De Gruyter Expositions in Mathematics 56 Editors Victor P. Maslov, Moscow, Russia Walter D. Neumann, New York, USA Marku...

Author:
Yakov Berkovich

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De Gruyter Expositions in Mathematics 56 Editors Victor P. Maslov, Moscow, Russia Walter D. Neumann, New York, USA Markus J. Pflaum, Boulder, USA Dierk Schleicher, Bremen, Germany Raymond O. Wells, Bremen, Germany

Yakov Berkovich Zvonimir Janko

Groups of Prime Power Order Volume 3

De Gruyter

Mathematical Subject Classification 2010: 20-02, 20D15, 20E07.

ISBN 978-3-11-020717-0 e-ISBN 978-3-11-025448-8 ISSN 0938-6572 Library of Congress Cataloging-in-Publication Data Berkovich, IA. G., 1938⫺ Groups of prime power order / by Yakov Berkovich, Zvonimir Janko. p. cm. ⫺ (De Gruyter expositions in mathematics ; ⬍ ⬎-56) Description based on v. 3, copyrighted c2011. Includes bibliographical references and index. ISBN 978-3-11-020717-0 (v. 3 : alk. paper) 1. Finite groups. 2. Group theory. I. Janko, Zvonimir, 1932⫺ II. Title. QA177.B469 2011 5121.23⫺dc22 2011004438

Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available in the Internet at http://dnb.d-nb.de. ” 2011 Walter de Gruyter GmbH & Co. KG, Berlin/New York Typesetting: Dimler & Albroscheit, Müncheberg Printing and binding: Hubert & Co. GmbH & Co. KG, Göttingen ⬁ Printed on acid-free paper Printed in Germany www.degruyter.com

Contents

List of deﬁnitions and notations . . . . . . . . . . . . . . . . . . . . . . . . . .

ix

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xv Prerequisites from Volumes 1 and 2 . . . . . . . . . . . . . . . . . . . . . . . . xvii 93 94

Nonabelian 2-groups all of whose minimal nonabelian subgroups are metacyclic and have exponent 4 . . . . . . . . . . . . . . . . . . . . . . . . .

1

Nonabelian 2-groups all of whose minimal nonabelian subgroups are nonmetacyclic and have exponent 4 . . . . . . . . . . . . . . . . . . . . . .

8

95

Nonabelian 2-groups of exponent 2e which have no minimal nonabelian subgroups of exponent 2e . . . . . . . . . . . . . . . . . . . . . . . . . . 10

96

Groups with at most two conjugate classes of nonnormal subgroups

97

p-groups in which some subgroups are generated by elements of order p

98

Nonabelian 2-groups all of whose minimal nonabelian subgroups are isomorphic to M2nC1 , n 3 ﬁxed . . . . . . . . . . . . . . . . . . . . . . . 31

99

2-groups with sectional rank at most 4 . . . . . . . . . . . . . . . . . . . 34

. . . 12 24

100 2-groups with exactly one maximal subgroup which is neither abelian nor minimal nonabelian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 101 p-groups G with p > 2 and d.G/ D 2 having exactly one maximal subgroup which is neither abelian nor minimal nonabelian . . . . . . . . . . 66 102 p-groups G with p > 2 and d.G/ > 2 having exactly one maximal subgroup which is neither abelian nor minimal nonabelian . . . . . . . . . . 77 103 Some results of Jonah and Konvisser . . . . . . . . . . . . . . . . . . . . 93 104 Degrees of irreducible characters of p-groups associated with ﬁnite algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 105 On some special p-groups . . . . . . . . . . . . . . . . . . . . . . . . . 102 106 On maximal subgroups of two-generator 2-groups . . . . . . . . . . . . . 110

vi

Groups of prime power order

107 Ranks of maximal subgroups of nonmetacyclic two-generator 2-groups

. 113

108 p-groups with few conjugate classes of minimal nonabelian subgroups . . 120 109 On p-groups with metacyclic maximal subgroup without cyclic subgroup of index p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 110 Equilibrated p-groups

. . . . . . . . . . . . . . . . . . . . . . . . . . . 125

111 Characterization of abelian and minimal nonabelian groups . . . . . . . . 134 112 Non-Dedekindian p-groups all of whose nonnormal subgroups have the same order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 113 The class of 2-groups in 70 is not bounded . . . . . . . . . . . . . . . . 148 114 Further counting theorems . . . . . . . . . . . . . . . . . . . . . . . . . 152 115 Finite p-groups all of whose maximal subgroups except one are extraspecial 157 116 Groups covered by few proper subgroups

. . . . . . . . . . . . . . . . . 162

117 2-groups all of whose nonnormal subgroups are either cyclic or of maximal class . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176 118 Review of characterizations of p-groups with various minimal nonabelian subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 119 Review of characterizations of p-groups of maximal class

. . . . . . . . 185

120 Nonabelian 2-groups such that any two distinct minimal nonabelian subgroups have cyclic intersection . . . . . . . . . . . . . . . . . . . . . . . 192 121 p-groups of breadth 2

. . . . . . . . . . . . . . . . . . . . . . . . . . . 197

122 p-groups all of whose subgroups have normalizers of index at most p

. . 204

123 Subgroups of ﬁnite groups generated by all elements in two shortest conjugacy classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237 124 The number of subgroups of given order in a metacyclic p-group . . . . . 239 125 p-groups G containing a maximal subgroup H all of whose subgroups are G-invariant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269 126 The existence of p-groups G1 < G such that Aut.G1 / Š Aut.G/

. . . . 272

127 On 2-groups containing a maximal elementary abelian subgroup of order 4 275 128 The commutator subgroup of p-groups with the subgroup breadth 1

. . . 277

Contents

vii

129 On two-generator 2-groups with exactly one maximal subgroup which is not two-generator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 130 Soft subgroups of p-groups

. . . . . . . . . . . . . . . . . . . . . . . . 287

131 p-groups with a 2-uniserial subgroup of order p . . . . . . . . . . . . . . 292 132 On centralizers of elements in p-groups . . . . . . . . . . . . . . . . . . 295 133 Class and breadth of a p-group . . . . . . . . . . . . . . . . . . . . . . . 300 134 On p-groups with maximal elementary abelian subgroup of order p 2 . . . 304 135 Finite p-groups generated by certain minimal nonabelian subgroups . . . 315 136 p-groups in which certain proper nonabelian subgroups are two-generator 328 137 p-groups all of whose proper subgroups have its derived subgroup of order at most p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338 138 p-groups all of whose nonnormal subgroups have the smallest possible normalizer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343 139 p-groups with a noncyclic commutator group all of whose proper subgroups have a cyclic commutator group . . . . . . . . . . . . . . . . . . . . . . 355 140 Power automorphisms and the norm of a p-group . . . . . . . . . . . . . 363 141 Nonabelian p-groups having exactly one maximal subgroup with a noncyclic center . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368 142 Nonabelian p-groups all of whose nonabelian maximal subgroups are either metacyclic or minimal nonabelian . . . . . . . . . . . . . . . . . . . . . 370 143 Alternate proof of the Reinhold Baer theorem on 2-groups with nonabelian norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373 144 p-groups with small normal closures of all cyclic subgroups

. . . . . . . 376

A.27 Wreathed 2-groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384 A.28 Nilpotent subgroups

. . . . . . . . . . . . . . . . . . . . . . . . . . . . 393

A.29 Intersections of subgroups

. . . . . . . . . . . . . . . . . . . . . . . . . 405

A.30 Thompson’s lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416 A.31 Nilpotent p 0 -subgroups of class 2 in GL.n; p/ . . . . . . . . . . . . . . . 428 A.32 On abelian subgroups of given exponent and small index . . . . . . . . . 434

viii

Groups of prime power order

A.33 On Hadamard 2-groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 437 A.34 Isaacs–Passman’s theorem on character degrees . . . . . . . . . . . . . . 440 A.35 Groups of Frattini class 2 . . . . . . . . . . . . . . . . . . . . . . . . . . 446 A.36 Hurwitz’ theorem on the composition of quadratic forms . . . . . . . . . 449 A.37 On generalized Dedekindian groups . . . . . . . . . . . . . . . . . . . . 452 A.38 Some results of Blackburn and Macdonald . . . . . . . . . . . . . . . . . 457 A.39 Some consequences of Frobenius’ normal p-complement theorem . . . . 460 A.40 Varia

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472

A.41 Nonabelian 2-groups all of whose minimal nonabelian subgroups have cyclic centralizers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 514 A.42 On lattice isomorphisms of p-groups of maximal class

. . . . . . . . . . 516

A.43 Alternate proofs of two classical theorems on solvable groups and some related results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 519 A.44 Some of Freiman’s results on ﬁnite subsets of groups with small doubling 527 Research problems and themes III

. . . . . . . . . . . . . . . . . . . . . . . . 536

Author index

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 630

Subject index

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 632

List of deﬁnitions and notations

Set theory jM j is the cardinality of a set M (if G is a ﬁnite group, then jGj is called its order). x 2 M (x 62 M ) means that x is (is not) an element of a set M . N M (N 6 M ) means that N is (is not) a subset of the set M ; moreover, if M ¤ N M , we write N M. ¿ is the empty set. N is called a nontrivial subset of M if N ¤ ¿ and N M . If N M , we say that N is a proper subset of M . M \ N is the intersection and M [ N is the union of sets M and N . If M; N are sets, then N M D ¹x 2 N j x 62 M º is the difference of N and M .

Number theory and general algebra p is always a prime number. is a set of primes; 0 is the set of all primes not contained in . m, n, k, r, s are, as a rule, natural numbers. .m/ is the set of prime divisors of m; then m is a -number if .m/ . np is the p-part of n, n is the -part of n. GCD.m; n/ is the greatest common divisor of m and n. m j n should be read as: m divides n. GF.p m / is the ﬁnite ﬁeld containing p m elements. F is the multiplicative group of a ﬁeld F. L.G/ is the lattice of all subgroups of a group G. If n D p1˛1 : : : pk˛k is the standard prime decomposition of n, then .n/ D

Pk

i D1 ˛i .

x

Groups of prime power order

Groups We consider only ﬁnite groups which are denoted, with a few exceptions, by uppercase Latin letters. If G is a group, then .G/ D .jGj/. G is a p-group if jGj is a power of p; G is a -group if .G/ . G is, as a rule, a ﬁnite p-group. H G means that H is a subgroup of G. H < G means that H G and H ¤ G (in that case, H is called a proper subgroup of G). ¹1º denotes the group containing only one element. H is a nontrivial subgroup of G if ¹1º < H < G. H is a maximal subgroup of G if H < G and it follows from H M < G that H D M. If H is a proper normal subgroup of G, then we write H G G. Expressions ‘normal subgroup of G’ and ‘G-invariant subgroup’ are synonyms. A normal subgroup of G is nontrivial provided G > H > ¹1º. H is a minimal normal subgroup of G if (a) H is normal in G; (b) H > ¹1º; (c) N GG and N < H implies N D ¹1º. Thus the group ¹1º has no minimal normal subgroup. G is simple if it is a minimal normal subgroup of G (so jGj > 1). H is a maximal normal subgroup of G if H < G and G=H is simple. The subgroup generated by all minimal normal subgroups of G is called the socle of G and denoted by Sc.G/. We put, by deﬁnition, Sc.¹1º/ D ¹1º. NG .M / D ¹x 2 G j x 1 M x D M º is the normalizer of a subset M in G. CG .x/ is the centralizer of an element x in G: CG .x/ D ¹z 2 G j zx D xzº. T CG .M / D x2M CG .x/ is the centralizer of a subset M in G. If A B and A; B are normal in G, then CG .B=A/ D H , where H=A D CG=A .B=A/. A wr B (or A o B) is the wreath product of the ‘passive’ group A and the transitive permutation group B (in what follows we assume that B is regular); B is called the active factor of the wreath product). Then the order of that group is jAjjBj jBj. Aut.G/ is the group of automorphisms of G (the automorphism group of G). Inn.G/ is the group of all inner automorphisms of G. Out.G/ D Aut.G/=Inn.G/ is the outer automorphism group of G.

List of deﬁnitions and notations

xi

N .G/ is the norm of G, the intersection of normalizers of all subgroups of G. If a; b 2 G, then ab D b 1 ab. An element x 2 G inverts a subgroup H G if hx D h1 for all h 2 H . If M G, then hM i D hx j x 2 M i is the subgroup of G generated by M . M x D x 1 M x D ¹y x j y 2 M º for x 2 G and M G. Œx; y D x 1 y 1 xy D x 1 x y is the commutator of elements x; y of G. If M; N G, then ŒM; N D hŒx; y j x 2 M; y 2 N i is a subgroup of G. o.x/ is the order of an element x of G. An element x 2 G is a -element if .o.x// . G is a -group, if .G/ . Obviously, G is a -group if and only if all of its elements are -elements. G 0 is the subgroup generated by all commutators Œx; y, x; y 2 G (i.e., G 0 D ŒG; G), G .2/ D ŒG 0 ; G 0 D G 00 D .G 0 /0 , G .3/ D ŒG 00 ; G 00 D .G 00 /0 and so on. G 0 is called the commutator (or derived) subgroup of G. T Z.G/ D x2G CG .x/ is the center of G. Zi .G/ is the i -th member of the upper central series of G; in particular, Z0 .G/ D ¹1º, Z1 .G/ D Z.G/. Ki .G/ is the i-th member of the lower central series of G; in particular, K2 .G/ D G 0 . We have Ki .G/ D ŒG; : : : ; G (i 1 times). We set K1 .G/ D G. If G is nonabelian, then .G/=K3 .G/ D Z.G=K3 .G//. M.G/ D hx 2 G j CG .x/ D CG .x p /i is the Mann subgroup of a p-group G. Sylp .G/ is the set of p-Sylow subgroups of an arbitrary ﬁnite group G. Sn is the symmetric group of degree n. An is the alternating group of degree n. †pn is a Sylow p-subgroup of Spn . GL.n; F/ is the set of all nonsingular n n matrices with entries in a ﬁeld F, the n-dimensional general linear group over F, SL.n; F/ D ¹A 2 GL.n; F/ j det.A/ D 1 2 Fº, the n-dimensional special linear group over F. T If H G, then HG D x2G x 1 H x is the core of the subgroup H in G and H G , the intersection of all normal subgroups of G containing H , is the normal closure or normal hull of H in G. Obviously, HG is normal in G. If G is a p-group, then p b.x/ D jG W CG .x/j; b.x/ is said to be the breadth of x 2 G, where G is a p-group; b.G/ D max¹b.x/ j x 2 Gº is the breadth of G.

xii

Groups of prime power order

If H G and jG W NG .H /j D p sb.H / , then sb.H / is said to be the subgroup breadth of H . Next, sb.G/ D max¹sb.H / j H Gº. ˆ.G/ is the Frattini subgroup of G (D the intersection of all maximal subgroups of G), ˆ.¹1º/ D ¹1º, p d.G/ D jG W ˆ.G/j i D ¹H < G j ˆ.G/ H; jG W H j D p i º, i D 1; : : : ; d.G/, where G > ¹1º. If H < G, then 1 .H / is the set of all maximal subgroups of H . exp.G/ is the exponent of G (the least common multiple of the orders of elements of G). If G is a p-group, then exp.G/ D max¹o.x/ j x 2 Gº. k.G/ is the number of conjugacy classes of G (D G-classes), the class number of G. Kx is the G-class containing an element x (sometimes we also write cclG .x/). Cm is the cyclic group of order m. G m is the direct product of m copies of a group G. A B is the direct product of groups A and B. A B is a central product of groups A and B, i.e., A B D AB with ŒA; B D ¹1º. Epm D Cpm is the elementary abelian group of order p m . G is an elementary abelian p-group if and only if it is a p-group > ¹1º and G coincides with its socle. Next, ¹1º is elementary abelian for each prime p. A group G is said to be homocyclic if it is a direct product of isomorphic cyclic subgroups (obviously, elementary abelian p-groups are homocyclic). ES.m; p/ is an extraspecial group of order p 1C2m (a p-group G is said to be extraspecial if G 0 D ˆ.G/ D Z.G/ is of order p). Note that for each positive integer m, there are exactly two nonisomorphic extraspecial groups of order p 2mC1 . S.p 3 / is a nonabelian group of order p 3 and exponent p > 2. A special p-group is a nonabelian p-group G such that G 0 D ˆ.G/ D Z.G/ is elementary abelian. Direct products of extraspecial p-groups are special. D2m is the dihedral group of order 2m, m > 2. Some authors consider E22 as the dihedral group D4 . Q2m is the generalized quaternion group of order 2m 23 . SD2m is the semidihedral group of order 2m 24 . Mpm is a nonabelian p-group containing exactly p cyclic subgroups of index p. cl.G/ is the nilpotence class of a p-group G. dl.G/ is the derived length of a p-group G. CL.G/ is the set of all G-classes.

List of deﬁnitions and notations

xiii

A p-group of maximal class is a nonabelian group G of order p m with cl.G/ D m 1. m .G/ D hx 2 G j o.x/ p m i, m .G/ D hx 2 G j o.x/ D p m i and Ãm .G/ D m hx p j x 2 Gi. A p-group G is said to be regular if for any x; y 2 G there exists z 2 hx; yi0 such that .xy/p D x p y p z p . A p-group is absolutely regular if jG=Ã1 .G/j < p p . A p-group is thin if it is either absolutely regular or of maximal class. G D A B is a semidirect product with kernel B and complement A. A group G is an extension of a normal subgroup N by a group H if G=N Š H . A group G splits over N if G D H N with H G and H \ N D ¹1º (in that case, G is a semidirect product of H and N with kernel N ). H # D H ¹eH º, where eH is the identity element of the group H . If M G, then M # D M ¹eG º. An automorphism ˛ of G is regular (D ﬁxed-point-free) if it induces a regular permutation on G # (a permutation is said to be regular if it has no ﬁxed points). An involution is an element of order 2 in a group. A group G is said to metacyclic if it contains a normal cyclic subgroup C such that G=C is cyclic. A group G is said to be minimal nonmetacyclic if it is nonmetacyclic but all its proper subgroups are metacyclic. A subgroup A of a group G is said to be soft if CG .A/ D A and jNG .A/ W Aj D p. A section of a group G is an epimorphic image of some subgroup of G. If F D GF.p n /, then we may write GL.m; p n /; SL.m; p n /; : : : instead of GL.m; F/, SL.m; F/; : : : . cn .G/ is the number of cyclic subgroups of order p n in a p-group G. sn .G/ is the number of subgroups of order p n in a p-group G. en .G/ is the number of subgroups of order p n and exponent p in G. A group G is said to be minimal nonabelian if it is nonabelian but all its proper subgroups are abelian. An -group is a p-group G all of whose subgroups of index p n are abelian but G contains a nonabelian subgroup of index p n1 . In particular, A1 -group is a minimal nonabelian p-group for some p. ˛n .G/ is the number of An -subgroups in a p-group G.

xiv

Groups of prime power order

MA.G/ is the set of minimal nonabelian subgroups of a p-group G. MAk .G/ D ¹H 2 MA.G/ j k .H / D H º. Dk .G/ D hMAk .G/i D hH j H 2 MAk .G/i. Ln D j¹x 2 G j x n D 1ºj.

Characters and representations Irr.G/ is the set of all irreducible characters of G over complex numbers. A character of degree 1 is said to be linear. Lin.G/ is the set of all linear characters of G (obviously, Lin.G/ Irr.G/). Irr1 .G/ D Irr.G/ Lin.G/ is the set of all nonlinear irreducible characters of G; n.G/ D jIrr1 .G/j. .1/ is the degree of a character of G. H is the restriction of a character of G to H G. G is the character of G induced from the character of some subgroup of G. N is a character of G deﬁned as follows: .x/ N D .x/ (here wN is the complex conjugate of a complex number w). Irr. / is the set of irreducible constituents of a character of G. 1G is the principal character of G. Irr# .G/ D Irr.G/ ¹1G º. If is a character of G, then ker. / D ¹x 2 G j .x/ D .1/º is the kernel of a character . Z. / D ¹x 2 G j j .x/j D .1/º is the quasikernel of . If N is normal in G, then Irr.G j N / D ¹ 2 Irr.G/ j N 6 ker. /º. P h ; i D jGj1 x2G .x/ .x 1 / is the inner product of characters and of G. IG ./ D hx 2 G j x D i is the inertia subgroup of 2 Irr.H / in G, where H G G. 1G is the principal character of G (1G .x/ D 1 for all x 2 G). M.G/ is the Schur multiplier of G. cd.G/ D ¹ .1/ j 2 Irr.G/º. mc.G/ D k.G/=jGj is the measure of commutativity of G. P T.G/ D 2Irr.G/ .1/, f.G/ D T.G/=jGj.

Preface

This is the third volume of the book devoted to elementary parts of p-group theory. Sections 93–95, 98–102, 113, 117, 118, 120–123, 125–133, 137, 139, 140–142, 144 are written by the second author, Sections 107, 138 and Appendix 41 are jointly written by both authors, all other material, apart from Appendices 33 and 44, is written by the ﬁrst author. All exercises and about all problems are due to the ﬁrst author. All material of this part is appeared in the book form for the ﬁrst time. Some interesting problems of elementary p-group theory are solved in this volume: (i) classiﬁcation of p-groups containing exactly one maximal subgroup which is neither abelian nor minimal nonabelian, (ii) classiﬁcation of groups all of whose nonnormal subgroups have the same order (independently, this result was obtained by Guido Zappa), (iii) classiﬁcation of p-groups all of whose nonnormal subgroups have normalizers of index p, (iv) computation of the number of subgroups of given order in metacyclic p-groups (for p > 2 this was done by Avinoam Mann), (v) computation of the order of the derived subgroup of a group with subgroup breadth 1, (vi) classiﬁcation of p-groups all of whose subgroups have derived subgroups of order p (so-called A2 -groups satisfy this condition), (vii) classiﬁcation of p-groups G all of whose maximal abelian subgroups containing a non-G-invariant cyclic subgroup of minimal order, say p , have order p C1 . Some results proved in this part have no analogs in existing books devoted to ﬁnite p-groups. We list only a few such results: (a) study the p-groups G all of whose minimal nonabelian subgroups have exponent < exp.G/, (b) study the groups admitting an irredundant covering by few proper subgroups, (c) study of some 2-groups with sectional rank 4, (d) study the p-groups which do not generated by certain minimal nonabelian subgroups, (e) study the p-groups, p > 3, in which certain nonabelian subgroups are generated by two elements, (f) study the p-groups with nonnormal maximal elementary abelian subgroup of order p 2 (this is a continuation of the paper of Glauberman–Mazza), (g) classiﬁcation of p-groups with exactly one maximal subgroup which is neither abelian nor minimal nonabelian,

xvi

Groups of prime power order

(h) classiﬁcation of p-groups all of whose proper subgroups have derived subgroups of orders p, (i) classiﬁcation of p-groups all of whose nonnormal cyclic subgroups of minimal possible order have index p is their normalizers, (j) classiﬁcation of p-groups G all of whose maximal abelian subgroups containing non-G-invariant cyclic subgroup of minimal order, say p , have order p nC1 , (k) classiﬁcation of p-groups all of whose maximal subgroups have cyclic derived subgroups, (l) characterizations of abelian and minimal nonabelian groups, (m) describing all possible sets of numbers of generators of maximal subgroups of two-generator 2-groups, (n) study the equilibrated p-groups, (o) classiﬁcation of nonabelian 2-groups in which any two distinct minimal nonabelian subgroups have cyclic intersection, (p) study the p-groups of breadth 2, (q) study groups containing a soft subgroup, (r) proof of the Schenkman theorem on the norm of a ﬁnite group and a new proof of Baer’s theorem on an arbitrary 2-group with nonabelian norm. For further information, see the Contents. Essential part of this volume is devoted to investigation of impact of minimal nonabelian subgroups on the structure of a p-group. Some appendices are devoted to nilpotent subgroups of nonnilpotent groups. The section ‘Research problems and themes III’, written by the ﬁrst author, contains about 900 research problems and themes some of which are solved by the second author. There are in the text approximately 200 exercises most of which are solved. Avinoam Mann (Hebrew University of Jerusalem) analyzed the list of problems and made a great number of constructive comments and corrections. He also read a number of sections and made numerous useful remarks. Moshe Roitman (University of Haifa) wrote Appendix 44 and helped with LATEX. Noboru Ito wrote Appendix 33. We are indebted to these three mathematicians. We are grateful to the publishing house of Walter de Gruyter and all its workers for supporting and promoting the publication of our book and especially to Simon Albroscheit and Kay Dimler.

Prerequisites from Volumes 1 and 2

In this section we state some results from Volumes 1 and 2 which we use in what follows. If we formulate Lemma 1.4 from Volume 1, then it is named below also as Lemma 1.4. Lemma INTR. (a) If M; N G G, then the quotient group G=.M \ N / is isomorphic to a subgroup of G=M G=N . (b) If Z is a cyclic subgroup of maximal order in an abelian p-group, then Z is a direct factor of G. In particular, an abelian p-group is a direct product of cyclic subgroups. (c) (Fitting’s lemma) If M and N are nilpotent normal subgroups of a group G, then cl.MN / cl.M / C cl.N /. (d) If G is a p-group, then G=ˆ.G/ is elementary abelian of order, say d . Every minimal set of generators of G contains exactly d members. (e) If G is a p-group, then ˆ.G/ D G 0 Ã1 .G/. Lemma 1.1. If A is an abelian subgroup of index p in a nonabelian p-group G, then jGj D pjG 0 jjZ.G/j. Theorem 1.2. Suppose that a nonabelian p-group has a cyclic subgroup of index p. Then one of the following holds: D b 2 D 1; ab D a1 i Š D2n is of class n 1, all elements of (a) G D ha; b j a2 the set G hai are involutions. n1

(b) G D ha; b j a2 D 1; a2 D b 2 ; ab D a1 i Š Q2n is of class n 1, all elements of the set G hai have the same order 4. n1

n2

(c) G D ha; b j a2 D 1; a2 D b 2 ; ab D a1 i Š SD2n is of class n 1, 1 D ¹D Š D2n1 ; Q Š Q2n1 ; hai Š C2n1 º, 1 .G/ D D, 2 .G/ D Q. n1

n1

n2

n2

(d) G D ha; b j ap D b p D 1; ab D a1Cp i Š Mpn is of class 2, where n > 3 n2 if p D 2, Z.G/ D hap i, 1 .G/ D hap ; bi Š Ep2 . The groups D2n , Q2n , SD2n are called dihedral, generalized quaternion, semidihedral, respectively. Proposition 1.3. If a p-group has only one subgroup of order p, then it is either cyclic or a generalized quaternion group.

xviii

Groups of prime power order

Lemma 1.4. Let N be a normal subgroup of a p-group G. If N has no G-invariant abelian subgroup of type .p; p/, then it is either cyclic or a 2-group of maximal class. It follows from Lemma 1.4 that if Z2 .G/ is cyclic, then G is either cyclic or a 2-group of maximal class. Lemma 1.6 (Taussky). If a nonabelian 2-group G satisﬁes jG W G 0 j D 4, then it is a 2-group of maximal class. Exercise 1.6(a). The number of abelian maximal subgroups in a nonabelian p-group G is either 0 or 1 or p C 1. Exercise P1. If a nonabelian p-group G has two distinct abelian maximal subgroups, then jG 0 j D p. Proposition 1.8 (Suzuki). If a nonabelian p-group G has a self-centralizing abelian subgroup of order p 2 , then G is of maximal class. (For the converse assertion, see Theorem 9.6(c).) Exercise 1.8(a). If a p-group G is minimal nonabelian, then jG 0 j D p;

d.G/ D 2;

jG W Z.G/j D p 2 ;

and one of the following holds: m

n

(a) G D ha; b j ap D b p D 1; ab D a1Cp pm

m1

i is metacyclic.

pn

(b) G D ha; b j a D b D 1; Œa; b D c; c p D 1; Œa; c D Œb; c D 1i is nonmetacyclic. In that case, Ã1 .G/ D hap i hb p i, G=Ã1 .G/ is nonabelian of order p 3 and exponent p unless p D 2 (if p D 2, then G=Ã1 .G/ Š E4 ). (c) G Š Q8 . Next, a group G is nonmetacyclic if and only if G 0 is a maximal cyclic subgroup of G. The group G is metacyclic if and only if j1 .G/j p 2 . No noncentral cyclic subgroup is normal in G. Theorems 1.10 and 1.17. Suppose that a p-group G is neither cyclic nor a 2-group of maximal class. Then (a) c1 .G/ 1 C p .mod p 2 /. (b) If k > 1, then ck .G/ 0 .mod p/. Theorem 1.20. If all subgroups of a nonabelian p-group G are normal, then we have G D Q E, where Q Š Q8 and exp.E/ 2. Exercise 1.69(a). Let G be a p-group and let H ¤ K be two distinct maximal subgroups of G. Then jG 0 W H 0 K 0 j p. In particular, if G 0 is cyclic, there is A 2 1 such that jG 0 W A0 j p.

Prerequisites from Volumes 1 and 2

xix

Lemma 4.2. Let G be a p-group with jG 0 j D p. Then G D .A1 A2 As /Z.G/, the central product, where A1 ; : : : ; As are minimal nonabelian, so G=Z.G/ is elementary abelian of even rank. In particular, if G=G 0 is elementary abelian, then we have jA1 j D D jAs j D p 3 , E D A1 As is extraspecial and G D EZ.G/. Lemma 4.3. Let E be a subgroup of a p-group G with jE 0 j D p and Z.E/ D ˆ.E/. If ŒG; E D E 0 , then G D E CG .E/. Theorem 5.2 (Hall’s enumeration principle). Let G be a p-group and M be a set of proper subgroups of G. Given H G, let ˛.H / be the number of members of the set M contained in H . Then X ˛.G/

˛.H / .mod p/: H 21

Theorems 5.3 and 5.4. Suppose that a p-group G of order p m is neither cyclic nor a 2-group of maximal class and 1 n < m. Then sn .G/ 1 C p .mod p 2 /. Theorem 5.8. Let G be a group of order p m > p 3 and exponent p, 2 < n < m. Let M denote the set of all 2-generator subgroups of order p n in G and ˛.K/ the number of elements of the set M contained in K G. (a) If n D m 1, then ˛.G/ 2 ¹0; p; p 2 º. (b) p divides ˛.G/. Theorem 7.1. Let G be a p-group. (a) Regularity is inherited by sections. (b) If cl.G/ < p or jGj p p or exp.G/ D p, then G is regular. (c) If Kp1 .G/ is cyclic, then G is regular. Theorem 7.2. Suppose that G is a regular p-group. (b) exp.n .G// p n . n

(c) Ãn .G/ D ¹x p j x 2 Gº. (d) jn .G/j D jG W Ãn .G/j. Theorem 9.5. Let G be a group of maximal class and order p m , m pC1. Then ˆ.G/ and G=Z.G/ have exponent p. If m D p C 1, then G is irregular and jÃ1 .G/j D p. Theorem 9.6. Let G be a group of maximal class and order p m , p > 2, m > p C 1. Then G is irregular and (a) jG W Ã1 .G/j D p p . In particular, Ã1 .G/ D Kp .G/. (b) There is G1 2 1 such that jG1 W Ã1 .G1 /j D p p1 .

xx

Groups of prime power order

(c) G has no normal subgroups of order p p and exponent p (here the condition that m > p C 1 is essential). Moreover, if N G G and jG W N j > p, then N is absolutely regular since N < G1 . Next, if N G G of order p p1 , then exp.N / D p. In particular, G has no normal cyclic subgroups of order p 2 . (d) Let Z2 D Z2 .G/ be a normal subgroup of order p 2 in G, G0 D CG .Z2 /. Then G0 is regular such that j1 .G0 /j D p p1 . (e) Let 1 D ¹M1 D G0 ; M2 ; : : : ; MpC1 º, where G0 is deﬁned in (d). Then the subgroups M2 ; : : : ; MpC1 are of maximal class (and so irregular; see Theorem 9.5). Thus the subgroups G1 from (b) and G0 from (d) coincide. In what follows we call G1 the fundamental subgroup of G. (f) In this part, m 3 (i.e., we do not assume as in other parts that m > p C 1). The group G has an element a such that jCG .a/j D p 2 , i.e., CG .a/ D ha; Km1 .G/i. Theorem 9.7. A nonabelian p-group G is of maximal class if and only if G=KpC1 .G/ is of maximal class. Theorem 9.8. (a) Absolutely regular p-groups G (i.e., G with jG=Ã1 .G/j < p p ) are regular. (b) If a p-group G is such that jG 0 =Ã1 .G 0 /j < p p1 , then G is regular. (c) Any irregular p-group G has a characteristic subgroup R of order p p1 and exponent p such that R G 0 . Theorem 9.11. A p-group G, p > 2, is metacyclic if and only if jG=Ã1 .G/j p 2 . Exercise 9.13. Let G be a p-group of maximal class, p > 2, and H < G. Then d.H / p. If d.H / D p, then G Š †p2 2 Sylp .Sp2 /. In particular (Blackburn), if G is a p-group of maximal class, p > 2, N G G and G=N Š †p2 , then N D ¹1º. Theorem 10.1. Let p n > 2 and let A < G be abelian of exponent p n . Then the number of abelian subgroups B G of order pjAj that contain A is congruent to 1 modulo p. Corollary 10.2. Let N be a normal subgroup of a p-group G and let A < N be a maximal G-invariant abelian subgroup of exponent p n , where p n > 2. Then we have n .CN .A// D A. Theorems 10.4 and 10.5. Suppose that G is a p-group, p > 2. Let n .G/ be the number of elementary abelian subgroups of order p n in G. If k 2 ¹3; 4º and k .G/ > 0, then we have k .G/ 1 .mod p/. Exercise P2. Let G and k be such as in the previous theorem. (a) If Zk .G/ has no G-invariant elementary abelian subgroup of order p k , then we have k .G/ D 0. (b) If a G-invariant subgroup N of G has no G-invariant elementary abelian subgroup of order p k , then k .N / D 0.

Prerequisites from Volumes 1 and 2

xxi

Lemma 10.8. Suppose that G is a minimal nonnilpotent group. Then G D PQ, where P 2 Sylp .G/, Q D G 0 2 Sylq .G/ and (a) P is cyclic, jP W .P \ Z.G/j D p. (b) Q is either elementary abelian or special, jQ=ˆ.Q/j D q b , where b is the order of q modulo p. (c) If Q is special, then b is even and ˆ.Q/ D Z.Q/ Z.G/, jˆ.Q/j p b=2 . Theorem (Frobenius’ normal p-complement theorem). If a group G has no p-closed minimal nonnilpotent subgroup of order divisible by p, then G has a normal p-complement (D p-nilpotent). Theorem (Burnside’s normal p-complement theorem). If a Sylow p-subgroup of a group G is contained in the center of its normalizer, then G is p-nilpotent. Proposition 10.17. If B G is a nonabelian subgroup of order p 3 in a p-group G and CG .B/ < B, then G is of maximal class. Remark 10.5. Let H < G and assume that NG .H / is of maximal class. Then G is also of maximal class. Proposition 10.19. Suppose that H is a nonabelian subgroup of order p 3 in a metacyclic p-group G. If p > 2, then G D H . If p D 2, then G is of maximal class. Proposition 10.28. A nonabelian p-group G is generated by minimal nonabelian subgroups. In particular, if, in addition, G is not minimal nonabelian, it contains two nonconjugate minimal nonabelian subgroups. Theorem 10.33. If all minimal nonabelian subgroups of a nonabelian 2-group G are generated by involutions, then G D hxi A, where A 2 1 is abelian and all elements in G A are involutions (such G is said to be generalized dihedral). Theorem 12.1. (a) If a p-group G has no normal subgroup of order p p and exponent p, then G is either absolutely regular or irregular of maximal class. (b) If an irregular p-group G has an absolutely regular maximal subgroup H , then it is either of maximal class or G D H 1 .G/, where 1 .G/ is of order p p (and, of course, of exponent p). Exercise P3. If p .G/ has no G-invariant subgroup of order p p and exponent p, then G is either absolutely regular or of maximal class. Theorem 12.12. Let a group G of order p m be neither absolutely regular nor of maximal class and suppose that H 2 1 is of maximal class. Then (a) d.G/ D 3. (b) Set D m 2 if m p C 1 and D p if m > p C 1. Then G=K .G/ is of order p C1 and, if m > 4, it is of exponent p.

xxii

Groups of prime power order

(c) Exactly p 2 maximal subgroups of G are of maximal class. If, in addition, p > 2 and m > 4, then the remaining p C 1 maximal subgroups of G have no two generators and their intersection .G/ has index p 2 in G. Theorem 13.2. Suppose that a p-group G is neither absolutely regular nor of maximal class. Then (a) c1 .G/ 1 C p C C p p1 .mod p p /. (b) If k > 1, then ck .G/ 0 .mod p p1 /. Corollary 13.3. Suppose that an irregular p-group G is neither absolutely regular nor of maximal class and k < p. Then it has a normal subgroup M of order p k and exponent p, and the number of subgroups in G of order p kC1 and exponent p containing M is 1 .mod p/. Theorem 13.5. If a p-group G is neither absolutely regular nor of maximal class, then ep .G/ 1 .mod p/ (here en .G/ is the number of subgroups of order p n and exponent p in G). Exercise P4. Let N be a normal subgroup of a p-group G. If N has no G-invariant subgroup of order p p and exponent p, then it is either absolutely regular or of maximal class. Theorem 13.6. Let a group G of order p m be neither absolutely regular nor of maximal class, let n be a natural number with m > n p C1. Denote by ˛.G/ the number of subgroups of maximal class and order p n in G. Then p 2 divides ˛.G/. Theorem 13.7. Let p > 2 and suppose that a p-group has no normal elementary abelian subgroup of order p 3 . Then one of the following holds: (a) G is metacyclic. (b) G is a 3-group of maximal class not isomorphic to a Sylow 3-subgroup of the symmetric group of degree 9. (c) G D 1 .G/C , where j1 .G/j D p 3 and C is cyclic. Exercise P5. Let G be a p-group, p > 2, and suppose that 1 .G/ has no G-invariant elementary abelian subgroup of order p 3 . Then G is one of the groups from Theorem 13.7. Exercise 13.10(a). Let H < G, where G is a p-group. If every subgroup of G of order pjH j containing H is of maximal class, then G is also of maximal class. Proposition 13.18. Let G be a p-group, and let M < G be of maximal class. (a) Set D D ˆ.M /, N D NG .M / and C D CN .M=D/. Let t be the number of subgroups K G of maximal class such that M < K and jK W M j D p. Then t D c1 .N=M / c1 .C =M /. If G is not of maximal class, then t 0 .mod p/.

xxiii

Prerequisites from Volumes 1 and 2

(b) Suppose, in addition, that M is irregular and G is not of maximal class and a positive integer k is ﬁxed. Then the number t of the subgroups L < G of maximal class and order p k jM j such that M < L is a multiple of p. Theorem 36.1. Let G be a nonabelian p-group and R < G 0 be a G-invariant subgroup of index p. Then G is metacyclic if and only if G=R is metacyclic. Corollary 36.6 (Blackburn). Suppose that a nonabelian p-group G and all its maximal subgroups are two-generator. Then G is either metacyclic or p > 2 and K3 .G/ D Ã1 .G/ has index p 3 in G (in the last case, jG W G 0 j D p 2 ). Theorem 36.16. Suppose that a p-group G is such that G=Ã2 .G/ is of maximal class. Then G is also of maximal class. Exercise P6. If a p-group G is such that G=Ã2 .G/ is of maximal class, then G is also of maximal class. Lemma 42.1. Let G be a group of order p m satisfying j2 .G/j p pC1 < jGj. Then one of the following holds: (a) G is an Lp -group. (b) G is absolutely regular. (c) p D 2, G is metacyclic and G D ha; b j a2

m2

D b 8 D 1; ab D a1 ; a2

m3

D b 4 ; m > 4i:

Here Z.G/ D hb 2 i, G 0 D ha2 i and G=G 0 is abelian of type .4; 2/, ˆ.G/ D m4 ha2 ; b 2 i, 2 .G/ D ha2 ; b 4 i. Corollary 44.6. A p-group is metacyclic if and only if one of the following quotient groups is metacyclic: G=ˆ.G 0 /;

G=K3 .G/;

G=Ã1 .G 0 /;

and, if p > 2, then G=Ã1 .G/:

Corollary 44.9. If G=Ã2 .G/ is a metacyclic 2-group, then G is also metacyclic. Theorem 44.12. Suppose that N is a two-generator normal subgroup of a p-group G. If N ˆ.G/, then N is metacyclic. Theorem 50.1. Let G be a 2-group which has no normal elementary abelian subgroup of order 8. Then G has a normal metacyclic subgroup N such that G=N is isomorphic to a subgroup of D8 . Lemma 57.1. Let G be a nonabelian p-group and let A be a maximal abelian normal subgroup of G. Then for any x 2 GA, there is a 2 A such that Œa; x ¤ 1, Œa; xp D 1, and Œa; x; x D 1 which implies that ha; xi is minimal nonabelian. Therefore G is generated by its minimal nonabelian subgroups. In particular, if all minimal nonabelian subgroups of a nonabelian p-group G, p > 2, have exponent p, then Hp .G/ is abelian. (For a stronger result, see Mann’s commentary to Problem 115.)

xxiv

Groups of prime power order

Lemma 57.2. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are of exponent 4 and let A be a maximal normal abelian subgroup of G. Then all elements in G A are of order 4 and so either exp.A/ D 2 or exp.A/ D exp.G/. If x 2 G A with x 2 2 A, then x inverts each element in Ã1 .A/ and in A=1 .A/. If exp.G/ > 4, then either G=A is cyclic of order 4 or G=A Š Q8 . Lemma 65.2(a). If G is a nonabelian two-generator p-group and G 0 1 .Z.G//, then G is minimal nonabelian. Theorems 66.1 and 69.1. If G is a minimal nonmetacyclic p-group, then one of the following holds: (a) G is of order p 3 and exponent p. (b) G is a 3-group of maximal class and order 34 . (c) G Š D8 C4 D Q8 C4 is of order 16. (d) G D Q8 C2 . (e) G is special of order 25 with jZ.G/j D 4, exactly one maximal subgroup of G is abelian. It follows from the previous theorem that a p-group G is metacyclic in any of the following cases: (i) every subgroup of G of exponent p 2 is metacyclic (in particular, if 2 .G/ is metacyclic), (ii) every three-generator subgroup of G is metacyclic. Exercise P7. A p-group is metacyclic if one of the following holds: (a) 2 .G/ is metacyclic. (b) G=Ã2 .G/ is metacyclic. (c) G=K3 .G/ˆ.G 0 / is metacyclic. Exercise P8. If a 2-group G and all its maximal subgroups are two-generator, then G is metacyclic. Propositions 71.3–71.5. If a two-generator p-group is a nonmetacyclic A2 -group, then p > 2. Theorems 82.1–82.3. Let G be a 2-group with exactly three involutions and assume that Z.G/ is noncyclic. Then G has a normal metacyclic subgroup M such that G=M is elementary abelian of order at most 4. Theorem 90.1. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to D8 or Q8 . Then G is one of the following groups: (a) G is generalized dihedral (i.e., jG W H2 .G/j D 2). (b) G D H Z.G/, where H is of maximal class and Ã1 .Z.G// Z.H /. (c) G D H Z.G/, where H is extraspecial and Ã1 .Z.G// Z.H /.

Prerequisites from Volumes 1 and 2

xxv

Corollary 90.2. Let G be a nonabelian 2-group in which any two noncommuting elements generate a subgroup of maximal class. Then G is one of the groups (a), (b) or (c) from Theorem 90.1. Conversely, each group in (a), (b) or (c) of Theorem 90.1 satisﬁes the assumption of our corollary. Exercise P9. Let H D ha; bi be a two-generator p-group with H 0 of order p. Then ˆ.H / D hap ; b p ; Œa; bi and H is minimal nonabelian. Exercise P10. Let G be a p-group with jG 0 j D p. If H is a minimal nonabelian subgroup of G, then G D H CG .H /. Exercise P11. All p 2 C p C 1 subgroups of order p 2 in an elementary abelian group E D ha; b; ci of order p 3 are: ha; bi, ha; b i ci (p C 1 subgroups containing hai) and haj b; ak ci (p 2 subgroups not containing hai), where i; j; k are any integers modulo p. Corollary 92.7. Let G be a non-Dedekindian p-group and let R.G/ be the intersection of all nonnormal subgroups. If R.G/ > ¹1º, then p D 2, jR.G/j D 2 and G is one of the following groups: (a) G Š Q8 C4 E2s , s 0. (b) G Š Q8 Q8 E2s , s 0. (c) G has an abelian maximal subgroup A of exponent > 2 and an element x 2 G A of order 4 which inverts each element in A.

93

Nonabelian 2-groups all of whose minimal nonabelian subgroups are metacyclic and have exponent 4

We shall determine the structure of nonabelian 2-groups all of whose minimal nonabelian subgroups are metacyclic and have exponent 4, i.e., they are isomorphic to D8 , Q8 or H2 D ha; b j a4 D b 4 D 1; ab D a1 i. In Theorem 90.1 we have determined the nonabelian 2-groups all of whose minimal nonabelian subgroups are isomorphic to D8 or Q8 . The nonabelian 2-groups all of whose minimal nonabelian subgroups are isomorphic to Q8 or H2 are identiﬁed in Theorem 92.6. Here we determine the title 2-groups G which possess as subgroups both D8 and H2 (and possibly Q8 ). It turns out, as a surprise, that such 2-groups G must be of exponent > 4. More precisely, we prove the following result. Theorem 93.1. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are metacyclic and have exponent 4 and assume that D8 and H2 D ha; b j a4 D b 4 D 1; ab D a1 i (and possibly Q8 ) actually appear as subgroups of G. Then G has a unique abelian maximal subgroup A of exponent > 4 and an involution t 2 G A such that t inverts each element in A=1 .A/ and in 2 .A/ and there is an element a 2 A of order > 4 such that at D a1 with 1 ¤ 2 1 .A/, where in case that Ã1 .A/ is cyclic we have 62 Ã1 .A/. Conversely, all these 2-groups satisfy the assumptions of the theorem. Proof. Let G be a 2-group satisfying the assumptions of Theorem 93.1 and suppose that A is a maximal normal abelian subgroup of G. We shall freely use Lemmas 57.1, 57.2 and 65.2(a). It follows that all elements in G A are of order 4. (i) We show that G=A is elementary abelian. Suppose that this is false. Then there is an element v of order 4 in G A such that v 2 D t 62 A. Set T D Ahvi, A0 D CA .v/ and T0 D A0 hvi so that A ¤ A0 and T ¤ T0 . Let T1 T be such that T1 > T0 and jT1 W T0 j D 2. Set A1 D T1 \A so that jA1 W A0 j D 2. Note that T10 A\T0 D A0 Z.T1 / and so cl.T1 / D 2. Let a be any element in A1 A0 so that a2 2 A0 . We get 1 ¤ Œa; v 2 A0 and Œa; v2 D Œa2 ; v D 1 and so Œa; v is an involution in Z.T1 /. This implies that ha; vi0 D hŒa; vi and so ha; vi is minimal nonabelian. Set ha; vi \ A D A hai so that ha; vi D A hvi satisfying A \ hvi D ¹1º. If jA j D 2, then ha; vi D A hvi is abelian, a contradiction.

2

Groups of prime power order

Hence jA j D 4 so that ha; vi Š H2 and A D hai Š C4 with av D a1 . We have proved that all elements in A1 A0 are of order 4 and they are inverted by v. Since hA1 A0 i D A1 , v inverts A1 so that A0 D CA .v/ is elementary abelian and t D v 2 centralizes A1 . Set A2 D CA .t/ so that A2 A1 , A2 ¤ A and A2 is hvi-invariant which gives that T2 D A2 hvi is a subgroup of T , T2 ¤ T and T2 \ A D A2 . Let T3 be a subgroup of T such that T3 > T2 and jT3 W T2 j D 2 and set A3 D T3 \ A so that jA3 W A2 j D 2. Let b be any element in A3 A2 so that b 2 2 A2 D CA .t/. It follows that 1 ¤ Œb; t 2 A \ T2 D A2

and

1 D Œb 2 ; t D Œb; tb Œb; t D Œb; t2

so that hb; ti is minimal nonabelian with a noncentral involution t which yields that hb; ti Š D8 and o.b/ 4. All elements in the set A3 A2 are of order 4 so that A3 D hA3 A2 i is of exponent 4. If o.b/ D 2, then o.ba/ D 4, where a 2 A1 A0 and o.a/ D 4. In any case, there is an element w 2 A3 A2 of order 4 which together with hw; ti Š D8 gives w t D w 1 . Suppose that there is an involution u in A2 A1 . Then 1 ¤ Œu; v 2 A and we get 1 D Œu2 ; v D Œu; vu Œu; v D Œu; v2

and 1 D Œu; v 2 D Œu; vŒu; vv

so that Œu; v is a central involution in hu; vi. Hence hu; vi is minimal nonabelian with a noncentral involution u which gives hu; vi Š D8 . But then v u D v 1 and so Œu; v 62 A, a contradiction. We have proved that all element in A2 A0 are of order 4. In particular, w 2 2 A0 D CA .v/. We compute 1 D Œw 2 ; v D Œw; vw Œw; v D Œw; v2 and so Œw; v 2 T2 \A D A2 is an involution in A2 and therefore Œw; v 2 A0 D CA .v/. It follows that Œw; v 2 Z.hw; vi/ and thus hw; vi is minimal nonabelian satisfying hw; vi \ A hwi Š C4 so that hw; vi Š H2 and hw; vi \ A D hwi which im2 plies w v D w 1 . But then w t D w v D w, which contradicts a result from the previous paragraph. (ii) We prove that exp.G/ > 4. Suppose that this is false. In what follows we assume that exp.G/ D 4. In that case, G 0 Ã1 .G/ 1 .A/ and so Ã1 .G/ is elementary abelian. Since H2 is a subgroup of G, we have jÃ1 .G/j 4. (ii1) If a 2 A and x 2 G with Œa; x ¤ 1, then ha; xi is minimal nonabelian. In that case, ha; xi Š D8 if and only if a or x (or both) is an involution. Indeed, 1 D Œa; x 2 D Œa; xŒa; xx and so Œa; x 2 1 .A/ is an involution which commutes with a and x. Hence ha; xi0 D hŒa; xi and so, by Lemma 65.2(a), ha; xi is minimal nonabelian. (ii2) We have Ã1 .A/ ¤ ¹1º and Ã1 .A/ Z.G/. Assume that A is elementary abelian. Let H D ha; b j a4 D b 4 D 1; ab D a1 ; a2 D z; b 2 D ui be a subgroup of G so that H \ A D hu; zi Š E4 . Let t be an involution in the set

93

2-groups with metacyclic minimal nonabelian subgroups of exponent 4

3

A .CA .a/ [ CA .b//. By (ii1), we have ht; ai Š ht; bi Š D8 and so at D a1 D az and b t D b 1 D bu. We get .ab/t D .ab/zu and so Œt; ab ¤ 1. Again by (ii1), we obtain ht; abi Š D8 and so .ab/t D .ab/1 D .ab/.ab/2 D .ab/u, a contradiction. We have proved that Ã1 .A/ ¤ ¹1º. Let a 2 A and assume that a2 62 Z.G/. Then there is g 2 G such that Œa2 ; g ¤ 1. But then Œa; g ¤ 1 and so (ii1) implies that ha; gi is minimal nonabelian. In that case, a2 2 ˆ.ha; gi/ D Z.ha; gi/ and so Œa2 ; g D 1, a contradiction. We have proved that Ã1 .A/ Z.G/. (ii3) We have Ã1 .G/ Z.G/ which implies that each maximal abelian subgroup of G is also normal in G. Whenever x; y 2 G with Œx; y ¤ 1, then hx; yi is minimal nonabelian. Suppose that this is false. Then there is an element v 2 G A such that v 2 62 Z.G/ and v 2 2 1 .A/. In that case, there is w 2 G .Ahvi/ such that Œw; v 2 ¤ 1. By (ii1), hw; v 2 i Š D8 and so (replacing w with wv 2 if necessary), we see that there is an involution t 2 G.Ahvi such that ht; v 2 i Š D8 . We study the two-generator subgroup H D ht; vi and set A0 D H \ A. Since H=A0 Š E4 , we have ˆ.H / D A0 and A0 is elementary abelian. Hence Œt; v is an involution in A0 and 1 D Œt 2 ; v D Œt; vt Œt; v shows that t commutes with Œt; v. If Œt; v also commutes with v or tv, then we get Œt; v 2 Z.H /, ht; vi0 D hŒt; vi and so ht; vi is minimal nonabelian. But in that case, we conclude v 2 2 ˆ.H / D Z.H / and Œt; v 2 D 1, a contradiction. We have proved that CH .A0 / D A0 and so A0 (being elementary abelian) is a maximal normal abelian subgroup of H . We have already seen that D8 Š ht; v 2 i is a subgroup of H . Suppose that H2 is not a subgroup of H . In that case, we are in a position to use Theorem 90.1. If H is quasidihedral, then hvi is normal in H in which case v 2 2 Z.H /, a contradiction. It follows that H D H0 Z.H /, where H0 is extraspecial and Ã1 .Z.H // Z.H0 /. But in the last case, A0 D Ã1 .H / Z.H /, a contradiction. We have shown that H has both D8 and H2 as subgroups. By (ii2) applied to H and A0 , we get Ã1 .A0 / ¤ ¹1º, contrary to the fact that A0 is elementary abelian. We have proved that Ã1 .G/ Z.G/ and so G is of class 2 which implies that each maximal abelian subgroup of G is also normal in G. Let x; y 2 G with Œx; y ¤ 1. Then Œx; y 2 Ã1 .G/ Z.G/ and Ã1 .G/ 1 .A/ so that hx; yi0 D hŒx; yi is of order 2 and so hx; yi is minimal nonabelian. (ii4) D8 C4 is not a subgroup of G. Indeed, assume that ha; t j a4 D t 2 D 1; at D a1 i hbi, where hbi Š C4 , is a subgroup of G. Then Œt; ab D Œt; a ¤ 1 and (ii3) implies ht; abi Š D8 . It follows .ab/t D .ab/1 D a1 b 1 and .ab/t D at b t D a1 b, a contradiction. (ii5) Each maximal abelian subgroup A is of type .4; 2; : : : ; 2/. In particular, C4 C4 is not a subgroup of G. By (ii2) and (ii3), A is not elementary abelian, A is normal in G, and assume that A has a subgroup hai hbi Š C4 C4 . Since D8 is a subgroup of G, there is an involution t 2 G A. By (ii1) and (ii4), t either centralizes ha; bi or t inverts each element of ha; bi. Suppose that t centralizes ha; bi and let x be any element of order 4

4

Groups of prime power order

in A. By (ii1), t normalizes hxi (since Œt; x ¤ 1 implies that ht; xi Š D8 ) and again by (ii4), t centralizes x. But elements of order 4 in A generate A and so t centralizes A, a contradiction. Hence t inverts ha; bi and by (ii4), t inverts each element of order 4 in A. It follows that t inverts A. This implies that each minimal nonabelian subgroup of Ahti is isomorphic to D8 and so Ahti ¤ G and there are no involutions in G .Aht i/. We have G D .Aht i/S with .Aht i/ \ S D A, where S ¤ A and all elements in S A are of order 4. Let v 2 S A. If Œv; t ¤ 1, then (ii3) implies that hv; ti Š D8 . But then vt 2 G .Ahti/ and vt is an involution, a contradiction. It follows that t commutes with each element in S A and so ŒS; t D 1 and t centralizes A, a contradiction. We have proved that any maximal abelian subgroup A of G is of type .4; 2; : : : ; 2/. (ii6) If v 2 G A is such that v 2 62 Ã1 .A/, then v centralizes 1 .A/. Suppose that a is an element of order 4 in A so that (by (ii5)) A D hai1 .A/ with hai \ 1 .A/ D ha2 i D Ã1 .A/. Further, suppose that there exists t 2 1 .A/ such that Œv; t ¤ 1. Then hv; ti Š D8 and so v t D v 1 and t 0 D vt is an involution in Av. By (ii5), we get Œa; v ¤ 1 and so ha; vi Š H2 . We have Œav; t D Œv; t D v 2 ¤ 1 and so hav; ti Š D8 which gives v 2 D Œav; t D .av/2 D a2 v 2 Œv; a;

Œv; a D a2 ;

av D a1

and .av/2 D v 2 . We compute 0

.av/t D .av/vt D .a1 v/t D a1 v 1 D .av/a2 v 2 ¤ av since a2 ¤ v 2 and so hav; t 0 i Š D8 which implies 0

.av/t D .av/1 D .av/.av/2 D .av/v 2 ; a contradiction with the above. (ii7) If t is any involution in G A, then t does not centralize 1 .A/. Suppose that t centralizes 1 .A/ so that CA .t/ D 1 .A/. In that case, t inverts each element in A. If v is an element of order 4 in A, then A D hvi1 .A/ with hvi \ 1 .A/ D hv 2 i D Ã1 .A/ and v t D v 1 : Since jÃ1 .G/j 4, there is an element g 2 G .Aht i/ such that g 2 ¤ v 2 . By (ii6), g centralizes 1 .A/ and so Œg; v ¤ 1 which implies that hg; vi Š H2 . If v g D v 1 , then v gt D .v 1 /t D v and gt centralizes A, a contradiction. It follows that v g D vg 2 and so v gt D .vg 2 /t D v 1 g 2 D v.v 2 g 2 /: If Œg; t ¤ 1, then we get hg; ti Š D8 , gt is an involution which does not normalize hvi, a contradiction. Hence Œg; t D 1, o.gt/ D 4, .gt/2 D g 2 and so hv; gti Š H2 with hv; gti0 D hv 2 g 2 i which implies that v 2 g 2 is a square in hv; gti. This is a contradiction since g 2 and v 2 are the only involutions in hv; gti which are squares in hv; gti.

93

2-groups with metacyclic minimal nonabelian subgroups of exponent 4

5

(ii8) We consider a ﬁnal conﬁguration which will produce a contradiction. Let H D hg; v j g 4 D v 4 D 1; v g D v 1 i be a subgroup of G. Let A be a maximal abelian subgroup of G containing hv; g 2 i Š C4 C2 . Since g centralizes 1 .A/ (by (ii6)), g inverts on A, all elements in Ag are of order 4, and Ahgi Š H2 E2s , s 0. Since D8 is a subgroup of G, there is an involution t 2 G .Ahgi/. By (ii7), t does not centralize 1 .A/ and we see that also gt and vgt do not centralize 1 .A/. If Œg; t D 1, then .gt/2 D g 2 62 Ã1 .A/ and then (ii6) gives a contradiction. Similarly, if Œvg; t D 1, then .vgt/2 D .vg/2 D g 2 62 Ã1 .A/ and again (ii6) gives a contradiction. We have proved that Œg; t ¤ 1;

Œvg; t ¤ 1;

and so hg; ti Š hvg; ti Š D8

which gives g t D g 1 and .vg/t D .vg/1 . We compute .vg/t D .vg/1 D .vg/.vg/2 D .vg/g 2 D vg 1 D v t g t D v t g 1 ; and so v t D v. Let s 2 1 .A/ C1 .A/ .t/ so that hs; ti Š D8 , w D ts is an element of order 4 in the coset At and w t D w 1 . Since w commutes with v, (ii5) implies that w 2 D v 2 and hw; gi Š H2 with g w D g t s D g 1 . Hence gw is an element of order 4 and .gw/t D g 1 w 1 D .gg 2 /.ww 2 / D .gw/.g 2 v 2 / ¤ gw: By (ii3), hgw; ti Š D8 which implies .gw/t D .gw/1 . On the other hand, .gw/t D .gw/1 D .gw/.gw/2 D .gw/w 2 D .gw/v 2 ; which contradicts the above result .gw/t D .gw/.g 2 v 2 /. We have thus proved that exp.G/ > 4. (iii) It remains to determine the structure of G in case exp.G/ > 4. By (i) and Lemma 57.2, A is a maximal subgroup of G, all elements in G A are of order 4, exp.A/ 8 and for each g 2 G A, g inverts A=1 .A/ and on Ã1 .A/. If a 2 A with o.a/ D 8 and g 2 G A, then ag D a1 , 2 1 .A/ and so Œa; g D a1 ag D a2 , where o.a2 / D 4 and therefore jG 0 j > 2. This gives together with jGj D 2jZ.G/jjG 0 j (Lemma 1.1) that A is a unique abelian maximal subgroup of G. Since D8 is a subgroup of G, there is an involution t 2 G A. Let v be an element of order 4 in A. Then v t D v 1 , 2 1 .A/, and so Œv; t D v 2 2 1 .A/ and 1 D Œv; t 2 D Œv; tŒv; tt which gives Œv; tt D Œv; t. If Œv; t ¤ 1, then Œv; t is an involution commuting with v and t and so hv; ti0 D hŒv; ti which implies that hv; ti is minimal nonabelian (Lemma 65.2(a)). Hence in that case, hŒv; ti Š D8 and so v t D v 1 . We have proved that t either centralizes or inverts each element of order 4 in A. Assume that A0 D CA .t/ D Z.G/ is not elementary abelian in which case we have exp.A0 / D 4. Since t inverts Ã1 .A/, there is an element v of order 4 in A A0 such

6

Groups of prime power order

that v t D v 1 and v 2 2 A0 . Let w be an element of order 4 in A0 . If w 2 ¤ v 2 , then vw is an element of order 4 in A A0 and so .vw/t D .vw/1 D v 1 w 1 D v t w t D v 1 w t which implies w t D w 1 , a contradiction. Hence w 2 D v 2 , A0 is of type .4; 2; : : : ; 2/ and Ã1 .A0 hvi/ D hv 2 i. Suppose that there is an element y of order 4 in A .A0 hvi/ so that y t D y 1 and y 2 2 A0 . If y 2 D v 2 , then yv is an involution in A .A0 hvi/ centralized by t, a contradiction. If there exists an involution u 2 A .A0 hvi/, then o.uv/ D 4 and uv 2 A .A0 hvi/. Hence in case that 2 .A/ > .A0 hvi/ we have an element y of order 4 in 2 .A/ .A0 hvi/ with y 2 2 A0 and y 2 ¤ v 2 . For each l 2 A0 hvi we obtain .yl/2 D y 2 l 2 with l 2 2 hv 2 i, and so all elements in y.A0 hvi/ are of order 4 and they are inverted by t . Since hy.A0 hvi/i D A0 hy; vi, t inverts A0 hvi. This is a contradiction because for an element w of order 4 in A0 , we have w t D w. We have proved that 2 .A/ D A0 hvi and A0 hv; ti D A0 hv; ti with hv; ti Š D8 ;

A0 \ hv; ti D hv 2 i D Z.hv; ti/ and Ã1 .A0 / D hv 2 i

so that each minimal nonabelian subgroup in A0 hv; ti is isomorphic to D8 or Q8 . Since H2 is a subgroup of G, there is a 2 A .A0 hvi/ such that .at/2 2 A0 hv 2 i and v at D v 1 so that hat; vi Š H2 with hat; vi0 D hv 2 i. Note that vw (with w 2 A0 and w 2 D v 2 ) is an involution in .A0 hvi/ A0 and .vw/at D .vw/t D .vw/v 2 and so hat; vwi is the nonmetacyclic minimal nonabelian of order 24 , a contradiction. We have proved that A0 D CA .t/ D Z.G/ is elementary abelian. Then all elements of order 4 in 2 .A/ are inverted by t and so t inverts 2 .A/ and therefore we have CA .t/ D Z.G/ D 1 .A/. Since H2 is not a subgroup of 2 .A/hti, there are elements a 2 A 2 .A/ and v 2 2 .A/ 1 .A/ such that .at /2 D ¤ 1;

2 1 .A/;

v at D v 1

and ¤ v 2 :

We have atat D and so at D a1 . If Ã1 .A/ is cyclic, then v 2 2 Ã1 .A/ and so we must have in that case 62 Ã1 .A/. (iv) We have obtained 2-groups stated in Theorem 93.1. It is easy to check that all these 2-groups satisfy the assumptions of our theorem. Indeed, let G be a 2-group described in the second paragraph of Theorem 93.1. Since t inverts each element in 2 .A/, we have CA .t/ D 1 .A/ D Z.G/. Let xt be any element in G A, where x 2 A. Then xt also inverts 2 .A/ and on A=1 .A/. We get .xt/2 D xtxt D xx t D x.x 1 s/ D s, where s 2 1 .A/ so that o.xt/ 4. Let w be an element of order 4 in Ã1 .A/ such that 62 hwi. From at D a1 we deduce .at/2 D ¤ 1 and so hwi \ hat i D ¹1º together with w at D w t D w 1 implies that

93

2-groups with metacyclic minimal nonabelian subgroups of exponent 4

7

hat; wi Š H2 . Also, hw; ti Š D8 and therefore we have proved that both D8 and H2 actually appear as subgroups of G. Let H be any minimal nonabelian subgroup of G so that H 6 A. Let c 2 H A and d 2 .H \ A/ ˆ.H / so that H D hc; d i and o.c/ 4. Suppose that o.d / 8. In that case, d c D d 1 r with r 2 1 .A/ so that Œd; c D d 2 r is of order 4, a contradiction. Hence o.d / 4 and in fact o.d / D 4 since 1 .A/ Z.G/. Since c inverts 2 .A/, we have d c D d 1 . If o.c/ D 2, then H D hc; d i Š D8 . In case o.c/ D 4 and c 2 D d 2 , we get H Š Q8 . If o.c/ D 4 and c 2 ¤ d 2 , then H Š H2 and we are done.

94

Nonabelian 2-groups all of whose minimal nonabelian subgroups are nonmetacyclic and have exponent 4

We consider here nonabelian 2-groups G all of whose minimal nonabelian subgroups are nonmetacyclic and have exponent 4. Then it is easy to prove that exp.G/ D 4. In fact, we prove the following result. Theorem 94.1. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are nonmetacyclic and have exponent 4. Then exp.G/ D 4, 1 .G/ is elementary abelian and if A is a maximal normal abelian subgroup of G containing 1 .G/, then ˆ.G/ 1 .A/ is elementary abelian and Ã1 .A/ Z.G/. Proof. Since D8 is not a subgroup of G, 1 .G/ is elementary abelian. Let A be a maximal normal abelian subgroup of G containing 1 .G/. Suppose that exp.G/ > 4. By Lemma 57.2, jG W Aj D 2, all elements in G A are of order 4 and if x is one of them, then x inverts Ã1 .A/. Let y be an element of order 4 in Ã1 .A/. Then y x D y 1 which implies that hx; yi Š Q8 or hx; yi Š H2 , a contradiction. We have proved that exp.G/ D 4 and so exp.A/ 4 and G=A is elementary abelian. Then ˆ.G/ D Ã1 .G/ 1 .A/ and so ˆ.G/ is elementary abelian. If a 2 A and g 2 G with Œa; g ¤ 1, then ha; gi is minimal nonabelian. Indeed, Œa; g is an involution in 1 .A/ and 1 D Œa; g 2 D Œa; gŒa; gg so that Œa; g commutes with a and g and therefore ha; gi0 D hŒa; gi. By Lemma 65.2(a), ha; gi is minimal nonabelian. We prove that Ã1 .A/ Z.G/. Indeed, let a 2 A be such that a2 62 Z.G/. Then there is g 2 G with Œa2 ; g ¤ 1. But then Œa; g ¤ 1 and so, by the previous paragraph, ha; gi is minimal nonabelian. In that case, a2 2 ˆ.ha; gi/ D Z.ha; gi/ and so Œa2 ; g D 1, a contradiction. Remark. It is easy to see that the group H16 C4 contains both nonmetacyclic minimal nonabelian 2-groups of exponent 4, namely, H16 D ha; b j a4 D b 2 D 1; Œa; b D z; z 2 D Œa; z D Œb; z D 1i (of order 16) and H32 D ha; b j a4 D b 4 D 1; Œa; b D z; z 2 D Œa; z D Œb; z D 1i (of order 32) and each minimal nonabelian subgroup of H16 C4 is nonmetacyclic.

94

2-groups with nonmetacyclic minimal nonabelian subgroups of exponent 4

9

We prove a lemma for p D 2 which might be useful for future similar investigations. The corresponding case for p > 2 was treated in Lemma 28.2. Lemma 94.2. Let G be a 2-group such that all subgroups of ˆ.G/ are normal in G. Then ˆ.G/ is abelian and jG W CG .ˆ.G//j 2. If jG W CG .ˆ.G//j D 2, then each element x 2 G .CG .ˆ.G// is of order 4 and x inverts each element in ˆ.G/. Proof. By assumption, ˆ.G/ is Dedekindian. Suppose that ˆ.G/ is nonabelian. Then ˆ.G/ possesses a subgroup Q Š Q8 . Since Q is normal in G and each cyclic subgroup of Q is normal in G, it follows that no element in G induces an outer automorphism on Q and so G D QCG .Q/ with Q \ CG .Q/ D Z.Q/. If M is a maximal subgroup of G containing CG .Q/, then Q 6 M , which contradicts the fact that Q ˆ.G/. We have proved that ˆ.G/ is abelian. Suppose that ˆ.G/ is cyclic. Let P be a cyclic subgroup of G which contains ˆ.G/ with jP W ˆ.G/j D 2. Let x 2 G P so that x 2 2 ˆ.G/ and P is a maximal cyclic subgroup of index 2 in P hxi. By the structure of such groups, x either centralizes ˆ.G/ or inverts ˆ.G/. Now assume that ˆ.G/ is noncyclic so that ˆ.G/ D Z1 Zn , where Zi are cyclic and n > 1. Set Zi D hai i and

K D Z1 Zi 1 ZiC1 Zn

and consider G=K. By the above, for each x 2 G we have aix D ai k or aix D ai1 k for some k 2 K. But Zi is normal in G and so k D 1. We have proved that each x 2 G either centralizes or inverts each Zi , i D 1; 2; : : : ; n. Suppose that ˆ.G/ possesses a subgroup hv; wi Š C4 C4 such that for an element x 2 G we have v x D v 1 and w x D w. In that case, .vw/x D v 1 w D v 2 .vw/, which contradicts the fact that hvwi is normal in G. We have proved that each element x 2 G either centralizes ˆ.G/ or inverts each element in ˆ.G/. This gives jG W CG .ˆ.G//j 2. Suppose that jG W CG .ˆ.G//j D 2 and let x 2 G .CG .ˆ.G//. Since x inverts ˆ.G/ and x 2 2 ˆ.G/, it follows that o.x 2 / 2 which gives o.x/ 4. The proof is complete.

95

Nonabelian 2-groups of exponent 2e which have no minimal nonabelian subgroups of exponent 2e

Here we solve a general problem (Nr. 1475 for p D 2) by classifying the title groups. More precisely, we prove the following result. Theorem 95.1. Let G be a nonabelian 2-group of exponent 2e (e 3) which does not possess a minimal nonabelian subgroup of exponent 2e . Then G has a unique maximal normal abelian subgroup A. The factor group G=A is either cyclic or generalized quaternion and each element in G A is of order < 2e . Also, G possesses at least one metacyclic minimal nonabelian subgroup and jCG .1 .A// W Aj 2. Proof. Let G be a 2-group satisfying the assumptions of Theorem 95.1 and let A be any maximal normal abelian subgroup of G and g 2 G A. Set 2m D max¹exp.H / j H < G is minimal nonabelianº so that m 2 and m < e. Further, set T D Ahgi, A0 D CA .g/ so that A0 < A and T0 D A0 hgi is abelian, T0 < T and T =A is cyclic. Let T1 T with T1 > T0 and jT1 W T0 j D 2. Set A1 D T1 \ A so that T1 D A1 hgi. Let a 2 A1 A0 so that a2 2 A0 and Œa; g 2 A \ T0 D A0 Z.T1 /. We have 1 D Œa2 ; g D Œa; ga Œa; g D Œa; g2 so that Œa; g is an involution commuting with a and g and therefore ha; gi is minimal nonabelian (Lemma 65.2(a)). In particular, we get o.a/ 2m and o.g/ 2m . Since A1 D hA1 A0 i and A1 is abelian, we have exp.A1 / 2m . We have proved that for each g 2 G A, o.g/ 2m and A0 D CA .g/ is of exponent 2m . Suppose at the moment that g 2 2 A and let x be an element in A. Then we have also .xg/2 2 A. Set x g D x 1 w for some w 2 A. In that case, w D xx g D x.g 1 xg/ D xg 2 gxg D g 2 .xg/2 is an element of order 2m1 in A. We have shown that an element g 2 G A with g 2 2 A inverts A=m1 .A/. Thus for each x 2 A, .x 2

m1

/g D .x g /2

m1

D .x 1 w/2

m1

D x 2

m1

95

2-groups of exponent 2e with no minimal nonabelian subgroup of exponent 2e 11

and so g inverts Ãm1 .A/. Since exp.A/ D exp.G/ D 2e and e > m, it follows that Ãm1 .A/ contains elements of order 4. Hence G=A contains only one subgroup of order 2 which implies that G=A is either cyclic of order 2m or generalized quaternion of order 2mC1 (m 2). Since g 2 G A with g 2 2 A inverts a cyclic subgroup of order 4 in Ãm1 .A/, it follows that G possesses a metacyclic minimal nonabelian subgroup. Suppose that G possesses another maximal normal abelian subgroup B, B ¤ A. Then all elements in G A are of order < 2e , all elements in A B D A .A \ B/ are of order < 2e and since an element b 2 B .A \ B/ centralizes A \ B, we have also exp.A \ B/ < 2e . But then exp.G/ < 2e , a contradiction. Suppose that A satisﬁes jCG .1 .A// W Aj 4. Let H=A Š C4 be a subgroup of order 4 in CG .1 .A//=A and let g be an element in H A such that hgi covers H=A. Set A0 D CA .g/, T0 D A0 hgi so that T0 is abelian and A0 1 .A/. Further, set S D 2 .Ãm1 .A// so that S is normal in H , exp.S/ D 4 and g 2 inverts S . Therefore S \A0 D 1 .S / and S T0 is a subgroup of H . Let T be a subgroup of ST0 containing T0 such that jT W T0 j D 2. Suppose that w is an element in .T \ S/ 1 .S/ so 2 that o.w/ D 4 and w 2 2 1 .S/ A0 and w g D w 1 . We now study the subgroup hw; gi, where 1 ¤ Œw; g 2 T0 \ A D A0 . Also, 1 D Œw 2 ; g D Œw; gw Œw; g D Œw; g2 and so Œw; g is an involution commuting with both w and g. It now follows from Lemma 65.2(a) that hw; gi is minimal nonabelian. But then g 2 2 ˆ.hw; gi/ D Z.hw; gi/ and so g 2 centralizes w, a contradiction. The theorem is proved. Theorem 95.2. Let G be a nonabelian 2-group of exponent 2e (e 3) which does not possess a minimal nonabelian subgroup of exponent 2e . Let A be a unique maximal normal abelian subgroup of G (Theorem 95.1). Suppose that G possesses a cyclic subgroup Z with Z \ A D ¹1º. Then G also possesses a nonmetacyclic minimal nonabelian subgroup. Proof. Set Z D hvi, v 2 D t and A0 D CA .v/ so that the subgroup T0 D A0 hvi is abelian. By Theorem 95.1, 1 .A/ 6 A0 and set T1 D 1 .A/T0 . Let T2 be a subgroup of T1 containing T0 such that jT2 W T0 j D 2. We have T2 D .1 .A/ \ T2 /T0 and take an involution u 2 .1 .A/ \ T2 / 1 .A0 /. Then Œu; v 2 1 .A/ \ T0 D 1 .A0 / so that Œu; v is an involution commuting with u and v and therefore hu; vi is minimal nonabelian. We have t D v 2 2 Z.hu; vi/ so that hu; Œu; v; t i Š E8 and thus hu; vi is nonmetacyclic.

96

Groups with at most two conjugate classes of nonnormal subgroups

The non-Dedekindian p-groups all of whose nonnormal subgroups are conjugate were classiﬁed in 58. Below we offer another proof. Next we classify the p-groups with exactly two conjugate classes of nonnormal subgroups. Both these results are due to Otto Schmidt. If G is a minimal nonabelian p-group, then (see Exercise 1.8(a) or Lemma 65.1) G D ha; bi, where m

n

m

n

(MA1) ap D b p D c p D 1, Œa; b D c, Œa; c D Œb; c D 1, jGj D p mCnC1 , G D hbi .haihci/ D hai .hbihci/ is nonmetacyclic with kernels haihci, hbi hci, respectively. (MA2) ap D b p D 1, ab D a1Cp cyclic with kernel hai, m > 1.

m1

, jGj D p mCn and G D hbi hai is meta-

(MA3) G Š Q8 , the ordinary quaternion group. Let nc.G/ be the number of conjugate classes of nonnormal subgroups of a group G and let no.G/ be the number of orders of nonnormal subgroups of G. To facilitate the proof, we ﬁrst prove some lemmas. Lemma 96.1. Let G D ha; bi be a minimal nonabelian p-group as in (MA1) or (MA2). (a) All nonnormal cyclic subgroups of G have the same order if and only if one of the following holds: (a1) G is as in (MA2) with m n. (a2) G is as in (MA1) with m D n. (b) All nonnormal subgroups of G have the same order if and only if one of the following holds: (b1) G 2 ¹D8 ; S.p 3 /º. (b2) G is as in (MA2) with m n. (c) All nonnormal cyclic subgroups of G of same order are conjugate if and only if G Š Mpt . (d) All nonnormal subgroups of G are conjugate if and only if G Š Mpt .

96

Groups with at most two conjugate classes of nonnormal subgroups

13

Proof. We use the same notation as in (MA1) and (MA2). Write A D hai and B D hbi. Let H < G be nonnormal; then G 0 6 H so that H \ G 0 D ¹1º since jG 0 j D p. (a) Suppose that all nonnormal cyclic subgroups of G have the same order. Let G be metacyclic as in (MA2) and let H be cyclic. If n > m and m .B/ D hxi, m1 then .ax/b D .ax/ap 62 haxi hence B and D D haxi are not G-invariant, and they are cyclic of orders p n and p m ¤ p n , respectively, a contradiction. It remains to show that if n m, then G satisﬁes the hypothesis. We have H \A D ¹1º since G 0 < A has order p and A is cyclic, so H , being isomorphic to a subgroup of G=A, has order p n . Since n1 .G/ Z.G/, our G satisﬁes the hypothesis and so it is as in (a1). Suppose that G is nonmetacyclic as in (MA1). Since A and B are not G-invariant, we must have m D n and G is as in (a2). Conversely, if G is as in (a2) with m D n, then, since m1 .G/ Z.G/, our G satisﬁes the hypothesis. (b) Suppose that all nonnormal subgroups of G have the same order; then H is cyclic since it is not generated by its G-invariant maximal subgroups. Thus one can use (a). Let G be metacyclic as in (a1); then m n. It remains to show that such G satisﬁes the hypothesis. In view of H \ A D ¹1º, we obtain that jH j jG W Aj D p n . Since n1 .G/ Z.G/, we have jH j D p n , and G is as in (b2). Now let G be nonmetacyclic as in (a2) so m D n. Assume that m > 1. Then we see that B 1 .A/ is nonnormal and noncyclic, a contradiction. Thus G is as in (b1). (c) Suppose that all nonnormal cyclic subgroups of G of same order are conjugate; then these subgroups have the same order. Let G be nonmetacyclic as in (MA1) and let m; n > 1. Assume that m > n. Let n .A/ D hxi. Then the cyclic subgroups B and hbxi of the same order p n are not normal, and since BG ¤ hbxiG , we conclude that B and hbxi are not conjugate. Thus m n. Similarly, n m so m D n. Then the cyclic subgroups A and B of the same order p m are neither normal nor conjugate, a contradiction. Now let G be nonmetacyclic as in (MA1) and m > n D 1. Then G has exactly p 2 nonnormal subgroups of order p. Since any nonnormal subgroup has p conjugates, there are in G two nonconjugate subgroups of order p. Let G be metacyclic as in (MA2). If n D 1, then G Š MpmC1 satisﬁes the condition. Now let n > 1. If n < m, set n .A/ D hxi; then the cyclic subgroups B and hbxi of order p n are neither G-invariant (otherwise, G D A hbxi is abelian) nor conjugate since their cores in G are different. If n m, then the cyclic subgroups B and hap bi of order p n are neither G-invariant nor conjugate since their cores in G are different, a contradiction. Statement (d) follows from (c). Below we present an alternate proof of Theorem 1.23. Lemma 96.2 (= Theorem 1.23 (Passman)). Assume G is a non-Dedekindian p-group. Then there is a G-invariant subgroup R of index p in G 0 such that G=R is non-Dedekindian.

14

Groups of prime power order

Proof. If p > 2, the result is obvious as Dedekindian groups of odd order are abelian. Now let p D 2. Assume that for every G-invariant subgroup K of index 2 in G 0 , the quotient group G=K is Dedekindian. We want to prove that G is itself Dedekindian. One may assume that K > ¹1º. Assuming that G is a minimal counterexample, we get jKj D 2 (this is due to the fact that the derived subgroup of a nonabelian Dedekindian group is of order 2; see Theorem 1.20) so jG 0 j D 4. It follows that G is not minimal nonabelian (Lemma 65.1) and there is in G=K a subgroup Q=K Š Q8 (Theorem 1.20). Since Q is not of maximal class (Theorem 1.2), we get jQ0 j D 2 (Proposition 1.6), and so G 0 D Q0 K Š E4 and G 0 Z.G/. Let A < G be minimal nonabelian. Assume that jAj > 8; then K < A (otherwise, A is isomorphic to a subgroup of G=K, which is impossible since all minimal nonabelian subgroups of the nonabelian Dedekindian group G=K are isomorphic to Q8 by Theorem 1.20). Assume that A=K is abelian; then it is of type .4; 2/ (here we use Theorem 1.20 again), and hence A0 D K. In this case, G 0 =K D Ã1 .G=K/ < A=K, and we conclude that G 0 < A, and so A G G. Then A=Q0 is nonabelian in view of Q0 \ K D ¹1º, hence A=Q0 Š Q8 by hypothesis. Thus one can assume from the start that A=K Š Q8 . In this case, A Š H2;2 D hx; y j x 4 D y 4 D 1; x y D x 3 i since j1 .A/j D 4 and exp.A/ D 4 (see Lemma 65.1 or (MA1) and (MA2)). We know that all subgroups of order 2 are characteristic in A so normal in G. Then A=hy 2 i Š D8 so G=hy 2 i is not Dedekindian, contrary to the assumption. Thus jAj D 8 so that any minimal nonabelian subgroup of G has order 8. Assume that A Š D8 . Let L < G 0 of order 2 be such that L 6 A. Then AL=L Š D8 so G=L is not Dedekindian, a contradiction. Thus A Š Q8 . Then A G G (indeed, any abelian subgroup of type .2; 2/ of a Dedekindian group G=A0 is normal). Write C D CG .A/. Then G=C is a subgroup of a Sylow 2-subgroup of Aut.A/ Š S4 , which is isomorphic to D8 . It follows that G=C Š E4 , and we conclude that G D A C . Since jG 0 j D 4, C is nonabelian. Let B < C be minimal nonabelian; then, by the above, B Š Q8 . If A \ B > ¹1º, then A B is extraspecial so it contains a subgroup Š D8 (Appendix 16), contrary to what has been proved above. Thus we get A \ B D ¹1º. If U < A0 B 0 D G 0 is of order 2 and A0 ¤ U ¤ B 0 , then AB=U is an extraspecial subgroup of order 25 of the Dedekindian group G=U , a ﬁnal contradiction. The presented proof of Lemma 96.2 is longer than the original one. Our aim here was to show an application of minimal nonabelian subgroups. The following lemma is a partial case of Theorem 96.7. Lemma 96.3. Let G be a p-group with jG 0 j D p and nc.G/ D 1. Then G Š Mpn . Proof. We use induction on jGj. In view of Lemma 96.1(c), one may assume that G is not minimal nonabelian. Let H < G be minimal nonabelian. Then G D H CG .H /

96

Groups with at most two conjugate classes of nonnormal subgroups

15

(Lemma 4.3) so CG .H / 6 H . Since nc.H / nc.G/ D 1 (see the decomposition of G in the previous sentence), we get H 2 ¹Q8 ; Mpn º by induction. (i) H Š Q8 . Since G is non-Dedekindian, we get exp.CG .H // > 2. Suppose that L D hxi CG .H / is cyclic of order 4. If L\H D ¹1º and hai, hbi are distinct cyclic subgroups of order 4 in H , then the nonnormal subgroups haxi, hbxi do not belong to the same G-class1 , and this is a contradiction. Now assume that L \ H D Z.H /. Then the central product L H (of order 16) has exactly six nonnormal subgroups of order 2, and these subgroups do not belong to the same conjugate class since 6 is not a power of 2 (note that HL G G since G 0 < H < HL), a contradiction. Thus G has no subgroups Š Q8 . (ii) Let H Š Mpn . Assume that there is in G a subgroup L of order p with L 6 H . Since H has a nonnormal subgroup B of order p, nc.G/ D 1 and H GG, we get LGG. Then HL D H L, and this subgroup has exactly p 2 nonnormal subgroups of order p which do not belong to the same G-class (indeed, every nonnormal subgroup of G of order p has exactly p conjugates). We get a contradiction since HLGG. Thus we have 1 .G/ D 1 .H / Š Ep2 . If p > 2, then, by Theorem 13.7 or Theorem 69.4, G is metacyclic. By Lemma 65.2(a), since jG 0 j D p, the group G is minimal nonabelian, contrary to the assumption. Now let p D 2. By Lemma 65.2(a), G is nonmetacyclic (see the previous paragraph). Therefore, by Theorem 66.1, 2 .G/ 6 H . It follows that there is a cyclic L < G of order 4 such that L 6 H . Then L G G by hypothesis so L \ H D H 0 (recall that Z.H / is cyclic). Write F D L2 .H /. Then F=H 0 Š E8 . Since j1 .F /j D 4, it follows that F is nonabelian. Then it contains a minimal nonabelian subgroup of order 8, contrary to what has been proved above. If G is nilpotent and nc.G/ D 1, then, as is easily seen, G is primary. Therefore, in Theorem 96.4 we consider p-groups. Theorem 96.4 (O. Schmidt [Sch3]). Let G be a p-group. If nc.G/ D 1, then we have G Š MpnC1 . Proof. We proceed by induction on G. In view of Lemma 96.3, one may assume that jG 0 j > p; then G is neither Dedekindian nor minimal nonabelian. Therefore, by Lemma 96.2, there is in G 0 a G-invariant subgroup R of order p such that G=R is not Dedekindian. By induction, G=R Š Mpn . Let U=R and V =R be two distinct cyclic subgroups of index p in G=R. Then U; V 2 1 are distinct abelian so U \ V D Z.G/. Since R < G 0 ˆ.G/, we get d.G/ D d.G=R/ D 2 so G is minimal nonabelian (Lemma 65.2(a)), a ﬁnal contradiction. Below we freely use properties of minimal nonnilpotent groups which are listed in Theorem A.22.1. 1 We

have haxib D ha1 xi 62 ¹haxi; hbxiº, and the same is true for hbxia .

16

Groups of prime power order

Supplement to Theorem 96.4 (O. Schmidt [Sch3]). Let G be a nonnilpotent group. Then nc.G/ D 1 if and only if G is minimal nonabelian with derived subgroup of prime order. Proof. Let H G be minimal nonnilpotent. Then H D PQ, where P 2 Sylp .G/ is cyclic and H 0 D Q 2 Sylq .G/, p ¤ q. In this case, all nonnormal subgroups of G are conjugate with P . It follows that P is maximal in H and all subgroups of Q are normal in G so jQj D q. Since NG .P / is nonnormal in G, we get NG .P / D P so H D G as H is normal in G and, by Frattini’s lemma, G D H NG .P / D HP D H . Next we classify the p-groups G with nc.G/ D 2. Lemma 96.5. If G D ha; bi is a minimal nonabelian p-group and the number of conjugate classes of nonnormal subgroups of G is equal to nc.G/ D 2, then p D 2 and m m1 G 2 ¹D8 ; H2;m ; m 2º. Here H2;m D ha; b j a2 D b 4 D 1; ab D a1C2 i. Proof. Let G be as in (MA1) or (MA2). Write A D hai, B D hbi. (i) Let G be nonmetacyclic as in (MA1). The subgroups A and B are neither G-invariant nor conjugate. The normal closures AG D A G 0 , B G D B G 0 are of order pjAj, pjBj, respectively. Suppose that m > 1. Then H D B1 .G/ D B E, where E Š Ep2 . Let L < E be of order p such that B L ¤ B G 0 . Then B L is not normal in G (otherwise, B D .B L/ \ .B G 0 / G G), and we get nc.G/ > 2, a contradiction. Similarly, we consider the case n D 1 so that jGj D p 3 . But D8 is metacyclic and nc.S.p 3 // D p > 2 (recall that S.p 3 / is nonabelian of order p 3 and exponent p > 2). (ii) Now let G be metacyclic as in (MA2); n > 1 (otherwise, G Š MpmC1 and nc.G/ D 1). Since G 0 < 1 .G/ Š Ep2 , every nonnormal subgroup of G being abelian is cyclic. If H < G is cyclic of order p n such that G 0 6 H , then G D H A is a semidirect product so H is not normal in G. (ii1) Assume that n > m. Then B1 D habi of order p n is neither a-invariant (otherwise, G D haiB1 would be abelian) nor conjugate with B since B1 6 B G D B G 0 . Suppose that m .B/ D hyi; then A1 D hayi of order p m is not b-invariant and jA1 j < jBj D jB1 j so nc.G/ 3, a contradiction. (ii2) Let n D m. If U=G 0 < G=G 0 is cyclic of order p m , then U D G 0 V so the cyclic subgroup V is not normal in G since G 0 6 V so that A \ V D ¹1º. There are in G=G 0 exactly p 2m1 p 2m2 D p m1 p m1 .p 1/ distinct cyclic subgroups Ui =G 0 of order p m . If Ui D G 0 Vi , then the subgroups V1 ; : : : ; Vpm1 that have trivial intersections with A are not G-invariant and pairwise nonconjugate in G since all Ui =G 0 G G=G 0 are distinct. It follows that p m1 2 so p D 2 D m; then G Š H2;2 .

96

Groups with at most two conjugate classes of nonnormal subgroups

17

(ii3) Let n < m. As we have noticed, n > 1. The subgroup n .G/ is abelian of type .p n ; p n / so it contains exactly p 2n p 2n2 D p n1 .p C 1/ p n1 .p 1/ cyclic subgroups of order p n and exactly p n1 of them contain G 0 . It follows that G has exactly p n1 .p C1/p n1 D p n nonnormal cyclic subgroups of order p n which are partitioned in p n1 classes of conjugate subgroups. We have nc.G/ D 2 p n1 so n D 2 and p D 2, and we get G D H2;m . n1

n2

Let G D M C , where M D ha; b j ap D b p D 1; ab D a1Cp i Š Mpn n2 and C D hci. Set D ap . Then Z.G/ D h i hci contains exactly p C 1 subgroups of order p and c1 .G/ D p 2 C p C 1. Therefore exactly c1 .G/ .p C 1/ D p 2 subgroups of order p of G are non-G-invariant so they belong to p distinct G-classes. Also the subgroup hbihci of order p 2 is non-G-invariant so that nc.G/ 1Cp > 2. Lemma 96.6. If G is a p-group such that jG 0 j D p and nc.G/ D 2, then p D 2 and G 2 ¹D8 ; H2;m º. Proof. In view of Lemmas 96.3 and 96.5, one may assume that G is not minimal nonabelian. Let A < G be minimal nonabelian; then A G G since A0 D G 0 in view of jA0 j D p D jG 0 j, so G D A C , where C D CG .A/ (Lemma 4.3). By assumption, we have C 6 A. Two subgroups of A are G-conjugate if and only if they are A-conjugate. Therefore nc.A/ nc.G/ D 2 so A 2 ¹Q8 ; Mpn ; D8 ; H2;m º by Lemma 96.3 and Theorem 96.5. (i) Let A Š D8 . Then A contains two nonnormal subgroups L1 ; L2 of order 2 which are not conjugate in A, and so neither in G. If L of order 2 is such that L 6 A (L exists by Theorem 1.17(b) since G is not of maximal class), then L G G; then L3 D L1 L is not normal in G since A \ L3 D L1 , and we get nc.G/ > 2, a contradiction. Thus G has no subgroup Š D8 . (ii) Let A Š Q8 . Since G is not Dedekindian, exp.C / > 2. Let hli D L C be cyclic of order 4. If A \ L D Z.A/, then A L Š D8 L contains a subgroup Š D8 , contrary to (i). Thus A \ L D ¹1º hence H D AL D A L. Let hai, hbi and hci be distinct (cyclic) subgroups of order 4 in A. Then the subgroups hali, hbli are not c-invariant and hcli is not a-invariant; since these three subgroups are not G-conjugate in pairs, we get nc.G/ > 2, a contradiction. Thus G has no subgroup Š Q8 . (iii) Suppose that m

A D ha; b j a4 D b 2 D 1; ab D a3 i Š H2;m : Then B D hbi and B1 D hba2 i are neither conjugate nor A-invariant so are not G-invariant. Assume that there is L < G of order 2 such that L 6 A. Then L < Z.G/. Since B L and B1 L are not G-invariant (for example, A \ .B L/ D B is not A-invariant), we get nc.G/ > 2. Thus L does not exist so 1 .G/ D 1 .A/ Š E4 .

18

Groups of prime power order

Since C 6 A and A \ C D Z.A/ D ˆ.A/, it follows that d.G/ > 2 so G is nonmetacyclic. Let M < G be minimal nonmetacyclic. Since j1 .M /j j1 .G/j D 4 and all minimal nonabelian subgroups of G have order > 8, we obtain that jM j D 25 (Lemma 66.1(d)). Because of jM 0 j D 4 > 2 D jG 0 j, we get a contradiction. Next we assume that G has no subgroup Š H2;m . (iv) Now suppose that A is as in (MA2) with n D 1; then A Š MpmC1 , and B D hbi is not G-invariant. Let L < G be a subgroup of order p not contained in A. Write H D L1 .A/; then jH j D p 3 . By (i), we have exp.H / D p. Assume that H is nonabelian. Then p 2 C p noncentral subgroups of H of order p belong to p C 1 distinct G-classes so nc.G/ p C 1 > 2, a contradiction. Now assume that H Š Ep3 . Then we get jH \ Z.G/j 2 ¹p; p 2 º. In the ﬁrst case, we obtain a contradiction as above. Let jH \ Z.G/j D p 2 . Then p 2 subgroups of H of order p that are not contained in Z.G/ belong to p distinct G-classes, and we conclude that p D 2. As H 6 Z.G/, it contains also a non-G-invariant subgroup of order 4; then nc.G/ 2 C 1 > 2, a contradiction. Thus 1 .G/ D 1 .A/ Š Ep2 . Now it follows from (i) and (ii) that C is cyclic. Since A \ C D Z.A/ D Ã1 .A/, we get Ã1 .A/ < C . Let C0 C be (cyclic) of order o.a/ (note that hap i < C and C is cyclic). Write F D A0 C0 . Then F D A0 K, where jKj D p (basic theorem on abelian p-groups). It follows from F 6 A that K 6 A. Since jKj D p, we get a contradiction to the result of the previous paragraph. Theorem 96.7 (O. Schmidt [Sch4]). If a group G is nilpotent and nc.G/ D 2, then either p D 2 and G 2 ¹D8 ; Q16 ; H2;m º or G Š Mpn Cq , where q ¤ p are any primes. Proof. It is easily checked that the groups from the conclusion satisfy the hypothesis. Let G D P Q, where P 2 Sylp .G/ with nc.P / > 0 and Q > ¹1º is a p 0 -group. If ¹1º L Q and U < P is not P -invariant, then L1 D U L is not G-invariant since P \ L1 D U . It follows that jQj D q, a prime, and nc.P / D 1 so P Š Mpn by Theorem 96.4. If G D P Q is a p-group, Q > ¹1º, then nc.G/ > 2 (Lemma 96.6). Next we assume that G is a p-group; then G has no nontrivial direct factors by the previous paragraph. If G has a cyclic subgroup of index p, then G 2 ¹D8 ; Q16 º (note that nc.SD16 / D 3). Next we assume that G has no cyclic subgroup of index p. In view of Lemma 96.6, one may assume that jG 0 j > p. By Lemma 96.2, there exists in G 0 a G-invariant subgroup R of index p such that G=R is not Dedekindian. Then we have j.G=R/0 j D p and 1 nc.G=R/ nc.G/ D 2 so, by Theorem 96.4 and Lemma 96.6, G=R 2 ¹H2;m .p D 2/; Mpn ; D8 º. Since R < G 0 ˆ.G/, we get d.G/ D d.G=R/ D 2. Below we consider these three possibilities. (i) If G=R Š D8 , then jG W G 0 j D 4 so G is of maximal class by Taussky’s theorem, a contradiction since, by assumption, G has no cyclic subgroup of index 2. (ii) Let p D 2 and G=R Š H2;m , m 2. To obtain a contradiction, one may assume that jRj D 2. Then G=R has two nonnormal and not conjugate subgroups S1 =R and S2 =R of order 2m ; then S1 and S2 are not conjugate in G. Since, in view of nc.G/ D 2, all subgroups of order 2m are normal in G, S1 and S2 are cyclic of or-

96

Groups with at most two conjugate classes of nonnormal subgroups

19

der 2mC1 . As GN D G=.S1 /G Š D8 has exactly four nonnormal subgroups of order 2, and among of them SN1 and SN2 , the inverse images in G of these four subgroups must be cyclic. It follows that Z D .S1 /G D Z.G/ (if Z < Z.G/, then jG W Z.G/j D 4 hence jG 0 j D 2 by Lemma 1.1, a contradiction). Let S1 < M 2 1 . Then S1 is cyclic of index 2 in M so M 2 ¹C2 C2m ; M2mC1 º (Theorem 1.2). In the second case, M has a nonnormal subgroup of order 2 < 2mC1 so nc.G/ > 2, a contradiction. Thus M is abelian. Similarly, the remaining two members of the set 1 are abelian; then G is minimal nonabelian since d.G/ D 2. We have jG 0 j D 2, and this is a contradiction. (iii) Let G=R Š Mpn . To obtain a contradiction, it sufﬁces to assume that jRj D p. Let A=R and B=R be two distinct cyclic subgroups of index p in G=R. Then A and B are distinct abelian maximal subgroups of G so A \ B D Z.G/ has index p 2 in G and G is minimal nonabelian; then jG 0 j D p, contrary to the assumption. The 2-groups H2;m are not presented in [Sch4] so Schmidt’s proof is not full. O. Schmidt [Sch4] also classiﬁed the nonnilpotent groups G with nc.G/ D 2. His proof is very long and not full (at ﬁrst, this fact was noticed in [SU]). This question is the subject of the recent paper [M] as well. For related results see also [PR]. Below we offer another approach to prove that theorem. In the nonnilpotent case we show an essentially stronger result, namely, we classify the nonnilpotent groups G with no.G/ D 2, where no.G/ is the number of orders of non-G-invariant subgroups in G. Write .G/ D ¹jH j j H < G is nonnormal in Gº so that j.G/j D no.G/. (MNN) (O. Schmidt, Y. Gol’fand; see Lemma A.22.1) Let G be a minimal nonnilpotent group. Then jGj D p a q ˇ , where p and q are distinct prime numbers. Let P 2 Sylp .G/, Q 2 Sylq .G/. Then the following statements hold: (a) One of the subgroups P , Q (say Q) is normal and coincides with G 0 . (b) If Q is abelian, then Q \ Z.G/ D ¹1º and exp.Q/ D p. (c) P is cyclic and jP W .P \ Z.G//j D p. (d) Let Q be nonabelian. Then Q is special (i.e., Z.Q/ D Q0 D ˆ.Q/ is elementary abelian) and jQ=Q0 j D q b , where b is the order of q .mod p/, Q \ Z.G/ D Z.Q/. If L < Z.Q/ is of index p, then Q=L is extraspecial so that b is even. A group from (MNN) is said to be an S.p; q/-group. In Theorem 96.8, p; q are distinct primes. Theorem 96.8. Let G be a nonnilpotent group such that no.G/ D 2. Then G is solvable and one of the following holds: (a) G D H C , where either (a1) H is minimal nonabelian of order p a q and jC j D r ¤ q, .G/ D ¹p a; p a rº, or (a2) H is nonabelian of order pq and jC j D p, .G/ D ¹p; p 2 º.

20

Groups of prime power order

In what follows, G D P Q, where P 2 Sylp .G/ is non-G-invariant, Q 2 Sylq .G/ and Q G G. (b) G is minimal nonabelian of order pq 2 , .G/ D ¹p; qº. (c) We have jQj D q, P is cyclic of order p a , a > 1, G=Z.G/ is a Frobenius group of order p 2 q, .G/ D ¹p a ; p a1 º. (d) One has G D P Q with Q of order q 2 and cyclic P of order p a , Ã1 .P / D Z.G/, G=Z.G/ is a Frobenius group of order pq 2 , all subgroups of Q are G-invariant, .G/ D ¹p a ; p a qº. (e) One has P Š Q8 , jQj D q, G has exactly one cyclic subgroup of index 2 and .G/ D ¹4; 8º. Proof. By Burnside’s two-prime theorem, G is solvable since it has at most two nonnormal Sylow subgroups of distinct orders. Let H < G. Then no.H / no.G/ D 2. If, in addition, H is not G-invariant, then even no.H / 1. Suppose that G D P Q is an S.p; q/-group as in (MNN). Then jQj > q (otherwise, no.G/ D 1) and P is not G-invariant. If Q0 > ¹1º, then PK, where ¹1º < K Q0 , is not normal in G so jQ0 j q. Assume that jQ0 j D q; then jQ=Q0 j q 2 and PQ0 is not normal in G. If Q0 < Q1 < Q, then Q1 is not normal in G so no.G/ > 2, a contradiction. Thus Q0 D ¹1º so Q is elementary abelian and G is minimal nonabelian. Since every nontrivial subgroup of Q is not normal in G, we get jQj D q 2 . Assume that jP j > p. If ¹1º < Q1 < Q, then Ã1 .P /Q1 is not normal in G so no.G/ > 2, a contradiction. Thus G is an minimal nonabelian subgroup of order pq 2 as in (b). In what follows we assume that G is not minimal nonnilpotent. Next let P Q D H < G be an S.p; q/-subgroup as in (MNN). Then no.H / 2 hence H is minimal nonabelian by the above, so either no.H / D 1 with jQj D q (Supplement to Theorem 96.4) or no.H / D 2 and H is of order pq 2 , H G G. Suppose that r 2 .G/ .H / and let R 2 Sylr .G/. Since G is solvable, one may assume that PR < G. If R is not normal in G, then P , R and PR are not normal in G and have different orders hence no.G/ > 2, a contradiction. Thus R G G. If R1 < R is G-invariant, then PR1 is not normal in G since PR1 \ H D P . We conclude that R is a minimal normal subgroup of G. Since P and RP are not normal in G, we get H G G so G D H R, and we conclude that jRj D r. Thus G is as in (a). In what follows we assume that .G/ D .H / D ¹p; qº, where H here and below is a proper S.p; q/-subgroup of G chosen above. (i) Suppose that H is not normal in G; then no.H / D 1 so H 0 D Q G G is of order q and Ã1 .P / G G. All subgroups of G of order ¤ jP j; jH j are G-invariant hence P 2 Sylp .G/ and .G/ D ¹jP j; jP jqº. Then NG .P / is not G-invariant and Q0 2 Sylq .G/ is normal in the group G since jQ0 j > jQj D q in view of H < G. Next, jNG .P /j 2 ¹jP j; jP jqº. (i1) Assume that Q0 is noncyclic of order q c > q 2 . Then there exist in Q0 two distinct maximal subgroups Q1 and Q2 (of order q c1 > q D jQj). In this case,

96

Groups with at most two conjugate classes of nonnormal subgroups

21

Q1 ; Q2 G G and PQ1 and PQ2 are distinct subgroups of G of order jP jq c1 > jH j so these subgroups are G-invariant. It follows that P L D PQ1 \ PQ2 G G and jG W Lj D q 2 . By Frattini’s lemma, we obtain G D LNG .P / hence P < NG .P /, jNG .P /j > jH j so that no.G/ > 2 since NG .P / is nonnormal in G, a contradiction. (i2) Let Q0 be cyclic of order q c > q 2 ; then NG .P / is abelian so NG .P / D P since Q is a unique subgroup of order q in G. If Q1 < Q0 is of index q in Q, then PQ1 D NG .P /Q1 G G, a contradiction since NG .P / > PQ1 . Thus jQ0 j D q 2 . By hypothesis, all subgroups of order q are G-invariant. In this case, .G/ D ¹jP j; jP jqjº. (i3) Assume that P < NG .P /. Then NG .P / D P Q1 , where jQ1 j D q, Q1 G G and CG .Q1 / PQ0 D G so G D H Q1 , and hence H G G, contrary to the assumption. Thus NG .P / D P . We have Ã1 .P / G G so Ã1 .P / D Z.G/, and we conclude that G=Ã1 .P / is a Frobenius group of order pq 2 . As we have noticed, all subgroups of Q0 .2 Sylq .G// are G-invariant; then G is as in (d). (ii) Now suppose that H G G; then jH j 2 ¹p a q; pq 2 º. (ii1) Assume that jH j D pq 2 is minimal nonabelian with jH 0 j D q 2 ; then we have no.H / D 2 D no.G/ so .G/ D .H / D ¹p; qº and P 2 Sylp .G/ since a Sylow p-subgroup of G is not G-invariant (consider its intersection with H ). By Frattini’s lemma, G D NG .P /H D NG .P /PQ D NG .P / Q so P < NG .P /. Since NG .P / is not normal in G and jNG .P /j 62 ¹p; qº D .G/, we get a contradiction. (ii2) Let jH j D p a q and p j jG W H j. If P < P0 2 Sylp .G/, then P0 is not normal in G and jP0 W P j D p (otherwise, every subgroup lying between P and P0 is not G-invariant so no.G/ > 2); then .G/ D ¹jP j; jP0 jº D ¹jP j; pjP jº. Next, NG .P / is not normal in G since NG .P / \ H D P . It follows that NG .P / D P0 is of order pjP j D p aC1 . On the other hand, G D NG .P /H D NG .P / Q D P0 Q (Frattini’s lemma). It follows that jGj D pjH j D p aC1 q. If P0 is cyclic, then Ã2 .P / D Z.G/ and G=Z.G/ is a Frobenius group of order p 2 q so G is as in (c). In what follows we assume that P0 is noncyclic. There is L < P0 of order p such that L 6 P . Suppose that L is not normal in G. Then we have p D jLj D jP j so P0 Š Ep2 . In this case, G D H CP .Q/, where jCP .Q/j D p, contrary to the assumption. If L G G, then G D H L as above. Then .G/ D ¹jP j; pjP jº is as in (a). Now assume that 1 .P0 / P . In this case, P0 is generalized quaternion since it is noncyclic by assumption. Then P0 Š Q8 (if jP0 j > 8, then no.G/ > 2). Such G has a unique cyclic subgroup CG .Q/ of index 2 so it is as in (e) (indeed, G=CG .Q/ > ¹1º is cyclic of order dividing q 1). We have .G/ D ¹4; 8º. (ii3) Next assume that P 2 Sylp .G/, i.e., (ii2) does not hold. Then G D NG .P /H D NG .P /PQ D NG .P / Q; NG .P / is not normal in G and P < NG .P/. In this case, jNG .P / W P j D q. It follows

22

Groups of prime power order

that Q0 Š Eq 2 , where Q0 2 Sylq .G/, and .G/ D ¹jP j; jP jqº. Since q 62 .G/, every Q1 < Q0 is G-invariant. If Q1 ¤ Q, then G D H Q1 so Q1 Z.G/. It follows that Q0 Z.G/ since Q0 has q C 1 > 2 subgroups of order q, and then G is nilpotent, a ﬁnal contradiction. Corollary 96.9 (O. Schmidt [S2]). If G is a nonnilpotent group with nc.G/ D 2, then one of the following holds: (i) G Š A4 , the alternating group of degree 4. (ii) G D P Q, where P 2 Sylp .G/ and G 0 D Q 2 Sylq .G/ are cyclic satisfying Ã1 .P / D Z.G/, and G=Z.G/ is a Frobenius group of order pq 2 . (iii) G D H C , where H is a minimal nonabelian subgroup of order p a q with jH 0 j D q and jC j D r is a prime different from p and q. (iv) G D P Q (semidirect product with kernel Q), where jQj D q, P is cyclic, jG W CG .Q/j D p 2 and G=Z.G/ is a Frobenius group of order p 2 q. Proof. As 1 no.G/ nc.G/ D 2, G is one of the groups from Theorem 96.4, the Supplement to Theorem 96.4 and Theorem 96.8. Since the nonnilpotent groups G with no.G/ D 1 satisfy nc.G/ D 1 as well, by the Supplement to Theorem 96.4, we need only consider the groups from Theorem 96.8. As in the proof of Theorem 96.8, let H D PQ G be an S.p; q/-subgroup, P 2 Sylp .H / and Q D H 0 2 Sylq .H /. (a) Let G D H C be as in Theorem 96.8(a2), jC j D r ¤ q. If also r ¤ p, then nc.G/ D 2. Now let r D p. Then P0 D P C 2 Sylp .G/ is not G-invariant. Let Sylp .G/ D ¹P0 ; P1 ; : : : ; Pq º. For i D 1; : : : ; q, let Pi D C Si , where Si ¤ Pi \H , and set Hi D Si Q, S0 D P . The subgroups S0 ; : : : ; Sp are not normal in G. The pairwise distinct (by construction) subgroups H D H0 ; H1 ; : : : ; Hq , being direct factors of G, are G-invariant. Since SiG D Hi for all i , we get nc.G/ q C 1 > 2, a contradiction. Thus r ¤ p so G is as in (iii). (b) Let G D PQ be minimal nonabelian of order pq 2 as in Theorem 96.8(b). Since the q C 1 subgroups of order q must be conjugate under P , we get p D q C 1 so q D 2, p D 3 and therefore G Š A4 is as in (i). (c) The group of Theorem 96.8(c) satisﬁes the hypothesis, and G is as in (iv). (d) Let the group G D P Q be as in Theorem 96.8(d). Then, provided Q Š Eq 2 , G does not satisfy the hypothesis. Indeed, if Q1 ; : : : ; QqC1 are all the subgroups of order q in Q, then, for i ¤ j , we have PQi ¤ .PQi /G D Ã1 .P /Qi ¤ .PQj /G D Ã1 .P /Qj ¤ PQj so that the subgroups P; PQ1 ; : : : ; PQqC1 are not G-invariant and not conjugate to each other, i.e., nc.G/ q C 2 > 2, a contradiction. If Q is cyclic of order q 2 , then G satisﬁes the hypothesis, and G is as in (ii). (e) The group from Theorem 96.8(e) does not satisfy the hypothesis. Indeed, if M1 and M2 are distinct nonnilpotent maximal subgroups of G and Pi < Mi \P of order 4,

96

Groups with at most two conjugate classes of nonnormal subgroups

23

i D 1; 2, then P1G D M1 ¤ M2 D P2G so P; P1 ; P2 are neither normal nor conjugate in pairs so nc.G/ > 2. Problem 1. Classify the p-groups G which satisfy no.G/ D 2 (the p-groups G satisfying no.G/ D 1 are classiﬁed in 112). Problem 2. Classify the p-groups G, p > 2, satisfying nc.G/ D p. Problem 3. Classify the p-groups G with exactly two conjugate classes of nonnormal cyclic subgroups.

97

p-groups in which some subgroups are generated by elements of order p

The 2-groups all of whose nonmetacyclic subgroups are generated by involutions are classiﬁed in 84. Corollary 97.4 contains a stronger version of the main theorem from 84 for p > 2. Deﬁnition. A p-group G is said to be an Ln -group if 1 .G/ is of order p n and exponent p and G=1 .G/ is cyclic of order greater than p (see 17, 18). In what follows we consider Ln -groups only for n D p. Exercise. Let G be an Lp -group of exponent p e ; then e > 2 and jGj D p pCe1 . Prove that Ã1 .G/ is cyclic of order p e1 and G D hx 2 G j o.x/ D p e i. Solution. We have p e D exp.G/ exp.Ã1 .G// exp.G=Ã1 .G// D p exp.Ã1 .G// so jÃ1 .G/j exp.Ã1 .G// p e1. Therefore it sufﬁces to show that jÃ1 .G/j p e1. If G is regular, then jÃ1 .G/j D jG=1 .G/j D p e1 . In case that G is irregular, then, by Theorem 9.8(a), we have jG=Ã1 .G/j p p so jÃ1 .G/j p p jGj D p e1 , and we conclude that jÃ1 .G/j D p p1 , as required. Next, exp.e1 .G// D p e1 so hx 2 G j o.x/ D p e i D hG e1 .G/i D G: The main result of this section is the following Theorem 97.1 ([Ber40]). Let a p-group G of exponent p e > p be neither absolutely regular nor of maximal class. Then G contains a subgroup H of order p pCe1 such that j1 .H /j D p p , H=1 .H / is cyclic of order p e1 and 2 .H / is regular (so if e > 2, then H is an Lp -group). Our proof of Theorem 97.1 is based on some consequences of Blackburn’s theory of p-groups of maximal class (see 9, 12, 13; proofs of some of them are reproduced below; see Lemmas 97.J(d), 97.5 and 97.6). If a p-group G is either absolutely regular or of maximal class and order > p pC1 , then it has no subgroup H such as in conclusion of Theorem 97.1 (in the second case

97

p-groups in which some subgroups are generated by elements of order p

25

this follows from the description of members of the set 1 and some other subgroups given in Lemma 97.J(h)). Therefore absolutely regular p-groups and p-groups of maximal class are excluded from the hypothesis of Theorem 97.1. Corollary 97.2. Suppose that a p-group G of exponent greater than p is not absolutely regular. If all proper not absolutely regular subgroups of G are generated by elements of order p, then one and only one of the following holds: (a) jGj D p pC1 and j1 .G/j > p p1 so if G is regular, then j1 .G/j D p p . (b) jGj D p pC1 , cl.G/ D p, j1 .G/j D p p1 (in this case, all proper subgroups of G are absolutely regular). (c) G has maximal class, jGj > p pC1 and every irregular member of the set 1 has two distinct subgroups of order p p and exponent p. Corollary 97.3. Let G be an irregular p-group, p > 2. Suppose that whenever H < G is neither absolutely regular nor of maximal class, then 1 .H / D H . Then G is of maximal class. Corollary 97.4. Let p > 2 and suppose that M is a maximal metacyclic subgroup of a nonmetacyclic p-group G, where jM j > p 2 . Suppose that whenever M < N G and jN W M j D p, then 1 .N / D N . Then p D 3 and G is of maximal class. In particular, if p > 2 and a p-group G of exponent greater than p is nonmetacyclic and such that all proper nonmetacyclic subgroups of G are generated by elements of order p, then one and only one of the following assertions holds: (a) G is regular of order p 4 and j1 .G/j D p 3 . (b) G is of maximal class and order 34 . (c) p D 3, G is of maximal class, jGj > 34 and every irregular member of the set 1 has two distinct (nonabelian) subgroups of order 33 and exponent 3. Note that if a p-group of maximal class, p > 2, has an elementary abelian subgroup of order p p , then G is isomorphic to a Sylow p-subgroup of the symmetric group of degree p 2 (Exercise 9.13). Therefore a group of Corollary 97.4 has no abelian subgroups of order 33 and exponent 3 unless jGj D 34 . In Lemma 97.J we collect some known results which are used in what follows. Lemma 97.J. Let G > ¹1º be a p-group, p > 2. (a) (Theorems 7.1 and 7.2) If G is regular, then exp.1 .G// D p and j1 .G/j D jG=Ã1 .G/j. Absolutely regular p-groups, groups of exponent p and p-groups of class < p are regular. (b) (Exercise 9.1(b)) If G is of maximal class and order p m , then it has exactly one normal subgroup of order p i for all 1 i < m 1. (c) (Proposition 1.8) If H < G is of order p 2 and jCG .H /j D p 2 , then G is of maximal class.

26

Groups of prime power order

(d) (Exercise 13.10(a)) If A < G and all subgroups of G of order pjAj containing A are of maximal class (so that jAj > p), then G is also of maximal class. (e) (Theorem 12.12) Suppose that M < G is an irregular subgroup of maximal class and index p. If G is not of maximal class, then G=Kp .G/ is of order p pC1 and exponent p and 1 D ¹M D M1 ; : : : ; Mp2 ; T1 ; : : : ; TpC1 º, where all Mi ’s are of maximal class and all Tj ’s are not of maximal T class (Ti are also not absopC1 2 lutely regular since jG=Ã1 .G/j p pC1 ), .G/ D SpC1 i D1 Ti1D D has index p in G and G=.G/ is abelian of type .p; p/ so iD1 Ti D G. (f) (Theorem 12.1(b)) If G is neither absolutely regular nor of maximal class and H 2 1 is absolutely regular, then G D H 1 .G/, where 1 .G/ is of order p p and exponent p (in particular, j1 .H /j D p p1 ). (g) (Theorem 13.5) If G is neither absolutely regular nor of maximal class, then the numberep .G/ of subgroups of order p p and exponent p in G is 1 .mod p/. (h) (Theorem 9.6) If G is of maximal class and order > p pC1 , then 1 D ¹G1 ; G2 ; : : : ; GpC1 º; where G1 is absolutely regular with j1 .G1 /j D p p1 (G1 is called the fundamental subgroup of G), and the subgroups G2 ; : : : ; GpC1 are irregular of maximal class. All p-groups of maximal class and order p pC1 are irregular. A p-group G of maximal class and order > p pC1 has no normal subgroup of order p p and exponent p. Assume that this is false, and let R G G be of order p p and exponent p. Then, by Lemma 97.J(b), R ˆ.G/ < G1 , a contradiction since the fundamental subgroup G1 is absolutely regular. If a p-group G satisﬁes exp.1 .G// > p, then it is irregular (Lemma 97.J(a)). To facilitate the proof of Theorem 97.1, we prove the following three assertions which have been proved in previous sections of the book. Lemma 97.5. Suppose that H < G is such that N D NG .H / is of maximal class. Then the p-group G is also of maximal class. Proof. We use induction on jGj. One may assume that N < G; then H is not characteristic in N (otherwise, N D G). In this case, by Lemma 97.J(b), we have jN W H j D p hence jH j > p. As jZ.N /j D p and Z.G/ < N , we get Z.G/ D Z.N / so jZ.G/j D p and Z.G/ ˆ.N / < H . Set GN D G=Z.G/. If jHN j D p, then CGN .HN / D NN is of order p 2 so GN is of maximal class (Lemma 97.J(c)). Now let jHN j > p; then NN is of maximal class so GN is also of maximal class by induction, and we are done since jZ.G/j D p. Lemma 97.6. Suppose that A 2 1 is absolutely regular and M < G is irregular of maximal class. Then G is of maximal class. 1 It

follows from the last equality that exp.Ti / D exp.G/ for some i p C 1.

97

p-groups in which some subgroups are generated by elements of order p

27

Proof. Assume that G is not of maximal class. Then, by Lemma 97.J(d), one can take H G such that M < H G, where jH W M j D p and H is not of maximal class. In this case, by Lemma 97.J(e), H=Kp .H / is of order p pC1 and exponent p. It follows that H has no absolutely regular maximal subgroups. However, A \ H is an absolutely regular maximal subgroup of H , and this is a contradiction. Lemma 97.7. All proper subgroups of an Lp -group G are regular; in particular, the subgroup 2 .G/ is regular. Proof. It sufﬁces to show that all maximal subgroups of an Lp -group G are regular. This is true for p D 2 since then G is either abelian or Š Mpn which is minimal nonabelian. In what follows we assume that p > 2. Take M 2 1 . Suppose that 1 .G/ 6 M ; then 1 .M / D M \1 .G/ is of order p p1 and exponent p. Since M is not of maximal class (indeed, M=1 .M / Š G=1 .G/ is cyclic of order > p), it follows that M is absolutely regular so regular (Lemma 97.J(g,a)). Now let 1 .G/ < M . Let D < 1 .G/ be a G-invariant subgroup of index p 2 . Set C D CG .1 .G/=D/; then jG W C j p. Take in C =1 .G/ a subgroup H=1 .G/ which is maximal in G=1 .G/; then H=D is abelian so cl.H / < p, and hence H is regular (Lemma 97.J(a)). Since G=1 .G/ has only one maximal subgroup, we get H D M . It is clear now that 2 .G/.< G/ is regular. Remark. Suppose that p > 2 and G is an irregular Lp -group of exponent p e . Assume that there is in G a normal cyclic subgroup C of order p e . Set D D C \ 1 .G/; then C 1 .G/ D G and jDj D p by the product formula, and G=D D .C =D/ .1 .G/=D// so d.G/ > 2 (it is important that p > 2). By Lemma 97.7, all proper subgroups of G are regular. It follows that d.G/ D 2, contrary to what has just been proved. Thus all cyclic subgroups of order p e are not normal in G. Proof of Theorem 97.1. First consider the case p D 2. Note that absolutely regular 2-groups are cyclic. Since G is not of maximal class, there is in G a normal abelian subgroup R of type .2; 2/ (Lemma 97.J(g)). Put C D CG .R/; then jG W C j 2 so exp.C / 2e1 . Suppose that exp.C / D 2e ; then there is in C R an element x of order 2e . In this case, A D hx; Ri is abelian of type .2e ; 2/ or .2e ; 2; 2/. In any case, A contains an abelian subgroup H of type .2e ; 2/, and H is the desired subgroup. Now let exp.C / D 2e1 . Take y 2 G C of order 2e . Suppose that U D CG .y/ is cyclic; then CG .U / D U and Z.G/ is cyclic. If e D 2, then G is of maximal class (Lemma 97.J(c)), contrary to the hypothesis. Thus e > 2. In this case, U \ R is of order 2 so H D RU is an L2 -subgroup of exponent 2e . If CG .y/ is noncyclic, then there is an involution x 2 CG .y/ hyi; in this case, H D hx; yi D hxi hyi is abelian of type .2e ; 2/ so H is the desired subgroup. In what follows we assume that p > 2. We use induction on jGj. Take an element x 2 G of order p e . Let x 2 L < G, where L is either absolutely regular or irregular of maximal class such that if L < M G, then M is neither absolutely regular nor of maximal class (L exists by hypothesis and Lemma 97.J(d)). Let L < F G and

28

Groups of prime power order

jF W Lj D p; then F is neither absolutely regular nor of maximal class by the choice of L. Therefore if F < G, then F contains the desired subgroup H . Next assume that F D G; then jG W Lj D p. (i) Let L be absolutely regular. Then, by Lemma 97.J(f), we have G D L1 .G/, where 1 .G/.< G/ is of order p p and exponent p. Set H D hx; 1 .G/i, where x 2 L has order p e ; then 1 .H / D 1 .G/ is of order p p and jH j D p pCe1 by the product formula. If e > 2, then H is an Lp -group so 2 .H / is regular by Lemma 97.7. It remains to show that if e D 2, then H is regular. This is true provided that H D G since, by hypothesis, G of order p pC1 is not of maximal class, so cl.G/ < p (Lemma 97.J(a)). Now let H < G and assume, by way of contradiction, that H is irregular. Then H is of maximal class since jH j D p pC1 (Lemma 97.J(a) again) and, by assumption, L is absolutely regular of index p in G. It follows from Lemma 97.6 that G is of maximal class, contrary to the hypothesis. Thus H is the desired subgroup. (ii) Now let L be irregular of maximal class. If F > L is as in the paragraph preceding (i), then, by Lemma 97.J(e), 1 .F / D ¹L D L1 ; L2 ; : : : ; Lp2 ; T1 ; : : : ; TpC1 º; where L1 ; : : : ; Lp2 are of maximal class and T1 ; : : : ; TpC1 are neither absolutely regSpC1 ular nor of maximal class and G D iD1 Ti . It follows that one of the Ti ’s, say T1 , has exponent p e . Therefore, by induction, there is H T1 of exponent p e such that 1 .H / is of order p p and exponent p, H=1 .H / is cyclic of order p e1 and 2 .H / is regular. Lemma 97.8. Suppose that A is a proper absolutely regular subgroup of a p-group G, exp.A/ > p and whenever A < B G and jB W Aj D p, then 1 .B/ D B. Then G is of maximal class. Proof. By Lemma 97.J(a), B is irregular so G is also irregular. Assume that G is not of maximal class. Let jG W Aj D p; then 1 .G/ D G by hypothesis. However, by Lemma 97.J(f), G D A1 .G/, where j1 .G/j D p p < jGj D j1 .G/j, a contradiction. Now let jG W Aj > p. If A < B < G, where jB W Aj D p, then, by what has just been proved, B is of maximal class so G is also of maximal class by Lemma 97.J(d). Proof of Corollary 97.2. Suppose that G is regular and set L D 1 .G/; then jLj p p since G is not absolutely regular. However, G (of exponent > p) has a maximal subgroup of exponent > p so it is absolutely regular, and this implies jLj D p p . By Lemma 97.J(a), jG W Lj D p (otherwise, if L < M < G, then 1 .M / D L < M , contrary to the hypothesis) so G is as in (a). Next we assume that G is irregular. Since, by hypothesis, G has no regular subgroup H of order p pC1 and exponent p 2 such that 1 .H / < H is of order p p , it follows that G is of maximal class (Theorem 97.1). If all maximal subgroups of G are absolutely regular, then G is as in (b) by Lemma 97.J(h). Obviously, every group of maximal class and order p pC1 satisﬁes the hypothesis. Now let jGj > p pC1 . If M 2 1

97

p-groups in which some subgroups are generated by elements of order p

29

is irregular, then 1 .M / D M by hypothesis. If L D 1 .ˆ.M //, then L is of order p p1 and exponent p (Lemma 97.J(h,a,b)). Take an element x 2 M L of order p and set U1 D hx; Li. Pick an element y 2 M U1 of order p and set U2 D hy; Li. Then U1 and U2 are distinct of order p p and exponent p (Lemma 97.J(a)), and the proof is complete. Proof of Corollary 97.3. Assume that G is not of maximal class. Then there is in G a subgroup H such as in Theorem 97.1. Since H is neither absolutely regular nor of maximal class and 1 .H / ¤ H , we get a contradiction. Proof of Corollary 97.4. Let M and N be as in the statement of the corollary. Then M is absolutely regular since p > 2 so, by Lemma 97.J(f), N is of maximal class. By Lemma 97.J(d), G is also of maximal class. Since p p1 D j1 .M /j p 2 and p > 2, we get p D 3. Here we offer another proof of Theorem 97.1 in the case exp.G/ D p e > p 2 . We have to prove that G contains an Lp -subgroup of order p pCe1 . To this end, we use induction on jGj. By Lemma 97.J(g), there is M G G of order p p and exponent p. Take a cyclic X < G of order p e and set F D MX; then F of exponent p e > p 2 is neither absolutely regular nor of maximal class. Therefore if F < G, the result follows by induction. Now let F D G. Suppose that X \ M > ¹1º; then jGj D p pCe1 . We claim that G is an Lp -group. It sufﬁces to show that 1 .G/ D M . Assume that this is false. Since G=M is cyclic, we conclude that j1 .G/j p pC1 . Assume that j1 .G/j D p pC1 . Then X \ 1 .G/ is cyclic of order p 2 hence 1 .G/ is irregular and we conclude that it is of maximal class (Lemma 97.J(a)). Since the group G is not of maximal class, the number ep .G/ of subgroups of order p p and exponent p in G is 1 .mod p/ (Lemma 97.J(g)), and all these subgroups lie in 1 .G/. As ep .G/ > 1 and d.1 .G// D 2, it follows that all maximal subgroups of 1 .G/ have exponent p so exp.1 .G// D p and 1 .G/ is regular (Lemma 97.J(a)), a contradiction. Thus, in the case under consideration, G is an Lp -group. Now let X \M D ¹1º. Let R < M be a G-invariant subgroup of index p. Set H D RX . Let us show that H is an Lp -group. Indeed, H is not absolutely regular since 1 .X/R is of order p p and exponent p (Lemma 97.J(a)). Next, H=R is cyclic of order p e > p 2 so H is not of maximal class. If K=R < H=R is of order p, then 1 .H / K 1 .H / so 1 .H / D 1 .X/R D K is of order p p and exponent p whence H is an Lp -subgroup. By Lemma 97.7, 2 .H / is regular. Proposition 97.9. Let H be a normal absolutely regular subgroup of a p-group G, jH j > p p1 and 1 .G/ 6 H . (a) If for every z 2 G H of order p the subgroup V D hz; H i is of maximal class, then G is also of maximal class. (b) If for every z 2 G H of order p we have 1 .hz; H i/ D hz; H i, then G is of maximal class.

30

Groups of prime power order

Proof. By hypothesis, exp.H / > p. It follows from Lemma 97.J(a) and Theorem 9.5 that G is irregular. Assume, by way of contradiction, that G is not of maximal class. Let H H0 < G, where H0 is absolutely regular such that jH0 j is as large as possible. By Lemma 97.J(d), H0 < B G, where jB W H0 j D p and B is not of maximal class. Then, by Lemma 97.J(f), we have B D H0 1 .B/, where 1 .B/ is of order p p and exponent p so there exists an element x 2 1 .B/ H0 of order p. Set U D hH; xi; then, by hypothesis, U is of maximal class and order > p p so irregular. We have H0 ; U < B, H0 is absolutely regular of index p in B and U is irregular of maximal class. Therefore, by Lemma 97.6, B is of maximal class, contrary to its choice. The above argument also proves (b). It is possible to change the condition 1 .G/ 6 H in Proposition 97.9 to the following one: G is not absolutely regular. Indeed, assume that 1 .G/ < H . Then, by Lemma 97.J(f), G must be of maximal class.

Problems Below G is a p-group. Problem 1. Classify the p-groups all of whose proper nonabelian subgroups that are nonmetacyclic are generated by elements of order p. Problem 2. Study the p-groups G containing a proper abelian subgroup M of exponent > p such that whenever M < N G and jN W M j D p, then 1 .N / D N (2 .N / D N ). Problem 3. Let A be a subgroup of index > p k in a p-group G. Suppose that all subgroups of G containing A as a subgroup of index p k , are of maximal class. Is it true that G also of maximal class? (Compare this with Lemma 97.J(d).) Problem 4. Study the 2-groups G containing a metacyclic subgroup M and such that whenever M < N G and jN W M j D 2, then d.N / D 2. Problem 5. Study the p-groups G containing a minimal nonabelian subgroup M such that whenever M < N G and jN W M j D p, then 1 .N / D N .

98

Nonabelian 2-groups all of whose minimal nonabelian subgroups are isomorphic to M2nC1 , n 3 ﬁxed

Here we classify completely the title groups. This classiﬁcation for an arbitrary n 3 (ﬁxed) is quite involved and shows that our technique with minimal nonabelian subgroups is efﬁcient. Theorem 98.1. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to M2nC1 , n 3 ﬁxed. Then E D 1 .G/ is elementary abelian. Let A be a maximal normal abelian subgroup of G containing E. Then A D CG .E/, all elements in G A are of order 2n , G=A is elementary abelian of order 4, and if x 2 G A, then jE W CE .x/j D 2, A0 D CA .x/ D CE .x/hx 2 i, A=A0 is cyclic, hA; xi0 is cyclic of order jA=A0 j, hA; xi0 \ A0 D 1 .hxi/, x inverts A=A0 and Ãn2 .A/. We have two possibilities: (a) exp.A/ D 2m 2n in which case jG W Aj D 2 and A is of type .2m; 2n2; 2; : : : ; 2/. (b) exp.A/ 2n1 in which case A is of type .2n1; 2r ; 2; : : : ; 2/ with 1 r n 2 and either jG W Aj D 2 or jG W Aj D 4, where in the last case G=n2 .A/ Š Q8 . Conversely, if G is a 2-group of type (a) or (b), then each minimal nonabelian subgroup of G is isomorphic to M2nC1 , n 3 ﬁxed. Proof. Let G be a title group with n 3 ﬁxed. Since G has no dihedral subgroups, E D 1 .G/ is elementary abelian. Set A D CG .E/ so that A is normal in G. If A is nonabelian, then consider a minimal nonabelian subgroup X Š M2nC1 in A. But then all three involutions in X are central in X , which contradicts the structure of M2nC1 . Hence A is abelian and so A D CG .E/ is a unique maximal abelian subgroup of G containing E. Let x be any element in G A so that o.x/ 4 and set hzi D E \hxi, where z is an involution. Set T D Ehxi and E0 D CE .x/ so that E > E0 and T0 D CT .x/ D E0 hxi is abelian. Let T1 T be such that T1 > T0 and jT1 W T0 j D 2. Set E1 D T1 \ E so that T1 D E1 hxi, jE1 W E0 j D 2 and let e be an involution in E1 E0 . Then we have 1 ¤ Œe; x 2 T0 \ E D E0 and so Œe; x is an involution commuting with e and x and therefore he; xi0 D hŒe; xi which implies that he; xi is minimal nonabelian (Lemma 65.2(a)) and so he; xi Š M2nC1 . In particular, hŒe; xi D 1 .ˆ.he; xi// D hzi,

32

Groups of prime power order

e x D ez and hei normalizes hxi which gives o.x/ D 2n . We have proved that all elements in G A are of order 2n and so G=A is elementary abelian. In particular, x 2 2 A and so x induces an involutory automorphism on E. Hence x centralizes E=E0 and so T0 is normal in T with T =T0 Š E=E0 . Suppose that jE=E0 j > 2 and let f 2 E E1 . Then considering T2 D hf iT0 , we get in the same way as above that hf; xi Š M2nC1 and so f x D f z. But then ef 62 E0 and .ef /x D .ez/.f z/ D ef , a contradiction. We have proved that jE W E0 j D 2 and ŒE; x D hzi D 1 .hxi/. Let t 2 E E0 so that t x D t z and ht; xi Š M2nC1 . Let s be an arbitrary element in A. We may set s x D s 1 w for some w 2 A. Note that o.x/ D 2n , o.sx/ D 2n and t sx D t x D t z so that Œt; sx D z with z 2 Z.Ahxi/ and so ht; sxi Š M2nC1 . n1 n1 In particular, hsxi > hzi and so .sx/2 D .x/2 D z. We compute (noting that x 2 2 A and .sx/2 2 A) w D ss x D s.x 1 sx/ D sx 2 xsx D x 2 .sx/.sx/ D x 2 .sx/2 : Therefore w 2 D x 2 .sx/2 inverts A=n2 .A/. Since n2

n2

.s 2

n1

/x D .s x /2

n2

n1

D zz D 1 and so s x D s 1 w implies that x

D .s 1 w/2

n2

D s 2

n2

D .s 2

n2

/1 ;

we see that x also inverts each element in Ãn2 .A/. We shall determine the structure of A0 D CA .x/, where A0 E0 hx 2 i. Suppose that there is an element v of order 4 in A0 such that v 2 D z 0 ¤ z. Taking an involution t 2 E E0 , we get .vt/2 D z 0 and Œvt; x D Œv; xt Œt; x D Œt; x D z and so hvt; xi is minimal nonabelian. But ˆ.hvt; xi/ hz; z 0 i Š E4 , which contradicts the structure of M2nC1 . Hence A0 is of type .2s0 ; 2; : : : ; 2/ with s0 n 1 and 1 .A0 / D E0 . Suppose that A0 contains an element y of order 2n . Then Ãn2 .hyi/ Š C4 is inverted by x, a contradiction. Hence s0 D n 1 and so A0 D E0 hx 2 i. Also, A1 D EA0 is of type .2n1 ; 2; : : : ; 2/ with Ãn2 .A1 / D hzi. For each l 2 A1 A0 , l 2 2 hx 2 i, Œl; x D z and so hl; xi Š M2nC1 . For each y 2 Ax (x 2 G A), CA .y/ D A0 and so y 2 2 A0 . This implies that x inverts A=A0 . Let A =A0 be a subgroup of order 2 in A=A0 and let a be an element in A A0 . We have .a /x D a s with 1 ¤ s 2 A0 since x inverts A=A0 . Then we get 2 a D .a /x D .a s/x D a s 2 and so s 2 D 1 and s is an involution in E0 D 1 .A0 /. The involution s D Œa ; x commutes with a and x and so ha ; xi is minimal nonabelian. Hence ha ; xi Š M2nC1 with ˆ.ha ; xi/ D hx 2 i and so Œa ; x D s D z. If ha ; x 2 i is cyclic, then we obtain that o.a / D 2n and x inverts Ãn2 .ha i/ Š C4 with Ãn2 .ha i/ A0 , a contradiction. Hence ha ; x 2 i is noncyclic and so i D a y is an involution for a suitable y 2 hx 2 i, where i 2 E E0 . Thus E A and so A D Ehx 2 i D A1 . We have proved that Ehx 2 i=A0 is a unique subgroup of order 2 in A=A0 and so A=A0 is cyclic. The mapping a ! a1 ax (a 2 A) is a homomorphism from A onto hA; xi0 with the kernel A0 D Z.hA; xi/. Thus A=A0 Š hA; xi0 . Since x inverts the cyclic group A=A0 , we have jA W .hA; xi0 A0 /j D 2 and therefore hA; xi0 \ A0 D hzi noting that ŒA1 ; x D hzi.

98

2-groups with minimal nonabelian subgroups isomorphic to M2nC1 , n ﬁxed

33

(i) Suppose that A possesses an element a of order 2n . Then each element x 2 GA inverts Ãn2 .hai/ D hvi Š C4 which implies that jG W Aj D 2, v 62 A0 D CA .x/ and v 2 2 E0 . Since A1 D Ehx 2 i is of type .2n1 ; 2; : : : ; 2/, we have v 2 D z and v 2 A1 A0 . This implies that jA=A0 j 2n1 . Let y be an element in the set A A0 such that hyi covers A=A0 and so o.y/ 2n1 . Suppose that o.y/ D 2n1 in which n2 case jA=A0 j D 2n1 , A splits over A0 and y 2 D e 2 E E0 . But then x inverts Ãn2 .hyi/ D hei and so x centralizes e, contrary to CE .x/ D E0 . It follows that o.y/ D 2m 2n . By the above, we get hyi \ A0 D hzi. Since hyi covers A=A0 and hx 2 i covers A0 =E0 , we get A D Ehyihx 2 i, where hyi \ hx 2 i D hzi. Let y0 2 hyi be of order 2n1 . Then we obtain o.y0 x 2 / D 2n2 , 1 .hy0 x 2 i/] 2 E E0 and hy; x 2 i D hyi hy0 x 2 i so that A is of type .2m ; 2n2 ; 2; : : : ; 2/. We have obtained the groups of type (a) of our theorem. (ii) Suppose that exp.A/ D 2n1 . In that case, hx 2 i is a cyclic subgroup in A of the maximal possible order. Let C be a complement of hx 2 i in A so that A D C hx 2 i. Set C0 D A0 \ C , C1 D A1 \ C so that C1 is elementary abelian since A1 =hx 2 i is elementary abelian. Hence we have A1 D C1 hx 2 i, A0 D C0 hx 2 i, jC1 W C0 j D 2, E D C1 hzi and E0 D C0 hzi. We know that A=A0 Š C =C0 is cyclic of order 2r , r1 r 1. Let c 2 C be such that hci covers C =C0 . Then c 2 D e is an involution in the set C1 C0 so that o.c/ D 2r , r n 1. Suppose that r D n 1. Since x inverts Ãn2 .hci/ D hei, we infer that x centralizes e 62 E0 , a contradiction. We have proved that r n 2 and so A is of type .2n1 ; 2r ; 2; : : : ; 2/, where 1 r n 2. It follows that n2 .A/ D hc; E; x 4 i is of index 2 in A, n2 .A/ is normal in G and all elements in A n2 .A/ are of order 2n1 . Assume, in addition, that G=A is elementary abelian of order 4. Let B=n2 .A/ be a subgroup of order 2 in G=n2 .A/ distinct from A=n2 .A/. But then each element d 2 B n2 .A/ is of order 2n , but d 2 2 n2 .A/ is of order 2n2 , a contradiction. Hence A=n2 .A/ is a unique subgroup of order 2 in G=n2 .A/. Since G=n2 .A/ cannot be cyclic (noting that G=A is elementary abelian of order 4), it follows that G=n2 .A/ is generalized quaternion. But then d.G=n2 .A// D 2 and so jG=Aj D 4 and G=n2 .A/ Š Q8 . We have obtained the groups of type (b) of our theorem. It remains to show that all obtained groups G of type (a) or (b) have the property that each minimal nonabelian subgroup M of G is isomorphic to M2nC1 , n 3 ﬁxed. First suppose that jM W .M \ A/j D 2 and let x 2 M A. Then M \ A 6 A0 D CA .x/. Since A=A0 is cyclic, it follows that .M \ A/ A0 contains an element l 2 A1 A0 , where A1 D Ehx 2 i. But then we know that hl; xi Š M2nC1 and so hl; xi D M . Now suppose that jM W .M \ A/j > 2 in which case jG=Aj D 4 and G=n2 .A/ Š Q8 . In this case, M covers Q=A. But for any m 2 M A we obtain that o.m2 / D 2n1 and so m2 2 A n2 .A/. Thus the subgroup M covers also G=n2 .A/ and so M=.M \ n2 .A// Š Q8 . Hence M 0 covers .M \A/=.M \n2 .A//. But we have jM 0 j D 2 which gives a contradiction since all elements in the set A n2 .A/ are of order 2n1 4. Our theorem is proved.

99

2-groups with sectional rank at most 4

Here we give some results of K. Harada from [Har], where in case of 2-groups of sectional rank at most 4 we simplify some of his proofs. Recall that the sectional rank r.G/ of a p-group G is r.G/ D max¹d.H / j H Gº. The most impressive result is Theorem 99.9 stating that a 2-group which possesses a self-centralizing abelian subgroup of order 8 is of sectional rank at most 4. For related results see 51, where the 2-groups containing a self-centralizing subgroup isomorphic to E8 are studied. Lemma 99.1. Suppose that G D A B is a 2-group, where A Š E8 , A E G, B Š E4 , A \ B D ¹1º and CG .A/ D A. Then one of the following holds: (i) CA .B/ D ŒA; B Š C2 and G Š Q8 Q8 . (ii) CA .B/ D ŒA; B Š E4 and G Š E4 o C2 . Proof. We know that Aut.A/ Š GL.3; 2/ has exactly two conjugacy classes of foursubgroups. Therefore there exist at most two possible structures for A B. Indeed, Q8 Q8 (the extraspecial group of order 25 and “type +”) and E4 o C2 are those two possibilities. In the ﬁrst case, CA .B/ D ŒA; B Š C2 and in the second case, CA .B/ D ŒA; B Š E4 . Lemma 99.2. Let G be a 2-group with an elementary abelian subgroup A of order 16 and index 2. If jZ.G/j D 4, then G Š E4 o C2 . In particular, each involution in Z.G/ has exactly four square roots in G A so G contains exactly six cyclic subgroups of order 4 (it follows that G has exactly 19 involutions). Proof. Let g be an element of order 4 in G A. Then CA .g/ D Z Š E4 and g 2 2 Z which implies Z D Z.G/, ˆ.G/ Z and therefore G 0 Z and so hZ; gi E G. Hence all four conjugates of g in G lie in ¹gZº and so G 0 D ˆ.G/ D Z.G/ and G is special. Let g 0 2 G A be such that .g 0 /2 D g 2 . Then g 0 D ga for some a 2 A and g 2 D .g 0 /2 D .ga/2 D gaga D g 2 ag a which gives ag D a and so a 2 Z and g 0 2 ¹gZº. Hence g 2 2 Z has exactly four square roots and they are all conjugate to g in G. If i is an involution in G A, then we see (as above) that ¹iZº is the set of all involutions in G A and they form a single conjugate class of involutions in G A. It follows that G A must contain exactly four square roots of each of the three involutions in Z. But this gives exactly 12 elements of order 4 in G A and so the remaining four elements in G A must be involutions forming a single conjugate

99 2-groups with sectional rank at most 4

35

class. The structure of G is uniquely determined and so G Š E4 o C2 . Indeed, let t be an involution in G A and let X be a complement of Z in A. Then X \ X t E G so that X \ X t D ¹1º. Lemma 99.3. Let G be a 2-group containing a subgroup R Š Q8 Q8 of order 25 . If CG .R/ R, then r.G/ 4. Proof. Set R D Q1 Q2 , where Q1 and Q2 are the only two quaternion subgroups of R and Q1 \ Q2 D Z.R/ D hzi. Let U be a normal four-subgroup of G and assume that U 6 R. Then U \ R D hzi, j.RU / W Rj D 2 so that ŒR; U hzi and so if u 2 U hzi, then Q1u D Q2 is not possible. If u induces an outer automorphism on Qi (i 2 ¹1; 2º), then ŒQi ; u Š C4 , a contradiction. Thus u induces on both Q1 and Q2 inner automorphisms and so u induces an inner automorphism on R, contrary to CG .R/ R. Hence U R. If G does not have a normal elementary abelian subgroup of order 8, then r.G/ 4 (Theorem 50.3 ). Hence we may assume that G possesses a normal elementary abelian subgroup X of order 8. Assume that X 6 R. By the previous paragraph, X \ R D X1 Š E4 , X1 is a normal four-subgroup in G and z 2 X1 . We have j.RX/ W Rj D 2 and so ŒR; X X1 which shows that X cannot interchange Q1 and Q2 . But ŒR; X is elementary abelian and so X induces on both Q1 and Q2 inner automorphisms and so X induces an inner automorphism group on R, a contradiction. We have proved that X R with z 2 X . Set V D NG .R/. Since Q8 Q16 and Q8 SD16 do not possess an elementary abelian normal subgroup of order 8, no element of V R induces an inner automorphism on Q1 or on Q2 . Hence any element of V R must either interchange Q1 and Q2 or induce an outer automorphism on both Q1 and Q2 . Note that a Sylow 2-subgroup of the outer automorphism group of Q8 Q8 is isomorphic to D8 and so we get jV =Rj 4 since X R. Suppose that G > V . Set GN D G=hzi where Z.R/ D hzi D Z.G/. Then VN contains N where a normal elementary abelian subgroup TN of order 16 which is distinct from R, y T D R with an element y 2 G V which normalizes V. By the action of elements N we have C N .tN/ Š E4 for tN 2 TN R. N Therefore TN \ RN Š E4 and of VN RN on R, R N N N T =.T \ R/ Š E4 so that T \ R D X Š E8 since X is normal in G and T =X Š E4 . Therefore there is t 2 T R which induces an outer automorphism on both Q1 and Q2 and so Œt; R is not elementary abelian. On the other hand, Œt; R X and so Œt; R is elementary abelian, a contradiction. Hence we have V D G and so jGj 27 . By the action of the elements of G R on R, we see that r.G/ 4. Indeed, let A=B be a section of G such that A=B Š E25 . If B D ¹1º, then A Š E25 and so A covers G=R Š E4 and so jGj D 27 . Let u be an involution in A R which induces an involutory outer automorphism on both Q1 and Q2 in which case CR .u/ Š E4 . On the other hand, A \ R Š E8 , a contradiction. Hence B ¤ ¹1º. If jBj D 4, then A D G is of order 27 and jˆ.G/j 4. But in this case, there is v 2 G R such that Q1v D Q2 and so jˆ.G/j 8, a contradiction. Hence we may assume jBj D 2 and since G cannot

36

Groups of prime power order

have an elementary abelian subgroup of order 26 , we obtain that ˆ.A/ D B Š C2 . If G D A is of order 26 , then B D Z.G/ D Z.R/ and if x 2 A R, then x centralizes R=Z.R/, a contradiction. Hence we have jGj D 27 , A is a maximal subgroup of G with ˆ.A/ D B Z.G/ and so B D Z.R/ D Z.G/ D hzi. Since A > R is obviously not possible (as before), we conclude that A covers G=R Š E4 . But then considering again GN D G=hzi, we have for elements tN 2 AN RN that tN centralizes a subgroup of N contrary to the established fact that C N .tN/ Š E4 . We have proved that order 8 in R, R r.G/ 4. Lemma 99.4. Suppose that G is a 2-group which contains a subgroup R Š E4 o C2 . If CG .Z.R// D R, then r.G/ 4. Proof. Since r.R/ 4, we may assume R < G. Let A be a unique elementary abelian subgroup of order 16 in R and let D be a subgroup of G containing R satisfying jD W Rj D 2. Since A char R, we have A E D. Also, CG .A/ CG .Z.R// D R and so CG .A/ D A. We study V D NG .A/ and set VN D V =A so that VN is a subgroup of GL.4; 2/. Let a be an involution in R A and set XN D CVN .a/ N so that X D. Then X normalizes R and so Z.R/ E X. But CG .Z.R// D R and so X D D. Thus jCVN .a/j N D 4 which implies (noting that exp.VN / 4) VN Š C4 ; E4 or D8 . Suppose that N V Š C4 which gives V D D. But then A char D which gives D D G. If in this case H is a subgroup of G containing A, then either H D R and d.H / 4 or H D G and ˆ.H / Z.R/ D R0 and so d.H / 4. If H 6 A, then d.H \ A/ 3 and H=.H \ A/ is cyclic and so d.H / 4. It follows that we may assume that VN Š E4 or VN Š D8 and so in both cases D=A Š E4 since jCVN .a/j N D 4. Since R A has exactly four involutions and they are all conjugate in R to a, we conclude that T D CD .a/ covers D=R and T \ R D hai Z.R/. If x 2 T R, then T =Z.R/ Š E4 implies that x 2 2 Z.R/ and so hx; Z.R/i Š D8 . Therefore there is an involution t 2 T R so that ha; ti Š E4 and CZ.R/ .t/ D hzi Š C2 which gives Z.D/ D hzi and D splits over A. Let t 0 2 ¹t; taº so that CZ.R/ .t 0 / D hzi and suppose that jCA .t 0 /j D 8 in which case CA .t 0 / covers A=Z.R/. But then CA .t 0 / is hai-admissible and CCA .t 0 / .a/ > hzi, a contradiction. It follows that CA .t 0 / Š E4 so that by Lemma 99.2 we have hA; t 0 i Š E4 o C2 and this together with Z.D/ D hzi gives A char D because A is a unique elementary abelian subgroup of order 16 in D. Now set Q D ha; tihCA .a/; CA .t/i, where the four-group ha; ti acts faithfully on E D CA .a/CA .t/ Š E8 and Z.Q/ D hzi. By Lemma 99.1, we have Q Š Q8 Q8 . Suppose that CG .Q/ 6 Q. Then, as A normalizes CG .Q/, we can ﬁnd a subgroup B in CG .Q/ of order 4 with B > Z.Q/ D hzi and A normalizes B. It follows that ŒA; B Z.Q/ D Z.D/ D hzi and so B normalizes A. On the other hand, B also centralizes Q and so B normalizes R D Ahai, where a 2 Q. By the above, CVN .a/ N D DN and so B D. This is a contradiction since CD .Q/ D Z.Q/ D hzi. We have proved that CG .Q/ Q. From Lemma 99.3 it follows that r.G/ 4. Lemma 99.5. Let A Š C8 be self-centralizing in a 2-group G. Then r.G/ 4.

99 2-groups with sectional rank at most 4

37

Proof. We set A D hai and a4 D c. Assume that G possesses a normal elementary abelian subgroup E of order 8 (because otherwise, Theorem 50.3 implies r.G/ 4). We set E D hc; x; yi, where E \ Z.G/ D hci and Z.G/ < A. If Œa2 ; E D 1, then a would centralize a four-subgroup of E, contrary to CG .A/ D A. Hence Œa2 ; E ¤ 1 and so Z.G/ D hci. Since GL.3; 2/ has only one class of elements of order 4, we may assume that x a D xc and y a D yx. This yields that ax D ac D a5 and ay D ax and the structure of AE is uniquely determined. In particular, hA; xi Š M16 has exactly two cyclic subgroups of order 8: A D hai and haxi D Ay . We obtain .ay/2 D a2 x and AE has only two further cyclic subgroups of order 8: hayi and hay xi, where .ay/y D ay x. Also, 2 .AE/ D Eha2 i D hxi ha2 y; yi Š C2 D8 . We know that Aut.A/ Š E4 and we have NAE .A/ D hA; xi since y 62 NAE .A/. Moreover, hc; xi is a unique normal four-subgroup of G which is contained in E. Since exp.AE/ D 8, we may assume that jG W .AE/j 4 (because otherwise, obviously r.G/ 4). Assume that A1 D NG .A/ D hA; xi is of order 16. Then A2 D NG .A1 / D AE and set A3 D NG .A2 / so that jA3 W A2 j D 2. Since 2 .A2 / char A2 , we get 2 .A2 / D Eha2 i E A3 . If CA3 .E/ D E, then A3 =E Š D8 and so C4 Š A2 =E char A3 =E which implies that an element in G A3 normalizing A3 also normalizes E and A2 , contrary to NG .A2 / D A3 . Hence CA3 .E/=E Š C2 so that A3 =E Š C4 C2 is abelian. Set E D CA3 .E/ so that E is abelian of type .2; 2; 2; 2/ or .4; 2; 2/. Since hai normalizes E and CE .a/ D hci, it follows that Ã1 .E / hci. Take an element t 2 E E. Then t centralizes E and t 2 2 hci. We have ˆ.A2 / D ha2 ; xi and A3 =Eha2 i Š E4 and so ˆ.A3 / D Eha2 i since our element t 2 A3 A2 fuses hai to hayi or hayxi. There are three maximal subgroups of A3 (containing Eha2 i): A2 , Eha2 ; ati and Eha2 ; ti. Since .Eha2 i/=E and .Eht i/=E are two distinct subgroups of order 2 in .Eha2 ; ti/=E, it follows that .Eha2 ; ti/=E Š E4 which gives that exp.Eha2 ; ti/ D 4. As all elements in A2 hE; a2 i are of order 8, Al2 D hE; a2 ; ati for some l 2 GA3 normalizing A3 . Hence all elements in hE; a2 ; atihE; a2 i are of order 8 and, in particular, the element at is of order 8. Since A3 D A2 E , A2 E A3 , E E A3 and A2 \ E D E, we have Œa; t 2 A2 \ E D E, where t 2 2 hci D Z.G/ and E is abelian. We get a1 t 1 at D e 2 E and so at D ae. On the other hand, t 2 A3 A2 and so at 62 A1 D hA; xi which implies that e D yc i x j , i; j D 0; 1. We compute .at/2 D atat D at 2 .t 1 at/ D at 2 ayc i x j D a2 t 2 yc i x j and so .at/4 D a2 t 2 yc i x j a2 t 2 yc i x j D a2 ya2 x j yx j D a2 ya2 y D a2 y ya2 c D a4 c D cc D 1; a contradiction. It remains to treat the case where A1 D NG .A/ is of order 32. We have A1 =A Š E4 and x 2 A1 . Choose an element t 2 A1 hA; xi such that at D a1 . Then we see that t satisﬁes t 2 2 CG .A/ D A and so t 2 2 hci. It follows that hA; ti Š D16 or Q16

38

Groups of prime power order

and hA; txi Š SD16 with .tx/2 2 hci. Also, A is a unique cyclic subgroup of order 8 in hA; ti and in hA; txi. Thus A1 contains exactly two cyclic subgroups of order 8: hai and haxi D Ay . This forces that jA2 W A1 j D 2, where A2 D NG .A1 / and so A2 D A1 E. Since t inverts A, A2 =E Š D8 . Moreover, CA2 .E/ D E as ŒE; a2 ¤ 1. By the structure of GL.3; 2/, A2 =E contains a four-subgroup D=E with CE .D/ Š C2 and so CE .D/ D hci. We have D D Eha2 ; ti or D D Eha2 ; ati, where t 2 2 hci and .at/2 2 hci. Since ŒD; E D hci, we get also ˆ.D/ D hci and so D Š Q8 Q8 . If CG .D/ 6 D, then we can ﬁnd a group V of CG .D/ of order 4 which contains hci and is normalized by A. But then Œa; V hci and so V normalizes A and so V A2 . On the other hand, CA2 .D/ CA2 .E/ D E and so CA2 .D/ D hci, a contradiction. Hence CG .D/ D and then Lemma 99.3 gives r.G/ 4. Lemma 99.6. Let G be a 2-group with a self-centralizing abelian subgroup A of order 8. If r.G/ > 4, then jGj 27 , G has a normal elementary abelian subgroup E of order 8 and A is not normal in G. Proof. Obviously, jGj 26 . Suppose that jGj D 26 . Then G has subgroups H > K with K E H and H=K Š E25 . If K D ¹1º, then H is a maximal subgroup of G and H Š E25 . But then jZ.G/j 8 and Z.G/ A, a contradiction. Hence H D G and K Š C2 so that ˆ.G/ D G 0 D K. Since G cannot be extraspecial, we have Z.G/ > K. But Z.G/ < A and so jZ.G/j D 4. Let a 2 A Z.G/. On the other hand, jG 0 j D 2 implies that jG W CG .a/j D 2 so that CG .A/ > A, a contradiction. We get jGj 27 and Theorem 50.3 implies that G has a normal elementary abelian subgroup E of order 8. Also, an S2 -subgroup of Aut.A/ is of order at most 8 which together with CG .A/ D A implies that A is not normal in G. Lemma 99.7. Suppose that G is a 2-group which contains a self-centralizing subgroup A Š E8 . Then r.G/ 4. Proof. Suppose that r.G/ > 4. Then Lemma 99.6 implies that jGj 27 , A is not normal in G, G possesses a normal elementary abelian subgroup E of order 8 and A ¤ E. (i) We show that jA \ Ej D 4 which implies E NG .A/. Since Z.G/ < A, A \ E is a nontrivial central subgroup of AE. Suppose that A\E Š C2 . Then A contains a four-subgroup A1 with A1 \E D ¹1º. If x 2 A1 is an involution which centralizes E, then taking y 2 A1 hxi we get jCE .y/j 4, contrary to CG .A/ D A. This yields that A1 acts faithfully on E. Since CE .A1 / D A\E Š C2 , Lemma 99.1(i) yields that AE Š Q8 Q8 . But then CG .AE/ CG .A/ D A and so Lemma 99.3 gives a contradiction. (ii) Let B be any elementary abelian subgroup of order 8 in NG .A/. Then we have jA \ Bj 4. As CG .A/ D A, we have A \ B ¤ ¹1º and assume that A \ B Š C2 . Let B1 Š E4 be a complement of A\B in B. Obviously, B1 acts faithfully on A and so Lemma 99.1 implies AB Š Q8 Q8 or AB Š E4 o C2 . Clearly, CG .AB/ CG .A/ D A in both

99 2-groups with sectional rank at most 4

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cases. By Lemma 99.3, we get AB Š E4 oC2 and let V be a unique elementary abelian subgroup of order 16 in AB so that AB D AV with V \ A D Z D Z.AB/ Š E4 . Assume that CG .Z/ > AB and let T be a subgroup of CG .Z/ containing AB with jT W .AB/j D 2. Then V is normal in T and if a 2 A Z, then .AB/ V contains exactly four involutions and they are all conjugate to a in AB since CAB .a/ D A. Hence CG .a/ covers T =.AB/, contrary to CG .A/ D A. We have proved that CG .Z/ D AB. But then Lemma 99.4 implies that r.G/ 4, a contradiction. (iii) NG .A/=A is not isomorphic to D8 . Suppose that NG .A/=A Š D8 . Since G > NG .A/, NG .A/ contains a conjugate B D At ¤ A of A, where t 62 NG .A/ and .NG .A//t D NG .A/. Since B E NG .A/ and A \ B Š E4 , B maps into the center of NG .A/=A. Hence AE D AB and note that AE Š C2 D8 and so A and E are the only two elementary abelian subgroups of order 8 in AE. Then B D A or B D E and this is a contradiction in both cases. (iv) GN D G=E is isomorphic to D2n , n 3, or to SD2m , m 4. We investigate the structure of GN D G=E. Obviously, NG .AE/ normalizes A since E E G and A and E are the only two elementary abelian subgroups of order 8 in AE. N D 4. This fact together with Hence jNG .AE/=Ej D 4 which means that jCGN .A/j N N G > NG .AE/ implies that G Š D2n or G Š SD2m . (v) Now we obtain a ﬁnal contradiction. Since r.G/ > 4, G contains subgroups G0 and U with G0 =U Š E32 . If E 6 G0 , then d.E \G0 / 2 and (by (iv)) d.G0 =E \G0 / 2, a contradiction. Hence E G0 . Since d.G0 =E/ 2, we must have E \ U D ¹1º. Hence ŒE; G0 E \ U D ¹1º and so E Z.G0 /. Since jG0 =Ej 4, G0 contains the inverse image K of Z.G=E/ in G. As K G0 CG .E/, we have K Š E16 or K Š C4 C2 C2 . The latter case does not occur as ˆ.G0 / \ E U \ E D ¹1º. Thus K Š E16 . Let a 2 A .A \ E/ so that a 62 K and CK .a/ Š E4 . Hence Lemma 99.2 implies that ha; Ki Š E4 o C2 . As A D ha; CK .a/i is normal in ha; Ki, we can ﬁnd an elementary abelian subgroup B of K of order 8 with jA \ Bj D 2 and B NG .A/. This contradicts (ii) and our lemma is proved. Lemma 99.8. Let G be a 2-group with a self-centralizing subgroup A Š C4 C2 . Then r.G/ 4. Proof. Suppose that r.G/ > 4. Then Lemma 99.6 implies that jGj 27 , A is not normal in G and G possesses a normal elementary abelian subgroup E of order 8. We set A D ha; b j a4 D b 2 D 1; Œa; b D 1; a2 D ci. It is well known that Aut.A/ Š D8 , CA .Aut.A// D hci and if is the unique central involution in Aut.A/, then inverts A. (i) We have jA \ Ej D 4. Suppose that jA \ Ej 2. Then A \ E Š C2 and A \ E D hci, hbci or hbi. Suppose ﬁrst that A \ E D hci. As A=hci Š E4 , each element x in A hci induces an involutory automorphism on E with jCE .x/j D 4 and hE; xi0 D ŒE; x D hci.

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Groups of prime power order

Indeed, if x centralizes E, then for an element y 2 A hc; xi we have jCE .y/j 4, which contradicts CG .A/ D A. If hE; xi0 D hei with e 2 E hci, then hei is normal in AE and A centralizes hei, a contradiction. Hence we get .AE/0 D ŒA; E D hci and .AE/=hci Š E16 . Then we obtain that Z.AE/ D .AE/0 D ˆ.AE/ D hci and so AE is extraspecial with an elementary abelian subgroup of order 8 which implies that AE Š Q8 Q8 . As CG .AE/ CG .A/ D A AE, Lemma 99.3 yields that r.G/ 4, a contradiction. We next treat the case A \ E D hbci or hbi. Without loss of generality we may assume that A \ E D hbi. We have .AE/=E Š hai Š C4 . If Œc; E D 1, then a centralizes a four-subgroup in E, a contradiction. Therefore hai acts faithfully on E and this forces Z.AE/ D Z.G/ D hbi. Set E D hb; x; yi. Without loss of generality we may set x a D xb, y a D yx and so ax D ab, Œc; x D 1, ay D ax, and x 2 NG .A/. Clearly hb; xi is a unique normal four-subgroup of G contained in E. Hence x maps into the center of NG .A/=A. Since x does not invert A, NG .A/=A Š C2 or NG .A/=A Š E4 . First suppose that NG .A/=A Š E4 , where NAE .A/ D hA; xi. Then NG .A/ .AE/ contains an element t which inverts A. We have NG .A/ D hA; x; t i and t 2 2 hb; ci so that ht; Ai=hbi Š D8 or Q8 . Setting F D NG .A/E, we see that F=E Š D8 or Q8 and F=E is faithful on E, inasmuch as c maps onto the center of F=E and does not centralize E. In that case, F=E Š D8 and so ht; Ai=hbi Š D8 , t 2 2 hbi and .ta/2 2 hbi. By the structure of GL.3; 2/, F=E must contain a four-subgroup F1 =E with CE .F1 / Š C2 . Interchanging t and ta if necessary, we may assume that F1 D hE; c; t i with Z.F1 / D hbi. But then Œht; ci; E D F10 D hbi and F1 =hbi Š E16 so that F1 is extraspecial and F1 Š Q8 Q8 . Suppose that CG .F1 / 6 F1 . Since a normalizes CG .F1 /, we can ﬁnd a subgroup V of order 4 in CG .F1 / which contains hbi and is normalized by hai. But then Œa; V hbi and so V normalizes A which implies that V F . On the other hand, CF .F1 / D Z.F1 / D hbi. This contradiction forces CG .F1 / F1 and then Lemma 99.3 implies r.G/ 4, a contradiction. It remains to treat the case A1 D NG .A/ D hA; xi, where A1 is the nonmetacyclic minimal nonabelian group of order 16. Also, we set A2 D NG .A1 / and A3 D NG .A2 /. As ˆ.A1 / D hb; ci, A1 has exactly three maximal subgroups: A, Ay D hax; bi Š C4 C2 and hc; b; xi Š E8 and A2 permutes A and Ay and so A2 D AE D hA; x; yi is of order 25 . Also, A3 > A2 since jGj 27 . On the other hand, as ˆ.A2 / D hb; c; xi, A2 has exactly three maximal subgroups: A1 , hc; Ei Š C2 D8 and hay; b; c; xi, where A1 cannot be conjugate to hc; Ei. Hence jA3 W A2 j D 2, jA3 j D 26 and A3 contains an element t 2 A3 A2 such that At1 D hay; b; c; xi, .ay/2 D a.yay/ D a ax D cx. If CA3 .E/ D E, then A3 =E Š D8 and A2 =E Š C4 is characteristic in A3 =E. Therefore if s 2 G A3 is such that s 2 2 A3 and As3 D A3 , then s normalizes A2 , contrary to A3 D NG .A2 /. Thus E satisﬁes CA3 .E/ > E and since A2 =E is faithful on E, we get CA3 .E/ D hE; ti, where ŒE; t D 1, t 2 2 E and t 2 A3 A2 . We have A3 D A2 hE; ti, where A2 and the abelian subgroup hE; ti (of order 16) are normal in

99 2-groups with sectional rank at most 4

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A3 and A2 \ hE; ti D E so that A3 =E Š C4 C2 and so 1 .A3 / hE; c; t i: We shall show that B D hb; c; xi is a unique normal self-centralizing elementary abelian subgroup of order 8 in A3 . Since B D ˆ.A2 /, we see that B is normal in A3 . Since t normalizes A2 but does not normalize A1 , we have at ay .mod B/, where A2 =B Š E4 and so A3 =B Š D8 . The element y maps onto Z.A3 =B/ and y does not centralize B and so CA3 .B/ D B. Suppose that hE; ti Š E16 . Then the four-group hy; ti acts faithfully on B and hy; ti centralizes hb; xi so that ChE;ti .c/ D hb; xi Š E4 and so by Lemma 99.1, we obtain hE; c; t i Š E4 o C2 . Since all involutions in hE; c; t i hE; ti are contained in B, it follows that in this case B is a unique normal self-centralizing elementary abelian subgroup of order 8 in A3 . Now suppose that hE; ti Š C4 C2 C2 . Since a centralizes Ã1 .hE; ti/ D ht 2 i, we get t 2 D b. Suppose that c inverts an element t1 2 hE; tiE in which case t12 D b and c t1 D cb. Hence t1 normalizes hb; ci so t1 normalizes CA2 .hb; ci/ D hA; xi D A1 . As t1 62 A2 D NG .A1 /, this is a contradiction. Thus all involutions in hE; c; t ihE; ti are contained in B and so B has the required property also in this case. Now we get a ﬁnal contradiction which will ﬁnally prove our statement (i). Let l be an element in G A3 such that Al3 D A3 and l 2 2 A3 . Since B is characteristic in A3 , B is normal in hA3 ; li and so hA3 ; li=B Š D8 C2 . Set F D ChA3 ;li .B/ so that F=B Š C2 and F is normal hA3 ; li D A3 F with A3 \ F D B. Let l 0 2 F B so that l 0 62 A3 , l 02 2 B and Œl 0 ; A3 A3 \ F D B. Hence l 0 centralizes B and A2 =B and so l 0 normalizes A2 , contrary to A3 D NG .A2 /. (ii) We have jNG .A/j 25 . By (i), jA \ Ej D 4 and so we may set E D hb; c; xi, where A \ E D hb; ci. Suppose that jNG .A/j D 26 . Since Ã1 .A/ D hci, we get hci D Z.NG .A// D Z.G/. We have E NG .A/ and E E G and so x maps into the center of NG .A/=A Š D8 . Thus x inverts A. In particular, x a D xc. Let e be an element of NG .A/ such that ae D a and b e D bc. Then e 2 2 CG .A/ D A. First suppose that o.e/ 4. Then e 2 2 hb; ci and so he; b; ci Š D8 . Thus we may assume that o.e/ D 2. As jCE .e/j D 4, we may also assume that Œe; x D 1. We claim that hA; e; xi Š Q8 Q8 . Indeed, we have hA; e; xi0 D hci and ˆ.hA; e; xi/ D hci D Z.hA; e; xi/ and so hA; e; xi is extraspecial of order 25 containing E Š E8 and so hA; e; xi Š Q8 Q8 . Since hA; e; xi is self-centralizing in G, Lemma 99.3 yields that r.G/ 4, a contradiction. Now suppose that o.e/ D 8 so that we may assume e 2 D a. There is an element d 2 NG .A/ with ad D ab and b d D b and so ˆ.NG .A// A. On the other hand, we have NG .A/=A Š D8 and so x maps into ˆ.NG .A/=A/ which implies that ˆ.NG .A// D hA; xi D AE: As x inverts A, hA; xi Š C2 D8 and so A is a unique abelian maximal subgroup of

42

Groups of prime power order

hA; xi which is not elementary abelian. It follows that A is characteristic in NG .A/ which forces NG .A/ D G, contrary to the fact that jGj 27 (Lemma 99.6). (iii) Set GN D G=E. Then jCGN .a/j N > 4. N 24 . This implies that GN Š D2n or that Suppose that jCGN .a/j N 4, where jGj N G Š SD2n , n 4. It follows that CG .E/ > E and so if K is the inverse image of N then K is an abelian normal subgroup of order 16. Z.G/, Since r.G/ > 4, G contains subgroups T > U with U E T and T =U Š E25 . Clearly, E T because otherwise d.T / 4. In this case, d.T =E/ 2 which implies E \ U D ¹1º. Hence ŒT; E E \ U D ¹1º and so E is central in T . But jT =Ej is N of order at least 4 and this implies that T also contains the inverse image K of Z.G/. Also, ˆ.T / U and ˆ.K/ E and so K Š E16 . Since CK .a/ D hb; ci, we get (Lemma 99.2) ha; Ki Š E4 o C2 with Z.ha; Ki/ D hb; ci Š E4 . By Lemma 99.4, CG .hb; ci/ > ha; Ki and let X be a subgroup of CG .hb; ci/ containing ha; Ki as a subgroup of index 2. Note that (Lemma 99.2) ha; Ki K contains exactly four square roots of c and they form a single conjugate class in ha; Ki which is also X-invariant (since X centralizes hb; ci). This shows that CG .a/ must cover X=ha; Ki, contrary to CG .A/ D A. (iv) We are now in a position to get a ﬁnal contradiction. N is of order 26 . We set again GN D G=E. By (iii), the inverse image D of CGN .a/ 5 Since D normalizes hA; xi D AE and (by (ii)) jNG .A/j 2 , it follows that hA; xi 6Š C2 D8 (because otherwise, A would be characteristic in AE). Therefore x does not invert A. Without loss of generality we may assume (replacing b with bc if necessary) ax D ab and so ˆ.AE/ D hb; ci. We have .ax/2 D a.xax/ D a.ab/ D a2 b D cb and so A D ha; bi and hax; bi are the only subgroups of AE which are isomorphic to C4 C2 . This implies that jDj D 26 , jNG .A/j D 25 , NG .A/=A Š E4 and if t 2 D NG .A/, then At D hax; bi and c t D cb which gives Z.G/ D Z.D/ D hbi. Let d be an element in NG .A/ which inverts A. Then d 2 NG .A/ .AE/ and since d 2 centralizes A, we have d 2 2 A and so d 2 2 hb; ci. Since jGj 27 (Lemma 99.6), we obtain CG .E/ > E. Hence aN (which does not centralize E) centralizes a subgroup E0 =E of order 2 in CG .E/=E, where E0 is normal in G. This in turn implies that CD .E/ > E. Since any element of D NG .A/ sends c onto cb, we have CD .E/ NG .A/. Thus CD .E/ D hE; d i or CD .E/ D hE; ad i. Noting that both d and ad inverts A, we may assume without loss of generality that CD .E/ D hE; d i. Setting K D hE; d i, we ﬁrst assume that K satisﬁes K Š E16 . As CK .a/ D hb; ci, ha; Ki Š E4 o C2 with Z.ha; Ki/ D hb; ci (Lemma 99.2). If CG .Z.ha; Ki// > ha; Ki, then take a subgroup Y in CG .Z.ha; Ki// containing ha; Ki with jY W ha; Kij D 2. Since ha; Ki contains exactly four square roots of c and they are all conjugate to a in ha; Ki (Lemma 99.2), we get that CG .a/ covers Y =ha; Ki, contrary to CG .A/ D A. Hence we have CG .Z.ha; Ki// D ha; Ki and then Lemma 99.4 gives that r.G/ 4, a contradiction.

99 2-groups with sectional rank at most 4

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It remains to treat the case K Š C4 C2 C2 . Since Ã1 .K/ is central in D, we get d 2 D b. We compute .adx/2 D .adx/.adx/ D ax.dad /x D axd 2 ad x D axba1 x D ab.xa1 x/ D ab.ab/1 D 1: Setting C D hb; c; adxi, we shall argue that C is a normal self-centralizing elementary abelian subgroup of order 8 in D. There are three nontrivial cosets in NG .A/ modulo E: aE, dE, adxE. All elements in aE are of order 4 (noting that ˆ.ha; Ei/ D hb; ci and the three maximal subgroups of ha; Ei are A Š C4 C2 , hax; bi Š C4 C2 , and E, where.ax/2 D cb) and all elements in dE are of order 4 since d 2 D b and d centralizes E. Finally, in the coset adxE there are exactly four involutions forming the set ¹adxhb; ciº since CE .adx/ D hb; ci. Hence all involutions in NG .A/ are contained in E or C . As E E D and NG .A/ E D, we have C E D. Since each t 2 D NG .A/ has the property c t D cb, we get CD .C / NG .A/ D CD .hb; ci/ and no one of d; x; dx centralizes adx, where dC , xC and dxC are three nontrivial cosets of C in NG .A/. Hence CD .C / D C , as required. Suppose that D contains another such subgroup C1 , namely C1 Š E8 , C1 E D, and CD .C1 / D C1 . Then b 2 C1 since hbi D Z.D/. Suppose that C \ C1 D hbi. Let C1 Š E4 be a complement of hbi in C1 so that C1 acts faithfully on C and CC .C1 / D hbi D Z.C C1 /. By Lemma 99.1, C C1 Š Q8 Q8 . We have a 62 C C1 since a2 D c and Ã1 .C C1 / D hbi. Since jD W .C C1 /j D 2, we conclude that a normalizes C C1 and D D C C1 hai. Suppose that CG .C C1 / 6 C C1 . Since hai normalizes CG .C C1 /, there is an hai-invariant subgroup V of order 4 in CG .C C1 / which contains hbi. But then Œhai; V hbi and so V normalizes ha; bi D A and therefore V D, where .C C1 / \ V D hbi. In that case, V centralizes C , contrary to the fact that C is selfcentralizing in D. Thus CG .C C1 / C C1 and so Lemma 99.3 implies r.G/ 4, a contradiction. Hence we may assume that jC \ C1 j D 4. Since hb; ci D C \ E is a normal four-subgroup of D and D=C Š D8 , it follows that four involutions in C hb; ci are conjugate in D and so hb; ci is a unique normal four-subgroup of D contained in C . This implies C \C1 D hb; ci. Consequently, C1 CD .hb; ci/ D NG .A/. Thus C1 D C , contrary to C1 ¤ C . Hence C is the unique subgroup in D with the required property which implies that C is characteristic in D. Let F G be a subgroup of G containing D so that jF W Dj D 2. Then C is normal in F and D=C Š D8 . Let H D CF .C / so that H E F , jH W C j D 2 and H is abelian. We get F D DH with D \ H D C and ŒD; H D \ H D C . Let h 2 H C so that h2 2 C and h centralizes C and D=C . We get ah D ay with y 2 C . If y 2 hb; ci, then h normalizes A D ha; bi, contrary to NG .A/ D. We get y 62 hb; ci and so y D adx b i c j with i; j D 0; 1. Then ah D a adxb i c j D a2 dxb i c j D cdxb i c j D dxb i c j C1 2 K and h centralizes hb; ci. Thus Ah K, where K is abelian of order 16. This is a ﬁnal contradiction since CG .Ah / D Ah . Our lemma is proved.

44

Groups of prime power order

Now Lemmas 99.5, 99.7 and 99.8 give the following remarkable result. Theorem 99.9 (K. Harada). Let G be a 2-group with a self-centralizing abelian subgroup of order 8. Then each subgroup of G is generated by at most four elements, i.e., the sectional rank of G is at most 4. Theorem 99.10 (K. Harada). Let G be a 2-group containing a subgroup T D U W , where U Š E4 and W Š E4 , D2n , n 3, or SD2m , m 4. Suppose that for each involution x 2 U , CG .x/ D T . Then the sectional rank r.G/ of G is 4. Proof. Since r.T / D 4, we may assume that G > T . We have Z.T / D U Z.W / Š E8 or E16 . Let x 2 NG .T / T with x 2 2 T . If U \ U x ¤ ¹1º, then x normalizes U \ U x and so x centralizes an involution in U \ U x , a contradiction. Hence we have U \ U x D ¹1º which implies Z.W / Š E4 and so W Š E4 and T Š E16 . For any element s 2 NG .T / T with s 2 2 T , U \ U s D ¹1º and so hs; T i Š E4 o C2 (Lemma 99.2). In particular, hs; T i T contains an involution. Let k D jNG .T / W T j and R D NG .T /. Then any element u 2 U has exactly k conjugates in T by the action of R. This gives 2 k 8. Suppose that k D 2. Then R Š E4 o C2 . This forces G D R D NG .T / as T is characteristic in R. Thus r.G/ D 4 by Lemma 99.4. Hence we may assume that k > 2. Suppose next that k D 8. Since there cannot be two orbits of length 8 (under the action of R), all involutions in U are conjugate in R. Let a be an involution in R T which maps into the center of R=T . Since all involutions of ha; T i T are conjugate in ha; T i (as ha; T i Š E4 o C2 ), we have R D CR .a/T . Hence CT .a/ is normal is R. Moreover, U \ CT .a/ D ¹1º. Let L be an R-invariant subgroup with CT .a/ < L < T and jL W CT .a/j D 2. Then jL \ U j D 2 and so all involutions in U cannot be conjugate in R, a contradiction. We have proved that k D 4. If G D R D NG .T /, then jGj D 26 and it is easy to see that in that case r.G/ D 4. Indeed, let A=B Š E25 be a section of G. If B D ¹1º, then A is an elementary abelian subgroup of index 2 in G. But then U \ A ¤ ¹1º, contrary to jCG .U \ A/j D 24 . Hence we have B Š C2 and A D G. But in this case, G 0 D B Š C2 and so an involution in U cannot have k D 4 conjugates in R D G. Hence we have r.G/ D 4 in this case. It follows that we may assume that G > R. This implies that using an element h 2 G NG .R/ with h2 2 R, we get another elementary abelian subgroup T1 D T h of order 16. As CG .u/ D T for any involution in U , we have T1 \U D ¹1º. Hence R D T1 T as jRj D 26 and T1 \T D Z.R/ Š E4 . This in turn implies that R is a split extension of E16 by E4 . Then U1 D U h is a complement of T in R and CT .b/ D T1 \ T for any involution b in U1 and, in addition, hb; T i Š E4 o C2 which shows that all elements in R .T [ T1 / are of order 4 and R0 D ˆ.R/ D Z.R/ D T \ T1 . Hence R has exactly two elementary abelian subgroups T an T1 of order 16. This forces that jNG .R/ W Rj D 2. Setting P D NG .R/, we shall argue that P D G. It sufﬁces to show that T and T1 are the only subgroups of P which are isomorphic to E16 . Indeed, let t 2 P R so that T t D T1 and therefore

99 2-groups with sectional rank at most 4

45

1 .CR .t// T \ T1 . This implies that any elementary abelian subgroup of P which is not contained in R is of order at most 8. Thus G D P , as required. It remains to be shown that r.G/ 4. For this, it sufﬁces to prove that G does not possess a normal elementary abelian subgroup of order 8. Suppose that this is false and let A be such a subgroup. If A R, then A T or A T1 . However, as T t D T1 (t 2 G R) and A 6 T \ T1 , A cannot be t -invariant. Hence A 6 R and A \ R is a normal four-subgroup of G D P which gives A \ R D T \ T1 . Let a 2 A R. Then T a D T1 and so G=T \ T1 Š E4 o C2 which shows that a does not map into the center of G=.T \ T1 /. This contradicts the fact that A=.T \ T1 / E G=.T \ T1 /. Our theorem is proved. It is interesting to notice that if a p-group G has a nonabelian subgroup B of order p 3 such that CG .B/ < B, then G is of maximal class (Proposition 10.17(a)); then every subgroup of G is generated by p elements.

100

2-groups with exactly one maximal subgroup which is neither abelian nor minimal nonabelian

Suppose that G is a p-group all of whose maximal subgroups are metacyclic except one (which is nonmetacyclic). If p > 2 and jGj > p 4 , then Y. Berkovich has shown (with a short and elegant proof) that G must be a so-called L3 -group, i.e., 1 .G/ is of order p 3 and exponent p and G=1 .G/ is cyclic of order p 2 (see Proposition A.40.12). However, if p D 2, then the problem of determination of such groups is much more difﬁcult and this was done in 87. All these results will be used very heavily in this section. Here we continue with this idea of classifying p-groups all of whose maximal subgroups, but one, have a certain strong property. In this section we determine up to isomorphism the 2-groups G all of whose maximal subgroups, but one, are abelian or minimal nonabelian. We begin with the case d.G/ D 2 (Theorems 100.1, 100.2 and 100.3). Actually, a detailed investigation of such groups has already begun in Lemma 76.5. Then we determine such groups satisfying d.G/ > 2 (Theorems 100.4, 100.5 and 100.6). All resulting groups will be presented in terms of generators and relations, but we shall also describe all important characteristic subgroups of these groups for two reasons. One reason is that only the knowledge of the subgroup structure of these groups will make our theorems useful for applications. Another reason is that with this knowledge we see that 2-groups appearing in distinct theorems are nonisomorphic. Conversely, it is easy to check that all groups given in these theorems indeed possess exactly one maximal subgroup which is neither abelian nor minimal nonabelian. The corresponding problem for p > 2 is open, but we think that this problem is within the reach of the present methods in ﬁnite p-group theory. Theorem 100.1. Let G be a two-generator 2-group with exactly one maximal subgroup H which is neither abelian nor minimal nonabelian. If G has an abelian maximal subgroup A, then 1 D ¹A; M; H º is the set of the maximal subgroups of G, where M is minimal nonabelian, A and M are both metacyclic, d.H / D 3 and we have more precisely: G D ha; x j Œa; x D v; v 4 D 1; v 2 D z; v x D v 1 ; v a D v 1 ; x 2 2 hzi; a2 2 hzi; m 2i; m

100

2-groups with exactly one exceptional maximal subgroup

47

where G 0 D hvi Š C4 ; E E G;

K3 .G/ D ŒG; G 0 D hzi Š C2 ;

G D Ehai;

G=E Š C2m ;

H D Eha2 i;

E D hv; xi Š D8 or Q8 ;

ˆ.G/ D G 0 ha2 i is abelian;

Z.G/ D ha2 ; zi;

A D hax; vi is an abelian maximal subgroup of G, M D hv; ai is metacyclic minimal nonabelian of order 2mC2 and jGj D 2mC3 , m 2. Proof. If G has more than one abelian maximal subgroup, then all three maximal subgroups of G are abelian, a contradiction. Hence A is a unique abelian maximal subgroup of G. It follows that 1 D ¹A; M; H º, where M is minimal nonabelian. The subgroup A \ M D ˆ.G/ is abelian, ˆ.M / < ˆ.G/ and jˆ.G/ W ˆ.M /j D 2. Also, ˆ.M / D Z.M / Z.G/ so that for an element m 2 M A, Cˆ.G/ .m/ D Z.M /. If CA .m/ > Z.M /, then G D M C with C D CG .M / and M \ C D Z.M /, contrary to d.G/ D 2. It follows that CA .m/ D Z.M / D Z.G/ and so jG W Z.G/j D 8. From jGj D 2jZ.G/jjG 0 j (Lemma 1.1) follows that jG 0 j D 4. For each x 2 G A, CA .x/ D Z.M / and so x 2 2 Z.M / D Z.G/ which implies that x inverts A=Z.M /. If A=Z.M / Š E4 , then ˆ.G/ D Ã1 .G/ Z.M /, a contradiction. Hence we obtain that A=Z.M / Š C4 and since m 2 M A inverts A=Z.M /, we have G=Z.M / Š D8 . Let a 2 A ˆ.G/. Then hai covers A=Z.M / and so v D Œa; m 2 ˆ.G/ Z.M /. We get 1 D Œa; m2 D Œa; mŒa; mm D vv m and so v m D v 1 which implies that o.v/ D 4. Indeed, if o.v/ D 2, then Œv; m D 1, which contradicts the fact that Cˆ.G/ .m/ D Z.M /. We get G 0 D hvi Š C4 and Œv; m D v 2 implies that M 0 D hv 2 i. Since v 2 is a square in M , it follows that M is metacyclic (Lemma 65.1). In particular, j1 .ˆ.G//j 4 which together with A=Z.M / Š C4 gives 1 .A/ ˆ.G/ and so A is also metacyclic. Here H D ˆ.G/hami is our third maximal subgroup of G. From Cˆ.G/ .am/ D Z.M / follows that H is nonabelian with Z.H / D Z.M / and so Lemma 1.1 yields jH 0 j D 2. If d.H / D 2, then Lemma 65.2 gives that H would be minimal nonabelian, a contradiction. Hence d.H / 3 and so H is the only maximal subgroup of G which is nonmetacyclic. In fact, d.H / D 3 since ˆ.G/ is metacyclic. We are in a position to use Theorem 87.12 for n D 2 since G 0 Š C4 . This gives the generators and relations described in our theorem, where we have used the notation from Theorem 87.12. Theorem 100.2. Let G be a 2-group with d.G/ D 2 which has exactly one maximal subgroup H which is neither abelian nor minimal nonabelian. If the other two maximal subgroups H1 and H2 are minimal nonabelian with H10 D H20 , then one of the following holds: (a) G is one of the groups given in Theorem 87.10. (b) G is the group of order 25 given in Theorem 87.14.

48

Groups of prime power order n

(c) G D hh; x j h2 D 1; n 2; Œh; x D s; s 2 D 1; Œs; h D z; z 2 D Œz; h D Œz; x D Œx; s D 1; x 2 2 hzii, where jGj D 2nC3 ;

G 0 D hz; si Š E4 ;

K3 .G/ D ŒG; G 0 D hzi Š C2 ;

ˆ.G/ D G 0 hh2 i is abelian and the maximal subgroups of G are H1 D hG 0 ; hi, H2 D hG 0 ; xhi (both are nonmetacyclic minimal nonabelian) and H D hx; s; h2 i with d.H / D 3 and H10 D H20 D H 0 D hzi. Proof. Here A D H1 \ H2 D ˆ.G/ is a maximal normal abelian subgroup of G. Set H10 D H20 D hzi A so that H1 =hzi and H2 =hzi are two distinct abelian maximal subgroups in G=hzi. It follows that H=hzi is also abelian and so H 0 D hzi since H is nonabelian. If d.H / D 2, then, by Lemma 65.2(a), H would be minimal nonabelian, a contradiction. Thus d.H / 3. If G=hzi is abelian, then G 0 D hzi and so G D H1 CG .H1 / which gives d.G/ D 3, a contradiction. Hence G=hzi is nonabelian and so .G=hzi/0 Š C2 since G=hzi has three distinct abelian maximal subgroups. Thus jG 0 j D 4 and G 0 A D ˆ.G/. Taking h1 2 H1 A and h2 2 H2 A, we get hh1 ; h2 i D G and so s D Œh1 ; h2 2 G 0 hzi. If G 0 Š C4 , then z is a square in H1 and H2 and so both H1 and H2 are metacyclic (Lemma 65.1). Since d.H / 3, G has exactly one nonmetacyclic maximal subgroup. But then Theorem 87.12 for n D 2 implies that G has an abelian maximal subgroup, a contradiction. Thus G 0 D hs; zi Š E4 , where s is an involution. If s 2 Z.G/, then G=hsi would be abelian (because hh1 ; h2 i D G and s D Œh1 ; h2 ), a contradiction. Hence s 62 Z.G/ and so s 62 Z.H1 / or s 62 Z.H2 /. Without loss of generality we may assume that s 62 Z.H1 /. Suppose that z is a square in H1 , i.e., there is v 2 H1 such that v 2 D z. Suppose at the moment that v 2 H1 A in which case hv; G 0 i D hv; si Š D8 since s v D sz. It follows that hv; G 0 i D H1 . Since CG .H1 / H1 (otherwise, d.G/ D 3), G would be of maximal class (Proposition 10.17), a contradiction (noting that 2-groups of maximal class have a cyclic maximal subgroup). Thus v 2 A D ˆ.G/. In that case, both H1 and H2 are metacyclic (Lemma 65.1) which together with d.H / 3 allows us to use 87, part 2o . If G has a normal elementary abelian subgroup of order 8, then we get the groups in part (a) of our theorem. If G has no normal elementary abelian subgroup of order 8, then we get the group of order 25 given in part (b) of our theorem. Now we assume that z is not a square in H1 which implies that H1 is nonmetacyclic (Lemma 65.1). If h1 is an involution, then s h1 D sz shows that hh1 ; si Š D8 and so n1 hh1 ; si D H1 is metacyclic, a contradiction. Hence o.h1 / D 2n , n 2. Set u D h21 so that u 2 Z.H1 / and u 62 G 0 since z is not a square in H1 and s h1 D sz. We have E D 1 .H1 / D hz; s; ui Š E8 ;

EEG

and

EA

which implies that H2 is nonmetacyclic and therefore z is also not a square in H2 . Since H1 D Ehh1 i D hh1 ; si, we have jH1 j D 2nC2 , jGj D 2nC3 and A D hh21 ; s; zi is abelian of order 2nC1 and type .2n1 ; 2; 2/. Also, H1 is a splitting extension of G 0

100

2-groups with exactly one exceptional maximal subgroup

49

by hh1 i Š C2n , n 2. As d.G=G 0 / D 2, we get that G=G 0 is abelian of type .2n ; 2/. We obtain G D FH1 , where F \ H1 D G 0 and jF W G 0 j D 2 so that F D hG 0 ; xi with o.x/ 4. In fact, we have x 2 2 hzi. Indeed, if F is not elementary abelian, then Ã1 .F / Š C2 and Ã1 .F / G 0 . But F E G and so Ã1 .F / Z.G/ which implies that Ã1 .F / D hzi as G 0 6 Z.G/. Since hxh1 iG 0 =G 0 is another cyclic subgroup of index 2 in G=G 0 (distinct from H1 =G 0 ), M D hG 0 ; xh1 i is a maximal subgroup of G distinct from H1 and M 0 D hzi (since H20 D H 0 D hzi). If G 0 Z.M /, then M would be abelian, a contradiction. We get s xh1 D sz and so M D hxh1 ; si is minimal nonabelian which yields that M D H2 . We may set h2 D xh1 , where Œh1 ; h2 D s and G 0 D hs; zi. From s xh1 D sz it follows 1

1

1

s x D .s xh1 /h1 D .sz/h1 D s h1 z D .sz/z D s and so F is abelian. As s D Œh1 ; h2 , we get s D Œh1 ; h2 D Œh1 ; xh1 D Œh1 ; h1 Œh1 ; xh1 D Œh1 ; xh1 and so Œh1 ; x D sz: We conclude that ˆ.G/ D G 0 hh21 i is abelian and H D F hh21 i, where Œh21 ; x D Œh1 ; xh1 Œh1 ; x D .sz/h1 .sz/ D .szz/.sz/ D z: Since H=hzi D H=H 0 is abelian of type .2; 2; 2n1 /, we have d.H / D 3. Replacing s with s 0 D sz, we get Œh1 ; x D s 0 and writing again s instead of s 0 , we may write Œh1 ; x D s 2 G 0 hzi. Also, writing h instead of h1 , we obtain the relations of part (c) of our theorem. Theorem 100.3. Let G be a 2-group with d.G/ D 2 which has exactly one maximal subgroup H which is neither abelian nor minimal nonabelian. If the other two maximal subgroups H1 and H2 are minimal nonabelian with H10 ¤ H20 , then one of the following holds: (a) G is a group of order 26 with n D 2 given in Theorem 87.15(a). (b) G is a group of order 2mC4 .m 2/ with n D 2 given in Theorem 87.16. (c) G D hh; x j Œh; x D v; v 4 D 1; v h D vz1 ; v x D v 1 v 2 D z1 z2 ; z12 D z22 D 1; m Œz1 ; h D Œz1 ; x D Œz2 ; h D Œz2 ; x D 1; x 2 2 hz1 ; z2 i; h2 D .z1 z2 / i, where D 0; 1;

m 2;

jGj D 2mC4 ;

K3 .G/ D ŒG; G 0 D hz1 ; z2 i Š E4

G 0 D hv; z1 i Š C4 C2 ; and

hz1 ; z2 i Z.G/:

Moreover, the maximal subgroups of the group G are H1 D G 0 hhi, H2 D G 0 hhxi and H D G 0 hx; h2 i, where H1 and H2 are both nonmetacyclic minimal nonabelian with H10 D hz1 i, H20 D hz2 i and H 0 D hz1 ; z2 i Š E4 (d.H / D 3).

50

Groups of prime power order

Proof. Set hz1 i D H10 and hz2 i D H20 so that W D hz1 ; z2 i Š E4 , W Z.G/ and W A D H1 \ H2 D ˆ.G/, where A is abelian. Here ¹H1 =hz1 i; H2 =hz1 i; H=hz1 iº is the set of maximal subgroups of G=hz1 i and H1 =hz1 i is abelian and two-generated, H2 =hz1 i is minimal nonabelian and so H=hz1 i must be nonabelian. If H=hz1 i is minimal nonabelian, then by a result of N. Blackburn (Theorem 44.5), G=hz1 i would be metacyclic. But then, by another result of N. Blackburn (Lemma 44.1 and Corollary 44.6), G is also metacyclic, contrary to W G 0 and W Š E4 . Hence H=hz1 i is neither abelian nor minimal nonabelian. By Theorem 100.1, d.H=hz1 i/ D 3;

.H=hz1 i/0 Š C2 ;

G 0 =hz1 i Š C4 :

Similarly, considering G=hz2 i, we obtain that .H=hz2 i/0 Š C2 and G 0 =hz2 i Š C4 . It follows that G 0 is abelian of type .4; 2/ with Ã.G 0 / D hz1 z2 i. On the other hand, ¹H1 =W; H2 =W; H=W º is the set consisting of the maximal subgroups of the nonabelian group G=W , where both H1 =W and H2 =W are abelian. Hence H=W is also abelian and so H 0 W . By the above, H 0 is distinct from hz1 i and hz2 i and so either H 0 D W or H 0 D hz1 z2 i. Suppose that H 0 D hz1 z2 i. Then ¹H1 =hz1 z2 i; H2 =hz1 z2 i; H=hz1 z2 iº is the set of maximal subgroups of G=hz1 z2 i, where H1 =hz1 z2 i and H2 =hz1 z2 i are minimal nonabelian and H=hz1 z2 i is abelian. By Theorem 106.3 (or by results of 71), we deduce that G=hz1 z2 i is metacyclic, contrary to G 0 =hz1 z2 i Š E4 . We have thus proved that H 0 D W D hz1 ; z2 i. Take h1 2 H1 A, h2 2 H2 A so that we have hh1 ; h2 i D G. If v D Œh1 ; h2 2 W , then v is an involution in Z.G/ and G=hvi is abelian, G 0 hvi, a contradiction. Hence v 62 W and so v 2 G 0 W is of order 4 with v 2 D z1 z2 and hvi is not normal in G (and so also hvz1 i is not normal in G). Indeed, if hvi is normal in G, then G=hvi would be abelian since Œh1 ; h2 D v and hh1 ; h2 i D G. In particular, hvi cannot be normal in both H1 and H2 and so we may assume without loss of generality that hvi is not normal in H1 . Hence Œh1 ; v D z1 which gives v h1 D vz1 and H1 D hh1 ; vi. Suppose that H1 is metacyclic. Then there exists h01 2 H1 such that .h01 /2 D z1 . 0 If h01 2 H1 A, then v h1 D vz1 and so H1 D hh01 ; vi is of order 24 . But then jGj D 25 and jG 0 j D 23 imply (using a result of O. Taussky) that G is of maximal class, a contradiction. It follows that h01 2 A and then .h01 v/2 D .h01 /2 v 2 D z1 .z1 z2 / D z2 which implies that H2 is also metacyclic. We are in a position to use 87, 2o . By Theorems 87.9 and 87.10, G has no normal elementary abelian subgroup of order 8 (since jG 0 j D 8). We have ˆ.G 0 / ¤ ¹1º and Z.G/ W is noncyclic. If G=ˆ.G 0 / has no normal elementary abelian subgroup of order 8, then G is isomorphic to a group of

100

2-groups with exactly one exceptional maximal subgroup

51

order 26 given in Theorem 87.15(a) for n D 2. If G=ˆ.G 0 / has a normal elementary abelian subgroup of order 8, then G is a group of order 2mC4 , m 2, given in Theorem 87.16 for n D 2. Suppose that H1 is nonmetacyclic. If there exists an element l 2 H1 A such that l 2 2 G 0 , then v l D vz1 gives that H1 D hl; vi D G 0 hli is nonmetacyclic minimal nonabelian of order 24 . But in that case, jGj D 25 and jG 0 j D 23 imply (using a result of O. Taussky) that G is of maximal class, a contradiction. It then follows that 1 .H1 / D 1 .A/ Š E8 and so H2 is also nonmetacyclic minimal nonabelian. Since h22 2 H1 , we have Œh1 ; h22 D z1 ; D 0; 1: We compute

z1 D Œh1 ; h22 D Œh1 ; h2 Œh1 ; h2 h2 D vv h2 ; and so

v h2 D v 1 z1 D v.z1 z2 /z1 D vz1

C1

z2 :

If D 0, then v h2 D v.z1 z2 /, contrary to H20 D hz2 i. Thus D 1 and so v h2 D vz2 which implies H2 D hh2 ; vi D G 0 hh2 i. Also, v h1 D vz1 gives H1 D hh1 ; vi D G 0 hh1 i and since h21 62 G 0 , we have H1 =G 0 Š C2m , m 2, and then also H2 =G 0 Š C2m . Since d.G=G 0 / D 2, we see that G=G 0 is abelian of type .2m ; 2/, m 2. We may set G D FH1 with F \H1 D G 0 and jF W G 0 j D 2. Since hh1 i covers H1 =G 0 Š C2m , v h1 D vz1 and neither z1 nor z2 are squares of any element in A D G 0 hh21 i D ˆ.G/, m we get h21 D .z1 z2 / , D 0; 1. We may set h2 D h1 x with x 2 F G 0 so that from v h2 D vz2 follows vz2 D v h2 D .v h1 /x D .vz1 /x D v x z1 and so v x D v.z1 z2 / D v 1 which gives x 2 2 hz1 ; z2 i Z.G/. It then follows from v D Œh1 ; h2 that v D Œh1 ; h1 x D Œh1 ; xŒh1 ; h1 x D Œh1 ; x: Finally, we have H2 D G 0 hh1 xi and H D F hh21 i, where F 0 D hŒv; xi D hz1 z2 i and Œh21 ; x D Œh1 ; xh1 Œh1 ; x D v h1 v D .vz1 /v D z1 v 2 D z1 .z1 z2 / D z2 and so indeed H 0 D hz1 ; z2 i Š E4 which shows that H is neither abelian nor minimal nonabelian. Writing h instead of h1 , we have obtained the relations given in part (c) of our theorem. We turn now to the case d.G/ 3. Since G possesses at least one minimal nonabelian maximal subgroup, it follows that in this case d.G/ D 3. It is well known that the number of abelian maximal subgroups in a nonabelian 2-group G is 0; 1 or 3. According to this fact, we shall subdivide our study of the title groups with d.G/ D 3.

52

Groups of prime power order

Theorem 100.4. Let G be a 2-group with d.G/ D 3 which has exactly one maximal subgroup which is neither abelian nor minimal nonabelian. If G possesses more than one abelian maximal subgroup, then one of the following holds: (a) G D Q Z, where Q Š Q8 , Z Š C2n , n 3 and Q \ Z D Z.Q/. (b) G D Q Z, where Q Š Q8 and Z Š C2n , n 2. (c) G D D Z, where D Š D8 and Z Š C2n , n 2. Proof. By our assumption, G has exactly three abelian maximal subgroups. This implies jG 0 j D 2 and G possesses exactly three maximal subgroups which are minimal nonabelian. Suppose that H is a minimal nonabelian maximal subgroup of G. Since H 0 D G 0 Š C2 , we get G D H Z.G/, where Z.G/ \ H D Z.H / D ˆ.H / D ˆ.G/: All three maximal subgroups of G, containing Z.G/, are abelian. Let G be a title group with jG 0 j D 2. Then G possesses a maximal subgroup H which is minimal nonabelian. From H 0 D G 0 Š C2 follows that G D H Z.G/ with jG W Z.G/j D 4. In that case, d.G/ D 3 and all three maximal subgroups of G containing Z.G/ are abelian. It follows that for a title group G the assumption G 0 Š C2 is equivalent with the assumption that G has more than one abelian maximal subgroup. In what follows H will denote a ﬁxed maximal subgroup of G which is minimal nonabelian. Suppose that there is an involution c 2 Z.G/ Z.H /. Then G D H hci and so each maximal subgroup of G which does not contain hci is isomorphic to G=hci Š H and so is minimal nonabelian, a contradiction. Hence there are no involutions in Z.G/ H which implies that 1 .Z.H // D 1 .Z.G// so that d.Z.H // D d.Z.G//. It follows that for each x 2 Z.G/ H , x 2 2 Z.H / ˆ.Z.H //. Suppose that jGj D 24 . Then each nonabelian maximal subgroup of G is isomorphic to D8 or Q8 and so is minimal nonabelian, a contradiction. Hence jGj 25 and, in particular, H is not isomorphic to Q8 or D8 . (i) First assume that H is metacyclic. Since H is not isomorphic to Q8 , it follows that H is a “splitting” metacyclic group and so we may set m

n

H D ha; b j a2 D b 2 D 1; ab D az; z D a2

m1

i;

where m 2;

n 1;

m C n 4;

H 0 D hzi;

jH j D 2mCn ;

jGj D 2mCnC1 :

We see that Z.H / D ha2 i hb 2 i D ˆ.H / D ˆ.G/ and, for each x 2 Z.G/ H , x 2 2 ha2 ; b 2 i ha4 ; b 4 i since ha4 ; b 4 i D ˆ.Z.H //. Suppose that n D 1 so that b is an involution, m 3, H Š M2mC1 , Z.H / D ha2 i. Hence Z.G/ Š C2m is cyclic and therefore we may choose c 2 Z.G/ H such that c 2 D a2 which gives .ca/2 D c 2 a2 D 1. As Œca; b D z, we get D D hca; bi Š D8 ,

100

2-groups with exactly one exceptional maximal subgroup

53

hca; bi \ Z.G/ D hzi which together with jG W Z.G/j D 4 gives G D DZ.G/. But D 2 .Z.G// contains a subgroup Q Š Q8 and so G D QZ.G/ with Z.G/ Š C2m , m 3, Q \ Z.G/ D Z.Q/ D hzi and we have obtained the groups stated in part (a) of our theorem. It remains to treat the case n 2. Suppose that there is an involution x 2 G H . We know that x 62 Z.G/ and so Œa; x ¤ 1 or Œb; x ¤ 1. Obviously, ha; b; xi D G. Suppose that Œa; x ¤ 1. As hai E G, ha; xi is minimal nonabelian of order 2mC1 . Because jGj D 2mCnC1 and n 2, ˆ.G/ha; xi D ha; x; b 2 i is a maximal subgroup of G which is neither abelian nor minimal nonabelian. Assume at the moment that Œb; x ¤ 1. In that case, hbi hzi is normal in G and so hb; xi is a nonmetacyclic minimal nonabelian subgroup of G of order 2nC2 . It follows that hb; xi must be maximal in G with jGj D 2nC3 (and so m D 2). But the case of a nonmetacyclic minimal nonabelian maximal subgroup in G will be studied in part (ii) of this proof. Hence we may assume Œx; b D 1 so that Œx; ab D Œx; aŒx; b D z which implies that hx; abi is minimal nonabelian and so hx; abi must be maximal in G. Now, habi covers H=hai Š C2n , n 2, and so o.ab/ 2n . We get n

n

n

n1 .2n 1/

.ab/2 D a2 b 2 Œb; a2 n

n

D a2 :

If n m, then we have .ab/2 D 1 and so o.ab/ D 2n and habi \ hai D ¹1º. Since hab; zi E G, we see that hab; xi is a nonmetacyclic minimal nonabelian subgroup of order 2nC2 . In that case, hab; xi must be maximal in G (with m D 2) and again this will be studied in part (ii) of this proof. It follows that we may assume n < m and we n n n set in that case s D m n 1. From .ab/2 D a2 and o.a2 / D 2s we infer that o.ab/ D 2nCs D 2m and habi hzi so that habi E G. Hence hab; xi is metacyclic minimal nonabelian of order 2mC1 and so hab; xi must be maximal in the group G. From jGj D 2mCnC1 follows n D 1, contrary to our assumption. We may assume Œa; x D 1 and so we must have Œb; x ¤ 1. Since hbi hzi E G, hb; xi is nonmetacyclic minimal nonabelian of order 2nC2 . If hb; xi is maximal in G, then this case will be treated in part (ii) of this proof. Thus we may assume that hb; xi is not maximal in G and so M D ˆ.G/hb; xi is maximal in G and M is neither abelian nor minimal nonabelian. It then follows that the subgroup hab; xi (with Œab; x D z), which is minimal nonabelian, must be also a maximal subgroup in G. Since habi n n covers H=hai, we have o.ab/ 2n , n 2 and .ab/2 D a2 . If n m, then we get o.ab/ D 2n and habi\hai D ¹1º and so hab; xi is nonmetacyclic minimal nonabelian of order 2nC2 . In that case, hab; xi is maximal in G (with m D 2) and again this will be treated in part (ii) of this proof. We may assume that n < m and then o.ab/ D 2m , habi hzi and so hab; xi is metacyclic minimal nonabelian of order 2mC1 . But then jGj D 2mCnC1 implies n D 1, contrary to our assumption. We have proved that we may assume that there are no involutions in G H . If there is c 2 Z.G/ H such that c 2 D h2 for some h 2 H , then the abelian subgroup hh; ci is noncyclic since hhi and hci are two distinct cyclic subgroups of hh; ci of the same order. But hh; ci\H D hhi and so there is an involution in hh; ciH , a contradiction.

54

Groups of prime power order

It follows that not every element in Ã1 .H / D Z.H / is a square of an element in H . By Proposition 26.23, H is not a powerful 2-group which implies H 0 D hzi 6 Ã2 .H /. This forces m D 2 and c 2 is not a square in H for any c 2 Z.G/ H . We compute for any integers i; j .ai b j /2 D a2i b 2j Œb j ; ai D a2i b 2j z ij : We get that a2i b 2j 2 Ã1 .H / D Z.H / is a square in H if and only if i or j is even. Therefore for any c 2 Z.G/ H , c 2 D a2i b 2j , where both i and j are odd, and then (since m D 2 and so a2 D z) c 2 D zb 2j , where j is odd. Consider the nonabelian subgroup S D ha; b j ci, where .b j c/2 D b 2j c 2 D b 2j zb 2j D z; and so S Š Q8 . Hence we conclude that G D hS; ci D S hci Š Q8 C2n with n 2, where S hb 2 i Š Q8 C2n1 is a unique maximal subgroup of G which is neither abelian nor minimal nonabelian. We have obtained the groups stated in part (b) of our theorem. (ii) It remains to consider the case where H is nonmetacyclic minimal nonabelian. We may set m

n

H D ha; b j a2 D b 2 D 1; Œa; b D z; z 2 D Œa; z D Œb; z D 1i; where we may assume m 2, n 1 as jH j 24 . Here H 0 D hzi, jH j D 2mCnC1 and so jGj D 2mCnC2 . Also, z is not a square in H , Z.H / D ha2 i hb 2 i hzi D ˆ.H / D ˆ.G/ and for each x 2 Z.G/ H , x 2 2 Z.H / ˆ.Z.H //. (ii1) First assume n D 1 so that Z.H / D ha2 i hzi and for a c 2 Z.G/ H , 2 c D a2i z j . Suppose that i is even and then j must be odd and so we may set in that 0 0 case c 2 D a4i z and compute for an element c 0 D a2i c 2 Z.G/ H 0

0

0

0

.c 0 /2 D .a2i c/2 D a4i c 2 D a4i a4i z D z: This gives G D H hc 0 i with .c 0 /2 D z, where hzi D H 0 , and it is easy to see that in that case G is an A2 -group (see Proposition 71.1), a contradiction. We have proved that i must be odd. The subgroup D D hai c; bi is minimal nonabelian since Œai c; b D Œa; bi D z. We have also .ai c/2 D a2i c 2 D a2i a2i z j D z j which shows that D Š D8 . But hci \ hzi D ¹1º, where hzi D Z.D/, and so hD; ci D hai c; b; ci D G D D hci with o.c/ D 2m , m 2. The subgroup D hc 2 i is a unique maximal subgroup of G which is neither abelian nor minimal nonabelian and we have obtained the groups stated in part (c) of our theorem. (ii2) It remains to consider the case n 2. In this case, for an element c 2 Z.G/H we have c 2 D a2i b 2j z k , where at least one of the integers i; j; k is odd.

100

2-groups with exactly one exceptional maximal subgroup

55

Suppose that both i and j are even so that in this case k is odd and we may set 0 0 0 0 c 2 D a4i b 4j z. For the element c 0 D a2i b 2j c, we get 0

0

0

0

0

0

.c 0 /2 D a4i b 4j c 2 D a4i b 4j a4i b 4j z D z; and so G D H hc 0 i with .c 0 /2 D z and hzi D H 0 which gives that G is an A2 -group from Proposition 71.1, a contradiction. Now assume that one of the integers i; j is even and the other one is odd. Note that i; j occur symmetrically and so we may assume that i is odd and j is even. In that case, the subgroup T D hai b j c; bi is minimal nonabelian since Œai b j c; b D Œa; bi D z. Using the fact that b j 2 Z.G/, we get .ai b j c/2 D a2i b 2j c 2 D a2i b 2j a2i b 2j z k D z k : Since ˆ.T / D hb 2 i hzi, we have jT j D 2nC2 . On the other hand, jGj D 2mCnC2 with m 2 and so ˆ.G/T is a maximal subgroup of G which is neither abelian nor minimal nonabelian. Consider now the minimal nonabelian subgroup U D hab; aci, where Œab; ac D z. We have .ab/2 D a2 b 2 z;

.ac/2 D a2 c 2 D a2 a2i b 2j z k D a2.iC1/ b 2j z k ;

where both i C 1 and j are even, and ˆ.U / D ha2 b 2 z; a2.i C1/ b 2j z k ; zi ha2 b 2 ; ziˆ.Z.H // since a2.i C1/ b 2j 2 ˆ.Z.H //. But Z.H / D ha2 i hb 2 i hzi and so d.Z.H // D 3 which gives ˆ.U / < Z.H / D ˆ.G/. This shows that ˆ.G/U is another maximal subgroup of G which is neither abelian nor minimal nonabelian, a contradiction. It remains to consider the possibility that both i and j are odd. Then we consider the minimal nonabelian subgroup V D hai b j c; bi, where Œai b j c; b D Œa; bi D z. We get .ai b j c/2 D .ai b j /2 c 2 D a2i b 2j z ij c 2 D a2i b 2j z a2i b 2j z k D z kC1 which shows that jV j D 2nC2 . But jGj D 2mCnC2 with m 2 and so ˆ.G/V is a maximal subgroup of G which is neither abelian nor minimal nonabelian. Now we consider a minimal nonabelian subgroup W D ha; bci, where Œa; bc D z. We compute .bc/2 D b 2 c 2 D b 2 a2i b 2j z k D a2i b 2.1Cj / z k ;

56

Groups of prime power order

where i is odd and 1 C j is even. Then we have ˆ.W / D ha2 ; .bc/2 ; zi D ha2 ; a2i b 2.1Cj / z k ; zi D ha2 ; ziˆ.Z.H // < Z.H / D ˆ.G/ since b 2.1Cj / 2 ˆ.Z.H //. Hence ˆ.G/W is another maximal subgroup of G which is neither abelian nor minimal nonabelian, a contradiction. In the rest of this section we shall assume that G is a title group with d.G/ D 3 which possesses at most one abelian maximal subgroup. We know that in that case jG W Z.G/j 8 and jG 0 j > 2. Let H be a maximal subgroup of G which is minimal nonabelian. Then ˆ.H / D Z.H / ˆ.G/ and jH W ˆ.H /j D 4. As jG W ˆ.H /j D 8, we must also have ˆ.H / D ˆ.G/. Let K ¤ H be another maximal subgroup of G which is minimal nonabelian. Then Z.K/ D ˆ.K/ D ˆ.G/ which implies ˆ.G/ Z.G/ and so ˆ.G/ D Z.G/. Let M be the unique maximal subgroup of G which is neither abelian nor minimal nonabelian. Since jM W ˆ.G/j D 4 and ˆ.G/ D Z.G/, we have M D S ˆ.G/, where S is minimal nonabelian, S \ˆ.G/ D ˆ.S/ < ˆ.G/ and so M 0 D S 0 Š C2 with d.M / 3. Theorem 100.5. Let G be a 2-group with d.G/ D 3 which has exactly one maximal subgroup M which is neither abelian nor minimal nonabelian. If G possesses exactly one abelian maximal subgroup A, then ˆ.G/ D Z.G/;

G 0 Š E4 ;

M 0 Š C2 ;

d.M / 3

and one of the following holds: (a) If G has no normal elementary abelian subgroup of order 8, then G is one of the groups given in Theorem 87.8(e). (b) If G has a normal subgroup E Š E8 but 1 .G/ > E, then G D ht; t 0 ; c j t 2 D t 02 D c 4 D 1; Œt; t 0 D c 2 D z; Œc; t D u; u2 D Œc; t 0 D Œu; t D Œu; t 0 D Œu; c D Œz; t D Œz; t 0 D 1i; where

jGj D 25 ;

G 0 D ˆ.G/ D Z.G/ D hz; ui Š E4 ;

1 .G/ D M D ht; t 0 iG 0 Š C2 D8 ; the subgroup A D ht 0 ; ciG 0 is abelian of type .4; 2; 2/ and the other ﬁve maximal subgroups of G are nonmetacyclic minimal nonabelian. (c) If G has a normal subgroup E Š E8 , 1 .G/ D E and E 6 A, then mC1

G D ha; b; t j a2

m

D b 4 D t 2 D 1; a2 D z; b 2 D u;

Œa; t D u; Œb; t D z; Œu; a D Œu; t D Œa; b D Œz; t D 1i;

100

2-groups with exactly one exceptional maximal subgroup

57

where jGj D 2mC4 ; m 2;

G 0 D hz; ui Š E4 ;

ˆ.G/ D Z.G/ D ha2 ; ui Š C2m C2 ; A D ha; bi is abelian of type .2mC1 ; 4/;

E 6 Z.G/;

M D hb; ti ha2 i;

where hb; ti is the nonmetacyclic minimal nonabelian group of order 24 and the other ﬁve maximal subgroups of G are minimal nonabelian. (d) If G has a normal subgroup E Š E8 , 1 .G/ D E and E A, then G D ha; b; d j a4 D b 2 D d 4 D 1; a2 D d 2 D z; Œa; d D z; Œa; b D c; c 2 D Œc; d D Œc; a D Œc; b D Œb; d D 1i; where jGj D 25 ;

G 0 D ˆ.G/ D Z.G/ D hz; ci Š E4 ; M D ha; d; ci Š Q8 C2 ;

A D hb; d iˆ.G/;

E 6 Z.G/

and the other ﬁve maximal subgroups of G are minimal nonabelian. Proof. Let 1 D ¹M; A; H1 ; : : : ; H5 º be the set of maximal subgroups of G, where H1 ; : : : ; H5 are minimal nonabelian. By Exercise 1.69(a), we have jG 0 W .A0 H10 /j D jG 0 W H10 j 2 and so jG 0 j D 4 (since jG 0 j > 2). If G 0 D hvi Š C4 , then we get H10 D D H50 D hv 2 i. But then G=hv 2 i is a nonabelian group with at least ﬁve abelian maximal subgroups Hi =hv 2 i, i D 1; : : : ; 5, a contradiction. Hence we obtain that G 0 Š E4 . Since A=M 0 and M=M 0 are two abelian maximal subgroups of the nonabelian group G=M 0 , it follows that there is exactly one minimal nonabelian maximal subgroup of G, say H5 , such that H50 D M 0 . With similar arguments we see that we may assume that H10 D H20 , H30 D H40 , H50 D M 0 are three pairwise distinct subgroups of order 2 in G 0 Š E4 . Suppose that the group G has no normal elementary abelian subgroup of order 8. Then A; H1 ; : : : ; H5 are metacyclic and so M is the only maximal subgroup of G which is nonmetacyclic. Since d.G/ D 3 and G 0 Š E4 , we see that G is isomorphic to one of the groups stated in Theorem 87.8(e) which gives part (a) of our theorem. From now on we assume that G has a normal elementary abelian subgroup E of order 8. Suppose at the moment that G possesses an elementary abelian subgroup F of order 16. Obviously, F is a maximal elementary abelian subgroup in G. Indeed, if X is an elementary abelian subgroup of order 32 in G, then jX \H1 j D 16, a contradiction. Since G 0 Z.G/, we have G 0 F and so F E G. If G=F is noncyclic, then there are at least three distinct maximal subgroups of G containing F and so at least one of them is minimal nonabelian, a contradiction. Hence G=F is cyclic and let a 2 G F be such that hai covers G=F . Suppose that jG W F j D 2 holds so that F is an abelian

58

Groups of prime power order

maximal subgroup in G. Since CF .a/ D Z.G/ D ˆ.G/ and jG=ˆ.G/j D 8, we get CF .a/ D G 0 Š E4 . By Lemma 99.2, G Š E4 o C2 and so we may assume that a is an involution. Let f1 ; f2 2 F G 0 so that F D hf1 ; f2 i G 0 and G D hf1 ; f2 ; ai. We have ha; f1 i Š ha; f2 i Š D8 and ha; f1 iG 0 and ha; f2 iG 0 are two distinct maximal subgroups of G which are isomorphic to D8 C2 (and so they are neither abelian nor minimal nonabelian), a contradiction. We have proved that G=F Š C2m , m 2. Since a2 2 ˆ.G/ D Z.G/, a induces an involutory automorphism on F which together with jG W Z.G/j D 8 implies CF .a/ D G 0 . As a2 62 F and 1 .hai/ Z.G/, we must have 1 .hai/ F (because E32 is not a subgroup of G). Hence hai \ F D hai \ G 0 D hzi Š C2 and so o.a/ D 2mC1 ;

m 2;

jGj D 2mC4 ;

Z.G/ D ˆ.G/ D G 0 ha2 i:

We may set F D hx; y; u; zi, where G 0 D hu; zi, Œa; x D u, Œa; y D z, G D hx; y; ai and so the structure of G is completely determined. Now, ha; yi Š hax; yi Š M2mC2 so that hui ha; yi and hui hax; yi are two distinct maximal subgroups of G which are neither abelian nor minimal nonabelian, a contradiction. We have proved that G does not possess an elementary abelian subgroup of order 16. Since G 0 Z.G/, we have G 0 < E Š E8 . Next suppose that E < 1 .G/ so that there is an involution t 2 G E with CE .t/ D G 0 and S D hE; ti E G. If G=S is noncyclic, then S is contained in a maximal subgroup of G which is minimal nonabelian, a contradiction (to the structure of minimal nonabelian 2-groups). Hence G=S is cyclic and let c 0 2 G S be such that hc 0 i covers G=S. Let t 0 2 E G 0 so that 1 ¤ Œt; t 0 D z 2 G 0 and ht; t 0 i D D Š D8 . Also, E1 D hG 0 ; ti is another elementary abelian normal subgroup of order 8 in G. All elements in S .E [ E1 / are of order 4 and v D t t 0 is one of them. We infer that v 2 D z, S D D hui Š D8 C2 , where u 2 G 0 hzi, and also Z.S/ D G 0 with Z.G/ D ˆ.G/ D G 0 hc 02 i so that G D ht; t 0 ; c 0 i and M D S hc 02 i must be a unique maximal subgroup of G which is neither abelian nor minimal nonabelian. If x 2 G M , then x is either an involution or 1 .hxi/ G 0 . If x is an involution, then Œt; x ¤ 1 (because E16 is not a subgroup of G) and so ht; xi Š D8 , jG W S j D 2, jGj D 25 , and ht; xiG 0 would be another maximal subgroup of G which is neither abelian nor minimal nonabelian, a contradiction. We have proved that x is not an involution and so 1 .hxi/ G 0 . Indeed, 1 .hxi/ Z.G/ and so 1 .hxi/ S which implies that 1 .hxi/ Z.S/ D G 0 . Now, A \ M is equal to one of three abelian maximal subgroups of M (containing ˆ.G/ D Z.G/): hE1 ; c 02 i, hE; c 02 i, hG 0 hvi; c 02 i, where A is the unique abelian maximal subgroup of G. We choose c 2 A M (instead of c 0 ), where hci also covers G=S , o.c/ 4, 1 .hci/ G 0 , ˆ.G/ D Z.G/ D G 0 hc 2 i, and c centralizes exactly one of the elements in the set ¹t; t 0 ; v D t t 0 º. Indeed, otherwise, G=hzi would be abelian since c generates G together with any two elements in the above set. But then G 0 D hzi, a contradiction. Interchanging t and t 0 (if necessary), we may assume that Œc; t 0 D 1 or Œc; t t 0 D 1. In that case, Œc; t ¤ 1 and if Œc; t D z, then again G=hzi would be abelian, a contradiction. It follows that we may set Œc; t D u 2 G 0 hzi.

100

2-groups with exactly one exceptional maximal subgroup

59

First assume Œc; t t 0 D 1 which gives Œc; t 0 D u. We have M D ht; t 0 iˆ.G/ and A D hc; t t 0 iˆ.G/. It follows that the other ﬁve maximal subgroups ˆ.G/T of G must be minimal nonabelian, where T is one of the following minimal nonabelian subgroups: ht; ci; ht 0 ; ci; ht; t 0 ci; ht 0 ; tci; ht t 0 ; tci: We have to show that in each of these cases ˆ.T / ˆ.G/ D hG 0 ; c 2 i D hu; z; c 2 i. Note that Œt; c D u and so ˆ.ht; ci/ D hc 2 ; ui which implies 1 .hci/ 2 ¹hzi; huziº. We have Œt; t 0 c D zu and so ˆ.ht; t 0 ci/ D h.t 0 c/2 D c 2 u; zui. Here if o.c/ 8, then 1 .ht 0 ci/ D 1 .hci/ and then 1 .hci/ 2 ¹hzi; huiº which together with the above result gives 1 .hci/ D hzi, and if o.c/ D 4, then c 2 D uz. We have Œt t 0 ; tc D z and so ˆ.ht t 0 ; tci/ D h.t t 0 /2 D z; .t c/2 D c 2 u; zi D hc 2 u; zi: If o.c/ D 4, then by the above c 2 D uz and then ˆ.ht t 0 ; tci D hzi, a contradiction. In case o.c/ 8, then by the above 1 .hci/ D hzi and as 1 .hc 2 ui/ D 1 .hci/ D hzi, we get again ˆ.ht t 0 ; tci/ D hc 2 ui 6 ˆ.G/, a contradiction. Now assume Œc; t 0 D 1 and from before we know that Œt; t 0 D z and Œc; t D u. We have here M D ht; t 0 iˆ.G/ and A D hc; t 0 iˆ.G/ so that the other ﬁve maximal subgroups must be minimal nonabelian. Since Œt; c D u, we get ˆ.ht; ci/ D hc 2 ; ui which gives 1 .hci/ 2 ¹hzi; huziº. Further, Œt; t 0 c D zu and so we conclude that ˆ.ht; t 0 ci/ D h.t 0 c/2 D c 2 ; zui which implies that 1 .hci/ 2 ¹hui; hziº. This fact together with our previous result gives 1 .hci/ D hzi. We have Œt 0 ; tc D z and so ˆ.ht 0 ; tci/ D h.tc/2 D c 2 u; zi. If o.c/ 8, then .c 2 u/2 D c 4 and hc 4 i hzi since 1 .hci/ D hzi. In this case, ˆ.ht 0 ; tci/ 6 ˆ.G/, a contradiction. Hence o.c/ D 4 and so c 2 D z. We have obtained a uniquely determined group of order 25 given in part (b) of our theorem. From now on we may assume that 1 .G/ D E Š E8 . (i) Assume that 1 .G/ D E 6 Z.G/ D ˆ.G/ and E 6 A, where A is the unique abelian maximal subgroup of G. Then A \ E D G 0 , A covers G=E and A is metacyclic. Since there are three maximal subgroups of G containing E, there exists at least one of them, denoted by H , which is minimal nonabelian. If H=E is noncyclic, then there are two distinct maximal subgroups X1 ¤ X2 of H containing E. In that case, E X1 \ X2 D ˆ.H / D ˆ.G/ D Z.G/, a contradiction. Hence H=E is cyclic. Since d.G=E/ D 2, G=E is abelian of type .2m ; 2/, m 1. Therefore A=G 0 Š G=E is of type .2m ; 2/, where A \ H=G 0 Š C2m . Let a be an element in A \ H such that hai covers A \ H=G 0 . Noting that 1 .G/ D E, we have o.a/ D 2mC1 and 1 .hai/ D hzi G 0 , where m z D a2 . If t 2 E G 0 , then Œt; a D u 2 G 0 hzi because H is nonmetacyclic and therefore u is not a square in H . Since A=G 0 Š C2m C2 , there is an element b 2 A H such that 1 ¤ b 2 2 G 0 and b 2 ¤ z. Indeed, if b 2 D z, then taking an element v of order 4 in hai, we get .bv/2 D b 2 v 2 D z 2 D 1, where bv 2 A H , a contradiction. Hence we get that b 2 2 ¹u; uzº. We have ˆ.H / D ˆ.G/ D ha2 ; ui

60

Groups of prime power order

and G D ha; b; t i. If Œb; t 2 hui, then G=hui is abelian, a contradiction. Hence we obtain Œb; t 2 ¹z; uzº, o.b/ D 4 and A is abelian of type .4; 2mC1 /. We set Œb; t D zu and b 2 D uz , ; D 0; 1. First suppose that o.a/ > 4 so that ha4 i hzi. In that case, ˆ.hb; ti/ D hb 2 ; Œb; ti G 0 < ˆ.G/ D ha2 ; ui and so M D hb; tiˆ.G/. The fact that ˆ.hab; ti/ D h.ab/2 D a2 uz ; Œab; t D zuC1 i D ˆ.G/ gives D 0. We may assume that b 2 D u, i.e., D 0. Indeed, if b 2 D uz D u0 , then we replace H D ha; ti with H1 D ha0 D ab; ti, where o.a0 / D o.a/, ha0 i hzi and Œa0 ; t D uz D u0 and so writing again a and u instead of a0 and u0 , respectively, we have obtained the relations for groups G of order 2mC4 given in part (c) of our theorem. It remains to examine the case o.a/ D 4. In this case, m D 1, a2 D z, G is a special group of order 25 , where ˆ.G/ D hu; zi. We have A D ha; bi Š C4 C4 and ha; ti D H is the nonmetacyclic minimal nonabelian group of order 24 . Further, Œb; t D zu

with

ˆ.hb; ti/ D huz ; zu i;

Œab; t D zuC1 with

ˆ.hab; ti/ D huz C1 ; zuC1 i;

Œb; at D zu

ˆ.hb; ati/ D huz ; uz; zu i

with

and Œab; at D zuC1 with

ˆ.hab; ati/ D huz C1 ; uz; zuC1 i:

If D D 0, then ˆ.hab; ti/ D huzi and ˆ.hab; ati/ D huzi. In case D D 1, we get ˆ.hb; ti/ D huzi and ˆ.hb; ati/ D huzi. It follows that in the above two cases our group G has two distinct maximal subgroups which are neither abelian nor minimal nonabelian, a contradiction. Therefore we must have ¤ in which case we may set D C 1. But in this case, we check that each nonabelian maximal subgroup of G is minimal nonabelian and so G would be an A2 -group, a contradiction. (ii) Assume that 1 .G/ D E 6 Z.G/ D ˆ.G/ and E A, where A is the unique abelian maximal subgroup of G. Since there are three maximal subgroups of G containing E, there is a maximal subgroup H of G containing E which is minimal nonabelian. Then H is nonmetacyclic with Z.H / \ E D G 0 and H=E is cyclic. Taking an element b 2 E G 0 , we may set ˛

H D ha; b j a2 D b 2 D 1; ˛ 2; Œa; b D c; c 2 D Œa; c D Œb; c D 1i; where hci D H 0 , Z.H / D hci ha2 i, jGj D 2˛C3 , and setting a2 D z we have G 0 D hz; ci Š E4 since c is not a square in H . Here ha; ci (containing G 0 ) is an abelian normal subgroup of type .2˛ ; 2/ in G having exactly two cyclic subgroups hai and haci of order 2˛ . Since ab D ac, we obtain that NH .hai/ D ha; ci which implies that ˛1

100

2-groups with exactly one exceptional maximal subgroup

61

N D NG .hai/ covers G=H and N \ H D ha; ci. It follows that N is a nonabelian maximal subgroup of G (because A E), where N=G 0 Š G=E is noncyclic abelian and so N=G 0 is of type .2; 2˛ /. Hence there is d 2 N H with 1 ¤ d 2 2 G 0 and so o.d / D 4. But d normalizes hai and therefore Œd; a 2 hai \ G 0 D hzi which gives Œd; a D z. There exist exactly three maximal subgroups of G containing E: H D ha; bi, hd; biˆ.H /, had; biˆ.H /, where exactly one of two last subgroups is abelian. It follows that either Œd; b D 1 or Œad; b D 1 in which case Œd; b D c (since Œa; b D c). We may set Œd; b D c , where D 0; 1, and note that G D ha; b; d i. First assume that ˛ 3. If d 2 D z, then hd; ai Š M2˛C1 and there are involutions ˛2 in hd; aihai, a contradiction. Thus d 2 2 G 0 hzi in which case hd; a2 i Š C4 C4 ˛2 ˛2 since a2 2 Z.G/. Hence, replacing d with da2 (if necessary), we may assume that d 2 D c. If D 0, then A D hd; biˆ.G/ is an abelian maximal subgroup of G and we check that all other six maximal subgroups of G are minimal nonabelian and so G is an A2 -group, a contradiction. Hence we must have D 1. We have Œb; d D c and ˆ.hb; d i/ D hci < ˆ.G/. Also, Œab; ad D z and ˆ.hab; ad i/ D ha2 c; a2 cz; zi D ha2 ci < ˆ.G/ since .a2 c/2 D a4 and ha4 i hzi. Hence hb; d iˆ.G/ and hab; ad iˆ.G/ are two distinct maximal subgroups of G which are neither abelian nor minimal nonabelian, a contradiction. We have proved that we must have ˛ D 2 so that a2 D z, G is special satisfying ˆ.G/ D hz; ci and jGj D 25 . We have Œd; b D c and if D 1, then we replace d with d 0 D ad so that Œd 0 ; b D 1 and Œa; d 0 D z. Writing again d instead of d 0 , we may assume that Œd; b D 1 and Œa; d D z (as before). If d 2 D z, then we obtain the group of order 25 given in part (d) of our theorem. It remains to analyze the cases d 2 2 ¹c; czº. If d 2 D c, then we infer that hb; ad i Š D8 since Œb; ad D c and .ad /2 D a2 d 2 Œd; a D zcz D c. This is a contradiction since 1 .G/ Š E8 . Suppose that d 2 D cz. In that case, we replace a with a0 D ab, z with z 0 D zc and d with d 0 D db. Then we get a02 D .ab/2 D zc D z 0 ;

d 02 D .db/2 D d 2 D cz D z 0 ;

Œa0 ; d 0 D Œab; db D zc D z 0 ;

Œa0 ; b D Œab; b D c;

Œb; d 0 D Œb; db D 1;

and so, writing again a; z; d instead of a0 ; z 0 ; d 0 , respectively, we obtain again the group given in part (d) of our theorem. (iii) We turn now to the difﬁcult case, where 1 .G/ D E Z.G/ D ˆ.G/. Let H1 D H be a maximal subgroup of G which is minimal nonabelian and such that H 0 ¤ M 0 . Since Z.H / D Z.G/ E, H is nonmetacyclic and we may set ˛

ˇ

H D ha; b j a2 D b 2 D 1; Œa; b D c; c 2 D Œa; c D Œb; c D 1i; where ˛ 2, ˇ 2, hci D H 0 , Z.H / D ˆ.G/ D hci ha2 i hb 2 i is abelian of type ˛1 ˇ1 .2˛1 ; 2ˇ 1 ; 2/, and jGj D 2˛Cˇ C2 . We have G 0 < E D ha2 ; a2 ; ci Š E8 .

62

Groups of prime power order

We consider the group G=M 0 , where .G=M 0 /0 D G 0 =M 0 Š C2 , d.G=M 0 / D 3, G=M 0 has exactly three abelian maximal subgroups A=M 0 , M=M 0 and H5 =M 0 (since H50 D M 0 ) and the other four maximal subgroups Hi =M 0 , i D 1; : : : ; 4, are minimal nonabelian. Thus G=M 0 is an A2 -group of Proposition 71.1 which implies that there is an element d 2 G H such that Œd; G D M 0 and 1 ¤ d 2 2 G 0 . Since there are exactly three maximal subgroups of G containing hd i Š C4 , at least one of them H is minimal nonabelian, where H E and so H is nonmetacyclic. Thus .H /0 is a maximal cyclic subgroup in H and so .H /0 ¤ hd 2 i which gives G 0 D h.H /0 ; d 2 i. Taking an element a 2 .H \ H / ˆ.G/, we conclude that H D ha ; d i and so ˆ.H / D h.a /2 ; d 2 ; .H /0 i D h.a /2 ; G 0 i D ˆ.G/ and E D h1 .ha i/; G 0 i. Set o.a / D 2 , 2, so that ˆ.H / D ˆ.G/ is of type .2 1 ; 2; 2/. On the other hand, ˆ.G/ is of type .2˛1 ; 2ˇ 1 ; 2/. Interchanging the elements a and b (if necessary), we may assume that ˇ D 2 and then D ˛ so that ˆ.G/ is of type .2˛1 ; 2; 2/ and o.b/ D 4. Since Œd; G D M 0 , we have hŒd; a i D M 0 and so .H /0 D M 0 and therefore H D H5 and d 2 2 G 0 M 0 . Because H =E is abelian of type .2˛1 ; 2/, it follows that H \ H =E is cyclic of order 2˛1 and so ha i covers H \ H =E. Since H \ H D ha i G 0 , it follows that H=G 0 is abelian of type .2˛ ; 2/ which implies that H D H0 .H \H / with H0 \.H \H / D G 0 and jH0 W G 0 j D 2. Hence there is an element b 2 H H with 1 ¤ .b /2 2 G 0 , ha ; b i D H , .b /2 ¤ c (since H 0 D hci and so c is not a square in H ) and so Œa ; b D c and o.b / D 4. In addition, from Œd; G D M 0 follows that either Œd; b D m with hmi D M 0 or Œd; b D 1. Also d 2 D cm and .b /2 D c m, where ; D 0; 1. Assume that Œd; b D 1 in which case d 2 ¤ .b /2 (because if d 2 D .b /2 , then db is an involution in G H , a contradiction) and so hd; b i Š C4 C4 and hd 2 ; .b /2 i D G 0 . In that case, Œb a ; d D m since Œa ; d D m and we get ˆ.hb a ; d i/ D h.b a /2 D c m .a /2 c D .a /2 c C1 m; d 2 D cm ; mi D G 0 h.a /2 i D ˆ.G/; and so hb a ; d i is a minimal nonabelian maximal subgroup of G which satisﬁes hb a ; d i0 D hmi and hb a ; d i ¤ H . This is a contradiction since H D H5 is the only maximal subgroup of G which is minimal nonabelian and .H /0 D M 0 D hmi. We have proved that Œd; b D m which together with hb ; d i ¤ H implies that hb ; d iˆ.G/ D M must be the unique maximal subgroup of G which is neither abelian nor minimal nonabelian. If D 0 and D 1, then .b /2 D cm, d 2 D c and therefore .b d /2 D cm c m D 1 and b d would be an involution in G H , a contradiction. Hence we have either D or D 1 and D 0. In this case, Œa ; b d D cm and ˆ.ha ; b d i/ D h.a /2 ; c C1 m ; cmi D ˆ.G/ implies that D 1 and D 0 is not possible and so D . Further, we have Œb ; a d D cm and so ˆ.hb ; a d i/ D hc m; .a /2 cmC1 ; cmi forces that D D 0. But then Œa b ; a d D c shows together with ˆ.ha b ; a d i/ D h.a /2 cm; ci < ˆ.G/ that ha b ; a d iˆ.G/ is another maximal subgroup of G which is neither abelian nor minimal nonabelian, a contradiction. Our theorem is proved.

100

2-groups with exactly one exceptional maximal subgroup

63

Theorem 100.6. Let G be a 2-group with d.G/ D 3 which has exactly one maximal subgroup M which is neither abelian nor minimal nonabelian. If G has no abelian maximal subgroups, then we get n

n1

G D ha; b; c j a4 D b 4 D c 2 D 1; a2 D x; b 2 D y; c 2

D z;

Œa; b D z; Œa; c D y; Œb; c D xy; Œx; b D Œx; c D Œy; a D Œy; c D Œz; a D Œz; b D 1i; where jGj D 2nC4 , n 3, G 0 D hx; y; zi Š E8 , Z.G/ D ˆ.G/ D G 0 hc 2 i is abelian of type .2n1 ; 2; 2/, M D ˆ.G/ha; bi D hc 2 iha; bi with hc 2 i\ha; bi D hzi D ha; bi0 , hc 2 i Š C2n1 and ha; bi is the nonmetacyclic minimal nonabelian group of exponent 4 and order 25 and the other six maximal subgroups of G are nonmetacyclic minimal nonabelian. Proof. We put 1 D ¹H1 ; H2 ; : : : ; H6 ; M º to be the set of maximal subgroups of G, where H1 ; : : : ; H6 are minimal nonabelian. We know that jM 0 j D 2 and d.M / 3 (see the remark preceding Theorem 100.5). By a result of A. Mann, jG 0 W .H10 H20 /j 2 and so jG 0 j 8. However, if jG 0 j D 2, then we know that G has three abelian maximal subgroups (see the second paragraph of the proof of Theorem 100.4), a contradiction. Hence jG 0 j D 4 or jG 0 j D 8. Suppose that for some Hi ¤ Hj we have Hi0 D Hj0 . Then by a result of A. Mann, we get jG 0 j D 4 and moreover G 0 Š E4 . Indeed, if G 0 Š C4 , then the nonabelian group G=1 .G 0 / would possess at least six abelian maximal subgroups Hi =1 .G 0 /, i D 1; : : : ; 6, a contradiction. The group G=M 0 is obviously an A2 -group satisfying .G=M 0 /0 Š C2 . By Proposition 71.1, G=M 0 has exactly three abelian maximal subgroups so that we may set H50 D H60 D M 0 D hui. In that case, Hi0 , i D 1; : : : ; 4, cannot be all pairwise distinct and so we may set H20 D H30 D H40 D hvi with hu; vi D G 0 and G=hvi has exactly three abelian maximal subgroups Hi =hvi, i D 2; 3; 4. It follows that we must have H10 D huvi so that G=huvi with .G=huvi/0 Š C2 has exactly one abelian maximal subgroup H1 =huvi. If d.M=huvi/ D 2, then M=huvi is minimal nonabelian so that G=huvi would be an A2 -group. But in that case (Proposition 71.1), G=huvi would have three abelian maximal subgroups, a contradiction. Hence we must have d.M=huvi/ 3 in which case M=huvi is a unique maximal subgroup of G=huvi which is neither abelian nor minimal nonabelian. By Theorem 100.5, we must have .G=huvi/0 Š E4 , a contradiction. We have proved that all Hi0 are pairwise distinct subgroups of order 2 in G 0 . This implies that G 0 Š E8 . If M 0 D Hi0 for some i 2 ¹1; 2; : : : ; 6º, then considering G=M 0 we see that there must exist a maximal subgroup Hj , j ¤ i , such that M 0 D Hi0 D Hj0 , a contradiction. Hence ¹H10 ; : : : ; H60 ; M 0 º is the set of seven pairwise distinct subgroups of order 2 in G 0 . Since G 0 Hi , all Hi (i D 1; : : : ; 6 ) are nonmetacyclic minimal nonabelian. The group G=G 0 is abelian of rank 3. Suppose that there is an involution t 2 G G 0 . Then F D G 0 ht i Š E16 and G=F is noncyclic. But then there is a maximal subgroup H of G such that H F and H is minimal nonabelian,

64

Groups of prime power order

a contradiction. We have proved that G 0 D 1 .G/. Set T =G 0 D 1 .G=G 0 / Š E8 . If G=T is noncyclic, then there is a maximal subgroup K of G such that K T and K is minimal nonabelian. But then d.K=G 0 / D 3, a contradiction. Hence G=T is cyclic and so G=G 0 is abelian of type .2m ; 2; 2/, m 1. (i) First assume m D 1, i.e., T D G, G=G 0 Š E8 and G is a special group with 0 G D 1 .G/ Š E8 . We shall determine the structure of M > G 0 . We have M D G 0 S , where S D ha; bi is minimal nonabelian and G 0 \ S D ˆ.S/ < G 0 . Set hzi D S 0 D M 0 Š C2 . Suppose at the moment that ˆ.S/ D hzi so that S Š Q8 . Then G=hzi is an A2 -group, where M=hzi Š E16 is a unique abelian maximal subgroup of G=hzi and E4 Š .G=hzi/0 Z.G=hzi/. But then Proposition 71.4(b) implies that 1 .G=hzi/ Š E8 , a contradiction. We have proved that ˆ.S/ Š E4 and 1 .S/ D ˆ.S/. Hence S is the metacyclic minimal nonabelian group of order 16 and exponent 4. We choose a; b 2 S ˆ.S/ so that a2 D z, b 2 D y, Œa; b D z and hy; zi D ˆ.S/ D ˆ.M /. Since G 0 D ˆ.G/, there is c 2 G M such that c 2 D x 2 G 0 hy; zi. We obtain hx; y; zi D G 0 and ha; b; ci D G. All other six maximal subgroups (distinct from M ) are nonmetacyclic minimal nonabelian. We have ˆ.ha; ci/ D hz; x; Œa; ci D G 0 so that Œa; c D x ˛ yz ˇ . Also, ˆ.hb; ci/ D hy; x; Œb; ci D G 0 gives Œb; c D x y ı z. Further, ˆ.hab; ci/ D hy; x; Œab; c D x ˛C y ıC1 z ˇ C1 i D G 0 which implies ˇ D 0. From ˆ.hb; aci/ D hy; x ˛C1 yz; Œb; ac D x y ı i D G 0 we get D 1, and from ˆ.hab; aci/ D hy; x ˛C1 yz; Œab; ac D x ˛C1 y ıC1 i D G 0 it follows ˛ D 0. Finally, ˆ.ha; bci/ D hz; y ıC1 z; Œa; bc D zyi D hy; zi < G 0 gives a contradiction since G 0 ha; bci is another maximal subgroup of G (distinct from the subgroup M ) which is neither abelian nor minimal nonabelian. (ii) Suppose that T < G, where T =G 0 D 1 .G=G 0 / Š E8 and ¹1º ¤ G=T is cyclic so that G=G 0 is abelian of type .2m ; 2; 2/, m 2. The unique maximal subgroup of G containing T must be equal to M . There are normal subgroups U and V of G such that G D U V , U \ V D G 0 , U=G 0 Š E4 and V =G 0 is cyclic of order 2m , m 2. Let c be an element in V G 0 such that hci covers V =G 0 . We have o.c/ D 2n , n 3, where n D mC1 (noting that 1 .G/ D G 0 ). n1 Set hzi D 1 .hci/ so that z D c 2 and z 2 G 0 . Then we conclude that M D U hc 2 i, 0 2 ˆ.G/ D Z.G/ D G hc i is abelian of type .2n1 ; 2; 2/ and jGj D 2nC4 . Suppose that a; b 2 U G 0 satisfy U D G 0 ha; bi, where a2 ; b 2 2 G 0 and G D ha; b; ci. Since each maximal subgroup Hi (i D 1; : : : ; 6) is nonmetacyclic and contains ˆ.G/ and z

100

2-groups with exactly one exceptional maximal subgroup

65

is a square in ˆ.G/, it follows that Hi0 ¤ hzi for all i D 1; : : : ; 6. This implies that M 0 D hzi and therefore Œa; b D z. Now, G=hzi has the unique maximal abelian subgroup M=hzi and six minimal nonabelian maximal subgroups Hi =hzi, i D 1; : : : ; 6, and so G=hzi is an A2 -group with the following properties. We have d.G=hzi/ D 3 and so G=hzi is nonmetacyclic of order 2nC3 > 24 since n 3, .G=hzi/0 Š E4 , G 0 =hzi Z.G=hzi/ (as G 0 Z.G/) and G=hzi has a normal elementary abelian subgroup hG 0 ; 2 .hci/i=hzi of order 8. Hence G=hzi is an A2 -group from Proposition 71.4(b) which implies the fact that hG 0 ; 2 .hci/i=hzi D 1 .G=hzi/. Set a2 D x and b 2 D y and consider the abelian group M=hzi. If the abelian subgroup U=hzi of order 16 and exponent 4 has rank greater than 2, then we get 1 .U=hzi/ > G 0 =hzi, which contradicts the above fact. Hence we must have U=hzi Š C4 C4 which implies that G 0 D hx; y; zi. Since all Hi are minimal nonabelian (containing ˆ.G/ D hc 2 ; x; yi), we get Œa; c D x ˛ yz ˇ and Œb; c D xy z ı , where ˛; ˇ; ; ı D 0; 1. From ˆ.hab; ci/ D h.ab/2 D xyz; c 2 ; Œab; c D x ˛C1 y C1 z ˇ Cı i D ˆ.G/ and the fact that 1 .hc 2 i/ D hzi it follows that ˛ C 1 ¤ C 1 which gives D ˛ C 1 and Œb; c D xy ˛C1 z ı . Interchanging a and b respectively x and y, i.e., writing a0 D b, b 0 D a, x 0 D y, 0 y D x, we get a02 D b 2 D y D x 0 ;

b 02 D a2 D x D y 0 ;

Œa0 ; b 0 D Œb; a D z; Œa0 ; c D Œb; c D xy ˛C1 z ı D .x 0 /˛C1 y 0 z ı ; Œb 0 ; c D Œa; c D x ˛ yz ˇ D x 0 y 0˛ z ˇ : Writing again a; b; x; y instead of a0 ; b 0 ; x 0 ; y 0 , respectively, we obtain a2 D x; Œa; c D x ˛C1 yz ı ;

b 2 D y;

Œa; b D z;

Œb; c D xy ˛ z ˇ ;

ˇ; ı D 0; 1;

which are the old relations in which ˛ is replaced with ˛ C 1. This shows that we may assume ˛ D 0 and so we get Œa; c D yz ˇ and Œb; c D xyz ı , ˇ; ı D 0; 1. Finally, replacing c with c 0 D caı b ˇ , we get c 02 D c 2 .aı b ˇ /2 Œaı b ˇ ; c D c 2 l with l 2 G 0 and so c 04 D c 4 , hc 0 i covers G=U and hc 0 i \ G 0 D hzi. In addition, we have Œa; c 0 D Œa; caı b ˇ D Œa; cŒa; bˇ D yz ˇ z ˇ D y; Œb; c 0 D Œb; caı b ˇ D Œb; cŒb; aı D xyz ı z ı D xy: Writing again c instead of c 0 , we obtain the relations stated in our theorem.

102

p-groups G with p > 2 and d.G / > 2 having exactly one maximal subgroup which is neither abelian nor minimal nonabelian

In this section we determine up to isomorphism the title groups (Theorems 102.1, 102.3 and 102.5). It is obvious that for such groups G we have d.G/ D 3. All resulting groups will be presented in terms of generators and relations. But we shall also state all important characteristic subgroups of these groups so that the results could be useful for applications. In 100, 101 and 102 the difﬁcult problem Nr. 861 stated by the ﬁrst author is completely solved. Theorem 102.1. Let G be a p-group, p > 2, which possesses exactly one maximal subgroup which is neither abelian nor minimal nonabelian. Suppose that d.G/ D 3 and G has more than one abelian maximal subgroup. Then we have one of the following possibilities: (a) G D U Z, where U Š S.p 3 /, Z Š Cpm , m 3, and U \ Z D Z.U /. Here G is an L3 -group. (b) G D U Z, where U Š S.p 3 / or U Š Mp3 and Z Š Cpm , m 2. Conversely, the groups in (a) and (b) satisfy the assumptions of our theorem. Proof. Our assumptions imply that G=Z.G/ Š Ep2 and G has exactly p C 1 abelian maximal subgroups (Exercise 1.6(a)). By Lemma 1.1, jGj D pjG 0 jjZ.G/j gives that jG 0 j D p. Conversely, assume that G is a title group with jG 0 j D p. Since G has at most pC1 abelian maximal subgroups, there is a minimal nonabelian maximal subgroup H . From jG 0 j D p we conclude that G D H CG .H / (Exercise P10), where H \CG .H / D Z.H / D ˆ.H / D ˆ.G/ and CG .H / D Z.G/. We have G=Z.G/ Š Ep2 and so all p C 1 maximal subgroups of G containing Z.G/ are abelian. In what follows H will denote a ﬁxed maximal subgroup of G which is minimal nonabelian. Suppose that there exists an element c 2 Z.G/ H of order p. Then we have G D H hci and so each maximal subgroup of G which does not contain hci is isomorphic to G=hci Š H and so is minimal nonabelian, a contradiction. Hence there are no elements of order p in Z.G/H which implies 1 .Z.H // D 1 .Z.G// so that d.Z.H // D d.Z.G//. It follows that for each x 2 Z.G/ H , x p 2 Z.H / ˆ.Z.H //.

78

Groups of prime power order

Obviously, jGj p 5 since the exceptional maximal subgroup M (which is neither abelian nor minimal nonabelian) is of order p 4 . (i) First assume that H is metacyclic. We may set m

n

ha; b j ap D b p D 1; ab D az; z D ap

m1

i;

where m 2, n 1, mCn 4, H 0 D hzi D G 0 , jH j D p mCn , and jGj D p mCnC1 . We see that Z.H / D hap i hb p i D ˆ.H / D ˆ.G/ and for each x 2 Z.G/ H , 2 2 2 2 x p 2 hap ; b p i hap ; b p i since hap ; b p i D ˆ.Z.H //. Suppose that n D 1 so that o.b/ D p, m 3, H Š MpmC1 and Z.H / D hap i. Here Z.G/ Š Cpm is cyclic and so there is c 2 Z.G/ H such that c p D ap which gives .ca/p D c p ap D 1 and o.ca/ D p. Since Œca; b D z, it follows that U D hca; bi Š S.p 3 /, U \ Z.G/ D hzi which together with jG W Z.G/j D p 2 gives G D U Z.G/. We have obtained the groups stated in part (a) of our theorem. Here M D U hc p i, all p C 1 maximal subgroups of G containing hci are abelian and we have to show that all subgroups hc i a; c j bi are minimal nonabelian maximal subgroups of G for all integers i; j .mod p/ unless i 1 .mod p/ and j 0 .mod p/. Indeed, Œc i a; c j b D Œa; b D z and so, by Exercise P9, hc i a; c j bi is minimal nonabelian and ˆ.hc i a; c j bi/ D h.c i a/p D c pi ap D api ap D ap.i C1/ ; .c j b/p D apj ; zi D hap i D ˆ.G/ if and only if either i 6 1 .mod p/ or j 6 0 .mod p/. It remains to treat the case n 2. Suppose, in addition, that there exists an element x 2 G H of order p. We know that x 62 Z.G/ and so Œa; x ¤ 1 or Œb; x ¤ 1. Obviously, ha; b; xi D G. First suppose Œa; x ¤ 1. Since hai E G, ha; xi Š MpmC1 is minimal nonabelian of order p mC1 . But jGj D p mCnC1 and n 2 which implies that M D ˆ.G/ha; xi D ha; xi hb p i is a maximal subgroup of G which is neither abelian nor minimal nonabelian. Assume at the moment that also Œx; b ¤ 1. In that case, we get hbihzi E G and so hb; xi is nonmetacyclic minimal nonabelian of order p nC2 . Then it follows that hb; xi must be a maximal subgroup of G with jGj D p nC3 (and so m D 2). But this case will be studied in part (ii) of this proof (where G possesses a nonmetacyclic minimal nonabelian maximal subgroup). Hence we may assume Œx; b D 1. This gives Œx; ab D Œx; a D z r , r 6 0 .mod p/ and so hx; abi is minimal nonabelian which forces that hx; abi must be a maximal subgroup of G. Now, habi covers H=hai Š Cpn n n n n and so o.ab/ p n , where n 2. We have .ab/p D ap b p D ap . If n m, then o.ab/ D p n and habi \ hai D ¹1º. Since hab; zi E G, we see that hab; xi is a nonmetacyclic minimal nonabelian subgroup of order p nC2 . In that case, hab; xi must be a maximal subgroup of G (with m D 2) and again this will be studied in part (ii) of the proof. It follows that we may assume n < m. In that case, o.ab/ D p m and habi hzi so that habi E G. Hence hab; xi is metacyclic minimal nonabelian of order p mC1 and so hab; xi must be a maximal subgroup of G. From jGj D p mCnC1 follows n D 1, contrary to our assumption.

102

p-groups G with p > 2, d.G/ > 2, exactly one exceptional maximal subgroup 79

We may assume that Œa; x D 1 and then Œb; x ¤ 1. Since hbi hzi E G, hb; xi is nonmetacyclic minimal nonabelian of order p nC2 . If hb; xi is a maximal subgroup of G, then this case will be treated in part (ii) of this proof. Thus we may assume that hb; xi is not a maximal subgroup of G and so M D ˆ.G/hb; xi with m > 2. It follows that hab; xi, which is minimal nonabelian, must be a maximal subgroup of G. n n Since habi covers H=hai, we obtain that o.ab/ p n , n 2, and .ab/p D ap . If n m, then o.ab/ D p n and habi\hai D ¹1º and so hab; xi is nonmetacyclic minimal nonabelian of order p nC2 . In that case, hab; xi must be a maximal subgroup of G with m D 2, a contradiction. We may assume that n < m and then o.ab/ D p m with habi hzi and so hab; xi is metacyclic minimal nonabelian of order p mC1 . But then jGj D p mCnC1 implies n D 1, contrary to our assumption. Therefore we may assume that there are no elements of order p in G H . We know that for an element c 1 2 Z.G/ H , c p 2 Z.H / ˆ.Z.H // and so c p is not a p-th power of any element in Z.H /. But Ã1 .H / D Z.H / H 0 D hzi and so H is a powerful group. Then c p D hp for some element h 2 H Z.H / (see Proposition 26.10). It follows hc 2 G H and .hc/p D hp c p D 1 and so hc is of order p, a contradiction. (ii) It remains to consider the case where H is nonmetacyclic minimal nonabelian. We may set m

n

ha; b j ap D b p D 1; Œa; b D z; z p D Œa; z D Œb; z D 1i; where we may assume m 2, n 1 since jGj p 5 . Here we have H 0 D hzi, jH j D p mCnC1 , and jGj D p mCnC2 . Also, hzi is a maximal cyclic subgroup in H , Z.H / D hap i hb p i hzi D ˆ.H / D ˆ.G/ and for each x 2 Z.G/ H we have x p 2 Z.H / ˆ.Z.H //. (ii1) First assume n D 1 so that Z.H / D hap ihzi and for an element c 2 Z.G/H we have c p D apr z s , where both integers r; s are not divisible by p. Suppose that r 0 .mod p/ and then s 6 0 .mod p/. Replacing c with another suitable gener2 0 ator of hci, we may assume that c p D ap r z for some integer r 0 . Take the element 0 2 0 c 0 D apr c 2 Z.G/ H ; then we get .c 0 /p D ap r c p D z. Thus G D H hc 0 i with .c 0 /p D z and hzi D H 0 and so we have obtained a p-group from Proposition 71.1(ii) which is an A2 -group, a contradiction. We have proved that c p D apr z s with r 6 0 .mod p/. Set a0 D ar so that o.a0 / D p m ;

Œa0 ; b D z r

and

.ca0 /p D c p .a0 /p D z s ;

where ha0 ; bi D H . Consider the subgroup U D hca0 ; bi. Since Œca0 ; b D z r and o.b/ D p, we have in case s 6 0 .mod p/ that U Š Mp3 and in case s 0 .mod p/ that U Š S.p 3 /. However, c p D .a0 /p z s , G D ha0 ; b; ci, o.c/ D p m , m > 1, and hci \ U D ¹1º and so we get G D U hci which are the groups stated in part (b) of our theorem. Here M D U hc p i and all p C 1 maximal subgroups containing hci are abelian.

80

Groups of prime power order

We have to show that all subgroups hc i a0 ; c j bi are minimal nonabelian maximal subgroups of G (not containing hci) for all integers i; j (mod p) unless i 1 .mod p/ and j 0 .mod p/ holds. Indeed, Œc i a0 ; c j b D Œa0 ; b D z r and ˆ.hc i a0 ; c j bi/ D hc pi .a0 /p D apri z si apr D apr.i1/ z si ; c pj D aprj z sj ; z r i D ˆ.G/ D ˆ.H / if either i 6 1 .mod p/ or j 6 0 .mod p/. (ii2) It remains to consider the case n 2. In this case, for an element c 2 Z.G/H we have c p D api b pj z k , where at least one of the integers i; j; k is 6 0 .mod p/. First suppose that i 0 .mod p/ and j 0 .mod p/ so that k 6 0 .mod p/. 2 0 2 0 We may deﬁne c p D ap i b p j z k for some integers i 0 ; j 0 . Considering the element 0 0 c 0 D api b pj c 2 Z.G/ H , we get .c 0 /p D ap

2i 0

b p

2j 0

cp D zk ;

k 6 0

.mod p/;

and so G D H hc 0 i with hc 0 i \ H D hzi D H 0 and this is an A2 -group from Proposition 71.1(ii), a contradiction. Now assume that one and only one of the integers i; j is 0 .mod p/. Because of the symmetry, we may assume i 6 0 (mod p) and j 0 .mod p/. We have ˆ.hai b j c; bi/ D hapi b pj c p D z k ; b p ; Œai b j c; b D z i ¤ 1i < ˆ.G/ and ˆ.hab; ar ci/ D hap b p ; apr c p D ap.iCr/ b pj z k ; Œab; ar c D z r ¤ 1i < ˆ.G/ for some suitable r 6 0 .mod p/ such that i C r 0 .mod p/. Hence we deduce that ˆ.G/hai b j c; bi and ˆ.G/hab; ar ci are two distinct maximal subgroups of G which are neither abelian nor minimal nonabelian, a contradiction. Finally, we consider the case where both i and j are 6 0 .mod p/. We have ˆ.hai b j c; bi/ D hapi b pj c p D z k ; b p ; Œai b j c; b D z i ¤ 1i < ˆ.G/ and ˆ.ha; b r ci/ D hap ; b pr c p D api b p.j Cr/ z k ; Œa; b r c D z r ¤ 1i < ˆ.G/ for some suitable r 6 0 .mod p/ such that j C r 0 .mod p/. Hence we deduce that ˆ.G/hai b j c; bi and ˆ.G/ha; b r ci are two distinct maximal subgroups of G which are neither abelian nor minimal nonabelian, a contradiction. Our theorem is proved. Lemma 102.2. Let G be a p-group, p > 2, which possesses exactly one maximal subgroup M which is neither abelian nor minimal nonabelian. Suppose that d.G/ D 3 and G has at most one abelian maximal subgroup. Then ˆ.G/ D Z.G/ D ˆ.H / for each minimal nonabelian maximal subgroup H of G. Also, jG 0 j > p, jM 0 j D p and d.M / 3 which implies jGj p 5 .

102

p-groups G with p > 2, d.G/ > 2, exactly one exceptional maximal subgroup 81

Proof. From the ﬁrst two paragraphs of the proof of Theorem 102.1 we know that jG W Z.G/j p 3 and jG 0 j > p. Let H be a maximal subgroup of G which is minimal nonabelian. Then ˆ.H / D Z.H / ˆ.G/ and jH W ˆ.H /j D p 2 which gives that ˆ.H / D ˆ.G/. Let K ¤ H be another maximal subgroup of G which is minimal nonabelian. Then Z.K/ D ˆ.K/ D ˆ.G/ which implies CG .ˆ.G// hH; Ki D G and so ˆ.G/ D Z.G/. Since jM W ˆ.G/j D p 2 and ˆ.G/ D Z.G/, we have M D S ˆ.G/, where S is minimal nonabelian and S \ ˆ.G/ D ˆ.S/ < ˆ.G/ D Z.M /. This implies M 0 D S 0 Š Cp and d.M / 3. Theorem 102.3. Let G be a p-group, p > 2, satisfying d.G/ D 3 which has exactly one maximal subgroup M which is neither abelian nor minimal nonabelian. Suppose that G has exactly one abelian maximal subgroup A. Then we have ˆ.G/ D Z.G/, G 0 Š Ep2 , jM 0 j D p, d.M / 3, 1 .G/ D E Š Ep3 , and E 6 ˆ.G/. If E 6 A, then we have the following possibilities: mC1

2

m

D b p D t p D 1; Œb; t D z; ap D z n ; b p D u; Œt; a D u; (a) G D ha; b; t j ap Œu; a D Œu; t D Œa; b D Œz; t D 1i, where m 2 and n 6 0 .mod p/. We have jGj D p mC4 ;

G 0 D hu; zi Š Ep2 ;

ˆ.G/ D Z.G/ D hap ; ui Š Cpm Cp ;

1 .G/ D E D hu; z; t i Š Ep3 ;

A D ha; bi is abelian of type .p mC1 ; p 2 /, M D hb; ti hap i, where hb; ti is a nonmetacyclic minimal nonabelian group of order p 4 , and all other p 2 C p 1 maximal subgroups of G are minimal nonabelian. 2

2

(b) G D ha; b; t j ap D b p D t p D 1; ap D z; u D Œt; a; Œb; t D ul z; b p D un z s ; Œu; a D Œu; t D Œa; b D Œz; t D 1i; where l; n; s are integers mod p, n 6 0 .mod p/, n is a square (in GF.p/) and .l C s/2 4n .mod p/. The group G is a special group of order p 5 with G 0 D hu; zi Š Ep2 ;

1 .G/ D E D hu; z; t i Š Ep3 ;

A D ha; bi Š Cp2 Cp2 , M D haj b; tiG 0 , where j .1=2/.l s/ .mod p/, and all other maximal subgroups of G are minimal nonabelian. If E A, then we have: ˛

2

˛1

(c) G D ha; b; d j ap D b p D d p D 1; ap D z; c D Œa; b; z D Œd; a; d p D n r s p c z ; Œd; b D c ; c D Œc; d D Œc; a D Œc; b D 1i, where ˛ 2, n; r; s are integers mod p, n 6 0 .mod p/. We have jGj D p ˛C3 ;

G 0 D hz; ci Š Ep2 ;

ˆ.G/ D Z.G/ D hap ; ci Š Cp˛1 Cp

and A D ˆ.G/hb; as d i. If ˛ 3, then M D ˆ.G/hb; d i with s 6 0 .mod p/. If ˛ D 2, then M D ˆ.G/hb; a r d i with r 6 s .mod p/. All other maximal subgroups of G (distinct from A and M ) are minimal nonabelian.

82

Groups of prime power order

Proof. From Lemma 102.2 we conclude that ˆ.G/ D Z.G/ (and so G is of class 2), jM 0 j D p, d.M / 3, jG 0 j > p and jGj p 5 . By Exercise 1.69(a), we get that jG 0 W .A0 H 0 /j D jG 0 W H 0 j p, where H is a minimal nonabelian maximal subgroup of G. Hence jG 0 j p 2 and so jG 0 j D p 2 . If G 0 D hvi Š Cp2 , then all p 2 C p C 1 maximal subgroups of the nonabelian group G=hv 2 i are abelian, a contradiction (see Exercise 1.6(a)). Hence G 0 Š Ep2 . We see also that each of the p C 1 subgroups of order p in G 0 is the commutator group of exactly p nonabelian maximal subgroups of G. p For each x; y 2 G we have .xy/p D x p y p Œy; x. 2 / D x p y p and so G is regular. By Theorem 7.2, 1 .G/ is of exponent p and j1 .G/j D jG W Ã1 .G/j. Suppose that j1 .G/j p 5 . Let H be a minimal nonabelian maximal subgroup of G. Then we obtain that jH \ 1 .G/j p 4 , contrary to the structure of H . We have proved that j1 .G/j p 4 . Suppose that G has no normal elementary abelian subgroup of order p 3 . Then we get j1 .A/j p 2 and so A is metacyclic. If H is any minimal nonabelian maximal subgroup of G, then jH j p 4 and the fact that 1 .H / Š Ep3 is not possible imply that H is metacyclic. In that case, M (with d.M / 3) is the only maximal subgroup of G which is not metacyclic. By Proposition A.40.12 of Berkovich, G is an L3 -group. In this case, 1 .G/ Š S.p 3 / (the nonabelian group of order p 3 and exponent p) and G=1 .G/ is cyclic of order p 2 . But then Ep2 Š G 0 1 .G/, contrary to the fact that G 0 Z.G/. We have proved that G possesses a normal elementary abelian subgroup E Š Ep3 of order p 3 . Suppose that G has an elementary abelian subgroup F of order p 4 . Since G 0 Š Ep2 and G 0 Z.G/, we have G 0 F and so F E G. Also, j1 .G/j p 4 implies that F D 1 .G/. If G=F is noncyclic, then there is a minimal nonabelian maximal subgroup K of G containing F , contrary to the structure of K. Hence G=F is cyclic of order p. Let a 2 G F be such that hai covers G=F . Then o.a/ D p m , m 2, which implies that 1 .hai/ D hzi ˆ.G/ D Z.G/ and so z 2 F . On the other hand, G=G 0 is abelian of rank 3 which forces z 2 G 0 . Set F D G 0 hf1 ; f2 i for some f1 ; f2 2 F G 0 so that ha; f1 ; f2 i D G and ˆ.G/ D G 0 hap i. As ŒF; hai D G 0 , we may choose f1 2 F G 0 so that Œf1 ; a D z and thus hf1 ; ai Š MpmC1 . This gives p p hf1 ; aiG 0 Š Cp MpmC1 . Since Œf1 ; f2 a D z and .f2 a/p D f2 ap Œa; f2 . 2 / D ap , it follows that 1 .hf2 ai/ D 1 .hai/ D hzi and hf1 ; f2 ai Š MpmC1 . Hence we conclude that hf1 ; f2 aiG 0 Š Cp MpmC1 is another maximal subgroup of G (distinct from hf1 ; aiG 0 ) which is neither abelian nor minimal nonabelian, a contradiction. We have proved that G does not possess an elementary abelian subgroup of order p 4 . Assume that 1 .G/ D S is a nonabelian subgroup of order p 4 and exponent p. Then we have Z.S/ D G 0 . If G=S is noncyclic, then there is a minimal nonabelian maximal subgroup H of G containing S, which contradicts the structure of H . Hence G=S is cyclic of order p. Let t; t 0 2 S G 0 so that S D G 0 ht; t 0 i and then we have 1 ¤ Œt 0 ; t D z 2 G 0 , ht; t 0 i Š S.p 3 / and S Š S.p 3 / Cp . Let M be the unique maximal subgroup of G containing S so that M is neither abelian nor minimal

102

p-groups G with p > 2, d.G/ > 2, exactly one exceptional maximal subgroup 83

nonabelian and let A be the unique abelian maximal subgroup of G. Then A > G 0 and A \ S Š Ep3 so that we may assume that A \ S D G 0 ht 0 i. Also, A covers G=S and so if c 2 A M , then hci covers A=.A \ S / Š G=S and o.c/ p 2 which implies 1 .hci/ ˆ.G/ \ S D Z.G/ \ S D Z.S/ D G 0 : We have ˆ.G/ D G 0 hc p i, and so G D hc; t; t 0 i and Œc; t 0 D 1. If Œc; t 2 hzi, then G=hzi is abelian, a contradiction. Hence Œc; t D u 2 G 0 hzi so that G 0 D hu; zi. The other p 2 C p 1 maximal subgroups of G (which are distinct from A and M ) are of the form T ˆ.G/, where T is one of the following minimal nonabelian subgroups: hc; .t 0 /i ti and hc j t 0 ; c k ti, where i; j; k are integers mod p and both j and k cannot be congruent 0 .mod p/ (see Exercise P9 and P11). Here T ˆ.G/ must be minimal nonabelian and this will be the case if and only if ˆ.T / ˆ.G/ D G 0 hc p i. It is enough to consider hc; ti and hc j t 0 ; cti for any integer j mod p. We have ˆ.hc; ti/ D hc p ; Œc; t D ui D ˆ.G/ if and only if 1 .hci/ ¤ hui which gives 1 .hci/ D hul zi for some integer l mod p. Further, we get ˆ.hc j t 0 ; cti/ D hc pj .t 0 /p D c pj ; .ct/p D c p t p D c p ; Œc j t 0 ; ct D uj zi D hc p ; uj zi: But 1 .hci/ D 1 .hc p i/ D hul zi and so for j D l, ˆ.hc l t 0 ; cti/ D hc p i < ˆ.G/ which shows that hc l t 0 ; ctiˆ.G/ is another maximal subgroup of G (distinct from M ) which is neither abelian nor minimal nonabelian, a contradiction. We have proved that 1 .G/ D E Š Ep3 . (i) Assume that Ep3 Š E D 1 .G/ 6 Z.G/ D ˆ.G/ and E 6 A, where A is the unique abelian maximal subgroup of G. Then we see that A \ E D G 0 , A covers G=E and A is metacyclic. Since G=G 0 is abelian of rank 3, we have d.G=E/ D 2 and so there exists at least one maximal subgroup H of G which is minimal nonabelian. If H=E is noncyclic, then E ˆ.H / D ˆ.G/ D Z.G/, a contradiction. Hence we have H=E Š .H \ A/=G 0 Š Cpm , m 1. Since d.G=E/ D 2, G=E Š A=G 0 is abelian of type .p m ; p/. Let a 2 H \ A be such that hai covers .H \ A/=G 0 . Noting m that 1 .G/ D E, we have o.a/ D p mC1 and 1 ¤ z D ap 2 G 0 . Let t 2 E G 0 so that Œt; a D u 2 G 0 hzi and so G 0 D hu; zi since H is nonmetacyclic and so H 0 D hui is a maximal cyclic subgroup in H . Since A=G 0 Š Cpm Cp , there is an element b 2 A H such that 1 ¤ b p 2 G 0 , b p 2 G 0 hzi and Œa; b D 1. Indeed, if b p 2 hzi, then the subgroup hb; 2 .hai/i contains elements of order p in A H , a contradiction. We have ˆ.H / D ˆ.G/ D hap ; ui, G D ha; b; t i and A D hai hbi is abelian of type .p mC1 ; p 2 /, m 1. If Œb; t 2 hui, then G=hui would be abelian, a contradiction. Hence Œb; t 2 G 0 hui and so replacing b with b 0 D b s (with some s 6 0 .mod p/), we may assume from the start that Œb; t D ul z for some integer l mod p.

84

Groups of prime power order

We have A D ha; bi and so we have to check the p 2 Cp other maximal subgroups of G which are of the form T ˆ.G/, where T is one of the minimal nonabelian subgroups: ha; b i t i, haj b; ak ti, where i; j; k are any integers mod p (see Exercise P11). (i1) First assume m 2 so that ˆ.G/ > G 0 . Consider T D hb; ti, where ˆ.hb; ti/ D hb p ; Œb; ti G 0 < ˆ.G/ and so M D hb; tiˆ.G/ must be our unique maximal subgroup of G which is neither abelian nor minimal nonabelian. All other maximal subgroups of G (which are distinct from A and M ) must be minimal nonabelian. Indeed, ˆ.ha; b i ti/ D hap ; b ip t p D b ip ; Œa; b i t D u1 i D ˆ.G/: Further, for j 6 0 .mod p/ we have ˆ.haj b; ti/ D hapj b p ; Œaj b; t D uj ul zi: Here hapj b p i covers ˆ.G/=G 0 and 1 .hapj b p i/ D hzi. Hence if l 6 0 .mod p/, then ˆ.hal b; ti/ D hapl b p i < ˆ.G/, a contradiction (since hal b; tiˆ.G/ will be another maximal subgroup which is neither abelian nor minimal nonabelian). Therefore we conclude that l 0 .mod p/ and so Œb; t D z. Finally, if both j and k are not congruent 0 .mod p/, then we have ˆ.haj b; ak ti/ D hapj b p ; apk ;

Œaj b; ak t D uj zi D ˆ.G/:

We have b p 2 G 0 hzi and so b p D un z s , where n and s are some integers mod p with n 6 0 .mod p/. Then we replace a with a0 D an b s and u with u0 D un z s and get m .a0 /p D z n ; b p D u0 ; Œt; a0 D Œt; an b s D un z s D u0 : Writing again a instead a0 and u instead u0 , we see that we have obtained the relations stated in part (a) of our theorem. (i2) Now assume m D 1 so that jGj D p 5 and G 0 D Z.G/ D ˆ.G/ D hu; zi and therefore G is a special p-group. In this case, we have ap D z;

Œt; a D u;

Œa; b D 1;

Œb; t D ul z;

b p D un z s ;

where l; n; s are some integers mod p with n 6 0 .mod p/. Also, A is abelian of type .p 2 ; p 2 /. For all integers i mod p we have ˆ.ha; b i ti/ hap D z; Œa; b i t D u1 i D G 0 D ˆ.G/ and so all subgroups ha; b i ti are minimal nonabelian maximal subgroups of G. Hence among the rest of the p 2 nonabelian maximal subgroups haj b; ak tiG 0 (j; k are any integers mod p) there is exactly one which is not minimal nonabelian. We have ˆ.haj b; ak t i/ D hun z sCj ; z k ; ulj zi;

102

p-groups G with p > 2, d.G/ > 2, exactly one exceptional maximal subgroup 85

and so if k 6 0 .mod p/, then ˆ.haj b; ak t i/ D hu; zi D G 0 . It remains to examine the case k 0 .mod p/, where we must have ˆ.haj b; ti/ D hun z sCj ; ulj zi ¤ G 0 D hu; zi for exactly one j . It follows that the quadratic congruence ˇ ˇ ˇ n ˇ s C j ˇ ˇ D j 2 C j.s l/ C .n sl/ 0 ˇl j 1 ˇ

.mod p/

must have exactly one solution in j . This occurs if and only if .s l/2 4.n sl/ 0

.mod p/

or, equivalently, .s C l/2 4n .mod p/. In this case, j .1=2/.l s/ .mod p/ and for that integer j the maximal subgroup M D haj b; tiG 0 is the only one which is neither abelian nor minimal nonabelian. We have obtained the groups from part (b) of our theorem. (ii) Assume that Ep3 Š E D 1 .G/ 6 Z.G/ D ˆ.G/ and E A, where A is the unique abelian maximal subgroup of G. Since d.G=E/ D 2, there exists (at least one) maximal subgroup H containing E which is minimal nonabelian. Then H is nonmetacyclic, H=E ¤ ¹1º is cyclic and Z.H / \ E D G 0 . Indeed, if H=E is noncyclic, then we have E ˆ.H / D ˆ.G/ D Z.G/, contrary to our assumption. Taking an element a 2 H E such that hai covers H=E and an element b 2 EG 0 , we get 1 .hai/ G 0 and ˛ H D ha; b j ap D b p D 1; c D Œa; b; c p D Œa; c D Œb; c D 1i; where ˛ 2, H 0 D hci, Z.H / D ˆ.H / D hap i hci, and jGj D p ˛C3 . Setting ˛1 ap D z, we have G 0 D hc; zi, E D hbi G 0 Š Ep3 because hci is a maximal cyclic subgroup in H and therefore hci ¤ hzi. Now, ha; ci is an abelian normal subgroup of G of type .p ˛ ; p/ which possesses exactly p cyclic subgroups hac i i (i any integer mod p) of order p ˛ . But Œa; b i D c i and so we get NH .hai/ D ha; ci and all subgroups hac i i are conjugate in H . Therefore N D NG .hai/ covers G=H and so G D NH with N \H D ha; ci. As N=G 0 Š G=E is abelian of rank 2, it follows that N=G 0 is abelian of type .p ˛ ; p/. Hence there exists an element d 2 N H such that 1 ¤ d p 2 G 0 and hd i normalizes hai. But N is a maximal subgroup of our group G which does not contain E and so N is nonabelian (noting that in our case E A). This gives 1 ¤ Œd; a 2 hzi and so replacing d with a suitable power d j (j 6 0 .mod p/), we may assume from the start that Œd; a D z. If d p 2 hzi, then we get hd; ai Š Mp˛C1 in which case there are elements of order p in hd; aiH , a contradiction. We have proved that d p 2 G 0 hzi so that hd; ai is a metacyclic minimal nonabelian maximal subgroup of G D ha; b; d i. Now, H D ha; bi is a minimal nonabelian maximal subgroup of G containing E and the other p maximal subgroups of G containing E are hb; ai d iˆ.G/, where i is any integer mod p and

86

Groups of prime power order

ˆ.G/ D G 0 hap i. For exactly one i , hb; ai d iˆ.G/ is the unique abelian maximal subgroup A of G, i.e., Œai d; b D 1 and then c i Œd; b D 1 and so we may set Œd; b D c s for some integer s mod p. Since d p 2 G 0 hzi, we may set d p D c n z r for some integers n; r mod p with n 6 0 .mod p/. All p 2 maximal subgroups of G which do not contain E are hb j a; b k d iˆ.G/, where j; k are any integers mod p. They are all metacyclic minimal nonabelian since ˆ.hb j a; b k d i/ D hap ; d p D c n z r ; Œb j a; b k d D c sj Ck z 1 ¤ 1i D ˆ.G/; where we have used the facts hap i hzi and n 6 0 .mod p/. We obtain that A D hb; as d iˆ.G/ is the unique abelian maximal subgroup of G and in the set of the p 1 nonabelian maximal subgroups hb; ai d iˆ.G/ for i 6 s .mod p/ there is exactly one which is not minimal nonabelian. We compute for all i 6 s .mod p/ ˆ.hb; ai d i/ D hb p D 1; .ai d /p D api c n z r ; Œb; ai d D c i s ¤ 1i D hapi z r ; ci: If ˛ 3, then hz r i hap i and so in this case ˆ.hb; ai d i/ ¤ ˆ.G/ if and only if i 0 .mod p/. Then we have M D hb; d iˆ.G/ and in this case s 6 0 .mod p/. If ˛ D 2, then ˆ.hb; ai d i/ D hz i Cr ; ci since ap D z. Hence in this case we have ˆ.hb; ai d i/ ¤ ˆ.G/ if and only if i r .mod p/. Then M D hb; ar d iˆ.G/ and in this case we must have r 6 s .mod p/. We have obtained the groups stated in part (c) of our theorem. (iii) It remains to consider the case Ep3 Š E D 1 .G/ Z.G/ D ˆ.G/. We shall show that this difﬁcult case cannot occur. If Hi is any minimal nonabelian maximal subgroup of G, then ˆ.Hi / D ˆ.G/ E Š Ep3 and so jHi j p 5 which implies jGj p 6 . By the ﬁrst paragraph of this proof, we know that there are exactly p nonabelian maximal subgroups of G whose commutator subgroup is equal to M 0 D hmi. Obviously, G=M 0 is an A2 -group of order p 5 with .G=M 0 /0 D G 0 =M 0 Š Cp and G=M 0 has exactly pC1 abelian maximal subgroups. By Proposition 71.1, there is a minimal nonabelian maximal subgroup H=M 0 of G=M 0 and an element d 2 G H with 1 ¤ d p 2 G 0 and Œhd i; G D M 0 . We have H 0 D hci with c 2 G 0 M 0 and G 0 D hc; mi which implies that H is a minimal nonabelian maximal subgroup of G (noting that H ¤ M since H 0 ¤ M 0 ). There are exactly p C 1 maximal subgroups Xi of G (i D 1; 2; : : : ; p C 1) which contain hd i. Then Xi \ H is an abelian maximal subgroup of H . Further, we obtain that Xi0 D Œhd i; .Xi \ H / M 0 and so Xi is either abelian or Xi0 D hmi. It follows that ¹X1 ; : : : ; XpC1 º D ¹A; M; H1 ; : : : ; Hp1 º, where Hj (j D 1; 2; : : : ; p 1) are 0 0 minimal nonabelian with .Hj / D hmi D M . But Hj (containing E) is nonmetacyclic and so M 0 is a maximal cyclic subgroup in Hj which implies d p 2 G 0 M 0 and we may set d p D c s mt , where s; t are some integers (mod p) with s 6 0 .mod p/.

102

p-groups G with p > 2, d.G/ > 2, exactly one exceptional maximal subgroup 87

Consider H D H1 so that H is a minimal nonabelian maximal subgroup of G containing hd i and .H /0 D hmi. Choose an element a 2 .H \ H / ˆ.G/ so that H D hd; a i and Œd; a D m. By Exercise P9, ˆ.H / D hd p ; Œd; a D mih.a /p i D G 0 h.a /p i; and we know that ˆ.H / D ˆ.G/. If 1 .ha i/ G 0 , then E 6 ˆ.G/, a contradiction. Hence we have 1 .ha i/ 6 G 0 which implies ˆ.H / D ˆ.G/ D h.a /p i G 0 and this is an abelian group of type .p; p; p 1 /, where o.a / D p , 2 and H \ H D ha i G 0 . It follows that .H \ H /=G 0 Š Cp and since H=G 0 is noncyclic abelian, there exists b 2 H .H \ H / such that 1 ¤ .b /p 2 G 0 and ha ; b i D H . We may set Œa ; b D c, where hci D H 0 . Also, hci is a maximal cyclic subgroup in H which gives .b /p 2 G 0 hci and we have G D ha ; b ; d i, jGj D p C4 , 2. We may set .b /p D c v mw , where v; w are some integers (mod p) with w 6 0 .mod p/. Suppose that Œd; b D 1 so that A D hd; b iˆ.G/. Also, hd; a i D H and so we investigate other maximal subgroups hd; .a /i b iˆ.G/ containing hd i, where i 6 0 .mod p/. We get ˆ.hd; .a /i b i/ D hd p 2 G 0 hmi; Œd; .a /i b D mi ; .a /pi .b /p i D ˆ.G/: But then there is no subgroup M in the set of maximal subgroups of G containing hd i, a contradiction. Hence we have Œd; b D mr with r 6 0 .mod p/. Since the maximal subgroups A and M are contained in the set of the p C 1 maximal subgroups of G which contain hd i, it follows that all p 2 maximal subgroups hd i a ; d j b iˆ.G/ of G (i; j are any integers mod p) which do not contain hd i must be minimal nonabelian. For each integer i mod p we must have ˆ.hd i a ; b i/ D hd pi .a /p ; .b /p D c v mw ; Œd i a ; b D cmri i D ˆ.G/ and this will be the case if and only if hc v mw ; cmri i D G 0 D hc; mi or, equivalently, ˇ ˇ ˇv w ˇ ˇ ˇ ˇ1 ri ˇ D vri w 6 0 .mod p/: But r 6 0 .mod p/ and so if v 6 0 .mod p/, then the congruence vri w 0 .mod p/ would have a solution in i , a contradiction. Hence v 0 .mod p/ and so .b /p D mw . For each integer i mod p we must have ˆ.hd i a ; db i/ D hd pi .a /p ; d p .b /p D c s mt Cw ; Œd i a ; db D cmri 1 i D ˆ.G/

88

Groups of prime power order

and this will be the case if and only if hc s mt Cw ; cmri1 i D G 0 D hc; mi or, equivalently, ˇ ˇ ˇs t C w ˇ ˇ ˇ ˇ1 ri 1ˇ D .sr/i s t w 6 0 .mod p/: But sr 6 0 .mod p/ and so the congruence .sr/i s t w 0 .mod p/ would have a solution in i , a ﬁnal contradiction. Our theorem is proved. Lemma 102.4. Let P be a p-group with jP 0 j D p. If P possesses a minimal nonabelian maximal subgroup H , then P has exactly p C 1 abelian maximal subgroups. Proof. By Exercise P.10, P D H CP .H / and in our case CP .H / is abelian so that CP .H / D Z.P /. But then all p C 1 maximal subgroups of P containing Z.P / are abelian and so we are done (see Exercise 1.6(a)). Theorem 102.5. Let G be a p-group, p > 2, with d.G/ D 3 which has exactly one maximal subgroup M which is neither abelian nor minimal nonabelian. Suppose that G has no abelian maximal subgroups. Then ˆ.G/ D Z.G/, 1 .G/ D G 0 Š Ep3 , jM 0 j D p, d.M / 3 and we have one of the following possibilities: 2

2

2

(a) G D ha; b; c j ap D b p D c p D 1; ap D z; b p D y; c p D x; Œa; b D z; Œa; c D yz ˇ ; Œb; c D x 1 y ı z ; Œx; b D Œx; a D Œy; a D Œy; c D Œz; b D Œz; c D 1i, where ˇ; ı; are integers mod p, 6 0 .mod p/ and .ˇ ı/2 C 4 is not a square in GF.p/. We have jGj D p 6 , 1 .G/ D G 0 D ˆ.G/ D Z.G/ D hx; y; zi Š Ep3 and so G is a special p-group. Also, M D ha; biG 0 and all other maximal subgroups of G are nonmetacyclic minimal nonabelian. 2

2

n

n1

D z; Œa; b D z; (b) G D ha; b; c j ap D b p D c p D 1; ap D x; b p D y; c p Œc; a D x ˛ y z ˇ ; Œc; b D x y z ı ; Œx; b D Œx; c D Œy; a D Œy; c D Œz; a D Œz; b D 1i, where n 3, ˛; ˇ; ; ı; ; are integers mod p, 6 0 .mod p/, 6 0 .mod p/, and .˛/2 C4 is not a square in GF.p/. In this case, we have jGj D p nC4 , 1 .G/ D G 0 D hx; y; zi Š Ep3 and ˆ.G/ D Z.G/ D hc p ; x; yi is abelian of type .p n1 ; p; p/. Also, M D ha; biˆ.G/ and all other maximal subgroups of G are nonmetacyclic minimal nonabelian. Proof. Let 1 D ¹H1 ; H2 ; : : : ; Hp2 Cp ; M º be the set of maximal subgroups of G, where Hi (i D 1; : : : ; p 2 C p) are minimal nonabelian and M is neither abelian nor minimal nonabelian. From Lemma 102.2 we get ˆ.G/ D Z.G/ D Z.Hi / D Z.M /, d.M / 3, jM 0 j D p and jG 0 j > p. By a result of A. Mann (see Exercise 1.69(a)), jG W .H10 H20 /j p and so p 2 jG 0 j p 3 . Suppose that for some Hi ¤ Hj we have Hi0 D Hj0 or Hi0 D M 0 . Then, by Exercise 1.69(a), jG 0 j p 2 and so jG 0 j D p 2 . If G 0 Š Cp2 , then the nonabelian group G=1 .G 0 / has p 2 C p C 1 abelian maximal subgroups, which is a contradiction by Exercise 1.6(a). Hence G 0 Š Ep2 . Let X be any ﬁxed subgroup of order p in G 0 . Then there is a minimal nonabelian maximal subgroup Hi (i 2 ¹1; : : : ; p 2 C pº) such that Hi0 ¤ X so that Hi =X is minimal nonabelian. By Lemma 102.4, G=X has exactly

102

p-groups G with p > 2, d.G/ > 2, exactly one exceptional maximal subgroup 89

p C 1 abelian maximal subgroups. Hence there are exactly p C 1 maximal subgroups of G whose commutator subgroup is equal X. But G 0 has p C 1 subgroups of order p and so G must have .p C 1/2 D p 2 C 2p C 1 maximal subgroups, a contradiction. We have proved that ¹H10 ; H20 ; : : : ; Hp0 2 Cp ; M 0 º is the set of p 2 C p C 1 pairwise distinct subgroups of order p in G 0 and so, in particular, G 0 Š Ep3 . Assume that there is an element t 2 G G 0 of order p. Since G=G 0 is abelian of rank 3, G=.G 0 ht i/ is noncyclic. But then there is a maximal subgroup Y of G containing G 0 ht i which is minimal nonabelian, a contradiction (with the structure of Y ). We have proved that 1 .G/ D G 0 Š Ep3 and all minimal nonabelian maximal subgroups of G are nonmetacyclic. Set T =G 0 D 1 .G=G 0 / Š Ep3 . If G=T is noncyclic, then there is a maximal subgroup K of G containing T which is minimal nonabelian. Since K > T , T is abelian of type .p 2 ; p 2 ; p 2 /. But K 0 < G 0 and so K 0 is not a maximal cyclic subgroup in K, contrary to the fact that K is nonmetacyclic minimal nonabelian. We have proved that G=T is cyclic. (i) First assume that T D G, i.e., G=G 0 Š Ep3 and so in this case G is a special group of order p 6 with G 0 D 1 .G/ Š Ep3 . We determine the structure of M . We have M D G 0 S , where S D ha; bi is minimal nonabelian and G 0 \ S D ˆ.S/ < G 0 since d.M / 3. Set S 0 D M 0 D hzi Š Cp . If ˆ.S/ D hzi, then hzi is a unique subgroup of order p in S which implies that S would be cyclic, a contradiction. Hence ˆ.S/ D 1 .S/ Š Ep2 and so S is metacyclic of order p 4 and exponent p 2 . We may choose a; b 2 S G 0 so that ap D z, b p D y 2 ˆ.S/ hzi and Œa; b D z. Since Ã1 .M / D Ã1 .S/ D hy; zi M 0 D hzi, it follows that M is a powerful group. By Proposition 26.10, each element in hy; zi is a p-th power of an element in M . Let c be an element in G M . Suppose that c p 2 hy; zi. Then there is m 2 M such that mp D c p . We get p .mc/p D mp c p Œc; m. 2 / D 1;

contrary to the fact that 1 .G/ D G 0 . Hence we have c p D x 2 G 0 hy; zi and so G 0 D hx; y; zi and G D ha; b; ci. All p 2 C p maximal subgroups of G which are distinct from M must be minimal nonabelian. We obtain ˆ.ha; ci/ D hz; x; Œa; ci D G 0 and so Œa; c D y r y 0 , where r 6 0 0 .mod p/ and y 0 2 hx; zi. Replacing c with c 0 D c r , where r 0 6 0 .mod p/ is such 0 that rr 1 .mod p/, we get 0

0

0

0

0

Œa; c 0 D Œa; c r D Œa; cr D y rr .y 0 /r D y.y 0 /r ; 0

0

0

where c 0 2 G M , .y 0 /r 2 hx; zi and .c 0 /p D .c p /r D x r D x 0 2 G hy; zi. Writing again c and x instead of c 0 and x 0 , respectively, we see that we may assume from the start Œa; c D yy 00 and c p D x, where y 00 2 hx; zi and so we may set Œa; c D x ˛ yz ˇ for some integers ˛; ˇ mod p. From ˆ.hb; ci/ D hy; x; Œb; ci D G 0 follows that Œb; c D x y ı z , where ; ı; are some integers mod p with 6 0 .mod p/.

90

Groups of prime power order

Maximal subgroups of G containing hci are ha; ciG 0 and hai b; ciG 0 . Therefore we must have for all integers i mod p ˆ.hai b; ci/ D hyz i ; x; Œai b; c D x ˛i C y i Cı z ˇ i C i D G 0 which is equivalent to ˇ ˇ ˇ 0 1 i ˇˇ ˇ ˇ 1 0 0 ˇˇ D i 2 C .ı ˇ/i 6 0 ˇ ˇ˛i C i C ı ˇi C ˇ

.mod p/:

Hence the quadratic congruence i 2 C .ı ˇ/i 0 .mod p/ should not have any solution in i which is equivalent to the requirement that .ˇ ı/2 C 4 is not a square in GF(p). We have to examine the p 2 maximal subgroups hc j a; c k biG 0 of G (j; k are integers mod p) which do not contain hci. For all k 6 0 .mod p/ we get ˆ.ha; c k bi/ D hz; x k y; Œa; c k b D x k˛ y k z kˇ C1 i D G 0 which is equivalent to ˇ ˇ ˇ0 0 1 ˇˇ ˇ ˇk 1 0 ˇˇ m D k 2 k˛ D k.k ˛/ 6 0 ˇ ˇk˛ k kˇ C 1ˇ

.mod p/:

Hence we must have ˛ 0 .mod p/ and so Œa; c D yz ˇ . For all j 6 0 .mod p/ we obtain ˆ.hc j a; bi/ D hx j z; y; Œc j a; b D x j y ıj z j C1 i D G 0 which is equivalent to ˇ ˇ j 0 ˇ ˇ 0 1 ˇ ˇj ıj

ˇ ˇ 1 ˇ ˇ D j.j C C 1/ 6 0 0 ˇ j C 1ˇ

.mod p/:

Since 6 0 .mod p/, we must have 1 .mod p/ and so Œb; c D x 1 y ı z . We have obtained the groups of order p 6 stated in part (a) of our theorem. It remains to check that all maximal subgroups hc j a; c k biG 0 are minimal nonabelian unless j k 0 .mod p/ in which case ha; biG 0 D M . Indeed, ˆ.hc j a; c k bi/ D hx j z; x k y; Œc j a; c k b D x j y ıj Ck z j Cˇ kC1 i < G 0 if and only if ˇ ˇ ˇj ˇ 0 1 ˇ ˇ ˇk ˇ D k 2 C k.ˇj ıj / j 2 0 1 0 ˇ ˇ ˇj ıj C k j C ˇk C 1ˇ

.mod p/:

102

p-groups G with p > 2, d.G/ > 2, exactly one exceptional maximal subgroup 91

The quadratic congruence in k k 2 C kj.ˇ ı/ j 2 0

.mod p/

(where j 2 GF.p/ is ﬁxed) has a solution in k if and only if the discriminant ()

j 2 .ˇ ı/2 C 4j 2 D ..ˇ ı/2 C 4/j 2

is a square in GF(p). But we know that .ˇ ı/2 C 4 is not a square in GF(p) and so we must have j 0 .mod p/. From () we then get k 2 0 .mod p/ and so also k 0 .mod p/ and we are done. (ii) Now assume that T < G, where T =G 0 D 1 .G=G 0 / Š Ep3 and T =G 0 is cyclic. Hence G=G 0 is abelian of type .p m ; p; p/, m 2, and the unique maximal subgroup of G containing T is obviously equal to M . There are normal subgroups U and V of G such that G D U V , U \ V D G 0 , U=G 0 Š Ep2 and V =G 0 Š Cpm , m 2. Let c be an element in V G 0 such that hci covers V =G 0 . We have o.c/ D p n , n 3, where n1 D z, where z 2 G 0 . Then M D n D m C 1 (noting that 1 .G/ D G 0 ). Set c p p 0 p hc iU , ˆ.G/ D Z.G/ D G hc i is abelian of type .p n1 ; p; p/ and jGj D p nC4 . Let a; b 2 U G 0 be such that U D G 0 ha; bi, where ap ; b p 2 G 0 and G D ha; b; ci. Since each minimal nonabelian maximal subgroup Hi of G (i D 1; : : : ; p 2 C p) is nonmetacyclic and contains ˆ.G/ and hzi is not a maximal cyclic subgroup in ˆ.G/, it follows that Hi0 ¤ hzi (for all i ) and so M 0 D hzi. Therefore 1 ¤ Œa; b 2 hzi and so we may assume Œa; b D z. Now, G=hzi has the unique abelian maximal subgroup M=hzi and all other maximal subgroups of the factor group G=hzi are minimal nonabelian. Hence G=hzi is an A2 -group with d.G=hzi/ D 3 and of order p nC3 > p 4 (since n 3), G 0 =hzi Š Ep2 , G 0 =hzi Z.G=hzi/ and G=hzi has a normal elementary abelian subgroup of order p 3 , namely hG 0 ; 2 .hci/i=hzi. Therefore G=hzi is an A2 -group from Proposition 71.4(b) which implies 1 .G=hzi/ D hG 0 ; 2 .hci/i=hzi. Set ap D x and b p D y and consider the abelian group M=hzi. If the abelian subgroup U=hzi of order p 4 and exponent p 2 has rank > 2, then 1 .U=hzi/ > G 0 =hzi, which contradicts the above fact. Hence we get U=hzi Š Cp2 Cp2 which implies G 0 D hx; y; zi. Since ˆ.hc; ai/ D hc p ; x; Œc; ai D ˆ.G/ and ˆ.hc; bi/ D hc p ; y; Œc; bi D ˆ.G/, we must have Œc; a D x ˛ y z ˇ ; Œc; b D x y z ı for some integers ˛; ˇ; ; ı; ; mod p with 6 0 .mod p/ and 6 0 .mod p/. The maximal subgroups of G containing hci are ha; ciˆ.G/ and hai b; ciˆ.G/. Therefore we must have for all integers i mod p ˆ.hc; ai bi/ D hc p ; x i y; Œc; ai b D x ˛i C y i C z ˇ iCı i D ˆ.G/ which is equivalent to ˇ ˇ 0 0 ˇ ˇ i 1 ˇ ˇ˛i C i C

ˇ 1 ˇˇ 0 ˇˇ D i 2 C . ˛/i 6 0 ˇi C ı ˇ

.mod p/:

92

Groups of prime power order

Hence the quadratic congruence i 2 C . ˛/i 0 .mod p/ should not have any solution in i which is equivalent to the requirement that . ˛/2 C 4 is not a square in GF(p). We have obtained the groups stated in part (b) of our theorem. It remains to check that all maximal subgroups hc j a; c k biˆ.G/ are minimal nonabelian unless j k 0 .mod p/ in which case ha; biˆ.G/ D M . Note that ˆ.G/ D hc p i hxi hyi and

2

ˆ.ˆ.G// D hc p i hzi;

where ˆ.G/=ˆ.ˆ.G// Š Ep3 . We have ˆ.hc j a; c k bi/ D hc pj x; c pk y; Œc j a; c k b D x j ˛k y j k z ıj ˇ kC1 i < ˆ.G/ if and only if ˇ ˇ ˇj 1 0 ˇˇ ˇ ˇk 0 1 ˇˇ D k 2 C .˛ /j k j 2 0 ˇ ˇ 0 j ˛k j k ˇ

.mod p/:

The quadratic congruence in k ()

k 2 C .˛ /j k j 2 0

.mod p/

(where j 2 GF.p/ is ﬁxed) has a solution in k if and only if the discriminant j 2 .˛ /2 C 4j 2 D ..˛ /2 C 4/j 2 is a square in GF(p). But we know that .˛ /2 C 4 is not a square in GF(p) and so we must have j 0 .mod p/. From () we then get k 2 0 .mod p/ and so (noting that 6 0 .mod p/) k 0 .mod p/ and we are done.

101

p-groups G with p > 2 and d.G / D 2 having exactly one maximal subgroup which is neither abelian nor minimal nonabelian

Here we determine up to isomorphism the title groups (Theorems 101.1 and 101.2). As a result we get a complete classiﬁcation of p-groups (which are not A2 -groups) from Lemma 76.5 (Theorems 100.1 and 101.1). Also, p-groups of Lemma 76.13 are completely determined (Theorems 100.2, 100.3 and 101.2). Theorem 101.1. Let G be a two-generator p-group, p > 2, with exactly one maximal subgroup M which is neither abelian nor minimal nonabelian. If G has an abelian maximal subgroup A, then we have G D hh; k j Œh; k D v; Œv; k D z; Œv; h D z ; v p D z p D Œz; h D Œz; k D 1; hp D z ; k p

nC1

D z i;

where n 1 and ; ; are integers mod p with 6 0 .mod p/. We have G 0 D hv; zi Š Ep2 ;

jGj D p nC4 ;

Z.G/ D hk p ; zi;

G 0 \ Z.G/ D hzi Š Cp ;

ˆ.G/ D Z.G/G 0 ;

ŒG 0 ; G D hzi

and so G is of class 3. Also, S D hv; hi Š S.p 3 / (if 0 .mod p/) or S Š Mp3 (if 6 0 .mod p/), S is normal in G, G D Shki;

S \ hki hzi; M D Shk p i;

A D CG .G 0 /;

G=S Š CpnC1 ; d.M / D 3;

G=Z.G/ Š S.p 3 /;

M 0 D hzi;

1 D ¹A; M; M1 ; : : : ; Mp1 º;

0 D hzi. Finally, G is an where all Mi are minimal nonabelian with M10 D D Mp1 L3 -group if and only if 6 0 .mod p/ and in that case 1 .G/ Š S.p 3 /, G=1 .G/ is cyclic of order p nC1 (n 1) and Z.G/ D hk p i Š CpnC1 is cyclic.

Proof. Obviously, A is a unique abelian maximal subgroup of G (otherwise, by Exercise 1.6(a), all p C 1 maximal subgroups of G would be abelian). By Exercise 1.69(a),

101

p-groups G with p > 2, d.G/ D 2, exactly one exceptional maximal subgroup 67

jG 0 W .A0 M10 /j p, where M1 is a minimal nonabelian maximal subgroup of G and so jG 0 j p 2 . But if jG 0 j D p, then this fact together with d.G/ D 2 would imply (Exercise P9) that G is minimal nonabelian, a contradiction. Hence jG 0 j D p 2 . Then it follows from jGj D pjG 0 jjZ.G/j (Lemma 1.1) that jG W Z.G/j D p 3 . Set 1 D ¹A; M; M1 ; : : : ; Mp1 º, where all Mi .i D 1; : : : ; p 1/ are minimal nonabelian. We have Z.G/ Mi (otherwise d.G/ D 3) and so Z.G/ D Z.Mi / D ˆ.Mi / for all i D 1; : : : ; p 1. Also, ˆ.Mi / < ˆ.G/ < Mi and so ˆ.G/ is abelian. For each x 2 G A, CA .x/ D Z.G/ and so x p 2 Z.G/. Hence G=Z.G/ is generated by its elements of order p and so G=Z.G/ Š S.p 3 / because d.G/ D 2 and so G=Z.G/ cannot be elementary abelian. This implies G 0 \ Z.G/ Š Cp ;

ˆ.G/ D Z.G/G 0 ;

cl.G/ D 3:

Also, Mi0 D M 0 D G 0 \ Z.G/ for all i D 1; : : : ; p 1. If d.M / D 2, then M 0 Š Cp would imply that M is minimal nonabelian, a contradiction. Hence we get d.M / 3. In particular, jM j p 4 and so jGj p 5 . (i) First assume that G 0 D hvi Š Cp2 is cyclic. Since hv p i D Mi0 is not a maximal cyclic subgroup in Mi > ˆ.G/ D Z.G/G 0 , it follows that all Mi (i D 1; : : : ; p 1) are metacyclic. In particular, j1 .ˆ.G//j p 2 . Suppose that A is also metacyclic so that M (with d.M / 3) is the only nonmetacyclic maximal subgroup of G. By a result of Y. Berkovich (see Proposition A.40.12), G is an L3 -group. But then we have G 0 1 .G/, where 1 .G/ is of order p 3 and exponent p and so G 0 Š Ep2 , a contradiction. It follows that A must be nonmetacyclic in which case 1 .A/ 6 ˆ.G/. Let a be an element of order p in Aˆ.G/ and let k 2 GA be such that hˆ.G/; ki D M1 . Since Œk; v ¤ 1, we may replace k with another generator of hki so that we may assume that Œk; v D v p . As hk; vi0 D hv p i Z.G/, it follows that hk; vi is minimal nonabelian and so hk; vi D M1 . By Exercise P9, we have Z.G/ D ˆ.M1 / D hk p ; v p ; Œk; v D v p i D hk p ; v p i: All maximal subgroups of G distinct from A D ˆ.G/hai are ˆ.G/hai ki D Z.G/hai k; vi; where i is any integer mod p. Since Œai k; v D Œai ; vk Œk; v D v p 2 Z.G/, it follows that hai k; vi is minimal nonabelian. Again, by Exercise P9, ˆ.hai k; vi/ D h.ai k/p ; v p ; Œai k; v D v p i D h.ai k/p ; v p i: The factor group G=hv p i is minimal nonabelian (since we have d.G=hv p i/ D 2 and .G=hv p i/0 Š Cp ) and so computing in G=hv p i, we get p

.ai k/p D aip k p Œk; ai . 2 / x;

where x 2 hv p i:

68

Groups of prime power order p

But aip D 1 and Œk; ai 2 hvi so that Œk; ai . 2 / 2 hv p i which gives .ai k/p D k p y for some y 2 hv p i. By the above, ˆ.hai k; vi/ D hk p y; v p i D hk p ; v p i D Z.G/ and so we conclude that ˆ.G/hai ki D Z.G/hai k; vi D hai k; vi is minimal nonabelian for all i D 1; : : : ; p 1. It follows that G is an A2 -group, a contradiction. (ii) We have proved that G Š Ep2 . Since ŒG; G 0 D G 0 \ Z.G/ Š Cp , we get by the Hall–Petrescu formula (Appendix 1) for any x; y 2 G, .xy/p D x p y p l for some l 2 G 0 \ Z.G/. We have A D CG .G 0 / and we take an element k 2 G A such that ˆ.G/hki D M1 is a minimal nonabelian maximal subgroup of G. Then for any element v 2 G 0 Z.G/, we get Œv; k D z, where hzi D G 0 \ Z.G/. Since hk; vi0 D hzi, hk; vi is minimal nonabelian and so hk; vi D M1 . In particular, ˆ.hk; vi/ D hk p ; v p ; Œv; ki D hk p ; zi D ˆ.M1 / D Z.G/: Thus hk p i covers Z.G/=hzi, where jZ.G/j p 2 . Set Z.G/=hzi Š ˆ.G/=G 0 Š Cpn with n 1 so that jGj D p nC4 . Consider the abelian group G=G 0 of rank 2. Since .G 0 hki/=G 0 Š CpnC1 , there is a subgroup S=G 0 of order p such that G D S hki and S \ hki hzi. Let h 2 S G 0 so that hp 2 hzi since hp 2 Z.G/ \ G 0 D hzi. Assume that S A in which case h 2 A ˆ.G/ and G D hh; ki. We may assume that Œh; k D v and examine all maximal subgroups ˆ.G/hhi ki of G (i is any integer mod p) which are distinct from A. We have Œv; hi k D Œv; kŒv; hi k D Œv; k D z and so hv; hi ki is minimal nonabelian. On the other hand, ˆ.hv; hi ki/ D hv p D 1; .hi k/p D hip k p l; Œv; hi k D zi D hk p ; zi D Z.G/ (where l 2 hzi) since .hi k/p D k p l 0 for some l 0 2 hzi. This means that ˆ.G/hhi ki D hv; hi ki and so all these p maximal subgroups of G are minimal nonabelian. But then G is an A2 -group, a contradiction. We have proved that S 6 A D CG .G 0 / and so 1 ¤ Œv; h 2 hzi. Since G D hh; ki, we may set Œh; k D v and Œv; k D z, where v 2 G 0 Z.G/ and hzi D G 0 \ Z.G/. nC1 Also, Œv; h D z , hp D z , and k p D z , where ; ; are integers mod p with 6 0 .mod p/. Here S D hv; hi Š S.p 3 / or Mp3 , S is normal in G, G D S hki with hki \ S hzi and M D S Z.G/ D S hk p i with d.M / D 3. It remains to examine all p maximal subgroups ˆ.G/hhi ki (i D 0; 1; : : : ; p 1) of G which are distinct from M D ˆ.G/hhi D S hk p i. We compute Œv; hi k D Œv; kŒv; hi D zz i D z i C1 ; where the congruence i C 1 0 .mod p/) has exactly one solution i 0 for i (noting 0 that 6 0 .mod p/). Hence A D ˆ.G/hhi ki is an abelian maximal subgroup of G

101

p-groups G with p > 2, d.G/ D 2, exactly one exceptional maximal subgroup 69

and for all other i 6 i 0 .mod p/ we see that hv; hi ki is minimal nonabelian and moreover, ˆ.hv; hi ki/ D hv p D 1; .hi k/p D hip k p l; Œv; hi k D z i C1 ¤ 1i for some l 2 hzi. Hence ˆ.hv; ai ki/ D hk p ; zi D Z.G/ and so ˆ.G/hhi ki D hv; hi ki is a minimal nonabelian maximal subgroup of G. Our theorem is proved. Theorem 101.2. Let G be a two-generator p-group, p > 2, with exactly one maximal subgroup H which is neither abelian nor minimal nonabelian. If G has no abelian maximal subgroup, then 1 D ¹H; H1 ; : : : ; Hp º, where Hi (i D 1; : : : ; p) are nonmetacyclic minimal nonabelian, G 0 Š Ep3 , W D ŒG; G 0 Š Ep2 , W Z.G/ (and so G is of class 3) and CG .G 0 / D ˆ.G/ is abelian. Moreover, ¹H 0 ; H10 ; : : : ; Hp0 º is the set of p C 1 subgroups of order p in W and the following holds. (a) If jGj p 6 , then we have G D hh; x j hp

mC1

m

D 1; Œh; x D v; hp D u; Œv; h D z;

Œv; x D u˛ ; v p D z p D Œu; x D Œz; h D Œz; x D 1; x p 2 hu; zii; where m 2 and ˛ is an integer mod p with ˛ 6 0 .mod p/. Here jGj D p mC4 ;

G 0 D hu; z; vi Š Ep3 ;

Z.G/ D hhp i hzi Š Cpm Cp ;

W D ŒG; G 0 D hu; zi Z.G/; ˆ.G/ D Z.G/ hvi:

Finally, H D ˆ.G/hxi, where in case x p 2 W hui we have d.H / D 3 and in case x p 2 hui we have d.H / D 4 and Hi D hv; x i hi (i D 1; : : : ; p) is the set of p nonmetacyclic minimal nonabelian maximal subgroups of G. (b) If jGj D p 5 , then 2

G D hh; x j hp D 1; Œh; x D v; hp D u˛ ; Œv; h D z; Œv; x D u; v p D z p D Œu; x D Œz; h D Œz; x D 1; hp D uˇ z i; where ˛; ˇ; are integers mod p with ˇ 6 0 .mod p/. We have ˆ.G/ D G 0 D hu; z; vi Š Ep3 and W D ŒG; G 0 D hu; zi D Z.G/ Š Ep2 . If p 5, then we have ˛ .mod p/. In that case, ˛ 0 .mod p/ implies Ã1 .G/ D hui and 1 .G/ D H Š S.p 3 / Cp and ˛ 6 0 .mod p/ implies Ã1 .G/ D W , 1 .G/ D G 0 and H Š Mp3 Cp . Also, all p maximal subgroups Hi D G 0 hx i hi (i integer mod p) are nonmetacyclic minimal nonabelian. If p D 3, then either ˇ D 1 and 6 ˛ .mod 3// or ˇ D 1 and ˛ .mod 3/. In that case, H D G 0 hxi Š S.27/ C3 or H Š M27 C3 and all three maximal subgroups Hi D G 0 hx i hi (i integer mod p) are nonmetacyclic minimal nonabelian.

70

Groups of prime power order

Proof. We set 1 D ¹H; H1 ; : : : ; Hp º, where Hi (i D 1; : : : ; p) are minimal nonabelian. Since H is neither abelian nor minimal nonabelian, we have jH j p 4 and so jGj p 5 . First suppose that two distinct minimal nonabelian maximal subgroups of G have the same commutator subgroup, say, H10 D H20 . Then considering G=H10 (see Exercise 1.6(a)), we see that all maximal subgroups of G=H10 are abelian and so we get H 0 D H10 D D Hp0 D hzi Š Cp : By Exercise 1.69(a), we have jG 0 W .H10 H20 /j D jG 0 W H10 j p, and so jG 0 j p 2 . But if jG 0 j D p, then this fact together with d.G/ D 2 implies (see Exercise P9) that G is minimal nonabelian, a contradiction. Hence jG 0 j D p 2 . Also, d.H / 3 and so H is nonmetacyclic. Indeed, if d.H / D 2, then (noting that jH 0 j D p) H would be minimal nonabelian, a contradiction. Suppose for a moment that G 0 D hvi Š Cp2 is cyclic. Then H10 D D Hp0 D hv 2 i and G 0 D hvi Hi so that Hi0 is not a maximal cyclic subgroup in Hi and therefore Hi is metacyclic for all i D 1; : : : ; p. By Proposition A.40.12, G is an L3 -group. But in that case, G 0 is of exponent p, a contradiction. We have proved that G 0 Š Ep2 . We notice that ˆ.G/ D H1 \ H2 is a maximal normal abelian subgroup of G. Taking elements h1 2 H1 ˆ.G/ and h2 2 H2 ˆ.G/, we have hh1 ; h2 i D G and so s D Œh1 ; h2 2 G 0 hzi and s 62 Z.G/. Indeed, if s 2 Z.G/, then G=hsi would be abelian, a contradiction. In particular, ŒG; G 0 D hzi D H10 and so G is of class 3. Since s 62 Z.G/, we have s 62 Z.H1 / or s 62 Z.H2 / and we may assume without loss of generality that s 62 Z.H1 /. Suppose that there is an element x 2 H1 ˆ.G/ such that x p 2 hzi. Then G 0 hxi is minimal nonabelian of order p 3 and so G 0 hxi D H1 , contrary to jGj p 5 . Assume that hzi D H10 is not a maximal cyclic subgroup in H1 . Then there is v 2 ˆ.G/ such that v 2 D z. This implies that all Hi , i D 1; : : : ; p, are metacyclic. Again by Proposition A.40.12, G is an L3 -group. This means that the subgroup U D 1 .G/ is of order p 3 and exponent p and G=U is cyclic of order p 2 . We have G 0 U and so if U is nonabelian, then CG .G 0 / covers G=U and CG .G 0 / is an abelian maximal subgroup of G, a contradiction. If U is elementary abelian, then jG W CG .U /j D p since a Sylow p-subgroup of GL3 .p/ is isomorphic to S.p 3 / and so is of exponent p. But in that case, CG .U / is an abelian maximal subgroup of G, a contradiction. Hence hzi D H10 is a maximal cyclic subgroup in H1 which yields that H1 is nonmetacyclic. Noting that jH1 j p 4 , we get E D 1 .H1 / D 1 .ˆ.G// Š Ep3 which also implies that all Hi are nonmetacyclic. By the previous paragraph, 1 .hh1 i/ D hui E and u 2 E G 0 . Further, we have H1 D Ehh1 i and so H1 is a splitting extension of G 0 by hh1 i, where o.h1 / D p n , n 2. Since G=G 0 is abelian of rank 2, we get G D H1 F with H1 \ F D G 0 and jF W G 0 j D p. We have G=F Š H1 =G 0 Š Cpn . If F is nonabelian, then CG .G 0 /

101

p-groups G with p > 2, d.G/ D 2, exactly one exceptional maximal subgroup 71

covers G=F and so CG .G 0 / is an abelian maximal subgroup of G, a contradiction. Hence F is abelian. Assume that Ã1 .F / 6 hzi. Then we obtain that jÃ1 .F /j D p and G 0 D hzi Ã1 .F / Z.G/, a contradiction. Hence Ã1 .F / hzi and so for an element x 2 F G 0 we have x p 2 hzi. Since G satisﬁes G D hh1 ; xi, we may set Œx; h1 D s 2 G 0 hzi and Œs; h1 D z, i where H10 D hzi Z.G/. Then x h1 D xs, s h1 D sz and s h1 D sz i for all i 1. We get 2 x h1 D .xs/h1 D .xs/.sz/ D xs 2 z i

i

and claim that we have x h1 D xs i z .2/ for all i 2. Indeed, by induction on i, iC1

x h1

i

i

i

i

i C1 2

D .x h1 /h1 D .xs/h1 D .xs i z .2/ /.sz i / D xs iC1 z .2/Ci D xs i C1 z .

/:

p p p Our formula gives x h1 D xs p z . 2 / D x and so F hh1 i is an abelian maximal subgroup of G, a contradiction. We have proved that H10 D hz1 i, H20 D hz2 i; : : : ; Hp0 D hzp i are pairwise distinct subgroups of order p in G 0 \ Z.G/. By Exercise 1.69(a), we get jG 0 W .H10 H20 /j p and so jG 0 j p 3 . Set W D hz1 ; : : : ; zp i so that W is an elementary abelian subgroup of order p 2 contained in G 0 \ Z.G/ which implies that G 0 is abelian of exponent p 2 . We have G D hx; yi for some x; y 2 G. If Œx; y 2 Z.G/, then G=hŒx; yi is abelian which implies that G 0 D hŒx; yi is cyclic, contrary to the fact that W G 0 . Thus Œx; y 2 G 0 W which gives

jG 0 j D p 3 ;

W Š Ep2 ;

¹1º ¤ ŒG; G 0 W Z.G/

and so G is of class 3. Let hzpC1 i be the subgroup of order p in W with hzpC1 i ¤ hzi i for all i D 1; : : : ; p. For any ﬁxed i 2 ¹1; : : : ; pº we consider G=hzi i, where Hi =hzi i is abelian (and two-generated) and Hj =hzi i is minimal nonabelian for all j ¤ i, j 2 ¹1; : : : ; pº. This implies that H=hzi i must be nonabelian (Exercise 1.6(a)). If G=hzi i is metacyclic, then Theorem 36.1 gives that G is also metacyclic, contrary to Ep2 Š W G 0 . Hence G=hzi i is nonmetacyclic. Suppose that H=hzi i is minimal nonabelian. Then G=hzi i is a nonmetacyclic A2 -group. If jG=hzi ij > p 4 , then Theorem 65.7(a) implies that G 0 =hzi i Š Ep2 . Suppose that jG=hzi ij D p 4 and G 0 =hzi i Š Cp2 . In that case, each maximal subgroup of G=hzi i is metacyclic and so G=hzi i is minimal nonmetacyclic. By Theorem 69.1, G=hzi i is a group of maximal class and order 34 . But in that case, G 0 =hzi i cannot be cyclic (see Exercise 9.1(c)). We have proved that in any case G 0 =hzi i Š Ep2 . Assume now that H=hzi i is not minimal nonabelian and we know already that H=hzi i is nonabelian. By Theorem 101.1, we have again G 0 =hzi i Š Ep2 . As a consequence we get Œx; yp 2 hzi i for each i D 1; : : : ; p which implies that Œx; yp D 1 and so G 0 Š Ep3 is elementary abelian. By Lemma 36.5(b), G 0 =ŒG; G 0 is cyclic and so ŒG; G 0 D W (since ŒG; G 0 W ). Since .G=W /0 Š Cp and d.G=W / D 2, G=W is minimal nonabelian (Exercise P9)

72

Groups of prime power order

and so H=W is abelian which implies ¹1º ¤ H 0 W . Suppose that H 0 D hzj i for some j 2 ¹1; : : : ; pº. Then G=hzj i is nonabelian with at least two distinct abelian maximal subgroups Hj =hzj i and H=hzj i. But then .G=hzj i/0 Š Cp (Exercise P1), a contradiction. We have proved that H 0 D hzpC1 i or H 0 D W . Suppose that H 0 D W Z.G/. In that case, d.H / 3. Indeed, if d.H / D 2, then H is a two-generator group of class 2 in which case H 0 must be cyclic (Proposition 36.5(a)), a contradiction. Consider G=hzpC1 i with d.G=hzpC1 i/ D 2 and having minimal nonabelian maximal subgroups Hi =hzpC1 i for all i D 1; : : : ; p. The remaining maximal subgroup H=hzpC1 i is neither abelian nor minimal nonabelian since d.H=hzpC1 i/ 3. But .Hi =hzpC1 i/0 D W=hzpC1 i for all i D 1; : : : ; p, contrary to the ﬁrst part of this proof. Hence we must have H 0 D hzpC1 i. We have proved that H 0 ; H10 ; : : : ; Hp0 are p C 1 pairwise distinct subgroups of order p in W . Since H is not minimal nonabelian, we have d.H / 3 and so jH j p 4 and jGj p 5 . Also, 1 .Hi / D G 0 ˆ.G/ for all i D 1; : : : ; p, where ˆ.G/ is abelian. Therefore we have either CG .G 0 / D ˆ.G/ or CG .G 0 / is a maximal subgroup of G. In any case, there exist two minimal nonabelian maximal subgroups of G, say H1 and H2 , such that G 0 6 Z.H1 / and G 0 6 Z.H2 /. Then H1 \ H2 D ˆ.G/ and taking some elements h1 2 H1 ˆ.G/ and h2 2 H2 ˆ.G/, we have hh1 ; h2 i D G and so v D Œh1 ; h2 2 G 0 W . Indeed, if v 2 W , then G=hvi is abelian, a contradiction. We may set Œv; h1 D z1 and Œv; h2 D z2 so that H10 D hz1 i, H20 D hz2 i and W D hz1 ihz2 i. All maximal subgroups of G are H1 D ˆ.G/hh1 i and ˆ.G/hhi1 h2 i, where i is any integer mod p. We compute Œv; hi1 h2 D Œv; h2 Œv; hi1 h2 D z2 .z1i /h2 D z1i z2 ¤ 1 which shows CG .v/ D ˆ.G/ and so CG .G 0 / D ˆ.G/. As hv; h1 i (with Œv; h1 D z1 ) is minimal nonabelian, we have hv; h1 i D H1 D G 0 hh1 i. Hence H1 =G 0 Š Cpm is pm m cyclic of order p , m 1, and h1 2 W hz1 i. The abelian group G=G 0 is of rank 2 and so G=G 0 is of type .p m ; p/ and jGj D p mC4 . Finally, ˆ.G/ D G 0 hh1 i D hh1 i hvi hz1 i p

p

is abelian of type .p m ; p; p/. pm

(i) First suppose that m 2. Set u D h1 so that u 2 W H10 , o.h1 / D p mC1 p 3 and ˆ.G/=G 0 is cyclic of order p m1 p since H1 =G 0 Š Cpm . Consider any Hi for 2 i p so that Hi \ H1 D ˆ.G/. Let hi 2 Hi ˆ.G/ and v 2 G 0 W so that 1 ¤ Œhi ; v D zi and Hi0 D hzi i. Since hhi ; vi is minimal nonabelian, we have hhi ; vi D Hi D G 0 hhi i and so Hi =G 0 is also cyclic of order p m . Further, we have p p p ıp hi 2 ˆ.G/ G 0 and hhi i covers ˆ.G/=G 0 . It follows hi D h1 k for some k 2 G 0 and ı 6 0 .mod p/. Then p2

hi

ıp 2

D h1

and so

pm

hhi i D hui;

pm

where u D h1 ;

101

p-groups G with p > 2, d.G/ D 2, exactly one exceptional maximal subgroup 73

and this implies that u 2 W Hi0 . We have proved that u 62 Hi0 for all i D 1; : : : ; p which forces hui D H 0 . Since G=G 0 is abelian of type .p m ; p/, m 2, we get G D H1 F with H1 \F D G 0 and jF W G 0 j D p. For the maximal subgroup ˆ.G/F of G we have .ˆ.G/F /=G 0 D .ˆ.G/=G 0 / .F=G 0 / Š Cpm1 Cp and so .ˆ.G/F /=G 0 is not cyclic which implies that ˆ.G/F D H . Taking the elements h1 D h 2 H1 ˆ.G/ and x 2 F ˆ.G/, we have o.x/ p 2 , G D hh; xi and m so v D Œh; x 2 G 0 W . We set again u D hp and we know that H 0 D hui. Also j set Œv; h D z1 D z, where H10 D hzi, W D hui hzi, v h D vz and v h D vz j for j 1. Since CG .G 0 / D CG .v/ D ˆ.G/, we have x p 2 W D hu; zi and Œv; x D u˛ with ˛ 6 0 .mod p/. We have obtained all relations stated in part (a) of our theorem. From Œh; x D v we get Œh2 ; x D Œh; xh Œh; x D v h v D .vz/v D v 2 z: j

We prove by induction on j 2 that Œhj ; x D v j z . 2 / . Indeed, j

j

Œhj C1 ; x D Œhhj ; x D Œh; xh Œhj ; x D v h .v j z . 2 / / D .vz j /.v j z . 2 / / j

j

j C1 2

j

D v j C1 z j C. 2 / D v j C1 z .

/:

p

In particular, Œhp ; x D v p z . 2 / D 1 and so hp 2 Z.G/ since G D hh; xi. We get Z.G/ D hhp i hzi Š Cpm Cp

and ˆ.G/ D Z.G/ hvi:

Further, we have H D F hhp i with F \ hhp i D hui. If x p 2 W hui, then F is minimal nonabelian and so d.H / D 3. If x p 2 hui, then F D hv; xi hzi, where hui D Z.hv; xi/ and hv; xi Š S.p 3 / or Mp3 and so d.H / D 4. Finally, we have to check all p maximal subgroups Hi D ˆ.G/hx i hi of G which are distinct from H and we have to show that they are minimal nonabelian. We have Œv; x i h D Œv; hŒv; x i h D z.u˛i /h D zu˛i ; where hzu˛i i are pairwise distinct subgroups of order p in W for i D 1; : : : ; p since ˛ 6 0 .mod p/). Thus hv; x i hi is minimal nonabelian with hv; x i hi0 D hzu˛i i ¤ hui. By the Hall–Petrescu formula (Appendix 1), we get for all r; s 2 G, m

m

p m m m . / .p / .rs/p D r p s p c2 2 c3 3 ;

where c2 2 G 0 and c3 2 W D ŒG; G 0 . But m 2 and so .rs/p D r p s p . We get m

m

m

m

m

m

m

.x i h/p D .x i /p hp D hp D u and so o.x i h/ D p mC1 which together with hv; x i hi D G 0 hx i hi and G 0 \hx i hi D hui implies that jhv; x i hij D p mC3 . But jGj D p mC4 and so hv; x i hi D Hi is minimal nonabelian and we are done.

74

Groups of prime power order

(ii) Suppose that m D 1. Then jGj D p 5 and G 0 D ˆ.G/ with CG .G 0 / D G 0 . We have H1 \ H D G 0 and since 1 .H1 / D G 0 , we get 1 ¤ hp 2 W D ŒG; G 0 D Z.G/ Š Ep2 for an element h 2 H1 G 0 . Also, we have Ã1 .G/ W . Take an element x 2 H G 0 so that x p 2 W , G D hh; xi and v D Œh; x D G 0 W . Set Œv; x D u so that 1 ¤ u 2 W and hui D H 0 . Then Œv; h D z 62 hui from which we conclude that H10 D hzi and W D huihzi. Since hv; xi is minimal nonabelian, we have hv; xi ¤ H which implies x p D u˛ (for some integer ˛ mod p) and H D hv; xi hzi, where hv; xi Š S.p 3 / or Mp3 . Since H1 is nonmetacyclic minimal nonabelian, hzi is a maximal cyclic subgroup in H1 which implies hp D uˇ z with ˇ 6 0 .mod p/. All p maximal subgroups Hi D G 0 hx i hi (i is any integer mod p) of G which are distinct from H must be minimal nonabelian. Since Œv; x i h D Œv; hŒv; x i h D z.ui /h D zui ¤ 1 and hv; x i hi is minimal nonabelian with hv; x i hi0 D hui zi and so we must have that Hi D hv; x i hi, we get h.x i h/p i ¤ hui zi or, equivalently, ()

h.x i h/p ; ui zi D hu; zi

for all integers i .mod p/. (ii1) First we assume p 5 in which case G is regular. By the Hall–Petrescu formula (Appendix 1), we have in our case for all r; s 2 G, .p/ .p/ .rs/p D r p s p c2 2 c3 3 ; where c2 2 G 0 and c3 2 W D ŒG; G 0 and so .rs/p D r p s p . Hence .x i h/p D x pi hp D u˛i .uˇ z / D u˛iCˇ z : Our condition () is equivalent to ˇ ˇ ˇ˛i C ˇ ˇ ˇ ˇ D .˛ /i C ˇ 6 0 ˇ i 1ˇ

.mod p/

for all integers i .mod p/, where we know that ˇ 6 0 .mod p/. This is equivalent to ˛ 0 .mod p/ and so ˛ .mod p/. If ˛ 0 .mod p/, then Ã1 .G/ D hui and

1 .G/ D H Š S.p 3 / Cp :

In case ˛ 6 0 .mod p/ we get Ã1 .G/ D W and 1 .G/ D G 0 so that H Š Mp3 Cp . (ii2) Finally, we suppose p D 3. In that case, the Hall–Petrescu formula gives for all r; s 2 G, .rs/3 D r 3 s 3 c3 , where c3 2 W . This is not a sufﬁcient information because

101

p-groups G with p > 2, d.G/ D 2, exactly one exceptional maximal subgroup 75

we have to know exactly the element c3 . Using the usual commutator identities (see 7, p. 98) together with xy D yxŒx; y, we compute exactly .rs/3 . We get .rs/2 D r.sr/s D r.rsŒs; r/s D r 2 s.sŒs; rŒs; r; s/ D r 2 s 2 Œs; rŒs; r; s; .rs/3 D r 2 s 2 Œs; rŒs; r; s rs D r 2 s 2 rŒs; rsŒs; r; rŒs; r; s D r 2 .s 2 r/sŒs; rŒs; r; sŒs; r; rŒs; r; s: But we have r 2 .s 2 r/sŒs; r D r 2 .rs 2 Œs 2 ; r/sŒs; r D r 3 s 3 Œs 2 ; rŒs 2 ; r; sŒs; r D r 3 s 3 Œs; rŒs; rs Œs 2 ; r; sŒs; r D r 3 s 3 Œs; r.Œs; rŒs; r; s/Œs 2 ; r; sŒs; r D r 3 s 3 Œs; r; sŒs 2 ; r; s and so .rs/3 D r 3 s 3 Œs; r; sŒs 2 ; r; sŒs; r; sŒs; r; rŒs; r; s D r 3 s 3 Œs 2 ; r; sŒs; r; r:

(1)

Also, we get Œs 2 ; r D Œs; rs Œs; r D Œs; rŒs; r; sŒs; r D Œs; r2 Œs; r; s and then Œs 2 ; r; s D ŒŒs; r2 Œs; r; s; s D ŒŒs; r2 ; s D Œs; r; sŒs;r Œs; r; s D Œs; r; s2 which together with (1) yields .rs/3 D r 3 s 3 Œs; r; s2 Œs; r; r D r 3 s 3 Œs; r; s1 Œs; r; r D r 3 s 3 Œs; Œs; rŒŒs; r; r: Thus we have obtained the formula ()

.rs/3 D r 3 s 3 Œs; Œs; rŒŒs; r; r:

Since hhp ; zi D huˇ z ; zi D hu; zi (noting that ˇ 6 0 .mod 3/),we have to use our condition () only for i D 1; 2. By (), .xh/3 D x 3 h3 Œh; Œh; xŒŒh; x; x D u˛ uˇ z Œh; vŒv; x D u˛Cˇ C1 z 1 ; and so from hu˛Cˇ C1 z 1 ; uzi D hu; zi we get ˇ ˇ ˇ˛ C ˇ C 1 1ˇ ˇ ˇ D ˛ C ˇ 1 6 0 (2) ˇ 1 1 ˇ

.mod 3/:

We compute Œh; x 2 D Œh; xŒh; xx D vv x D v.vu/ D v 2 u and so by (), .x 2 h/3 D x 6 h3 Œh; v 2 uŒv 2 u; x 2 D u2˛ .uˇ z /z 2 u D u˛Cˇ C1 z

C1

:

76

Groups of prime power order

From our condition () for i D 2 we get hu˛Cˇ C1 z C1 ; u2 zi D hu; zi or, equivalently, ˇ ˇ ˇ˛ C ˇ C 1 C 1ˇ ˇ ˇ D ˛ C ˇ C 1 6 0 .mod 3/: (3) ˇ 1 1 ˇ

Now, (2) and (3) hold if and only if .˛ C ˇ 1/.˛ C ˇ C 1/ 6 0

.mod 3/:

This is equivalent to ..ˇ 1/ C .˛ //..ˇ 1/ .˛ // 6 0

.mod 3/

.ˇ 1/2 .˛ /2 6 0

.mod 3/:

or

Hence if ˇ D 1, then 6 ˛ .mod 3/ and if ˇ D 1, then ˛ .mod 3/. We have obtained the groups stated in part (b) of our theorem which is now completely proved.

103

Some results of Jonah and Konvisser

In this section we will present another approach to counting theorems due to Jonah and Konvisser [KonJ, JKon]. Deﬁnition 1. Let G be a p-group. The line M1 M2 determined by two distinct maximal subgroups M1 and M2 of G is the set of all maximal subgroups of G containing M1 \ M2 . Obviously, jM1 M2 j D p C 1, i.e., every line contains exactly p C 1 points. Proposition 103.1. Let … be a ﬁnite projective space in which each line contains p C1 points and let f be a function from the points of … to the set of integers. Suppose that there is a point m0 2 … with the following property: If m1 and m2 are two points different from m0 on the same line through m0 , then f .m1 / f .m2 / .mod p/. P Then m2… f .m/ f .m0 / .mod p/. Proof. We have X

f .m/ D f .m0 / C

m2…

XX

f .m/ ;

L0 … m2L0

where L runs over all lines of … through m0 and L0 D L ¹m0 º. Furthermore, X f .m/ jL0 jf .m/ 0 .mod p/ m2L0

since f is constant (mod p) on the points of L0 and jL0 j D p by hypothesis. The two displayed formulas yield the result. Applying these ideas to p-groups, we get the following Theorem 103.2. Let G be a p-group, C be a nonempty set of proper subgroups of G and ˛.H / be the number of elements of C contained in H G. Suppose that: (i) For each proper normal subgroup M of G which contains an element of C, one has ˛.M / 1 .mod p/.

94

Groups of prime power order

(ii) There is a maximal subgroup M0 of G with the property: if a maximal subgroup M1 ¤ M0 contains an element of C, then so do all maximal subgroups on the line M0 M1 . Then ˛.G/ 1 .mod p/ or, what is the same, jCj 1 .mod p/ since ˛.G/ D jC j. Proof. By Hall’s enumeration principle and Proposition 103.1, we have X ˛.M / ˛.M0 /j1 j 1 .mod p/ ˛.G/

M 21

since, by hypothesis, if M 2 1 ¹M0 º, then ˛.M / ˛.M0 / 1 .mod p/ and j1 j 1 .mod p/. At this point it is appropriate to give the following Deﬁnition 2. Let C be a nonempty set of proper subgroups of a p-group G. (a) A maximal subgroup M0 of G which satisﬁes Lemma 103.2(ii) is called an origin (for C). (b) A proper subgroup L of G is called a local origin if for each M 2 1 the intersection L \ M contains an element of C . (c) A pair E1 ; E2 of distinct elements of the set C is said to support good lines if each maximal subgroup M < G containing E1 \ E2 also contains an element of C . Proposition 103.3 (Method of locating origins). Let C be a nonempty set, closed under conjugation, of proper subgroups of a p-group G satisfying ˛.M / D 0 or ˛.M / 1 .mod p/ for each M 2 1 . Then G contains an origin M0 if any at least one of the following three conditions is satisﬁed: (a) G contains a local origin L, in which case any maximal subgroup M L is an origin. (b) Each pair of distinct G-invariant elements E1 ; E2 2 C supports good lines, in which case any maximal subgroup M of G containing an element of C is an origin. (c) A maximal subgroup M of G contains a family E1 ; : : : ; Em of G-invariant elements of the set C satisfying: if E is a G-invariant element of the set C , then for some i 2 ¹1; : : : ; mº, the pair E, Ei supports good lines, in which case M is an origin. Proof. Statement (a) follows from the deﬁnitions. Indeed, let L M0 2 1 and let M 2 1 ¹M0 º. Then ˛.M0 / 1 .mod p/ and ˛.M / 1 .mod p/ since L is a local origin. It follows that M0 is an origin. (b) Let M0 2 1 contain an element of the set C . Then M0 contains a G-invariant element E0 2 C since ˛.M0 / 1 .mod p/ in view of ˛.M0 / > 0. For the same

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Some results of Jonah and Konvisser

95

reasons, each M1 2 1 contains a G-invariant element E1 of the set C whenever M1 contains an element of the set C . Thus each M on the line M0 M1 contains an element of the set C for M > M0 \ M1 E0 \ E1 , which in turn implies that M contains a G-invariant element of C because the G-invariant pair E0 ; E1 supports good lines by hypothesis. Statement (c) follows as (b). Exercise 1 coincides with the case n D 2 of Theorem 1.10(b). Exercise 1. Let G be a noncyclic p-group, p > 2. Then the number of noncyclic subgroups of order p 2 in G is congruent to 1 .mod p/. Solution. We use induction on jGj. Let C be the set of all abelian subgroups of type .p; p/ in G; then C is nonempty (Lemma 1.4) and G-invariant. By induction, if Ep2 Š E M1 2 1 , then ˛.M1 / 1 .mod p/ so that M1 contains a G-invariant abelian subgroup E1 of type .p; p/. One may assume that ˛.G/ > 1. Let E2 ¤ E1 be a G-invariant element of the set C (if E2 does not exist, the result is obvious). Then the product E1 E2 is a group of order p 4 and exponent p since p > 2. Since .E1 E2 /\M contains an abelian subgroup of type .p; p/ for all M 2 1 , it follows that E1 E2 is a local origin; then M is an origin. By Lemma 103.2, ˛.G/ 1 .mod p/. Exercise 2 (Sylow). Let G be a group of order p m and n < m. Then the number of subgroups of order p n in G is congruent to 1 .mod p/. Hint. Let C be the set of all subgroups of order p n in G. As above, C contains a G-invariant element E1 . If E2 ¤ E1 is another G-invariant element of the set C (if E2 does not exist, the result is obvious), then, obviously, E1 E2 is a local origin (indeed, M \ E1 E2 contains an element from C for M 2 1 ). Now the result follows from Lemma 103.2. In the previous two exercises one can take as C the set of all G-invariant subgroups of the set C deﬁned in these exercises. The following result is also known (see Theorem 10.4). Proposition 103.4. Let G be a p-group, p > 2. If G contains an elementary abelian subgroup of order p 3 , then the number of such subgroups in G is congruent to 1 .mod p/. Proof. We use induction on jGj. Let C be the set of all elementary abelian subgroups of order p 3 in G. Then, as above, there exists a G-invariant element E1 2 C . Let E2 ¤ E1 be a G-invariant element in C (see Exercises 1 and 2). Set H D E1 E2 ; then exp.H / D p since p > 2 and cl.H / 2 (see Theorems 7.1(b) and 7.2(b)). Then one of the following assertions holds: (i) jH j D p 4 . Then E1 \ E2 Z.H /. Let a maximal subgroup M of G contain E1 \ E2 . If H M , then H and so M contains an element of C. If H 6 M , then

96

Groups of prime power order

H \M is an element of C since jH \M j D p 3 and E1 \E2 < H \M < H so H \M is elementary abelian of order p 3 since E1 \ E2 Z.H / and exp.H \ M / D p. We see that the pair E1 ; E2 supports good lines. (ii) jH j p 5 . If M 2 1 , then jM \ H j p 4 so M \ H contains an abelian subgroup of order p 3 which is an element of the set C hence the pair E1 ; E2 supports good lines. Now the result follows from Proposition 103.3 and Lemma 103.2. Exercise 3. Suppose that a p-group G has order > p 3 , p > 2. Then the number of abelian subgroups of order p 3 in G is congruent to 1 .mod p/. Using the above approach, the following results are established in [KonJ]. Theorem 103.5. Let G be a p-group, p > 2, and let C be any one of the following classes of abelian subgroups of G: (i) elementary abelian subgroups of order p k for some ﬁxed k 5. (ii) abelian subgroups of order p k for some ﬁxed k 5. (iii) abelian subgroups of ﬁxed index p or p 2 . Suppose that the set C is nonempty. Then jCj 1 .mod p/ except in the index p 2 case (case (iii)) where the number can also be exactly 2. Corollary 103.6 ([Alp2]). If G is a p-group, p > 2, having an abelian subgroup of index p 3 , then there is a normal abelian subgroup of the same index in G. Proof. Indeed, let A be an abelian subgroup of index p 3 in G, and let A < M 2 1 ; then jM W Aj D p 2 . Then, by Theorem 103.5, the number of subgroups of index p 2 in M is not divisible by p since p > 2, and so one of these subgroups is normal in G. Corollary 103.7. Suppose that a p-group G possesses an elementary abelian subgroup Ek of order p k , p > 2, k < 6. Then the normal closure EkG of Ek in G contains a G-invariant elementary abelian subgroup of order p k . Indeed, the number of elementary abelian subgroups of order p k in EkG is not divisible by p so one of them is G-invariant. Exercise 4. Suppose that a 2-group G contains an abelian subgroup E of type .2; 2/. Then one of the following holds: (a) E G is dihedral or semidihedral. (b) E G contains a G-invariant abelian subgroup of type .2; 2/. (Hint. Use Theorem 1.17(b).)

104

Degrees of irreducible characters of p-groups associated with ﬁnite algebras

All results of this section are due to Isaacs [Isa11] In what follows q is a power of a prime p, F D GF.q/ (the ﬁeld of cardinality q, where q is a power of a prime), R a ﬁnite dimensional F-algebra (as always, with the identity element). Let J D J.R/ be the Jacobson radical of R (= the maximal nilpotent two-sided ideal of R; see [Lang, p. 658]); then 1 C J D ¹1 C x j x 2 J º is a p-subgroup of R , the group of units in R (Theorem 34.6). We refer to the group 1 C J arising in this way from an F-algebra R as an F-algebra group G. Let x 2 G; then x D 1 C a for some a 2 J . Observe that jGj D jJ j so G is a p-group. The upper unitriangular group UT.n; q/ is a Sylow p-subgroup of GL.n; q/ and it has the form 1 C J , where J is the space of strictly n n upper-triangular matrices over F; then J is the radical of the F-algebra F 1 C J . Thompson conjectured that the degrees of complex irreducible characters of the group UT.n; q/ are degrees of q. This conjecture is a corollary of the following general Isaacs’ theorem which is the main result of this section: Theorem 104.1 ([Isa11, Theorem A]). Let G be an F-algebra group. Then the degrees of all irreducible complex characters of G are powers of q. The theorem is a consequence of the following six lemmas. Lemma 104.2. Let N be a normal subgroup of a group G and suppose that G=N is abelian. Let 2 Irr.G/ and suppose that N D 2 Irr.N /. Then every irreducible constituent of G has the form for some 2 Lin.G=N /.1 Proof. Note that .1N /G D G=N , the regular character of G=N , and so the irreducible constituents of .1N /G are exactly the characters as in the statement of the lemma. Since (see [BZ, Theorem 5.1.2]) Irr. G / Irr.. N /G / and . N /G D .1N N /G D .1N /G ; the result follows. particular, all irreducible constituents of G are extensions of to G. The lemma is a particular case of a known result of Gallagher. 1 In

98

Groups of prime power order

Lemma 104.3. Let G; N and G=N be such as in Lemma 104:2. Let 2 Irr.N / be G-invariant and suppose that 2 Irr. G /. Then there exists a subgroup M of G with N M G and 2 Irr.M / such that N D and M D e, where e 2 D jG W M j. Furthermore, vanishes on the set G M . Proof. Take M to be the unique smallest subgroup of G containing N such that vanishes on G M and let 2 Irr. M / be arbitrary. To see that M D e for some integer e > 0, let 2 Irr. M / be arbitrary. Then ; 2 Irr. M / and so, by Lemma 104.2, D for some 2 Lin.M=N / since M=N is abelian. Let K D ker./; then N K M . Since G=N is abelian, extends to a character of G=N and 2 Irr../G / D Irr. G / (consider . /M ). Now, is an irreducible constituent of M D . /M , and Lemma 104.2 then tells us that D for some 2 Lin.G=M /. For the elements x 2 M K we see that .x/ D .x/ ¤ 1 D .x/ and thus .x/ D 0. By the deﬁnition of M , then K D M and hence D 1M . We conclude that D , i.e., M D e, as claimed. Now e 2 D he; ei D h M ; M i D jG W M j; where the third equality holds since vanishes on G M . It remains to show that N is irreducible. If this is false, it follows from the fact that M=N is abelian that is induced from a character of some subgroup T with N T < M [BZ, Theorem 7.2.2]. Next, T is normal in M , and thus vanishes on M T . It follows that vanishes on G T , a contradiction since T < M . For an arbitrary positive integer q we shall say that G is a q-power-degree group if all of its irreducible characters have degrees that are powers of q. Lemma 104.4. Fix a positive integer q and let G be a ﬁnite group. Let M G G and suppose that H is a collection of subgroups of G. Assume the following: (1) M is the intersection of any two distinct members of the set H . S S (2) G D H 2H H (i.e., G=M D H 2H H=M is a nontrivial partition). (3) jG W H j D q for each H 2 H . (4) G=M is abelian of order q 2 . (5) M and all members of the set H are q-power-degree groups. Then G is a q-power-degree group. 1 Proof. By (1)–(4), jH j D qq1 D q C 1 and if H and K are any distinct members of the set H , we have HK D G by the product formula. To show that G is a q-power-degree group, let 2 Irr.G/ be arbitrary. We choose 2 Irr. M / and let T D IG ./, the inertia subgroup of in G. Let H 2 H and ˛ 2 Irr. H /. Then 2

.1/ ˛.1/ H .1/ D jH W M j.1/ D q.1/:

104

Degrees of irreducible characters of p-groups associated with ﬁnite algebras

99

Since M and H are q-power-degree groups, however, and so .1/ j ˛.1/, it follows that ˛.1/ 2 ¹.1/; q.1/º: In the ﬁrst case, ˛M D and thus H T . Otherwise, we get H D ˛ and this forces H \ T D M. If M < T , then since T is partitioned by subgroups T \ H (H 2 H ), we must have T \ H > M for some H 2 H . Therefore, by the result of the previous paragraph, H T . If, in fact, H < T , then T \K > M for some other member K 2 H different from H , and in this case T HK D G. We see, therefore, that the only possibilities are T D M , T D G or T 2 H . Suppose that T < G. By the Clifford Correspondence Theorem 7.2.2, we know that D ˛ G for some ˛ 2 Irr.T /. Since either T D M or T 2 H , it follows that T is a q-power-degree group, and so ˛.1/ is a q-power. As .1/ D jG W T j˛.1/ and jG W T j 2 ¹q; q 2 º; the result follows in this case. We are left with the situation where T D G so is G-invariant. In this case, Lemma 104.3 yields the existence of a subgroup Q with M Q G and a character 2 Irr.Q/ such that Q D e and M D . Since .1/ D .1/ is a q-power and .1/ D e.1/, it sufﬁces to show that e is either 1 or q. To this end, we shall use the additional information (from Lemma 104.3) that e 2 D jG W Qj and that vanishes on the set G Q. Let H 2 H and consider ˛ 2 Irr. H /. Since is H -invariant and both and ˛ have q-power degrees, we have seen that ˛.1/ < jH W M j.1/ D q.1/ so we get ˛.1/ D .1/. We know that .1/ D e.1/ and it follows that H is a sum of exactly e irreducible constituents and so h H ; H i e. Write S D H \ Q and note that H vanishes on H S since it vanishes on H Q. Also, S is irreducible since M D 2 Irr.M / and M S. We thus have ejH W S j h H ; H ijH W Sj D h S ; S i D heS ; eS i D e 2 : It follows that e jH W S j and so jS=M j D q=jH W Sj q=e. Now Q=M is covered by subgroups .Q \ H /=M as H runs over H , and we have just seen that each of these subgroups has order at least qe . Since the pairwise intersections of these subgroups are trivial and jG=Qj D e 2 , we have jQ=M j D

q2 q2 jQjq 2 D D 2 jGj jG W Qj e

and so X q2 1 D jQ=M j 1 D .j.Q \ H /=M j 1/ 2 e H 2H q q 1 D .q C 1/ 1 : jH j e e

100

Groups of prime power order

If e < q, we can divide by qe 1 to get qe C 1 q C 1 and this yields e D 1. We have now shown that either e D 1 or e D q, and the proof is complete. Note that G=M in Lemma 104.4 has a prime exponent since this quotient group is equally partitioned (see 68). The following lemma resembles the known Nakayama’s lemma. Lemma 104.5. Let J D J.R/ and suppose that U J is a multiplicatively closed subspace. If J D U C J 2 , then U D J . Proof. We show by induction on m that J D U CJ m for any positive integer m. Since J is nilpotent, the result will follow. We can assume that m > 2 and J D U C J m1 . Then, since U J is a multiplicatively closed subspace, we get J 2 D J.U C J m1 / J U C J m ; J U D .U C J m1 /U U C J m and so J 2 U C J m . It follows that J D U C J 2 U C J m U C J m1 D J and we conclude that J D U C J m . Lemma 104.6. Let J D J.R/ and suppose that U J is an ideal of R. Set G D 1CJ and N D 1 C U . Then N is a normal algebra subgroup of G and G=N is naturally isomorphic to the algebra group corresponding to R=U . Proof. Write RN to denote the F-algebra R=U and, as usual, let rN 2 RN denote the image of r 2 R. Observe that GN D 1 C JN and so GN is an F-algebra group and the image of G under the group homomorphism g 7! g. N The result now follows from the observation that N D 1 C U is the kernel of this homomorphism. Before stating the next lemma, we introduce a bit more notation. If G D 1 C J is an F-algebra group, it is clear that the collection of the F-algebra subgroups of G is closed under intersection. It follows that for each element x 2 G there is a unique smallest F-algebra subgroup containing x, and we write A.x/ to denote this subgroup. In fact, it is easy to construct A.x/ explicitly. Write x D 1 C u and observe that the smallest multiplicatively closed F-subspace of J containing u is uFŒu. It follows that A.x/ D 1 C uFŒu. The key observation here is that A.x/ is abelian for all x 2 G. This follows since the multiplication in uFŒu is clearly commutative. Lemma 104.7. Let G be an F-algebra group for R that is not of the form A.x/ for any element x 2 G. (In particular, this holds if G is nonabelian.) Then there exists an F-algebra subgroup M G G and a family H of F-algebra subgroups of G satisfying properties (1)–(4) of Lemma 104:4.

104

Degrees of irreducible characters of p-groups associated with ﬁnite algebras 101

Proof. We have G D 1CJ , where J D J.R/ for some ﬁnite dimensional F-algebra R, and we can assume that R D F 1 C J . If J 2 has codimension 1 in J , we can write J D J 2 C Fu for some element u 2 J . It follows that J D J 2 C uFŒu and hence, by Lemma 104.6, we have J D uFŒu and A.u/ D A.1 C u/, a contradiction. We conclude that the codimension of J 2 in J is at least 2. Now let U be an F-subspace of J having codimension 2 and containing J 2 . Note that U and all F-subspaces of J containing U are ideals of R and that the multiplication in J.R=U / is trivial. Let M D 1 C U and let H be the collection of the subgroups 1 C V , where V runs over all hyperplanes of J that contain U . By Lemma 104.6, M is normal in G and G=M is isomorphic to 1 C J.R=U /, which is abelian. All of the assertions are now clear. We mention that Lemma 104.7 essentially continues to hold even if the ﬁeld F is inﬁnite. Obviously, we need to delete the assertion about the indices of the members H 2 H and M , but everything else goes through without change. Proof of Theorem 104.1. We are given an F-algebra group G, where jFj D q (q being a power of a prime), and we show by induction on jGj that G is a q-power-degree group. We can certainly assume that G is nonabelian and thus, by Lemma 104.7, we get a normal algebra subgroup M and a collection of algebra subgroups H satisfying certain properties. In particular, all of these subgroups are proper in G and thus each is a q-power-degree group by the inductive hypothesis. It follows from Lemma 104.4 that G is a q-power-degree group, as required. It follows from Theorem 104.1 that UT.n; q/ is a q-power-degree group. The important paper [Isa11] contains a number of related results.

105

On some special p-groups

This section was inspired by a letter of Isaacs in which he disproved one problem posed by the ﬁrst author on character degrees (see below). Most main results of this section are also due to Isaacs. 1o . Here we repeat some considerations from 46. Let s > 2 be a positive integer, F D GF.p s /, F0 be the prime subﬁeld of F and S D Gal.F=F0 / D hi; the set of all automorphisms of F. It is known that o./ D s and if .x/ D x for some x 2 F, then x 2 F0 . We will construct a special p-group P D Ap .s; / (see 46) and study its structure. For x; y 2 F, write .x; y/ to denote the matrix .aij / with ai i D 1; i D 1; 2; 3; a12 D x; a13 D y; a23 D .x/; a21 D a31 D a32 D 0: The set of matrices .x; y/ (x; y 2 F) is a group (denote it by P D Ap .s; /) with respect to matrix multiplication; obviously, jP j D jFj2 D p2s . The identity element of P is .0; 0/. It is easy to check that (i) We have (1) .x; y/.x1 ; y1 / D .x C x1 ; y C y1 C x.x1 //;

.x; y/1 D .x; y C x.x//

and the commutator (2)

Œ.x; y/; .x1 ; y1 / D .0; x.x1 / x1 .x//:

If .x1 ; y1 /.x; y/ D .x; y/.x1 ; y1 /, then by (2) we get x.x1 / D x1 .x/ or, if x1 ¤ 0, x11 x D .x11 x/. In that case, x D x1 x0 , where x0 2 F0 . Thus (ii) If x ¤ 0, then (3) CP ..x; y// D ¹.xx0 ; y 0 /; x0 2 F0 ; y 0 2 Fº;

jCP ..x; y//j D jF0 jjFj D p sC1 :

Obviously, Z.P / D ¹.0; y/; y 2 Fº, and so jZ.P /j D jFj D p s and Z.P / Š FC , the additive group of the ﬁeld F. In particular, Z.P / Š Eps . It follows that G is a group with small abelian subgroups (see 20).

105

On some special p-groups

103

By induction on n, we get (4)

1 .x; y/n D nx; ny C n.n 1/x.x/ : 2

Setting in the above formula n D p, we obtain (5)

.x; y/p D .0; p.p 1/=2 x.x//:

If p > 2, then .x; y/p D .0; 0/ is the identity element of P . In case p D 2, we get .x; y/2 D .0; x.x//. If p D 2 and .x; y/ is an involution, then we obtain x D 0 so .x; y/ 2 Z.P /. Thus (iii) exp.P / D p if p > 2 and exp.P / D 4 if p D 2; in the last case, we have Z.P / D 1 .P /. It follows from (3) and (iii) that if .x; y/ 2 P Z.P /, then CP ..x; y// D h.x; y/; Z.P /i is isomorphic to Cp2 Eps1 if p D 2 and is isomorphic to EpsC1 if p > 2. Therefore, sC1 2s if p D 2 and o..x; y// D 4, then .x; y/2 is contained in exactly 22.21/ D 2s1 cyclic subgroups of order 4. It follows from (iii) that if p D 2, then P has exactly 22s 2s D 2s1 .2s 1/ 2.2 1/ cyclic subgroups of order 22 and 2s 1 subgroups of order 2. Therefore (iv) If p D 2, then Ã1 .P / D Z.P /. In particular, ˆ.P / D Ã1 .P /P 0 D Z.P / (by (i), P 0 Z.P /). It follows from (ii) that jP 0 j p s1 (since jP 0 j is at least the size of a conjugacy class). We will prove that jP 0 j D p s . Let x1 ; : : : ; xs be a normal basis of the extension F=F0 consisting of primitive elements of F.1 Then, by (ii), the commutators ci D Œ.xi ; 0/; .1; 0/ D .0; xi .xi /;

i D 1; : : : ; s:

We claim that c1 ; : : : ; cs are linearly independent. Indeed, assume c1a1 : : : csas D .0; 0/ for some a1 ; : : : ; as 2 ¹0; : : : ; p 1º. Set ui D xi .xi /, i D 1; : : : ; s. Then a2 u2 C C as us D 0 or, what is the same, .a1 x1 C C as xs / D a1 x1 C C as us : 1 The elements x ; : : : ; x constitute a normal basis of the extension F=F if .x / 2 ¹x ; : : : ; x º s s 1 0 1 i for all i and 2 Gal.F=F0 /. A normal basis consisting of primitive elements does always exists; see [Ward, Kapitel 8].

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Groups of prime power order

This implies a1 x1 C C as xs 2 F0 and so a1 D D as D 0, proving our claim. Therefore jhc1 ; : : : ; cs ij D p s and so P 0 D Z.P /. Thus (v) P is special, P 0 D Z.P / D ˆ.P / is elementary abelian of order p s . By (3) and (v), we obtain (vi) The class number of P is k.P / D jZ.P /j C

jP j jZ.P /j D p s C p sC1 p: p s1

Next, jLin.P /j D jP W P 0 j D p s

and

jIrr1 .P /j D k.G/ jLin.P /j D p sC1 p;

where Irr1 .P / is the set of nonlinear characters of P . Let s be odd. If 2 Irr.P /, then .1/2 jP W Z.P /j D p s and so .1/2 p s1 . In this case, X .1/2 p s1 .p sC1 p/ D p 2s p s p 2s p s D jP j jP W P 0 j D 2Irr1 .P /

and so (vii) If s is odd, then the degree of any nonlinear irreducible character of P is equal 1 1 to p 2 .s1/ , i.e., cd.P / D ¹1; p 2 .s1/ º. It is easy to check that if s is even, then jcd.P /j > 2. According to 46, in this case, jcd.G/j D 3. An automorphism of a group G is said to be central or normal, if it centralizes Inn.G/, the group of inner automorphisms of G. In this case, for any x; g 2 G one has .x/1 .g/.x/ D .x 1 gx/ D x 1 .g/x and so

.g/..x/x 1 / D ..x/x 1 /.g/

hence, since im./ D G, we get .x/ D xzx , where zx 2 Z.G/ for all x 2 G, and we conclude that induces the identity automorphism on G=Z.G/. Conversely, if 2 Aut.G/ induces the identity automorphism of G=Z.G/, it is central. The set Autc .G/ of central automorphisms of a group G is a normal subgroup of Aut.G/.2 (viii) (L. Kazarin) Let p D 2 and let s > 1 be odd, P D A2 .s; /, where is a generator of Gal.F=F0 /, 2 Autc .P /. Then .a; b/ D .a; b C .a// for all .a; b/ 2 P , where D is a linear operator of the s-dimensional vector space GF.2s /=GF.2/. 2 Indeed,

if ˛ 2 Aut.G/, 2 Autc .G/ and g 2 G, then .˛ 1 ˛/.g/ D ˛ 1 ..˛.g// D ˛ 1 .˛.g/z/ D g˛.z/

for some z 2 Z.G/, and we conclude that ˛ 1 ˛ 2 Autc .G/.

105

On some special p-groups

105

2

We claim that jAutc .P /j D 2s . Indeed, .a; b/ D .a; 0/.0; b/. Let us show that ¹a.a/ j a 2 GF.2s /º D GF.2s /: In fact, if a.a/ D a1 .a1 / (a; a1 2 GF.2s //, then .a1 a1 / D .a1 a1 /1 so that a D a1 in view of o./ D s is odd. It follows that for each b 2 GF.2s / there exists a unique a 2 GF.2s / such that b D a.a/, and so .0; b/ D .a; 0/2 . Therefore is deﬁned uniquely by its action on elements of the form .a; 0/. Let .a; 0/ D .a; .a//, where W GF.2s / ! GF.2s /. Since .a; 0/2 D .0; a.a//, then for b D a.a/ we have .0; b/ D ..a; 0/2 / D ..a; 0/ /2 D .a; .a//2 D .0; a.a// D .0; b/; i.e., ﬁxes all elements of Z.P /. In this case, .a; b/ D ..a; 0/.0; b// D .a; .a//.0; b/ D .a; b C .a//: If .a; b/ D .0; 0/ and D is linear, then a D 0 and b D .a/ D .0/ D 0, and so is an injection so a surjection. We claim that is linear. We have .a; 0/ .c; 0/ D .a; .a/.c; .c// D .a C c; .a/ C .c/ C a.c//: On the other hand, .a; 0/.c; 0/ D .a C c; a.c// and ..a; 0/.c; 0// D .a C c; .a C c/ C a.c//: Since is an automorphism, we get .a C c/ D .a/ C .c/. It is easy to check that the mapping W .a; b/ 7! .a; b C .a// is an automorphism of P for every linear map W GF.2s /=GF.2/ ! GF.2s /=GF.2/: Thus, .a; b/ D .a; b C .a// D .a; b/.0; .a//; and 2 Autc .P / since .0; .a// 2 Z.P /. It is easy to see that for every there is a unique central automorphism of P corresponding to . Since the number of ’s is equal to the number of s s matrices over GF.2/ (every such matrix contains s 2 entries equal 0 or 1), the claim (viii) follows. 2o . Let P , F, and p s be as in the previous subsection. Set bD

ps 1 D 1 C p C C p s1 : p1

The multiplicative cyclic group F of order p s 1 has a unique subgroup C of order b since b j p s 1. We deﬁne an action of C on P as follows: .x; y/c D .xc; yc pC1 / .c 2 C /:

106

Groups of prime power order

Note that .c/ D c p for all c 2 F ( is a generator of Aut.F/). We have .x; y/c .u; v/c D .xc; yc 1Cp /.uc; vc 1Cp / D ..x C u/c; .y C v C x.u//c 1Cp / D ..x; y/.u; v//c ; and we conclude that .x; y/ 7! .xc; yc pC1 / is an action. If .x; y/c D .x; y/, then c D 1, i.e., every element of C induces a ﬁxed-point-free automorphism of P . Therefore CP D .C; P / is a Frobenius group with kernel P and cyclic complement C of order b. We will construct a p-solvable group G such that b.Q/ does not divide .1/ for all 2 Irr.G/, where Q 2 Sylp .G/ and b.Q/ D max¹.1/; 2 Irr.Q/º. This example is taken from Isaacs’ unpublished note. We retain the above notation. In what follows we will assume that s D p > 2. Then o./ D p. We deﬁne an action of S D hi on P D Ap .p; / as follows: .x; y/ D ..x/; .y//: Set G D .C S/ P . If .x; y/ D .x; y/, then x; y 2 F0 , and so jCP ./j D jF0 j2 D p 2 ;

jCZ.P / ./j D p:

Thus ﬁxes exactly p classes of Z.P / and p 2 classes of P and p classes of P =P 0 D P =Z.P /. By Brauer’s permutation Lemma 10.3.1(c), ﬁxes exactly p linear characters of both groups Z.P / and P . If ¤ 1Z.P / is one of them in Lin.Z.P // and H D ker./, then ﬁxes all linear characters of Z.P / with kernel H . It follows that ﬁxes exactly one maximal subgroup of Z.P /. We choose so that .c/ D c p for all c 2 F. If c 2 C and .c/ D c, then we have c D 1 since .jC j; p/ D 1. Therefore S C D .S; C / is a Frobenius group with kernel C and complement S D h i. The group .S; C / has exactly jC j D b subgroups of order jSj D s D p and all of them are conjugate in SC (Sylow). If T is a subgroup of order p in .S; C / D .T; C /, then T ﬁxes exactly one maximal subgroup HT of Z.P /. Since b is equal to the number of maximal subgroups of Z.P / and to the number of subgroups of order p in .S; C /, there is a one-to-one correspondence T $ HT , where T and HT are as above. By the result of the previous paragraph, S ﬁxes exactly p 2 p D p.p 1/ nonlinear irreducible characters of P (these characters are irreducible constituents of ˛ P , where ˛ runs over all p 1 nonprincipal linear characters of Z.P /=HS and HS is the unique maximal subgroup of Z.P / ﬁxed by S). It follows that ˛ SP is the sum of p 2 irreducible characters of SP of degree p .p1/=2 . In particular, the inertia subgroup IG .˛/ D SP , and so all irreducible constituents of ˛ G have degree bp .p1/=2 . If T is a subgroup of order p in SC , then it is conjugate to S in SC . Therefore, if HT is the unique maximal subgroup of Z.P / ﬁxed by T and ˇ is a nonprincipal linear character of Z.P /=HT , then IG .ˇ/ D TP . If is a nonlinear irreducible character of G=Z.P /, then .1/ divides jG W P j D bp. Therefore, the p-part of the degree of every irreducible character of G does not exceed p .p1/=2 .

105

On some special p-groups

107

The Sylow p-subgroup SP of G has an irreducible character of degree p .pC1/=2 , however. This is because S deﬁnitely does not stabilize all nonlinear irreducible characters of P . (It stabilizes only those that have a particular ﬁxed maximal subgroup of Z.P / as kernel.) This proves that b.SP / does not divide .1/ for all 2 Irr.G/. The same question is open for p D 2. Isaacs [Isa4] proved that for every set S of powers of a prime p containing p 0 D 1 there exists a p-group P such that cd.P / D S (see Theorem 105.2 below). Next, he showed that for every p-group P there exists a group G with cyclic Sylow p-subgroup C such that cd.G/ D cd.P /. We will prove this result. More exactly, we will show that for every set S of powers of a prime p containing 1 D p 0 and with maximum member p a there exists a group G having a cyclic Sylow p-subgroup C of order p a such that cd.G/ D S. For each member p e 2 S, let Ve be an elementary abelian q-group on which a cyclic group of order p e acts faithfully and irreducibly. Let C act on Ve with kernel of order p ae (in that case, C acts on Ve irreducibly), and let W be the direct product of the groups Ve for all p e 2 S so that C acts on W . Let G D C W be the semidirect product of W with C . We claim that cd.G/ D S. Obviously, it is enough to prove that p a 2 cd.G/ (by Ito’s theorem (see Introduction, Theorem 17), the set cd.D/ contains only powers of p not exceeding jC j D p a ). Let W D Ve1 Ves ;

where S D ¹1; p e1 ; : : : ; p es D p a º:

Let xi 2 Ve#i , i D 1; : : : ; s, x D x1 : : : xs . Then the normal closure of hxi in G is equal to W , the socle of G. Therefore, by [BZ, Corollary 9.6] (Gasch¨utz’ theorem), Irr.G/ has a faithful character . We claim that .1/ D p es D p a . Note that C Va is a Frobenius group. As Va 6 ker. /, it follows that Va has a nonlinear constituent, and so .1/ p a in view of cd.C Va / D ¹1; p a º. Since .1/ divides p a by Ito’s theorem (see Introduction, Theorem 17), we get .1/ D p a , as desired. Of course, if cd.G/ D S, where S is as in the previous paragraph, then G has an abelian normal p-complement. Problem 1. Let P be a Sylow p-subgroup of a p-solvable group G. Study cd.P / in the case where p 2 does not divide .1/ for all 2 Irr.G/. Problem 2. Let P D Ap .s; /, where is a generator of Aut.GF.p s //, s > 2 is even. Find cd.P /. Exercise. Let G be a group of class two of maximal order generated by n > 1 elements of order p (for the matrix construction of this group, see [Macd3]). Prove that (a) exp.G/ D p if p > 2 and exp.G/ D 4 if p D 2, jGj D p n.nC1/=2 . (b) G is special and jZ.G/j D p n.n1/=2 . (c) If x 2 G Z.G/, then jG W CG .x/j D p n1 , i.e., G is a p-group with small abelian subgroups; see 20. (d) k.G/ D p n.n1/=2 C p n.nC1/=2.n1/ p n.n1/=2.n1/ .

108

Groups of prime power order

(e) For any positive integer n such that 2r n, Irr.G/ has a character of degree p r . P (f) Find T.G/ D 2Irr.G/ .1/. (g) (Mann) If p > 2, then we have Aut.G/=Autc .G/ Š Aut.G=G 0 / Š GL.n; p/, jAutc .G/j D jG 0 jn and Autc .G/ is elementary abelian. (h) (Mann) If p D 2, then Aut.G/=Autc .G/ Š Sn . (i) If n > 5, then G 0 has non-commutators. (In fact, this is true in the case n > 3. Moreover, if n D 4, then G has an epimorphic image H of order p 8 such that not all elements of H 0 are commutators [Macd3]. Numerous examples of groups, in which not all elements of G 0 are commutators, have been published; see, for example, [Fit], [Isa6] and [Macd3].) 3o . Let X be a set of powers of p such that 1 2 X. We will prove that there exists a p-group G such that cd.G/ D X . As we know (see the exercise, this is the case if X D ¹1; p; : : : ; p e º. Isaacs [Isa4] has shown that such a G exists for any X; moreover, G may be taken to be of class 2. Let U be an abelian group which acts on an abelian group A. Set b A D Lin.A/, the group of linear characters of A. As we know (see [BZ, 1.9]), there exists a natural isomorphism of b A onto A. If ˛ 2 b A, u 2 U and a 2 A, then, setting ˛ u .a/ D ˛.uau1 /; we can deﬁne an action of U on b A. The following lemma is of independent interest. Lemma 105.1 ([Isa4]). Let U be an abelian group which acts on an abelian group A and S be the set of the sizes of the orbits in this action. Write G D U b A, where the b semidirect product is constructed with respect to the action of U on A, induced by the given action of U on A. Then cd.G/ D S. Furthermore, if ŒA; U; U D ¹1º, then G has the nilpotence class 2. Proof. Let 2 Lin.b A/ and let T be the stabilizer of in U . Since the cyclic group b A= ker./ is centralized by an abelian subgroup T , it follows that b AT = ker./ is abelAT / is linear. Obviously, b ian and thus every 2 Irr.b AT D IG ./, the inertia subgroup of in G, and thus every 2 Irr.G / has degree jG W b AT j D jU W T j. Therefore, cd.G/ D ¹jU W T j j T is a stabilizer in U of some 2 Lin.b A/º: However, there is a natural correspondence between Lin.b A/ and A, and this deﬁnes a permutation isomorphism of the actions of U on A and Lin.b A/, respectively, and we conclude that cd.G/ D S, as required. Now suppose that ŒA; U; U D ¹1º. If ˛ 2 b A, we have for u 2 U and a 2 A that Œ˛; u.a/ D .˛ 1 ˛ u /.a/ D ˛.a1 /˛.uau1 / D ˛.Œa; u1 /:

105

On some special p-groups

109

Therefore, if v 2 U , we get by the above displayed formula, Œ˛; u; v.a/ D Œ˛; u.Œa; v 1 / D ˛.Œa; v 1 ; u1 / D ˛.1/ D 1 and so Œb A; U; U D ¹1º. Since b A is abelian and Œb A; U b A, this yields CG .Œb A; U / b b b b b U A D G, i.e., ŒA; U Z.G/. As G=ŒA; U D U A=ŒA; U is abelian, it follows that cl.G/ 2, completing the proof. Now we are ready to prove the main result of this section. Theorem 105.2 ([Isa4]). Let p be a prime and let 0 D e0 < e1 < < em be integers. Then there exists a p-group that is generated by elements of order p and has nilpotence class 2 such that cd.G/ D ¹p ei j 1 i mº. Proof. Suppose that U is an elementary abelian p-group of rank em with generators u1 ; : : : ; uem . Let A be an elementary abelian p-group with basis a1 ; z1;1 ; : : : ; z1;e1 ; : : : ; am ; zm;1 ; : : : ; zm;em (m blocks of sizes e1 ; : : : ; em , respectively) and deﬁne an action of U on A as follows. Put .zi; /u D zi; for all i; ; and .ai /u D ai if > ei and .ai /u D ai zi; if

ei . Since the automorphisms of A deﬁned this way all have order p and commute pairwise, this does deﬁne an action of U on A. Let us compute the sizes of the orbits of this action. Write Z D hzi; j ei i A and let a 2 A. Obviously, Z Z.G/, where G D U b A is deﬁned as in Lemma 105.1. Assume that a 2 A Z. Then there exists a unique subscript i such that we can write a D bcz, where b 2 haj j j < ii;

1 ¤ c 2 hai i;

z 2 Z:

Suppose that u 2 U . If u involves the generator u with ei , the exponents of zi; in a and in au will not be equal, and u does not centralize a. Thus we obtain that CU .a/ hu j > ei i. Since the reverse inclusion is obvious, we ﬁnally obtain CG .a/ D hu j > ei i. We now have jU W CU .a/j D p ei and we see that the orbit sizes of the action of U on A are precisely the numbers p ei for 0 i m. Since ŒA; U Z Z.G/, we have ŒA; U; U D ¹1º and the result follows by Lemma 105.1.

106

On maximal subgroups of two-generator 2-groups

To facilitate the proof of Theorem 106.3, we ﬁrst prove Lemmas 106.1 and 106.2. The lemma below follows from Propositions 71.3, 71.4 and 71.5, but we prefer to give an independent proof. Lemma 106.1. If a nonmetacyclic A2 -group G is two-generator, then p > 2.1 Proof (Berkovich). Assume that this is false, i.e., there exists a nonmetacyclic twogenerator 2-group G which is an A2 -group. Since G is not metacyclic, it has a maximal subgroup, say A, such that d.A/ > 2 (Corollary 36.6); then d.A/ D 3 (Schreier; see Appendix 25). From Lemma 65.1 we deduce that A is not an A1 -group so it is abelian. Let 1 D ¹M; N; Aº. If one of the subgroups M; N , say M , is abelian, then A \ M D Z.G/ so Z.G/ D ˆ.G/ since G=Z.G/ Š E4 Š G=ˆ.G/, and we conclude that G is an A1 -group, contrary to the hypothesis. Thus M and N are A1 -subgroups. We have Z.M / D ˆ.M / < ˆ.G/ < A so CG .Z.M // AM D G, and we conclude that Z.M / D Z.G/ since Z.G/ < ˆ.G/ (otherwise, G is an A1 -group). Similarly, Z.N / D Z.G/. Since the two-generator group G=Z.G/ of order 8 has two noncyclic subgroups M=Z.G/ and N=Z.G/ of order 4, it is generated by involutions. It follows that G=Z.G/ Š D8 . Suppose that R G G satisﬁes G=R Š D8 . We claim that then R D Z.G/. Indeed, it follows from d.G/ D 2 that R < ˆ.G/. Since G=R has exactly one cyclic subgroup of order 4, one may assume, without loss of generality, that M=R Š E4 holds so R D ˆ.M / D Z.M /. In that case, CG .R/ AM D G so that R D Z.G/ (compare indices). Thus there is in G only one normal subgroup, namely Z.G/, such that the corresponding quotient group is isomorphic to D8 . N D Set GN D G=Ã1 .A/. Then GN is nonabelian of order 16 since AN Š E8 and d.G/ N D 4, GN is not of maximal class. Then, by Proposition 10.17, d.G/ D 2. As exp.G/ GN has no nonabelian subgroup of order 8 (otherwise, d.G/ > 2) so it is an A1 -group. N of order 2 such that N Š E4 . In this case, there exists RN < Z.G/ It follows that Z.G/ 0 N RN Š D8 since a nonabelian group G= N RN has a subgroup A= N RN Š E4 . RN ¤ GN ; then G= By the ﬁrst paragraph, R D Z.G/. Let B=Z.G/ < G=Z.G/ be cyclic of order 4; then a 2-group G which is a two-generator A2 -group is metacyclic; this will be the case if and only if jG 0 j D p 2 by Corollary 65.3. 1 Thus

106

On maximal subgroups of two-generator 2-groups

111

B is abelian. As B ¤ A (indeed, A=Z.G/ Š E4 ), we get a contradiction since, by the above, there is in G only one abelian maximal subgroup. Lemma 106.2 ([BJ3, Lemma 2.1]). Let G be a nonmetacyclic p-group and let R < G 0 be G-invariant of order p. If G=R is minimal nonabelian and all nonabelian maximal subgroups of G are two-generator, then G is an A2 -group and p > 2. Proof. By hypothesis, jG 0 j > p so G is not an A1 -group (Lemma 65.1). We also have d.G/ D d.G=R/ D 2 since R < G 0 ˆ.G/. Let M 2 1 be nonabelian. Then d.M / D 2 by hypothesis, and M=R, as a maximal subgroup of the A1 -group G=R, is abelian which implies in view of jRj D p that M 0 D R. Hence M is an A1 -group (Lemma 65.2(a)) and so G is an A2 -group. By Lemma 106.1, p > 2. Note that if R < G 0 is G-invariant of order p and G=R is a metacyclic A1 -group, then G is a metacyclic A2 -group. Indeed, G is metacyclic by Theorem 36.1. Since jG 0 j D p 2 , G is an A2 -group by Corollary 65.3. Now we are ready to prove our main result. Theorem 106.3 ([BJ3, Theorem 2.2]). If a 2-group G and all its nonabelian maximal subgroups are two-generator, then G is either metacyclic or minimal nonabelian. Proof. Assume that G is a counterexample of minimal order. Then G is nonabelian (otherwise, G is metacyclic since, by hypothesis, d.G/ 2). Assume that jG 0 j D 2. Then, since we have d.G=G 0 / D 2, G is an A1 -group by Lemma 65.2(a), contrary to the assumption. It follows that jG 0 j > 2. Let R < G 0 be G-invariant of order 2; then GN D G=R is nonabelian. Since GN satN < jGj, it is either metacyclic or minimal nonabelian by isﬁes the hypothesis and jGj induction. If GN is metacyclic, then G is also metacyclic by Theorem 36.1, contrary to the assumption. Thus GN D G=R is a nonmetacyclic A1 -group. By hypothesis, G and all its nonabelian maximal subgroups are two-generator. Then, by Lemma 106.2, G is an A2 -group and p > 2, a ﬁnal contradiction. Let a nonabelian 2-group be neither metacyclic nor minimal nonabelian. If all nonabelian maximal subgroups of G are two-generator, then d.G/ D 3 (Theorem 106.3). If G is a 2-group with d.G/ > 2 and all maximal subgroups of G are two-generator, then the quotient group G=K4 .G/, where Kn .G/ is the n-th member of the lower central series of G, is described up to isomorphism in 71 (the order of such quotient group does not exceed 29 ). However, see 113 which shows that there exist such groups of arbitrary class. Remark. Let G be a two-generator p-group, p > 2, possessing an abelian maximal subgroup, say A, and suppose that all nonabelian maximal subgroups of G are two-generator. Suppose that jG 0 j > p (if jG 0 j D p, then G is minimal nonabelian by Lem-

112

Groups of prime power order

ma 65.2(a)). Let R < G 0 be a G-invariant subgroup of order p and let G=R be of maximal class. We claim that then G is also of maximal class. Indeed, since R < ˆ.G/, we get p 2 D j.G=R/ W .G 0 =R/j D jG W G 0 j D pjZ.G/j by Lemma 1.1, i.e., Z.G/ is of order p so coincides with R. It follows that G is of maximal class, as was to be shown. Conversely, if a p-group G of maximal class and order > p 3 has an abelian maximal subgroup, say A, then all nonabelian subgroups of G are of maximal class so two-generator. Indeed, if H 2 1 is nonabelian, then H is of maximal class by Fitting’s lemma. Next, if F < G is nonabelian, then F H 2 1 , and since H \ A is maximal abelian in H , we conclude that F is of maximal class by induction in H , completing the proof. If all nonabelian maximal subgroups of a p-group G are twogenerator and jG=Ã1 .G/j p 4 , then G has an abelian maximal subgroup. Indeed, G=Ã1 .G/ has a maximal subgroup, say A=Ã1 .G/, which is not generated by two elements (Theorem 5.8(b)); then d.A/ > 2 so A is abelian by hypothesis.

107

Ranks of maximal subgroups of nonmetacyclic two-generator 2-groups

Let G be a nonmetacyclic two-generator 2-group. By Theorem 44.5, there is in G a maximal subgroup C such that d.C / > 2 (in fact, by Schreier’s theorem on the number of generators of subgroups, d.C / D 3; see Appendix 25). Let A; B; C be all maximal subgroups of G. It is important to know all possibilities for d.A/ and d.B/ and also on the structure of A and B. As Theorem 107.1 asserts, there are exactly two possibilities: d.A/ D d.B/ D 2 and d.A/ D d.B/ D 3. Theorem 107.1 ([BJ3, Theorem 5.1]). Suppose that G is a nonmetacyclic two-generator 2-group. Then the number of two-generator maximal subgroups of G is even. Proof. By hypothesis, G is nonabelian. If A 2 1 , then exp.G=Ã1 .A// 4 so that Ã2 .G/ Ã1 .A/ D ˆ.A/ and hence d.A=Ã2 .G// D d.A/. Therefore, without loss of generality, one may assume that Ã2 .G/ D ¹1º; then exp.G/ D 4. It follows that G=G 0 is abelian either of type .4; 2/ or .4; 4/. Since G is not metacyclic, at least one maximal subgroup of G is not two-generator by Corollary 36.6. Let 1 D ¹A; B; C º and d.C / D 3 (see the paragraph preceding the theorem). Assume that the statement of the theorem is false. Then one may assume d.A/ D 2 and d.B/ D 3.D d.C //. Let R be a G-invariant subgroup of index 2 in G 0 . Then G=R is minimal nonabelian and nonmetacyclic (Theorem 36.1). By what has just been said, jG=Rj 25 . If jG=Rj D 25 , then all maximal subgroups of G=R are not generated by two elements (see Lemma 65.1), contrary to the assumption. Thus jG=Rj D 24 . In this case, G=G 0 is abelian of type .4; 2/ and j1 .G=R/j D 23 (Lemma 65.1). All maximal subgroups of G=R which are different from C =R D 1 .G=R/ are abelian of type .4; 2/. Hence B=R is abelian of type .4; 2/ and so it is two-generator. By assumption, we have on the one hand d.B/ D 3 implying that R ¤ ˆ.B/ (note that jRj D jˆ.B/j), and on the other hand that ˆ.B/ is a G-invariant subgroup of index 23 in B. Write GN D G=ˆ.B/; then GN (of order 24 ) is not of maximal class since it contains a subgroup BN Š E8 . It follows that jGN W GN 0 j D 8 D jG=G 0 j so ˆ.B/ < G 0 (this also follows from Taussky’s theorem), and we conclude that jG 0 W ˆ.B/j D 2. Since G 0 has only one G-invariant subgroup of index 2 (Lemma 36.5), we conclude that ˆ.B/ D R. This is a contradiction since, by the above, R ¤ ˆ.B/. Thus d.B/ D 2 so that the number of two-generator maximal subgroups of G is even, as was to be

114

Groups of prime power order

shown. (There exists an A2 -group G of order 26 and d.G/ D 3 all of whose seven maximal subgroups are A1 -groups so two-generator (see 71).) Thus, if the number of two-generator maximal subgroups of a 2-group G is odd, then either G is metacyclic or d.G/ D 3. Exercise. Let G be a p-group such that jG=G 0 j D p 2 . Then G has no maximal subgroup of rank p C 1. If A 2 1 with d.A/ D p, then G=ˆ.A/ Š †p2 , a Sylow p-subgroup of the symmetric group of degree p 2 . Solution. Let A 2 1 . Then G=ˆ.A/ is of maximal class (to prove, use Lemma 1.1 and induction) so that d.A/ < p C 1 (Theorem 9.5). Now let A 2 1 with d.A/ D p. Then G=ˆ.A/ Š †p2 (Exercise 9.13). Theorem 107.2 ([BJ3, Theorem 5.2]). Suppose that G is a nonmetacyclic two-generator 2-group. Then one of the following holds: (a) Every maximal subgroup of G is not generated by two elements if and only if G=G 0 has no cyclic subgroup of index 2. (b) Exactly one maximal subgroup of G is not generated by two elements if and only if G=G 0 has a cyclic subgroup of index 2. Proof. (a) Suppose that all maximal subgroups of a given nonmetacyclic two-generator 2-group G are not two-generator (then their ranks are equal to 3 by Appendix 25). We claim that, in this case, G=G 0 has no cyclic subgroup of index 2. Assume that this is false. As in the proof of Theorem 107.1, one may assume that Ã2 .G/ D ¹1º hence exp.G/ D 4 and G is nonabelian. Let R be a G-invariant subgroup of index 2 in G 0 . Then G=R is a nonmetacyclic A1 -group (Theorem 36.1 and Lemma 65.2(a)). As in the proof of Theorem 107.1, 23 < jG=Rj 25 and R is the unique G-invariant subgroup of index 2 in G 0 . Assume that jG=Rj D 24 ; then G=G 0 is abelian of type .4; 2/ and the maximal subgroups of G=R are abelian of types .4; 2/, .4; 2/ and .2; 2; 2/. Let B=R < G=R be abelian of type .4; 2/. By hypothesis, we have B=ˆ.B/ Š E8 . Since G=ˆ.B/ is not of maximal class (indeed, it contains a subgroup Š E8 , namely B=ˆ.B/), it follows that jG=ˆ.B/ W .G=ˆ.B//0 j D 8 (Taussky’s theorem) and so G 0 \ ˆ.B/ D R which yields ˆ.B/ > R, and this is a contradiction (recall that d.B=R/ D 2). As in the proof of Theorem 107.1, we get R D ˆ.B/, and this is a contradiction. Thus jG=Rj D 25 so G=G 0 is abelian of type .4; 4/. It follows that, in the case under consideration, G=G 0 has no cyclic subgroup of index 2. Now suppose that G=G 0 has no cyclic subgroup of index 2. Let R be (the unique) G-invariant subgroup of index 2 in G 0 . Write GN D G=R. Then GN is a nonmetacyclic N Š E8 (Lemma 65.1). Since A1 -group by Theorem 36.1 and Lemma 65.2(a) so 1 .G/ G=G 0 has no cyclic subgroup of index 2, it is abelian of type .2m ; 2n /, m; n > 1, m1 ; p n1 / hence noncyclic. N N therefore it follows that G= 1 .G/ is abelian of type .p N N N N 1 .G/ N is maximal Therefore 1 .G/ ˆ.G/ since d.G/ D d.G/ D 2. Thus, if A=

107

Ranks of maximal subgroups of nonmetacyclic two-generator 2-groups

115

N N N in G= 1 .G/, then d.A=R/ D 3 since A=R is abelian and contains 1 .G/ Š E8 , and we conclude that d.A/ D 3. Since A 2 1 is arbitrary, the proof of (a) is complete. Statement (b) follows from (a) and Theorem 107.1. Suppose that a nonmetacyclic two-generator p-group G contains a metacyclic abelian maximal subgroup A. If G has a normal elementary abelian subgroup E of order p 3 , then A has a cyclic subgroup of index p. Indeed, G D AE and G=.A \ E/ Š .A=.A \ E// .E=.A \ E//. Since d.G/ D 2, we conclude that A=.A \ E/ is cyclic, and our claim follows in view of A \ E D 1 .A/. We have G 0 A \ E. In the following two theorems we shall consider the nonmetacyclic two-generator 2-groups G containing exactly one non-two-generator maximal subgroup. Theorem 107.3 ([BJ3, Theorem 5.3]). Suppose that G is a nonmetacyclic two-generator 2-group which possesses exactly one maximal subgroup K which is not two-generator. If 1 D ¹M; N; Kº is the set of maximal subgroups of G, then jG 0 W K 0 j D 2, G=K 0 is nonmetacyclic minimal nonabelian, both M and N are either metacyclic or nonmetacyclic and G=G 0 has a cyclic subgroup of index 2. Proof. By Theorem 107.2, G=G 0 has a cyclic subgroup of index 2. We know that d.M / D 2 D d.N / and d.K/ D 3. If K 0 D ¹1º, then Theorem 106.3 implies that G is minimal nonabelian in which case both M and N are metacyclic since, in the case under consideration, 1 .M / Š 1 .N / Š E4 and M; N are abelian. Therefore one may assume that G is not minimal nonabelian and hence jG 0 j > 2 (Lemma 65.2(a)); then K 0 > ¹1º. We have K=ˆ.K/ Š E8 so that d.G=ˆ.K// D 2 implies that G=ˆ.K/ is nonmetacyclic minimal nonabelian of order 24 (this follows from the previous paragraph; this also follows from Proposition 10.17). Hence we get K 0 G 0 \ ˆ.K/ and jG 0 W .G 0 \ ˆ.K//j D 2. Assume that K 0 < G 0 \ ˆ.K/. In this N where G, N MN , NN are two-generator groups and KN is abelian case, consider G=K 0 D G, N with d.K/ D 3. By Theorem 106.3, GN must be minimal nonabelian. This is a contraN 0 is of order > 2. We have proved that K 0 D G 0 \ ˆ.K/, and diction since GN 0 D .G/ 0 so, by the above, jG W K 0 j D 2 and therefore GN D G=K 0 is nonmetacyclic minimal N D 3). nonabelian (recall that d.K/ Now assume that M is nonmetacyclic but N is metacyclic. Then M 0 ¤ ¹1º since d.M / D 2, and let R < M 0 be a G-invariant subgroup with jM 0 W Rj D 2. In this case, we consider GN D G=R so that MN is nonmetacyclic minimal nonabelian (Theorem 36.1 and Lemma 65.2(a)). Note that NN is metacyclic and we set EN D 1 .MN / Š E8 (LemN EN is noncyclic, then EN ˆ.G/ N since d.G/ N D 2, and so EN Š E8 ma 65.1). If G= N N EN is cyclic of oris contained in metacyclic subgroup N , a contradiction. Hence G= N N N N der 4 since M > E in view of d.M / D 2 < 3 D d.E/. Let aN 2 GN MN be such N EN and hNzi D hai that hai N is cyclic of order 8, hai N covers G= N \ EN Š C2 so that, in N since ha; N D GN (recall that GN is not split over E). N We have particular, hNz i 2 Z.G/ N Ei 2 0 2 N N N N N M D haN iE and M D ŒaN ; E D hui, N where uN 2 E hzN i since uN is not a square N (recall that MN is minimal nonabelian). in MN by Lemma 65.1, and so hzN ; ui N Z.G/

116

Groups of prime power order

N But then Let vN 2 EN hNz ; ui N so that vN aN D vN N with N 2 hz; N ui N hence N 2 Z.G/. 2 N aN D .vN / N N aN D vN N 2 D vN vN aN D .vN /

N D ¹1º, a contradiction since MN D haN 2 ; Ei N is nonabelian. The proof is and so ŒaN 2 ; E complete. It follows from Theorem 107.3 and Corollary 36.6 that the number of metacyclic maximal subgroups in any nonmetacyclic two-generator 2-group is even. Let us show that provided X is a nonmetacyclic two-generator 2-group of minimal possible order, then it is an A1 -group of order 16 which is uniquely determined. Since X=Ã2 .X/ is nonmetacyclic two-generator (Corollary 44.9), we get Ã2 .X/ D ¹1º so that exp.X/ D 4. By Taussky’s theorem and Theorem 36.1, X=X 0 is abelian of type .4; 2/. If R < X 0 is X-invariant of index 2, then X=R is nonmetacyclic and minimal nonabelian (Theorem 36.1) so that R D ¹1º. It follows that X is nonmetacyclic minimal nonabelian of order 16 so it is determined uniquely (Lemma 65.1). Theorem 107.4 ([BJ3, Theorem 5.4]). Let G be a smallest group from Theorem 107.3, where maximal subgroups M and N of G are both nonmetacyclic. Then G is uniquely determined as follows: G D ha; b j a2 D b 4 D 1; Œa; b D c; Œb; c D m; c 2 D m2 D Œm; a D Œm; b D Œa; c D 1i; where jGj D 25 ;

G 0 D hc; mi Š E4 ;

K3 .G/ D Z.G/ D hmi

and so cl.G/ D 3. The three maximal subgroups of G are M D hG 0 ; bi;

N D hG 0 ; abi

(which are both nonmetacyclic minimal nonabelian of order 24 ) and K D hG 0 ; a; b 2 i Š C2 D8 :1 Our group G exists as a subgroup of the alternating group A8 . Proof. By the paragraph preceding the theorem, a smallest possibility for two-generator nonmetacyclic subgroups M and N is the nonmetacyclic minimal nonabelian group H16 D hx; y j x 2 D y 4 D 1; Œx; y D z; z 2 D Œz; x D Œz; y D 1i of order 24 . Therefore we try to construct our two-generator group G under the assumption that jGj D 25 and the set of maximal subgroups ¹M; N; Kº is such that 1 Thus

G is an A3 -group.

107

Ranks of maximal subgroups of nonmetacyclic two-generator 2-groups

117

M Š N Š H16 and K is a nonabelian group of order 24 with d.K/ D 3 (indeed, K must be nonabelian by Theorem 106.3). Since G is neither of maximal class (and so jG 0 j < 23 by Taussky’s theorem) nor minimal nonabelian (and so jG 0 j > 2 by Lemma 65.2(a)), we have jG 0 j D 4. As d.G/ D 2, we conclude that G=G 0 is abelian of type .4; 2/ so G D AB, where A \ B D G 0 , A=G 0 Š C2 and B=G 0 Š C4 . Let a 2 A G 0 and b 2 B G 0 be such that A D hG 0 ; ai and B D hG 0 ; bi. We have ˆ.G/ D G 0 hb 2 i so that G D ha; b; G 0 i D ha; bi, and we set c D Œa; b ¤ 1. By Proposition 10.17, since K is neither of maximal class nor minimal nonabelian, we get K D U Z.K/, where U is nonabelian of order 8. Note that K=ˆ.K/ Š E8 . We obtain G DM [N [K

and

exp.M / D exp.N / D exp.K/ D 4

so

exp.G/ D 4

since G D M [N [K. It follows from jˆ.G/j D 8 that ˆ.G/ is abelian. Assume that ˆ.G/ is abelian of type .4; 2/; then every set of generators of ˆ.G/ contains an element of order 4 so, since ˆ.G/ D Ã1 .G/, we get exp.G/ D 8, which is a contradiction. Thus ˆ.G/ Š E8 so G 0 Š E4 and K Š D8 C2 since Z.K/ Š E4 . It follows that jG 0 W K 0 j D 2 so G=K 0 is nonmetacyclic A1 -group (of order 24 and hence we get G=K 0 Š H16 by Lemmas 65.1 and 65.2(a)); in particular, M 0 D N 0 D K 0 since M=K 0 and N=K 0 are abelian. Since jK 0 j D 2, we have K 0 D hmi Z.G/ and so G 0 D hc; mi.Š E4 /, where o.c/ D 2. As we have shown, G=K 0 is a nonmetacyclic A1 -group. Therefore, by Lemma 65.1, the generator of G 0 =K 0 is not a square in G=K 0 , and we conclude that a2 2 hmi. Since K=G 0 Š E4 and G=G 0 is abelian of type .4; 2/, we get K=G 0 D 1 .G=G 0 / so that M=G 0 Š C4 Š N=G 0 . It follows from ˆ.G/ Š E8 that ˆ.G/ D hb 2 i G 0 so that Ahb 2 i=hmi Š E8 and so we must have K D Ahb 2 i with ˆ.K/ D hmi. The other two maximal subgroups are G 0 hbi and G 0 habi so that we may set M D B D G 0 hbi and N D G 0 habi. We have M D hb; c; mi D hb; ci since m 2 ˆ.M /, and similarly N D hab; ci, so that Œb; c D Œab; c D m since M 0 D N 0 D K 0 D hmi. We compute m D Œab; c D Œa; cb Œb; c D Œa; cb m which gives Œa; c D 1 and therefore A D ha; c; mi is abelian of order 8. Further, Œa; b 2 D Œa; bŒa; bb D cc b D c.cm/ D m: If a2 D m, then ha; b 2 i Š D8 , and then a0 D ab 2 is an involution. Replacing a with a0 (if necessary), we may assume from the start that a is an involution, and then A D ha; c; mi Š E8 (see the above two displayed formulas; recall that A is abelian since Œa; c D 1 D Œa; m D Œc; m). We get K D hci ha; b 2 i Š C2 D8 and the structure of G is uniquely determined as given in the statement of the theorem. It is easy to see that Z.G/ D hmi D K3 .G/ so cl.G/ D 3. The existence of G as a subgroup of A8 is established if we set a D .2; 7/.3; 8/

and b D .1; 2; 3; 4/.5; 6; 7; 8/:

118

Groups of prime power order

It is enough to check that the permutations a and b satisfy all the above relations. Since we get m D .1; 6/.2; 7/.3; 8/.4; 5/ ¤ 1G and hmi D Z.G/, our permutation representation is faithful. Theorem 107.5 ([BJ3, Theorem 5.5]). Suppose that G is a nonmetacyclic two-generator 2-group and let M; N; K be all its maximal subgroups such that M; N are nonmetacyclic two-generator and d.K/ D 3 (see Theorem 107.3). Then there is R G G such that G=R is isomorphic to the group from Theorem 107.4. Proof. We may assume that jGj > 25 (otherwise, in view of Theorem 107.4, there is nothing to prove). Since two-generator abelian 2-groups are metacyclic, the (nonmetacyclic two-generator) subgroups M and N are nonabelian. It follows that G is neither abelian nor minimal nonabelian; in particular, jG 0 j > 2 (Lemma 65(a)). In view of Theorem 106.3, K is nonabelian. Let L1 be a G-invariant subgroup of index 2 in M 0 (since d.M / D 2, the quotient group M 0 =K3 .M / is cyclic so L1 is uniquely determined). By Theorem 36.1, M=L1 is not metacyclic and neither is G=L1 . We also have L1 < ˆ.M / < ˆ.G/ D N \ K. Since d.N=L1 / D 2 D d.G=L1 /, it follows that d.K=L1 / D 3 (Corollary 36.6). By Theorem 107.3, N=L1 is nonmetacyclic so nonabelian. Thus the group G=L1 satisﬁes the hypothesis. Therefore, without loss of generality, one may assume that L1 D ¹1º. Let L2 be a G-invariant subgroup of index 2 in N 0 . As above, N=L2 is not metacyclic so is M=L2 , and d.K=L2 / D 3, so again G=L2 satisﬁes the hypothesis, and we may assume that L2 D ¹1º. By Lemma 65.2(a), M and N are A1 -groups since, by construction, jM 0 j D 2 D jN 0 j. Now set L3 D Ã2 .M /. Then, by Supplement to Corollary 36.6, M=L3 is not metacyclic and, as above, G=L3 satisﬁes the hypothesis so, arguing as above, one may assume that L3 D ¹1º. Similarly, one may assume that Ã2 .N / D ¹1º. In that case, we get exp.M / D exp.N / D 4. Since M and N are nonmetacyclic two-generator, they are nonabelian so A1 -groups as epimorphic images of A1 -groups. From Lemma 65.1 we deduce jM j 25 (indeed, an A1 -group of exponent 2n has order 22nC1 ). By assumption, we must have jGj > 25 so jM j D 25 and M=M 0 is abelian of type .4; 4/, jGj D 26 . In that case, by Lemma 65.1, Z.M / D ˆ.M / D 1 .M / Š E8 Š 1 .N / D Z.N / D ˆ.N /: Since Z.M / D ˆ.M / < ˆ.G/ < N

and

Z.N / < ˆ.G/ < M;

it follows that Z.M / D Z.N / D Z.G/.Š E8 / (indeed, we have Z.G/ < M in view of d.G/ D 2). The quotient group G=Z.G/ of order 8 is not elementary abelian since d.G/ D 2. It follows that G=Z.G/ contains a cyclic subgroup, say C =Z.G/, of order 4. In that case, C is abelian so d.C / D 3. Since K is a unique maximal subgroup of G not generated by two elements, we get C D K so that K is abelian. This is a ﬁnal contradiction: by the above, G has no abelian maximal subgroups.

107

Ranks of maximal subgroups of nonmetacyclic two-generator 2-groups

119

Let G be a two-generator 2-group all of whose maximal subgroups are not twogenerator (in that case all those subgroups are three-generator by Appendix 25); then G is nonabelian. Denote by H32 the unique A1 -group of order 32 and exponent 4 (see Lemma 65.1; all maximal subgroups of H32 are abelian of type .4; 2; 2/). Supplement to Theorem 107.5 ([BJ3]). If all maximal subgroups of a two-generator 2-group G are not two-generator, then there is R G G such that G=R Š H32 . Proof. Indeed, if M < G is maximal, then ˆ.M / D Ã1 .M / Ã2 .G/ so that, as we have noticed, d.M=Ã2 .G// D d.M / D 3. Thus G=Ã2 .G/ satisﬁes the hypothesis so one may assume that Ã2 .G/ D ¹1º holds. By Theorem 107.2(a), G=G 0 is abelian of type .4; 4/. If R is G-invariant of index 2 in G 0 , then G=R Š H32 (Lemmas 65.2(a) and 65.1), as desired. It follows that if a group G from the supplement has minimal possible order, then G Š H32 . Remark 1. Suppose that a nonabelian two-generator p-group G, p > 2, has an abelian subgroup A of index p. (i) If 1 .G/ D G, then exp.G=G 0 / D p so jG=G 0 j D p 2 and G is of maximal class (to prove this, we use Lemma 1:1 and induction). (ii) If 1 .G/ 6 A, then jG W G 0 Ã1 .A/j D jG W ˆ.G/j D p 2 (indeed, A1 .G/ D G, G 0 Ã1 .A/ G 0 Ã1 .G/ D ˆ.G/ and G=G 0 Ã1 .A/ is elementary abelian since it is abelian of rank 2 and generated by elements of order p as an epimorphic image of G=Ã1 .A/ D A1 .G/=Ã1 .A/). Thus jG=Ã1 .A/ W .G=Ã1 .A//0 j D p 2 . It follows from (i) that G=Ã1 .A/ is of maximal class; then d.A/ D d.A=Ã1 .A/ p by Exercise 9.13. Remark 2. Let p > 2 and let G be a two-generator p-group containing a maximal subgroup M such that d.M / D p C 1 (see Appendix 25). Let us study the structure of G=ˆ.M /. To simplify our task, one may assume that ˆ.M / D ¹1º; then M Š EppC1 so G is not of maximal class (Theorems 9:5 and 9:6). It follows that jZ.G/j > p (if jZ.G/j D p, then jG W G 0 j D pjZ.G/j D p 2 by Lemma 1.1, so G is of maximal class contrary to what has just been said). Thus we have jG=G 0 j D pjZ.G/j p 3 . If G=G 0 j > p 3 , then, in view of M Š EppC1 , p C 1 4, M=G 0 < G=G 0 and d.M=G 0 / > 2 D d.G=G 0 / D d.G/, we get a contradiction. Thus jG=G 0 j D p 3 hence exp.G=G 0 / D p 2 since d.G=G 0 / D d.G// D 2, and we conclude that G=G 0 is abelian of type .p 2 ; p/. Write H=G 0 D 1 .G=G 0 /; then M 1 .G/ H so that M D H D 1 .G/ since M 2 1 . Let x 2 G M ; then G D hx; M i, o.x/ D p 2 and N D G, N the group x p 2 Z.G/ since M is abelian. Write GN D G=hx p i. Since 1 .G/ N G is of maximal class (Remark 1(i)), isomorphic to †p2 , a Sylow p-subgroup of the symmetric group of degree p 2 by Exercise 9.13. If KN < GN is maximal and nonabelian, then KN is of maximal class (Fitting’s lemma). However, K is not of maximal class since d.G/ D 2 (here we use Theorem 12:12(a)), so jZ.K/j > p. We have Z.K/ < M (otherwise, G D M Z.K/ so cl.G/ D 2, a contradiction). Thus CG .Z.K// MK D G so Z.K/ D Z.G/ since Z.G/ < ˆ.G/ < K in view of d.G/ D 2.

108

p-groups with few conjugate classes of minimal nonabelian subgroups

Let 1 .G/ be the number of conjugate classes of A1 -subgroups (= minimal nonabelian subgroups) of a p-group G (see 76). If 1 .G/ 1, then G is either abelian or an A1 -group (Theorem 10.28). Therefore only the case where 1 .G/ > 1 is interesting. It follows from Remark 76.1 that if a p-group G is neither abelian nor minimal nonabelian, then 1 .G/ p. Proposition 108.1. Suppose that a p-group G contains a normal nonabelian subgroup N of index p n such that G=N is noncyclic. Then we have 1 .G/ n C p. If, in addition, 1 .G/ D n C p, then all proper G-invariant subgroups of N are abelian. Proof. By hypothesis, n 2. Let N D N0 < N1 < < Nn D G be a chain of G-invariant subgroups such that G=Nn2 is noncyclic; then all indices of this chain are equal to p. Since a nonabelian p-group is generated by its minimal nonabelian subgroups (Theorem 10.28), there is a minimal nonabelian subgroup Si < Ni such Si 6 Ni 1 , i D 1; : : : ; n 2. We have G=Nn2 Š Ep2 . Let S0 N0 D N be minimal nonabelian and Nn1 D Nn1;1 ; : : : ; Nn1;pC1 all (nonabelian) maximal subgroups of G containing Nn2 . Let Sn1;i < Nn1;i be minimal nonabelian not contained in Nn2 , i D 1; 2; : : : ; p C 1. Since .n 2/ C 1 C .p C 1/ D n C p minimal nonabelian subgroups S0 ; S1 ; : : : ; Sn2 ; Sn1;1 ; : : : ; Sn1;pC1 are not conjugate in pairs (indeed, normal closures of these subgroups in G are pairwise distinct), we get 1 .G/ n C p. Now suppose that 1 .G/ D n C p. Assume that there is in N a proper nonabelian G-invariant subgroup R; then G=R is noncyclic of order p nC1 . By the previous paragraph, 1 .G/ .n C 1/ C p > n C p, a contradiction. Thus all proper G-invariant subgroups of N are abelian. Proposition 108.2. Suppose that a p-group G of order p m > p 3 is of maximal class with abelian A 2 1 . Then 1 .G/ D p. Proof. It is well known that all nonabelian subgroups of G are of maximal class so all minimal nonabelian subgroups of G have the same order p 3 , and the number of minimal nonabelian subgroups in G equals p m3 . Indeed, 1 D ¹H1 ; : : : ; Hp ; Aº, where H1 ; : : : ; Hp are of maximal class (Fitting’s lemma). Then, by induction, Hi contains

108

p-groups with few conjugate classes of minimal nonabelian subgroups

121

p m13 D p m4 minimal nonabelian subgroups, and Hi \ Hj D ˆ.G/.< A/ is abelian for all i ¤ j , so the number of minimal nonabelian subgroups in G is equal to p p m13 D p m3 , as asserted. If Si Hi is minimal nonabelian, i D 1; : : : ; p, then S1 ; : : : ; Sp are pairwise nonconjugate since SiG D Hi for i D 1; : : : ; p (indeed, if a G-invariant subgroup has index > p in G, then it is contained in ˆ.G/; see Exercise 9.1(b)). We have jG W NG .Si /j D p m4 for all i D 1; : : : ; p. Indeed, NG .Si / is nonabelian hence of maximal class. Since all normal subgroups of NG .Si / of index > p are abelian, we get jNG .Si /j D pjSi j D p 4 , proving our claim. Thus all minimal nonabelian subgroups of Hi are conjugate in G for i D 1; : : : ; p and do not lie in Gj (j ¤ i ) hence 1 .G/ D p. Proposition 108.3. Suppose that G is a group of exponent p. Then 1 .G/ D p if and only if G is of maximal class and order > p 3 with an abelian subgroup of index p. In this case, jGj p p , and so we must have p > 3. Proof. By hypothesis, jGj > p 3 . If G is of maximal class with an abelian subgroup of index p, then 1 .G/ D p (Proposition 108.2). Now let 1 .G/ D p. Then, by Theorem 76.17, d.G/ D 2 and G has an abelian maximal subgroup. As ˆ.G/ D G 0 , we get jG W G 0 j D p 2 . It follows that G is of maximal class so jGj p p (Theorem 9.5). Since G is not of maximal class, we get p > 3. Let .G/ be the number of conjugate classes of maximal cyclic subgroups of a p-group G. Then .G/ k D 1 C p C C p d.G/1 . Indeed, if x 2 G ˆ.G/, then hxi is a maximal cyclic subgroup of G. The group G=ˆ.G/ has exactly k subgroups of order p, say M1 =ˆ.G/; : : : ; Mk =ˆ.G/. Let Ci Mi be cyclic such that Ci 6 ˆ.G/ for all i . Then C1 ; : : : ; Ck are maximal cyclic subgroups in G. Since we have CiG Ci ˆ.G/ D Mi for all i and Mi \ Mj D ˆ.G/ (i ¤ j ), it follows that C1 ; : : : ; Ck are pairwise nonconjugate in G so that .G/ k. It is interesting to classify the p-groups G satisfying .G/ D k (all 2-groups of maximal class satisfy this condition). Exercise. If G is a nonabelian p-group, then 1 .G/ d.G/1. Improve this estimate in the case d.G/ > 3. (Hint. If A is minimal nonabelian, then the subgroup Aˆ.G/ is nonabelian of index p d.G/2 in G and G-invariant. Apply Proposition 108.1.)

Problems Problem 1. Estimate the maximal possible number of conjugate classes of minimal nonabelian subgroups in a p-group of maximal class and order p m . Problem 2. Estimate the maximal possible number of conjugate classes of minimal nonabelian subgroups in representation groups of elementary abelian group Epn . Problem 3. Estimate the maximal possible number of conjugate classes of minimal nonabelian subgroups in metacyclic groups of order p m .

109

On p-groups with metacyclic maximal subgroup without cyclic subgroup of index p

According to [MS, Lemma 4.9], if p > 3 is a prime and if a is an automorphism of order p of the abelian group L of type .p 2 ; p 2 /, then a centralizes 1 .L/ (the proof of this result is also reproduced in [AS, Lemma A.1.30]). We offer a generalization of this result. Our proof is based on entirely other ideas. Theorem 109.1. Suppose that p > 2 and L is a metacyclic p-group without cyclic subgroup of index p. An element a 2 Aut.L/ of order p does not centralize 1 .L/ if and only if p D 3, the partial holomorph G D hai L is a 3-group of maximal class and one of the following holds: (a) L is abelian of type .3m ; 3n /, m > 1; n > 1 and jm nj 1. m

n

m1

(b) L D ha; b j a3 D b 3 D 1; ab D a1C3

i is minimal nonabelian, jmnj 1.

Proof. Suppose that G D hx; G1 i is a 3-group of maximal class and order > 34 with regular subgroup G1 of index 3 such that o.x/ D 3. Then G1 , the fundamental subgroup of G, is metacyclic so d.G1 / D 2 (see Theorems 9.6(b) and 9.11). If G1 is abelian of type .3m ; 3n /, then jm nj 1 since G has no normal cyclic subgroup of order 32 (indeed, G has a normal abelian subgroup of type .3; 3/ and it is a unique G-invariant subgroup of order 9 by Exercise 9,1; if k D min¹m; nº, then Ãk .G1 / is cyclic of order p jmnj and characteristic in L so normal in G). Next, we obtain that CG .1 .G1 // D G1 so x does not centralize 1 .G1 /. Now suppose that G1 is nonabelian. Then jG10 j D 3 since G10 is G-invariant cyclic so G1 is minimal nonabelian (Lemma 65.2(a)). In this case, by Lemma 65.1 (see also Exercise 1.8(a)), G1 has the same deﬁning relations as L in part (b) of our theorem. Since Ãjmnj .G1 / is cyclic, we get again jm nj 1. As, by hypothesis, G1 =1 .G1 / is noncyclic, we also have 1 .G1 / ˆ.G1 / D Z.G1 / since d.G1 / D 2. Since again CG .1 .G1 // D G1 , the element x does not centralize 1 .G1 /. Next we assume that G is not a 3-group of maximal class. Now suppose that L and a are as in the statement of our theorem. We have to describe the structure of L. Put G D ha; Li. Since L is regular, the subgroup 1 .L/ is abelian of type .p; p/; clearly, 1 .L/ G G. As j1 .G/j p 3 , G is nonmetacyclic. Since L has no cyclic subgroup of index p, the quotient group L=1 .L/ is noncyclic.

109

Metacyclic maximal subgroup

123

Suppose that a does not centralize 1 .L/. It follows that H D ha; 1 .L/i is nonabelian of order p 3 and exponent p since p > 2. We have G D LH since H 6 L and L 2 1 . Assume that G is regular. Then exp.1 .G// D p (Theorem 7.2(b)) so, considering the intersection L \ 1 .G/, we conclude that j1 .G/j D pj1 .L/j D p 3 D jH j whence 1 .G/ D H (H is deﬁned in the previous paragraph). It follows that G has no elementary abelian subgroups of order p 3 . Since G is neither metacyclic nor a 3-group of maximal class (Theorems 9.5 and 9.6), we can conclude from Theorem 13.7 that G=1 .G/ is cyclic, a contradiction since G=1 .G/ Š L1 .G/=1 .G/ Š L.L \ 1 .G// D L=1 .L/; and the last quotient group, as we have noticed, is noncyclic. Now let G be irregular (by assumption in the ﬁrst paragraph of the proof, G is not a 3-group of maximal class). It follows from Theorems 9.5 and 9.6 that G is not of maximal class for p > 3 as well. Therefore, by Theorem 12.1(b), G D L1 .G/, where 1 .G/ is of order p p and exponent p. Since L \ 1 .G/ D 1 .L/ is elementary abelian of order p 2 , we get p D 3. It follows that 1 .G/ D H D ha; 1 .L/i is nonabelian of order 33 and exponent 3 so G has no elementary abelian subgroups of order 33 . In this case, since G=H is noncyclic, the group G is of maximal class (Theorem 13.7), a ﬁnal contradiction. As we have noticed, the regular maximal subgroup of a 3-group of maximal class and order > 34 is such as in conclusion of Theorem 109.1. Corollary 109.2. Suppose that p > 2 and L is an abelian group of type .p m ; p n /, m > 1, n > 1. An element a 2 Aut.G/ of order p does not centralize 1 .L/ if and only if p D 3, G D ha; Li is a 3-group of maximal class and jm nj 1. Remark. Now we consider a similar but more complicated situation for p D 2. Suppose that a metacyclic 2-group L without cyclic subgroup of index 2 is maximal in a 2-group G and 1 .L/ Z.L/; then 1 .L/ is a four-subgroup. In this case, G is not of maximal class. Suppose that there is an involution a 2 GL which does not centralize 1 .L/. We claim that G is not metacyclic. Assume that this is false. Then the nonabelian subgroup hx; 1 .L/ Š D8 so G is of maximal class (Proposition 10.19), contrary to what has just been said. By hypothesis, CG .1 .L// D L. Now we use Theorem 66.1 classifying minimal nonmetacyclic 2-groups. Let H be a minimal nonmetacyclic subgroup of G. If H Š E8 , then L \ H D 1 .L/ so CG .1 .L// HL D G, a contradiction. In case H Š Q8 C2 , we have Z.H / D 1 .H / D 1 .L/ < H \ L so we get CG .1 .L// HL D G, a contradiction. Now assume that jH j D 25 (this H is special with center of order 4). As exp.H / D 4, it follows that 1 .L/ D 1 .H /, and since 1 .H / D Z.H /, we get CG .1 .L// HL D G, a contradiction. Thus H Š D8 C4 is of order 16. Similarly, if F < G is minimal nonabelian, then either jF j D 8 or F Š M2n (Lemma 65.1). As we know (see Theorem 10.28), it is possible to choose F so that F 6 L.

124

Groups of prime power order

If a noncyclic p-group L which contains a cyclic subgroup Z of index p is not a 2-group of maximal class (then 1 .L/ is abelian of type .p; p/), is maximal in a p-group G D ha; Li and an element a 2 G L of order p does not centralize 1 .L/, then H D ha; 1 .L/i is nonabelian of order p 3 and G D HZ with H \ Z D Z.H / and 1 .H / D H . If, in addition, p D 2, then all minimal nonmetacyclic subgroups of G have order 24 and are isomorphic to D8 C4 (Theorem 66.1). Let L be a maximal subgroup of a p-group G D hx; Li, where p > 2 and 1 .L/ is abelian of type .p; p/ (then L is either metacyclic or an irregular 3-group of maximal class by Theorem 13.7) and o.x/ D p. Suppose that x does not centralize 1 .L/. Then H D hx; 1 .L/i is nonabelian of order p 3 and exponent p. Clearly, G has no subgroup of order p 4 and exponent p (if such a subgroup, say F , exists, then we get j1 .F \ L/ D p 3 > p 2 D j1 .L/j, a contradiction). Then either G D H C , where H D 1 .G/ and C is cyclic, or G is a 3-group of maximal class (Theorem 12.1(a)). In the ﬁrst case, jL=Ã1 .L/j D j1 .L/j D p 2 (Theorem 7.2(d)) so L is metacyclic (Theorem 9.11), and this case we have already considered. Now assume that G and L are 3-groups of maximal class and jGj > 34 . Then CG .1 .L// D G1 is metacyclic and x does not centralize 1 .G1 /. Such a G1 , by Theorem 109.1, is either abelian or minimal nonabelian so G1 ¤ L; we see that L \ G1 D ˆ.G/ is abelian of index 3 in L. Exercise 1. Let G be a p-group, p > 2, and suppose that a noncyclic M 2 1 is metacyclic. If CG .1 .M // D M , then either M contains a cyclic subgroup of index p or G is a 3-group of maximal class. (Here we do not assume that G splits over M .) Exercise 2. Let G be a 2-group and suppose that M 2 1 is metacyclic but neither cyclic nor of maximal class. Suppose that CG .1 .M // D M . Classify the minimal nonabelian subgroups of G which are not contained in M . Answer. Q8 , D8 and M2n .

110

Equilibrated p-groups

We begin with the following Deﬁnition (compare with [BDM]). A p-group G is said to be weakly equilibrated (in short WE-group) if it follows from G D AB with A; B < G that one of the factors A; B is G-invariant. A group G is said to be equilibrated (brieﬂy: E-group) if every of its subgroup is weakly equilibrated. If, in addition, G is a p-group, then it is called a WEp -group and an Ep -group, respectively. Obviously, the properties WE, E, WEp and Ep are inherited by quotient groups. It is not true that the property WEp is inherited by subgroups (for example, if the fundamental subgroup G1 of a 3-group G of maximal class and order 35 is nonabelian, then G1 is not an E3 -group which follows from Exercise 5 below). However, as Lemma 110.1 shows, this group G is a WE3 -group. A minimal nonabelian WEp -group is an Ep -group since abelian p-groups are Ep -groups. Next, we follow [BDM]. We consider p-groups only. Note that in [BDM] the nonnilpotent groups are also studied. In what follows G is a non-Dedekindian WEp -group of order p m (Dedekindian p-groups are Ep -groups). Since groups of order p 3 are Ep -groups, one may assume that m > 3. As all subgroups of Mpm of order > p are normal, Mpm is an Ep -group. Moreover, all groups from Theorem 1.25 are Ep -groups. Exercise 1. (a) A nonmetacyclic minimal nonabelian p-group G of order p m > p 3 is not a WEp -group. (b) All two-generator WE2 -groups are metacyclic. (c) If G is a nonmetacyclic two-generator WEp -group, p > 2, then jG W G 0 j D p 2 . (d) If a metacyclic p-group G is a non-WEp -group with jG 0 j D p and G D AB, where A; B are not G-invariant, then A and B are cyclic. Solution. (a) We ﬁrst assume that m D 4. In this case, by Lemma 65.1, 2

G D ha; b j ap D b p D 1; c D Œa; b; c p D Œa; c D Œb; c D 1i: Set A D hai and B D hap c; bi. Then jAj D jBj D p 2 and A \ B D ¹1º so we have G D AB by the product formula. Since G 0 D hci 6 A and G 0 6 B, neither A nor B

126

Groups of prime power order

are G-invariant, and our claim follows. Now let m > 4. In this case, r

s

G D ha; b j ap D b p D 1; c D Œa; b; c p D 1; Œa; c D Œb; c D 1i: 2

We also assume that r s; then r > 1. Set T D hap ; b p i. Then the quotient group G=T is a nonmetacyclic minimal nonabelian p-group of order p 4 so, by the above, it is not a WEp -group. In this case, G is also not a WEp -group since the property WEp is inherited by epimorphic images. (b) Suppose that a two-generator WE2 -group G is nonmetacyclic; then it is nonabelian and jG W G 0 j > 4 (Taussky’s theorem). By Theorem 36.1, if R < G 0 is G-invariant of index 2, then G=R is nonmetacyclic and minimal nonabelian. Therefore, by part (a), G=R is not a WE2 -group since jG=Rj > 23 , and our claim follows. (c) Assume, by way of contradiction, that jG W G 0 j > p 2 . If R < G 0 of G-invariant of index p, then G=R is nonmetacyclic and minimal nonabelian (Theorem 36.1) so it is not a WEp -group by (a) since jG=Rj > p 3 ; then G is not a WEp -group as well. (d) A noncyclic subgroup of G contains 1 .G/ > G 0 and a subgroup containing 1 .G/ is normal. If C < G is cyclic such that G=C is cyclic, then A \ C D ¹1º D B \ C , and the claim follows. (In the case under consideration, G is minimal nonabelian by Lemma 65.2(a).) Exercise 1. A 2-group G of maximal class is an E2 -group. Solution. Assume that G D AB, where A; B < G, and jAj jBj, A and B are not normal in G; then jAj 4, jBj 4. Let C < G be cyclic of index 2. Then, since jB \ C j jA \ C j, we get B \ C A \ B. As B \ C G G, one may assume that B \ C D ¹1º. In this case, jG W Aj jBj D 2 so A G G, contrary to the assumption. It follows that G is also an E2 -group since all its proper subgroups are either cyclic or of maximal class. Exercise 2. Suppose that G is a two-generator nonmetacyclic p-group of order p 4 , p > 2. Then G is a WEp -group if and only if G is of maximal class and has no subgroup isomorphic to Ep3 . Solution. Let G be a WEp -group. We have to prove that G has no subgroup Š Ep3 . By Exercise 1(a), G is not minimal nonabelian. Let B < G be minimal nonabelian. Since d.G/ D 2, we get CG .B/ < B so G is of maximal class (Proposition 10.17). Assume that G has a subgroup U Š Ep3 . Then U D Z.G/ A, where A Š Ep2 , so AG D ¹1º, i.e., A is not G-invariant. Let V < G be of order p 2 such that V 6 U ; then V is not normal in G (otherwise, V D ˆ.G/ < U ).1 In this case, CG .V \ U / D U V D G so U \ V D Z.G/, and we get V \ A D ¹1º. Then G D AB by the product formula, and we conclude that G is not a WEp -group, contrary to the assumption. Now suppose that G has no subgroup Š Ep3 . We have to prove that then G is a WEp -group. Assume that G D AB, where A and B are not normal in G; then jAj D p 2 D jBj. If A does not exist, then U is generated by all subgroups of G of order p 2 so G has a cyclic subgroup of index p, a contradiction (Theorem 1.2). 1 If V

110

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Equilibrated p-groups

is noncyclic, then Z.G/ < A (otherwise, Z.G/A Š Ep3 , contrary to the assumption). If A is cyclic, then Z.G/ D Ã1 .G/ < A since G=Ã1 .G/ is of order p 3 (Theorem 9.11) and exponent p. Thus, in any case, Z.G/ is contained in A and B so, by the product formula, jABj < p 4 D jGj, a contradiction. Now it is clear that G is an Ep -group if it has no subgroup Š Ep3 . Exercise 3. Suppose that G is a p-group of maximal class and order > p 4 , p > 3. Then G is not a WEp -group. Solution. By Theorems 9.5 and 9.6, jG=Ã1 .G/j p 4 . Let Ã1 .G/ N G G be such that jG=N j D p 4 ; then N < ˆ.G/, exp.G=N / D p and G=N (of maximal class and exponent p) has a subgroup Š Ep3 . By Exercise 3, G=N is not a WEp -group so is G. 2

2

Exercise 4. If p > 2 and G D ha; b j ap D b p D 1; ab D a1Cp i is nonabelian metacyclic of order p 4 and exponent p 2 , then G is not a WEp -group. Solution. We have Z.G/ D hap i hb p i D 1 .G/ D Ã1 .G/ and G 0 D hap i. Set B D hbi; then B is not G-invariant. The quotient group G=hap b p i is nonabelian so it has a nonnormal subgroup A=hap b p i of order p. It follows that A is cyclic (indeed, all noncyclic subgroups of G are normal). In this case, A \ B D ¹1º, and so we get G D AB by the product formula. Thus G is not a WEp -group. n

n

n1

Remark. Moreover, if p > 2, then G D ha; b j ap D b p D 1; ab D a1Cp i is not a WEp -group. The group G is metacyclic minimal nonabelian (Lemma 65:2(a)). We have Ãn1 .G/ D 1 .G/ Z.G/. If L is of order p and L 62 ¹1 .hbi/; G 0 º, then n1 there is x 2 G such that L D hx p i (Theorem 7:2(c)). Then, by the product formula, G D U V , where U D hbi, V D hxi are not G-invariant. Indeed, V \ G 0 D ¹1º, and if V G G, then G D hai V is abelian, a contradiction. From this it also follows that m n n1 G D ha; b j ap D b p D 1; ab D a1Cp i, where m < n, is not a WEp -group nm (consider the quotient group G=hb p i). It is easy to show that a nonabelian metacyclic group G of order 24 and exponent 4 is a WE2 -group. To prove this, it sufﬁces to notice that all subgroups of order 2 of G are characteristic and exactly one of them is maximal cyclic. Indeed, let G D AB, where A; B < G are not G-invariant. We have G Š H2 D hx; y j x 4 D y 4 D 1; x y D x 3 i: Since G=hx 2 y 2 i Š Q8 , we get AG ¤ hx 2 y 2 i ¤ BG . Then one of subgroups AG ; BG contains G 0 D ha2 i so is G-invariant. Exercise 5. If G is a two-generator WEp -group, p > 2, of order > p 3 that satisﬁes jG=G 0 j > p 2 , then G is metacyclic. (Hint. Use Exercise 1(c). This also follows from Exercise 1(a) (take in G 0 a G-invariant subgroup of index p).) Note that if G is a 3-group of maximal class not isomorphic to †32 2 Syl3 .S32 /, then all subgroups of G are two-generator (see Exercise 9.13).

128

Groups of prime power order

Lemma 110.1 (compare with [BDM, 4, Example 4]). A 3-group G of maximal class and order > 34 is a WE3 -group. Proof. We use induction on jGj. Write Z D Z.G/. Assume that G D HK, where H; K < G are not G-invariant. By the paragraph preceding the lemma, Exercises 9.13, 3 and induction, in the factorization G=Z D .HZ=Z/.KZ=Z/ one of the factors, say HZ=Z, is G-invariant; then HZ G G. By assumption, Z 6 H so that HZ D H Z. Since H 6 ˆ.G/, we get H Z 2 1 (every normal subgroup of index > p in a p-group X of maximal class is contained in ˆ.X/). All members of the set 1 , apart from the fundamental subgroup G1 , are of maximal class. It follows that H Z D G1 . Since G1 is metacyclic, the subgroup H is cyclic. Then ˆ.H / D ˆ.G1 /, a cyclic subgroup of order > 3, is G-invariant, a ﬁnal contradiction (see Exercise 9.1(c)).2 It follows that if G is a 3-group of maximal class and order 3m > 34 , then G is an E3 -group provided its fundamental subgroup G1 is either abelian or m is even (see the remark after Exercise 5).3 Indeed, our group G is an E3 -group if and only if G1 is a WE3 -group since if G1 is a WE3 -group, then every proper subgroup of G is also a WE3 -group so G is an E3 -group (if H 2 1 ¹G1 º, then H is of maximal class and H \ G1 is abelian). Now we are ready to prove the following Theorem 110.2 ([BDM, Theorem 4.1]). Suppose that G is a nonabelian two-generator Ep -group of order > p 3 and p > 2. Then one of the following holds: (a) G is metacyclic (see Exercise 5). (b) G is a 3-group of maximal class (all such groups are described in Lemma 110.3 below). (c) G is of maximal class and order p 4 without subgroup Š Ep3 ; see Exercise 3. Proof. Suppose that G is neither metacyclic nor of maximal class. Then jGj > p 4 (Exercise 3). Next, by Exercise 3 and Theorem 9.11, jG=Ã1 .G/j D p 3 . By Exercise 6, jG=G 0 j D p 2 . From Theorem 36.16 and Corollary 44.9 it follows that G=Ã2 .G/ is neither metacyclic nor of maximal class, where Ã2 .G/ D Ã1 .Ã1 .G//. Therefore one may assume that Ã2 .G/ D ¹1º; then exp.G/ D p 2 . Let R Ã1 .G/ be G-invariant of order p. Then, by induction, either G=R is of maximal class and order p 4 or a 3-group of maximal class and order 35 . In both cases, GN D G=R has no subgroup Š Ep3 so all its maximal subgroups are two-generator. (i) Let GN be of maximal class and order p 4 . Then GN has an abelian subgroup T =R of type .p 2 ; p/ (Exercise 3). By Lemma 65.2(a), Exercises 1(a) and 5, T is abelian of type .p 2 ; p 2 / or .p 2 ; p; p/ since exp.G/ D p 2 . Since jG=G 0 j D p 2 , we get jZ.G/j D 2 Thus, if a p-group of maximal class and order > p 4 is a WE -group, then p 2 ¹2; 3º; see Exerp cises 2, 4 and Lemma 110.3(b), below. 3 However, we do not assert that if m is even, then G is an E -group; see Lemma 110.3. 3

110

Equilibrated p-groups

129

1 0 p jG=G j

D p so that Z.G/ D R, and we conclude that G is of maximal class. In the case under consideration, since jG=Ã1 .G/j D p 3 , we get p D 3 (Theorem 9.5). (ii) Let GN be a 3-group of maximal class and order 35 (recall that exp.G0 / D 32 ). By Theorem 13.7, there is in G a normal subgroup E Š E27 . As G=R is also of maximal class, we conclude that E D Ã1 .D/, where D=R is the fundamental subgroup of the quotient group G=R. Let H=E D Z.G=E/; then H G G is of order 34 . Since jH j D 34 > 33 D jEj D j1 .D/j, it follows that exp.H / D 9 so that K D Ã1 .H / is G-invariant of order 3. By induction, G=K is of maximal class and of order 35 with normal subgroup E=K of order 33 and exponent 3, contrary to Theorem 9.6(c). Lemma 110.3 (see [BDM, Example 4]). A 3-group G of maximal class and order 3m > 34 is an E3 -group if and only if its fundamental subgroup G1 is either abelian or sC1 s s G1 D ha; b j a3 D b 3 D 1; ab D a1C3 i: Proof. All subgroups of G are two-generator (see Exercise 9.13). By Lemma 110.1, G is a WE3 -group. If the fundamental subgroup G1 of G is abelian, it is a WE3 -group. Then all proper subgroups of G which have (metacyclic) abelian fundamental subgroups are WE3 -groups. Thus G is an E3 -group if G1 is a WE3 -group. Assume that G1 is nonabelian. Then G1 , as we know, is minimal nonabelian (this follows from Exercise 9.1(b) and Lemma 65.2(a)). As we know, G is an E3 -group of and only if G1 is a WE3 -group (so an E3 -group). Next, by Exercise 5 and the remark following it, G1 is not a WE3 -group if m 1 is even; then m is odd. Now let m D 2k. By Lemma 65.1 and Theorems 9.5 and 9.6, we have r

s

G1 D ha; b j a3 D b 3 D 1; ab D a1C3

r1

i;

jr sj D 1; r Cs D m1 D 2k 1:

Since E9 Š 1 .G1 / < G1 , all noncyclic subgroups of G1 are G1 -invariant. Let s D r C 1. Then G1 =Ãr .G1 / is not a WE3 -group (Exercise 5 and the remark following it) so G1 and G are not E3 -groups. Now let r D s C 1. We claim that then G is a WE3 -group. Assume that this is false. Then G D AB, where A and B are nonnormal in G so A and B do not contain G 0 hence they have trivial intersections with hai. It follows that jAj p s , jBj p s so that jGj D jABj p 2s < p rCs D jGj, a contradiction. Thus, in the case under consideration, G is a WE3 -group. Exercise 6. Let G be an extraspecial group of order p 2mC1 , m > 1. (a) Prove that G is a WEp -group. (b) If, in addition, exp.G/ D p, then G is not an Ep -group. (c) If m > 2 and p > 2, then G is not an Ep -group. Solution. (a) Assume that G D AB, where A; B < G are not G-invariant. Then we have G 0 6 A; B so A; B are abelian of orders p m (see 4). By the product formula, jABj < p 2mC1 D jGj, a contradiction.

130

Groups of prime power order

(b) It sufﬁces to assume that m D 2. We have G D D D1 , where D and D1 are nonabelian of order p 3 . Let L < D1 be of order p such that L ¤ G 0 . Set H D D L. It sufﬁces to show that H is not an E3 -group. Let L1 < D of order p be such that L1 ¤ G 0 and set A D L L1 . Then A Š Ep2 is not normal in H since D \ A D L is not normal in D. Let L2 < Z.H / of order p be such that L2 62 ¹L; G 0 º and let L3 < D of order p be such that L3 62 ¹L1 ; G 0 º. Set B D L2 L3 . Then B is not normal in H since B \ D D L2 is not normal in D. Since A \ B D ¹1º, we get H D AB is not a WEp -group so G is not an Ep -group. (c) Let G D D1 Dm1 Dm , where D1 ; : : : ; Dm1 are nonabelian of order p 3 and exponent p. Then D1 D2 is not an Ep -group by (b) so that G is not an Ep -group. m

n

Exercise 7. Prove that the metacyclic p-group G D ha; b j ap D b p D 1; ab D m1 a1Cp ; n mi, p > 2, is not a WEp -group. (Hint. See the last paragraph of the proof of Lemma 110.3.) r

s

Exercise 8. Let G D ha; b j ap D b p D 1; ab D a1Cp Then G is a WEp -group if and only if s < r.

r1

i, p > 2, r C s > 4.

Solution. Using the argument in the proof of Theorem 110.3, which is also applied to minimal nonabelian metacyclic p-groups with p > 3, we have to prove that if s < r, then G is a WEp -group. Assume that this is false. Then G D AB, where A and B are not G-invariant. Since G 0 < 1 .G/, we get j1 .A/j D j1 .B/j D p so A and B are cyclic and A \ hai D ¹1º D B \ hai. It follows that jAj p s , jBj p s so that jGj jAjjBj p 2s < p rCs D jGj, a contradiction. Thus G is a WEp -group if and only if s < r. Exercise 9. Let a WE-group G be nonnilpotent and solvable. Then we have G D D P , a semidirect product, where P 2 Sylp .G/ for some prime p is G-invariant and D is Dedekindian. Solution. Let H < G be a maximal nonnormal subgroup of G; then jG W H j D p a for some prime p. Let P 2 Sylp .G/. Since G D HP and H is not G-invariant, we get P G G. Let F=P < G=P . Since G D HP HF , we obtain G D HF so F G G. It follows that G=P is Dedekindian. It remains to apply Hall’s Theorem A.28.4 on solvable groups. Exercise 10. Let a nonabelian WEp -group G D AB be such that jG=G 0 j D p 2 and A G G, B < G. Prove that jG W Aj D p. Solution. If jG W Aj > p, then A G 0 D ˆ.G/ so G D AB D B, a contradiction. Exercise 11. Let a nonabelian two-generator WEp -group G D AB, where A G G and B < G. Prove that G=A is cyclic. (Hint. If G=A is noncyclic, then A ˆ.G/.) Exercise 12. Let a nonabelian two-generator non-WEp -group G D AB, where A and B are maximal nonnormal subgroups of G. Then G=AG and G=B G are cyclic. (Hint. If G=AG is noncyclic, then A < AG ˆ.G/.)

110

Equilibrated p-groups

131

Exercise 13. Let G be a 2-group such that G=1 .G/ Š Q8 . Is it true that G is not a WE2 -group? (Hint. See Lemma 42.1(c) about the structure of G. By that lemma, G is metacyclic.) Exercise 14. Suppose that G is a metacyclic p-group. Then one of the following holds: (a) jGj D p 3 . (b) G is a 2-group of maximal class. (c) G has a subgroup isomorphic to Mpn , n > 3. (d) 1 .G/ Z.G/. Solution. Suppose that (a), (b) and (c) do not hold; then G is non-Dedekindian and 1 .G/ Š Ep2 . Let A G be minimal nonabelian. If jAj D p 3 , then either p D 2 and G is of maximal class or G D A (Proposition 10.19), contrary to the assumption. It follows that 1 .A/ Z.A/ (Lemma 65.1). Since all such A generate G (Theorem 10.28) and 1 .A/ D 1 .G/, we conclude that 1 .G/ Z.G/, i.e., (d) holds. Lemma 110.4 ([BDM, Theorem 4.2]). Let G be a metacyclic p-group, p > 2, let hai D M G G and G=M D hbM i. Then G D hbihbai. Proof. One has G D ha; bi. Suppose that ab D ak ; then we get k 1 .mod p/ since o.b/ is a power of p and p j .k o.b/ 1/. Let us show that hbihbai D hbaihbi. We have ˇ1 ˇ1 b ˛ .ba/ˇ D b ˛Cˇ ab : : : ab a D b ˛Cˇ a1CkCCk : Thus we must show that every integer can be expressed modulo p n D o.a/ in the form 1 C k C C k ˇ 1 . If this is not so, there exist integers ˇ; with 0 ˇ < < p n such that 1 C k C C k ˇ 1 D 1 C k C C k 1 .mod p n /: Hence, since p does not divide k, we get .1/

1 C k C C k ˇ 1 0

.mod p n /;

and 0 < ˇ < p n . It follows that k 6 1 .mod p n /. But k 1 .mod p/; thus, if k D 1 C p m s with GCD.s; p/ D 1, then 1 m < n. It follows from (1) that k ˇ 1 k ˇ 1 D

0 k1 sp m

.mod p n /

so k ˇ 1 .mod p mCn /. We have .2/

k ˇ D .1 C sp m / ˇ D 1 C . ˇ/sp m C Cp mCn ;

where C is a positive integer. By (2), we obtain that p n j . ˇ/ so ˇ p n , a contradiction.

132

Groups of prime power order

Theorem 110.5 ([BDM, Theorem 4.2]). Let G be a metacyclic WEp -group, p > 2. Then cl.G/ 2. Proof. We retain the same meaning for M, a and b as in the statement of Lemma 110.4. Since G D hbihbai is a WEp -group, either hbi G G or hbai G G; hence Œb; a, a generator of G 0 , is contained either in hbi or hbai. But also Œb; a 2 M D hai. It follows that Œb; a commutes with b and a hence Œb; a commutes with hb; ai D G, and we conclude that G 0 D hŒb; ai Z.G/. Exercise 15 ([BDM, Lemma 4.3(a)]). Suppose that a nonabelian Ep -group G has exponent p. Then jGj D p 3 . Solution. By hypothesis, p > 2. Assume that jGj D p 4 . Then G contains an abelian subgroup A of index p so, by Exercise 3, cl.G/ D 2. By Lemma 65.1, G is not minimal nonabelian. Therefore it follows from Proposition 10.17 that G D D L, where L Z.G/ is of order p. As in the solution of Exercise 6(b), G is not a WEp -group. Now let jGj > p 4 . Since G is not minimal nonabelian (Lemma 65.1), it contains a nonabelian subgroup H of order p 4 . By what has just been proved, the subgroup H is not a WEp -group so G is not an Ep -group. Thus jGj D p 3 . Exercise 16. Let G be an Ep -group of order > p 4 , p > 2, containing a nonabelian subgroup A of order p 3 and exponent p. Prove that one of the following holds: (a) 1 .G/ D A and G D AC , where C is cyclic. (b) G is a 3-group of maximal class. Solution. Assume that G has a normal subgroup U Š Ep3 . Let U1 < U be a G-invariant subgroup which is as small as possible and such that U1 6 A. Set H D AU1 . If A \ U1 D ¹1º, then jU1 j D p and H D A U1 is not a WEp -group (see solution of Exercise 6(b)). Thus A \ U1 > ¹1º. Moreover, all minimal normal subgroups of G contained in U are contained in A. It follows that jH j D p 4 . Assume that U1 D U . Then H is not of maximal class (Exercise 3) so, by Proposition 10.17, CH .A/ 6 A. Since 1 .H / D H , we get H D A C for some C of order p, and then H is not a WEp -group. Thus jU1 j D p 2 and U1 6 Z.H / (otherwise, A is a direct factor of H ). In this case, we have CH .U1 / Š Ep3 . Then, by Exercise 3, H is not of maximal class so cl.H / D 2, and we conclude that exp.H / D p since p > 2. Then H D A C again, a contradiction. Thus U does not exist and the result now follows from Theorem 13.7. Theorem 110.6 ([BDM, Lemma 4.3(c)]). If an Ep -group G contains a proper subgroup H of maximal class and order > p 3 , p > 2, then p D 3 and G is of maximal class. Proof. Assume that G is not of maximal class. By Theorems 9.5 and 9.6, one may assume that jH j D p 4 (indeed, H contains a subgroup of maximal class and order p 4 ); then exp.H / D p 2 and H has no subgroup Š Ep3 (Exercise 3). Theorem 13.7 yields

110

Equilibrated p-groups

133

that either H is a 3-group of maximal class or H D C 1 .H /, where C is cyclic of order p 2 and 1 .H / is nonabelian of order p 3 and exponent p. By Exercise 16, the second case is impossible since a group from Exercise 16(a) has no such a subgroup as H (it is important that jGj > p 4 ). Thus H is of maximal class and order 34 . Again H has no subgroups of order 33 and exponent 3. Indeed, if H has a nonabelian subgroup A of order 33 and exponent 3, then, by Exercise 16, G D AC , where A D 1 .G/ and C is cyclic of order > 32 . Such a G, however, has no subgroup of maximal class and order 34 . By Exercise 13.10(a), we have H < M G, where jM W H j D p and M is not of maximal class. One may assume without loss of generality that G D M ; then G=G 0 Š E33 (Theorem 12.12(a)). By Theorem 12.12(b), GN D G=K3 .G/ is of order 34 and exponent 3 since jGj > 34 . As above, GN D SN CN , where SN is nonabelian of order 33 and jCN j D 3. As we know, such a GN is not an E3 -group, a contradiction. Thus G is a 3-group of maximal class. For further details on Ep -groups, see [BDM] and [Sil]. In view of Theorem 110.2, the classiﬁcation of Ep -groups, p > 2, will follow from the solution of the following Problem 1. Classify the p-groups all of whose nonabelian two-generator subgroups are either metacyclic or of maximal class. Problem 2. Classify the metacyclic WEp -groups.

111

Characterization of abelian and minimal nonabelian groups

If G is abelian, then G 0 D ˆ.G/0 and it appears that this property characterizes abelian groups (Theorem 111.1). If G is either abelian or minimal nonabelian, then H 0 D ˆ.G/0 for all H 2 1 , where 1 is the set of all maximal subgroups of G, and Theorem 111.3 shows that this property is characteristic for groups all of whose maximal subgroups are abelian. In this section we consider also nonnilpotent groups. In Lemma J we collect some known results cited in what follows. Lemma J. Let G be a ﬁnite group. (a) (R. Baer [Bae6]) If P 2 Syl.G/ is G-invariant, then ˆ.P / D P \ ˆ.G/. (b) (Redei; see Exercise 1.8(a)) If G is a nilpotent minimal nonabelian group, then G is a p-group, jG 0 j D p, Z.G/ D ˆ.G/ is of index p 2 in G and one of the following holds: (i) p D 2 and G is the ordinary quaternion group, m

n

m1

(ii) G D ha; b j ap D b p D 1; ab D a1Cp ; m > 1i is metacyclic of order p mCn , m n (iii) G D ha; b j ap D b p D c p D 1; c D Œa; b; Œa; c D Œb; c D 1i is nonmetacyclic of order p mCnC1 . In particular, if exp.G/ D p, then p > 2 and jGj D p 3 . (c) (Miller–Moreno; see Lemma 10.8) If G is a nonnilpotent minimal nonabelian group, then we have jGj D p a q b , G D P Q, where P 2 Sylp .G/ is cyclic, Q D G 0 2 Sylq .G/ is a minimal normal subgroup of G, Z.G/ D ˆ.G/ has index p in P . (d) (Wielandt) A group G is nilpotent if and only if G 0 ˆ.G/ (or, what is the same, G=ˆ.G/ is nilpotent). (e) If G is nilpotent and noncyclic, then G=G 0 is also noncyclic. (f) [Gas1] If a Sylow p-subgroup is normal in G=ˆ.G/, then a Sylow p-subgroup is normal in G. If H is normal in G, then ˆ.H / ˆ.G/. (g) .G=ˆ.G// D .G/, where .G/ is the set of prime divisors of jGj.

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Characterization of abelian and minimal nonabelian groups

135

(h) [BG] If NF .F \ H / ¤ F \ H ¤ NH .F \ H / for any two distinct F; H 2 1 , then either G is nilpotent or G=ˆ.G/ is (nonnilpotent) minimal nonabelian. The converse assertion is also true. (i) If G is a nonabelian p-group such that G 0 Z.G/ has exponent p, then we have ˆ.G/ Z.G/. (j) (Fitting; see Corollary 6.5) Let Q be a normal abelian Sylow q-subgroup of a group G. Then Q \ Z.G/ is a direct factor of G. Let us prove Lemma J(a). Since ˆ.P / D D ˆ.G/ \ P (Lemma J(f)), it sufﬁces to prove the reverse containment. To this end, one may assume that ˆ.P / D ¹1º; then P is elementary abelian. We have to show that p 62 .ˆ.G//. Let H be a p 0 -Hall subgroup of G. Then P D D L, where L is H -admissible (Maschke). In this case, G D HP D .HL/ D, a semidirect product with kernel D, so D D ¹1º since D ˆ.G/. Thus, in general case, ˆ.P / D D. Let us prove Lemma J(i). Take x; y 2 G. Since cl.G/ D 2, we have 1 D Œx; yp D Œx; y p so that Ã1 .G/ Z.G/, and we obtain ˆ.G/ D G 0 Ã1 .G/ Z.G/. Let 1 (1n ) be the set of maximal (nonnormal maximal) subgroups of a group G. If G is not nilpotent, then the set 1n is not empty (Lemma J(d)). Theorem 111.1. The following conditions for a group G are equivalent: (a) G 0 D ˆ.G/0 . (b) G is abelian. Proof. As we have noticed, (b) ) (a) so it remains to prove the reverse implication. We have G 0 D ˆ.G/0 ˆ.G/ so G is nilpotent (Lemma J(d)); then G D P1 Pk , where P1 ; : : : ; Pk are Sylow subgroups of G. We have (1) (2) (3)

ˆ.G/ D ˆ.P1 / ˆ.Pk /; G 0 D P10 Pk0 ; ˆ.G/0 D ˆ.P1 /0 ˆ.Pk /0 :

Equality (1) follows from Lemma J(a), and (2), (3) are known properties of direct products. Since ˆ.G/0 D G 0 , it follows from (2) and (3) that ˆ.Pi /0 D Pi0 so Pi satisﬁes the hypothesis for i D 1; : : : ; k. To complete the proof, it sufﬁces to show that Pi is abelian for i D 1; : : : ; k so one may assume that k D 1, i.e., G is a p-group. Assume that G is nonabelian of the least possible order. We get ˆ.G/0 D G 0 > ¹1º so that ˆ.G/ is nonabelian. Let R < ˆ.G/0 be G-invariant of index p. Since R < ˆ.G/0 D G 0 < ˆ.G/, we get ˆ.G=R/0 D .ˆ.G/=R/0 D ˆ.G/0 =R D G 0 =R D .G=R/0 whence the nonabelian group G=R satisﬁes the hypothesis so we must have R D ¹1º by induction. In that case, we get jˆ.G/0 j D jG 0 j D p so G 0 Z.G/. By Lemma J(i), ˆ.G/ Z.G/ so that ˆ.G/ is abelian, contrary to what has been said above.

136

Groups of prime power order

Lemma 111.2. Let G be a noncyclic p-group and D D hˆ.H / j H 2 1 i. Then D ˆ.G/ and jˆ.G/ W Dj p with equality only for p > 2. Proof. Since all members of the set 1 are G-invariant, we have D ˆ.G/ (Lemma J(f)). As all maximal subgroups of the quotient group G=D are elementary abelian, G=D is either elementary abelian or nonabelian of order p 3 and exponent p (Lemma J(d)). In the ﬁrst case, D D ˆ.G/; in the second case, p > 2 and jˆ.G/ W Dj D jˆ.G=D/j D p. Thus, in both cases, jˆ.G/ W Dj p, as desired. If H 2 1 , then ˆ.G/ < H so ˆ.G/0 H 0 . Below we consider the extreme case when ˆ.G/0 D H 0 for all H 2 1 . Theorem 111.3. The following assertions for a group G are equivalent: (a) ˆ.G/0 D H 0 for all H 2 1 . (b) G is either abelian or minimal nonabelian. Proof. Obviously, groups all of whose maximal subgroups are abelian satisfy condition (a) hence implication (b) ) (a) is true. Therefore it remains to prove the reverse implication. In what follows one may assume that G is nonabelian. (i) Suppose that G is a nonabelian p-group satisfying condition (a). Note that ˆ.G/, ˆ.G/0 and H 0 are G-invariant for all H 2 1 . If ˆ.G/ is abelian, then all maximal subgroups of G are also abelian by hypothesis, and G is minimal nonabelian so (b) holds. Next we assume that ˆ.G/0 > ¹1º. Let R < ˆ.G/0 be G-invariant of index p N is nonabelian and jˆ.G/ N 0 j D p. Take H 2 1 . We have and set GN D G=R; then ˆ.G/ .4/

N 0 D ˆ.G=R/0 D .ˆ.G/=R/0 D ˆ.G/0 =R D H 0 =R D HN 0 ˆ.G/

N since is of order p so, by Lemma J(i), ˆ.HN / Z.HN / and hence ˆ.HN / Z.ˆ.G// N N N N N N N ˆ.H / ˆ.G/ H . Setting D D hˆ.H / j H 2 1 i, we conclude D Z.ˆ.G// N is abelian since jˆ.G/ N W Dj N p by Lemma 111.2, contrary to what has so ˆ.G/ already been said. Thus, if G is a p-group, then (a) ) (b). In what follows we assume that G is not a prime-power group. Write T D ˆ.G/0 ; then T D H 0 for all H 2 1 by hypothesis. Therefore all maximal subgroups of G=T are abelian so the quotient group G=T is either abelian or minimal nonabelian. As in (i), one may assume that ˆ.G/ is nonabelian. (ii) Suppose that G is nilpotent; then G D P1 Pk , where Pi 2 Sylpi .G/, i D 1; : : : ; k, k > 1. It follows from (2) and (3) which are true for any nilpotent group that ˆ.Pi /0 D Hi0 for all maximal subgroups Hi of Pi so Pi is either abelian or minimal nonabelian by (i). Then ˆ.Pi / is abelian for all i so ˆ.G/ is abelian in view of (1), contrary to the assumption. Thus the theorem is true provided G is nilpotent. Next we assume that G is not nilpotent. In this case, by Lemma J(d), G=T is nonnilpotent (and minimal nonabelian). (iii) It remains to consider the case where G=T is minimal nonabelian and nonnilpotent. It follows from the structure of G=T (Lemma J(c)) that ˆ.G=T / D ˆ.G/=T is

111

Characterization of abelian and minimal nonabelian groups

137

cyclic. Remembering that ˆ.G/ is nilpotent and T D ˆ.G/0 , we conclude that ˆ.G/ is cyclic (Lemma J(e)) so abelian, a ﬁnal contradiction. Corollary 111.4. Suppose that a group G is neither abelian nor minimal nonabelian. Then there exists a nonabelian H 2 1 such that ˆ.G/0 < H 0 . Proof. We have ˆ.G/0 H 0 for all H 2 1 . Assume that ˆ.G/0 D H 0 for all nonabelian H 2 1 ; then ˆ.G/ is nonabelian (take a nonabelian H 2 1 ) so the set 1 has no abelian members. In that case, ˆ.G/0 D H 0 for all H 2 1 by hypothesis, so that G is minimal nonabelian (Theorem 111.3), contrary to the hypothesis. We claim that the following conditions for a group G are equivalent: (a) H 0 ˆ.G/ for all H 2 1 , (b) either G is nilpotent or G=ˆ.G/ is nonnilpotent minimal nonabelian, (c) NF .F \ H / ¤ F \ H ¤ NH .H \ F / for all distinct F; H 2 1 . Obviously, (a) , (b) and (b) ) (c). Next, (b) follows from (c) by Lemma J(h). Below we use some known results on Frobenius groups (see [BZ, 10.2]). Recall that G is said to be a Frobenius group if there is a subgroup H < G, H > ¹1º, such that H \H x D ¹1º for all x 2 G H . In that case, G D H N is a semidirect product with kernel N , Sylow subgroups of H are either cyclic or generalized quaternion. Lemma 111.5. The following conditions for a nonnilpotent group G are equivalent: (a) All members of the set 1n are abelian. (b) G=Z.G/ D .U=Z.G// .Q1 =Z.G// is a Frobenius group with elementary abelian kernel Q1 =Z.G/ D .G=Z.G//0 and a cyclic complement U=Z.G/, U 2 1 . Proof. If U; V are two distinct abelian maximal subgroups of a nonabelian group G, then U \ V D Z.G/. As we have noticed, the set 1n is nonempty. T Suppose that G satisﬁes condition (a). Given H 2 1n , set HG D x2G H x ; then HG D Z.G/ by the previous paragraph. Set GN D G=Z.G/; then GN D HN QN 1 is a Frobenius group with kernel QN 1 and complement HN . Since HN is abelian, it is cyclic. There is in QN 1 an HN -invariant Sylow subgroup by Sylow’s theorem. Therefore, since N the subgroup QN 1 is a p-group; moreover, QN 1 is a minimal normal HN is maximal in G, subgroup of GN so it is elementary abelian. All nonnormal maximal subgroups of GN are conjugate with HN (Schur–Zassenhaus) so all members of the set 1n are conjugate in G (indeed, all members of the set 1n , being abelian, contain Z.G/). Thus (a) ) (b). Conversely, every group as in (b) satisﬁes condition (a). In fact, if H 2 1n , then Z.G/ < H . By (b), H=Z.G/ is cyclic so H is abelian. The proof of Lemma 111.5 shows that if the set 1n has at least one abelian member, then the group G has the same structure as in Lemma 111.5(b). Corollary 111.6. A nonnilpotent group G satisﬁes ˆ.G/0 D H 0 for all H 2 1n if and only if GN D G=ˆ.G/0 is as in Lemma 111.5(b).

138

Groups of prime power order

Proof. Suppose GN D G=ˆ.G/0 is as in Lemma 111.5(b) and HN is a nonnormal maxiN then HN is cyclic by hypothesis, so H 0 ˆ.G/0 . As ˆ.G/ < H , mal subgroup of G; 0 we get ˆ.G/ H 0 so H 0 D ˆ.G/0 and whence G satisﬁes the hypothesis. Now suppose that ˆ.G/0 D H 0 for all H 2 1n . Then all nonnormal maximal subgroups of G=ˆ.G/0 are abelian so G=ˆ.G/0 is as in Lemma 111.5(b). Proposition 111.7. Suppose that a nonnilpotent group G is not minimal nonabelian and such that M 0 D ˆ.G/ for all nonabelian M 2 1 . Then (a) G D P Q is a semidirect product with kernel Q D G 0 2 Sylq .G/, P 2 Sylp .G/ is of order p, ˆ.G/ D ˆ.Q/ > ¹1º. Set H D P ˆ.G/. The set 1 consists of two conjugacy classes of subgroups with representatives H and Q. (b) P G D G, where P G is the normal closure of P in G. (c) G=ˆ.G/ is minimal nonabelian so that Q=ˆ.G/ is a minimal normal subgroup of G=ˆ.G/. (d) H 0 < ˆ.G/ if and only if G is minimal nonnilpotent. (e) If ˆ.G/ is abelian, then G is either minimal nonnilpotent or a Frobenius group. Proof. By hypothesis, all maximal subgroups of G=ˆ.G/ are abelian so it is nonnilpotent and minimal nonabelian by Lemma J(d,c). We obtain that ˆ.G=ˆ.G// D ¹1º so that jG=ˆ.G/j D pq b , where a subgroup of order p is not normal in G=ˆ.G/ (Lemma J(b,c)). By hypothesis, the set 1 has a nonabelian member so ˆ.G/ > ¹1º. Lemma J(g) yields .G/ D .G=ˆ.G// D ¹p; qº. Due to Lemma J(f), Q 2 Sylq .G/ is normal in G. We have G D P Q, where P 2 Sylp .G/. From Lemma J(a) we deduce ˆ.Q/ D Q \ ˆ.G/. By the above, ˆ.G/ D P1 ˆ.Q/, where P1 is a maximal subgroup of P . The subgroup H D P ˆ.Q/ 2 1 and all nonnormal maximal subgroups of G are conjugate with H . Since P1 Q=Q is the unique normal maximal subgroup of G=Q, it follows that G=Q Š P is cyclic. Let, as above, P1 < P be maximal; then, as we have noticed, P1 ˆ.G/. The subgroup P1 Q D P1 Q D M 2 1 . Assume that P1 > ¹1º and M is nonabelian. We have P1 6 M 0 D ˆ.G/, a contradiction. Thus, if P1 > ¹1º, then M is abelian. Similarly, P1 6 .P ˆ.G//0 D H 0 so that H is also abelian. It follows that G is minimal nonabelian, contrary to the hypothesis. Thus we get P1 D ¹1º so that jP j D p and Q 2 1 . This completes the proof of (a). By Lemma J(a), ˆ.G/ D ˆ.Q/. Assume that there is L G G of index q. Since L 2 1 , we get ˆ.G/ < L, and this is a contradiction since G=ˆ.G/ has no normal subgroup of index q. It follows that P G D G, and the proof of (b) is complete. Suppose that H 0 < ˆ.G/. Then H is abelian by hypothesis. In this case, we have CG .ˆ.G// H G P G D G so that ˆ.G/ D Z.G/ since Z.G=ˆ.G// D ¹1º, and we conclude that G is minimal nonnilpotent since G=ˆ.G/ is minimal nonabelian. The proof of (d) is complete. Assume that ˆ.Q/.D ˆ.G// is abelian and G is not minimal nonnilpotent. Then, by (d), H D P ˆ.Q/ is nonabelian so H 0 D ˆ.Q/ by hypothesis. In this case, due to

111

Characterization of abelian and minimal nonabelian groups

139

Lemma J(j), NH .P / D P so that H is a Frobenius group as jP j D p. Since G=ˆ.Q/ is a Frobenius group, it follows that G is also a Frobenius group (in the case under consideration, p > 2, by Burnside). This completes the proof of (e) and, thereby, of the proposition. Let a nonabelian group G be such that H 0 D ˆ.H / for all nonabelian H G. Then G has no minimal nonnilpotent subgroups by Lemma J(d). It follows that the group G is nilpotent. Let A G be minimal nonabelian. Then A is a p-subgroup for some prime p and A0 D ˆ.A/ has order p and index p 2 in A (Lemma J(b)) so that jAj D jA0 jp 2 D p 3 . Thus all minimal nonabelian subgroups of G have order equal to the cube of a prime. Suppose that P 2 Syl2 .G/ is nonabelian. Then P is among 2-groups described in 90. Now suppose that G D Q H , where Q 2 Sylq .G/ is nonabelian and H > ¹1º. Assume that Z H is cyclic of order p 2 . Then K D QZ does not satisfy the hypothesis since ˆ.K/ 6 K 0 . Thus either G is a prime power or G D Q H , where Q 2 Sylq .G/ is nonabelian and exp.H / > 1 is square free. It follows that if H is also nonabelian, then exp.Q/ D q and we conclude that q > 2.

Problems Problem 1. Study the nonabelian p-groups G of exponent > p all of whose subgroups H G satisfy Ã1 .ˆ.H // D Ã1 .H /. (Obviously, we must have p > 2 since for any 2-group X we have ˆ.X/ D Ã1 .X/.) Problem 2. Classify the p-groups G satisfying H 0 D ˆ.G/0 for all those H G that are neither abelian nor minimal nonabelian. Problem 3. Study the p-groups G which satisfy H 0 D ˆ.G/0 for all H 2 i , where i 2 ¹1; : : : ; d.G/ 1º is ﬁxed. (For i D 1, see Theorem 111.3.)

112

Non-Dedekindian p-groups all of whose nonnormal subgroups have the same order

In this section we classify the groups G with the title property. If a subgroup H < G is nonnormal, it is cyclic since it is not generated by its (G-invariant) maximal subgroups. The p-groups all of whose nonnormal subgroups are cyclic are classiﬁed in Theorem 16.2. Therefore Theorems 112.3 and 112.4 are partial cases of Theorem 16.2. However, our proofs of the theorems are essentially shorter and not so involved. Moreover, our approach is entirely different. Therefore we decided to present the proofs of Theorems 112.3 and 112.4. We also classify the nonnilpotent groups with the title property. Theorem 112.3 was proved by D. Passman [Pas], and Theorem 112.4 was also proved by G. Zappa [Zap] (his proof is essentially longer). If G is a non-Dedekindian minimal nonabelian p-group, then (see Exercise 1.8(a) or Lemma 65.1) G D ha; bi, where m

n

m

n

(MA1) ap D b p D c p D 1, Œa; b D c, Œa; c D Œb; c D 1, jGj D p mCnC1 , G D hbi .hai hci/ D hai .hbi hci/ is nonmetacyclic. (MA2) ap D b p D 1, ab D a1Cp cyclic.

m1

, jGj D p mCn and G D hbi hai is meta-

If a minimal nonabelian p-group is Dedekindian, then it is isomorphic to Q8 . If G is a nonnilpotent minimal nonabelian group, then (Miller–Moreno [MM]) (MNA) G D P Q, where P 2 Sylp .G/ is cyclic, G 0 D Q 2 Sylq .G/ is an elementary abelian minimal normal subgroups (p; q are distinct primes), Z.G/ D Ã1 .P /, jQj D q b , where b is a minimal positive integer such that p j q b 1 hence b is the order of q modulo p. The group from (MNA) is said to be an A.p; q/-group. Below we freely use the following fact. The diagonal subgroup X of the direct product G D A B of two isomorphic nonabelian groups A and B is not G-invariant. Indeed, assume that this is false; then A \ X D B \ X D ¹1º so CG .X/ AB D G and X 2 Z.G/ is abelian, a contradiction since X Š A and A is nonabelian. The following lemma may be useful in proofs of such results as Theorems 112.2 and 112.4. For its proof, see Lemma 96.1.

112

Nonnormal subgroups have the same order

141

Lemma 112.1 (= Lemma 96.1). Let G D ha; bi be a minimal nonabelian p-group as in (MA1) or (MA2). (a) All nonnormal cyclic subgroups of G have the same order if and only if one of the following holds: (a1) G is as in (MA2) with m n. (a2) G is as in (MA1) with m D n. (b) All nonnormal subgroups of G have the same order if and only if one of the following holds: (b1) G 2 ¹D8 ; S.p 3 /º. (b2) G is as in (MA2) with m n. (c) All nonnormal cyclic subgroups of G of same order are conjugate if and only if G Š Mpt . (d) All nonnormal subgroups of G are conjugate if and only if G Š Mpt . Let nc.G/ denote the number of conjugate classes of nonnormal subgroups of a group G. If G is nilpotent and nc.G/ D 1, then G is primary. Therefore, in Theorem 112.2, we conﬁne to p-groups. Theorem 112.2 (O. Schmidt [Sch3]). Let G be a p-group. If nc.G/ D 1, then we have G Š MpnC1 . Proof. See Lemma 96.4. Theorem 112.3 (Passman [Pas, Proposition 2.4] = Theorem 1.25). If all nonnormal subgroups of a non-Dedekindian p-group G of order > p 3 have the same order p, then one and only one of the following holds: (a) G Š Mpn . (b) G D Z.G/1 .G/, where Z.G/ is cyclic and 1 .G/ is nonabelian of order p 3 so 1 .G/ 2 ¹S.p 3 /; D8 º (note, that D8 Z.G/ Š Q8 Z.G/). (c) The group G D Q D is extraspecial of order 25 , where Q Š Q8 and D Š D8 . Proof. It is easy to check that the groups (a)–(c) satisfy the hypothesis. By Proposition 1.23 (see also Lemma 96.2), jG 0 j D p (otherwise, there is in G 0 a G-invariant subgroup R of index p such that G=R contains a nonnormal subgroup K=R; then K is not G-invariant of order > p, contrary to the hypothesis). If G is minimal nonabelian, then either jGj D p 3 or G Š Mpn , n > 3, by Lemma 112.1(a). Next we assume that G is not minimal nonabelian. Let U < G be minimal nonabelian; then U 0 D G 0 so U G G. In view of Lemma 4.3, G D U C , where C D CG .U /. Lemma 112.1(b) yields U 2 ¹Q8 ; D8 ; MpnC1 ; S.p 3 /º since, if U is non-Dedekindian, it satisﬁes the hypothesis of that lemma. By assumption, C 6 U . Let K < G be nonnormal of order p; then K is contained in exactly one subgroup of G of order p 2 (otherwise, K G G). Considering K1 .Z.G//, we con-

142

Groups of prime power order

clude that Z.G/ is cyclic and so G has no subgroup Š Ep3 (indeed, if such a subgroup exists, it must lie in Z.G/). Next we consider all four possibilities for U . (i) Let U Š D8 . Assume that C has a subgroup, say V , of order 2 that satisﬁes V ¤ Z.U /. If A < U is nonnormal, then A V Š E4 is not G-invariant (indeed, .A V / \ U D A is not U -invariant), a contradiction. Thus C is either cyclic or Š Q8 (recall that jG 0 j D 2) so G D U C is a group from (b) or (c). In what follows we assume that G has no subgroups Š D8 . (ii) Let U Š Q8 . Since G is not Dedekindian, we get exp.C / > 2 by Theorem 1.20. Let L C be cyclic of order 4; write H D U L. Assume that U \ L D ¹1º; then H D U L. Since 1 .H / < Z.H /, exp.H / D 4 and H is non-Dedekindian (Theorem 1.20), it has a nonnormal cyclic subgroup of order 4, a contradiction. Thus jH j D 16. Then H D U L has a subgroup Š D8 (see Appendix 16), contrary to the assumption. Next we assume that G has no subgroups Š Q8 . (iii) Let U Š S.p 3 /, p > 2. As in (i), C is cyclic (of order > p with U \C D Z.U / since jGj > p 3 ), and we get a group from (b) since then U D 1 .G/ by the regularity of G. In what follows we assume that G has no subgroups Š S.p 3 /. (iv) Suppose that U Š MpnC1 . Assume that there is K < G of order p such that K 6 U . Put H D 1 .U /K; then H of order p 3 is generated by elements of order p. It follows from (i) and (iii) that H Š Ep3 . In this case, H has a non-G-invariant subgroup of order p 2 (otherwise, H Z.G/, which is a contradiction). Thus we obtain 1 .G/ D 1 .U / Š Ep2 and C is cyclic by (ii). There is S C such that S 6 U but Ã1 .S/ < U . Let T < U be cyclic of index p. Then U \ S Z.U / D ˆ.U / < T , jT j jS j so (the abelian group) T S D T L, where L of order p, and L is not contained in U (otherwise, S < U ), a ﬁnal contradiction. Let G be a non-Dedekindian p-group such that jH=HG j p for all H < G. If a subgroup H < G is nonnormal of maximal order, then G=HG is one of groups of Theorem 112.3. Let us show how to prove Theorem 112.2 making use of Theorem 112.3. Suppose that H < G is nonnormal; then H is cyclic. Set jH j D p s . If s D 1, then G Š MpnC1 (Theorem 112.3). If s > 1, then G=1 .H / Š Mpm by induction, and the proof is ﬁnished as in Theorem 96.4. D. Passman [Pas, Theorem 2.9] considered the p-groups without nonnormal subgroups U < V such that jV W U j D p. Suppose that, as in [Pas, Theorem 2.9], a non-Dedekindian p-group G has no two nonnormal subgroups U < V with jV W U j D p. Let H < G be nonnormal of maximal order. Then H is cyclic since all its maximal subgroups are G-invariant, and G=HG is as in Theorem 112.2.1 1 As

Passman [Pas] has noticed, the converse is also true, i.e., if all nonnormal subgroups of G are cyclic, then it has no chain U < V with nonnormal U; V such that jV W U j D p. Therefore Theorem 2.9

112

Nonnormal subgroups have the same order

143

We consider an important partial case of [Pas, Theorem 2.9]. Note that the statement of Theorem 112.4 is neither stated in [Pas] nor follows from [Pas, Theorem 2.9] (since in the last theorem groups of order < 28 were not considered). For a complete classiﬁcation of groups from [Pas, Theorem 2.9], see Theorem 16.2. Now we are ready to prove the main result of this section. Theorem 112.4. Suppose that all nonnormal subgroups of a p-group G have the same order p > p; then 1 .G/ D 1 .Z.G//, j1 .G/j p 2 . If j1 .G/j D p, then G Š Q16 , D 2. Now we let 1 .G/ Š Ep2 ; then G=1 .G/ is abelian. Let H < G be nonnormal; then H is cyclic, G=HG is as in Theorem 112.3 and one and only one of the hollowing holds: m

n

(a) G D ha; b j ap D b p D 1; ab D a1Cp m n D > 1.

m1

i is metacyclic minimal nonabelian,

(b) G D Q L, where Q Š Q8 , L Š C4 , and D 2. (c) G is minimal nonmetacyclic and special of order 25 , D 2.2 (d) G is special of order 26 with Z.G/ D G 0 D ˆ.G/ D 1 .G/ Š E4 , D 2.3 Proof. By hypothesis, jGj > p 3 . The hypothesis is inherited by non-Dedekindian subgroups. Note that all nonnormal subgroups of any proper non-Dedekindian epimorphic image of G are also of same order (which, however, is < p ). We have 1 .G/ Z.G/ since > 1. Let H < G be nonnormal; then all maximal subgroups of H are G-invariant so they do not generate H , and we infer that H is cyclic. If U=HG < G=HG is not normal, then we have jU j D jH j so jU=HG j D p, and hence G=HG is as in Theorem 112.3. Next, H is contained in exactly one (G-invariant) subgroup of G of order p C1 so NG .H /=H has exactly one subgroup of order p. We assume that j1 .G/j > p (if j1 .G/j D p, then G Š Q16 ). It follows that H 1 .Z.G//=H is cyclic so j1 .Z.G//j D p 2 since H is cyclic and j1 .G/j > p. If G is minimal nonabelian, it is metacyclic as in Lemma 112.3(b2) with m n > 1. Next we assume that G is not minimal nonabelian. If G is of maximal class, then j1 .G/j D p since 1 .G/ Z.G/; then, as we know, G Š Q16 . Next we assume that G is not of maximal class; then 1 .G/ D 1 .Z.G// Š Ep2 and so G=1 .G/ is Dedekindian since all subgroups of G containing 1 .G/ are noncyclic so G-invariant. from [Pas] yields an almost full classiﬁcation of p-groups all of whose nonnormal subgroups are cyclic. See 16. 2 Deﬁning relations of G, namely, G D ha; b; c j a4 D b 4 D Œa; b D 1; c 2 D a2 ; ac D ab 2 ; b c D ba2 i; are given in Theorem 66.1. 3 Deﬁning relations of G, namely, G D ha; b; c; d j a4 D b 4 D Œa; b D 1; c 2 D a2 b 2 ; ac D a1 ; b c D a2 b 1 ; d 2 D a2 ; ad D a1 b 2 ; b d D b 1 ; Œc; d D 1i; are given in Theorem 16.2(ix).

144

Groups of prime power order

Assume that G=1 .G/ is nonabelian. By the previous paragraph, there exists Q8 Š Q=1 .G/ G=1 .G/ for some Q G and all maximal subgroups of Q containing 1 .G/ are abelian since 1 .G/ Z.G/; then jZ.Q/j D 8 and jQ0 j D 2 (Lemma 1.4). In this case, Q0 < 1 .G/, a contradiction since Q=1 .G/ Š Q8 is nonabelian. Thus G=1 .G/ is abelian so G 0 1 .G/. We conclude that G 0 is of order p 2 and exponent p. (a) By the previous paragraph, G 0 1 .G/ Š Ep2 , and since 1 .G/ Z.G/, we get cl.G/ D 2. If d.G/ D 2, then G is minimal nonabelian (Lemma 65.2(a)), contrary to the assumption. Thus we have d.G/ > 2 so G is not metacyclic. If p > 2, then, since .Ep2 Š/1 .G/ Z.G/, G has no minimal nonmetacyclic subgroups (Lemma 69.1(a,b)) so it is metacyclic, d.G/ D 2, contrary to what has just been said. Next we assume that p D 2 and d.G/ > 2. (b) Suppose that Q < G is nonabelian of order 8; then Q Š Q8 as 1 .Q/ Z.G/, and QGG since Q is noncyclic. Write C D CG .Q/; then G=C is isomorphic to a noncyclic subgroup of D8 , a Sylow 2-subgroup of Aut.Q/ Š S4 , the symmetric group of degree 4. As G 0 Z.G/ C , we get G=C Š E4 so G D Q C is a central product by the product formula since Q\C D Z.Q/ has index 4 in Q. Since G is not Dedekindian, there is a cyclic L C of order > 2. If Q \ L > ¹1º, then Q L has a nonnormal subgroup of order 2 < 2 , a contradiction. Thus Q \ L D ¹1º. Let elements u; v 2 Q generate different cyclic subgroups of order 4 and let x be a generator of L. If o.x/ > 4, then huxi and hux 2 i of different orders are not v-invariant, a contradiction. Thus exp.C / D 4 and so D 2. Suppose that C contains a minimal nonabelian subgroup, say D; then we obtain j1 .D/j j1 .G/j D 4 so D is metacyclic by Lemma 65.1; then jDj 16 since exp.D/ exp.C / D 4. If jDj D 8, then D Š Q8 and QD D Q D since Q \ L D ¹1º for a cyclic L < C of order 4. Then the diagonal subgroup of Q D (of order 8 > 4) is not normal in Q D, a contradiction. Thus D Š H2;2 D ha; b j a4 D b 4 ; ab D a3 i, the unique nonabelian metacyclic group of order 16 and exponent 4. Then Q \ D D Z.Q/ D hzi is a maximal cyclic subgroup of D of order 2 since j1 .G/j D 4; we have z D a2 b 2 . To obtain a contradiction, one may assume that G D Q D. Let u and v be generators of distinct cyclic subgroups of Q of order 4. Put H D hub; ai. We have .ub/2 D u2 b 2 D zb 2 D a2 b 2 b 2 D a2 ;

Œub; a D Œb; a D a2

so H Š Q8 . Since Œub; v D Œu; v D z ¤ a2 (indeed, Q \ hai D ¹1º), we get z 62 H , so H \ Q D ¹1º. As Q; H G G, we get F D QH D Q H . Then the diagonal subgroup X of F is not normal in F , a contradiction since jXj D 8. Thus D does not exist so C is abelian and d.C / 2 as 1 .C / 1 .Z.G// Š E4 . If C is abelian of type .4; 4/, there is in C a cyclic subgroup L of order 4 containing Q \ C D Z.Q/, contrary to what has already been proved. Thus C must be abelian

112

Nonnormal subgroups have the same order

145

of type .4; 2/ so, if C4 Š L1 < C , then L1 \ Q D ¹1º and G D Q L1 (indeed, by the product formula, jGj D 25 D jQ L1 j). In what follows we assume that G has no nonabelian subgroups of order 8. Let us show that D 2. Indeed, by (a), d.G/ > 2. Then G=G 0 contains a subgroup E=G 0 Š E8 . Since d.E/ > 2, E is neither abelian (since j1 .G/j D 4) nor minimal nonabelian. Let T < E be minimal nonabelian. We claim that T is metacyclic of order 16 and exponent 4. Indeed, 8 < jT j < jEj D jG 0 jjE=G 0 j 25

and 1 .T / 1 .G/ Š E4

so T is metacyclic (see (MA1) and (MA2)) and jT j D 16 and hence T Š H2;2 since exp.T / D 4. Since T has a nonnormal (cyclic) subgroup of order 22 , we get D 2. Then H Š C4 , jHG j D 2 and C2 Š HG ˆ.G/. By the above, GN D G=HG is a group from Theorem 112.3. (c) Let GN D G=HG D QN CN , where QN Š Q8 , QN \ CN D 1 .CN / and CN Š C2n , n > 1. Suppose that M < G is minimal nonmetacyclic. Since M has no nonabelian subgroup of order 8 and 1 .G/ Š E4 , we have jM j D 25 (Theorem 66.1). In this case, N D QN 2 .CN / has order 16 so 2 .G/ L and jLj D jHG jjLj N D 25 . LN D 2 .G/ 5 Since 2 .M / D M has order 2 (Theorem 66.1), we get L D M so 2 .G/ D M . Suppose that .2 .G/ D/M < G. In this case, there is in G a cyclic subgroup V of order 8. Then V G G (since D 2 by (b)) and Z D V \ M is cyclic of order 4 and G-invariant. It follows from the hypothesis and G 0 D M 0 Š E4 that G=Z is nonabelian and Dedekindian since D 2 and E4 Š G 0 6 Z; next, by assumption, we have jG=Zj > 8 (indeed, jGj > jM j D 25 ). Therefore M=Z Š Q8 so there are exactly four maximal subgroups of M that does not contain Z (indeed, M contains seven maximal subgroups and only three of them contain Z). Let K be one of such subgroups that is nonabelian (K exists since M has at most one abelian maximal subgroup in view of jM 0 j D 4). Then K \ V D 1 .V / as Z < V . Since M=1 .V / is nonabelian, it contains at most three abelian maximal subgroups by Exercise 1.6(a). Therefore one may assume that K is chosen so that K=1 .V / is nonabelian. We have K V =1 .V / D .K=1 .V // .V =1 .V //, and this group is non-Dedekindian by Theorem 1.20 since jV =1 .V /j D 4. Therefore since K=1 .V / 2 ¹D8 ; Q8 º, the group K V =1 .V / (of exponent 4) possesses a nonnormal subgroup, say T =1 .V /, of order 4.4 Since jT j D 8 > 4, we get a contradiction. Thus G D M is minimal nonmetacyclic as in part (c) of the theorem. (d) Suppose that G=HG D GN D DN QN is extraspecial of order 25 , where DN Š D8 and QN Š Q8 . Since HG < Z.G/ and GN is a monolith, we get G=Z.G/ Š E16 so jˆ.G/j 4 and exp.G/ D 4. Suppose that M < G is minimal nonmetacyclic. As in 4 This is clear if K= .V / Š Q (in that case, all subgroups of order 2 are normal there). In case 1 8 K=1 .V / Š D8 and R=1 .V / < K=1 .V / is noncentral so of order 2, R=1 .V / 1 .V =1 .V // of order 4 is not normal in K=1 .V /.

146

Groups of prime power order

part (c), we obtain jM j D 25 and M is special. Then Z.M / D M 0 D ˆ.M / Š E4 . It follows that ˆ.G/ D ˆ.M / so Z.G/ D ˆ.G/ D G 0 D M 0 and hence G is special of order 26 . If G is as in (c) or (d), then every subgroup of G of order > 4 contains 1 .G/ D G 0 so is G-invariant. This means that this G satisﬁes the hypothesis.5 Note that the central product G D U C , where U D ha; b j a4 D b 4 D 1; ab D a3 i Š H2;2

and C D hci

such that a2 D c 2 , has nonnormal subgroups of order 4 and 2 (for example, hbi and haci), however G=ha2 b 2 i Š D8 C4 . For completeness, we prove the following Proposition 112.5. Let G be a nonnilpotent group and a positive integer n be ﬁxed. Suppose that .H / D n for all nonnormal H < G.6 Then G D PQ is minimal nonabelian as in (MNA), where P 2 Sylp .G/ is cyclic, G 0 D Q 2 Sylq .G/, and one of the following holds: (a) jP j D p n , jQj D q. (b) n D 1, jP j D p, Q Š Eq 2 . Proof. Groups (a) and (b) satisfy the hypothesis. Now let M G be minimal nonnilpotent. Then M D P Q, where P 2 Sylp .M / is cyclic and Q D G 0 2 Sylq .M / (see Lemma 10.8). Since P is not normal in M so neither in G and .M / ¤ .P /, we get jP j D p n and so M G G. Then Q G G since Q is characteristic in M . Since a maximal subgroup of M containing P is not normal in M , it follows that P is maximal in M so Q is a minimal normal subgroup of M and M is minimal nonabelian. By Frattini’s lemma, G D NG .P /M D NG .P / Q is a semidirect product with kernel Q. As NG .P / is not normal in G, we get NG .P / D P so G D NG .P /Q D PQ D M . Thus G is a minimal nonabelian group. Let n > 1. Then jQj D q (otherwise, G also has a nonnormal subgroup of order q and .q/ D 1 < n) so jGj D p n q. Let n D 1 and jQj > q. Then Q has no proper subgroup of order q 2 so jQj D q 2 . Also, G of order pq 2 satisﬁes the hypothesis (this group is as in (b)). Proposition 112.5 is a generalization of the nonnilpotent part of the following. Corollary 112.6 (O. Schmidt [Sch3]). If G is a nonnilpotent group with nc.G/ D 1, then G is minimal nonabelian as in Proposition 112.5(a). 5 It is noticed in Theorem 16.2(ix) that all maximal subgroups of the group G from part (d) are minimal nonmetacyclic. Let us prove this. If S 2 1 , then S (of order 25 ) is nonmetacyclic since exp.G/ D 4. If M S is minimal nonmetacyclic, then, as in the ﬁrst paragraph of (d), we get jM j D 25 D jS j so M D S. 6 Recall that .H / is the number of prime factors of n, i.e., ƒ.H / D a C C a provided that 1 k Q jH j D kiD1 p ai is the standard decomposition.

112

Nonnormal subgroups have the same order

147

The following proposition is also a particular case of Theorem 16.2. Proposition 112.7. Let G be a non-Dedekindian two-generator p-group of order > p 3 with cyclic derived subgroup G 0 . If all nonnormal subgroups of G are cyclic, then one of the following holds: (a) G is minimal nonabelian and metacyclic as in (MA2). (b) p D 2 and either G Š Q16 or G D ha; b j a8 D 1; a4 D b 4 ; ab D a3 i is of order 25 . Conversely, all nonnormal subgroups of groups from (a) and (b) are cyclic. Proof. First we check the last assertion. Let G be minimal nonabelian metacyclic as in (MA2). Then G 0 < 1 .G/ Š Ep2 so all nonnormal subgroups of G do not contain 1 .G/. If H < G is nonnormal, then j1 .H /j D p hence H , being abelian, is cyclic (Proposition 1.3) so the groups from (a) satisfy the hypothesis. Now assume that G is as in (b). Clearly, Q16 satisﬁes the hypothesis. Now let G be the second group in (b). If H < G is nonnormal and noncyclic, H 6 2 .G/ as 1 .G/ D R G G is the unique proper noncyclic subgroup of 2 .G/ Š C4 C2 . Thus exp.H / D 8.D exp.G// so H 2 1 , a contradiction. Next we ﬁnd the groups satisfying the hypothesis. Suppose that G is not a 2-group of maximal class (otherwise, G Š Q16 ). In this case, there is in G a normal subgroup R Š Ep2 ; then all subgroups of G=R are normal so G=R is Dedekindian. If p > 2, then G=R is abelian. Since G 0 R and G 0 is cyclic, we get jG 0 j D p so G is minimal nonabelian (Lemma 65.2(a)). In this case, G is metacyclic (otherwise, G has a nonnormal abelian subgroup of type .p; p/ which is noncyclic). Next we assume that p D 2 and G is not minimal nonabelian (see the ﬁrst paragraph); then we have jG 0 j > 2 (Lemma 65.2(a)). It remains to show that then G is as the second group in (b). We have G 0 6 R so G=R is nonabelian Dedekindian, and since d.G=R/ D 2, we get G=R Š Q8 , jGj D 25 and R < ˆ.G/. Assume that G has a subgroup E Š E8 ; then E Z.G/ since all subgroups of E of order 4 are G-invariant. In this case, G is minimal nonabelian, a contradiction. Thus G has no subgroup Š E8 so j2 .G/j D 8 and G (of order 25 ) is as in Lemma 42.1(c) and hence has deﬁning relations given in (b). Remark. Let p > 2 and let G be a nonabelian p-group all of whose nonnormal subgroups are cyclic. As in the proofs of Theorems 112.3 and 112.4, G has no subgroup Š Ep3 so G is a group from Theorem 13.7. If R G G is abelian of type .p; p/, then G=R is abelian by Theorem 1.20. If G is metacyclic, then, being cyclic, jG 0 j D p and G is minimal nonabelian by Lemma 65.2(a). If G is a 3-group of maximal class and jGj > 33 , then jGj D 34 , 1 .G/ Š E9 . If G D EC , where E D 1 .G/ Š S.p 3 / and C is cyclic, then all subgroups of G of type .p; p/ are normal so jG 0 j D p. Then, by Lemma 4.3, G D E CG .E/, where CG .E/ is cyclic. Problem. Classify the p-groups that have no three nonnormal subgroups of pairwise distinct orders.

113

The class of 2-groups in 70 is not bounded

Suppose that all maximal subgroups of a nonabelian p-group G are generated by two elements but d.G/ D 3. By Theorem 70.2, if such a group G is of class > 2, then p D 2, jGj 27 , and G=K3 .G/ is isomorphic to the Suzuki group of order 26 given in Theorem 70.1(d). If G is of class 3, then these groups are completely determined in 70, where in particular for p > 2, jGj p 5 and G is of class 2 and for p D 2, jGj 28 . Theorem 113.1 (E. Crestani). Let G be a 2-group all of whose maximal subgroups are generated by two elements but d.G/ D 3. Then the nilpotence class of G is not bounded. Proof. Eleonora Crestani from University of Padova has invented the following ingenious construction of a 2-group Gn of order 24nC4 for each ﬁxed n 2 such that d.Gn / D 3, each maximal subgroup of Gn is generated by two elements and the class of Gn is at least n. Therefore the nilpotence class of such groups is not bounded. In fact, E. Crestani has also proved that the nilpotence class of Gn is exactly 2n, but that would require further tedious computations. In order to construct our 2-group Gn for each ﬁxed n 2, we start with the homocyclic group H D hg4 ; g5 ; g6 ; g7 i Š C2n C2n C2n C2n of order 24n . Let K be the splitting extension of H by the cyclic group hg3 i Š C4 of order 4, where g3 induces the automorphism of order 4 on H given by (1) g34 D 1;

g4g3 D g41 g62 ;

g5g3 D g51 g72 ;

g6g3 D g6 g41 ;

g7g3 D g7 g51 :

g2

We compute gi 3 D gi1 for i D 4; 5; 6; 7 and so g32 inverts H which implies that all elements in Hg32 are involutions. Let L D Khg2 i be the cyclic extension of order 2 of K, where (2)

g22 D g6 g72 ;

g4g2 D g41 g62 g74 ; g5g2 D g51 g62 g74 ;

g6g2 D g6 g42 g52 ; g7g2 D g7 g4 g5 ;

g3g2 D g31 g7 :

Here we have to show that g2 induces an automorphism on K D hg3 ; g4 ; g5 ; g6 ; g7 i. It is enough to see that gig2 , i D 3; 4; : : : ; 7, generate K (which is obvious) and then

113

149

The class of 2-groups in 70 is not bounded

we have to show that the relations (1) for K remain invariant when we replace gi with gig2 , i D 3; : : : ; 7. For example, .g3g2 /2 D .g31 g7 /2 2 Hg32 (since K=H Š C4 ) and we know that all elements in Hg32 are involutions and so o.g31 g7 / D 4. Then we g2 verify .g4g2 /g3 D .g4g2 /1 .g6g2 /2 . Indeed, g2

1

.g4g2 /g3 D .g41 g62 g74 /g3

g7

2

D .g41 g62 g74 /.g3 g3 g7 /

D .g4 g62 g74 /g3 g7 D g41 g62 g62 g42 g74 g54 D g43 g54 g64 g74 ; and for the right-hand side we get the same result: .g4g2 /1 .g6g2 /2 D g4 g62 g74 g62 g44 g54 D g43 g54 g64 g74 : In the same way we show that the last three relations in (1) remain invariant under the above replacement. Then we have to show that g22 acts the same way on K as the inner automorphism of K induced by g6 g72 (since g22 D g6 g72 ). First we compute that g22 ﬁxes each of the elements g4 ; g5 ; g6 ; g7 in H and then we show that g2

g g72

g 3 2 D g3 6

:

Finally, Gn D Lhg1 i, where (3)

g12 D g32 g61 g73 ; g4g1 D g41 g62 g76 ; g5g1 D g51 g62 g74 ; g6g1 D g6 g42 g53 ; g7g1 D g7 g4 g5 ;

g2g1 D g2 g32 ;

g3g1 D g31 g6 :

We proceed in exactly the same way as in the previous paragraph. First we show that g1 induces an automorphism on L D hg2 ; g3 ; g4 ; g5 ; g6 ; g7 i, where obviously gig1 , i D 2; : : : ; 7, generate L. We only check that the relations (1) and (2) for L remain invariant when we replace all gi with gig1 , i D 2; 3; : : : ; 7. Then we check that g12 acts the same way on the generators g2 ; g3 ; : : : ; g7 of L as the inner automorphism of L induced by g32 g61 g73 (since g12 D g32 g61 g73 ). The group Gn is constructed. Our group Gn exists because it is obtained with three consecutive cyclic extensions (of orders 4, 2, 2) of the homocyclic group H Š C2n C2n C2n C2n , n 2. From our relations (1), (2) and (3) we see that the subgroups H; K; L are normal in Gn . It is easy to see that the nilpotence class of Gn is at least n. Indeed, the involution g32 inverts the cyclic subgroup hg4 i of order 2n and so hg4 ; g32 i Š D2nC1 is a dihedral group of order 2nC1 and class n. Hence Gn is of class at least n. We show that Gn0 D hg32 ; g4 ; g5 ; g6 ; g7 i D H hg32 i and since G=Gn0 Š E8 , we also have ˆ.Gn / D Gn0 , d.Gn / D 3 and Gn D hg1 ; g2 ; g3 i. Indeed, H D hg4 ; g5 ; g6 ; g7 i and K D H hg3 i are normal in Gn and so H hg32 i is also normal in G as K=H Š C4 . From our relations (1), (2), (3) we get g12 ; g22 ; g32 2 H hg32 i and Œg2 ; g1 D g32 ; Œg3 ; g1 D g32 g6 ; Œg3 ; g2 D g32 g7 ; Œg3 ; g6 D g4 ; Œg3 ; g7 D g5 ; which shows that Gn =.H hg32 i/ Š E8 and Gn0 D H hg32 i D ˆ.G/. Thus d.Gn / D 3 and Gn D hg1 ; g2 ; g3 i.

150

Groups of prime power order

It remains to prove that each maximal subgroup of Gn is generated by two elements. Our seven maximal subgroups of Gn are Hi D Ti Gn0 , where Ti (i D 1; 2; : : : ; 7) is one of the subgroups: hg1 ; g2 i; hg1 ; g3 i; hg2 ; g3 i; hg1 g2 ; g3 i; hg1 ; g2 g3 i; hg2 ; g1 g3 i; hg1 g2 ; g1 g3 i: Here we shall freely use the relations (1), (2), (3) (without quoting). (i) H1 D hg1 ; g2 iGn0 D hg1 ; g2 i since hg1 ; g2 i Gn0 . Indeed, we have Œg1 ; g2 D g32 ;

g12 D g32 g61 g73 ;

g22 D g6 g72

and so Œg1 ; g2 g12 g22 D g71 which shows that g32 ; g6 ; g7 2 hg1 ; g2 i. Furthermore, we have g6g1 .g72 /g1 D g6 g72 g51 and so g5 2 hg1 ; g2 i. Finally, g7g1 D g7 g5 g4 and so g4 2 hg1 ; g2 i and we are done. (ii) H2 D hg1 ; g3 iGn0 D hg1 ; g3 i since hg1 ; g3 i Gn0 which shows the following relations (in this order): Œg3 ; g1 D g32 g6 ;

g12 D g32 g61 g73 ;

g6g1 .g72 /g1 D g6 g72 g51 ;

g7g1 D g7 g5 g4 :

(iii) H3 D hg2 ; g3 iGn0 D hg2 ; g3 i since hg2 ; g3 i Gn0 which show the relations Œg3 ; g2 D g32 g7 ;

g7g3 D g7 g51 ;

g7g2 D g7 g4 g5 ;

g22 D g6 g72 :

(iv) H4 D hg1 g2 ; g3 iGn0 D hg1 g2 ; g3 i since hg1 g2 ; g3 i Gn0 . Indeed, we have .g1 g2 /2 D g5 g7 , .g5 g7 /g3 D g52 g73 and so g32 ; g5 ; g7 2 hg1 g2 ; g3 i. Further, we get Œg3 ; g1 g2 D g42 g6 .g51 g71 / and so g42 g6 2 hg1 g2 ; g3 i. Finally, we conclude .g42 g6 /g3 D g4 g63 2 hg1 g2 ; g3 i and so g4 ; g6 2 hg1 g2 ; g3 i. (v) H5 D hg1 ; g2 g3 iGn0 D hg1 ; g2 g3 i D T5 as T5 Gn0 . Indeed, Œg2 g3 ; g1 D g6 which implies that g6 2 T5 . Further, g12 D g32 g61 g73 implies that g32 g73 2 T5 , .g2 g3 /2 D g73 g51 g6 gives g73 g51 2 T5 and then also g32 g51 2 T5 . From the relations g6g1 D g6 g42 g53 and g6g2 g3 D g4 g52 g74 g63 it follows g42 g53 2 T5 , g4 g52 g74 2 T5 and then also g41 g51 g74 2 T5 . In view of .g4 g5 g74 /g1 D g43 g53 g72 2 T5 , we obtain that .g41 g51 g74 /3 .g43 g53 g72 / D g710 and so g72 2 T5 which also forces g4 g5 2 T5 . Then by the above, the relation .g4 g5 /2 .g42 g53 / D g51 gives that g5 2 T5 . But from the above facts it also follows that g4 ; g32 ; g7 2 T5 and we are done. (vi) H6 D hg2 ; g1 g3 iGn0 D hg2 ; g1 g3 i D T6 as T6 Gn0 . Indeed, Œg2 ; g1 g3 D g71 and so g7 2 T6 . From g22 D g6 g72 it follows that g6 2 T6 and .g1 g3 /2 D g32 g41 g73 yields g32 g41 2 T6 . Further, the relation g7g2 D g7 g4 g5 implies g4 g5 2 T6 . From g6g1 g3 D .g4 g53 /g63 g76 it follows g4 g53 2 T6 which together with a previous result gives g52 2 T6 . Finally, g7g1 g3 D g41 g52 g62 g73 implies with the previous results that g4 ; g5 ; g32 2 T6 and we are done. (vii) H7 D hg1 g2 ; g1 g3 iGn0 D hg1 g2 ; g1 g3 i D T7 since T7 Gn0 . Indeed, from .g1 g2 /2 D g5 g7 and .g1 g3 /2 D g32 g41 g73 we conclude that t1 D g5 g7 2 T7 and

113

The class of 2-groups in 70 is not bounded

151

t2 D g32 g41 g73 2 T7 . We compute Œg1 g2 ; g1 g3 D g32 g42 g51 g6 g71 and this gives t3 D g32 g42 g51 g6 g71 2 T7 . Further, .g5 g7 /g1 g3 D g43 g55 g64 g75 and so t4 D g43 g55 g64 g75 2 T7 . Finally, we obtain .g32 g41 g73 /g1 g2 D g32 g53 g62 g75 which implies that t5 D g32 g53 g62 g75 2 T7 . Now, the ﬁve elements t1 ; t2 ; t3 ; t4 ; t5 also lie in Gn0 and note that we have ˆ.Gn0 / D 2 2 2 2 hg4 ; g5 ; g6 ; g7 i so that Gn0 =ˆ.Gn0 / Š E25 . We claim that ht1 ; t2 ; t3 ; t4 ; t5 i D Gn0 and so Gn0 T7 , as required. Indeed, considering the cosets ti ˆ.Gn0 / (i D 1; : : : ; 5), we see that t1 ˆ.Gn0 / contains g5 g7 , t2 ˆ.Gn0 / contains g32 g4 g7 , t3 ˆ.Gn0 / contains g32 g5 g6 g7 , t4 ˆ.Gn0 / contains g4 g5 g7 and t5 ˆ.Gn0 / contains g32 g5 g7 and these new ﬁve elements are obviously linearly independent mod ˆ.Gn0 /.

114

Further counting theorems

In what follows let e.X/ be the number of solutions of x p D 1 in a p-group X and let 1 .X/ be the set of all maximal subgroups of a p-group X. In this section we closely follow [HupB, VIII.5]. Theorem 114.1 ([Bla10] = [HupB, Theorem VIII.5.1]). Let G be a p-group and let ¹1º < N 1 .Z.G//. Then X e.G=M / D .j1 .N /j 1/e.G=N / C e.G/: M 21 .N /

Proof. Let k D j1 .N /j; then jN j D .p 1/k C 1 since N is elementary abelian. If Z N is of order p, then j1 .N=Z/j D

jN j p .p 1/k C 1 p k1 jN W Zj 1 D D D : p1 p.p 1/ p.p 1/ p

Let Y D ¹.x; M / j x 2 G N; M 2 1 .N /; x p 2 M º: For x 2 G N and x p 2 N , let Yx D ¹M 2 1 .N / j x p 2 M º: Clearly, jYx j is the number of maximal subgroups of the quotient group N=hx p i. Thus p jYx j D k if x p D 1 and jYx j D k1 p if x ¤ 1 (see the ﬁrst displayed formula in the proof). Hence X X X X k1 X k1 k1 D ; jYx j D kC jYj D k C p p p p p p x x x D1

x ¤1

x D1

where x runs through the set of all elements of G N for which x p 2 N , i.e., all x with .xN /p D N . Thus the number of x 2 G N for which x p 2 N is equal to .e.G=N / 1/jN j. Hence k1 k1 jYj D .e.G/ jN j/ k C .e.G=N / 1/jN j p p (1) k1 k1 kjN j: D e.G/ k C e.G=N /jN j p p

114

153

Further counting theorems

On the other hand, suppose that M 2 1 .N /. The number of elements x 2 G N for which x p 2 M is .e.G=M / p/jM j, so X

jYj D

.2/

.e.G=M / p/

M 21 .N /

jN j : p

Comparing the two expressions for jYj in (1) and (2), we obtain e.G/

.p 1/k C 1 k1 C e.G=N /jN j kjN j D p p

X

.e.G=M / p/

M 21 .N /

jN j ; p

or, since jN j D .p 1/k C 1, k D j1 .N /j, we get e.G/

k1 jN j C e.G=N /jN j kjN j D p p

X

e.G=M / kjN j;

M 21 .N /

and we conclude that e.G/ C e.G=N /.j1 .N /j 1/ D

X

e.G=M /;

M 21 .N /

as was to be shown. Thus e.G/ D

X

e.G=M / .j1 .N /j 1/e.G=N /:

M 21 .N /

Lemma 114.2 ([HupB, Lemma VIII.5.2]). Suppose that a p-group G D AB (central product). Let D D A \ B, jDj D p and exp.A=D/ D exp.B=D/ D p. Then 1 .jAj e.A//.jBj e.B// e.G/ D e.A/e.B/ C : p p1 Proof. We have G Š .A B/=h.x; x 1 /i for a generator x of D. Thus pe.G/ D j¹.a; b/ 2 A B j ap D x i ; b p D x i for some i 2 ¹0:1; : : : ; p 1ººj: If i D 0, the number of solutions of ap D x i .D 1/ is e.A/; otherwise, it is equal to jAje.A/ since exp.A=D/ D p. Similar statements hold for B. Hence p1 jAj e.A/ jBj e.B/ p1 p1 .jAj e.A//.jBj e.B// ; D e.A/e.B/ C p1

pe.G/ D e.A/e.B/ C .p 1/

completing the proof.

154

Groups of prime power order

We have, by Lemma 114.2, 1 e.Q8 Q8 / D .2 2 C .8 2/2 / D 20; 2 1 e.D8 D8 / D .62 C .8 6/2 / D 20; 2 1 e.Q8 D8 / D .2 6 C 6 2/ D 12: 2 Theorem 114.3 ([HupB, Lemma VIII.5.3]). Suppose that G is a 2-group, d.G/ D d and jˆ.G/j D 2. Then e.G/ 2 ¹2d ; 2d ˙2d r º, where r is a positive integer satisfying 1 2r d . Further, if e.G/ D 2d ˙ 2d 2 d , then G is extraspecial. Proof. Clearly, exp.G/ D 4. If G is abelian, then e.G/ D 2d . Now let G be nonabelian. Then, by Lemma 4.2, G D EZ.G/, where E is extraspecial. Let E D ES.m; 2/ and d.Z.G// D s; then d D 2m C s , where D 0 if exp.Z.G// D 4 and D 1 if exp.Z.G// D 2. We have cd.E/ D ¹ .1/ j 2 Irr.G/º D ¹1; 2m º: By the Frobenius–Schur formula (see [BZ, 4.6]), we get e.E/ D 22m ˙ 2m (indeed, Irr1 .E/ D ¹ º, where .1/ D 2m ). Suppose that exp.Z.G// D 2. Then G D E T , where T Š E2s1 , and so (3)

e.G/ D e.E/e.T / D .22m ˙ 2m /2s1 D 22mCs1 ˙ 2mCs1 :

In this case, d D 2m C s 1;

m C s 1 D .2m C s 1/ m D d m;

2m d:

Now suppose that exp.Z.G// D 4. Then Z.G/ D C4 E2s1 , d.Z.G// D s and e.Z.G// D 2s . Let us apply Lemma 114.2 with A D E and B D Z.G/. We have (4)

2e.G/ D 2s Œ22m ˙ 2m C 22mC1 .22m ˙ 2m / D 22mCsC1

so e.G/ D 22mCs D 2d , and we obtain again the desired result. 1 Now assume that e.G/ D 2d ˙ 2d 2 d . Then, by (3) and (4), exp.Z.G// D 2 and d D 2m C s 1;

1 d DmCs1 2

(by (3))

so that 2m C s 1 D 2.m C s 1/. It follows that s D 1 so G D EZ.G/ D E is extraspecial, completing the proof. Theorem 114.4 ([HupB, Theorem VIII.5.4]). Suppose that G is a 2-group of class at most 2 and exponent at most 4. Suppose that ¹x 2 G j x 2 D 1º D N is a subgroup of G. Then jGj jN j3 ; indeed, jGj < jN j3 if G is not special.

114

Further counting theorems

155

Proof. Since the theorem obviously holds if G is abelian (in this case, jGj jN j2 j), we will assume, in what follows, that G is nonabelian. We prove the theorem by induction on jN j. If jN j D 2, then G is isomorphic to Q8 (Proposition 1.3), and the assertion is clear. Suppose then that jN j > 2. Clearly, ˆ.G/ N as exp.G=N / < exp.G/ D 4. First suppose that G 0 < N . We have N L where L=G 0 D 1 .G=G 0 /. Since L=G 0 is elementary abelian, we get L=G 0 D .N=G 0 / .K=G 0 / for some K L. As G 0 < N , we get K < L so induction may be applied to K. Thus jKj j1 .K/j3 D jN \ Kj3 D jG 0 j3 so jLj D jN jjKj=jG 0 j jN jjG 0 j2 . But jL W G 0 j D j1 .G=G 0 /j D jG=G 0 W Ã1 .G=G 0 /j D jG W ˆ.G/j so jG W Lj D jˆ.G/ W G 0 j. Hence jGj D jG W LjjLj D jˆ.G/ W G 0 jjLj jˆ.G/ W G 0 jjN jjG 0 j2 D jˆ.G/jjN jjG 0 j jN jjN jjG 0 j D jN j2 jG 0 j < jN j3 since G 0 < N by assumption. Suppose that G 0 D N ; then ˆ.G/ D N also in view of G 0 ˆ.G/ N D G 0 . Since G is of class 2, we obtain that N Z.G/. Suppose next that N < Z.G/. Choose z 2 Z.G/ N . Let M be a maximal subgroup of G not containing z (M exists since z 62 ˆ.G/). Thus G D M hzi; M \ hzi D Z; where Z D hz 2 i ¤ ¹1º (indeed, 1 .G/ D N < Z.G/ so o.z/ > 2). We apply induction to M=Z. If x 2 M and x 2 2 Z, we have x 2 D .z 2 /m D z 2m for some m and xz m 2 N since .xz m /2 D x 2 z 2m D x 2 x 2 D 1). Hence z m 2 M \ hzi D Z j3 and xZ 2 N=Z. Thus jM=Zj jN=Zj3 , and this implies jM j jN ; then jZj2 jGj D 2jM j

2jN j3 1 D jN j3 < jN j3 2 jZj 2

since jZj D 2 (recall that exp.G/ D 4). Thus one may assume that G 0 D ˆ.G/ D Z.G/ D N ; then G is special. We have e.G/ D jN j D 2n , and e.G=N / D 2d , say. Next, j1 .N /j D 2n 1. Hence, by Theorem 114.1, X (5) e.G=M / D .j1 .N / 1/e.G=N / C e.G/ D .2n 2/2d C 2n : M 21 .N /

By Theorem 114.3, e.G=M / 2 ¹2d ; 2d ˙ 2d r º, where the positive integer r is such that 2r d . Substituting in (5), we obtain an equation of the form X 2nCd 2d C1 C 2n D .˙2si /; i

where si 12 d . Suppose that n < 12 d . Then si > n for is divisible by 2nC1 , which is a contradiction since 2nC1 does not divide 2n . Hence n 12 d and jGj D 2d Cn

all i, so the right-hand side j .2nCd 2d C1 /, but 2nC1 3n 2 D jN j3 .

156

Groups of prime power order

Theorem 114.5 ([HupB, Theorem VIII.5.5]). Let G be a nonidentity 2-group and suppose that Aut.G/ permutes the set Inv.G/ of involutions of G transitively. Further, let j1 .Z.G//j D q. Then (a) 1 .Z.G// D 1 .G/. (b) (Gross [Gro]) jGj D q m for some positive integer m. (c) (Shaw) If G is nonabelian and of exponent 4, then G is special and jGj 2 ¹q 2 ; q 3 º. Proof. Statement (a) is obvious. k1

(b) Given an involution t 2 G, write mk D j¹y 2 G j y 2 D tºj (k 1). Since Aut.G/ permutes the set of involutions of G transitively, mk is independent of t , and the number of elements of order 2k in G is .q 1/mk (k 1). Hence jGj 1 D .q 1/.m1 C m2 C / and so q 1 divides jGj 1. Let jGj D q m 2s , where 2s < q. Since 2s q m 1 2s .q m 1/ 2s 1 jGj 1 D D C q1 q1 q1 q1 is an integer, we get s D 0 so jGj D q m , proving (b). (c) Since exp.G/ D 4, we get exp.G=1 .Z.G/// D 2 hence G=Z.G/ is elementary abelian. Thus G 0 1 .Z.G//. As G is nonabelian, we have G 0 > ¹1º. By hypothesis, 1 .Z.G// is a minimal characteristic subgroup of G so G 0 D 1 .Z.G//. Therefore G=G 0 is elementary abelian so G 0 D ˆ.G/. If G 0 < Z.G/, then Z.G/ contains an element z of order 4. Now, if y 2 G Z.G/, y 2 is an involution. Since z 2 is also an involution, we obtain y 2 D .z 2 /˛ for some ˛ 2 Aut.G/. But then y 2 D .z ˛ /2 and y.z ˛ /1 2 Z.G/ so y 2 Z.G/, which is a contradiction. Hence G 0 D Z.G/ D 1 .G/. By (b), jGj D q m for some positive integer m, and by Theorem 114.4, m 3. Since G is nonabelian, m > 1 so m D 2 or 3.1

1 Both

values of m are possible.

115

Finite p-groups all of whose maximal subgroups except one are extraspecial

In a letter Z. Janko reported that he proved that if all maximal subgroups of a 2-group G except one are extraspecial, then G is of maximal class and order 24 . Below we consider the similar problem for all p. Recall that all maximal abelian subgroups of any extraspecial group of order p 2mC1 have the same order p mC1 . In particular, if this group possesses an abelian subgroup of index p, then m D 1. Theorem 115.1. All nonabelian maximal subgroups of a nonabelian p-group G except one are extraspecial if and only if G is either minimal nonabelian or of order p 4 . Proof. All nonabelian maximal subgroups of a nonabelian group G of order p 4 are extraspecial unless G is minimal nonabelian. In what follows we assume that G is not minimal nonabelian and jGj > p 4 (clearly, jGj D p 2m for some m > 2 by Theorem 4.7(a)). Then there is a nonabelian E 2 1 . Since, by Exercise 1.6(a), the number of nonabelian members of the set 1 is at least p > 1, one may assume that E is extraspecial. Assume that there is an abelian A 2 1 . Since A \ E is an abelian maximal subgroup of the extraspecial group E, we get jEj D p 3 (Theorem 4.7(d)); then jGj D pjEj D p 4 ; contrary to the assumption. Thus the set 1 has no abelian members. Since j1 j p C 1 3, there is an extraspecial F 2 1 ¹Eº. Note that E 0 G G and E 0 D ˆ.E/ ˆ.G/ < F so E 0 D F 0 since F 0 is a unique normal subgroup of order p in F . Thus all extraspecial members of the set 1 have the same derived subgroup E 0 . By what has been said in the previous paragraph, all members of the set 1 except one are extraspecial. Since G has a maximal subgroup whose center is of order > p and this subgroup is neither abelian nor extraspecial, it follows that the number of extraspecial members in the set 1 is divisible by p (indeed, we have j1 j D 1 C p C C p d.G/1 ). Set GN D G=E 0 . We consider the following two cases. (i) Suppose that GN is nonabelian. Since GN has at most one nonabelian member and the number of abelian maximal subgroups of GN is at least p, we conclude that G=E 0

158

Groups of prime power order

is minimal nonabelian (see the previous paragraph and Exercise 1.6(a)). As EN and FN N D GN so GN 2 ¹D8 ; S.p 3 /º, where S.p 3 / is are elementary abelian, we get 1 .G/ 3 N D p 4 , contrary nonabelian of order p and exponent p. In this case, jGj D jE 0 jjGj to the assumption. (ii) Thus GN is abelian. As GN D EN FN , it follows that GN is elementary abelian. Since N D p 2mC1 so GN Š Ep2mC1 and jGj D 22mC2 N jEj D p 2m for some m > 1, we get jGj 0 0 with G D E . Therefore G is not extraspecial, and we conclude that jZ.G/j > p. Since G=Z.G/ is noncyclic, it contains two distinct maximal subgroups M=Z.G/ and N=Z.G/. As, by assumption, M and N are nonabelian and, obviously, M and N are not extraspecial (indeed, jZ.G/j > p and Z.G/ M \ N ), we get a contradiction. Exercise 1. Classify the p-groups all of whose maximal subgroups except one are elementary abelian. Answer. The dihedral group D8 is the unique group with that property. Corollary 115.2. If all maximal subgroups of a nonabelian p-group G except one are extraspecial, then G is of maximal class and order p 4 . A p-group G is said to be a CZ-group if G 0 is its unique minimal normal subgroup. In this case, Z.G/ is cyclic. If x; y 2 G, then we infer that 1 D Œx; yp D Œx; y p so Ã1 .G/ Z.G/ and hence ˆ.G/ D G 0 Ã1 .G/ Z.G/, and we conclude that ˆ.G/ is also cyclic (see 4 on such G). A minimal nonabelian p-group X is a CZ-group if and only if its center is cyclic (in this case, X is either of order p 3 or X Š Mpn ). Obviously, extraspecial p-groups are CZ-groups. Let G be a CZ-group. By Lemma 4.2, G D .A1 An /Z.G/, where A1 ; : : : ; An are minimal nonabelian. In our case, Z.G/ is cyclic so Z.Ai / is cyclic for all i. Then it follows that Ai 2 ¹D8 ; Q8 ; S.p 3 /; Mpk º. At least one member of the set 1 is not a CZ-group. Theorem 115.3. Suppose that all maximal subgroups of a nonabelian p-group G of order p m > p 4 except one are CZ-groups and let F 2 1 be an arbitrary CZ-subgroup. Then Z.G/ is cyclic. (a) If G=F 0 is abelian, then G is a CZ-group.1 Next we assume that G=M 0 is nonabelian for all CZ-members M 2 1 . Then we have F 0 D M 0 for all such M since Z.G/ is cyclic. (b) If the quotient group G=F 0 is minimal nonabelian, then d.G/ D 2 and jG 0 j D p 2 . Also, G=Z.F / is abelian of type .p 2 ; p/ and Z.F / < ˆ.G/. (c) Suppose that G=Z.F / is abelian (or, what is the same, G 0 Z.F /). Then G=Z.F / is abelian of type .p 2 ; p/ and d.F / 3. If M 2 1 ¹F º is a CZ-subgroup, then Z.M / ¤ Z.F / and M=Z.F / is cyclic; moreover, M Š Mpm1 . 1 But

we do not assert that all CZ-groups satisfy the hypothesis.

115

Finite p-groups all of whose maximal subgroups except one are extraspecial 159

Proof. Assume that Z.G/ is noncyclic. Then all members of the set 1 that contain Z.G/ are not CZ-groups. Since G=Z.G/ is noncyclic, the set 1 has at least p C 1 > 1 members which are not CZ-groups, and this is a contradiction. Thus (i) Z.G/ is cyclic. By hypothesis, the set 1 has two distinct members F and H that are CZ-groups. By (i), F 0 D H 0 since F 0 ; H 0 Z.G/. Then, by Exercise 1.6(a), the set 1 .G=F 0 / contains at least p C 1 abelian members. Assume that the set 1 .G=F 0 / has a nonabelian member M=F 0 . Then M is not a CZ-group (otherwise, M 0 D F 0 so M=F 0 is abelian), and we conclude that d.G/ 3. Therefore, by Exercise 1.6(a), the set 1 .G=F 0 / has at least p 2 nonabelian members so the set 1 has at least p 2 members that are not CZ-groups, and this is a contradiction. Thus (ii) G=F 0 is either abelian or minimal nonabelian for any CZ-subgroup F 2 1 , D F 0 for all CZ-subgroups H 2 1 ¹F º.

H0

We consider two cases, mentioned in (ii), separately. A. Suppose that G=F 0 is abelian; then G 0 D F 0 . Since Z.G/ is cyclic by (i), G 0 is a unique minimal normal subgroup of G so G is a CZ-group. B. Suppose that G=F 0 is minimal nonabelian. Since F 0 ˆ.F / ˆ.G/, we get d.G/ D d.G=F 0 / D 2. By Lemma 65.1, d.F / D d.F=F 0 / 3, and this is true for any CZ-member of the set 1 . Due to Lemma 4.2 (or Lemma 4.3), F is a product of minimal nonabelian subgroup and Z.F / so F=Z.F / Š Ep2 hence jG=Z.F /j D p 3 . Since G=Z.F / is noncyclic and d.G/ D 2, we obtain Z.F / < ˆ.G/. Suppose that the quotient group G=Z.F / is abelian; then G=Z.F / (of order p 3 ) is abelian of type .p 2 ; p/ since d.G/ D 2. Let H 2 1 ¹F º be a CZ-subgroup. We have Z.H / ¤ Z.F / (otherwise, Z.F / D Z.G/ and so G is minimal nonabelian, a contradiction). Then the quotient group H=Z.F / (recall that Z.F / ˆ.G/ < H ) is cyclic of order p 2 (indeed, F=Z.F / is the unique noncyclic maximal subgroup of G=Z.F / which is abelian of type .p 2 ; p/) so H is metacyclic since Z.F / is cyclic (indeed, F is a CZ-group). In this case, H is minimal nonabelian (Lemma 65.2(a)). Since Z.H / is cyclic and jH j > p 3 , we get H Š Mpm1 . Remark 1. Suppose that a p-group G is neither abelian nor minimal nonabelian. Let 10 be the set of nonabelian members of the set 1 . It follows from Exercise 1.6(a) that j10 j p. We suppose that the set 10 contains p 1 non-extraspecial members. One may assume that jGj D p m > p 4 (groups of order p 4 satisfy the hypothesis). Then there is an extraspecial E 2 1 . Since jEj 25 , it follows that E has no abelian subgroup of index p, and we conclude, as above, that 10 D 1 so the set 1 contains p 1 non-extraspecial members. Because of j1 j p C 1, there is an extraspecial F 2 1 ¹Eº. As in the proof of Theorem 115.1, Z.G/ is cyclic. Since E 0 ; F 0 Z.G/, N of maximal subgroups of GN conwe get E 0 D F 0 . Set GN D G=E 0 . Then the set 1 .G/ tains two distinct (elementary) abelian members, therefore, by Exercise 1.6(a), it conN D EN FN D GN hence, if GN is abelian, tains at least p C 1 abelian members. Next, 1 .G/

160

Groups of prime power order

it has exponent p. Assume that GN is neither abelian nor minimal nonabelian. Then the N contains at least p nonabelian members whose inverse images in G are not set 1 .G/ extraspecial, contrary to the hypothesis. Thus, if GN is nonabelian, it is minimal nonN D p 3 (Lemma 65.1). (i) Assume that GN is elementary abelian; then abelian and so jGj G is not extraspecial since jGj D pjEj D p 2.kC1/ , where jEj D p 2kC1 . It follows that jZ.G/j > p. Since G=Z.G/ is noncyclic, all p C 1 members of the set 1 containing Z.G/, are not extraspecial, contrary to the hypothesis (recall that the set 1 has no abelian members). (ii) Now let GN is minimal nonabelian of order p 3 . Then we get 0 j D p 4 , contrary to the assumption. N jGj D jGjjE Recall that a p-group G is said to be an An -group if it has a nonabelian subgroup of index p n1 but all its subgroups of index p n are abelian. The p-groups which are A2 -groups are classiﬁed in 71. Remark 2. Let a p-group G be neither abelian nor minimal nonabelian. Suppose that all nonabelian members of the set 1 are either extraspecial or minimal nonabelian. Then G is an A2 -group. Indeed, by hypothesis, jGj > p 3 . If jGj D p 4 , it satisﬁes the hypothesis. In what follows we assume that jGj > p 4 . If all nonabelian maximal subgroups are minimal nonabelian, then G is an A2 -group. In the following we assume that there is an extraspecial E 2 1 . If A 2 1 is either abelian or minimal nonabelian, then E \ A is abelian of index p in E so jEj D p 3 . Then jGj D p 4 , contrary to the assumption. Thus all maximal subgroups of G are extraspecial. This is impossible since there is in G a maximal subgroup with center of order > p. Theorem 115.4. Let a p-group G be neither abelian nor minimal nonabelian. Suppose that all nonabelian members of the set 1 are either minimal nonabelian or of maximal class. Then one and only of the following holds: (a) G is an A2 -group. (b) G is a p-group of maximal class with abelian subgroup of index p. (c) G is a 3-group of maximal class with nonabelian fundamental subgroup G1 (in this case, G1 is minimal nonabelian). Proof. Suppose that G is not an A2 -group; then jGj > p 4 . In this case, there is in the set 1 a member M of maximal class. Then, by Theorem 12.12(a), G is either of maximal class or d.G/ D 3; so, in the second case, the set 1 contains exactly p C 1 members that are either abelian or minimal nonabelian (Theorem 13.6). (i) Suppose that G is of maximal class. One may assume that p > 3 as all p-groups of maximal class and order > p 4 , p 3, satisfy the hypothesis (see 9). If the set 1 has an abelian member, then all nonabelian members of the set 1 are of maximal class (Fitting’s lemma) so G satisﬁes the hypothesis. Next we assume that 1 has no abelian members. If jGj > p pC1 , then the fundamental subgroup G1 of G must be minimal nonabelian since it is not of maximal class. In this case, as p p1 D j1 .G1 /j p 3 (Lemma 65.1), we get p D 3, contrary to the assumption.

115

Finite p-groups all of whose maximal subgroups except one are extraspecial 161

Now let jGj D p m p pC1 ; then p > 3 since m > 4. In this case, the set 1 has at most two distinct members that are not of maximal class. Let A 2 1 be not of maximal class. Then A is minimal nonabelian. We have A0 D Z.G/. Since cl.G/ > 3, we conclude that A is a unique member of the set 1 which is not of maximal class (otherwise, cl.G=Z.G// D 2). As j1 .A/j p 3 , it follows that G has no subgroup of order p 5 and exponent p. Since m > 4, we get exp.A/ > p. Next, the metacyclic abelian subgroup A=A0 is the fundamental subgroup of G=A0 (recall that A0 D Z.G/). It follows that jG=A0 j D p 4 so jGj D p 5 and A=A0 is abelian of type .p 2 ; p/. Since exp.G=A0 / D p, we get a contradiction. (ii) Suppose that 1 contains exactly p C 1 members, say A1 ; : : : ; ApC1 , that are either abelian or minimal nonabelian. Then all intersections M \Ai (i D 1; : : : ; pC1) are abelian and have index p in M (here M 2 1 is of maximal class). Since m > 4, M contains at most one abelian subgroup of index p. It follows that M \ A1 D D M \ ApC1 so G=.M \ A1 / contains p C 2 distinct subgroups M=.M \ A1 /; A1 =.M \ A1 /; : : : ; ApC1 =.M \ ApC1 / of order p, a contradiction since the abelian group G=.M \A1 / of type .p; p/ contains exactly p C 1 subgroups of order p. Exercise 2. Classify the p-groups all of whose nonabelian maximal subgroups are either of maximal class or have derived subgroup of order p. (Hint. Use 137.) Exercise 3. Classify the p-groups all of whose nonabelian maximal subgroups are either absolutely regular or of maximal class. (Hint. Use Theorem 12.1(b).) Exercise 4. Classify the p-groups all of whose nonabelian maximal subgroups are either metacyclic or of maximal class. Exercise 5. Classify the p-groups all of whose nonabelian members of the set 2 are either minimal nonabelian or of maximal class. Exercise 6. Classify the p-groups containing p extraspecial maximal subgroups.

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Groups covered by few proper subgroups

1o Introduction. We say that a group G is covered by proper subgroups A1 ; : : : ; An if (1)

G D A1 [ [ An :

We have, in (1), G > ¹1º and n > 1. A group is covered by its proper subgroups if and only if it is noncyclic (check!). Every noncyclic group is covered by (proper) cyclic subgroups. A group is not covered by two proper subgroups. Covering (1) is said to be irredundant if every proper subset of the set ¹A1 ; : : : ; An º does not cover G. In what follows we assume that (1) is an irredundant covering of G by proper subgroups. Remark 1. If, in (1), jA1 j jAn j, then jGj jA1 jn .n 1/ < njA1 j and hence jG W A1 j < n. Let M be a maximal subset (with respect to inclusion) of pairwise noncommuting elements of a nonabelian group G. We denote the set of all such subsets by ƒ.G/. Write .G/ D max¹jMj j M 2 ƒ.G/º: For any abelian group G we set .G/ D 1. If H is a subgroup of G, then we have .H / .G/. Recall that two groups G and G1 are lattice isomorphic if there is a bijective mapping of the set of subgroups of G onto the set of subgroups of G1 such that, provided F; H G, we have .F \ H / D F \ H and hF; H i D hF ; H i. If groups G and G1 are lattice isomorphic, then the inequality .G/ ¤ .G1 / is possible owing to the fact that some nonabelian groups are lattice isomorphic to abelian groups (indeed, n the group MpnC1 is lattice isomorphic to the abelian group S of type .p ; p/). As Lemma 116.2(a) shows, if M 2 ƒ.G/, then G D x2M CG .x/ and this covering is irredundant. Every nonabelian group contains two noncommuting elements. If a; b 2 G satisfy ab ¤ ba, then the elements a; b; ab are pairwise noncommuting, i.e., .G/ 3. For p-groups one can prove a stronger result. Lemma 116.1. Let G be a nonabelian p-group. Then (a) If G is minimal nonabelian, then .G/ D p C 1. (b) .G/ p C 1.

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Groups covered by few proper subgroups

163

Proof. (a) Since all maximal subgroups of G are abelian, any two noncommuting elements of G are contained in distinct maximal subgroups of G. Thus .G/ p C 1. If M1 ; : : : ; MpC1 are all the maximal subgroups of G and xi 2 Mi ˆ.G/ for all i , then, for i ¤ j , hxi ; xj i D G is nonabelian, so xi xj ¤ xj xi . Hence x1 ; : : : ; xpC1 are pairwise noncommuting elements so .G/ p C 1, completing the proof. (b) Let H be a minimal nonabelian subgroup of G. Then .H / D p C 1 by (a), and so .G/ .H / D p C 1. The following lemma establishes a connection between members of the set ƒ.G/ with some irredundant coverings of a nonabelian group G. Part (b) of this lemma also shows that the members of ƒ.G/ of cardinality .G/ have a special property. If G is a minimal nonabelian p-group of order p n , then jƒ.G/j D .p n1 p n2 /pC1 ; but G has only one irredundant covering (see Lemma 116.3(b) and Theorem 116.5). Lemma 116.2. Let G be a nonabelian group and M 2 ƒ.G/. Then (a) We have

[

(2) If N M1 2 ƒ.G/ and an irredundant covering.

S

CG .x/ D G:

x2M

x2N

CG .x/ D G, then N D M1 ; in particular, (2) is

S (b) Suppose, in addition, that jMj D .G/. If x 2 M, then hG y2M¹xº CG .y/i is abelian. S Proof. (a) Assume that there is g 2 G x2M CG .x/. Since M M [ ¹gº, it follows from the maximality of M that S gx D xg for some x 2 M; then g 2 CG .x/, contrary to the choice of g. Thus x2M CG .x/S D G. Now assume that there is u 2 M such that x2M¹uº CG .x/ D G. Then there exists v 2 M ¹uº such that u 2 CG .v/, so that u; S v are distinct commuting members of the set M, a contradiction. Thus the covering x2M CG .x/ D G is irredundant. S (b) Given x 2 M, set D D G y2M¹xº CG .y/. The set D is nonempty. Assume that there are noncommuting u; v 2 D. By the choice, every element of the set M¹xº does not commute with u and v. It follows that .M ¹xº/ [ ¹u; vº M2 2 ƒ.G/, a contradiction since jM2 j > jMj D .G/. Thus all elements of the set D commute so the subgroup hDi is abelian. It follows from Lemma 116.2(a) that if H < G is such that .H / D .G/ and if we have M D ¹x1 ; : : : ; x .G/ º 2 ƒ.H /, then there are i ¤ j with CG .xi / 6 H and CG .xj / 6 H . Indeed, there is an index i such that CG .xi / 6 H since .G/ [ kD1

CG ..xk / D G > H

(Lemma 116.2(a)).

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Groups of prime power order

If for all j ¤ i we have CG .xj / H , then H [ CG .xi / D G, which is impossible since H and CG .xj / are nonincident. 2o p-groups. In this section we study the p-groups containing a maximal subset (with respect to inclusion) of pairwise noncommuting elements of cardinality p C 1. Some related results are also established and discussed. Next, we study the p-groups which are covered by 2p proper subgroups. It is proved that if a p-group G admits an irredundant covering by p C 2 subgroups, then p D 2. A noncyclic p-group G admits an irredundant covering by pC1 maximal subgroups (indeed, if T G G is such that G=T is abelian of type .p; p/, then p C 1 maximal subgroups of G containing T cover G). Moreover, Lemma 2.1 showsTthat if a p-group pC1 G is covered by p C 1 proper subgroups A1 ; : : : ; ApC1 , then jG W i D1 Ai j D p 2 , i.e., all Ai are maximal in G. Lemma 116.3 is known; however, we prove it to make our exposition self-contained. Lemma 116.3. Suppose that a noncyclic p-group G of order p m is covered by n proper subgroups A1 ; : : : ; An as in (1). Then (a) n p C 1. (b) If n D p C 1, then the covering (1) is irredundant and jG W In particular, the Ai are maximal in G.

TpC1 i D1

Ai j D p 2 .

Proof. (a) If n p, then ˇ ˇ n ˇX # ˇ m1 ˇ ˇ A 1/ D p m p < jG # j; i ˇ p.p ˇ iD1

a contradiction. (b) Now let n D p C 1. Then the covering (1) is irredundant by (a). First assume that A1 ; : : : ; ApC1 are maximal in G; then jAi \ Aj j D p m2 for i ¤ j . We have ! p [ .Ai ApC1 / : (3) G D ApC1 [ iD1

Since Ai Aj D Ai .Ai \ Aj / for i ¤ j , the right-hand side of (3) contains at most p m1 C p.p m1 p m2 / D p m D jGj elements so (3) is a partition of G. Then it follows that Ai \ ApC1 D Aj \ ApC1 and .Ai ApC1 / \ .Aj ApC1 / D ¿ for all distinctTi; j < p C 1 (indeed, one pC1 can take in (3) Ak instead of ApC1 ). We conclude that i D1 Ai D A1 \ ApC1 has 2 index p in G. It follows from the above computation (see the displayed formula after (3)) that, in fact, all subgroups A1 ; : : : ; ApC1 must be maximal in G (otherwise, we obtain that SpC1 j i D1 Ai j < jGj).

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165

It follows from Lemmas 116.2 and 116.3 that if G is a nonabelian p-group, then we have .G/ pC1. In Theorem 116.5(b), the p-groups G which satisfy .G/ D pC1 are classiﬁed. Lemma 116.4. Let H be a minimal nonabelian subgroup of a p-group G. Then the intersection ƒ.H / \ ƒ.G/ is not empty if and only if G D H CG .H / (central product); in this case, ƒ.H / ƒ.G/. Proof. (i) Let M 2 ƒ.H / \ ƒ.G/; then we have S jMj D p C 1 (Lemma 116.1(a)). By hypothesis and Lemma 116.2(a), we get G D x2M CG .x/ so, by Lemma 116.3(b), T T jG W x2M CG .x/j D p 2 . Since x2M CG .x/ D CG .M/ and hMi D H , we get CG .H / D CG .M). As CG .H / \ H D Z.H / has index p 2 D jG W CG .M/j in H , we obtain G D H CG .H / by the product formula. (ii) Now suppose that we are given an (arbitrary) p-group G D H CG .H /, where H is minimal nonabelian, S and let M D ¹x1 ; : : : ; xpC1 º 2 ƒ.H /. Then we conclude G D H CG .H / x2M CG .x/, so M 2 ƒ.G/ by Lemmas 116.2(a) and 116.3(a). Thus ƒ.H / ƒ.G/ since M is arbitrary.1 Theorem 116.5. Let G be a nonabelian p-group. (a) If M 2 ƒ.G/ has cardinality p C 1, then jG W CG .x/j D p for all x 2 M and jG W CG .M/j D p 2 . (b) .G/ D p C 1 if and only if G D H Z.G/, where H is an arbitrary minimal nonabelian subgroup of G, H \ Z.G/ D Z.H /. If, in addition, G is of exponent p, then G D H E, where H is nonabelian of order p 3 and E is abelian. Proof. Statement (a) follows from Lemma 116.3(b). (b) Suppose that .G/ D p C 1. Let H G be minimal nonabelian. Then we see that M 2 ƒ.H / has cardinality p C 1 (Lemma 116.1(a)) so that M 2 ƒ.G/ by hypothesis. By Lemma 116.4, G D H CG .H /. We claim that CG .H / D Z.G/. It sufﬁces to show that CG .H / is abelian (indeed, then CG .CG .H // H CG .H / D G). Assume that this is false. Then CG .H / contains two noncommuting elements b; b1 . Let M D ¹a1 ; : : : ; apC1 º 2 ƒ.H /. Take a 2 L¹Z.H /[¹a1 ºº, where L is an (abelian) maximal subgroup of H containing a1 (such an a exists since jL Z.H /j > 1). Then, as Œab; ai D Œa; ai ¤ 1 for i > 1 (indeed, for i > 1 the subgroup ha; ai i D H is nonabelian), we see that the p C 1 elements ab; a2 ; : : : ; apC1 are pairwise noncommuting. Note that Œab; ab1 D Œb; b1 ¤ 1 and for i > 1 we have Œab1 ; ai D Œa; ai ¤ 1. It follows that the p C 2.> .G// elements ab; ab1 ; a2 ; : : : ; apC1 are pairwise noncommuting, which is a contradiction. Thus CG .H / is abelian so coincides with Z.G/. Let us show that for our group G D H Z.G/ we have .G/ < p C 2 (we have ƒ.H / ƒ.G/, but our assertion is stronger). Indeed, assume that g1 ; : : : ; gpC2 2 G do not assert that, in the case under consideration, .G/ D .H / (however, this equality holds by Theorem 116.5). 1 We

166

Groups of prime power order

are pairwise noncommuting. Then we have gi D hi zi , where hi 2 H and zi 2 Z.G/ (i D 1; 2; : : : ; p C 2). Let i ¤ j . Then Œhi ; hj D Œhi zi ; hj zj D Œgi ; gj ¤ 1 so the minimal nonabelian p-group H contains the p C 2 pairwise noncommuting elements h1 ; : : : ; hpC2 , contrary to Lemma 116.1(a). Now suppose that G D H Z.G/ is of exponent p (here H is of order p 3 as minimal nonabelian group of exponent p by Lemma 65.1, and Z.G/ is elementary abelian). In this case, H \ Z.G/ D Z.H / is of order p so Z.G/ D Z.H / E, where E is elementary abelian. Then G D H E, and this completes the proof of (b). Theorem 116.5(b), in particular, classiﬁes the nonabelian p-groups possessing exactly p C 1 distinct centralizers of noncentral elements.2 Proposition 116.6. The following assertions for a nonabelian p-group G are equivalent: (a) If H G is minimal nonabelian, then ƒ.H / ƒ.G/. (b) G D .B1 Bk /Z.G/, where B1 ; : : : ; Bk are minimal nonabelian. Proof. (a) ) (b). We proceed by induction on jGj. Let B1 G be minimal nonabelian. Then G D B1 CG .B1 / by Lemma 116.4. If CG .B1 / is abelian, we are done. If CG .B1 / is nonabelian, the result follows by induction applied to CG .B1 /. (b) ) (a). Let G be as in (b) and H G be minimal nonabelian. Since jG 0 j D p, then, by [B1, Lemma 4.3(a)], we obtain G D H CG .H / so ƒ.H / ƒ.G/ by Lemma 116.4. Remark 2. The argument in part (ii) of the proof of Lemma 116.4 shows that if H is a nonabelian subgroup of an arbitrary group G D H CG .H /, then ƒ.H / ƒ.G/. 3o Nonnilpotent groups. In this subsection G is a nonnilpotent group. We consider coverings of nonnilpotent groups by a few proper subgroups. Minimal nonabelian and minimal nonnilpotent groups play a crucial role in what follows. Let p be a prime divisor of jGj such that G has no normal p-complement. Then there is in G a minimal nonnilpotent subgroup H D Q P , where P D H 0 2 Sylp .H / and Q 2 Sylq .H / is cyclic (this follows from Frobenius’ normal p-complement theorem; see, for example, [Isa5, Theorem 9.18]). We have jP j D p bCc , where b is the order of p modulo q and p c D jP \ Z.H /j (see [BZ, Lemma 11.2]). In this case, there are in H exactly p b Sylow q-subgroups, say Q1 D hx1 i; : : : ; Qpb D hxpb i: Then x1 ; : : : ; xpb are pairwise noncommuting elements (indeed, if i ¤ j, then hxi ; xj i is nonnilpotent since it has two distinct Sylow q-subgroups Qi and Qj so this twogenerator subgroup coincides with H ). If ¹y1 ; : : : ; ys º is a maximal subset of pairwise 2 Note

that [P] yields an estimate of the index jG W Z.G/j in terms of .G/.

116

Groups covered by few proper subgroups

167

noncommuting elements of P , then ¹y1 ; : : : ; ys ; x1 ; : : : ; xpb º is a maximal subset (with respect to inclusion) of pairwise noncommuting elements of H of cardinality p b C s p b C 1 (note that s D 1 if and only if P is abelian). If P is nonabelian, then s pC1 (Lemma 116.1(b)). Thus .G/ .H / D p b Cs p b C1. Theorem 116.7. Let G be a nonabelian group and p be a prime divisor of jGj. (a) If G has no normal p-complement, then .G/ p C 1. If, in addition, p is the minimal prime divisor of jGj, then .G/ p 2 C 1. (b) Suppose that G has a normal p-complement however a Sylow p-subgroup is not a direct factor of G. Then .G/ p C 2. If, in addition, .G/ D p C 2, then either p D 2 and q D 3 or p is a Mersenne prime. (c) If G D P A, where P is nonabelian, A is abelian and .G/ < p C 2, then P is such as in Theorem 116.5(b) and A is abelian. Proof. Statement (a) was proved in the paragraph preceding the theorem.3 (b) Now assume that G has a normal p-complement H but P 2 Sylp .G/ is not a direct factor of G. In this case, the p-solvable group G contains a nonnilpotent subgroup PQ, where Q 2 Sylq .H /; then we get Q D PQ \ H G PQ. Therefore PQ contains a minimal nonnilpotent subgroup F D P1 Q1 , where P1 2 Sylp .F / is cyclic and Q1 D F 0 2 Sylq .F /. Then jQ1 j D q bCc , where b is the order of q modulo p and q c D jQ1 \ Z.F /j. As above, there is M 2 ƒ.F / of cardinality q b C 1. Since q b p C 1, we get jMj p C 2. Now assume that jMj D p C 2; then q b D p C 1 so either p D 2 and q D 3 or q D 2 and p is a Mersenne prime.4 Statement (c) now follows from Remark 2 and Theorem 116.5(b). Proposition order of a group G and let SpC1 116.8. Let p be a minimal prime divisor ofTthe pC1 G D i D1 Ai be an irredundant covering. Then jG W iD1 Ai j D p 2 . In particular, jG W Ai j D p for i D 1; : : : ; p C 1. Proof. It follows from Remark 1 that if p is the minimal prime divisor of jGj, then it is not covered by p proper subgroups (otherwise, there is in G a proper subgroup of index < p, contrary to Lagrange’s theorem). One may assume that jA1 j jApC1 j. Then, by Remark 1, jG W A1 j < p C 1 so that jG W A1 j D p and hence A1 G G. First assume that all Ai are maximal in G. Set jGj D g and jG W Ai j D ki , where i D 2; : : : ; p C 1. Note that ki p for all i . We have ! pC1 [ .Ai A1 / : (4) G D A1 [ iD2 3 If 4 If

G Š A4 , the alternating group of degree 4, p D 2, then .G/ D 22 C 1. G Š A4 and p D 3, then .G/ D 5 D 3 C 2.

168

Groups of prime power order

Further, G D A1 Ai for i > 1. Since Ai A1 D Ai .Ai \A1 / and jAi W .Ai \A1 /j D jG W A1 j D p so that jG W .Ai \ A1 /j D jG W Ai jjAi W .Ai \ A1 /j D pki we obtain jAi A1 j D jAi .Ai \ A1 /j D

g g g D ki pki ki

for i > 1; 1

1 : p

The right-hand side of formula (4) contains v elements, where pC1 pC1 1 X g 1 X g g g g 1 g D C C 1 1 p D g: v C 1 p p ki p p p p p p iD2

i D2

SinceTv D g, it follows that (4) is a partition of G and ki D p for all i . In this case, pC1 jG W i D1 Ai j D p 2 . SpC1 Now let Ai Bi < G, where Bi are maximal in G for all i . Then G D i D1 Bi is an irredundant covering of G by the ﬁrst sentence of the proof, and so jG W Bi j D p for all i by the previous paragraph. If Ai < Bi for some i , then, taking in the above displayed formula SpC1 Aj D Bj2 for j ¤ i , we get a contradiction. Thus Bi D Ai for all i and so jG W i D1 Ai j D p by the previous paragraph. Lemma 116.3(b) is a partial case of Proposition 116.8. Let G be a non-p-nilpotent group. Then, using Theorem 116.7, one can show the following results: (a) If p D 2, then .G/ 5. (b) If p > 2, then .G/ p C 1. (c) If p > 2 is a minimal prime divisor of jGj, then .G/ p 2 C 1. 4o On the number of maximal subgroups appearing in some coverings of p-groups. In this subsection we consider irredundant coverings of a p-group by k proper subgroups, where p C 1 < k 2p. It is impossible to avoid some repetition in computations (otherwise, the proofs would be unreadable). Remark 3. We claim that if a p-group G of order p 3 is neither cyclic nor Q8 , it admits an irredundant covering by 2p subgroups. Indeed, let T G G be such that G=T is abelian ofStype .p; p/. Let A1 =T; : : : ; ApC1 =T be all subgroups of order p in G=T . pC1 Then G D i D1 Ai is an irredundant covering. Since G is neither cyclic nor isomorphic to Q8 , one may assume that A1 is noncyclic (here we use Theorem 1.2 which implies that if a p-group contains > p cyclic subgroups of index p, it is isomorphic to Q8 ). In this case, there exists in T an A1 -invariant subgroup T0 such that A1 =T0

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Groups covered by few proper subgroups

169

is abelian of type .p; p/. Let T D T1 ; T2 ; : : : ; TpC1 be all maximal subgroups of A1 containing T0S . Then G is covered by 2p subgroups A2 ; : : : ; ApC1 ; T2 ; : : : ; TpC1 since pC1 A1 A2 [ . i D2 Ti /, and this covering is irredundant. Lemma 116.9. If a p-group G admits an irredundant covering by p C 2 subgroups A1 ; : : : ; ApC2 , then (a) If p > 2, then at least p C 1 of the Ai ’s are maximal in G. (b) If p D 2, then at least two of the Ai ’s are maximal in G. Proof. Let jA1 j jApC2 j and jGj D p n . By Remark 1.1, we get jG W A1 j D p. Assume that jG W ApC1 j > p. Then ˇpC2 ˇ p pC2 X X ˇ[ ˇ ˇ ˇ Ai ˇ jA1 j C jAi A1 j C jAi A1 j p Dˇ n

i D1

Dp

n1

i D2

C .p 1/.p

n1

i DpC1

p

n2

/ C 2.p n2 p n3 /

D p n p n3 .p 2 3p C 2/ D p n p n3 .p 1/.p 2/: If p > 2, then p n p n p n3 .p 1/.p 2/ < p n , which is a contradiction. Thus, if p > 2, then at least p C 1 subgroups Ai are maximal in G, completing this case. Now let p D 2 and assume that jA2 j < 2n1 . Then ˇX ˇ 4 X ˇ 4 ˇ ˇ Ai ˇˇ jA1 j C jAi A1 j D 2n1 C 3.2n2 2n3 / 2 Dˇ n

i D1

D2

n1

i D2

n3

C3 2

D 7 2n3 < 2n ;

a contradiction. Thus, if p D 2, then at least two Ai are maximal in G. S Let G D 4iD1 Ai be an irredundant covering of a 2-group G of order 2n > 23 , jA1 j jA2 j jA3 j jA4 j. We claim that if jG W A3 j D 2, then jG W A4 j D 2 so all Ai are maximal in G. Then we have (5)

G D A1 [ .A2 A1 / [ .A3 A1 A2 / [ .A4 A1 A2 A3 /:

Assume that jG W A4 j > 2. We have jG W .A1 \ A2 \ A3 /j D 23 (Lemma 116.3) hence jA3 A1 A2 j D 2n3 . Further, A3 A1 A2 D .A3 A1 / .A3 A1 / \ A2 D A3 .A3 A1 / .A3 \ A2 / .A1 \ A2 / D A3 .A3 A1 / .A3 \ A3 / .A3 \ A1 \ A2 / :

170

Groups of prime power order

It follows that jA3 A1 A2 j D 2n1 2n2 .2n2 2n3 / D 2n3 : Now it is clear that jA4 A1 A2 A3 j < 2n3 . Therefore the right-hand side of formula (5) contains v elements, where v < 2n1 C 2n2 C 2n3 C 2n3 D 2n ; which is a contradiction. Thus, in the case under consideration, either exactly two or four of the Ai ’s are maximal in G. Let G be a 2-group which is not generated by two elements. We claim that then G S admits an irredundant covering G D 4iD1 Ai , where Ai 2 1 for all i . Without loss of generality one may assume that ˆ.G/ D ¹1º. Let A1 ; A2 2 1 be distinct, and set T D A1 \ A2 . Let T < A3 2 1 ¹A1 ; A2 º and let S < T be of index 2; then A3 =S Š E4 . Let T =S; T1 =S; T2 =S < A3 =S be of index 2. As G=Ti Š E4 , there are distinct B1 ; B2 2 1 ¹A3 º such that Bi \ A3 D Ti , i D 1; 2. Since A3 is a subset of the set A1 [B1 [B2 (indeed, A3 D T [T1 [T2 is a subset of A1 [B1 [B2 ), it follows that G D A1 [ A2 [ B1 [ B2 is a covering (indeed, by the above, G D A1 [ A2 [ A3 is a subset of A1 [ A2 [ B1 [ B2 ). Due to the fact the intersection of any three distinct elements of the set ¹A1 ; A2 ; B1 ; B2 º has index p 3 in G, our covering is irredundant (Lemma 116.3(b)). Similarly, if p > 2 and a p-group G is not generated by two elements, then it admits an irredundant covering by 2p maximal subgroups. Lemma 116.10. Suppose that A; B; C; D are pairwise distinct maximal subgroups of a p-group G of order p n such that jG W .A \ B \ C /j D p 3 . Then (6)

jA [ B [ C j D 3p n1 3p n2 C p n3 ;

(7)

jD .A [ B [ C /j p n1 3p n2 C 3p n3 p n4 :

Proof. Note that if distinct U; V < G are maximal, then jG W .U \ V /j D p 2 . By the inclusion-exclusion identity, jA [ B [ C j D .jAj C jBj C C j/ .jA \ Bj C jB \ C j C jC \ Aj/ C jA \ B \ C j D 3p n1 3p n2 C p n3 ; proving (6). By hypothesis, A \ B ¤ B \ C ¤ C \ A (if, for example, A \ B D A \ C , then A \ B D .A \ B/ \ .A \ C / D A \ B \ C , a contradiction since jA \ Bj D p n2 > p n3 D jA \ B \ C j). We have (8)

D .A [ B [ C / D D .D \ .A [ B [ C //;

116

and so

Groups covered by few proper subgroups

171

D .A [ B [ C / D D .D \ A/ [ .D \ B/ [ .D \ C / :

By the inclusion-exclusion identity, j.D \ A/ [ .D \ B/ [ .D \ C /j D jD \ Aj C jD \ Bj C jD \ C j jD \ A \ Bj C jD \ A \ C j C jD \ B \ C j C jD \ A \ B \ C j: If A \ B D, then D \ A \ B \ C D D \ C has order p n2 , a contradiction since A \ B \ C D \ A \ B \ C has order p n3 by hypothesis. Thus A \ B 6 D, and the same is true for A \ C and B \ C . Then, by the product formula and the previous displayed formula, we have jD \ A \ Bj D jD \ A \ C j D jD \ B \ C j D p n3 ; jD \ .A [ B [ C /j D j.D \ A/ [ .D \ B/ [ .D \ C /j D 3p n2 3p n3 C jD \ A \ B \ C j: Since jD \ A \ B \ C j 2 ¹p n3 ; p n4 º, we obtain jD \ .A [ B [ C /j 3p n2 3p n3 C p n4 : Now (7) follows from (8). Theorem 116.11. If a group of order p n admits an irredundant covering by p C 2 subgroups, then p D 2. Proof. Assume that a group G of order p n admits an irredundant covering by p C 2 proper subgroups A1 ; : : : ; ApC2 and p > 2. By Lemma 116.9(a), oneT may assume that SpC1 pC1 A1 ; : : : ; ApC1 are maximal in G. Since G ¤ iD1 Ai , we have jG W i D1 Ai j p 3 . One may assume, without loss of generality, that jG W .A1 \ A2 \ A3 /j D p 3 . We may also assume that jG W ApC2 j D p. Then, by (6), we have jA1 [ A2 [ A3 j D 3p n1 3p n2 C p n3

(9) and, for i > 3, (10)

jAi .A1 [ A2 [ A3 /j < p n1 3p n2 C 3p n3 D p n3 .p 2 3p C 3/

by (7). Set A1 [ A2 [ A3 D U . We have (11)

GDU [

pC2 [ iD4

! .Ai U / :

172

Groups of prime power order

Therefore, taking into account (9) and (10), we obtain jGj D p n .3p n1 3p n2 C p n3 / C .p 1/p n3 .p 2 3p C 3/ D p n p n3 .p 2 3p C 2/ D p n p n3 .p 1/.p 2/ < p n since p > 2, a ﬁnal contradiction. Thus we must have p D 2. Proposition 116.12. If a p-group G is covered by at most k 2p proper subgroups A1 ; : : : ; Ak (we do not assume that this covering is irredundant), then at least p of these subgroups are maximal in G. If p > 3 and p C 2 < k < 2p, then at least p C 1 summands in our covering are maximal in G. Proof. (i) In view of Lemma 116.9, one may assume that p > 2. Let jA1 j jAk j. Then jG W A1 j D p as p > 2 (Remark 1). Since we do not assume that our covering is irredundant, one can add new summands to obtain k D 2p. We also may assume, by way of contradiction, that A1 ; : : : ; Ap1 are maximal in G and Ap ; : : : ; A2p have index p 2 in G. Indeed, if, for example, jG W Ai j > p 2 (i > p 1), one can replace Ai by a subgroup that contains Ai and has index p 2 in G. If, for example, jG W Ai j > p (i < p), one can replace Ai by a maximal subgroup of G that contains Ai . We have (12)

G D Ap1 [

p2 [

! .Ai Ap1 / [

i D1

2p [

! .Ai Ap1 / :

iDp

The right-hand side of formula (12) contains v elements, where v p n1 C .p 2/.p n1 p n2 / C .p C 1/.p n2 p n3 / D p n p n3 .p 1/2 < p n D jGj; a contradiction since v D jGj D p n . Thus at least p subgroups Ai (i 2p) are maximal in G. (ii) To prove the last assertion, one may assume, by way of contradiction, that A1 ; : : : ; Ap are maximal in G and jG W Ai j D p 2 for i > p (see (i)). (Here p > 3 since 3 C 2 D 2 3 1.) We may also assume that k D 2p 1 (if k < 2p 1, one can add to our union 2p 1 k new summands of order p n2 ). Then, as above, we obtain jGj D p n p n1 C .p 1/.p n1 p n2 / C .p 1/.p n2 p n3 / D p n1 C .p 1/.p n1 p n3 / D p n p n3 .p 1/ < p n ; a contradiction.

116

Groups covered by few proper subgroups

173

Proposition 116.13. Suppose that a p-group G of order p n p 4 , p > 2, is covered by k proper subgroups, say A1 ; : : : ; Ak , where p C 2 < k 2p. Let, in addition, any p C 2 subgroups Ai do not cover G, jG W Ai j D p for i p and jG W Ai j > p for i > p. Then (a) k D 2p and our covering is irredundant. Tp (b) j i D1 Ai j D p n2 . (c) jAi j D p n2 for i > p. T2p (d) j i DpC1 Ai j D p n3 . Proof. By the second assertion of Proposition 116.12, we get k D 2p so (a) is true. We have ! ! p1 2p [ [ .Ai Ap / [ .Ai Ap / : (13) G D Ap [ i D1

i DpC1

(c) Assume that jApC1 j jA2p j and jA2p j < p n2 . Then the right-hand side of (13) contains v elements, where v p n1 C .p 1/.p n1 p n2 / C .p 1/.p n2 p n3 / C .p n3 p n4 / D p n1 C .p 1/.p n1 p n3 / C .p n3 p n4 / D p n p n4 .p 1/2 < p n D jGj; which is a contradiction. This proves (c). (b, d) We have jG W Ai j D p 2 for i > p by (c). In this case, the right-hand side of (13) contains v elements, where v p n1 C .p 1/.p n1 p n2 / C p.p n2 p n3 / D p n D jGj; so (13) is a partition. This implies (b) and (d). A similar argument shows that if a p-group G is covered by p 2 proper subgroups, then at least two of these subgroups are maximal in G. Indeed, if only one summand of our covering is maximal in G (see Remark 1), we obtain p n p n1 C .p 2 1/.p n2 p n3 / D p n p n3 .p 1/ < p n ; a contradiction. Corollary 116.14. If a nonabelian p-group G has at most 2p pairwise noncommuting elements, then the centralizers of at least p of these elements are maximal in G. Remark 4. Let G be a group of maximal class and order p nC2 , n 2, with abelian subgroup A of index p. In this case, every x 2 G A satisﬁes jCG .x/j D p 2 (indeed,

174

Groups of prime power order

CA .y/ D Z.G/ is of order p and y p 2 Z.G/ for all y 2 G A), and the number of maximal abelian subgroups of order p 2 not contained in A is equal to jG Aj D pn p.p 1/ (indeed, if B is such a subgroup, then jB Aj D p.p 1/). These p n subgroups together with A cover G and this covering is irredundant. The so-obtained set of cardinality p n C 1 is contained in ƒ.G/. It is easy to see that .G/ D p n C 1. Remark 5. Let G be a nonabelian p-group. If x 2 G and Ax is a maximal abelian subgroup of CG .x/, then Ax is also a maximal abelian subgroup of G. Let M 2 ƒ.G/. Take in CG .x/ a maximal abelian subgroup Ax for every x 2 M (indeed, if B > Ax is abelian, then, by choice, B 6 CG .x/, a contradiction). Then j¹Ax j x 2 Mºj D jMj. It follows that G has at least .G/ maximal abelian subgroups. If the group G has exactly p C 1 maximal abelian subgroups, say A1 ; : : : ; ApC1 , they cover G. In this case, A1 ; : : : ; ApC1 are maximal in G (Lemma 116.2(b)) and G has the structure described in Theorem 116.6(b). Remark 6. Let B1 ; : : : ; Bn beSall maximal abelian subgroups of a nonabelian group G. Then .G/ n since G D niD1 Bi (it is possible that this covering may be redundant). If Bi \ Bj D Z.G/ for all i ¤ j (in this case, the considered covering is irredundant), then .G/ D n. Indeed, take xi 2 Bi Z.G/ for all i . We claim that ¹x1 ; : : : ; xn º 2 ƒ.G/ (indeed, if ¹x; x1 ; : : : ; xn º G, then x 2 Bi for some i so xxi D xi x). For example, let G be a Sylow 2-subgroup of the simple Suzuki group Sz.q/, where q D 22mC1 . Then Z.G/ D ˆ.G/ has index 22mC1 in G. If A < G is maximal abelian, then jA W Z.G/j D 2. It follows that there is M 2 ƒ.G/ of cardinality 22mC1 1; moreover, all members of the set ƒ.G/ have the same cardinality. Remark 7. Let G D AB be a central product of nonabelian groups A and B. Further, let M D ¹a1 ; : : : ; am º 2 ƒ.A/ and N D ¹b1 ; : : : ; bn º 2 ƒ.B/. Then m 1 C n elements of the set M1 D ¹a2 ; : : : ; am ; a1 b1 ; a1 b2 ; : : : ; a1 bn º are pairwise noncommuting. We claim that M1 2 ƒ.G/. Assume that there exists x 2 G M1 such that all elements of the set M1 [ ¹xº are pairwise noncommuting. We have x D ab, where S a 2 A and b 2 B. For i > 1 we have 1 ¤ Œab; ai D Œa; ai so that a 2 D D A niD2 CA .ai /. By Lemma 116.1(b), the subgroup hDi is abelian so Œa; a1 D 1 since a1 2 D. For i D 1; : : : ; n we have 1 ¤ Œab; a1 bi D Œab; bi D Œb; bi . We conclude that n C 1 > .B/ elements b; b1 ; : : : ; bn 2 B are pairwise noncommuting, which is a contradiction. It is easy to deduce from this by induction that if G is an extraspecial group of order p 2mC1 , then .G/ mp C 1. It follows from Remark 7 and Lemma 116.1(a) that if G is a nonabelian p-group such that .G/ 2p, then CG .H / is abelian for all minimal nonabelian H < G.

116

Groups covered by few proper subgroups

175

Problems Below we state some related problems. Problem 1. Classify the 2-groups which do not contain ﬁve pairwise noncommuting elements. (See Lemma 1.3 and Theorem 116.11.) Problem 2. Does there exist a p-group G admitting an irredundant covering by n subgroups, where p C 1 < n < 2p? If ‘yes’, classify such the groups. Problem 3. Describe the set of positive integers n such that there is an elementary abelian p-group admitting an irredundant covering by n maximal subgroups. Problem 4. Let M; N be groups and .M / D m, .N / D n. Then .M N / D mn. (i) Estimate .M N / in terms of M , N and M \ N . Consider the case where M; N are p-groups of maximal class. (ii) Find .G/, where G is an extraspecial group of order p 2mC1 . Problem 5. Classify the pairs of groups N G G such that .G=N / D .G/. (The set of such pairs is inﬁnite: let G be a minimal nonabelian p-group with noncyclic center and N < G with G 0 6 N .) Problem 6. Find .†pn /, where †pn is a Sylow p-subgroup of the symmetric group Spn of degree p n . Deal with the same problem for UT.m; p n /, a Sylow p-subgroup of the general linear group GL.m; p n /. Problem 7. Study the nonabelian p-groups G such that CG .H / is abelian for all minimal nonabelian H G. (See the paragraph following Remark 7.) Problem 8. Study the nonnilpotent groups G such that ƒ.H / ƒ.G/ for all minimal nonnilpotent H G. (Compare this with Lemma 2.2(b).) Problem 9. Study the groups that are covered by (i) minimal nonnilpotent subgroups, (ii) minimal nonabelian subgroups, (iii) Frobenius subgroups. Problem 10. Classify the p-groups that are covered by subgroups of maximal class. Problem 11. Let H be a proper subgroup of maximal class of a p-group G such that ƒ.H / ƒ.G/. Study the structure of G. Problem 12. Find .G/, where G 2 ¹An ; Sn º (for example, .A5 / D 21; see also [Br].)

117

2-groups all of whose nonnormal subgroups are either cyclic or of maximal class

The 2-groups all of whose nonnormal subgroups are cyclic are completely determined in 16. The ﬁrst step in classifying the title groups was done by the ﬁrst author in Theorem A.40.25. We ﬁnish the classiﬁcation of the title groups by proving the following result which solves Problem 1978. Theorem 117.1. Let G be a 2-group all of whose nonnormal subgroups are either cyclic or of maximal class. Assume, in addition, that G has a nonnormal subgroup of maximal class. Then we have one of the following possibilities: (a) G is generalized quaternion Q2n , n 5. (b) G D hy; b j y 8 D b 4 D 1; b y D bu; y 2 D ua; uy D uz; ay D a1 ; a2 D b 2 D Œa; b D z; Œu; a D Œu; b D 1i, where G is of order 25 and class 3 with 2 .G/ D hui ha; bi Š C2 Q8 , Z.G/ D hzi Š C2 , G 0 D hu; zi Š E4 and ˆ.G/ D hu; ai Š C2 C4 . (c) G is any 2-group of class 2 with 1 .G/ D G 0 Š E4 having a nonnormal quaternion subgroup. Conversely, each group in (a), (b) and (c) satisﬁes the assumptions of our theorem. Proof. Let H be any nonnormal subgroup of maximal class. Suppose that H is not generalized quaternion. Let V be a four-subgroup in H . Then V E G and so H Š D8 . But H has another four-subgroup W ¤ V and W is normal in G. It then follows that H D hV; W i E G, a contradiction. Therefore H is generalized quaternion. If G is of maximal class, then G Š Q2n , n 5 (part (a) of our theorem), and so we assume in the sequel that G is not of maximal class. This implies that G possesses a normal four-subgroup U and G=U is Dedekindian. Let Q be a nonnormal subgroup of maximal possible order in G such that Q Š Q2m , m 3. Set N D NG .Q/ and hzi D Z.Q/ so that Q < N < G, N E G and z 2 Z.N /. Then each subgroup X of N with X > Q is normal in G and so N=Q is Dedekindian. If X1 =Q and X2 =Q are two distinct subgroups of order 2 in N=Q, then X1 E G, X2 E G and X1 \X2 D Q, a contradiction. We have proved that N=Q ¤ ¹1º is either cyclic or N=Q Š Q8 . Let t be any involution in G Q. Since U D hz; ti E G and U \Q D hzi, we have j.UQ/ W Qj D 2 which implies t 2 N . On the other hand, Q=hzi Š D2m1 (if m 4)

117 2-groups with nonnormal subgroups being cyclic or of maximal class

177

or Q=hzi Š E4 (if m D 3) and G=U is Dedekindian. Hence m D 3 and Q Š Q8 . Set L D QU so that L E G and L=Q D 1 .N=Q/. Each involution in G is contained in N and so each involution in G is also contained in L. If CL .Q/ Q, then L would be of maximal class (and order 24 ), contrary to the fact that U is a normal four-subgroup in G. Hence L D V Q with V \ Q D hzi. If V Š C4 , then Q is a unique quaternion subgroup in L. In that case, Q E G, a contradiction. Hence V Š E4 , L D hti Q and 1 .G/ D U D hz; ti. Set C D CG .Q/ so that C \ L D ht; zi and C E G. If C does not cover N=Q, then N=C Š D8 (since Aut.Q8 / Š S4 ), contrary to the fact that G=U is Dedekindian. Thus we get N D Q C with Q \ C D hzi and C =hzi Š N=Q is either cyclic or quaternion. Set Q D hv; wi, where hvi, hwi, hvwi are three cyclic subgroups of index 2 in Q. Now, hv; ti, hw; ti and hvw; ti are normal in G and so hzi Z.G/, jG W NG .hvi/j 2, jG W NG .hwi/j 2 and NG .hvi/ \ NG .hwi/ D N . Hence either jG W N j D 2 or jG W N j D 4 and in the second case G=N Š E4 , where NG .hvi/=N , NG .hwi/=N and NG .hvwi/=N are three distinct subgroups of order 2 in G=N . In any case, there is an element v of order 4 in Q and an element x 2 G N with x 2 2 N so that v x D v t, where 2 ¹1; 1º. Hence Œv; x D t or zt and so, in particular, hz; ti G 0 . Clearly, z is not a square in C . Indeed, if there is y 2 C with y 2 D z, then o.vy/ D 2, where v 2 Q hzi and y 2 C L and so vy 2 N L, contrary to 1 .G/ D ht; zi. (i) First assume C > ht; zi. Then C =hzi is either cyclic of order 4 or C =hzi Š Q8 . In the ﬁrst case, there is c 2 C such that hci covers C =hzi and then (since z is not a square in C ) C D hci hzi, Z.N / D C and 1 .hci/ is characteristic in C . It follows that 1 .hci/ Z.G/ and U D ht; zi Z.G/. Suppose that C =hzi Š Q8 . Then we get ¹1º ¤ C 0 U D ht; zi. If C 0 D U , then, by a result of O. Taussky (Lemma 1.6), C is of maximal class and order 24 , a contradiction. Hence jC 0 j D 2 and since C 0 covers U=hzi, we have U D C 0 hzi. Then it follows that C 0 Z.G/ and so again U D ht; zi Z.G/. We know that G=U is Dedekindian. Assume that G=U is nonabelian, i.e., G=U is Hamiltonian. In that case, G=U has a quaternion subgroup V =U Š Q8 . Set W=U D Z.V =U / D .V =U /0 so that W is abelian of type .4; 2/ and therefore all elements in W U are of order 4. Let S1 , S2 , S3 be three maximal subgroups of V which contain U . Then Si =U Š C4 and since U Z.G/, each Si is an abelian maximal subgroup of V , i D 1; 2; 3. By Exercise P1, jV 0 j D 2. On the other hand, V 0 covers W =U and so there are involutions in W U , a contradiction. Hence G=U is abelian and so G 0 D U D 1 .G/ Z.G/. We have obtained the groups from part (c) of our theorem. (ii) Now assume C D ht; zi D U so that N D L D ht i Q Š C2 Q8 and jGj D 25 or 26 . Suppose, in addition, that 2 .G/ D N . In that case, Theorem 52.1(d) implies that G is a unique group of order 25 with Z.G/ D hzi, G 0 D ht; zi and ˆ.G/ Š C4 C2 which appears in part (b) of our theorem. From now on we may assume that there is an element x 2 G N of order 4 and so 1 ¤ x 2 2 U . In this case,

178

Groups of prime power order

we obtain j.G=U / W 1 .G=U /j 2 which implies that the Dedekindian group G=U cannot be Hamiltonian and so G=U is abelian which gives G 0 D U . We have also hzi Z.G/ U . Since 1 .G/ D U , the case U hxi Š D8 is not possible and so x centralizes U . Because x normalizes N D L and does not normalize Q, there is an element v of order 4 in Q with v 2 D z such that x does not normalize hvi. On the other hand, hv; ti E G and so we may set v x D vt for a suitable element t 2 U hzi which gives Œv; x D t. But U D hz; ti Z.N hxi/ and so hv; xi is minimal nonabelian with .hv; xi/0 D hŒv; xi D ht i. Also, hv; xi ht; zi D U and so hv; xi E G. This implies that .hv; xi/0 D ht i Z.G/ and so Z.G/ D ht; zi. We have again obtained some groups from part (c) of our theorem.

118

Review of characterizations of p-groups with various minimal nonabelian subgroups

Recall that a group G is said to be minimal nonabelian if it is nonabelian but all its maximal subgroups are abelian. A p-group H is minimal nonabelian if and only if H D ha; bi is two-generated and jH 0 j D p, i.e., Œa; bp D Œa; b; a D Œa; b; b D 1. In that case, ˆ.H / D Z.H / so that jH W Z.H /j D p 2 . Exercise 1.8(a) ([Red] and Lemma 65.1). Let G be a minimal nonabelian p-group. Then jG 0 j D p and G=G 0 is abelian of rank two and G is one of the following groups: m

n

(a) G D ha; b j ap D b p D 1; ab D a1Cp (G is metacyclic). m

m1

i, m 2, n 1, jGj D p mCn

n

(b) G D ha; b j ap D b p D c p D 1; Œa; b D c; Œa; c D Œb; c D 1i is nonmetacyclic of order p mCnC1 and if p D 2, then m C n > 2. Next, G 0 is a maximal cyclic subgroup of G. (c) G Š Q23 . Our group G is nonmetacyclic if and only if G 0 is a maximal cyclic subgroup of G. Proposition 10.19. Let G be a metacyclic p-group containing a nonabelian subgroup B of order p 3 . Then (a) if p D 2, then G is of maximal class, (b) if p > 2, then jGj D p 3 , i.e., G D B. Proposition 10.28. A nonabelian p-group is generated by its minimal nonabelian subgroups. Theorem 10.33. All minimal nonabelian subgroups of a 2-group are dihedral if and only if G D hxi A, where ax D a1 for all a 2 A. Theorem 10.34(a). A p-group G, p > 2, is generated by minimal nonabelian subgroups of exponent p unless it contains a subgroup Š †p2 2 Sylp .Sp2 /. The following result generalizes Theorem 10.28 and plays an important role in what follows. Lemma 57.1. Let G be a nonabelian p-group and let A be any maximal abelian normal subgroup of G. Then for each x 2 G A there is a 2 A such that ha; xi is minimal nonabelian.

180

Groups of prime power order

It follows that minimal nonabelian subgroups of G cover the set G A and so they generate G (see also Theorem 10.28). We have the following open problem. Classify ﬁnite p-groups which are covered by its minimal nonabelian subgroups. We have solved only the following two special cases of this problem. Lemma 65.2(a). Let G be a nonabelian p-group such that G 0 1 .Z.G// and d.G/ D 2. Then G is minimal nonabelian. Theorem 76.A. If G is a nonabelian p-group containing p 2 C p C 1 minimal nonabelian subgroups, then all its subgroups of index p 3 are abelian. Theorem 92.1. Let G be a nonabelian p-group such that for each minimal nonabelian subgroup H of G and each x 2 H Z.G/ we have CG .x/ H . Then G is one of the following groups: (a) G is minimal nonabelian. (b) p D 2, d.G/ D 3 and G D ha; b; c j a4 D b 4 D c 4 D 1; Œa; b D c 2 ; Œa; c D b 2 c 2 ; Œb; c D a2 b 2 ; Œa2 ; b D Œa2 ; c D Œb 2 ; a D Œb 2 ; c D Œc 2 ; a D Œc 2 ; b D 1i; where G is a special 2-group of order 26 with ha2 ; b 2 ; c 2 i D G 0 D Z.G/ D ˆ.G/ D 1 .G/ Š E8 and G is isomorphic to an S2 -subgroup of the simple group Sz.8/. (c) p > 2, d.G/ D 2, G is of order p 5 and 2

2

G D ha; x j ap D x p D 1; Œa; x D b; Œa; b D y1 ; Œx; b D y2 ; p

p

b p D y1 D y2 D Œa; y1 D Œx; y1 D Œa; y2 D Œx; y2 D 1; ˇ

ap D y1˛ y2 ; x p D y1 y2ı i; where in case p > 3, 4ˇ C .ı ˛/2 is a quadratic non-residue (mod p). Here ˆ.G/ D G 0 D hb; y1 ; y2 i D 1 .G/ Š Ep3 ; Z.G/ D K3 .G/ D Ã1 .G/ D hy1 ; y2 i Š Ep2 : Conversely, all the above groups satisfy the assumptions of the theorem. Theorem 92.2. Let G be a nonabelian p-group such that whenever A is a maximal subgroup of any minimal nonabelian subgroup H in G, then A is also a maximal abelian subgroup of G. Then each abelian subgroup of G is contained in a minimal nonabelian subgroup and one of the following holds: (a) G is minimal nonabelian. (b) G is metacyclic. (c) G is isomorphic to the group of order 26 deﬁned in Theorem 92.1(b). (d) G is isomorphic to a group of order p 5 deﬁned in Theorem 92.1(c).

118

Review of p-groups with various minimal nonabelian subgroups

181

Theorem 92.2 is in fact a proper generalization of Theorem 92.1 since the assumption of Theorem 92.1 implies the assumption of Theorem 92.2 but not vice versa. The assumptions of both Theorems 92.1 and 92.2 imply that each abelian subgroup of G is contained in a minimal nonabelian subgroup and so in both cases G is indeed covered by its minimal nonabelian subgroups. If we know something about the structure of minimal nonabelian subgroups of a 2-group G, then we are able in some cases to determine the structure of G. In studying 2-groups all of whose minimal nonabelian subgroups are of exponent 4 (which, in general, is still an unsolved problem) the following lemma is crucial. Lemma 57.2. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are of exponent 4 and let A be a maximal normal abelian subgroup of G. Then all elements in G A are of order 4 and so either exp.A/ D 2 or exp.A/ D exp.G/. If x 2 G A with x 2 2 A, then x inverts each element in Ã1 .A/ and in A=1 .A/. If exp.G/ > 4, then either G=A is cyclic of order 4 or G=A Š Q8 . Theorem 90.1. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to D8 or Q8 . Then G is one of the following groups: (a) G is generalized dihedral (i.e., jG W H2 .G/j D 2). (b) G D H Z.G/, where H is of maximal class and Ã1 .Z.G// Z.H /. (c) G D H Z.G/, where H is extraspecial and Ã1 .Z.G// Z.H /. Theorem 57.3. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to H2 D ha; b j a4 D b 4 D 1; ab D a1 i. Then the following holds: (a) If G is of exponent 8, then G has a unique abelian maximal subgroup A. We have exp.A/ 8 and E D 1 .A/ D 1 .G/ D Z.G/ is of order 4. All elements in G A are of order 4 and if v is one of them, then CA .v/ D E and v inverts ˆ.A/ and on A=E. (b) If G is of exponent 4, then G D K V , where exp.V / 2 and for the group K we have one of the following possibilities: (b1) K Š H2 is of order 24 . (b2) K is the minimal nonmetacyclic group of order 25 (see Theorem 66.1(d)). (b3) K is a unique special group of order 26 with Z.K/ Š E4 in which every maximal subgroup is minimal nonmetacyclic of order 25 (from (b2)): K D ha; b; c; d j a4 D b 4 D 1; c 2 D a2 b 2 ; Œa; b D 1; ac D a1 ; b c D a2 b 1 ; d 2 D a2 ; ad D a1 b 2 ; b d D b 1 ; Œc; d D 1i: (b4) K is a splitting extension of B D B1 Bm , m 2, with a cyclic group hbi of order 4, where Bi Š C4 , i D 1; 2; : : : ; m, and b inverts each element of B (and b 2 centralizes B).

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Theorem 57.4. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to H16 D ha; t j a4 D t 2 D 1; Œa; t D z; z 2 D Œa; z D Œt; z D 1i. Then 1 .G/ is a self-centralizing elementary abelian subgroup and G is of exponent 4. Moreover, the centralizer of any element of order 4 is abelian of type .4; 2; : : : ; 2/. Theorem 57.5. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to H32 D ha; b j a4 D b 4 D 1; Œa; b D z; z 2 D Œa; z D Œb; z D 1i. Then 1 .G/ Z.G/ and G is of exponent 4 and class 2. Theorem 57.6. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to M16 D ha; t j a8 D t 2 D 1; at D a5 i. Then E D 1 .G/ is elementary abelian and if A is a maximal normal abelian subgroup of G containing E, then A is of type .2s ; 2; : : : ; 2/, s 2, jG W Aj 4, each element x in G A is of order 8, jE W CE .x/j D 2, x inverts each element in A=E and in Ã1 .A/ and we have the following possibilities: (a) s > 2 in which case jG W Aj D 2 and G=E Š Q2s is generalized quaternion of order 2s . (b) s D 2 in which case either G Š M16 V with exp.V / 2 or G=E Š Q8 and jG W Aj D 4. In Theorem 92.6 we classify nonabelian 2-groups all of whose minimal nonabelian subgroups are isomorphic to Q8 or H2 D ha; b j a4 D b 4 D 1; ab D a1 i. This theorem generalizes a result of N. Blackburn concerning p-groups G which possess nonnormal subgroups and such that the intersection of all nonnormal subgroups is nontrivial (Corollary 92.7). Namely, it is easy to see that in such p-groups G we must have p D 2 and each minimal nonabelian subgroup of G is isomorphic to Q8 or H2 . Theorem 92.6. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to Q8 or H2 D ha; b j a4 D b 4 D 1; ab D a1 i. Then the following holds: (a) If G is of exponent > 4, then G has a unique abelian maximal subgroup A, jG 0 j > 2 and all elements in G A are of order 4. We have 1 .A/ D 1 .G/ Z.G/ and if x 2 G A, then x inverts each element of A=1 .A/. (b) If G is of exponent 4, then G D K V , where exp.V / 2 and for the group K we have one of the following possibilities: (b1) K Š Q8 or K Š H2 . (b2) K D ha; b; c j a4 D b 4 D c 4 D Œa; b D 1; c 2 D a2 ; Œa; c D b 2 ; Œb; c D a2 i is the minimal nonmetacyclic group of order 25 . (b3) K is a unique special group of order 26 with Z.K/ Š E4 given in Theorem 57.3(b3) in which every maximal subgroup is isomorphic to the minimal nonmetacyclic group of order 25 (from (b2)). (b4) K Š Q8 C4 .

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183

(b5) K Š Q8 Q8 . (b6) G D K V has an abelian maximal subgroup B of exponent 4 and an element v 2 G B of order 4 which inverts each element of B. (b7) K D Q C is a central product of Q D ha; bi Š Q8 and C D hc; d j c 4 D d 4 D 1; c d D c 1 i Š H2 with Q \ C D hc 2 d 2 i D Z.Q/, where K is special of order 26 and Z.K/ D 1 .K/ Š E4 . Conversely, in each of the above groups in part (b) every minimal nonabelian subgroup is isomorphic to Q8 or H2 . Corollary 92.7 ([Bla7]). Let G be a ﬁnite p-group which possesses nonnormal subgroups and let R.G/ be the intersection of all nonnormal subgroups. If R.G/ > ¹1º, then p D 2, jR.G/j D 2 and G is one of the following groups: (a) G Š Q8 C4 E2s , s 0. (b) G Š Q8 Q8 E2s , s 0. (c) G has an abelian maximal subgroup A of exponent > 2 and an element x 2 G A of order 4 which inverts each element in A. Theorem 93.1. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are metacyclic and have exponent 4 and assume that D8 and H2 D ha; b j a4 D b 4 D 1; ab D a1 i (and possibly Q8 ) actually appear as subgroups of G. Then G has a unique abelian maximal subgroup A of exponent > 4 and an involution t 2 G A such that t inverts each element in A=1 .A/ and in 2 .A/ and there is an element a 2 A of order > 4 such that at D a1 with 1 ¤ 2 1 .A/, where in case that Ã1 .A/ is cyclic we have 62 Ã1 .A/. Conversely, all these 2-groups satisfy the assumptions of the theorem. Theorem 94.1. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are nonmetacyclic and have exponent 4. Then exp.G/ D 4, 1 .G/ is elementary abelian and if A is a maximal normal abelian subgroup of G containing 1 .G/, then ˆ.G/ 1 .A/ is elementary abelian and Ã1 .A/ Z.G/. Theorem 98.1. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to M2nC1 , n 3 ﬁxed. Then E D 1 .G/ is elementary abelian. Let A be a maximal normal abelian subgroup of G containing E. Then A D CG .E/, all elements in G A are of order 2n , G=A is elementary abelian of order 4, and if x 2 G A, then jE W CE .x/j D 2, A0 D CA .x/ D CE .x/hx 2 i, A=A0 is cyclic, hA; xi0 is cyclic of order jA=A0 j, hA; xi0 \ A0 D 1 .hxi/, x inverts A=A0 and Ãn2 .A/. We have two possibilities: (a) exp.A/ D 2m 2n in which case jG W Aj D 2 and A is of type .2m; 2n2; 2; : : : ; 2/. (b) exp.A/ 2n1 in which case A is of type .2n1; 2r ; 2; : : : ; 2/ with 1 r n 2 and either jG W Aj D 2 or jG W Aj D 4, where in the last case G=n2 .A/ Š Q8 . Conversely, if G is a 2-group of type (a) or (b), then each minimal nonabelian subgroup of G is isomorphic to M2nC1 , n 3 ﬁxed.

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Finally, we consider the following general problem. Theorem 95.1. Let G be a nonabelian 2-group of exponent 2e (e 3) which does not possess a minimal nonabelian subgroup of exponent 2e . Then G has a unique maximal normal abelian subgroup A. The factor group G=A is either cyclic or generalized quaternion and each element in G A is of order < 2e . Also, G possesses at least one metacyclic minimal nonabelian subgroup and jCG .1 .A// W Aj 2. Proposition (see commentary of Mann to Problem 115). If p > 2 and all minimal nonabelian subgroups of a p-group G are of exponent p and exp.G/ > p, then we have jG W Hp .G/j D p. Exercise A40.79. Given a positive integer n and a p-group G of order p n , put Dn .G/ D hH 0 j H < G; jG W H j D p n i. Obviously, Dn .G/ is a characteristic subgroup of G. If G is nonabelian, then D1 .G/ < G if and only if G=D1 .G/ is minimal nonabelian. Exercise 124.7. Let A be a proper minimal nonabelian subgroup of a p-group G. Suppose that all A-containing subgroups of G of order pjAj are two-generator. Then A=A0 is a maximal abelian subgroup of NG .A/=A0 . Theorem 111.3. The following assertions for a group G are equivalent: (a) ˆ.G/0 D H 0 for all H 2 1 . (b) G is either abelian or minimal nonabelian. Theorem 124.30. Let H Š H2;p be a proper subgroup of a metacyclic p-group G. Here H2;p is the unique nonabelian metacyclic group of order p 4 and exponent p 2 . Then one of the following holds: (a) If p > 2, then jG W H j D p, jG 0 j D p 2 and either 3

2

3

2

2

G D hz; u j z p D up D 1; z u D z 1Cp i or G D hz; u j z p D up D 1; z u D z 1Cp i: (b) If p D 2, H G G, then jG W H j D 2. (c) If p D 2 and H is not normal in G, then 1 .G/ D 1 .H / and G=1 .G/ is either dihedral or semidihedral. Proposition 124.31. Suppose that H Š Mpn , n > 3, is a proper subgroup of a metacyclic p-group G. Then NG .H /=H is cyclic. Next assume that n D 4. (a) If p > 2, then jNG .H /=H j D p. (b) If p D 2, then jNG .H /=H j 4.

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In this section we review characterizations of p-groups of maximal class obtained in the book. We hope that this section will help the readers. Recall that a group G of order p m is called a p-group of maximal class if m > 2 and cl.G/ D m 1. Thus p-groups G of maximal class are nonabelian. Proposition 1.3. Suppose that a p-group G of order p m has only one subgroup of order p n , 0 < n < m. Then either G is cyclic or p D 2, n D 1 and G Š Q2m . Lemma 1.4. Let N be a normal subgroup of a p-group G. If N has no G-invariant abelian subgroup of type .p; p/, then N is either cyclic or a 2-group of maximal class. In particular, if N and G are as in Lemma 1.4 and Z2 .G/ \ N is cyclic, then N is either cyclic or a 2-group of maximal class. For N D G we get a known result of P. Roquette [Roq]: A p-group without normal abelian subgroup of type .p; p/ is either cyclic or 2-group of maximal class. Proposition 1.6 (O. Taussky). If G is a nonabelian 2-group such that jG W G 0 j D 4, then G 2 ¹D2m ; Q2m ; SD2m º. It follows from Proposition 1.6 that the groups D2m , Q2m and SD2m exhaust all groups of maximal class and order 2m (m > 2). Therefore some results stated below are trivial for p D 2. Proposition 1.8 (Suzuki). Let G be a nonabelian p-group and let A < G be of order p 2 . If CG .A/ D A, then G is of maximal class. In particular, if x 2 G # is such that jCG .x/j D p 2 , then G is of maximal class. By Theorem 9.6(f), if G is a p-group of maximal class, there is x 2 G such that jCG .x/j D p 2 . Exercise 9.1. Let G be a p-group of maximal class and order p m . Then (a) jG W G 0 j D p 2 , jZ.G/j D p, nonabelian epimorphic images of G are of maximal class. (b) If 1 i < m 1, then G has only one normal subgroup of order p i . (c) If p > 2 and m > 3, then G has no cyclic normal subgroups of order p 2 .

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Note that Exercise 9.1(b) and Lemma 1.4 imply Exercise 9.1(c). Lemma 9.1. Let G be a noncyclic subgroup of order p m , m > 2. If G contains, for any i D 1; : : : ; m 2, only one normal subgroup of order p i , then G is of maximal class. The converse is trivial. A number of most important properties of p-groups of maximal class is listed in the following two theorems. Theorem 9.5. Let G be a group of maximal class and order p m , m p C 1. Then ˆ.G/ and G=Z.G/ have exponent p. If m D p C 1, then the group G is irregular and jÃ1 .G/j D p. It follows from Theorem 9.5 that a p-group of maximal class and order > p pC1 is irregular. Theorem 9.6. Let G be a group of maximal class and order p m , p > 2, m > p C 1. Then (a) jG W Ã1 .G/j D p p . In particular, Ã1 .G/ D Kp .G/. (b) There is G1 2 1 such that jG1 W Ã1 .G1 /j D p p1 . This subgroup G1 is absolutely regular. (c) G has no normal subgroups of order p p and exponent p (this follows from (b) and Exercise 9.1(b); here the condition m > p C 1 is essential). Moreover, if N G G and jG W N j > p, then N is absolutely regular since N < G1 . Next, if N G G of order p p1 , then exp.N / D p, N D 1 .G1 / D 1 .G 0 /. In particular, G has no normal cyclic subgroups of order p 2 (see Exercise 9.1(c)). (d) Let Z2 D Z2 .G/ be a normal subgroup of order p 2 in G, G0 D CG .Z2 /. Then G0 is regular such that j1 .G0 /j D p p1 (see (b)). (e) Let 1 D ¹M1 D G0 ; M2 ; : : : ; MpC1 º, where G0 is deﬁned in (d). Then the subgroups M2 ; : : : ; MpC1 are of maximal class (and so irregular; see Theorem 9.5). Thus the subgroups G1 from (b) and G0 from (d) coincide. The subgroup G1 is called the fundamental subgroup of G (in general, G1 is deﬁned for all m 4). (f) In this part, m 3 (i.e., we do not assume as in other parts that m > p C 1). The group G has an element a such that jCG .a/j D p 2 , i.e., CG .a/ D ha; Z.G/i (this is trivial for m 2 ¹3; 4º). (g) ŒKi .G/; Kj .G/ Ki Cj C1 .G/ for all i; j 2 ¹1; : : : ; m 1º. Here K1 .G/ D G1 .1 By Exercise 9.1(b) for jGj > p 3 , G > G1 D K1 .G/ > K2 .G/ D G 0 > > Km1 .G/ D Z.G/ > Km .G/ D ¹1º is a chain of characteristic subgroups; all indices of this chain are equal to p. 1 In

what follows we do not use this property.

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Let G1 be the fundamental subgroup of a p-group G of maximal class, jGj > p pC1 , and exp.G1 / D p nC1 . If k n, then jk .G/j D p k.p1/ . If p D 3, then G1 is metacyclic: G1 is either abelian or minimal nonabelian (this follows from Theorem 9.11, Exercise 9.1(c) and Lemma 65.2(a)). Let G be a p-group of maximal class and order p m > p pC1 and p C 1 n < m. Then the number of subgroups of maximal class and order p n in G is equal to p mn . This follows from Theorem 9.6(e). Indeed, by that theorem, our claim follows in case n D m 1. To prove this for remaining values n, one should apply induction. Theorem 9.7. If a p-group G is such that G=KpC1 .G/ is of maximal class, then G is also of maximal class.2 Theorem 9.8. Let G be a p-group, p > 2. (a) (1st Hall regularity criterion) If G is absolutely regular, it is regular.3 (b) (Hall) If G has no normal subgroups of order p p1 and exponent p, it is regular.4 (c) (2nd Hall regularity criterion) If jG 0 W Ã1 .G 0 /j < p p1 , then G is regular. In particular, if G 0 is cyclic, then G is regular. (d) If G is irregular, then G 0 has a characteristic subgroup of exponent p and order p p1 . Theorem 9.11 ([Bla2, Theorem 2.6(i)]; Huppert; see also Corollary 36.7). If p > 2, then G is metacyclic if and only if jG W Ã1 .G/j p 2 . Exercise 9.13. If a p-group of maximal class, p > 2, possesses a subgroup H such that d.H / > p 1, then G Š †p2 , a Sylow p-subgroup of the symmetric group of degree p 2 .5 In particular, d.H / p for all H G (this is also true for p D 2). Remark 10.5. If A is a subgroup of a p-group G such that NG .A/ is of maximal class, so is G. Proposition 10.17. Let B be a nonabelian subgroup of order p 3 of a p-group G. If B satisﬁes CG .B/ < B, then G is of maximal class. It follows that if a 2-group G has at most two proper subgroups isomorphic to D8 , then G 2 ¹SD16 ; D16 ; SD32 º: Note that we do not know all 2-groups which contain only one subgroup isomorphic to Q8 . 2 For

a more general result, see Theorem 12.9. is also true for p D 2 since absolutely regular 2-groups are cyclic. 4 This is a partial case of Theorem 12.1(a). 5 This is not true for p D 2 as the dihedral group D 16 shows. 3 This

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Proposition 10.19. If a metacyclic p-group G contains a nonabelian subgroup of order p 3 , then either jGj D p 3 or G is a 2-group of maximal class.6 Proposition 10.24. Let R be a subgroup of order p of a nonabelian p-group G. If there is only one maximal chain connecting R with G, then either CG .R/ Š Ep2 (in this case, G is of maximal class by Proposition 1.8) or G Š Mpn . Two very important characterizations of p-groups of maximal class are contained in the following theorem which is due to Blackburn. These results are most frequently cited in our book. Theorem 12.1. (a) If a p-group G has no normal subgroup of order p p and exponent p, it is either absolutely regular or irregular of maximal class. (b) Suppose that a p-group G is not absolutely regular. If G contains an absolutely regular subgroup H of index p, then either G is of maximal class or G D H 1 .G/, where 1 .G/ is of order p p and exponent p. In particular, if all regular subgroups of an irregular p-group G are absolutely regular, then G is of maximal class and j1 .G/j D p p1 . The most important assertion of Lemma 1.4 follows from Theorem 12.1(a). Theorem 1.2 follows from Theorem 12.1(b) and Proposition 1.6. It follows from Theorem 12.1(a) that if N is a normal subgroup of a p-group G and N \ Zp .G/ is absolutely regular, then N is either absolutely regular or of maximal class. Recall that ck .G/ is the number of cyclic subgroup of order p k of a p-group G. Lemma 12.3. Let a p-group G be of maximal class and order p m > p pC1 , p > 2. Then (a) c1 .G/ 1 C p C C p p2 .mod p p /. (b) c2 .G/ p p2 .mod p p1 /. (c) If n > 2, then p p1 j cn .G/ D cn .G1 /. (d) The number of subgroups of order p p1 and exponent p in G is 1 .mod p mp /. Condition m > p C 1 is important in Theorem 12.1. Note that the classiﬁcation of groups of order p pC1 is one of the most important problems in p-group theory. Proposition 12.6. Suppose that a group G of order p m is not of maximal class and m > n > p C 1. Let N be the set of normal subgroups D of G such that G=D is of maximal class and order p n . Then jN j 0 .mod p/. The following theorem generalizes Theorem 9.7. Theorem 12.9. Suppose that a p-group G has only one normal subgroup L of index p pC1 . If G=L is of maximal class, then G is also of maximal class.7 6 This

follows from Proposition 10.17. is easily seen that this theorem follows from Proposition 12.6 (this was not noticed in 12 so that theorem was proved independently). 7 It

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Theorem 12.12. (a) If a two-generator p-group G contains a subgroup of maximal class and index p, then G itself is of maximal class. If d.G/ D 3, then G 0 D ˆ.G/, i.e., G=G 0 Š Ep3 . (b) Suppose that a group G of order p m , m > 3, contains a subgroup H of maximal class and index p. If G is not of maximal class, then exactly p 2 maximal subgroups of G are of maximal class. If, in addition, p > 2 and m > 4, then the remaining p C 1 maximal subgroups of G have no two generators and their intersection .G/ has index p 2 in G. If, in addition, H is irregular, then G=KpC1 .G/ is of order p pC1 and exponent p. Proposition 12.13. Let G be a p-group. If A 2 1 is absolutely regular and M < G is irregular of maximal class, then G is also of maximal class. Theorem 13.2. If G is a group of order p m which is neither absolutely regular nor of maximal class, then (a) c1 .G/ 1 C p C C p p1 .mod p p /. In particular, the number of solutions of x p D 1 in G is divisible by p p . (b) If k > 1, then p p1 divides ck .G/. Theorem 13.2 implies Theorem 10.17. Recall that ek .G/ is the number of subgroups of order p k and exponent p in a p-group G. The following theorem is the most important counting result. Theorem 13.5 and 13.6. If a p-group G is neither absolutely regular nor of maximal class, then (a) ep .G/ 1 .mod p/. (b) Let jGj D p m > p n p pC1 . Then the number of subgroups of maximal class and order p n in G is divisible by p 2 . It follows that if N is a normal subgroup of a p-group G and N has no G-invariant subgroup of order p p and exponent p, then it is either absolutely regular or of maximal class. Next, if Zp .G/ \ N is absolutely regular, then N is either absolutely regular or of maximal class. Proposition 13.14. (a) If 1 .G/ is either absolutely regular or irregular of maximal class, so is the p-group G. (b) A p-group G is of maximal class if and only if 2 .G/ is of maximal class. If an irregular p-group G is of maximal class, then one of the following holds: (i) j1 .G/j D p p1 , (ii) jG W 1 .G/j p. If G is of maximal class, then 2 .G/ D G.

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Groups of prime power order

Proposition 13.16. Suppose that a nonabelian p-group G has a cyclic subgroup U of order p 2 such that CG .U / is cyclic. Then G is of maximal class. It is important in the hypothesis of Proposition 13.16 that jU j D p 2 ; for jU j > p 2 the assertion does not hold. For a proof, see Appendix 40. Remark 13.4. Let G be of order p m > p k p pC1 , and M 2 1 . If H < G is of maximal class and order p k such that H 6 M and all such H are of maximal class, then G is also of maximal class. Remark 13.6. Let a p-group G satisfy the following conditions: (i) G contains a proper abelian subgroup A of order p 3 . (ii) Whenever A < H G and jH W Aj D p, then jZ.H /j D p. Then: (a) G is of maximal class, (b) jG W Aj D p. To justify (b), it sufﬁces to take A in CG .R/, where R G G is of order p 2 . Exercise 13.10. (a) Let H < G, where G is a p-group. If every subgroup of G of order pjH j containing H is of maximal class, then G is also of maximal class. (b) Let A be a proper absolutely regular subgroup of a p-group G, p > 2, exp.A/ > p, such that whenever A < B G with jB W Aj D p, then 1 .B/ D B. Then G is of maximal class. If, in addition, jAj > p p , then jG W Aj D p. Proposition 13.18. Let M < G be of maximal class, G is a p-group. (a) Set D D ˆ.M /, N D NG .M / and C D CN .M=D/. Let t be the number of subgroups K G of maximal class such that M < K and jK W M j D p. Then t D c1 .N=M / c1 .C =M /. If G is not of maximal class, then t 0 .mod p/. (b) Suppose, in addition, that M is irregular and G is not of maximal class and a positive integer k is ﬁxed. Then the number t of subgroups L < G of maximal class and order p k jM j such that M < L is a multiple of p. Theorem 13.19. Let G be a group of maximal class of order p m and exponent p e , m > p C 1. Then (a) If 3 k e, then k .G/ G1 , where k .G/ D hx 2 G j o.x/ D p k i, i.e., every element in G G1 has order p 2 . If m > p C 1, then exp.G1 / D p e . (b) If x 2 G G1 , then x p 2 Z.G/ so Hp .G=Z.G// D G1 =Z.G/. Here Hp .G/ D hx 2 G j o.x/ > pi is the Hughes subgroup of G. (c) If H < G, H 6 G1 and jH j > p p , then H is of maximal class. (d) If H < G is irregular, then 1 .G1 / < H . (e) If H < G is of order p p and H 6 G1 , then 1 .G1 / < H . (f) If p > 2 and L < G is of order p p1 , then either L < G1 or Z.G/ < L. (g) If L < G is of order p p1 and L 6 G1 , then 1 .G/ normalizes L. It follows from Theorems 9.5 and 13.19(b) that if G is of maximal class, then we have 2 .G/ D G.

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Theorem 36.12. Let G be a p-group of exponent p e > p 2 and 1 < k < e. Let U be a maximal member of the set of subgroups of G with exponent p k . (a) If U is absolutely regular, then G is also absolutely regular, U D k .G/ and the subgroup U is not of maximal class. (b) If U is irregular of maximal class, then G is also of maximal class. Lemma 36.13. Let G be a p-group of order > p pC1 with j2 .G/j D p pC1 . Then one of the following holds: (a) G is absolutely regular. (b) G is an Lp -group. n

(c) p D 2 and G D ha; b j a2 D 1; a2

n1

D b 4 ; ab D a1C2

n2

i.

Theorem 36.14. Let G be a p-group and suppose that H < G is a maximal member of the set of subgroups of G of exponent p 2 . Suppose, in addition, that jH j D p pC1 . Then one of the following holds: (a) H D 2 .G/ so G is as in Lemma 36.13. (b) G is a 2-group of maximal class (in this case, H is also of maximal class). Recall that Ã2 .G/ D Ã1 .Ã1 .G//. We have Ã2 .G/ Ã2 .G/ (strict inequality is possible). Theorem 36.16. Suppose that a p-group G is such that G=Ã2 .G/ is of maximal class. Then G is also of maximal class. Proposition 25.8(c). If G is an irregular p-group of maximal class and W G ! G0 is a lattice isomorphism, then G is also of maximal class.

120

Nonabelian 2-groups such that any two distinct minimal nonabelian subgroups have cyclic intersection

Our technique with minimal nonabelian subgroups of p-groups gives also the following rather deep result. This solves a problem stated by the ﬁrst author for p D 2. Theorem 120.1. Let G be a nonabelian 2-group such that any two distinct minimal nonabelian subgroups of G have cyclic intersection. Then we have one of the following possibilities: (a) G is minimal nonabelian. (b) G is a 2-group of maximal class. (c) G D Q V , where Q Š Q2n , n 3, is generalized quaternion and V ¤ ¹1º is elementary abelian. (d) G D ha; h j a8 D h4 D 1; Œa; h D v; v 4 D 1; Œv; h D v 2 ; a2 D vh2 i, where jGj D 25 ;

G 0 D hvi Š C4 ;

Z.G/ D 1 .G/ D hh2 ; v 2 i Š E4 ;

ˆ.G/ D hh2 ; vi Š C2 C4 ;

cl.G/ D 3;

and maximal subgroups of G are hh; vi Š H2 (minimal nonabelian), hai hh2 i (abelian of type .8; 2/) and hah; vi hh2 i Š Q8 C2 . Conversely, each of the above groups satisﬁes the assumption of our theorem. Proof. Since minimal nonabelian 2-groups and 2-groups of maximal class satisfy the assumption of our theorem, we may assume that G is neither minimal nonabelian nor a 2-group of maximal class. Suppose that G possesses noncommuting involutions. Since any two noncommuting involutions generate a dihedral subgroup, it follows that D D hz; u; t j z 2 D u2 D t 2 D 1; Œu; z D Œt; z D 1; Œu; t D zi Š D8 is a subgroup of G, where Z.D/ D D 0 D hzi and hu; zi and ht; zi are two four-subgroups of D. If CG .D/ D, then G is of maximal class (Proposition 10.17), a contradiction. Hence CG .D/ 6 D and let v 2 CG .D/ D be such that v 2 2 hzi. Then

120

2-groups with cyclic intersections of minimal nonabelian subgroups

193

we get D1 D hz; u; vt i Š D8 with D \ D1 D hz; ui Š E4 , contrary to our assumption. We have proved that 1 .G/ is elementary abelian of order 4 (since G is not generalized quaternion). Assume that E D 1 .G/ 6 Z.G/. Let t 2 E be a noncentral involution and set H D CG .t/ so that E H and H < G. Let S be a subgroup of G such that S > H 2 and jS W H j D 2 and take an element x 2 S H . Then t 0 D t x ¤ t and .t 0 /x D t x D t which gives W D ht; t 0 i Š E4 , CG .W / D H and S D NG .W /. Set z D t t 0 so that z x D z, Œx; t D z and so M D hx; ti D W hxi is minimal nonabelian with M 0 D hzi. We have either 1 .hxi/ D hzi (in which case M Š M2n , n 4, since M Š D8 is not possible because there are no noncommuting involutions) or 1 .hxi/ D hui with u 2 E W (in which case M is nonmetacyclic minimal nonabelian with 1 .M / D hui W Š E8 ). Suppose that M D hx; ti ¤ S . In this case, M \ H < H and take an element h 2 H M so that y D hx 2 S H and y 62 M . We get t y D t hx D t x D t 0 ;

.t 0 /y D t y D t 2

and

z y D .t t 0 /y D z:

Thus M1 D hy; ti is minimal nonabelian with M \ M1 ht; t 0 i Š E4 , contrary to our assumption. We have proved that S D M D hx; ti D W hxi is minimal nonabelian. If, in addition, 1 .hxi/ D hzi, then E D W E G. But we know that NG .W / D S and so S D G is minimal nonabelian, a contradiction. Hence E D 1 .hxi/ W D 1 .G/ Š E8 and CG .E/ D H so that G=H is isomorphic to a subgroup of D8 and S < G. Since H is a maximal normal abelian subgroup of G, we may use Lemma 57.1. For any element g 2 G S , there is h 2 H such that hg; hi is minimal nonabelian. By our assumption, hg; hi \ 1 .S/ is cyclic and so 1 .hg; hi/ Š C2 which forces hg; hi Š Q8 . In particular, o.g/ D 4 and g 2 2 E so that g induces an involutory automorphism on E (noting that CG .E/ D H < S). Let e 2 E be such that e 0 D e g ¤ e which together with 2 .e 0 /g D e g D e shows that .ee 0 /g D ee 0 ¤ 1. Hence hg; ei is minimal nonabelian with hg; ei0 D hee 0 i. But hg; ei \ S he; e 0 i Š E4 ; contrary to our assumption. We have proved that E D 1 .G/ Z.G/. Suppose that all minimal nonabelian subgroups of G are quaternion. In that case, Theorem 90.1 implies that G D Q V , where Q Š Q2n , n 3, is generalized quaternion and V ¤ ¹1º is elementary abelian, and so we have obtained the groups stated in part (c) of our theorem. In what follows we assume that the group G possesses a minimal nonabelian subgroup H D ha; bi which is not quaternion. Then j1 .H /j 4 and so 1 .H / Š E4 or 1 .H / Š E8 . Since 1 .H / Z.G/, it follows by the assumption of our theorem that H E G and 1 .H / Z.H / D ˆ.H /. Also (Exercise P9), ˆ.H / D ha2 ; b 2 ; Œa; bi. Suppose that there is an involution t 2 GH . As t 2 Z.G/, we get Œat; b D Œa; b and H1 D hat; bi is minimal nonabelian with ˆ.H1 / D h.at/2 D a2 ; b 2 ; Œa; bi D ˆ.H /

194

Groups of prime power order

so that H \ H1 1 .H / is noncyclic, a contradiction. Hence we have proved that 1 .H / D 1 .G/ D E and so 4 jEj 8. Let K be any minimal nonabelian subgroup of G which is distinct from H . If K is not quaternion, then j1 .K/j 4 and 1 .K/ 1 .H / D 1 .G/ so that H \ K is noncyclic, a contradiction. We have proved that each minimal nonabelian subgroup of G which is distinct from H is quaternion. Suppose that G=H is not elementary abelian. Then there is a subgroup L > H such that L=H Š C4 . Let L0 be a unique subgroup of index 2 in L which contains H . If x 2 L L0 , then hxi covers L=H and so o.x/ 8 since there are no involutions in L0 H (noting that 1 .H / D 1 .G/). On the other hand, Lemma 57.1 implies that L is generated by its minimal nonabelian subgroups which are equal to H and quaternion subgroups of L. This shows that L possesses a quaternion subgroup which is not contained in L0 . But this contradicts the above fact that all elements in L L0 are of order 8. We have proved that G=H ¤ ¹1º is elementary abelian. In what follows we shall determine the structure of any subgroup T > H with jT =H j D 2. (i) First we show that T possesses at least one abelian maximal subgroup A. Assume that this is false. Let ¹X; Y; Zº be the set of maximal subgroups of H so that X; Y and Z are abelian and at least one of them, say X, is normal in T . In that case, X is a maximal normal abelian subgroup of T . For each x 2 T H , there is a 2 X such that hx; ai is minimal nonabelian (Lemma 57.1) and so hx; ai Š Q8 . In particular, we obtain o.x/ D 4 and 1 ¤ x 2 2 1 .H / ˆ.H / and x 2 2 Z.T /. It follows that ˆ.T / D ˆ.H / which implies that Y and Z are also maximal normal abelian subgroups of T . Let x0 be a ﬁxed element in T H so that X hx0 i is nonabelian and each minimal nonabelian subgroup of Xhx0 i is isomorphic to Q8 . By Theorem 90.1, X hx0 i D R V , where R Š Q2n , n 3, and exp.V / 2. Since x0 62 ˆ.X hx0 i/ and o.x0 / D 4 with x02 2 Z.T /, it follows that x0 inverts X. Considering in the same way the nonabelian subgroups Y hx0 i and Zhx0 i, we conclude that x0 also inverts Y and Z. But X [ Y [ Z D H and so x0 inverts on H . This implies that H is abelian, a contradiction. Hence we have proved that T possesses at least one abelian maximal subgroup A. (ii) We have 1 .H / D ˆ.H / D Z.T / so that exp.H / D 4 and either jH j D 24 if 1 .H / Š E4 or jH j D 25 if 1 .H / Š E8 . Indeed, let x 2 T .A [ H /. By Lemma 57.1, there is a 2 A such that hx; ai is minimal nonabelian. Since hx; ai ¤ H , we have hx; ai Š Q8 . In particular, we obtain o.x/ D o.a/ D 4, x 2 D a2 D z 2 Z.T / and x inverts a. Suppose that x centralizes an element b of order 4 in A. If b 2 D z, then ab is an involution and .ab/x D a1 b D .ab/z, contrary to the fact that 1 .G/ Z.G/. Hence b 2 ¤ z and we consider the subgroup hab; xi. We have Œab; x D Œa; xb Œb; x D z b D z

and so hab; xi0 D hzi

which implies that hab; xi is minimal nonabelian. But .ab/2 D a2 b 2 D zb 2 ¤ z and x 2 D z so that hab; xi 6Š Q8 , contrary to the fact that hab; xi ¤ H (since, by choice,

120

2-groups with cyclic intersections of minimal nonabelian subgroups

195

x 2 T .A [ H /). We have proved that x does not centralize any element of order 4 in A and so CA .x/ D 1 .H / D 1 .T / which implies that 1 .H / D Z.T /. On the other hand, we conclude that ˆ.H / D Z.H / and ˆ.H / A so that ˆ.H / Z.T /. This forces ˆ.H / D 1 .H /, exp.H / D 4, jH j D 24 if 1 .H / Š E4 and jH j D 25 if 1 .H / Š E8 . (iii) We have jT 0 j D 4 and so A is the unique abelian maximal subgroup of T . Each maximal subgroup X of T which is distinct from the subgroups A and H is of the form X D R1 .H / with R Š Q8 and R \ 1 .H / D Z.R/. We have jH j D 4j1 .H /j and so jT j D 8j1 .H /j as 1 .H / D ˆ.H /. In view of Lemma 1.1, we get jT j D 2jZ.T /jjT 0 j, where Z.T / D 1 .H / so that jT 0 j D 4. By Exercise P1, T has exactly one abelian maximal subgroup. Let X be any maximal subgroup of T distinct from A and H . Then X is nonabelian and each minimal nonabelian subgroup of X is isomorphic to Q8 . By Theorem 90.1, we have X D R V , where R Š Q2n , n 3, and exp.V / 2. On the other hand, 1 .H / D Z.T / D ˆ.H / ˆ.T / and so 1 .H / X which gives Z.R/ V D 1 .X/ D 1 .H /: Since jX j D jH j D 4j1 .H /j, it follows that R Š Q8 . (iv) We have d.T / D 2. Suppose that d.T / > 2 in which case we have ˆ.T / D ˆ.H / D 1 .H / D Z.T /, exp.T / D 4, T 0 Z.T / and d.T / D 3. (iv1) Assume, in addition, that 1 .H / Š E4 so that H D ha; b j a4 D b 4 D 1; Œa; b D z; z 2 ¹a2 ; b 2 ºi Š H2 ; where a 2 A and A Š C4 C4 . Let c 2 A H , where 1 ¤ c 2 2 1 .H / D ha2 ; b 2 i, c 2 ¤ a2 and T D ha; b; ci. If Œc; b D 1, then T =hzi is abelian, contrary to jT 0 j D 4. Hence Œc; b ¤ 1 and hc; bi is minimal nonabelian so that hc; bi Š Q8 (as hc; bi ¤ H ). It follows Œc; b D b 2 and c 2 D b 2 . Consider the subgroup hcb; ai with Œcb; a D z so that hcb; ai Š Q8 since hcb; ai ¤ H . We get z D a2

and

.cb/2 D c 2 b 2 Œb; c D c 2 b 2 b 2 D c 2 D b 2 ¤ a2 ;

a contradiction. (iv2) It remains to consider the case 1 .H / Š E8 so that H D ha; b j a4 D b 4 D 1; Œa; b D d; d 2 D Œd; a D Œd; b D 1i; where a 2 A and 1 .H / D ha2 ; b 2 ; d i Š E8 . Let c 2 AH , where 1 ¤ c 2 2 1 .H / and T D ha; b; ci. If Œc; b D 1, then T =hd i is abelian, contrary to jT 0 j D 4. Hence we get Œc; b ¤ 1 and hc; bi is minimal nonabelian so that hc; bi Š Q8 (since hc; bi ¤ H ). It follows that c 2 D b 2 D Œc; b. We have Œc; ab D Œc; b D b 2 so that hc; abi Š Q8 since hc; abi is minimal nonabelian distinct from H . But .ab/2 D a2 b 2 d ¤ c 2 D b 2 , a contradiction.

196

Groups of prime power order

(v) Now we determine the structure of T . As d.T / D 2, we have A \ H D ˆ.T /;

ˆ.H / D 1 .H / D Z.T / < ˆ.T /;

1 .H / D 1 .T /:

Since jˆ.T / W 1 .H /j D 2, it follows that ˆ.T / is abelian of type .4; 2/ or .4; 2; 2/ (depending on the possibilities 1 .H / Š E4 or 1 .H / Š E8 ). Take some elements h 2 H A and a 2 A H so that ha; hi D T . If Œa; h 2 1 .H / D Z.T /, then T =hŒa; hi is abelian and so T 0 D hŒa; hi, contrary to jT 0 j D 4. Hence we obtain that Œa; h D v 2 ˆ.T / 1 .H / which implies o.v/ D 4, T 0 D hvi Š C4 and T is of class 3. Set v 2 D z so that H 0 D hzi. Since H 0 is not a maximal cyclic subgroup in H , it follows that H is metacyclic, 1 .H / Š E4 and so jT j D 25 . We have H D hh; v j h4 D v 4 D 1; Œh; v D v 2 D zi Š H2 ; where 1 .H / D hh2 ; zi Š E4 . For any element x 2 T .A [ H / there is an element k 2 A such that hx; ki is minimal nonabelian (Lemma 57.1) and so hx; ki Š Q8 which implies that o.x/ D 4 and x 2 2 1 .H /. Also, ˆ.H / D 1 .H / and so all elements in T A are of order 4. If any element in A H is of order 4, then Ã1 .T / D 1 .H /, contrary to A\H D ˆ.T /. Hence o.a/ D 8 and a2 2 .A\H /1 .H /. All elements in A H are of order 8 and so exp.T / D 8. By (iii), we get hahiˆ.T / Š Q8 C2 . Since Œah; v D Œa; vh Œh; v D Œh; v D z, we must have hah; vi Š Q8 and therefore v 2 D z D .ah/2 D ahah D a2 a1 h1 h2 ah D a2 h2 Œa; h D a2 h2 v and so a2 D vh2 . The structure of T is uniquely determined: T D ha; h j a8 D h4 D 1; Œa; h D v; v 4 D 1; Œv; h D v 2 ; a2 D vh2 i; where all elements in AH are of order 8 and A D haihh2 i is abelian of type .8; 2/. It is now easy to determine the structure of G. We know that G=H ¤ ¹1º is elementary abelian. Also, all minimal nonabelian subgroups of G are equal to H Š H2 and some quaternion subgroups. Suppose that jG=H j > 2. Then there exists a subgroup G0 > H such that G0 =H Š E4 . If T =H is a subgroup of order 2 in G0 =H , then the structure of T is uniquely determined in (v) and for any other subgroup L=H of order 2 in G0 =H , we have L Š T . In particular, exp.G0 / D exp.T / D 8. By Theorem 92.6, G0 has a unique abelian maximal subgroup A0 and A0 \ T D A is the unique abelian maximal subgroup of T and is of type .8; 2/. All elements in A0 H are of order 8 and 2 .A0 / D A0 \ H D A \ H which is abelian of type .4; 2/. There is no such abelian group A0 of order 25 . We have proved that G D T and so we have obtained the group of order 25 stated in part (d) of our theorem.

121

p-groups of breadth 2

In this section we prove a result of Parmeggiani and Stellmacher [PS1] on p-groups of breadth 2. We recall that the breadth bG .x/ of an element x in a p-group G is deﬁned as p bG .x/ D jG W CG .x/j and the breadth b D b.G/ is max¹bG .x/ j x 2 Gº. We know that for a p-group G we have b.G/ D 1 if and only if jG 0 j D p (see Exercise 2.7 and Proposition 121.9). This result is also proved in [Kno], where we additionally ﬁnd the result that jG 0 j p 3 for p-groups with b.G/ D 2. Theorem 121.1. Let G be a p-group. Then b.G/ D 2 if and only if one of the following holds: (a) jG 0 j D p 2 , or (b) jG 0 j D p 3 and jG W Z.G/j D p 3 . Moreover, if b.G/ D 2 and jG 0 j D p 3 , then jA=Z.G/j D p for every maximal normal abelian subgroup A of G. We shall prove Theorem 121.1 in a series of lemmas. For that purpose we need the following deﬁnitions. Let A be a normal abelian subgroup in a p-group G and x 2 G. Then p bA .x/ D jA W CA .x/j;

bA D bA .G/ D max¹bA .x/ j x 2 Gº;

BA D ¹x 2 G j bA .x/ D bA .G/º; Mx D ACG .x/;

Gx D hMax

TA D ¹x 2 G j bA .x/ D 1º; \ j a 2 Ai; Dx D Max : a2A

Lemma 121.2. For every x 2 G, ŒA; x D ŒA; hxi D ¹Œa; x j a 2 Aº. In particular, jŒA; xj D p bA .x/ D jA W CA .x/j. Proof. The map a ! Œa; x 2 A (a 2 A) is a homomorphism from A into A since 0 for any a; a0 2 A we have Œaa0 ; x D Œa; xa Œa0 ; x D Œa; xŒa0 ; x. The kernel of this homomorphism is CA .x/ and so A=CA .x/ Š ¹Œa; x j a 2 Aº D A0 , which is a subgroup of A. Suppose that Œa; x i 2 A0 for some i 1. Then Œa; x i C1 D Œa; x i x D Œa; xŒa; x i x D Œa; xŒa; x i ŒŒa; x i ; x 2 A0 and so ŒA; x D ŒA; hxi which gives jŒA; hxij D p bA .x/ .

198

Groups of prime power order

Lemma 121.3. Let u; v 2 G be such that ŒA; u \ ŒA; v D ¹1º. Then CA .u/ CA .v/ or bA .u/ < bA .uv/. Proof. Suppose that CA .u/ 6 CA .v/. Then there is a 2 CA .u/ CA .v/. Thus we get auv D av ¤ a and 1 ¤ Œa; uv D Œa; vŒa; uv D Œa; v so that ŒA; uv \ ŒA; v ¤ 1. We have ()

ŒA; u ŒA; v D ŒA; uvŒA; v:

Indeed, for any a 2 A, we obtain Œa; uv D Œa; vŒa; uv D Œa; vŒa; uŒŒa; u; v 2 ŒA; u ŒA; v; and Œa; u D Œa; .uv/v 1 D Œa; v 1 Œa; .uv/v

1

D Œa; v 1 Œa; .uv/ŒŒa; .uv/; v 1 2 ŒA; uvŒA; v; which proves (). From () and ŒA; uv\ŒA; v ¤ ¹1º it then follows (using also Lemma 121.2) that p bA .u/CbA .v/ < p bA .uv/CbA .v/ which gives bA .u/ < bA .uv/ and we are done. Lemma 121.4. Let x; y 2 TA , T D hx; yi and TN D .T CG .A//=CG .A/. Then one of the following holds: (a) T CG .A/ TA and ŒA; x D ŒA; T ; moreover, CA .x/ ¤ CA .y/ if TN is not cyclic. (b) T CG .A/ TA and CA .x/ D CA .T /; moreover, ŒA; x ¤ ŒA; y if TN is not cyclic and in this case ŒA; T D ŒA; x ŒA; y. (c) CA .T / D CA .x/ \ CA .y/ and ŒA; T D ŒA; xy D ŒA; x ŒA; y. Proof. If TN is cyclic, then jT =TG .A/j D p and so CA .x/ D CA .y/ D CA .T /, and (b) holds. Indeed, we conclude that j.Ahxi/ W .CA .x/hxi/j D p so that if a 2 A CA .x/, then Œa; x 2 A \ .CA .x/hxi/ D CA .x/. Hence 1 ¤ Œa; x 2 Z.ha; xi/ and therefore ha; xi0 D hŒa; xi is of order p. This implies that ha; xi is minimal nonabelian and so Œa; x p D 1 and x p 2 CG .A/. Similarly, we get y p 2 CG .A/ and therefore if TN is cyclic, then jT =CT .A/j D p. In what follows we may assume that the quotient group TN is not cyclic. Assume that ŒA; x D ŒA; y D hsi (where o.s/ D p) and CA .x/ D CA .y/. If a 2 A CA .x/, then we may set ax D as and ay D as j with j 6 0 .mod p/. We have seen in the previj ous paragraph that s 2 Z.Ahxi/ and s 2 Z.Ahyi/. Hence we have ax D as j and j so ax y D as j Cj D a and x j y 2 CG .A/, contrary to our assumption that TN is not cyclic. We have proved that both ŒA; x D ŒA; y and CA .x/ D CA .y/ cannot hold. Assume ﬁrst that xy 2 TA . If, in addition, ŒA; x D ŒA; y D ŒA; T , then by the above CA .x/ ¤ CA .y/ and (a) holds since for each v 2 T CG .A/, ŒA; v D ŒA; T

121 p-groups of breadth 2

199

is of order p and so v 2 TA . If ŒA; x ¤ ŒA; y, then we obtain ŒA; x \ ŒA; y D ¹1º and Lemma 121.3 implies that CA .x/ D CA .y/ D CA .T / and (b) holds since for each v 2 T CG .A/ we conclude that CA .v/ D CA .T / is of index p in A and so v 2 TA . We have jA W CA .T /j D p, CA .T / Z.AT / and A0 D ŒA; x ŒA; y CA .T /. Considering .AT /=A0 , we see that both x and y centralize A=A0 and so T D hx; yi centralizes A=A0 which gives ŒA; T A0 and ŒA; T D ŒA; x ŒA; y. Assume now that xy 62 TA . Since xy 62 CG .A/ (since TN is noncyclic), we have CA .xy/ D CA .x/ \ CA .y/ and so bA .xy/ D 2 and jŒA; xyj D p 2 . On the other hand, for each a 2 A Œa; xy D Œa; yŒa; xy D Œa; yŒa; xŒŒa; x; y 2 ŒA; xŒA; y: Thus ŒA; xy D ŒA; x ŒA; y D ŒA; T and (c) holds. Lemma 121.5. Let x; u; v 2 G be such that uv 1 62 CG .A/ and bA .x/ > 1. Then at least one of the elements u; v; uv; ux; vx; uvx is not in TA . Proof. Assume that D ¹u; v; uv; ux; vx; uvxº TA : Consider

T1 D huv 1 ; ui;

T2 D huv 1 ; vxi;

T3 D hu; vxi:

Then case (a) or (b) of Lemma 121.4 applies to each of the subgroups T1 ; T2 ; T3 . Indeed, considering T3 D hu; vxi, we have u; vx; u.vx/ 2 TA and so (a) or (b) of Lemma 121.4 holds. Consider T1 D huv 1 ; ui, where uv 1 62 CG .A/. We have v 2 T1 and so T1 D huv 1 ; ui D hu; vi, where u; v; uv 2 TA and so (a) or (b) of Lemma 121.4 holds for T1 . In particular, T1 CG .A/ TA and so uv 1 2 TA . Finally, consider T2 D huv 1 ; vxi. Since uv 1 ; vx; uv 1 vx D ux 2 TA , again (a) or (b) of Lemma 121.4 holds for T3 . By Lemma 121.4, we have jŒA; Ti j D p or p 2 for each i 2 ¹1; 2; 3º. Then ﬁx i; j 2 ¹1; 2; 3º such that i ¤ j and jŒA; Ti j D jŒA; Tj j. Since i ¤ j , we have hTi ; Tj i D hi and

Ti \ Tj 6 CG .A/:

Indeed, each Tk hi, k D 1; 2; 3. Also, hT1 ; T2 i D huv 1 ; u; vxi D hu; v; xi D hi; hT1 ; T3 i D huv 1 ; u; vxi D hT1 ; T2 i D hi; hT2 ; T3 i D huv 1 ; u; vxi D hT1 ; T2 i D hi: Finally, uv 1 2 T1 \ T2

and uv 1 62 CG .A/;

u 2 T 1 \ T3

and u 2 TA ;

vx 2 T2 \ T3

and vx 2 TA :

Hence Ti \ Tj 6 CG .A/.

200

Groups of prime power order

Assume that jŒA; Ti j D jŒA; Tj j D p 2 . Then case (b) of Lemma 121.4 applies to Ti and Tj , and every element in hi CG .A/ has the same centralizer on A. Since x 2 hi, it follows that x 2 TA , a contradiction. Assume now that jŒA; Ti j D jŒA; Tj j D p. Then, due to Lemma 121.4(a) and (b), every element in hiCG .A/ has the same commutator with A. But x 2 hiCG .A/ and so jŒA; xj D p and x 2 TA , a contradiction. Lemma 121.6. Let x 2 G and a 2 A. Then the following hold: (a) Œa; Max \ Mx ŒA; x. (b) ŒA; Dx ŒA; x D ŒDx ; x. (c) ŒGx ; x ŒA; G. Proof. Let y 2 Max D CG .ax/A so that y D cb for some c 2 CG .ax/ and b 2 A. It follows ay x y D .ax/y D .ax/cb D .ax/b D ax b and this gives Œa; y D a1 ay D a1 ax b .x 1 /y D x b .x 1 /y D .x b x 1 /.x.x 1 /y / D Œb; x 1 Œx 1 ; y: In particular, Œx 1 ; y 2 ŒA; G and so (noting that ŒA; G E G) Œy; x 2 ŒA; G which gives ŒGx ; x ŒA; G, and this is (c). Assume that y 2 Mx and so y D a0 d , where a0 2 A and d 2 CG .x/. We get Œx 1 ; y D Œx 1 ; a0 d D Œx 1 ; d Œx 1 ; a0 d D Œx 1 ; .a0 /d 2 ŒA; x 1 D ŒA; x: If, in addition, y 2 Max , then we know from the previous paragraph that Œa; y can be written as Œa; y D Œb; x 1 Œx 1 ; y, where b 2 A and so Œa; y 2 ŒA; x, which is (a). We note that ŒMx ; x D ŒA; x. Indeed, if y 2 Mx , then y D a0 d with a0 2 A, d 2 CG .x/ and so Œa0 d; x D Œa0 ; xd Œd; x D Œ.a0 /d ; x 2 ŒA; x: This result together with (a) yields Œa; Dx Œa; Max \ Mx ŒA; x ŒDx ; x ŒMx ; x D ŒA; x and thus ŒA; Dx ŒA; x D ŒDx ; x and this is (b). Lemma 121.7. Let bA > 1. Suppose that ŒG; z ŒA; G for all z 2 G satisfying () Then G 0 D ŒCG .A/; G.

bA .z/ > 1 and

2bA .z/ bA :

121 p-groups of breadth 2

201

Proof. Let x 2 BA and y 2 G. Then we have ŒA; x ŒA; yŒA; yx. Indeed, we obtain that ŒA; y D ¹Œa; y j a 2 Aº and for each a 2 A, Œa; yx D Œa; xŒa; yx D Œa; xŒa; yŒŒa; y; x D Œa; yŒaŒa; y; x D Œa; yŒay ; x which gives ŒA; yx D ¹Œa; yŒay ; x j a 2 Aº. Thus, Œay ; x 2 ŒA; yŒA; yx for all a 2 A and ¹Œay ; x j a 2 Aº D ŒA; x which proves our assertion. It follows that we have one of the following two cases (considering the subgroup ŒA; yŒA; yx which contains ŒA; x, where jŒA; xj D p bA ): (1) y or yx satisfy () for all y 2 G, or (2) y and yx 2 TA for some y 2 G, and bA D 2. Assume that we are in case (1). Then we have ŒG; y ŒA; G or ŒG; yx ŒA; G for all y 2 G. Since x satisﬁes (), we also obtain ŒG; x ŒA; G. Thus, if for an element y 2 G, ŒG; yx ŒA; G, we have for all g 2 G, Œg; yx D Œg; xŒg; yx and so Œg; yx 2 ŒA; G which implies Œg; y 2 ŒA; G since ŒA; G E G and therefore ŒG; y ŒA; G for all g 2 G. Thus G 0 D ŒA; G D ŒCG .A/; G. Suppose that we are in case (2). Then ŒG; z ŒA; G for all z 2 G .TA [ CG .A// and thus ŒG; z ŒCG .A/; G for every z 2 G TA . Assume that there exist u; v 2 G such that Œu; v 62 ŒCG .A/; G, i.e., ŒG; u 6 ŒCG .A/; G and ŒG; v 6 ŒCG .A/; G. Then also Œuv; v; Œux; v; Œvx; u; Œuvx; v 62 ŒCG .A/; G. Indeed, Œuv; v D Œu; vv 62 ŒCG .A/; G; Œux; v D Œu; vx Œx; v 62 ŒCG .A/; G since Œx; v 2 Œx; G ŒA; G, Œu; v 62 ŒCG .A/; G, and Œvx; u D Œv; ux Œx; u 62 ŒCG .A/; G; Œuvx; v D Œuv; vx Œx; v 62 ŒCG .A/; G since Œx; v 2 ŒA; G and Œuv; vx 62 ŒCG .A/; G. Hence u; v; uv; ux; vx; uvx 2 TA . Moreover, Œu; vv

1

1

D Œuv ; v D Œv.uv 1 /; v/ D Œuv 1 ; v 62 ŒCG .A/; G

and so uv 1 62 CG .A/. But then Lemma 121.5 gives a contradiction. Lemma 121.8. Suppose bA D 1. Then either (i) jA W .A \ Z.G//j D p and CA .x/ D CA .y/ for every x; y 2 G CG .A/, or (ii) jŒA; Gj D p. If, in addition, A is a maximal normal abelian subgroup of G, then in case (i) jG W Z.G//j p 1Cb and in case (ii) b.G=ŒA; G/ < b.G/.

202

Groups of prime power order

Proof. If G=CG .A/ is cyclic, then there is g 2 G such that G D CG .A/hgi. It follows that jA=CA .g/j D p, CA .g/ D Z.G/ \ A and ŒA; g D ŒA; G is of order p so (i) and (ii) hold. Hence we may assume that there exist x; y 2 G such that for T D hx; yi the quotient group T CG .A/=CG .A/ is not cyclic. Then cases (a) or (b) of Lemma 121.4 hold for T . Assume that (b) holds for T , i.e., CA .x/ D CA .y/ and ŒA; x ¤ ŒA; y. Then for every z 2 G CG .A/ either ŒA; z ¤ ŒA; x or ŒA; z ¤ ŒA; y. Hence Lemma 121.4 applied to hz; xi or hz; yi gives CA .x/ D CA .y/ D CA .z/ and thus CA .x/ D A \ Z.G/. This is (i). Assume that (a) holds for T , i.e., CA .x/ ¤ CA .y/ and ŒA; x D ŒA; y. Thus for every z 2 G CG .A/ either CA .z/ ¤ CA .x/ or CA .z/ ¤ CA .y/. Hence another application of Lemma 121.4 gives ŒA; x D ŒA; y D ŒA; z and thus ŒA; G D ŒA; x is of order p. This is (ii). Let A be a maximal normal abelian subgroup of G. Suppose that (i) holds and let a 2 A Z.G/. Then CG .a/ D A

and jG W Z.G/j D jG W AjjA W Z.G/j D jG W CG .a/jp p bC1 :

Assume that (ii) holds. Then D D ŒA; G is a normal subgroup of G of order p such that b.G=D/ < b D b.G/. Proposition 121.9 (Knoche). Suppose that b D b.G/ D 1. Then jG 0 j D p. Proof. We suppose, in addition, that A is a maximal normal abelian subgroup of G so that CG .A/ D A. For every x 2 G A we have G D CG .x/A D Dx D Gx . Hence Lemma 121.6(b) gives jŒG; Aj D p and Lemma 121.6(c) yields ŒG; x ŒG; A and so G 0 D ŒG; A is of order p. Proof of Theorem 121.1. Let A be a maximal normal abelian subgroup of G and suppose that b D b.G/ D 2. If bA D 2 for some maximal normal abelian subgroup A of G, then for each x 2 G such that bA .x/ D 2 we have Mx D CG .x/A D G so that Gx D G and Lemma 121.6(c) gives ŒG; x ŒA; G. Hence the condition () of Lemma 121.7 is satisﬁed and so G 0 D ŒA; G. Since G D Max D CG .ax/A for every a 2 A, we have G D Dx and so Lemma 121.6(b) gives ŒA; G ŒA; x. Thus ŒA; G D ŒA; x is of order p 2 and so jG 0 j D jŒA; xj D p 2 . Hence we may assume that bA D 1 for every maximal normal abelian subgroup A of G. We discuss the two cases of Lemma 121.8. Assume that case (i) of Lemma 121.8 holds. For a 2 A Z.G/ we get CG .a/ D A and thus jG=Z.G/j p 3 . Now, b D 2 gives jG=Z.G/j D p 3 and jŒA; Gj D p 2 since a has exactly p 2 conjugates in G. Consider GN D G=ŒA; G so that GN is either N G= N AN Š Ep2 and so GN has more than one abelian maximal abelian or AN D Z.G/, subgroup which implies jGN 0 j D p. Thus jG 0 =ŒA; Gj p and so we get (a) or (b) of our theorem.

121 p-groups of breadth 2

203

Assume that case (ii) of Lemma 121.8 holds, i.e., jŒA; Gj D p. Set GN D G=ŒA; G. N and b.G/ N 1. Hence by Proposition 121.9, jGN 0 j p. As jŒA; Gj D p Then AN Z.G/ and b D 2, we conclude (by Proposition 121.9) that jG 0 j D p 2 and we get case (a) of our theorem. Suppose that G satisﬁes (a) or (b) of our theorem. Then clearly b 2 and Proposition 121.9 gives b D 2 since jG 0 j > p. Theorem 121.1 is proved.

122

p-groups all of whose subgroups have normalizers of index at most p

This section is written by the second author. Here we give a proof of the following result which was proved for p > 2 and predicted for p D 2 by J. P. Bohanon in [Boh]. Theorem 122.1. Let G be a p-group such that for each subgroup H of G we have jG W NG .H /j p. Then the following holds: (a) If p D 2, then jG W Z.G/j 24 . (b) If p > 2, then jG W Z.G/j p 3 . Our proof is elementary but very involved. In particular, we do not use any computer results for 2-groups of orders 28 which have the title property and also we do not use the representation theory for such groups in case p > 2. We recall that the breadth bG .x/ of an element x in a p-group G is deﬁned as p bG .x/ D jG W CG .x/j; and the (element) breadth b.G/ of G is max¹bG .x/ j x 2 Gº: Also, for an abelian normal subgroup A in a p-group G and an element x 2 G we set p bA .x/ D jA W CA .x/j and deﬁne bA D bA .G/ D max¹bA .x/ j x 2 Gº; BA D ¹x 2 G j bA .x/ D bA .G/º; TA D ¹x 2 G j bA .x/ D 1º: We deﬁne the subgroup breadth sb.G/ of a p-group G by p sb.G/ D max¹jG W NG .H /j j H Gº: Note that the title groups are either Dedekindian or have the subgroup breadth 1.

122

p-groups all of whose subgroups have normalizers of index at most p

205

We shall ﬁrst prove some preliminary results (Propositions 122.2 to 122.18 ) about p-groups with the subgroup breadth 1. Some of these results are of independent interest since they give additional information about such groups which is not contained in Theorem 122.1. In particular, Propositions 122.2, 122.7, 122.8, 122.13 and 122.16 give important properties of p-groups with the subgroup breadth 1. Proposition 122.2. Let G be a p-group with sb.G/ D 1. Then ˆ.G/ is abelian. Proof. Obviously, ˆ.G/ is Dedekindian and so either ˆ.G/ is abelian or p D 2 and ˆ.G/ is Hamiltonian, i.e., ˆ.G/ D Q V , where Q Š Q8 and exp.V / 1. Suppose that we are in the second case, where V ¤ ¹1º (by a classical result of Burnside). Set U D Z.Q/ V so that U D Z.ˆ.G// is normal in G and jU W V j D 2. If V E G, then ˆ.G=V / D ˆ.G/=V Š Q8 , contrary to a classical result of Burnside. Thus V is not normal in G and set T D NG .V / so that ˆ.G/ < T and jG W T j D 2. Take an element g 2 G T . We have V g ¤ V and V g < U . It follows that V0 D V \ V g is normal in G and jV W V0 j D 2. Hence ˆ.G=V0 / D ˆ.G/=V0 Š Q8 C2 and such groups G=V0 are classiﬁed in Theorem 85.1. In particular, G=V0 contains a subgroup H of order 26 given by H D hx; y j x 8 D y 8 D 1; x 4 D y 4 D z; x 2 D a; y 2 D b; Œa; b D z; ay D at; t 2 D Œt; a D Œt; b D 1; t x D t y D t z; b x D btz; .xy/2 D tz ; D 0; 1i; where ha; bi Š Q8 ;

ˆ.H / D ha; bi ht i Š Q8 C2 ;

Z.H / D Z.ha; bi/ D hzi:

We have ay D at;

b x D bt z D b 1 t;

ayx D .at/x D at x D atz D a1 t

and so NH .ha; bi/ D ˆ.H / is of index 4 in H , a contradiction. Proposition 122.3. Let G be a p-group with sb.G/ D 1. If A G is a maximal normal abelian containing ˆ.G/ (noting that ˆ.G/ is abelian by Proposition 122.2), then 1 bA .G/ 2, i.e., for each x 2 G A we have p jA W CA .x/j p 2 . Proof. We infer that that G=A ¤ ¹1º is elementary abelian. Let x 2 G A so that x p 2 A and j.Ahxi/ W Aj D p. Set M D NAhxi .hxi/ and A0 D M \ A so that j.Ahxi/ W M j p and jA W A0 j p. We have M 0 A0 \ hxi D hx p i Z.Ahxi/; and so M is of class 2. Suppose that M is of class 2. For any a 2 A0 we have

206

Groups of prime power order

Œa; x 2 hx p i ¤ ¹1º since M 0 hx p i and M 0 ¤ ¹1º. We get Œa; xp D Œa; x p D 1 and so ŒA0 ; hxi D 1 .hxi/ D M 0 is of order p (see Lemma 121.2) which implies jA0 W CA0 .x/j D p. Since CG .A/ D A, we obtain p jA W CA .x/j p 2 , and we are done. Proposition 122.4. If G D Q1 Q2 , where ŒQ1 ; Q2 D ¹1º and Q1 Š Q2 Š Q8 , then we have sb.G/ > 1. Proof. First suppose that Q1 \ Q2 D Z.Q1 / D Z.Q2 / D G 0 D hzi so that G is extraspecial of order 25 and “type +”. In this case, G possesses a foursubgroup S such that S \ hzi D ¹1º. We have ŒNG .S/; S S \ hzi D ¹1º and so NG .S/ D CG .S / D S hzi which gives jG W NG .S/j D 4. Assume now that G D Q1 Q2 , and set Q1 D ha1 ; b1 i with Q10 D hz1 i and Q2 D ha2 ; b2 i with Q20 D hz2 i. Considering the subgroup U D ha1 a2 ; b1 b2 i with U 0 D hz1 z2 i, we have G D Q1 U with Q1 \ U D ¹1º and so NG .U / D U NQ1 .U /, where NQ1 .U / D CQ1 .U /. But .b1 b2 /a1 D b1 b2 z1 ;

.a1 a2 /b1 D a1 a2 z1 ;

.b1 b2 /b1 a1 D b1 b2 z1

and so CQ1 .U / D hz1 i which gives NG .U / D U hz1 i and we are done. Proposition 122.5 (J. Shareshian). Let G be a 2-group with sb.G/ D 1 and we assume that G possesses two involutions which do not commute. Then jG W Z.G/j 24 and jG 0 j 4 so that b.G/ 2. Proof. Let s; t be involutions in G which do not commute. Then hs; ti Š D2n is a dihedral subgroup of order 2n , n 3. If n > 3, then sb.D2n / > 1 and so D D hs; ti Š D8 . Set Z.D/ D hzi and note that ¹s; szº and ¹t; tzº are already complete conjugate classes in G of s and t, respectively. This implies hs; zi E G, ht; zi E G, D E G, G D D C , where C D CG .D/ and D \ C D hzi. If C is Dedekindian, then the fact that Z.C / D Z.G/ implies jG W Z.G/j 24 and jG 0 j 4. Suppose that C has a nonnormal subgroup H such that hzi 6 H and let K ¤ H be a conjugate of H in C . Then the four subgroups ht; H i D hti hH i;

ht; Ki;

htz; H i;

htz; Ki

are pairwise distinct and conjugate subgroups in G, a contradiction. Hence, assuming that C is not Dedekindian, each nonnormal subgroup of C contains hzi. Such groups C are completely determined by N. Blackburn (see Corollary 92.7) and so we have one of the following three possibilities: (1) C Š Q8 C4 E2s ; (2) C Š Q8 Q8 E2s (s 0); (3) C has an abelian normal subgroup A of exponent > 2 with jC W Aj D 2 and an element g of order 4 in C A such that g 2 D z and g inverts each element in A.

122

p-groups all of whose subgroups have normalizers of index at most p

207

In case (1), jG W Z.G/j D 24 and jG 0 j 4. In case (2), sb.Q8 Q8 / > 1 (by Proposition 122.4), a contradiction. Assume that we are in case (3). Note that .st/2 D z so that stg is an involution that inverts on A. As sb.D16 / > 1, we must have exp.A/ D 4. Suppose that there exists b 2 A such that o.b/ D 4 and b 2 ¤ z D g 2 . We obtain that hstg; bi Š D8 so that .stg/b D stgb 2 . Also, .stg/s D stgz and .stgb 2 /s D stgzb 2 . It follows that the four involutions stg, stgz, stgb 2 , stgzb 2 are pairwise distinct and conjugate, a contradiction. Therefore we have an element h 2 A such that h2 D z, hg D h1 , hh; gi Š Q8 and C Š Q8 E2s . But then C is Dedekindian, contrary to our assumption. Proposition 122.6 (J. Shareshian). Let G be a 2-group with sb.G/ D 1, Z.G/ is cyclic and all involutions of G commute with each other. Then G contains at most three involutions. Proof. Suppose that our proposition is false. Since 1 .G/ is elementary abelian of order 8, G possesses a normal elementary abelian subgroup E of order 8. Let z be the unique involution in E which lies in Z.G/, and let t be any involution in E hzi. Since jG W CG .t/j D 2, there are exactly two conjugates ¹t; t 0 º of t in G and so the four-subgroup ht; t 0 i is normal in G which implies z 2 ht; t 0 i and so t 0 D tz. This gives ŒG; E D hzi. Set E D hz; s; t i and consider the four-subgroup S D hs; ti. We have ŒNG .S/; S S \ hzi D ¹1º and so NG .S/ D CG .S/ D CG .s/ \ CG .t/, where CG .s/ and CG .t/ are subgroups of index 2 in G. Since jG W NG .S/j D 2, we get CG .s/ D CG .t/ D CG .E/ with jG W CG .E/j D 2. Let g 2 G CG .E/ so that s g D sz and t g D t z. But then .st/g D st and so st 2 Z.G/, a contradiction. Proposition 122.7. Let G be a metacyclic p-group with jG 0 j D p 2 . If sb.G/ D 1, then p D 2 and either G Š Q16 or G D ha; b j a8 D 1; b 2 D a4 ; ab D a1C4 i; n

where n 2 and D 0; 1. Proof. First suppose jGj D p 4 . Then G has a cyclic subgroup H of index p. If p > 2, then G Š Mp4 and jG 0 j D p, a contradiction. We have p D 2 and G is of maximal class. If G is dihedral or semidihedral, then there exists an involution i 2 G H and jG W CG .i/j D 4, a contradiction. Hence we obtain in this case that G Š Q16 . Now suppose that jGj > p 4 . In view of Corollary 65.3, a metacyclic p-group G is an A2 -group if and only if jG 0 j D p 2 . According to Proposition 71.2, we have two possibilities for the structure of G: (a)

m

n

G D ha; b j ap D 1; b p D ap

m1

; ab D a1Cp

m2

i;

where m 3, n 2, D 0; 1, and in case p D 2, m 4; (b)

p D 2;

G D ha; b j a8 D 1; ; b 2 D a4 ; ab D a1C4 i;

where n 2, ; D 0; 1.

n

208

Groups of prime power order

Suppose that G is a 2-group in (b). If D 0, then jG W NG .hbi/j D 2 implies that b centralizes ha2 i, a contradiction. Hence D 1 and we have obtained all 2-groups stated in our proposition. It remains to be shown that all groups in (a) are not of subgroup breadth 1. Suppose that G is a p-group from (a). If D 0, then jG W NG .hbi/j D p gives that b centralizes ap , a contradiction. Hence D 1 and so we may set G D ha; b j ap D b p D z; z p D 1; ap m

n

m1

D v; ab D avi;

where m 2, n 2, and if p D 2, then m 3. We have jGj D p mCnC1 , Œa; b D v, 2 Œap ; b D z, Œap ; b D 1, G 0 D hvi and G is of class 2 if and only if m 3 since in 2 that case hvi hap i Z.G/. (i) First assume m 3 so that G is of class 2. In that case, we also have Œa; b p D z, 2 Œa; b p D 1. (i1) Suppose m D n. Since hbai covers G=hai Š Cpn , it follows that o.ba/ p n . On the other hand, n

p n n n .ba/p D b p ap Œa; b. 2 / D z 1 z D 1

and so G D haihbai with hai \ hbai. But ap normalizes hbai and so we conclude that hap ihbai D hap i hbai which gives CG .ap / D G, a contradiction. n (i2) Suppose m > n. Then there is a0 2 hai such that ha0 i < hai and .a0 /p D z. Here again hba0 i covers G=hai Š Cpn and so o.ba0 / p n . On the other hand, n

p n n n .ba0 /p D b p .a0 /p Œa0 ; b. 2 / D z 1 z D 1

since a0 2 hap i and so Œa0 ; b 2 hzi. Thus G is a splitting extension of hai by hba0 i. It follows that ap normalizes hba0 i and so ap centralizes hba0 i, ap 2 Z.G/, a contradiction. m (i3) Suppose n > m. Then there is b 0 2 hbi such that hb 0 i < hbi and .b 0 /p D z 1 . Since b 0 2 hb p i, it follows that Œa; b 0 2 hzi. We get pm 2

.b 0 a/p D .b 0 /p ap Œa; b 0 . m

m

m

and .b 0 a/p

m1

D .b 0 /p

m1

ap

m1

/ D z 1 z D 1

p m1 2

Œa; b 0 .

/ D .b 0 /pm1 v ¤ 1;

and so o.b 0 a/ D p m and hbi \ hb 0 ai D ¹1º since .b 0 /p v is an element of order p m1 which is not contained in hzi (noting that v 2 hai and .b 0 /p 62 hai). Now, we get 0 0 nC1 0 m G D hb; b ai D hbihb ai since o.b/ D p , o.b a/ D p , hbi \ hb 0 ai D ¹1º, and jGj D p mCnC1 . But b p normalizes hb 0 ai and so Œb 0 a; b p 2 hvi \ hb 0 ai D ¹1º which gives b p 2 Z.G/, contrary to Œa; b p D z. m1

122

p-groups all of whose subgroups have normalizers of index at most p

209

(ii) Assume that m D 2 so that G is of class 3, G 0 D hvi, ŒG; G 0 D hzi and p > 2. (ii1) Suppose, in addition, that m D n D 2 so that jGj D p 5 . Since hbai covers G=hai Š Cp2 , we have o.ba/ p 2 . By the Hall–Petrescu formula (Appendix A.1), 2

2

p 2 2 2 . / .p / .ba/p D b p ap c2 2 c3 3 ;

where c2 2 hvi, c3 2 hzi and so .ba/p D z 1 z D 1 which gives o.ba/ D p 2 . Hence G D haihbai with hai \ hbai D ¹1º. On the other hand, ap D v normalizes hbai and so v centralizes ba. We get CG .v/ ha; bai D G, contrary to v b D vz. 2 (ii2) Suppose n > 2. Then there is b 0 2 hbi such that hb 0 i < hbi and .b 0 /p D z 1 . Note that ha; b 0 i is a proper subgroup of the A2 -group G and so ha; b 0 i is abelian or minimal nonabelian which gives Œa; b 0 2 hzi. We get 2

p2

.b 0 a/p D .b 0 /p ap Œa; b 0 . 2 / D z 1 z D 1 2

2

2

and so o.b 0 a/ p 2 . Also, we obtain p

.b 0 a/p D .b 0 /p ap Œa; b 0 . 2 / D .b 0 /p v ¤ 1 since .b 0 /p v is an element of order p which is not contained in hzi (indeed, hvi hai and .b 0 /p 62 hai, and so .b 0 /p v 62 hai ). Hence G D hbihb 0 ai with hbi \ hb 0 ai D ¹1º. Then b p normalizes hb 0 ai and so Œb p ; hb 0 ai hb 0 ai \ hvi D ¹1º. Thus we obtain that CG .b p / hb; b 0 ai D G and so b p 2 Z.G/. But then hbi induces on hai an automorphism of order p which implies that jG 0 j D p, a contradiction. Our proposition is proved. Proposition 122.8. Let G be a p-group with sb.G/ D 1. Then b.G/ 2 if p > 2 and b.G/ 3 if p D 2. Proof. Let A be a maximal normal abelian subgroup of G which contains ˆ.G/ (noting that ˆ.G/ is abelian by Proposition 122.2). By Proposition 122.3, 1 bA .G/ 2. Let x 2 A so that NG .hxi/ A and jG W NG .hxi/j p. We have CG .x/ A and NG .hxi/=CG .x/ is elementary abelian acting faithfully on hxi. Hence, if p > 2, then jNG .hxi/=CG .x/j p and if p D 2, then jNG .hxi/=CG .x/j 4. In the ﬁrst case bG .x/ 2 and in the second case bG .x/ 3. Let x 2 G A and assume that jA W CA .x/j D p. First suppose, in addition, that the subgroup A does not normalize hxi in which case NG .hxi/ covers G=A and we have NG .hxi/ \ A D CA .x/. Since NG .hxi/=CG .x/ is elementary abelian of order p in case p > 2 and elementary abelian of order 4 if p D 2, we get again bG .x/ 2 for p > 2 and bG .x/ 3 for p D 2. Now suppose that A normalizes hxi so that NG .hxi/ A and jG W NG .hxi/j p. Assume that p > 2 and that CG .x/ does not cover NG .hxi/=A. Then jNG .hxi/=.ACG .x//j D p and NG .hxi/=CG .x/ Š Cp2 . Let y 2 NG .hxi/CG .x/ be such that hyi covers NG .hxi/=CG .x/ so that y p 2 ACA .x/ 2 and y p 2 CA .x/. In that case, we have o.x/ p 3 and x y D xv, where v 2 hx p i and o.v/ D p 2 . Consider the metacyclic subgroup H D hxihyi with H 0 D hvi Š Cp2 .

210

Groups of prime power order

Since sb.H / D 1, Proposition 122.7 gives a contradiction. Hence if p > 2, CG .x/ must cover NG .hxi/=A and so in this case bG .x/ 2. Now assume that p D 2 and also NG .hxi/=.ACG .x// Š E4 which implies that NG .hxi/=CG .x/ Š C4 C2 . Then there exists again an element y 2 NG .hxi/ such that y 2 2 A CA .x/ and x y D xv, v 2 hx 2 i, o.v/ D 4 and o.x/ 24 . Considering the metacyclic subgroup K D hxihyi with K 0 D hvi Š C4 , Proposition 122,7 gives again a contradiction. Hence we obtain that jNG .hxi/=.ACG .x//j 2 which implies in this case bG .x/ 3. We consider now an element g 2 GA such that jA W CA .g/j D p 2 . Set P D Ahgi, where g p 2 CA .g/. By Lemma 121.2, jŒA; gj D p 2 . If hgi E P , then P 0 A \ hgi D hg p i Z.P /;

ŒA; g hg p i;

and so P is of class 2 and for each a 2 A, Œa; gp D Œa; g p D 1 and this implies ŒA; g D 1 .hgi/, contrary to jŒA; gj D p 2 . Hence jP W NP .hgi/j D p which yields that NG .hgi/ covers G=A. Set A1 D NA .hgi/ so that A1 E G. Suppose that in case p > 2, jNG .hgi/ W .A1 CG .g//j D p and in case p D 2, jNG .hgi/ W .A1 CG .g//j D 4. Then in both cases there is y 2 NG .hgi/ .A1 CG .g// such that y p 2 A1 CA .g/ and g y D gv, where v 2 hg p i, o.v/ D p 2 , o.g/ p 3 and in case p D 2, o.g/ 24 . Considering the metacyclic subgroup L D hgihyi with L0 D hvi Š Cp2 , Proposition 122.7 gives a contradiction. Hence in case p > 2, CG .g/ covers NG .hgi/=A1 and so bG .g/ D 2, and in case p D 2, jNG .hgi/=.A1 CG .g//j 2 so that in this case bG .g/ 3 and we are done. Proposition 122.9. Let G be a p-group with sb.G/ D 1. If G is a minimal counterexample to Theorem 122.1, then 1 .Z.G// G 0 . Proof. Let G be a minimal counterexample to Theorem 122.1 and suppose that there is i 2 1 .Z.G// with i 62 G 0 . Set Z=hi i D Z.G=hi i/ so that for p > 2, jG=Zj p 3 and for p D 2, jG=Zj 24 . Since ŒG; Z hii and i 62 G 0 , we get ŒG; Z D ¹1º and so Z D Z.G/, contrary to the fact that G was a minimal counterexample to Theorem 122.1. Proposition 122.10. Let G be a p-group with sb.G/ D 1. Let A be a maximal normal abelian subgroup of G containing ˆ.G/ such that bA .G/ D 1 (i.e., for each x 2 GA, jA W CA .x/j D p). Then Theorem 122.1 holds. Proof. Let G be a minimal counterexample to this proposition. Using Lemma 121.8, we have either jA W Z.G/j D p or jŒG; Aj D p. (i) First assume jA W Z.G/j D p. Let a 2 A Z.G/ so that CG .a/ D A. In case jG=Aj p 2 , we get jG=Z.G/j p 3 , which contradicts the fact that G is a minimal counterexample. Hence jG=Aj p 3 and then Proposition 122.8 forces p D 2 and bG .a/ D 3 so that jG=Aj D 23 . But then jG=Z.G/j D 24 , a contradiction. (ii) Now assume ŒG; A D hzi is of order p. We may also assume jA=Z.G/j > p which forces jG=Aj > p. For any x; y 2 G A such that hx; yiA=A Š Ep2 we have, by Lemma 121.4, CA .x/ ¤ CA .y/. We show that ˆ.A/ Z.G/. Indeed, if a 2 A

122

p-groups all of whose subgroups have normalizers of index at most p

211

and g 2 G are such that Œa; g ¤ 1, then Œa; g 2 hzi Z.G/ so that ha; gi0 D hzi and therefore ha; gi is minimal nonabelian. Then ap 2 ˆ.ha; gi/ D Z.ha; gi/ and so Œap ; g D 1 which gives ap 2 Z.G/. Due to Proposition 73.8, A possesses a basis ¹a1 ; a2 ; : : : ; an º, n 2, (noting that Ã1 .A/ Z.G/) such that z 2 han i. Set A0 D ha1 ; : : : ; an1 i and G0 D NG .A0 / so p that jG W G0 j p. We get ŒG0 ; A0 A0 \hzi D 1 so that A D A0 han i D Z.G0 /, where jA W A j D p. If jG0 W Aj > p, then there are elements x; y 2 G0 A such that hx; yiA=A Š Ep2 and CA .x/ D CA .y/ D A , contrary to the above remark. Hence jG0 =Aj p which together with jG=Aj > p gives G=A Š Ep2 and jG W G0 j D p. Let g1 ; g2 2 G A so that hg1 ; g2 i covers G=A. Then CA .g1 / ¤ CA .g2 / (by the above remark) yields CA .g1 /\CA .g2 / D Z.G/, A=Z.G/ Š Ep2 and for each x 2 AZ.G/, jG W CG .x/j D p. If p D 2, then jG=Z.G/j D 24 , a contradiction. We must have p > 2 and then Proposition 122.8 implies b.G/ 2. By Proposition 73.8, Z.G/ has a basis ¹z1 ; z2 ; : : : ; zr º, r 1, such that z 2 hz1 i. In case r > 1, we consider the subgroup S D hz2 ; : : : ; zr i ¤ ¹1º and the factor group G=S. Let g 2 G A and assume that Œg; A S. But then Œg; A S \ hzi D 1 and so g centralizes A, a contradiction. Hence A=S is a maximal normal abelian subgroup of G=S . Let a 2 A Z.G/ and assume ŒG; a S . Then ŒG; a S \ hzi D 1 and so a 2 Z.G/, a contradiction. Hence Z.G/=S D Z.G=S/ and so jG=S W Z.G=S/j D p 4 . This contradicts the fact that G was a minimal counterexample to our proposition. It follows that S D ¹1º and so Z.G/ D hz1 i is cyclic. Suppose that z1 is not a p-th power of any element in the subgroup A. Then for each p x 2 A Z.G/ we have x p 2 hz1 i and so hz1 i is a cyclic subgroup of the maximal possible order in A. Then there is a complement U Š Ep2 of hz1 i in A. We obtain that jNG .U / W Aj p and ŒNG .U /; U U \ hzi D ¹1º and so NG .U / centralizes A, a contradiction. Hence we may set A D hai hbi, where ap D z1 , hap i D Z.G/, z 2 hap i, and o.b/ D p. We shall prove that there are no elements of order p in G A. Assume that there is an element c of order p in G A. First suppose Œb; c ¤ 1 so that hŒb; ci D hzi and T D hb; ci Š S.p 3 / is the nonabelian group of order p 3 and exponent p with Z.T / D hzi. All p conjugates of b in G are contained in hb; zi and all p conjugates of c in G are contained in hc; zi which implies hb; zi E G, hc; zi E G, hT i E G, and G D T H , where H D CG .T / with T \ H D hzi. We have Z.H / D Z.G/ D hz1 i and H=Z.H / Š Ep2 . Since all p C 1 maximal subgroups of H containing hz1 i are abelian, we obtain that jH 0 j D p, H 0 hz1 i and so H 0 D hzi and G 0 D hzi. Due to the fact that ˆ.G/ D ˆ.T /ˆ.H / hz1 i and ˆ.A/ D hz1 i, we get ˆ.H / D hz1 i and so H is minimal nonabelian. It follows that either H Š S.p 3 / or H Š Mpn , n 3. In any case, there is an element d of order p in H Z.H /. We have ŒNG .hb; d i/; hb; d i hb; d i \ hzi D ¹1º; and so NG .hb; d i/ D CG .hb; d i/ D CG .b/ \ CG .d / D .H hbi/ \ .T hz1 ; d i/ D hz1 ihb; d i

212

Groups of prime power order

which is of index p 2 in G, a contradiction. Thus we have proved that Œb; c D 1. Since .Ahci/0 D hzi, we get NAhci .hb; ci/ D CAhci .hb; ci/ D CAhci .c/ which is of index p in Ahci. Due to jG W NG .hb; ci/j D p, it follows that NG .hb; ci/ covers G=A. But A E G and hb; ci \ A D hbi and so NG .hbi/ D CG .hbi/ also covers G=A and so b 2 Z.G/, a contradiction. We have proved 1 .G/ D 1 .A/ D hb; zi Š Ep2 : Since jG=Z.G/j D p 4 , we have jGj p 5 . If b.G/ D 1, then Proposition 121.9 (Knoche) implies jG 0 j D p. In case b.G/ D 2, Theorem 121.1 yields that jG 0 j D p 2 . By Theorem 13.7, G is metacyclic (since G cannot be a 3-group of maximal class because jG 0 j p 2 and jGj p 5 ). If jG 0 j D p, then G is minimal nonabelian in which case jG=Z.G/j D p 2 , a contradiction. Finally, if jG 0 j D p 2 , then Proposition 122.7 gives a contradiction. Proposition 122.11. Let G be a 2-group with jG 0 j D 2 such that G=Z.G/ Š E16 . Suppose that G possesses an abelian subgroup S which satisﬁes G 0 \ S D ¹1º and S=.S \ Z.G// Š E4 . Then jG W NG .S/j D 4 and so sb.G/ > 1. Proof. We have ŒNG .S/; S S \ G 0 D ¹1º and so NG .S/ D CG .S/ Z.G/S, where jG W .Z.G/S/j D 4. If CG .S/ > Z.G/S , then G has an abelian maximal subgroup which implies (Lemma 1.1) jGj D 2jZ.G/jjG 0 j and so jG=Z.G/j D 4, a contradiction. Hence CG .S/ D Z.G/S and we are done. Proposition 122.12. Let G be a 2-group of order 25 possessing a maximal subgroup M Š E16 . If there is an element x 2 G M such that CM .x/ Š E4 , then sb.G/ > 1. Proof. By Lemma 99.2, there is an involution i 2 G M so that CM .i/ D CM .x/. But then jG W CG .i/j D 4 and we are done. Proposition 122.13. Let G be a p-group with sb.G/ D 1. If jG 0 j D p, then we have G=Z.G/ Š Ep2 unless p D 2, G=Z.G/ Š E16 and G Š .Q8 D/ E2n , n 0, Q8 \ D D Z.Q8 / D D 0 , where D Š D8 or D Š P is the nonmetacyclic minimal nonabelian group of order 24 . Proof. Since jG 0 j D p, Lemma 4.2 gives G D A1 A2 Ak Z.G/ with k 1, where A1 ; : : : ; Ak are minimal nonabelian and A01 D A02 D D A0k D G 0 D hzi being of order p. If k D 1, then G=Z.G/ Š Ep2 and we are done. We may assume k > 1 and we set G D .A1 A2 /C , where C D CG .A1 A2 / and .A1 A2 / \ C D Z.A1 A2 / D Z.A1 /Z.A2 / so that jG=C j D p 4 and Z.G/ C . Note that jG 0 j D p implies b.G/ D 1 so that we may use Proposition 122.10. It follows that C D Z.G/ must be abelian and p D 2. Thus it remains to study a 2-group G D .A1 A2 /Z.G/, where A1 and A2 are minimal nonabelian with A01 D A02 D G 0 D hzi. We have G 0 D hzi A1 \ A2 Z.A1 A2 / D Z.A1 /Z.A2 / and G=Z.G/ Š E16 :

122

p-groups all of whose subgroups have normalizers of index at most p

213

(i) First assume that there is an involution t 2 G Z.G/. Let b 2 G Z.G/ be such that Œb; t D z and then B1 D hb; ti is minimal nonabelian. Set C D CG .B1 / so that G D B1 C with B1 \ C D Z.B1 /. Since G=Z.G/ Š E16 , C is nonabelian. Let B2 be a minimal nonabelian subgroup of C and then C D B2 Z.G/. We set G1 D B1 B2 so that G D G1 Z.G/. Suppose that B2 possesses a nonnormal cyclic subgroup hb 0 i. Set hui D 1 .hb 0 i/ and S D ht i hb 0 i so that u ¤ z, where hzi D G 0 D B20 D B10 . Then 1 .S/ D ht; ui does not contain z. But then Proposition 122.11 gives a contradiction. Hence B2 is Dedekindian which implies that B2 Š Q8 , B1 \ B2 D hzi and Z.B1 / D Z.G1 /. Set B2 D hr; si. (i1) Suppose that B1 is metacyclic. Then either B1 Š D8 or B1 Š M2m , m 4. In the second case, Z.B1 / is cyclic of order 4 so that there is v 2 Z.B1 / such that v 2 D z and then i D vr is a noncentral involution in G1 .B1 [ B2 /. We obtain S D hi; ti Š E4 , S \ Z.G/ D ¹1º and z 62 S so that Proposition 122.11 gives a contradiction. Hence in this case we must have G D .Q8 D8 /Z.G/. (i2) Now assume that B1 is nonmetacyclic so that we may set m

B1 D ht; b j t 2 D b 2 D 1; Œt; b D z; z 2 D Œz; t D Œz; b D 1i; where m 2 and Z.B1 / D hb 2 i hzi. Assume that m 3 in which case there is an element w of order 4 in hb 2 i Z.B1 / so that w 2 ¤ z and .wr/2 D w 2 z. Considering the abelian subgroup S D hti hwri, we have 1 .S/ D ht; w 2 zi 6 hzi, and again Proposition 122.11 gives a contradiction. Hence in this case we have m D 2 and so G D .Q8 P/Z.G/, where P is the nonmetacyclic minimal nonabelian group of order 24 . (ii) Assume that there exist no involutions in G Z.G/. We have seen above that G D .A1 A2 /Z.G/, where we may assume that A1 6Š Q8 (noting that Q8 Q8 possesses noncentral involutions). There is an element d 2 A1 Z.A1 / such that z 62 hd i and 1 ¤ d 2 2 Z.G/, where hzi D G 0 . Let b 2 G be given such that Œd; b D z and B1 D hd; bi is minimal nonabelian, where b; d 2 B1 Z.B1 /. Note that hd i=hd 2 i is a noncentral subgroup of order 2 in G=hd 2 i. We have G D B1 C , where C D CG .B1 / and B1 \C D Z.B1 /. If C =hd 2 i is abelian, then we obtain that C 0 hd 2 i\hzi D ¹1º and C D Z.G/ is abelian, contrary to G=Z.G/ Š E16 . Let B2 =hd 2 i be a minimal nonabelian subgroup of C =hd 2 i, where B2 hd 2 ihzi and B20 D hzi. If B2 =hd 2 i 6Š Q8 , then there is b 0 2 B2 Z.B2 / such that hb 0 i \ hd 2 ; zi hd 2 i. Working in the factor group .B1 B2 /=hd 2 i, we get (as in (i)) that sb..B1 B2 /=hd 2 i/ > 1, a contradiction. Hence B2 =hd 2 i Š Q8 so that B1 \ B2 D hd 2 i hzi. By our results in (i) (applied to .B1 B2 /=hd 2 i), we get B1 =hd 2 i Š D8 or P. Set 1 .hd 2 i/ D hui and since B2 =hd 2 i Š Q8 , we have 1 .B2 / D hu; zi Š E4 . Since Z.B2 / D hd 2 ; zi, any minimal nonabelian subgroup X of B2 covers B2 =hd 2 ; zi and z 2 X. As j1 .X/j 4, X is metacyclic and z is a square in X so that there is an element x 2 X hd 2 ; zi such that x 2 D z, hzi D X 0 , hxi E G and hxi is a maximal cyclic subgroup of G. If there is y 2 B1 such that y 2 D z, then xy is a noncentral in-

214

Groups of prime power order

volution in G, a contradiction. Hence z is not a square in B1 and therefore B1 is nonmetacyclic minimal nonabelian. If B1 =hd 2 i Š D8 , then we have Z.B1 / D hd 2 i hzi and there are no involutions in B1 hd 2 ; zi so that 1 .B1 / D hu; zi, contrary to the fact that B1 is nonmetacyclic. We have proved that B1 =hd 2 i Š P. Set E D 1 .B1 / so that E Š E8 , Z.B1 / D hd 2 iE and E \ hd 2 ; zi D hu; zi. By the structure of B1 (being nonmetacyclic minimal nonabelian), there exists e 2 B1 Z.B1 / such that e 2 D l is an involution in E, where E D hu; z; li and B1 D hd; ei. Let y 2 B2 hd 2 ; zi be such that hx; yi covers the factor group B2 =hd 2 ; zi. Then hx; yi is minimal nonabelian with y 2 2 hd 2 ; zihd 2 i. Assume that o.d 2 / 4 and let d 0 2 hd 2 i be an element of order 4 so that .d 0 /2 D u. Consider the subgroup S D he; xd 0 i Š C4 C4 with 1 .S/ D hl; uzi Š E4 and z 62 S. By Proposition 122.11, we get a contradiction (noting that S \Z.G/ D 1 .S/). Hence o.d 2 / D 2 and so d 2 D u. If y 2 D z, then hx; yi Š Q8 and B2 D huihx; yi. In case y 2 D uz, we conclude that B2 D hx; yi Š H2 is minimal nonabelian, where H2 D hx; y j x 4 D y 4 D 1; x y D x 1 i, x 2 D z, y 2 D uz and u is not a square in H2 . We have proved that B1 D hd; e j d 4 D e 4 D 1; Œd; e D z; z 2 D Œz; d D Œz; e D 1i; where d 2 D u, e 2 D l, 1 .B1 / D hu; l; zi, and B1 B2 Š B1 Q8 or B1 B2 Š B1 H2 . If B1 B2 Š B1 Q8 with B1 \ Q8 D hzi, then consider B1 =huz; lzi Š Q8 so that .B1 Q8 /=huz; lzi Š Q8 Q8 , which is a contradiction by Proposition 122.4. Hence we must have B2 Š H2 D hx; y j x 4 D y 4 D 1; x y D x 1 i; where x 2 D z, y 2 D uz and u is not a square in B2 . In this case, we consider the subgroup S D hy; dei Š C4 C4 . Since S \ Z.G/ D 1 .S / D huz; ulzi 6 hzi; Proposition 122.11 gives a contradiction. (iii) We have proved that G D .D8 Q8 /Z.G/ or G D .P Q8 /Z.G/. It remains to be proved that Z.G/ is elementary abelian. We may set G D .A1 A2 /Z.G/, where A1 D hb; t j b 4 D t 2 D 1; b 2 D z; b t D bzi Š D8

(a) or

A1 D hb; t j b 4 D t 2 D 1; b 2 D u; Œb; t D z; z 2 D Œz; b D Œz; t D 1i Š P

(b) and

A2 D hx; y j x 2 D y 2 D z; z 2 D 1; Œx; y D zi Š Q8

with A1 \ A2 D hzi:

122

p-groups all of whose subgroups have normalizers of index at most p

215

First assume that Z.G/ has an element v of order 4 such that v 2 D s 62 A1 A2 . In both cases (a) and (b), we consider the abelian subgroup S D hvx; ti of G which satisﬁes 1 .S/ D ht; szi 6 hzi so that Proposition 122.11 gives a contradiction. Now assume that Z.G/ has an element v of order 4 such that v 2 2 Z.A1 A2 /. In case (a), we must have v 2 D z and then we consider the subgroup S D hvx; ti Š E4 , where vx is an involution in G .A1 A2 / commuting with t and so z 62 S and Proposition 122.11 gives a contradiction. In case (b), we have either v 2 D z or v 2 D u since there is an outer automorphism ˛ of A1 induced by t ˛ D t and b ˛ D bt, where .bt/2 D uz and so u˛ D uz. If v 2 D z, we consider the subgroup S D hvx; ti Š E4 with z 62 S. In case v 2 D u, we consider the subgroup S D hvx; ti Š C4 C2

with 1 .S/ D ht; uzi 6 hzi:

In both cases, we get a contradiction with the help of Proposition 122.11. Our proposition is proved. Proposition 122.14. Let G be a 2-group with sb.G/ D 1. Then b.G/ 3 and let a be an element in G with eight conjugates. Then we have o.a/ D 8, jG W NG .hai/j D 2 and NG .hai/=CG .a/ Š E4 . Proof. We have b.G/ 3 by Proposition 122.8. Let a be an element in G with eight conjugates. Set C D CG .a/ so that jG W C j D 8. Also set H D NG .hai/ so that jG W H j 2. Since jH=C j 4, we have o.a/ 8. Suppose that o.a/ D 2n with n 4. Assume that there is an element x 2 H C such that ax D a1 z , where n1 z D a2 and D 0; 1. This implies hxi \ hai hzi. On the other hand, a2 normalizes hxi and so Œa2 ; x hxi \ hai hzi. But .a2 /x D .a1 z /2 D a2 and so Œa2 ; x D a4 with o.a4 / 4, a contradiction. It follows that H=C contains only the involutory automorphism induced by a ! az. Hence H=C is cyclic. Assume that H C contains an element y which induces on hai an automorphism of order 4 given n2 by ay D av, where v D a2 . Then ha; yi is metacyclic with ha; yi0 D hvi Š C4 . But then Proposition 122.7 gives a contradiction. We get H=C Š C4 , jG W H j D 2, and if y 2 H C is such that hyi covers H=C , then ay D a1 v, where v is an element of order 4 in hai and v 2 D z with 1 .hai/ D hzi. Note that .a4 /y D .a1 v/4 D a4 and so v y D v 1 and ay D .a1 v/y D .a1 v/1 v y D az 2

which implies hyi \ hai hzi. On the other hand, a2 normalizes hyi and so Œa2 ; y hyi \ hai hzi. But .a2 /y D .a1 v/2 D a2 z and so Œa2 ; y D a4 z which is of order 4, a contradiction. We have proved that o.a/ D 8 and then jG W H j D 2 and H=C Š E4 . Proposition 122.15. Let G be a 2-group with sb.G/ D 1. Let a be an element in G with eight conjugates. By Proposition 122.14, we have o.a/ D 8, jG W NG .hai/j D 2, and NG .hai/=CG .a/ Š E4 . If b; c 2 NG .hai/ CG .a/ are such that ab D a1 and

216

Groups of prime power order

ac D a1 z with z D a4 , then o.b/ D o.c/ D 8 and 1 .hbi/ D 1 .hci/ D hbi \ hai D hci \ hai D hzi: We have

L D hb; a2 i Š M16 , where b 4 D a4 D z, z 2 D 1, Œa2 ; b D z, L E G, t D a2 b 2 is an involution and 1 .L/ D ht; zi.

b a D ba2 , .a2 /a D a2 and so a induces on L an outer automorphism of order 4 2 as b a D ba4 D bz, where a2 induces on L an inner automorphism of order 2.

Aut.L/ D D h˛i, where D Š D8 and ˛ is an outer involutory automorphism of L induced by b ˛ D b 1 and .a2 /˛ D a2 .

Proof. Let a be an element in G with eight conjugates so that o.a/ D 8 and then set C D CG .a/, H D NG .hai/, where jG W H j D 2 and H=C Š E4 . Let b; c 2 H C be such that hb; ci covers H=C and ab D a1 , ac D a1 z with z D a4 . This implies that hbi\hai hzi and hci\hai hzi. On the other hand, a2 normalizes hbi and hci. If hbi \ hai D ¹1º, then a2 centralizes hbi, a contradiction. In case hci \ hai D ¹1º, a2 centralizes hci, a contradiction. Hence we have hbi \ hai D hci \ hai D hzi; m

l

and so we may set b 2 D z and c 2 D z, where m 1 and l 2. Indeed, if c 2 D z, then ha; ci Š SD16 , and so ha; ci possesses a noncentral involution i with Cha;ci .i/ D hi; zi Š E4 , a contradiction. We shall study the structure of the subgroup M D ha; b; ci, where M 0 ha2 i since Œa; b D a2 , Œa; c D a2 . If b has only four conjugates in M , then b c D ba2i (for some integer i ) since b has already four conjugates in hb; ai with hb; ai0 D ha2 i and this gives Œb; c 2 ha2 i and M 0 D ha2 i so that b.M / D 2. Similarly, if c has only four conjugates in M , then c b D ca2j (for some integer j ) and this gives Œc; b 2 ha2 i and M 0 D ha2 i so that b.M / D 2. Hence b has only four conjugates in M if and only if c has only four conjugates in M and in that case M 0 D ha2 i and b.M / D 2. Conversely, if M 0 D ha2 i, then b.M / D 2 and so both b and c have exactly four conjugates in M . We aim to show that o.b/ D o.c/ D 8. First assume that b or c has eight conjugates in M . By the previous paragraph, both b and c have exactly eight conjugates in M . From Proposition 122.14 we then conclude that o.b/ D o.c/ D 8. In the sequel we assume that both b and c have exactly four conjugates in M and so M 0 D ha2 i. Set Œb; c D a2i (i integer) and compute Œb 2 ; c D Œb; cb Œb; c D .a2i /b a2i D a2i a2i D 1; Œb; c 2 D Œb; cŒb; cc D a2i .a2i /c D a2i a2i D 1; and so hb 2 ; c 2 i Z.M /.

122

p-groups all of whose subgroups have normalizers of index at most p

217

Assume m 3, where b 2 D z and set b 0 D b 2 so that o.b 0 / D 8, .b 0 /4 D z 0 0 0 and b 2 Z.M /. It follows that hb ; ai D hb i hai with hb 0 i \ hai D hzi. Because of .b 0 a/2 D .b 0 /2 a2 ¤ 1 and .b 0 a/4 D .b 0 /4 a4 D zz D 1, we obtain that o.b 0 a/ D 4. Set S D hb 0 ai and we claim that S has more than two conjugates in M . Indeed, we have S b D hb 0 a1 i and S c D hb 0 a1 zi. If S D S b , then either b 0 a D b 0 a1 (and then a2 D 1) or b 0 a D .b 0 a1 /1 D .b 0 /1 a (and then .b 0 /2 D 1), a contradiction. In case S D S c , either b 0 a D b 0 a1 z (and then a2 D z) or b 0 a D .b 0 /1 az (and then .b 0 /2 D z), a contradiction. If S b D S c , then either b 0 a1 D b 0 a1 z (and then z D 1) or b 0 a1 D .b 0 /1 az (and then .b 0 /2 D a2 z), a contradiction. Thus m 2. l l2 so that o.c 0 / D 8, .c 0 /4 D z and Assume l 3, where c 2 D z and set c 0 D c 2 c 0 2 Z.M /. It follows that hc 0 ; ai D hc 0 i hai with hc 0 i \ hai D hzi. Then we have T D hc 0 ai Š C4 and we claim that T has more than two conjugates in M . Indeed, we obtain T b D hc 0 a1 i and T c D hc 0 a1 zi. If T D T b , then either c 0 a D c 0 a1 or c 0 a D .c 0 /1 a, a contradiction. In case T D T c , we conclude either c 0 a D c 0 a1 z or c 0 a D .c 0 /1 az, a contradiction. Finally, if T b D T c , then either c 0 a1 D c 0 a1 z or c 0 a1 D .c 0 /1 az, a contradiction. Thus l 2 and so l D 2 and o.c/ D 8. Suppose o.b/ D 4 so that b 2 D z, ha; bi Š Q16 since ab D a1 . Then m

Q D ha2 ; bi Š Q8 ;

m2

c 2 2 Z.M /;

o.c 2 / D 4;

and so c 2 62 Q. But we have c 4 D z, where hzi D Z.Q/ and so hQ; c 2 i D Q hc 2 i with Q \ hc 2 i D hzi and therefore hQ; c 2 i contains a subgroup isomorphic to D8 . By Proposition 122.5, jG 0 j 4 and so b.G/ 2, contrary to the fact that a has eight conjugates in G. Hence o.b/ D 8. We have proved that in any case o.b/ D o.c/ D 8, as required. We have L D hb; a2 i D hb; a2 b 2 D t j b 8 D t 2 D 1; b t D bz; z D b 4 D a4 i Š M16 : Since b a D ba2 , it follows that the only two distinct cyclic subgroups hbi and hba2 i of order 8 in L are conjugate under a and so hb; b a D ba2 i D L is normal in G. From the fact that hb 2 i D Z.L/ is normal in G and 2 .L/ D hb 2 ; a2 i Š C4 C2 , we infer that ha2 i is also normal in G (noting that ha2 i and hb 2 i are the only two cyclic subgroups of order 4 in 2 .L/). Each automorphism ˛ 2 Aut.L/ sends b onto one of the eight elements of order 8 in the set .hbi hb 2 i/ [ .hba2 i hb 2 i/ and ˛ sends a2 onto one of the two elements in ¹a2 ; a2 D a2 zº and so jAut.L/j D 8 2 D 16. Since Inn.L/ Š L=Z.L/ Š E4 , we have jAut.L/=Inn.L/j D 4. Also, Inn.L/ D hia2 ; ib i, where ix is the inner automorphism of L induced by conjugation with the element x 2 L Z.L/. We deﬁne the outer automorphisms ˇa (induced by conjugation with the element a 2 G) and ˛ by b ˇa D b a D ba2 ;

.a2 /ˇa D .a2 /a D a2

and b ˛ D b 1 ;

.a2 /˛ D a2 ;

where .ˇa /2 D ia2 and ˛ 2 D 1. We check that Inn.L/hˇa i Š D8 and that ˛ com-

218

Groups of prime power order

mutes with ˇa and ib and so Aut.L/ D .Inn.L/hˇa i/ h˛i Š D8 C2 : Our proposition is completely proved. The following proposition is the key result of the whole section. Proposition 122.16. Let G be a 2-group with sb.G/ D 1. Then b.G/ 2, i.e., the element breadth of G is at most 2. Proof. Let G be a minimal counterexample to this proposition. Note that in this case Proposition 122.8 implies that b.G/ 3. Then G possesses an element a which has exactly eight conjugates in G. We use Proposition 122.15 together with the notation and all details stated in that proposition. Set C D CG .L/ so that L \ C D Z.L/ D hb 2 i, a 2 G .LC / and ¹1º ¤ G=.LC / is elementary abelian of order at most 4. We aim to show that C is abelian. Suppose that C is nonabelian. Note that C cannot be Hamiltonian since C has a central subgroup hb 2 i Š C4 . Let hxi be a nonnormal cyclic subgroup of C such that z 62 hxi and 0 0 let c 0 2 C be such that hx c i ¤ hxi and then z 62 hx c i. Now, t D a2 b 2 is a noncentral involution in L and t ¤ t b D t z. Set S D hx; ti D hxi hti and we claim that S D hx; ti;

0

S1 D hx c ; ti;

0

S2 D hx; tzi and S3 D hx c ; tzi

are pairwise distinct conjugates of S. If S D S1 , then 0

S \ C D hxi D S1 \ C D hx c i; a contradiction. In case S D S2 , we get S \ L D hti D S2 \ L D htzi; a contradiction. If S1 D S2 , then 0

S1 \ C D hx c i D S2 \ C D hxi; a contradiction. Thus we have proved that each nonnormal subgroup of C contains hzi. By a result of N. Blackburn (see Corollary 92.7), we have three possibilities for the structure of C . (1) C Š Q8 Q8 E2s (s 0) which is not possible since C has a central subgroup hb 2 i which is cyclic of order 4. (2) C Š Q8 C4 E2s (s 0) and let Q Š Q8 be a quaternion subgroup of C . If z 2 Q, then the subgroup hb 2 ; Qi contains a subgroup isomorphic to D8 . In that case, Proposition 122.5 implies jG 0 j 4 which gives b.G/ 2, a contradiction. Hence Q D hx; yi with Z.Q/ D hz 0 i ¤ hzi. But then hb 2 xi is nonnormal in C and z 62 hb 2 xi, a contradiction.

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p-groups all of whose subgroups have normalizers of index at most p

219

(3) C possesses an abelian maximal subgroup A with exp.A/ > 2 and an element v 2 C A of order 4 such that x v D x 1 for all x 2 A. But C has a central subgroup hb 2 i Š C4 , a contradiction. We have proved that C is abelian. (i) First assume that bG .b/ D 3, i.e., b has exactly eight conjugates in G. By Proposition 122.14, G possesses an element d such that b d D b 1 and so, by Proposition 122.15, our group G induces on L Š M16 the full automorphism group of L (which is isomorphic to the group D8 C2 ) and so we may choose our element d so that b d D b 1 , .a2 /d D a2 , d 2 2 C , o.d / D 8, d 4 D z and G D .LC /ha; d i. Also, the automorphism of L induced by d commutes with all automorphisms of L and so f D Œd; a 2 C . Consider the subgroup H D hb; d j b 8 D d 8 D 1; b 4 D d 4 D z; b d D b 1 i; where Z.H / D hd 2 i. Since .bd /2 D bdbd D bd 2 b d D bd 2 b 1 D d 2 ;

Œb; bd D Œb; d D b 2 ;

it follows that the cyclic subgroup S D hbd i of order 8 is not normal in H . Thus two conjugates of S in G are already contained in H . On the other hand, .bd /a D b a d a D ba2 df D .bd /a2 f; which forces a2 f 2 H . Since a2 f 2 LC and H \ .LC / D hb; d 2 i, we deduce that 2 a2 f must be contained in the abelian subgroup hb; d 2 i. But b a f D .bz/f D bz ¤ b, a contradiction. (ii) Now assume that bG .b/ D 2, i.e., there is no element in G which inverts hbi. We get G D .LC /hai, where C D CG .L/ is abelian and jG W .LC /j D 2. Since b inverts hai, b 2 centralizes hai and so CG .hb 2 i/ hL; C; ai D G so that b 2 2 Z.G/. We have NG .hai/ D haiNLC .hai/ and noting that L D ha2 ; bi normalizes hai, we obtain that NLC .hai/ D LC0 , where C0 D NC .hai/, C0 hb 2 i and jC W C0 j D 2 (Proposition 122.14). Let x 2 C C0 so that C E G implies Œa; x D f0 2 C and ax D af0 . Hence we have f0 62 hai and so f0 2 C hzi since hai \ C D hzi. Consider the subgroup K D ha; b j a8 D b 8 D 1; a4 D b 4 D z; ab D a1 i; where Z.K/ D hb 2 i. Since .ab/2 D abab D ab 2 ab D ab 2 a1 D b 2

and Œab; a D Œb; a D a2 ;

the cyclic subgroup habi of order 8 is not normal in K and so two conjugates of habi in G are already contained in K. On the other hand, .ab/x D af0 b D .ab/f0 which

220

Groups of prime power order

implies f0 2 K. But K \ .LC / D ha2 ; bi D L and so f0 2 .L \ C / hzi. Thus f0 D .b 2 /˙1 and therefore replacing b with b 1 (if necessary), we may set f0 D b 2 and so ax D ab 2 . Then we get 2

ax D .ab 2 /x D .ab 2 /b 2 D ab 4 D az and so x 2 2 C0 but x 2 62 C1 , where C1 D CC .a/ hb 2 i. On the other hand, C centralizes ha2 i and so in C0 C1 we cannot have elements which invert hai. This gives jC0 W C1 j D 2;

x 2 2 C0 C1 ;

C1 D Z.G/:

For any y 2 C0 C1 we have ay D az and ayb D a1 z. By Proposition 122.15, we have o.yb/ D 8 and .yb/4 D z. Then we get y 4 b 4 D z and y 4 D 1 so that C0 is of exponent 4. Suppose that Z.G/ D C1 D hb 2 i is cyclic. If, in addition, G possesses noncommuting involutions, then Proposition 122.5 implies that jG 0 j 4 which gives b.G/ 2, a contradiction. By Proposition 122.6, G has at most three involutions and so 1 .G/ D 1 .L/ D hz; ti Š E4 : But exp.C0 / D 4 and jC0 W hb 2 ij D 2 and so there exist involutions in C0 hb 2 i, a contradiction. Hence we get C1 D Z.G/ > hb 2 i so that there is a central involution u 2 C1 hb 2 i. Considering the factor group GN D G=hui, then the fact that G was a N 2. On the other hand, from minimal counterexample to our proposition gives b.G/ 2 b 1 x 2 x a D a , a D ab , a D az we deduce that aN has in GN eight conjugates, a ﬁnal contradiction. Our proposition is proved. Proposition 122.17. Let G be a 2-group with sb.G/ D 1 which is a minimal counterexample to Theorem 122.1. Then jG 0 j D 4, and let z 2 G 0 be a central involution in G. If R=hzi D Z.G=hzi/, then G=R Š E16 . Proof. Let G be a 2-group with sb.G/ D 1 which is a minimal counterexample to Theorem 122.1. If jG 0 j D 2, then Proposition 122.13 implies jG=Z.G/j 24 , a contradiction. Hence jG 0 j > 2 and so b.G/ > 1 by Proposition 121.9 (Knoche). On the other hand, Proposition 122.16 gives b.G/ 2 and so b.G/ D 2. Since jG=Z.G/j 25 , Theorem 121.1 implies jG 0 j D 4. N 1 and Let z 2 G 0 be a central involution in G and set GN D G=hzi so that sb.G/ 0 N N Š E4 or G=Z. N N Š E16 . Assume that jGN j D 2. By Proposition 122.13, G=Z. G/ G/ N N Š E4 and so we have only to rule out this case. Set Z.G/ N D R=hzi, where G=Z. G/ G=R Š E4 and r 2 R if and only if ŒG; r hzi. We have ŒG; R D hzi. Indeed, if ŒG; R D ¹1º, then we get R D Z.G/, a contradiction. Also, ŒG; r hzi implies that jG W CG .r/j 2. Let A be a maximal normal abelian subgroup of G which contains ˆ.G/ so that ¹1º ¤ G=A is elementary abelian. In view of Propositions 122.3 and 122.10, there is g 2 G A such that jA W CA .g/j D 4 and then Lemma 121.2 implies jŒA; gj D 4 so

122

p-groups all of whose subgroups have normalizers of index at most p

221

that ŒA; G D G 0 . If jG W Aj D 2, then we have jG=Z.G/j D 8, a contradiction. Hence jG W Aj 4. Since ŒG; G 0 hzi, we have G 0 R. As ŒA; g 6 hzi, it follows that A 6 R. For each x 2 .AR/ A, x 2 TA , i.e., jA W CA .x/j D 2. Therefore R does not cover G=A because g 62 TA . Hence jA W .A\R/j D 2 and jG W .AR/j D 2 which gives G D .AR/hgi. We have R0 hzi and ŒA; R D hzi (noting that jR W .A \ R/j 2) and so .AR/0 D hzi. Since jA W CA .g/j D 4, g does not centralize A\R and so we get ŒA \ R; g D hzi which gives jA \ R W CA\R .g/j D 2 and therefore CA .g/ is a subgroup of index 2 in A \ R and so g 2 2 CA\R .g/. Also, R0 hzi and ŒR; g D hzi imply .Rhgi/0 D hzi. Since G is a minimal counterexample to Theorem 122.1, we have jG=Z.G/j 25 and so Propositions 122.5 and 122.9 imply that 1 .G/ is elementary abelian and 1 .Z.G// G 0 . In particular, D8 is not a subgroup of G. First suppose jG=Aj D 4. If R is abelian, then CG .CA .g// hR; g; Ai D G, CA .g/ Z.G/ and jG W CA .g/j D 24 , a contradiction. Hence R is nonabelian with R0 D hzi and A \ R is an abelian maximal subgroup of R. Then Lemma 1.1 gives jRj D 2jZ.R/jjR0 j and so Z D Z.R/ is a subgroup of index 2 in A\R. If CA .g/ D Z, we get CG .Z/ hR; g; Ai D G and jG W Zj D 24 , a contradiction. It follows that Z0 D CA .g/ \ Z is of index 4 in A \ R and Z0 D Z.Rhgi/ with j.Rhgi/=Z0 j D 24 . Since .Rhgi/0 D hzi and sb.Rhgi/ D 1, we may apply Proposition 122.13 and get (noting that D8 is not a subgroup of G) Rhgi Š .P Q8 / E2s , s 0, where P is the nonmetacyclic minimal nonabelian subgroup of order 24 and P \ Q8 D hzi. Now assume jG=Aj > 4 so that j.AR/=Aj 4. Note that we have .AR/0 D hzi and Z.AR/ < A which gives j.AR/=Z.AR/j 23 . Applying Proposition 122.13, we get AR Š .P Q8 / E2s , s 0. We have shown that in any case G possesses a maximal subgroup M D .P Q/S, where P D hb; t j b 4 D t 2 D 1; Œb; t D z; z 2 D Œb; z D Œt; z D 1i; Q D hr; si Š Q8 ;

P \ Q D hzi and S Š E2s ; s 0:

Set b 2 D u so that Z.P / D hu; zi Š E4

and 1 .P / D hu; z; t i Š E8 :

Since hbi and hbzi are two distinct conjugates in G of the cyclic subgroup hbi Š C4 , it follows that hb; bzi D hbi hzi is normal in G and so Ã1 .hb; bzi/ D hui is also normal in G. Thus u 2 Z.G/ and since 1 .Z.G// G 0 , we get G 0 D Z.P / Z.G/ which shows that G is of class 2. Suppose that S ¤ ¹1º and let x be an involution in S. We have ht; xi Š E4 and ht; xi \ G 0 D ¹1º. Then ŒNG .ht; xi/; ht; xi ht; xi \ G 0 D ¹1º; and so NG .ht; xi/ D CG .ht; xi/ D CG .x/ \ CG .t/:

222

Groups of prime power order

But 1 .Z.G// D G 0 D hu; zi whence x 62 Z.G/ and so CG .x/ D M . Because of jM W CM .t/j D 2, we have jG W .CG .x/\CG .t//j D 4, a contradiction. Hence we get S D ¹1º and so M D P Q, Z.M / D Z.P / D G 0 Z.G/. Since jG=Z.G/j 25 , we have Z.P / D G 0 D Z.G/. We shall show that there are no involutions in G M . Indeed, let i be an involution in G M and note that ht; ii Š E4 and ht; ii \ Z.G/ D ¹1º so that E D hz; u; t; ii is isomorphic to E16 . Also, b 2 D u and E E G and therefore jEhbij D 25 . We have ŒNG .ht; ii/; ht; ii ht; ii \ G 0 D ¹1º

and so

NG .ht; ii/ D CG .ht; ii/

which together with jG W NG .ht; ii/j D 2 gives CG .t/ D CG .i/ D CG .ti/ D EQ, where b 62 EQ and jG W .EQ/j D 2. In particular, CE .b/ D hu; zi Š E4 and so Proposition 122.12 implies sb.Ehbi/ > 1, a contradiction. We have proved that there are no involutions in G M and so 1 .G/ D 1 .M / D 1 .P / D hz; u; t i Š E8 : Note that P =hui Š D8 and so M=hui Š D8 Q8 and therefore, applying Proposition 122.13 on the factor group G=hui, we see that there is an element v 2 G M such that v 2 D u and Œv; M hui. If b v D b, then v 2 D b 2 D u gives that vb is an involution in G M , a contradiction. Hence b v D bu D b 1 and so hb; vi Š Q8 . Since t and t b D t z are the only conjugates of t in G, we cannot have t v D tu and so Œt; v D 1. If v centralizes Q D hr; si, then hhb; vi; Qi D hb; vi Q Š Q8 Q8 ; and then Proposition 122.4 gives a contradiction. In case r v D ru and s v D su, we have .rs/v D rs. Hence we may choose generators r; s of Q so that Œr; v D 1 and Œs; v D u. The structure of G (of order 27 ) is uniquely determined. We compute .bt s/2 D .bt/2 s 2 D uz z D u and

Œv; bts D u u D 1

and so v bts is an involution in G M , a ﬁnal contradiction. Our proposition is proved. Proposition 122.18. Let G be a 2-group with sb.G/ D 1 which is a minimal counterexample to Theorem 122.1. Then jG 0 j D 4 and let z 2 G 0 be a central involution in G. We have G D AB, where A and B are normal subgroups of G, A \ B D hzi;

A Š Q8 ; C4 ; E4 or C2

and if A Š Q8 , then ŒA; B D 1, and G 0 D B 0 , B=hzi Š Q8 D8 with Q8 \ D8 D Z.Q8 / or B=hzi Š Q8 P with Q8 \ P D Z.Q8 / D P0 , where P is the nonmetacyclic minimal nonabelian group of order 24 .

122

p-groups all of whose subgroups have normalizers of index at most p

223

Proof. By Proposition 122.17, we have jG 0 j D 4 and if z 2 G 0 is a central involution in G, then G=hzi is isomorphic to one of the groups stated in the second half of Proposition 122.13. More precisely, we have G D AB, where A and B are normal subgroups of G, A \ B D hzi, A=hzi Š E2n , n 0, B=hzi Š Q8 D with Q8 \ D D Z.Q8 / D D 0 , D Š D8 or P, where P is the nonmetacyclic minimal nonabelian group of order 24 . Further, G 0 B and set T =hzi D Z.B=hzi/ so that T D G 0 (in case B=hzi Š Q8 D8 ) or T =hzi Š E4 (in case B=hzi Š Q8 P ) and in both cases B=T Š E16 and Z.B/ T . If B < G, then the fact that G is a minimal counterexample to Theorem 122.1 gives jB=Z.B/j 24 and so in this case Z.B/ D T . Suppose that A is nonabelian so that in this case A0 D hzi and A=A0 is elementary abelian. We may also assume that A is not Hamiltonian and so sb.A/ D 1. By Proposition 122.5, D8 is not a subgroup of A. Also, P cannot be a subgroup of A since P=P0 is not elementary abelian. Then, by Proposition 122.13, we get A=Z.A/ Š E4 . Let A0 be a minimal nonabelian subgroup of A so that A D A0 Z.A/. But we have A00 D hzi D ˆ.A0 / D ˆ.A/ and so A0 Š Q8 and A0 \ Z.A/ D hzi with Z.A/=hzi being elementary abelian. If there is v 2 Z.A/ such that v 2 D z, then A0 hvi contains a subgroup isomorphic to D8 , a contradiction (Proposition 122.5). It follows that Z.A/ is elementary abelian and so A is Hamiltonian after all. We have proved that A is either abelian with Ã1 .A/ hzi or A is Hamiltonian with A0 D hzi. Suppose that A possesses a central involution t satisfying t ¤ z and A > ht; zi. Set G0 D Bht i so that G0 < G. As B < G, we get T D Z.B/ and T =hzi D Z.B=hzi/. We have ŒB; t hzi and G0 =hzi D .ht; zi=hzi/ .B=hzi/ Š C2 .B=hzi/: Hence Z.G0 =hzi/ D .ht iT /=hzi and so Z.G0 / ht iT . On the other hand, we have G0 < G and so (by the minimality of G) Z.G0 / D ht iT . In particular, t centralizes B and so CG .t/ hA; Bi D G. Hence t is a central involution in G which is not contained in G 0 (since G 0 B), contrary to Proposition 122.9. We have proved that A Š Q8 , C4 , E4 or C2 . Assume that A Š Q8 and let hvi be any cyclic subgroup of order 4 in A. As v 2 D z and Œv; B hzi, we have hvi E G. Set again G0 D hviB so that G0 < G which gives T D Z.B/ and T =hzi D Z.B=hzi/. Due to the fact that G0 =hzi Š C2 .B=hzi/, we have Z.G0 =hzi/ D .hviT /=hzi and so Z.G0 / hviT . But jG0 =.hviT /j D 24 and G0 < G and so Z.G0 / D hviT . In particular, v centralizes B. But hvi was any cyclic subgroup of order 4 in A Š Q8 and so ŒA; B D 1. It remains to prove that G 0 D B 0 . Assume jB 0 j D 2 so that G 0 D hzi B 0 Z.G/ and therefore G is of class 2. Also, B < G and so A > hzi which gives (because of the minimality of G) T D Z.B/ and B=Z.B/ Š E16 . By Proposition 122.13, we obtain that B D .Q8 P/ E2s , s 0, with B 0 D Z.Q8 / D hz 0 i D P0 D Q8 \ P

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Groups of prime power order

since D8 is not a subgroup of G. If z 2 Q8 P, then z 2 Z.P/ hz 0 i and B=hzi Š .Q8 D8 / E2s and so s D 0. In case z 62 Q8 P, we get B=hzi Š .Q8 P/ E2s1 and so s D 1. Thus it follows that we have either B D hzi .Q8 P/ or B D Q8 P. If A Š Q8 , then ŒA; B D 1 and so hA; Q8 i D AQ8 , which is a contradiction (by Proposition 122.4). Hence A Š E4 or A Š C4 and ŒA; B D hzi since G 0 D hz; z 0 i. First assume B D hzi .Q8 P/, where we set P D hb; t j b 4 D t 2 D 1; b 2 D u; Œb; t D z 0 ; .z 0 /2 D Œb; z 0 D Œt; z 0 D 1i and Q8 D hr; s j r 2 D s 2 D z 0 ; Œr; s D z 0 i. Since ¹t; tz 0 º are the only two conjugates of t in G and ¹hbi; hbz 0 iº are the only two conjugates of hbi Š C4 in G, we have ht; tz 0 i E G and hb; bz 0 i E G and so P E G. This gives ŒA; P A \ P D ¹1º and so A centralizes P. Since A E G and ŒA; B D hzi, we have jG W CG .A/j D 2 and so we may set ŒA; r D 1 and ŒA; s D z. Consider the cyclic subgroup hsbi Š C4 with .sb/2 D z 0 u and .sb/t D .sb/z 0 so that ¹hsbi; hsbz 0 iº are the only two conjugates of hsbi in G (and they are contained in Q8 P). For a 2 A hzi, .sb/a D szb D .sb/z and .sb/z 62 Q8 P, a contradiction. Finally, suppose that B D Q8 P, where we may set P D hb; t j b 4 D t 2 D 1; b 2 D z; Œb; t D z 0 ; .z 0 /2 D Œb; z 0 D Œt; z 0 D 1i (noting that both z and zz 0 are squares in P) and Q8 D hr; s j r 2 D s 2 D z 0 ; Œr; s D z 0 i. Also, ŒA; B D hzi and since ¹t; t z 0 º are the only two conjugates of t in G, we have ŒA; t D 1 (as ŒA; t D z is not possible) and so A centralizes t . Assume that A Š E4 and let i be an involution in A hzi. We have ŒNG .hi; ti/; hi; ti hi; ti \ G 0 D ¹1º since G 0 D hz; z 0 i. Hence NG .hi; ti/ D CG .hi; ti/ and since jG W NG .hi; ti/j D 2 and hi; ti \ Z.G/ D ¹1º (because 1 .Z.G// G 0 by Proposition 122.9), we obtain CG .t/ D CG .i/ D CG .it/ D hi; t; zi Q8 : As ŒA; B D hzi, we have Œi; b D z and so b i D bz D b 1 which gives hb; ii Š D8 , a contradiction. We have proved that A D hvi with v 2 D z and Œv; t D 1 (because A centralizes t ). Since A E G, we get jG W CG .v/j D 2 and so jB W CB .v/j D 2. If v centralizes Q8 , then Œv; b D z and so hv; bi Š Q8 centralizes Q8 D hr; si, which is a contradiction (by Proposition 122,4). Hence we may assume Œv; r D 1 and Œv; s D z. Suppose that v centralizes b. Then i 0 D vb is an involution and so Œi 0 ; t D 1 by Proposition 122.5. But v centralizes t and so also b centralizes t , a contradiction. Hence Œv; b D z and the structure of G is uniquely determined.

122

p-groups all of whose subgroups have normalizers of index at most p

225

We see that vt brs is an involution in G B. Indeed, .vt brs/2 D .vt/2 .brs/2 Œbrs; vt D 1 and then we must have Œt; vtbrs D 1. On the other hand, computing in G we get Œt; vtbrs D Œt; vŒt; tŒt; bŒt; rŒt; s D z 0 which is a ﬁnal contradiction. Our proposition is proved. Proof of Theorem 122.1 for p D 2. Let G be a 2-group with sb.G/ D 1 (i.e., the subgroup breadth of G is equal to 1) and assume that G is a minimal counterexample to Theorem 122.1, i.e., jG=Z.G/j 25 but for each proper subgroup or factor group X of G, we have jX=Z.X/j 24 . Then Proposition 122.5 gives that 1 .G/ is elementary abelian and Proposition 122.9 yields 1 .Z.G// G 0 . The starting point in our proof is Proposition 122.18 which gives the following facts. We have jG 0 j D 4, and let z be ﬁxed involution in G 0 \ Z.G/. Then G D AB, where A and B are normal subgroups of G with A \ B D hzi, A Š Q8 , C4 , E4 or C2 and if A Š Q8 , then ŒA; B D ¹1º. Further, B D CD with C \ D D G 0 D B 0;

ŒC; D hzi;

C =hzi Š Q8

and D=hzi Š D8 or P;

where P is the nonmetacyclic minimal nonabelian subgroup of order 24 . Hence C and D are also normal subgroups of G. At this point it is possible to determine the structure of C . For each x 2 C G 0 , 2 x 2 G 0 hzi and so hz; xi D G 0 hxi is abelian. It follows that all three maximal subgroups of C which contain G 0 are abelian and so jC 0 j D 2 (Exercise P1). But C 0 covers G 0 =hzi and so G 0 D hziC 0 Š E4 and since C 0 Z.G/, we get G 0 Z.G/ and so G is of class 2 whose commutator subgroup is a four-subgroup. We have 1 .C / D G 0 and z is not a square in C . Set hc 0 i D C 0 so that G 0 D B 0 D hz; c 0 i. If ˆ.C / D hc 0 i, then there is a maximal subgroup Q of C which does not contain hzi in which case C D hziQ with Q Š Q8 and Q0 D hc 0 i. In case ˆ.C / D G 0 , we infer that C is minimal nonabelian of order 24 and exponent 4. Since 1 .C / D G 0 Š E4 , it follows that C is metacyclic and so c 0 is a square in C . There are r; s 2 C G 0 such that r 2 D c 0 , s 2 D c 0 z and Œr; s D c 0 . In this case, C D hr; s j r 4 D s 4 D 1; r 2 D c 0 ; s 2 D zc 0 ; Œr; s D c 0 i Š H2 ; where H2 is deﬁned by H2 D hx; y j x 4 D y 4 D 1; x y D x 1 i. In any case, there are elements r; s 2 C G 0 of order 4 such that r 2 D c 0 , s 2 D c 0 z , D 0; 1 and Œr; s D c 0 . We see that .rs/2 D r 2 s 2 Œs; r D c 0 c 0 z c 0 D c 0 z . Fix this notation in the rest of the proof which we split into two parts: Part 1, where D=hzi Š D8 , and Part 2, where D=hzi Š P with P being the nonmetacyclic minimal nonabelian group of order 24 . It is interesting to note that Part 1 will be more difﬁcult than Part 2.

226

Groups of prime power order

Part 1, where D=hzi Š D8 . Each of the three maximal subgroups of D containing G 0 are abelian and so D 0 D hd 0 i is of order 2 and d 0 2 G 0 hzi and so we may set d 0 D c 0 z , D 0; 1. If ˆ.D/ < G 0 , then ˆ.D/ D D 0 D hd 0 i and D D hzi D0 , where D0 Š D8 , a contradiction (since dihedral groups cannot be subgroups in G). Hence we have ˆ.D/ D G 0 and so D is minimal nonabelian of order 24 and exponent 4. If 1 .D/ D G 0 , then D Š H2 and if 1 .D/ Š E8 , then D Š P. In case D Š H2 , the fact that D=hzi Š D8 shows that z is a square in D. If D Š P, then d 0 is the only involution in Z.D/ which is not a square in D. In any case, there is y 2 D such that y 2 D z. Also, G 0 D B 0 D Z.B/ Z.G/ and jB=Z.B/j D 24 so that we must have G > B which implies A > hzi. If A Š Q8 , then we know that ŒA; B D ¹1º and so A hyi (with A \ hyi D hzi) contains a subgroup isomorphic to D8 , a contradiction. It follows that A Š E4 ; or C4 . In view of jG=Z.G/j 25 , we have A 6 Z.G/ and so ŒA; B D hzi and G 0 D B 0 D Z.B/ D Z.G/ D ˆ.G/. (i) First assume D D hb; t j b 4 D 1; b 2 D z; t 2 D 1; Œb; t D d 0 ; .d 0 /2 D Œd 0 ; b D Œd 0 ; t D 1i Š P; where ¹t; t b D t d 0 º are the only two conjugates of t in G. On the other hand, we obtain ŒC; t ŒC; D hzi and so we must have ŒC; t D ¹1º which gives CB .t/ D hti C . Since ŒC; D hzi, we may set Œb; r D z 1 and Œb; s D z 2 , where 1 ; 2 2 ¹0; 1º. We compute Œt; tbr D d 0 , Œt; tbs D d 0 and Œt; tbrs D d 0 , which shows that no one of the elements ¹t br; t bs; t brsº could be an involution (since 1 .G/ is elementary abelian). On the other hand, ..tb/r/2 D .tb/2 r 2 Œr; tb D zd 0 c 0 z 1 D zz z 1 D z 1CC1 which gives C 1 0 .mod 2/, ..tb/s/2 D zd 0 c 0 z z 2 D zz z z 2 D z 1CCC2 which gives C C 2 0 .mod 2/, ..tb/.rs//2 D zd 0 c 0 z z 1 C2 D z 1CCC1 C2 which gives C C 1 C 2 0 .mod 2/. From the above relations we get 1 D 0 and D 0. But then we have d 0 D c 0 and so D 0 D C 0 which gives ŒC; D D hzi since B 0 D G 0 . This implies 2 D 1 and then the above relations also give D 1 so that C Š H2 . The structure of B is uniquely determined. Consider the abelian subgroup S D ht; si D ht i hsi, where s 2 D zc 0 . Due to the fact that ŒNB .S/; S S \ G 0 D hzc 0 i, an element x 2 B lies in NB .S/ if and only if Œt; x 2 hzc 0 i and Œs; x 2 hzc 0 i. We have CB .t/ D ht i C and t b D tc 0 (where c 0 ¤ zc 0 ) and so NB .S/ ht i C . Also, r 2 CB .t/ but Œs; r D c 0 ¤ zc 0 and so NB .S/ is a proper subgroup of CB .t/ which shows jB W NB .S/j 4, a contradiction. (ii) Now assume D D hb; f j b 4 D 1; f 4 D 1; b 2 D z; f 2 D d 0 ; Œb; f D d 0 i Š H2 ;

122

p-groups all of whose subgroups have normalizers of index at most p

227

where 1 .D/ D 1 .C / D G 0 D B 0 D hz; c 0 i D hz; d 0 i D Z.B/ D Z.G/ D ˆ.G/: Here we have two possibilities for c 0 d 0 D z , namely, D 0 or 1. (ii1) First assume D 0, i.e., d 0 D c 0 which implies ŒC; D D hzi since B 0 D G 0 . We set Œb; r D z 1 ; Œb; s D z 2 ; Œf; r D z 3 ; Œf; s D z 4 ; where 1 ; 2 ; 3 ; 4 2 ¹0; 1º, and compute .f r/2 D c 0 c 0 z 3 D z 3 ; .f s/2 D c 0 c 0 z z 4 D z C4 ; .f .rs//2 D c 0 c 0 z z 3 C4 D z C3 C4 : Assume 3 D 0 so that Œf; r D 1 and i D f r is an involution. Then i b D f b r b D f c 0 rz 1 D ic 0 z 1

and i s D f s r s D f z 4 rc 0 D ic 0 z 4

so that we have 4 D 1 ;

i bs D i

and

CB .i/ D G 0 hf; r; bsi:

Suppose, in addition, that 1 D 1 and consider the abelian subgroup S D hi i hf i Š C2 C4 with f 2 D c 0 . Since ŒNB .S/; S S \ G 0 D hc 0 i, an element x 2 B lies in NB .S/ if and only if Œi; x 2 hc 0 i and Œf; x 2 hc 0 i. As i b D i.c 0 z/, we conclude that NB .S/ CB .i/. But bs 2 CB .i/ and f bs D .f c 0 /s D f z 4 c 0 D f .zc 0 / and so NB .S/ is a proper subgroup of CB .i/ and so jB W NB .S/j 4, a contradiction. Thus we have proved that 1 D 4 D 0 and so Œb; r D Œf; s D 1. But ŒC; D D hzi and so we must have 2 D 1 and Œb; s D z. If D 0, then hr; si Š Q8 and hr; si hf i (with hr; si \ hf i D hc 0 i) contains a subgroup isomorphic to D8 , a contradiction. Hence D 1 and so s 2 D .rs/2 D c 0 z and therefore C Š H2 so that the structure of B is in this case uniquely determined. Let us consider the abelian subgroup S D hi i hbsi Š C2 C4 with .bs/2 D c 0 z. Since ŒNB .S /; S S \ G 0 D hc 0 zi, an element x 2 B lies in NB .S / if and only if Œi; x 2 hc 0 zi and Œbs; x 2 hc 0 zi. As i b D ic 0 , we have NB .S / CB .i/. But r 2 CB .i/ and Œbs; r D c 0 ¤ c 0 z and so NB .S / is a proper subgroup of CB .i/ and therefore jB W NB .S /j 4, a contradiction. We have proved that 3 D 1 and so Œf; r D z. If Œf; s D z, then Œf; rs D 1 and so interchanging s and rs (if necessary) and noting that s 2 D .rs/2 D c 0 z , we may assume from the start that Œf; s D 1 and so 4 D 0. Assume that D 1. Then j D f rs is an involution in B G 0 , j r D f z r sc 0 D jzc 0 , j s D jc 0 and ¹j; jzc 0 ; jc 0 º are three pairwise distinct conjugate involutions in B, a contradiction. Hence we obtain D 0 so that r 2 D s 2 D .rs/2 D c 0 , hr; si Š Q8 and u D f s is an involution in

228

Groups of prime power order

B G 0 . We have ur D f z sc 0 D uzc 0 and so ¹u; uzc 0 º are the only two conjugates of u in G. Also, ub D f c 0 sz 2 D uc 0 z 2 which gives 2 D 1, Œb; s D z and ubr D u so that CB .u/ D G 0 hf; s; bri. Let us consider the abelian subgroup S0 D hui hf i Š C2 C4 with f 2 D c 0 . Since ŒNB .S0 /; S0 S0 \ G 0 D hc 0 i, an element x 2 B lies in NB .S0 / if and only if Œu; x 2 hc 0 i and Œf; x 2 hc 0 i. As ub D uc 0 z, we infer that NB .S0 / CB .u/. But br 2 CB .u/ and Œf; br D c 0 z ¤ c 0 and so NB .S0 / is a proper subgroup of CB .u/ and so jB W NB .S0 /j 4, a contradiction. (ii2) We have proved that D 1, i.e., d 0 D c 0 z. Since ŒD; C hzi, we may set again Œb; r D z 1 ; Œb; s D z 2 ; Œf; r D z 3 ; Œf; s D z 4 ; where 1 ; 2 ; 3 ; 4 2 ¹0; 1º, and we look for involutions in B .C [ D/. We compute .f r/2 D c 0 z c 0 z 3 D z 1C3 ; .f s/2 D c 0 z c 0 z z 4 D z 1CC4 ; .f .rs//2 D c 0 z c 0 z z 3 C4 D z 1CC3 C4 : If 3 D 1, then .f r/2 D 1, .f s/2 D z 1C.C4 / , .f .rs//2 D z C4 and so f r is an involution and either f s or f rs is an involution. But in that case, we conclude that Œf r; f s D c 0 z 1C4 ¤ 1 and Œf r; f rs D c 0 z 4 ¤ 1, contrary to the fact that 1 .G/ must be elementary abelian. Hence 3 D 0 and then .f r/2 D z;

.f s/2 D z 1C.C4 / ;

.f rs/2 D z 1C.C4 / :

If C 4 1 .mod 2/, then both f s and f rs are involutions which is not possible since Œf s; f rs D c 0 ¤ 1. Hence C 4 0 .mod 2/ and so 4 D . We have proved that Œf; r D 1 and Œf; s D z which gives .f r/2 D .f s/2 D .f rs/2 D z. Also note that b 2 D .bf /2 D z and so if there is an involution i 2 A hzi, then i commutes with b (otherwise, Œi; b D z and then hi; bi Š D8 ), bf , f r, f s, f rs. But we have hb; bf; f r; f s; f rsi D B and so ŒA; B D 1, contrary to the fact that we know that ŒA; B D hzi. Hence A D hai Š C4 with a2 D z. Suppose that there are no involutions in G B. Then Œa; b D Œa; bf D Œa; f r D Œa; f s D Œa; f rs D z: From Œa; b D Œa; bf D z follows Œa; f D 1. But then Œa; r D Œa; s D Œa; rs D z, a contradiction. Hence it follows that a commutes with at least one element x in the set L D ¹b; bf; f r; f s; f rsº in which case ax is an involution in G B. We check that no two distinct elements in the set L commute. If a would commute with two distinct elements x; y 2 L, then ax and ay are involutions and then 1 D Œax; ay D Œx; y, a contradiction. We have proved that a must commute with exactly one of the ﬁve elements in L. But we may interchange s and rs and so it is enough to consider the following four cases: (a) Œa; b D 1;

(b) Œa; bf D 1;

(c) Œa; f r D 1;

(d) Œa; f s D 1:

122

p-groups all of whose subgroups have normalizers of index at most p

229

(ii2a) We assume Œa; b D 1 so that i D ab is an involution in G B and Œa; bf D Œa; f r D Œa; f s D Œa; f rs D z: This implies Œa; f D z, Œa; r D Œa; s D 1. Since i f D .ab/f D az bc 0 z D ic 0 , ¹i; ic 0 º are the only two conjugates of i in G. Then i r D .ab/r D a bz 1 D iz 1 gives 1 D 0 and i s D .ab/s D a bz 2 D iz 2 gives 2 D 0. We have CB .i/ D hbihr; si D hbiC and i f D ic 0 . Consider the abelian subgroup S D hi i hbri Š C2 C4 with .br/2 D zc 0 . Since ŒNG .S/; S S \ G 0 D hzc 0 i, an element x 2 G lies in NG .S/ if and only if Œi; x 2 hzc 0 i and Œbr; x 2 hzc 0 i. As i f D i c 0 , we have NG .S/ CG .i/. But s 2 CG .i/ and .br/s D br s D .br/c 0 and so NG .S/ is a proper subgroup of CG .i/ and therefore jG W NG .S/j 4, a contradiction. (ii2b) We assume that Œa; bf D 1 so that i D abf is an involution in G B and Œa; b D Œa; f r D Œa; f s D Œa; f rs D z. This gives Œa; f D z, Œa; r D Œa; s D 1. Since i f D .abf /f D az bc 0 z f D ic 0 ; i r D .abf /r D a bz 1 f D iz 1 ; i s D .abf /s D a bz 2 f z D iz C2 ; we get 1 D 0 and 2 D . If D 0, then we consider the abelian subgroup S1 D hi i hasi Š C2 C4 with .as/2 D zc 0 . As ŒNG .S1 /; S1 S1 \ G 0 D hzc 0 i, an element x 2 G lies in NG .S1 / if and only if Œi; x 2 hzc 0 i and Œas; x 2 hzc 0 i. Since i f D ic 0 and jG W CG .i/j D 2 (noting that 1 .Z.G// G 0 ), we obtain NG .S1 / CG .i/. But we have r 2 CG .i/ and .as/r D .as/c 0 and so NG .S1 / is a proper subgroup of CG .i/ and therefore we get jG W NG .S1 /j 4, a contradiction. If D 1, then we consider the abelian subgroup S2 D hi i hsi Š C2 C4 with s 2 D zc 0 . Since ŒNG .S2 /; S2 S2 \ G 0 D hzc 0 i, an element x 2 G lies in NG .S2 / if and only if Œi; x 2 hzc 0 i and Œs; x 2 hzc 0 i. Because i f D ic 0 and jG W CG .i/j D 2, we have NG .S2 / CG .i/. But r 2 CG .i/ and s r D sc 0 and so NG .S2 / is a proper subgroup of CG .i/ and therefore jG W NG .S2 /j 4, a contradiction. (ii2c) We assume Œa; f r D 1 so that i D af r is an involution in G B and Œa; b D Œa; bf D Œa; f s D Œa; f rs D z: This gives Œa; f D 1, Œa; r D 1, Œa; s D z. We have i b D .af r/b D az f c 0 z rz 1 D ic 0 z 1 ; i s D .af r/s D az f z rc 0 D ic 0 z C1 ; and so 1 D C1. As i r D .af r/r D i and Œi; sb D 1, we have CG .i/ D ha; f; r; sbi.

230

Groups of prime power order

If D 0, then we consider the abelian subgroup S1 D hi i hri Š C2 C4 with r 2 D c 0 . Since ŒNG .S1 /; S1 S1 \ G 0 D hc 0 i, an element x 2 G lies in NG .S1 / if and only if Œi; x 2 hc 0 i and Œr; x 2 hc 0 i. Because i b D ic 0 z and jG W CG .i/j D 2, we have NG .S1 / CG .i / < G. But sb 2 CG .i/ and Œr; sb D zc 0 and so NG .S1 / is a proper subgroup of CG .i/ and therefore jG W NG .S1 /j 4, a contradiction. If D 1, then we consider the abelian subgroup S2 D hi i hari Š C2 C4 with .ar/2 D zc 0 . Since ŒNG .S2 /; S2 S2 \ G 0 D hzc 0 i, an element x 2 G lies in NG .S2 / if and only if Œi; x 2 hzc 0 i and Œar; x 2 hzc 0 i. Because i b D .af r/b D ic 0 and jG W CG .i/j D 2, we get NG .S2 / CG .i/ < G. But sb 2 CG .i/ and Œar; sb D c 0 and so NG .S2 / is a proper subgroup of CG .i/ and therefore jG W NG .S2 /j 4, a contradiction. (ii2d) We assume Œa; f s D 1 so that i D af s is an involution in G B and Œa; b D Œa; bf D Œa; f r D Œa; f rs D z: This gives Œa; f D 1, Œa; r D z, Œa; s D 1. We have i b D .af s/b D az f c 0 z sz 2 D ic 0 z 2 ; and so D 0. Also,

i s D .af s/s D a f z s D iz ;

i r D .af s/r D az f sc 0 D izc 0

which gives 2 D 1 and i f D i , i br D i and therefore CG .i/ D ha; f; s; bri. We consider the abelian subgroup S D hi i hsi Š C2 C4 with s 2 D c 0 . Since ŒNG .S/; S S \G 0 D hc 0 i, an element x 2 G lies in NG .S/ if and only if Œi; x 2 hc 0 i and Œs; x 2 hc 0 i. As i b D ic 0 z and jG W CG .i/j D 2, we have NG .S/ CG .i/ < G. But br 2 CG .i/ and Œs; br D zc 0 and so NG .S/ is a proper subgroup of CG .i/ and therefore jG W NG .S/j 4, a contradiction. Part 2, where D=hzi Š P. Here P is the nonmetacyclic minimal nonabelian subgroup of order 24 . We set W=hzi D Z.D=hzi/ D ˆ.D=hzi/ so that W > G 0;

jW W G 0 j D 2;

W=hzi Š E4 ;

W=hzi D Z.B=hzi/

and hence Z.B/ W . There is y 2 D such that y 2 2 W G 0 (noting that ˆ.D/ covers W=hzi). For each g 2 G, Œy 2 ; g D Œy; g2 D 1 and so y 2 2 Z.G/ and therefore W Z.G/, W D Z.B/, and B=Z.B/ Š E24 . Since ŒA; B hzi and A=hzi is elementary abelian, we get W D ˆ.G/. Each of three maximal subgroups of D containing W are abelian and so D 0 D hd 0 i is of order 2, where d 0 2 G 0 hzi. First suppose that z 62 ˆ.D/ and then D D hzi P0 , where P0 Š P so that we may set P0 D hb; t j b 4 D t 2 D 1; Œb; t D d 0 ; .d 0 /2 D Œd 0 ; t D Œd 0 ; b D 1i:

122

p-groups all of whose subgroups have normalizers of index at most p

231

Since ¹t; t b D td 0 º are the only two conjugates of t in G and ŒC; D hzi, we must have Œt; r D Œt; s D 1. Similarly, hbi and hb t i D hbd 0 i are the only two conjugates of hbi in G and so Œb; r D Œb; s D 1. We have proved that ŒC; D D ¹1º. If D 0 D C 0 , then B 0 D C 0 D D 0 is of order 2, contrary to B 0 D G 0 . Hence we have d 0 ¤ c 0 and so d 0 D c 0 z. We claim that hbri Š C4 , h.br/t i D hbr c 0 zi and h.br/s i D hbr c 0 i are three distinct cyclic subgroups of order 4 and this gives a contradiction. Indeed, the elements of order 4 in hbri are br and .br/.br/2 D br b 2 c 0 , the elements of order 4 in hbr c 0 zi are br c 0 z and br b 2 z, and the elements of order 4 in hbr c 0 i are br c 0 and br b 2 . We see that the above six elements of order 4 are pairwise distinct and we are done. We have proved that z 2 ˆ.D/ and so W D ˆ.D/ with D=W Š E4 which together with jD 0 j D 2 implies that D is minimal nonabelian of order 25 . Since D=hzi Š P is nonmetacyclic of exponent 4, it follows that D is nonmetacyclic of exponent 4 or 8. If D is of exponent 4, then D D hb; f j b 4 D f 4 D 1; Œb; f D d 0 ; .d 0 /2 D Œd 0 ; b D Œd 0 ; f D 1i; where

1 .D/ D W D ˆ.D/ D Z.D/ D hb 2 ; f 2 ; d 0 i Š E8 :

In case that z is not a square in D, we get 1 .D=hzi/ D W=hzi Š E4 which implies that D=hzi is metacyclic minimal nonabelian, contrary to D=hzi Š P. Hence z is a square in D so that we may choose the generators b and f of D so that b 2 D z. If D is of exponent 8, then D D hb; t j b 8 D t 2 D 1; Œb; t D d 0 ; .d 0 /2 D Œd 0 ; b D Œd 0 ; t D 1i; where b 4 D z since D=hzi Š P is of exponent 4. We have proved that in any case z is a square in D. Suppose that A Š Q8 . Then we know that ŒA; B D ¹1º and let b 0 2 D be such that .b 0 /2 D z. But then the subgroup A hb 0 i with A \ hb 0 i D hzi possesses a subgroup isomorphic to D8 , a contradiction. If A Z.G/, then jG=Z.G/j D 24 , a contradiction. Hence we have A Š C4 or E4 and ŒA; B D hzi. (i) First suppose that D is of exponent 4, i.e., D D hb; f j b 4 D f 4 D 1; Œb; f D d 0 ; .d 0 /2 D Œd 0 ; b D Œd 0 ; f D 1i; where b 2 D z and d 0 2 G 0 hzi. Now, hf i (where f and ff 2 are its elements of order 4) and hf b i D hf d 0 i (where f d 0 and ff 2 d 0 are its elements of order 4) are the only conjugates of hf i in G. Hence if Œf; C D hzi, then hf i is conjugate under C to hf zi (where f z and ff 2 z are its elements of order 4) and so hf zi is distinct from hf i and hf d 0 i, a contradiction. Similarly, hf bi (where f b and f bf 2 zd 0 are its elements of order 4) and h.f b/b i D hf bd 0 i (where f bd 0 and f bf 2 z are its elements of order 4) are the only conjugates of hf bi in G. Hence if Œf b; C D hzi, then hf bi

232

Groups of prime power order

is conjugate under C to hf bzi (where f bz and f bf 2 d 0 are its elements of order 4) and so hf bzi is distinct from hf bi and hf bd 0 i, a contradiction. Since D D hf; f bi, we get ŒC; D D ¹1º which implies D 0 ¤ C 0 (since B 0 D G 0 ) and so d 0 D c 0 z. We claim that hf ri Š C4 has more than two conjugates in G which gives a contradiction. Indeed, hf ri (where f r and f rf 2 c 0 are its elements of order 4), h.f r/b i D hf rc 0 zi (where f rc 0 z and f rf 2 z are its elements of order 4), and h.f r/s i D hf rc 0 i (where f rc 0 and f rf 2 are its elements of order 4) are three pairwise distinct conjugates of hf ri. (ii) Now assume that D is of exponent 8, i.e., D D hb; t j b 8 D t 2 D 1; Œb; t D d 0 ; .d 0 /2 D Œd 0 ; b D Œd 0 ; t D 1i; where b 4 D z and d 0 2 G 0 hzi. We have here W D hb 2 i hc 0 i D hb 2 i hd 0 i Š C4 C2 and W D Z.B/ D Z.G/ D ˆ.G/. Since ¹t; t b D td 0 º is the complete conjugate class in G, ŒC; t D hzi is not possible and so we get Œt; r D Œt; s D 1. We have r 2 D c 0 and s 2 D .rs/2 D c 0 z , D 0; 1. Assume at the moment that D 1 so that C Š H2 . Then replace s with s 0 D sb 2 , where o.b 2 / D 4 and b 2 2 Z.B/. We obtain .s 0 /2 D .sb 2 /2 D s 2 b 4 D c 0 z z D c 0 and Œs 0 ; r D Œs; r D c 0 so that hr; s 0 i Š Q8 . Replacing C D hr; si with C D hzihr; s 0 i Š C2 Q8 , we may assume (writing again s instead of s 0 and C instead of C ) from the start that D 0, i.e., s 2 D .rs/2 D c 0 and C D hzi hr; si, where hr; si Š Q8 . Since Œb; hr; si hzi, we may choose the generators r; s of hr; si so that Œb; r D 1. Assume that d 0 D c 0 in which case we must have Œb; s D z since B 0 D G 0 . Consider the abelian subgroup S D ht i hb 2 si Š C2 C4 , where .b 2 s/2 D zc 0 . Note that ŒNB .S/; S S \ B 0 D hzc 0 i. We conclude that jB W CB .t/j D 2 and t b D tc 0 so that NB .S/ CB .t/. But r centralizes t and .b 2 s/r D .b 2 s/c 0 and so r 62 NB .S/ and therefore jB W NB .S/j 4, a contradiction. We have proved that d 0 D c 0 z and so we have the following two possibilities: Œb; s D 1 or Œb; s D z. But we shall show that these two possibilities give isomorphic groups. Indeed, if Œb; s D z, then replace b with b D br and replace s with s D ts. We have .b /4 D .br/4 D z; Œb ; t D Œbr; t D zc 0 ; where hc 0 i D hs ; ri0 and Q D hs ; ri Š Q8 since .s /2 D s 2 D c 0 , r 2 D c 0 , and Œs ; r D Œts; r D c 0 . In addition, Œhb ; ti; Q D ¹1º. Indeed, t commutes with Q , Œb ; r D Œbr; r D Œb; r D 1 and Œb ; s D Œbr; ts D Œb; tŒb; sŒr; tŒr; s D zc 0 z 1 c 0 D 1: Thus we may assume from the start that Œb; s D 1. The structure of B is uniquely determined.

122

p-groups all of whose subgroups have normalizers of index at most p

233

It is easy to see that there are involutions in G B. Indeed, if A Š E4 , then there is an involution i 2 A hzi D A B. Suppose that A D hvi Š C4 with v 2 D z. Since b 2 2 Z.G/ and b 4 D z, i D vb 2 is an involution in G B. Let i be an involution in G B from the previous paragraph. Then we deduce that hi; ti Š E4 , hi; ti \ G 0 D ¹1º and hi; ti \ Z.G/ D ¹1º. From ŒNG .hi; ti/; hi; ti hi; ti \ G 0 D ¹1º it follows that NG .hi; ti/ D CG .hi; ti/ D CG .i/ \ CG .t/: But jG W NG .hi; ti/j D 2 and so we must have CG .i/ D CG .t/ D hi; ti hb 2 ; r; si, which is a subgroup of index 2 in G. As Œt; b D c 0 z ¤ 1, we have also Œi; b ¤ 1. Note that ŒA; b hzi and so if i 2 A, then Œi; b D z. If i D vb 2 (in case A D hvi Š C4 ), then Œi; b D Œvb 2 ; b D Œv; b D z. Hence in any case Œi; b D z. Consider the abelian subgroup S D hit i hb 2 si Š C2 C4 , where .b 2 s/2 D b 4 s 2 D zc 0 . Note that ŒNG .S/; S S \ G 0 D hzc 0 i. In view of .it /b D iz tc 0 z D it c 0 , we must have NG .S/ CG .it / D CG .t/. On the other hand, r centralizes it but .b 2 s/r D b 2 s c 0 and so r 62 NG .S/. We obtain that jG W NG .S/j 4, a contradiction. Theorem 122.1 is completely proved for p D 2. Proof of Theorem 122.1 for p > 2. Let G be a p-group, p > 2, with sb.G/ D 1 which is a minimal counterexample to Theorem 122.1 and so jG=Z.G/j > p 3 . By Proposition 122.9, 1 .Z.G// G 0 . Due to Proposition 122.8, b.G/ 2, i.e., the element breadth of G is at most 2. If b.G/ D 1, then Proposition 121.9 implies that jG 0 j D p. But then Proposition 122.13 gives jG=Z.G/j D p 2 , a contradiction. Hence we obtain b.G/ D 2. In that case, Theorem 121.1 implies jG 0 j D p 2 (because jG 0 j D p 3 would imply jG=Z.G/j D p 3 ). Also, we note that G cannot possess an abelian maximal subgroup because in that case Lemma 1.1 gives jGj D pjZ.G/jjG 0 j and then we conclude jG=Z.G/j D p 3 , a contradiction. N D 1 and sb.G/ N D 1. Let 1 ¤ z 2 G 0 \ 1 .Z.G// so that for GN D G=hzi, b.G/ N Set R=hzi D Z.G/ and then Proposition 122.13 gives G=R Š Ep2 . We have r 2 R if and only if ŒG; r hzi in which case jG W CG .r/j p. Also, ŒG; R D hzi because if ŒG; R D 1, then jG=Z.G/j D p 2 , a contradiction. As ŒG; G 0 hzi, we have G 0 R. Let A be any maximal normal abelian subgroup in G which contains ˆ.G/, where ˆ.G/ is abelian (Proposition 122.2). By Proposition 122.10, we have sbA .G/ D 2, and we ﬁx an element g 2 G such that jA W CA .g/j D p 2 and then jŒA; gj D p 2 (Proposition 121.2) so that ŒA; g D G 0 . This implies that A 6 R and R does not cover G=A so that jA W .A \ R/j D p and jG W .AR/j D p, where g 2 G .AR/ so that G D .AR/hgi. For each x 2 .AR/ A, x 2 TA , i.e., jA W CA .x/j D p (noting that for each r 2 R we have jG W CG .r/j p). Since jA W CA .g/j D p 2 , g does not centralize R \ A and so ŒR \ A; g D hzi which implies j.R \ A/ W CR\A .g/j D p and therefore CR\A .g/ D CA .g/ so that g p 2 CR\A .g/. As ŒR; R hzi and Œg; R D hzi, we have .Rhgi/0 D hzi. Also, R0 hzi and ŒA; R D hzi (noting that R 6 A since G has no abelian maximal subgroup) and so .AR/0 D hzi.

234

Groups of prime power order

Assume that jG=Aj > p 2 so that j.AR/=Aj p 2 . We have Z.AR/ < A and so j.AR/ W Z.AR/j p 3 , which contradicts Proposition 122.13 applied to AR. Hence we get G=A Š Ep2 and then jR W .R \ A/j D p. Since .Rhgi/0 D hzi, we may apply Proposition 122.3 to Rhgi and we get j.Rhgi/ W Z.Rhgi/j D p 2 . Suppose that R is abelian. Then CG .CA .g// hA; R; gi D G which gives Z.G/ D CA .g/ and jG W Z.G/j D p 4 . Now assume that R is nonabelian so that R0 D hzi. In that case, R \ A is an abelian maximal subgroup of R and then we get from jRj D pjZ.R/jjR0 j (Lemma 1.1) that jR W Z.R/j D p 2 , where Z D Z.R/ is a subgroup of index p in R \ A. But we infer that j.Rhgi/ W Z.Rhgi/j D p 2 , and so there is an element h 2 .Rhgi/ R such that h 2 Z.Rhgi/. We get CG .Z/ hA; R; hi D G and so again Z D CA .g/ D Z.G/ and jG W Z.G/j D p 4 . Thus we have proved that for any maximal normal abelian subgroup A of G containing ˆ.G/, G=A Š Ep2 and CA .g/ D Z.G/ is of index p 4 in G, where g 2 G A is such that jA W CA .g/j D p 2 . We shall ﬁx this notation for A, R and g in the rest of the proof. (i) We prove that 1 .G/ is elementary abelian. Suppose that x; y are some elements of order p in the group G such that Œx; y ¤ 1. Then C D CG .x/ D NG .hxi/ implies that jG W C j D p and y 2 G C so that the set p1 ¹x; x y ; : : : ; x y º contains p conjugates of x in G and they are all central elements p1 in C which implies that N D hx; x y ; : : : ; x y i is an elementary abelian normal subgroup in G, N C , and jN j p 2 . Set N0 D CN .y/ so that jN W N0 j D p and x 2 N N0 since jG W CG .y/j D p. We have N0 D Z.hx; yi/ and hx; yi=N0 Š Ep2 so that 1 ¤ Œx; y D z 0 2 N0 and hx; yi0 D hz 0 i is of order p which implies that hx; yi is minimal nonabelian. Hence S D hx; yi Š S.p 3 / which is the minimal nonabelian group of order p 3 and exponent p and so N0 D hz 0 i and N Š Ep2 . All p conjugates of x in S are contained in hx; z 0 i and they are all conjugates of x in G and so hx; z 0 i E G. Similarly, all p conjugates of y in S are contained in hy; z 0 i and they are all conjugates of y in G and so hy; z 0 i E G. This fact implies that S D hx; yi E G, G D S P , where P D CG .S/ and S \ P D hz 0 i D Z.S/. If P is abelian, then G 0 D S 0 D hz 0 i is of order p, a contradiction. Hence P is nonabelian and so P possesses nonnormal subgroups (Lemma 1.18). If all nonnormal subgroups of P contain hz 0 i, then Corollary 92.7 (N. Blackburn) implies p D 2, a contradiction. Let P1 be a nonnormal subgroup of P such that z 0 62 P1 and set U D hP1 ; xi D P1 hxi. We have U \ P D P1 and U \ S D hxi. If h 2 G normalizes U , then h normalizes U \ S D hxi and so h centralizes x. But h also normalizes U \ P D P1 and so we get NG .U / D CG .x/ \ NG .P1 /. On the other hand, we have CG .x/ D P hxi and jG W CG .x/j D p. But P1 is not normal in P hxi (since P1 is a nonnormal subgroup of P1 ) and so NG .U / is a proper subgroup of P hxi which gives jG W NG .U /j p 2 , a contradiction.

122

p-groups all of whose subgroups have normalizers of index at most p

235

(ii) 1 .G/ is elementary abelian of order p 3 . First suppose that G possesses a normal elementary abelian subgroup F of order p 4 . Set F0 D F \ G 0 so that jF0 j p 2 . Let x; y be any elements in F F0 such that hx; yi Š Ep2 and hx; yi \ G 0 D ¹1º. Then ŒNG .hx; yi/; hx; yi hx; yi \ G 0 D ¹1º and so NG .hx; yi/ D CG .hx; yi/ D CG .x/ \ CG .y/: Note that for each x 2 F F0 we have jG W CG .x/j D p since 1 .Z.G// G 0 . As jG W NG .hx; yi/j D p, we have C D CG .x/ D CG .y/ and so C D CG .F / is a subgroup of index p in G. In particular, ˆ.G/ centralizes F and so we may choose a maximal normal abelian subgroup A of G so that A F ˆ.G/. Recalling our results about A and g 2 G A with jA W CA .g/j D p 2 (from the beginning of our proof), we know that G=A Š Ep2 and CA .g/ D Z.G/. Since CF .g/ 1 .Z.G// G 0 and jG 0 j D p 2 , we see that jCF .g/j p 2 and so F covers A=CA .g/. But C D CG .F / is a subgroup of index p in G and so C > A and C centralizes hZ.G/; F i D A, contrary to CG .A/ D A. Now assume that 1 .G/ Š Ep2 . In view of Theorem 13.7, G is either metacyclic (with G 0 Š Cp2 ) or a 3-group of maximal class. In case that G is metacyclic, Proposition 122.7 implies that sb.G/ > 1, a contradiction. If G is a 3-group of maximal class, then jG=G 0 j D 32 and so jGj D 34 . But then jG W Z.G/j 33 , a contradiction. Thus we have proved that 1 .G/ Š Ep3 . (iii) We have Ep3 Š O D 1 .G/ ˆ.G/. Suppose that this is false. Let N be a maximal subgroup of G which does not contain O so that O \ N D 1 .N / Š Ep2 . Also we know that N is nonabelian. By Theorem 13.7, N is either metacyclic or a 3-group of maximal class. Suppose that N is metacyclic. If N 0 Š Cp2 , then Proposition 122.7 implies that sb.N / > 1, a contradiction. Hence we have jN 0 j D p which gives that N is minimal nonabelian and then ˆ.N / D Z.N / and jN W Z.N /j D p 2 . Let y 2 O N so that jN W CN .y/j D p and thus y centralizes ˆ.N / D Z.N / which gives Z.N / D Z.G/ and jG W Z.G/j p 3 , a contradiction. We have proved that N is a 3-group of maximal class. In that case, jN W N 0 j D 32 . If jN 0 j D 3, then jN j D 33 , jGj D 34 and jG W Z.G/j 33 , a contradiction. Hence jN 0 j D 32 and jGj D 35 . We have G 0 N \ O and so G 0 D N 0 D N \ O Š E9 : Since N=N 0 Š E9 and O=N 0 Š C3 , we get ˆ.G/ D ˆ.N / D N \ O D G 0 . As G has no abelian maximal subgroup, we have CG .O/ D O so that O D A is a maximal normal abelian subgroup of G containing ˆ.G/. We recall our results about A, R and g from the beginning of the proof. Since R G 0 , we obtain R \ A D G 0 , g 3 2 G 0 , and N D Rhgi is another maximal subgroup of G which does not contain O since N \A D G 0 D 1 .N / Š E9 and j.N /0 j D 3. By Theorem 13.7, N is metacyclic and so N is minimal nonabelian with Z.N / D ˆ.N / and jN W Z.N /j D 32 .

236

Groups of prime power order

Let v 2 A G 0 so that v centralizes a maximal subgroup of N and so v centralizes ˆ.N / D Z.N / which forces Z.N / Z.G/. But then jG W Z.G/j 33 , a contradiction. It is now easy to complete the proof of Theorem 122.1 for p > 2. To do this, we recall again our results about A ˆ.G/, R, g and z from the beginning of the proof. We have ˆ.G/ .Rhgi/ \ A D R \ A, where G=A Š Ep2 , jA W .R \ A/j D p, j.R \ A/ W CR\A .g/j D p, and CA .g/ D CR\A .g/ D Z.G/. By (iii), we conclude that Ep3 Š O D 1 .G/ ˆ.G/ and so O R \ A. Since 1 .Z.G// G 0 , we have O 6 Z.G/ and so Ep2 Š G 0 D O \ Z.G/ and O covers .R \ A/=Z.G/. In particular, the group G is of class 2 with an elementary abelian commutator subgroup of order p 2 . For each x; y 2 G we have Œx p ; y D Œx; yp D 1 and so x p 2 Z.G/ which implies that ˆ.G/ D Ã1 .G/G 0 Z.G/. But then O ˆ.G/ Z.G/, a ﬁnal contradiction. Theorem 122.1 is completely proved. It follows from Theorem 122.1 that if G is a metacyclic group of order p 2mC1 , where m > 1 if p > 2 and n > 2 if p D 2, then there is in G a subgroup H such that jG W NG .H /j > p.

123

Subgroups of ﬁnite groups generated by all elements in two shortest conjugacy classes

Here we present some results of I. M. Isaacs (Subgroups generated by small classes in ﬁnite groups, Proc. Amer. Math. Soc. 136 (2008), 2299–2301) which generalize the results of A. Mann about subgroups of p-groups which are generated by all elements of the ﬁrst two class sizes. Let M.G/ be the subgroup of a ﬁnite group G generated by all elements that lie in conjugacy classes of the two smallest sizes. The following simple result is the key for all following arguments. Lemma 123.1. Let A be an abelian normal subgroup of a group G. Let x be a noncentral element of G and a 2 A. Then jCG .Œa; x/j > jCG .x/j, and so the G-class of Œa; x is smaller than that of x. Proof. Consider the map u ! Œu; x (u 2 A) from A into A. Since for any u; v 2 A we have Œuv; x D Œu; xv Œv; x D Œu; xŒv; x, it follows that our map is a homomorphism onto ŒA; x D ¹Œu; x j u 2 Aº with kernel CA .x/. Thus ŒA; x is a subgroup of A and jŒA; xj D jA=CA .x/j. Set H D ACG .x/ so that jH j D .jAjjCG .x/j/=jCA .x/j and jAj=jCA .x/j D jH j=jCG .x/j: Obviously, ŒA; x E H . Indeed, if c 2 CG .x/ and Œu; x 2 ŒA; x with u 2 A, then Œu; cc D Œuc ; x c D Œuc ; x 2 ŒA; x. We may assume that Œa; x ¤ 1 because x is noncentral (and so in case Œa; x D 1, our result is trivial). Hence the conjugacy class of Œa; x in H has jH j=jCH .Œa; x/j elements so that jH j=jCH .Œa; x/j < jŒA; xj D jAj=jCA .x/j D jH j=jCG .x/j which gives jCH .Œa; x/j > jCG .x/j and we are done. Lemma 123.2. For an abelian normal subgroup A of a group G, ŒA; M.G/ Z.G/. Proof. If x 2 M.G/ is contained in a conjugacy class of G of size m, where m is the smallest class size exceeding 1, then, by Lemma 123.1, we have Œa; x 2 Z.G/ for each a 2 A. Hence all generators of M.G/ centralize A=.A \ Z.G//. Thus M.G/ centralizes A=.A \ Z.G// and so ŒA; M.G/ Z.G/.

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Groups of prime power order

Theorem 123.3. Let G be a group that contains a self-centralizing abelian normal subgroup A. Then M.G/ is nilpotent of class at most 3. Proof. Set M D M.G/ so that Lemma 123.2 yields ŒA; M Z.G/ and therefore ŒA; M; M D ¹1º. Since ŒM; A; M D ¹1º, the three-subgroups lemma implies that ŒM; M; A D ¹1º and then M 0 CG .A/ D A. But in that case we get ŒM 0 ; M; M ŒA; M; M D ¹1º, and so M is nilpotent of class at most 3. A maximal normal abelian subgroup in a supersolvable group G is self-centralizing and so we get the following result. Corollary 123.4. Let G be a supersolvable group. Then M.G/ is nilpotent of class at most 3. In particular, if G is a p-group, then M.G/ is of class at most 3. Theorem 123.5. Let G be a group and assume that M.G/ has a nonidentity solvable normal subgroup. Then Z.M.G// is nontrivial. Proof. Write M D M.G/. By our assumption, the Fitting subgroup F.M / of M is nontrivial. Let Z D Z.F.M // so that Z is a nontrivial characteristic abelian subgroup of M , and therefore Z E G. Due to Lemma 123.2, ŒZ; M Z.G/. If ŒZ; M ¤ ¹1º, then Z.G/ is nontrivial. By the deﬁnition of M D M.G/, Z.G/ M and so Z.M / is nontrivial and we are done in this case. If ŒZ; M D ¹1º, then ¹1º ¤ Z Z.M /. In any case, Z.M / is nontrivial. Finally, we prove a result about arbitrary ﬁnite group G (without any assumptions). Theorem 123.6. Let G be any ﬁnite group. Then the Fitting subgroup of M.G/ has the nilpotence class at most 4. Proof. Write again M D M.G/ and set F D F.M /. Let n be the nilpotence class of F so that Kn .F / ¤ ¹1º and KnC1 .F / D ¹1º. We want to show that n 4 and so we may assume that n 3 so that Kn2 .F / is deﬁned. If this normal subgroup is abelian, Lemma 123.2 yields ŒKn2 .F /; M Z.G/ and then Kn .F / D ŒKn2 .F /; M; M D ¹1º, a contradiction. Hence Kn2 .F / is nonabelian and then ¹1º ¤ ŒKn2 .F /; Kn2 .F / K2n4 .F /: Since KnC1 .F / D ¹1º, we must have 2n 4 < n C 1 and so n < 5, as required.

124

The number of subgroups of given order in a metacyclic p-group

In this section we compute the number of subgroups of given order in a metacyclic p-group and also prove some structure results. In the following lemma we gather some known results which we use in what follows. Lemma J. Let G be a nonabelian metacyclic p-group. (a) Let G be of order p 4 and exponent p 2 ; then G is minimal nonabelian. (i) If p D 2, then G Š H2 D ha; b j a4 D b 4 D 1; ab D a3 i, all subgroups of order 2 are characteristic in G, exactly one maximal cyclic subgroup of G has order 2. 2 2 (ii) If p > 2, then G D ha; b j ap D b p D 1; ab D a1Cp i, all maximal cyclic subgroups of G have order p 2 . (b) (Proposition 10.19) If G has a nonabelian subgroup of order p 3 , then it is of maximal class. In particular, if p > 2, then jGj D p 3 . (c) If p D 2 and L G G is such that G=L is nonabelian of order 8, then L is characteristic in G. (d) (Theorem 1.2 and Lemma 1.6) There are four and only four series of nonabelian 2-groups with cyclic subgroup of index 2, namely D2n , Q2n , SD2n , M2n . (e) j1 .G/j ¤ p 2 if and only if G is a 2-group of maximal class. Let us prove Lemma J(a). There is in G a normal cyclic subgroup A D hai such that G=A is cyclic; then we have jAj D p 2 . If B < G is cyclic and such that AB D G, then jBj D p 2 ; let B D hbi. Since jG W CG .A/j D p and the group of automorphisms of A is generated by the automorphism a ! a1Cp of order p, one can write ab D a1Cp hence the group G has the same deﬁning relations as in (a). Therefore, if p > 2, then G is regular, and (ii) follows easily. Now let p D 2. The subgroup G 0 D ha2 i is characteristic in G. Next, G has exactly three subgroups of order 2: G 0 D ha2 i, U D hb 2 i and V D ha2 b 2 i, and all these subgroups are central. Since G=V Š Q8 and V is the unique subgroup of order 2 not contained in a cyclic subgroup of order 4, it follows that V is characteristic in G, and hence U is also characteristic in G.

240

Groups of prime power order

Let us prove Lemma J(c). We have Ã2 .G/ L. If Ã2 .G/ D L, then it is nothing to prove. Now let Ã2 .G/ < L. Then jG W Ã2 .G/j D 24 , so G=Ã2 .G/ is nonabelian metacyclic of order 24 and exponent 4, and now the result follows from Lemma J(a). In what follows we freely use the following fact: If G is a metacyclic p-group, then 1 .G/ Š Ep2 unless G is either cyclic or a 2-group of maximal class. 1o Mann’s theorem on the number of subgroups of given order in certain metacyclic p-groups. Let G be a metacyclic group of order p n and exponent p e and let f D ne; clearly, f e. We also write e D e.G/ and f D f .G/. This notation is applicable to subgroups and epimorphic images of G as well. We begin with the following Deﬁnition. A p-group G is said to be quasi-regular if it is metacyclic and (QR1) if p D 2, then G has no nonabelian sections of order 8. (QR2) if H is a section of G, then Ã1 .H / D ¹x p j x 2 H º, i.e., all elements of Ã1 .H / are p-th powers. A more general condition than (QR2) was considered by Mann (see 11) for all p-groups, where p is an arbitrary prime. All metacyclic p-groups, p > 2, are quasi-regular (Theorem 7.2(c)). If G is a quasiregular p-group, then exp.m .G// p m for all positive integers m (this is proved by induction on jGj). The group H2 (see Lemma J(a)) is not quasi-regular since it has two nonabelian sections of order 8. The groups M2n are quasi-regular. Exercise. Metacyclic quasi-regular p-groups G are powerful (see 26). Solution. This is obvious for p > 2 (indeed, G 0 ˆ.G/ D Ã1 .G/). Now let p D 2 and write GN D G=Ã2 .G/. We have to prove that GN is abelian. Assume that this is N 2 ¹23 ; 24 º. Then jGj N ¤ 23 by (QR1). By Lemma J(a), there exists false. We have jGj only one nonabelian metacyclic group of order 24 and exponent 4, namely H2 . For this 2 i Š D so G N N does not satisfy condition (QR1), which is a congroup we have G=hb 8 tradiction. Let sm .G/ (s0m .G/, cm .G/) be the number of subgroups (noncyclic subgroups, cyclic subgroups) of order p m in a p-group G. In what follows, c; e; f; m; n; t are positive integers. The next theorem was inspired by Mann’s letter to the ﬁrst author at June 28, 2009. Theorem 124.1 (Mann [M33] for p > 2; see also 5o ). Let G be a quasi-regular metacyclic group of order p n and exponent p e < p n , m n, and set f .G/ D f D n e. Then one of the following holds: (a) If m f , then sm .G/ D

p mC1 1 p1 .

(b) If f < m e, then sm .G/ D (c) If m > e, then sm .G/ D

p f C1 1 p1

p nmC1 1 p1

.

so that sm .G/ is independent of m.

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The number of subgroups of given order in a metacyclic p-group

241

As we know, this is the unique result presenting exact values for sm .G/ for such a general class of p-groups. Of course, Theorem 124.1 is not true for general metacyclic 2-groups as 2-groups of maximal class show. We need the following Lemma 124.2. Suppose that G is a quasi-regular metacyclic group of order p n and exponent p e . Given a cyclic A G of order p e , there is a cyclic B G such that G D AB and A \ B D ¹1º. Proof. We use induction on jGj. One may assume that the group G is nonabelian and f D n e > 1 (otherwise, the result is either trivial for n D e or follows from Introduction, Exercise 4, and Theorem 1.2). Then the subgroup Ã1 .G/ is noncyclic of exponent p e1 by (QR2)). We have jA \ Ã1 .G/j D p e1 . Therefore, by induction, Ã1 .G/ D U V , where U D A \ Ã1 .G/, V is cyclic and U \ V D ¹1º. It follows that A \ V D ¹1º. Let V D hvi. Then, by (QR2), there is b 2 G such that v D b p . Set B D hbi. Then A \ B D ¹1º and G D AB. (Clearly, jBj D p f .) Note that for general abelian p-groups cyclic subgroups of maximal order are complemented (Introduction, Exercise 4). As Lemma 124.2 shows, this property also holds for quasi-regular metacyclic p-groups so for all metacyclic p-groups, p > 2. Generalized quaternion groups show that this is not true, in general, for p D 2. However, it is not true that if p > 2 and A G G is cyclic and such that G=A is cyclic, then A is complemented in G. Example: G is abelian of type .p; p n /, n > 2, and A is a cyclic subgroup of order p 2 not contained in ˆ.G/. Indeed, then G=A is cyclic, but A is not complemented in G (otherwise, we have j2 .G/j D p 4 , a contradiction since, in fact, j2 .G/j D p 3 ). Proof of Theorem 124.1. One may assume that 1 < m < n. If f D 1, then G is either abelian of type .p e ; p/ or G Š Mpn (see Theorem 1.2). In both these cases, we get sm .G/ D p C 1, and the same result is obtained from (a–c). In what follows we assume that f > 1; then G has no cyclic subgroup of index p. By Lemma 124.2, G D AB, where A; B < G are cyclic of orders p e , p f , respectively. Taking into account that G is quasi-regular and f e, we get ()

jk .G/j D p 2k .k f / and

G=f .G/ Š Cpef D Cpn2f :

We have (1)

sm .G/ D s0m .G/ C cm .G/:

If H G is noncyclic, then 1 .G/ H since j1 .G/j D p 2 D j1 .H /j (Lemma J(e)), and so (2)

s0m .G/ D sm2 .G=1 .G//:

One can rewrite (1) as follows: (3)

sm .G/ D sm2 .G=1 .G// C cm .G/:

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Groups of prime power order

(A) We ﬁrst compute cm .G/ for m e (if m > e, then cm .G/ D 0). Since G is quasi-regular and p m1 .p 1/ D '.p m / is the number of elements of order p m in Cpm , it follows that cm .G/ D

(4)

jm .G/ m1 .G/j : p m1 .p 1/

(i) Suppose that m f . Then we get jm .G/j D p 2m , jm1 .G/j D p 2m2 so that, by (4), cm .G/ D

(5)

p 2m p 2m2 D p m1 .p C 1/: p m1 .p 1/

(ii) Suppose that f < m e. By (), we have jm .G/j D p 2f Cmf D p mCf

and jm1 .G/j D p 2f Cm1f D p mCf 1 :

Therefore, by (4), (6)

cm .G/ D

p mCf p mCf 1 D pf : p m1 .p 1/

(B) In view of (3), it remains to compute sm2 .G=1 .G//. Here we use induction on m. We have p n1 D jG=1 .G/j D p n2 ; e1 D e.G=1 .G// D e 1;

(7)

f1 D f .G=1 .G// D f 1 (for example, f1 D n1 e1 D .n 2/ .e 1/ D n e 1 D f 1). (j) Let m f ; then m 2 f 2 < f 1 D f1 D f .G=1 .G//. By induction (see part (a) of the statement), sm2 .G=1 .G// D

p m1 1 p m2C1 1 D p1 p1

so that, by (3) and (5), we get sm .G/ D

p m1 1 p mC1 1 C p m1 .p C 1/ D ; p1 p1

completing this case. (jj) Let f < m e; then f1 D f .G=1 .G// D f 1 m 2 e 2 < e 1 D e.G=1 .G// D e1 : If m 2 D f 1 or, what is the same, m 1 D f , then, by induction (see part (a)), sm2 .G=1 .G// D

pf 1 p m2C1 1 D p1 p1

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243

so, by (3) and (6), we obtain sm .G/ D p f C

p f C1 1 pf 1 D ; p1 p1

as in (b), and we are done in this case. Now let m 2 > f 1 or, what is the same, m > f C 1. Then, by induction (see part (b)), pf 1 p f 1C1 1 sm2 .G=1 .G// D D p1 p1 so, by (3) and (6), we have sm .G/ D p f C

p f C1 1 pf 1 D ; p1 p1

completing case (jj). (jjj) Now let m e C 1 or, what is the same, m 2 e 1 D e.G=1 .G// D e1 . Then sm .G/ D sm2 .G=1 .G// since all subgroups of G of order p m are noncyclic hence contain 1 .G/. Let m 2 D e 1 D e1 , i.e., e C 1 D m; then, by induction (see part (b)), sm2 .G=1 .G// D

pf 1 p ne 1 p nmC1 1 p .f 1/C1 1 D D D ; p1 p1 p1 p1

and this coincides with the formula in (c). Now let m 2 > e 1, i.e., m > e C 1. Then, by induction (see part (c)), sm2 .G=1 .G// D

p nmC1 1 p .n2/.m2/C1 1 D ; p1 p1

and this coincides with the formula in (c). Since all possibilities for m are considered, the proof is complete. Corollary 124.3 (Mann [M33] for p > 2). If G and m; n; f; e are as in Theorem 124.1 (i.e., G is quasi-regular), then sm .G/ D snm .G/. Proof. One may assume that G is noncyclic and 1 < m < n. (i) Let m f ; then n m n f D e. If n m D e, then m D n e D f . We have, by Theorem 124.1(b,a), snm .G/ D se .G/ D

p mC1 1 p f C1 1 D D sm .G/: p1 p1

If n m > e, then m < n e D f . We have, by Theorem 124.1(c,a), snm .G/ D

p mC1 1 p n.nm/C1 1 D D sm .G/: p1 p1

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Groups of prime power order

(ii) Let f < m e; then f D n e n m < n f D e. If n m D f , then m D n f D e. We have, by Theorem 124.1(a,b), snm D sf .G/ D

p f C1 1 D se .G/ D sm .G/: p1

If f < n m < e, then, by Theorem 124.1(b), snm .G/ D

p f C1 1 D sm .G/: p1

(iii) Let m > e, then n m < n e D f so, by Theorem 124.1(a,c), snm .G/ D

p nmC1 1 D sm .G/; p1

and the proof is complete. The group G D H2 with n D 4, e D f D 2 is not quasi-regular, however, sm .G/ is such as in Theorem 124.1. 2o The number of cyclic subgroups of given order in a metacyclic 2-group. In this subsection we shall ﬁnd the number of cyclic subgroups of given order in a metacyclic 2-group. In this and the following subsections we suppose that G is a metacyclic group of order 2n . Set w D w.G/ D max¹i j ji .G/j D 22i º but jwC1 .G/j ¤ 22.wC2/ : In this case, we write R.G/ D w .G/. Then we get jR.G/j D 22w and G=R.G/ is either cyclic or a 2-group of maximal class (Lemma J(e)). We shall retain this notation in what follows. If w D 0, then either G is cyclic or of maximal class (Lemma J(e)). Part A of the following known theorem completes the case w D 0. Theorem 124.4. Suppose that G is a 2-group of maximal class and n > 3. A. If n > m > 2, then cm .G/ D 1. (a) If G Š D2n , then c1 .G/ D 2n1 C 1 and c2 .G/ D 1. (b) If G Š Q2n , then c1 .G/ D 1, c2 .G/ D 2n2 C 1. (c) If G Š SD2n , then c1 .G/ D 2n2 C 1 and c2 .G/ D 2n3 C 1. B. We have sm .G/ D 2nm C 1 for 2 m < n. If G is a group from Theorem 124.4, then its maximal subgroups are known (for example, if G D SD2nC1 , then 1 D ¹C; D; Qº, where C is cyclic, D is dihedral and Q is generalized quaternion). This allows us, using induction and Hall’s enumeration principle, to prove that theorem. In what follows w D w.G/ is a positive integer.

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Theorem 124.5. Suppose that G satisﬁes R.G/ D G; then jGj D 2n D 22w . In that case, we have G D AB, where A G G and B are cyclic of the same order 2w and cm .G/ D 3 2m1 for all m 2 ¹1; : : : ; wº. Proof. Since G is metacyclic, there is a cyclic AGG such that G=A is cyclic. It follows from exp.G/ D 2w and jAj 2w that jG W Aj 2w D exp.G/ whence jAj D 2w . If B < G is cyclic such that AB D G, then, since exp.G/ D 2w , we get jBj D 2w , A \ B D ¹1º, and the ﬁrst assertion of the theorem is proven. Since j t .G/j D 22t for all nonnegative integers t w, we get jm .G/ m1 .G/j 22m 22m2 D D 3 2m1 ; 2m1 2m1

cm .G/ D completing the proof.

In view of Theorem 124.5, we suppose in what follows that .w .G/ D/R.G/ < G. By Theorem 124.5, if m w, then cm .G/ D 3 2m1 . Therefore in the following two theorems we assume that m > w and R.G/ < G. Theorem 124.6. Suppose that G=R.G/ is nonidentity cyclic and m > w. Then we have cm .G/ D 2w . Proof. We obtain jm .G/j D 22wC.mw/ D 2wCm ; jm1 .G/j D 22wC.m1w/ D 2wCm1 : Therefore cm .G/ D

jm .G/ m1 .G/j 2wCm 2wCm1 D D 2w ; 2m1 2m1

and the proof is complete. In what follows we assume that the quotient group G=R.G/ is of maximal class. If L=R.G/ < G=R.G/ is cyclic of order 2s , s > 0, then we infer that cm .L/ D 2w by Theorem 124.6. In case that L1 =R.G/ < G=R.G/ is another cyclic subgroup of order 2s , we conclude that L \ L1 has no cyclic subgroups of order 2wCs . If L < G is cyclic of order 2m > 2w , then, by the product formula, LR.G/=R.G/ is cyclic of order 2mw since jL \ R.G/j D 2w . Therefore we get Theorem 124.7. Suppose that G=R.G/ is a 2-group of maximal class and m > w. Then cm .G/ D 2w cmw .G=R.G//. Since cmw .G=R.G// is known (Theorem 124.4.A), we have completed the computation of cm .G/ if G=R.G/ is of maximal class. Thus the number cm .G/ is computed for all metacyclic 2-groups.

246

Groups of prime power order

3o The number of subgroups of given order in a metacyclic 2-group. In this subsection we compute the number sm .G/ of subgroups of order 2m in G, where G > ¹1º is a metacyclic 2-group. We ﬁrst consider a metacyclic 2-group G D R.G/.D w .G// (see the second paragraph of 2o ) for which this problem is solved without difﬁculty (however, this is a key result as we shall see in the sequel). In what follows G is a metacyclic 2-group of order 2n with R.G/ > ¹1º or, what is the same, w 1 (see Theorem 124.4). Theorem 124.8. If G is a metacyclic 2-group such that G D R.G/, w 1 (in this case, jGj D 22w ) and 1 m 2w, then one of the following holds: (a) If m w, then sm .G/ D 2mC1 1. (b) If 1 t w, then swCt .G/ D 2wtC1 1. Proof. It is easily checked that the theorem is true for w D 1 and t D w. Next we assume that w > 1 and t < w. As in the proof of Theorem 124.1, formulas (1), (2) and (3) hold. Therefore, in view of Theorem 124.5, it remains to ﬁnd the number of noncyclic subgroups of order 2m . If H < G is noncyclic, then 1 .G/ H so the number of noncyclic subgroups of order 2m in G is equal to sm2 .G=1 .G//. We proceed by induction on w. Obviously, R.G= t .G// D G= t .G/ for t w. (a) Suppose that m w. In our case, we obtain that w 0 D w.G=1 .G// D w 1, jG=1 .G/j D 22w2 . We have m 2 < w 1 D w 0 . Let H < G be of order 2m . If H is noncyclic, then the number of such H is G is equal to sm2 .G=1 .G// D 2.m2/C1 1 D 2m1 1 by induction (see the previous paragraph). If H < G is cyclic, then the number of such H is equal to cm .m .G// D 3 22m1 (Theorem 124.5) so, by (3), we obtain sm .G/ D 3 2m1 C .2m1 1/ D 2mC1 1; completing the proof of (a). (b) In what follows we assume that m D wCt , where 1 t < m. As exp.G/ D 2w , we obtain that exp.G= t .G// D 2wt . If H G is of order 2wCt , then t .G/ < H and jH= t .G/j D 2wt , and so swCt .G/ D swt .G= t .G//. As we know, we have R.G= t .G// D G= t .G/; therefore, using the formula from the previous sentence and (a) for G= t .G/, we get swCt .G/ D swt .G= t .G// D 2wtC1 1 by induction, and the proof is complete. It is easy to show that the groups from Theorem 124.8 coincide with metacyclic 2-groups G of order 22w and exponent 2w . Indeed, if G is such a group, then 1 .G/ is abelian of type .2; 2/ by Lemma J(e), and now, by induction, ji .G=1 .G//j D 22i for all i m 1, and our claim follows.

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The number of subgroups of given order in a metacyclic p-group

247

It follows that the number sm .G/ from Theorem 124.8 is computed by the same formulas as in Theorem 124.1. The group H2 is a group from Theorem 124.8 but not from Theorem 124.1. Note that a group G D AB from Theorem 124.8 has a noncyclic center. This is the case if one of the subgroups A; B, say A, is normal in G since 1 .A/ Z.G/ and, in view of jAut.A/j D 2m1 , 1 .B/ centralizes A so contained in Z.G/. Now suppose that A and B are not normal in G. Then there exists x 2 GNG .A/ such that Ax ¤ A. By Ore (see Lemma A.28.8), AAx ¤ G soT A \ Ax 1 .A/ > ¹1º for all x 2 G by the product formula. It follows that AG D x2G Ax > ¹1º so Z.G/ 1 .A/ > ¹1º. Similarly, Z.G/ 1 .B/ > ¹1º, and our claim follows since 1 .A/1 .B/ Z.G/ is noncyclic. Suppose that a metacyclic 2-group G is such that G=R.G/ is nonidentity cyclic and R.G/ > ¹1º. We claim that in this case G D AB, where A; B < G are cyclic and A \ B D ¹1º. Indeed, suppose that exp.G/ D 2e and, as above, exp.R.G// D p w ; then w < e. Let A < G be cyclic of order 2e ; then A \ R.G/ is cyclic of order 2w . By Theorem 124.5, R.G/ D .A \ R.G//B, where B < R.G/ is cyclic of order 2w and .A \ R.G// \ B D ¹1º. It follows that A \ B D ¹1º, so AB D G by the product formula. Let 1 D ¹A; B; C º be the set of all maximal subgroups of a noncyclic metacyclic 2-group G. Then, supposing 2m < jGj, we have, by Hall’s enumeration principle, (8)

sm .G/ D sm .A/ C sm .B/ C sm .C / 2sm .ˆ.G//:

We may assume that G has no cyclic subgroup of index 2 (see Theorem 124.4). In case 2mC2 D jGj D 2n , we get sm .G/ D sn2 .G/ D 3 3 2 1 D 7 D 22 C 3 by (8). We shall freely use this fact in what follows. Next we retain the notation introduced in this paragraph. Now we compute sm .G/ for the case where G=R.G/ is nonidentity cyclic. The following lemmas will help us to state the inductive hypothesis. If m w, then we have sm .G/ D sm .R.G//, and this number is computed in Theorem 124.8(a). Therefore in what follows we consider case m > w only. Lemma 124.9. Suppose that G is a metacyclic 2-group that satisﬁes jG=R.G/j D 2. If 1 t w C 1, then swCt .G/ D 2wtC2 1. Proof. If t D w C 1, then w C t D 2w C 1 D n, where 2n D jGj; then 1 D swCt .G/ D sn .G/ D 1 D 2w.wC1/C2 1; as in the statement. Next we assume that t < w C 1. We have exp.G/ D 2wC1 . If w D 1, then t D 1, G is abelian of type .4; 2/, and so swCt .G/ D s2 .G/ D 3 D 22 1 D 211C2 1, as in statement. Therefore one may also assume that w > 1.

248

Groups of prime power order

In view of Theorem 124.8(b), swCt .R.G// D 2wtC1 1 so it sufﬁces to ﬁnd the number of subgroups H < G of order 2wCt such that H 6 R.G/. Let H be such a subgroup; then exp.H / D 2wC1 D exp.G/. If H is cyclic, then m D w C 1 so t D 1, hence the number of such H is equal to cwC1 .G/ D 2w (Theorem 124.6), and we get swC1 .G/ D swC1 .R.G// C cwC1 .G/ D .2w1C1 1/ C 2w D 2wC1 1; completing this case. Next we assume that jH j D 2.wC1/Ct , where 0 < t < w. In this case, H is noncyclic and t .G/ < H . Suppose that H 6 R.G/. Since exp.H / D 2wC1 D exp.G/, it follows that H= t .G/ is of order 2wCtC12t D 2wtC1 and exponent 2wtC1 , and we conclude that H= t .G/ is cyclic. We have w.G= t .G// D w t. Therefore the number of such H is equal to cwtC1 .G= t .G// D 2wt (Theorem 124.6). Since swC1Ct .R.G// D 2w.tC1/C1 1 D 2wt 1 (Theorem 124.8), we obtain swCtC1 .G/ D 2wt C .2wt 1/ D 2wtC1 1: Replacing t C1 by t in the last equality, we get swCt .G/ D 2wtC2 1, as required. Lemma 124.10. Suppose that a metacyclic 2-group G is such that G=R.G/ is cyclic of order 4. (a) We have swC1 .G/ D 2wC1 1. (b) If 2 t m, then swCt .G/ D 2wtC3 1. Proof. (a) Since jwC1 .G/=R.G/j D 2, we have, by Lemma 124.9, swC1 .G/ D swC1 .wC1 .G// D 2w1C2 1 D 2wC1 1; and (a) is proven. (b) If t D w C 2, then 2wCt D jGj so swC.wC2/ .G/ D 1 D 2w.wC2/C3 1; as in the statement. In what follows we assume that t < w C 2. If w D 1, then jGj D 24 has a cyclic subgroup of index 2, t D 2 (by hypothesis), s1Ct .G/ D s3 .G/ D 3 D 22 1 D 212C3 1; and this coincides with the required result. In the sequel we assume that w > 1. Set U D wC1 .G/; then jU W R.G/j D 2 D jG W U j. As swCt .U / D 2wtC2 1 (Lemma 124.9), it sufﬁces to compute the number of H < G of order 2wCt such that H 6 U . Let H be such a subgroup. Then exp.H / D 2wC2 D exp.G/. If H is cyclic, then jH j D 2wC2 (this holds if and only if t D 2), and so the number of such H in G

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249

is equal to cwC2 .G/ D 2w (Theorem 124.6), and we get swC2 .G/ D cwC2 .G/ C swC2 .U / D 2w C .2w2C2 1/ D 2wC1 1 D 2w2C3 1; and this coincides with the required result for t D 2. (Note that if t D 2 and H 6 U , then H is cyclic.) Now we suppose that jH j D 2.wC2/Ct , where 0 < t < w. In this case, H is noncyclic of exponent 2wC2 D exp.G/, t .G/ < H and H= t .G/ is cyclic of order 2wCtC22t D 2wtC2 (cyclic since jH= t .G/j D 2wtC2 D exp.H= t .G/). As exp.R.G= t .G// D 2wt , the number of such H equals cwt C2 .G= t .G// D 2wt (Theorem 124.6). Since swCtC2 .U / D 2w.tC2/C2 1 D 2wt 1 (Lemma 124.9), we get swCt C2 .G/ D 2wt C .2wt 1/ D 2wtC1 1: Replacing in the last equality t C 2 by t , we get swCt .G/ D 2wtC3 1, completing the proof. Lemma 124.11. Suppose that G is a metacyclic 2-group such that G=R.G/ is cyclic of order 23 , w > 0. (a) If t D 1, then swC1 .G/ D 2wC1 1. (b) If t D 2, then swC2 .G/ D 2wC1 1. (c) If 3 t < w C 3, then swCt .G/ D 2wtC4 1. Proof. Statements (a) and (b) follow from Lemmas 124.9 and 124.10, respectively. Indeed, let t D 1. Then swC1 .G/ D swC1 .wC1 .G// D 2w1C2 1 D 2wC1 1

(Lemma 124.9):

Now let t D 2. Then swC2 .G/ D swC2 .wC2 .G// D 2w2C3 1 D 2wC1 1

(Lemma 124.10):

(c) One may assume that t < w C 3. If w D 1, then t D 3, by hypothesis, G has a cyclic subgroup of index 2 so s1C3 .G/ D 3 D 213C4 1. Next we assume that w > 1. Set U D t C2 .G/. Let H < G be of order 2wC3 . Since swC3 .U / D 2w3C3 1 D 2w 1 (Lemma 124.10), it sufﬁces to count the number of those subgroups H that are not contained in U . In this case, we get exp.H / D 2wC3 so H is cyclic. Then the number of such H is equal to cwC3 .G/ D 2w (Theorem 124.6). We get swC3 .G/ D 2w C .2w 1/ D 2wC1 1 D 2w3C4 1; and this coincides with the required result for t D 3.

250

Groups of prime power order

Suppose that jH j D 2wC3Ct , where 0 < t < w, and H < G, H 6 U . Then we conclude that t .G/ < H , H= t .G/ is cyclic of order 2wtC3 . The number of such H is equal to cwtC3 .G= t .G// D 2wt since exp.R.G= t .G/// D 2wt (Theorem 124.6). All such H are not contained in U as exp.H / D 2wC3 > 2wC2 D exp.U /. Since swC3Ct .U / D 2w.tC3/C3 1 D 2wt 1 (Lemma 124.10), we get swC3Ct .G/ D 2wt C .2wt 1/ D 2wtC1 1: Replacing in the last equality t C 3 by t , we get swCt .G/ D 2wtC4 1, completing the proof. The proofs of the previous three lemmas are similar. Our goal there was to state inductive hypothesis and show as to attain this. In any case, Lemma 124.9 must be proved (it is the basis of induction). Now we are ready to prove the following Theorem 124.12. Suppose that G is a metacyclic 2-group of order 2n such that w > 1 and G=R.G/ is cyclic of order 2c (in this case, n D 2w C c/.1 (a) If 1 t < c, then swCt .G/ D 2wC1 1. (b) If c t w C c, then swCt .G/ D 2wtCcC1 1. Proof. We proceed by induction on c. As in previous three lemmas, one may assume that t < w C c. (i) First we consider case t c. Since w > 0, G is noncyclic. Set U D wCc1 .G/; then exp.U / D 2wCc1 D 1 c1 > 1. 2 exp.G/, jG W U j D 2 and U=R.U / D UR.G/=R.G/ is cyclic of order 2 The theorem holds for c D 1; 2; 3 (Lemmas 124.9–124.11) so one may assume that c > 3. We have exp.G/ D 2wCc . By induction, swCt .U / D 2wtC.c1/C1 1 D 2wtCc 1 since t > c 1 D log2 .jUR.G/=R.G/j/. Therefore it sufﬁces to ﬁnd the number of those H < G of order 2wCt that are not contained in U . All such subgroups have the same exponent 2wCc D exp.G/. If such an H is cyclic, then jH j D 2wCc (this holds if and only if t D c), and the number of such H in G is equal to cwCc .G/ D 2w (Theorem 124.6). Therefore we obtain swCc .G/ D cwCc .G/ C swCc .U / D 2w C .2wcCc 1/ D 2wC1 1; and this coincides with the required result for t D c. Next we assume that jH j D 2wCcCt , where 0 < t < w. Then t .G/ < H and H= t .G/ is of order 2wCcCt2t D 2wtCc D exp.H= t .G// whence H= t .G/ is cyclic. We have swCcCt .U / D 2w.cCt /C.c1/C1 1 D 2wt 1 by induction. w D 1, then the number of proper subgroups of given order in G is equal to 22 1 since G has a cyclic subgroup of index 2 and cl.G/ 2. 1 If

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The number of subgroups of given order in a metacyclic p-group

251

Since R.G= t .G// D R.G/= t .G/ and exp.R.G= t .G/// D 2wt , the number of such H equals cwtCc .G= t .G// D 2wt (Theorem 124.6). Therefore swCcCt .G/ D 2wt C .2wt 1/ D 2wtC1 1: Replacing in the displayed formula t C c by t , we get swCt D 2wtCcC1 1, and the proof of (b) is complete. (ii) It remains to prove (a); then t < c. In this case, H < V D wCt .G/. The subgroup V =R.G/ is cyclic of order 2t . By part (b), applied to V , we obtain that swCt .V / D 2wtCt C1 1 D 2wC1 1. Since swCt .G/ D swCt .V /, the proof of (a) is complete. To complete the computation of the value sm .G/, it remains to consider the case where G=R.G/ is a 2-group of maximal class and order 2c . Note that the factor group G=R.G/ contains a cyclic subgroup of index 2 (Theorem 1.2). In the notation of (8) (namely, 1 D ¹A; B; C º is the set of maximal subgroups of G) we assume in the sequel that C =R.G/ is cyclic. These three cases will be covered in Theorems 124.16, 124.18 and 124.19 and in the supplements to them. In those theorems we consider the case w > 1, t c 1. In the supplements we consider the cases w D 1 and t D 1 separately. In the case under consideration, R.G/ < ˆ.G/ as d.G=R.G// D 2 D d.G/. The following two lemmas are the bases of induction for Theorems 124.16 and 124.18, respectively. Lemma 124.13. Suppose that G is a metacyclic 2-group such that G=R.G/ Š Q8 and w C t > w > 1, t w C 3. (a) If t D 1, then swC1 .G/ D 2wC1 1. (b) If 2 t w C 3, then swCt .G/ D 2wtC4 1. Proof. If t D w C 3, then we have w C t D n, where 2n D jGj, so we obtain that swCt .G/ D 1 D 2w.wC3/C4 1, as in (b). If t D w C 2, then w C t D n 1 so swCt .G/ D 3 D 22 1 D 2w.wC2/C4 1; as in (b). In what follows we assume that t < w C 2. We have R.A/ D R.B/ D R.C / D R.ˆ.G// D R.G/ and the quotient groups A=R.G/, B=R.G/ and C =R.G/ are cyclic of order 4, jˆ.G/=R.ˆ.G/j D 2. If t D 1, then, by Lemma 124.9, swC1 .G/ D s.wC1 .G// D 2w1C2 1 D 2wC1 1; and the proof of (a) is complete. Now let t > 1. Then swCt .A/ D swCt .B/ D swCt .C / D 2wtC3 1 (Lemma 124.10); swCt .ˆ.G// D 2wtC2 1

(Lemma 124.9)

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Groups of prime power order

so, by (8), swCt .G/ D 3 .2wtC3 1/ 2 .2wtC2 1/ D 2wtC4 1; and the proof is complete. Lemma 124.14. Suppose that G is a metacyclic 2-group such that G=R.G/ Š D8 and w C t > w > 1, 1 t w C 3. (a) If t D 1, then swC1 .G/ D 2wC3 2wC1 1 D 3 2wC1 1. (b) If t > 1, then swCt .G/ D 2wtC4 1. Proof. In the notation of identity (8), we have R.A/ D A, R.B/ D B so we obtain w.A/ D w.B/ D w C 1, R.C / D R.ˆ.G// D R.G/ so w.C / D w.ˆ.G// D w. In case t D w C 3, we get w C t D n, and, as in the previous lemma, we infer that swCt .G/ D 1 D 2w.wC3/C4 1, as in (b). If t D w C 2, then swCt .G/ D 3 D 22 1 D 2w.wC2/C4 1; as in (b). Next we assume that t < w C 2. For t D 1, we have swC1 .A/ D swC1 .B/ D 2.wC1/C1 1 D 2wC2 1 (Theorem 124.8(a)); swC1 .C / D swC1 .wC1 .C // D swC1 .ˆ.G// D 2w1C2 1 D 2wC1 1

(Lemma 124.9)

so that, by (8), swC1 .G/ D 2 .2wC2 1/ C .2wC1 1/ 2.2wC1 1/ D 2wC3 2wC1 1 D 3 2wC1 1; as in (a). If t > 1, then w C t D .w C 1/ C .t 1/ so that swCt .A/ D swCt .B/ D 2.wC1/.t1/C1 1 D 2wtC3 1 swCt .C / D 2wtC3 1 swCt .ˆ.G// D 2wtC2 1

(Theorem 124.8(b)); (Lemma 124.10); (Lemma 124.9)

so that, by (8), swCt .G/ D 3.2wtC3 1/ 2.2wtC2 1/ D 2wtC4 1; completing the proof.

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Lemma 124.15. Let G be a metacyclic 2-group of order 2n such that G=R.G/ Š Q24 , w > 1 and 1 t w C 4. Then (a) swC1 .G/ D 2wC1 1. (b) If 2 t w C 4, then swCt .G/ D 2wtC5 1. Proof. If t D w C 4, then swC.wC4/ .G/ D sn .G/ D 1 D 2w.wC4/C5 1; as in (b). In case t D w C 3, we get swC.wC3/ .G/ D 3 D 22 1 D 2w.wC3/C5 1; as in (b). If t D 1, then, by Lemma 124.9, swC1 .G/ D swC1 .wC1 .G// D 2w1C2 1 D 2wC1 1; as in (a). We have w.M / D w for M 2 ¹A; B; C; ˆ.G/º. Now let t D 2. Then swC2 .A/ D swC2 .B/ D 2w2C4 1 D 2wC2 1 (Lemma 124.13); swC2 .C / D 2w2C4 1 D 2wC2 1 swC2 .ˆ.G// D 2

w2C3

1D2

wC1

1

(Lemma 124.11); (Lemma 124.10) :

Therefore, by (8), we get swC2 .G/ D 3.2wC2 1/ 2.2wC1 1/ D 2wC3 1 D 2w2C5 1; as in (b). Next we assume that 2 < t < w C 3. By Lemma 124.13, Theorem 124.12 and identity (8), swCt .G/ D 2.2wtC4 1/ C .2wtC4 1/ 2.2wtC3 1/ D 2wtC5 1; and the proof is complete. Now we are ready to prove the following Theorem 124.16. Suppose that G is a metacyclic 2-group such that G=R.G/ Š Q2c and w C t > w > 1, c 1 t w C c. Then swCt .G/ D 2wtCcC1 1. Proof. We proceed by induction on c. The statement of the theorem is true for c D 3; 4 (Lemmas 124.13, 124.15) and t 2 ¹w C c 1; w C cº (direct checking; see Lemma 124.15). Next assume that c > 4 and t < w C c 1. As in Lemma 124.15, we

254

Groups of prime power order

have w.M / D w for M 2 ¹A; B; C; ˆ.G/º. We obtain A=R.G/ Š Q2c1 Š B=R.G/;

C =R.G/ Š C2c1 ;

ˆ.G/=R.G/ Š C2c2 :

Then, by induction, swCt .A/ D swCt .B/ D 2wtCc 1: By Theorem 124.12, swCt .C / D 2wt Cc 1;

swCt .ˆ.G// D 2wtCc1 1:

Therefore, by (8), we obtain swCt .G/ D 3.2wt Cc 1/ 2.2wtCc1 1/ D 2wtCcC1 1; and the proof is complete. There is no problem to compute swCt .G/ for t < c 1. Our method works also in this case. Now suppose that w D 1 and G=R.G/ is of maximal class and order 2c . Taking into account that C and ˆ.G/ have cyclic subgroups of index 2, cl.C / 2 and ˆ.G/ is abelian (indeed, CG .R.G// ˆ.G/), we obtain s1Ct .C / D s1Ct .ˆ.G// D 3. Therefore, by (8), we get (9)

s1Ct .G/ D s1Ct .A/ C s1Ct .B/ C 3 2 3 D s1Ct .A/ C s1Ct .B/ 3:

Supplement 1 to Theorem 124.16. Let G=R.G/ Š Q2c , w D 1, jGj D 2n D 22Cc , 1 t < c. Then (a) s2 .G/ D 3 (in that case, t D 1). (b) If 1 < t < c, then s1Ct .G/ D 21tCc C 3. Proof. One may assume that 1 C t < n 2.D c/ (we have sn2 .G/ D 22 C 3 D 21tCc C 3, where t D n 3 and c D n 2; see the displayed formula following (8)). If t D 1, then s1C1 .G/ D s2 .2 .G// D s2 .ˆ.G// D 3 since 2 .G/. ˆ.G// is abelian of type .4; 2/ and ˆ.G/ has a cyclic subgroup of index 2. Next we also assume that t > 1. By the same displayed formula following (8), if c D 3 and t D 2, then we get s3 .G/ D 22 C 3 D 212C3 C 3. If c D 4, then t D 2 (by assumption, 1 < t < n 3 D c 1) so, by (9) and the previous paragraph, s1C2 .G/ D 2.22 C 3/ 3 D 23 C 3 D 212C4 C 3; as in (b). In case c D 5, we get t 2 ¹2; 3º so, by (9), s1C2 .G/ D 2.23 C 3/ 3 D 24 C 3 D 212C5 C 3

(by the previous paragraph),

s1C3 .G/ D 2 .22 C 3/ 3 D 23 C 3 D 213C5 C 3 (by the ﬁrst paragraph).

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The number of subgroups of given order in a metacyclic p-group

255

If c D 6, then t 2 ¹2; 3; 4º. We have, by (9) and the previous paragraphs, s1C2 .G/ D 2.24 C 3/ 3 D 25 C 3 D 212C6 C 3; s1C3 .G/ D 2.23 C 3/ 3 D 24 C 3 D 213C6 C 3; s1C4 .G/ D 2 .22 C 3/ 3 D 23 C 3 D 214C6 C 3: Now let jG=R.G/j D 2c , where t 2 ¹2; : : : ; c1º, c > 3. We claim that s1Ct .G/ D We prove this by induction on c. By the above, this is true for c D 3; 4; 5; 6 so one may assume that c > 6. Applying induction and (9), we obtain 21tCc C3.

s1Ct .G/ D 2.21tCc1 C 3/ 3 D 21tCc C 3; and we are done. We see that the cases w D 1 and w > 1 are essentially different. Supplement 2 to Theorem 124.16. Let G=R.G/ Š Q2c , w > 1 and t D 1. Then we have swC1 .G/ D 2wC1 1. Proof. For arbitrary c 3 we have, by Lemma 124.9, swC1 .G/ D swC1 .wC1 .G// D 2w1C2 1 D 2wC1 1; as required. Lemma 124.17. Suppose that G is a metacyclic 2-group such that G=R.G/ Š D24 , w C t > w > 1 and 1 < t < w C 3. Then swCt .G/ D 2wtC5 1 D 2wtC.4C1/ 1. Proof. It follows from the structure of the maximal subgroups of G that w.M / D w for M 2 ¹A; B; C; ˆ.G/º. By Lemmas 124.14, 124.12 and (8), we have swCt .G/ D 2.2wtC4 1/ C .2wtC4 1/ 2.2wtC3 1/ D 2wtC5 1; and we are done. Now we are ready to prove the following Theorem 124.18. Suppose that G is a metacyclic 2-group such that G=R.G/ Š D2c , w C t > w > 1, c 1 t < w C c 1. Then swCt .G/ D 2wt CcC1 1. Proof. We proceed by induction on c. The theorem holds for c D 3; 4 (Lemmas 124.14 and 124.17) so one may assume that c > 4. As above, we have w.M / D w for each M 2 ¹A; B; C; ˆ.G/º. By induction, Theorem 124.12 and (8), we obtain swCt .G/ D 2.2wtCc 1/ C .2wtCc 1/ 2.2wtCc1 1/ D 2wtCcC1 1; completing the proof.

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Now we consider the case G=R.G/ Š D2c , w D 1. As above, jGj D 2n . Supplement 1 to Theorem 124.18. Suppose that G=R.G/ Š D2c , w D 1 (in this case, n D c C 2) and 1 t < c. Then we have s1Ct .G/ D 21tCc C 3. (If t D c, then s1Cc .G/ D j1 j D 3.) Proof. By the paragraph containing (8), sn2 .G/ D 22 C 3 D 21.n3/Cn2 C 3 (here t D n 3 and c D n 2), as in the statement. Therefore one may assume in the sequel that t < n 3 D .c C 2/ 3 D c 1. Let t D 1. If H < G is of order 4 and H ¤ R.G/, then HR.G/ of order 8 contains exactly two ¤ R.G/ subgroups of order 4 since HR.G/ is abelian of type .22 ; 2/ by Lemma J(b). Since s1 .G=R.G// D 2c1 C 1, we get s1C1 .G/ D 1 C 2.2c1 C 1/ D 2c C 3 D 211Cc C 3; as in the statement. In what follows we assume that t > 1; then c > 3. Let c D 4. Then t D 2. Using (9), we obtain s1C2 .G/ D 2 .22 C 3/ 3 D 23 C 3 D 212C4 C 3 by the displayed formula following (8). Let c D 5. Then t 2 ¹2; 3º and s1C2 .G/ D 2.23 C 3/ 3 D 24 C 3 D 212C5 C 3; s1C3 .G/ D 2 .22 C 3/ 3 D 23 C 3 D 213C5 C 3 (the ﬁrst equality follows from the previous paragraph and the second one follows from the displayed formula following (8)). Now we will prove by induction on c that s1Ct .G/ D 21tCc C 3. This is true for c D 3; 4; 5 and t D 1. By induction, Theorem 124.12 and (9), we get s1Ct .G/ D 2.21tCc1 C 3/ 3 D 21tCc C 3; as required. Supplement 2 to Theorem 124.18. Suppose that G=R.G/ Š D2c , w > 1. Then we have swC1 .G/ D 2wCc1 C 2wC1 1. Proof. Let X1 =R.G/; : : : ; X2c1 =R.G/; X2c1 C1 =R.G/ D Z.G=R.G// be all subgroups of order 2 in G=R.G/. Then R.Xi / D R.G/;

swC1 .Xi / D 2w1C2 1 D 2wC1 1 for all i

(Lemma 124.9)

and swC1 .R.G// D 2w1C1 1 D 2w 1

(Theorem 124.8(b)).

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257

We get Xi \ Xj D R.G/ for i ¤ j and, given H < G of order 2wC1 not contained in R.G/, there is exactly one i 2c1 C 1 such that H < Xi .D HR.G//. Therefore swC1 .G/ D

2c1 XC1

swC1 .Xi / 2c1 swC1 .R.G//

i D1

D .2c1 C 1/.2wC1 1/ 2c1 .2w 1/ D 2wCc1 C 2wC1 1; and the proof is complete. (It is easy to check that, for t D 1 and c D 3, the obtained result coincides with Lemma 124.14(a).) In the following supplement we consider the groups G such that G=R.G/ Š D2c , w > 1 and t D 2. Note that the cases t D 3; : : : ; c 2 are more difﬁcult and, to treat them, one has to use the enumeration principle and induction on c for each value of t . Supplement 3 to Theorem 124.18. Suppose that G is a metacyclic 2-group such that G=R.G/ Š D2c , w > 1 and t D 2. Then swC2 .G/ D 2wCc2 C 2wC1 1. Proof. Suppose that X1 =R.G/; : : : ; X2c2 =R.G/ < G=R.G/ are abelian of type .2; 2/ and X2c2 C1 =R.G/ D Y =R.G/ < G=R.G/ cyclic of order 4. For i ¤ j we have Xi \ Xj D wC1 .Y /. Given H < G of order 2wC2 , there is i 2 ¹1; : : : ; 2c2 C 1º such that H < Xi since jH \ R.G/j 2w . Therefore (10)

swC2 .G/ D

2c2 XC1

swC2 .Xi / 2c2 swC2 .wC1 .Y //:

i D1

We have R.Xi / D Xi for i 2c2 and R.Y / D R.G/. By Theorem 124.8(b), swC2 .Xi / D s.wC1/C1 .Xi / D 2.wC1/1C1 1 D 2wC1 1;

i 2c2 ;

and, by Lemmas 124.10, 124.9, respectively (recall that Y D X2c2 C1 ), swC2 .Y / D 2w2C3 1 D 2wC1 1;

swC2 .wC1 .Y // D 2w2C2 1 D 2w 1:

Therefore we get, by (10), swC2 .G/ D 2c2 .2wC1 1/ C .2wC1 1/ 2c2 .2w 1/ D 2wCc2 C 2wC1 1; and the proof is complete. The cases G=R.G/ 2 ¹Q2c ; SD2c º, where w > 1 and t D 2, are considered similarly. Theorem 124.19. Suppose that G is a metacyclic 2-group such that G=R.G/ Š SD2c and w C t > w > 1, c 1 t w C c. Then swCt .G/ D 2wtCcC1 1.

258

Groups of prime power order

Proof. It is easily seen that the theorem holds for t 2 ¹w C c 1; w C cº. Next we assume that t < w C c 1. By Theorems 124.12, 124.16, 124.18 and (8), we have swCt .G/ D 3.2wtCc 1/ 2.2wtCc1 1/ D 2wt CcC1 1; and the proof is complete. Supplement 1 to Theorem 124.19. Let G=R.G/ Š SD2c , w D 1 and t 1. Then (a) If t D 1, then s1C1 .G/ D s2 .G/ D 2c1 C 3. (b) If 1 < t c 1, then s1Ct .G/ D 21tCc C 3. Proof. In case t D c 1, the result follows from the paragraph containing (8). Now let t < c 1. By Supplements 1 to Theorems 124.16 and 124.18 and (9), s1C1 .G/ D s2 .G/ D 3 C .211Cc1 C 3/ 3 D 2c1 C 3; s1Ct .G/ D 2.21tCc1 C 3/ 3 D 21tCc C 3

for 1 < t < c;

completing the proof. Supplement 2 to Theorem 124.19. Suppose that G=R.G/ Š SD2c , w > 1. Then we have swC1 .G/ D 2wCc2 C 2wC1 1. Proof. Let X1 =R.G/; : : : ; X2c2 =R.G/; X2c2 C1 =R.G/ D Z.G=R.G// be all subgroups of order 2 in G=R.G/. Then swC1 .Xi / D 2w1C2 1 D 2wC1 1 for all i (Lemma 124.9), swC1 .R.G// D 2w1C1 1 D 2w 1 (Theorem 124.8) and Xi \ Xj D R.G/ for all i ¤ j . We obtain (see Supplement 2 to Theorem 124.18) swC1 .G/ D .2c2 C 1/.2wC1 1/ 2c2 .2w 1/ D 2wCc2 C 2wC1 1; and we are done. The lower restrictions for t in Theorems 124.16, 124.18 and 124.19 are made for technical reasons (otherwise, the statements of these theorems would be more complicated). However, if necessary, it is not difﬁcult, using the above approach, consider the c 3 cases for which t 2 ¹2; : : : ; c 2º. If G is a noncyclic p-group of order p n with sk .G/ D 1 for some k 2 ¹1; : : : ; n1º, then we have k D 1, p D 2 and G is a generalized quaternion group (Proposition 1.3). By Sylow’s theorem, sk .G/ 1 .mod p/ for the same k. It follows that if G is neither cyclic nor generalized quaternion and k < n, then sk .G/ 1 C p. Therefore it is natural to classify the noncyclic p-groups G of order p n > p 3 satisfying sk .G/ D 1 C p for some ﬁxed k 2 ¹2; : : : ; n 2º (sn1 .G/ D 1 C p if and only if d.G/ D 2).

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The number of subgroups of given order in a metacyclic p-group

259

Proposition 124.20. Let G be a group of order p n > p 3 satisfying sk .G/ D 1 C p for some ﬁxed k 2 ¹2; : : : ; n 2º. Then one of the following holds: (a) G is abelian of type .p n1 ; p/. (b) G Š Mpn . n2

(c) p D 2, k D 2 and G D ha; b j a2 is metacyclic as in Lemma 42.1(c).

n3

D 1; n > 4; a2

D b 2 D 1; ab D a1 i

Proof. It is easy to see that G is not a 2-group of maximal class (otherwise, by Theorem 124.4, sk .G/ 1 .mod 4/ so sk .G/ ¤ 1 C 2). In that case, G has a normal abelian subgroup R of type .p; p/ (Lemma 1.4). (i) Suppose that k D 2. Assume that G has a subgroup E of order p 3 and exponent p; then E is nonabelian (otherwise, we have s2 .E/ D 1 C p C p 2 > 1 C p), s2 .E/ D 1 C p D s2 .G/. It follows that all subgroups of G of order p 2 are contained in E so exp.G/ D p, s2 .G/ 1Cp C2p 2 .mod p 3 / (Theorem 5.9), a contradiction. Thus E does not exist. Assume that G=R has two distinct subgroups X=R and Y =R of order p. Then X and Y contain together at least s2 .X/ C s2 .Y / 1 D 2p C 1 > p C 1 distinct subgroups of order p 2 , a contradiction. Thus G=R is either cyclic or generalized quaternion. In the ﬁrst case, G is one of groups (a, b). If G=R is generalized quaternion, then we obtain that j2 .G/j D 23 so G is as in (c) by Lemma 42.1(c). (ii) Suppose that k > 2. Then, as in part (i), G=R contains exactly one (proper since k < n 1) subgroup of order p k1 . In that case, G=R is cyclic. The subgroup CG .R/ is abelian with cyclic subgroup of index p and jG W CG .R/j p. Then 1 .G/ D R so G is either as in (a) or in (b). Now let s1 .G/ D 1 C p. If p > 2, then G is either metacyclic or a 3-group of maximal class (Theorem 13.7). Such G are also described very well for p D 2 in 82. Proposition 124.21. Let M be a metacyclic p-group and let G be a p-group of order p n > p 3 , p > 2. If s2 .G/ D s2 .M /, then one of the following holds: (a) G is metacyclic. (b) G D EC , where E D 1 .G/ is nonabelian of order p 3 and exponent p and C is cyclic of order > p. (c) G is of maximal class and order p 4, all maximal subgroups of G are two-generator. Proof. One may assume that the p-group M is noncyclic; then G is also noncyclic. We have s2 .G/ D s2 .M / D ¹1 C p; 1 C p C p 2 º (Theorem 124.1). Assume that G is nonmetacyclic. Then s2 .G/ D 1 C p C p 2 by Theorem 124.20, and so 2 .G/ is nonmetacyclic by Remark 41.2. (i) Assume that G has a subgroup E of order p 4 and exponent p. Then we obtain that s2 .E/ D 1 C p C 2p 2 C sp 3 for some nonnegative integer (Theorem 5.9) so s2 .E/ > 1 C p C p 2 D s2 .G/, a contradiction.

260

Groups of prime power order

(ii) Assume that G has an elementary abelian subgroup E of order p 3 . Then we get s2 .E/ D 1 C p C p 2 , so, by the ﬁrst paragraph, all subgroups of G of order p 2 are contained in E. It follows that exp.G/ D p, and so, by (i), G D E is of order p 3 , contrary to the hypothesis. (iii) It follows from (ii) that either G is a 3-group of maximal class or G D EC , where E D 1 .G/ is nonabelian of order p 3 and exponent p and C is cyclic (Theorem 13.7). The group G D EC satisﬁes the hypothesis (note that one of such groups is of maximal class; then its order is equal to p 4 ). Suppose that G is a 3-group of maximal class and order > 34 . Let G1 be the fundamental subgroup of G (see 9); then G1 is metacyclic without cyclic subgroup of index 3 (Theorems 9.6 and 9.11) so that s2 .G1 / D 1 C 3 C 32 D s2 .M / D s2 .G/: It follows that all subgroups of G of order 32 are contained in G1 so that 2 .G/ G1 . However (see Proposition 13.14(b)), 2 .G/ D G > G1 , and this is a contradiction. Thus, if G is a 3-group of maximal class, then n D 4. By (ii), G has no subgroup of type .3; 3; 3/, and this holds if and only if G is not isomorphic to a Sylow 3-subgroup of the symmetric group of degree 9. 4o A property of metacyclic 2-groups with nonabelian section of order 8. In this subsection G is a metacyclic 2-group containing subgroups L G M such that M=L is nonabelian of order 8 (M=L is said to be a section of G). In what follows we freely use the following known fact: There is only one nonabelian metacyclic group of order 16 and exponent 4. Before we denoted this group by H2 . If G D R.G/, then R.Ã1 .G// D Ã1 .G/. We use this obvious fact in the proof of the following Theorem 124.22. Suppose that a metacyclic 2-group G has a nonabelian section of order 8. If G is not of maximal class, then the following hold: (a) There is R G G such that G=R Š Q8 and G=Ã2 .G/ Š H2 . (b) Ã1 .G/ has no nonabelian section of order 8. Proof. (a) We use induction on jGj. Let L G M G be such that M=L is nonabelian of order 8. One may assume that M < G (otherwise, there is nothing to prove). Let M H 2 1 . Then, by induction, there is S G H such that H=S is nonabelian of order 8. By Lemma J(c), S is characteristic in H so normal in G. Due to Lemma J(b), G=S is of maximal class. Let R=S D Z.G=S/; then G=R is nonabelian of order 8, as was to be shown. (b) Assume that Ã1 .G/ D ˆ.G/ has a nonabelian section of order 8. Then, by (a), there is S G Ã1 .G/ such that Ã1 .G/=S is nonabelian of order 8. Due to Lemma J(c), S is characteristic in Ã1 .G/ so normal in G. By Lemma J(b), the quotient group G=S is of maximal class (of order 25 ). Thus ˆ.G=S / D Ã1 .G=S/ is nonabelian of order 8, contrary to Lemma 1.4.

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The number of subgroups of given order in a metacyclic p-group

261

5o Mann’s proof of Theorem 124.1 for p > 2. The content of this subsection is taken from [Man33] (only in one place we made a change using Lemma 124.2). We retain the notation introduced in 1o . Proof of Theorem 124.1. Below G is a metacyclic p-group, p > 2. We use the same notation as in 1o . By Lemma 124.2, G is the product of two cyclic subgroups with trivial intersection, say G D hxihyi D AB; A\B D ¹1º. By regularity, p e is the maximum of the orders of x and y, and thus we may assume that jAj D p e , and then jBj D p f . Hence e f . Since G 0 Ã1 .G/, we have Ãe1 .G 0 / D ¹1º. In a regular group, by Theorem 7.2(f), ŒG; Ãk .G/ D Ãk .G 0 /. Taking k D e 1, we get ŒG; Ãe1 .G/ D Ãe1 .G 0 / D ¹1º, e1 and hence Ãe1 .G/ Z.G/, and in particular N D hx p i is a central subgroup of order p. Write zm .G/ for the number of subgroups of G of order p m that do not contain N . It is clear that (11)

sm .G/ D sm1 .G=N / C zm .G/:

Note that in the quotient group G=N the invariants corresponding to e and f are e 1 and f (arranged in order). We now evaluate zm .G/. Let H be any subgroup of order p m not containing N . Since 1 .G/ D N Ãf 1 .B/ has order p 2 , we see that j1 .H /j D jH \ 1 .G/j D p, and thus H is cyclic since p > 2. If m > f , then we have jH j > jG W Aj, therefore H \ A ¤ ¹1º, and H N so that zm .G/ D 0. Now let m f . Then jm .G/j D jm .A/m .B/j D p 2m ;

jm1 .G/j D p 2.m1/ ;

and G contains p 2m p 2.m1/ elements of order p m and so p 2m p 2.m1/ D p m1 .p C 1/ .p 1/p m1 cyclic subgroups of order p m . Let z be a generator of N . If a cyclic subgroup H conm1 tains N , then H is generated by some element t such that t p D z. Then the elem1 powers are equal to z are the elements of the coset t m1 .G/, ments of G whose p their number is p 2.m1/ , and H contains p m1 of them. It follows that there are p m1 cyclic subgroups of order p m containing N , and zm .G/ D p m . Now the ﬁrst formula of the theorem follows by induction, noting that m 1 f .G=N /. Next, let f m e. If m D f , then the result follows from the previous case, so we assume that m > f . Then we get sm .G/ D sm1 .G=N /. Here e > f , implying e.G=N / D e 1 and f .G=N / D f , and thus e.G=N / m 1 f .G=N /, and we apply induction. Finally, assume that m e. Again sm .G/ D sm1 .G=N /, and once more the case of equality m D e is covered by the previous case, therefore we may assume that m > e, implying m 1 e.G=N /, and we apply induction.

262

Groups of prime power order

6o The p-groups with the same number of subgroups of some order as in a metacyclic p-group. Mann reported that, given a metacyclic p-group G, p > 2, he classiﬁed the p-groups H such that s1 .H / D s1 .G/ and s2 .H / D s2 .G/ (see also [Man33, Proposition 2] and Theorem 124.31). First we offer another proof of this result. Then we shall prove a related result. Proposition 124.23. Suppose that G and H are p-groups of exponent > p, p > 2, and G is metacyclic. Write Lpk .G/ D ¹x 2 G j o.x/ p k º. If jLp2 .H /j D jLp2 .G/j, then one of the following holds: (a) H is metacyclic. (b) p D 3 and H is of maximal class and order 34 ; moreover, H is minimal nonmetacyclic. (c) j1 .H /j D p 3 and H=1 .H / > ¹1º is cyclic. In particular (Mann), if s1 .H / D s1 .G/ and s2 .H / D s2 .G/, then H is as in parts (a–c) with nonabelian 1 .H / of order p 3 . Proof. One may assume that H is nonmetacyclic. Then jLp2 .H /j D jLp2 .G/j 2 ¹p 3 ; p 4 º: Write R D 1 .G/; then R is abelian of type .p; p/. Let M=R D 1 .G=R/. Then we get M D 2 .G/ D Lp2 . If jM=Rj D p, then j2 .H /j D p 3 and so H has a cyclic subgroup of index p so it is as in (a). Next we assume that jM j D p 4 . Then we obtain j2 .H /j D p 4 .

If j1 .H /j D p 2 , then H is a 3-group of maximal class. Since 2 .H / D H , we get jH j D 34 so H is as in (b).

If j1 .H /j D p 3 , then H is as in (c).

Now let s1 .H / D s1 .G/.D 1 C p/ and s2 .H / D s2 .G/ 2 ¹1 C p; 1 C p C p 2 º. In this case, jLp2 .H /j D j2 .H /j D p 4 so that, by the above, H is as in (a–c) (with 1 .H / is nonabelian of order p 3 in (c)). The analogous problem for p D 2 is surprisingly complicated. Denote by sŒk .G/ the number of subgroups of index p k in a p-group G. Proposition 124.24. Suppose that G and H be p-groups of exponent > p and assume that G is metacyclic. If sŒ1 .H / D sŒ1 .G/ and sŒ2 .H / D sŒ2 .G/, then one of the following holds: (a) H is metacyclic. (b) p > 2 and K3 .H / D Ã1 .H / has index p 3 in G. Proof. One may assume that the group G is noncyclic. Then we have sŒ1 .G/ D 1Cp, i.e., 2 D d.G/ D d.H /. Using Hall’s enumeration principle, we further conclude that sŒ2 .G/ 2 ¹1 C p; 1 C p C p 2 º. If sŒ2 .G/ D 1 C p, then G has a cyclic subgroup of

124

The number of subgroups of given order in a metacyclic p-group

263

index p. Therefore, if sŒ2 .H / D pC1, then, by what has just been said, H has a cyclic subgroup of index p so metacyclic. Now let sŒ2 .H / D p 2 C p C 1. In this case, using Hall’s enumeration principle, we conclude that the ranks of all maximal subgroups of H are equal to 2. Then, by Proposition 36.3, if p > 2, then we get case (b), if p D 2, then H is metacyclic. Corollary 124.25. Suppose that G and H are p-groups and assume that G is metacyclic. If sk .H / D sk .G/ holds for all positive integers k, then H is either metacyclic or minimal nonmetacyclic of order 34 . Proposition 124.26. Let G be an absolutely regular p-group of exponent > p, p > 2. If H is such a p-group that c1 .H / D c1 .G/ and c2 .H / D c2 .G/, then one of the following holds: (a) H is absolutely regular. (b) H is irregular of maximal class and order p 2.p1/ and j1 .H /j D p p1 . Proof. Let j1 .G/j D p r and j2 .G/j D p rCs ; then, since G is absolutely regular, we have r; s p 1; c2 .G/ D

c1 .G/ D 1 C p C C p r1 D c1 .H /;

p rCs p r D p r1 .1 C p C C p s1 / D c2 .H /: p.p 1/

By Theorems 13.2(a) and 9.5, H is either absolutely regular or irregular of maximal class. Assume that the group H is irregular of maximal class. Then we obtain r D p 1, j1 .H /j D j1 .G/j D p p1 . Let jH j D p n and let F be an absolutely regular maximal subgroup of H ; then n p C 1 (see Theorems 9.5 and 9.6). By Theorem 13.19, all elements of the set H F have the same order p 2 . Therefore there are in H exactly p n p n1 D p n2 p.p 1/ cyclic subgroups of order p 2 not contained in F . Let j2 .F /j D p p1Ct ; then we get t p 1 since F is absolutely regular, and we have c2 .F / D

p p1Ct p p1 D p p2 .1 C p C C p t 1 / p.p 1/

so that c2 .H / D p p2 .1 C p C C p t 1 / C p n2 D p r1 .1 C p C C p s1 / D p p2 .1 C p C C p s1 /: As n p C 1, we get n D 2 C .r 1/ C .s 1/ D r C s .p 1/ C .s 1/ D 2.p 1/:

264

Groups of prime power order

7o On subgroups of metacyclic p-groups. We begin with the following exercises. Exercise 1. Let G be a nonabelian metacyclic p-group that is not generalized quaternion. Then the following conditions are equivalent: (a) All minimal nonabelian subgroups of G have no cyclic subgroups of index p. (b) 1 .G/ Z.G/. Solution. (a) ) (b). If a subgroup H G is minimal nonabelian, then we obtain that 1 .G/ D 1 .H / Z.H /. Since G is generated by its minimal nonabelian subgroups (Theorem 10.28), we get 1 .G/ Z.G/. (b) ) (a). Assume that a minimal nonabelian F G has a cyclic subgroup of index p. Then we obtain 1 .G/ 6 F (otherwise, F would be abelian). It follows that p D 2 so F Š Q8 (see Exercise 1.8(a) or Lemma 65.1). In this case, G is of maximal class (Lemma J(b)) so G is generalized quaternion, contrary to the hypothesis. Exercise 2. Let a metacyclic p-group G be neither abelian nor minimal nonabelian. If 1 .G/ 6 Z.G/, then G contains at least k p 1 minimal nonabelian subgroups with a cyclic subgroup of index p. If k D p 1, then jG 0 j D p 2 so all subgroups of index p 2 in G are abelian. Solution. One may assume that G has no cyclic subgroup of index p; then G has no nonabelian subgroup of order p 3 . If p D 2, then k 1 (Exercise 1). Now let p > 2. Then CG .1 .G// D C is maximal in G and the minimal nonabelian subgroups of C have no cyclic subgroups of index p. If A < G is minimal nonabelian without cyclic subgroup of index p, then A 6 C and we obtain that 1 .G/ D 1 .A/ Z.A/ so that CG .1 .G// CA D G, contrary to the hypothesis. Thus every minimal nonabelian subgroup of G not contained in C has a cyclic subgroup of index p. It follows from 76 that k p 1 in all cases. Now suppose that k D p 1. Then, by Lemma 76.5(a), 1 D ¹C; A1 ; : : : ; Ap1 ; Aº, where A1 ; : : : ; Ap1 are minimal nonabelian and A is abelian. In view of Lemma 64.1(u), we get jG 0 j pjA01 A0 j D p 2 . If jG 0 j D p, then G is minimal nonabelian (Lemma 65.2(a)), contrary to the hypothesis. Thus jG 0 j D p 2 . In this case, all subgroups of G of index p 2 are abelian (Corollary 65.3), as asserted. Exercise 3. Let G and k be such as in Exercise 2. Study the structure of G provided k D p. (Hint. See Lemma 76.13.) Lemma 124.27. Let G > ¹1º be a metacyclic p-group, A < B G. Then A0 < B 0 unless B is abelian. Proof. Assume that B is nonabelian and A0 D B 0 . Take in B 0 a subgroup L of index p. Then B=L is minimal nonabelian (Lemma 65.2(a)) so A=L is abelian. In this case, we get A0 L < B 0 , as was to be shown. Theorem 124.28. Suppose that G is a metacyclic p-group with derived subgroup of order p n > p and let t be the number of maximal subgroups of G having derived subgroup of order p n1 . Then t 2 ¹p; p C 1º.

124

The number of subgroups of given order in a metacyclic p-group

265

Proof. By Lemma 64.1(u), there exists H 2 1 such that jH 0 j D p n1 so that t > 0. Assume that there are distinct A; B 2 1 such that jA0 j jB 0 j < p n1 . Since G 0 is cyclic, we get A0 B 0 . Then, by Lemma 64.1(u), we have jG 0 j pjA0 B 0 j D pjB 0 j < p p n1 D p n ; contrary to the hypothesis. Thus there is in 1 at most one member whose derived subgroup has order < p n1 . Since G is noncyclic so j1 j D p C 1 and G has no maximal subgroup whose derived subgroup has order p n (Lemma 124.27), we conclude that t p. We are done since t p C 1. Using induction and Hall’s enumeration principle, one obtains Corollary 124.29. Suppose that G is a metacyclic p-group with derived subgroup of order p n > p 2 . If t denotes the number of subgroups H of G with jH 0 j D p n2 and jG W H j D p 2 , then p 2 p t .p C 1/2 . Proof. Let M1 ; : : : ; Ms be all members of the set 1 whose derived subgroups have order p n1 ; then s 2 ¹p; p C 1º (Theorem 124.28). Let t.Mi / be the number of those maximal subgroups of Mi whose derived subgroups have order p n2 and let D 0 if jˆ.G/0 j < p n2 and D 1 if jˆ.G/0 j D p n2 ; then we have t.Mi / 2 ¹p; p C 1º (Theorem 124.28). By Hall’s enumeration principle, tD

s X

t .Mi / p :

i D1

It follows that

p2

p t .p C 1/2 .

Exercise 4. Suppose that G is a metacyclic 2-group of order > 22k , k > 1. If all minimal nonabelian subgroups of G are isomorphic to k

k

k1

Hk D ha; b j a2 D b 2 D 1; ab D a1C2

i;

then k1 .G/ D Z.G/ is abelian of type .2k1 ; 2k1 / and G=Z.G/ is dihedral. Solution. By Proposition 10.28, the group G is generated by minimal nonabelian subgroups. It follows that all minimal nonabelian subgroups are not normal in G (otherwise, they would be coincide). If A < G is minimal nonabelian, then we obtain that k1 .G/ D k1 .A/ D Z.A/ and thus we get k1 .G/ Z.G/ since such subgroups A generate G (Theorem 10.28). Moreover, since A \ Z.G/ D Z.A/, we get k1 .G/ D Z.G/. Therefore the nonabelian metacyclic quotient group G=k1 .G/ is generated by abelian subgroups of type .2; 2/, and we conclude that it is dihedral since jG=1 .G/j 8 (see Theorem 1.2). Exercise 5. If A metacyclic 2-group G contains two distinct minimal nonabelian subgroups A and B of order 22k and exponent 2k , then G=k1 .G/ is either dihedral or semidihedral. If G=k1 .G/ is dihedral, then k1 .G/ D Z.G/.

266

Groups of prime power order

Solution. We have k1 .G/ D k1 .A/ D k1 .B/. As A=k1 .G/, B=k1 .G/ are nonnormal four-subgroups of G=k1 .G/, it follows that G=k1 .G/ is of maximal class (Lemma J(e)); moreover, G=k1 .G/ is either dihedral or semidihedral. Suppose that G=k1 .G/ is dihedral. If H=k1 .G/ is abelian of type .2; 2/, then k k k1 either H is abelian or H Š ha; b j a2 D b 2 D 1; ab D a1C2 i. In both cases, we obtain k1 .G/ D Z.H /. As such subgroups H generate G, we get k1 .G/ Z.G/: Now the equality of these two subgroups follows since Z.G/ \ H D k1 .G/. Exercise 6. Suppose that G is a metacyclic p-group and B < G minimal nonabelian. Then B=B 0 is a maximal abelian subgroup of G=B 0 . Solution. Assume that B=B 0 < A=B 0 , where A=B 0 is abelian. Then A0 D B 0 , contrary to Lemma 124.27. Exercise 7. Let A be a proper minimal nonabelian subgroup of a p-group G. Suppose that all subgroups of G containing A of order pjAj are two-generator. Then A=A0 is a maximal abelian subgroup of NG .A/=A0 . Hint. Let A < T NG .A/ be such that jT W Aj D p and T =A0 is abelian. Since d.T / D 2 and T 0 D A0 is of order p, it follows from Lemma 65.2(a) that T is minimal nonabelian, which is a contradiction. Exercise 8 (= Proposition 10.17). Let B be a nonabelian subgroup of order p 3 of a p-group G. If CG .B/ < G, then G is of maximal class. Solution. Set N D NG .B/. Then, by N=C -Theorem, jN j D p 4 so N is a unique subgroup of order p 4 in G containing B. We claim that d.N / D 2 (otherwise, N contains at least two distinct abelian subgroups of order p 3 so jZ.N /j D p 2 ; then CG .B/ 6 B, a contradiction). Clearly, Z.G/ D Z.B/ D B 0 . Then, by Exercise 7, B=B 0 is a maximal abelian subgroup of G=B 0 so G=B 0 is of maximal class (Proposition 1.8). Since Z.G/ D B 0 is of order p, it follows that G is of maximal class as well. Exercise 9. Let H < G, where G is a metacyclic p-group. If H > R.H /, then we have R.H / G G. (Hint. We have R.H / D e .G/, where exp.R.H // D p e .) Exercise 10. Let G be a metacyclic 2-group and let hn .G/ be the number of subgroups H < G such that H D R.H / is of order 22n . Then one of the following holds: (a) hn .G/ D 1. (b) hn .G/ is even. In that case, if hn .G/ > 0, then G=n1 .G/ is either dihedral or semidihedral. Solution. Let H < G be such that H D R.H / is of order 22n . If H G G, then we have H D n .G/ and hn .G/ D 1. Now let H be nonnormal in G; then all such H

124

The number of subgroups of given order in a metacyclic p-group

267

are nonnormal in G. In this case, hn .G/ is even. Then n1 .H / D n1 .G/ and H=n1 .H / Š E4 is non-G-invariant. By Lemma J(e), It follows that G=n1 .G/ is either dihedral or semidihedral. Exercise 11. Prove that there exists a metacyclic p-group G such that G contains a maximal normal abelian subgroup A with nonabelian G=A. Solution (Mann). Let G be a semidirect product of a cyclic group A of order p n by a k cyclic group B whose generator induces on A the automorphism x ! x 1Cp . The order of B is the order of the automorphism. Choose n D 2k C 2. Then G 0 is the subgroup Ãk .A/ and Z.G/ D Ãnk .A/. Let L be the unique subgroup between G 0 and Z.G/, and let D D CB .L/. Then LD is a maximal normal abelian subgroup with nonabelian factor group. Many more examples can be produced by choosing other values of n and k to make the distance between G 0 and Z.G/ bigger. 2

2

Let H2;p D ha; b j ap D b p D 1; ab D a1Cp i. Then H2;p is the unique nonabelian metacyclic group of order p 4 and exponent p 2 . For p D 2 we have H2;2 D H2 . Theorem 124.30. Let H Š H2;p be a proper subgroup of a metacyclic p-group G. Then one of the following holds: (a) If p > 2, then jG W H j D p, jG 0 j D p 2 , G=G 0 is abelian of type .p 2 ; p/ and G=Z is cyclic of order p 2 for some cyclic Z G G. (b) If p D 2 and H G G, then jG W H j D 2, jG 0 j D 4, G=G 0 is abelian of type .4; 2/ and G=Z is cyclic of order 4 for some cyclic Z G G. (c) If p D 2 and the subgroup H is not normal in G, then we have 1 .G/ D 1 .H / and jG W CG .1 .G//j 2. If, in addition, jGj > 25 , then G=1 .G/ is either dihedral or semidihedral. Proof. Let G have a proper normal nonabelian subgroup H of order p 4 and exponent p 2 . Then we have R D 1 .H / D 1 .G/ G G. Set C D CG .R/. Let L < R be of order p, L ¤ H 0 . Then H=L C =L is nonabelian of order p 3 so H D C if p > 2 and C =L is of maximal class if p D 2 (Lemma J(b)). In the case under consideration, jG W C j p. Let p > 2; then H D C . In this case, G 0 < C so jG 0 j D p 2 since G 0 is cyclic and G is not minimal nonabelian (see Lemma 65.2(a). There is a cyclic Z G G such that G=Z is cyclic. Since G 0 < Z, it follows that G=G 0 is abelian of type .p 2 ; p/. Now let p D 2 and H G G; then L is characteristic in H (Lemma J(a)). Then C D G since L G G so G=L is of maximal class (Lemma J(b)). Since L 6 G 0 , it follows that jG W G 0 j D 8 so G=G 0 is abelian of type .4; 2/. As above, there is a cyclic Z G G such that G=Z is cyclic of order 4. Let H be not normal in G and C as above; then p D 2. Suppose that jGj > 25 . Take L < R.D 1 .G// of order 2, L ¤ H 0 . Then H=L C =L is nonabelian of order 8 so C =L is of maximal class. In this case, C =1 .G/ G=1 .G/ is of maximal class with nonnormal abelian subgroup H=1 .G/ of type .2; 2/, and we conclude that G=1 .G/ is either dihedral or semidihedral.

268

Groups of prime power order

Proposition 124.31. Suppose that H Š Mpn , n > 3, is a proper subgroup of a metacyclic p-group G. Then NG .H /=H is cyclic. Next, assume that n D 4. (a) If p > 2, then jNG .H /=H j D p. (b) If p D 2, then jNG .H /=H j 4. Proof. Assume that NG .H /=H is noncyclic. Write N D NG .H /. In this case, we have H 6 ˆ.N /. Indeed, otherwise 1 .H / D 1 .N / Z.N / (see the proof of Lemma 1.4). Then 1 .H / Z.H / so H is abelian, a contradiction. Thus NG .H /=H is cyclic. Next we assume that n D 4. By Exercise 8, H=H 0 is a maximal abelian subgroup of N=H 0 . If p > 2, then a Sylow p-subgroup of Aut.H=H 0 / is nonabelian of order p 3 and exponent p so N=H , being cyclic, has order p. Now let p D 2. Then Aut.H=H 0 / Š D8 so N=H is a cyclic subgroup of D8 , and we conclude that N=H is cyclic of order 4.

Problems Below we state some related problems. Problem 1. Study the metacyclic p-groups, p > 2, possessing a nonabelian section of order p 3 . (See Theorem 124.22.) Problem 2. Find the number of subgroups of given order in a nonmetacyclic minimal nonabelian p-group. Problem 3. Let G be an abelian group of type .p a1 ; : : : ; p ak /. Find the types of all maximal subgroups of G counting multiplicities. Problem 4. Find the number of subgroups of given order in an abelian p-group of rank 3. Problem 5. Let G be a homocyclic p-group of rank d . (i) Find the number of subgroups of given order in G. (ii) Find the number of subgroups of rank 2 and given order in G. Problem 6. Given a positive integer k > 1, classify the p-groups G such that there exists a metacyclic p-group M satisfying sk .G/ D sk .M /. Problem 7. Study the p-groups G such that sk .G/ D sk .A/ for all positive integers k and some abelian p-group A. Problem 8. Given p > 2 and n > p, does there exist an absolutely regular p-group A (see 9) and an irregular p-group G of the same order p n satisfying sk .G/ D sk .A/ for all k 2 ¹2; : : : ; n 1º?

125

p-groups G containing a maximal subgroup H all of whose subgroups are G -invariant

Here we solve a problem stated by the ﬁrst author. Theorem 125.1. Let G be a nonabelian p-group containing a maximal subgroup H such that all subgroups of H are G-invariant. Then there is an element g 2 G H such that one of the following holds: (i) p D 2, H is Hamiltonian, i.e., H D Q V , where Q Š Q8 , exp.V / 1, and g 2 Z.G/, o.g/ 4. (ii) p D 2, H is abelian of exponent 2e , e 2, and either g inverts each element e1 in H or e 3 and g satisﬁes hg D h1C2 for all h 2 H . In both cases, Z.G/ D CH .g/ D 1 .H / is elementary abelian and o.g/ 4. e1

for all h 2 H ,

e1

for all h 2 H ,

(iii) p D 2, H is abelian of exponent 2e , e 3, and hg D h1C2 where Z.G/ D CH .g/ D e1 .H /.

(iv) p > 2, H is abelian of exponent p e , e 2, and hg D h1Cp where Z.G/ D CH .g/ D e1 .H /.

Proof. (i) First assume that H is nonabelian. Then p D 2 and H D Q V , where Q Š Q8 and exp.V / 1. Set U D Z.Q/ V D 1 .H / so that U Z.G/ since each subgroup of order 2 in U is normal (and so central) in G. But each subgroup of Q is also normal in G and so G D Q C , where C D CG .Q/, Q \ C D Z.Q/ and H \ C D U . Thus C is abelian and so C D Z.G/. We take an element g 2 C H so that g 2 Z.G/, g 2 2 U and o.g/ 4. (ii) Now assume that H is abelian, p D 2, and 2 .H / 6 Z.G/. Let g be an element in G H . We have CH .g/ D Z.G/ and g 2 2 Z.G/. Let v be an element of order 4 in H Z.G/. Then v g D v 1 D vz, where z D v 2 2 Z.G/. Suppose that Z.G/ possesses an element w of order 4. If w 2 D z, then vw is an involution in H Z.G/ and vw is central in G, a contradiction. Hence we get w 2 D u ¤ z and hv; wi Š C4 C4 . But then we obtain .vw/g D .vz/w D .vw/z and z 62 hvwi as .vw/2 D zu which implies that hvwi is not normal in the group G, a contradiction. We have proved that Z.G/ D 1 .H / D CH .g/ is elementary abelian. For any x 2 H Z.G/ we obtain that o.x/ 4 and Chxi .g/ D 1 .hxi/ and so either x g D x 1 or o.x/ > 4 and x g D x 1 x 0 , where hx 0 i D 1 .x/. Also, g 2 2 Z.G/ and so o.g/ 4.

270

Groups of prime power order

Suppose that g does not invert each element in H . Then exp.H / D 2e with e 3. If g inverts each element of order 2e in H , then g inverts each element in H since elements of order 2e generate H , a contradiction. Hence there is an element h1 2 H of e1 order 2e such that g does not invert hh1 i and so hg1 D h1C2 . There exists a basis 1 ¹h1 ; : : : ; hr º, r 1, of H which contains h1 (since h1 is of the maximal possible order in H ). Assume that there is hi 2 ¹h1 ; : : : ; hr º such that i > 1, o.hi / D 2e and hgi D h1 i . But then .h1 hi /g D h1C2 1 e1

e1

e1

1 2 h1 i D .h1 hi / h1

e1

e1

¤ h21 which shows that hh1 hi i is not normal in G, and 1 .hh1 hi i/ D h21 h2i e1 a contradiction. Hence we must have hgi D h1C2 . Suppose that there is an elei 0 ment hj 2 ¹h1 ; : : : ; hr º such that j > 1, o.hj / D 2e , e 0 < e, but hjg ¤ hj1

e 0 1

e 0 3 and hjg D hj1C2

and so

:

0

Let h01 2 hh1 i be such that o.h01 / D 2e which implies that .h01 /g D .h01 /1 . But then .h01 hj /g D .h01 /1 hj1C2

e 0 1

and

1 .hh01 hj i/ D .h01 /2

e 0 1

hj2

e 0 1

¤ hj2

e 0 1

which shows that hh01 hj i is not normal in G, a contradiction. Hence we must have hjg D hj1 D hj1C2

e1

since o.hj / < 2e . We have proved that hgi D h1C2 for all 1 i r and so for i e1 all h 2 H , hg D h1C2 and we are done. (iii) Assume that H is abelian, p D 2, and 2 .H / Z.G/. Then H is of exponent 2e , e 3, and let g 2 G H so that g centralizes 2 .H /. If x 2 H Z.G/, then o.x/ 8, Chxi .g/ D hx 2 i and so x g D xx 0 , where hx 0 i D 1 .hxi/. Since all e1 elements of order 2e generate H , there is h1 2 H of order 2e such that hg1 D h11C2 . Let ¹h1 ; : : : ; hr º, r 1, be a basis of H containing h1 . Suppose that there is an element hi 2 ¹h1 ; : : : ; hr º such that i > 1, o.hi / D 2e and hgi D hi . But then e1

.h1 hi /g D .h1 hi /h21 e1

a contradiction. Hence hgi D h1C2 i 0 o.hj / D 2e , e 0 3, e 0 < e, and

e1

e1

and h21

62 hh1 hi i;

. Assume that there is hj 2 ¹h1 ; : : : ; hr º, j > 1, e 0 1

hjg D hj1C2

:

0

Let h01 2 hh1 i with o.h01 / D 2e which gives .h01 /g D h01 . But then e 0 1

.h01 hj /g D .h01 hj /hj2

125

p-groups G with a maximal subgroup having only G-invariant subgroups e 0 1

so that hj2

271

62 hh01 hj i, a contradiction. Hence hjg D hj D hj1C2 e1

It follows that for each h 2 H , hg D h1C2

e1

:

, Z.G/ D e1 .H /, and we are done.

(iv) Finally, assume p > 2 so that H is abelian of exponent p e , e 2. We have Z.G/ D CH .g/, where g is an element in G H and g satisﬁes g p 2 Z.G/. For each x 2 H Z.G/, g induces on hxi an automorphism of order p and so x g D xx 0 , where 1 ¤ x 0 2 1 .hxi/. Since all elements of order p e generate H , there is h1 2 H of order p e such that, replacing g with g l for some l 6 0 .mod p/ if necessary, we may set 1Cp e1 hg1 D h1 : Let ¹h1 ; : : : ; hr º, r 1, be a basis of H containing h1 . Let hi 2 ¹h1 ; : : : ; hr º be such that i > 1, o.hi / D p e but hgi D hi . Then p e1

.h1 hi /g D .h1 hi /h1

;

p e1

1Cp e1 0 hi ,

62 hh1 hi i, a contradiction. Hence we may set hgi D hi where h1 h0i 2 1 .hhi i/. Then 1Cp e1 1Cp e1 0 hi hi

.h1 hi /g D h1 and .h1 hi /p

e1

p e1

h0i D h1

p e1 0 hi /

.hi

D .h1 hi /1Cp

e1

where

h0i

p e1 p e1 hi i

2 1 .hh1 hi i/ D hh1

1Cp e1

which gives h0i D 1. Thus hgi D hi . Let hj 2 ¹h1 ; : : : ; hr º be such that j > 1, 0 g 0 e 0 0 o.hj / D p , e 2, e < e but hj D hj hj , where 1 ¤ hj0 2 1 .hhj i/. Let h01 2 hh1 i 0 be such that o.h01 / D p e which gives that .h01 /g D h01 . But then .h01 hj /g D .h01 hj /hj0 ; where hj0 62 hh01 hj i, a contradiction. Hence 1Cp e1

hjg D hj D hj and so for each h 2 H , hg D h1Cp

e1

, e1 D Z.G/, and we are done.

126

The existence of p-groups G1 < G such that Aut.G1/ Š Aut.G /

Benjamin Sambale has constructed a 2-group G of order 32 which possesses a maximal subgroup G1 such that Aut.G1 / Š Aut.G/. We give here a simple proof of this fact which solves Problem 211 (for p D 2). Theorem 126.1. There exist 2-groups G1 < G with jGj D 32 and Aut.G1 / Š Aut.G/. Proof. We construct our 2-group G as a splitting and faithful extension of its abelian normal subgroup H D ha; b j a4 D b 2 D Œa; b D 1i of type .4; 2/ with a cyclic subgroup hci Š C4 of order 4, where ac D a1 b and b c D ba2 . Then ha2 i D Z.G/ and c 2 inverts each element in H and so all elements in Hc 2 are involutions. Thus all elements in G .H hc 2 i/ D Hc [Hc 1 are of order 4 and so G1 D H hc 2 i D 1 .G/ is a characteristic subgroup of G. Since H D 2 .G1 /;

ha2 ; bi D 1 .H /;

ha2 i D Ã1 .H / D Z.G/;

we conclude that H , ha2 ; bi Š E4 and ha2 i are also characteristic subgroups of G. As b D aac , we have G D ha; ci and so G is given in terms of generators and relations as follows: ()

G D ha; c j a4 D c 4 D .aac /2 D Œa; aac D 1; ac D a1 i: 2

Note that hai and habi are the only two cyclic subgroups of order 4 in H and since haic D hac i D ha1 bi D habi, we have NG .hai/ D G1 . Hence for each x 2 G G1 and y 2 H 1 .H /, hyi and hy x i are two distinct cyclic subgroups of order 4 in H , which gives that hx; yi D G. Each automorphism of G sends a to one of ¹a; a1 ; ab; a1 bº and c is sent to one of 16 elements in G G1 which shows that jAut.G/j 4 16 D 64. Conversely, take a 2 ¹a; a1 ; ab; a1 bº (4 possibilities) and c 2 G G1 D Hc [ Hc 1 (16 possibilities). We know from the previous paragraph that ha ; c i D G and we verify that a and c satisfy all relations in () (in place of a and c, respectively) and so a ! a , c ! c induces an automorphism of G, hence jAut.G/j D 4 16 D 64. Indeed, a and c are elements of order 4. Also, a and .a /c are two elements in H 1 .H / so that a .a /c 2 1 .H / and therefore .a .a /c /2 D 1. Obviously, a and .a /c (lying in the abelian subgroup H ) commute. Finally, .c /2 2 G1 H

126 The existence of p-groups G1 < G such that Aut.G1 / Š Aut.G/

273

and so .c /2 is an involution which inverts each element in H and therefore we obtain 2 .a /.c / D .a /1 . It is easy to show that jAut.G1 /j D 64. We have G1 D ha; b; d D c 2 j a4 D b 2 D Œa; b D d 2 D 1; ad D a1 ; b d D bi;

()

and for each 2 Aut.G1 / we conclude that a 2 ¹a; a1 ; ab; a1 bº (4 possibilities), b 2 ¹b; ba2 º (2 possibilities) and d 2 G1 H (8 possibilities) which implies that jAut.G1 /j 4 2 8 D 64. For the converse, if we consider a 2 ¹a; a1 ; ab; a1 bº, b 2 ¹b; ba2 º and d 2 G1 H , then obviously ha ; b ; d i D G1 and a ; b ; d satisfy all relations in () since each element in G1 H is an involution which inverts each element in H . We have proved that jAut.G1 /j D 64. It remains to show Aut.G1 / Š Aut.G/. First we determine the structure of Aut.G/. The restriction of Aut.G/ to H provides a homomorphism f W Aut.G/ ! Aut.H /. If 2 ker.f /, then a D a;

bDb

D .ac 1 ac/ D a.c /1 ac

which gives ac D a1 b D ac and so c 2 Hc. We may set D h with a h D a and c h D hc (h 2 H ) so that ker.f / D ¹ h j h 2 H º. The map h ! h (h 2 H ) is an isomorphism from ker.f / onto H and so ker.f / Š C4 C2 and Aut.G/=ker.f / Š Aut.H / Š D8 and therefore f is a homomorphism from Aut.G/ onto Aut.H /. Now we construct a complement of ker.f / in Aut.G/. Deﬁne L D h˛; ˇ 2 Aut.G/; where a˛ D ab; c ˛ D c and aˇ D a; c ˇ D c 1 i: Then a˛ D .ab/˛ D .a ac 1 ac/˛ D .a2 c 1 ac/˛ 2

D a2 .c 1 ab c/ D a2 a1 b ba2 D a1 2

and c ˛ D c so that o.˛/ D 4 and obviously o.ˇ/ D 2. Then we show that .˛ˇ/2 D 1. Indeed, a˛ˇ D .ab/ˇ D ab ˇ D a.ac 1 ac/ˇ D a acac 1 D a2 cac 1 and c ˛ˇ D c 1 . This gives (noting that a2 2 Z.G/) a.˛ˇ / D .a2 cac 1 /˛ˇ D .a2 cac 1 /2 c 1 .a2 cac 1 /c 2

D a2 cac 1 a2 cac 1 c 1 a2 ca D a2 ca2 c 1 a D a 2

and c .˛ˇ / D c which shows .˛ˇ/2 D 1. We have proved that L D h˛; ˇ j ˛ 4 D ˇ 2 D .˛ˇ/2 D 1i Š D8 . Since Z.L/ D h˛ 2 i and ˛ 2 62 ker.f /, we get ker.f / \ L D ¹1º. Thus we have shown so far that Aut.G/ is a semidirect product of ker.f / Š H and

274

Groups of prime power order

L Š Aut.H / Š D8 . It is easy to justify that CAut.G/ .ker.f // D ker.f /. Indeed, if we have CAut.G/ .ker.f // > ker.f /, then ˛ 2 centralizes ker.f /. In particular, we obtain 2 2 2 that ˛ 2 a D a ˛ 2 . But c ˛ a D c a D ac and c a ˛ D .ac/˛ D a1 c, a contradiction. We have proved that Aut.G/ is isomorphic to the holomorph of H Š C4 C2 . Now we determine the structure of Aut.G1 /, where G1 is given in (). The restriction of Aut.G1 / to H gives rise to a homomorphism f0 W Aut.G1 / ! Aut.H / with ker.f0 / D ¹h j h 2 H º; where ah D a; b h D b; d h D hd: The map h ! h .h 2 H / gives an isomorphism ker.f0 / Š H . Thus we conclude Aut.G1 /=ker.f0 / Š Aut.H / Š D8 . The automorphisms 2 Aut.G1 / with d D d form a complement L0 Š D8 of ker.f0 / in Aut.G1 /. Let 2 L0 with a D a1 b, b D ba2 so that 2 inverts each element in H and therefore h 2 i D Z.L0 /. In case CAut.G1 / .ker.f0 // > ker.f0 /, we infer that 2 centralizes ker.f0 /. In particular, we 2 2 2 have a 2 D 2 a . But then d a D .ad / D a1 d and d a D d a D ad , a contradiction. Hence CAut.G1 / .ker.f0 // D ker.f0 / and so Aut.G1 / is also isomorphic to the holomorph of H Š C4 C2 and we are done. A similar example was presented in [Li4].

127

On 2-groups containing a maximal elementary abelian subgroup of order 4

In the following theorem Problem 1591 is solved. Theorem 127.1 (Janko). Let G be a 2-group which contains a maximal elementary abelian subgroup A of order 4. If G contains also an elementary abelian subgroup E of order 16, then G is isomorphic to one of the groups of Theorem 49.3.1 Proof. By Exercise 51 in Appendix 40, G has no normal elementary abelian subgroup of order 8. Due to Exercise 52 in Appendix 40, every subgroup of G is generated by four elements. In particular, G has no elementary abelian subgroup of order > 16. Obviously, G is neither abelian nor a 2-group of maximal class. Then G possesses a normal four-subgroup U . If A Z.G/, then AE is elementary abelian of order > 16, contrary to what has just been said. Therefore, since A is a maximal elementary abelian subgroup of G, it follows that 1 .Z.G// < A and so, in particular, Z.G/ is cyclic. Set hzi D 1 .Z.G//; then we have hzi A \ U and jG W CG .U /j D 2. Note that jE \ CG .U /j 8 and so A ¤ U which gives A \ U D hzi. But then A 6 CG .U / and so A \ CG .U / D hzi. Let t 2 A CG .U / so that CG .t/ D ht i CCG .U / .t/ by the modular law, and therefore z is the only involution in L D CCG .U / .t/ (indeed, if z 2 F < CCG .U / .t/, where F is abelian of type .2; 2/, then A < ht; F i Š E8 , contrary to the hypothesis). Hence L is either cyclic of order 2m , m 2, or L is generalized quaternion. Indeed, if L is cyclic of order 2, then G would be of maximal class (Proposition 1.8), a contradiction. Suppose that L Š C2m , m 2. By Theorem 48.1, in that case G does not possess an elementary abelian subgroup of order 16. Hence we have L Š Q2m , m 3, and then G is isomorphic to one of the groups from Theorem 49.3 and we are done. Thus the groups from Theorem 127.1 are determined up to isomorphism. 1 In Theorem 49.3 the 2-groups G containing an involution t such that C .t/ D hti Q, where Q G is a generalized quaternion group, are classiﬁed if G also possesses an elementary abelian subgroup of order 16.

276

Groups of prime power order

Remark. Let t be an element of order p of a p-group G such that CG .t/ D hti Q, where j1 .Q/j D p is a generalized quaternion group. Then 1 .CG .t// Š Ep2 is a maximal elementary abelian subgroup of G. In that case, G has no normal subgroup of order p pC1 and exponent p (see Exercise 51 in Appendix 40). Now, if p D 2, by the proof of the theorem, G has no elementary abelian subgroup of order 25 by Exercise 52 in Appendix 40. G. Glauberman and N. Mazza have proved that if p > 2 and a p-group G possesses a maximal elementary abelian subgroup of order p 2 , then it has no elementary abelian subgroup of order p pC1 . As the symmetric group Sp2 , p > 2, shows, this estimate is best possible. Exercise. Suppose that a p-group G, p > 2, possesses a maximal elementary abelian subgroup A of order p 2 . If there is in G an elementary abelian subgroup E of order p 3 , then there is a 2 A# such that CG .a/ D hai Q, where Q has exactly one subgroup of order p. Next, G has no normal subgroup of order p pC1 and exponent p (compare with Glauberman–Mazza’s result from the paragraph preceding the exercise). In view of Theorem 13.7, one may assume that E GG. In that case, AE is of maximal class and order p 4 . Assume, in addition, that p > 3; then exp.AE/ D p. Let AE H G, where H is as large as possible among all subgroups of exponent p. As CH .A/ D A, it follows that H is of maximal class (Proposition 1.8), so jH j p p (Theorem 9.5).

128

The commutator subgroup of p-groups with the subgroup breadth 1

Recall that the subgroup breadth sb.G/ of a p-group G is deﬁned by p sb.G/ D max¹jG W NG .H /j j H Gº: This section is a continuation of 122 and here we prove some results which are crucial on the way of classifying completely p-groups G with sb.G/ D 1. Theorem 128.1. If a p-group G has the subgroup breadth 1, then jG 0 j p 2 . Proof. By Propositions 122.8 and 122.16, the element breadth b.G/ of G is at most 2. Recall that the element breadth b.G/ of a p-group G is deﬁned by p b.G/ D max¹jG W CG .x/j j x 2 Gº: If b.G/ D 1, then by Proposition 121.9 (Knoche), jG 0 j D p. We may assume in the sequel that b.G/ D 2. Then we use Theorem 121.1 (Parmeggiani–Stellmacher) and conclude that we must be in part (b) of that theorem (otherwise, jG 0 j D p 2 and we are ﬁnished) which implies that jG 0 j D p 3 and jG W Z.G/j D p 3 . By Proposition 122.2, ˆ.G/ is abelian. Let A be a maximal normal abelian subgroup of G containing ˆ.G/ so that G=A ¤ ¹1º is elementary abelian. If jG=Aj D p, then Lemma 1.1 gives jGj D pjZ.G/jjG 0 j, which contradicts jG 0 j D jG W Z.G/j D p 3 . Hence jG=Aj p 2 . But Z.G/ < A and so G=A Š Ep2 and jA W Z.G/j D p. For each x 2 G A, x p 2 Z.G/ and so G=Z.G/ is of order p 3 and exponent p. Also, for each x 2 G A, CA .x/ D Z.G/ and so jŒA; xj D p (Lemma 121.2) and ŒA; x Z.G/ since A=Z.G/ Z.G=Z.G//. (i) First assume that G=Z.G/ is nonabelian. Then p > 2 and G=Z.G/ Š S.p 3 / (the nonabelian group of order p 3 and exponent p). Since G 0 covers A=Z.G/, we get jG 0 \ Z.G/j D p 2 . For any x; y 2 G A such that Ahx; yi D G we have ŒA; x ¤ ŒA; y. Indeed, if ŒA; x D ŒA; y, then we obtain A=ŒA; x D Z.G=ŒA; x/ which implies that G=ŒA; x has (at least) two distinct abelian maximal subgroups. But then j.G=ŒA; x/0 j D p and then jG 0 j D p 2 , a contradiction. Hence if Gi =A, i D 1; 2; : : : ; p C 1, are p C 1 pairwise distinct subgroups of order p in G=A, then ŒA; Gi D ŒG 0 ; Gi (i D 1; 2; : : : ; p C 1) are p C 1 pairwise distinct subgroups of order p in G 0 \ Z.G/ so that G 0 \ Z.G/ Š Ep2 . On the other hand,

278

Groups of prime power order

for each x 2 A Z.G/ we have CG .x/ D A which shows that there are no elements of order p in AZ.G/. In particular, all elements in the set G 0 Z.G/ are of order p 2 . Let a 2 G 0 Z.G/ and then set z D ap so that Ã1 .G 0 / D hzi and G 0 is abelian of type .p 2 ; p/. Let M be any maximal subgroup of G which contains Z.G/. Since G 0 M , we must have M > A. This simple remark will be used often in the following arguments. Let g 2 G A be such that Œa; g D z. Then g p 2 Z.G/ and CG .g/ D Z.G/hgi. Indeed, we have CA .g/ D Z.G/ and so if CG .g/ > Z.G/hgi, then CG .g/ covers G=A and so CG .g/ would be a maximal subgroup of G containing Z.G/ but not containing A, contrary to our simple remark from the previous paragraph. On the other hand, jNG .hgi/ W CG .g/j D p and so o.g/ p 2 and NG .hgi/ D Ahgi. Indeed, if hgi E G, then A=Z.G/ and .Z.G/hgi/=Z.G/ are two distinct normal subgroups of order p in the factor group G=Z.G/, contrary to G=Z.G/ Š S.p 3 /. Also, if NG .hgi/ ¤ Ahgi, then NG .hgi/ is a maximal subgroup of G which contains Z.G/ but does not contain A, contrary to our simple remark. We have proved that hai normalizes hgi. Since Œa; g D z, we have 1 .hgi/ D hzi. If o.g/ p 3 , then there is an element h 2 hg p i Z.G/ such that o.h/ D p 2 and hp D z 1 . But then ah 2 AZ.G/ and .ah/p D ap hp D zz 1 D 1, a contradiction. Hence o.g/ D p 2 and 1 ¤ g p 2 hzi. We get ha; gi D U Š Mp3 with U \ A D hai. Since 1 .U / Š Ep2 , there is an element g 0 of order p in U A. But then we must have CA .g 0 / D Z.G/ and jG W CG .g 0 /j D p which shows that CG .g 0 / covers G=A. Hence CG .g 0 / is a maximal subgroup of G containing Z.G/ but not containing A, contrary to our simple remark. (ii) Assume that G=Z.G/ is abelian. In that case, we get G=Z.G/ Š Ep3 and so ˆ.G/ Z.G/. For any x; y 2 G we have Œx; yp D Œx p ; y D 1 and so G 0 Š Ep3 . If there is an element x 2 G Z.G/ of order p, then we infer that jG W CG .x/j D p and so CG .x/ is an abelian maximal subgroup of G, a contradiction. Hence 1 .G/ Z.G/. If x; y; z 2 G Z.G/ are such that G D Z.G/hx; y; zi, then G 0 D hŒx; y; Œx; z; Œy; zi. Let h be any element in G Z.G/. Then o.h/ p 2 and Œhhi; G Š Ep2 . We have CG .h/ D Z.G/hhi with jG W CG .h/j D p 2 because G has no abelian maximal subgroup. If hhi E G, then Œhhi; G hhi and so Œhhi; G is cyclic of order p 2 , a contradiction. It follows that jG W NG .hhi/j D p. Then we get Œhhi; NG .hhi/ hhi \ Œhhi; G and so 1 .hhi/ Œhhi \ G: We have proved the following fact which will be used often in the following arguments: () For each h 2 G Z.G/ we have 1 .hhi/ Œhhi; G G 0 .

128

The commutator subgroup of p-groups with the subgroup breadth 1

279

(ii1) Suppose that p > 2. Amongst all elements in G Z.G/ let a be one of smallest possible order p m , m1 m 2. Then we set t1 D ap so that (by ./) t1 2 G 0 . Further, let b be any element in G .Z.G/hai/ such that Œa; b D t1 . From () we conclude that such an element b exists. Let c 2 G .Z.G/ha; bi/ so that setting t2 D Œa; c and t3 D Œb; c, we obtain ht1 ; t2 ; t3 i D G 0 , where Z.G/ha; b; ci D G. Since Œb; a D t11 and Œb; c D t3 , we get Œhbi; G D ht1 ; t3 i. As Œab; a D t11 and Œab; c D t2 t3 , we have Œhabi; G D ht1 ; t2 t3 i: First suppose that o.b/ > o.a/ D p m . Then, using (), we get ˇ

1 .hbi/ D 1 .habi/ D ht1˛ t3 i D ht1 .t2 t3 /ı i D ht1 t2ı t3ı i; where not both ˛ and ˇ are 0 .mod p/ and also not both and ı are 0 .mod p/. We get ı 0 .mod p/ and then ˇ 0

˛ 6 0

.mod p/;

6 0

.mod p/;

.mod p/:

But then b p 2 Z.G/;

o.b p / p m ;

hb p i ht1 i

which implies that there is an element y 2 hb p i such that y p .ay/p

m1

D ap

m1

yp

m1

m1

D t11 . We obtain

D t1 t11 D 1

and so o.ay/ p m1 and ay 2 G Z.G/, which is a contradiction. We have proved that for each b 2 G Z.G/ with Œa; b D t1 we have o.b/ D p m . It follows that for each x 2 Z.G/, o.xb/ D p m and so o.x/ p m which gives that exp.Z.G// p m . Hence all elements in .Z.G/hai/ Z.G/ are also of order p m . m1 ˇ Since Œhbi; G D ht1 ; t3 i, the fact stated in () implies b p D t1˛ t3 , where not both ˛ and ˇ are 0 .mod p/. Then .ab/p

m1

D ap

m1

bp

m1

D t1 t1˛ t3 D t1˛C1 t3 2 G 0 ˇ

ˇ

ˇ

and so the minimality of o.b/ D p m yields that o.ab/ D p m and t1˛C1 t3 ¤ 1. On the other hand, the fact that Œhabi; G D ht1 ; t2 t3 i gives together with () that .ab/p

m1

ˇ

D t1˛C1 t3 D t1 .t2 t3 /ı D t1 t2ı t3ı ;

where not both and ı are 0 .mod p/. We conclude that ı 0 .mod p/ and then ˇ 0 .mod p/ and ˛C1

.mod p/

with

˛ 6 0

.mod p/

and

˛ C 1 6 0

.mod p/:

280

Groups of prime power order

Hence b p compute

m1

D t1˛ and so, replacing b with b 0 D b , where ˛ 1 .mod p/, we

.ab 0 /p

m1

D ap

m1

.b 0 /p

m1

D t1 .b p

m1

/ D t1 t1 D t1 t11 D 1: ˛

Since ab 0 62 Z.G/ and o.ab 0 / p m1 , we get a contradiction to the minimality of o.b/ D p m . (ii2) Assume that p D 2. Amongst all elements in G Z.G/ let a be one of maximal possible order 2m , m1 m 2. Then set t1 D a2 so that () implies t1 2 Œa; G. Let b be any element in G .Z.G/hai/ such that Œa; b D t1 . Take an element c 2 G .Z.G/ha; bi/ and set t2 D Œa; c and t3 D Œb; c, where G 0 D ht1 ; t2 ; t3 i. Since Œb; c D t3 and Œb; a D t1 , ˇ () implies 1 .hbi/ ht1 ; t3 i and therefore 1 .hbi/ D ht1˛ t3 i, where ˛; ˇ 2 ¹0; 1º and not both ˛ and ˇ are equal to 0. As Œc; b D t3 and Œc; a D t2 , it follows from () that 1 .hci/ ht2 ; t3 i and therefore 1 .hci/ D ht2 t3 i, where ; 2 ¹0; 1º and not both and are equal to 0. First assume m D 2 so that all elements in G Z.G/ are of order 4. We have ˇ

ˇ

.ab/2 D a2 b 2 Œb; a D t1 t1˛ t3 t1 D t1˛ t3 ;

Œab; a D t1 ;

Œab; c D t2 t3 ;

ˇ

and so Œab; G D ht1 ; t2 t3 i. Using (), we get t1˛ t3 2 ht1 ; t2 t3 i and so ˇ D 0, ˛ D 1, .ab/2 D t1 , and b 2 D t1 . We have

.ac/2 D t1 t2 t3 t2 D t1 t2C1 t3 ;

Œac; a D t2 ;

Œac; b D t1 t3 ;

and so Œac; G D ht2 ; t1 t3 i. Again using (), we get t1 t2C1 t3 2 ht2 ; t1 t3 i and so D 1. We have .bc/2 D t1 t2 t3 t3 D t1 t2 ;

Œbc; b D t3 ;

Œbc; a D t1 t2

and so Œbc; G D ht3 ; t1 t2 i. Applying (), we obtain that t1 t2 2 ht3 ; t1 t2 i, and so D 1, c 2 D t2 t3 . Finally, we have .ab c/2 D .ab/2 c 2 Œc; ab D t1 t2 t3 t2 t3 D t1 ;

Œabc; a D t1 t2 ;

Œabc; b D t1 t3

and so Œabc; G D ht1 t2 ; t1 t3 i. Using (), we get t1 2 ht1 t2 ; t1 t3 i, a contradiction. We have proved that m 3, where a is an element of maximal possible order 2m in G Z.G/. Now suppose that for each b 2 G .Z.G/hai/ such that Œa; b D t1 we m1 ˇ have always o.b/ D 2m . We get b 2 D t1˛ t3 , where not both ˛ and ˇ are equal to 0. m1 Since Œab; a D t1 , Œab; c D t2 t3 , we obtain (using ()) .ab/2 D t1 .t2 t3 /ı , where not both and ı are equal 0. We note that Œa; ab D t1 and so, by our assumption, we have o.ab/ D 2m . On the other hand, .ab/2

m1

ˇ

ˇ

D t1 t1˛ t3 D t1˛C1 t3 :

128

The commutator subgroup of p-groups with the subgroup breadth 1 ˇ

ˇ

281

Hence t1˛C1 t3 D t1 t2ı t3ı and this implies ı D 0 and then t1˛C1 t3 D t1 gives ˇ D 0 and ˛ C 1 .mod 2/. Since ı D 0, we must have D 1 and then ˛ D 0. But then both ˛ and ˇ are equal to 0, a contradiction. By the previous paragraph, we may take an element b in G .Z.G/hai/ such that m1 m1 Œa; b D t1 and o.b/ < 2m . In that case, we have .ab/2 D a2 D t1 . It is easy to see that exp.Z.G// D 2m1 . Indeed, in any case, exp.Z.G// 2m since all elements in G Z.G/ are of order at most 2m . Suppose that there is z 2 Z.G/ with o.z/ D 2m . Then b 0 D bz is of order 2m and since Œb 0 ; G D Œb; G D ht1 ; t3 i, we have (by ()) 1 .hb 0 i/ 2 ht1 ; t3 i and so .b 0 /2

m1

D .b/2

m1

m1

z2

m1

D z2

m1

2 ht1 ; t3 i:

m1

2 ¹t3 ; t1 t3 º, then .az/2 2 ¹t3 ; t1 t3 º, a contradiction as Œaz; G D ht1 ; t2 i. If z 2 m1 Therefore z 2 D t1 . However, Œc; G D Œcz; G D ht2 ; t3 i and so 1 .hci/ D t2 t3 , where ; 2 ¹0; 1º and not both and are equal to 0 and 1 .hcz/ ht2 ; t3 i. In case m1 o.c/ < 2m , we get .cz/2 D t1 , a contradiction. If o.c/ D 2m , then c2

m1

D t2 t3

m1

and .cz/2

D t2 t3 t1 ;

a contradiction. Hence exp.Z.G// D 2m1 and so all elements in Z.G/a and Z.G/ab are of order 2m and all elements in Z.G/b are of order < 2m . Consider ac, where Œac; b D t1 t3 , Œac; a D t2 so that Œac; G D ht2 ; t1 t3 i and therefore (by ()) 1 .haci/ ht2 ; t1 t3 i: On the other hand, .ac/2 2m1

m1

m1

D t1 c 2

:

2m1

D 1, then .ac/ D t1 , a contradiction. Hence we have o.c/ D 2m and so If c m1 m1 2 c D t2 t3 , where not both and are equal 0. Then .ac/2 D t1 t2 t3 2 ht2 ; t1 t3 i m1 2 and so D 1 and c D t 2 t3 . Consider .ab/c, where Œabc; a D t1 t2 , Œabc; c D t2 t3 so that, by (), 1 .habci/ ht1 t2 ; t2 t3 i: On the other hand, ..ab/c/2

m1

m1

D t1 t2 t3 2 ht1 t2 ; t2 t3 i; m1

D t3 and .abc/2 D t1 t3 . and so D 0 and we get c 2 Finally, we consider bc, where Œbc; b D t3 , Œbc; a D t1 t2 so that (by ()) 1 .hbci/ ht1 t2 ; t3 i: On the other hand, .bc/2

m1

D t3 .

282

Groups of prime power order

We see that hbi normalizes the subgroups hai;

habi;

hci;

haci;

hbci;

habci;

and all elements in the cosets Z.G/a;

Z.G/ab;

Z.G/c;

Z.G/ac;

Z.G/bc;

Z.G/abc

are of order 2m and they are also normalized by hbi. Let b0 be an element of smallest possible order 2s in Z.G/b, where 2 s m 1, s1 ˇ m 3. Then b02 D t1˛ t3 , where neither ˛ nor ˇ are equal to 0. Since a2

m1

D t1 ;

c2

m1

D t3 ;

m1

.ac/2

D t1 t3 ;

there is an element x of order 2s in one of the subgroups ha2 i; s1

so that x 2 Z.G/ and x 2

hc 2 i;

h.ac/2 i

ˇ

D t1˛ t3 . Then we get s1

.xb0 /2

ˇ

ˇ

D t1˛ t3 t1˛ t3 D 1

and so o.xb0 / 2s1 and xb0 2 Z.G/b, a ﬁnal contradiction. Thus our theorem is proved. Theorem 128.2. Let G be a p-group with the subgroup breadth 1 and let p > 2. Then G possesses an abelian maximal subgroup and G 0 (of order p 2 ) is elementary abelian. Proof. By Theorem 122.1(b), jG W Z.G/j p 3 . Due to Theorem 128.1, jG 0 j p 2 . If jG W Z.G/j D p 2 , then G=Z.G/ Š Ep2 and this implies jG 0 j D p. In case jG 0 j D p, Proposition 122.13 yields G=Z.G/ Š Ep2 . In what follows we may assume that jG=Z.G/j D p 3 and jG 0 j D p 2 . Also, we assume that G does not possess any abelian maximal subgroup. Let A be a maximal normal abelian subgroup of G containing ˆ.G/ (which is abelian) so that G=A is elementary abelian of order p 2 . As Z.G/ < A and jG=Z.G/j D p 3 , we have G=A Š Ep2 and jA W Z.G/j D p. For any g 2 G A, g p 2 Z.G/ and so G=Z.G/ is of order p 3 and exponent p. (i) First assume that G=Z.G/ is nonabelian and so G=Z.G/ Š S.p 3 / (the nonabelian group of order p 3 and exponent p). Since G 0 covers A=Z.G/, we infer that G 0 \ Z.G/ Š Cp . For any x 2 A Z.G/, CG .x/ D A which implies o.x/ p 2 . Suppose that there is an element y 2 GA of order p. As CA .y/ D Z.G/, it follows that that CG .y/ must cover G=A, which contradicts the structure of G=Z.G/ Š S.p 3 /. We have proved that 1 .G/ Z.G/. In particular, G 0 D hai Š Cp2 and set ap D z. Let g 2 GA so that o.g/ p 2 and 1 ¤ g p 2 Z.G/. As Œg; a ¤ 1 and Œg; a 2 Z.G/, we have hŒg; ai D hzi, hg; ai0 D hzi and hg; ai is minimal nonabelian. Then replacing g with g i for some i 6 0 .mod p/ (if necessary), we may assume that Œg; a D z.

128

The commutator subgroup of p-groups with the subgroup breadth 1

283

If hgi E G, then A=Z.G/ and .hgiZ.G//=Z.G/ would be two distinct normal subgroups of order p in G=Z.G/ Š S.p 3 /, a contradiction. Hence jG W NG .hgi/j D p. If NA .hgi/ D Z.G/, then NG .hgi/ must cover G=A and NG .hgi/ \ A D Z.G/, contrary to the structure of G=Z.G/. It follows that NG .hgi/ D Ahgi so that a normalizes hgi and so Œg; a D z implies that 1 .hgi/ D hzi. If o.g/ p 3 , then there is an element l 2 hg p i Z.G/ such that o.l/ D p 2 and l p D z 1 . But then .al/p D ap l p D zz 1 D 1 and so o.al/ D p and al 2 A Z.G/, a contradiction. Hence we have o.g/ D p 2 and ha; gi Š Mp3 with ha; gi \ A D hai. On the other hand, 1 .ha; gi/ Š Ep2 and so there are elements of order p in ha; gi A, a contradiction. (ii) Now assume that G=Z.G/ is abelian and so G=Z.G/ Š Ep3 . Hence ˆ.G/ Z.G/ and so for any x; y 2 G we have Œx; yp D Œx p ; y D 1 which implies that G 0 Š Ep2 . If x; y 2 G Z are such that .hx; yiZ.G//=Z.G/ Š Ep2 , then Œx; y ¤ 1 since G has no abelian maximal subgroups. Let a; b 2 G Z.G/ be such that .ha; biZ.G//=Z.G/ Š Ep2 which implies that Œa; b D t1 ¤ 1. Let c be an element in G .ha; biZ.G// so that ha; b; ciZ.G/ D G. Set Œa; c D t2 ¤ 1 and we claim that ht1 ; t2 i D G 0 . Indeed, suppose t2 D Œa; c D t1i for some i 6 0 .mod p/. Then Œa; b i c D Œa; bi Œa; c D t1i t1i D 1 and .ha; b i ciZ.G//=Z.G/ Š Ep2 , a contradiction. j Assume that Œb; c D t1 for some j 6 0 .mod p/. Then we get j j

Œb; aj c D Œb; aj Œb; c D t1 t1 D 1 which together with .hb; aj ciZ.G//=Z.G/ Š Ep2 gives a contradiction. Thus we have 1 ˇ proved that Œb; c D t1˛ t2 , where ˇ 6 0 .mod p/. As haˇ b; b ˛ˇ c; biZ.G/ D G, we have 1 .haˇ b; b ˛ˇ ciZ.G//=Z.G/ Š Ep2 : On the other hand, Œaˇ b; b ˛ˇ

1

c D Œa; bˇ ˛ˇ

1

ˇ

Œa; cˇ Œb; c D t1˛ t2

ˇ

t1˛ t2 D 1;

a contradiction. We have proved that G possesses an abelian maximal subgroup A. Then Z.G/ < A and jA=Z.G/j D p 2 . For each x 2 G A, CA .x/ D Z.G/ and so x p 2 Z.G/. Hence G=Z.G/ is generated by its elements of order p and therefore exp.G=Z.G// D p. In particular, A=Z.G/ Š Ep2 . For a ﬁxed element g 2 G A the map a ! Œa; g

.a 2 A/

284

Groups of prime power order

is a homomorphism from A into A with the kernel CA .g/ D Z.G/ and so A=Z.G/ Š ¹Œa; g j a 2 Aº D G 0 which implies that G 0 Š Ep2 . Our theorem is proved. Theorem 128.3. Let G be a 2-group with the subgroup breadth 1. Then G possesses a normal abelian subgroup of index 4. Proof. Suppose that this result is false. Then we get jG=Z.G/j 24 and so, by Theorem 122.1(a), jG=Z.G/j D 24 . Let A be a maximal normal abelian subgroup of G so that G=A is of order 23 . Since Z.G/ < A and jG=Z.G/j D 24 , we have jG=Aj D 23 and jA=Z.G/j D 2. By Proposition 122.16, the element breadth of G is at most 2. But if a 2 A Z.G/, then CG .a/ D A and so jG W CG .a/j D 23 , a contradiction.

129

On two-generator 2-groups with exactly one maximal subgroup which is not two-generator

By Theorem 71.6, a nonmetacyclic two-generator 2-group G contains an even number of two-generator maximal subgroups. We begin a study of the structure of that G provided it contains exactly two two-generator maximal subgroups (Problem 1969). Theorem 129.1 (Janko). Let G be a two-generator 2-group with exactly one maximal subgroup which is not two-generator. Then the other two maximal subgroups are either both metacyclic (and such groups G are completely determined in 87) or both are nonmetacyclic. Proof. Set 1 D ¹A; B; C º, where d.C / 3. By Schreier’s inequality (Appendix 25), we have d.C / D 3. Since G is nonmetacyclic, G is nonabelian and d.A/ D d.B/ D 2. If both A and B are metacyclic, then such groups G are completely determined in 87. Therefore we may assume that A is nonmetacyclic and so A is nonabelian. In what follows we assume (by way of contradiction) that B is metacyclic. Let H 2 ¹A; B; C º. Then we have ˆ.H / D Ã1 .H / Ã2 .G/. On the other hand, we infer that Ã2 .A/ Ã2 .G/ and Ã2 .A/ E G so that A=Ã2 .A/ is two-generator and nonmetacyclic (Corollary 44.9). Also, B=Ã2 .A/ is metacyclic (but noncyclic) and d.C =Ã2 .A// D 3. Finally, G=Ã2 .G/ is nonmetacyclic (Corollary 44.9) and therefore G=Ã2 .A/ is nonmetacyclic with d.G=Ã2 .A// D 2. It follows that we may assume that Ã2 .A/ D ¹1º. Hence exp.A/ D 4 and so exp.G/ 8. Since B is metacyclic of exponent 8, we get jBj 26 . Suppose that jBj D 26 . Then B \ A is a metacyclic subgroup of exponent 4 and order 25 , a contradiction. Hence B is metacyclic of order 25 and so jGj 26 . If jGj D 24 , then jAj D 23 which together with d.A/ D 2 implies that A is metacyclic, a contradiction. It follows that jGj D 25 or jGj D 26 . (i) First assume that jGj D 25 . If jA0 j D 4, then, by a result of O. Taussky (Lemma 1.6), A is of maximal class and so A is metacyclic, a contradiction. Hence jA0 j D 2 which together with d.A/ D 2 implies that A is the nonmetacyclic minimal nonabelian group of order 24 . Then we get E D 1 .A/ Š E8 and E E G. If G=E Š E4 , then E D ˆ.G/ (since d.G/ D 2) and so E B, a contradiction (since B is metacyclic). Hence G=E Š C4 and let g 2 G A be such that hgi covers G=E. In that case, g 2 2 A E is of order 4 and so o.g/ D 8 and hgi \ E Š C2 . Since CA .E/ D E and

286

Groups of prime power order

G=E Š C4 , it follows that CG .E/ D E and 1 .G/ D E so that G is a faithful and nonsplitting extension of E Š E8 by C4 . As A D Ehg 2 i is a unique maximal subgroup of G containing E, it follows that C does not contain E. Hence C \ E D U Š E4 and C covers G=E so that C =U Š C4 . But 1 .C / D U and so if h 2 C U is such that hhi covers C =U , then o.h/ D 8. It follows that C has a cyclic subgroup of index 2, contrary to d.C / D 3. (ii) It remains to consider the case jGj D 26 . In this case, jˆ.G/j D 24 and ˆ.G/ is a maximal subgroup of A. On the other hand, ˆ.G/ < B and B is metacyclic and so ˆ.G/ is a metacyclic group of exponent 4. Hence we have either ˆ.G/ Š C4 C4 or ˆ.G/ Š H2 D hx; y j x 4 D y 4 D 1; x y D x 1 i. In any case, 1 .ˆ.G// Š E4 and Ã1 .ˆ.G// D 1 .ˆ.G//. Since A does not possess elements of order 8, we have for each x 2 A ˆ.G/, x 2 2 1 .ˆ.G// which gives ˆ.A/ D Ã1 .A/ D 1 .ˆ.G//, contrary to d.A/ D 2. Our theorem is proved.

130

Soft subgroups of p-groups

We present here some work of L. Hethelyi on soft subgroups of p-groups. A subgroup A of a nonabelian p-group G is called a soft subgroup if it satisﬁes CG .A/ D A and jNG .A/ W Aj D p. Note that a nonabelian p-group P is of maximal class if and only if P possesses a soft subgroup of order p 2 . Our induction arguments will be based on the following: Lemma 130.1. Let A be a soft subgroup in a nonabelian p-group G, N D NG .A/, N and let K be the core of A in G, GN D G=K. Then NN is soft in GN and NN D AN Z.G/. 0 Moreover, jN j D p if jG W Aj > p. N AZ. N G/. N Now we see Proof. We may assume that G > N . Obviously, NN D NGN .A/ N is normal in G, N hence trivial. Then jNN W Aj N D p implies the second asthat AN \ Z.G/ N We infer that C N .NN / C N .A/ N N N .A/ N D NN , i.e., NN is sertion: NN D AN Z.G/. G G G N N self-centralizing. Let us denote N2 D NG .N /, so NGN .N / D N2 . For x 2 N2 N , Ax ¤ A is a maximal subgroup in N , hence Ax \ A D Z.N /, jN W Z.N /j D p 2 , jN 0 j D p. Counting the conjugate subgroups Ax , x 2 N2 , we obtain jN2 W N j D p N (noting that N has at most p C 1 abelian maximal subgroups), thus NN is soft in G. Lemma 130.2. Suppose that A is a soft subgroup in a nonabelian p-group G and set jG W Aj D p n , where n 1. Then the subgroups of G containing A form a chain A D N0 < N1 < < Nn1 D M < Nn D G; where NG .Ni1 / D Ni and jNi W Ni1 j D p, i D 1; : : : ; n. Proof (see also Appendix 40, Exercise 72). By induction, Lemma 130.1 yields at once jNi W Ni 1 j D p. Now if A X G, then let Nj be the largest subgroup of our chain contained in X . Suppose Nj < X. Since Nj C1 D NG .Nj / D NX .Nj /, we have Nj C1 X , a contradiction. Hence Nj D X. In what follows we shall use the notation introduced in Lemma 130.2. Moreover, for X Y , core.X W Y / denotes the largest normal subgroup of Y contained in X and X Y stands for the smallest normal subgroup of Y containing X. Corollary 130.3. For any v 2 G M we have hA; Av i D M , where M is the unique maximal subgroup of G containing a soft subgroup A of G.

288

Groups of prime power order

Proof. By Lemma 130.2, we have hA; Av i D Nj for some j n 1. Since A Nj 1 and A Njv , it follows that v normalizes Nj . But from NG .Nj / D Nj C1 we infer v 2 Nj C1 and so Nj D M . Corollary 130.4. There exists a 2 A such that CG .a/ D A, where A is a soft subgroup of G. T Proof. As A D CG .A/ D a2A CG .a/, the assertion follows from Lemma 130.2. Proposition 130.5. Let M be the unique maximal subgroup of a p-group G containing a soft subgroup A of G. For the upper central series of M we have: (i) Zi .M / D core.Ni 1 W G/ D Ni 1 \ Niv1 for any v 2 G M , i D 1; : : : ; n. (ii) Zi C1 .M /=Zi .M / is elementary abelian of order p 2 , i D 1; : : : ; n 1. (iii) Zi .G/ Zi .M /, i D 1; : : : ; n. Proof. (i) For i D 1 we have N0 D A and let v 2 G M be ﬁxed. By Corollary 130.3, hA; Av i D M . Now, Z.M / D CM .A/ \ CM .Av / D A \ Av . Finally, Lemma 130.1 proves (i) by induction. v (ii) We have Zi .M / D Ni 1 \ Ni1 Ni 1 \ Niv Ni \ Niv D Zi C1 .M /. The index in both steps is 1 or p, so jZi C1 .M /=Zi .M /j p 2 : If the index is p 2 , inserting also Ni \ Niv1 in the middle, we see that the factor is elementary abelian. Indeed, .Ni 1 \ Niv /=Zi .M / and .Ni \ Niv1 /=Zi .M / are two distinct subgroups of order p in Zi C1 .M /=Zi .M /. (iii) For i D 1 we have Z1 .G/ core.A W G/ D Z1 .M /. Lemma 130.1 is again applicable and gives our result. Now, as a corollary to Proposition 130.5 (i) and (iii), we obtain: Theorem 130.6. Let A be a soft subgroup of a p-group G with jG W Aj D p n , where n 1. Then there is a unique maximal subgroup M of G containing A. The nilpotence class of M is n and the nilpotence class of G is at least nC1. Since A is also soft in Ni , i D 1; : : : ; n 1, every above statement is also true for Ni . Corollary 130.7. The nilpotence class of Ni is i C 1 for i D 0; : : : ; n 1. Corollary 130.8. We have jA W core.A W G/j p n1 . Proof. Making use of Proposition 130.5 (i) and (ii), we obtain jA W core.A W G/j D p nC1 jM W core.A W G/j D p nC1 jZn .M / W Z1 .M /j p nC1 p 2.n1/ D p n1 :

130

Soft subgroups of p-groups

289

Proposition 130.9. Let M be the unique maximal subgroup of a p-group G containing a soft subgroup A of G. For the lower central series of M we have: 0 0 0v W G/ D hNnC1i ; NnC1i i for any v 2 G M , where (i) Ki .M / D n:cl:.NnC1i i D 2; : : : ; n;

(ii) Ki .M /=Ki C1 .M / is elementary abelian of order p 2 , i D 2; : : : ; n. Proof. (i) Let i D n and we may assume that n > 1. Since Kn .M / is normal in G, Kn .M / Z1 .M / A and cl.M=Kn .M // D n1, Lemma 130.1 implies that the factor group N1 =Kn .M / is soft in G=Kn .M /. Indeed, if N1 =Kn .M / is nonabelian, then AKn .M / would be soft in G=Kn .M / and then cl.M=Kn .M // D n. Hence we obtain that N10 Kn .M /. Conversely, for T D hN10 ; N10v i we have Z2 .M /=T D .N1 \ N1v /=T Z.hN1 ; N1v i=T / D Z.M=T /; hence cl.M=T / < cl.M /, so T Kn .M / also holds and therefore Kn .M / D hN10 ; N10v i D n:cl:.N10 W G/: Now, for i < n, (i) follows by backwards induction. (ii) Suppose that i D n. Since we may assume that n > 1, Lemma 130.1 implies jKn .M /j D jhN10 ; N10v ij jN10 j2 p 2 : The result then follows by induction. Proposition 130.10. Suppose that M is the unique maximal subgroup of a p-group G containing a soft subgroup A of G. For the series of commutator subgroups Ni0 we have Ni0 A D Ni1 , i D 1; : : : ; n and so N10 < N20 < < Nn0 D G 0 and so, in particular, G 0A D M . Proof. Clearly, we have Ni0 A Ni 1 and Ni0 A E Ni . Now Lemma 130.2 yields that Ni0 A D Ni 1 . Theorem 130.11. Suppose that A and B are soft subgroups of a p-group G contained in the same maximal subgroup M of G. Then B is conjugate to A. Proof. Suppose that jG W Aj D p 2 . Then jG=Z.M /j D p 3 , A=Z.M / 6 Z.G=Z.M //, B=Z.M / 6 Z.G=Z.M //, and jA=Z.M /j D jB=Z.M /j D p. However, we obtain that jM=Z.M /j D p 2 and so A=Z.M / is conjugate to B=Z.M /. Thus, for jG W Aj D p 2 , A is conjugate to B. Suppose that jG W Aj > p 2 . Set N1 D NG .A/ and GN D G=Z.M / with bar conN D NG .B/ are soft subgroups of GN contained in MN (by vention. Then NN1 and NGN .B/ N is conjugate to NN1 and Proposition 130.5(i) and Lemma 130.1). By induction, NGN .B/ so NG .B/ is conjugate to N1 . Thus we can assume B < N1 . However, B is conjugate to A in N2 D NG .N1 / by the previous paragraph.

290

Groups of prime power order

Theorem 130.12. Suppose that A is a maximal normal abelian subgroup of a nonabelian p-group G. Assume that G=A is cyclic and that G possesses a soft subgroup B distinct from A. Then G D AB. Proof. Suppose that this theorem is false. Since A ¤ B, neither A nor B is of index p in G. As AB < G and AB E G, we get jG W .AB/j D p and so M D AB is the unique maximal subgroup of G containing B. We have A\B D Z.M / E G. Now set N1 D NG .B/ so that jN1 W Bj D p and N1 M . Let x 2 G N1 be such that x normalizes N1 . Then B\B x D Z.N1 / Z.M / and jB W Z.N1 /j D p. Set R D AZ.N1 /; then R E G and G=R Š Cp2 . Since CG .B/ D B, we have for each b 2 B Z.N1 /, CG .b/ D B. Let jBj D p k and jM j D p n so that jB Z.N1 /j D p k p k1 D p k1 .p 1/ and jM Rj D p n p n1 D p n1 .p 1/: Note that all conjugates in G of the subset B Z.N1 / lie in M R and there are exactly jG W NG .B/j D jG W N1 j D p nC1 =p kC1 D p nk such distinct conjugates. For any g 2 G N1 we have .B Z.N1 // \ .B Z.N1 //g D ¿ since for each b 2 B Z.N1 /, CG .b/ D B. Hence the number of elements contained in the union of all these conjugates is p nk p k1 .p 1/ D p n1 .p 1/, which is the number of all elements contained in M R. It follows that each element y 2 M R is conjugate in G to an element b 2 B Z.N1 / and so CG .y/ M . On the other hand, G=R Š Cp2 and so if v 2 G M , then v p 2 M R but CG .v p / 6 M , a contradiction. Our theorem is proved. Theorem 130.13. Suppose that A is a maximal normal abelian subgroup of a nonabelian p-group G. Assume that G=A is cyclic and that G possesses a soft subgroup B distinct from A. Then we have: (i) d.G=Z.G// D 2, and if jG W Aj D p ˛ , then G=.G 0 Z.G// is of type .p ˛ ; p/. (ii) G is a CF-group, i.e., the index of any term of the lower central series of G beyond G 0 in its predecessor is at most p. Proof. (i) By Theorem 130.12, we have G D AB and so Z.G/ D A \ B. Let M be the unique maximal subgroup of G containing B. We have G 0 A \ M . On the other hand, G 0 B is a normal subgroup of G containing B and so G 0 B D M which implies A \ M D G 0 Z.G/. Since G=.G 0 Z.G// D A=.G 0 Z.G// M=.G 0 Z.G//; we infer that G=.G 0 Z.G// is of type .p ˛ ; p/. Take an element b 2 B Z.G/ so that hbi covers B=Z.G/ and take an element a 2 A M . Then hbi covers G=A and G D ha; bi.A \ M / D ha; biG 0 Z.G/ D ha; biZ.G/ which implies d.G=Z.G// D 2.

130

Soft subgroups of p-groups

291

(ii) Since ap 2 G 0 Z.G/, we have Œap ; b 2 K3 .G/. Now Œap ; b D ap .b 1 ap b/ D ap .ap /b D .a1 ab /p D Œa; bp ; so Œa; bp 2 K3 .G/. Since G 0 D hŒa; b; K3 .G/i, jG 0 W K3 .G/j D p. It follows by an easy induction that jKi .G/ W KiC1 .G/j D p for Ki .G/ ¤ ¹1º: in fact, if i 3 and Ki 1 .G/ D hu; Ki .G/i with up 2 Ki .G/, then Ki .G/ D hŒa; u; Œb; u; Ki C1 .G/i and Œa; u D 1 as u 2 G 0 A and a 2 A. Also, we obtain that Œb; up 2 Ki C1 .G/ since Œb; up D .ub /p up D Œb; up 2 Ki C1 .G/. Indeed, Œb; up D .b 1 up b/up D .b 1 u1 b/p up D ..u1 /b u/p D Œb; up since u 2 G 0 ; .u1 /b 2 G 0 and G 0 A. Thus G is a CF-group. Theorem 130.14. Suppose that G is a p-group of class greater than 2 and A is an abelian subgroup of G of index p. Then the following are equivalent: (a) Every maximal abelian subgroup of G is soft. (b) G has a soft subgroup distinct from A. (c) G=Z.G/ is a p-group of maximal class. (d) jZ2 .G/ W Z.G/j D p. Proof. The implication (a) ) (b) is trivial. (b) ) (c). By Theorem 130.13, G is a CF-group and G=.G 0 Z.G// is of type .p; p/. N GN 0 is of type .p; p/ and GN is a CF-group, so GN is of maximal If GN D G=Z.G/, then G= class. The implication (c) ) (d) is trivial. (d) ) (a). Let B ¤ A be an abelian maximal subgroup of G so that AB D G and A \ B D Z.G/. Set N D NG .B/ so that N D .N \ A/B. We have ŒN \ A; B B \ A D Z.G/ so N \ A Z2 .G/. Hence jN W Bj D j.N \ A/ W Z.G/j D p and B is soft. Theorem 130.15. Suppose that G is a nonabelian p-group which possesses a maximal normal abelian subgroup A such that G=A is cyclic. If jZ.G/ \ G 0 j D p, then G has a soft subgroup distinct from A. Proof. Note that G D Ahbi for some element b 2 G A so that CA .b/ D Z.G/ and CG .b/ D Z.G/hbi D B. We prove that B is soft in G. First of all, B is abelian and self-centralizing and set N D NG .B/. Further, we have jN W Bj D j.N \ A/ W Z.G/j and j.N \ A/ W Z.G/j D jŒN \ A; hbij D p by Lemma 121.2.

131

p-groups with a 2-uniserial subgroup of order p

Let us say that a subgroup H of a p-group G is n-uniserial (n 1) if for each index i D 1; 2; : : : ; n there is a unique subgroup Ki such that H Ki and jKi W H j D p i . In case the subgroups of G containing H form a chain, we say that H is uniserially embedded in G. We shall classify here all p-groups G which possess a subgroup of order p which is 2-uniserial and so extend some results for p > 2 of N. Blackburn and L. Hethelyi to the case of 2-groups. Theorem 131.1 (Janko). Let G be a p-group and let L be a 2-uniserial subgroup of order p. Then L is uniserially embedded in G (and so G is one of the groups in Exercise A40.77 of Appendix 40) or p D 2 and m

G D ha; t; x j a2 D t 2 D Œa; t D 1; u D a2 vDa

2m2

m1

;

; x 2 D tv; ax D a1 ; t x D tui;

where m 3;

jGj D 2mC2 ;

Z.G/ D htvi Š C4 ;

ˆ.G/ D hti ha2 i Š C2 C2m1 ;

G 0 D ha2 i Š C2m1

and here L D hti Š C2 is 2-uniserial but not 3-uniserial in G (and so ht i is not uniserially embedded in G). Proof. Let L be a 2-uniserial subgroup of order p in G. (i) First assume p > 2. Set N D NG .L/ D CG .L/. If jN=Lj D p, then N is a self-centralizing abelian subgroup of order p 2 and jNG .N / W N j D p. This means that N is a soft subgroup in G and so Lemma 130.2 implies that all subgroups of G containing N form a chain. Hence in this case L is uniserially embedded in G. We may assume jN=Lj > p. Since there is only one subgroup of order p in N=L, it follows that N=L is cyclic. If N D G, then L is uniserially embedded in G and we are done. Thus we may assume N < G. In this case, L is not characteristic in N and so N is not cyclic. Set R D 1 .N / Š Ep2 and N1 D NG .R/ so that N < N1 since R is characteristic in N . As there is only one subgroup of order p in N1 =R, we see that N1 =R is cyclic. Since R < N , we infer that R D 1 .N1 / is characteristic in N1 , and we conclude

131 p-groups with a 2-uniserial subgroup of order p

293

that N1 D G. In that case, 1 .G/ D R Š Ep2 and G possesses a cyclic subgroup of index p which does not contain L. Thus L is uniserially embedded in G and we are done in case p > 2. (ii) Assume now p D 2 and set L D hti. Since the involution t is contained in exactly one subgroup of order 4 in G, Theorem A.14.1 implies that we have one of the following possibilities: (a) G is cyclic. (b) CG .t/ D ht i C2m , m 1. (c) CG .t/ D ht i Q2m , m 3. (d) G D ha; bi with deﬁning relations m

a2 D b 4 D 1; t 2 D 1;

m 2;

u D a2

ab D a1 t ;

m1

;

b 2 D ut;

D 0; 1:

Here Z.G/ D ht; ui Š E4 , G=hti Š Q2mC1 , G 0 D ha2 t i, and G is metacyclic if and only if D 0. In case (a), ht i is uniserially embedded in G. In cases (c) and (d), CG .t/=ht i is a generalized quaternion group and so there are two distinct cyclic subgroups S=hti and T =hti of order 4 contained in CG .t/=ht i. But then S and T are two distinct subgroups of order 8 which contain hti and so ht i is not 2-uniserial in G, a contradiction. It remains to consider the above case (b), where CG .t/ D hti C with C D hai and o.a/ D 2m , m 1. If CG .t/ D G, then hti is uniserially embedded in G. From now on we assume CG .t / < G and so G is nonabelian. We may apply Theorem 48.1 in which such groups G are classiﬁed. In cases (a) and (b) of that theorem, G is either of maximal class or G Š M2n , n 4, and in these cases hti is uniserially embedded in G. In cases (d–h) of that theorem, we have m 2 and t is contained in a subgroup D Š D8 . On the other hand, t is also contained in ht i 2 .hai/ which is an abelian subgroup of type .2; 4/. Hence in all these cases ht i is not 2-uniserial in G, a contradiction. We have proved that we must be in case (c) of Theorem 48.1. From here we know that T D CG .t/ D hti hai is of index 2 in G, o.a/ D 2m with m 3 and t 2 ˆ.G/ so that ˆ.G/ D ht i ha2 i, where ha2 i D ˆ.T /. Hence there is an element m1 x 2 G T such that x 2 2 ht; a2 i ha2 i. Set u D a2 and 2 .hai/ D hvi Š C4 2 so that v D u, u 2 Z.G/ and 1 .T / D ht; ui Š E4 . Also, we have t x D tu and so x 2 62 ht; ui and ˆ.G/ D ht i ha2 i D hx 2 ; a2 i. Obviously, x does not centralize a2 . Indeed, if Œx; a2 D 1, then x centralizes ˆ.G/ D hx 2 ; a2 i and so x would centralize t, a contradiction. Thus we have proved that x induces on T an involutory automorphism such that x t D tu and x does not centralize ha2 i. In Theorem 34.8(d) the structure of Aut.T / is determined. In particular, we get ax D a1 u , where D 0; 1. If D 1, then we

294

Groups of prime power order

replace a with a0 D at and obtain .a0 /x D .at/x D ax t x D .a1 u/ .tu/ D a1 t D .at/1 D .a0 /1 : Hence writing again a instead of a0 , we may assume from the start that ax D a1 . We see that CT .x/ D ht vi D Z.G/ Š C4 since .tv/2 D u. Also, ha2 i Z.G/ and o.x 2 / 4 and so x 2 D .tv/˙1 D t v ˙1 . Replacing v with v 1 if necessary, we may assume that x 2 D t v and so the structure of G is uniquely determined. Let xt r as (r; s are integers) be any element in G T . Then we compute .xt r as /2 D xt r as xt r as D x 2 .t r as /x t r as D t v .tu/r as t r as D tvur D .tv/˙1 ; and see that all elements in G T are of order 8. It follows that ht ihvi is the only subgroup of order 8 containing ht i which shows that ht i is 2-uniserial in G. But G is not m3 3-uniserial since ht i ha2 i and ht; xi Š M24 are two distinct subgroups of or4 der 2 containing ht i, and the theorem is proved.

132

On centralizers of elements in p-groups

We present here some results of N. Blackburn [Bla12] about the centralizers of elements in p-groups. Suppose that K is a p-group and s is an element of order p in the center of K. We are concerned with the question: What are the p-groups G for which K G and K D CG .s/? In particular, is there a bound on the order or class of G? In fact, we show by Theorem 132.1 that these two last questions are the same. Theorem 132.1. Let G be a p-group of class k. Suppose that s 2 G and jCG .s/j D p r . Then jGj p rk . Proof. Let G D K1 .G/ > K2 .G/ > > Kk .G/ > KkC1 .G/ D ¹1º be the lower central series of G. For 1 i k, set H D hs; Ki .G/i. Every conjugate of s in H is of the form s h D sŒs; h (h 2 H ) so the number of conjugates of s in H is at most jH 0 j. Thus jH 0 j jH W CH .s/j jH j=p r jKi .G/j=p r : But we see that Ki .G/=KiC1 .G/ Z.G=Ki C1 .G// and H=Ki .G/ is cyclic and so H=Ki C1 .G/ is abelian. Then we get H 0 Ki C1 .G/ and so, by the above inequality, jKi C1 .G/j jKi .G/j=p r which implies jKi .G/ W Ki C1 .G/j p r for i D 1; : : : ; k. Hence jGj p rk . If s is an element of order p in a p-group G, then set C D CG .s/. Then, as a rule, we have inﬁnitely many p-groups G with such a ﬁxed centralizer C . For example, if C Š Ep2 , then there are inﬁnitely many p-groups of maximal class which all have such a small centralizer of s. However, there are examples for the structure of C which is small in a rather strong sense, where there exist only ﬁnitely many p-groups G with such a ﬁxed centralizer C . This will be seen in the next two theorems. Theorem 132.2. Let s be an element of order p in a p-group G and set C D CG .s/. m1 If there exists a positive integer m such that s D x p for some x 2 C and if C has mC1 no elementary abelian subgroup of order p , then jG W C j < p m . Proof. If this theorem is false, then there is a subgroup H of G such that H C and jH W C j D p m . Thus s has p m conjugates s1 ; s2 ; : : : ; spm in H . If si D s ui (ui 2 H ),

296

Groups of prime power order p m1

put xi D x ui ; thus x1 ; : : : ; xpm are distinct conjugates of x in H since xi D si . If M is a maximal subgroup of H containing C , then xi 2 M as x 2 M and M is normal in H . Since jM W C j D p m1 , it follows that p m1

si D xi

2 C;

i D 1; : : : ; p m :

Indeed, if si 62 C , then C \ hxi i D ¹1º and so, by the product formula, we obtain jM j jC jjhxi ij D jC jp m , a contradiction (because jH W M j D p and jH j D jC jp m ). But then si ; sj commute for any i; j 2 ¹1; 2; : : : ; p m º. Hence hs1 ; : : : ; spm i is elementary abelian of order p mC1 , a contradiction. Theorem 132.3. Let s be an element of order p in a p-group G and set C D CG .s/. Suppose that s is the p-th power of some element x 2 C and jC j D p l , where l 3 and 2l p C 4. Then jG W C j < p l2 . Proof. If l D 3, then, by the structure of groups of order p 3 , hsi D 1 .ˆ.C // and so hsi is characteristic in C . But then NG .C / centralizes hsi which implies G D C and we are done. Now we assume l > 3 so that p is odd and 2l p C 3. If our assertion is false, there exists a subgroup H of G such that H C and jH W C j D p l2 . Let M be a maximal subgroup of H such that M C . Thus jM j D p 2l3 p p and so M is regular (Theorem 7.1(b)). It follows that the elements of M of order at most p form a characteristic subgroup E of M (Theorem 7.2(b)). We have M E H and s 2 E. As jH W C j D p l2 , s has p l2 conjugates s1 ; : : : ; spl2 in H . If si D s ui (ui 2 H ), p p put xi D x ui . Since x 2 C , we have xi 2 M . If i ¤ j , we get xi ¤ xj . By Theo1 l2 rem 7.2(a), we infer that xi xj 62 E. Thus jM W Ej > p and so jM W Ej p l1 . Similarly, jEj > p l2 since si 2 E. Hence jEj p l1 which implies jM j p 2l2 , a contradiction. Our theorem is proved. In his paper [Bla12], N. Blackburn has studied 2-groups G which possess an involution s such that CG .s/ is the nonmetacyclic minimal nonabelian group of order 16. We shall describe these groups G more closely by using the results from 51 and 130. Theorem 132.4. Suppose that G is a 2-group of order 25 which possesses an involution s such that C D CG .s/ is the nonmetacyclic minimal nonabelian group of order 24 . Then E D 1 .C / is a self-centralizing elementary abelian group of order 8. Also, E D ˆ.G/ and so G is isomorphic to a unique group of order 25 given in Theorem 51.3. Proof. We may set C D CG .s/ D hr; t j r 4 D t 2 D 1; Œr; t D z; z 2 D Œz; r D Œz; t D 1i; where

hzi D C 0 ;

Z.C / D ˆ.C / D hr 2 ; zi

and E D 1 .C / D hr 2 ; z; t i Š E8

with CG .E/ D E:

132

On centralizers of elements in p-groups

297

Since G centralizes hzi D C 0 , it follows that s 2 ¹r 2 ; r 2 zº and G (normalizing Z.C /) fuses r 2 and r 2 z. We may set s D r 2 and for any v 2 G C , s v D sz. Note that for any t 0 2 E Z.C / we have .rt 0 /2 D r 2 .t 0 /2 Œt 0 ; r D sz and so we may put r v D rt 0 with some t 0 2 E Z.C /. It follows that t 0 2 G 0 and so E ˆ.G/ gives E D ˆ.G/. Then Theorem 51.3 applies and such a group G is uniquely determined. Theorem 132.5. Let G be a 2-group of order > 25 which possesses an involution s such that C D CG .s/ is the nonmetacyclic minimal nonabelian group of order 24 . Then we have jGj D 2nC3 , n 3, jG 0 j D 2n , Z.G/ D hzi D C 0 is of order 2, G is of class n C 1 (coclass 2) and G is a CF-group (i.e., the index of any term of the lower central series of G beyond G 0 D K2 .G/ in its predecessor is at most 2). Moreover, Z.C / D hz; si Š E4 , G 0 \C D W ¤ Z.C / is a normal four-subgroup of G, K3 .G/ is abelian of order 2n1 and rank 2, s inverts K3 .G/, L D hsiK3 .G/ is normal in G, G=L Š D8 , s 62 G 0 , ˆ.G/ D hsiG 0 , and so d.G/ D 2. Finally, E D 1 .C / is a selfcentralizing elementary abelian subgroup of order 8, E < ˆ.G/, and we have one of the following possibilities: (a) n D 3, E is normal in G and G is isomorphic to one of the four distinct groups of order 26 given in Theorem 51.4. (b) n > 3, E is not normal in G, G has no normal elementary abelian subgroup of order 8, W is the unique normal four-subgroup in G, and G is one of the groups described in (c) and (d) of Theorem 51.6. Proof. We may set C D CG .s/ D hr; t j r 4 D t 2 D 1; Œr; t D z; z 2 D Œz; r D Œz; t D 1i; where hzi D C 0 ;

Z.C / D ˆ.C / D hr 2 ; zi

and E D 1 .C / D hr 2 ; z; t i Š E8

with CG .E/ D E:

C 0 , it follows that s

Since NG .C / > C centralizes hzi D 2 ¹r 2 ; r 2 zº and NG .C / (normalizing Z.C /) fuses r 2 and r 2 z. This gives jNG .C / W C j D 2 and we may assume s D r 2 (noting that .rt/2 D r 2 t 2 Œt; r D r 2 z). Also, Z.G/ Z.C / and so Z.G/ D hzi is of order 2. Set A D hr; zi D hrihzi Š C4 C2 so that CG .r/ C and therefore CG .r/ D A, which implies that A is a self-centralizing abelian subgroup of type .4; 2/. As NG .A/ centralizes hsi D Ã1 .A/, it follows that NG .A/ D C which together with jC W Aj D 2 gives that A is a soft subgroup in G. We may apply all results from 130 about soft subgroups. Set jG W Aj D 2n so that jGj D 2nC3 with n 3. By Lemma 130.2, all subgroups of G containing A form a chain. As CG .r/ D A, the conjugacy class of r has 2n elements and so jG 0 j 2n . However, if jG 0 j > 2n , then jG 0 j D 2nC1 and so jG=G 0 j D 4.

298

Groups of prime power order

By a result of O. Taussky (Lemma 1.6), G would be of maximal class, a contradiction (since G has an abelian subgroup A of type .4; 2/). We have proved that jG 0 j D 2n . Let M be a unique maximal subgroup of G containing the soft subgroup A. By Theorem 130.6, the nilpotence class of M is equal to n and the class of G is at least n C 1. However, if the class of G is greater than n C 1, then the class of G is n C 2 and so G would be of maximal class, a contradiction. We have proved that G is of class n C 1. But jG=G 0 j D 23 and so the index of any term of the lower central series of G beyond G 0 in its predecessor is at most 2 which implies that G is a CF-group. By Proposition 130.10, we have G 0 A D M . On the other hand, we have hzi D C 0 and so hzi G 0 \ A. But jM j D 2nC2 and so, by the product formula, G 0 \ A D hzi. In particular, s D r 2 62 G 0 and so hsiG 0 D ˆ.G/, jG W ˆ.G/j D 4, and d.G/ D 2. The subgroup A is also a soft subgroup in M and let K ( C ) be a unique maximal subgroup of M containing A. By Proposition 130.10 (applied to M ), K D AM 0 with A \ M 0 D hzi, M 0 is normal in G, and jG 0 W M 0 j D 2. On the other hand, we have K3 .G/ D ŒG; G 0 M 0 and so the fact that G is a CF-group gives K3 .G/ D M 0 . Since s D r 2 2 ˆ.M / and M 0 ˆ.M /, we get ˆ.M / D hsiM 0 ;

where jˆ.M / W M 0 j D 2:

Note that ˆ.M / is normal in G and so ˆ.M /M 0 is a normal subset of G with exactly jM 0 j D 2n1 elements. Hence the involution s and all its 2n1 D jG W C j conjugates must be contained in ˆ.M / M 0 . It follows that all elements in sM 0 D sK3 .G/ are involutions. This implies that K3 .G/ is abelian. Also, L D hsiK3 .G/ D ˆ.M / and so L is normal in the group G. We have jLj D 2n so that G=L is nonabelian of order 8 since G 0 6 K > L. But M=ˆ.M / D M=L Š E4 and so G=L Š D8 . As AM 0 D K with A\M 0 D hzi and A < C K with jC W Aj D 2, it follows that C \ M 0 D W is of order 4. We know that s inverts W . If W Š C4 , then W hsi Š D8 , contrary to the fact that C is minimal nonabelian. Hence we infer that W Š E4 and so E D hsi W D 1 .C / Š E8 . Since CM 0 .s/ D W and s inverts M 0 , we obtain W D 1 .M 0 /. It follows that M 0 is abelian of rank 2 and W is a normal four-subgroup in G with W ¤ Z.C / D hz; si since s 62 G 0 . Also, s 2 ˆ.G/ and W M 0 ˆ.G/ gives E D hsi W D 1 .C / ˆ.G/. But G 0 > M 0 and so G 0 6 E and therefore E < ˆ.G/. Also, E is self-centralizing in G and so we are in a position to apply the results from 51. If n D 3, then M 0 D K3 .G/ is of order 4 and so K3 .G/ D W . All four conjugates of s in G lie in sW and so E D hsi W is normal in G. Theorem 51.4 is applicable and we get exactly four groups of order 64. If n > 3, then M 0 D K3 .G/ is of order > 4 and so all 2n1 conjugates of s in G (forming the set sK3 .G/) do not lie in E. This gives that E is not normal in G. In this case, Theorem 51.6 is applicable. Our group G has no normal elementary abelian subgroup of order 8 and W is the unique normal four-subgroup in G. The group G is then described in parts (c) and (d) of Theorem 51.6.

132

On centralizers of elements in p-groups

299

Exercise. Suppose that x is a noncentral element of a p-group G, o.x/ D p, such that C D CG .x/ is minimal nonabelian such that C =1 .G/ is nonidentity cyclic. Prove that (a) 1 .C / Š Ep3 . (b) jZ.G/j D p. (c) If x 2 R < 1 .C / is of order p 2 , then we have jNG .R/ W C j D p. In particular, jNG .C / W C j D p so that NG .C / D NG .R/. Study the structure of NG .C / in detail.

133

Class and breadth of a p-group

Let G be a p-group and let x be an element of G. We recall that the breadth of x is the nonnegative integer b.x/ such that p b.x/ D jG W CG .x/j: The breadth of the group G, denoted by b.G/ or just b, is the maximum of the breadths of its elements. Thus p b is the size of the largest of the conjugacy classes of G. It was shown by C. R. Leedham-Green, P. M. Neumann, and J. Wiegold in [LGNW] that there is a close relationship between the breadth b of a p-group G and its nilpotence class c D cl.G/. In particular, we present here a (technically somewhat simpler) proof of their result: Theorem 133.1. Let G be a p-group of the breadth b and class c. Then c<

p b C 1: p1

Since p=.p 1/ D 1 C 1=.p 1/ 2, we get a bound c < 2b C 1 or c 2b which is independent of p. A slightly stronger result was proved by M. Cartwright (Bull. London Math. Soc. 19 (1987), 425–430) in case of 2-groups, where he has shown that c 53 b C 1. The Class-Breadth Conjecture. For any p-group G we have c b C 1. This bound is attained by any group G of order p n (n 3) and maximal class n 1 since such a group always possesses an element x such that jCG .x/j D p 2 and so we have b D b.G/ D b.x/ D n 2. However, the class-breadth conjecture was shown in [ENOB] to be false at least for p D 2. But the conjecture remains open for p-groups of odd order. The class-breadth conjecture has been established under various conditions. For example, if b < p, then c bC1 (Theorem 133.5) and if G is metabelian, then c bC1 (Theorem 133.6). In order to prove Theorem 133.1, we have to prove ﬁrst the following two technical lemmas. If G D K1 .G/ > K2 .G/ > > Kc .G/ > KcC1 .G/ D ¹1º

133

Class and breadth of a p-group

301

is the lower central series of a p-group G of class c, then we deﬁne the two-step centralizer Ci (i D 1; 2; : : : ; c 1) by Ci D CG .Ki .G/=Ki C2 .G// D ¹g 2 G j Œg; x 2 Ki C2 .G/ for all x 2 Ki .G/º: It is clear that Ci is normal in G, Ci Ki C2 .G/ and Ci is a proper subgroup of G for all i D 1; 2; : : : ; c 1. Lemma 133.2. Let G be a p-group and x 2 G. Then b.x/ s.x/, where s.x/ denotes the number of indices i (i D 1; 2; : : : ; c 1) for which x 62 Ci . Proof. We use induction on the length c of the lower central series of G. The lemma is certainly correct for p-groups with the lower central series of length 1 (in that case, G is abelian), for then b.x/ and s.x/ are both 0 whatever x may be. Let us suppose that the lemma is known to be right for p-groups with the lower central series of length c 1. We set S.x/ D ¹i j 1 i c 1 and x 62 Ci º so that s.x/ D jS.x/j. Put GN D G=Kc .G/ and embellish other symbols with a bar to N or simply to denote the cordenote images under canonic epimorphism of G onto G, N responding notion in G. Notice that if 1 i c 2, then Ci D CGN .Ki .G/=.Ki C2 .G// D Ci =Kc .G/ so that our convention is unambiguous; and if 1 i c 2, then xN 2 Ci if and only N x/ if x 2 Ci . An instance for our inductive hypothesis is that b. N sN .x/. N There are two cases to consider. Suppose ﬁrst that x 2 Cc1 which means that x centralizes Kc1 .G/. Then we have N x/, N x/ S.x/ D S. N so that s.x/ D sN .x/; N and since certainly b.x/ b. N the inductive hypothesis gives N x/ b.x/ b. N sN .x/ N D s.x/; as required. The other possibility is x 62 Cc1 , i.e., x does not centralize Kc1 .G/. In this case, N x/[¹c1º, N S.x/ D S. N and so s.x/ D sN .x/C1. N On the other hand, as Kc1 .G/ Z.G/ and therefore Kc1 .G/ is centralized by x, N we have (noting that Kc1 .G/ CGN .x/ N and CG .x/ CGN .x/) N N jGN W .CG .x/Kc1 .G//j D jG W .CG .x/Kc1 .G//j < jG W CG .x/j; jGN W CGN .x/j the last inequality is strict because Kc1 .G/ 6 CG .x/ (by our assumption). This gives N x/ N x/ b. N < b.x/ and so b. N b.x/ 1, and therefore by the above, N x/ b.x/ b. N C 1 sN .x/ N C 1 D s.x/: This case also yields the desired inequality and induction completes the proof.

302

Groups of prime power order

Lemma 133.3. Let G be a p-group of class c 2. Then there exists an element g 2 G which lies in t.g/ two-step centralizers Ci (1 i c 1) and t.g/ < p1 .c 1/. Proof.PFor x 2 G, let ti .x/ D 1 if x 2 Ci and ti .x/ D 0 if x 62 Ci , 1 i c 1. Thus c1 i D1 ti .x/ D t .x/ is the number of two-step centralizers Ci in which x lies. We estimate the average of ti .x/ over G: 1 1 1 X jCi j ; ti .x/ D jGj jGj p x2G

the latter inequality arises from the fact that Ci is a proper subgroup of G and hence has index at least p. It follows that c1 1 XX 1 X c1 : ti .x/ D t.x/ jGj jGj p i D1 x2G

x2G

In this inequality the summands t .x/ are nonnegative, and one of them, t.1/ D c 1, c1 exceeds c1 p . Therefore at least one summand t.g/ for some g 2 G is less than p , and this is what the lemma states. Proof of Theorem 133.1. We combine both Lemmas 133.2 and 133.3 together with the notation introduced there. Since s.g/ C t .g/ D c 1, we get b b.g/ s.g/ D .c 1/ t.g/ > .c 1/ 1 p1 : D .c 1/ 1 D .c 1/ p p This gives c <

p p1 b

1 .c 1/ p

C 1 and we are done.

Lemma 133.4. If Ci ,S i D 1; 2; : : : ; c 1, are the two-step centralizers in a p-group G of class c 2 and if c1 iD1 Ci ¤ G, then c bC1 and so the class-breadth conjecture holds. S Proof. If x 2 G, x 62 Ci , then we have s.x/ D c 1 and, using Lemma 133.2, we get c 1 b.x/ b and c b C 1. Theorem 133.5. Let G be a p-group with b D b.G/ < p. Then c D cl.G/ b C 1 and so the class-breadth conjecture holds. Proof. From Theorem 133.1 follows 1 b p bC1D 1C bC1DbC1C c< p1 p1 p1 p1 bC1C D b C 2; p1 and since both c and b are integers, we get c b C 1.

133

Class and breadth of a p-group

303

Theorem 133.6. Let G be a p-group such that Kp .G/ is contained in Z.G 0 /. Then the class-breadth conjecture holds. In particular, if G is metabelian, then the class-breadth conjecture holds. Proof. Let Ci , 1 i c 1, be the two-step centralizers of the lower central series of length c of G. That is, Ci D CG .Ki .G/=Ki C2 .G//. We note that G cannot be covered by less than p C1 proper subgroups. Indeed, considering a normal subgroup G0 of G such that G=G0 Š Ep2 , we see that Ep2 is partitioned by exactly p C 1 subgroups of order p. We may assume that c p C 2. Indeed, if c p C 1, then the two-step centralizers C1 ; C2 ; : : : ; Cc1 contain at most p proper subgroups of G and so they cannot cover G. But then, by Lemma 133.4, the class-breadth conjecture holds. We ﬁrst use the fact that Kp .G/ Z.K2 .G// to show Ci CiC1 whenever we have p i < c 1. To do this, suppose that g 2 Ci . Now Ki C1 .G/ is generated by the set of commutators ¹Œx; a j x 2 Ki .G/; a 2 Gº, and so we need only to prove that ŒŒx; a; g 2 KiC3 .G/ whenever x 2 Ki .G/ and a 2 G. From the Hall–Witt identity (Introduction, Exercise 13) Œx; y 1 ; zy Œy; z 1 ; xz Œz; x 1 ; yx D 1 we get by setting y D a1 and z D g 1

Œx; a; ga Œa1 ; g 1 ; xg Œg; x 1 ; a1 x D 1: As x 2 Ki .G/ Kp .G/ while Œa1 ; g 1 2 K2 .G/, we have Œa1 ; g 1 ; x D 1 by our assumption. Hence 1

Œx; a; ga Œg; x 1 ; a1 x D 1: Since g 2 Ci and x 1 2 Ki .G/, it follows that Œg; x 1 2 Ki C2 .G/ and so we obtain Œg; x 1 ; a1 2 KiC3 .G/ which implies that Œg; x 1 ; a1 x 2 Ki C3 .G/. Thus we get 1 Œx; a; ga 2 Ki C3 .G/ and so Œx; a; g 2 Ki C3 .G/ and g 2 CiC1 . We have shown that Cp CpC1 Cc1 : Therefore

c1 [ i D1

Ci D Cc1 [

p1 [ i D1

Ci :

S This exhibits ic1 D1 Ci as a union of p subgroups of index at most p in G; it therefore cannot be the whole of G (see 116). Lemma 133.4 then completes the proof.

134

On p-groups with maximal elementary abelian subgroup of order p 2

1o . Glauberman and Mazza [GM] have proved that if a p-group G, p > 2, possesses a maximal elementary abelian subgroup R of order p 2 (i.e., R is not contained in an elementary abelian subgroup of G of order p 3 ; in general, E is a maximal elementary abelian subgroup of a p-group G provided 1 .CG .E// D E), then G has no elementary abelian subgroup of order p pC1 . The proof of this deep result is not elementary. The result above is not true for p D 2 (see 127). Some information on the structure of the groups with title property is contained in Theorem 134.6. In particular, if we have exp.RG / D p, then G has no elementary abelian subgroup of order p pC1 . The proof of this result is fairly involved. However, we cannot present an elementary proof of the theorem of Glauberman–Mazza. In this section we continue to study the structure of groups from [GM] clearing the structure of CG .R/ also in case p D 2. In that case, there is G containing an elementary abelian subgroup of order 24 , and all such G are determined in Theorem 127.1. If a 2-group G has a maximal elementary abelian subgroup R of order 4, then every subgroup of G is generated by four elements so G has no elementary abelian subgroup of order 25 . Indeed, in view of a theorem of MacWilliams (see 50), it sufﬁces to show that G has no normal elementary abelian subgroup of order 8. Assume that E Š E8 is a normal subgroup of G. Then, since RE is not of maximal class, we get CRE .R/ > R (Proposition 1.8), and CRE .R/ is elementary abelian by the modular law, and this is a contradiction. The same MacWilliams’ result shows that every subgroup of G is generated by four elements. Note that the wreath product G D Q2n wr C2 has a maximal elementary abelian subgroup of order 4 and a maximal subgroup B (the base of this wreath product) with d.B/ D 4. We begin with the following auxiliary result. Proposition 134.1. Suppose that a p-group G contains a maximal elementary abelian subgroup R of order p 2 and assume that R is not normal in G. If x 2 R Z.G/, then CG .x/ D CG .R/ has no metacyclic subgroup of order p 4 and exponent p 2 . Proof. By hypothesis, G is nonabelian and 1 .Z.G// < R. Write C D CG .R/ and N D NG .R/; then jN W C j D p by N/C-Theorem (see Proposition 12 in Introduction). Obviously, 1 .C / D R. Assume, by way of contradiction, that L C is meta-

134

On p-groups with maximal elementary abelian subgroup of order p 2

305

cyclic of order p 4 and exponent p 2 . Clearly, Z.G/ \ R D U has order p so Z.G/ is cyclic. If y 2 RU , then CG .y/ D C.D CG .R// since R D hyiU and U Z.G/. Since exp.L/ D p 2 and L is metacyclic, we have L D AB, where A and B are cyclic of order p 2 such that A\B D ¹1º. At least one of the subgroups 1 .A/, 1 .B/ is different from U.D 1 .Z.G//; denote that subgroup by hxi; then, as we have noticed, CG .x/ D C . Let, for deﬁniteness, x 2 A. We have hxi U D R. If G has no normal abelian subgroup of type .p; p/, it is a 2-group of maximal class (Lemma 1.4), and such a group G has no subgroup isomorphic to L, a contradiction. Let E G G be abelian of type .p; p/; then we have R ¤ E by hypothesis, and U < E. Write F D hx; Ei, where x is chosen in the previous paragraph. Clearly, we get x 62 E (otherwise, E D hxi U D R). Since R < F and 1 .F / D F , it follows that F is nonabelian of order p 3 . Recall that A < L is the cyclic subgroup of order p 2 containing x. Write W D A E (the natural semidirect product with kernel E); then F < W and F=E is a unique subgroup of order p in the cyclic group W=E of order p 2 . Since the centralizer CW .E/ has index p in W , it contains F so F is abelian, contrary to what has just been proved. Thus L does not exist. Theorem 134.2. Suppose that a nonabelian p-group G, p > 2, possesses a maximal elementary abelian subgroup R of order p 2 . Set C D CG .R/. Then one of the following holds: (a) G is metacyclic; then R D 1 .G/. (b) G is a p-group of maximal class. If, in addition, R G G, then p D 3.1 (c) C has a cyclic subgroup of index p so it is abelian of type .p n ; p/, n > 1. Proof. If R Z.G/, then C D G has no elementary abelian subgroup of order p 3 so it is metacyclic (Theorem 13.7). In what follows we assume that R 6 Z.G/; then Z.G/ is cyclic. It follows that, in any case, C is metacyclic. Suppose that R G G. If G contains an elementary abelian subgroup A of order p 3 , then CRA .R/ > R is elementary abelian, contrary to the hypothesis. Then, by Theorem 13.7, one of the following holds: (i) G is metacyclic. (ii) G D 1 .G/Z, where 1 .G/ is nonabelian of order p 3 and exponent p and Z is cyclic. (iii) G is a 3-group of maximal class. In case (i), there exist no further restrictions on the structure of G. In case (ii), CG .R/ has a cyclic subgroup of index p. Finally, in case (iii), if jGj > 34 , we have R D 1 .ˆ.G// and CG .R/ D C has no cyclic subgroups of index 3 by Exercise 9.1(c); in this case, C is either abelian or minimal nonabelian. In what follows we assume that R is not G-invariant; then G is not metacyclic. We also assume that G is not of maximal class. Then jC j > p 2 (Proposition 1.8). Then, by Proposition 134.1, the (metacyclic) centralizer C has no metacyclic subgroup of order p 4 and exponent p 2 . We claim that in the case under consideration C 1 If

a 3-group G of maximal class is not isomorphic to a Sylow 3-subgroup of the symmetric group of degree 32 , then all maximal elementary abelian subgroups of G have order 32 (Exercise 9.13). If p > 3, then there is a p-group G of maximal class and order > p 4 that has no such subgroup as R (this is the case if 1 .G/ ˆ.G/).

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has a cyclic subgroup of index p. One may assume that jC j > p 4 . Since C is regular (Theorem 7.1(c)), we get j2 .C /j p 4 and exp.2 .C // D p 2 so that j2 .C /j D p 3 by what has just been said. It follows that C =R has only one subgroup of order p, and hence it is cyclic. If Z < C is maximal such that R 6 Z (Z exists since R 6 ˆ.C /), then Z is cyclic of index p in C . Since R Z.C /, the subgroup C is abelian of type .p n ; p/ as in (c). Note that the p-groups G, p > 2, such that CG .x/ is abelian of type .p n ; p/ for some x 2 G of order p were studied in great detail in the rarely cited important Blackburn’s result (see Theorem A.38.10); that paper yields essential additional information on groups in part (c) of Theorem 134.2. A subgroup A of a p-group G is said to be soft in G, if it satisﬁes CG .A/ D A and jNG .A/ W Aj D p (see 130). Thus soft subgroups are abelian. A subgroup C from Theorem 134.2(c) is soft. Moreover, if a nonnormal R < G is of order p 2 , then we get jNG .R/ W CG .R/j D p, and if, in addition, CG .R/ is abelian, then it is soft in G. Soft subgroups have a number of remarkable properties (see [Het4] and further papers of L. Hethelyi listed in MathSciNet; see also 130). One of such properties is proved in the following remark (note that this proof is distinct from the original one due to L. Hethelyi [Het4]). Remark 1 (the result of this remark coincides with Lemma 130.2). Let A be a nonnormal maximal abelian subgroup of a p-group G, jNG .A/ W Aj D p. Let us prove that if A < H < G, then jNG .H / W H j D p (it follows that there is only one maximal chain connecting A with G). Set N0 D NG .A/; then jN0 W Aj D p and N0 is nonabelian. Set N1 D NG .N0 /. Then N0 contains jN1 W N0 j > 1 conjugates of A under N1 . Since N0 is nonabelian, the number of abelian subgroups of index p in N0 is equal to p C 1 (see Exercise 1.6(a)), therefore we get jN1 W N0 j D p. The intersection of all abelian subgroups of index p in N0 coincides with Z.N0 / D Z G N1 . The quotient group N1 =Z is nonabelian since its subgroup A=Z (of index p 2 ) is not normal. Since CG .A/ D A, we obtain Z.G/ Z < A. Let R Z.G/ be of order p. Then either A=R or N0 =R is a maximal abelian subgroup of G=R since N1 =R, having a nonabelian epimorphic image of N1 =Z.G/, is nonabelian.2 Clearly, the pair K=R < G=R, where K=R is the above chosen maximal abelian subgroup of G=R containing A=R, satisﬁes the identity jNG=R .K=R/ W K=Rj D p by the above as K 2 ¹A; N0 º. Thus K=R is soft in G=R. By induction, there is only one maximal chain connecting K=R and G=R so there is only one maximal chain connection A and G. Indeed, there is nothing to prove in case K D A. If K > A, then K D N0 so the result also holds since N0 is a unique subgroup of G of order pjAj containing A. Similarly, by induction, we obtain the second assertion on indices. Remark 2. If R and G are as in Theorem 134.2, then every subgroup H G such that R < H and exp.H / D p has order p p . Indeed, we have jCH .R/j D p 2 so H is 2 In

fact (see Lemma 130.1), N0 =R is abelian, but we do not use this fact.

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of maximal class (Proposition 1.8), and now the claim follows from Blackburn’s theory of p-groups of maximal class (see Theorems 9.5, 9.6). Assume that there is in G a normal subgroup L of order p pC1 and exponent p. Then CRL .R/ D R by the modular law, so RL is of maximal class (Proposition 1.8). This is a contradiction since a p-group of maximal class cannot contain such a subgroup as L (Theorem 9.6). The case p D 2 is considered in the following theorem. If a 2-group G contains a maximal elementary abelian subgroup R G G of order 4, then G has no normal elementary abelian subgroup of order 23 ; the structure of such G is in some detail described in 50. Note that a minimal nonmetacyclic group X of order 25 satisﬁes j1 .X/j D 4 and d.X/ D 3 (the group X is special). Theorem 134.3. Suppose that a nonabelian 2-group G contains a maximal elementary abelian subgroup R of order 4 and assume that R is not normal in G. Then one of the following holds: (a) The centralizer CG .R/ has a cyclic subgroup of index 2 (so it is abelian). (b) The centralizer CG .R/ D Q Z, where Q is a generalized quaternion group and jZj D 2.3 Proof. Set C D CG .R/; then 1 .C / D R. As R is not normal in G, the subgroup C has no metacyclic subgroup of order 16 and exponent 4 by Proposition 134.1, and this allows us to describe the structure of C . If C is abelian, we get case (a) since C has no abelian subgroup of type .4; 4/ and so j2 .C /j 23 (see the proof of Theorem 134.2). Suppose that C is nonabelian. Then C contains a minimal nonabelian subgroup A. As 1 .A/ 1 .C / D R, it follows that A is metacyclic (Lemma 65.1). Assume that jAj > 8. Then R < A since 1 .A/ Š E4 Š R D 1 .C /, so R D 1 .A/ Z.A/, and we infer that A has no cyclic subgroup of index 2 (otherwise, A will be abelian). Since, by Proposition 134.1, A has no metacyclic subgroup of order 16 and exponent 4, we get a contradiction. Therefore jAj D 8. As A 6Š D8 , we obtain A Š Q8 . Thus all minimal nonabelian subgroups of C are isomorphic to Q8 . It follows that C D Q Z, where Q is a generalized quaternion group and jZj D 2 (see Corollary A.17.3), and the proof is complete. Proposition 134.4 ([GM, Lemma 2.5] for p > 2). Suppose that a p-group G that is not a 2-group of maximal class contains a non-G-invariant maximal elementary abelian subgroup R of order p 2 . Then G has only one normal elementary abelian subgroup of order p 2 unless p D 2, and G contains a proper subgroup of order 24 that is isomorphic to the group D8 C4 of order 16. 3 In this case, Z.G/ D .Q/ has order 2 since this subgroup is characteristic in C and Z.G/ is cyclic. 1 It follows that CG .Z/ D C . The 2-groups G containing an involution x such that CG .x/ D Q hxi, where Q is either cyclic or a generalized quaternion group, are described in 48, 49.

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Proof. Assume that E and F are distinct G-invariant abelian subgroups of type .p; p/ in G. Since Z.G/ is cyclic, we get E \F D U , where U D 1 .Z.G// so jE \F j D p and the subgroup H D EF has order p 3 by the product formula. For the subgroups E=U and F=U we get E=U; F=U Z.G=U /. If H is abelian, it is elementary, and so R 6 H . In case that H is nonabelian, it is either of exponent p > 2 or isomorphic to D8 (this follows from the description of groups of order p 3 ). In that case, all noncyclic subgroups of index p in H are normal in G since H=U Z.G/ and U D ˆ.H /. It follows that R 6 H , again. Write D D HR; then 1 .D/ D D and, as U D H \ R has order p, we get jDj D p 4 by the product formula. Since E=U and F=U are central subgroups of G=U , it follows that D=U Š Ep3 so that d.D/ D 3 and cl.D/ D 2. Suppose that H is abelian. Then CD .R/ is of exponent p so it coincides with R by hypothesis, and it follows from Proposition 1.8 that cl.D/ D 3 > 2, contrary to the last sentence of the previous paragraph. Now let H be nonabelian. By Proposition 10.17, we have CD .H / 6 H since D is not of maximal class, and so Z.D/ has order p 2 . It follows that Z.D/ is cyclic (otherwise, we obtain R < RZ.D/ Š Ep3 , contrary to the hypothesis). In that case, we have p D 2 (if p > 2, then D D 1 .D/ is of exponent p, a contradiction). It follows that D Š D8 C4 has order 24 (note that D8 C4 Š Q8 C4 ). In particular, if, in Proposition 134.4, p > 2, then G has only one normal abelian subgroup of type .p; p/, as asserted in [GM, Lemma 2.5]. Deﬁnition. A proper subgroup A of a p-group G is said to be generalized soft if, whenever A H < G, then jNG .H / W H j D p. (In that case, there is only one maximal chain connecting A and G, but the converse is not true.) In the following proposition we consider the p-groups containing a subgroup of order p that is, as a rule, generalized soft. Proposition 134.5. Suppose that a p-group G contains a subgroup L of order p such that there is only one maximal chain connecting L and G. Then one of the following holds: (a) G is abelian with cyclic subgroup of index p. n

(b) G D ha; b j ap D b p D 1; b a D a1Cp (c) G is a p-group of maximal

n1

i Š MpnC1 .

class.4

Proof. Write N D NG .L/; then N=L is cyclic. If N D G, we have case (a). Next we assume that N < G. If jN=Lj D p, then G is of maximal class by Proposition 1.8. Now assume that jN=Lj > p. Since L is not G-invariant, it is not characteristic in N so N is not cyclic. Let R D 1 .N / and N1 D NG .R/. Since R is characteristic in N , we get N < N1 . By hypothesis, N1 =R is cyclic. Since R < N < N1 , it follows that 4 Not

all p-groups of maximal class contain such a subgroup as L (for example, a p-group G of maximal class, p > 3, such that j1 .G/j D p p1 has no such subgroup).

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R D 1 .N1 / is characteristic in N1 , and we conclude that N1 D G. In that case, G=R is cyclic so G possesses a cyclic subgroup of index p hence G Š MpnC1 by Theorem 1.2. Remark 3. Below we describe the pairs L < G of 2-groups such that L Š E4 , L is not G-invariant and there is only one maximal chain connecting L with G. We write C D CG .L/; then C < G. One may assume that L < C (otherwise, G is of maximal class by Proposition 1.8). In that case, C =L > ¹1º is cyclic so C is a maximal abelian subgroup of G of rank 2 or 3. If jC =Lj D 2, then we obtain that C 2 ¹E8 ; C4 C2 º. Such G are described in 50, 77. Next assume that jC =Lj > 2. Let d.C / D 3. Then T D 1 .C / Š E8 is a proper characteristic subgroup in C . In that case, we see that NG .T /=T > C =T > ¹1º is cyclic by hypothesis, and so T D 1 .NG .T // is characteristic in NG .T /, and we conclude that NG .T / D G hence 1 .G/ D T . It follows that G=1 .G/ is cyclic and 1 .G/ Š E8 . Then G has a cyclic subgroup of index 4 (such G are described in 74). Now let C be abelian of rank 2; then L D 1 .C / so C has a cyclic subgroup of index 2 by Proposition 134.1. In that case, NG .L/=L is cyclic by hypothesis. Therefore it follows from L < C < NG .L/ that L D 1 .NG .L// is characteristic in NG .L/ so NG .L/ D G, i.e., L G G, contrary to the hypothesis. 2o . In this subsection we prove two theorems. We suppose that the p-groups R, E and G appearing in these theorems satisfy the following Hypothesis. (1) p > 2, R is a nonnormal maximal elementary abelian subgroup of order p 2 of a p-group G. (2) E is a maximal elementary abelian subgroup of G, E ¤ R. (3) G is not of maximal class. Let us prove that if R and G satisfy the Hypothesis, then Z.G/ is cyclic. Indeed, Z.G/ < CG .R/ and 1 .CG .R// D R so that 1 .Z.G// < R, and our claim follows. Theorem 134.6. Let p; G; R; E be as in the Hypothesis. Suppose that exp.RG / D p and let RG M G G, where M is as large as possible subjecting exp.M / D p. Then one of the following holds: (a) The subgroups RG and M are of maximal class, jM j p p . (b) If G is irregular, then jM j D p p . (c) If j1 .G/j D p p , then G=1 .G/ is cyclic. (d) If jM j D p p , then ME is of maximal class and order p pC1 so that jEj p p . Theorem 134.7. Let p; G; R; E be as in the Hypothesis. Suppose that exp.RG / > p. Then the following hold: (a) There is M G G of order p p and exponent p. Write H D MR and F D M C , where C D CG .R/; then H is of maximal class and order p pC1 , H < F .

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(b) Let H < T G, where jT W H j D p; then the subgroup T is not of maximal class and T D M2 .C / is the unique subgroup of G of order pjH j containing H so that NG .H /=H is cyclic. (c) F=M is abelian of type .p n1 ; p/, H G F and F=H is cyclic of order p n1 . (d) If R < B < H with jB W Rj D p, then B is nonabelian and CG .B/ is cyclic. (e) The subgroup H is not normal in G. (f) 1 .Z.G// Ã1 .C /. (g) If jME W M j > p, then R 6 ME. (h) If A G G is of order p k < p p and exponent p, then A < M . Our proofs of Theorems 134.6 and 134.7 are essentially based on Blackburn’s theory of p-groups of maximal class (see 9) and Hall’s results on regular p-groups (see 7). In both theorems G is nonabelian. Remark 4. If R is as in the Hypothesis, then it is not contained in ˆ.G/. Assume, by way of contradiction, that R ˆ.G/; then ˆ.G/ is noncyclic. In this case, there is in ˆ.G/ a G-invariant abelian subgroup L of type .p; p/ (Lemma 1.4). As L Z.ˆ.G// (consider CG .L/!), the subgroup RL is elementary abelian properly containing R, a contradiction. It follows from Z.G/ < CG .R/ that Z.G/ is cyclic (see the paragraph following the Hypothesis. Note that if G satisﬁes the Hypothesis, it has no normal subgroup of order p pC1 and exponent p. Indeed, assume that B G G is of order p pC1 and exponent p. Write H D BR. By Theorem 9.6(c), H is not of maximal class as B H so that CH .R/ > R (Proposition 1.8), and we conclude that CB .R/ 6 R. If L < CB .R/ is of order p is such that L 6 R, then RL D R L is an elementary abelian p-group properly containing R, which is a contradiction. Thus B does not exist. Proof of Theorem 134.6. By Proposition 1.8, if R < K G and exp.K/ D p, then K is of maximal class and order p p . Therefore, by assumption, exp.G/ > p. Obviously, G is nonmetacyclic. In this theorem exp.RG / D p. Since R is not normal in G, we get R < RG . Let M be a maximal G-invariant subgroup of exponent p containing RG . Then, by Proposition 1.8, the subgroups RG and M are of maximal class as CM .R/ D R D CRG .R/ (see part (1) of the Hypothesis). By Proposition 1.8 and Theorem 9.5, we obtain that jM j p p . Set Q D M C , where C D CG .R/. Since, by Theorem 134.2(c), C is abelian of type .p n ; p/, n > 1, and R D M \ C , then Q=M Š C =.C \ M / D C =R is cyclic of order p n1 . (i) By Theorem 7.2(b), if G is regular, then M D 1 .G/. There is no further restriction on the structure of G in this case. We claim that if G is irregular, then jM j D p p . Indeed, by Theorem 12.1(a), there is N G G of order p p and exponent p since, by assumption, G is not of maximal class.

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As we know, M p p . One may assume that M 6 N (otherwise, there is nothing to prove). Let N1 be a G-invariant subgroup of N satisfying N1 6 M and such that N1 is as small as possible. Then jMN1 j D pjM j p pC1 hence exp.MN1 / D p 2 by the choice of M . Therefore MN1 is irregular hence jMN1 j p pC1 (Theorem 7.1(b)), and we conclude that MN1 is of maximal class and order p pC1 (Theorem 7.2(b)). Since M 6 ˆ.MN1 / (otherwise, we have MN1 D N1 , contrary to the proved equality jMN1 j D p pC1 ), we infer that jMN1 W M j D p (Exercise 9.1(b)); then jM j D p p , as asserted. It is important in what follows that R < M . (ii) Let T =M G=M be of order p such that T 6 Q.D M C / (see the second paragraph of the proof; such a T exists if and only if G=M is noncyclic). Since CT .R/ D T \ C D R; the subgroup T is of maximal class (Proposition 1.8). (iii) In what follows we suppose that jM j D p p (this is the case provided G is irregular by (i), however, we do not assume here that G is irregular). Set W D ME, where E < G is as in part (2) of the Hypothesis. We claim that W is of maximal class. This is the case if E < M . Now let E 6 M ; then M < W . To prove our claim, take M < U ME such that jU W M j D p. By the modular law, we get U D M.U \ E/ so that 1 .U / D U . Assume that exp.U / D p. Then U is not of maximal class since jU j D p pC1 (Theorem 9.5) so that CU .R/ > R (Proposition 1.8), which is a contradiction. Thus exp.U / > p so that U is irregular (Theorem 7.2(b)). As jU j D p pC1 , we conclude that U is of maximal class (Theorem 7.1(b)). Thus all subgroups of order p pC1 lying between M and W D ME are of maximal class. It follows from Exercise 10.10 that W is also of maximal class. By Exercise 9.1(b), jW j D pjM j D p pC1 since M 6 ˆ.W /, and we conclude that jEj < jW j D p pC1 . (iv) Suppose that j1 .G/j D p p ; then M D 1 .G/ (see the deﬁnition of M in the second paragraph of the proof). Assume that G=M is noncyclic. Assume, in addition, that p D 3. Then G has no elementary abelian subgroup of order p 3 . In this case, however, G=M is cyclic by Theorem 13.7, contrary to the assumption. Thus p > 3. Since Q=M D M C =M is cyclic, there is T =M < G=M of order p such that T =M 6 Q=M (Proposition 1.3); then T 6 Q. It follows that T \Q D M so that CT .R/ D R, and Proposition 1.8 implies that T is of maximal class. Since jT j D p pC1 , we infer that T is irregular (Theorem 9.5). Let T < L G be such that jL W T j D p. The subgroup L is not of maximal class since M G L, jLj > p pC1 and M is of order p p and exponent p (see Theorem 9.6(c)). Therefore, by Theorem 12.12(b), L=Kp .L/ is of order p pC1 and exponent p. It follows that L has no absolutely regular subgroup of index p (otherwise, jL=Ã1 .L/ p p ). Let D < M be G-invariant of order p 2 and set V D CG .D/; then V is maximal in G as D 6 Z.M / in view of jZ.M /j D p < jDj. The intersection M \ V D 1 .V / has order p p1 hence V is absolutely regular since it is not of maximal class in view of D Z.V / (Theorem 12.1(b)). Then L \ V , as

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a subgroup of the absolutely regular group V , is absolutely regular of index p in L, contrary to what has been said above. Thus G=M is cyclic, as was to be shown. The proof of our theorem is completed. Proof of Theorem 134.7. By hypothesis, in this case exp.RG / > p hence RG is irregular in view of 1 .RG / D RG (Theorem 7.2(b)). Since G is not of maximal class, there is M G G of order p p and exponent p (Theorem 12.1(a)). We have R 6 M since exp.RG / > p D exp.M /. (i) We claim that H D MR is of maximal class and order p pC1 . Indeed, we have jM \ Rj D p since Z.G/ is cyclic, and so jMRj D p pC1 by the product formula. Next, CH .R/ D H \ CG .R/ D R since CH .R/ D R CM .R/ D R by the modular law, and our claim follows from Proposition 1.8 (in the case under consideration, exp.CH .R// D p by Theorems 7.1(b) and 7.2(b) since cl.CH .R// < p). (ii) Let R < S < H , where jH W S j D p; then we get S D R.S \M / by the modular law, so exp.S/ D p since 1 .S/ D S and jSj D p p (Theorems 7.1(b) and 7.2(b)). If H G G, then, considering the action of G on maximal subgroups of H via conjugation, we obtain S G G (indeed, since exp.H / D p 2 and d.H / D 2, H has at most p maximal subgroups of exponent p and one of them, namely M , is normal in G), and this is a contradiction since RG S and exp.RG / > exp.S/. Thus H is not normal in G. (iii) Write F D M C ; then we obtain MR D H < F since jF j p 2 jM j > jH j (recall that R 6 M and exp.C / > p; see Theorem 134.2(c)). Let H < T G, where jT W H j D p; then T is not of maximal class since M G T (Theorem 9.6(c)). Then we have CT .R/ > R (Proposition 1.8) so that 2 .C / < T . By the product formula, we obtain T D M2 .C / M C D F so that T is the unique subgroup of NG .H / of order pjH j containing H , and we infer that NG .H /=H is cyclic since p > 2 (Proposition 1.3). By Theorem 12.12(b), T =Kp .T / is of order p pC1 and exponent p so that Ã1 .T / D Kp .T / D Kp .H / D Z.M / D Ã1 .H / D Ã1 .2 .C // D R \ Z.G/ D 1 .Z.G//: It follows that F=M is abelian of type .p n1 ; p/ so that H GF and hence F=H is cyclic of order p n1 . If R < B < H with jB W Rj D p, then B is nonabelian of order p 3 and exponent p, and CT .B/ D 2 .C /; moreover, CG .B/ is cyclic since R < B. (iv) Let jM j D p p and jME W M j > p. We have to prove that R 6 ME. Assume that this is false. Since R \ M D Z.M / is of order p, we have jMRj D p pC1 by the product formula. Next, CMR .R/, being of exponent p, coincides with R, and we conclude that MR is of maximal class (Proposition 1.8). Let M < U < ME be such that jU W M j D p and U ¤ MR. As U D M.U \ E/ by the modular law, we obtain that 1 .U / D U . Assume that exp.U / D p. Write H D UR (note that U G ME). Since H is not of maximal class (Theorem 9.6(c)), we get CH .R/ > R. Again by the modular law, we conclude that CH .R/ D M.U \ CH .R// so that exp.CH .R// D p,

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313

contrary to Theorem 134.2(c) since H is neither metacyclic nor of maximal class (note that MR is also of maximal class). Hence we obtain that exp.U / > p so U is irregular (Theorem 7.2(b)). Since jU j D pjM j D p pC1 , it follows that U is of maximal class (Theorem 7.1(b)). Thus all subgroups of G of order pjM j lying between M and ME are of maximal class. Therefore ME is also of maximal class (Exercise 10.10). Since M 6 ˆ.ME/, we get jME W M j D p (Exercise 9.1(b)), contrary to the assumption. (v) Now let A G G be of order p k < p p and exponent p. We have to prove that A satisﬁes A < M , where M is as in part (a) of the statement of our theorem. This is true for k D 1 since Z.G/ is cyclic. Now assume that we have proved that any normal subgroup of G of order < p k and exponent p is contained in M . Set J D AM . Then, by assumption, jA \ M j D p k1 . By the product formula, we infer that jJ j D p pC1 . Due to Exercise 9.1(b), J is not of maximal class since it has two nonincident proper normal subgroups of different orders. It follows that cl.J / < p, and we conclude that J is regular (Theorem 7.1(b)). Since 1 .J / D J , it follows from Theorem 7.2(b) that exp.J / D p. This is a contradiction since, as we have noticed (see the paragraph preceding the proof of Theorem 134.6), G has no normal subgroup of order p pC1 and exponent p. Thus we get A < M . Exercise 1. Theorem 134.7(h) also holds under the condition of Theorem 134.6. (Hint. Argue as in part (v) of the proof of Theorem 134.7.) Exercise 2. Suppose that a pair of p-groups R < G satisfy the Hypothesis. If M and N are two distinct G-invariant subgroups of order p p and exponent p, then MN is of maximal class and order p pC1 . Solution. Let A < M be a G invariant subgroup of index p. By Theorem 134.7(h) and Exercise 1, we get A < N so that MN is of order p pC1 . As we know, exp.MN / D p 2 so MN is irregular. It follows that MN is of maximal class. Exercise 3. Suppose that R is a nonnormal maximal elementary abelian subgroup of a p-group G, p > 2. Then C D CG .R/ is abelian with cyclic subgroup of index p unless NG .R/ is either metacyclic or a 3-group of maximal class. Solution. By Theorem 13.7, C is metacyclic. Set N D NG .R/; then jN W C j D p. Since N has no normal elementary abelian subgroup of order p 3 , the result follows from Theorem 13.7. Remark 5. Suppose that G is a p-group and M G G is of order p p and exponent p such that every subgroup of order p pC1 between M and G has an exponent p 2 . Then G has no elementary abelian subgroup of order p pC1 . Indeed, let E G be elementary abelian and let E 6 M (otherwise, there is nothing to prove). Write H D ME and let H < U G, where jU W H j D p. Then, by hypothesis, exp.U / D p 2 . By the modular law, U D M.U \ E/ so that 1 .U / D U , and hence U is of maximal class (Theorems 7.1(b) and 7.2(b)) and order p pC1 . Since U is arbitrary, we conclude as

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above that the group G is of maximal class and order p pC1 (Exercise 10.10). It follows that jEj p p , as asserted. (This argument is the same as in part (iv) of the proof of Theorem 134.7.)

Problems Problem 1. Suppose that G, p > 2, possesses a maximal elementary abelian subgroup of order p 2 and H G. (i) Is it true that d.H / p? (ii) Is it true that jH j < p pC1 provided exp.H / D p? Problem 2. Suppose that G, p > 2, possesses a maximal elementary abelian subgroup n1 of order p n . Is it true that G has no elementary abelian subgroup of order p 1Cp ? Problem 3. Study the p-groups all of whose minimal nonabelian (so all nonabelian) subgroups are generalized soft. Problem 4. Study the p-groups containing a cyclic generalized soft subgroup of order p n . (The problem is nontrivial even for n D 2.) In the following three problems R is a nonnormal maximal elementary abelian subgroup of order p 2 in a p-group G. Problem 5. Study the structure of RG , where exp.R/ > p. Consider in detail the case when RG is of maximal class. Problem 6. Study the structure of G provided RG is of maximal class. Problem 7. Study the structure of G provided CG .R/ is metacyclic. Problem 8. Study the p-groups G containing a nonabelian subgroup B of order p 3 such that CG .B/ is cyclic. (See the statement of Theorem 134.7.)

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Finite p-groups generated by certain minimal nonabelian subgroups

Recall that a group is said to be minimal nonabelian if it is nonabelian but all its proper subgroups are abelian. A nonabelian group contains a minimal nonabelian subgroup. Therefore it is natural to believe that minimal nonabelian groups essentially impact on the structure of a nonabelian group. As many places of our book show, knowledge of all minimal nonabelian subgroups of a nonabelian p-group G allows us to do strong conclusions on its structure. This is analogous to impact of minimal nonnilpotent subgroups on the structure of a nonnilpotent group (this is true in view of Frobenius’ normal p-complement theorem). To justify our thought, we present the following two examples (other examples are given after the proof of Lemma 135.5): (i) If a p-group G possesses at most p 2 C p C 1 minimal nonabelian subgroups, then all its subgroups of index p 3 are abelian (see 76). (ii) If all minimal nonabelian subgroups of a 2-group G have the same order 8, then either the Hughes subgroup has index 2 in G or G D H Z.G/, where H is either of maximal class or extraspecial and Ã1 .Z.G// Z.H / (see Theorem 90.1). In what follows all considered groups have prime power order. Let MA.G/ be the set of all minimal nonabelian subgroups of a p-group G. If G is nonabelian, we have, by Lemma 135.7 below, G D hMA.G/i, i.e., G is generated by all members of the set MA.G/ (this result is important in the proof of assertion (i) above). Since minimal nonabelian subgroups are small, that result shows that in a complicated p-group G the set MA.G/ is large. However, in some cases, certain welldescribed subsets of that set also generate G, and in what follows we present some of such subsets. There is, for p > 2, only one, up to isomorphism, minimal nonabelian p-group generated by elements of order p and its order is p 3 (Exercise 1.8(a)). We denote this group by S.p 3 /. The dihedral group D8 of order 8 is the only minimal nonabelian 2-group generated by involutions. Let k be a positive integer and let MAk .G/ be the set of minimal nonabelian subgroups A of a p-group G such that A D k .A/. Since for any p-group X > ¹1º we

316

Groups of prime power order

have exp.X=1 .X// < exp.X/, it follows that if A 6Š D8 is as above and k > 1, then ()

exp.A/ exp.1 .A// exp.A=1 .A// p p k1 D p k ;

since, in view of jA0 j D p, the quotient group A=1 .A/ is abelian of exponent p k1 and exp.1 .A// D p. In the following paragraph we shall use the following fact. Let G be a nonabelian p-group. Further, assume that all containing ˆ.G/ subgroups of G of order pjˆ.G/j are abelian. Then CG .ˆ.G// possesses all subgroups T of G of order pjˆ.G/j such that ˆ.G/ < T . Since all such T generate G, it follows that ˆ.G/ Z.G/. Let G be a nonabelian p-group. Suppose that ˆ.G/ 6 Z.G/. In this case, there is a sequence ˆ.G/ D T0 < T1 < < Td.G/ D G of subgroups and A1 ; : : : ; Ad.G/ of minimal nonabelian subgroups such that T1 is nonabelian and, for i D 1; : : : ; d.G/, one has Ai Ti but Ai 6 Ti1 (Lemma 135.7 below); then jTi W Ti1 j D p and Ai Ti 1 D Ti . It follows that G D Td.G/ D hA1 ˆ.G/; A2 ; : : : ; Ad.G/ i D hA1 ; A2 ; : : : ; Ad.G/ i: Thus, in the case under consideration, G is generated by d.G/ minimal nonabelian subgroups. If, however, ˆ.G/ Z.G/, the same argument shows that G is generated by d.G/ 1 minimal nonabelian subgroups (note that there is a containing ˆ.G/ subgroup of G of order pjˆ.G/j that is not contained in Z.G/). The obtained estimate is not best possible (for example, an extraspecial group G of order p 1C2m is generated by m < 2m D d.G/ minimal nonabelian subgroups since it is a central product of m nonabelian subgroups of order p 3 ). Let k D ı.G/ denote the minimal number of minimal nonabelian subgroups, say A1 ; : : : ; Ak , generating a nonabelian p-group G. Set T0 D ˆ.G/;

T1 D ˆ.G/A1 ;

T2 D T1 A2 ;

:::;

Tk D Tk1 Ak D G:

Since, for all i , ˆ.Ai / ˆ.G/ has index p 2 in Ai , we get jTi W Ti 1 j p 2 . It follows that p d.G/ D jG W ˆ.G/j D jTk W Tk1 j : : : jT1 W T0 j p 2k D p 2ı.G/ ; and we obtain ı.G/ d.G/ 2ı.G/: As we have noticed, the lower estimate is attained for extraspecial p-groups. This estimate can be slightly improved in the case ˆ.G/ Z.G/. In this section, however, we study the generation of a nonabelian p-group G by minimal nonabelian subgroups satisfying certain additional properties. Write Dk .G/ D hMAk .G/i;

Dk .G/ D hA j A 2 MAk .G/; exp.A/ D p k i:

Obviously, Dk .G/; Dk .G/ k .G/.

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317

In what follows we establish criteria guaranteeing the equality Dk .G/ D G and also study the p-groups G satisfying Dk .G/ < G for a positive integer k (see Theorem 135.1(b) and Remark 2). Let us prove that if G is a nonabelian regular group of exponent p e , then we have G D De .G/. Assume that G is a counterexample of minimal order. Then we conclude that e > 1 (Lemma 135.7) and G is not minimal nonabelian. Setting H D e1 .G/, we have exp.H / D p e1 (Theorem 7.2(b)). By Lemma 135.7, there exists a minimal nonabelian subgroup A < G such that A 6 H ; then exp.A/ D p e . If M 2 1 is nonabelian of exponent p e , then we get De .M / D M by induction (such M exist: take M containing A). By Theorem 7.2(b), there exists in G at most one maximal subgroup of exponent p e1 (otherwise, H D G). Therefore all other members of the set 1 have exponent p e . Since the number of abelian maximal subgroups in G is 0, 1 or p C1 and j1 j 1 .mod p/, there are at least p nonabelian members of the set 1 . It follows that there are in 1 at least p 1 nonabelian members of exponent p e . Since G is nonabelian regular, we have p > 2 so p 1 > 1. It follows that there is a nonabelian N 2 1 ¹M º of exponent p e . By induction, we see that De .N / D N . Therefore De .G/ De .M /De .N / D MN D G, contrary to the assumption. It is proved in Lemma 30.3 that if a nonabelian 2-group G D 1 .G/, then we have G D D1 .G/. To clear up our path, we offer another proof of this equality. Since G is nonabelian, it contains two noncommuting involutions a and b. Then ha; bi is dihedral so it contains a subgroup A Š D8 . Set H D D1 .G/. Let x 2 G A be an involution. If x 2 CG .A/, we get U D Ahxi D A hxi. There are in U exactly four maximal subgroups not containing x, and these subgroups generate U and are isomorphic to D8 , and we conclude that x 2 U H . Now assume that x does not centralize A. In this case, there is in A D 1 .A/ an involution y such that xy ¤ yx. Then hx; yi is dihedral so it is generated by subgroups isomorphic to D8 , and again x 2 hx; yi H . Thus all involutions are contained in H so H 1 .G/ D G. Considering the group S.p 3 / as an analog of D8 , we note that a result similar to the result of the previous paragraph does not hold for p > 2 as a Sylow p-subgroup †p2 of the symmetric group of degree p 2 shows. Indeed, †p2 has exactly two maximal subgroups, say E and M , both of exponent p, E is abelian and M is nonabelian,1 and so D1 .G/ D M < G (Lemma 135.7) since M is nonabelian in view of cl.†p2 / D p > 2, and any L 2 1 ¹M:Eº satisﬁes D1 .L/ D L \ E hence MA1 .L/ D ¿. As Theorem 30.1(d) shows, if a nonabelian p-group G D 1 .G/, p > 2, has no subgroup Š †p2 , then G is generated by subgroups isomorphic to S.p 3 /. However, the proof of this result was omitted there. Here (see Theorem 135.1(a)) we offer its 1 The group G D † p 2 is the wreath product of two groups of order p. Let E be the base of G; then E is elementary abelian of order p p and G D hxi E is a semidirect product, o.x/ D p. Next, we have CG .x/ D hxi Z.G/, where jZ.G/j D p, so jCG .x/j D p 2 , and hence G is of maximal class (Proposition 1.8). Set M D hxi ˆ.G/; then M D 1 .M / is regular so exp.M / D p. The set E [ M contains exactly 2p p p p1 1 elements of order p, and this number coincides with the number of elements of order p in G. Thus G has exactly two maximal subgroups E and M of exponent p.

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Groups of prime power order

proof independent of Theorem 30.1. Next, we study in Theorem 135.1(b) the p-groups G D 1 .G/, p > 2, such that D1 .G/ < G. Remark 1. Let G D 1 .G/, p > 2, be a nonabelian p-group. Suppose that whenever x; y 2 G are noncommuting elements of order p, then D1 .hx; yi/ D hx; yi. We claim that then G D D1 .G/. By hypothesis, G, being nonabelian, contains a subgroup L Š S.p 3 /. Let H D D1 .G/; then H > ¹1º. Assume that H < G. Then there is x 2 G H of order p. If x 2 CG .L/, then U D hx; Li D hxi L. Since U is generated by p 2 its maximal subgroups not containing x, and all these subgroups are isomorphic to L Š S.p 3 /, we get x 2 H , a contradiction. Now assume that x 62 CG .L/. Then there is y 2 L# (of order p) such that xy ¤ yx. In this case, by hypothesis, the nonabelian subgroup hx; yi D D1 .hx; yi/ hence x 2 H , a contradiction. Thus we have H D G. Theorem 135.1. Let G D 1 .G/ be a nonabelian p-group, p > 2. Then (a) (Theorem 30.1(d)) If G has no subgroup isomorphic to †p2 , then G D D1 .G/. (b) (Locating of elements of order p) Suppose that H D D1 .G/ < G and E is a maximal by inclusion normal elementary abelian subgroup of G. Then the set H [ E contains all elements of G of order p. Next, E is a unique maximal normal elementary abelian subgroup of G, jEj p p and G D HE; in particular, ˆ.H / H . Corollary 135.2. The result of Remark 1 follows from Theorem 135.1(a). Proof. Our group G has no subgroup isomorphic to †p2 since †p2 is generated by two noncommuting elements of order p and, as we have noticed, D1 .G/ < G. Therefore the result follows from Theorem 135.1(a). The next lemma is useful in proofs by induction (see the proofs of Lemma 135.8 and Theorem 135.1). Lemma 135.3 (= Lemma 30.2). Let k be a positive integer and A a proper subgroup of a p-group G D k .G/ such that k .A/ D A. Then we have A M 2 1 , where M D k .M /. In particular, there are two distinct U; V 2 1 such that k .U / D U and k .V / D V . Lemma 135.4. Suppose that a nonabelian p-group G has an abelian subgroup A of index p. If jG W G 0 j D p 2 , then G is of maximal class. Proof. We proceed by induction on jGj. One may assume that jGj > p 3 . By Lemma 1.1, jZ.G/j D p1 jG W G 0 j D p so Z.G/ is a unique minimal normal subgroup of G. Since Z.G/ < G 0 and jG=Z.G/ W G 0 =Z.G/j D jG W G 0 j D p 2 ; the quotient group G=Z.G/ is of maximal class by induction, and the result follows since jZ.G/j D p.

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319

Lemma 135.5. Suppose that A < G is a normal abelian subgroup of a p-group G such that A 6 Z.G/. Then for every x 2 G CG .A/ there is a 2 A such that hx; ai is a minimal nonabelian subgroup.2 Proof. Let x 2 G CG .A/. Bringing to mind our aim, one may assume, without loss of generality, that G D Ahxi; then CA .x/ D U DW A \ Z.G/. One may also assume that jA W U j D p (if not, let us consider instead of A an x-invariant subgroup A1 A of order pjU j such that U < A1 ). Since G is nonabelian, the quotient group G=U is noncyclic. It follows that G=U D .hx; U i=U / .A=U / is abelian of type .p k ; p/, where p k is the order of x modulo U . In this case, G=U contains two distinct cyclic subgroups M=U and N=U of order p k so maximal in G=U . Then M; N are abelian maximal subgroups of G, and we see that Z.G/ D M \N has index p 2 in G. By Lemma 1.1, jG 0 j D p1 jG W Z.G/j D p. Let a 2 A U.D A CG .x// and set H D hx; ai; then H is nonabelian, jH 0 j D jG 0 j D p and d.H / D 2. From Lemma 65.2(a) it follows that H is minimal nonabelian. Below we offer some applications of Lemma 135.5. Suppose that all minimal nonabelian subgroups of a nonabelian 2-group G have the same order 8. Let A be a maximal abelian normal subgroup of G. By Lemma 135.5, all elements of the set G A have order 2 or 4. Assume that there is x 2 G A of order 4. Then there is a 2 A such that H D ha; xi has order 8. Since ˆ.G/ H \ A, it follows that x 2 2 A, and we conclude that ˆ.G/ A. Such groups are classiﬁed in Theorem 90.1. Let p > 2 and A be a maximal abelian normal subgroup of a nonabelian p-group G of exponent > p. If all minimal nonabelian subgroups of G are isomorphic to S.p 3 /, then A is equal to the Hughes subgroup Hp .G/ of G. Indeed, by Lemma 135.5, all elements of the set G A have order p, and we conclude that Hp .G/ A as exp.A/ > p. Since A is generated by elements of maximal order, we get, in fact, Hp .G/ D A.3 Suppose that all minimal nonabelian subgroups M of a nonabelian p-group G satisfy k .M / Z.M /, where k is a ﬁxed positive integer. We claim that if A is a normal abelian subgroup of G such that exp.A/ p k and jAj is as large as possible, then k .G/ A. Assume that there is an element x 2 G A of order p k . Then hx; Ai is nonabelian (Corollary 10.2) so A 6 Z.G/. In this case, there is in A an element a such that H D ha; xi is minimal nonabelian. By hypothesis, x; y 2 Z.H / so H is abelian, a contradiction. Thus x does not exist, and we conclude that k .G/ A; moreover, k .G/ D A. Lemma 135.6. Suppose that G D k .G/ is a nonabelian p-group, p k > 2. Then the set MAk .G/ is not empty. 2 In particular, if A < G is a maximal normal abelian, then for each x 2 G A there is a 2 A such that the subgroup hx; ai is minimal nonabelian. This important result, which is due to Z. Janko, coincides with Lemma 57.1. 3 According to Mann’s commentary to item 115 in Research problems and themes I, one has the equality jG W Aj D p.

320

Groups of prime power order

Proof (compare with the proof of Lemma 30.4(a)). Let E < G be a G-invariant abelian subgroup of exponent p k of maximal order. Take x 2 G E of order p k . Then H D hx; Ei is nonabelian by Corollary 10.2. By Lemma 135.5, there is a 2 E such that the subgroup B D hx; ai is minimal nonabelian. Since o.x/ D o.a/ p k , we get k .B/ D B. By (), exp.B/ p k so that B 2 MAk .G/. Lemma 135.7 (= Theorem 10.28). A nonabelian p-group is generated by minimal nonabelian subgroups. We present an application of Lemma 135.7. Let G D D C , where D Š D8 and C Š C2n (n > 2). At ﬁrst sight, the set MA.G/ has no member of exponent 2n . But this is false. Indeed, we have n1 .G/ D Dn1 .C / < G. By Lemma 135.7, there is M 2 MA.G/ such that M 6 n1 .G/ hence exp.M / D 2n . Note that Lemma 135.7 follows from Lemma 135.5. Indeed, if, in Lemma 135.5, A is a maximal normal abelian subgroup of G, then x 2 G A is contained in some minimal nonabelian subgroup of G. Since hG Ai D G, our claim follows. The following lemma covers a partial case of Theorem 135.1(a) (but it does not follow from that theorem). Lemma 135.8. Suppose that a nonabelian p-group G D 1 .G/ has no subgroup isomorphic to †p2 . If G possesses an elementary abelian subgroup V of index p, then we have exp.G/ D p. Proof. Since G is nonabelian, we get p > 2. By hypothesis, there is x 2 G V of order p; then G D hxi V is a semidirect product with kernel V . Let x 2 F 2 1 ; then F D hxi .F \ V / by the modular law, so 1 .F / D F . By induction, exp.F / D p. Thus every maximal subgroup of G containing x has exponent p. If jGj p p , it is regular so of exponent p (Theorems 7.1(b), 7.2(b)). Next we assume that jGj p pC1 . If jG W G 0 j D p 2 , then G is of maximal class (Lemma 135.4) so we get G Š †p2 by Exercise 9.13, a contradiction. Thus jG W G 0 j > p 2 . Since 1 .G/ D G, it follows that G=G 0 is elementary abelian. Set H D hxi G 0 ; then G=H is elementary abelian of order > p. Let T1 =H; : : : ; Tn =H be all maximal subgroups of G=H ; thenSn p C 1. By the above, exp.Ti / D p for all i n. Therefore it follows from G D niD1 Ti that exp.G/ D p. Proof of Theorem 135.1. (a) Let G be a counterexample of minimal order; then we get jGj > p 3 and exp.G/ > p by Exercise 1.8(a) and Lemma 135.7 so G is irregular and jGj p pC1 . All proper nonabelian subgroups H D 1 .H / of G satisfy D1 .H / D H by induction. Due to Lemmas 135.6 and 135.3, there is a nonabelian U 2 1 such that 1 .U / D U ; then D1 .U / D U by induction. By the last assertion of Lemma 135.3, there exists V 2 1 ¹U º such that 1 .V / D V . One may assume that V ¤ D1 .V / (otherwise, D1 .G/ D D1 .U V / D1 .U /D1 .V / D U V D G). Then V is elementary abelian by induction. If x 2 GV is of order p, then G D hxi V , a semidirect product with kernel V , and now the result follows from Lemmas 135.8 and 135.7 since G has no subgroup isomorphic to †p2 .

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Finite p-groups generated by certain minimal nonabelian subgroups

321

(b) By hypothesis, we get H D D1 .G/ < G. Let E be a normal elementary abelian subgroup of G such that jEj is as large as possible. Assume that there is an element x 2 G .H [ E/ of order p. Set L D hx; Ei. By Corollary 10.2, L is nonabelian. If L is regular, then it follows from 1 .L/ D L that exp.L/ D p by Theorem 7.2(b), and so L D D1 .L/ by Exercise 1.8(a) and Lemma 135.7, a contradiction since L 6 H by the choice of x. Thus L is irregular so jEj p p (Theorem 7.1(b)). Assume that jZ.L/j D p. It follows from Lemma 135.4 that L is of maximal class (indeed, we have jG W G 0 j D pjZ.G/j D p 2 by Lemma 1.1) so L Š †p2 by Exercise 9.13). In that case, U D hx; ˆ.L/i is nonabelian (by Fitting’s lemma) of exponent p (Theorem 7.2(b)) so U D D1 .U /; then x 2 U H , contrary to the choice of x. Thus jZ.L/j > p. Note that Z.L/ D CE .x/. Let T =Z.L/ be an L-invariant subgroup of order p in E=Z.L/ so T is elementary abelian. Then V D hx; T i is nonabelian since x does not centralize T , and 1 .V / D V . We have cl.V / D 2 since jV =Z.L/j D p 2 . It follows from this and p > 2 that exp.V / D p (Theorems 7.1(b) and 7.2(b)). Then, by Lemma 135.7 and Exercise 1.8(a), we have V D D1 .V /, and we obtain x 2 V H , a contradiction. Thus the set G.H [E/ has no elements of order p, as was to be shown. Since G D 1 .G/ and the subgroup HE . H [ E/ contains all elements of G .D 1 .G// of order p, we conclude that G D HE. Because G=H Š E=.E \ H / is elementary abelian, we get ˆ.G/ H . In particular, E 6 H since, by assumption, H < G. Assume that E1 < G is another maximal G-invariant elementary abelian subgroup. Then EE1 is nonabelian, by Corollary 10.2, so it is of class 2 (Fitting’s lemma). It follows that exp.EE1 / D p (Theorems 7.1(b) and 7.2(b)) so D1 .EE1 / D EE1 by Exercise 1.8(a) and Lemma 135.7, and we conclude that E < EE1 H , contrary to the last sentence of the previous paragraph. The subgroup E from Theorem 135.1(b) has the following property: If x 2 G E is of order p, then the subgroup Hx D hx; Ei contains a subgroup isomorphic to †p2 . Indeed, since E 6 D1 .G/, it follows that D1 .Hx / < Hx , and the claim follows from Theorem 135.1(a). Theorem 135.1(a) can be restated as follows: If p > 2, then for a nonabelian p-group G one of the following holds: (i) 1 .G/ is elementary abelian. (ii) G contains a subgroup isomorphic to †p2 . (iii) D1 .G/ D 1 .G/. Assume that p > 2. The group †p2 satisﬁes the hypothesis of Theorem 135.1(b). Let G D †p2 L, where L > ¹1º is elementary abelian p-group. Let M; E < †p2 be maximal of exponent p, M be nonabelian and E be abelian. Write W D D1 .G/. We claim that W D M L. By Lemma 135.7 and Exercise 1.8(a), we get M L W . It remains to prove the reverse inclusion W M L. Let S.p 3 / Š A < G and set F D LA; then exp.F / D p, F is nonabelian so D1 .F / D F . We have to prove that A < M L. By the modular law, F D L.F \†p2 / so that T D F \†p2 is nonabel-

322

Groups of prime power order

ian of exponent p since exp.F / D p and L Z.F /. We get T Š S.p 3 / as jT j D p 3 . It follows that T D1 .†p2 / D M . Thus A < LT LM so that W D LM . Therefore the set of p-groups satisfying the hypothesis of Theorem 135.1(b) is inﬁnite. Remark 2. Let k > 1 and G D k .G/ be a nonabelian p-group. Suppose that G is such that Dk .G/ < G. Then G possesses only one maximal G-invariant abelian subgroup A of exponent p k . The set A [ Dk .G/ contains all elements of orders p k so that G D ADk .G/. In particular, A 6 Dk .G/ and G 0 Dk .G/. Indeed, we have MAk .G/ ¤ ¿ (Lemma 135.6) so Dk .G/ D H > ¹1º. By hypothesis, H < G. Let A be a G-invariant abelian subgroup of exponent p k such that jAj is as large as possible. We claim that the set H [A contains all elements of orders p k . Assume that this is false. Then there exists an element x 2 G .H [A/ of order p k . Set F D hx; Ai; then F is nonabelian by Corollary 10.2 since p k > 2. By Lemma 135.5, there is an element a 2 A such that L D hx; ai is minimal nonabelian. As o.x/; o.a/ p k , it follows that k .L/ D L, and we conclude that x 2 L 2 MAk .G/ D H by (), contrary to the choice of x. Thus x does not exist. Since HA . H [ A/ is normal in G and the set H [ A contains all elements of G of orders p k , we get G D k .G/ HA so that G D HA. In particular, A 6 H and G 0 H since G=H Š A=.H \ A/ is abelian. Assume that there is another maximal G-invariant abelian subgroup A1 of exponent p k . By the above, A1 6 H . Therefore, since A1 H [ A, it follows that A1 D .A1 \ H / [ .A1 \ A/ is a union of two nonincident subgroups, which is impossible. The groups from Theorem 135.1(b) and Remark 2 deserve further investigation. In particular, we do not even know if a group G from Remark 2 satisfying Dk .G/ > G does exist. Write k .G/ D jMAk .G/j. Clearly, k .G/ D k .Dk .G//. Proposition 135.9. Let G be a 2-group with nonabelian 1 .G/. Then (a) 1 .G/ D 1 if and only if G 2 ¹D8 ; SD16 º. (b) 1 .G/ D 2 if and only if G 2 ¹D16 ; SD32 º. (c) 1 .G/ D 3 if and only if 1 .G/ D D C , the central product, where D Š D8 , C is cyclic of order 4 and D \ C D Z.D/. Next, GN D G=CG .D/ 2 ¹E4 ; D8 º. If GN Š E4 , then G D D CG .D/, where CG .D/ is cyclic. If GN Š D8 , then we have G D DQ, where D \ Q D Z.D/ and Q is either cyclic or a generalized quaternion group, Q is not G-invariant, jQ W CG .D/j D 2. In the second case, G is as in Theorem 43.9 (in this case, G contains exactly seven involutions). Proof. Note that 1 .G/ D 1 .1 .G//. Since 1 .G/ is nonabelian, we conclude that 1 .G/ 1. As we know, all members of the set MA1 .G/ are isomorphic to D8 . Let D 2 MA1 .G/. If CG .D/ < D, then we see that G is of maximal class by Proposition 10.17. In this case, if jGj D 2n , then we get 1 .G/ D 2n3 if G Š D2n and 1 .G/ D 2n4 if G Š SD2n (clearly, 1 .Q2n / D 0). Therefore, if 1 .G/ D 1, then

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G 2 ¹D8 ; SD16 º, and if 1 .G/ D 2, then we have G 2 ¹D16 ; SD32 º. If CG .D/ 6 D, let L CG .D/ be cyclic of minimal order such that L 6 D; then we get jLj 4. In case jLj D 2, we obtain that 1 .DL/ D 1 .D L/ D 4 > 3, a contradiction. If jLj D 4, then DL D DL is the central product of order 16, and then 1 .DL/ D 3. The cases (a), (b) are completed. Now let 1 .G/ D 3; then G is not of maximal class. One may assume from the start that DGG. If L1 ¤ L is another cyclic subgroup of order 4 in CG .D/ (see the previous paragraph), then 1 .DL1 / D 3 and then 1 .G/ 5, a contradiction. Thus L is the unique cyclic subgroup of order 4 in CG .D/. Then we have 1 .CG .D// D 0. Indeed, if D1 2 MA1 .CG .D//, then, since 1 .CG .D// 2, CG .D/ is of maximal class by (a) and (b), so DD1 D D D1 is extraspecial of order 25 . If L0 < D1 is cyclic of order 4, then 1 .DL0 / D 3 by the above. As D1 6 DL0 , we get 1 .DD1 / > 3, a contradiction. It follows that CG .D/ is cyclic by Theorem 1.17(b). In this case, we have 1 .G/ D D L, where L is cyclic of order 4 since 1 .G/ D D1 .G/. The quotient group G=CG .D/ 2 ¹E4 ; D8 º is a subgroup of a Sylow 2-subgroup of Aut.D/ Š S4 containing D=Z.D/ Š E4 . If G=CG .D/ Š E4 , then G D D CG .D/, where CG .D/ is cyclic. Let GN D G=CG .D/ Š D8 . Let xN 2 GN DN be an involution and Q D hx; CG .D/i. Then Q has only one subgroup of order 2 coinciding with Z.D/ so Q is either cyclic or a generalized quaternion group. In that case, G is as in Theorem 43.9(b, c). Let G be a 2-group with 1 .G/ D 4. Take D8 Š D < G. If CG .D/ < D, then G is of maximal class so G 2 ¹D32 ; SD64 º. Next we assume that CG .D/ 6 D. Assume that CG .D/ D C > ¹1º is cyclic. Then 1 .DC / D 3. Let D8 Š D1 6 CD. Since G is not of maximal class, CG .D1 / 6 D1 . In this case, we obtain 1 .D1 CG .D1 // 3 so 1 .G/ 5 > 4, a contradiction. Thus CG .D/ is noncyclic. If G D D R, where R > ¹1º is either cyclic or generalized quaternion, then 1 .G/ D 4. It is easy to show that if a 2-group G contains exactly one subgroup Š Q8 , then we have G 2 ¹Q8 ; SD16 ; Q8 C2n º (the last group has order 2nC2 24 ). It is not easy to classify the 2-groups with exactly two or three subgroups isomorphic to Q8 . Proposition 135.10. Suppose that p > 2 and G is a p-group with nonabelian 1 .G/. If 1 .G/ D 1, then one of the following holds: (a) G D C 1 .G/, where C is cyclic and 1 .G/ Š S.p 3 /, jGj D p 2 jC j. (b) G is a group of maximal class and order 34 . If, in addition, G D 1 .G/, then we have G Š †32 . Proof. By hypothesis, we have D1 .G/ Š S.p 3 /. One may assume that D1 .G/ < G. If D1 .G/ D 1 .G/, then G has no normal elementary abelian subgroup of order p 3 so G is as in (a) or a 3-group of maximal class. In the second case, jGj D 34 so G is as in (b).4 4 According to a report of Mann, the set of groups X of maximal class and order p pC1 , p > 2, satisfying j1 .G/j D p p is nonempty.

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Groups of prime power order

Now let D1 .G/ < 1 .G/. Take x 2 1 .G/ D1 .G/ and write H D hx; D1 .G/i; then 1 .H / D H . If exp.H / D p, then we get 1 .G/ p > 1 by Theorem 5.8(b), a contradiction. Thus exp.H / > p so H is irregular, and we obtain p D 3 by Theorem 12.1(a). If H D 1 .G/, then G is of maximal class by Theorem 13.5 since H has at most two subgroups of order 33 and exponent 3. In this case, in view of D1 .G/ G G, we get jGj D 34 (Theorem 9.6(c)) so G D H . Thus we conclude that (the irregular 3-group) G D 1 .G/ > D1 .G/ has only one nonabelian subgroup D1 .G/ of order 33 and exponent 3 so G Š †32 by Theorem 30.6. Set ˛1 .G/ D jMA.G/j. It is known (see 76) that ˛1 .G/ p unless G is either abelian or minimal nonabelian. Recall that a p-group is said to be an A2 -group if all its subgroups of index p 2 are abelian but G has a nonabelian maximal subgroup (see 76). Given a noncyclic p-group G, set 2 D ¹H < G j ˆ.G/ < H; jG W H j D p 2 º. Proposition 135.11. If a p-group G possesses a proper A2 -subgroup, then there exist ˛1 .G/ .2p 3/ pairwise distinct minimal nonabelian subgroups generating G. Proof. Let H 2 1 be neither abelian nor minimal nonabelian (such an H exists by hypothesis). Then ˛1 .H / p by Remark 76.1. The subgroup H contains a member T of the set 2 ; then T ¤ Z.G/. Let H1 =T D H=T; H2 =T; : : : ; HpC1 =T be all subgroups of order p in G=T ; then we have H1 D H; H2 ; : : : ; HpC1 2 1 . One may assume that H D H1 ; : : : ; Hp are nonabelian. By Remark 76.1, H contains p 1 pairwise distinct minimal nonabelian subgroups, say S1 ; : : : ; Sp1 , that are not contained in T . For i 2 ¹2; : : : ; pº, let Mi be a minimal nonabelian subgroup of Hi that is not contained in T (if Hi is minimal nonabelian, then Mi D Hi ). By Lemma 135.7, we have G D hHp ; S1 i D hMA.Hp / [ ¹S1 ºi. Therefore G D hMA.G/ ¹S2 ; : : : ; Sp ; M2 ; : : : ; Mp1 ºi if p > 2; G D hMA.G/ ¹S2 ºi

if p D 2:

The proof is complete. Given a subgroup N of a nonabelian p-group X, denote by N .X/ the subgroup generated by all those minimal nonabelian subgroups of X which are not contained in N . If N is abelian, then N .X/ D X (Lemma 135.7), and N .X/ D ¹1º if and only if N D X is minimal nonabelian. Proposition 135.12. Suppose that N is a proper subgroup of a nonabelian p-group G. If N .G/ < G, then p D 2. Proof. By the paragraph preceding the proposition, N is neither abelian nor minimal nonabelian. Assume, by way of contradiction, that p > 2. We proceed by induction on jGj. First suppose that N 2 1 . Since j1 j 1 C p .mod p 2 / and, by Exercise 1.6(a), the set 1 contains 0, 1 or p C 1 abelian members, it follows that the set 1 contains

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325

at least p nonabelian members, say N D M1 ; M2 : : : ; ; Mp . Let i > 1. If A Mi is minimal nonabelian and such that A 6 N \ Mi , then A 6 N so that A N .G/. By induction, N \Mi .Mi / D Mi for i > 1. Therefore Mi N .G/ for i > 1. Since p 3, we get N .G/ M2 M3 D G, contrary to the hypothesis. Now suppose that a nonabelian N 62 1 . Then we have N < M1 2 1 , and M1 is nonabelian as well. By the previous paragraph, we see that M1 .G/ D G. However, G D M1 .G/ N .G/. Since, by hypothesis, N .G/ < G, this is a ﬁnal contradiction. Thus, if G is not minimal nonabelian, then p D 2. Let a nonabelian 2-group G be such that 1 D ¹N; M; Aº, where N; M are nonabelian and A is abelian. Then N .G/ M < G. Thus the set of nonabelian 2-groups satisfying the hypothesis of Proposition 135.12 is inﬁnite. However, some useful information on the pairs of 2-groups N < G satisfying N .G/ < G is contained in the following Proposition 135.13. Suppose that N is a proper subgroup of a nonabelian 2-group G. If N .G/ < G, then the following hold: (a) N 2 1 . (b) N .G/ 2 1 . (c) If N \ N .G/ D T , then there is an abelian A 2 1 such that T < A. (d) A is the unique abelian member of the set 1 . In what follows we assume that d.G/ > 2. (e) The set 1 has no minimal nonabelian members. (f) If a nonabelian K < G is such that K 6 N .G/ and K 6 N , then K \ N is nonabelian and N \K .K/ D K \ N .G/. (g) If a nonabelian L < G is such that L 6 N , L 6 N .G/ and L is minimal by inclusion subjecting the above conditions, then d.L/ D 2 and L\N .G/ .L/ D L \ N : Proof. As we have noticed, N is nonabelian. (i) First suppose that N 2 1 . Assume that the set 1 has no abelian members. Let M 2 1 ¹N º be arbitrary and let K 2 1 ¹N; M º be such that N \ M < K. Let S K be minimal nonabelian such that S 6 N \K.D N \M /. Then S 6 M . As S N .G/, we get N .G/ 6 M . Since M is arbitrary, it follows that N .G/ is not contained in any maximal subgroup of G, and we conclude that N .G/ D G, contrary to the hypothesis. Thus there is an abelian A 2 1 . Let N .G/ M 2 1 and let N \ M < L 2 1 ¹M; N º. We claim that L is abelian. Assume that this is false. Take S L such that S 6 N \L.D N \M /. Then

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S 6 M so that S 6 N .G/. Since S 6 N , we get a contradiction. Thus L is abelian, and so one may assume that L D A. (ii) Now we show that N; N .G/ 2 1 . Assume that N [ N .G/ [ A ¤ G. Take x 2 G .N [M [A/, where N .G/ M 2 1 . Then, by Lemma 135.5, there exists a 2 A such that the subgroup U D ha; xi is minimal nonabelian. By the choice of x, the subgroup U is not contained in N and N .G/, which is a contradiction. Thus we get G D N [ N .G/ [ A. Then it follows from Lemma 116.3 that N; N .G/ 2 1 , completing the proof of (a) and (b). (iii) Assume that there exists an abelian A1 2 1 ¹Aº. Set T D N \ A1 and let H; N; A1 be all maximal subgroups of G containing T . Clearly, T ¤ N \ A. Take x 2 H T . Then there is a1 2 A1 such that M D ha1 ; xi is minimal nonabelian. By the choice of x, we have M 6 N whence M N .G/. Such M cover the set H T . Since hH T i D H , it follows that H N .G/. Because H and N .G/ are distinct maximal subgroups of G, we get a contradiction. Thus A1 does not exist, completing the proof of (d). (iv) Supposing that d.G/ > 2, we will prove that the set 1 has no minimal nonabelian members. Assume that this is false and let H 2 1 be minimal nonabelian. Then, by hypothesis, either H D N or H D N .G/. (iv1) Assume that H D N .G/. Then H is the unique minimal nonabelian subgroup of G not contained in N , i.e., in the notation of 76, ˇ1 .G; N / D 1. Then, by Lemma 76.5, d.G/ D 2, contrary to the hypothesis. (iv2) Assume that H D N . Then N is the unique minimal nonabelian subgroup of G not contained in N .G/, and again, as in (iv1), ˇ1 .G; N .G// D 1 and d.G/ D 2, a contradiction. Thus H does not exist, so (e) is true. In what follows we assume that d.G/ > 2. (v) Let a nonabelian K < G be as in (f) (K exists since d.G/ > 2, for example, K 2 1 ¹N; N .G/; Aº). By hypothesis, K is not minimal nonabelian. Due to Lemma 135.7, K \ N .G/ is nonabelian (otherwise, all minimal nonabelian subgroups of K are contained in K \ N so K N by Lemma 135.7). Let S < K be minimal nonabelian such that S 6 K \ N .G/; then S 6 N .G/. By hypothesis, S < N so S K \ N . It follows that K\N .G/ .K/ D K \ N . Since one of the proper subgroups K \ N and ıK\N .K/ of K is nonabelian, we infer from Lemma 135.7 that the second is also nonabelian. This completes the proof of (f). (vi) Let L be such as in (g). Write T D N .G/ \ N \ A. Take x 2 N .G/ T , y 2 N T such that xy ¤ yx (since hN T i D N , such x; y exist). Let H D hx; yi; then H is nonabelian and H 6 N .G/, H 6 N and H < G as d.H / D 2 < d.G/. Then we have H \N .G/ .H / D H \ N by (f). If a nonabelian L H is minimal by inclusion such L 6 N .G/ and L 6 N , then the above argument shows that L is the desired subgroup.

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Remark 3. Suppose that p > 2 and G is a p-group such that 1 < 1 .G/ < p (see Proposition 11), say MA1 .G/ D ¹S1 ; : : : ; Sk º, 1 < k < p. Then all subgroups Si are normal in G. Set H D S1 S2 . If exp.H / D p, then, by Lemma 30.5(a), we have 1 .H / p > 1 .G/, a contradiction. Thus we get exp.H / > p so H is irregular (Theorem 7.2(b)), and we infer that jH j D p 4 , p D 3 by Fitting’s lemma and Theorem 7.1(b). By Theorems 30.7 and 9.6, the group G is of maximal class and order 34 . (In this case, G 6Š †32 .)

Problems Problem 1. Study the p-groups G D k .G/ containing exactly two distinct maximal subgroups U; V such that k .U / D U and k .V / D V . Problem 2. Study the nonabelian p-groups G D k .G/.D hx 2 G j o.x/ D p k i), k > 1, that are not generated by minimal nonabelian subgroups of exponent p k . (See Remark 2 and Theorem 135.9.) Problem 3. Let k > 1. Study the p-groups G > k .G/ such that H D k .H / for all H < G with exp.H / p k . Problem 4. Study the p-groups G such that 1 .ˆ.G// < ˆ.G/ however 1 .ˆ.H // D ˆ.H / .1 .G 0 / < G 0 however 1 .H 0 / D H 0 / for all H < G (two problems). Problem 5. Given k > 1, classify the nonabelian p-groups G D k .G/ containing exactly one minimal nonabelian subgroup of exponent p k . (Compare this with Proposition 135.10.) Problem 6. Classify the p-groups G, p > 2, with 1 .G/ p (see Remark 3). Problem 7. Classify the 2-groups G such that 1 .G/ Š D8 C2 .

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We study the p-groups all of whose nonabelian subgroups of a ﬁxed order are two-generator. A number of important results of this kind for abelian subgroups were proved by Blackburn and are used essentially in the Odd Order paper [FT]. Blackburn has classiﬁed the p-groups, p > 2, all of whose normal abelian subgroups are two-generator (see Theorem 13.7 and Theorem 69.4); for p D 2 this was done by the second author in 50. In contrast, in this section we study the p-groups, p > 2, in which certain nonabelian subgroups are two-generator. All our proofs are fairly short but involved. It is known a comparatively small number of structure results for groups of exponent p. One of such results is Theorem 136.1. Theorem 136.1. Suppose that G is a nonabelian group of order p m and exponent p and n is a ﬁxed integer such that 3 < n < m. Then the following assertions are equivalent: (a) All nonabelian subgroups of G of order p n are two-generator. (b) G is of maximal class with abelian subgroup of index p. In particular, m p. Since all nonabelian groups of order p 3 are two-generator, the hypothesis of Theorem 136.1 is fulﬁlled in case n D 3 for all nonabelian groups of exponent p. Therefore we assume there that n > 3. Note that p > 3 in this theorem (indeed, groups of exponent 3 are of class at most 2). It is impossible to state an analog of Theorem 136.1 for nonabelian p-groups of exponent > p since many such groups need not necessarily contain a nonabelian subgroup of order p n for a ﬁxed n > 2. For groups of exponent p the existence of such subgroups is due to the fact that minimal nonabelian subgroups of exponent p have order p 3 (see Lemma 136.2(ii)). We collect in Lemma 136.2 most results used in what follows. Some results are not stated in full generality. Lemma 136.2. Suppose that G is a nonabelian group of order p m . (i) (Theorem 5.8(b)) If G is of exponent p and n 2 ¹3; : : : ; m 1º, then the number of two-generator subgroups of order p n in G is divisible by p.

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p-groups in which certain proper nonabelian subgroups are two-generator 329

(ii) (Redei; see Lemma 65.1) Suppose that G is minimal nonabelian. Then d.G/ D 2;

jG 0 j D p;

jG W Z.G/j D p 2 :

If exp.G/ D p, then jGj D p 3 (in this case p > 2).1 Next, j1 .G/j p 3 and jG=Ã1 .G/j p 3 . (iii) (Exercise 13.10(a)) Let H < G. If all subgroups of G of order pjH j containing H are of maximal class, then G is also of maximal class. (iv) (Tuan; Lemma 1.1) If A < G is abelian of index p, then jGj D pjG 0 jjZ.G/j. (v) (Proposition 10.17) If B G is nonabelian of order p 3 such that CG .B/ < B, then G is of maximal class. (vi) (Exercise 1.6(a)) The number of abelian subgroups of index p in G is equal to one of the numbers 0, 1 or p C 1. (vii) (Theorem 12.12(a)) Suppose that G contains a subgroup H of maximal class and index p. If d.G/ D 2, then G is also of maximal class (otherwise, we have d.G/ D 3). (viii) (Blackburn; see Theorem 9.5) If G of maximal class is regular (in particular, of exponent p), then jGj p p . (ix) (Hall’s regularity criterion; see Theorem 9.8(a)) If G is irregular, then we have jG=Ã1 .G/j p p . In particular, absolutely regular p-groups are regular. (x) (Theorem 44.12) Suppose that N is a two-generator normal subgroup of G. If N ˆ.G/, then N is metacyclic. (xi) (Lemma 30.3(a, b)) Suppose that a p-group G D k .G/ contains a subgroup E D k .E/. Then we have E U 2 k with k .U / D U . There is also V 2 1 ¹U º with k .V / D V . (xii) (Blackburn; see Theorem 12.1(a)) If a group G is neither absolutely regular nor of maximal class, it has a normal subgroup of order p p and exponent p. (xiii) (Feit–Thompson [FT, Lemma 8.3]; see also Corollary 10.2) If E < G is a maximal normal elementary abelian subgroup of G, p > 2, then E satisﬁes 1 .CG .E// D E. (xiv) (Lemma 57.1) Let A < G be a maximal normal abelian subgroup. Then for any x 2 G A there is a 2 A such that the subgroup hx; ai is minimal nonabelian. (xv) (Theorem 10.28) Any nonabelian p-group G is generated by minimal nonabelian subgroups. (xvi) If G contains an abelian subgroup of index p and jG W G 0 j D p 2 , then G is of maximal class. 1 It

follows from this the fact important in what follows: a nonabelian group of exponent p contains a nonabelian subgroup of order p 3 .

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Groups of prime power order

(xvii) If G is of order p 4 and exponent p and d.G/ D 2, then G is of maximal class. (xviii) If G is a p-group of maximal class with abelian subgroup of index p, then all nonabelian subgroups of G are of maximal class so two-generator. Proof of Lemma 136.2(xvi). We proceed by induction on jGj. By Lemma 136.2(iv), jZ.G/j D p1 jG W G 0 j D p. Write GN D G=Z.G/. Then jGN W GN 0 j p 2 D jG W G 0 j so that jGN W GN 0 j D p 2 . Therefore, if jGj D p 3 , we are done. If jGj > p 3 , then GN is of maximal class by induction. Then G is also of maximal class since jZ.G/j D p. Proof of Lemma 136.2(xvii). Obviously, the group G has an abelian subgroup of index p. Since G 0 D ˆ.G/ and, by hypothesis, jG W G 0 j D p 2 , the result follows from Lemma 136.2(xvi). Proof of Lemma 136.2(xviii). One may assume that jGj > p 3 . We proceed by induction on jGj. Let H 2 1 be nonabelian and A 2 1 be abelian. As G D AH , we conclude that cl.H / D cl.G/ 1 (Fitting’s lemma) hence H is of maximal class. Now let U < G be nonabelian. Let U H 2 1 . By the above, H is of maximal class, and A \ H is abelian of index p in H . Therefore, by induction, U is of maximal class. Proof of Theorem 136.1. We ﬁrst have to prove that (a) ) (b). Assume that G is a counterexample of minimal order; then G is not of maximal class. Since exp.G/ D p, we have G 0 D ˆ.G/. (i) Suppose that n D m 1. Since j1 j 1 .mod p/, there is in G a non-two-generator subgroup, say A, of order p n (Lemma 136.2(i)). By hypothesis, A is abelian. Due to Lemma 136.2(ii), G is not minimal nonabelian since m > 4. Therefore there is a nonabelian M 2 1 . We have d.G/ d.M / C 1 D 3 since, by hypothesis, d.M / D 2. Next, M \ A is an abelian subgroup of index p in M . As jM W M 0 j D jM W ˆ.M /j D p 2 , the subgroup M is of maximal class (Lemma 136.2(xvi)). Then, by Lemma 136.2(vii), d.G/ D 3 since, by assumption, G is not of maximal class. Assume that m D 5 so n D 4. Let B < G be minimal nonabelian; then jBj D p 3 (Lemma 136.2(ii)). If CG .B/ < B, then G is of maximal class (Lemma 136.2(v)), contrary to the assumption. Thus there exists x 2 CG .B/ B. Then K D B hxi is nonabelian of order p 4 D p n and d.K/ D 3, contrary to the hypothesis. Thus m > 5. Now we are ready to complete the proof of our implication in case n D m 1. Take N G G such that N < G 0 and jG=N j D p 5 (recall that jG W G 0 j D p 3 ); then N > ¹1º, and G=N is neither abelian nor of maximal class. Therefore there is in G=N a nonabelian maximal subgroup M=N (Lemma 136.2(ii)); then M 2 1 is nonabelian. By hypothesis, d.M / D 2 hence d.M=N / D 2. Thus all nonabelian maximal sub-

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p-groups in which certain proper nonabelian subgroups are two-generator 331

groups of G=N are two-generator. By induction, G=N is of maximal class, which is a contradiction since d.G=N / D 3 > 2. Thus, if n D m 1, then G is of maximal class with and abelian subgroup of index p. (ii) Now let 3 < n < m 1. Let B < G be nonabelian of order p n (B exists by Lemma 136.2(ii)), and let B < H < G, where jH j D p nC1 . By hypothesis, all nonabelian maximal subgroups of H are two-generator so H is of maximal class by (i). Thus all B-containing subgroups of G of order pjBj D p nC1 are of maximal class. By Lemma 136.2(iii), the group G is of maximal class as well. Take L 2 1 such that d.L/ > 2 (L exists by Lemma 136.2(i)). Then either by hypothesis (provided m D 5) or by induction (provided m > 5), we conclude that the subgroup L is abelian. The implication (b) ) (a) follows from Lemma 136.2(xviii). Finally, the inequality m p follows from Lemma 136.2(viii). Remark 1. Suppose that a p-group G of order p m > p 4 is neither abelian nor minimal nonabelian. We claim that all proper nonabelian subgroups of G have centers of order p if and only if G is of maximal class with abelian subgroup of index p. Indeed, all minimal nonabelian subgroups have the same order p 3 (Lemma 136.2(ii)). If B < G is minimal nonabelian, then CG .B/ D Z.B/ < B is of order p. Therefore G is of maximal class (Lemma 136.2(v)). If R GG is of order p 2 , then CG .R/ has no minimal nonabelian subgroups by the above, so CG .R/ is abelian and maximal in G. Conversely, if G is of maximal class with abelian subgroup of index p, then all proper nonabelian subgroups of G have the same center of order p (Lemma 136.2(xviii)). The simple argument above was the starting point of some subsequent considerations. Theorem 136.3. Suppose that G is a nonabelian group of order p m > p 4 and assume that n 2 ¹4; : : : ; m1º is ﬁxed. If a p-group G has a nonabelian subgroup of order p n and all nonabelian subgroups of order p n have centers of the same order p, then G is of maximal class. Proof. By hypothesis, G is neither abelian nor minimal nonabelian. (i) Suppose that n D m 1. Let R G G be of order p 2 . Then jG W CG .R/j p. In this case, there is a maximal subgroup M of G such that R < M CG .R/. Since R Z.M / is of order p 2 > p, the subgroup M is abelian by hypothesis. Let H 2 1 be nonabelian; then jZ.H /j D p. In this case, H has an abelian subgroup H \ M of index p. By Lemma 136.2(iv), we have jH W H 0 j D pjZ.H /j D p 2 and so, by Lemma 136.2(xvi), H is of maximal class. In this case, all nonabelian members of the set 1 are of maximal class and d.G/ d.H / C 1 D 2 C 1 D 3. Assume, by way of contradiction, that G is not of maximal class. Then it follows from Theorem 12.12(c) that d.G/ D 3 and the number of subgroups of maximal class and index p in G is equal to p 2 . As we have noticed, all nonabelian members of the set 1 are of maximal class. It follows that G has exactly .p 2 C p C 1/ p 2 D p C 1 > 1 pairwise distinct abelian maximal subgroups so cl.G/ D 2. In this case, cl.G/ D 2

332

Groups of prime power order

hence cl.H / D 2, and we conclude that jH j D p 3 . Then jGj D pjH j D p 4 < p m , which is a contradiction. Thus, if n D m 1, then G is of maximal class. (ii) Now suppose that jGj D p m > p nC1 p 5 . Let H < G be nonabelian of order p n and H < F < G, where jF W H j D p. Then, by hypothesis, all nonabelian maximal subgroups of F have centers of the same order p so F is of maximal class by (i). Thus all subgroups of G containing H and having order pjH j are of maximal class. In this case, by Lemma 136.2(iii), the group G is also of maximal class, completing case (ii) and thereby the proof of our theorem. Theorem 106.3 asserts that if a 2-group G and all its nonabelian maximal subgroups are two-generator, then G is either metacyclic or minimal nonabelian. The following theorem extends this result to p-groups, p > 3. Note that in Theorem 136.4 we do not assume that d.G/ D 2 (we deduce this from another additional condition). Theorem 136.4. Suppose that all nonabelian maximal subgroups of a non-absolutely regular p-group G, p > 3, are two-generator. If, in addition, there is M 2 1 such that jM W M 0 j D p 2 , then G is of maximal class with abelian subgroup of index p. Proof. One may assume that exp.G/ > p (otherwise, the result follows from Theorem 136.1); then Ã1 .G/ > ¹1º. We have jG=Ã1 .G/j p p p 5 (the last inequality holds in view of p 5). Then, by Lemma 136.2(ii), there is in G=Ã1 .G/ a nonabelian maximal subgroup H=Ã1 .G/ hence d.H / D 2 by hypothesis, and we obtain that d.G/ d.H / C 1 D 2 C 1 D 3. Therefore, since Ã1 .G/ Ã1 .G/G 0 D ˆ.G/ and jG=Ã1 .G/j > p 3 , it follows that G=Ã1 .G/ is nonabelian. If K=Ã1 .G/ < G=Ã1 .G/ is nonabelian of index p (such a K=Ã1 .G/ exists since G=Ã1 .G/ is not minimal nonabelian in view of Lemma 136.2(ii)), then d.K/ D 2 by hypothesis, whence we get d.K=Ã1 .G// D 2. Thus any nonabelian maximal subgroup of the group G=Ã1 .G/ (of exponent p) is two-generator. It follows that G=Ã1 .G/ is of maximal class by Theorem 136.1. In particular, d.G/ D 2 since Ã1 .G/ ˆ.G/. By Lemma 136.2(i), there is A=Ã1 .G/ < G=Ã1 .G/ of index p such that d.A=Ã1 .G// > 2; then d.A/ > 2 hence A is abelian by hypothesis. Next, there exists M 2 1 such that jM W M 0 j D p 2 by hypothesis. In this case, M \ A is an abelian maximal subgroup of M , and it follows from Lemma 136.2(xvi) that M is of maximal class. As the two-generator p-group G contains a subgroup M of maximal class and index p, we infer that G is also of maximal class by Lemma 136.2(vii). Theorem 136.4 is an extension of Theorem 136.1. Supplement to Theorem 136.4. Let a p-group G of order > p 4 be neither abelian nor minimal nonabelian. If for all nonabelian M 2 1 we have jM W M 0 j D p 2 , then G is of maximal class. Proof. Let N G G be of index p 4 . Then G=N possesses an abelian subgroup A=N of index p. As jA W A0 j > p 2 , it follows by hypothesis that A 2 1 is abelian. Since G

136

p-groups in which certain proper nonabelian subgroups are two-generator 333

is not minimal nonabelian, there is a nonabelian M 2 1 . Then M \ A is abelian of index p in M so, by Lemma 136.2(xvi), M is of maximal class since jM W M 0 j D p 2 . Now, repeating word for word the last part of the proof of Theorem 136.3, we complete the proof. All maximal subgroups of a 3-group of maximal class and order > 34 are two-generator. The same is true for 2-groups of maximal class. However, there are inﬁnitely many 2-groups G with d.G/ D 3 all of whose maximal subgroups are two-generator; see 70, 113. In the proof of Theorem 136.7 we use the following two lemmas. Lemma 136.5. Let G be a p-group of order > p p , p > 2. If G contains a subgroup R of order p p and exponent p and if all such subgroups are (elementary) abelian, then 1 .G/ is also (elementary) abelian. Proof. Bringing to mind our aim, one may assume that G D 1 .G/. It remains to prove that G is abelian. We proceed by induction on jGj. We have R U 2 1 such that 1 .U / D U and V 2 1 ¹U º such that 1 .V / D V (Lemma 136.2(xi)). By induction, U and V are elementary abelian. Then we obtain that U V D G so cl.G/ 2 (Fitting’s lemma). Since G is not minimal nonabelian, we conclude that it is abelian (Theorem 136.1). Lemma 136.6. Let a non-absolutely regular nonabelian group G have order p pC1 , p > 3. Suppose that all nonabelian maximal subgroups of G are two-generator. Then one of the following holds: (a) 1 .G/ is elementary abelian of order p p and G=Ã1 .G/ is of maximal class so d.G/ D 2. Next, G is regular, jZ.G/j D p 2 and G=G 0 is abelian of type .p 2 ; p/. If B 2 1 ¹1 .G/º, then B=B 0 is abelian of type .p 2 ; p/. (b) G is of maximal class with abelian subgroup of index p. Proof. By Lemma 136.2(ii), G is not minimal nonabelian as jG=Ã1 .G/j p p > p 3 . Let a nonabelian K 2 1 ; then d.K/ D 2. By Theorem 136.1, G=Ã1 .G/ is nonabelian of maximal class with an abelian subgroup of index p. It follows that d.G/ D 2 since Ã1 .G/ ˆ.G/. If the quotient group A=Ã1 .G/ is maximal in G=Ã1 .G/ of rank > 2 (see Lemma 136.2(i)), then A is abelian. Suppose that G is not of maximal class; then jZ.G/j D p 2 so that cl.G/ D p 1 4: In this case, G is regular (Theorem 7.1(b)), and we obtain that 1 .G/ D M 2 1 (Theorem 7.2(d)) and exp.M / D p. Assume that M ¤ A. Then, since cl.G/ > 2, the subgroup M is nonabelian. The intersection A \ M is an abelian subgroup of index p in M . Since exp.M / D p and d.M / D 2, we get M 0 D ˆ.M / so that jM W M 0 j D p 2 , and Lemma 136.2(xvi)

334

Groups of prime power order

implies that M is of maximal class. As d.G/ D 2, it follows that G is of maximal class (Lemma 136.2(vii)), contrary to the assumption. Now let M D A; then M.D 1 .G// is elementary abelian of order p p . As we have noticed, jZ.G/j D p 2 . In this case, by Lemma 136.2(iv), jG W G 0 j D pjZ.G/j D p 3 so G=G 0 is abelian of type .p 2 ; p/ since d.G/ D 2. Let R < G 0 be G-invariant of index p. By Lemma 65.2(a), the quotient group G=R is minimal nonabelian. As R < M and M=R Š Ep3 , it follows that 1 .G=R/ Š Ep3 . Let B 2 1 ¹1 .G/º. Since B=R is abelian of type .p 2 ; p/, we get cl.B/ p 2. We conclude from BM D G that cl.B/ p 2 (Fitting’s lemma). Thus we obtain cl.B/ D p 2, and we infer that R D B 0 so that B=B 0 is abelian of type .p 2 ; p/. Theorem 136.7. Let a p-group G of order > p p , p > 3, be not absolutely regular, and let a ﬁxed natural number n be such that 3 < n < p. Suppose that all nonabelian subgroups of G of order p n and exponent p and all nonabelian subgroups of G of order p p and exponent p 2 are two-generator. Then one of the following holds: (a) 1 .G/ is elementary abelian. (b) G is of maximal class; in that case, 1 .ˆ.G// is (elementary) abelian. Proof. Suppose that 1 .G/ is nonabelian. By hypothesis, Theorems 9.5, 9.6, 9.8 and Lemma 136.2(xii), we get j1 .G/j p p1 with equality if and only if G is of maximal class. Suppose that G is of maximal class. Then G is irregular and j1 .ˆ.G//j D p p1 (Theorems 9.5, 9.6). Since, in view of p > 3, R D 1 .ˆ.G// is not metacyclic, we obtain d.R/ > 2 (Lemma 136.2(x)) so it is elementary abelian by Theorem 136.1 if n < p 1 and hypothesis if n D p 1. Next we assume that G is not of maximal class. In this case, there is in G a nonabelian subgroup R of order p p and exponent p (Lemma 136.5), and all such subgroups are of maximal class as n < p (Theorem 136.1). Take R < S G, where jS j D p pC1 . Assume that exp.S/ D p. Then all nonabelian subgroups of index p in S are twogenerator so S is of maximal class (Theorem 136.1), contrary to Lemma 136.2(viii). Thus we get exp.S/ D p 2 . By Lemma 136.6, S is of maximal class as 1 .S/. R/ is nonabelian. Since S runs over the set of all subgroups of G that contain R and have order pjRj, it follows, in view of Lemma 136.2(iii), that G is of maximal class. All p-groups of maximal class with an abelian subgroup of index p satisfy the hypothesis of Theorem 136.7; however, there is a group of maximal class and order p pC1 , p > 3, that does not satisfy this hypothesis [Bla3]. Theorem 136.8. Let G be a nonabelian p-group of order > p 4 and suppose that all its nonabelian subgroups of order p 4 are generated by two elements. Then one of the following holds: (a) G is of maximal class. (b) The subgroup 1 .G/ is abelian.

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p-groups in which certain proper nonabelian subgroups are two-generator 335

Proof. Assume that H D 1 .G/ is nonabelian. (i) We claim that if G contains a nonabelian subgroup B of order p 3 , then G is of maximal class. This is the case if CG .B/ < B by Lemma 136.2(v). Suppose that there is in CG .B/ B an element y of minimal possible order. Since y p 2 Z.B/, we get o.y/ p 2 and the subgroup K D hy; Bi has order p 4 and d.K/ D 3, contrary to the hypothesis since K is nonabelian. By Theorem 9.6(f), a p-group G of maximal class contains an element x such that CG .x/ is of order p 2 . Therefore, if CG .x/ < A < G with jA W CG .x/j D p, then A is nonabelian of order p 3 . Next we assume, in addition, that G is not of maximal class. Then, by the above, all subgroups of G of order p 3 are abelian. Since H D 1 .G/ is assumed to be nonabelian, we get jH j p 4 . Assume that G has no nonabelian subgroup of order p 4 . If G is regular, then we have exp.H / D p. Therefore, since H is nonabelian by assumption, we get jH j D p 3 (Lemma 136.2(ii)). But, by assumption, all subgroups of G of order p 4 which contain H are abelian, a contradiction. Now let G be irregular. As G is not of maximal class, it follows, by Lemma 136.2(xii), that there exists R G G of order p p and exponent p. Assume that p > 2. We claim that R is abelian. Indeed, if p D 3 and R < R1 < G, where jR1 j D 34 , then R1 is abelian by assumption; then R is also abelian. If p > 3, then R is abelian since it has no minimal nonabelian subgroup. Then, by Lemma 136.5, 1 .G/ is abelian so G is as in case (a). Now let p D 2. Then there are in the nonabelian subgroup H two noncommuting involutions x; y. In this case, hx; yi is dihedral so it contains a dihedral subgroup B of order 8. Then B is contained in a nonabelian subgroup of order 24 , contrary to the assumption. In what follows we assume that G has a nonabelian subgroup of order p 4 . By the above, all subgroups of G of order p 3 are abelian. It follows that all nonabelian subgroups of G of order p 4 are minimal nonabelian. Therefore, provided p D 2, G has a nonabelian (dihedral) subgroup of order 8 which is contained in some overgroup of order 16 which is not minimal nonabelian. Thus we must have p > 2. (ii) To obtain a contradiction, it remains to prove that the subgroup H.D 1 .G// is abelian. Therefore, bringing to mind our aim, one may assume, without loss of generality, that G D H.D 1 .G//. Let E < G be a G-invariant elementary abelian subgroup such that jEj is as large as possible. Let x 2 1 .G/ E be of order p. Then we infer that K D hx; Ei is nonabelian (Lemma 136.2(xiii)). If jEj < p 3 , then jEj D p 2 and K is nonabelian of order p 3 and exponent p, contrary to the assumption. Thus we get jEj > p 2 . In this case, by Lemma 136.2(xiv), there is a 2 E such that B D ha; xi is minimal nonabelian; then 1 .B/ D B. Therefore, by Lemma 136.2(ii), B is nonabelian of order p 3 and exponent p, a contradiction. Theorem 136.9. Let G be a nonabelian group of order p m > p 4 . (a) If exp.G/ D p and G possesses only one subgroup, say H , of order p 4 such that d.H / > 2, then G is of maximal class and m D 5.

336

Groups of prime power order

(b) Let p > 3 and let G be an irregular p-group. If G has only one subgroup, say H , of order p 4 and exponent p such that d.H / > 2, then one of the following holds: (b1) G is 5-group of maximal class, all subgroups of G of order 55 and exponent 5 are of maximal class. (b2) G D C 1 .G/, where 1 .G/ is of maximal class and of order 55 and exponent 5, C is absolutely regular. (b3) G is a group of maximal class, j1 .G/j D 54 . Proof. (a) Assume that G is of order p m > p 5 and exponent p. Let L G G be of order p 2 ; then C D CG .L/ has index p in G. If L < B < C is of order p 4 , then d.B/ > 2 (Lemma 136.2(xvii)). Since C of order > p 4 has only one non-two-generator subgroup of order p 4 , it follows that C is nonabelian. By Lemma 136.2(xv), there exists a nonabelian T < C of order p 3 with T 6 B. If U < L is of order p such that U 6 T , then D D T U ¤ B, jDj D p 4 , d.D/ > 2 and D ¤ B, contrary to the hypothesis. Thus m D 5. Due to Lemma 136.2(i), there is B 2 1 satisfying d.B/ > 2. By hypothesis, all members of the set 1 ¹Bº are two-generator so they are of maximal class (Lemma 136.2(xiii)). Assume that G is not of maximal class. Then the set 1 has exactly p 2 members of maximal class (Lemma 136.2(xxi)), and all other p C 1 > 1 members of 1 are not of maximal class so they are not two-generator (Lemma 136.2(xiii)), contrary to the hypothesis. Thus G is of maximal class and order p 5 , as was to be shown. (b) We have exp.G/ > p (Theorem 7.1(b)). If the subgroup 1 .G/ is abelian, then we see that j1 .G/j D p 4 by hypothesis. In this case, G is a 5-group of maximal class (Lemma 136.2(xii)) so G is as in (b3). Next we assume that 1 .G/ is nonabelian. Assume that G is not of maximal class. Then there is in G a subgroup R of order p p and exponent p (Lemma 136.2(xii)). Since p > 3, R is nonabelian (otherwise, R contains more than one subgroup isomorphic to Ep4 ). Then, by part (a), R is of maximal class and p D 5. Thus all subgroups of G of order 55 and exponent 5 have maximal class. Then, by Lemma 136.2(i) or Theorem 136.1, G has no subgroup of order 56 and exponent 5. Let L G G be abelian of type .5; 5/. It is easy to see that C D CG .L/ has index 5 in G (otherwise, LR is of order > 55 and exponent p). Assume that C has a G-invariant subgroup M of order 55 and exponent 5. Since exp.LM / D 5, we get L Z.M /, a contradiction since M , by the above, is of maximal class. Thus M does not exist. Therefore, by Lemma 136.2(xv), C is absolutely regular. Lemma 136.2(xi) then yields R D 1 .G/. Now G D CR is as in (b2). Now let G be of maximal class. Then K D 1 .ˆ.G// is of order p p1 and exponent p. Assume that p > 5; then K is nonabelian by hypothesis. Since K ˆ.G/, it follows that jZ.K/j > p (see Lemma 1.4). Then, by (a), K has two distinct nonabelian non-two-generator subgroups of order p 4 , contrary to the hypothesis. It follows that p D 5. If G has no subgroup of order 55 and exponent 5, it satisﬁes the hypothesis. Now assume that G has a subgroup R of order 55 and exponent 5. Then, by the above, R is of maximal class, completing the proof.

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p-groups in which certain proper nonabelian subgroups are two-generator 337

Remark 2. We present another application of Lemma 136.2(xiv). Let G D k .G/ be a p-group of exponent p k such that G is nonabelian but k .H / is abelian for all H 2 1 . We claim that then G is minimal nonabelian. In view of Lemma 136.2(xi), there is A 2 1 such that k .A/ D A. Then A is abelian by hypothesis. Let x 2 GA be of order p k . By Lemma 2(xiv), there is a 2 A such that H D ha; xi is minimal nonabelian. Since k .H / D H is nonabelian, it follows H D G, and we are done.

Problems Problem 1. Suppose that a group G is neither abelian nor minimal nonabelian with jGj D p m and p k D min¹jAj j A < G is minimal nonabelianº. Let a ﬁxed integer n be such that k < n < m. Study the structure of G provided all its nonabelian subgroups of order p n are two-generator. If p D 2 and n D m 1, then 70 shows that the set of 2-groups satisfying the condition of Problem 1 is inﬁnite. Problem 2. Study the p-groups of order > p 3 all of whose maximal subgroups except one have derived quotient groups of index p 2 . This problem is not solved even for groups of exponent p.2 Problem 3. Study the p-groups all of whose nonabelian subgroups of index p 2 are two-generator. (Note that the p-groups all of whose subgroups of index p 2 are abelian, are classiﬁed; see Lemma 65.1 and 71. All p-groups of maximal class with abelian subgroup of index p and 3-groups of maximal class satisfy this condition.) Problem 4. Study the p-groups all of whose nonabelian subgroups of order p 5 are two-generator. (See Theorem 136.8.)3 Problem 5. Study the p-groups of order p m and exponent p all of whose nonabelian subgroups of order p n , 2 < n < m 1 (n is ﬁxed), are nonnormal. (All p-groups of maximal class and exponent p > 5 with abelian subgroup of index p satisfy this condition.) Problem 6. Classify the p-groups all of whose nonabelian subgroups of class 2 are two-generator. Problem 7. Classify the p-groups G with exactly one nonabelian subgroup H of order p 4 such that d.H / D 3. (See Theorem 136.9.) Some close results are contained in 30, 135. p D 2, then G is of maximal class. Indeed, by Taussky’s theorem (Lemma 1.6), all nonabelian maximal subgroups of G are of maximal class (one may assume that jGj > 8). Assume that G is not of maximal class. Then, by Theorem 5.4, the number of members of maximal class in the set 1 is divisible by 4. Therefore the number of remaining members, say M , of the set 1 is equal to 3 since d.G/ D 3, and we get a contradiction since for these M we have jM W M 0 j > 4, again by Taussky’s theorem. 3 If G is a 2-group of order 25 and all its subgroups of orders 24 and 25 are two-generator, then G is metacyclic. This Blackburn’s result follows from Theorem 66.1. 2 If

137

p-groups all of whose proper subgroups have its derived subgroup of order at most p

Here we give a characterization of the title groups. In the proofs we partly use some ´ ideas of J. Q. Zhang, X. H. Li (Proposition 137.3) and V. Cepuli´ c, O. C. Piljavska (Proposition 137.5). To facilitate the proof of the main result (Theorem 137.7), we shall ﬁrst prove some auxiliary results. Proposition 137.1. Let G be a title group. Then for all x; y 2 G such that hx; yi < G we have o.Œx; y/ p and Œx; y 2 Z.G/. Proof. Suppose that Œx; y ¤ 1. Let X be a maximal subgroup of G containing hx; yi. Then X 0 D hŒx; yi E G with o.Œx; y/ D p and so Œx; y 2 Z.G/. Proposition 137.2. If G is a title group, then G 0 is abelian of order p 3 . Proof. We may assume that G is nonabelian. Let X ¤ Y be two maximal subgroups of G. Then jX 0 j p and jY 0 j p. By Exercise 1.69(a), we get jG 0 W .X 0 Y 0 /j p and so jG 0 j p 3 . By Burnside, G is abelian. Proposition 137.3 (Zhang and Li). If G is a title group and jG 0 j p 2 , then d.G/ 3. Proof. Since jG 0 j p 2 , G is not minimal nonabelian (Lemma 65.1) so there exists a maximal subgroup A with jA0 j D p; then A0 GG. Suppose that M 0 A0 for each maximal subgroup M of G. Then G=A0 is minimal nonabelian, and we get d.G=A0 / D 2 (Lemma 65.1), A0 ˆ.G/ and so d.G/ D 2, completing this case. We may assume that G has a maximal subgroup B such that B 0 6 A0 . In that case, we get jA0 j D jB 0 j D p and A0 \ B 0 D ¹1º. Let a1 ; a2 2 A and a3 ; a4 2 B be such that A0 D hŒa1 ; a2 i and B 0 D hŒa3 ; a4 i. Since jha1 ; a2 ; a3 ; a4 i0 j p 2 , we obtain ha1 ; a2 ; a3 ; a4 i D G by hypothesis, and so d.G/ 4. We assume, by way of contradiction, that d.G/ D 4. Then we have o.Œx; y/ p and Œx; y 2 Z.G/ for any x; y 2 G by hypothesis, and we conclude that G 0 is elementary abelian and G 0 Z.G/. In particular, cl.G/ D 2. For any k 2 ¹1; 2º and l 2 ¹3; 4º we have ha1 ; a2 ; al i < G and hak ; a3 ; a4 i < G hence ha1 ; a2 ; al i0 D hŒa1 ; a2 i and hak ; a3 ; a4 i0 D hŒa3 ; a4 i: It follows that

Œak ; al 2 hŒa1 ; a2 i \ hŒa3 ; a4 i D ¹1º:

137

p-groups G with jH 0 j p for all proper subgroups H of G

339

This implies Œa1 ; a2 a3 D Œa1 ; a2 Œa1 ; a3 D Œa1 ; a2 and Œa2 a3 ; a4 D Œa2 ; a4 Œa3 ; a4 D Œa3 ; a4 : But then ha1 ; a2 a3 ; a4 i is a proper subgroup of G and we have jha1 ; a2 a3 ; a4 i0 j p 2 , contrary to the hypothesis. The proof is complete. Proposition 137.4 (Berkovich). Let G be a nonabelian p-group. If d.G/ D 2, then we have H 0 < G 0 for each H < G. Proof. Let R < G 0 be a G-invariant subgroup of index p in G 0 . Then j.G=R/0 j D p and d.G=R/ D 2. This implies that G=R is minimal nonabelian. For each maximal subgroup H of G we have H 0 R < G 0 , and we are done. ´ Proposition 137.5 (Cepuli´ c and Piljavska). Let G be a title p-group with p > 2. Then for any a; b 2 G we have Œap ; b D Œa; b p D Œa; bp . Proof. Set g D Œa; b. If g commutes with a, then for each n 1 we prove by induction that Œan ; b D Œa; bn . Indeed, for n > 1, Œan ; b D Œaan1 ; b D Œa; ba

n1

Œan1 ; b D Œa; bŒan1 ; b

D Œa; bŒa; bn1 D Œa; bn : In particular, Œap ; b D Œa; bp . Now assume that Œg; a D z ¤ 1. Since hg; ai < G, Proposition 137.1 implies that o.z/ D p and z 2 Z.G/. We note that g a D a1 ga D g.g 1 a1 ga/ D gŒg; a D gz;

and so

i

g a D gz i for all i 1:

We have Œap ; b D Œa ap1 ; b D Œa; ba D Œa; ba

p1

Œa; ba

p2

p1

Œap1 ; b D Œa; ba

p1

Œa ap2 ; b

Œap2 ; b

and so continuing we ﬁnally get Œap ; b D Œa; ba D ga

p1

p1

ga

Œa; ba

p2

p2

: : : Œa; ba Œa; b

: : : g a g D .gz p1 /.gz p2 / : : : .gz/g p

D g p z .p1/C.p2/CC1 D g p z .p1/ 2 D g p D Œa; bp ; where we have used the fact that p > 2. Thus we have proved that in any case we get Œap ; b D Œa; bp . Now, Œa; b p D Œb p ; a1 D Œb; ap D Œa; bp ; and the proof is complete.

340

Groups of prime power order

Proposition 137.6. Suppose that G is a p-group which has one of the following properties: (a) jG 0 j p. (b) d.G/ D 2, jG 0 j D p 2 . (c) p > 2, d.G/ D 2, cl.G/ D 3, G 0 Š Ep3 , Ã1 .G/ Z.G/. (d) d.G/ D 3, cl.G/ D 2, G 0 Š Ep3 or Ep2 . Then G has the title property, i.e., jH 0 j p for each proper subgroup H of G. Proof. If G is a p-group with property (a), then obviously G has the title property. Suppose that G is a p-group in (b). By Proposition 137.4, for each H < G we have H 0 < G 0 and so G has the title property. Now assume that G is a p-group with property (c). Since cl.G/ D 3, it follows that ¹1º ¤ K3 .G/ Z.G/. But d.G/ D 2 and so ¹1º ¤ G 0 =K3 .G/ is cyclic and therefore K3 .G/ Š Ep2 . By assumption, Ã1 .G/ Z.G/ and so ˆ.G/ D Ã1 .G/G 0 is abelian and G=ˆ.G/ Š Ep2 . Also, Ã1 .G/K3 .G/ Z.G/ and in fact Ã1 .G/K3 .G/ D Z.G/. Indeed, if Ã1 .G/K3 .G/ < Z.G/, then G=Z.G/ Š Ep2 . But in that case, G has p C 1 abelian maximal subgroups and this implies (Exercise P1) jG 0 j D p, a contradiction. Let M be any maximal subgroup of G so that jM W ˆ.G/j D p. Then M is either abelian or satisﬁes Z.G/ D Z.M / and M=Z.M / Š Ep2 . In the second case, we may use Lemma 1.1 because ˆ.G/ is an abelian maximal subgroup of M . Then we conclude from jM j D pjZ.M /jjM 0 j that jM 0 j D p. We have proved that in this case G has the title property. Suppose that G is a p-group with property (d). For any x; y 2 G we have Œx p ; y D Œx; yp D 1 and so Ã1 .G/ Z.G/. It follows that ˆ.G/ D Ã1 .G/G 0 Z.G/ and G=ˆ.G/ Š Ep3 . Let M be any maximal subgroup of G satisfying M=ˆ.G/ Š Ep2 . Then p C 1 maximal subgroups of M containing ˆ.G/ are abelian. This implies that jM 0 j p, and we are done. Theorem 137.7. A p-group G has the property that each proper subgroup of G has its derived subgroup of order at most p if and only if one of the following holds: (a) jG 0 j p. (b) d.G/ D 2, jG 0 j D p 2 . (c) p > 2, d.G/ D 2, cl.G/ D 3, G 0 Š Ep3 , Ã1 .G/ Z.G/.1 (d) d.G/ D 3, cl.G/ D 2, G 0 Š Ep3 or Ep2 . Here we have ˆ.G/ D Z.G/. Proof. If G is a p-group in (a)–(d), then Proposition 137.6 implies jH 0 j p for each subgroup H < G. Suppose that G is a p-group all of whose proper subgroups have its derived subgroup of order p. If jG 0 j p, then we obtain the groups in part (a) of our theorem. 1 Such

p-groups G exist. See for example A2 -groups of order p 5 , p > 2, in Proposition 71.5(b).

137

p-groups G with jH 0 j p for all proper subgroups H of G

341

In what follows we assume that jG 0 j p 2 . By Proposition 137.2, G 0 is abelian of order p 2 or p 3 . Due to Proposition 137.3, we have d.G/ 3. (i) First assume that d.G/ D 2. If jG 0 j D p 2 , then we have obtained the groups in part (b) of our theorem. In the sequel we shall assume jG 0 j D p 3 . By Exercise 1.69(a), all p C 1 maximal subgroups Mi .i D 1; 2; : : : ; p C 1/ of G are nonabelian, jMi0 j D p and for any i ¤ j we have Mi0 \ Mj0 D ¹1º so that Mi0 Mj0 Š Ep2

and Mi0 Mj0 Z.G/:

If cl.G/ D 2, then d.G/ D 2 would imply that G 0 is cyclic, contrary to the existence of the subgroup Mi0 Mj0 Š Ep2 . Hence cl.G/ 3. But ¹1º ¤ K3 .G/ D ŒG; G 0 Mi0 Mj0 Z.G/ and so cl.G/ D 3. Set E D Mi0 Mj0 D G 0 \ Z.G/ Š Ep2 . Whenever a; b 2 G are such that ha; bi D G, then Œa; b 2 G 0 E. Indeed, if 1 ¤ Œa; b 2 E, then we have o.Œa; b/ D p and Œa; b 2 Z.G/. But then G=hŒa; bi is abelian and so G 0 D hŒa; bi is of order p, a contradiction. Let g D Œa; b, where ha; bi D G and g 2 G 0 E. For any x 2 G we have g x D ge with some e 2 E. Then i

g x D ge i

and so

p

g x D g:

It follows that Ã1 .G/ centralizes G 0 and so ˆ.G/ D Ã1 .G/G 0 centralizes G 0 . (i1) Now assume p > 2. Suppose, in addition, that G 0 is not elementary abelian. Then we get E D 1 .G 0 / and set ¹1º ¤ Ã1 .G 0 / D hsi < E so that G 0 =hsi Š Ep2 . If K3 .G/ D ŒG; G 0 D hsi, then cl.G=hsi/ D 2 so that d.G=hsi/ D 2 would imply that G 0 =hsi D .G=hsi/0 is cyclic, a contradiction. Hence there is an element c 2 G ˆ.G/ such that g c D gl with l D Œg; c 2 E hsi. Let d 2 G ˆ.G/ be such that hc; d i D G so that Œc; d 2 G 0 E. By Proposition 137.5, we get Œc; d p D Œc; d p D s j , where j 6 0 (mod p). Consider the maximal subgroup C D hˆ.G/; ci. Since g; c; d p 2 C , we have C 0 hŒg; c; Œc; d p i D hl; s j i D E Š Ep2 ; which is a contradiction. We have proved that G 0 Š Ep3 . For any x; y 2 G we get, by Proposition 137.5, Œx p ; y D Œx; yp D 1 and so Ã1 .G/ Z.G/. We have obtained the groups given in part (c) of our theorem. (i2) It remains to consider the case p D 2. Assume, in addition, that ¹1º ¤ K3 .G/ D ŒG; G 0 < E and set ŒG; G 0 D hui, where u is an involution in E Z.G/. Note that ˆ.G/ centralizes G 0 and for each x 2 G ˆ.G/ and y 2 G 0 E we have y x D yu0 with u0 2 hui. Set G0 D CG .G 0 / so that jG W G0 j D jG0 W ˆ.G/j D 2. Since G=hui is of class 2 and d.G=hui/ D 2, we conclude that G 0 =hui is cyclic. Hence if g 2 G 0 E, then g 2 D v

342

Groups of prime power order

is an involution in E hui and therefore E D 1 .G 0 / D hu; vi and Ã1 .G 0 / D hvi. Take some elements a 2 G0 ˆ.G/ and b 2 G G0 . Then ha; bi D G and therefore Œa; b D h 2 G 0 E with h2 D v, ha D h and hb D hu. Consider the maximal subgroup H D hˆ.G/; bi. Since Œa2 ; b D Œa; ba Œa; b D ha h D h2 D v

and

Œh; b D u;

we get H 0 hu; vi D E Š E4 , a contradiction. We have proved that K3 .G/ D ŒG; G 0 D E D G 0 \ Z.G/ Š E4 : Let a; b 2 G ˆ.G/ be such that ha; bi D G. Then g D Œa; b 2 G 0 E, Œg; a D c1 , Œg; b D c2 , where hc1 ; c2 i D E D K3 .G/. We set c3 D c1 c2 and get Œg; ab D Œg; bŒg; ab D Œg; bŒg; a D c2 c1 D c3 : We compute the commutator subgroups of our three nonabelian maximal subgroups X1 D hˆ.G/; ai;

X2 D hˆ.G/; bi

and X3 D hˆ.G/; abi;

where we note that we must have jXi0 j D 2 for i D 1; 2; 3. Since Œg; a D c1 and Œa; b 2 D Œa; bŒa; bb D gg b D g gc2 D g 2 c2 ; we infer that X10 D hc1 i and so we must have g 2 c2 2 hc1 i. This forces either g 2 D c2 or g 2 D c3 . As Œg; b D c2 and Œa2 ; b D Œa; ba Œa; b D g a g D gc1 g D g 2 c1 ; we obtain X20 D hc2 i and so we must have g 2 c1 2 hc2 i. This forces either g 2 D c1 or g 2 D c3 . With the above we get exactly g 2 D c3 . Since Œg; ab D c3 and Œa2 ; ab D Œa; aba Œa; ab D g a g D gc1 g D g 2 c1 (where we have used the fact that Œa; ab D Œa; b), we infer that X30 D hc3 i and so we must have g 2 c1 2 hc3 i. But we know that g 2 D c3 and so g 2 c1 D c3 c1 D c2 2 hc3 i, a contradiction. We have proved that such 2-groups do not exist! (ii) Finally, assume that d.G/ D 3. For any x; y 2 G we have hx; yi < G and so Proposition 137.1 implies that o.Œx; y/ p and Œx; y 2 Z.G/. But then G 0 is elementary abelian (of order p 2 or p 3 ) and G 0 Z.G/ and so we have obtained the groups from part (d) of our theorem. For any a; b 2 G we have Œap ; b D Œa; bp D 1 and so ˆ.G/ Z.G/. If Z.G/ 6 ˆ.G/, then there is a maximal subgroup M of G such that G D hM; xi, where x 2 Z.G/. But then G 0 D M 0 and so jG 0 j D 2, a contradiction. Hence ˆ.G/ D Z.G/. Theorem 137.7 is proved.

138

p-groups all of whose nonnormal subgroups have the smallest possible normalizer

1o . A p-group G is called an S1 -group if jNG .H / W H j D p for all nonnormal subgroups H of G. We determine here up to isomorphism all non-Dedekindian S1 -groups by using the results (but not the proofs) from [ZG]. This solves Problem 116(i) stated by the ﬁrst author. Lemma 138.1. Assume that G is a minimal nonabelian p-group of order > p 3 . If G is an S1 -group, then 2

2

G Š ha; b j ap D b p D 1; ab D a1Cp i D Mp .2; 2/: Proof. Suppose that G has an elementary abelian subgroup E of order p 3 . Then each subgroup of order p in E must be normal in G and so E Z.G/. Let X be a cyclic nonnormal subgroup in G. But then NG .X/ E and so jNG .X/ W X j p 2 , a contradiction. We have proved that G is metacyclic and so n

m

G Š ha; b j ap D b p D 1; ab D a1Cp

n1

i;

where n 2, m 1, n C m 4 and G 0 D hŒa; bi D hap i. Obviously, hbi is not normal in G and NG .hbi/ D hbi hap i which forces n D 2 and so m 2. Assume m2 m2 m2 m 3 so that b p 2 Z.G/, o.ab p / D p 2 and hab p i is not normal in G. m2 m2 m2 Indeed, if hab p i C G, then we have ŒG; hab p i hab p i \ G 0 D ¹1º and m2 p so ab 2 Z.G/ which forces a 2 Z.G/, a contradiction. But n1

NG .hab p and therefore jNG .hab p we are done.

m2

i/ W hab p

m2

m2

i/ D hai hb p i

ij p 2 , a contradiction. Hence m D 2 and

Theorem 138.2 ([ZG]). Assume that G is a nonabelian p-group, p > 2, and jGj > p 3 . Then G is an S1 -group if and only if 2

2

G Š ha; b j ap D b p D 1; ab D a1Cp i D Mp .2; 2/: Proof (Janko). Suppose that G is a nonabelian p-group, p > 2, jGj > p 3 and G is an S1 -group.

344

Groups of prime power order

First assume that G is nonabelian of order p 4 . Suppose, in addition, that G is not minimal nonabelian and let H be a minimal nonabelian subgroup of order p 3 in G. Then 1 .H / contains a G-invariant subgroup E Š Ep2 . Let E0 be a noncentral subgroup of order p of H with E0 < E. Then we must have NG .E0 / D E. On the other hand, CH .E/ D E and so CG .E/ 6 H , a contradiction. Hence G is minimal nonabelian and so Lemma 138.1 implies that G Š Mp .2; 2/. Now we use induction on jGj. Let G be a nonabelian S1 -group of order p 5 , and let N be a subgroup of order p in G 0 \ Z.G/. Then G=N is an S1 -group and so (by induction) G=N is either abelian or G=N Š Mp .2; 2/. (i) First assume that G=N is abelian. Then G 0 D N is of order p. Let K be a minimal nonabelian subgroup of G so that K 0 D G 0 D N and K satisﬁes either jKj D p 3 or K Š Mp .2; 2/ (Lemma 138.1). We get G D KCG .K/, where K \ CG .K/ D Z.K/ and CG .K/ 6 K (since jGj p 5 ). If Y is a nonnormal cyclic subgroup of K, then jNG .Y / W Y j p 2 , a contradiction. (ii) Now suppose that G=N Š Mp .2; 2/. Since G=N is metacyclic, it follows from Theorem 36.1 that G is also metacyclic. As G 0 Š Cp2 , we conclude that G is a metacyclic A2 -group of order p 5 . By Proposition 71.2(a), we have 3

2

2

G Š ha; b j ap D 1; b p D ap ; ab D a1Cp i; where D 0; 1. Consider the maximal subgroup L D haihb p i of G. Since hai is a cyclic subgroup of index p in L (and L is obviously noncyclic), there is an element x of order p in Lhai. But G=hai Š Cp2 acts faithfully on hai and so x does not centralize hai which implies that hxi is not normal in G. On the other hand, L is minimal nonabelian and so x centralizes hap i and therefore jNG .hxi/ W hxij p 2 , a contradiction. Since Mp .2; 2/ is obviously an S1 -group, our theorem is proved. Theorem 138.3. Let G be a non-Dedekindian 2-group. Then G is an S1 -group if and only if one of the following holds: (a) G is of maximal class distinct from Q8 . (b) G Š ha; b j a2 D b 4 D 1; ab D a1C2 D 1, then n 3. n

n1

i, where n 2, D 0; 1, and if

Proof (Janko). Suppose that G is a non-Dedekindian 2-group and G is an S1 -group. By Lemma 138.1, each minimal nonabelian subgroup H of G is isomorphic to D8 , Q8 or M2 .2; 2/ D H2 . (i) Suppose that G possesses a subgroup H Š D8 . Let t be a noncentral involution in H . Since jCH .t/ W htij D 2, we have CG .t/ < H and so CG .H / H . By Proposition 10.17, G is of maximal class. In what follows we may assume that D8 is not a subgroup of G. (ii) Now suppose that G possesses a subgroup H Š Q8 so that in this case we must have H < G. Set H D ha; bi and Z.H / D hzi. First assume that H has a cyclic subgroup C of order 4 such that C is not normal in G. Then NG .C / D H which implies

138

Smallest normalizers of nonnormal subgroups

345

that CG .H / H and so again (Proposition 10.17) G is of maximal class. In what follows we may assume that each cyclic subgroup of H is normal in G. This gives that H E G and G D HK, where K D CG .H / and H \ K D hzi. Suppose that K has a cyclic subgroup K1 D hki of order 4. If k 2 D z, then we have HK1 D H K1 with H \ K1 D hzi. But we know that in this case HK1 possesses a subgroup isomorphic to D8 , a contradiction. Hence we must have HK1 D H K1 , where we set k 2 D z 0 . In this case, we consider haki Š C4 , where .ak/2 D zz 0 and .ak/b D ak z, which implies that haki is not normal in HK1 . But ha; ki Š C4 C4 centralizes haki and so HK1 is not an S1 -group, a contradiction. We have proved that K is elementary abelian. But then G is Dedekindian, contrary to our assumption. (iii) In the sequel we assume that each minimal nonabelian subgroup of G is isomorphic to H2 . Then we use Theorem 57.3, where such 2-groups are classiﬁed. If G is a group from part (b) of Theorem 57.3, then G is of exponent 4 and by inspection we see that the only S1 -group there is the group H2 which appears in part (b1) of that theorem. We have obtained a group from part (b) of our theorem for the special case n D 2 and D 0. Indeed, in case (b2) of Theorem 57.3, G contains the unique minimal nonmetacyclic group H of order 25 given in Theorem 66.1(d) with H D ha; b; c j a4 D b 4 D Œa; b D 1; c 2 D a2 ; ac D ab 2 ; b c D ba2 i; and here hai is not normal in H , but ha; bi centralizes hai and so H is not an S1 -group. In case (b3) of Theorem 57.3 the groups there contain a minimal nonmetacyclic group of order 25 as a subgroup. Finally, in case (b4) of Theorem 57.3, we have a subgroup K D Bhbi, where o.b/ D 4, hbi \ B D ¹1º, b inverts each element of B, hbi is not normal in K, but 1 .B/ is elementary abelian of order 4 and 1 .B/ centralizes hbi and so such a group K is not an S1 -group. It remains to consider the groups G in Theorem 57.3(a) which are of exponent 8. Here G has a unique abelian maximal subgroup A. We have exp.A/ 8, E D 1 .A/ D 1 .G/ D Z.G/ is of order 4. All elements in G A are of order 4 and if b is one of them, then CA .b/ D E and b inverts each element of ˆ.A/ and of A=E. Since hbi is not normal in G and E centralizes hbi, we get E Š E4 and therefore A is of rank 2. If C D hci is any cyclic subgroup of order 4 in A, then C C G and so c b D c 1 . In case c 2 D b 2 , we infer that hb; ci Š Q8 , a contradiction. Thus we have proved that b 2 D u is not a n1 square in A and so A D hui hai with o.a/ D 2n , n 3, and we set a2 D z so that E D 1 .A/ D 1 .G/ D hu; zi. Since b inverts each element of A=E, we have ab D a1 with 2 E. If D u, then .ba/2 D ba ba D b 2 ab a D u a1 u a D 1; a contradiction. In case D uz, we obtain .ba/2 D ba ba D b 2 ab a D u a1 uz a D z:

346

Groups of prime power order n2

But ba inverts v D a2 (of order 4) and so hba; vi Š Q8 , a contradiction. We have proved that ab D a1 z , D 0; 1, and so we have obtained all groups stated in part (b) of our theorem. (iv) Conversely, if G is a 2-group of maximal class (distinct from Q8 ), then G is obviously a non-Dedekindian S1 -group. Assume that G is a 2-group from part (b) of our theorem. If n D 2, then D 0 and G Š H2 , which is an S1 -group. Suppose that n 3 n1 and set b 2 D u and a2 D z so that ab D a1 z and 1 .G/ D Z.G/ D hu; zi. i j Let x D b a with 2 1 .G/ (i; j are integers) be any element in G .hui hai/. Then we compute x 2 D .b i aj /2 D b i aj b i aj D b 2 .b 1 aj b/aj D u.ab /j aj D u .a1 z /j aj D uaj z j aj D uz j ; and we see that x 2 2 1 .G/ hzi. For any y 2 1 .G/ hzi we infer that G=hyi is dihedral or semidihedral and so G=hyi is an S1 -group. Let M be any subgroup of G. If M contains an element y 2 1 .G/hzi, then M=hyi is a subgroup of G=hyi and so either M E G or jNG .M / W M j D 2. Suppose that M does not contain any element in 1 .G/ hzi. Then, by the above, M hui hai and so M E G since every subgroup of huihai is normal in G. Hence G is an S1 -group. Our theorem is proved. 2o . Below we prove a more general result. Let G be a non-Dedekindian p-group and C < G be nonnormal cyclic of minimal possible order. Write jC j D p .G/ D p . Denote by C the set of all nonnormal cyclic subgroups of G of order p . Clearly, if H < G is non-Dedekindian, then we obtain that .H / .G/. If G satisﬁes the hypothesis of Theorem 138.4, then the members of the set C are maximal cyclic subgroups of G. Our main aim is to prove the following Theorem 138.4 (Berkovich). Let G be a non-Dedekindian p-group of order > p 3 such that any nonnormal cyclic subgroup of order p D p .G/ has index p in its normalizer. Then one and only one of the following holds: (a) If p > 2, then G is the unique nonabelian metacyclic group of order p 4 and exponent p 2 , D 2. (b) G is a 2-group of maximal class. In this case, D 1 unless G is generalized quaternion and D 2. (c) G D hx; y j x 2 D y 4 D 1; x y D x 1C2 i is metacyclic, where 2 ¹0; 1º, n 2, and if D 1, then n > 2. Here jGj D 2nC2 , G D hyi hxi (semidirect n1 product with kernel hxi), Z.G/ D hx 2 ; y 2 i D 1 .G/, ˆ.G/ D hx 2 ; y 2 i D hy 2 i G 0 is abelian of type .2n1 ; 2/, G=G 0 ia abelian of type .4; 2/, G=hy 2 i is of maximal class but not generalized quaternion. In this case, D 2. n

n1

Proof. If G is a 2-group of maximal class satisfying the hypothesis of the theorem, then all its abelian subgroups of order > 4 are normal cyclic and all its nonnormal sub-

138

Smallest normalizers of nonnormal subgroups

347

groups of order > 4 are of maximal class. If K < G is nonabelian, then a normal subgroup of K of index > 2 is contained in ˆ.K/ < ˆ.G/ and so are normal in G. Therefore, if U 2 C , then jU j 4 hence 2, and jNG .U / W U j D 2, and so G satisﬁes the hypothesis. This proves (b). Next we assume that our group G is not a 2-group of maximal class. (i) If we have U 2 C , then any abelian subgroup A of G containing U has order at most pjU j.1 Proof. Indeed, we have A NG .U / so that jA W U j jNG .U / W U j D p. (ii) G has no elementary abelian subgroup of order p 3 . Proof. Assume, by way of contradiction, that a subgroup E < G is elementary abelian of order p 3 . Then, by (i), all subgroups of order p contained in E are normal in G, and we conclude that E Z.G/. Let U 2 C . Then jE \ U j p hence, by the product formula, EU is abelian of order p 2 jU j, contrary to (i). (iii) All subgroups of order p are normal in G and j1 .G/j D p 2 (recall that we have assumed that G is not a 2-group of maximal class). In particular, > 1. Proof. Assume that a subgroup T < G is nonnormal of order p. Then D 1 hence jNG .T /j D pjT j D p 2 by hypothesis. Since NG .T / D CG .T /, it follows that G is of maximal class by Proposition 1.8 (then, by assumption, p > 2). This argument shows that any subgroup of G properly containing a nonnormal subgroup of order p is either of order p 2 or of maximal class. Then, since p > 2, G possesses a G-invariant abelian subgroup R of type .p; p/ (Lemma 1.4). As Z.G/ is cyclic, it follows that there is in R a non-G-invariant subgroup L of order p. Write E D CG .R/; then jEj D p1 jGj p 3 . Let B=R E=R be of order p. Then B is abelian of order p 3 D p 2 jLj (see (ii)), contrary to (i). Thus the group G has no nonnormal subgroups of order p, and we conclude that 1 .G/ Z.G/ so that j1 .G/j D p 2 by (ii). It follows from (iii) that G is not of maximal class also for p > 2. (iv) The group G possesses a minimal nonabelian subgroup H such that .H / D

.D .G//. Moreover, H is metacyclic of order p 4 and exponent p 2 and D 2. Proof. Let T 2 C and write N D NG .T /; then jN W T j D p so that jN j D pjT j D p C1 p 3 since > 1 by (iii). Assume that N Š Q8 . Then D 2 and CG .T / D T so G is of maximal class, a contradiction. Thus we have N 6Š Q8 so that N is abelian of type .p ; p/ by Theorem 1.2. Denote by N1 the normalizer of N in G; then N1 is nonabelian. The subgroup N GN1 has exactly c .N / D p cyclic subgroups of index p (here c .N / is the number of cyclic subgroups of order p in N ). Let N1 act via conjugation on cyclic subgroups of index p in N . Since the stabilizer of T in N1 coincides 1 By

hypothesis, if jNG .U / W U j > p, then U G G.

348

Groups of prime power order

with NN1 .T / D N , we get jN1 W N j D c .N / D p. Therefore the normalizer of any cyclic subgroup of N of order jT j in N1 (even in G by hypothesis) coincides with N . Assume that p D 2 and all minimal nonabelian subgroups of N1 are Š Q8 (G has no dihedral subgroup of order 8 by (iii)). Then, by Corollary A.17.3 and hypothesis, we get N1 D QE, where Q is generalized quaternion and exp.E/ 2. By construction, N1 ¤ Q (otherwise, all abelian subgroups of Q are cyclic but N < Q is noncyclic abelian) so that E > ¹1º. If jQj > 8, then there exists in Q a nonnormal cyclic subgroup M of order 4 (then D 2), and the subgroup NN1 .M / D NQ .M / E has order 8jEj > 2jM j, contrary to the hypothesis. Thus Q Š Q8 so that N1 is Dedekindian. In this case, T G N1 , a contradiction since NG .T / D N < N1 . Now let p be arbitrary. By the previous paragraph, there is in N1 a minimal nonabelian subgroup H 6Š Q8 ; then we have jH j > p 3 since 1 .H / Z.H / by (iii). By (ii) and Lemma 65.1, H is metacyclic. Assume that .H / > . Then jH j p .H /C2 > p C2 D jN1 j, which is a contradiction. Thus .H / D . Let L < H be nonnormal cyclic of order p . By hypothesis, jH j D p jNH .L/j D p C2 . Let Z G H be cyclic such that H=Z is cyclic; then H 0 < Z. It follows that L \ Z D ¹1º, and we conclude that H D L Z, a semidirect product with kernel Z, and jZj D jH W Lj D p 2 . Let 2 .L/ D hbi and Z D hzi. Then the cyclic subgroup U D hbzi of order p 2 is not normal in H so that .H / D 2 .G/ > 1 hence D 2 and jH j D jLjjZj D p 4 so H is metacyclic of exponent 4. As a by-product, we have also obtained N1 D H . (v) The theorem is true if p > 2. Proof. Indeed, by (ii), (iii) and Proposition 10.19, G is metacyclic. Take in G a minimal nonabelian subgroup H such that .H / D (H exists in view of (iv)). By (iv), H is of order p 4 and exponent p 2 . Let V be of order p such that V ¤ H 0 ; then we obtain V G G by (iii). In this case, H=V is a nonabelian subgroup of order p 3 in the metacyclic group G=V . It follows from Proposition 10.19 that H=V D G=V so G D H , as was to be shown. In what follows we assume that p D 2. Let H G be minimal nonabelian with

.H / D .D .G//. By (iv), H is metacyclic of order 16 and exponent 4 and D 2. (vi) The group G is as in (c). Proof. If G is minimal nonabelian, then it coincides with H so G is metacyclic of order 16 and exponent 4 by the paragraph preceding (vi). Next we assume that G is not minimal nonabelian; then we get H < G so jGj > 16. Set jGj D 2nC2 ; then n > 2. As we know, H D ha; b j a4 D b 4 D 1; ab D a3 i. Let L D hbi < H be nonnormal cyclic of order 4; then L is not normal in G so that

D 2 by (iii). In this case, by hypothesis, we obtain NH .L/ D ha2 ; bi D NG .L/ (the last equality holds by hypothesis). Therefore NG=Ã1 .L/ .L=Ã1 .L// D NG .L/=Ã1 .L/ is abelian of type .2; 2/. It follows from Proposition 1.8 that G=Ã1 .L/ is of maximal class. We claim that G is metacyclic. Assume that this is false. If Ã1 .L/ < G 0 , then

138

Smallest normalizers of nonnormal subgroups

349

jG W G 0 j D 4 so G is of maximal class by Proposition 1.6, contrary to the assumption. Thus Ã1 .L/ 6 G 0 so that G 0 \ L D ¹1º. In this case, G 0 Š .G=Ã1 .L//0 hence G 0 is cyclic of order 2n1 . As Ã1 .L/ D ˆ.L/ < ˆ.G/, we get d.G/ D d.G=Ã1 .L// D 2 and so G=G 0 is abelian of type .4; 2/. The subgroup F D L G 0 (semidirect product with kernel G 0 ) is a metacyclic maximal subgroup of G. Also, we have 1 .G/ < F as j1 .G/j D 4 D j1 .G 0 / 1 .L/j. We obtain G=G 0 D .F=G 0 / .K=G 0 /, where F=G 0 Š L is cyclic of order 4 and jK=G 0 j D 2. Since F \K D G 0 is cyclic, it follows that K has only one subgroup of order 2 since it does not contain Ã1 .L/, and we infer that K is either cyclic or generalized quaternion. We also have G D L K (semidirect product with kernel K) and jKj D 2n . Let M=G 0 < G=G 0 be cyclic of order 4 such that M ¤ F (note that G=G 0 contains exactly two cyclic subgroups of order 4 and F=G 0 is one of them); then M is metacyclic. Since d.G/ D 2, we have 1 D ¹F; M; K Ã1 .L/º; where F and M are metacyclic. Assume that KÃ1 .L/ is metacyclic. Then, by Theorem 66.1, we get jGj D 25 and exp.G/ D 4, a contradiction since, in this case, K is cyclic of order 8. Thus K Ã1 .L/ is nonmetacyclic so K is a generalized quaternion group of order 2n , where n > 2, by the above. If n > 3, then K possesses a nonnormal cyclic subgroup D of order 4 and NK .D/ Ã1 .L/ NG .D/ has order 16 D 4jDj > 2jDj, contrary to the hypothesis. Hence n D 3, K Š Q8 and b induces on K an involutory automorphism since b 2 centralizes K but b does not centralizes K. Let G 0 ; U; V be pairwise distinct cyclic subgroups of order 4 in K. Since G 0 G G, it follows that jG W NG .U /j 2 so U G G. Similarly, V G G. Then b inverts G 0 ; U; V so K, and we conclude that K is abelian (Burnside), a contradiction. The obtained contradiction shows that G is metacyclic. Let Z G G be cyclic such that G=Z is cyclic. Then we get G 0 < Z so jG=Zj < jG=G 0 j D 8. Since G has no cyclic subgroup of index 2 (Theorem 1.2), we obtain jG=Zj D 4. We have G D L Z (semidirect product since L \ G 0 D ¹1º). All the above allows us to write out deﬁning relations for G in the following form: G D ha4 D z 2 D 1; z a D z 1C2 n

n1

i;

where 2 ¹0; 1º (Theorem 1.2), and if D 1, then n > 2, so G is as in (c). (vii) It remains to check that the groups (a)–(c) listed in the conclusion of our theorem satisfy the hypothesis. The group from (a), obviously, satisﬁes the hypothesis. As we have noticed, 2-groups of maximal class satisfy the hypothesis (see the ﬁrst paragraph of the proof). Now let G be a group from (c) and jGj D 2nC2 > 24 . In this case, .G/ D D 2 since 1 .G/ D Z.G/ and the set C2 contains a nonnormal cyclic subgroup hai of

350

Groups of prime power order

order 4 (see the deﬁning relations for G). Let L < G be nonnormal cyclic of order 4 and Z D hzi. We have to prove that jNG .L/ W Lj D 2. Since cl.G/ > 2, CG .Z/ is the unique abelian maximal subgroup of G and LZ D G since L 6 CG .Z/, and so we obtain L\Z D ¹1º by the product formula. It follows that CG .L/ D LZ.G/ is abelian of type .4; 2/. Since NG .L/ D CG .L/, we ﬁnally get jNG .L/ W Lj D 2. The proof is complete. To prove that Theorem 138.3 follows from Theorem 138.4, it sufﬁces to show that all groups (a)–(c) from the conclusion of the latter theorem satisfy the hypothesis of Theorem 138.3. It sufﬁces to check only the groups from Theorem 138.4(c). We omit this routine checking. Note only that if H < G is nonnormal and contains 1 .G/, then jNG .H / W H j D 2 since G=1 .G/ is of maximal class; therefore it sufﬁces to assume that H is cyclic (see Proposition 10.19). Let p be a prime and n 4 be a natural number. Denote by tp .n/ the number of pairwise nonisomorphic groups of order p n satisfying the hypothesis of Theorem 138.4. If p > 2, then we have tp .n/ D 1 for n D 4 and tp .n/ D 0 for n > 4. Next, t2 .n/ D 4 for n D 4 and t2 .n/ D 5 for n > 4. This follows from Theorem 138.4. Corollary 138.5. If any nonnormal subgroup (cyclic subgroup) of a non-Dedekindian p-group G has index p in its normalizer, then G is one of the groups (a)–(c) from Theorem 138.4.

Problems Problem 1. Classify the non-Dedekindian p-groups G such that whenever C A < G, where C 2 C and A is abelian, then jA W C j p. Problem 2. Classify the non-Dedekindian p-groups such that whenever C < G is nonnormal cyclic, then jNG .C / W C j p 2 . Problem 3. Let H be a subgroup of a p-group G. Study the structure of G if for every maximal cyclic subgroup C of G not contained in H we have jNG .C / W C j D p. Problem 4. Let H be a maximal subgroup of a non-Dedekindian 2-group G. Describe the structure of G provided all minimal nonabelian subgroups of G not contained in H (see Theorem 10.28) are isomorphic to Q8 . Problem 5. Classify the non-Dedekindian p-groups G such that jNG .C / W C j D p for all nonnormal maximal cyclic C < G. Problem 6. Classify the non-Dedekindian p-groups G such that jNG .H / W H j D p for all maximal nonnormal H < G. Problem 7. Suppose that H < G is noncyclic. Study the structures of H and G provided all maximal abelian (cyclic) subgroups of H are maximal abelian (cyclic) in G.

138

Smallest normalizers of nonnormal subgroups

351

In conclusion we solve Problem 1 (see Theorem 138.7 and 138.8). Lemma 138.6. Let G be a non-Dedekindian metacyclic p-group of order > p 3 satisfying the hypothesis of Problem 1. Then G is one of the groups from Theorem 138.4. Proof. One may assume that G satisﬁes jGj > p 4 . Also assume that G is not a 2-group of maximal class. Then R D 1 .G/ is abelian of type .p; p/ (Lemma 1.4 and Proposition 10.19). Since CG .R/ contains an abelian subgroup A of type .p 2 ; p/, it follows that all subgroups of order p are normal in G whence R Z.G/. Let H G be minimal nonabelian. By what has just been said and Proposition 10.19, we get jH j > p 3 . As in the proof of Theorem 138.4, we deduce that jH j D p 4 and exp.H / D p 2 . Let K < 1 .G/ be of order p such that K ¤ H 0 . Then H=K is a nonabelian subgroup of order p 3 in G=K so, since jG=Kj > p 3 , it follows that G=K is a 2-group of maximal class (Proposition 10.19). By Taussky’s Proposition 1.6, we have jG W G 0 j > 4 so, by assumption, K 6 G 0 , and we conclude that G 0 is cyclic of index 8 in G hence G=G 0 is abelian of type .4; 2/. As above, there is a cyclic subgroup Z > G such that G=Z is cyclic of order 4. Let L be a cyclic subgroup of order 4 in H such that L \ H 0 D ¹1º. Then we have L 2 C ( D 2) and L \ Z D L \ G 0 D ¹1º. Hence G D LZ and CZ .L/ D 1 .Z/ shows that G is a group from Theorem 138.4(c). Theorem 138.7. If a non-Dedekindian p-group G of order > p 3 , p > 2, is a group of Problem 1, then G is metacyclic of order p 4 and exponent p 2 . Proof. In view of Lemma 138.6, it sufﬁces to show that G is metacyclic. Assume, by way of contradiction, that G is not metacyclic. As in the proof of Theorem 138.4, we have (i) 1 .G/ Z.G/ is of order p 2 . Then G is not a group of maximal class. (ii) If H G is minimal nonabelian, it is metacyclic of order p 4 and exponent p 2 . By assumption, H < G. Let R H be of order p such that R ¤ H 0 ; then we have R G G by (i). In this case, H=R is a nonabelian subgroup of order p 3 in G=R. Write N=R D NG=R .H=R/; then jN j > p 4 . By (i), N has no elementary abelian subgroup of order p 3 . Then N is a group from Theorem 13.7. Therefore, by (i), N is metacyclic of order > p 4 , contrary to Lemma 138.6. Not all arguments in the proof of Theorem 138.4 work under hypothesis of the following theorem (for example, one cannot assert that G=Ã1 .L/ is a 2-group of maximal class). Theorem 138.8. If G is a non-Dedekindian nonmetacyclic 2-group of Problem 1 and jGj > 25 , then D 2. Suppose that G 6Š Q2n C2 . Then (a) There is in G an abelian subgroup, say A, of index 2. Let G D hx; Ai. (b) 1 .G/ D Z.G/ is of order 4 and all elements of the set G A have order 4. Also, we have d.A/ D 2.

352

Groups of prime power order

(c) All subgroups of A of order 4 are normal in G. (d) Any cyclic subgroup of G of order 4 not contained in A is nonnormal in G. Conversely, any group G satisfying conditions (a)–(d) also satisﬁes the condition of Problem 1. Proof. Let RGG be abelian of type .2; 2/ (Lemma 1.4) and let B=R < CG .R/=R be of order 2; then B is abelian of order 8. By hypothesis, all subgroups of R of order 2 are normal in G so that R Z.G/. As in part (ii) of the proof of Theorem 138.4, we obtain j1 .G/j D 4 so 1 .G/ D R Z.G/. Arguing as in the proof of Theorem 138.4, one proves that any minimal nonabelian subgroup of G is either isomorphic to Q8 or metacyclic of order 16 and exponent 4. As, by hypothesis, G 6Š Q2m C2 , there is in G a metacyclic minimal nonabelian subgroup H of order 16 and exponent 4 (H exists in view of Corollary A.17.3 and what has been said in the previous paragraph). Since H has a nonnormal cyclic subgroup, say L, of order 4, we have D 2. By hypothesis, CH .L/ of order 8 is maximal abelian in G so that CG .L/ D CH .L/. As 1 .G/ Z.G/ < CG .L/, it follows that Z.G/ D 1 .G/. Let A be a maximal normal abelian subgroup of G; then 1 .G/ < A so that A is metacyclic but noncyclic since j1 .G/j D 4. For any x 2 G A there exists an element a 2 A such that U D ha; xi is minimal nonabelian (Lemma 57.1). Since exp.U / D 4, we get o.x/ D 4 so that all elements of the set G A have the same order 4. Because x 2 2 1 .G/ D Z.G/ centralizes A, it follows that x 2 2 A, and we conclude that G=A is elementary abelian. Therefore, since jGj > 25 , we get jAj > 8 (otherwise, G=A Š Aut.A/ Š D8 ). In this case, all cyclic subgroups of A of order 4 are normal in G by hypothesis. Let R < A be cyclic of order 4. Then jG W CG .R/j D 2 since R 6 Z.G/ in view of exp.Z.G// D 2. Assume that jG=Aj > 2. Then we infer that CG .R/ > A hence there is b 2 CG .R/ A of order 4 (recall that all elements of the set GA have order 4). Clearly, hb; Ai is nonabelian (indeed, A is maximal abelian in G) of order 2jAj and R < Z.hb; Ai/. By Corollary A.17.3, hb; Ai has a minimal nonabelian subgroup H1 6Š Q8 since hb; Ai 6Š Q2n E2s : the center of the last group has exponent 2 < 4 D exp.R/. By the above, H1 is metacyclic of order 16 and exponent 4. Since R centralizes H1 and R 6 H1 , there is in H1 R (of order 32) an abelian subgroup of order 16 containing a nonnormal cyclic subgroup of H1 of order 4 D 2 , contrary to the hypothesis. Thus jG W Aj D 2. Let D < G be cyclic of order 4 such that D 6 A. Then we have G D AD. Assume that D G G. Then G=.A \ D/ is abelian so A \ D D G 0 is of order 2. In this case, by Lemma 1.1, jGj D 2jG 0 jjZ.G/j D 16; contrary to the hypothesis. Thus all cyclic subgroups of G of order 4 not contained in A are nonnormal in G. It remains to show that our group G satisﬁes the condition of Problem 1 if it satisﬁes conditions (a)–(d) in the statement of the theorem. Note that the set C coincides with the set of all cyclic subgroups of G that are not contained in A.

138

Smallest normalizers of nonnormal subgroups

353

Let D 2 C .D C2 /. Assume that D < B < G, where B is abelian of order 16. As jB \ Aj D 8, the subgroup B \ A has a cyclic subgroup L of order 4. Then CG .L/ AD D G so that L < Z.G/, a contradiction since exp.Z.G// D 2 < 4 D exp.L/. We claim that if the 2-group G D hxiA of order 26 satisfying conditions (a)–(d) of Theorem 138.8 is metacyclic, then it is as in Theorem 138.4(c). Let x 2 G A and write X D hxi. We want to prove that then A D 1 .X/ C , where C is cyclic. If this is false, there is in A an element y of order 4 such that y 2 D x 2 (indeed, if all invariants of an abelian group, say B, are > p, then any element of B of order p is contained in a cyclic subgroup of B of order p 2 ; this follows from Theorem 6.1 with G D Q). Since hyi G G by (d), one has y x D x 1 (otherwise, y 2 Z.G/ D 1 .G/), and we conclude that hx; yi Š Q8 so that G is of maximal class (Proposition 10.19), which is a contradiction. Thus (e) A D 1 .X/ C , where C is cyclic (of index 2 in A). (f) We claim that the quotient group GN D G=1 .X/ is of maximal class. Indeed, GN has a cyclic subgroup AN Š C of index 2 by (e). Assume that GN is not of maximal class. Then, by Theorem 1.2, GN is either abelian of type .2n ; 2/ or isomorphic to M2nC1 . In the ﬁrst case, we get X G G, a contradiction. Thus GN Š M2nC1 , n 4, as jGj 26 . N then XN < B. N In this case, we get XN G BN Let BN be a noncyclic maximal subgroup of G; N D 2n 24 . It then follows that B is nonabelian by hypothsince BN is abelian, and jBj esis. The quotient group B=X is cyclic of order 2n1 23 . Then CB .X/ is abelian of order 2n 24 , contrary to the hypothesis. Thus GN is of maximal class. As in the proofs of Theorems 138.4 and 138.8, we get (recall that G is not of maximal class) (g) G 0 is cyclic and G=G 0 is abelian of type .4; 2/. (h) G is as in Theorem 138.4(c). Therefore, if A has no cyclic subgroup of index 2, then G is nonmetacyclic. Now we present a nonmetacyclic 2-group G satisfying conditions (a)–(d) of Theorem 138.8. Let m

n

G D hx; a; b j a2 D b 2 D 1; m; n 2; m C n 5; x 2 D a2

m1

; ab D ba; ax D a1 ; b x D b 1 i:

We have ha; xi Š Q2mC1 and hb; xi Š D2nC1 . Further, we have jGj D 2mCnC1 > 25 j and c D Œx; ai b j D a2i b 2 2 G 0 with c D 1 if and only if i m 1 and j n 1. m1 n1 It follows that Z.G/ D ha2 i hb 2 i is abelian of type .2; 2/. Clearly, we have 0 G D Œhxi; A. Since the noncyclic subgroup ha2 ihb 2 i of index 8 in G is contained in G 0 and jG W G 0 j > 4 (Proposition 1.6), it coincides with G 0 so that G 0 D ha2 i hb 2 i is abelian of type .2m1 ; 2n1 /, and we conclude that G is nonmetacyclic. If y 2 G A

354

Groups of prime power order

is arbitrary, then y inverts A since CG .y/ D hy; Z.G/i is abelian of type .4; 2/. It follows that G satisﬁes the condition of Problem 1. It is interesting to classify the nonnilpotent groups G such that whenever H < G is nonnormal, then either NG .H / D H or jNG .H / W H j is a prime. There is a nonsolvable group satisfying this condition, for example, PSL.2; 5/. Using Theorem 138.4, Frobenius’ normal p-complement theorem for p D 2 (see Lemma 10.8), the Odd Order Theorem and the classiﬁcation of simple groups with abelian Sylow 2-subgroup of type .2; 2/, it is easy to prove that PSL.2; 5/ is the unique nonsolvable group satisfying the above condition.

139

p-groups with a noncyclic commutator group all of whose proper subgroups have a cyclic commutator group

This section is written by the second author. Here we give a characterization of the title groups (Theorem 139.A). The surprising and deep result is that if G is a title group, then G 0 is elementary abelian of order p 2 or p 3 and for each H < G we have jH 0 j p. We use here a corresponding characterization result of A. Leone (see Theorem B in [Leo]), but most of our proofs are simpler. At the beginning we prove two very useful lemmas, and then the main result (Theorem 139.A) will be proved in a series of propositions concerning the title groups. Theorem 139.A. A p-group G has a noncyclic commutator group G 0 and each proper subgroup H of G has a cyclic commutator group H 0 if and only if for each H < G we have jH 0 j p and exactly one of the following holds: (a) d.G/ D 2, cl.G/ D 3, and G 0 Š Ep2 . (b) p > 2, d.G/ D 2, cl.G/ D 3, G 0 Š Ep3 and Ã1 .G/ Z.G/. (c) d.G/ D 3, cl.G/ D 2, G 0 Š Ep3 or Ep2 and ˆ.G/ D Z.G/. A p-group G is said be a KC-group if the commutator group G 0 is cyclic. Then the title groups are actually p-groups which are minimal non-KC-groups. Lemma 139.1. Let G D hx; yi be a nonabelian two-generator KC-group. If p > 2 or p D 2 and ŒG 0 ; G Ã2 .G 0 /, then Ã1 .G 0 / D hŒx; yp i D hŒx; y p i. Proof. By hypothesis, G 0 D hŒx; yi. If Ã1 .G 0 / D ¹1º, then jG 0 j D p and G is minimal nonabelian (Lemma 65.2(a)), and the result follows from Lemma 65.1. Next we assume that Ã1 .G 0 / > ¹1º and consider the quotient group GN D G=Ã2 .G 0 /. Then we N D 2 and GN 0 Š Cp2 . As G=Ã N 1 .GN 0 / is minimal nonabelian (Lemma 65.2(a)), get d.G/ for each H 2 1 we have jHN 0 j p. N (i) Assume that p > 2. In this case, we may apply Proposition 137.5 for the group G. p p 0 We get Œx; y D Œx; y , where 2 Ã2 .G /. But this gives hŒx; y p i D hŒx; yp i and we are done.

356

Groups of prime power order

(ii) Suppose that p D 2. Then we compute Œx; y 2 D Œx; yŒx; yy D Œx; yŒx; y.Œx; y1 Œx; yy / D Œx; y2 ŒŒx; y; y: By our assumption, we have ŒŒx; y; y 2 ŒG 0 ; G Ã2 .G 0 / and so we obtain again hŒx; y 2 i D hŒx; y2 i. Lemma 139.2. Let G D ha; b; ci be a p-group of Theorem 137.7(d), i.e., d.G/ D 3;

cl.G/ D 2;

G 0 Š Ep3 or Ep2

and

ˆ.G/ D Z.G/:

Then for each subgroup X of order p which is contained in G 0 there is a maximal subgroup L of G such that L0 D X. Proof. If G 0 Š Ep3 , then all p 2 C p C 1 maximal subgroups of G must have pairwise distinct commutator subgroups of order p. Indeed, if L ¤ K are two distinct maximal subgroups of G such that L0 D K 0 , then Exercise 1.69(a) would imply jG 0 j p 2 , a contradiction. Hence in this case each subgroup of order p in G 0 is a commutator subgroup of a maximal subgroup of G. Assume now that G 0 Š Ep2 . We may choose the generators a; b; c of G so that setting Œa; b D k and Œb; c D l we have G 0 D hk; li. But then for any i D 0; 1; : : : ; p 1 we obtain Œac i ; b D Œa; bŒc; bi D kl i and so again each subgroup of order p in G 0 is a commutator subgroup of one of the maximal subgroups ˆ.G/hb; ci and ˆ.G/hac i ; bi (i D 0; 1; : : : ; p 1) of G. Proposition 139.3. Let G be a p-group which is a minimal non-KC-group. Then we have: (i) Z.G/ ˆ.G/. (ii) G is metabelian. (iii) d.G/ 3. Proof. (i) Suppose Z.G/ 6 ˆ.G/. Then there is a maximal subgroup M of G such that G D hM; xi, where x 2 Z.G/. But then G 0 D M 0 is cyclic, a contradiction. (ii) Let M be a maximal subgroup of G so that M 0 is cyclic and M 0 E G. Since Aut.M 0 / is abelian, G=CG .M 0 / is abelian and so G 0 centralizes M 0 . Set H D hM 0 j M is any maximal subgroup in Gi so that H is characteristic in G and H Z.G 0 /. Each maximal subgroup M of G contains H and M=H is abelian. Thus G=H is either abelian or minimal nonabelian. This gives jG 0 =H j p and so G 0 is abelian. (iii) We consider G=ˆ.G 0 / so that G 0 =ˆ.G 0 / is elementary abelian of order p 2 . Each maximal subgroup M=ˆ.G 0 / of G=ˆ.G 0 / has its derived subgroup of order p and so, by Proposition 137.3, d.G=ˆ.G 0 // 3 and therefore d.G/ 3.

139

p-groups G with noncyclic G 0 and cyclic H 0 for all subgroups H < G

357

Proposition 139.4. A p-group G is a minimal non-KC-group with H D hM 0 j M is any maximal subgroup in Gi being cyclic if and only if G 0 Š Ep2 and d.G/ D 2. Here we have cl.G/ D 3. Proof. Suppose that G is a p-group which is a minimal non-KC-group and H is cyclic. Then G 0 > H and each maximal subgroup of G=H is abelian. It follows that G=H is minimal nonabelian. This gives d.G=H / D 2 and so d.G/ D 2 and jG 0 W H j D p so that G 0 is abelian of type .p n ; p/, n 1, where jH j D p n and exp.G 0 / D p n . We set G D ha; bi, where Œa; b 2 G 0 H . The commutator group G 0 is generated by all conjugates of Œa; b in G and since G 0 is abelian, we have exp.G 0 / D p n D o.Œa; b/. Thus H and hŒa; bi are two distinct cyclic subgroups of order p n (and index p) in G 0 . Suppose, by way of contradiction, that n > 1. Then cn .G 0 / D p and since one of the cyclic subgroups, H , of order p n is normal in G, it follows that all of them are normal in G. But then hŒa; bi is normal in G and so G 0 D hŒa; bi is cyclic, a contradiction. It follows that n D 1, jH j D p and o.Œa; b/ D p so that G 0 Š Ep2 . Conversely, assume that G is a p-group with d.G/ D 2 and G 0 Š Ep2 . Let N < G 0 be a G-invariant subgroup of order p contained in G 0 . Then we infer that d.G=N / D 2 and .G=N /0 D G 0 =N is of order p and so G=N is minimal nonabelian. Hence, if M is any maximal subgroup of G, then M G 0 and M=N is abelian, which gives M 0 N . It is easy to see that such a p-group G is of class 3. Indeed, if G 0 Z.G/, then considering a subgroup K < G 0 of order p with K ¤ N , we conclude that G=K is minimal nonabelian and so for any maximal subgroup M of G we get also M 0 K and so M 0 N \ K D ¹1º. In this case, G would be minimal nonabelian and so jG 0 j D p, a contradiction. Hence we obtain that G 0 6 Z.G/ and so G is of class 3. Our proposition is proved. Proposition 139.5. Let G be a p-group which is a minimal non-KC-group with H D hM 0 j M is any maximal subgroup in Gi being noncyclic and d.G/ D 2. Then p > 2;

cl.G/ D 3;

G 0 Š Ep3 ;

and

Ã1 .G/ Z.G/:

Proof. Suppose that G is a p-group which is a minimal non-KC-group, H is noncyclic and d.G/ D 2. Then there exist maximal subgroups M and N such that H0 D M 0 N 0 is noncyclic. Then the factor group G=H0 has two distinct abelian maximal subgroups M=H0 and N=H0 and this gives jG 0 W H0 j p. Suppose for a moment that G 0 D H0 . Let X be a G-invariant subgroup such that M 0 X < H0 and jH0 W Xj D p. Since d.G=X/ D 2 and .G=X/0 D H0 =X is of order p, it follows that G=X is minimal nonabelian. But then N=X is abelian and so N 0 X , a contradiction. We have proved that jG 0 W H0 j D p and since d.G=H0 / D 2 and .G=H0 /0 D G 0 =H0 is of order p, it follows that G=H0 is minimal nonabelian. For each maximal subgroup L of G, L0 H0

358

Groups of prime power order

and so H0 D H and d.H / D 2. Also, for any two distinct maximal subgroups K and L of G we have K 0 L0 D H . Indeed, by Exercise 1.69(a), jG 0 W .K 0 L0 /j p and so K 0 L0 D H . Consider the factor group G=ˆ.H / so that H=ˆ.H / Š Ep2 and jG 0 =ˆ.H /j D p 3 . Using Theorem 137.7 for G=ˆ.H /, we see that G=ˆ.H / is a group from part (c) of that theorem. In particular, we have p > 2, cl.G=ˆ.H // D 3 and G 0 =ˆ.H / Š Ep3 so that d.G 0 / D 3. Let Mi (i D 1; 2; : : : ; p C 1) be the set of all maximal subgroups of G and so we infer that G 0 =Mi0 is noncyclic. For any ﬁxed i 2 ¹1; 2; : : : ; p C 1º we consider the factor group G=Mi0 . We see that d.G=Mi0 / D 2, .G=Mi0 /0 D G 0 =Mi0 is noncyclic, but for each maximal subgroup Mj =Mi0 (j D 1; 2; : : : ; p C 1) of G=Mi0 we have .Mj =Mi0 /0 .Mj0 Mi0 /=Mi0 H=Mi0 and so each maximal subgroup of G=Mi0 has its commutator group contained in the same cyclic subgroup H=Mi0 . We may apply Proposition 139.4 to conclude that G 0 =Mi0 Š Ep2 . Thus Mi0 (i D 1; 2; : : : ; p C 1) are pairwise distinct cyclic maximal subgroups of H . Set jMi0 j D p m , m 1, so that H is abelian of type .p m ; p/. If m > 1, then such a group H has exactly p cyclic maximal subgroups (of order p m ) and a noncyclic maximal subgroup (of type .p m1 ; p/). This is a contradiction (because we need in H exactly p C 1 pairwise distinct cyclic maximal subgroups) and so m D 1 and jG 0 j D p 3 . But we have d.G 0 / D 3 and so

ˆ.G 0 / D ˆ.H / D ¹1º:

Hence G is a group from part (c) of Theorem 137.7 and we are done. The following result is crucial for the whole investigation. Proposition 139.6. Let G be a p-group which is a minimal non-KC-group. Suppose that there exist nonabelian maximal subgroups M; N of G such that M 0 \ N 0 D ¹1º. If p > 2 or p D 2 and ŒL0 ; L Ã2 .L0 / for all L 2 1 , then H D hM 0 j M 2 1 i is elementary abelian and central in G. Proof. As M \N is abelian, there are m 2 M \N and n 2 N M with N 0 D hŒm; ni. Indeed, if N 0 D hŒn0 ; ni with n; n0 2 N M , then n0 D ni m for some m 2 M \ N and Œn0 ; n D Œni m; n D Œm; n. We apply Lemma 139.1 on the group hm; ni which is a nonabelian two-generator group with a cyclic commutator group hŒm; ni. It follows that hŒm; np i D hŒm; np i. But np 2 M and so Œm; np 2 N 0 \ M 0 D ¹1º and therefore Ã1 .N 0 / D ¹1º which gives jN 0 j D p. In an analogous way, we also get jM 0 j D p. Let L be any maximal subgroup of G with L0 ¤ ¹1º. Then either L0 \ M 0 D ¹1º or L0 \ N 0 D ¹1º since M 0 \ N 0 D ¹1º. This gives jL0 j D p and so L0 Z.G/. Hence H is elementary abelian and central in G. In what follows we may assume that d.G/ D 3. This implies that H D hM 0 j M is any maximal subgroup in Gi D G 0 ; i.e., G 0 is generated with commutator subgroups of all maximal subgroups of G.

139

p-groups G with noncyclic G 0 and cyclic H 0 for all subgroups H < G

359

Proposition 139.7. Let G be a p-group which is a minimal non-KC-group satisfying d.G/ D 3. Suppose that there exist maximal subgroups M; N of G such that M 0 ¤ ¹1º, N 0 ¤ ¹1º and M 0 \ N 0 D ¹1º. If p > 2 or p D 2 and ŒL0 ; L Ã2 .L0 / for all maximal subgroups L of G, then G is a p-group in Theorem 137.7(d), i.e., d.G/ D 3;

cl.G/ D 2;

G 0 Š Ep3 or Ep2 ;

and ˆ.G/ D Z.G/. Proof. By Proposition 139.6, G 0 is elementary abelian (noncyclic) and central in G. Due to Theorem 137.7, G is a p-group in Theorem 137.7 (d) and we are done. Proposition 139.8. Let G be a p-group which is a minimal non-KC-group satisfying d.G/ D 3. If p > 2 or p D 2 and ŒL0 ; L Ã2 .L0 / for all maximal subgroups L of G, then there exist maximal subgroups M; N of G such that M 0 ¤ ¹1º;

N 0 ¤ ¹1º and

M 0 \ N 0 D ¹1º:

Proof. Suppose, by way of contradiction, that L0 \ T 0 ¤ ¹1º for any maximal subgroups L; T of G with L0 ¤ ¹1º, T 0 ¤ ¹1º. Since G 0 is noncyclic, there exist maximal subgroups M and N of G such that hM 0 ; N 0 i is noncyclic and we have M 0 \N 0 ¤ ¹1º. Applying Proposition 139.6 on the factor group G=.M 0 \ N 0 /, we get jM 0 =.M 0 \ N 0 /j D p

and

jN 0 =.M 0 \ N 0 /j D p

so that jM 0 j D jN 0 j D p n , n > 1. We may set M 0 D hhi, N 0 D hki so that hp D k p . Then we have hk 1 ¤ 1 and .hk 1 /p D 1 so that hhk 1 i is a subgroup of order p in G 0 with hhk 1 i 6 M 0 and hhk 1 i 6 N 0 and we get M 0 \ N 0 D hhp i D hk p i. Let L be any maximal subgroup of G. Then we have either L0 M 0 \ N 0 or jL0 j D p n , L0 > M 0 \ N 0 and jL0 =.M 0 \ N 0 /j D p. We study the factor group G=.M 0 \ N 0 / which is a minimal non-KC-group (noting that hM 0 ; N 0 i=.M 0 \ N 0 / Š Ep2 ), where each maximal subgroup of G=.M 0 \ N 0 / has its commutator group of order at most p and d.G=.M 0 \ N 0 // D 3. It follows that G=.M 0 \ N 0 / is a group from Theorem 137.7(d) and so we may apply Lemma 139.2 on this group. Hence there is a maximal subgroup L of G with L0 D hhk 1 ; M 0 \ N 0 i. But then L0 is noncyclic, a contradiction. Lemma 139.9. Suppose that G is a 2-group which is a minimal non-KC-group such that exp.G 0 / D 4 and d.G/ D 3. Assume, in addition, that there is a maximal subgroup M of G such that M 0 Š C4 and ŒM 0 ; M ¤ ¹1º. If there exists another maximal subgroup N of G with N 0 ¤ ¹1º and M 0 \ N 0 D ¹1º, then the following statements hold: (i) M 0 \ L0 ¤ ¹1º for each maximal subgroup L of G with L ¤ N . (ii) N D CG .G 0 / D CG .M 0 /. (iii) jN 0 j D 2.

360

Groups of prime power order

Proof. For any a 2 M 0 and n 2 N we have Œa; n 2 M 0 \ N 0 D ¹1º

and so

CG .M 0 / D N :

Hence, if L is any maximal subgroup of G with L ¤ N , then L 6 CG .M 0 /. Therefore ¹1º ¤ ŒM 0 ; L M 0 \ L0 and this gives (i). If L is any maximal subgroup of G with L ¤ N , then M 0 \ L0 ¤ ¹1º together with M 0 \ N 0 D ¹1º implies that L0 \ N 0 D ¹1º. Hence ŒN 0 ; L N 0 \ L0 D ¹1º and so N 0 Z.G/. Also, ŒL0 ; N L0 \ N 0 D ¹1º and so CG .L0 / N which together with CG .M 0 / D N gives CG .G 0 / D N and (ii) is proved. We may choose m 2 M \ N and n 2 N M so that N 0 D hŒm; ni (noting that M \ N is abelian). As N 0 Z.G/, we may apply Lemma 139.1 on the group hm; ni. We get hŒm; n2 i D hŒm; n2 i D ¹1º, which gives jN 0 j D 2 and this is (iii). Proposition 139.10. Let G be a 2-group which is a minimal non-KC-group satisfying d.G/ D 3. Then ŒM 0 ; M Ã2 .M 0 / for all maximal subgroups M of G. Proof. Suppose, by way of contradiction, that there exists a maximal subgroup X of G such that ŒX 0 ; X 6 Ã2 .X 0 /. In particular, X 0 is cyclic of order 4 and so G 0 is of exponent 4. Note that the factor group G=Ã2 .G 0 / is a minimal non-KC-group such that d.G=Ã2 .G 0 // D 3 and .G=Ã2 .G 0 //0 is noncyclic of exponent 4. We want to show that such a 2-group G=Ã2 .G 0 / does not exist and so we may assume in the sequel that Ã2 .G 0 / D ¹1º. In other words, we may assume that G is 2-group which is a minimal non-KC-group with d.G/ D 3 and G 0 is noncyclic of exponent 4. First suppose that for each maximal subgroup L of G we have ŒL0 ; L Ã2 .L0 /. Then Proposition 139.8 implies that there exist maximal subgroups M; N of G such that M 0 ¤ ¹1º, N 0 ¤ ¹1º and M 0 \ N 0 D ¹1º. By Proposition 139.7, G 0 is elementary abelian of order 4 or 8, contrary to the fact that G 0 is noncyclic of exponent 4. Thus we have proved that there exists a maximal subgroup M of G that satisﬁes M 0 Š C4 and ŒM 0 ; M 6 Ã2 .M 0 /, i.e., ŒM 0 ; M ¤ ¹1º. (a) Assume that there is a maximal subgroup N of G such that N 0 ¤ ¹1º

and M 0 \ N 0 D ¹1º:

In this case, all assumptions of Lemma 139.9 are satisﬁed and so the statements (i), (ii) and (iii) of that lemma hold. Let L be any maximal subgroup of G with L ¤ N . Then L0 \ M 0 ¤ ¹1º, L0 \ N 0 D ¹1º, and either L0 Z.G/ or CG .L0 / D N . Assume that L0 Z.G/. Here N \L is abelian and so there are elements l 2 N \L and l 0 2 L N such that L0 D hŒl; l 0 i. By Lemma 139.1 applied to hl; l 0 i, we obtain that hŒl; l 0 2 i D hŒl; l 02 i D ¹1º (since .l 0 /2 2 N \ L) which implies jL0 j D 2 and so L0 D 1 .M 0 /. Suppose that CG .L0 / D N so that in this case L0 Š C4 and L0 \ M 0 1 .M 0 /. We see that G D hN; ti, where t 2 G N and t inverts L0 . Since CG .G 0 / D N and t inverts G 0 (noting that G 0 is generated by all X 0 , where X is any maximal subgroup of G), it follows that each subgroup of G 0 is normal in G. Set 1 .M 0 / D hzi so that

139

p-groups G with noncyclic G 0 and cyclic H 0 for all subgroups H < G

361

Ã1 .G 0 / D hzi. Then we obtain G 0 D 1 .G 0 /M 0 with 1 .G 0 / \ M 0 D hzi. We get G 0 D M 0 N 0 K with K 1 .G 0 /. Considering the factor group G=K, it follows that we may assume from the start that G 0 D M 0 N 0 is abelian of type .4; 2/ (since we want to show that G=K does not exist). Set N 0 D hui so that E D 1 .G 0 / D hu; zi Z.G/. We know that N D CG .G 0 /. Also set N D ˆ.G/ha; bi so that Œa; b 2 hui. Consider G=hzi so that G=hzi is a group from Theorem 137.7(d). This implies that ˆ.G=hzi/ D ˆ.G/=hzi D Z.G=hzi/. It follows that .ˆ.G//0 hzi \ hui D ¹1º and for any x 2 N ˆ.G/, Œˆ.G/; x hzi \ hui D ¹1º so that the three maximal subgroups of N containing ˆ.G/ are abelian. This also forces Œa; b D u. Take an element c 2 G N so that G D ha; b; ci. If Œa; c 2 E and Œb; c 2 E, then G=E would be abelian and so G 0 E, a contradiction. Hence (interchanging a and b if necessary), we may assume that Œa; c 2 G 0 E so that Œa; c2 D z. We may assume that Œb; c D z with D 0; 1. Indeed, suppose that Œb; c 62 hzi. Set L D ˆ.G/hb; ci so that L is a maximal subgroup of G which is distinct from N . By Lemma 139.9, L0 \ M 0 ¤ ¹1º and so L0 hzi. It follows that Œb; c 2 G 0 E. Then we have (since ˆ.G/hbi is abelian) Œab; c D Œa; cb Œb; c D Œa; cŒb; c 2 E: The maximal subgroup L1 D ˆ.G/hab; ci of G is distinct from N and so, in view of Lemma 139.9, L01 \ M 0 ¤ ¹1º and so L01 hzi and Œab; c 2 hzi. As G D ha; ab; ci and Œa; ab D Œa; b D u, we replace in this case b with ab so that we may assume from the start that Œb; c D z with D 0; 1. We consider the proper subgroup K D hac; b; a2 i of G (noting that a2 2 ˆ.G/). Then we have Œac; b D Œa; bc Œc; b D uc z D uz and Œa2 ; ac D Œa2 ; c D Œa; ca ŒŒa; c D Œa; c2 D z as a 2 N commutes with Œa; c (because N D CG .G 0 /). But then K 0 huz ; zi D E is noncyclic, a contradiction. (b) Assume that for all maximal subgroups L of G with L0 ¤ ¹1º, L0 \ M 0 ¤ ¹1º, where G 0 is noncyclic of exponent 4, M 0 Š C4 and ŒM 0 ; M ¤ ¹1º. For each maximal subgroup L of G we have either L0 M 0 or L0 Š C4 and L0 \ M 0 D hi i, where we set hi i D Ã1 .M 0 /. Thus Ã1 .G 0 / D hii and G 0 =hi i is elementary abelian of order 4 (since G 0 is noncyclic). It follows that G=hi i is a group from Theorem 137.7(d) and so G 0 =hii Š 1 .G 0 / Š E4 or E8 . Let F be a four-subgroup contained in 1 .G 0 / such that F > hii. Then Lemma 139.2 (applied on the group G=hi i) gives that G possesses a maximal subgroup L1 such that L01 covers F=hi i. But L01 is cyclic and so jL01 j D 2 and therefore F D L01 hi i, contrary to our assumption in (b). Our proposition is proved.

362

Groups of prime power order

We can now prove our main result. Proof of Theorem 139.A. If G is a p-group from (a), (b) or (c) of Theorem 139.A, then Proposition 137.6 implies that jH 0 j p for each subgroup H < G. Conversely, if G is a p-group with G 0 being noncyclic but for each H < G we have H 0 is cyclic, then Propositions 139.4, 139.5, 139.7, 139.8 and 139.10 show that jH 0 j p and we have one of the possibilities stated in parts (a), (b) or (c) of Theorem 139.A.

140

Power automorphisms and the norm of a p-group

An automorphism of a ﬁnite group G is called a power automorphism if it maps each subgroup of G onto itself. This will be the case if and only if maps each element x 2 G to some power x i of x (therefore such a is called a power automorphism). We denote by A.G/ the group of all power automorphisms of G and we shall prove here some elementary facts about A.G/ (Theorem 140.7). Then, in Corollary 140.8, we obtain a result of E. Schenkman [Sc] asserting that the norm of a p-group G (which is the intersection of the normalizers of all subgroups of G) is contained in the second center Z2 .G/ of G. Lemma 140.1. If 2 A.G/ and g 2 G, the inner automorphism induced by g 1 g is a power automorphism. Proof. Let H G and g 2 G. Then

H g D .H g / D .H /g D H g

whence (noting that g commutes with g ) H D Hg

g 1

D Hg

1 g

and so g 1 g induces a power automorphism of G. Lemma 140.2. If 2 A.G/ and g; h 2 G, then Œg 1 g ; hŒh1 h ; g D 1 and Œg 1 g ; h 2 hgi \ hhi: Proof. Since g h D .gh/ 2 hghi, it follows that g h commutes with gh. Hence (noting that Œg; g D 1 and Œh; h D 1) Œg 1 g ; hŒhh ; g 1 D gg h1 g 1 .g h /.gh/h g 1 D gg h1 g 1 .gh/.g h /h g 1 D 1: We have proved ()

Œg 1 g ; hŒhh ; g 1 D 1:

364

Groups of prime power order

By Lemma 140.1, g 1 g and h1 h induce inner power automorphisms on G and so Œg 1 g ; h 2 hhi and Œhh ; g 1 2 hgi. Hence for all g; h 2 G we get Œg 1 g ; h 2 hgi \ hhi: Interchanging g and h, we also have Œh1 h ; g 2 hgi \ hhi which shows that Œh1 h ; g D Œh1 h ; ghh

D Œh1 h ; h1 h ghh

D hh h h1 g 1 h h h1 h h1 h ghh

D g 1 .h1 h /g.hh / D g 1 g hh

D g hh

g 1 D Œhh ; g 1 :

Putting this into (), we ﬁnally get the required relation Œg 1 g ; hŒh1 h ; g D 1: Given an automorphism of a group G and g; h 2 G, we set Œg; D g 1 g and Œg; ; h D ŒŒg; ; h. Lemma 140.3. If 2 A.G/ and Œgi ; ; hj D 1 for i D 1; 2; : : : ; n, j D 1; 2; : : : ; m, then Œg1 g2 : : : gn ; ; h1 h2 : : : hm D 1: Proof. By Lemma 140.2 which asserts that Œg; ; hŒh; ; g D 1 (g; h 2 G), we may assume m n. We prove this lemma by induction on m. If m D 1, then n D 1 and Œg1 ; ; h1 D 1 by assumption. In case m > 1, we get Œg1 g2 : : : gn ; ; .h1 h2 : : : hm1 /hm D Œg1 g2 : : : gn ; ; hm Œg1 g2 : : : gn ; ; h1 h2 : : : hm1 hm D 1 by induction hypothesis. Lemma 140.4. If 2 A.hg; hi/ and Œg; ; h D 1, then is central in hg; hi, i.e., for each g1 2 hg; hi we have g1 D g1 z for some z 2 Z.hg; hi/. Proof. Set G D hg; hi and let g1 ; h1 2 G. Then g1 D a1 a2 : : : an ;

h1 D b1 b2 : : : bm ;

where each ai and bj is equal to g or h (noting that G is assumed to be ﬁnite). We claim that for each i; j we have Œai ; ; bj D 1. Indeed, Œg; ; g D Œg 1 g ; g D 1 since g 1 g is a power of g. Similarly, Œh; ; h D 1 and by our assumption Œg; ; h D 1. Finally, by Lemma 140.2, Œg; ; hŒh; ; g D 1 and so also Œh; ; g D 1. Using Lemma 140.3, we get Œg1 ; ; h1 D Œa1 a2 : : : an ; ; b1 b2 : : : bm D 1: If we set g11 g1 D z, then we get Œz; h1 D 1 and so z 2 Z.hg; hi/. Hence is central in hg; hi and we are done.

140

365

Power automorphisms and the norm of a p-group

Lemma 140.5. Let G be a p-group which is either abelian or minimal nonabelian with p > 2 and x; y 2 G such that o.x/ o.y/. Then for some integer u we have hxi \ hx u yi D ¹1º and

o.x u / D o.y/:

Proof. If hxi hyi, then there is an integer u with x u D y 1 so that o.x u / D o.y/ and x u y D 1 and we are done. We may assume that hxi 6 hyi. Since o.x/ o.y/, we have jhxi=.hxi \ hyi/j jhyi=.hxi \ hyi/j D p r with r 1. There exists an element x u D x 0 2 hxi hyi of order o.y/ such that r r .x 0 /p D y p , where jhx 0 i=.hxi \ hyi/j D p r . Since G is either abelian or minimal nonabelian with p > 2, we have for each a; b 2 G and s 1, s

s

1

s

.ab/p D ap b p Œb; a 2 p

s .p s 1/

Hence .x 0 y/p D 1 and so o.x 0 y/ p r . If .x 0 y/p r

.x 0 /p

r1

D yp

r1

r1

s

s

D ap b p : D 1, then

2 hxi \ hyi;

a contradiction. Hence o.x 0 y/ D p r and the element .x 0 y/p D .x 0 /p 0 0 of order p contained in hx yi. Suppose that hxi \ hx yi ¤ ¹1º. Then r1

.x 0 /p

r1

yp

r1

2 hxi

and so y p

r1

r1

yp

r1

is

2 hxi \ hyi;

contrary to jhyi=.hxi \ hyi/j D p r . Thus we have proved that hxi \ hx u yi D ¹1º and o.x u / D o.y/. Lemma 140.6. For a; b 2 G and 2 A.G/ we have Œab; D Œa; b Œb; . Proof. We have Œa; b Œb; D b 1 .a1 a / b .b 1 b / D b 1 a1 a b D .ab/1 .ab/ D Œab; : Theorem 140.7. Every power automorphism of a p-group G is central, i.e., for each g 2 G we have g D gz for some z 2 Z.G/. In particular, ﬁxes each element of G 0 . Proof. Let G be a p-group with a noncentral power automorphism . Then for some g; h 2 G we have Œg; ; h ¤ 1. By Lemma 140.2, d D Œg; ; h 2 hgi \ hhi Z.K/, where K D hg; hi. Since is noncentral in K=hd p i, we may assume d p D 1. We consider the subgroup H D hŒg; ; Œh; i of K. By Lemma 140.2, Œg; ; hŒh; ; g D 1

366

Groups of prime power order

and so we may assume that o.Œg; / o.Œh; /. Now, Œh; D hr for some integer r. Hence ŒŒg; ; Œh; D ŒŒg; ; hr D d r 2 hd i Z.H / since Œg; ; h D d 2 Z.K/ and so Œg; ; hr D Œg; ; hr . This shows that H is either abelian or minimal nonabelian with H 0 D hd i. In case p D 2, we can conclude that Œh; D h1 h D h1Ci D hr , where i is odd and so r D 1 C i is even, which gives d r D 1 and so in this case H is abelian. We have proved that the subgroup H D hŒg; ; Œh; i is either abelian or minimal nonabelian with p > 2, where H 0 D hd i and d D Œg; ; h. As o.Œg; / o.Œh; /, we may use Lemma 140.5 for the group H . It follows that there is an integer u such that hŒg; i \ hŒg; u Œh; i D ¹1º

()

and o.Œg; u / D o.Œh; /:

Now, d and Œg; are both contained in hgi and so one is a power of the other. However, d commutes with h while Œg; does not. Hence d is a power of Œg; , say d D Œg; t . Then setting v D u.1 t /, we get using Lemma 140.6 Œg v h; D Œg ut g u h; D Œg ut ; g

uh

Œg u h;

(since Œg ut ; is a power of g) D Œg ut ; h Œg u h; (and since Œg ut ; h D .g ut .g /ut /h D ...g 1 g /t /u /h D ..Œg; t /u /h D .d u /h D d u because d 2 Z.K/) D d u Œg u h; D d u Œg u ; h Œh; D d u Œg u ; Œg u ; ; hŒh; (since

Œg u ; D g u .g /u D .g 1 g /u D Œg; u

and Œg u ; ; h D ŒŒg u ; ; h D ŒŒg; u ; h D Œg; u .h1 Œg; h/u D Œg; u .Œg; h /u D .Œg; 1 Œg; h /u D Œg; ; hu D d u because Œg; commutes with d D Œg; 1 Œg; h 2 hgi and so Œg; also commutes with Œg; h ) D d u Œg; u d u Œh; D Œg; u Œh; :

140

Power automorphisms and the norm of a p-group

367

We have proved that Œg v h; D Œg; u Œh; : By (), we get hŒg; i \ hŒg v h; i D ¹1º. Since d D Œg; t is of order p, we have 1 ¤ Œg; 2 hgi. On the other hand, Œg v h; is a power of g v h. Hence we have either Œg v h; D 1 or hgi \ hg v hi D ¹1º. By Lemma 140.2, Œg v h; ; g D ŒŒg v h; ; g hg v hi \ hgi: Thus, in any case, Œg v h; ; g D 1. We get hg v h; gi D hg; hi and so, by Lemma 140.4, is central in hg; hi. But this implies that Œg; ; h D 1, a contradiction. Our theorem is proved. The norm N .G/ of a group G is the intersection of the normalizers of all subgroups of G. Corollary 140.8. The norm N .G/ of a p-group G is contained in the second center Z2 .G/ of G. Proof. Let a 2 N .G/. Then a induces in G an inner power automorphism. By Theorem 140.7, for all g 2 G we have g a D gz with some z 2 Z.G/. This implies that a 2 Z2 .G/ and we are done. Exercise 1. If M Š Mpn , then N .G/ is a noncyclic maximal subgroup of G. Exercise 2. If G is a nonabelian metacyclic group of order p 4 and exponent p 2 , then N .G/ D Z.G/. Exercise 3. If G 6Š Q8 is a 2-group of maximal class, then we have N .G/ D Z.G/ unless G Š Q2n , n > 3 (then N .G/ D Z2 .G/. Exercise 4. Find N .G/, where (a) G is minimal nonabelian p-group. (b) G is extraspecial. (c) G is special. (d) G is metacyclic. (e) jˆ.G/j D p. (f) jG 0 j D p. (g) G is a p-group of maximal class, p > 2. (h) G is a Sylow p-subgroup of the holomorph of a cyclic p-group. (i) G is an A2 -group. (j) G is minimal nonmetacyclic 2-group. Exercise 5. Find N .A B/ is terms of A and B.

141

Nonabelian p-groups having exactly one maximal subgroup with a noncyclic center

Theorem 141.1 solves a problem posed by the ﬁrst author. Note that if G is a nonabelian p-group all of whose maximal subgroups have a cyclic center, then p D 2 and G is a 2-group of maximal class. Theorem 141.1. A nonabelian p-group G has exactly one maximal subgroup with a noncyclic center if and only if Z.G/ is cyclic and G has exactly one normal abelian subgroup of type .p; p/. Proof. Suppose that G is a nonabelian p-group with exactly one maximal subgroup M such that Z.M / is noncyclic. If Z.G/ is noncyclic, then G=Z.G/ is cyclic since M is the only maximal subgroup of G containing Z.G/. But then G is abelian, a contradiction. Hence Z.G/ must be cyclic. Suppose that G has no normal abelian subgroup of type .p; p/. Then p D 2 and G is of maximal class (Lemma 1.4). But in that case, either each maximal subgroup of G has a cyclic center or G Š D8 . In the last case, G has two distinct maximal subgroups with a noncyclic center, a contradiction. Thus we have proved that G has a normal abelian subgroup U of type .p; p/. As jG W CG .U /j D p (because Z.G/ is cyclic), we get CG .U / D M . Also, 1 .Z.G// D hzi < U . Assume that G possesses another normal abelian subgroup V distinct from U of type .p; p/. Since CG .V / is a maximal subgroup of G with a noncyclic center (recall that Z.G/ is cyclic), we have CG .V / D M . In particular, we get V M , U \V D hzi and U V Š Ep3 with U V Z.M /. Take x 2 G M . We may choose u 2 U hzi and v 2 V hzi so that ux D uz and v x D vz 1 . Indeed, if ux D uz i and v x D vz j , where i; j are some integers with i 6 0 .mod p/ and j 6 0 .mod p/, we ﬁnd integers i 0 ; j 0 so that i i 0 1 .mod p/ and jj 0 1 .mod p/. Then, replacing u with 0 0 u0 D ui and v with v 0 D v j , we get (noting that i 0 6 0 .mod p/, j 0 6 0 .mod p/) 0

0

0

0

0

0

.u0 /x D .ui /x D .ux /i D .uz i /i D ui z i i D ui z D u0 z; and

0

0

0

0

0

0

.v 0 /x D .v j /x D .v x /j D .vz j /j D v j z jj D v j z 1 D v 0 z 1 :

Thus, writing again u instead of u0 and v instead of v 0 , we can assume from the start that ux D uz and v x D vz 1 . It follows that .uv/x D ux v x D uz vz 1 D uv:

141 p-groups with exactly one maximal subgroup having a noncyclic center

369

But then hz; uvi Š Ep2 and hz; uvi Z.G/, which contradicts the fact that Z.G/ is cyclic. We have proved that U is the unique normal elementary abelian subgroup of order p 2 in G. Suppose that G is a nonabelian p-group with a cyclic center Z.G/ which possesses a unique normal abelian subgroup U of type .p; p/. Since Z.G/ is cyclic, we conclude that jG W CG .U /j D p and so M D CG .U / is a maximal subgroup of G with a noncyclic center. Let N be any maximal subgroup of G with a noncyclic center Z.N /. Since Z.N / is normal in G, it follows that 1 .Z.N // contains an abelian subgroup V of type .p; p/ which is normal in G. Because Z.G/ is cyclic, we have CG .V / D N . By our assumption, V D U and so CG .U / D M D CG .V / D N , and we are done. Theorem 141.2. Let G be a nonabelian p-group all of whose nonabelian maximal subgroups have a cyclic center. Then one of the following holds: (a) G is minimal nonabelian. (b) G is a 2-group of maximal class. (c) G D LZ.G/, where L Š D8 ; Q8 or Mpn , n 3 (if p D 2, then n 4), and either Z.G/ is cyclic with Z.G/ > Z.L/ or G D Lht i, where t is an element of order p. (d) G has exactly one abelian maximal subgroup A of rank 2, Z.G/ is cyclic, and G possesses exactly one normal abelian subgroup of type .p; p/ (contained in A). Proof. We may assume that G is not minimal nonabelian. First assume that G contains more than one abelian maximal subgroup. Then we get G=Z.G/ Š Ep2 and jG 0 j D p. All p C 1 maximal subgroups of G containing Z.G/ are abelian and therefore each maximal subgroup M of the group G which does not contain Z.G/ is nonabelian. Set Z1 D M \ Z.G/ D Z.M / so that (by our assumption) Z1 is cyclic and so Z.G/ is of rank 2. Now let L be a minimal nonabelian subgroup of M so that M D LZ1 , G D LZ.G/ and L \ Z1 D Z.L/ ˆ.G/. It follows that L has a cyclic subgroup of index p and so L Š D8 ; Q8 or Mpn , n 3 (if p D 2, then n 4 ). Suppose that Z.G/ is not cyclic. Let t be an element of order p in Z.G/Z1 . If Lht i < G, then a maximal subgroup of G containing Lhti has a noncyclic center, a contradiction. Hence in this case G D L ht i. Assume that Z.G/ is cyclic. Then Z1 D ˆ.Z.G// D ˆ.G/ and so each maximal subgroup M1 of G which does not contain Z.G/ contains Z1 D ˆ.G/ and Z.M1 / D Z1 is cyclic. Now assume that G has exactly one abelian maximal subgroup A. If A is cyclic, then all maximal subgroups of G have cyclic center and then p D 2 and G is a 2-group of maximal class. We may assume that A is of rank 2 and so, by Theorem 141.1, Z.G/ is cyclic and G has exactly one normal abelian subgroup of type .p; p/ (contained in 1 .A/). Finally, assume that all maximal subgroups of G are nonabelian. Then all maximal subgroups of G have cyclic center. But then p D 2 and G is a 2-group of maximal class and such groups have a cyclic subgroup of index 2, a contradiction.

142

Nonabelian p-groups all of whose nonabelian maximal subgroups are either metacyclic or minimal nonabelian

We solve here Problem 1535(i) and prove the following result. Theorem 142.1. Let G be a nonabelian p-group all of whose nonabelian maximal subgroups are either metacyclic or minimal nonabelian. Then one of the following holds: (a) G is minimal nonabelian (= A1 -group). (b) G is an A2 -group, i.e., G is not minimal nonabelian but all nonabelian maximal subgroups of G are minimal nonabelian (and such groups are completely determined in 71). (c) G is metacyclic. Proof. Let G be a title group. If all maximal subgroups of G are abelian, then G is minimal nonabelian and we have obtained case (a) of our theorem. We may assume that G has nonabelian maximal subgroups and they are all either metacyclic or minimal nonabelian. It follows that d.G/ 3. First assume that G has more than one abelian maximal subgroup. Then G contains exactly p C 1 abelian maximal subgroups, d.G/ D 3 and jG 0 j D p. Let H be any nonabelian maximal subgroup of G. Then jH 0 j D p and H is either minimal nonabelian or H is metacyclic. But in the second case we have d.H / D 2 which together with jH 0 j D p implies that H is also minimal nonabelian. Hence in this case all nonabelian maximal subgroups of G are minimal nonabelian and so G is an A2 -group from case (b) of our theorem. From now on we may assume that G has at most one abelian maximal subgroup. If, in addition, all nonabelian maximal subgroups of G are minimal nonabelian, then again G is an A2 -group from part (b) of our theorem. Now suppose that all nonabelian maximal subgroups of G are metacyclic. We may also assume that G is nonmetacyclic (because of the possibility (c) of our theorem). If G is minimal nonmetacyclic, then, by the results of 66 and 69, G is either the minimal nonmetacyclic 2-group of order 25 or G is a 3-group of order 34 and maximal class. But these both groups are also A2 -groups and so they are included in part (b) of our theorem. Assume now that G has exactly one nonmetacyclic maximal subgroup M

142 p-groups with abelian, metacyclic or minimal nonabelian maximal subgroup 371

in which case M is the unique abelian maximal subgroup of G. If p D 2, then such groups are completely determined in 87. Considering all results of that section (Theorems 87.8, 87.10, 87.12, 87.14 to 87.17, and 87.19 to 87.21), we see in the case where G is not minimal nonabelian that the unique nonmetacyclic maximal subgroup of G is always nonabelian, a contradiction. Now suppose that p > 2. Then these groups are determined by Y. Berkovich in Proposition A.40.12. It follows that assuming jGj > p 4 (noting that in case jGj p 4 , G is an A2 -group ), G is an L3 -group, i.e., 1 .G/ D E is of order p 3 and exponent p and G=E is cyclic of order > p. Let M be the unique maximal subgroup of G containing E. Since M is nonmetacyclic, M is abelian and M D CG .E/. Let a 2 G M so that G D hE; ai and E \hai ¤ ¹1º. If CG .a/ > hai, then jCG .a/ W haij D p and CG .a/ is another abelian maximal subgroup of G (distinct from M ), a contradiction. Hence we get CG .a/ D hai. Let X be any maximal subgroup of G distinct from M . Then X covers G=E and X \ E Š Ep2 . Let a0 2 X be such that X D hX \ E; a0 i and ha0 i \ .X \ E/ ¤ ¹1º. Since ha0 i is a cyclic subgroup of index p in X and X is nonabelian, we get X Š Mpn , n 4, and so X is minimal nonabelian. Hence G is again an A2 -group from part (b) of our theorem. We have reduced our problem to a classiﬁcation of the title groups with the following properties: (i) G has at most one abelian maximal subgroup. (ii) G has a nonabelian maximal subgroup H which is minimal nonabelian but not metacyclic. (iii) G has a nonabelian maximal subgroup K which is metacyclic but not minimal nonabelian. In particular, K 0 is cyclic of order p 2 . We have E D 1 .H / Š Ep3 . If E ˆ.H /, then E ˆ.G/ and then each maximal subgroup of G would be nonmetacyclic, which contradicts our assumption (iii). Hence E D 1 .H / 6 ˆ.H / and we may set n

H D ha; b j ap D b p D 1; n 2; Œa; b D z; z p D Œz; a D Œz; b D 1i; where E D hap ; z; bi, ˆ.H / D Z.H / D hap ; zi and H 0 D hzi is a maximal cyclic subgroup of H . In particular, H=E is cyclic of order p n1 . If G=E is abelian, then G 0 E and so G 0 is elementary abelian. But this contradicts our assumption (iii), where K 0 is cyclic of order p 2 . It follows that G=E is nonabelian with a cyclic subgroup of index p. In particular, we have n 3 and so H=E is cyclic of order p 2 . All p C 1 maximal subgroups of G containing E are nonmetacyclic and so (by our assumption (iii)) d.G/ D 3 which forces ˆ.H / D ˆ.G/ since jG W ˆ.H /j D p 3 . Suppose that G possesses an abelian maximal subgroup A. By a result of A. Mann, we have jG 0 W .A0 H 0 /j p and so jG 0 W H 0 j p which implies that jG 0 j p 2 and so G 0 D K 0 Š Cp2 . On the other hand, G 0 ˆ.G/ D ˆ.H / and G 0 > H 0 , which contradicts the fact that H 0 D hzi is a maximal cyclic subgroup in H . n1

372

Groups of prime power order

Thus we have proved that all maximal subgroups of G are nonabelian. In particular, all p C 1 maximal subgroups of G containing E are nonabelian and nonmetacyclic and so they all must be nonmetacyclic minimal nonabelian. Let H1 ¤ H be another nonmetacyclic minimal nonabelian maximal subgroup of the group G containing E. Since H1 > ˆ.G/ D ˆ.H / and jH1 W ˆ.H1 /j D p 2 with ˆ.H1 / ˆ.G/, we have ˆ.H1 / D ˆ.H / D ˆ.G/ D Z.H / D Z.H1 /: This implies that ˆ.G/ Z.G/ and so G is of class 2. For any x; y 2 G we have Œx; yp D Œx p ; y D 1 and so G 0 is elementary abelian. But this contradicts our assumption (iii), where K 0 is cyclic of order p 2 . This is a ﬁnal contradiction and our theorem is proved.

143

Alternate proof of the Reinhold Baer theorem on 2-groups with nonabelian norm

Recall that the norm N .G/ of a group G is the intersection of the normalizers of all subgroups of G. Clearly, the subgroup N .G/ is characteristic in G and Dedekindian, i.e., N .G/ D Q E A, where Q 2 ¹¹1º; Q8 º, exp.E/ 2 and A is abelian of odd order. In this section we offer another proof of the following remarkable result due to Reinhold Baer. Theorem 143.1. If the norm N .G/ of a 2-group G is nonabelian, then N .G/ D G. Proof (Berkovich). Assume, by way of contradiction, that N .G/ < G. By Theorem 1.20, there is in N .G/ a subgroup Q Š Q8 . We have, by assumption, Q < G so that jGj 16. Let A D hai and B D hbi be distinct cyclic subgroups of Q of order 4. (i) If Q H G, then Q N .H /. This is obvious. (ii) If Q < H G and jH W Qj D 2, then H D Q C is Dedekindian. Assume that this is false. Then H has a nonnormal cyclic subgroup L of order 4 such that L 6 Q. As QL D H , it follows that Q does not normalize L, contrary to the hypothesis. This H is Dedekindian, and our claim follows. (iii) CG .Q/ has no cyclic subgroup of order 4. Assume, however, that the subgroup L D hxi CG .Q/ is cyclic of order 4. Set H D Q L; then we get 16 jH j 32. If jH j D 16, our claim follows from (ii). Now let jH j D 32; then H D Q L and haxi is not b-invariant, a contradiction. (iv) By (ii), jG W Qj > 2. (v) It follows from hypothesis and (iii) that if D D hd i < G is cyclic of order 4, then Q \ D > ¹1º. Indeed, assume that Q \ D D ¹1º. Then, by (iii), Q is not normal in F D QD so F=QF Š D8 . But the norm of D8 coincides with its center, and this is a contradiction. (vi) We claim that exp.G/ D 4. Assume that T < G is cyclic of order 8. Since Q normalizes T , we get H D QT G. Bringing to mind our aim, one may assume that G D QT . By (v), one has Q \ T > ¹1º so that 16 jGj 32. From (iv) we deduce jGj > 16. Let jGj D 32. Write H1 D AT and H2 D BT . Since, by (i), A N .H1 /,

374

Groups of prime power order

it follows that H1 is not of maximal class in view of A 6 ˆ.H1 /. Similarly, H2 is not of maximal class. Then, by Theorem 1.2, Ã1 .T / D ˆ.Hi / Z.Hi /, i D 1; 2. Thus Ã1 .T /, a cyclic subgroup of order 4, centralizes hA; Bi D Q, contrary to (iii). Now we are ready to complete the proof of our theorem. By (v) and (vi), Ã1 .G/ D Ã1 .Q/ so G=Ã1 .G/, being of exponent 2, is abelian, and we conclude that G 0 D Ã1 .G/ has order 2 since G is nonabelian. Then G=CG .Q/ is isomorphic to a subgroup of D8 2 Syl2 .Aut.Q//, and G=CG .Q/ contains a four-subgroup isomorphic to Q=Z.Q/. As Ã1 .Q/ CG .Q/, we get G=CG .Q/ Š Q=Z.Q/ since Q \ CG .Q/ D Z.Q/, and we infer that G D QCG .Q/. Since exp.CG .Q// D 2 by (iii), we deduce that CG .Q/ D Z.Q/ E, where E < CG .Q/. In that case, we obtain G D QCG .Q/ D Q.Z.Q/ E/ D QE; where Q \ E D ¹1º. Thus G D Q E so G is Dedekindian as exp.E/ D 2. Remark. In the proof of Theorem 143.3 we do not use the ﬁniteness of G. Thus that theorem is also true for inﬁnite 2-groups. The following theorem is a partial case of Schenkman’s result (see Corollary 140.8). Theorem 143.2. Suppose that G is an arbitrary ﬁnite group with nonabelian N .G/. Then P 2 Syl2 .N .G// centralizes all elements of G of odd order and P Z2 .G/. Proof. Let Q8 Š Q N .G/, let a 2 Q# and let x 2 G be of order p k , where p > 2 is a prime. Set H D ha; xi. Let y 2 hxi be of order p and F D ha; yi. Assume that F is nonabelian. In case o.a/ D 2, we see that F is a nonabelian group of order 2p so its norm equals ¹1º, a contradiction since a 2 N .G/. If o.a/ D 4, then F is minimal nonabelian with norm of order 2, a contradiction again. Thus F is abelian. Hence Q centralizes all subgroups of G of odd order. As, by Theorem 10.28, P 2 Syl2 .N .G// is generated by its minimal nonabelian subgroups all of which are isomorphic to Q8 , it follows that P centralizes all subgroups of G of odd order. Now let P P1 2 Syl2 .G/. By Theorem 143.1, the subgroup P1 is Dedekindian. Therefore Z.P / Z.P1 / (Theorem 1.20). Since, by the above, Z.P / centralizes all elements of G of odd order, we conclude that Z.P / Z.G/. As P =Z.P / P1 =Z.P /, P1 =Z.P / is abelian and P =Z.P / centralizes all elements of G=Z.P / of odd order, we get P =Z.P / Z.G=Z.P // so that P Z2 .G/, proving the last assertion. Exercise 1. Let N1 .G/ be the intersection of the normalizers of all cyclic subgroups of a 2-group G of order 8. If N1 .G/ contains a subgroup isomorphic to Q8 , then we have G D N1 .G/. (Hint. See the proof of Theorem 143.1.) Exercise 2. Let G 6Š SD16 be a 2-group, D < G and D Š D8 . Then there exists in G a subgroup L of order 2 such that L 6 D and D 6 NG .L/.

143

Alternate proof of Baer’s theorem on 2-groups with nonabelian norm

375

Exercise 3. Let M be a proper nonabelian subgroup of order p 3 of a p-group G, p > 2. Then, if 1 .G/ > 1 .M /, there is a subgroup L < G of order p such that L 6 M and M 6 NG .L/. Schenkman [Sc] has proved that N .G/ Z2 .G/ and ŒG 0 ; N .G/ D ¹1º for arbitrary groups G (ﬁnite and inﬁnite).

144

p-groups with small normal closures of all cyclic subgroups

A p-group G is called a BI.p k /-group (G 2 BI.p k / for short) with a ﬁxed integer k 0 if jhxiG W hxij p k for all x 2 G. Obviously, each p-group is a BI.p k /-group for some integer k and so we can only investigate such groups for some small k. Using to some extent the results of Heng Lv, Wei Zhou and Dapeng Yu [LZY], we shall establish here some remarkable properties of BI.p 2 /-groups for p > 2 (Theorems 144.6, 144.7 and 144.9). The corresponding problem for p D 2 is completely open. Look at similar problems stated in Problem 148. Lemma 144.1. Let G 2 BI.p k /. Then jG W NG .hai/j p k for all a 2 G. Proof. Let o.a/ D p n , n 1. Since G is a BI.p k /-group, we have jhaiG j p nCk . Hence there are at most .p nCk p n1 /=.p n p n1 / D .p kC1 1/=.p 1/ < p kC1 cyclic subgroups of order p n in haiG . This gives jG W NG .hai/j p k . Lemma 144.2. Suppose that G D ha; bi is a nonabelian p-group. If jhaiG W haij p and jhbiG W hbij p with hai \ hbi D ¹1º, then G 0 is of order p. Proof. If one of hai or hbi, say hai, is normal in G, then G is a semidirect product of hai and hbi. In that case, we see that hbi is not normal in G and so jhbiG W hbij D p and jhai \ hbiG j D p. But G 0 hai \ hbiG and so jG 0 j D p. Hence we may assume that jhaiG W haij D jhbiG W hbij D p: Set A D haiG and B D hbiG which yields jA W haij D p, G=A ¤ ¹1º and Ahbi D G. Similarly, jB W hbij D p, G=B ¤ ¹1º and Bhai D G. Hence we obtain G 0 A \ B. If hbi \ A D ¹1º or hai \ B D ¹1º, then jA \ Bj D p and so jG 0 j D p and we are done. It follows that we may assume hbi\A ¤ ¹1º and hai\B ¤ ¹1º. As hai\hbi D ¹1º, we see that hbi\A D hb1 i is of order p and A D haihb1 i and similarly, hai\B D ha1 i is of order p and so A \ B D ha1 i hb1 i Š Ep2 . In particular, o.a/ p 2 , o.b/ p 2 ,

144 p-groups with small normal closures of all cyclic subgroups

377

and G D haihbi is a product of the two cyclic subgroups hai and hbi with G 0 ha1 i hb1 i: For any x 2 G, haix \ hai ¤ ¹1º and so haiG ha1 i which implies ha1 i E G and so a1 2 Z.G/. Similarly, b1 2 Z.G/ and so G 0 ha1 i hb1 i Z.G/. In this case, hŒa; bi D G 0 is of order p and we are done. Lemma 144.3. If jGj D p mC2 and exp.G/ D p m with m 3 and p > 2, then we have jG 0 j p 2 . Proof. By Theorem 74.1, G is either metacyclic or an L3 -group. In the second case, G 0 < 1 .G/ and so jG 0 j p 2 . Let G be metacyclic, and let R be a normal elementary m1 abelian subgroup of order p 2 in G and a 2 G of order p m . Since ap centralizes R and G is metacyclic, we have jhai \ Rj D p and so .Rhai/=R is a cyclic subgroup of index p in G=R and G=R is noncyclic. Therefore there exists b 2 G .Rhai/ with b p 2 R so that o.b/ p 2 . Also, G=.Rhap i/ is elementary abelian of order p 2 which gives Rhap i D ˆ.G/ and so G D ha; bi. Since metacyclic p-groups for p > 2 are regular, we obtain that exp.hbiG / p 2 . Then G 0 hbiG gives exp.G 0 / p 2 . On the other hand, G 0 is cyclic and so jG 0 j p 2 and we are done. Lemma 144.4. Let G D hx; yi 2 BI.p 2 /, p > 2. If hxi \ hyi D ¹1º, then the following hold: (i) cl.G/ 4. In particular, cl.G/ 3 if o.x/ o.y/ p 3 . (ii) jG 0 j p 3 and exp.G 0 / p 2 . (iii) Ã2 .G/ Z.G/. Proof. We may assume that G is nonabelian. Let o.x/ D p m , o.y/ D p n . Since G is a BI.p 2 /-group, we have jhxiG W hxij p 2 . Then from jhxijjhx y ij=jhxi \ hx y ij D jhxihx y ij jhxiG j p mC2 2

we get jhxi \ hx y ij p m2 and x p 2 hxi \ hx y i. Hence 2

2

2

hx p i D h.x y /p i D h.x p /y i: 2

2

2

Let M D hx p ; yi. Then hx p i E M . By Lemma 144.1, we have x p 2 NG .hyi/ and 2 2 so hyi E M . But hxi\hyi D ¹1º and so Œx p ; y D 1 and x p 2 Z.G/. Similarly, we 2 get y p 2 Z.G/. Set H D hxiG and K D hyiG so that G 0 H \K. If jH \Kj p 3 , then we obtain cl.G/ 4. Hence if o.y/ p 2 , then jKj p 4 and thus H \ K < K which yields that cl.G/ 4. In this case, we also conclude that exp.G 0 / p 2 . Indeed, assume that H \ K D G 0 is cyclic of order p 3 . Then the group G is regular and K D hyiG is of

378

Groups of prime power order

exponent p 2 , a contradiction. We have a similar situation if o.x/ p 2 . So we need only consider the case where o.x/; o.y/ p 3 . Without loss of generality we may assume that m n 3. We shall prove that under this condition cl.G/ 3 which will complete the proof of (1). Since G is a BI.p 2 /-group and hxi \ hyi D ¹1º, we see that jH \ Kj p 4 . If jH \ Kj D p 3 , then xp m1

m1

yp

2 H \ K;

n1

2H \K

n1

and, by the above, x p ; yp 2 Z.G/. This implies that H \ K 2 Z2 .G/ and then 0 cl.G/ 3 and exp.G / p 2 . So we need to consider the case jH \ Kj D p 4 m2

where x p

m2

n2

; yp

n2

m1

2 H \K

n1

and so H \ K D hx p ihy p i. Set A D hx p ; y p i and so, by the above, we N get A Z.G/ since m; n 3. Now consider G D G=A and we use the “bar convention”. Then GN D hx; N yi N and we obtain N

N

jhxi N G W hxij N D jhyi N G W hyij N D p: By Lemma 144.2, jGN 0 j p and so jG 0 j p 3 . If jG 0 j p 2 , then cl.G/ 3. In case jG 0 j D p 3 , we have A G 0 . Since A Z.G/, we conclude that ŒG; G 0 A Z.G/ and so cl.G/ 3. Obviously, in any of the above cases, jG 0 j p 3 and exp.G 0 / p 2 . We have proved (1) and (2). Let g1 D x i y j 2 G. By the Hall–Petrescu formula (see Appendix 1), 2

2

2

.x i y j /p D .x i /p .y j /p a2l2 a3l3 a4l4 ; 2 where ai 2 Ki .G/, li D pi , i D 2; 3; 4. If p > 3, then p 2 j li for i D 2; 3; 4. Since exp.G 0 / p 2 , it follows that p2

2

2

2

g1 D .x i y j /p D .x i /p .y j /p 2 Z.G/: For any g D x k1 y m1 : : : x kr y mr 2 G, we see that 2

2

2

2

2

g p D .x k1 /p .y m1 /p : : : .x kr /p .y mr /p 2 Z.G/: Now we consider the case p D 3. We shall prove in the following four cases that exp.K3 .G// 3. (i) Suppose that o.y/ D 3. Then K D hyiG is of order 33 . Since K is generated by elements of order 3, it follows that exp.K/ 3. But G 0 K and so we are done. (ii) Suppose that o.x/ D o.y/ D 32 . Then jH j; jKj 34 and so jH \ Kj 33 . Obviously, exp.K3 .G// 3 if jH \ Kj 32 and so we need only consider the case jH \ Kj D 33 . Now we have jH j D jKj D 34 , jGj D 35 and jG 0 j jH \ Kj D 33 . If jG 0 j 32 , then exp.K3 .G// 3. It remains to consider the case that G 0 D H \ K

144 p-groups with small normal closures of all cyclic subgroups

379

is of order 33 . If G 0 is cyclic, then G is regular and so G D hx; yi is of exponent 32 , a contradiction. Therefore G 0 is noncyclic. By a result of Burnside, G 0 is abelian. Indeed, if G 0 were nonabelian, then Z.G 0 / is cyclic and so G 0 would also be cyclic, a contradiction. It follows that j1 .G 0 /j 32 , jG=1 .G 0 /j 33 and K3 .G/ 1 .G 0 /, and we are done. (iii) Suppose that o.x/ D 3m 33 and o.y/ D 32 . (We argue similarly in the case o.y/ D 3n 33 and o.x/ D 32 .) As above, we need only consider jH \ Kj D 33 and G 0 D H \ K. If G 0 is cyclic, then G is regular and so exp.K/ D 32 , a contradiction. Hence G 0 is noncyclic and then G 0 is noncyclic abelian with j1 .G 0 /j 32 . In that case, K3 .G/ 1 .G 0 /, and we are done. (iv) Suppose that o.x/; o.y/ 33 . By (2), we get jG 0 j 33 and we see by the above m1 n1 arguments that K3 .G/ hx 3 i y3 and so exp.K3 .G// 3. Since 32 j l2 ; l4 and 3 j l3 and exp.K3 .G// 3, it follows from the Hall–Petrescu formula that Ã2 .G/ Z.G/. Lemma 144.5. Suppose that G D hx; yi 2 BI.p 2 / is noncyclic, p > 2, o.x/ D p m and o.y/ D p n . If m n and m 3, then there exists y1 2 G such that G D hx; y1 i p and hxi \ hy1 i D ¹1º or hxi \ hy1 i D hy1 i with o.y1 / D p 2 , where p D 3. Proof. We proceed by induction on m C n. Let H D hxiG and K D hyiG . We ﬁrst prove our lemma in the special cases where n D 2 and m D n D 3, which illustrate the idea in the general case. (a) n D 2 and hxi \ hyi D hy p i. If p D 3, then y1 D y satisﬁes all the requirem2 ments. Hence we need only consider p > 3. Consider the subgroup M D hx p ; yi. m2 m2 Since jhx p iG j p 4 and y p 2 hx p i, we have jM j p 5 and so M is regular. m2 m1 p As y p D x dp for some integer d (by the assumption), we get .yx dp / D1 m1 by the regularity of M (Theorem 7.2(a)). Set y1 D yx dp . Then G D hx; y1 i and hxi \ hy1 i D ¹1º. (b) m D n D 3. We shall ﬁrst prove that here G satisﬁes cl.G/ 5, exp.G 0 / p 2 and exp.K3 .G// p in the following subcases. (b1) hxi \ hyi D hx p i. Then x p 2 Z.G/ and x p D .x p /y D .x y /p . Since G is a BI.p 2 /-group, we have jH j; jKj p 5 . By Theorem 74.1, either H is metacyclic (if jH j p 4 , then H is certainly metacyclic) or jH j D p 5 , j1 .H /j D p 3 and exp.1 .H // D p. If H is metacyclic, then H is regular. Since x; x y 2 H , we see that .x 1 x y /p D 1 (Theorem 7.2(a)) and exp.G 0 / p as G 0 1 .H /. Because H is metacyclic, j1 .H /j p 2 and so cl.G/ 3. So we need only consider the case that jH j D jKj D p 5

and j1 .H /j D j1 .K/j D p 3

with exp.1 .H // D exp.1 .K// D p. Set GN D G=1 .H / and then o.x/ N D p 2 and hxi N E GN so that GN D hxih N yi N is meta0 0 N N cyclic and so G < hxi. N Therefore jG j p and so K3 .G/ 1 .H /. We have proved that in this case cl.G/ 5, exp.G 0 / p 2 and exp.K3 .G// p.

380

Groups of prime power order 2

(b2) hxi \ hyi D hx p i. If H or K is an L3 -group, then the argument in the second paragraph of (b1) can be applied to see that G satisﬁes cl.G/ 5, exp.G 0 / p 2 and exp.K3 .G// p. So we need only consider the case where both H and K are metacyclic. (b2.1) H and K are metacyclic and H \ K is cyclic. Since G 0 H \ K , we obtain 2 2 2 G 0 D hŒx; yi. As x; x y 2 H and x p D .x p /y D .x y /p , we see that the regularity 2 2 of H implies .x 1 x y /p D 1 and so Œx; yp D 1. Hence jG 0 j p 2 , jK3 .G/j p and cl.G/ 3. (b2.2) H and K are metacyclic and H \ K is noncyclic. Since p is odd, we have j1 .H \ K/j p 2 . But H and K are metacyclic and so 1 .H / D 1 .H \ K/ D 1 .K/ is elementary abelian of order p 2 . Set GN D G=1 .H / so that GN D hx; N yi; N

hxi N \ hyi N D ¹1º;

N

jhxi N G W hxij N p;

N

jhyi N G W hyij N p:

By Lemma 144.2, we have jGN 0 j p and so K3 .G/ 1 .H /. Therefore cl.G/ 4, exp.G 0 / p 2 and exp.K3 .G// p. Now, 2

hxi \ hyi D hx p i or hxi \ hyi D hx p i; 2

2

and so x p D y kp , where 1 k p 1. Since cl.G/ 5, using the Hall–Petrescu formula we get 2 2 2 .xy k /p D x p y kp c2n2 c3n3 c4n4 c5n5 ; 2 where ci 2 Ki .G/, ni D pi , i D 1; : : : ; 5, and p 2 j n2 , p j n3 , p j n4 , p j n5 since 2 p > 2. Hence .xy k /p D 1. Let y0 D xy k so that G D hx; y0 i and o.y0 / p 2 . By (a), there exists y1 2 G such that hxi \ hy1 i D ¹1º or hxi \ hy1 i D hy13 i with o.y1 / D 9. (c) m D n 4. 2 2 3 (c1) hx p i hxi \ hyi. Hence we obtain x p 2 Z.G/ and hx p i E G. Consider 3 p 3 GN D G=hx i so that GN D hx; N yi N and o.x/ N D o.y/ N D p . By (b), there exists y1 2 G 2 N such that G D hx; N yN1 i and o.yN1 / p . Thus we get G D hx; y1 i and o.y1 / < o.x/. Let G0 D hx p ; y1 i. By induction, there exists y2 2 G0 such that G0 D hx p ; y2 i and p hx p i \ hy2 i D ¹1º or hx p i \ hy2 i D hy2 i with p D 3 and o.y2 / D 9. Therefore p G D hx; G0 i D hx; y2 i and hxi \ hy2 i D ¹1º or hxi \ hy2 i D hy2 i with o.y2 / D 9. (c2) hxi\hyi D hx p i. We shall prove in this part that cl.G/ 6, exp.G 0 / p 3 , exp.K3 .G// p 2 and exp.K4 .G// p. By Theorem 74.1, H is either metacyclic or H is an L3 -group and the same holds for K. (c2.1) First suppose that H (or K) is an L3 -group. Now consider the factor group N G D G=1 .H / D hx; N yi, N where hxi N \ hyi N D ¹1º. By Lemma 144.4, exp.GN 0 / p 2 . 0 2 N Then G is a cyclic group of order p as GN 0 HN D hxi. N Therefore exp.G 0 / p 3 , 2 0 exp.K3 .G// p and exp.K4 .G// p. Since jG j p 5 , we have cl.G/ 6. (c2.2) H and K are metacyclic and H \K is cyclic. Since G 0 H \K, we see that m1 0 G is cyclic and G 0 D hŒx; yi. Consider the factor group GN D G=hx p i D hx; N yi, N m1

144 p-groups with small normal closures of all cyclic subgroups

381

where hxi N \ hyi N D ¹1º. By Lemma 144.4, oŒx; N y N p 2 and so oŒx; y p 3 . Hence 0 3 2 exp.G / p , exp.K3 .G// p , exp.K4 .G// p and cl.G/ 4. (c2.3) H and K are metacyclic and H \ K is noncyclic. We have 1 .H / D 1 .K/ D 1 .H \ K/ Š Ep2 : N

N yi, N where hxi N \ hyi N D ¹1º and jhxi N G W hxij N p, Consider GN D G=1 .H / D hx; N 0 0 G jhyi N W hyij N p. By Lemma 144.2, we have jGN j p and so jG j p 3 which gives cl.G/ 4, exp.G 0 / p 3 , exp.K3 .G// p 2 and exp.K4 .G// p. m1 m1 m1 Since hxi \ hyi D hx p i, we have x p D y lp for some positive integer l p 1. By the Hall–Petrescu formula, we get m1

m1

m1

.xy l /p D xp y lp c2n2 c3n3 c4n4 c5n5 c6n6 ; m1 where ci 2 Ki .G/, ni D p i , i D 1; : : : ; 6, and p 3 j n2 , p 2 j nj , j D 3; : : : ; 6 m1 since p > 2. It follows that .xy l /p D 1 and so G D hx; xy l i with o.x/ > o.xy l /. By induction, there is an element y1 2 G such that G D hx; y1 i and hxi \ hy1 i D ¹1º p or hxi \ hy1 i D hy1 i with o.y1 / D p 2 , where p D 3. mk

(c3) hxi\hyi D hx p i for some k 2. In this case, we consider the factor group mkC1 N G D G=hx p i so that GN D hx; N yi. N Then we apply (c2) to ﬁnd g0 2 G such that GN D hx; N yN0 i with o.yN0 / < o.y/. N Hence we obtain G D hx; y0 i with o.y0 / < o.y/. By induction, there is y1 2 G satisfying all the requirements. mn (d) m > n 3. Let G1 D hx p ; yi. By induction, there exists an element y1 mn mn mn p p p such that G1 D hx ; y1 i and hx i \ hy1 i D ¹1º or hx p i \ hy1 i D hy1 i with o.y1 / D p 2 , where p D 3. Hence G D hx; G1 i D hx; y1 i satisﬁes all the requirements. Theorem 144.6. Let G 2 BI.p 2 / with p > 2. Then Ã2 .G/ Z.G/ and exp.G 0 / p 2 . Proof. We ﬁrst prove that Ã2 .G/ Z.G/. Let x 2 G be such that o.x/ D p m p 3 . For any y 2 G we consider G1 D hx; yi. 2 It sufﬁces to prove that Œx p ; y D 1. If hxi \ hyi D ¹1º, then Lemma 144.4 implies 2 2 that Ã2 .G1 / Z.G1 / and so Œx p ; y D Œx; y p D 1: (i) o.y/ p 2 . By the above argument, we need only consider the case o.y/ D p 2 2 2 and hxi \ hyi D hy p i. If m D 3, then x p 2 Z.G1 / and so Œx p ; y D 1. Let m > 3 p and x0 2 hxi be such that x0 D y p . By Lemma 144.1, jhxi W Nhxi .hyi/j p 2 and so hx0 i normalizes hyi and therefore Œx0 ; y 2 hy p i so that hx0 ; yi is of order p 3 and p p class 2. Now set y1 D x0 y so that y1 D x0 y p D 1 and G1 D hx; yi D hx; y1 i. 2 By Lemma 144.4, x p 2 Z.G1 / and we are done. (ii) o.y/ p 3 . Without loss of generality we may assume o.x/ o.y/. If there is y1 such that G1 D hx; yi D hx; y1 i and hxi \ hy1 i D ¹1º, then, by Lemma 144.4, we 2 have Ã2 .G1 / Z.G1 / which gives Œx p ; y D 1. Due to Lemma 144.5, there is only one exceptional case where p D 3, G1 D hx; yi D hx; y1 i, hxi \ hy1 i D hy13 i and 2 o.y1 / D 9. By (i) applied to G1 D hx; y1 i, we see that x 3 2 Z.G1 / and we are done.

382

Groups of prime power order

Now we prove exp.G 0 / p 2 . We claim that o.Œx; y/ p 2 for all x; x 2 G. If o.y/ p 2 , then jhyiG j p 4 . Suppose that exp.hyiG / p 3 . Then hyiG is metacyclic and so regular which implies that exp.hyiG / D exp.hyi/ p 2 , a contradiction. Therefore we get exp.hyiG / p 2 and so o.Œx; y/ p 2 since Œx; y 2 hyiG for each x 2 G. If o.x/ o.y/ p 3 , then consider G1 D hx; yi so that, by Lemma 144.5, there exists an element y1 such that G1 D hx; yi D hx; y1 i and hxi \ hy1 i D ¹1º or hxi \ hy1 i D hy13 i and o.y1 / D 9. If hxi \ hy1 i D ¹1º, then, by Lemma 144.4, o.Œx; y/ p 2 . In case hxi \ hy1 i ¤ ¹1º, we get o.y1 / D 32 and so we also see that o.Œx; y/ 32 . Indeed, by the above argument, we infer that jhy1 iG1 j 34 and exp.hy1 iG1 / 32 and since G10 hy1 iG1 , we get o.Œx; y/ 32 . It sufﬁces to show that exp.ha; bi/ p 2 for any a; b 2 G with o.a/; o.b/ p 2 . Set A D ha; bi, H D haiA and K D hbiA . Hence we get jH j p 4 , jKj p 4 and jAj p 6 . If jAj D p 6 , then jH \ Kj p 2 and then jA0 j p 2 , jK3 .A/j p and cl.A/ 3. By the Hall–Petrescu formula, exp.A/ p 2 . So it remains to consider the case jAj p 5 . Suppose that exp.A/ p 3 . By Theorem 74.1, A is either metacyclic or jAj D p 5 and A is an L3 -group so that j1 .A/j D p 3 and exp.1 .A// D p. If A is metacyclic, then A is regular, and therefore exp.A/ p 2 . In the second case, we get exp.2 .A// p 2 . But a; b 2 2 .A/, a contradiction. Hence exp.A/ p 2 . We have proved that exp.G 0 / p 2 . Theorem 144.7. Let G 2 BI.p 2 / with p > 2. Then cl.G/ 4. Proof. It sufﬁces to prove that Œx; y 2 Z3 .G/ for any x; y 2 G. We set o.x/ D p m , o.y/ D p n , H D hxiG and K D hyiG . If jH \ Kj p 3 , then Œx; y 2 Z3 .G/ since Œx; y 2 H \ K E G. (i) m 2 and n 2. Hence jH j; jKj p 4 and jH 0 j; jK 0 j p 2 . If y 2 H , then Œx; y 2 H 0 and so Œx; y 2 Z2 .G/ Z3 .G/. Similarly, Œx; y 2 Z3 .G/ if x 2 K. Now assume that x 62 K and y 62 H . Then jH \ Kj p 3 and then Œx; y 2 Z3 .G. (ii) m 2 and n 3 (and similarly for m 3 and n 2 ). Since jH j p 4 , we need only consider jH \ Kj D p 4 and then H K. By Lemma 144.3, jK 0 j p 2 and therefore Œx; y 2 Z3 .G/. (iii) m 3 and n 3. If hxi \ hyi D ¹1º, then jH \ Kj p 4 . As before, we need m1 n1 only consider jH \ Kj D p 4 . Then x p ; yp 2 H \ K. By Theorem 144.6, m1 n1 hx p ; y p i Z.G/. Hence we get H \ K Z3 .G/. We consider now the case hxi \ hyi ¤ ¹1º. Set G1 D hx; yi, where we can assume m n. By Lemma 144.5, p there exists y1 2 G1 such that G1 D hx; y1 i and hxi\hy1 i D ¹1º or hxi\hy1 i D hy1 i with o.y1 / D p 2 , where p D 3. If hxi \ hy1 i D ¹1º, we see that G10 Z3 .G/ as above and then Œx; y 2 Z3 .G/. In the second case, o.y1 / D 32 and so, by (ii) above, we also have Œx; y 2 G10 Z3 .G/. Lemma 144.8. Let G 2 BI.p 2 / with p > 2. Suppose that x; y 2 G, o.x/ D p m p 3 and o.y/ D p n p 3 . If hxi \ hyi D ¹1º, then Œx; y 2 Z2 .G/.

144 p-groups with small normal closures of all cyclic subgroups m1

383

n1

Proof. By Theorem 144.6, x p ; yp 2 Z.G/. Set H D hxiG and K D hyiG . 2 Since G is a BI.p /-group, we have jH \ Kj p 4 . If jH \ Kj p 2 , then we have m1 n1 ; yp 2H \K Œx; y 2 H \ K Z2 .G/. In case jH \ Kj D p 3 , we get x p m1 n1 p p and x ;y 2 Z.G/ and so H \ K Z2 .G/. We need only consider the case jH \ Kj D p 4 in which case H \ K D hx p

m2

ihy p

n2

i:

m2

n1

2 Z.G/. Since also y p 2 Z.G/, we get If m 4, then, by Theorem 144.6, x p Œx; y 2 H \ K Z2 .G/. Similarly, we have Œx; y 2 H \ K Z2 .G/ in case n 4. We may assume that jH \ Kj D p 4 and m D n D 3. Since hxi \ hyi D ¹1º, we have jH \ hyij D p 2 and so y p 2 H . Thus H D hx; y p i D hxihy p i is of order p 5 and HK E G is of order p 6 since HK D H hyi with y p 2 H . But jhxihyij D jhxijjhyij D p 6 and so HK D hxihyi is a product of two cyclic subgroups. By Corollary 36.8, H is metacyclic and so H is also regular. Thus HK is metacyclic of exponent p 3 which implies that .HK/0 is cyclic of order p 2 . Since Œx; y 2 .HK/0 and .HK/0 Z2 .G/, we are done. Theorem 144.9. Let G 2 BI.p 2 / with p > 2. Then we have cl.G/ 3 if and only if Œ2 .G/; G Z2 .G/. Proof. Obviously, we need only prove that our condition is sufﬁcient. Let x; y 2 G. If o.y/ p 2 (or o.x/ p 2 ), then Œx; y 2 Z2 .G/ by assumption. Hence we may assume that o.x/ o.y/ p 3 . Let G1 D hx; yi. By Lemma 144.5, there is y1 2 G1 such that G1 D hx; y1 i and hxi \ hy1 i D ¹1º or hxi \ hy1 i D hy13 i with o.y1 / D 9. If o.y1 / p 2 , we have Œx; y1 2 Z2 .G/ by assumption and Œx; y 2 G10 Z2 .G/. In case o.y1 / p 3 , we get hxi \ hy1 i D ¹1º and, by Lemma 144.8, Œx; y1 2 Z2 .G/ and Œx; y 2 G10 Z2 .G/. Therefore cl.G/ 3.

Appendix 27

Wreathed 2-groups

In this section we study the subgroup structure of wreath products of some small 2-groups with a group of order 2. k Let solk .G/ be the number of solutions of x 2 D 1 in a group G. Exercise 1. Let A be a ﬁnite group and G D A wr C , where C D hc j c 2 D 1i Š C2 . Then sol1 .G/ D jAj C sol1 .A/2 : (We have c1 .G/ D sol1 .G/ 1.) It follows that the set G .A Ac / contains exactly jAj involutions. Since CG .c/ D hci , where is the diagonal of the base A Ac , all involutions of the set G .AAc / are conjugate in G in view of jG W CG .c/j D jAj. Solution. Let B D A Ac be the base of G D C B (a semidirect product with kernel B). An element g 2 G B can uniquely be represented in the form g D xy c c for x; y 2 A. We have g 2 D xy c cxy c c D xcyxcy D x.yx/c y D .xy/.yx/c : If g 2 D 1, then xy D 1, i.e., y D x 1 . It follows that G B contains exactly jAj involutions and sol1 .G/ D jAj C sol1 .B/ D jAj C sol1 .A/2 : In particular, if A Š C2n , then sol1 .G/ D 2n C 4 so c1 .G/ D 2n C 3. Next, c1 .D8 wr C2 / D 8 C 62 1 D 43;

c1 .Q8 wr C2 / D 8 C 22 1 D 11:

1o . Let G D A wr C , where n

A D ha j a2 D 1i Š C2n ;

C D hc j c 2 D 1i Š C2 :

Then jGj D 22nC1 and n

n

G D ha; b; c j a2 D b 2 D c 2 D 1; Œa; b D 1; ac D bi: The group B D hai hac i is the base of the wreath product G. Every element of B can be written uniquely in the form ai .ac /j , where i; j D 1; 2; : : : ; 2n . If n D 1, then we have G Š D8 . So, in what follows, we assume that n > 1. The abelian subgroup B

A.27

Wreathed 2-groups

385

of type .2n ; 2n / is maximal in G so we get Z.G/ D CB .c/ < B. Let ai .aj /c 2 Z.G/, 0 i; j < p n . Then ai .aj /c D .ai .aj /c /c D aj .ai /c : It follows that ai D aj . It is easy to see that .aac /i D ai .ai /c 2 Z.G/ for all i . Thus we have proved assertion (a) of the following Lemma A.27.1. (a) Z.G/ D haac i Š C2n . (b) G is not metacyclic unless n D 1. Proof. It remains to prove (b). Due to (a), h1 .B/; ci is nonabelian so it is isomorphic to D8 . By Proposition 10.19, if a metacyclic 2-group contains a nonabelian subgroup of order 8, it is of maximal class. Since G is not of maximal class unless n D 1, it is not metacyclic for n > 1. Remark. Let G D A wr B be a p-group. Then d.G/ D d.A/ C d.B/. We claim that if ¹x1 ; : : : ; xr º and ¹y1 ; : : : ; ys º are minimal bases of the groups A and B, respectively, then B D ¹x1 ; : : : ; xr ; y1 ; : : : ; ys º is a minimal basis of G. Indeed, we have hBi D G, hB ¹xi ºi < G and hB ¹yj ºi < G for all i r and j s. In particular, jG W ˆ.G/j D jA W ˆ.A/jjB W ˆ.B/j D p rCs : It follows from Lemma 1.1 that 22nC1 D jGj D 2jZ.G/jjG 0 j D 2nC1 jG 0 j so that jG 0 j D 2n . As G D ha; ci, we have d.G/ D 2. Since ˆ.G/ is a subgroup of index 2 in an abelian group B of type .2n ; 2n /, we get the following Lemma A.27.2. (a) jG 0 j D 2n . (b) ˆ.G/ is abelian of type .2n ; 2n1 /. Lemma A.27.3. G=Z.G/ Š D2nC1 . In particular, cl.G/ D n C 1. Proof. We have .aZ.G//c D ac Z.G/ D a1 .aac Z.G// D a1 Z.G/ (Lemma A.27.1(a)). Since ha; Z.G/i D B and G D hB; ci, the result follows. Since G=ŒB; C is abelian, we get G 0 D ŒB; C . Lemma A.27.4. Let n > 1. (a) G 0 D ha1 ac i Š C2n and G=G 0 is abelian of type .2n ; 2/.1 (b) D D hc; G 0 i Š D2nC1 and G=D is cyclic; then j1 .G/j D 2jDj D 2nC2 . 1 This

agrees with the Ito–Ohara result; see Corollary 36.11 (indeed, as we shall prove, our nonmetacyclic group G is a product of two cyclic subgroups).

386

Groups of prime power order

(c) 1 .G/ D D1 .B/ is of order 2nC2 , D \ 1 .B/ D 1 .Z.G//. (d) D < 2 .G/, jG W n1 .G/j D 2. Proof. We have a1 ac D Œa; c 2 G 0 and o.a1 ac / D 2n D jG 0 j which proves (a) (Lemma A.27.2(a)). Since .a1 ac /c D .a1 ac /1 , we see that D D hc; G 0 i Š D2nC1 . As d.G/ D 2, D G G and Z.D/ is cyclic, it follows that D 6 ˆ.G/ (Lemma 1.4) so that the quotient group G=D is cyclic. By Exercise 1, c1 .G/ D 2n C 3 D c1 .D/ C 2. It follows from n > 1 that jD \ 1 .B/j D 2 so 1 .G/ D D1 .B/ (see Proposition 1.17(a)) is of order 2jDj D 2nC2 , completing the proof of (b) and (c). Note that D D 2 .D/ and 1 .B/ 6 D so we obtain 2 .G/ > D. It follows from the cyclicity of G=1 .G/ as an epimorphic image of G=D and j1 .G/j D 2nC2 that jn1 .G/j D 22n hence jG W n1 .G/j D 2, completing the proof of (d). Lemma A.27.5. The class number of G is equal to k.G/ D 22n1 C 2nC1 2n1 . Proof. We have cd.G/ D ¹1; 2º (Introduction, Theorem 17) so that jGj jG W G 0 j 22 2nC1 2 2nC1 D 2nC1 C D 22n1 C 2nC1 2n1 ; 22

k.G/ D jG W G 0 j C

and the proof is complete. Lemma A.27.6. We have o.ac/ D 2nC1 hence exp.G/ D 2nC1 . Proof. We obtain exp.G/ exp.B/ exp.C / D 2nC1 . The element .ac/2 D aac has order 2n hence o.ac/ D 2nC1 . Lemma A.27.7. The subgroup G 0 \ Z.G/ D 1 .G 0 / is of order 2 and ˆ.G/ D ha2 i haac i D ha2 i Z.G/ D G 0 Z.G/ is abelian of type .2n ; 2n1 /. Proof. Clearly, we see that G 0 \ Z.G/ D 1 .G 0 / since G 0 < D and D G G is of maximal class. Further, we get ˆ.G/ D Ã1 .G/ and so a2 ; aac 2 ˆ.G/ as aac D a2 Œa; c. Since ha2 i \ haac i D ¹1º, the ﬁrst equality in the displayed line follows. Because of G 0 \ ha2 i D ¹1º, B D G 0 hai and a2 2 ˆ.G/, we get ˆ.G/ D ha2 i G 0 (compare orders). Since G 0 Z.G/ ˆ.G/ and G 0 \ Z.G/ D 1 .G 0 /, we obtain G 0 Z.G/ D ˆ.G/ by the product formula. Let us compute ck .G/ for all positive integers k, k n C 1 D log2 .exp.G//. Since ck .B/ D

jk .B/ k1 .B/j 22k 22k2 D D 3 2k1 2k1 2k1

A.27

387

Wreathed 2-groups

for k n, it remains to compute the number solk .G B/ of elements of order 2k > 2 in the set G B; then ck .G/ D 3 2k1 C

solk .G B/ 2k1

(here 2k1 D '.2k /, where './ is Euler’s totient function). Let x D ai .ac /j c 2 G B, i; j D 0; 1; 2; : : : ; 2n 1. We have x 2 D ai Cj .ac /i Cj D .aac /i Cj 2 Z.G/: Suppose that x 2 D 1. Then ai Cj D 1 so i C j 0 .mod 2n /. We see that there are in G B exactly 2n elements of order 2, i.e., c1 .G/ D 2n C c1 .B/ D 2n C 3 (this coincides with the result of Exercise 1). Now suppose that x 4 D 1. In that case, we have 1 D x 4 D a2.i Cj / .ac /2.iCj / so 2.iCj / D 1. It follows that i C j 0 .mod 2n1 /. If i D 0, then j 2 ¹0; 2n1 º. a Similarly, for arbitrary i we have j 2 ¹2n1 i; 2n i º. Thus G B contains exactly 2 2n elements of order 4 so exactly 2 2n 2n D 2n elements of order 4. Now we prove for k 2 ¹2; : : : ; nº that G B contains exactly 2nC.k2/ elements of k order 2k . By the previous paragraph, this is true for k D 2. Let x 2 D 1 (x 2 G B). Then a.iCj /2

k1

.ac /.iCj /2

k1

D1

so that i C j 0

.mod 2n.k1/ /:

It follows that for arbitrary i we have j D s 2n.k1/ i;

where s D 1; 2; : : : ; 2k1 :

Thus for a ﬁxed i there are in G B exactly 2k1 elements ai .ac /j c of order 2k whence the set G B contains exactly 2k1 2n D 2nC.k1/ elements of order 2k , k < n C 1. By induction, the set G B has 2nCk2 elements of order 2k1 . Thus the set G B possesses exactly 2nCk1 2nCk2 D 2nCk2 elements of order 2k . Now let o.x/ D 2nC1 ; then o.x 2 / D 2n , where x 2 D ai Cj .ac /i Cj D .aac /i Cj . It follows that i C j is odd, and we conclude that G B has exactly 2n1 2n D 22n1 elements of order 2nC1 . Since 2n C 2n C 2nC1 C 2nC2 C C 22n1 D 22n D jG Bj; we have found all elements of the set G B. Let Nk .X/ be the number of elements of order 2k in a set X; then, with k D 3; : : : ; n, Nk .B/ D 3 22.k1/ ;

N1 .G B/ D N2 .G/ D 2n ;

Nk .G B/ D 2nCk2 :

We obtain the number ck .G/ from the following equality: Nk .G/ D Nk .G B/ C Nk .B/ D 2k1 ck .G/:

388

Groups of prime power order

Thus we have c1 .G/ D 2n C 3;

c2 .G/ D 3 2 C 2n1 ;

ck .G/ D 3 2k1 C 2n1 cn .G/ D 3 2

n1

C2

n1

for k D 3; : : : ; n; D 2nC1 ;

cnC1 .G/ D 22n1 =2n D 2n1 :

Since .ac/2 D aac 2 Z.G/ is of order 2n , we infer that Z.G/ has index 2 in haci so Z.G/ is not a maximal cyclic subgroup of G. It follows from the proof of Lemma 27.3 n1 n1 that Z.G=Z.G// D ha2 Z.G/i. However, the subgroup ha2 ; Z.G/i is abelian of type .2n ; 2/. Since G=Z.G/ Š D2nC1 has only one minimal normal subgroup, it follows that haci is not normal in G; moreover, it has exactly 2n1 conjugates in G. We claim that all cyclic subgroups of the group G that are not contained in ˆ.G/ are not G-invariant. Indeed, assume that X GG is cyclic and X 6 ˆ.G/. Since d.G/ D 2, it follows that G=X is cyclic, and hence G is metacyclic, a contradiction. Let cG 0 , uG 0 and cuG 0 be all involutions in the quotient group G=G 0 . Then we get c1 .hc; G 0 i/ D 2n C 1 (Lemma A.27.4(b)). Since c1 .G/ D 2n C 3, it follows that the subgroups L1 D hu; G 0 i and L2 D hcu; G 0 i contain together exactly four involutions (one of them is an element of G 0 ). Therefore one may assume that the subgroup L1 contains exactly one involution so L1 is cyclic or generalized quaternion. Since G 0 is contained in an even number of subgroups of maximal class and order 2jG 0 j D 2nC1 and D D hc; G 0 i Š D2nC1 is dihedral so of maximal class, we obtain L1 Š Q2nC1 . As CG .G 0 / > G 0 , it follows that L2 is abelian of type .2n ; 2/, and so G 0 is a maximal cyclic subgroup of G. Then 1 .G/ D DL1 D D1 .B/ (compare the number of involutions and use Theorem 1.17(a); DL1 is not of maximal class since 1 .B/ G DL1 ). Now we shall ﬁnd all three members of the set 1 . We have B 2 1 . Since G satisﬁes cl.G/ D n C 1 > 2, B is the unique abelian maximal subgroup of G and two remaining members of the set 1 have class at least n (Fitting’s lemma). The subgroup ˆ.G/ D ha2 i h.ac/2 i D ha2 i haac i D ha2 i Z.G/ is abelian of type .2n ; 2n1 /. Since c 62 B, the subgroup M D hc; ˆ.G/i D hc; a2 ; aac i is another maximal subgroup of G. It then follows from Lemma A.27.3 that M=Z.G/ is abelian of type .2; 2/ if n D 2 and isomorphic to D2n if n > 2. By the above, we have cl.M / D n; then Z.M / D Z.G/. Since hc; G 0 i \ Z.G/ D 1 .G 0 /, we obtain M D hc; G 0 iZ.G/ by the product formula so d.M / D 3. As .ac/2 D aac 2 Z.G/, we get G 0 \ haci D 1 .G 0 / so N D haciG 0 is the third (metacyclic) member of the set 1 . Indeed, we have N ¤ B since exp.N / D 2nC1 > 2n D exp.B/.2 2 This

agrees with the known result that a nonmetacyclic two-generator 2-group has an even number of metacyclic subgroups of index 2 (see Theorem 107.1).

A.27

389

Wreathed 2-groups

As haci \ hai D ¹1º, we get G D hacihai by the product formula. Since all members of the set 1 are pairwise nonisomorphic, they are characteristic in G. It follows that Aut.G/ is a 2-group.3 Let D ¹xx c j x 2 Aº be the diagonal subgroup of B. Then D Z.G/. Assume that G has an elementary abelian subgroup E of order 8. Then G D BE so E \B D 1 .B/ Š E4 , CG .E \B/ BE D G hence E \B Z.G/, a contradiction since Z.G/ D haac i is cyclic. Thus E does not exist. 2o . Let G D M wr C , where M is a 2-group of maximal class and order > 8 and C D hci Š C2 . Set B D M M c (B is the base of G). Since B 0 D M 0 .M 0 /c G 0 ˆ.G/ is G-invariant and 1 .G=B 0 / D G=B 0 , we deduce that G 0 D ˆ.G/. Therefore, since d.G/ D 3 (see the remark following Lemma A.27.1), we obtain G=G 0 Š E8 hence B 0 < G 0 (compare indices!). Next, D ¹xx c j x 2 M º is the diagonal subgroup of B which is isomorphic to M and < ˆ.G/ D G 0 so that M 0 ˆ.G/ D G 0 . Clearly, B 0 \ has index 4 in and jG W B 0 j D 25 . It follows that jG W B 0 j D 8 D jG W G 0 j whence B 0 D G 0 . Since M 0 \ D ¹1º, the subgroup M 0 is a semidirect product with kernel M 0 and has index 8 D jG W G 0 j in G. Thus G 0 D M 0 . Assume that G possesses a normal subgroup E Š E8 . Clearly, E 6 B since B has no normal elementary abelian subgroup of order 8 in view of jM j > 8. Then we have E \ B D Z.B/ so that CG .Z.B// BE D G, and hence Z.B/ D Z.G/ is noncyclic. Since CB .c/ D , it follows that Z.G/ < hence Z.G/ is cyclic, and this is a contradiction. Thus E does not exist. In the case under consideration, the subgroup Z Z c , where Z < M is cyclic of index 2, is a G-invariant metacyclic subgroup, and we have G=.Z Z c / Š D8 since Z Z c 6 G 0 and 1 .G=.Z Z c // D G=.Z Z c /. Thus the result of Theorem 50.1 is best possible. Also, the result of Theorem 50.3 is also best possible. 3o . In this subsection G D Q wr C , where n1

Q D ha; b j a2

D 1; a2

n2

D b 2 ; ab D a1 i Š Q2n ;

C D hc j c 2 i Š C2 :

We have jGj D 22nC1 ;

Z.G/ D hb 2 .b 2 /c i;

jZ.G/j D 2;

G D ha; b; ci

and d.G/ D 3 by the remark following Lemma A.27.1. The subgroup B D Q Qc is the base of the wreath product G. Let A D hai; then A1 D AAc is an abelian normal subgroup of G (indeed, A1 is B-invariant and c-invariant and G D hB; ci). It is easy to proof: If 2 Aut.G/ is of odd order, then D idG for n D 1; if n > 1, then stabilizes the chain G > Z.G/ > ¹1º so D idG again. 3 Another

390

Groups of prime power order

see that G=A1 Š D8 so A1 6 ˆ.G/. Since .ac/2 D aac is of order 2n1 D exp.B/, we have o.ac/ D 2n D exp.G/. We obtain that .bc/2 D bb c 62 A1 , .bb c /c D bb c so hbb c A1 i D Z.G=A1 /. Set T =A1 D hbb c A1 i. Then T G G and G=T is abelian of type .2; 2/. By Exercise 1, c1 .G/ D 2n C 3 and all involutions of the set G B are conjugate in G. By the previous paragraph, N1 .G B/ D 2n ;

N1 .G/ D 2n C 3;

sol1 .G/ D 2n C 4:

Every element of G B can be presented uniquely in the form z D xy c c for some x; y 2 Q. Note that xy c D y c x for all x; y 2 Q. We have .xy c c/2 D xy c cxy c c D xy c x c y D .xy/.yx/c ;

(1)

k

k1

.xy c c/2 D .xy/2

(1a)

..yx/c /2

k1

:

Note that sol1 .Q/ D 2

(2)

and, for k > 1;

solk .Q/ D 2n1 C 2k :

It follows from (2) that (3)

N2 .B/ D sol2 .Q/2 sol1 .Q/2 D .2n1 C 4/2 4 D 22.n1/ C 2nC2 C 12

and, for 2 < k < n, (4)

Nk .B/ D .2n1 C 2k /2 .2n1 C 2k1 /2 D .2n C 3 2k1 /2k1 :

Now let z D xy c c 2 G B, x; y 2 Q and z 4 D 1. Then, by (1), .xy/2 D 1 so x D y 1 u, where u2 D 1 and u 2 Q, and we infer that sol2 .G B/ D 2jQj D 2nC1 and hence N2 .G B/ D sol2 .G B/ sol1 .G B/ D 2n : It follows that N2 .G/ D 22.n1/ C 2nC2 C 2n C 12

(5) so that

1 c2 .G/ D N2 .G/ D 22n3 C 2nC1 C 2n1 C 6: 2

Let 2 < k < n, z 2 G B with z 2 D 1. Then we get x D y 1 u, where u 2 Q and o.u/ 2k1 (see (1a)). It follows that k

solk .G B/ D jQjsolk1 .Q/ D 2n .2n1 C 2k1 / (see (2)). Then (6)

Nk .G B/ D 2n .2n1 C 2k1 / 2n .2n1 C 2k2 / D 2nCk2

A.27

Wreathed 2-groups

391

and we obtain Nk .G/ D Nk .B/ C Nk .G B/ D .2n C 3 2k1 /2k1 C 2nCk2 (7)

D 3.2n1 C 2k1 /2k1 ; ck .G/ D 3.2n1 C 2k1 /

.2 < k n/:

Since exp.G/ D 2n , all elements of G of order 2n lie in the set G B in view of exp.B/ D 2n1 . If o.xy c c/ D 2n , then y D x 1 u, where u 2 Q with o.u/ D 2n1 . It follows that the number of elements of order 2n in G is equal to jQj 2n2 D 22n2 so that 22n2 D 2n1 : cn .G/ D '.2n / We state the obtained results in the following Proposition A.27.8. Suppose that G D Q wr C2 , where Q Š Q2n . Then G 0 D ˆ.G/ is of index 8 in G and (a) c1 .G/ D 2n C 3. (b) c2 .G/ D 22n3 C 2nC1 C 2n1 C 6. (c) If 2 < k < n, then ck .G/ D 3 .2n1 C 2k1 /. (d) cn .G/ D 2n1 . Exercise 2. Let G D D wr C2 be a 2-group, where D has a cyclic subgroup of index 2. Find ck .G/ and describe all maximal subgroups of G. Exercise 3. Let G D A wr C2 , where A is an arbitrary (ﬁnite) group, C2 D hci. Is it true that all involutions in the set G .A Ac / are conjugate in G? Exercise 4. Let G D M wr C2 be a 2-group, where M has a cyclic subgroup of index 2. Find cl.G/. Exercise 5. Consider the standard wreath product G D E1 wr E2 , where E1 ; E2 are elementary abelian groups of order 2n , 2m , respectively. Find c1 .G/. Exercise 6. Let A be a 2-group, G D A wr C2m . Express c1 .G/ in terms of A. Exercise 7. Let G D A wr Cp be a p-group. Express jG W G 0 j in terms of A. Exercise 8. Let G D M1 M2 , where M1 and M2 are 2-groups of maximal class. Find the number of subgroups of maximal class and given order in G. Exercise 9. Let G D M wr C2 , where M is a 2-group of maximal class. Find the number of nonabelian subgroups of order 8 in G.

392

Groups of prime power order

Exercise 10. Study the subgroup structure of the standard wreath product G D E2n wr M; where M is a 2-group of maximal class. Exercise 11. Study the subgroup structure of G D C2m wr C2n . Exercise 12. Study the subgroup structure of the standard wreath product G D C2n wr M; where M is a 2-group of maximal class. Exercise 13. Let G D A wr B be the standard wreath product. Is it true that G satisﬁes exp.G/ D exp.A/ exp.B/? Solution. Let a 2 A, b 2 B, o.a/ D m, o.b/ D n. Then .ab/n D aab

1

ab

2

ab

.n1/

is of order o.a/ D m. Exercise 14. Describe the structure of Aut(C2n wr C2 /. Exercise 15. Describe the structure of Aut.Q wr C2 /, where Q is a 2-group of maximal class. Exercise 16. Let G D Cpn wr Cp . Describe the structure of 1 .G/. Exercise 17. Prove an analog of Proposition A.27.8 for the groups G D D2n wr C2 and G D SD2n wr C2 .

Appendix 28

Nilpotent subgroups

1o Theorems of Sylow, Hall, Carter etc. We prove Sylow’s theorem in the following form: Theorem A.28.1 (Sylow). All maximal p-subgroups of a group G are conjugate and have order jGjp , and their number is 1 .mod p/. A subgroup P G is a Sylow p-subgroup of G provided jP j is a power of p and p does not divide jG W P j. According to Theorem A.28.1, the Sylow p-subgroups coincide with the maximal p-subgroups of G and are conjugate in G. Let Sylp .G/ denote the set of all Sylow p-subgroups of G and .G/ be the set of the prime divisors of jGj. Lemma A.28.2 (Cauchy). If p 2 .G/, then G contains a subgroup of order p. In particular, a maximal p-subgroup of G is > ¹1º. Proof. Suppose that G is a counterexample of minimal order. Then G has no proper subgroup C of order divisible by p and jGj ¤ p. If G has only one maximal subgroup, say M , then it is cyclic. Indeed, if x 2 G M , then hxi is not contained in M so we get hxi D G. Then hx o.x/=p i < G is of order p, a contradiction. If G is abelian and A ¤ B are maximal subgroups of G, then G D AB so jGj D

jAjjBj ; jA \ Bj

and p does not divide P jGj, a contradiction. In case that G is nonabelian, we obtain that jGj D jZ.G/j C kiD1 hi , where h1 ; : : : ; hk are sizes of noncentral G-classes. Since the hi are indices of proper subgroups in G, we see that p divides hi for all i . Then p divides jZ.G/j so Z.G/ contains an element of order p, and hence G is not a counterexample. The set S D ¹Ai ºniD1 of subgroups of a group G is said to be G-invariant if Axi 2 S for all i n and x 2 G. All members of a G-invariant set S are conjugate if and only if S has no nonempty proper G-invariant subset. Lemma A.28.3. Let S ¤ ¿ be a G-invariant set of subgroups of a group G. Suppose that whenever N ¤ ¿ is a G-invariant subset of S, then jN j 1 .mod p/. Then all members of the set S are conjugate in G.

394

Groups of prime power order

Proof. Assume that S has a proper G-invariant subset N ¤ ¿. Then S N ¤ ¿ is G-invariant so jSj D jN j C jS N j 1 C 1 6 1

.mod p/;

a contradiction. Thus S is a minimal G-invariant subset so all its members are conjugate in G. Let P be a maximal p-subgroup of a group G and P1 be a p-subgroup of G. In case PP1 G, we see that PP1 is a p-subgroup by the product formula, and we conclude that P1 P . If P1 is a maximal p-subgroup of G, then NP .P1 / D P \ P1 . Proof of Theorem A.28.1. One may assume that p divides jGj. Let S0 D ¹P D P0 ; P1 ; : : : ; Pr º be a G-invariant set of all maximal p-subgroups of G; then we have Pi > ¹1º for all i (Lemma A.28.2). Let r > 0 and let P act on the set S0 ¹P º via conjugation. Then the size of every P -orbit on the set S0 ¹P º is a power of p greater than 1 because the P -stabilizer of a ‘point’ Pi is equal to P \ Pi < P . Thus p divides r hence jS0 j D r C 1 1 .mod p/; and the ﬁrst assertion follows (Lemma A.28.3). Then p does not divide jG W NG .P /j. Since P is a maximal normal p-subgroup of NG .P / D N , the prime p does not divide jN=P j (Lemma A.28.2) whence p does not divide jG W N jjN W P j D jG W P j, i.e., jP j D jGjp . A standard statement of Sylow’s theorem is as follows: (a) A group G contains a subgroup of order jGjp . (b) The number of subgroups from (a) is 1 .mod p/. (c) All maximal p-subgroups of G have order jGjp . Remark 1. (1) (Frobenius) Let P be a p-subgroup of a group G and suppose that the set M D ¹P1 ; : : : ; Pr º Sylp .G/ is P -invariant and P 6 Pi for all i . Then we obtain that jMj 0 .mod p/ (let P act on M via conjugation) so, by Theorem A.28.1, the number of Sylow p-subgroups of G containing P is 1 .mod p/ (indeed, take as M the set of all Sylow p-subgroups not containing P ). Similarly, if P 2 Sylp .G/, then the number of p-subgroups of G of given order that are not contained in P is divisible by p (let P act on the set of the above p-subgroups!). (2) (Frobenius) Let M D ¹M1 ; : : : ; Ms º be the set of all subgroups of order p k in a group G of order p m , k < m. We claim that s D jMj 1 .mod p/. To prove this, we use induction on m. Let 1 D ¹G1 ; : : : ; Gr º be the set of all maximal subgroups of G. Because the number of subgroups of index p in the elementary abelian p-group

A.28

395

Nilpotent subgroups

G=ˆ.G/ of order, say p d , is equal to pd 1 D 1 C p C C p d 1 1 p1

.mod p/;

we get r D j1 j 1 .mod p/, so one may assume that k < m1. Let ˛i be the number of members of the set M contained in Gi and ˇj the number of members of the set 1 containing Mj for all i; j . Then, by double counting, ˛1 C C ˛r D ˇ1 C C ˇs ; By the above, r 1 .mod p/, and by induction, ˛i 1 .mod p/ for all i . Next, ˇj is the number of maximal subgroups in G=Mj ˆ.G/ so ˇj 1 .mod p/ for all j . By the displayed formula, jMj D s r 1 .mod p/. (3) (Frobenius; see also [Bur, Theorem 9.II]) It follows from (1) and (2) that the number of p-subgroups of order p k in a group G of order p k m is 1 .mod p/. (4) Suppose that the subsets S1 ; S2 Sylp .G/ are nonempty and disjoint. Further, let Pi 2 Si be such that the set Si is Pi -invariant, i D 1; 2; then jSi j 1 .mod p/, i D 1; 2 (see the proof of Theorem A.28.1). It follows that S2 is not P1 -invariant (otherwise, considering the action of P1 on S2 via conjugation, we get jS2 j 0 .mod p/ since P1 62 S2 ). (5) (Burnside) Let G be a non-p-closed group (i.e., we have Op .G/ 62 Sylp .G/) and P 2 Sylp .G/. Suppose that Q 2 Sylp .G/ ¹P º is such that D D P \ Q is a maximal, by inclusion, intersection of Sylow p-subgroups of G. Assume that D > ¹1º. We claim that N D NG .D/ is not p-closed. Assume that this is false. Then P1 2 Sylp .N / is nonnormal in N . It follows from the properties of p-groups that P \ P1 > D and Q \ P1 > D so P1 is not contained in P . Therefore, if P1 U 2 Sylp .G/, then we get U ¤ P . However, P \ Q D D < P \ P1 P \ U , a contradiction. The following assertion is easily checked. If R is an abelian minimal normal subgroup of a group G and G D HR, where H < G, then H is maximal in G and we have H \ R D ¹1º. Theorem A.28.4 (P. Hall [Hal1]). If is a set of primes, then all maximal -subgroups of a solvable group G are conjugate. By Theorems A.28.1 and A.28.4, a maximal -subgroup of a solvable group G is its -Hall subgroup, and this gives the standard form of Hall’s theorem. Let Hall .G/ be the set of all -Hall subgroups of G. Given H G, the subgroup \ HG D Hx x2G

is said to be the core of H in G.

396

Groups of prime power order

The proofs of Theorems A.28.4, A.28.6 and A.28.7 are based on the following Lemma A.28.5 ([Ore]). If distinct maximal subgroups F and H of a solvable group G > ¹1º have equal cores, then they are conjugate in G. Proof. One may assume that FG D ¹1º; then G is not nilpotent (otherwise, jGj is a prime and F D H D ¹1º). Let R be a minimal normal, say a p-subgroup, of G; then we have RF D G D RH , R \ F D ¹1º D R \ H . Let K=R be a minimal normal, say a q-subgroup, of G=R, q is a prime. In this case, K is nonnilpotent (otherwise, we get NG .K \ F / > F and so ¹1º < K \ F FG D ¹1º, contrary to the assumption) whence q ¤ p. Then, by the product formula, F \ K and H \ K are nonnormal Sylow q-subgroups of K hence they are conjugate in K (Theorem A.28.1) so are in G. It follows that then NG .F \ K/ D F and NG .H \ K/ D H are also conjugate in G, as was to be shown. Proof of Theorem A.28.4.1 We use induction on jGj. Let F and H be maximal -subgroups of G and R a minimal normal, say p-subgroup, of G. If p 2 , then R F and R H by the product formula, and F=R, H=R are maximal -subgroups of G=R so they are conjugate by induction; then F and H are also conjugate. Now assume that O .G/ D ¹1º; then p 2 0 . Let F1 =R, H1 =R be maximal -subgroups of G=R containing FR=R, HR=R, respectively; then we get F1x D H1 for some x 2 G and F1 =R; H1 =R 2 Hall .G=R/ by induction. Assume that F1 < G. Then, by induction, we obtain F 2 Hall .F1 / as a maximal -subgroup of F1 so F1 D F R. Similarly, H1 D H R. Therefore F x 2 Hall .H1 /. Then H D .F x /y D F xy for some y 2 H1 by induction, and we are done. Now let F1 D G; then G=R is a -group. Suppose that K=R is a minimal normal, say a q-subgroup, of G=R; then q 2 hence q ¤ p.2 0 /. Let Q 2 Sylq .K/; then we have K D Q R. By the Frattini argument, G D NG .Q/K D NG .Q/QR D NG .Q/R hence NG .Q/ 2 Hall .G/ is maximal in G. Assume FR < G; then F 62 Hall .G/. In this case, NG .Q/ \ FR, as a -Hall subgroup of FR (product formula!), is conjugate with F , contrary to the choice of F . Thus FR D HR D G and FG D ¹1º D HG by assumption; then F and H are maximal in G (see the paragraph preceding the statement of the theorem). In this case, by Lemma A.28.5, F and H are conjugate in G. Let us prove, for completeness, by induction on jGj that a group G is solvable if it has a p 0 -Hall subgroup for all p 2 .G/ (Hall–Chunikhin). Let Gp0 2 Hallp0 .G/, where p 0 D .G/ ¹pº, and let q 2 p 0 . If Gq 0 2 Hallq 0 .G/, then we conclude that Gp0 \Gq 0 2 Hallq 0 .Gp0 / by the product formula, so Gp0 is solvable by induction. Let R be a minimal normal, say an r-subgroup, of Gp0 ; then r 2 p 0 . In view of Burnside’s two-prime theorem, we may assume that j.G/j > 2, so there exists s 2 .G/¹p; rº. Let Gs 0 2 Halls 0 .G/; then we have G D Gp0 Gs 0 . By Theorem A.28.1, one may assume 1 This

proof is independent of the Schur–Zassenhaus theorem. See also Theorem A.43.5.

A.28

Nilpotent subgroups

397

that R < Gs 0 ; then we get R .Gs 0 /G . Next, .Gs 0 /G is solvable as a subgroup of Gs 0 . Thus G has a minimal normal, say an r-subgroup, which we denote by R again. In case R 2 Sylr .G/, we conclude that G=R Š Gr 0 is solvable so is G. If R 62 Sylr .G/, then Gr 0 R=R 2 Hallr 0 .G=R/. Let q 2 .G/¹rº; then R < Gq 0 so Gq 0 =R 2 Hallq 0 .G=R/. In this case, by induction, G=R is solvable, and the proof is complete. A subgroup K is said to be a Carter subgroup (= C-subgroup) of G if it is nilpotent and coincides with its normalizer in G. Theorem A.28.6 (Carter [Car]). A solvable group G possesses a C-subgroup and all C-subgroups of G are conjugate. Proof. We use induction on jGj. One may assume that G is not nilpotent. Let R be a minimal normal, say p-subgroup, of G. Existence. By induction, G=R contains a C-subgroup S=R so NG .S/ D S. Let T be a p 0 -Hall subgroup of S (Theorem A.28.4); then TR is normal in S since TR=R is a p 0 -Hall subgroup in the nilpotent subgroup S=R, and S D T P , where P 2 Sylp .S/. One may assume that T > ¹1º (otherwise, we see that S D P is a C-subgroup of G). Set K D NS .T /; then K D T NP .T / is nilpotent and KR D S (Theorem A.28.4 and the Frattini argument). If y 2 NS .K/, then y 2 NS .T / D K since T is characteristic in K, and so NS .K/ D K. If x 2 NG .K/, then x 2 NG .KR/ D NG .S/ D S so x 2 NS .K/ D K, and hence K is a C-subgroup of G. Conjugacy. Let K, L be C-subgroups of G; then KR=R, LR=R are C-subgroups of G=R so, by induction, LR D .KR/x for some x 2 G, and we have K x LR. One may assume that R is not contained in K (otherwise, K x D LR so R < L D NG .L/ since LR is nilpotent, and we obtain K x D L, as desired); then RK is nonnilpotent whence R is not contained in L. If LR < G, then .K x /y D L for y 2 LR by induction, and we are done. Now let G D LR; then G D KR (recall that .KR/x D LR) so G=R is nilpotent, and this is true for each choice of R. Thus R is the unique minimal normal subgroup of G so K and L are maximal in G and KG D LG . Then K and L are conjugate in G (Lemma A.28.5). Theorem A.28.7. Let G D N1 : : : Nk be a nonnilpotent group, where N1 ; : : : ; Nk are proper pairwise permutable and nilpotent subgroups. Then there exists an index i k such that the normal closure NiG < G. Lemma A.28.8 ([Ore]). If G D AB, where A; B < G, then A and B are not conjugate. Proof. Assume that B D Ag for g 2 G. Then g D ab, a 2 A and b 2 B, so we obtain 1 B D Aab D Ab and A D B b D B, G D A, a contradiction. Proof of Theorem A.28.7.2 Assume that G is a minimal counterexample. Then G is nonnilpotent. By [Wiel2] and [Keg1], G is solvable. Let R < G be a minimal normal 2 For

k D 2, Theorem A.28.7 was proved by Janko and Kegel [Keg1], independently.

398

Groups of prime power order

subgroup, say a p-subgroup; then we have G=R D .N1 R=R/ : : : .Nk R=R/, where all Ni R=R are nilpotent. If G=R is not nilpotent, we get .Ni R/G < G for some i by induction so NiG < G. Now let G=R be nilpotent. Then, by hypothesis, we get Ni R D G for all i so Ni is maximal in G and .Ni /G D ¹1º for all i. Therefore N1 N2 D G. By Lemma A.28.5, N1 and N2 are conjugate in G, contrary to Lemma A.28.8. Exercise 1. If K; L are distinct Carter subgroups of a solvable group G, then we have KL ¤ LK. Solution. Write M D hK; Li. Then K; L, being Carter subgroups of M , are conjugate in M (Theorem 28.6) so they are not permutable by Theorem 28.8. Exercise 2. (a) Let K; L < G, where G is an arbitrary group. If K and L are conjugate in hK; Li, then KL ¤ LK. (This is restatement of Lemma 28.8.) (b) Suppose that K H < G, where K is a Carter subgroup of a solvable group G and x 2 G H . Is it true that HH x ¤ H x H ? Exercise 3. Let G be solvable and H < G be such that all subgroups of G containing H coincide with their normalizers and x 2 G H . Is it true that HH x ¤ H x H ? Supplement 1 to Lemma A.28.5. If for arbitrary maximal subgroups F and H of a group G with equal cores we have .jG W F j/ D .jG W H j/, then G is solvable. Proof. Assume that G is a minimal counterexample. As the hypothesis is inherited by epimorphic images, G has only one minimal normal subgroup R, and G=R is solvable so R is nonsolvable (the equality R D G is possible). Let p 2 .R/, P 2 Sylp .R/ and NG .P / F < G, where F is maximal in G. By Frattini’s lemma, G D RF so jG W F j D jR W .F \ R/j and p 62 .jG W F j/. Let q 2 .jR W .F \ Rj/; then q ¤ p. Take Q 2 Sylq .R/ and let NG .Q/ H < G, where H is maximal in G. Then we get FR D G D HR so FG D ¹1º D HG . However, q 2 .jG W F j/ and q 62 .jG W H j/, a contradiction. A group G is said to be p-solvable, if every its composition factor is either a p- or p 0 -number. Supplement 2 to Lemma A.28.5. Suppose that G is a p-solvable group with minimal normal p-subgroup R. Assume that G possesses a maximal subgroup H that satisﬁes HG D ¹1º. Then all maximal subgroups of G with core ¹1º are conjugate. Proof. By hypothesis, we have j.G/j > 1. Let F be another maximal subgroup of G with FG D ¹1º; then G D FR D HR and F \ R D ¹1º D H \ R. If R 2 Sylp .G/, we are done (Schur–Zassenhaus). Now let p 2 .G=R/; then G=R is not simple. Let K=R be a minimal normal subgroup of G=R. Then, as in the proof of Lemma A.28.5, j.K/j > 1 so, taking into account that HG D ¹1º and G=R is p-solvable, we conclude that K=R is a p 0 -subgroup. In that case, F \K and H \K as p 0 -Hall subgroups

A.28

Nilpotent subgroups

399

of K are conjugate in K so in G (Schur–Zassenhaus) whence F D NG .F \ K/ and H D NG .H \ K/ are also conjugate in G. Supplement 3 to Lemma A.28.5 ([Gas4]). Let M be a minimal normal subgroup of a solvable group G. Suppose that K 1 and K 2 are complements of M in G such that K 1 \ CG .M / D K 2 \ CG .M /. Then K 1 and K 2 are conjugate in G. Proof (compare with [Weh, Theorem 3.7]). From K i M D G we conclude that K i is maximal in G, i D 1; 2. We have K i \ M D ¹1º so that .K i /G CG .M /, i D 1; 2. By the modular law, we get CG .M / D M CK i .M /; then K i \ CG .M / D CK i .M / so CK 1 .M / D CK 2 .M / by hypothesis, and hence CK i .M / D .K i /G , i D 1; 2. Then .K 1 /G D .K 2 /G so K 1 and K 2 are conjugate in G by Lemma A.28.5. 2o Groups with a cyclic Sylow p-subgroup. Here we prove two results on groups with a cyclic Sylow p-subgroup. Theorem A.28.9 (compare with [Wiel3]). Let P 2 Sylp .G/ be cyclic. If H G G and p divides GCD.jH j; jG W H j/, then H is p-nilpotent and G is p-solvable. Remark 2. Suppose that P 2 Sylp .G/ is cyclic. (1) Suppose, in addition, that p j jZ.G/j. Set N D NG .P /. Since N has no minimal nonnilpotent subgroup S satisfying S 0 2 Sylp .S/ (Lemma 10.8), it follows that N is p-nilpotent so P Z.N /. In this case, G is p-nilpotent (Burnside’s normal p-complement theorem). (2) Suppose that G, p, P and H are given as in Theorem A.28.9 and assume that the subgroup P1 D P \ H.2 Sylp .H // is normal in H so in G. Let T 2 Hallp0 .H / (Schur–Zassenhaus); then, by Schur–Zassenhaus and Frattini, H D P1 T;

G D H NG .T / D P1 T NG .T / D P1 NG .T /:

It follows from the last equality, the cyclicity of P and P1 < P that NG .T / D G. Thus T G G so H D P1 T is p-decomposable. (3) If ¹1º < H < G are arbitrary and R ˆ.H / is normal in G, then R ˆ.G/. Indeed, assuming that this is false, we get G D RM for some maximal subgroup M of G. Then, by the modular law, we get H D R.H \ M / so H D H \ M and R < H M , a contradiction. (4) If H G G are arbitrary and G D AH , where A is as small as possible, then we have A \ H ˆ.A/. Indeed, if this is false, then A D B.A \ H /, where B < A is maximal. Then G D AH D B.A \ H /H D BH , contrary to the choice of A. Proof of Theorem A.28.9. By the product formula, we have P1 D P \ H 2 Sylp .H /. Set N D NG .P1 /; then P N . Further, set N1 D NH .P1 / D N \ H . Then, by Remark 2(2) applied to the pair N1 < N , we get N1 D P1 T , where T 2 Hallp0 .N1 /, and so H is p-nilpotent (Burnside’s normal p-complement theorem).3 By Frattini’s 3 Since H \P

ˆ.P /, it follows, by the deep Tate’s theorem [Hup, Satz 4.4.7], that H is p-nilpotent.

400

Groups of prime power order

lemma, G D HN so G=H Š N=.N \ H / D N=N1 . Therefore it remains to prove that N=N1 is p-solvable. To this end, we may assume that N D G; then P1 is normal in G. In this case, G=CG .P1 / as a p 0 -subgroup of Aut.P1 / is cyclic of order dividing p 1. By Remark 2(1), we see that CG .P1 / is p-nilpotent, and we conclude that G is p-solvable. Theorem A.28.10 ([Hal3, Theorem 4.61]). If P 2 Sylp .G/ is cyclic of order p m and k m, then ck .G/ 1 .mod p mkC1 /. Proof. Let C D ¹Z1 ; : : : ; Zr º be the set of all subgroups of order p k in G not contained in P . Let P act on the set C via conjugation. Then the P -stabilizer of Zi equals P \ Zi which is of order p k1 at most. Thus r D jC j 0 .mod p m.k1/ /. 3o Subgroup generated by some minimal nonabelian subgroups. Let p 2 .G/ and let Ap .G/ be the set of all minimal nonabelian subgroups A of an arbitrary (ﬁnite) group G such that A0 is a p-subgroup (in this case, A is either a p-group or p-closed and minimal nonnilpotent). Set Y Lp .G/ D hH j H 2 Ap .G/i and L.G/ D Lp .G/: p2.G/

It is known (Exercise 10.6 and Theorem 10.28) that L.G/ D G if G is a nonabelian p-group and G 0 L.G/ for arbitrary G. Theorem A.28.11. Given an arbitrary group G, the quotient group G=Lp .G/ is p-nilpotent. Moreover, if p j jG=Lp .G/j, then the Sylow p-subgroups of G are abelian. Proof. Take P 2 Sylp .G/. One may assume that P is not contained in Lp .G/. Then P is abelian (Theorem 10.28), proving the last assertion. Assume that G=Lp .G/ is not p-nilpotent. T

Yakov Berkovich Zvonimir Janko

Groups of Prime Power Order Volume 3

De Gruyter

Mathematical Subject Classification 2010: 20-02, 20D15, 20E07.

ISBN 978-3-11-020717-0 e-ISBN 978-3-11-025448-8 ISSN 0938-6572 Library of Congress Cataloging-in-Publication Data Berkovich, IA. G., 1938⫺ Groups of prime power order / by Yakov Berkovich, Zvonimir Janko. p. cm. ⫺ (De Gruyter expositions in mathematics ; ⬍ ⬎-56) Description based on v. 3, copyrighted c2011. Includes bibliographical references and index. ISBN 978-3-11-020717-0 (v. 3 : alk. paper) 1. Finite groups. 2. Group theory. I. Janko, Zvonimir, 1932⫺ II. Title. QA177.B469 2011 5121.23⫺dc22 2011004438

Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available in the Internet at http://dnb.d-nb.de. ” 2011 Walter de Gruyter GmbH & Co. KG, Berlin/New York Typesetting: Dimler & Albroscheit, Müncheberg Printing and binding: Hubert & Co. GmbH & Co. KG, Göttingen ⬁ Printed on acid-free paper Printed in Germany www.degruyter.com

Contents

List of deﬁnitions and notations . . . . . . . . . . . . . . . . . . . . . . . . . .

ix

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xv Prerequisites from Volumes 1 and 2 . . . . . . . . . . . . . . . . . . . . . . . . xvii 93 94

Nonabelian 2-groups all of whose minimal nonabelian subgroups are metacyclic and have exponent 4 . . . . . . . . . . . . . . . . . . . . . . . . .

1

Nonabelian 2-groups all of whose minimal nonabelian subgroups are nonmetacyclic and have exponent 4 . . . . . . . . . . . . . . . . . . . . . .

8

95

Nonabelian 2-groups of exponent 2e which have no minimal nonabelian subgroups of exponent 2e . . . . . . . . . . . . . . . . . . . . . . . . . . 10

96

Groups with at most two conjugate classes of nonnormal subgroups

97

p-groups in which some subgroups are generated by elements of order p

98

Nonabelian 2-groups all of whose minimal nonabelian subgroups are isomorphic to M2nC1 , n 3 ﬁxed . . . . . . . . . . . . . . . . . . . . . . . 31

99

2-groups with sectional rank at most 4 . . . . . . . . . . . . . . . . . . . 34

. . . 12 24

100 2-groups with exactly one maximal subgroup which is neither abelian nor minimal nonabelian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 101 p-groups G with p > 2 and d.G/ D 2 having exactly one maximal subgroup which is neither abelian nor minimal nonabelian . . . . . . . . . . 66 102 p-groups G with p > 2 and d.G/ > 2 having exactly one maximal subgroup which is neither abelian nor minimal nonabelian . . . . . . . . . . 77 103 Some results of Jonah and Konvisser . . . . . . . . . . . . . . . . . . . . 93 104 Degrees of irreducible characters of p-groups associated with ﬁnite algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 105 On some special p-groups . . . . . . . . . . . . . . . . . . . . . . . . . 102 106 On maximal subgroups of two-generator 2-groups . . . . . . . . . . . . . 110

vi

Groups of prime power order

107 Ranks of maximal subgroups of nonmetacyclic two-generator 2-groups

. 113

108 p-groups with few conjugate classes of minimal nonabelian subgroups . . 120 109 On p-groups with metacyclic maximal subgroup without cyclic subgroup of index p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 110 Equilibrated p-groups

. . . . . . . . . . . . . . . . . . . . . . . . . . . 125

111 Characterization of abelian and minimal nonabelian groups . . . . . . . . 134 112 Non-Dedekindian p-groups all of whose nonnormal subgroups have the same order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 113 The class of 2-groups in 70 is not bounded . . . . . . . . . . . . . . . . 148 114 Further counting theorems . . . . . . . . . . . . . . . . . . . . . . . . . 152 115 Finite p-groups all of whose maximal subgroups except one are extraspecial 157 116 Groups covered by few proper subgroups

. . . . . . . . . . . . . . . . . 162

117 2-groups all of whose nonnormal subgroups are either cyclic or of maximal class . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176 118 Review of characterizations of p-groups with various minimal nonabelian subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 119 Review of characterizations of p-groups of maximal class

. . . . . . . . 185

120 Nonabelian 2-groups such that any two distinct minimal nonabelian subgroups have cyclic intersection . . . . . . . . . . . . . . . . . . . . . . . 192 121 p-groups of breadth 2

. . . . . . . . . . . . . . . . . . . . . . . . . . . 197

122 p-groups all of whose subgroups have normalizers of index at most p

. . 204

123 Subgroups of ﬁnite groups generated by all elements in two shortest conjugacy classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237 124 The number of subgroups of given order in a metacyclic p-group . . . . . 239 125 p-groups G containing a maximal subgroup H all of whose subgroups are G-invariant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269 126 The existence of p-groups G1 < G such that Aut.G1 / Š Aut.G/

. . . . 272

127 On 2-groups containing a maximal elementary abelian subgroup of order 4 275 128 The commutator subgroup of p-groups with the subgroup breadth 1

. . . 277

Contents

vii

129 On two-generator 2-groups with exactly one maximal subgroup which is not two-generator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 130 Soft subgroups of p-groups

. . . . . . . . . . . . . . . . . . . . . . . . 287

131 p-groups with a 2-uniserial subgroup of order p . . . . . . . . . . . . . . 292 132 On centralizers of elements in p-groups . . . . . . . . . . . . . . . . . . 295 133 Class and breadth of a p-group . . . . . . . . . . . . . . . . . . . . . . . 300 134 On p-groups with maximal elementary abelian subgroup of order p 2 . . . 304 135 Finite p-groups generated by certain minimal nonabelian subgroups . . . 315 136 p-groups in which certain proper nonabelian subgroups are two-generator 328 137 p-groups all of whose proper subgroups have its derived subgroup of order at most p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338 138 p-groups all of whose nonnormal subgroups have the smallest possible normalizer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343 139 p-groups with a noncyclic commutator group all of whose proper subgroups have a cyclic commutator group . . . . . . . . . . . . . . . . . . . . . . 355 140 Power automorphisms and the norm of a p-group . . . . . . . . . . . . . 363 141 Nonabelian p-groups having exactly one maximal subgroup with a noncyclic center . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368 142 Nonabelian p-groups all of whose nonabelian maximal subgroups are either metacyclic or minimal nonabelian . . . . . . . . . . . . . . . . . . . . . 370 143 Alternate proof of the Reinhold Baer theorem on 2-groups with nonabelian norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373 144 p-groups with small normal closures of all cyclic subgroups

. . . . . . . 376

A.27 Wreathed 2-groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384 A.28 Nilpotent subgroups

. . . . . . . . . . . . . . . . . . . . . . . . . . . . 393

A.29 Intersections of subgroups

. . . . . . . . . . . . . . . . . . . . . . . . . 405

A.30 Thompson’s lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416 A.31 Nilpotent p 0 -subgroups of class 2 in GL.n; p/ . . . . . . . . . . . . . . . 428 A.32 On abelian subgroups of given exponent and small index . . . . . . . . . 434

viii

Groups of prime power order

A.33 On Hadamard 2-groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 437 A.34 Isaacs–Passman’s theorem on character degrees . . . . . . . . . . . . . . 440 A.35 Groups of Frattini class 2 . . . . . . . . . . . . . . . . . . . . . . . . . . 446 A.36 Hurwitz’ theorem on the composition of quadratic forms . . . . . . . . . 449 A.37 On generalized Dedekindian groups . . . . . . . . . . . . . . . . . . . . 452 A.38 Some results of Blackburn and Macdonald . . . . . . . . . . . . . . . . . 457 A.39 Some consequences of Frobenius’ normal p-complement theorem . . . . 460 A.40 Varia

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472

A.41 Nonabelian 2-groups all of whose minimal nonabelian subgroups have cyclic centralizers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 514 A.42 On lattice isomorphisms of p-groups of maximal class

. . . . . . . . . . 516

A.43 Alternate proofs of two classical theorems on solvable groups and some related results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 519 A.44 Some of Freiman’s results on ﬁnite subsets of groups with small doubling 527 Research problems and themes III

. . . . . . . . . . . . . . . . . . . . . . . . 536

Author index

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 630

Subject index

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 632

List of deﬁnitions and notations

Set theory jM j is the cardinality of a set M (if G is a ﬁnite group, then jGj is called its order). x 2 M (x 62 M ) means that x is (is not) an element of a set M . N M (N 6 M ) means that N is (is not) a subset of the set M ; moreover, if M ¤ N M , we write N M. ¿ is the empty set. N is called a nontrivial subset of M if N ¤ ¿ and N M . If N M , we say that N is a proper subset of M . M \ N is the intersection and M [ N is the union of sets M and N . If M; N are sets, then N M D ¹x 2 N j x 62 M º is the difference of N and M .

Number theory and general algebra p is always a prime number. is a set of primes; 0 is the set of all primes not contained in . m, n, k, r, s are, as a rule, natural numbers. .m/ is the set of prime divisors of m; then m is a -number if .m/ . np is the p-part of n, n is the -part of n. GCD.m; n/ is the greatest common divisor of m and n. m j n should be read as: m divides n. GF.p m / is the ﬁnite ﬁeld containing p m elements. F is the multiplicative group of a ﬁeld F. L.G/ is the lattice of all subgroups of a group G. If n D p1˛1 : : : pk˛k is the standard prime decomposition of n, then .n/ D

Pk

i D1 ˛i .

x

Groups of prime power order

Groups We consider only ﬁnite groups which are denoted, with a few exceptions, by uppercase Latin letters. If G is a group, then .G/ D .jGj/. G is a p-group if jGj is a power of p; G is a -group if .G/ . G is, as a rule, a ﬁnite p-group. H G means that H is a subgroup of G. H < G means that H G and H ¤ G (in that case, H is called a proper subgroup of G). ¹1º denotes the group containing only one element. H is a nontrivial subgroup of G if ¹1º < H < G. H is a maximal subgroup of G if H < G and it follows from H M < G that H D M. If H is a proper normal subgroup of G, then we write H G G. Expressions ‘normal subgroup of G’ and ‘G-invariant subgroup’ are synonyms. A normal subgroup of G is nontrivial provided G > H > ¹1º. H is a minimal normal subgroup of G if (a) H is normal in G; (b) H > ¹1º; (c) N GG and N < H implies N D ¹1º. Thus the group ¹1º has no minimal normal subgroup. G is simple if it is a minimal normal subgroup of G (so jGj > 1). H is a maximal normal subgroup of G if H < G and G=H is simple. The subgroup generated by all minimal normal subgroups of G is called the socle of G and denoted by Sc.G/. We put, by deﬁnition, Sc.¹1º/ D ¹1º. NG .M / D ¹x 2 G j x 1 M x D M º is the normalizer of a subset M in G. CG .x/ is the centralizer of an element x in G: CG .x/ D ¹z 2 G j zx D xzº. T CG .M / D x2M CG .x/ is the centralizer of a subset M in G. If A B and A; B are normal in G, then CG .B=A/ D H , where H=A D CG=A .B=A/. A wr B (or A o B) is the wreath product of the ‘passive’ group A and the transitive permutation group B (in what follows we assume that B is regular); B is called the active factor of the wreath product). Then the order of that group is jAjjBj jBj. Aut.G/ is the group of automorphisms of G (the automorphism group of G). Inn.G/ is the group of all inner automorphisms of G. Out.G/ D Aut.G/=Inn.G/ is the outer automorphism group of G.

List of deﬁnitions and notations

xi

N .G/ is the norm of G, the intersection of normalizers of all subgroups of G. If a; b 2 G, then ab D b 1 ab. An element x 2 G inverts a subgroup H G if hx D h1 for all h 2 H . If M G, then hM i D hx j x 2 M i is the subgroup of G generated by M . M x D x 1 M x D ¹y x j y 2 M º for x 2 G and M G. Œx; y D x 1 y 1 xy D x 1 x y is the commutator of elements x; y of G. If M; N G, then ŒM; N D hŒx; y j x 2 M; y 2 N i is a subgroup of G. o.x/ is the order of an element x of G. An element x 2 G is a -element if .o.x// . G is a -group, if .G/ . Obviously, G is a -group if and only if all of its elements are -elements. G 0 is the subgroup generated by all commutators Œx; y, x; y 2 G (i.e., G 0 D ŒG; G), G .2/ D ŒG 0 ; G 0 D G 00 D .G 0 /0 , G .3/ D ŒG 00 ; G 00 D .G 00 /0 and so on. G 0 is called the commutator (or derived) subgroup of G. T Z.G/ D x2G CG .x/ is the center of G. Zi .G/ is the i -th member of the upper central series of G; in particular, Z0 .G/ D ¹1º, Z1 .G/ D Z.G/. Ki .G/ is the i-th member of the lower central series of G; in particular, K2 .G/ D G 0 . We have Ki .G/ D ŒG; : : : ; G (i 1 times). We set K1 .G/ D G. If G is nonabelian, then .G/=K3 .G/ D Z.G=K3 .G//. M.G/ D hx 2 G j CG .x/ D CG .x p /i is the Mann subgroup of a p-group G. Sylp .G/ is the set of p-Sylow subgroups of an arbitrary ﬁnite group G. Sn is the symmetric group of degree n. An is the alternating group of degree n. †pn is a Sylow p-subgroup of Spn . GL.n; F/ is the set of all nonsingular n n matrices with entries in a ﬁeld F, the n-dimensional general linear group over F, SL.n; F/ D ¹A 2 GL.n; F/ j det.A/ D 1 2 Fº, the n-dimensional special linear group over F. T If H G, then HG D x2G x 1 H x is the core of the subgroup H in G and H G , the intersection of all normal subgroups of G containing H , is the normal closure or normal hull of H in G. Obviously, HG is normal in G. If G is a p-group, then p b.x/ D jG W CG .x/j; b.x/ is said to be the breadth of x 2 G, where G is a p-group; b.G/ D max¹b.x/ j x 2 Gº is the breadth of G.

xii

Groups of prime power order

If H G and jG W NG .H /j D p sb.H / , then sb.H / is said to be the subgroup breadth of H . Next, sb.G/ D max¹sb.H / j H Gº. ˆ.G/ is the Frattini subgroup of G (D the intersection of all maximal subgroups of G), ˆ.¹1º/ D ¹1º, p d.G/ D jG W ˆ.G/j i D ¹H < G j ˆ.G/ H; jG W H j D p i º, i D 1; : : : ; d.G/, where G > ¹1º. If H < G, then 1 .H / is the set of all maximal subgroups of H . exp.G/ is the exponent of G (the least common multiple of the orders of elements of G). If G is a p-group, then exp.G/ D max¹o.x/ j x 2 Gº. k.G/ is the number of conjugacy classes of G (D G-classes), the class number of G. Kx is the G-class containing an element x (sometimes we also write cclG .x/). Cm is the cyclic group of order m. G m is the direct product of m copies of a group G. A B is the direct product of groups A and B. A B is a central product of groups A and B, i.e., A B D AB with ŒA; B D ¹1º. Epm D Cpm is the elementary abelian group of order p m . G is an elementary abelian p-group if and only if it is a p-group > ¹1º and G coincides with its socle. Next, ¹1º is elementary abelian for each prime p. A group G is said to be homocyclic if it is a direct product of isomorphic cyclic subgroups (obviously, elementary abelian p-groups are homocyclic). ES.m; p/ is an extraspecial group of order p 1C2m (a p-group G is said to be extraspecial if G 0 D ˆ.G/ D Z.G/ is of order p). Note that for each positive integer m, there are exactly two nonisomorphic extraspecial groups of order p 2mC1 . S.p 3 / is a nonabelian group of order p 3 and exponent p > 2. A special p-group is a nonabelian p-group G such that G 0 D ˆ.G/ D Z.G/ is elementary abelian. Direct products of extraspecial p-groups are special. D2m is the dihedral group of order 2m, m > 2. Some authors consider E22 as the dihedral group D4 . Q2m is the generalized quaternion group of order 2m 23 . SD2m is the semidihedral group of order 2m 24 . Mpm is a nonabelian p-group containing exactly p cyclic subgroups of index p. cl.G/ is the nilpotence class of a p-group G. dl.G/ is the derived length of a p-group G. CL.G/ is the set of all G-classes.

List of deﬁnitions and notations

xiii

A p-group of maximal class is a nonabelian group G of order p m with cl.G/ D m 1. m .G/ D hx 2 G j o.x/ p m i, m .G/ D hx 2 G j o.x/ D p m i and Ãm .G/ D m hx p j x 2 Gi. A p-group G is said to be regular if for any x; y 2 G there exists z 2 hx; yi0 such that .xy/p D x p y p z p . A p-group is absolutely regular if jG=Ã1 .G/j < p p . A p-group is thin if it is either absolutely regular or of maximal class. G D A B is a semidirect product with kernel B and complement A. A group G is an extension of a normal subgroup N by a group H if G=N Š H . A group G splits over N if G D H N with H G and H \ N D ¹1º (in that case, G is a semidirect product of H and N with kernel N ). H # D H ¹eH º, where eH is the identity element of the group H . If M G, then M # D M ¹eG º. An automorphism ˛ of G is regular (D ﬁxed-point-free) if it induces a regular permutation on G # (a permutation is said to be regular if it has no ﬁxed points). An involution is an element of order 2 in a group. A group G is said to metacyclic if it contains a normal cyclic subgroup C such that G=C is cyclic. A group G is said to be minimal nonmetacyclic if it is nonmetacyclic but all its proper subgroups are metacyclic. A subgroup A of a group G is said to be soft if CG .A/ D A and jNG .A/ W Aj D p. A section of a group G is an epimorphic image of some subgroup of G. If F D GF.p n /, then we may write GL.m; p n /; SL.m; p n /; : : : instead of GL.m; F/, SL.m; F/; : : : . cn .G/ is the number of cyclic subgroups of order p n in a p-group G. sn .G/ is the number of subgroups of order p n in a p-group G. en .G/ is the number of subgroups of order p n and exponent p in G. A group G is said to be minimal nonabelian if it is nonabelian but all its proper subgroups are abelian. An -group is a p-group G all of whose subgroups of index p n are abelian but G contains a nonabelian subgroup of index p n1 . In particular, A1 -group is a minimal nonabelian p-group for some p. ˛n .G/ is the number of An -subgroups in a p-group G.

xiv

Groups of prime power order

MA.G/ is the set of minimal nonabelian subgroups of a p-group G. MAk .G/ D ¹H 2 MA.G/ j k .H / D H º. Dk .G/ D hMAk .G/i D hH j H 2 MAk .G/i. Ln D j¹x 2 G j x n D 1ºj.

Characters and representations Irr.G/ is the set of all irreducible characters of G over complex numbers. A character of degree 1 is said to be linear. Lin.G/ is the set of all linear characters of G (obviously, Lin.G/ Irr.G/). Irr1 .G/ D Irr.G/ Lin.G/ is the set of all nonlinear irreducible characters of G; n.G/ D jIrr1 .G/j. .1/ is the degree of a character of G. H is the restriction of a character of G to H G. G is the character of G induced from the character of some subgroup of G. N is a character of G deﬁned as follows: .x/ N D .x/ (here wN is the complex conjugate of a complex number w). Irr. / is the set of irreducible constituents of a character of G. 1G is the principal character of G. Irr# .G/ D Irr.G/ ¹1G º. If is a character of G, then ker. / D ¹x 2 G j .x/ D .1/º is the kernel of a character . Z. / D ¹x 2 G j j .x/j D .1/º is the quasikernel of . If N is normal in G, then Irr.G j N / D ¹ 2 Irr.G/ j N 6 ker. /º. P h ; i D jGj1 x2G .x/ .x 1 / is the inner product of characters and of G. IG ./ D hx 2 G j x D i is the inertia subgroup of 2 Irr.H / in G, where H G G. 1G is the principal character of G (1G .x/ D 1 for all x 2 G). M.G/ is the Schur multiplier of G. cd.G/ D ¹ .1/ j 2 Irr.G/º. mc.G/ D k.G/=jGj is the measure of commutativity of G. P T.G/ D 2Irr.G/ .1/, f.G/ D T.G/=jGj.

Preface

This is the third volume of the book devoted to elementary parts of p-group theory. Sections 93–95, 98–102, 113, 117, 118, 120–123, 125–133, 137, 139, 140–142, 144 are written by the second author, Sections 107, 138 and Appendix 41 are jointly written by both authors, all other material, apart from Appendices 33 and 44, is written by the ﬁrst author. All exercises and about all problems are due to the ﬁrst author. All material of this part is appeared in the book form for the ﬁrst time. Some interesting problems of elementary p-group theory are solved in this volume: (i) classiﬁcation of p-groups containing exactly one maximal subgroup which is neither abelian nor minimal nonabelian, (ii) classiﬁcation of groups all of whose nonnormal subgroups have the same order (independently, this result was obtained by Guido Zappa), (iii) classiﬁcation of p-groups all of whose nonnormal subgroups have normalizers of index p, (iv) computation of the number of subgroups of given order in metacyclic p-groups (for p > 2 this was done by Avinoam Mann), (v) computation of the order of the derived subgroup of a group with subgroup breadth 1, (vi) classiﬁcation of p-groups all of whose subgroups have derived subgroups of order p (so-called A2 -groups satisfy this condition), (vii) classiﬁcation of p-groups G all of whose maximal abelian subgroups containing a non-G-invariant cyclic subgroup of minimal order, say p , have order p C1 . Some results proved in this part have no analogs in existing books devoted to ﬁnite p-groups. We list only a few such results: (a) study the p-groups G all of whose minimal nonabelian subgroups have exponent < exp.G/, (b) study the groups admitting an irredundant covering by few proper subgroups, (c) study of some 2-groups with sectional rank 4, (d) study the p-groups which do not generated by certain minimal nonabelian subgroups, (e) study the p-groups, p > 3, in which certain nonabelian subgroups are generated by two elements, (f) study the p-groups with nonnormal maximal elementary abelian subgroup of order p 2 (this is a continuation of the paper of Glauberman–Mazza), (g) classiﬁcation of p-groups with exactly one maximal subgroup which is neither abelian nor minimal nonabelian,

xvi

Groups of prime power order

(h) classiﬁcation of p-groups all of whose proper subgroups have derived subgroups of orders p, (i) classiﬁcation of p-groups all of whose nonnormal cyclic subgroups of minimal possible order have index p is their normalizers, (j) classiﬁcation of p-groups G all of whose maximal abelian subgroups containing non-G-invariant cyclic subgroup of minimal order, say p , have order p nC1 , (k) classiﬁcation of p-groups all of whose maximal subgroups have cyclic derived subgroups, (l) characterizations of abelian and minimal nonabelian groups, (m) describing all possible sets of numbers of generators of maximal subgroups of two-generator 2-groups, (n) study the equilibrated p-groups, (o) classiﬁcation of nonabelian 2-groups in which any two distinct minimal nonabelian subgroups have cyclic intersection, (p) study the p-groups of breadth 2, (q) study groups containing a soft subgroup, (r) proof of the Schenkman theorem on the norm of a ﬁnite group and a new proof of Baer’s theorem on an arbitrary 2-group with nonabelian norm. For further information, see the Contents. Essential part of this volume is devoted to investigation of impact of minimal nonabelian subgroups on the structure of a p-group. Some appendices are devoted to nilpotent subgroups of nonnilpotent groups. The section ‘Research problems and themes III’, written by the ﬁrst author, contains about 900 research problems and themes some of which are solved by the second author. There are in the text approximately 200 exercises most of which are solved. Avinoam Mann (Hebrew University of Jerusalem) analyzed the list of problems and made a great number of constructive comments and corrections. He also read a number of sections and made numerous useful remarks. Moshe Roitman (University of Haifa) wrote Appendix 44 and helped with LATEX. Noboru Ito wrote Appendix 33. We are indebted to these three mathematicians. We are grateful to the publishing house of Walter de Gruyter and all its workers for supporting and promoting the publication of our book and especially to Simon Albroscheit and Kay Dimler.

Prerequisites from Volumes 1 and 2

In this section we state some results from Volumes 1 and 2 which we use in what follows. If we formulate Lemma 1.4 from Volume 1, then it is named below also as Lemma 1.4. Lemma INTR. (a) If M; N G G, then the quotient group G=.M \ N / is isomorphic to a subgroup of G=M G=N . (b) If Z is a cyclic subgroup of maximal order in an abelian p-group, then Z is a direct factor of G. In particular, an abelian p-group is a direct product of cyclic subgroups. (c) (Fitting’s lemma) If M and N are nilpotent normal subgroups of a group G, then cl.MN / cl.M / C cl.N /. (d) If G is a p-group, then G=ˆ.G/ is elementary abelian of order, say d . Every minimal set of generators of G contains exactly d members. (e) If G is a p-group, then ˆ.G/ D G 0 Ã1 .G/. Lemma 1.1. If A is an abelian subgroup of index p in a nonabelian p-group G, then jGj D pjG 0 jjZ.G/j. Theorem 1.2. Suppose that a nonabelian p-group has a cyclic subgroup of index p. Then one of the following holds: D b 2 D 1; ab D a1 i Š D2n is of class n 1, all elements of (a) G D ha; b j a2 the set G hai are involutions. n1

(b) G D ha; b j a2 D 1; a2 D b 2 ; ab D a1 i Š Q2n is of class n 1, all elements of the set G hai have the same order 4. n1

n2

(c) G D ha; b j a2 D 1; a2 D b 2 ; ab D a1 i Š SD2n is of class n 1, 1 D ¹D Š D2n1 ; Q Š Q2n1 ; hai Š C2n1 º, 1 .G/ D D, 2 .G/ D Q. n1

n1

n2

n2

(d) G D ha; b j ap D b p D 1; ab D a1Cp i Š Mpn is of class 2, where n > 3 n2 if p D 2, Z.G/ D hap i, 1 .G/ D hap ; bi Š Ep2 . The groups D2n , Q2n , SD2n are called dihedral, generalized quaternion, semidihedral, respectively. Proposition 1.3. If a p-group has only one subgroup of order p, then it is either cyclic or a generalized quaternion group.

xviii

Groups of prime power order

Lemma 1.4. Let N be a normal subgroup of a p-group G. If N has no G-invariant abelian subgroup of type .p; p/, then it is either cyclic or a 2-group of maximal class. It follows from Lemma 1.4 that if Z2 .G/ is cyclic, then G is either cyclic or a 2-group of maximal class. Lemma 1.6 (Taussky). If a nonabelian 2-group G satisﬁes jG W G 0 j D 4, then it is a 2-group of maximal class. Exercise 1.6(a). The number of abelian maximal subgroups in a nonabelian p-group G is either 0 or 1 or p C 1. Exercise P1. If a nonabelian p-group G has two distinct abelian maximal subgroups, then jG 0 j D p. Proposition 1.8 (Suzuki). If a nonabelian p-group G has a self-centralizing abelian subgroup of order p 2 , then G is of maximal class. (For the converse assertion, see Theorem 9.6(c).) Exercise 1.8(a). If a p-group G is minimal nonabelian, then jG 0 j D p;

d.G/ D 2;

jG W Z.G/j D p 2 ;

and one of the following holds: m

n

(a) G D ha; b j ap D b p D 1; ab D a1Cp pm

m1

i is metacyclic.

pn

(b) G D ha; b j a D b D 1; Œa; b D c; c p D 1; Œa; c D Œb; c D 1i is nonmetacyclic. In that case, Ã1 .G/ D hap i hb p i, G=Ã1 .G/ is nonabelian of order p 3 and exponent p unless p D 2 (if p D 2, then G=Ã1 .G/ Š E4 ). (c) G Š Q8 . Next, a group G is nonmetacyclic if and only if G 0 is a maximal cyclic subgroup of G. The group G is metacyclic if and only if j1 .G/j p 2 . No noncentral cyclic subgroup is normal in G. Theorems 1.10 and 1.17. Suppose that a p-group G is neither cyclic nor a 2-group of maximal class. Then (a) c1 .G/ 1 C p .mod p 2 /. (b) If k > 1, then ck .G/ 0 .mod p/. Theorem 1.20. If all subgroups of a nonabelian p-group G are normal, then we have G D Q E, where Q Š Q8 and exp.E/ 2. Exercise 1.69(a). Let G be a p-group and let H ¤ K be two distinct maximal subgroups of G. Then jG 0 W H 0 K 0 j p. In particular, if G 0 is cyclic, there is A 2 1 such that jG 0 W A0 j p.

Prerequisites from Volumes 1 and 2

xix

Lemma 4.2. Let G be a p-group with jG 0 j D p. Then G D .A1 A2 As /Z.G/, the central product, where A1 ; : : : ; As are minimal nonabelian, so G=Z.G/ is elementary abelian of even rank. In particular, if G=G 0 is elementary abelian, then we have jA1 j D D jAs j D p 3 , E D A1 As is extraspecial and G D EZ.G/. Lemma 4.3. Let E be a subgroup of a p-group G with jE 0 j D p and Z.E/ D ˆ.E/. If ŒG; E D E 0 , then G D E CG .E/. Theorem 5.2 (Hall’s enumeration principle). Let G be a p-group and M be a set of proper subgroups of G. Given H G, let ˛.H / be the number of members of the set M contained in H . Then X ˛.G/

˛.H / .mod p/: H 21

Theorems 5.3 and 5.4. Suppose that a p-group G of order p m is neither cyclic nor a 2-group of maximal class and 1 n < m. Then sn .G/ 1 C p .mod p 2 /. Theorem 5.8. Let G be a group of order p m > p 3 and exponent p, 2 < n < m. Let M denote the set of all 2-generator subgroups of order p n in G and ˛.K/ the number of elements of the set M contained in K G. (a) If n D m 1, then ˛.G/ 2 ¹0; p; p 2 º. (b) p divides ˛.G/. Theorem 7.1. Let G be a p-group. (a) Regularity is inherited by sections. (b) If cl.G/ < p or jGj p p or exp.G/ D p, then G is regular. (c) If Kp1 .G/ is cyclic, then G is regular. Theorem 7.2. Suppose that G is a regular p-group. (b) exp.n .G// p n . n

(c) Ãn .G/ D ¹x p j x 2 Gº. (d) jn .G/j D jG W Ãn .G/j. Theorem 9.5. Let G be a group of maximal class and order p m , m pC1. Then ˆ.G/ and G=Z.G/ have exponent p. If m D p C 1, then G is irregular and jÃ1 .G/j D p. Theorem 9.6. Let G be a group of maximal class and order p m , p > 2, m > p C 1. Then G is irregular and (a) jG W Ã1 .G/j D p p . In particular, Ã1 .G/ D Kp .G/. (b) There is G1 2 1 such that jG1 W Ã1 .G1 /j D p p1 .

xx

Groups of prime power order

(c) G has no normal subgroups of order p p and exponent p (here the condition that m > p C 1 is essential). Moreover, if N G G and jG W N j > p, then N is absolutely regular since N < G1 . Next, if N G G of order p p1 , then exp.N / D p. In particular, G has no normal cyclic subgroups of order p 2 . (d) Let Z2 D Z2 .G/ be a normal subgroup of order p 2 in G, G0 D CG .Z2 /. Then G0 is regular such that j1 .G0 /j D p p1 . (e) Let 1 D ¹M1 D G0 ; M2 ; : : : ; MpC1 º, where G0 is deﬁned in (d). Then the subgroups M2 ; : : : ; MpC1 are of maximal class (and so irregular; see Theorem 9.5). Thus the subgroups G1 from (b) and G0 from (d) coincide. In what follows we call G1 the fundamental subgroup of G. (f) In this part, m 3 (i.e., we do not assume as in other parts that m > p C 1). The group G has an element a such that jCG .a/j D p 2 , i.e., CG .a/ D ha; Km1 .G/i. Theorem 9.7. A nonabelian p-group G is of maximal class if and only if G=KpC1 .G/ is of maximal class. Theorem 9.8. (a) Absolutely regular p-groups G (i.e., G with jG=Ã1 .G/j < p p ) are regular. (b) If a p-group G is such that jG 0 =Ã1 .G 0 /j < p p1 , then G is regular. (c) Any irregular p-group G has a characteristic subgroup R of order p p1 and exponent p such that R G 0 . Theorem 9.11. A p-group G, p > 2, is metacyclic if and only if jG=Ã1 .G/j p 2 . Exercise 9.13. Let G be a p-group of maximal class, p > 2, and H < G. Then d.H / p. If d.H / D p, then G Š †p2 2 Sylp .Sp2 /. In particular (Blackburn), if G is a p-group of maximal class, p > 2, N G G and G=N Š †p2 , then N D ¹1º. Theorem 10.1. Let p n > 2 and let A < G be abelian of exponent p n . Then the number of abelian subgroups B G of order pjAj that contain A is congruent to 1 modulo p. Corollary 10.2. Let N be a normal subgroup of a p-group G and let A < N be a maximal G-invariant abelian subgroup of exponent p n , where p n > 2. Then we have n .CN .A// D A. Theorems 10.4 and 10.5. Suppose that G is a p-group, p > 2. Let n .G/ be the number of elementary abelian subgroups of order p n in G. If k 2 ¹3; 4º and k .G/ > 0, then we have k .G/ 1 .mod p/. Exercise P2. Let G and k be such as in the previous theorem. (a) If Zk .G/ has no G-invariant elementary abelian subgroup of order p k , then we have k .G/ D 0. (b) If a G-invariant subgroup N of G has no G-invariant elementary abelian subgroup of order p k , then k .N / D 0.

Prerequisites from Volumes 1 and 2

xxi

Lemma 10.8. Suppose that G is a minimal nonnilpotent group. Then G D PQ, where P 2 Sylp .G/, Q D G 0 2 Sylq .G/ and (a) P is cyclic, jP W .P \ Z.G/j D p. (b) Q is either elementary abelian or special, jQ=ˆ.Q/j D q b , where b is the order of q modulo p. (c) If Q is special, then b is even and ˆ.Q/ D Z.Q/ Z.G/, jˆ.Q/j p b=2 . Theorem (Frobenius’ normal p-complement theorem). If a group G has no p-closed minimal nonnilpotent subgroup of order divisible by p, then G has a normal p-complement (D p-nilpotent). Theorem (Burnside’s normal p-complement theorem). If a Sylow p-subgroup of a group G is contained in the center of its normalizer, then G is p-nilpotent. Proposition 10.17. If B G is a nonabelian subgroup of order p 3 in a p-group G and CG .B/ < B, then G is of maximal class. Remark 10.5. Let H < G and assume that NG .H / is of maximal class. Then G is also of maximal class. Proposition 10.19. Suppose that H is a nonabelian subgroup of order p 3 in a metacyclic p-group G. If p > 2, then G D H . If p D 2, then G is of maximal class. Proposition 10.28. A nonabelian p-group G is generated by minimal nonabelian subgroups. In particular, if, in addition, G is not minimal nonabelian, it contains two nonconjugate minimal nonabelian subgroups. Theorem 10.33. If all minimal nonabelian subgroups of a nonabelian 2-group G are generated by involutions, then G D hxi A, where A 2 1 is abelian and all elements in G A are involutions (such G is said to be generalized dihedral). Theorem 12.1. (a) If a p-group G has no normal subgroup of order p p and exponent p, then G is either absolutely regular or irregular of maximal class. (b) If an irregular p-group G has an absolutely regular maximal subgroup H , then it is either of maximal class or G D H 1 .G/, where 1 .G/ is of order p p (and, of course, of exponent p). Exercise P3. If p .G/ has no G-invariant subgroup of order p p and exponent p, then G is either absolutely regular or of maximal class. Theorem 12.12. Let a group G of order p m be neither absolutely regular nor of maximal class and suppose that H 2 1 is of maximal class. Then (a) d.G/ D 3. (b) Set D m 2 if m p C 1 and D p if m > p C 1. Then G=K .G/ is of order p C1 and, if m > 4, it is of exponent p.

xxii

Groups of prime power order

(c) Exactly p 2 maximal subgroups of G are of maximal class. If, in addition, p > 2 and m > 4, then the remaining p C 1 maximal subgroups of G have no two generators and their intersection .G/ has index p 2 in G. Theorem 13.2. Suppose that a p-group G is neither absolutely regular nor of maximal class. Then (a) c1 .G/ 1 C p C C p p1 .mod p p /. (b) If k > 1, then ck .G/ 0 .mod p p1 /. Corollary 13.3. Suppose that an irregular p-group G is neither absolutely regular nor of maximal class and k < p. Then it has a normal subgroup M of order p k and exponent p, and the number of subgroups in G of order p kC1 and exponent p containing M is 1 .mod p/. Theorem 13.5. If a p-group G is neither absolutely regular nor of maximal class, then ep .G/ 1 .mod p/ (here en .G/ is the number of subgroups of order p n and exponent p in G). Exercise P4. Let N be a normal subgroup of a p-group G. If N has no G-invariant subgroup of order p p and exponent p, then it is either absolutely regular or of maximal class. Theorem 13.6. Let a group G of order p m be neither absolutely regular nor of maximal class, let n be a natural number with m > n p C1. Denote by ˛.G/ the number of subgroups of maximal class and order p n in G. Then p 2 divides ˛.G/. Theorem 13.7. Let p > 2 and suppose that a p-group has no normal elementary abelian subgroup of order p 3 . Then one of the following holds: (a) G is metacyclic. (b) G is a 3-group of maximal class not isomorphic to a Sylow 3-subgroup of the symmetric group of degree 9. (c) G D 1 .G/C , where j1 .G/j D p 3 and C is cyclic. Exercise P5. Let G be a p-group, p > 2, and suppose that 1 .G/ has no G-invariant elementary abelian subgroup of order p 3 . Then G is one of the groups from Theorem 13.7. Exercise 13.10(a). Let H < G, where G is a p-group. If every subgroup of G of order pjH j containing H is of maximal class, then G is also of maximal class. Proposition 13.18. Let G be a p-group, and let M < G be of maximal class. (a) Set D D ˆ.M /, N D NG .M / and C D CN .M=D/. Let t be the number of subgroups K G of maximal class such that M < K and jK W M j D p. Then t D c1 .N=M / c1 .C =M /. If G is not of maximal class, then t 0 .mod p/.

xxiii

Prerequisites from Volumes 1 and 2

(b) Suppose, in addition, that M is irregular and G is not of maximal class and a positive integer k is ﬁxed. Then the number t of the subgroups L < G of maximal class and order p k jM j such that M < L is a multiple of p. Theorem 36.1. Let G be a nonabelian p-group and R < G 0 be a G-invariant subgroup of index p. Then G is metacyclic if and only if G=R is metacyclic. Corollary 36.6 (Blackburn). Suppose that a nonabelian p-group G and all its maximal subgroups are two-generator. Then G is either metacyclic or p > 2 and K3 .G/ D Ã1 .G/ has index p 3 in G (in the last case, jG W G 0 j D p 2 ). Theorem 36.16. Suppose that a p-group G is such that G=Ã2 .G/ is of maximal class. Then G is also of maximal class. Exercise P6. If a p-group G is such that G=Ã2 .G/ is of maximal class, then G is also of maximal class. Lemma 42.1. Let G be a group of order p m satisfying j2 .G/j p pC1 < jGj. Then one of the following holds: (a) G is an Lp -group. (b) G is absolutely regular. (c) p D 2, G is metacyclic and G D ha; b j a2

m2

D b 8 D 1; ab D a1 ; a2

m3

D b 4 ; m > 4i:

Here Z.G/ D hb 2 i, G 0 D ha2 i and G=G 0 is abelian of type .4; 2/, ˆ.G/ D m4 ha2 ; b 2 i, 2 .G/ D ha2 ; b 4 i. Corollary 44.6. A p-group is metacyclic if and only if one of the following quotient groups is metacyclic: G=ˆ.G 0 /;

G=K3 .G/;

G=Ã1 .G 0 /;

and, if p > 2, then G=Ã1 .G/:

Corollary 44.9. If G=Ã2 .G/ is a metacyclic 2-group, then G is also metacyclic. Theorem 44.12. Suppose that N is a two-generator normal subgroup of a p-group G. If N ˆ.G/, then N is metacyclic. Theorem 50.1. Let G be a 2-group which has no normal elementary abelian subgroup of order 8. Then G has a normal metacyclic subgroup N such that G=N is isomorphic to a subgroup of D8 . Lemma 57.1. Let G be a nonabelian p-group and let A be a maximal abelian normal subgroup of G. Then for any x 2 GA, there is a 2 A such that Œa; x ¤ 1, Œa; xp D 1, and Œa; x; x D 1 which implies that ha; xi is minimal nonabelian. Therefore G is generated by its minimal nonabelian subgroups. In particular, if all minimal nonabelian subgroups of a nonabelian p-group G, p > 2, have exponent p, then Hp .G/ is abelian. (For a stronger result, see Mann’s commentary to Problem 115.)

xxiv

Groups of prime power order

Lemma 57.2. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are of exponent 4 and let A be a maximal normal abelian subgroup of G. Then all elements in G A are of order 4 and so either exp.A/ D 2 or exp.A/ D exp.G/. If x 2 G A with x 2 2 A, then x inverts each element in Ã1 .A/ and in A=1 .A/. If exp.G/ > 4, then either G=A is cyclic of order 4 or G=A Š Q8 . Lemma 65.2(a). If G is a nonabelian two-generator p-group and G 0 1 .Z.G//, then G is minimal nonabelian. Theorems 66.1 and 69.1. If G is a minimal nonmetacyclic p-group, then one of the following holds: (a) G is of order p 3 and exponent p. (b) G is a 3-group of maximal class and order 34 . (c) G Š D8 C4 D Q8 C4 is of order 16. (d) G D Q8 C2 . (e) G is special of order 25 with jZ.G/j D 4, exactly one maximal subgroup of G is abelian. It follows from the previous theorem that a p-group G is metacyclic in any of the following cases: (i) every subgroup of G of exponent p 2 is metacyclic (in particular, if 2 .G/ is metacyclic), (ii) every three-generator subgroup of G is metacyclic. Exercise P7. A p-group is metacyclic if one of the following holds: (a) 2 .G/ is metacyclic. (b) G=Ã2 .G/ is metacyclic. (c) G=K3 .G/ˆ.G 0 / is metacyclic. Exercise P8. If a 2-group G and all its maximal subgroups are two-generator, then G is metacyclic. Propositions 71.3–71.5. If a two-generator p-group is a nonmetacyclic A2 -group, then p > 2. Theorems 82.1–82.3. Let G be a 2-group with exactly three involutions and assume that Z.G/ is noncyclic. Then G has a normal metacyclic subgroup M such that G=M is elementary abelian of order at most 4. Theorem 90.1. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to D8 or Q8 . Then G is one of the following groups: (a) G is generalized dihedral (i.e., jG W H2 .G/j D 2). (b) G D H Z.G/, where H is of maximal class and Ã1 .Z.G// Z.H /. (c) G D H Z.G/, where H is extraspecial and Ã1 .Z.G// Z.H /.

Prerequisites from Volumes 1 and 2

xxv

Corollary 90.2. Let G be a nonabelian 2-group in which any two noncommuting elements generate a subgroup of maximal class. Then G is one of the groups (a), (b) or (c) from Theorem 90.1. Conversely, each group in (a), (b) or (c) of Theorem 90.1 satisﬁes the assumption of our corollary. Exercise P9. Let H D ha; bi be a two-generator p-group with H 0 of order p. Then ˆ.H / D hap ; b p ; Œa; bi and H is minimal nonabelian. Exercise P10. Let G be a p-group with jG 0 j D p. If H is a minimal nonabelian subgroup of G, then G D H CG .H /. Exercise P11. All p 2 C p C 1 subgroups of order p 2 in an elementary abelian group E D ha; b; ci of order p 3 are: ha; bi, ha; b i ci (p C 1 subgroups containing hai) and haj b; ak ci (p 2 subgroups not containing hai), where i; j; k are any integers modulo p. Corollary 92.7. Let G be a non-Dedekindian p-group and let R.G/ be the intersection of all nonnormal subgroups. If R.G/ > ¹1º, then p D 2, jR.G/j D 2 and G is one of the following groups: (a) G Š Q8 C4 E2s , s 0. (b) G Š Q8 Q8 E2s , s 0. (c) G has an abelian maximal subgroup A of exponent > 2 and an element x 2 G A of order 4 which inverts each element in A.

93

Nonabelian 2-groups all of whose minimal nonabelian subgroups are metacyclic and have exponent 4

We shall determine the structure of nonabelian 2-groups all of whose minimal nonabelian subgroups are metacyclic and have exponent 4, i.e., they are isomorphic to D8 , Q8 or H2 D ha; b j a4 D b 4 D 1; ab D a1 i. In Theorem 90.1 we have determined the nonabelian 2-groups all of whose minimal nonabelian subgroups are isomorphic to D8 or Q8 . The nonabelian 2-groups all of whose minimal nonabelian subgroups are isomorphic to Q8 or H2 are identiﬁed in Theorem 92.6. Here we determine the title 2-groups G which possess as subgroups both D8 and H2 (and possibly Q8 ). It turns out, as a surprise, that such 2-groups G must be of exponent > 4. More precisely, we prove the following result. Theorem 93.1. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are metacyclic and have exponent 4 and assume that D8 and H2 D ha; b j a4 D b 4 D 1; ab D a1 i (and possibly Q8 ) actually appear as subgroups of G. Then G has a unique abelian maximal subgroup A of exponent > 4 and an involution t 2 G A such that t inverts each element in A=1 .A/ and in 2 .A/ and there is an element a 2 A of order > 4 such that at D a1 with 1 ¤ 2 1 .A/, where in case that Ã1 .A/ is cyclic we have 62 Ã1 .A/. Conversely, all these 2-groups satisfy the assumptions of the theorem. Proof. Let G be a 2-group satisfying the assumptions of Theorem 93.1 and suppose that A is a maximal normal abelian subgroup of G. We shall freely use Lemmas 57.1, 57.2 and 65.2(a). It follows that all elements in G A are of order 4. (i) We show that G=A is elementary abelian. Suppose that this is false. Then there is an element v of order 4 in G A such that v 2 D t 62 A. Set T D Ahvi, A0 D CA .v/ and T0 D A0 hvi so that A ¤ A0 and T ¤ T0 . Let T1 T be such that T1 > T0 and jT1 W T0 j D 2. Set A1 D T1 \A so that jA1 W A0 j D 2. Note that T10 A\T0 D A0 Z.T1 / and so cl.T1 / D 2. Let a be any element in A1 A0 so that a2 2 A0 . We get 1 ¤ Œa; v 2 A0 and Œa; v2 D Œa2 ; v D 1 and so Œa; v is an involution in Z.T1 /. This implies that ha; vi0 D hŒa; vi and so ha; vi is minimal nonabelian. Set ha; vi \ A D A hai so that ha; vi D A hvi satisfying A \ hvi D ¹1º. If jA j D 2, then ha; vi D A hvi is abelian, a contradiction.

2

Groups of prime power order

Hence jA j D 4 so that ha; vi Š H2 and A D hai Š C4 with av D a1 . We have proved that all elements in A1 A0 are of order 4 and they are inverted by v. Since hA1 A0 i D A1 , v inverts A1 so that A0 D CA .v/ is elementary abelian and t D v 2 centralizes A1 . Set A2 D CA .t/ so that A2 A1 , A2 ¤ A and A2 is hvi-invariant which gives that T2 D A2 hvi is a subgroup of T , T2 ¤ T and T2 \ A D A2 . Let T3 be a subgroup of T such that T3 > T2 and jT3 W T2 j D 2 and set A3 D T3 \ A so that jA3 W A2 j D 2. Let b be any element in A3 A2 so that b 2 2 A2 D CA .t/. It follows that 1 ¤ Œb; t 2 A \ T2 D A2

and

1 D Œb 2 ; t D Œb; tb Œb; t D Œb; t2

so that hb; ti is minimal nonabelian with a noncentral involution t which yields that hb; ti Š D8 and o.b/ 4. All elements in the set A3 A2 are of order 4 so that A3 D hA3 A2 i is of exponent 4. If o.b/ D 2, then o.ba/ D 4, where a 2 A1 A0 and o.a/ D 4. In any case, there is an element w 2 A3 A2 of order 4 which together with hw; ti Š D8 gives w t D w 1 . Suppose that there is an involution u in A2 A1 . Then 1 ¤ Œu; v 2 A and we get 1 D Œu2 ; v D Œu; vu Œu; v D Œu; v2

and 1 D Œu; v 2 D Œu; vŒu; vv

so that Œu; v is a central involution in hu; vi. Hence hu; vi is minimal nonabelian with a noncentral involution u which gives hu; vi Š D8 . But then v u D v 1 and so Œu; v 62 A, a contradiction. We have proved that all element in A2 A0 are of order 4. In particular, w 2 2 A0 D CA .v/. We compute 1 D Œw 2 ; v D Œw; vw Œw; v D Œw; v2 and so Œw; v 2 T2 \A D A2 is an involution in A2 and therefore Œw; v 2 A0 D CA .v/. It follows that Œw; v 2 Z.hw; vi/ and thus hw; vi is minimal nonabelian satisfying hw; vi \ A hwi Š C4 so that hw; vi Š H2 and hw; vi \ A D hwi which im2 plies w v D w 1 . But then w t D w v D w, which contradicts a result from the previous paragraph. (ii) We prove that exp.G/ > 4. Suppose that this is false. In what follows we assume that exp.G/ D 4. In that case, G 0 Ã1 .G/ 1 .A/ and so Ã1 .G/ is elementary abelian. Since H2 is a subgroup of G, we have jÃ1 .G/j 4. (ii1) If a 2 A and x 2 G with Œa; x ¤ 1, then ha; xi is minimal nonabelian. In that case, ha; xi Š D8 if and only if a or x (or both) is an involution. Indeed, 1 D Œa; x 2 D Œa; xŒa; xx and so Œa; x 2 1 .A/ is an involution which commutes with a and x. Hence ha; xi0 D hŒa; xi and so, by Lemma 65.2(a), ha; xi is minimal nonabelian. (ii2) We have Ã1 .A/ ¤ ¹1º and Ã1 .A/ Z.G/. Assume that A is elementary abelian. Let H D ha; b j a4 D b 4 D 1; ab D a1 ; a2 D z; b 2 D ui be a subgroup of G so that H \ A D hu; zi Š E4 . Let t be an involution in the set

93

2-groups with metacyclic minimal nonabelian subgroups of exponent 4

3

A .CA .a/ [ CA .b//. By (ii1), we have ht; ai Š ht; bi Š D8 and so at D a1 D az and b t D b 1 D bu. We get .ab/t D .ab/zu and so Œt; ab ¤ 1. Again by (ii1), we obtain ht; abi Š D8 and so .ab/t D .ab/1 D .ab/.ab/2 D .ab/u, a contradiction. We have proved that Ã1 .A/ ¤ ¹1º. Let a 2 A and assume that a2 62 Z.G/. Then there is g 2 G such that Œa2 ; g ¤ 1. But then Œa; g ¤ 1 and so (ii1) implies that ha; gi is minimal nonabelian. In that case, a2 2 ˆ.ha; gi/ D Z.ha; gi/ and so Œa2 ; g D 1, a contradiction. We have proved that Ã1 .A/ Z.G/. (ii3) We have Ã1 .G/ Z.G/ which implies that each maximal abelian subgroup of G is also normal in G. Whenever x; y 2 G with Œx; y ¤ 1, then hx; yi is minimal nonabelian. Suppose that this is false. Then there is an element v 2 G A such that v 2 62 Z.G/ and v 2 2 1 .A/. In that case, there is w 2 G .Ahvi/ such that Œw; v 2 ¤ 1. By (ii1), hw; v 2 i Š D8 and so (replacing w with wv 2 if necessary), we see that there is an involution t 2 G.Ahvi such that ht; v 2 i Š D8 . We study the two-generator subgroup H D ht; vi and set A0 D H \ A. Since H=A0 Š E4 , we have ˆ.H / D A0 and A0 is elementary abelian. Hence Œt; v is an involution in A0 and 1 D Œt 2 ; v D Œt; vt Œt; v shows that t commutes with Œt; v. If Œt; v also commutes with v or tv, then we get Œt; v 2 Z.H /, ht; vi0 D hŒt; vi and so ht; vi is minimal nonabelian. But in that case, we conclude v 2 2 ˆ.H / D Z.H / and Œt; v 2 D 1, a contradiction. We have proved that CH .A0 / D A0 and so A0 (being elementary abelian) is a maximal normal abelian subgroup of H . We have already seen that D8 Š ht; v 2 i is a subgroup of H . Suppose that H2 is not a subgroup of H . In that case, we are in a position to use Theorem 90.1. If H is quasidihedral, then hvi is normal in H in which case v 2 2 Z.H /, a contradiction. It follows that H D H0 Z.H /, where H0 is extraspecial and Ã1 .Z.H // Z.H0 /. But in the last case, A0 D Ã1 .H / Z.H /, a contradiction. We have shown that H has both D8 and H2 as subgroups. By (ii2) applied to H and A0 , we get Ã1 .A0 / ¤ ¹1º, contrary to the fact that A0 is elementary abelian. We have proved that Ã1 .G/ Z.G/ and so G is of class 2 which implies that each maximal abelian subgroup of G is also normal in G. Let x; y 2 G with Œx; y ¤ 1. Then Œx; y 2 Ã1 .G/ Z.G/ and Ã1 .G/ 1 .A/ so that hx; yi0 D hŒx; yi is of order 2 and so hx; yi is minimal nonabelian. (ii4) D8 C4 is not a subgroup of G. Indeed, assume that ha; t j a4 D t 2 D 1; at D a1 i hbi, where hbi Š C4 , is a subgroup of G. Then Œt; ab D Œt; a ¤ 1 and (ii3) implies ht; abi Š D8 . It follows .ab/t D .ab/1 D a1 b 1 and .ab/t D at b t D a1 b, a contradiction. (ii5) Each maximal abelian subgroup A is of type .4; 2; : : : ; 2/. In particular, C4 C4 is not a subgroup of G. By (ii2) and (ii3), A is not elementary abelian, A is normal in G, and assume that A has a subgroup hai hbi Š C4 C4 . Since D8 is a subgroup of G, there is an involution t 2 G A. By (ii1) and (ii4), t either centralizes ha; bi or t inverts each element of ha; bi. Suppose that t centralizes ha; bi and let x be any element of order 4

4

Groups of prime power order

in A. By (ii1), t normalizes hxi (since Œt; x ¤ 1 implies that ht; xi Š D8 ) and again by (ii4), t centralizes x. But elements of order 4 in A generate A and so t centralizes A, a contradiction. Hence t inverts ha; bi and by (ii4), t inverts each element of order 4 in A. It follows that t inverts A. This implies that each minimal nonabelian subgroup of Ahti is isomorphic to D8 and so Ahti ¤ G and there are no involutions in G .Aht i/. We have G D .Aht i/S with .Aht i/ \ S D A, where S ¤ A and all elements in S A are of order 4. Let v 2 S A. If Œv; t ¤ 1, then (ii3) implies that hv; ti Š D8 . But then vt 2 G .Ahti/ and vt is an involution, a contradiction. It follows that t commutes with each element in S A and so ŒS; t D 1 and t centralizes A, a contradiction. We have proved that any maximal abelian subgroup A of G is of type .4; 2; : : : ; 2/. (ii6) If v 2 G A is such that v 2 62 Ã1 .A/, then v centralizes 1 .A/. Suppose that a is an element of order 4 in A so that (by (ii5)) A D hai1 .A/ with hai \ 1 .A/ D ha2 i D Ã1 .A/. Further, suppose that there exists t 2 1 .A/ such that Œv; t ¤ 1. Then hv; ti Š D8 and so v t D v 1 and t 0 D vt is an involution in Av. By (ii5), we get Œa; v ¤ 1 and so ha; vi Š H2 . We have Œav; t D Œv; t D v 2 ¤ 1 and so hav; ti Š D8 which gives v 2 D Œav; t D .av/2 D a2 v 2 Œv; a;

Œv; a D a2 ;

av D a1

and .av/2 D v 2 . We compute 0

.av/t D .av/vt D .a1 v/t D a1 v 1 D .av/a2 v 2 ¤ av since a2 ¤ v 2 and so hav; t 0 i Š D8 which implies 0

.av/t D .av/1 D .av/.av/2 D .av/v 2 ; a contradiction with the above. (ii7) If t is any involution in G A, then t does not centralize 1 .A/. Suppose that t centralizes 1 .A/ so that CA .t/ D 1 .A/. In that case, t inverts each element in A. If v is an element of order 4 in A, then A D hvi1 .A/ with hvi \ 1 .A/ D hv 2 i D Ã1 .A/ and v t D v 1 : Since jÃ1 .G/j 4, there is an element g 2 G .Aht i/ such that g 2 ¤ v 2 . By (ii6), g centralizes 1 .A/ and so Œg; v ¤ 1 which implies that hg; vi Š H2 . If v g D v 1 , then v gt D .v 1 /t D v and gt centralizes A, a contradiction. It follows that v g D vg 2 and so v gt D .vg 2 /t D v 1 g 2 D v.v 2 g 2 /: If Œg; t ¤ 1, then we get hg; ti Š D8 , gt is an involution which does not normalize hvi, a contradiction. Hence Œg; t D 1, o.gt/ D 4, .gt/2 D g 2 and so hv; gti Š H2 with hv; gti0 D hv 2 g 2 i which implies that v 2 g 2 is a square in hv; gti. This is a contradiction since g 2 and v 2 are the only involutions in hv; gti which are squares in hv; gti.

93

2-groups with metacyclic minimal nonabelian subgroups of exponent 4

5

(ii8) We consider a ﬁnal conﬁguration which will produce a contradiction. Let H D hg; v j g 4 D v 4 D 1; v g D v 1 i be a subgroup of G. Let A be a maximal abelian subgroup of G containing hv; g 2 i Š C4 C2 . Since g centralizes 1 .A/ (by (ii6)), g inverts on A, all elements in Ag are of order 4, and Ahgi Š H2 E2s , s 0. Since D8 is a subgroup of G, there is an involution t 2 G .Ahgi/. By (ii7), t does not centralize 1 .A/ and we see that also gt and vgt do not centralize 1 .A/. If Œg; t D 1, then .gt/2 D g 2 62 Ã1 .A/ and then (ii6) gives a contradiction. Similarly, if Œvg; t D 1, then .vgt/2 D .vg/2 D g 2 62 Ã1 .A/ and again (ii6) gives a contradiction. We have proved that Œg; t ¤ 1;

Œvg; t ¤ 1;

and so hg; ti Š hvg; ti Š D8

which gives g t D g 1 and .vg/t D .vg/1 . We compute .vg/t D .vg/1 D .vg/.vg/2 D .vg/g 2 D vg 1 D v t g t D v t g 1 ; and so v t D v. Let s 2 1 .A/ C1 .A/ .t/ so that hs; ti Š D8 , w D ts is an element of order 4 in the coset At and w t D w 1 . Since w commutes with v, (ii5) implies that w 2 D v 2 and hw; gi Š H2 with g w D g t s D g 1 . Hence gw is an element of order 4 and .gw/t D g 1 w 1 D .gg 2 /.ww 2 / D .gw/.g 2 v 2 / ¤ gw: By (ii3), hgw; ti Š D8 which implies .gw/t D .gw/1 . On the other hand, .gw/t D .gw/1 D .gw/.gw/2 D .gw/w 2 D .gw/v 2 ; which contradicts the above result .gw/t D .gw/.g 2 v 2 /. We have thus proved that exp.G/ > 4. (iii) It remains to determine the structure of G in case exp.G/ > 4. By (i) and Lemma 57.2, A is a maximal subgroup of G, all elements in G A are of order 4, exp.A/ 8 and for each g 2 G A, g inverts A=1 .A/ and on Ã1 .A/. If a 2 A with o.a/ D 8 and g 2 G A, then ag D a1 , 2 1 .A/ and so Œa; g D a1 ag D a2 , where o.a2 / D 4 and therefore jG 0 j > 2. This gives together with jGj D 2jZ.G/jjG 0 j (Lemma 1.1) that A is a unique abelian maximal subgroup of G. Since D8 is a subgroup of G, there is an involution t 2 G A. Let v be an element of order 4 in A. Then v t D v 1 , 2 1 .A/, and so Œv; t D v 2 2 1 .A/ and 1 D Œv; t 2 D Œv; tŒv; tt which gives Œv; tt D Œv; t. If Œv; t ¤ 1, then Œv; t is an involution commuting with v and t and so hv; ti0 D hŒv; ti which implies that hv; ti is minimal nonabelian (Lemma 65.2(a)). Hence in that case, hŒv; ti Š D8 and so v t D v 1 . We have proved that t either centralizes or inverts each element of order 4 in A. Assume that A0 D CA .t/ D Z.G/ is not elementary abelian in which case we have exp.A0 / D 4. Since t inverts Ã1 .A/, there is an element v of order 4 in A A0 such

6

Groups of prime power order

that v t D v 1 and v 2 2 A0 . Let w be an element of order 4 in A0 . If w 2 ¤ v 2 , then vw is an element of order 4 in A A0 and so .vw/t D .vw/1 D v 1 w 1 D v t w t D v 1 w t which implies w t D w 1 , a contradiction. Hence w 2 D v 2 , A0 is of type .4; 2; : : : ; 2/ and Ã1 .A0 hvi/ D hv 2 i. Suppose that there is an element y of order 4 in A .A0 hvi/ so that y t D y 1 and y 2 2 A0 . If y 2 D v 2 , then yv is an involution in A .A0 hvi/ centralized by t, a contradiction. If there exists an involution u 2 A .A0 hvi/, then o.uv/ D 4 and uv 2 A .A0 hvi/. Hence in case that 2 .A/ > .A0 hvi/ we have an element y of order 4 in 2 .A/ .A0 hvi/ with y 2 2 A0 and y 2 ¤ v 2 . For each l 2 A0 hvi we obtain .yl/2 D y 2 l 2 with l 2 2 hv 2 i, and so all elements in y.A0 hvi/ are of order 4 and they are inverted by t . Since hy.A0 hvi/i D A0 hy; vi, t inverts A0 hvi. This is a contradiction because for an element w of order 4 in A0 , we have w t D w. We have proved that 2 .A/ D A0 hvi and A0 hv; ti D A0 hv; ti with hv; ti Š D8 ;

A0 \ hv; ti D hv 2 i D Z.hv; ti/ and Ã1 .A0 / D hv 2 i

so that each minimal nonabelian subgroup in A0 hv; ti is isomorphic to D8 or Q8 . Since H2 is a subgroup of G, there is a 2 A .A0 hvi/ such that .at/2 2 A0 hv 2 i and v at D v 1 so that hat; vi Š H2 with hat; vi0 D hv 2 i. Note that vw (with w 2 A0 and w 2 D v 2 ) is an involution in .A0 hvi/ A0 and .vw/at D .vw/t D .vw/v 2 and so hat; vwi is the nonmetacyclic minimal nonabelian of order 24 , a contradiction. We have proved that A0 D CA .t/ D Z.G/ is elementary abelian. Then all elements of order 4 in 2 .A/ are inverted by t and so t inverts 2 .A/ and therefore we have CA .t/ D Z.G/ D 1 .A/. Since H2 is not a subgroup of 2 .A/hti, there are elements a 2 A 2 .A/ and v 2 2 .A/ 1 .A/ such that .at /2 D ¤ 1;

2 1 .A/;

v at D v 1

and ¤ v 2 :

We have atat D and so at D a1 . If Ã1 .A/ is cyclic, then v 2 2 Ã1 .A/ and so we must have in that case 62 Ã1 .A/. (iv) We have obtained 2-groups stated in Theorem 93.1. It is easy to check that all these 2-groups satisfy the assumptions of our theorem. Indeed, let G be a 2-group described in the second paragraph of Theorem 93.1. Since t inverts each element in 2 .A/, we have CA .t/ D 1 .A/ D Z.G/. Let xt be any element in G A, where x 2 A. Then xt also inverts 2 .A/ and on A=1 .A/. We get .xt/2 D xtxt D xx t D x.x 1 s/ D s, where s 2 1 .A/ so that o.xt/ 4. Let w be an element of order 4 in Ã1 .A/ such that 62 hwi. From at D a1 we deduce .at/2 D ¤ 1 and so hwi \ hat i D ¹1º together with w at D w t D w 1 implies that

93

2-groups with metacyclic minimal nonabelian subgroups of exponent 4

7

hat; wi Š H2 . Also, hw; ti Š D8 and therefore we have proved that both D8 and H2 actually appear as subgroups of G. Let H be any minimal nonabelian subgroup of G so that H 6 A. Let c 2 H A and d 2 .H \ A/ ˆ.H / so that H D hc; d i and o.c/ 4. Suppose that o.d / 8. In that case, d c D d 1 r with r 2 1 .A/ so that Œd; c D d 2 r is of order 4, a contradiction. Hence o.d / 4 and in fact o.d / D 4 since 1 .A/ Z.G/. Since c inverts 2 .A/, we have d c D d 1 . If o.c/ D 2, then H D hc; d i Š D8 . In case o.c/ D 4 and c 2 D d 2 , we get H Š Q8 . If o.c/ D 4 and c 2 ¤ d 2 , then H Š H2 and we are done.

94

Nonabelian 2-groups all of whose minimal nonabelian subgroups are nonmetacyclic and have exponent 4

We consider here nonabelian 2-groups G all of whose minimal nonabelian subgroups are nonmetacyclic and have exponent 4. Then it is easy to prove that exp.G/ D 4. In fact, we prove the following result. Theorem 94.1. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are nonmetacyclic and have exponent 4. Then exp.G/ D 4, 1 .G/ is elementary abelian and if A is a maximal normal abelian subgroup of G containing 1 .G/, then ˆ.G/ 1 .A/ is elementary abelian and Ã1 .A/ Z.G/. Proof. Since D8 is not a subgroup of G, 1 .G/ is elementary abelian. Let A be a maximal normal abelian subgroup of G containing 1 .G/. Suppose that exp.G/ > 4. By Lemma 57.2, jG W Aj D 2, all elements in G A are of order 4 and if x is one of them, then x inverts Ã1 .A/. Let y be an element of order 4 in Ã1 .A/. Then y x D y 1 which implies that hx; yi Š Q8 or hx; yi Š H2 , a contradiction. We have proved that exp.G/ D 4 and so exp.A/ 4 and G=A is elementary abelian. Then ˆ.G/ D Ã1 .G/ 1 .A/ and so ˆ.G/ is elementary abelian. If a 2 A and g 2 G with Œa; g ¤ 1, then ha; gi is minimal nonabelian. Indeed, Œa; g is an involution in 1 .A/ and 1 D Œa; g 2 D Œa; gŒa; gg so that Œa; g commutes with a and g and therefore ha; gi0 D hŒa; gi. By Lemma 65.2(a), ha; gi is minimal nonabelian. We prove that Ã1 .A/ Z.G/. Indeed, let a 2 A be such that a2 62 Z.G/. Then there is g 2 G with Œa2 ; g ¤ 1. But then Œa; g ¤ 1 and so, by the previous paragraph, ha; gi is minimal nonabelian. In that case, a2 2 ˆ.ha; gi/ D Z.ha; gi/ and so Œa2 ; g D 1, a contradiction. Remark. It is easy to see that the group H16 C4 contains both nonmetacyclic minimal nonabelian 2-groups of exponent 4, namely, H16 D ha; b j a4 D b 2 D 1; Œa; b D z; z 2 D Œa; z D Œb; z D 1i (of order 16) and H32 D ha; b j a4 D b 4 D 1; Œa; b D z; z 2 D Œa; z D Œb; z D 1i (of order 32) and each minimal nonabelian subgroup of H16 C4 is nonmetacyclic.

94

2-groups with nonmetacyclic minimal nonabelian subgroups of exponent 4

9

We prove a lemma for p D 2 which might be useful for future similar investigations. The corresponding case for p > 2 was treated in Lemma 28.2. Lemma 94.2. Let G be a 2-group such that all subgroups of ˆ.G/ are normal in G. Then ˆ.G/ is abelian and jG W CG .ˆ.G//j 2. If jG W CG .ˆ.G//j D 2, then each element x 2 G .CG .ˆ.G// is of order 4 and x inverts each element in ˆ.G/. Proof. By assumption, ˆ.G/ is Dedekindian. Suppose that ˆ.G/ is nonabelian. Then ˆ.G/ possesses a subgroup Q Š Q8 . Since Q is normal in G and each cyclic subgroup of Q is normal in G, it follows that no element in G induces an outer automorphism on Q and so G D QCG .Q/ with Q \ CG .Q/ D Z.Q/. If M is a maximal subgroup of G containing CG .Q/, then Q 6 M , which contradicts the fact that Q ˆ.G/. We have proved that ˆ.G/ is abelian. Suppose that ˆ.G/ is cyclic. Let P be a cyclic subgroup of G which contains ˆ.G/ with jP W ˆ.G/j D 2. Let x 2 G P so that x 2 2 ˆ.G/ and P is a maximal cyclic subgroup of index 2 in P hxi. By the structure of such groups, x either centralizes ˆ.G/ or inverts ˆ.G/. Now assume that ˆ.G/ is noncyclic so that ˆ.G/ D Z1 Zn , where Zi are cyclic and n > 1. Set Zi D hai i and

K D Z1 Zi 1 ZiC1 Zn

and consider G=K. By the above, for each x 2 G we have aix D ai k or aix D ai1 k for some k 2 K. But Zi is normal in G and so k D 1. We have proved that each x 2 G either centralizes or inverts each Zi , i D 1; 2; : : : ; n. Suppose that ˆ.G/ possesses a subgroup hv; wi Š C4 C4 such that for an element x 2 G we have v x D v 1 and w x D w. In that case, .vw/x D v 1 w D v 2 .vw/, which contradicts the fact that hvwi is normal in G. We have proved that each element x 2 G either centralizes ˆ.G/ or inverts each element in ˆ.G/. This gives jG W CG .ˆ.G//j 2. Suppose that jG W CG .ˆ.G//j D 2 and let x 2 G .CG .ˆ.G//. Since x inverts ˆ.G/ and x 2 2 ˆ.G/, it follows that o.x 2 / 2 which gives o.x/ 4. The proof is complete.

95

Nonabelian 2-groups of exponent 2e which have no minimal nonabelian subgroups of exponent 2e

Here we solve a general problem (Nr. 1475 for p D 2) by classifying the title groups. More precisely, we prove the following result. Theorem 95.1. Let G be a nonabelian 2-group of exponent 2e (e 3) which does not possess a minimal nonabelian subgroup of exponent 2e . Then G has a unique maximal normal abelian subgroup A. The factor group G=A is either cyclic or generalized quaternion and each element in G A is of order < 2e . Also, G possesses at least one metacyclic minimal nonabelian subgroup and jCG .1 .A// W Aj 2. Proof. Let G be a 2-group satisfying the assumptions of Theorem 95.1 and let A be any maximal normal abelian subgroup of G and g 2 G A. Set 2m D max¹exp.H / j H < G is minimal nonabelianº so that m 2 and m < e. Further, set T D Ahgi, A0 D CA .g/ so that A0 < A and T0 D A0 hgi is abelian, T0 < T and T =A is cyclic. Let T1 T with T1 > T0 and jT1 W T0 j D 2. Set A1 D T1 \ A so that T1 D A1 hgi. Let a 2 A1 A0 so that a2 2 A0 and Œa; g 2 A \ T0 D A0 Z.T1 /. We have 1 D Œa2 ; g D Œa; ga Œa; g D Œa; g2 so that Œa; g is an involution commuting with a and g and therefore ha; gi is minimal nonabelian (Lemma 65.2(a)). In particular, we get o.a/ 2m and o.g/ 2m . Since A1 D hA1 A0 i and A1 is abelian, we have exp.A1 / 2m . We have proved that for each g 2 G A, o.g/ 2m and A0 D CA .g/ is of exponent 2m . Suppose at the moment that g 2 2 A and let x be an element in A. Then we have also .xg/2 2 A. Set x g D x 1 w for some w 2 A. In that case, w D xx g D x.g 1 xg/ D xg 2 gxg D g 2 .xg/2 is an element of order 2m1 in A. We have shown that an element g 2 G A with g 2 2 A inverts A=m1 .A/. Thus for each x 2 A, .x 2

m1

/g D .x g /2

m1

D .x 1 w/2

m1

D x 2

m1

95

2-groups of exponent 2e with no minimal nonabelian subgroup of exponent 2e 11

and so g inverts Ãm1 .A/. Since exp.A/ D exp.G/ D 2e and e > m, it follows that Ãm1 .A/ contains elements of order 4. Hence G=A contains only one subgroup of order 2 which implies that G=A is either cyclic of order 2m or generalized quaternion of order 2mC1 (m 2). Since g 2 G A with g 2 2 A inverts a cyclic subgroup of order 4 in Ãm1 .A/, it follows that G possesses a metacyclic minimal nonabelian subgroup. Suppose that G possesses another maximal normal abelian subgroup B, B ¤ A. Then all elements in G A are of order < 2e , all elements in A B D A .A \ B/ are of order < 2e and since an element b 2 B .A \ B/ centralizes A \ B, we have also exp.A \ B/ < 2e . But then exp.G/ < 2e , a contradiction. Suppose that A satisﬁes jCG .1 .A// W Aj 4. Let H=A Š C4 be a subgroup of order 4 in CG .1 .A//=A and let g be an element in H A such that hgi covers H=A. Set A0 D CA .g/, T0 D A0 hgi so that T0 is abelian and A0 1 .A/. Further, set S D 2 .Ãm1 .A// so that S is normal in H , exp.S/ D 4 and g 2 inverts S . Therefore S \A0 D 1 .S / and S T0 is a subgroup of H . Let T be a subgroup of ST0 containing T0 such that jT W T0 j D 2. Suppose that w is an element in .T \ S/ 1 .S/ so 2 that o.w/ D 4 and w 2 2 1 .S/ A0 and w g D w 1 . We now study the subgroup hw; gi, where 1 ¤ Œw; g 2 T0 \ A D A0 . Also, 1 D Œw 2 ; g D Œw; gw Œw; g D Œw; g2 and so Œw; g is an involution commuting with both w and g. It now follows from Lemma 65.2(a) that hw; gi is minimal nonabelian. But then g 2 2 ˆ.hw; gi/ D Z.hw; gi/ and so g 2 centralizes w, a contradiction. The theorem is proved. Theorem 95.2. Let G be a nonabelian 2-group of exponent 2e (e 3) which does not possess a minimal nonabelian subgroup of exponent 2e . Let A be a unique maximal normal abelian subgroup of G (Theorem 95.1). Suppose that G possesses a cyclic subgroup Z with Z \ A D ¹1º. Then G also possesses a nonmetacyclic minimal nonabelian subgroup. Proof. Set Z D hvi, v 2 D t and A0 D CA .v/ so that the subgroup T0 D A0 hvi is abelian. By Theorem 95.1, 1 .A/ 6 A0 and set T1 D 1 .A/T0 . Let T2 be a subgroup of T1 containing T0 such that jT2 W T0 j D 2. We have T2 D .1 .A/ \ T2 /T0 and take an involution u 2 .1 .A/ \ T2 / 1 .A0 /. Then Œu; v 2 1 .A/ \ T0 D 1 .A0 / so that Œu; v is an involution commuting with u and v and therefore hu; vi is minimal nonabelian. We have t D v 2 2 Z.hu; vi/ so that hu; Œu; v; t i Š E8 and thus hu; vi is nonmetacyclic.

96

Groups with at most two conjugate classes of nonnormal subgroups

The non-Dedekindian p-groups all of whose nonnormal subgroups are conjugate were classiﬁed in 58. Below we offer another proof. Next we classify the p-groups with exactly two conjugate classes of nonnormal subgroups. Both these results are due to Otto Schmidt. If G is a minimal nonabelian p-group, then (see Exercise 1.8(a) or Lemma 65.1) G D ha; bi, where m

n

m

n

(MA1) ap D b p D c p D 1, Œa; b D c, Œa; c D Œb; c D 1, jGj D p mCnC1 , G D hbi .haihci/ D hai .hbihci/ is nonmetacyclic with kernels haihci, hbi hci, respectively. (MA2) ap D b p D 1, ab D a1Cp cyclic with kernel hai, m > 1.

m1

, jGj D p mCn and G D hbi hai is meta-

(MA3) G Š Q8 , the ordinary quaternion group. Let nc.G/ be the number of conjugate classes of nonnormal subgroups of a group G and let no.G/ be the number of orders of nonnormal subgroups of G. To facilitate the proof, we ﬁrst prove some lemmas. Lemma 96.1. Let G D ha; bi be a minimal nonabelian p-group as in (MA1) or (MA2). (a) All nonnormal cyclic subgroups of G have the same order if and only if one of the following holds: (a1) G is as in (MA2) with m n. (a2) G is as in (MA1) with m D n. (b) All nonnormal subgroups of G have the same order if and only if one of the following holds: (b1) G 2 ¹D8 ; S.p 3 /º. (b2) G is as in (MA2) with m n. (c) All nonnormal cyclic subgroups of G of same order are conjugate if and only if G Š Mpt . (d) All nonnormal subgroups of G are conjugate if and only if G Š Mpt .

96

Groups with at most two conjugate classes of nonnormal subgroups

13

Proof. We use the same notation as in (MA1) and (MA2). Write A D hai and B D hbi. Let H < G be nonnormal; then G 0 6 H so that H \ G 0 D ¹1º since jG 0 j D p. (a) Suppose that all nonnormal cyclic subgroups of G have the same order. Let G be metacyclic as in (MA2) and let H be cyclic. If n > m and m .B/ D hxi, m1 then .ax/b D .ax/ap 62 haxi hence B and D D haxi are not G-invariant, and they are cyclic of orders p n and p m ¤ p n , respectively, a contradiction. It remains to show that if n m, then G satisﬁes the hypothesis. We have H \A D ¹1º since G 0 < A has order p and A is cyclic, so H , being isomorphic to a subgroup of G=A, has order p n . Since n1 .G/ Z.G/, our G satisﬁes the hypothesis and so it is as in (a1). Suppose that G is nonmetacyclic as in (MA1). Since A and B are not G-invariant, we must have m D n and G is as in (a2). Conversely, if G is as in (a2) with m D n, then, since m1 .G/ Z.G/, our G satisﬁes the hypothesis. (b) Suppose that all nonnormal subgroups of G have the same order; then H is cyclic since it is not generated by its G-invariant maximal subgroups. Thus one can use (a). Let G be metacyclic as in (a1); then m n. It remains to show that such G satisﬁes the hypothesis. In view of H \ A D ¹1º, we obtain that jH j jG W Aj D p n . Since n1 .G/ Z.G/, we have jH j D p n , and G is as in (b2). Now let G be nonmetacyclic as in (a2) so m D n. Assume that m > 1. Then we see that B 1 .A/ is nonnormal and noncyclic, a contradiction. Thus G is as in (b1). (c) Suppose that all nonnormal cyclic subgroups of G of same order are conjugate; then these subgroups have the same order. Let G be nonmetacyclic as in (MA1) and let m; n > 1. Assume that m > n. Let n .A/ D hxi. Then the cyclic subgroups B and hbxi of the same order p n are not normal, and since BG ¤ hbxiG , we conclude that B and hbxi are not conjugate. Thus m n. Similarly, n m so m D n. Then the cyclic subgroups A and B of the same order p m are neither normal nor conjugate, a contradiction. Now let G be nonmetacyclic as in (MA1) and m > n D 1. Then G has exactly p 2 nonnormal subgroups of order p. Since any nonnormal subgroup has p conjugates, there are in G two nonconjugate subgroups of order p. Let G be metacyclic as in (MA2). If n D 1, then G Š MpmC1 satisﬁes the condition. Now let n > 1. If n < m, set n .A/ D hxi; then the cyclic subgroups B and hbxi of order p n are neither G-invariant (otherwise, G D A hbxi is abelian) nor conjugate since their cores in G are different. If n m, then the cyclic subgroups B and hap bi of order p n are neither G-invariant nor conjugate since their cores in G are different, a contradiction. Statement (d) follows from (c). Below we present an alternate proof of Theorem 1.23. Lemma 96.2 (= Theorem 1.23 (Passman)). Assume G is a non-Dedekindian p-group. Then there is a G-invariant subgroup R of index p in G 0 such that G=R is non-Dedekindian.

14

Groups of prime power order

Proof. If p > 2, the result is obvious as Dedekindian groups of odd order are abelian. Now let p D 2. Assume that for every G-invariant subgroup K of index 2 in G 0 , the quotient group G=K is Dedekindian. We want to prove that G is itself Dedekindian. One may assume that K > ¹1º. Assuming that G is a minimal counterexample, we get jKj D 2 (this is due to the fact that the derived subgroup of a nonabelian Dedekindian group is of order 2; see Theorem 1.20) so jG 0 j D 4. It follows that G is not minimal nonabelian (Lemma 65.1) and there is in G=K a subgroup Q=K Š Q8 (Theorem 1.20). Since Q is not of maximal class (Theorem 1.2), we get jQ0 j D 2 (Proposition 1.6), and so G 0 D Q0 K Š E4 and G 0 Z.G/. Let A < G be minimal nonabelian. Assume that jAj > 8; then K < A (otherwise, A is isomorphic to a subgroup of G=K, which is impossible since all minimal nonabelian subgroups of the nonabelian Dedekindian group G=K are isomorphic to Q8 by Theorem 1.20). Assume that A=K is abelian; then it is of type .4; 2/ (here we use Theorem 1.20 again), and hence A0 D K. In this case, G 0 =K D Ã1 .G=K/ < A=K, and we conclude that G 0 < A, and so A G G. Then A=Q0 is nonabelian in view of Q0 \ K D ¹1º, hence A=Q0 Š Q8 by hypothesis. Thus one can assume from the start that A=K Š Q8 . In this case, A Š H2;2 D hx; y j x 4 D y 4 D 1; x y D x 3 i since j1 .A/j D 4 and exp.A/ D 4 (see Lemma 65.1 or (MA1) and (MA2)). We know that all subgroups of order 2 are characteristic in A so normal in G. Then A=hy 2 i Š D8 so G=hy 2 i is not Dedekindian, contrary to the assumption. Thus jAj D 8 so that any minimal nonabelian subgroup of G has order 8. Assume that A Š D8 . Let L < G 0 of order 2 be such that L 6 A. Then AL=L Š D8 so G=L is not Dedekindian, a contradiction. Thus A Š Q8 . Then A G G (indeed, any abelian subgroup of type .2; 2/ of a Dedekindian group G=A0 is normal). Write C D CG .A/. Then G=C is a subgroup of a Sylow 2-subgroup of Aut.A/ Š S4 , which is isomorphic to D8 . It follows that G=C Š E4 , and we conclude that G D A C . Since jG 0 j D 4, C is nonabelian. Let B < C be minimal nonabelian; then, by the above, B Š Q8 . If A \ B > ¹1º, then A B is extraspecial so it contains a subgroup Š D8 (Appendix 16), contrary to what has been proved above. Thus we get A \ B D ¹1º. If U < A0 B 0 D G 0 is of order 2 and A0 ¤ U ¤ B 0 , then AB=U is an extraspecial subgroup of order 25 of the Dedekindian group G=U , a ﬁnal contradiction. The presented proof of Lemma 96.2 is longer than the original one. Our aim here was to show an application of minimal nonabelian subgroups. The following lemma is a partial case of Theorem 96.7. Lemma 96.3. Let G be a p-group with jG 0 j D p and nc.G/ D 1. Then G Š Mpn . Proof. We use induction on jGj. In view of Lemma 96.1(c), one may assume that G is not minimal nonabelian. Let H < G be minimal nonabelian. Then G D H CG .H /

96

Groups with at most two conjugate classes of nonnormal subgroups

15

(Lemma 4.3) so CG .H / 6 H . Since nc.H / nc.G/ D 1 (see the decomposition of G in the previous sentence), we get H 2 ¹Q8 ; Mpn º by induction. (i) H Š Q8 . Since G is non-Dedekindian, we get exp.CG .H // > 2. Suppose that L D hxi CG .H / is cyclic of order 4. If L\H D ¹1º and hai, hbi are distinct cyclic subgroups of order 4 in H , then the nonnormal subgroups haxi, hbxi do not belong to the same G-class1 , and this is a contradiction. Now assume that L \ H D Z.H /. Then the central product L H (of order 16) has exactly six nonnormal subgroups of order 2, and these subgroups do not belong to the same conjugate class since 6 is not a power of 2 (note that HL G G since G 0 < H < HL), a contradiction. Thus G has no subgroups Š Q8 . (ii) Let H Š Mpn . Assume that there is in G a subgroup L of order p with L 6 H . Since H has a nonnormal subgroup B of order p, nc.G/ D 1 and H GG, we get LGG. Then HL D H L, and this subgroup has exactly p 2 nonnormal subgroups of order p which do not belong to the same G-class (indeed, every nonnormal subgroup of G of order p has exactly p conjugates). We get a contradiction since HLGG. Thus we have 1 .G/ D 1 .H / Š Ep2 . If p > 2, then, by Theorem 13.7 or Theorem 69.4, G is metacyclic. By Lemma 65.2(a), since jG 0 j D p, the group G is minimal nonabelian, contrary to the assumption. Now let p D 2. By Lemma 65.2(a), G is nonmetacyclic (see the previous paragraph). Therefore, by Theorem 66.1, 2 .G/ 6 H . It follows that there is a cyclic L < G of order 4 such that L 6 H . Then L G G by hypothesis so L \ H D H 0 (recall that Z.H / is cyclic). Write F D L2 .H /. Then F=H 0 Š E8 . Since j1 .F /j D 4, it follows that F is nonabelian. Then it contains a minimal nonabelian subgroup of order 8, contrary to what has been proved above. If G is nilpotent and nc.G/ D 1, then, as is easily seen, G is primary. Therefore, in Theorem 96.4 we consider p-groups. Theorem 96.4 (O. Schmidt [Sch3]). Let G be a p-group. If nc.G/ D 1, then we have G Š MpnC1 . Proof. We proceed by induction on G. In view of Lemma 96.3, one may assume that jG 0 j > p; then G is neither Dedekindian nor minimal nonabelian. Therefore, by Lemma 96.2, there is in G 0 a G-invariant subgroup R of order p such that G=R is not Dedekindian. By induction, G=R Š Mpn . Let U=R and V =R be two distinct cyclic subgroups of index p in G=R. Then U; V 2 1 are distinct abelian so U \ V D Z.G/. Since R < G 0 ˆ.G/, we get d.G/ D d.G=R/ D 2 so G is minimal nonabelian (Lemma 65.2(a)), a ﬁnal contradiction. Below we freely use properties of minimal nonnilpotent groups which are listed in Theorem A.22.1. 1 We

have haxib D ha1 xi 62 ¹haxi; hbxiº, and the same is true for hbxia .

16

Groups of prime power order

Supplement to Theorem 96.4 (O. Schmidt [Sch3]). Let G be a nonnilpotent group. Then nc.G/ D 1 if and only if G is minimal nonabelian with derived subgroup of prime order. Proof. Let H G be minimal nonnilpotent. Then H D PQ, where P 2 Sylp .G/ is cyclic and H 0 D Q 2 Sylq .G/, p ¤ q. In this case, all nonnormal subgroups of G are conjugate with P . It follows that P is maximal in H and all subgroups of Q are normal in G so jQj D q. Since NG .P / is nonnormal in G, we get NG .P / D P so H D G as H is normal in G and, by Frattini’s lemma, G D H NG .P / D HP D H . Next we classify the p-groups G with nc.G/ D 2. Lemma 96.5. If G D ha; bi is a minimal nonabelian p-group and the number of conjugate classes of nonnormal subgroups of G is equal to nc.G/ D 2, then p D 2 and m m1 G 2 ¹D8 ; H2;m ; m 2º. Here H2;m D ha; b j a2 D b 4 D 1; ab D a1C2 i. Proof. Let G be as in (MA1) or (MA2). Write A D hai, B D hbi. (i) Let G be nonmetacyclic as in (MA1). The subgroups A and B are neither G-invariant nor conjugate. The normal closures AG D A G 0 , B G D B G 0 are of order pjAj, pjBj, respectively. Suppose that m > 1. Then H D B1 .G/ D B E, where E Š Ep2 . Let L < E be of order p such that B L ¤ B G 0 . Then B L is not normal in G (otherwise, B D .B L/ \ .B G 0 / G G), and we get nc.G/ > 2, a contradiction. Similarly, we consider the case n D 1 so that jGj D p 3 . But D8 is metacyclic and nc.S.p 3 // D p > 2 (recall that S.p 3 / is nonabelian of order p 3 and exponent p > 2). (ii) Now let G be metacyclic as in (MA2); n > 1 (otherwise, G Š MpmC1 and nc.G/ D 1). Since G 0 < 1 .G/ Š Ep2 , every nonnormal subgroup of G being abelian is cyclic. If H < G is cyclic of order p n such that G 0 6 H , then G D H A is a semidirect product so H is not normal in G. (ii1) Assume that n > m. Then B1 D habi of order p n is neither a-invariant (otherwise, G D haiB1 would be abelian) nor conjugate with B since B1 6 B G D B G 0 . Suppose that m .B/ D hyi; then A1 D hayi of order p m is not b-invariant and jA1 j < jBj D jB1 j so nc.G/ 3, a contradiction. (ii2) Let n D m. If U=G 0 < G=G 0 is cyclic of order p m , then U D G 0 V so the cyclic subgroup V is not normal in G since G 0 6 V so that A \ V D ¹1º. There are in G=G 0 exactly p 2m1 p 2m2 D p m1 p m1 .p 1/ distinct cyclic subgroups Ui =G 0 of order p m . If Ui D G 0 Vi , then the subgroups V1 ; : : : ; Vpm1 that have trivial intersections with A are not G-invariant and pairwise nonconjugate in G since all Ui =G 0 G G=G 0 are distinct. It follows that p m1 2 so p D 2 D m; then G Š H2;2 .

96

Groups with at most two conjugate classes of nonnormal subgroups

17

(ii3) Let n < m. As we have noticed, n > 1. The subgroup n .G/ is abelian of type .p n ; p n / so it contains exactly p 2n p 2n2 D p n1 .p C 1/ p n1 .p 1/ cyclic subgroups of order p n and exactly p n1 of them contain G 0 . It follows that G has exactly p n1 .p C1/p n1 D p n nonnormal cyclic subgroups of order p n which are partitioned in p n1 classes of conjugate subgroups. We have nc.G/ D 2 p n1 so n D 2 and p D 2, and we get G D H2;m . n1

n2

Let G D M C , where M D ha; b j ap D b p D 1; ab D a1Cp i Š Mpn n2 and C D hci. Set D ap . Then Z.G/ D h i hci contains exactly p C 1 subgroups of order p and c1 .G/ D p 2 C p C 1. Therefore exactly c1 .G/ .p C 1/ D p 2 subgroups of order p of G are non-G-invariant so they belong to p distinct G-classes. Also the subgroup hbihci of order p 2 is non-G-invariant so that nc.G/ 1Cp > 2. Lemma 96.6. If G is a p-group such that jG 0 j D p and nc.G/ D 2, then p D 2 and G 2 ¹D8 ; H2;m º. Proof. In view of Lemmas 96.3 and 96.5, one may assume that G is not minimal nonabelian. Let A < G be minimal nonabelian; then A G G since A0 D G 0 in view of jA0 j D p D jG 0 j, so G D A C , where C D CG .A/ (Lemma 4.3). By assumption, we have C 6 A. Two subgroups of A are G-conjugate if and only if they are A-conjugate. Therefore nc.A/ nc.G/ D 2 so A 2 ¹Q8 ; Mpn ; D8 ; H2;m º by Lemma 96.3 and Theorem 96.5. (i) Let A Š D8 . Then A contains two nonnormal subgroups L1 ; L2 of order 2 which are not conjugate in A, and so neither in G. If L of order 2 is such that L 6 A (L exists by Theorem 1.17(b) since G is not of maximal class), then L G G; then L3 D L1 L is not normal in G since A \ L3 D L1 , and we get nc.G/ > 2, a contradiction. Thus G has no subgroup Š D8 . (ii) Let A Š Q8 . Since G is not Dedekindian, exp.C / > 2. Let hli D L C be cyclic of order 4. If A \ L D Z.A/, then A L Š D8 L contains a subgroup Š D8 , contrary to (i). Thus A \ L D ¹1º hence H D AL D A L. Let hai, hbi and hci be distinct (cyclic) subgroups of order 4 in A. Then the subgroups hali, hbli are not c-invariant and hcli is not a-invariant; since these three subgroups are not G-conjugate in pairs, we get nc.G/ > 2, a contradiction. Thus G has no subgroup Š Q8 . (iii) Suppose that m

A D ha; b j a4 D b 2 D 1; ab D a3 i Š H2;m : Then B D hbi and B1 D hba2 i are neither conjugate nor A-invariant so are not G-invariant. Assume that there is L < G of order 2 such that L 6 A. Then L < Z.G/. Since B L and B1 L are not G-invariant (for example, A \ .B L/ D B is not A-invariant), we get nc.G/ > 2. Thus L does not exist so 1 .G/ D 1 .A/ Š E4 .

18

Groups of prime power order

Since C 6 A and A \ C D Z.A/ D ˆ.A/, it follows that d.G/ > 2 so G is nonmetacyclic. Let M < G be minimal nonmetacyclic. Since j1 .M /j j1 .G/j D 4 and all minimal nonabelian subgroups of G have order > 8, we obtain that jM j D 25 (Lemma 66.1(d)). Because of jM 0 j D 4 > 2 D jG 0 j, we get a contradiction. Next we assume that G has no subgroup Š H2;m . (iv) Now suppose that A is as in (MA2) with n D 1; then A Š MpmC1 , and B D hbi is not G-invariant. Let L < G be a subgroup of order p not contained in A. Write H D L1 .A/; then jH j D p 3 . By (i), we have exp.H / D p. Assume that H is nonabelian. Then p 2 C p noncentral subgroups of H of order p belong to p C 1 distinct G-classes so nc.G/ p C 1 > 2, a contradiction. Now assume that H Š Ep3 . Then we get jH \ Z.G/j 2 ¹p; p 2 º. In the ﬁrst case, we obtain a contradiction as above. Let jH \ Z.G/j D p 2 . Then p 2 subgroups of H of order p that are not contained in Z.G/ belong to p distinct G-classes, and we conclude that p D 2. As H 6 Z.G/, it contains also a non-G-invariant subgroup of order 4; then nc.G/ 2 C 1 > 2, a contradiction. Thus 1 .G/ D 1 .A/ Š Ep2 . Now it follows from (i) and (ii) that C is cyclic. Since A \ C D Z.A/ D Ã1 .A/, we get Ã1 .A/ < C . Let C0 C be (cyclic) of order o.a/ (note that hap i < C and C is cyclic). Write F D A0 C0 . Then F D A0 K, where jKj D p (basic theorem on abelian p-groups). It follows from F 6 A that K 6 A. Since jKj D p, we get a contradiction to the result of the previous paragraph. Theorem 96.7 (O. Schmidt [Sch4]). If a group G is nilpotent and nc.G/ D 2, then either p D 2 and G 2 ¹D8 ; Q16 ; H2;m º or G Š Mpn Cq , where q ¤ p are any primes. Proof. It is easily checked that the groups from the conclusion satisfy the hypothesis. Let G D P Q, where P 2 Sylp .G/ with nc.P / > 0 and Q > ¹1º is a p 0 -group. If ¹1º L Q and U < P is not P -invariant, then L1 D U L is not G-invariant since P \ L1 D U . It follows that jQj D q, a prime, and nc.P / D 1 so P Š Mpn by Theorem 96.4. If G D P Q is a p-group, Q > ¹1º, then nc.G/ > 2 (Lemma 96.6). Next we assume that G is a p-group; then G has no nontrivial direct factors by the previous paragraph. If G has a cyclic subgroup of index p, then G 2 ¹D8 ; Q16 º (note that nc.SD16 / D 3). Next we assume that G has no cyclic subgroup of index p. In view of Lemma 96.6, one may assume that jG 0 j > p. By Lemma 96.2, there exists in G 0 a G-invariant subgroup R of index p such that G=R is not Dedekindian. Then we have j.G=R/0 j D p and 1 nc.G=R/ nc.G/ D 2 so, by Theorem 96.4 and Lemma 96.6, G=R 2 ¹H2;m .p D 2/; Mpn ; D8 º. Since R < G 0 ˆ.G/, we get d.G/ D d.G=R/ D 2. Below we consider these three possibilities. (i) If G=R Š D8 , then jG W G 0 j D 4 so G is of maximal class by Taussky’s theorem, a contradiction since, by assumption, G has no cyclic subgroup of index 2. (ii) Let p D 2 and G=R Š H2;m , m 2. To obtain a contradiction, one may assume that jRj D 2. Then G=R has two nonnormal and not conjugate subgroups S1 =R and S2 =R of order 2m ; then S1 and S2 are not conjugate in G. Since, in view of nc.G/ D 2, all subgroups of order 2m are normal in G, S1 and S2 are cyclic of or-

96

Groups with at most two conjugate classes of nonnormal subgroups

19

der 2mC1 . As GN D G=.S1 /G Š D8 has exactly four nonnormal subgroups of order 2, and among of them SN1 and SN2 , the inverse images in G of these four subgroups must be cyclic. It follows that Z D .S1 /G D Z.G/ (if Z < Z.G/, then jG W Z.G/j D 4 hence jG 0 j D 2 by Lemma 1.1, a contradiction). Let S1 < M 2 1 . Then S1 is cyclic of index 2 in M so M 2 ¹C2 C2m ; M2mC1 º (Theorem 1.2). In the second case, M has a nonnormal subgroup of order 2 < 2mC1 so nc.G/ > 2, a contradiction. Thus M is abelian. Similarly, the remaining two members of the set 1 are abelian; then G is minimal nonabelian since d.G/ D 2. We have jG 0 j D 2, and this is a contradiction. (iii) Let G=R Š Mpn . To obtain a contradiction, it sufﬁces to assume that jRj D p. Let A=R and B=R be two distinct cyclic subgroups of index p in G=R. Then A and B are distinct abelian maximal subgroups of G so A \ B D Z.G/ has index p 2 in G and G is minimal nonabelian; then jG 0 j D p, contrary to the assumption. The 2-groups H2;m are not presented in [Sch4] so Schmidt’s proof is not full. O. Schmidt [Sch4] also classiﬁed the nonnilpotent groups G with nc.G/ D 2. His proof is very long and not full (at ﬁrst, this fact was noticed in [SU]). This question is the subject of the recent paper [M] as well. For related results see also [PR]. Below we offer another approach to prove that theorem. In the nonnilpotent case we show an essentially stronger result, namely, we classify the nonnilpotent groups G with no.G/ D 2, where no.G/ is the number of orders of non-G-invariant subgroups in G. Write .G/ D ¹jH j j H < G is nonnormal in Gº so that j.G/j D no.G/. (MNN) (O. Schmidt, Y. Gol’fand; see Lemma A.22.1) Let G be a minimal nonnilpotent group. Then jGj D p a q ˇ , where p and q are distinct prime numbers. Let P 2 Sylp .G/, Q 2 Sylq .G/. Then the following statements hold: (a) One of the subgroups P , Q (say Q) is normal and coincides with G 0 . (b) If Q is abelian, then Q \ Z.G/ D ¹1º and exp.Q/ D p. (c) P is cyclic and jP W .P \ Z.G//j D p. (d) Let Q be nonabelian. Then Q is special (i.e., Z.Q/ D Q0 D ˆ.Q/ is elementary abelian) and jQ=Q0 j D q b , where b is the order of q .mod p/, Q \ Z.G/ D Z.Q/. If L < Z.Q/ is of index p, then Q=L is extraspecial so that b is even. A group from (MNN) is said to be an S.p; q/-group. In Theorem 96.8, p; q are distinct primes. Theorem 96.8. Let G be a nonnilpotent group such that no.G/ D 2. Then G is solvable and one of the following holds: (a) G D H C , where either (a1) H is minimal nonabelian of order p a q and jC j D r ¤ q, .G/ D ¹p a; p a rº, or (a2) H is nonabelian of order pq and jC j D p, .G/ D ¹p; p 2 º.

20

Groups of prime power order

In what follows, G D P Q, where P 2 Sylp .G/ is non-G-invariant, Q 2 Sylq .G/ and Q G G. (b) G is minimal nonabelian of order pq 2 , .G/ D ¹p; qº. (c) We have jQj D q, P is cyclic of order p a , a > 1, G=Z.G/ is a Frobenius group of order p 2 q, .G/ D ¹p a ; p a1 º. (d) One has G D P Q with Q of order q 2 and cyclic P of order p a , Ã1 .P / D Z.G/, G=Z.G/ is a Frobenius group of order pq 2 , all subgroups of Q are G-invariant, .G/ D ¹p a ; p a qº. (e) One has P Š Q8 , jQj D q, G has exactly one cyclic subgroup of index 2 and .G/ D ¹4; 8º. Proof. By Burnside’s two-prime theorem, G is solvable since it has at most two nonnormal Sylow subgroups of distinct orders. Let H < G. Then no.H / no.G/ D 2. If, in addition, H is not G-invariant, then even no.H / 1. Suppose that G D P Q is an S.p; q/-group as in (MNN). Then jQj > q (otherwise, no.G/ D 1) and P is not G-invariant. If Q0 > ¹1º, then PK, where ¹1º < K Q0 , is not normal in G so jQ0 j q. Assume that jQ0 j D q; then jQ=Q0 j q 2 and PQ0 is not normal in G. If Q0 < Q1 < Q, then Q1 is not normal in G so no.G/ > 2, a contradiction. Thus Q0 D ¹1º so Q is elementary abelian and G is minimal nonabelian. Since every nontrivial subgroup of Q is not normal in G, we get jQj D q 2 . Assume that jP j > p. If ¹1º < Q1 < Q, then Ã1 .P /Q1 is not normal in G so no.G/ > 2, a contradiction. Thus G is an minimal nonabelian subgroup of order pq 2 as in (b). In what follows we assume that G is not minimal nonnilpotent. Next let P Q D H < G be an S.p; q/-subgroup as in (MNN). Then no.H / 2 hence H is minimal nonabelian by the above, so either no.H / D 1 with jQj D q (Supplement to Theorem 96.4) or no.H / D 2 and H is of order pq 2 , H G G. Suppose that r 2 .G/ .H / and let R 2 Sylr .G/. Since G is solvable, one may assume that PR < G. If R is not normal in G, then P , R and PR are not normal in G and have different orders hence no.G/ > 2, a contradiction. Thus R G G. If R1 < R is G-invariant, then PR1 is not normal in G since PR1 \ H D P . We conclude that R is a minimal normal subgroup of G. Since P and RP are not normal in G, we get H G G so G D H R, and we conclude that jRj D r. Thus G is as in (a). In what follows we assume that .G/ D .H / D ¹p; qº, where H here and below is a proper S.p; q/-subgroup of G chosen above. (i) Suppose that H is not normal in G; then no.H / D 1 so H 0 D Q G G is of order q and Ã1 .P / G G. All subgroups of G of order ¤ jP j; jH j are G-invariant hence P 2 Sylp .G/ and .G/ D ¹jP j; jP jqº. Then NG .P / is not G-invariant and Q0 2 Sylq .G/ is normal in the group G since jQ0 j > jQj D q in view of H < G. Next, jNG .P /j 2 ¹jP j; jP jqº. (i1) Assume that Q0 is noncyclic of order q c > q 2 . Then there exist in Q0 two distinct maximal subgroups Q1 and Q2 (of order q c1 > q D jQj). In this case,

96

Groups with at most two conjugate classes of nonnormal subgroups

21

Q1 ; Q2 G G and PQ1 and PQ2 are distinct subgroups of G of order jP jq c1 > jH j so these subgroups are G-invariant. It follows that P L D PQ1 \ PQ2 G G and jG W Lj D q 2 . By Frattini’s lemma, we obtain G D LNG .P / hence P < NG .P /, jNG .P /j > jH j so that no.G/ > 2 since NG .P / is nonnormal in G, a contradiction. (i2) Let Q0 be cyclic of order q c > q 2 ; then NG .P / is abelian so NG .P / D P since Q is a unique subgroup of order q in G. If Q1 < Q0 is of index q in Q, then PQ1 D NG .P /Q1 G G, a contradiction since NG .P / > PQ1 . Thus jQ0 j D q 2 . By hypothesis, all subgroups of order q are G-invariant. In this case, .G/ D ¹jP j; jP jqjº. (i3) Assume that P < NG .P /. Then NG .P / D P Q1 , where jQ1 j D q, Q1 G G and CG .Q1 / PQ0 D G so G D H Q1 , and hence H G G, contrary to the assumption. Thus NG .P / D P . We have Ã1 .P / G G so Ã1 .P / D Z.G/, and we conclude that G=Ã1 .P / is a Frobenius group of order pq 2 . As we have noticed, all subgroups of Q0 .2 Sylq .G// are G-invariant; then G is as in (d). (ii) Now suppose that H G G; then jH j 2 ¹p a q; pq 2 º. (ii1) Assume that jH j D pq 2 is minimal nonabelian with jH 0 j D q 2 ; then we have no.H / D 2 D no.G/ so .G/ D .H / D ¹p; qº and P 2 Sylp .G/ since a Sylow p-subgroup of G is not G-invariant (consider its intersection with H ). By Frattini’s lemma, G D NG .P /H D NG .P /PQ D NG .P / Q so P < NG .P /. Since NG .P / is not normal in G and jNG .P /j 62 ¹p; qº D .G/, we get a contradiction. (ii2) Let jH j D p a q and p j jG W H j. If P < P0 2 Sylp .G/, then P0 is not normal in G and jP0 W P j D p (otherwise, every subgroup lying between P and P0 is not G-invariant so no.G/ > 2); then .G/ D ¹jP j; jP0 jº D ¹jP j; pjP jº. Next, NG .P / is not normal in G since NG .P / \ H D P . It follows that NG .P / D P0 is of order pjP j D p aC1 . On the other hand, G D NG .P /H D NG .P / Q D P0 Q (Frattini’s lemma). It follows that jGj D pjH j D p aC1 q. If P0 is cyclic, then Ã2 .P / D Z.G/ and G=Z.G/ is a Frobenius group of order p 2 q so G is as in (c). In what follows we assume that P0 is noncyclic. There is L < P0 of order p such that L 6 P . Suppose that L is not normal in G. Then we have p D jLj D jP j so P0 Š Ep2 . In this case, G D H CP .Q/, where jCP .Q/j D p, contrary to the assumption. If L G G, then G D H L as above. Then .G/ D ¹jP j; pjP jº is as in (a). Now assume that 1 .P0 / P . In this case, P0 is generalized quaternion since it is noncyclic by assumption. Then P0 Š Q8 (if jP0 j > 8, then no.G/ > 2). Such G has a unique cyclic subgroup CG .Q/ of index 2 so it is as in (e) (indeed, G=CG .Q/ > ¹1º is cyclic of order dividing q 1). We have .G/ D ¹4; 8º. (ii3) Next assume that P 2 Sylp .G/, i.e., (ii2) does not hold. Then G D NG .P /H D NG .P /PQ D NG .P / Q; NG .P / is not normal in G and P < NG .P/. In this case, jNG .P / W P j D q. It follows

22

Groups of prime power order

that Q0 Š Eq 2 , where Q0 2 Sylq .G/, and .G/ D ¹jP j; jP jqº. Since q 62 .G/, every Q1 < Q0 is G-invariant. If Q1 ¤ Q, then G D H Q1 so Q1 Z.G/. It follows that Q0 Z.G/ since Q0 has q C 1 > 2 subgroups of order q, and then G is nilpotent, a ﬁnal contradiction. Corollary 96.9 (O. Schmidt [S2]). If G is a nonnilpotent group with nc.G/ D 2, then one of the following holds: (i) G Š A4 , the alternating group of degree 4. (ii) G D P Q, where P 2 Sylp .G/ and G 0 D Q 2 Sylq .G/ are cyclic satisfying Ã1 .P / D Z.G/, and G=Z.G/ is a Frobenius group of order pq 2 . (iii) G D H C , where H is a minimal nonabelian subgroup of order p a q with jH 0 j D q and jC j D r is a prime different from p and q. (iv) G D P Q (semidirect product with kernel Q), where jQj D q, P is cyclic, jG W CG .Q/j D p 2 and G=Z.G/ is a Frobenius group of order p 2 q. Proof. As 1 no.G/ nc.G/ D 2, G is one of the groups from Theorem 96.4, the Supplement to Theorem 96.4 and Theorem 96.8. Since the nonnilpotent groups G with no.G/ D 1 satisfy nc.G/ D 1 as well, by the Supplement to Theorem 96.4, we need only consider the groups from Theorem 96.8. As in the proof of Theorem 96.8, let H D PQ G be an S.p; q/-subgroup, P 2 Sylp .H / and Q D H 0 2 Sylq .H /. (a) Let G D H C be as in Theorem 96.8(a2), jC j D r ¤ q. If also r ¤ p, then nc.G/ D 2. Now let r D p. Then P0 D P C 2 Sylp .G/ is not G-invariant. Let Sylp .G/ D ¹P0 ; P1 ; : : : ; Pq º. For i D 1; : : : ; q, let Pi D C Si , where Si ¤ Pi \H , and set Hi D Si Q, S0 D P . The subgroups S0 ; : : : ; Sp are not normal in G. The pairwise distinct (by construction) subgroups H D H0 ; H1 ; : : : ; Hq , being direct factors of G, are G-invariant. Since SiG D Hi for all i , we get nc.G/ q C 1 > 2, a contradiction. Thus r ¤ p so G is as in (iii). (b) Let G D PQ be minimal nonabelian of order pq 2 as in Theorem 96.8(b). Since the q C 1 subgroups of order q must be conjugate under P , we get p D q C 1 so q D 2, p D 3 and therefore G Š A4 is as in (i). (c) The group of Theorem 96.8(c) satisﬁes the hypothesis, and G is as in (iv). (d) Let the group G D P Q be as in Theorem 96.8(d). Then, provided Q Š Eq 2 , G does not satisfy the hypothesis. Indeed, if Q1 ; : : : ; QqC1 are all the subgroups of order q in Q, then, for i ¤ j , we have PQi ¤ .PQi /G D Ã1 .P /Qi ¤ .PQj /G D Ã1 .P /Qj ¤ PQj so that the subgroups P; PQ1 ; : : : ; PQqC1 are not G-invariant and not conjugate to each other, i.e., nc.G/ q C 2 > 2, a contradiction. If Q is cyclic of order q 2 , then G satisﬁes the hypothesis, and G is as in (ii). (e) The group from Theorem 96.8(e) does not satisfy the hypothesis. Indeed, if M1 and M2 are distinct nonnilpotent maximal subgroups of G and Pi < Mi \P of order 4,

96

Groups with at most two conjugate classes of nonnormal subgroups

23

i D 1; 2, then P1G D M1 ¤ M2 D P2G so P; P1 ; P2 are neither normal nor conjugate in pairs so nc.G/ > 2. Problem 1. Classify the p-groups G which satisfy no.G/ D 2 (the p-groups G satisfying no.G/ D 1 are classiﬁed in 112). Problem 2. Classify the p-groups G, p > 2, satisfying nc.G/ D p. Problem 3. Classify the p-groups G with exactly two conjugate classes of nonnormal cyclic subgroups.

97

p-groups in which some subgroups are generated by elements of order p

The 2-groups all of whose nonmetacyclic subgroups are generated by involutions are classiﬁed in 84. Corollary 97.4 contains a stronger version of the main theorem from 84 for p > 2. Deﬁnition. A p-group G is said to be an Ln -group if 1 .G/ is of order p n and exponent p and G=1 .G/ is cyclic of order greater than p (see 17, 18). In what follows we consider Ln -groups only for n D p. Exercise. Let G be an Lp -group of exponent p e ; then e > 2 and jGj D p pCe1 . Prove that Ã1 .G/ is cyclic of order p e1 and G D hx 2 G j o.x/ D p e i. Solution. We have p e D exp.G/ exp.Ã1 .G// exp.G=Ã1 .G// D p exp.Ã1 .G// so jÃ1 .G/j exp.Ã1 .G// p e1. Therefore it sufﬁces to show that jÃ1 .G/j p e1. If G is regular, then jÃ1 .G/j D jG=1 .G/j D p e1 . In case that G is irregular, then, by Theorem 9.8(a), we have jG=Ã1 .G/j p p so jÃ1 .G/j p p jGj D p e1 , and we conclude that jÃ1 .G/j D p p1 , as required. Next, exp.e1 .G// D p e1 so hx 2 G j o.x/ D p e i D hG e1 .G/i D G: The main result of this section is the following Theorem 97.1 ([Ber40]). Let a p-group G of exponent p e > p be neither absolutely regular nor of maximal class. Then G contains a subgroup H of order p pCe1 such that j1 .H /j D p p , H=1 .H / is cyclic of order p e1 and 2 .H / is regular (so if e > 2, then H is an Lp -group). Our proof of Theorem 97.1 is based on some consequences of Blackburn’s theory of p-groups of maximal class (see 9, 12, 13; proofs of some of them are reproduced below; see Lemmas 97.J(d), 97.5 and 97.6). If a p-group G is either absolutely regular or of maximal class and order > p pC1 , then it has no subgroup H such as in conclusion of Theorem 97.1 (in the second case

97

p-groups in which some subgroups are generated by elements of order p

25

this follows from the description of members of the set 1 and some other subgroups given in Lemma 97.J(h)). Therefore absolutely regular p-groups and p-groups of maximal class are excluded from the hypothesis of Theorem 97.1. Corollary 97.2. Suppose that a p-group G of exponent greater than p is not absolutely regular. If all proper not absolutely regular subgroups of G are generated by elements of order p, then one and only one of the following holds: (a) jGj D p pC1 and j1 .G/j > p p1 so if G is regular, then j1 .G/j D p p . (b) jGj D p pC1 , cl.G/ D p, j1 .G/j D p p1 (in this case, all proper subgroups of G are absolutely regular). (c) G has maximal class, jGj > p pC1 and every irregular member of the set 1 has two distinct subgroups of order p p and exponent p. Corollary 97.3. Let G be an irregular p-group, p > 2. Suppose that whenever H < G is neither absolutely regular nor of maximal class, then 1 .H / D H . Then G is of maximal class. Corollary 97.4. Let p > 2 and suppose that M is a maximal metacyclic subgroup of a nonmetacyclic p-group G, where jM j > p 2 . Suppose that whenever M < N G and jN W M j D p, then 1 .N / D N . Then p D 3 and G is of maximal class. In particular, if p > 2 and a p-group G of exponent greater than p is nonmetacyclic and such that all proper nonmetacyclic subgroups of G are generated by elements of order p, then one and only one of the following assertions holds: (a) G is regular of order p 4 and j1 .G/j D p 3 . (b) G is of maximal class and order 34 . (c) p D 3, G is of maximal class, jGj > 34 and every irregular member of the set 1 has two distinct (nonabelian) subgroups of order 33 and exponent 3. Note that if a p-group of maximal class, p > 2, has an elementary abelian subgroup of order p p , then G is isomorphic to a Sylow p-subgroup of the symmetric group of degree p 2 (Exercise 9.13). Therefore a group of Corollary 97.4 has no abelian subgroups of order 33 and exponent 3 unless jGj D 34 . In Lemma 97.J we collect some known results which are used in what follows. Lemma 97.J. Let G > ¹1º be a p-group, p > 2. (a) (Theorems 7.1 and 7.2) If G is regular, then exp.1 .G// D p and j1 .G/j D jG=Ã1 .G/j. Absolutely regular p-groups, groups of exponent p and p-groups of class < p are regular. (b) (Exercise 9.1(b)) If G is of maximal class and order p m , then it has exactly one normal subgroup of order p i for all 1 i < m 1. (c) (Proposition 1.8) If H < G is of order p 2 and jCG .H /j D p 2 , then G is of maximal class.

26

Groups of prime power order

(d) (Exercise 13.10(a)) If A < G and all subgroups of G of order pjAj containing A are of maximal class (so that jAj > p), then G is also of maximal class. (e) (Theorem 12.12) Suppose that M < G is an irregular subgroup of maximal class and index p. If G is not of maximal class, then G=Kp .G/ is of order p pC1 and exponent p and 1 D ¹M D M1 ; : : : ; Mp2 ; T1 ; : : : ; TpC1 º, where all Mi ’s are of maximal class and all Tj ’s are not of maximal T class (Ti are also not absopC1 2 lutely regular since jG=Ã1 .G/j p pC1 ), .G/ D SpC1 i D1 Ti1D D has index p in G and G=.G/ is abelian of type .p; p/ so iD1 Ti D G. (f) (Theorem 12.1(b)) If G is neither absolutely regular nor of maximal class and H 2 1 is absolutely regular, then G D H 1 .G/, where 1 .G/ is of order p p and exponent p (in particular, j1 .H /j D p p1 ). (g) (Theorem 13.5) If G is neither absolutely regular nor of maximal class, then the numberep .G/ of subgroups of order p p and exponent p in G is 1 .mod p/. (h) (Theorem 9.6) If G is of maximal class and order > p pC1 , then 1 D ¹G1 ; G2 ; : : : ; GpC1 º; where G1 is absolutely regular with j1 .G1 /j D p p1 (G1 is called the fundamental subgroup of G), and the subgroups G2 ; : : : ; GpC1 are irregular of maximal class. All p-groups of maximal class and order p pC1 are irregular. A p-group G of maximal class and order > p pC1 has no normal subgroup of order p p and exponent p. Assume that this is false, and let R G G be of order p p and exponent p. Then, by Lemma 97.J(b), R ˆ.G/ < G1 , a contradiction since the fundamental subgroup G1 is absolutely regular. If a p-group G satisﬁes exp.1 .G// > p, then it is irregular (Lemma 97.J(a)). To facilitate the proof of Theorem 97.1, we prove the following three assertions which have been proved in previous sections of the book. Lemma 97.5. Suppose that H < G is such that N D NG .H / is of maximal class. Then the p-group G is also of maximal class. Proof. We use induction on jGj. One may assume that N < G; then H is not characteristic in N (otherwise, N D G). In this case, by Lemma 97.J(b), we have jN W H j D p hence jH j > p. As jZ.N /j D p and Z.G/ < N , we get Z.G/ D Z.N / so jZ.G/j D p and Z.G/ ˆ.N / < H . Set GN D G=Z.G/. If jHN j D p, then CGN .HN / D NN is of order p 2 so GN is of maximal class (Lemma 97.J(c)). Now let jHN j > p; then NN is of maximal class so GN is also of maximal class by induction, and we are done since jZ.G/j D p. Lemma 97.6. Suppose that A 2 1 is absolutely regular and M < G is irregular of maximal class. Then G is of maximal class. 1 It

follows from the last equality that exp.Ti / D exp.G/ for some i p C 1.

97

p-groups in which some subgroups are generated by elements of order p

27

Proof. Assume that G is not of maximal class. Then, by Lemma 97.J(d), one can take H G such that M < H G, where jH W M j D p and H is not of maximal class. In this case, by Lemma 97.J(e), H=Kp .H / is of order p pC1 and exponent p. It follows that H has no absolutely regular maximal subgroups. However, A \ H is an absolutely regular maximal subgroup of H , and this is a contradiction. Lemma 97.7. All proper subgroups of an Lp -group G are regular; in particular, the subgroup 2 .G/ is regular. Proof. It sufﬁces to show that all maximal subgroups of an Lp -group G are regular. This is true for p D 2 since then G is either abelian or Š Mpn which is minimal nonabelian. In what follows we assume that p > 2. Take M 2 1 . Suppose that 1 .G/ 6 M ; then 1 .M / D M \1 .G/ is of order p p1 and exponent p. Since M is not of maximal class (indeed, M=1 .M / Š G=1 .G/ is cyclic of order > p), it follows that M is absolutely regular so regular (Lemma 97.J(g,a)). Now let 1 .G/ < M . Let D < 1 .G/ be a G-invariant subgroup of index p 2 . Set C D CG .1 .G/=D/; then jG W C j p. Take in C =1 .G/ a subgroup H=1 .G/ which is maximal in G=1 .G/; then H=D is abelian so cl.H / < p, and hence H is regular (Lemma 97.J(a)). Since G=1 .G/ has only one maximal subgroup, we get H D M . It is clear now that 2 .G/.< G/ is regular. Remark. Suppose that p > 2 and G is an irregular Lp -group of exponent p e . Assume that there is in G a normal cyclic subgroup C of order p e . Set D D C \ 1 .G/; then C 1 .G/ D G and jDj D p by the product formula, and G=D D .C =D/ .1 .G/=D// so d.G/ > 2 (it is important that p > 2). By Lemma 97.7, all proper subgroups of G are regular. It follows that d.G/ D 2, contrary to what has just been proved. Thus all cyclic subgroups of order p e are not normal in G. Proof of Theorem 97.1. First consider the case p D 2. Note that absolutely regular 2-groups are cyclic. Since G is not of maximal class, there is in G a normal abelian subgroup R of type .2; 2/ (Lemma 97.J(g)). Put C D CG .R/; then jG W C j 2 so exp.C / 2e1 . Suppose that exp.C / D 2e ; then there is in C R an element x of order 2e . In this case, A D hx; Ri is abelian of type .2e ; 2/ or .2e ; 2; 2/. In any case, A contains an abelian subgroup H of type .2e ; 2/, and H is the desired subgroup. Now let exp.C / D 2e1 . Take y 2 G C of order 2e . Suppose that U D CG .y/ is cyclic; then CG .U / D U and Z.G/ is cyclic. If e D 2, then G is of maximal class (Lemma 97.J(c)), contrary to the hypothesis. Thus e > 2. In this case, U \ R is of order 2 so H D RU is an L2 -subgroup of exponent 2e . If CG .y/ is noncyclic, then there is an involution x 2 CG .y/ hyi; in this case, H D hx; yi D hxi hyi is abelian of type .2e ; 2/ so H is the desired subgroup. In what follows we assume that p > 2. We use induction on jGj. Take an element x 2 G of order p e . Let x 2 L < G, where L is either absolutely regular or irregular of maximal class such that if L < M G, then M is neither absolutely regular nor of maximal class (L exists by hypothesis and Lemma 97.J(d)). Let L < F G and

28

Groups of prime power order

jF W Lj D p; then F is neither absolutely regular nor of maximal class by the choice of L. Therefore if F < G, then F contains the desired subgroup H . Next assume that F D G; then jG W Lj D p. (i) Let L be absolutely regular. Then, by Lemma 97.J(f), we have G D L1 .G/, where 1 .G/.< G/ is of order p p and exponent p. Set H D hx; 1 .G/i, where x 2 L has order p e ; then 1 .H / D 1 .G/ is of order p p and jH j D p pCe1 by the product formula. If e > 2, then H is an Lp -group so 2 .H / is regular by Lemma 97.7. It remains to show that if e D 2, then H is regular. This is true provided that H D G since, by hypothesis, G of order p pC1 is not of maximal class, so cl.G/ < p (Lemma 97.J(a)). Now let H < G and assume, by way of contradiction, that H is irregular. Then H is of maximal class since jH j D p pC1 (Lemma 97.J(a) again) and, by assumption, L is absolutely regular of index p in G. It follows from Lemma 97.6 that G is of maximal class, contrary to the hypothesis. Thus H is the desired subgroup. (ii) Now let L be irregular of maximal class. If F > L is as in the paragraph preceding (i), then, by Lemma 97.J(e), 1 .F / D ¹L D L1 ; L2 ; : : : ; Lp2 ; T1 ; : : : ; TpC1 º; where L1 ; : : : ; Lp2 are of maximal class and T1 ; : : : ; TpC1 are neither absolutely regSpC1 ular nor of maximal class and G D iD1 Ti . It follows that one of the Ti ’s, say T1 , has exponent p e . Therefore, by induction, there is H T1 of exponent p e such that 1 .H / is of order p p and exponent p, H=1 .H / is cyclic of order p e1 and 2 .H / is regular. Lemma 97.8. Suppose that A is a proper absolutely regular subgroup of a p-group G, exp.A/ > p and whenever A < B G and jB W Aj D p, then 1 .B/ D B. Then G is of maximal class. Proof. By Lemma 97.J(a), B is irregular so G is also irregular. Assume that G is not of maximal class. Let jG W Aj D p; then 1 .G/ D G by hypothesis. However, by Lemma 97.J(f), G D A1 .G/, where j1 .G/j D p p < jGj D j1 .G/j, a contradiction. Now let jG W Aj > p. If A < B < G, where jB W Aj D p, then, by what has just been proved, B is of maximal class so G is also of maximal class by Lemma 97.J(d). Proof of Corollary 97.2. Suppose that G is regular and set L D 1 .G/; then jLj p p since G is not absolutely regular. However, G (of exponent > p) has a maximal subgroup of exponent > p so it is absolutely regular, and this implies jLj D p p . By Lemma 97.J(a), jG W Lj D p (otherwise, if L < M < G, then 1 .M / D L < M , contrary to the hypothesis) so G is as in (a). Next we assume that G is irregular. Since, by hypothesis, G has no regular subgroup H of order p pC1 and exponent p 2 such that 1 .H / < H is of order p p , it follows that G is of maximal class (Theorem 97.1). If all maximal subgroups of G are absolutely regular, then G is as in (b) by Lemma 97.J(h). Obviously, every group of maximal class and order p pC1 satisﬁes the hypothesis. Now let jGj > p pC1 . If M 2 1

97

p-groups in which some subgroups are generated by elements of order p

29

is irregular, then 1 .M / D M by hypothesis. If L D 1 .ˆ.M //, then L is of order p p1 and exponent p (Lemma 97.J(h,a,b)). Take an element x 2 M L of order p and set U1 D hx; Li. Pick an element y 2 M U1 of order p and set U2 D hy; Li. Then U1 and U2 are distinct of order p p and exponent p (Lemma 97.J(a)), and the proof is complete. Proof of Corollary 97.3. Assume that G is not of maximal class. Then there is in G a subgroup H such as in Theorem 97.1. Since H is neither absolutely regular nor of maximal class and 1 .H / ¤ H , we get a contradiction. Proof of Corollary 97.4. Let M and N be as in the statement of the corollary. Then M is absolutely regular since p > 2 so, by Lemma 97.J(f), N is of maximal class. By Lemma 97.J(d), G is also of maximal class. Since p p1 D j1 .M /j p 2 and p > 2, we get p D 3. Here we offer another proof of Theorem 97.1 in the case exp.G/ D p e > p 2 . We have to prove that G contains an Lp -subgroup of order p pCe1 . To this end, we use induction on jGj. By Lemma 97.J(g), there is M G G of order p p and exponent p. Take a cyclic X < G of order p e and set F D MX; then F of exponent p e > p 2 is neither absolutely regular nor of maximal class. Therefore if F < G, the result follows by induction. Now let F D G. Suppose that X \ M > ¹1º; then jGj D p pCe1 . We claim that G is an Lp -group. It sufﬁces to show that 1 .G/ D M . Assume that this is false. Since G=M is cyclic, we conclude that j1 .G/j p pC1 . Assume that j1 .G/j D p pC1 . Then X \ 1 .G/ is cyclic of order p 2 hence 1 .G/ is irregular and we conclude that it is of maximal class (Lemma 97.J(a)). Since the group G is not of maximal class, the number ep .G/ of subgroups of order p p and exponent p in G is 1 .mod p/ (Lemma 97.J(g)), and all these subgroups lie in 1 .G/. As ep .G/ > 1 and d.1 .G// D 2, it follows that all maximal subgroups of 1 .G/ have exponent p so exp.1 .G// D p and 1 .G/ is regular (Lemma 97.J(a)), a contradiction. Thus, in the case under consideration, G is an Lp -group. Now let X \M D ¹1º. Let R < M be a G-invariant subgroup of index p. Set H D RX . Let us show that H is an Lp -group. Indeed, H is not absolutely regular since 1 .X/R is of order p p and exponent p (Lemma 97.J(a)). Next, H=R is cyclic of order p e > p 2 so H is not of maximal class. If K=R < H=R is of order p, then 1 .H / K 1 .H / so 1 .H / D 1 .X/R D K is of order p p and exponent p whence H is an Lp -subgroup. By Lemma 97.7, 2 .H / is regular. Proposition 97.9. Let H be a normal absolutely regular subgroup of a p-group G, jH j > p p1 and 1 .G/ 6 H . (a) If for every z 2 G H of order p the subgroup V D hz; H i is of maximal class, then G is also of maximal class. (b) If for every z 2 G H of order p we have 1 .hz; H i/ D hz; H i, then G is of maximal class.

30

Groups of prime power order

Proof. By hypothesis, exp.H / > p. It follows from Lemma 97.J(a) and Theorem 9.5 that G is irregular. Assume, by way of contradiction, that G is not of maximal class. Let H H0 < G, where H0 is absolutely regular such that jH0 j is as large as possible. By Lemma 97.J(d), H0 < B G, where jB W H0 j D p and B is not of maximal class. Then, by Lemma 97.J(f), we have B D H0 1 .B/, where 1 .B/ is of order p p and exponent p so there exists an element x 2 1 .B/ H0 of order p. Set U D hH; xi; then, by hypothesis, U is of maximal class and order > p p so irregular. We have H0 ; U < B, H0 is absolutely regular of index p in B and U is irregular of maximal class. Therefore, by Lemma 97.6, B is of maximal class, contrary to its choice. The above argument also proves (b). It is possible to change the condition 1 .G/ 6 H in Proposition 97.9 to the following one: G is not absolutely regular. Indeed, assume that 1 .G/ < H . Then, by Lemma 97.J(f), G must be of maximal class.

Problems Below G is a p-group. Problem 1. Classify the p-groups all of whose proper nonabelian subgroups that are nonmetacyclic are generated by elements of order p. Problem 2. Study the p-groups G containing a proper abelian subgroup M of exponent > p such that whenever M < N G and jN W M j D p, then 1 .N / D N (2 .N / D N ). Problem 3. Let A be a subgroup of index > p k in a p-group G. Suppose that all subgroups of G containing A as a subgroup of index p k , are of maximal class. Is it true that G also of maximal class? (Compare this with Lemma 97.J(d).) Problem 4. Study the 2-groups G containing a metacyclic subgroup M and such that whenever M < N G and jN W M j D 2, then d.N / D 2. Problem 5. Study the p-groups G containing a minimal nonabelian subgroup M such that whenever M < N G and jN W M j D p, then 1 .N / D N .

98

Nonabelian 2-groups all of whose minimal nonabelian subgroups are isomorphic to M2nC1 , n 3 ﬁxed

Here we classify completely the title groups. This classiﬁcation for an arbitrary n 3 (ﬁxed) is quite involved and shows that our technique with minimal nonabelian subgroups is efﬁcient. Theorem 98.1. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to M2nC1 , n 3 ﬁxed. Then E D 1 .G/ is elementary abelian. Let A be a maximal normal abelian subgroup of G containing E. Then A D CG .E/, all elements in G A are of order 2n , G=A is elementary abelian of order 4, and if x 2 G A, then jE W CE .x/j D 2, A0 D CA .x/ D CE .x/hx 2 i, A=A0 is cyclic, hA; xi0 is cyclic of order jA=A0 j, hA; xi0 \ A0 D 1 .hxi/, x inverts A=A0 and Ãn2 .A/. We have two possibilities: (a) exp.A/ D 2m 2n in which case jG W Aj D 2 and A is of type .2m; 2n2; 2; : : : ; 2/. (b) exp.A/ 2n1 in which case A is of type .2n1; 2r ; 2; : : : ; 2/ with 1 r n 2 and either jG W Aj D 2 or jG W Aj D 4, where in the last case G=n2 .A/ Š Q8 . Conversely, if G is a 2-group of type (a) or (b), then each minimal nonabelian subgroup of G is isomorphic to M2nC1 , n 3 ﬁxed. Proof. Let G be a title group with n 3 ﬁxed. Since G has no dihedral subgroups, E D 1 .G/ is elementary abelian. Set A D CG .E/ so that A is normal in G. If A is nonabelian, then consider a minimal nonabelian subgroup X Š M2nC1 in A. But then all three involutions in X are central in X , which contradicts the structure of M2nC1 . Hence A is abelian and so A D CG .E/ is a unique maximal abelian subgroup of G containing E. Let x be any element in G A so that o.x/ 4 and set hzi D E \hxi, where z is an involution. Set T D Ehxi and E0 D CE .x/ so that E > E0 and T0 D CT .x/ D E0 hxi is abelian. Let T1 T be such that T1 > T0 and jT1 W T0 j D 2. Set E1 D T1 \ E so that T1 D E1 hxi, jE1 W E0 j D 2 and let e be an involution in E1 E0 . Then we have 1 ¤ Œe; x 2 T0 \ E D E0 and so Œe; x is an involution commuting with e and x and therefore he; xi0 D hŒe; xi which implies that he; xi is minimal nonabelian (Lemma 65.2(a)) and so he; xi Š M2nC1 . In particular, hŒe; xi D 1 .ˆ.he; xi// D hzi,

32

Groups of prime power order

e x D ez and hei normalizes hxi which gives o.x/ D 2n . We have proved that all elements in G A are of order 2n and so G=A is elementary abelian. In particular, x 2 2 A and so x induces an involutory automorphism on E. Hence x centralizes E=E0 and so T0 is normal in T with T =T0 Š E=E0 . Suppose that jE=E0 j > 2 and let f 2 E E1 . Then considering T2 D hf iT0 , we get in the same way as above that hf; xi Š M2nC1 and so f x D f z. But then ef 62 E0 and .ef /x D .ez/.f z/ D ef , a contradiction. We have proved that jE W E0 j D 2 and ŒE; x D hzi D 1 .hxi/. Let t 2 E E0 so that t x D t z and ht; xi Š M2nC1 . Let s be an arbitrary element in A. We may set s x D s 1 w for some w 2 A. Note that o.x/ D 2n , o.sx/ D 2n and t sx D t x D t z so that Œt; sx D z with z 2 Z.Ahxi/ and so ht; sxi Š M2nC1 . n1 n1 In particular, hsxi > hzi and so .sx/2 D .x/2 D z. We compute (noting that x 2 2 A and .sx/2 2 A) w D ss x D s.x 1 sx/ D sx 2 xsx D x 2 .sx/.sx/ D x 2 .sx/2 : Therefore w 2 D x 2 .sx/2 inverts A=n2 .A/. Since n2

n2

.s 2

n1

/x D .s x /2

n2

n1

D zz D 1 and so s x D s 1 w implies that x

D .s 1 w/2

n2

D s 2

n2

D .s 2

n2

/1 ;

we see that x also inverts each element in Ãn2 .A/. We shall determine the structure of A0 D CA .x/, where A0 E0 hx 2 i. Suppose that there is an element v of order 4 in A0 such that v 2 D z 0 ¤ z. Taking an involution t 2 E E0 , we get .vt/2 D z 0 and Œvt; x D Œv; xt Œt; x D Œt; x D z and so hvt; xi is minimal nonabelian. But ˆ.hvt; xi/ hz; z 0 i Š E4 , which contradicts the structure of M2nC1 . Hence A0 is of type .2s0 ; 2; : : : ; 2/ with s0 n 1 and 1 .A0 / D E0 . Suppose that A0 contains an element y of order 2n . Then Ãn2 .hyi/ Š C4 is inverted by x, a contradiction. Hence s0 D n 1 and so A0 D E0 hx 2 i. Also, A1 D EA0 is of type .2n1 ; 2; : : : ; 2/ with Ãn2 .A1 / D hzi. For each l 2 A1 A0 , l 2 2 hx 2 i, Œl; x D z and so hl; xi Š M2nC1 . For each y 2 Ax (x 2 G A), CA .y/ D A0 and so y 2 2 A0 . This implies that x inverts A=A0 . Let A =A0 be a subgroup of order 2 in A=A0 and let a be an element in A A0 . We have .a /x D a s with 1 ¤ s 2 A0 since x inverts A=A0 . Then we get 2 a D .a /x D .a s/x D a s 2 and so s 2 D 1 and s is an involution in E0 D 1 .A0 /. The involution s D Œa ; x commutes with a and x and so ha ; xi is minimal nonabelian. Hence ha ; xi Š M2nC1 with ˆ.ha ; xi/ D hx 2 i and so Œa ; x D s D z. If ha ; x 2 i is cyclic, then we obtain that o.a / D 2n and x inverts Ãn2 .ha i/ Š C4 with Ãn2 .ha i/ A0 , a contradiction. Hence ha ; x 2 i is noncyclic and so i D a y is an involution for a suitable y 2 hx 2 i, where i 2 E E0 . Thus E A and so A D Ehx 2 i D A1 . We have proved that Ehx 2 i=A0 is a unique subgroup of order 2 in A=A0 and so A=A0 is cyclic. The mapping a ! a1 ax (a 2 A) is a homomorphism from A onto hA; xi0 with the kernel A0 D Z.hA; xi/. Thus A=A0 Š hA; xi0 . Since x inverts the cyclic group A=A0 , we have jA W .hA; xi0 A0 /j D 2 and therefore hA; xi0 \ A0 D hzi noting that ŒA1 ; x D hzi.

98

2-groups with minimal nonabelian subgroups isomorphic to M2nC1 , n ﬁxed

33

(i) Suppose that A possesses an element a of order 2n . Then each element x 2 GA inverts Ãn2 .hai/ D hvi Š C4 which implies that jG W Aj D 2, v 62 A0 D CA .x/ and v 2 2 E0 . Since A1 D Ehx 2 i is of type .2n1 ; 2; : : : ; 2/, we have v 2 D z and v 2 A1 A0 . This implies that jA=A0 j 2n1 . Let y be an element in the set A A0 such that hyi covers A=A0 and so o.y/ 2n1 . Suppose that o.y/ D 2n1 in which n2 case jA=A0 j D 2n1 , A splits over A0 and y 2 D e 2 E E0 . But then x inverts Ãn2 .hyi/ D hei and so x centralizes e, contrary to CE .x/ D E0 . It follows that o.y/ D 2m 2n . By the above, we get hyi \ A0 D hzi. Since hyi covers A=A0 and hx 2 i covers A0 =E0 , we get A D Ehyihx 2 i, where hyi \ hx 2 i D hzi. Let y0 2 hyi be of order 2n1 . Then we obtain o.y0 x 2 / D 2n2 , 1 .hy0 x 2 i/] 2 E E0 and hy; x 2 i D hyi hy0 x 2 i so that A is of type .2m ; 2n2 ; 2; : : : ; 2/. We have obtained the groups of type (a) of our theorem. (ii) Suppose that exp.A/ D 2n1 . In that case, hx 2 i is a cyclic subgroup in A of the maximal possible order. Let C be a complement of hx 2 i in A so that A D C hx 2 i. Set C0 D A0 \ C , C1 D A1 \ C so that C1 is elementary abelian since A1 =hx 2 i is elementary abelian. Hence we have A1 D C1 hx 2 i, A0 D C0 hx 2 i, jC1 W C0 j D 2, E D C1 hzi and E0 D C0 hzi. We know that A=A0 Š C =C0 is cyclic of order 2r , r1 r 1. Let c 2 C be such that hci covers C =C0 . Then c 2 D e is an involution in the set C1 C0 so that o.c/ D 2r , r n 1. Suppose that r D n 1. Since x inverts Ãn2 .hci/ D hei, we infer that x centralizes e 62 E0 , a contradiction. We have proved that r n 2 and so A is of type .2n1 ; 2r ; 2; : : : ; 2/, where 1 r n 2. It follows that n2 .A/ D hc; E; x 4 i is of index 2 in A, n2 .A/ is normal in G and all elements in A n2 .A/ are of order 2n1 . Assume, in addition, that G=A is elementary abelian of order 4. Let B=n2 .A/ be a subgroup of order 2 in G=n2 .A/ distinct from A=n2 .A/. But then each element d 2 B n2 .A/ is of order 2n , but d 2 2 n2 .A/ is of order 2n2 , a contradiction. Hence A=n2 .A/ is a unique subgroup of order 2 in G=n2 .A/. Since G=n2 .A/ cannot be cyclic (noting that G=A is elementary abelian of order 4), it follows that G=n2 .A/ is generalized quaternion. But then d.G=n2 .A// D 2 and so jG=Aj D 4 and G=n2 .A/ Š Q8 . We have obtained the groups of type (b) of our theorem. It remains to show that all obtained groups G of type (a) or (b) have the property that each minimal nonabelian subgroup M of G is isomorphic to M2nC1 , n 3 ﬁxed. First suppose that jM W .M \ A/j D 2 and let x 2 M A. Then M \ A 6 A0 D CA .x/. Since A=A0 is cyclic, it follows that .M \ A/ A0 contains an element l 2 A1 A0 , where A1 D Ehx 2 i. But then we know that hl; xi Š M2nC1 and so hl; xi D M . Now suppose that jM W .M \ A/j > 2 in which case jG=Aj D 4 and G=n2 .A/ Š Q8 . In this case, M covers Q=A. But for any m 2 M A we obtain that o.m2 / D 2n1 and so m2 2 A n2 .A/. Thus the subgroup M covers also G=n2 .A/ and so M=.M \ n2 .A// Š Q8 . Hence M 0 covers .M \A/=.M \n2 .A//. But we have jM 0 j D 2 which gives a contradiction since all elements in the set A n2 .A/ are of order 2n1 4. Our theorem is proved.

99

2-groups with sectional rank at most 4

Here we give some results of K. Harada from [Har], where in case of 2-groups of sectional rank at most 4 we simplify some of his proofs. Recall that the sectional rank r.G/ of a p-group G is r.G/ D max¹d.H / j H Gº. The most impressive result is Theorem 99.9 stating that a 2-group which possesses a self-centralizing abelian subgroup of order 8 is of sectional rank at most 4. For related results see 51, where the 2-groups containing a self-centralizing subgroup isomorphic to E8 are studied. Lemma 99.1. Suppose that G D A B is a 2-group, where A Š E8 , A E G, B Š E4 , A \ B D ¹1º and CG .A/ D A. Then one of the following holds: (i) CA .B/ D ŒA; B Š C2 and G Š Q8 Q8 . (ii) CA .B/ D ŒA; B Š E4 and G Š E4 o C2 . Proof. We know that Aut.A/ Š GL.3; 2/ has exactly two conjugacy classes of foursubgroups. Therefore there exist at most two possible structures for A B. Indeed, Q8 Q8 (the extraspecial group of order 25 and “type +”) and E4 o C2 are those two possibilities. In the ﬁrst case, CA .B/ D ŒA; B Š C2 and in the second case, CA .B/ D ŒA; B Š E4 . Lemma 99.2. Let G be a 2-group with an elementary abelian subgroup A of order 16 and index 2. If jZ.G/j D 4, then G Š E4 o C2 . In particular, each involution in Z.G/ has exactly four square roots in G A so G contains exactly six cyclic subgroups of order 4 (it follows that G has exactly 19 involutions). Proof. Let g be an element of order 4 in G A. Then CA .g/ D Z Š E4 and g 2 2 Z which implies Z D Z.G/, ˆ.G/ Z and therefore G 0 Z and so hZ; gi E G. Hence all four conjugates of g in G lie in ¹gZº and so G 0 D ˆ.G/ D Z.G/ and G is special. Let g 0 2 G A be such that .g 0 /2 D g 2 . Then g 0 D ga for some a 2 A and g 2 D .g 0 /2 D .ga/2 D gaga D g 2 ag a which gives ag D a and so a 2 Z and g 0 2 ¹gZº. Hence g 2 2 Z has exactly four square roots and they are all conjugate to g in G. If i is an involution in G A, then we see (as above) that ¹iZº is the set of all involutions in G A and they form a single conjugate class of involutions in G A. It follows that G A must contain exactly four square roots of each of the three involutions in Z. But this gives exactly 12 elements of order 4 in G A and so the remaining four elements in G A must be involutions forming a single conjugate

99 2-groups with sectional rank at most 4

35

class. The structure of G is uniquely determined and so G Š E4 o C2 . Indeed, let t be an involution in G A and let X be a complement of Z in A. Then X \ X t E G so that X \ X t D ¹1º. Lemma 99.3. Let G be a 2-group containing a subgroup R Š Q8 Q8 of order 25 . If CG .R/ R, then r.G/ 4. Proof. Set R D Q1 Q2 , where Q1 and Q2 are the only two quaternion subgroups of R and Q1 \ Q2 D Z.R/ D hzi. Let U be a normal four-subgroup of G and assume that U 6 R. Then U \ R D hzi, j.RU / W Rj D 2 so that ŒR; U hzi and so if u 2 U hzi, then Q1u D Q2 is not possible. If u induces an outer automorphism on Qi (i 2 ¹1; 2º), then ŒQi ; u Š C4 , a contradiction. Thus u induces on both Q1 and Q2 inner automorphisms and so u induces an inner automorphism on R, contrary to CG .R/ R. Hence U R. If G does not have a normal elementary abelian subgroup of order 8, then r.G/ 4 (Theorem 50.3 ). Hence we may assume that G possesses a normal elementary abelian subgroup X of order 8. Assume that X 6 R. By the previous paragraph, X \ R D X1 Š E4 , X1 is a normal four-subgroup in G and z 2 X1 . We have j.RX/ W Rj D 2 and so ŒR; X X1 which shows that X cannot interchange Q1 and Q2 . But ŒR; X is elementary abelian and so X induces on both Q1 and Q2 inner automorphisms and so X induces an inner automorphism group on R, a contradiction. We have proved that X R with z 2 X . Set V D NG .R/. Since Q8 Q16 and Q8 SD16 do not possess an elementary abelian normal subgroup of order 8, no element of V R induces an inner automorphism on Q1 or on Q2 . Hence any element of V R must either interchange Q1 and Q2 or induce an outer automorphism on both Q1 and Q2 . Note that a Sylow 2-subgroup of the outer automorphism group of Q8 Q8 is isomorphic to D8 and so we get jV =Rj 4 since X R. Suppose that G > V . Set GN D G=hzi where Z.R/ D hzi D Z.G/. Then VN contains N where a normal elementary abelian subgroup TN of order 16 which is distinct from R, y T D R with an element y 2 G V which normalizes V. By the action of elements N we have C N .tN/ Š E4 for tN 2 TN R. N Therefore TN \ RN Š E4 and of VN RN on R, R N N N T =.T \ R/ Š E4 so that T \ R D X Š E8 since X is normal in G and T =X Š E4 . Therefore there is t 2 T R which induces an outer automorphism on both Q1 and Q2 and so Œt; R is not elementary abelian. On the other hand, Œt; R X and so Œt; R is elementary abelian, a contradiction. Hence we have V D G and so jGj 27 . By the action of the elements of G R on R, we see that r.G/ 4. Indeed, let A=B be a section of G such that A=B Š E25 . If B D ¹1º, then A Š E25 and so A covers G=R Š E4 and so jGj D 27 . Let u be an involution in A R which induces an involutory outer automorphism on both Q1 and Q2 in which case CR .u/ Š E4 . On the other hand, A \ R Š E8 , a contradiction. Hence B ¤ ¹1º. If jBj D 4, then A D G is of order 27 and jˆ.G/j 4. But in this case, there is v 2 G R such that Q1v D Q2 and so jˆ.G/j 8, a contradiction. Hence we may assume jBj D 2 and since G cannot

36

Groups of prime power order

have an elementary abelian subgroup of order 26 , we obtain that ˆ.A/ D B Š C2 . If G D A is of order 26 , then B D Z.G/ D Z.R/ and if x 2 A R, then x centralizes R=Z.R/, a contradiction. Hence we have jGj D 27 , A is a maximal subgroup of G with ˆ.A/ D B Z.G/ and so B D Z.R/ D Z.G/ D hzi. Since A > R is obviously not possible (as before), we conclude that A covers G=R Š E4 . But then considering again GN D G=hzi, we have for elements tN 2 AN RN that tN centralizes a subgroup of N contrary to the established fact that C N .tN/ Š E4 . We have proved that order 8 in R, R r.G/ 4. Lemma 99.4. Suppose that G is a 2-group which contains a subgroup R Š E4 o C2 . If CG .Z.R// D R, then r.G/ 4. Proof. Since r.R/ 4, we may assume R < G. Let A be a unique elementary abelian subgroup of order 16 in R and let D be a subgroup of G containing R satisfying jD W Rj D 2. Since A char R, we have A E D. Also, CG .A/ CG .Z.R// D R and so CG .A/ D A. We study V D NG .A/ and set VN D V =A so that VN is a subgroup of GL.4; 2/. Let a be an involution in R A and set XN D CVN .a/ N so that X D. Then X normalizes R and so Z.R/ E X. But CG .Z.R// D R and so X D D. Thus jCVN .a/j N D 4 which implies (noting that exp.VN / 4) VN Š C4 ; E4 or D8 . Suppose that N V Š C4 which gives V D D. But then A char D which gives D D G. If in this case H is a subgroup of G containing A, then either H D R and d.H / 4 or H D G and ˆ.H / Z.R/ D R0 and so d.H / 4. If H 6 A, then d.H \ A/ 3 and H=.H \ A/ is cyclic and so d.H / 4. It follows that we may assume that VN Š E4 or VN Š D8 and so in both cases D=A Š E4 since jCVN .a/j N D 4. Since R A has exactly four involutions and they are all conjugate in R to a, we conclude that T D CD .a/ covers D=R and T \ R D hai Z.R/. If x 2 T R, then T =Z.R/ Š E4 implies that x 2 2 Z.R/ and so hx; Z.R/i Š D8 . Therefore there is an involution t 2 T R so that ha; ti Š E4 and CZ.R/ .t/ D hzi Š C2 which gives Z.D/ D hzi and D splits over A. Let t 0 2 ¹t; taº so that CZ.R/ .t 0 / D hzi and suppose that jCA .t 0 /j D 8 in which case CA .t 0 / covers A=Z.R/. But then CA .t 0 / is hai-admissible and CCA .t 0 / .a/ > hzi, a contradiction. It follows that CA .t 0 / Š E4 so that by Lemma 99.2 we have hA; t 0 i Š E4 o C2 and this together with Z.D/ D hzi gives A char D because A is a unique elementary abelian subgroup of order 16 in D. Now set Q D ha; tihCA .a/; CA .t/i, where the four-group ha; ti acts faithfully on E D CA .a/CA .t/ Š E8 and Z.Q/ D hzi. By Lemma 99.1, we have Q Š Q8 Q8 . Suppose that CG .Q/ 6 Q. Then, as A normalizes CG .Q/, we can ﬁnd a subgroup B in CG .Q/ of order 4 with B > Z.Q/ D hzi and A normalizes B. It follows that ŒA; B Z.Q/ D Z.D/ D hzi and so B normalizes A. On the other hand, B also centralizes Q and so B normalizes R D Ahai, where a 2 Q. By the above, CVN .a/ N D DN and so B D. This is a contradiction since CD .Q/ D Z.Q/ D hzi. We have proved that CG .Q/ Q. From Lemma 99.3 it follows that r.G/ 4. Lemma 99.5. Let A Š C8 be self-centralizing in a 2-group G. Then r.G/ 4.

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Proof. We set A D hai and a4 D c. Assume that G possesses a normal elementary abelian subgroup E of order 8 (because otherwise, Theorem 50.3 implies r.G/ 4). We set E D hc; x; yi, where E \ Z.G/ D hci and Z.G/ < A. If Œa2 ; E D 1, then a would centralize a four-subgroup of E, contrary to CG .A/ D A. Hence Œa2 ; E ¤ 1 and so Z.G/ D hci. Since GL.3; 2/ has only one class of elements of order 4, we may assume that x a D xc and y a D yx. This yields that ax D ac D a5 and ay D ax and the structure of AE is uniquely determined. In particular, hA; xi Š M16 has exactly two cyclic subgroups of order 8: A D hai and haxi D Ay . We obtain .ay/2 D a2 x and AE has only two further cyclic subgroups of order 8: hayi and hay xi, where .ay/y D ay x. Also, 2 .AE/ D Eha2 i D hxi ha2 y; yi Š C2 D8 . We know that Aut.A/ Š E4 and we have NAE .A/ D hA; xi since y 62 NAE .A/. Moreover, hc; xi is a unique normal four-subgroup of G which is contained in E. Since exp.AE/ D 8, we may assume that jG W .AE/j 4 (because otherwise, obviously r.G/ 4). Assume that A1 D NG .A/ D hA; xi is of order 16. Then A2 D NG .A1 / D AE and set A3 D NG .A2 / so that jA3 W A2 j D 2. Since 2 .A2 / char A2 , we get 2 .A2 / D Eha2 i E A3 . If CA3 .E/ D E, then A3 =E Š D8 and so C4 Š A2 =E char A3 =E which implies that an element in G A3 normalizing A3 also normalizes E and A2 , contrary to NG .A2 / D A3 . Hence CA3 .E/=E Š C2 so that A3 =E Š C4 C2 is abelian. Set E D CA3 .E/ so that E is abelian of type .2; 2; 2; 2/ or .4; 2; 2/. Since hai normalizes E and CE .a/ D hci, it follows that Ã1 .E / hci. Take an element t 2 E E. Then t centralizes E and t 2 2 hci. We have ˆ.A2 / D ha2 ; xi and A3 =Eha2 i Š E4 and so ˆ.A3 / D Eha2 i since our element t 2 A3 A2 fuses hai to hayi or hayxi. There are three maximal subgroups of A3 (containing Eha2 i): A2 , Eha2 ; ati and Eha2 ; ti. Since .Eha2 i/=E and .Eht i/=E are two distinct subgroups of order 2 in .Eha2 ; ti/=E, it follows that .Eha2 ; ti/=E Š E4 which gives that exp.Eha2 ; ti/ D 4. As all elements in A2 hE; a2 i are of order 8, Al2 D hE; a2 ; ati for some l 2 GA3 normalizing A3 . Hence all elements in hE; a2 ; atihE; a2 i are of order 8 and, in particular, the element at is of order 8. Since A3 D A2 E , A2 E A3 , E E A3 and A2 \ E D E, we have Œa; t 2 A2 \ E D E, where t 2 2 hci D Z.G/ and E is abelian. We get a1 t 1 at D e 2 E and so at D ae. On the other hand, t 2 A3 A2 and so at 62 A1 D hA; xi which implies that e D yc i x j , i; j D 0; 1. We compute .at/2 D atat D at 2 .t 1 at/ D at 2 ayc i x j D a2 t 2 yc i x j and so .at/4 D a2 t 2 yc i x j a2 t 2 yc i x j D a2 ya2 x j yx j D a2 ya2 y D a2 y ya2 c D a4 c D cc D 1; a contradiction. It remains to treat the case where A1 D NG .A/ is of order 32. We have A1 =A Š E4 and x 2 A1 . Choose an element t 2 A1 hA; xi such that at D a1 . Then we see that t satisﬁes t 2 2 CG .A/ D A and so t 2 2 hci. It follows that hA; ti Š D16 or Q16

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Groups of prime power order

and hA; txi Š SD16 with .tx/2 2 hci. Also, A is a unique cyclic subgroup of order 8 in hA; ti and in hA; txi. Thus A1 contains exactly two cyclic subgroups of order 8: hai and haxi D Ay . This forces that jA2 W A1 j D 2, where A2 D NG .A1 / and so A2 D A1 E. Since t inverts A, A2 =E Š D8 . Moreover, CA2 .E/ D E as ŒE; a2 ¤ 1. By the structure of GL.3; 2/, A2 =E contains a four-subgroup D=E with CE .D/ Š C2 and so CE .D/ D hci. We have D D Eha2 ; ti or D D Eha2 ; ati, where t 2 2 hci and .at/2 2 hci. Since ŒD; E D hci, we get also ˆ.D/ D hci and so D Š Q8 Q8 . If CG .D/ 6 D, then we can ﬁnd a group V of CG .D/ of order 4 which contains hci and is normalized by A. But then Œa; V hci and so V normalizes A and so V A2 . On the other hand, CA2 .D/ CA2 .E/ D E and so CA2 .D/ D hci, a contradiction. Hence CG .D/ D and then Lemma 99.3 gives r.G/ 4. Lemma 99.6. Let G be a 2-group with a self-centralizing abelian subgroup A of order 8. If r.G/ > 4, then jGj 27 , G has a normal elementary abelian subgroup E of order 8 and A is not normal in G. Proof. Obviously, jGj 26 . Suppose that jGj D 26 . Then G has subgroups H > K with K E H and H=K Š E25 . If K D ¹1º, then H is a maximal subgroup of G and H Š E25 . But then jZ.G/j 8 and Z.G/ A, a contradiction. Hence H D G and K Š C2 so that ˆ.G/ D G 0 D K. Since G cannot be extraspecial, we have Z.G/ > K. But Z.G/ < A and so jZ.G/j D 4. Let a 2 A Z.G/. On the other hand, jG 0 j D 2 implies that jG W CG .a/j D 2 so that CG .A/ > A, a contradiction. We get jGj 27 and Theorem 50.3 implies that G has a normal elementary abelian subgroup E of order 8. Also, an S2 -subgroup of Aut.A/ is of order at most 8 which together with CG .A/ D A implies that A is not normal in G. Lemma 99.7. Suppose that G is a 2-group which contains a self-centralizing subgroup A Š E8 . Then r.G/ 4. Proof. Suppose that r.G/ > 4. Then Lemma 99.6 implies that jGj 27 , A is not normal in G, G possesses a normal elementary abelian subgroup E of order 8 and A ¤ E. (i) We show that jA \ Ej D 4 which implies E NG .A/. Since Z.G/ < A, A \ E is a nontrivial central subgroup of AE. Suppose that A\E Š C2 . Then A contains a four-subgroup A1 with A1 \E D ¹1º. If x 2 A1 is an involution which centralizes E, then taking y 2 A1 hxi we get jCE .y/j 4, contrary to CG .A/ D A. This yields that A1 acts faithfully on E. Since CE .A1 / D A\E Š C2 , Lemma 99.1(i) yields that AE Š Q8 Q8 . But then CG .AE/ CG .A/ D A and so Lemma 99.3 gives a contradiction. (ii) Let B be any elementary abelian subgroup of order 8 in NG .A/. Then we have jA \ Bj 4. As CG .A/ D A, we have A \ B ¤ ¹1º and assume that A \ B Š C2 . Let B1 Š E4 be a complement of A\B in B. Obviously, B1 acts faithfully on A and so Lemma 99.1 implies AB Š Q8 Q8 or AB Š E4 o C2 . Clearly, CG .AB/ CG .A/ D A in both

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cases. By Lemma 99.3, we get AB Š E4 oC2 and let V be a unique elementary abelian subgroup of order 16 in AB so that AB D AV with V \ A D Z D Z.AB/ Š E4 . Assume that CG .Z/ > AB and let T be a subgroup of CG .Z/ containing AB with jT W .AB/j D 2. Then V is normal in T and if a 2 A Z, then .AB/ V contains exactly four involutions and they are all conjugate to a in AB since CAB .a/ D A. Hence CG .a/ covers T =.AB/, contrary to CG .A/ D A. We have proved that CG .Z/ D AB. But then Lemma 99.4 implies that r.G/ 4, a contradiction. (iii) NG .A/=A is not isomorphic to D8 . Suppose that NG .A/=A Š D8 . Since G > NG .A/, NG .A/ contains a conjugate B D At ¤ A of A, where t 62 NG .A/ and .NG .A//t D NG .A/. Since B E NG .A/ and A \ B Š E4 , B maps into the center of NG .A/=A. Hence AE D AB and note that AE Š C2 D8 and so A and E are the only two elementary abelian subgroups of order 8 in AE. Then B D A or B D E and this is a contradiction in both cases. (iv) GN D G=E is isomorphic to D2n , n 3, or to SD2m , m 4. We investigate the structure of GN D G=E. Obviously, NG .AE/ normalizes A since E E G and A and E are the only two elementary abelian subgroups of order 8 in AE. N D 4. This fact together with Hence jNG .AE/=Ej D 4 which means that jCGN .A/j N N G > NG .AE/ implies that G Š D2n or G Š SD2m . (v) Now we obtain a ﬁnal contradiction. Since r.G/ > 4, G contains subgroups G0 and U with G0 =U Š E32 . If E 6 G0 , then d.E \G0 / 2 and (by (iv)) d.G0 =E \G0 / 2, a contradiction. Hence E G0 . Since d.G0 =E/ 2, we must have E \ U D ¹1º. Hence ŒE; G0 E \ U D ¹1º and so E Z.G0 /. Since jG0 =Ej 4, G0 contains the inverse image K of Z.G=E/ in G. As K G0 CG .E/, we have K Š E16 or K Š C4 C2 C2 . The latter case does not occur as ˆ.G0 / \ E U \ E D ¹1º. Thus K Š E16 . Let a 2 A .A \ E/ so that a 62 K and CK .a/ Š E4 . Hence Lemma 99.2 implies that ha; Ki Š E4 o C2 . As A D ha; CK .a/i is normal in ha; Ki, we can ﬁnd an elementary abelian subgroup B of K of order 8 with jA \ Bj D 2 and B NG .A/. This contradicts (ii) and our lemma is proved. Lemma 99.8. Let G be a 2-group with a self-centralizing subgroup A Š C4 C2 . Then r.G/ 4. Proof. Suppose that r.G/ > 4. Then Lemma 99.6 implies that jGj 27 , A is not normal in G and G possesses a normal elementary abelian subgroup E of order 8. We set A D ha; b j a4 D b 2 D 1; Œa; b D 1; a2 D ci. It is well known that Aut.A/ Š D8 , CA .Aut.A// D hci and if is the unique central involution in Aut.A/, then inverts A. (i) We have jA \ Ej D 4. Suppose that jA \ Ej 2. Then A \ E Š C2 and A \ E D hci, hbci or hbi. Suppose ﬁrst that A \ E D hci. As A=hci Š E4 , each element x in A hci induces an involutory automorphism on E with jCE .x/j D 4 and hE; xi0 D ŒE; x D hci.

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Groups of prime power order

Indeed, if x centralizes E, then for an element y 2 A hc; xi we have jCE .y/j 4, which contradicts CG .A/ D A. If hE; xi0 D hei with e 2 E hci, then hei is normal in AE and A centralizes hei, a contradiction. Hence we get .AE/0 D ŒA; E D hci and .AE/=hci Š E16 . Then we obtain that Z.AE/ D .AE/0 D ˆ.AE/ D hci and so AE is extraspecial with an elementary abelian subgroup of order 8 which implies that AE Š Q8 Q8 . As CG .AE/ CG .A/ D A AE, Lemma 99.3 yields that r.G/ 4, a contradiction. We next treat the case A \ E D hbci or hbi. Without loss of generality we may assume that A \ E D hbi. We have .AE/=E Š hai Š C4 . If Œc; E D 1, then a centralizes a four-subgroup in E, a contradiction. Therefore hai acts faithfully on E and this forces Z.AE/ D Z.G/ D hbi. Set E D hb; x; yi. Without loss of generality we may set x a D xb, y a D yx and so ax D ab, Œc; x D 1, ay D ax, and x 2 NG .A/. Clearly hb; xi is a unique normal four-subgroup of G contained in E. Hence x maps into the center of NG .A/=A. Since x does not invert A, NG .A/=A Š C2 or NG .A/=A Š E4 . First suppose that NG .A/=A Š E4 , where NAE .A/ D hA; xi. Then NG .A/ .AE/ contains an element t which inverts A. We have NG .A/ D hA; x; t i and t 2 2 hb; ci so that ht; Ai=hbi Š D8 or Q8 . Setting F D NG .A/E, we see that F=E Š D8 or Q8 and F=E is faithful on E, inasmuch as c maps onto the center of F=E and does not centralize E. In that case, F=E Š D8 and so ht; Ai=hbi Š D8 , t 2 2 hbi and .ta/2 2 hbi. By the structure of GL.3; 2/, F=E must contain a four-subgroup F1 =E with CE .F1 / Š C2 . Interchanging t and ta if necessary, we may assume that F1 D hE; c; t i with Z.F1 / D hbi. But then Œht; ci; E D F10 D hbi and F1 =hbi Š E16 so that F1 is extraspecial and F1 Š Q8 Q8 . Suppose that CG .F1 / 6 F1 . Since a normalizes CG .F1 /, we can ﬁnd a subgroup V of order 4 in CG .F1 / which contains hbi and is normalized by hai. But then Œa; V hbi and so V normalizes A which implies that V F . On the other hand, CF .F1 / D Z.F1 / D hbi. This contradiction forces CG .F1 / F1 and then Lemma 99.3 implies r.G/ 4, a contradiction. It remains to treat the case A1 D NG .A/ D hA; xi, where A1 is the nonmetacyclic minimal nonabelian group of order 16. Also, we set A2 D NG .A1 / and A3 D NG .A2 /. As ˆ.A1 / D hb; ci, A1 has exactly three maximal subgroups: A, Ay D hax; bi Š C4 C2 and hc; b; xi Š E8 and A2 permutes A and Ay and so A2 D AE D hA; x; yi is of order 25 . Also, A3 > A2 since jGj 27 . On the other hand, as ˆ.A2 / D hb; c; xi, A2 has exactly three maximal subgroups: A1 , hc; Ei Š C2 D8 and hay; b; c; xi, where A1 cannot be conjugate to hc; Ei. Hence jA3 W A2 j D 2, jA3 j D 26 and A3 contains an element t 2 A3 A2 such that At1 D hay; b; c; xi, .ay/2 D a.yay/ D a ax D cx. If CA3 .E/ D E, then A3 =E Š D8 and A2 =E Š C4 is characteristic in A3 =E. Therefore if s 2 G A3 is such that s 2 2 A3 and As3 D A3 , then s normalizes A2 , contrary to A3 D NG .A2 /. Thus E satisﬁes CA3 .E/ > E and since A2 =E is faithful on E, we get CA3 .E/ D hE; ti, where ŒE; t D 1, t 2 2 E and t 2 A3 A2 . We have A3 D A2 hE; ti, where A2 and the abelian subgroup hE; ti (of order 16) are normal in

99 2-groups with sectional rank at most 4

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A3 and A2 \ hE; ti D E so that A3 =E Š C4 C2 and so 1 .A3 / hE; c; t i: We shall show that B D hb; c; xi is a unique normal self-centralizing elementary abelian subgroup of order 8 in A3 . Since B D ˆ.A2 /, we see that B is normal in A3 . Since t normalizes A2 but does not normalize A1 , we have at ay .mod B/, where A2 =B Š E4 and so A3 =B Š D8 . The element y maps onto Z.A3 =B/ and y does not centralize B and so CA3 .B/ D B. Suppose that hE; ti Š E16 . Then the four-group hy; ti acts faithfully on B and hy; ti centralizes hb; xi so that ChE;ti .c/ D hb; xi Š E4 and so by Lemma 99.1, we obtain hE; c; t i Š E4 o C2 . Since all involutions in hE; c; t i hE; ti are contained in B, it follows that in this case B is a unique normal self-centralizing elementary abelian subgroup of order 8 in A3 . Now suppose that hE; ti Š C4 C2 C2 . Since a centralizes Ã1 .hE; ti/ D ht 2 i, we get t 2 D b. Suppose that c inverts an element t1 2 hE; tiE in which case t12 D b and c t1 D cb. Hence t1 normalizes hb; ci so t1 normalizes CA2 .hb; ci/ D hA; xi D A1 . As t1 62 A2 D NG .A1 /, this is a contradiction. Thus all involutions in hE; c; t ihE; ti are contained in B and so B has the required property also in this case. Now we get a ﬁnal contradiction which will ﬁnally prove our statement (i). Let l be an element in G A3 such that Al3 D A3 and l 2 2 A3 . Since B is characteristic in A3 , B is normal in hA3 ; li and so hA3 ; li=B Š D8 C2 . Set F D ChA3 ;li .B/ so that F=B Š C2 and F is normal hA3 ; li D A3 F with A3 \ F D B. Let l 0 2 F B so that l 0 62 A3 , l 02 2 B and Œl 0 ; A3 A3 \ F D B. Hence l 0 centralizes B and A2 =B and so l 0 normalizes A2 , contrary to A3 D NG .A2 /. (ii) We have jNG .A/j 25 . By (i), jA \ Ej D 4 and so we may set E D hb; c; xi, where A \ E D hb; ci. Suppose that jNG .A/j D 26 . Since Ã1 .A/ D hci, we get hci D Z.NG .A// D Z.G/. We have E NG .A/ and E E G and so x maps into the center of NG .A/=A Š D8 . Thus x inverts A. In particular, x a D xc. Let e be an element of NG .A/ such that ae D a and b e D bc. Then e 2 2 CG .A/ D A. First suppose that o.e/ 4. Then e 2 2 hb; ci and so he; b; ci Š D8 . Thus we may assume that o.e/ D 2. As jCE .e/j D 4, we may also assume that Œe; x D 1. We claim that hA; e; xi Š Q8 Q8 . Indeed, we have hA; e; xi0 D hci and ˆ.hA; e; xi/ D hci D Z.hA; e; xi/ and so hA; e; xi is extraspecial of order 25 containing E Š E8 and so hA; e; xi Š Q8 Q8 . Since hA; e; xi is self-centralizing in G, Lemma 99.3 yields that r.G/ 4, a contradiction. Now suppose that o.e/ D 8 so that we may assume e 2 D a. There is an element d 2 NG .A/ with ad D ab and b d D b and so ˆ.NG .A// A. On the other hand, we have NG .A/=A Š D8 and so x maps into ˆ.NG .A/=A/ which implies that ˆ.NG .A// D hA; xi D AE: As x inverts A, hA; xi Š C2 D8 and so A is a unique abelian maximal subgroup of

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Groups of prime power order

hA; xi which is not elementary abelian. It follows that A is characteristic in NG .A/ which forces NG .A/ D G, contrary to the fact that jGj 27 (Lemma 99.6). (iii) Set GN D G=E. Then jCGN .a/j N > 4. N 24 . This implies that GN Š D2n or that Suppose that jCGN .a/j N 4, where jGj N G Š SD2n , n 4. It follows that CG .E/ > E and so if K is the inverse image of N then K is an abelian normal subgroup of order 16. Z.G/, Since r.G/ > 4, G contains subgroups T > U with U E T and T =U Š E25 . Clearly, E T because otherwise d.T / 4. In this case, d.T =E/ 2 which implies E \ U D ¹1º. Hence ŒT; E E \ U D ¹1º and so E is central in T . But jT =Ej is N of order at least 4 and this implies that T also contains the inverse image K of Z.G/. Also, ˆ.T / U and ˆ.K/ E and so K Š E16 . Since CK .a/ D hb; ci, we get (Lemma 99.2) ha; Ki Š E4 o C2 with Z.ha; Ki/ D hb; ci Š E4 . By Lemma 99.4, CG .hb; ci/ > ha; Ki and let X be a subgroup of CG .hb; ci/ containing ha; Ki as a subgroup of index 2. Note that (Lemma 99.2) ha; Ki K contains exactly four square roots of c and they form a single conjugate class in ha; Ki which is also X-invariant (since X centralizes hb; ci). This shows that CG .a/ must cover X=ha; Ki, contrary to CG .A/ D A. (iv) We are now in a position to get a ﬁnal contradiction. N is of order 26 . We set again GN D G=E. By (iii), the inverse image D of CGN .a/ 5 Since D normalizes hA; xi D AE and (by (ii)) jNG .A/j 2 , it follows that hA; xi 6Š C2 D8 (because otherwise, A would be characteristic in AE). Therefore x does not invert A. Without loss of generality we may assume (replacing b with bc if necessary) ax D ab and so ˆ.AE/ D hb; ci. We have .ax/2 D a.xax/ D a.ab/ D a2 b D cb and so A D ha; bi and hax; bi are the only subgroups of AE which are isomorphic to C4 C2 . This implies that jDj D 26 , jNG .A/j D 25 , NG .A/=A Š E4 and if t 2 D NG .A/, then At D hax; bi and c t D cb which gives Z.G/ D Z.D/ D hbi. Let d be an element in NG .A/ which inverts A. Then d 2 NG .A/ .AE/ and since d 2 centralizes A, we have d 2 2 A and so d 2 2 hb; ci. Since jGj 27 (Lemma 99.6), we obtain CG .E/ > E. Hence aN (which does not centralize E) centralizes a subgroup E0 =E of order 2 in CG .E/=E, where E0 is normal in G. This in turn implies that CD .E/ > E. Since any element of D NG .A/ sends c onto cb, we have CD .E/ NG .A/. Thus CD .E/ D hE; d i or CD .E/ D hE; ad i. Noting that both d and ad inverts A, we may assume without loss of generality that CD .E/ D hE; d i. Setting K D hE; d i, we ﬁrst assume that K satisﬁes K Š E16 . As CK .a/ D hb; ci, ha; Ki Š E4 o C2 with Z.ha; Ki/ D hb; ci (Lemma 99.2). If CG .Z.ha; Ki// > ha; Ki, then take a subgroup Y in CG .Z.ha; Ki// containing ha; Ki with jY W ha; Kij D 2. Since ha; Ki contains exactly four square roots of c and they are all conjugate to a in ha; Ki (Lemma 99.2), we get that CG .a/ covers Y =ha; Ki, contrary to CG .A/ D A. Hence we have CG .Z.ha; Ki// D ha; Ki and then Lemma 99.4 gives that r.G/ 4, a contradiction.

99 2-groups with sectional rank at most 4

43

It remains to treat the case K Š C4 C2 C2 . Since Ã1 .K/ is central in D, we get d 2 D b. We compute .adx/2 D .adx/.adx/ D ax.dad /x D axd 2 ad x D axba1 x D ab.xa1 x/ D ab.ab/1 D 1: Setting C D hb; c; adxi, we shall argue that C is a normal self-centralizing elementary abelian subgroup of order 8 in D. There are three nontrivial cosets in NG .A/ modulo E: aE, dE, adxE. All elements in aE are of order 4 (noting that ˆ.ha; Ei/ D hb; ci and the three maximal subgroups of ha; Ei are A Š C4 C2 , hax; bi Š C4 C2 , and E, where.ax/2 D cb) and all elements in dE are of order 4 since d 2 D b and d centralizes E. Finally, in the coset adxE there are exactly four involutions forming the set ¹adxhb; ciº since CE .adx/ D hb; ci. Hence all involutions in NG .A/ are contained in E or C . As E E D and NG .A/ E D, we have C E D. Since each t 2 D NG .A/ has the property c t D cb, we get CD .C / NG .A/ D CD .hb; ci/ and no one of d; x; dx centralizes adx, where dC , xC and dxC are three nontrivial cosets of C in NG .A/. Hence CD .C / D C , as required. Suppose that D contains another such subgroup C1 , namely C1 Š E8 , C1 E D, and CD .C1 / D C1 . Then b 2 C1 since hbi D Z.D/. Suppose that C \ C1 D hbi. Let C1 Š E4 be a complement of hbi in C1 so that C1 acts faithfully on C and CC .C1 / D hbi D Z.C C1 /. By Lemma 99.1, C C1 Š Q8 Q8 . We have a 62 C C1 since a2 D c and Ã1 .C C1 / D hbi. Since jD W .C C1 /j D 2, we conclude that a normalizes C C1 and D D C C1 hai. Suppose that CG .C C1 / 6 C C1 . Since hai normalizes CG .C C1 /, there is an hai-invariant subgroup V of order 4 in CG .C C1 / which contains hbi. But then Œhai; V hbi and so V normalizes ha; bi D A and therefore V D, where .C C1 / \ V D hbi. In that case, V centralizes C , contrary to the fact that C is selfcentralizing in D. Thus CG .C C1 / C C1 and so Lemma 99.3 implies r.G/ 4, a contradiction. Hence we may assume that jC \ C1 j D 4. Since hb; ci D C \ E is a normal four-subgroup of D and D=C Š D8 , it follows that four involutions in C hb; ci are conjugate in D and so hb; ci is a unique normal four-subgroup of D contained in C . This implies C \C1 D hb; ci. Consequently, C1 CD .hb; ci/ D NG .A/. Thus C1 D C , contrary to C1 ¤ C . Hence C is the unique subgroup in D with the required property which implies that C is characteristic in D. Let F G be a subgroup of G containing D so that jF W Dj D 2. Then C is normal in F and D=C Š D8 . Let H D CF .C / so that H E F , jH W C j D 2 and H is abelian. We get F D DH with D \ H D C and ŒD; H D \ H D C . Let h 2 H C so that h2 2 C and h centralizes C and D=C . We get ah D ay with y 2 C . If y 2 hb; ci, then h normalizes A D ha; bi, contrary to NG .A/ D. We get y 62 hb; ci and so y D adx b i c j with i; j D 0; 1. Then ah D a adxb i c j D a2 dxb i c j D cdxb i c j D dxb i c j C1 2 K and h centralizes hb; ci. Thus Ah K, where K is abelian of order 16. This is a ﬁnal contradiction since CG .Ah / D Ah . Our lemma is proved.

44

Groups of prime power order

Now Lemmas 99.5, 99.7 and 99.8 give the following remarkable result. Theorem 99.9 (K. Harada). Let G be a 2-group with a self-centralizing abelian subgroup of order 8. Then each subgroup of G is generated by at most four elements, i.e., the sectional rank of G is at most 4. Theorem 99.10 (K. Harada). Let G be a 2-group containing a subgroup T D U W , where U Š E4 and W Š E4 , D2n , n 3, or SD2m , m 4. Suppose that for each involution x 2 U , CG .x/ D T . Then the sectional rank r.G/ of G is 4. Proof. Since r.T / D 4, we may assume that G > T . We have Z.T / D U Z.W / Š E8 or E16 . Let x 2 NG .T / T with x 2 2 T . If U \ U x ¤ ¹1º, then x normalizes U \ U x and so x centralizes an involution in U \ U x , a contradiction. Hence we have U \ U x D ¹1º which implies Z.W / Š E4 and so W Š E4 and T Š E16 . For any element s 2 NG .T / T with s 2 2 T , U \ U s D ¹1º and so hs; T i Š E4 o C2 (Lemma 99.2). In particular, hs; T i T contains an involution. Let k D jNG .T / W T j and R D NG .T /. Then any element u 2 U has exactly k conjugates in T by the action of R. This gives 2 k 8. Suppose that k D 2. Then R Š E4 o C2 . This forces G D R D NG .T / as T is characteristic in R. Thus r.G/ D 4 by Lemma 99.4. Hence we may assume that k > 2. Suppose next that k D 8. Since there cannot be two orbits of length 8 (under the action of R), all involutions in U are conjugate in R. Let a be an involution in R T which maps into the center of R=T . Since all involutions of ha; T i T are conjugate in ha; T i (as ha; T i Š E4 o C2 ), we have R D CR .a/T . Hence CT .a/ is normal is R. Moreover, U \ CT .a/ D ¹1º. Let L be an R-invariant subgroup with CT .a/ < L < T and jL W CT .a/j D 2. Then jL \ U j D 2 and so all involutions in U cannot be conjugate in R, a contradiction. We have proved that k D 4. If G D R D NG .T /, then jGj D 26 and it is easy to see that in that case r.G/ D 4. Indeed, let A=B Š E25 be a section of G. If B D ¹1º, then A is an elementary abelian subgroup of index 2 in G. But then U \ A ¤ ¹1º, contrary to jCG .U \ A/j D 24 . Hence we have B Š C2 and A D G. But in this case, G 0 D B Š C2 and so an involution in U cannot have k D 4 conjugates in R D G. Hence we have r.G/ D 4 in this case. It follows that we may assume that G > R. This implies that using an element h 2 G NG .R/ with h2 2 R, we get another elementary abelian subgroup T1 D T h of order 16. As CG .u/ D T for any involution in U , we have T1 \U D ¹1º. Hence R D T1 T as jRj D 26 and T1 \T D Z.R/ Š E4 . This in turn implies that R is a split extension of E16 by E4 . Then U1 D U h is a complement of T in R and CT .b/ D T1 \ T for any involution b in U1 and, in addition, hb; T i Š E4 o C2 which shows that all elements in R .T [ T1 / are of order 4 and R0 D ˆ.R/ D Z.R/ D T \ T1 . Hence R has exactly two elementary abelian subgroups T an T1 of order 16. This forces that jNG .R/ W Rj D 2. Setting P D NG .R/, we shall argue that P D G. It sufﬁces to show that T and T1 are the only subgroups of P which are isomorphic to E16 . Indeed, let t 2 P R so that T t D T1 and therefore

99 2-groups with sectional rank at most 4

45

1 .CR .t// T \ T1 . This implies that any elementary abelian subgroup of P which is not contained in R is of order at most 8. Thus G D P , as required. It remains to be shown that r.G/ 4. For this, it sufﬁces to prove that G does not possess a normal elementary abelian subgroup of order 8. Suppose that this is false and let A be such a subgroup. If A R, then A T or A T1 . However, as T t D T1 (t 2 G R) and A 6 T \ T1 , A cannot be t -invariant. Hence A 6 R and A \ R is a normal four-subgroup of G D P which gives A \ R D T \ T1 . Let a 2 A R. Then T a D T1 and so G=T \ T1 Š E4 o C2 which shows that a does not map into the center of G=.T \ T1 /. This contradicts the fact that A=.T \ T1 / E G=.T \ T1 /. Our theorem is proved. It is interesting to notice that if a p-group G has a nonabelian subgroup B of order p 3 such that CG .B/ < B, then G is of maximal class (Proposition 10.17(a)); then every subgroup of G is generated by p elements.

100

2-groups with exactly one maximal subgroup which is neither abelian nor minimal nonabelian

Suppose that G is a p-group all of whose maximal subgroups are metacyclic except one (which is nonmetacyclic). If p > 2 and jGj > p 4 , then Y. Berkovich has shown (with a short and elegant proof) that G must be a so-called L3 -group, i.e., 1 .G/ is of order p 3 and exponent p and G=1 .G/ is cyclic of order p 2 (see Proposition A.40.12). However, if p D 2, then the problem of determination of such groups is much more difﬁcult and this was done in 87. All these results will be used very heavily in this section. Here we continue with this idea of classifying p-groups all of whose maximal subgroups, but one, have a certain strong property. In this section we determine up to isomorphism the 2-groups G all of whose maximal subgroups, but one, are abelian or minimal nonabelian. We begin with the case d.G/ D 2 (Theorems 100.1, 100.2 and 100.3). Actually, a detailed investigation of such groups has already begun in Lemma 76.5. Then we determine such groups satisfying d.G/ > 2 (Theorems 100.4, 100.5 and 100.6). All resulting groups will be presented in terms of generators and relations, but we shall also describe all important characteristic subgroups of these groups for two reasons. One reason is that only the knowledge of the subgroup structure of these groups will make our theorems useful for applications. Another reason is that with this knowledge we see that 2-groups appearing in distinct theorems are nonisomorphic. Conversely, it is easy to check that all groups given in these theorems indeed possess exactly one maximal subgroup which is neither abelian nor minimal nonabelian. The corresponding problem for p > 2 is open, but we think that this problem is within the reach of the present methods in ﬁnite p-group theory. Theorem 100.1. Let G be a two-generator 2-group with exactly one maximal subgroup H which is neither abelian nor minimal nonabelian. If G has an abelian maximal subgroup A, then 1 D ¹A; M; H º is the set of the maximal subgroups of G, where M is minimal nonabelian, A and M are both metacyclic, d.H / D 3 and we have more precisely: G D ha; x j Œa; x D v; v 4 D 1; v 2 D z; v x D v 1 ; v a D v 1 ; x 2 2 hzi; a2 2 hzi; m 2i; m

100

2-groups with exactly one exceptional maximal subgroup

47

where G 0 D hvi Š C4 ; E E G;

K3 .G/ D ŒG; G 0 D hzi Š C2 ;

G D Ehai;

G=E Š C2m ;

H D Eha2 i;

E D hv; xi Š D8 or Q8 ;

ˆ.G/ D G 0 ha2 i is abelian;

Z.G/ D ha2 ; zi;

A D hax; vi is an abelian maximal subgroup of G, M D hv; ai is metacyclic minimal nonabelian of order 2mC2 and jGj D 2mC3 , m 2. Proof. If G has more than one abelian maximal subgroup, then all three maximal subgroups of G are abelian, a contradiction. Hence A is a unique abelian maximal subgroup of G. It follows that 1 D ¹A; M; H º, where M is minimal nonabelian. The subgroup A \ M D ˆ.G/ is abelian, ˆ.M / < ˆ.G/ and jˆ.G/ W ˆ.M /j D 2. Also, ˆ.M / D Z.M / Z.G/ so that for an element m 2 M A, Cˆ.G/ .m/ D Z.M /. If CA .m/ > Z.M /, then G D M C with C D CG .M / and M \ C D Z.M /, contrary to d.G/ D 2. It follows that CA .m/ D Z.M / D Z.G/ and so jG W Z.G/j D 8. From jGj D 2jZ.G/jjG 0 j (Lemma 1.1) follows that jG 0 j D 4. For each x 2 G A, CA .x/ D Z.M / and so x 2 2 Z.M / D Z.G/ which implies that x inverts A=Z.M /. If A=Z.M / Š E4 , then ˆ.G/ D Ã1 .G/ Z.M /, a contradiction. Hence we obtain that A=Z.M / Š C4 and since m 2 M A inverts A=Z.M /, we have G=Z.M / Š D8 . Let a 2 A ˆ.G/. Then hai covers A=Z.M / and so v D Œa; m 2 ˆ.G/ Z.M /. We get 1 D Œa; m2 D Œa; mŒa; mm D vv m and so v m D v 1 which implies that o.v/ D 4. Indeed, if o.v/ D 2, then Œv; m D 1, which contradicts the fact that Cˆ.G/ .m/ D Z.M /. We get G 0 D hvi Š C4 and Œv; m D v 2 implies that M 0 D hv 2 i. Since v 2 is a square in M , it follows that M is metacyclic (Lemma 65.1). In particular, j1 .ˆ.G//j 4 which together with A=Z.M / Š C4 gives 1 .A/ ˆ.G/ and so A is also metacyclic. Here H D ˆ.G/hami is our third maximal subgroup of G. From Cˆ.G/ .am/ D Z.M / follows that H is nonabelian with Z.H / D Z.M / and so Lemma 1.1 yields jH 0 j D 2. If d.H / D 2, then Lemma 65.2 gives that H would be minimal nonabelian, a contradiction. Hence d.H / 3 and so H is the only maximal subgroup of G which is nonmetacyclic. In fact, d.H / D 3 since ˆ.G/ is metacyclic. We are in a position to use Theorem 87.12 for n D 2 since G 0 Š C4 . This gives the generators and relations described in our theorem, where we have used the notation from Theorem 87.12. Theorem 100.2. Let G be a 2-group with d.G/ D 2 which has exactly one maximal subgroup H which is neither abelian nor minimal nonabelian. If the other two maximal subgroups H1 and H2 are minimal nonabelian with H10 D H20 , then one of the following holds: (a) G is one of the groups given in Theorem 87.10. (b) G is the group of order 25 given in Theorem 87.14.

48

Groups of prime power order n

(c) G D hh; x j h2 D 1; n 2; Œh; x D s; s 2 D 1; Œs; h D z; z 2 D Œz; h D Œz; x D Œx; s D 1; x 2 2 hzii, where jGj D 2nC3 ;

G 0 D hz; si Š E4 ;

K3 .G/ D ŒG; G 0 D hzi Š C2 ;

ˆ.G/ D G 0 hh2 i is abelian and the maximal subgroups of G are H1 D hG 0 ; hi, H2 D hG 0 ; xhi (both are nonmetacyclic minimal nonabelian) and H D hx; s; h2 i with d.H / D 3 and H10 D H20 D H 0 D hzi. Proof. Here A D H1 \ H2 D ˆ.G/ is a maximal normal abelian subgroup of G. Set H10 D H20 D hzi A so that H1 =hzi and H2 =hzi are two distinct abelian maximal subgroups in G=hzi. It follows that H=hzi is also abelian and so H 0 D hzi since H is nonabelian. If d.H / D 2, then, by Lemma 65.2(a), H would be minimal nonabelian, a contradiction. Thus d.H / 3. If G=hzi is abelian, then G 0 D hzi and so G D H1 CG .H1 / which gives d.G/ D 3, a contradiction. Hence G=hzi is nonabelian and so .G=hzi/0 Š C2 since G=hzi has three distinct abelian maximal subgroups. Thus jG 0 j D 4 and G 0 A D ˆ.G/. Taking h1 2 H1 A and h2 2 H2 A, we get hh1 ; h2 i D G and so s D Œh1 ; h2 2 G 0 hzi. If G 0 Š C4 , then z is a square in H1 and H2 and so both H1 and H2 are metacyclic (Lemma 65.1). Since d.H / 3, G has exactly one nonmetacyclic maximal subgroup. But then Theorem 87.12 for n D 2 implies that G has an abelian maximal subgroup, a contradiction. Thus G 0 D hs; zi Š E4 , where s is an involution. If s 2 Z.G/, then G=hsi would be abelian (because hh1 ; h2 i D G and s D Œh1 ; h2 ), a contradiction. Hence s 62 Z.G/ and so s 62 Z.H1 / or s 62 Z.H2 /. Without loss of generality we may assume that s 62 Z.H1 /. Suppose that z is a square in H1 , i.e., there is v 2 H1 such that v 2 D z. Suppose at the moment that v 2 H1 A in which case hv; G 0 i D hv; si Š D8 since s v D sz. It follows that hv; G 0 i D H1 . Since CG .H1 / H1 (otherwise, d.G/ D 3), G would be of maximal class (Proposition 10.17), a contradiction (noting that 2-groups of maximal class have a cyclic maximal subgroup). Thus v 2 A D ˆ.G/. In that case, both H1 and H2 are metacyclic (Lemma 65.1) which together with d.H / 3 allows us to use 87, part 2o . If G has a normal elementary abelian subgroup of order 8, then we get the groups in part (a) of our theorem. If G has no normal elementary abelian subgroup of order 8, then we get the group of order 25 given in part (b) of our theorem. Now we assume that z is not a square in H1 which implies that H1 is nonmetacyclic (Lemma 65.1). If h1 is an involution, then s h1 D sz shows that hh1 ; si Š D8 and so n1 hh1 ; si D H1 is metacyclic, a contradiction. Hence o.h1 / D 2n , n 2. Set u D h21 so that u 2 Z.H1 / and u 62 G 0 since z is not a square in H1 and s h1 D sz. We have E D 1 .H1 / D hz; s; ui Š E8 ;

EEG

and

EA

which implies that H2 is nonmetacyclic and therefore z is also not a square in H2 . Since H1 D Ehh1 i D hh1 ; si, we have jH1 j D 2nC2 , jGj D 2nC3 and A D hh21 ; s; zi is abelian of order 2nC1 and type .2n1 ; 2; 2/. Also, H1 is a splitting extension of G 0

100

2-groups with exactly one exceptional maximal subgroup

49

by hh1 i Š C2n , n 2. As d.G=G 0 / D 2, we get that G=G 0 is abelian of type .2n ; 2/. We obtain G D FH1 , where F \ H1 D G 0 and jF W G 0 j D 2 so that F D hG 0 ; xi with o.x/ 4. In fact, we have x 2 2 hzi. Indeed, if F is not elementary abelian, then Ã1 .F / Š C2 and Ã1 .F / G 0 . But F E G and so Ã1 .F / Z.G/ which implies that Ã1 .F / D hzi as G 0 6 Z.G/. Since hxh1 iG 0 =G 0 is another cyclic subgroup of index 2 in G=G 0 (distinct from H1 =G 0 ), M D hG 0 ; xh1 i is a maximal subgroup of G distinct from H1 and M 0 D hzi (since H20 D H 0 D hzi). If G 0 Z.M /, then M would be abelian, a contradiction. We get s xh1 D sz and so M D hxh1 ; si is minimal nonabelian which yields that M D H2 . We may set h2 D xh1 , where Œh1 ; h2 D s and G 0 D hs; zi. From s xh1 D sz it follows 1

1

1

s x D .s xh1 /h1 D .sz/h1 D s h1 z D .sz/z D s and so F is abelian. As s D Œh1 ; h2 , we get s D Œh1 ; h2 D Œh1 ; xh1 D Œh1 ; h1 Œh1 ; xh1 D Œh1 ; xh1 and so Œh1 ; x D sz: We conclude that ˆ.G/ D G 0 hh21 i is abelian and H D F hh21 i, where Œh21 ; x D Œh1 ; xh1 Œh1 ; x D .sz/h1 .sz/ D .szz/.sz/ D z: Since H=hzi D H=H 0 is abelian of type .2; 2; 2n1 /, we have d.H / D 3. Replacing s with s 0 D sz, we get Œh1 ; x D s 0 and writing again s instead of s 0 , we may write Œh1 ; x D s 2 G 0 hzi. Also, writing h instead of h1 , we obtain the relations of part (c) of our theorem. Theorem 100.3. Let G be a 2-group with d.G/ D 2 which has exactly one maximal subgroup H which is neither abelian nor minimal nonabelian. If the other two maximal subgroups H1 and H2 are minimal nonabelian with H10 ¤ H20 , then one of the following holds: (a) G is a group of order 26 with n D 2 given in Theorem 87.15(a). (b) G is a group of order 2mC4 .m 2/ with n D 2 given in Theorem 87.16. (c) G D hh; x j Œh; x D v; v 4 D 1; v h D vz1 ; v x D v 1 v 2 D z1 z2 ; z12 D z22 D 1; m Œz1 ; h D Œz1 ; x D Œz2 ; h D Œz2 ; x D 1; x 2 2 hz1 ; z2 i; h2 D .z1 z2 / i, where D 0; 1;

m 2;

jGj D 2mC4 ;

K3 .G/ D ŒG; G 0 D hz1 ; z2 i Š E4

G 0 D hv; z1 i Š C4 C2 ; and

hz1 ; z2 i Z.G/:

Moreover, the maximal subgroups of the group G are H1 D G 0 hhi, H2 D G 0 hhxi and H D G 0 hx; h2 i, where H1 and H2 are both nonmetacyclic minimal nonabelian with H10 D hz1 i, H20 D hz2 i and H 0 D hz1 ; z2 i Š E4 (d.H / D 3).

50

Groups of prime power order

Proof. Set hz1 i D H10 and hz2 i D H20 so that W D hz1 ; z2 i Š E4 , W Z.G/ and W A D H1 \ H2 D ˆ.G/, where A is abelian. Here ¹H1 =hz1 i; H2 =hz1 i; H=hz1 iº is the set of maximal subgroups of G=hz1 i and H1 =hz1 i is abelian and two-generated, H2 =hz1 i is minimal nonabelian and so H=hz1 i must be nonabelian. If H=hz1 i is minimal nonabelian, then by a result of N. Blackburn (Theorem 44.5), G=hz1 i would be metacyclic. But then, by another result of N. Blackburn (Lemma 44.1 and Corollary 44.6), G is also metacyclic, contrary to W G 0 and W Š E4 . Hence H=hz1 i is neither abelian nor minimal nonabelian. By Theorem 100.1, d.H=hz1 i/ D 3;

.H=hz1 i/0 Š C2 ;

G 0 =hz1 i Š C4 :

Similarly, considering G=hz2 i, we obtain that .H=hz2 i/0 Š C2 and G 0 =hz2 i Š C4 . It follows that G 0 is abelian of type .4; 2/ with Ã.G 0 / D hz1 z2 i. On the other hand, ¹H1 =W; H2 =W; H=W º is the set consisting of the maximal subgroups of the nonabelian group G=W , where both H1 =W and H2 =W are abelian. Hence H=W is also abelian and so H 0 W . By the above, H 0 is distinct from hz1 i and hz2 i and so either H 0 D W or H 0 D hz1 z2 i. Suppose that H 0 D hz1 z2 i. Then ¹H1 =hz1 z2 i; H2 =hz1 z2 i; H=hz1 z2 iº is the set of maximal subgroups of G=hz1 z2 i, where H1 =hz1 z2 i and H2 =hz1 z2 i are minimal nonabelian and H=hz1 z2 i is abelian. By Theorem 106.3 (or by results of 71), we deduce that G=hz1 z2 i is metacyclic, contrary to G 0 =hz1 z2 i Š E4 . We have thus proved that H 0 D W D hz1 ; z2 i. Take h1 2 H1 A, h2 2 H2 A so that we have hh1 ; h2 i D G. If v D Œh1 ; h2 2 W , then v is an involution in Z.G/ and G=hvi is abelian, G 0 hvi, a contradiction. Hence v 62 W and so v 2 G 0 W is of order 4 with v 2 D z1 z2 and hvi is not normal in G (and so also hvz1 i is not normal in G). Indeed, if hvi is normal in G, then G=hvi would be abelian since Œh1 ; h2 D v and hh1 ; h2 i D G. In particular, hvi cannot be normal in both H1 and H2 and so we may assume without loss of generality that hvi is not normal in H1 . Hence Œh1 ; v D z1 which gives v h1 D vz1 and H1 D hh1 ; vi. Suppose that H1 is metacyclic. Then there exists h01 2 H1 such that .h01 /2 D z1 . 0 If h01 2 H1 A, then v h1 D vz1 and so H1 D hh01 ; vi is of order 24 . But then jGj D 25 and jG 0 j D 23 imply (using a result of O. Taussky) that G is of maximal class, a contradiction. It follows that h01 2 A and then .h01 v/2 D .h01 /2 v 2 D z1 .z1 z2 / D z2 which implies that H2 is also metacyclic. We are in a position to use 87, 2o . By Theorems 87.9 and 87.10, G has no normal elementary abelian subgroup of order 8 (since jG 0 j D 8). We have ˆ.G 0 / ¤ ¹1º and Z.G/ W is noncyclic. If G=ˆ.G 0 / has no normal elementary abelian subgroup of order 8, then G is isomorphic to a group of

100

2-groups with exactly one exceptional maximal subgroup

51

order 26 given in Theorem 87.15(a) for n D 2. If G=ˆ.G 0 / has a normal elementary abelian subgroup of order 8, then G is a group of order 2mC4 , m 2, given in Theorem 87.16 for n D 2. Suppose that H1 is nonmetacyclic. If there exists an element l 2 H1 A such that l 2 2 G 0 , then v l D vz1 gives that H1 D hl; vi D G 0 hli is nonmetacyclic minimal nonabelian of order 24 . But in that case, jGj D 25 and jG 0 j D 23 imply (using a result of O. Taussky) that G is of maximal class, a contradiction. It then follows that 1 .H1 / D 1 .A/ Š E8 and so H2 is also nonmetacyclic minimal nonabelian. Since h22 2 H1 , we have Œh1 ; h22 D z1 ; D 0; 1: We compute

z1 D Œh1 ; h22 D Œh1 ; h2 Œh1 ; h2 h2 D vv h2 ; and so

v h2 D v 1 z1 D v.z1 z2 /z1 D vz1

C1

z2 :

If D 0, then v h2 D v.z1 z2 /, contrary to H20 D hz2 i. Thus D 1 and so v h2 D vz2 which implies H2 D hh2 ; vi D G 0 hh2 i. Also, v h1 D vz1 gives H1 D hh1 ; vi D G 0 hh1 i and since h21 62 G 0 , we have H1 =G 0 Š C2m , m 2, and then also H2 =G 0 Š C2m . Since d.G=G 0 / D 2, we see that G=G 0 is abelian of type .2m ; 2/, m 2. We may set G D FH1 with F \H1 D G 0 and jF W G 0 j D 2. Since hh1 i covers H1 =G 0 Š C2m , v h1 D vz1 and neither z1 nor z2 are squares of any element in A D G 0 hh21 i D ˆ.G/, m we get h21 D .z1 z2 / , D 0; 1. We may set h2 D h1 x with x 2 F G 0 so that from v h2 D vz2 follows vz2 D v h2 D .v h1 /x D .vz1 /x D v x z1 and so v x D v.z1 z2 / D v 1 which gives x 2 2 hz1 ; z2 i Z.G/. It then follows from v D Œh1 ; h2 that v D Œh1 ; h1 x D Œh1 ; xŒh1 ; h1 x D Œh1 ; x: Finally, we have H2 D G 0 hh1 xi and H D F hh21 i, where F 0 D hŒv; xi D hz1 z2 i and Œh21 ; x D Œh1 ; xh1 Œh1 ; x D v h1 v D .vz1 /v D z1 v 2 D z1 .z1 z2 / D z2 and so indeed H 0 D hz1 ; z2 i Š E4 which shows that H is neither abelian nor minimal nonabelian. Writing h instead of h1 , we have obtained the relations given in part (c) of our theorem. We turn now to the case d.G/ 3. Since G possesses at least one minimal nonabelian maximal subgroup, it follows that in this case d.G/ D 3. It is well known that the number of abelian maximal subgroups in a nonabelian 2-group G is 0; 1 or 3. According to this fact, we shall subdivide our study of the title groups with d.G/ D 3.

52

Groups of prime power order

Theorem 100.4. Let G be a 2-group with d.G/ D 3 which has exactly one maximal subgroup which is neither abelian nor minimal nonabelian. If G possesses more than one abelian maximal subgroup, then one of the following holds: (a) G D Q Z, where Q Š Q8 , Z Š C2n , n 3 and Q \ Z D Z.Q/. (b) G D Q Z, where Q Š Q8 and Z Š C2n , n 2. (c) G D D Z, where D Š D8 and Z Š C2n , n 2. Proof. By our assumption, G has exactly three abelian maximal subgroups. This implies jG 0 j D 2 and G possesses exactly three maximal subgroups which are minimal nonabelian. Suppose that H is a minimal nonabelian maximal subgroup of G. Since H 0 D G 0 Š C2 , we get G D H Z.G/, where Z.G/ \ H D Z.H / D ˆ.H / D ˆ.G/: All three maximal subgroups of G, containing Z.G/, are abelian. Let G be a title group with jG 0 j D 2. Then G possesses a maximal subgroup H which is minimal nonabelian. From H 0 D G 0 Š C2 follows that G D H Z.G/ with jG W Z.G/j D 4. In that case, d.G/ D 3 and all three maximal subgroups of G containing Z.G/ are abelian. It follows that for a title group G the assumption G 0 Š C2 is equivalent with the assumption that G has more than one abelian maximal subgroup. In what follows H will denote a ﬁxed maximal subgroup of G which is minimal nonabelian. Suppose that there is an involution c 2 Z.G/ Z.H /. Then G D H hci and so each maximal subgroup of G which does not contain hci is isomorphic to G=hci Š H and so is minimal nonabelian, a contradiction. Hence there are no involutions in Z.G/ H which implies that 1 .Z.H // D 1 .Z.G// so that d.Z.H // D d.Z.G//. It follows that for each x 2 Z.G/ H , x 2 2 Z.H / ˆ.Z.H //. Suppose that jGj D 24 . Then each nonabelian maximal subgroup of G is isomorphic to D8 or Q8 and so is minimal nonabelian, a contradiction. Hence jGj 25 and, in particular, H is not isomorphic to Q8 or D8 . (i) First assume that H is metacyclic. Since H is not isomorphic to Q8 , it follows that H is a “splitting” metacyclic group and so we may set m

n

H D ha; b j a2 D b 2 D 1; ab D az; z D a2

m1

i;

where m 2;

n 1;

m C n 4;

H 0 D hzi;

jH j D 2mCn ;

jGj D 2mCnC1 :

We see that Z.H / D ha2 i hb 2 i D ˆ.H / D ˆ.G/ and, for each x 2 Z.G/ H , x 2 2 ha2 ; b 2 i ha4 ; b 4 i since ha4 ; b 4 i D ˆ.Z.H //. Suppose that n D 1 so that b is an involution, m 3, H Š M2mC1 , Z.H / D ha2 i. Hence Z.G/ Š C2m is cyclic and therefore we may choose c 2 Z.G/ H such that c 2 D a2 which gives .ca/2 D c 2 a2 D 1. As Œca; b D z, we get D D hca; bi Š D8 ,

100

2-groups with exactly one exceptional maximal subgroup

53

hca; bi \ Z.G/ D hzi which together with jG W Z.G/j D 4 gives G D DZ.G/. But D 2 .Z.G// contains a subgroup Q Š Q8 and so G D QZ.G/ with Z.G/ Š C2m , m 3, Q \ Z.G/ D Z.Q/ D hzi and we have obtained the groups stated in part (a) of our theorem. It remains to treat the case n 2. Suppose that there is an involution x 2 G H . We know that x 62 Z.G/ and so Œa; x ¤ 1 or Œb; x ¤ 1. Obviously, ha; b; xi D G. Suppose that Œa; x ¤ 1. As hai E G, ha; xi is minimal nonabelian of order 2mC1 . Because jGj D 2mCnC1 and n 2, ˆ.G/ha; xi D ha; x; b 2 i is a maximal subgroup of G which is neither abelian nor minimal nonabelian. Assume at the moment that Œb; x ¤ 1. In that case, hbi hzi is normal in G and so hb; xi is a nonmetacyclic minimal nonabelian subgroup of G of order 2nC2 . It follows that hb; xi must be maximal in G with jGj D 2nC3 (and so m D 2). But the case of a nonmetacyclic minimal nonabelian maximal subgroup in G will be studied in part (ii) of this proof. Hence we may assume Œx; b D 1 so that Œx; ab D Œx; aŒx; b D z which implies that hx; abi is minimal nonabelian and so hx; abi must be maximal in G. Now, habi covers H=hai Š C2n , n 2, and so o.ab/ 2n . We get n

n

n

n1 .2n 1/

.ab/2 D a2 b 2 Œb; a2 n

n

D a2 :

If n m, then we have .ab/2 D 1 and so o.ab/ D 2n and habi \ hai D ¹1º. Since hab; zi E G, we see that hab; xi is a nonmetacyclic minimal nonabelian subgroup of order 2nC2 . In that case, hab; xi must be maximal in G (with m D 2) and again this will be studied in part (ii) of this proof. It follows that we may assume n < m and we n n n set in that case s D m n 1. From .ab/2 D a2 and o.a2 / D 2s we infer that o.ab/ D 2nCs D 2m and habi hzi so that habi E G. Hence hab; xi is metacyclic minimal nonabelian of order 2mC1 and so hab; xi must be maximal in the group G. From jGj D 2mCnC1 follows n D 1, contrary to our assumption. We may assume Œa; x D 1 and so we must have Œb; x ¤ 1. Since hbi hzi E G, hb; xi is nonmetacyclic minimal nonabelian of order 2nC2 . If hb; xi is maximal in G, then this case will be treated in part (ii) of this proof. Thus we may assume that hb; xi is not maximal in G and so M D ˆ.G/hb; xi is maximal in G and M is neither abelian nor minimal nonabelian. It then follows that the subgroup hab; xi (with Œab; x D z), which is minimal nonabelian, must be also a maximal subgroup in G. Since habi n n covers H=hai, we have o.ab/ 2n , n 2 and .ab/2 D a2 . If n m, then we get o.ab/ D 2n and habi\hai D ¹1º and so hab; xi is nonmetacyclic minimal nonabelian of order 2nC2 . In that case, hab; xi is maximal in G (with m D 2) and again this will be treated in part (ii) of this proof. We may assume that n < m and then o.ab/ D 2m , habi hzi and so hab; xi is metacyclic minimal nonabelian of order 2mC1 . But then jGj D 2mCnC1 implies n D 1, contrary to our assumption. We have proved that we may assume that there are no involutions in G H . If there is c 2 Z.G/ H such that c 2 D h2 for some h 2 H , then the abelian subgroup hh; ci is noncyclic since hhi and hci are two distinct cyclic subgroups of hh; ci of the same order. But hh; ci\H D hhi and so there is an involution in hh; ciH , a contradiction.

54

Groups of prime power order

It follows that not every element in Ã1 .H / D Z.H / is a square of an element in H . By Proposition 26.23, H is not a powerful 2-group which implies H 0 D hzi 6 Ã2 .H /. This forces m D 2 and c 2 is not a square in H for any c 2 Z.G/ H . We compute for any integers i; j .ai b j /2 D a2i b 2j Œb j ; ai D a2i b 2j z ij : We get that a2i b 2j 2 Ã1 .H / D Z.H / is a square in H if and only if i or j is even. Therefore for any c 2 Z.G/ H , c 2 D a2i b 2j , where both i and j are odd, and then (since m D 2 and so a2 D z) c 2 D zb 2j , where j is odd. Consider the nonabelian subgroup S D ha; b j ci, where .b j c/2 D b 2j c 2 D b 2j zb 2j D z; and so S Š Q8 . Hence we conclude that G D hS; ci D S hci Š Q8 C2n with n 2, where S hb 2 i Š Q8 C2n1 is a unique maximal subgroup of G which is neither abelian nor minimal nonabelian. We have obtained the groups stated in part (b) of our theorem. (ii) It remains to consider the case where H is nonmetacyclic minimal nonabelian. We may set m

n

H D ha; b j a2 D b 2 D 1; Œa; b D z; z 2 D Œa; z D Œb; z D 1i; where we may assume m 2, n 1 as jH j 24 . Here H 0 D hzi, jH j D 2mCnC1 and so jGj D 2mCnC2 . Also, z is not a square in H , Z.H / D ha2 i hb 2 i hzi D ˆ.H / D ˆ.G/ and for each x 2 Z.G/ H , x 2 2 Z.H / ˆ.Z.H //. (ii1) First assume n D 1 so that Z.H / D ha2 i hzi and for a c 2 Z.G/ H , 2 c D a2i z j . Suppose that i is even and then j must be odd and so we may set in that 0 0 case c 2 D a4i z and compute for an element c 0 D a2i c 2 Z.G/ H 0

0

0

0

.c 0 /2 D .a2i c/2 D a4i c 2 D a4i a4i z D z: This gives G D H hc 0 i with .c 0 /2 D z, where hzi D H 0 , and it is easy to see that in that case G is an A2 -group (see Proposition 71.1), a contradiction. We have proved that i must be odd. The subgroup D D hai c; bi is minimal nonabelian since Œai c; b D Œa; bi D z. We have also .ai c/2 D a2i c 2 D a2i a2i z j D z j which shows that D Š D8 . But hci \ hzi D ¹1º, where hzi D Z.D/, and so hD; ci D hai c; b; ci D G D D hci with o.c/ D 2m , m 2. The subgroup D hc 2 i is a unique maximal subgroup of G which is neither abelian nor minimal nonabelian and we have obtained the groups stated in part (c) of our theorem. (ii2) It remains to consider the case n 2. In this case, for an element c 2 Z.G/H we have c 2 D a2i b 2j z k , where at least one of the integers i; j; k is odd.

100

2-groups with exactly one exceptional maximal subgroup

55

Suppose that both i and j are even so that in this case k is odd and we may set 0 0 0 0 c 2 D a4i b 4j z. For the element c 0 D a2i b 2j c, we get 0

0

0

0

0

0

.c 0 /2 D a4i b 4j c 2 D a4i b 4j a4i b 4j z D z; and so G D H hc 0 i with .c 0 /2 D z and hzi D H 0 which gives that G is an A2 -group from Proposition 71.1, a contradiction. Now assume that one of the integers i; j is even and the other one is odd. Note that i; j occur symmetrically and so we may assume that i is odd and j is even. In that case, the subgroup T D hai b j c; bi is minimal nonabelian since Œai b j c; b D Œa; bi D z. Using the fact that b j 2 Z.G/, we get .ai b j c/2 D a2i b 2j c 2 D a2i b 2j a2i b 2j z k D z k : Since ˆ.T / D hb 2 i hzi, we have jT j D 2nC2 . On the other hand, jGj D 2mCnC2 with m 2 and so ˆ.G/T is a maximal subgroup of G which is neither abelian nor minimal nonabelian. Consider now the minimal nonabelian subgroup U D hab; aci, where Œab; ac D z. We have .ab/2 D a2 b 2 z;

.ac/2 D a2 c 2 D a2 a2i b 2j z k D a2.iC1/ b 2j z k ;

where both i C 1 and j are even, and ˆ.U / D ha2 b 2 z; a2.i C1/ b 2j z k ; zi ha2 b 2 ; ziˆ.Z.H // since a2.i C1/ b 2j 2 ˆ.Z.H //. But Z.H / D ha2 i hb 2 i hzi and so d.Z.H // D 3 which gives ˆ.U / < Z.H / D ˆ.G/. This shows that ˆ.G/U is another maximal subgroup of G which is neither abelian nor minimal nonabelian, a contradiction. It remains to consider the possibility that both i and j are odd. Then we consider the minimal nonabelian subgroup V D hai b j c; bi, where Œai b j c; b D Œa; bi D z. We get .ai b j c/2 D .ai b j /2 c 2 D a2i b 2j z ij c 2 D a2i b 2j z a2i b 2j z k D z kC1 which shows that jV j D 2nC2 . But jGj D 2mCnC2 with m 2 and so ˆ.G/V is a maximal subgroup of G which is neither abelian nor minimal nonabelian. Now we consider a minimal nonabelian subgroup W D ha; bci, where Œa; bc D z. We compute .bc/2 D b 2 c 2 D b 2 a2i b 2j z k D a2i b 2.1Cj / z k ;

56

Groups of prime power order

where i is odd and 1 C j is even. Then we have ˆ.W / D ha2 ; .bc/2 ; zi D ha2 ; a2i b 2.1Cj / z k ; zi D ha2 ; ziˆ.Z.H // < Z.H / D ˆ.G/ since b 2.1Cj / 2 ˆ.Z.H //. Hence ˆ.G/W is another maximal subgroup of G which is neither abelian nor minimal nonabelian, a contradiction. In the rest of this section we shall assume that G is a title group with d.G/ D 3 which possesses at most one abelian maximal subgroup. We know that in that case jG W Z.G/j 8 and jG 0 j > 2. Let H be a maximal subgroup of G which is minimal nonabelian. Then ˆ.H / D Z.H / ˆ.G/ and jH W ˆ.H /j D 4. As jG W ˆ.H /j D 8, we must also have ˆ.H / D ˆ.G/. Let K ¤ H be another maximal subgroup of G which is minimal nonabelian. Then Z.K/ D ˆ.K/ D ˆ.G/ which implies ˆ.G/ Z.G/ and so ˆ.G/ D Z.G/. Let M be the unique maximal subgroup of G which is neither abelian nor minimal nonabelian. Since jM W ˆ.G/j D 4 and ˆ.G/ D Z.G/, we have M D S ˆ.G/, where S is minimal nonabelian, S \ˆ.G/ D ˆ.S/ < ˆ.G/ and so M 0 D S 0 Š C2 with d.M / 3. Theorem 100.5. Let G be a 2-group with d.G/ D 3 which has exactly one maximal subgroup M which is neither abelian nor minimal nonabelian. If G possesses exactly one abelian maximal subgroup A, then ˆ.G/ D Z.G/;

G 0 Š E4 ;

M 0 Š C2 ;

d.M / 3

and one of the following holds: (a) If G has no normal elementary abelian subgroup of order 8, then G is one of the groups given in Theorem 87.8(e). (b) If G has a normal subgroup E Š E8 but 1 .G/ > E, then G D ht; t 0 ; c j t 2 D t 02 D c 4 D 1; Œt; t 0 D c 2 D z; Œc; t D u; u2 D Œc; t 0 D Œu; t D Œu; t 0 D Œu; c D Œz; t D Œz; t 0 D 1i; where

jGj D 25 ;

G 0 D ˆ.G/ D Z.G/ D hz; ui Š E4 ;

1 .G/ D M D ht; t 0 iG 0 Š C2 D8 ; the subgroup A D ht 0 ; ciG 0 is abelian of type .4; 2; 2/ and the other ﬁve maximal subgroups of G are nonmetacyclic minimal nonabelian. (c) If G has a normal subgroup E Š E8 , 1 .G/ D E and E 6 A, then mC1

G D ha; b; t j a2

m

D b 4 D t 2 D 1; a2 D z; b 2 D u;

Œa; t D u; Œb; t D z; Œu; a D Œu; t D Œa; b D Œz; t D 1i;

100

2-groups with exactly one exceptional maximal subgroup

57

where jGj D 2mC4 ; m 2;

G 0 D hz; ui Š E4 ;

ˆ.G/ D Z.G/ D ha2 ; ui Š C2m C2 ; A D ha; bi is abelian of type .2mC1 ; 4/;

E 6 Z.G/;

M D hb; ti ha2 i;

where hb; ti is the nonmetacyclic minimal nonabelian group of order 24 and the other ﬁve maximal subgroups of G are minimal nonabelian. (d) If G has a normal subgroup E Š E8 , 1 .G/ D E and E A, then G D ha; b; d j a4 D b 2 D d 4 D 1; a2 D d 2 D z; Œa; d D z; Œa; b D c; c 2 D Œc; d D Œc; a D Œc; b D Œb; d D 1i; where jGj D 25 ;

G 0 D ˆ.G/ D Z.G/ D hz; ci Š E4 ; M D ha; d; ci Š Q8 C2 ;

A D hb; d iˆ.G/;

E 6 Z.G/

and the other ﬁve maximal subgroups of G are minimal nonabelian. Proof. Let 1 D ¹M; A; H1 ; : : : ; H5 º be the set of maximal subgroups of G, where H1 ; : : : ; H5 are minimal nonabelian. By Exercise 1.69(a), we have jG 0 W .A0 H10 /j D jG 0 W H10 j 2 and so jG 0 j D 4 (since jG 0 j > 2). If G 0 D hvi Š C4 , then we get H10 D D H50 D hv 2 i. But then G=hv 2 i is a nonabelian group with at least ﬁve abelian maximal subgroups Hi =hv 2 i, i D 1; : : : ; 5, a contradiction. Hence we obtain that G 0 Š E4 . Since A=M 0 and M=M 0 are two abelian maximal subgroups of the nonabelian group G=M 0 , it follows that there is exactly one minimal nonabelian maximal subgroup of G, say H5 , such that H50 D M 0 . With similar arguments we see that we may assume that H10 D H20 , H30 D H40 , H50 D M 0 are three pairwise distinct subgroups of order 2 in G 0 Š E4 . Suppose that the group G has no normal elementary abelian subgroup of order 8. Then A; H1 ; : : : ; H5 are metacyclic and so M is the only maximal subgroup of G which is nonmetacyclic. Since d.G/ D 3 and G 0 Š E4 , we see that G is isomorphic to one of the groups stated in Theorem 87.8(e) which gives part (a) of our theorem. From now on we assume that G has a normal elementary abelian subgroup E of order 8. Suppose at the moment that G possesses an elementary abelian subgroup F of order 16. Obviously, F is a maximal elementary abelian subgroup in G. Indeed, if X is an elementary abelian subgroup of order 32 in G, then jX \H1 j D 16, a contradiction. Since G 0 Z.G/, we have G 0 F and so F E G. If G=F is noncyclic, then there are at least three distinct maximal subgroups of G containing F and so at least one of them is minimal nonabelian, a contradiction. Hence G=F is cyclic and let a 2 G F be such that hai covers G=F . Suppose that jG W F j D 2 holds so that F is an abelian

58

Groups of prime power order

maximal subgroup in G. Since CF .a/ D Z.G/ D ˆ.G/ and jG=ˆ.G/j D 8, we get CF .a/ D G 0 Š E4 . By Lemma 99.2, G Š E4 o C2 and so we may assume that a is an involution. Let f1 ; f2 2 F G 0 so that F D hf1 ; f2 i G 0 and G D hf1 ; f2 ; ai. We have ha; f1 i Š ha; f2 i Š D8 and ha; f1 iG 0 and ha; f2 iG 0 are two distinct maximal subgroups of G which are isomorphic to D8 C2 (and so they are neither abelian nor minimal nonabelian), a contradiction. We have proved that G=F Š C2m , m 2. Since a2 2 ˆ.G/ D Z.G/, a induces an involutory automorphism on F which together with jG W Z.G/j D 8 implies CF .a/ D G 0 . As a2 62 F and 1 .hai/ Z.G/, we must have 1 .hai/ F (because E32 is not a subgroup of G). Hence hai \ F D hai \ G 0 D hzi Š C2 and so o.a/ D 2mC1 ;

m 2;

jGj D 2mC4 ;

Z.G/ D ˆ.G/ D G 0 ha2 i:

We may set F D hx; y; u; zi, where G 0 D hu; zi, Œa; x D u, Œa; y D z, G D hx; y; ai and so the structure of G is completely determined. Now, ha; yi Š hax; yi Š M2mC2 so that hui ha; yi and hui hax; yi are two distinct maximal subgroups of G which are neither abelian nor minimal nonabelian, a contradiction. We have proved that G does not possess an elementary abelian subgroup of order 16. Since G 0 Z.G/, we have G 0 < E Š E8 . Next suppose that E < 1 .G/ so that there is an involution t 2 G E with CE .t/ D G 0 and S D hE; ti E G. If G=S is noncyclic, then S is contained in a maximal subgroup of G which is minimal nonabelian, a contradiction (to the structure of minimal nonabelian 2-groups). Hence G=S is cyclic and let c 0 2 G S be such that hc 0 i covers G=S. Let t 0 2 E G 0 so that 1 ¤ Œt; t 0 D z 2 G 0 and ht; t 0 i D D Š D8 . Also, E1 D hG 0 ; ti is another elementary abelian normal subgroup of order 8 in G. All elements in S .E [ E1 / are of order 4 and v D t t 0 is one of them. We infer that v 2 D z, S D D hui Š D8 C2 , where u 2 G 0 hzi, and also Z.S/ D G 0 with Z.G/ D ˆ.G/ D G 0 hc 02 i so that G D ht; t 0 ; c 0 i and M D S hc 02 i must be a unique maximal subgroup of G which is neither abelian nor minimal nonabelian. If x 2 G M , then x is either an involution or 1 .hxi/ G 0 . If x is an involution, then Œt; x ¤ 1 (because E16 is not a subgroup of G) and so ht; xi Š D8 , jG W S j D 2, jGj D 25 , and ht; xiG 0 would be another maximal subgroup of G which is neither abelian nor minimal nonabelian, a contradiction. We have proved that x is not an involution and so 1 .hxi/ G 0 . Indeed, 1 .hxi/ Z.G/ and so 1 .hxi/ S which implies that 1 .hxi/ Z.S/ D G 0 . Now, A \ M is equal to one of three abelian maximal subgroups of M (containing ˆ.G/ D Z.G/): hE1 ; c 02 i, hE; c 02 i, hG 0 hvi; c 02 i, where A is the unique abelian maximal subgroup of G. We choose c 2 A M (instead of c 0 ), where hci also covers G=S , o.c/ 4, 1 .hci/ G 0 , ˆ.G/ D Z.G/ D G 0 hc 2 i, and c centralizes exactly one of the elements in the set ¹t; t 0 ; v D t t 0 º. Indeed, otherwise, G=hzi would be abelian since c generates G together with any two elements in the above set. But then G 0 D hzi, a contradiction. Interchanging t and t 0 (if necessary), we may assume that Œc; t 0 D 1 or Œc; t t 0 D 1. In that case, Œc; t ¤ 1 and if Œc; t D z, then again G=hzi would be abelian, a contradiction. It follows that we may set Œc; t D u 2 G 0 hzi.

100

2-groups with exactly one exceptional maximal subgroup

59

First assume Œc; t t 0 D 1 which gives Œc; t 0 D u. We have M D ht; t 0 iˆ.G/ and A D hc; t t 0 iˆ.G/. It follows that the other ﬁve maximal subgroups ˆ.G/T of G must be minimal nonabelian, where T is one of the following minimal nonabelian subgroups: ht; ci; ht 0 ; ci; ht; t 0 ci; ht 0 ; tci; ht t 0 ; tci: We have to show that in each of these cases ˆ.T / ˆ.G/ D hG 0 ; c 2 i D hu; z; c 2 i. Note that Œt; c D u and so ˆ.ht; ci/ D hc 2 ; ui which implies 1 .hci/ 2 ¹hzi; huziº. We have Œt; t 0 c D zu and so ˆ.ht; t 0 ci/ D h.t 0 c/2 D c 2 u; zui. Here if o.c/ 8, then 1 .ht 0 ci/ D 1 .hci/ and then 1 .hci/ 2 ¹hzi; huiº which together with the above result gives 1 .hci/ D hzi, and if o.c/ D 4, then c 2 D uz. We have Œt t 0 ; tc D z and so ˆ.ht t 0 ; tci/ D h.t t 0 /2 D z; .t c/2 D c 2 u; zi D hc 2 u; zi: If o.c/ D 4, then by the above c 2 D uz and then ˆ.ht t 0 ; tci D hzi, a contradiction. In case o.c/ 8, then by the above 1 .hci/ D hzi and as 1 .hc 2 ui/ D 1 .hci/ D hzi, we get again ˆ.ht t 0 ; tci/ D hc 2 ui 6 ˆ.G/, a contradiction. Now assume Œc; t 0 D 1 and from before we know that Œt; t 0 D z and Œc; t D u. We have here M D ht; t 0 iˆ.G/ and A D hc; t 0 iˆ.G/ so that the other ﬁve maximal subgroups must be minimal nonabelian. Since Œt; c D u, we get ˆ.ht; ci/ D hc 2 ; ui which gives 1 .hci/ 2 ¹hzi; huziº. Further, Œt; t 0 c D zu and so we conclude that ˆ.ht; t 0 ci/ D h.t 0 c/2 D c 2 ; zui which implies that 1 .hci/ 2 ¹hui; hziº. This fact together with our previous result gives 1 .hci/ D hzi. We have Œt 0 ; tc D z and so ˆ.ht 0 ; tci/ D h.tc/2 D c 2 u; zi. If o.c/ 8, then .c 2 u/2 D c 4 and hc 4 i hzi since 1 .hci/ D hzi. In this case, ˆ.ht 0 ; tci/ 6 ˆ.G/, a contradiction. Hence o.c/ D 4 and so c 2 D z. We have obtained a uniquely determined group of order 25 given in part (b) of our theorem. From now on we may assume that 1 .G/ D E Š E8 . (i) Assume that 1 .G/ D E 6 Z.G/ D ˆ.G/ and E 6 A, where A is the unique abelian maximal subgroup of G. Then A \ E D G 0 , A covers G=E and A is metacyclic. Since there are three maximal subgroups of G containing E, there exists at least one of them, denoted by H , which is minimal nonabelian. If H=E is noncyclic, then there are two distinct maximal subgroups X1 ¤ X2 of H containing E. In that case, E X1 \ X2 D ˆ.H / D ˆ.G/ D Z.G/, a contradiction. Hence H=E is cyclic. Since d.G=E/ D 2, G=E is abelian of type .2m ; 2/, m 1. Therefore A=G 0 Š G=E is of type .2m ; 2/, where A \ H=G 0 Š C2m . Let a be an element in A \ H such that hai covers A \ H=G 0 . Noting that 1 .G/ D E, we have o.a/ D 2mC1 and 1 .hai/ D hzi G 0 , where m z D a2 . If t 2 E G 0 , then Œt; a D u 2 G 0 hzi because H is nonmetacyclic and therefore u is not a square in H . Since A=G 0 Š C2m C2 , there is an element b 2 A H such that 1 ¤ b 2 2 G 0 and b 2 ¤ z. Indeed, if b 2 D z, then taking an element v of order 4 in hai, we get .bv/2 D b 2 v 2 D z 2 D 1, where bv 2 A H , a contradiction. Hence we get that b 2 2 ¹u; uzº. We have ˆ.H / D ˆ.G/ D ha2 ; ui

60

Groups of prime power order

and G D ha; b; t i. If Œb; t 2 hui, then G=hui is abelian, a contradiction. Hence we obtain Œb; t 2 ¹z; uzº, o.b/ D 4 and A is abelian of type .4; 2mC1 /. We set Œb; t D zu and b 2 D uz , ; D 0; 1. First suppose that o.a/ > 4 so that ha4 i hzi. In that case, ˆ.hb; ti/ D hb 2 ; Œb; ti G 0 < ˆ.G/ D ha2 ; ui and so M D hb; tiˆ.G/. The fact that ˆ.hab; ti/ D h.ab/2 D a2 uz ; Œab; t D zuC1 i D ˆ.G/ gives D 0. We may assume that b 2 D u, i.e., D 0. Indeed, if b 2 D uz D u0 , then we replace H D ha; ti with H1 D ha0 D ab; ti, where o.a0 / D o.a/, ha0 i hzi and Œa0 ; t D uz D u0 and so writing again a and u instead of a0 and u0 , respectively, we have obtained the relations for groups G of order 2mC4 given in part (c) of our theorem. It remains to examine the case o.a/ D 4. In this case, m D 1, a2 D z, G is a special group of order 25 , where ˆ.G/ D hu; zi. We have A D ha; bi Š C4 C4 and ha; ti D H is the nonmetacyclic minimal nonabelian group of order 24 . Further, Œb; t D zu

with

ˆ.hb; ti/ D huz ; zu i;

Œab; t D zuC1 with

ˆ.hab; ti/ D huz C1 ; zuC1 i;

Œb; at D zu

ˆ.hb; ati/ D huz ; uz; zu i

with

and Œab; at D zuC1 with

ˆ.hab; ati/ D huz C1 ; uz; zuC1 i:

If D D 0, then ˆ.hab; ti/ D huzi and ˆ.hab; ati/ D huzi. In case D D 1, we get ˆ.hb; ti/ D huzi and ˆ.hb; ati/ D huzi. It follows that in the above two cases our group G has two distinct maximal subgroups which are neither abelian nor minimal nonabelian, a contradiction. Therefore we must have ¤ in which case we may set D C 1. But in this case, we check that each nonabelian maximal subgroup of G is minimal nonabelian and so G would be an A2 -group, a contradiction. (ii) Assume that 1 .G/ D E 6 Z.G/ D ˆ.G/ and E A, where A is the unique abelian maximal subgroup of G. Since there are three maximal subgroups of G containing E, there is a maximal subgroup H of G containing E which is minimal nonabelian. Then H is nonmetacyclic with Z.H / \ E D G 0 and H=E is cyclic. Taking an element b 2 E G 0 , we may set ˛

H D ha; b j a2 D b 2 D 1; ˛ 2; Œa; b D c; c 2 D Œa; c D Œb; c D 1i; where hci D H 0 , Z.H / D hci ha2 i, jGj D 2˛C3 , and setting a2 D z we have G 0 D hz; ci Š E4 since c is not a square in H . Here ha; ci (containing G 0 ) is an abelian normal subgroup of type .2˛ ; 2/ in G having exactly two cyclic subgroups hai and haci of order 2˛ . Since ab D ac, we obtain that NH .hai/ D ha; ci which implies that ˛1

100

2-groups with exactly one exceptional maximal subgroup

61

N D NG .hai/ covers G=H and N \ H D ha; ci. It follows that N is a nonabelian maximal subgroup of G (because A E), where N=G 0 Š G=E is noncyclic abelian and so N=G 0 is of type .2; 2˛ /. Hence there is d 2 N H with 1 ¤ d 2 2 G 0 and so o.d / D 4. But d normalizes hai and therefore Œd; a 2 hai \ G 0 D hzi which gives Œd; a D z. There exist exactly three maximal subgroups of G containing E: H D ha; bi, hd; biˆ.H /, had; biˆ.H /, where exactly one of two last subgroups is abelian. It follows that either Œd; b D 1 or Œad; b D 1 in which case Œd; b D c (since Œa; b D c). We may set Œd; b D c , where D 0; 1, and note that G D ha; b; d i. First assume that ˛ 3. If d 2 D z, then hd; ai Š M2˛C1 and there are involutions ˛2 in hd; aihai, a contradiction. Thus d 2 2 G 0 hzi in which case hd; a2 i Š C4 C4 ˛2 ˛2 since a2 2 Z.G/. Hence, replacing d with da2 (if necessary), we may assume that d 2 D c. If D 0, then A D hd; biˆ.G/ is an abelian maximal subgroup of G and we check that all other six maximal subgroups of G are minimal nonabelian and so G is an A2 -group, a contradiction. Hence we must have D 1. We have Œb; d D c and ˆ.hb; d i/ D hci < ˆ.G/. Also, Œab; ad D z and ˆ.hab; ad i/ D ha2 c; a2 cz; zi D ha2 ci < ˆ.G/ since .a2 c/2 D a4 and ha4 i hzi. Hence hb; d iˆ.G/ and hab; ad iˆ.G/ are two distinct maximal subgroups of G which are neither abelian nor minimal nonabelian, a contradiction. We have proved that we must have ˛ D 2 so that a2 D z, G is special satisfying ˆ.G/ D hz; ci and jGj D 25 . We have Œd; b D c and if D 1, then we replace d with d 0 D ad so that Œd 0 ; b D 1 and Œa; d 0 D z. Writing again d instead of d 0 , we may assume that Œd; b D 1 and Œa; d D z (as before). If d 2 D z, then we obtain the group of order 25 given in part (d) of our theorem. It remains to analyze the cases d 2 2 ¹c; czº. If d 2 D c, then we infer that hb; ad i Š D8 since Œb; ad D c and .ad /2 D a2 d 2 Œd; a D zcz D c. This is a contradiction since 1 .G/ Š E8 . Suppose that d 2 D cz. In that case, we replace a with a0 D ab, z with z 0 D zc and d with d 0 D db. Then we get a02 D .ab/2 D zc D z 0 ;

d 02 D .db/2 D d 2 D cz D z 0 ;

Œa0 ; d 0 D Œab; db D zc D z 0 ;

Œa0 ; b D Œab; b D c;

Œb; d 0 D Œb; db D 1;

and so, writing again a; z; d instead of a0 ; z 0 ; d 0 , respectively, we obtain again the group given in part (d) of our theorem. (iii) We turn now to the difﬁcult case, where 1 .G/ D E Z.G/ D ˆ.G/. Let H1 D H be a maximal subgroup of G which is minimal nonabelian and such that H 0 ¤ M 0 . Since Z.H / D Z.G/ E, H is nonmetacyclic and we may set ˛

ˇ

H D ha; b j a2 D b 2 D 1; Œa; b D c; c 2 D Œa; c D Œb; c D 1i; where ˛ 2, ˇ 2, hci D H 0 , Z.H / D ˆ.G/ D hci ha2 i hb 2 i is abelian of type ˛1 ˇ1 .2˛1 ; 2ˇ 1 ; 2/, and jGj D 2˛Cˇ C2 . We have G 0 < E D ha2 ; a2 ; ci Š E8 .

62

Groups of prime power order

We consider the group G=M 0 , where .G=M 0 /0 D G 0 =M 0 Š C2 , d.G=M 0 / D 3, G=M 0 has exactly three abelian maximal subgroups A=M 0 , M=M 0 and H5 =M 0 (since H50 D M 0 ) and the other four maximal subgroups Hi =M 0 , i D 1; : : : ; 4, are minimal nonabelian. Thus G=M 0 is an A2 -group of Proposition 71.1 which implies that there is an element d 2 G H such that Œd; G D M 0 and 1 ¤ d 2 2 G 0 . Since there are exactly three maximal subgroups of G containing hd i Š C4 , at least one of them H is minimal nonabelian, where H E and so H is nonmetacyclic. Thus .H /0 is a maximal cyclic subgroup in H and so .H /0 ¤ hd 2 i which gives G 0 D h.H /0 ; d 2 i. Taking an element a 2 .H \ H / ˆ.G/, we conclude that H D ha ; d i and so ˆ.H / D h.a /2 ; d 2 ; .H /0 i D h.a /2 ; G 0 i D ˆ.G/ and E D h1 .ha i/; G 0 i. Set o.a / D 2 , 2, so that ˆ.H / D ˆ.G/ is of type .2 1 ; 2; 2/. On the other hand, ˆ.G/ is of type .2˛1 ; 2ˇ 1 ; 2/. Interchanging the elements a and b (if necessary), we may assume that ˇ D 2 and then D ˛ so that ˆ.G/ is of type .2˛1 ; 2; 2/ and o.b/ D 4. Since Œd; G D M 0 , we have hŒd; a i D M 0 and so .H /0 D M 0 and therefore H D H5 and d 2 2 G 0 M 0 . Because H =E is abelian of type .2˛1 ; 2/, it follows that H \ H =E is cyclic of order 2˛1 and so ha i covers H \ H =E. Since H \ H D ha i G 0 , it follows that H=G 0 is abelian of type .2˛ ; 2/ which implies that H D H0 .H \H / with H0 \.H \H / D G 0 and jH0 W G 0 j D 2. Hence there is an element b 2 H H with 1 ¤ .b /2 2 G 0 , ha ; b i D H , .b /2 ¤ c (since H 0 D hci and so c is not a square in H ) and so Œa ; b D c and o.b / D 4. In addition, from Œd; G D M 0 follows that either Œd; b D m with hmi D M 0 or Œd; b D 1. Also d 2 D cm and .b /2 D c m, where ; D 0; 1. Assume that Œd; b D 1 in which case d 2 ¤ .b /2 (because if d 2 D .b /2 , then db is an involution in G H , a contradiction) and so hd; b i Š C4 C4 and hd 2 ; .b /2 i D G 0 . In that case, Œb a ; d D m since Œa ; d D m and we get ˆ.hb a ; d i/ D h.b a /2 D c m .a /2 c D .a /2 c C1 m; d 2 D cm ; mi D G 0 h.a /2 i D ˆ.G/; and so hb a ; d i is a minimal nonabelian maximal subgroup of G which satisﬁes hb a ; d i0 D hmi and hb a ; d i ¤ H . This is a contradiction since H D H5 is the only maximal subgroup of G which is minimal nonabelian and .H /0 D M 0 D hmi. We have proved that Œd; b D m which together with hb ; d i ¤ H implies that hb ; d iˆ.G/ D M must be the unique maximal subgroup of G which is neither abelian nor minimal nonabelian. If D 0 and D 1, then .b /2 D cm, d 2 D c and therefore .b d /2 D cm c m D 1 and b d would be an involution in G H , a contradiction. Hence we have either D or D 1 and D 0. In this case, Œa ; b d D cm and ˆ.ha ; b d i/ D h.a /2 ; c C1 m ; cmi D ˆ.G/ implies that D 1 and D 0 is not possible and so D . Further, we have Œb ; a d D cm and so ˆ.hb ; a d i/ D hc m; .a /2 cmC1 ; cmi forces that D D 0. But then Œa b ; a d D c shows together with ˆ.ha b ; a d i/ D h.a /2 cm; ci < ˆ.G/ that ha b ; a d iˆ.G/ is another maximal subgroup of G which is neither abelian nor minimal nonabelian, a contradiction. Our theorem is proved.

100

2-groups with exactly one exceptional maximal subgroup

63

Theorem 100.6. Let G be a 2-group with d.G/ D 3 which has exactly one maximal subgroup M which is neither abelian nor minimal nonabelian. If G has no abelian maximal subgroups, then we get n

n1

G D ha; b; c j a4 D b 4 D c 2 D 1; a2 D x; b 2 D y; c 2

D z;

Œa; b D z; Œa; c D y; Œb; c D xy; Œx; b D Œx; c D Œy; a D Œy; c D Œz; a D Œz; b D 1i; where jGj D 2nC4 , n 3, G 0 D hx; y; zi Š E8 , Z.G/ D ˆ.G/ D G 0 hc 2 i is abelian of type .2n1 ; 2; 2/, M D ˆ.G/ha; bi D hc 2 iha; bi with hc 2 i\ha; bi D hzi D ha; bi0 , hc 2 i Š C2n1 and ha; bi is the nonmetacyclic minimal nonabelian group of exponent 4 and order 25 and the other six maximal subgroups of G are nonmetacyclic minimal nonabelian. Proof. We put 1 D ¹H1 ; H2 ; : : : ; H6 ; M º to be the set of maximal subgroups of G, where H1 ; : : : ; H6 are minimal nonabelian. We know that jM 0 j D 2 and d.M / 3 (see the remark preceding Theorem 100.5). By a result of A. Mann, jG 0 W .H10 H20 /j 2 and so jG 0 j 8. However, if jG 0 j D 2, then we know that G has three abelian maximal subgroups (see the second paragraph of the proof of Theorem 100.4), a contradiction. Hence jG 0 j D 4 or jG 0 j D 8. Suppose that for some Hi ¤ Hj we have Hi0 D Hj0 . Then by a result of A. Mann, we get jG 0 j D 4 and moreover G 0 Š E4 . Indeed, if G 0 Š C4 , then the nonabelian group G=1 .G 0 / would possess at least six abelian maximal subgroups Hi =1 .G 0 /, i D 1; : : : ; 6, a contradiction. The group G=M 0 is obviously an A2 -group satisfying .G=M 0 /0 Š C2 . By Proposition 71.1, G=M 0 has exactly three abelian maximal subgroups so that we may set H50 D H60 D M 0 D hui. In that case, Hi0 , i D 1; : : : ; 4, cannot be all pairwise distinct and so we may set H20 D H30 D H40 D hvi with hu; vi D G 0 and G=hvi has exactly three abelian maximal subgroups Hi =hvi, i D 2; 3; 4. It follows that we must have H10 D huvi so that G=huvi with .G=huvi/0 Š C2 has exactly one abelian maximal subgroup H1 =huvi. If d.M=huvi/ D 2, then M=huvi is minimal nonabelian so that G=huvi would be an A2 -group. But in that case (Proposition 71.1), G=huvi would have three abelian maximal subgroups, a contradiction. Hence we must have d.M=huvi/ 3 in which case M=huvi is a unique maximal subgroup of G=huvi which is neither abelian nor minimal nonabelian. By Theorem 100.5, we must have .G=huvi/0 Š E4 , a contradiction. We have proved that all Hi0 are pairwise distinct subgroups of order 2 in G 0 . This implies that G 0 Š E8 . If M 0 D Hi0 for some i 2 ¹1; 2; : : : ; 6º, then considering G=M 0 we see that there must exist a maximal subgroup Hj , j ¤ i , such that M 0 D Hi0 D Hj0 , a contradiction. Hence ¹H10 ; : : : ; H60 ; M 0 º is the set of seven pairwise distinct subgroups of order 2 in G 0 . Since G 0 Hi , all Hi (i D 1; : : : ; 6 ) are nonmetacyclic minimal nonabelian. The group G=G 0 is abelian of rank 3. Suppose that there is an involution t 2 G G 0 . Then F D G 0 ht i Š E16 and G=F is noncyclic. But then there is a maximal subgroup H of G such that H F and H is minimal nonabelian,

64

Groups of prime power order

a contradiction. We have proved that G 0 D 1 .G/. Set T =G 0 D 1 .G=G 0 / Š E8 . If G=T is noncyclic, then there is a maximal subgroup K of G such that K T and K is minimal nonabelian. But then d.K=G 0 / D 3, a contradiction. Hence G=T is cyclic and so G=G 0 is abelian of type .2m ; 2; 2/, m 1. (i) First assume m D 1, i.e., T D G, G=G 0 Š E8 and G is a special group with 0 G D 1 .G/ Š E8 . We shall determine the structure of M > G 0 . We have M D G 0 S , where S D ha; bi is minimal nonabelian and G 0 \ S D ˆ.S/ < G 0 . Set hzi D S 0 D M 0 Š C2 . Suppose at the moment that ˆ.S/ D hzi so that S Š Q8 . Then G=hzi is an A2 -group, where M=hzi Š E16 is a unique abelian maximal subgroup of G=hzi and E4 Š .G=hzi/0 Z.G=hzi/. But then Proposition 71.4(b) implies that 1 .G=hzi/ Š E8 , a contradiction. We have proved that ˆ.S/ Š E4 and 1 .S/ D ˆ.S/. Hence S is the metacyclic minimal nonabelian group of order 16 and exponent 4. We choose a; b 2 S ˆ.S/ so that a2 D z, b 2 D y, Œa; b D z and hy; zi D ˆ.S/ D ˆ.M /. Since G 0 D ˆ.G/, there is c 2 G M such that c 2 D x 2 G 0 hy; zi. We obtain hx; y; zi D G 0 and ha; b; ci D G. All other six maximal subgroups (distinct from M ) are nonmetacyclic minimal nonabelian. We have ˆ.ha; ci/ D hz; x; Œa; ci D G 0 so that Œa; c D x ˛ yz ˇ . Also, ˆ.hb; ci/ D hy; x; Œb; ci D G 0 gives Œb; c D x y ı z. Further, ˆ.hab; ci/ D hy; x; Œab; c D x ˛C y ıC1 z ˇ C1 i D G 0 which implies ˇ D 0. From ˆ.hb; aci/ D hy; x ˛C1 yz; Œb; ac D x y ı i D G 0 we get D 1, and from ˆ.hab; aci/ D hy; x ˛C1 yz; Œab; ac D x ˛C1 y ıC1 i D G 0 it follows ˛ D 0. Finally, ˆ.ha; bci/ D hz; y ıC1 z; Œa; bc D zyi D hy; zi < G 0 gives a contradiction since G 0 ha; bci is another maximal subgroup of G (distinct from the subgroup M ) which is neither abelian nor minimal nonabelian. (ii) Suppose that T < G, where T =G 0 D 1 .G=G 0 / Š E8 and ¹1º ¤ G=T is cyclic so that G=G 0 is abelian of type .2m ; 2; 2/, m 2. The unique maximal subgroup of G containing T must be equal to M . There are normal subgroups U and V of G such that G D U V , U \ V D G 0 , U=G 0 Š E4 and V =G 0 is cyclic of order 2m , m 2. Let c be an element in V G 0 such that hci covers V =G 0 . We have o.c/ D 2n , n 3, where n D mC1 (noting that 1 .G/ D G 0 ). n1 Set hzi D 1 .hci/ so that z D c 2 and z 2 G 0 . Then we conclude that M D U hc 2 i, 0 2 ˆ.G/ D Z.G/ D G hc i is abelian of type .2n1 ; 2; 2/ and jGj D 2nC4 . Suppose that a; b 2 U G 0 satisfy U D G 0 ha; bi, where a2 ; b 2 2 G 0 and G D ha; b; ci. Since each maximal subgroup Hi (i D 1; : : : ; 6) is nonmetacyclic and contains ˆ.G/ and z

100

2-groups with exactly one exceptional maximal subgroup

65

is a square in ˆ.G/, it follows that Hi0 ¤ hzi for all i D 1; : : : ; 6. This implies that M 0 D hzi and therefore Œa; b D z. Now, G=hzi has the unique maximal abelian subgroup M=hzi and six minimal nonabelian maximal subgroups Hi =hzi, i D 1; : : : ; 6, and so G=hzi is an A2 -group with the following properties. We have d.G=hzi/ D 3 and so G=hzi is nonmetacyclic of order 2nC3 > 24 since n 3, .G=hzi/0 Š E4 , G 0 =hzi Z.G=hzi/ (as G 0 Z.G/) and G=hzi has a normal elementary abelian subgroup hG 0 ; 2 .hci/i=hzi of order 8. Hence G=hzi is an A2 -group from Proposition 71.4(b) which implies the fact that hG 0 ; 2 .hci/i=hzi D 1 .G=hzi/. Set a2 D x and b 2 D y and consider the abelian group M=hzi. If the abelian subgroup U=hzi of order 16 and exponent 4 has rank greater than 2, then we get 1 .U=hzi/ > G 0 =hzi, which contradicts the above fact. Hence we must have U=hzi Š C4 C4 which implies that G 0 D hx; y; zi. Since all Hi are minimal nonabelian (containing ˆ.G/ D hc 2 ; x; yi), we get Œa; c D x ˛ yz ˇ and Œb; c D xy z ı , where ˛; ˇ; ; ı D 0; 1. From ˆ.hab; ci/ D h.ab/2 D xyz; c 2 ; Œab; c D x ˛C1 y C1 z ˇ Cı i D ˆ.G/ and the fact that 1 .hc 2 i/ D hzi it follows that ˛ C 1 ¤ C 1 which gives D ˛ C 1 and Œb; c D xy ˛C1 z ı . Interchanging a and b respectively x and y, i.e., writing a0 D b, b 0 D a, x 0 D y, 0 y D x, we get a02 D b 2 D y D x 0 ;

b 02 D a2 D x D y 0 ;

Œa0 ; b 0 D Œb; a D z; Œa0 ; c D Œb; c D xy ˛C1 z ı D .x 0 /˛C1 y 0 z ı ; Œb 0 ; c D Œa; c D x ˛ yz ˇ D x 0 y 0˛ z ˇ : Writing again a; b; x; y instead of a0 ; b 0 ; x 0 ; y 0 , respectively, we obtain a2 D x; Œa; c D x ˛C1 yz ı ;

b 2 D y;

Œa; b D z;

Œb; c D xy ˛ z ˇ ;

ˇ; ı D 0; 1;

which are the old relations in which ˛ is replaced with ˛ C 1. This shows that we may assume ˛ D 0 and so we get Œa; c D yz ˇ and Œb; c D xyz ı , ˇ; ı D 0; 1. Finally, replacing c with c 0 D caı b ˇ , we get c 02 D c 2 .aı b ˇ /2 Œaı b ˇ ; c D c 2 l with l 2 G 0 and so c 04 D c 4 , hc 0 i covers G=U and hc 0 i \ G 0 D hzi. In addition, we have Œa; c 0 D Œa; caı b ˇ D Œa; cŒa; bˇ D yz ˇ z ˇ D y; Œb; c 0 D Œb; caı b ˇ D Œb; cŒb; aı D xyz ı z ı D xy: Writing again c instead of c 0 , we obtain the relations stated in our theorem.

102

p-groups G with p > 2 and d.G / > 2 having exactly one maximal subgroup which is neither abelian nor minimal nonabelian

In this section we determine up to isomorphism the title groups (Theorems 102.1, 102.3 and 102.5). It is obvious that for such groups G we have d.G/ D 3. All resulting groups will be presented in terms of generators and relations. But we shall also state all important characteristic subgroups of these groups so that the results could be useful for applications. In 100, 101 and 102 the difﬁcult problem Nr. 861 stated by the ﬁrst author is completely solved. Theorem 102.1. Let G be a p-group, p > 2, which possesses exactly one maximal subgroup which is neither abelian nor minimal nonabelian. Suppose that d.G/ D 3 and G has more than one abelian maximal subgroup. Then we have one of the following possibilities: (a) G D U Z, where U Š S.p 3 /, Z Š Cpm , m 3, and U \ Z D Z.U /. Here G is an L3 -group. (b) G D U Z, where U Š S.p 3 / or U Š Mp3 and Z Š Cpm , m 2. Conversely, the groups in (a) and (b) satisfy the assumptions of our theorem. Proof. Our assumptions imply that G=Z.G/ Š Ep2 and G has exactly p C 1 abelian maximal subgroups (Exercise 1.6(a)). By Lemma 1.1, jGj D pjG 0 jjZ.G/j gives that jG 0 j D p. Conversely, assume that G is a title group with jG 0 j D p. Since G has at most pC1 abelian maximal subgroups, there is a minimal nonabelian maximal subgroup H . From jG 0 j D p we conclude that G D H CG .H / (Exercise P10), where H \CG .H / D Z.H / D ˆ.H / D ˆ.G/ and CG .H / D Z.G/. We have G=Z.G/ Š Ep2 and so all p C 1 maximal subgroups of G containing Z.G/ are abelian. In what follows H will denote a ﬁxed maximal subgroup of G which is minimal nonabelian. Suppose that there exists an element c 2 Z.G/ H of order p. Then we have G D H hci and so each maximal subgroup of G which does not contain hci is isomorphic to G=hci Š H and so is minimal nonabelian, a contradiction. Hence there are no elements of order p in Z.G/H which implies 1 .Z.H // D 1 .Z.G// so that d.Z.H // D d.Z.G//. It follows that for each x 2 Z.G/ H , x p 2 Z.H / ˆ.Z.H //.

78

Groups of prime power order

Obviously, jGj p 5 since the exceptional maximal subgroup M (which is neither abelian nor minimal nonabelian) is of order p 4 . (i) First assume that H is metacyclic. We may set m

n

ha; b j ap D b p D 1; ab D az; z D ap

m1

i;

where m 2, n 1, mCn 4, H 0 D hzi D G 0 , jH j D p mCn , and jGj D p mCnC1 . We see that Z.H / D hap i hb p i D ˆ.H / D ˆ.G/ and for each x 2 Z.G/ H , 2 2 2 2 x p 2 hap ; b p i hap ; b p i since hap ; b p i D ˆ.Z.H //. Suppose that n D 1 so that o.b/ D p, m 3, H Š MpmC1 and Z.H / D hap i. Here Z.G/ Š Cpm is cyclic and so there is c 2 Z.G/ H such that c p D ap which gives .ca/p D c p ap D 1 and o.ca/ D p. Since Œca; b D z, it follows that U D hca; bi Š S.p 3 /, U \ Z.G/ D hzi which together with jG W Z.G/j D p 2 gives G D U Z.G/. We have obtained the groups stated in part (a) of our theorem. Here M D U hc p i, all p C 1 maximal subgroups of G containing hci are abelian and we have to show that all subgroups hc i a; c j bi are minimal nonabelian maximal subgroups of G for all integers i; j .mod p/ unless i 1 .mod p/ and j 0 .mod p/. Indeed, Œc i a; c j b D Œa; b D z and so, by Exercise P9, hc i a; c j bi is minimal nonabelian and ˆ.hc i a; c j bi/ D h.c i a/p D c pi ap D api ap D ap.i C1/ ; .c j b/p D apj ; zi D hap i D ˆ.G/ if and only if either i 6 1 .mod p/ or j 6 0 .mod p/. It remains to treat the case n 2. Suppose, in addition, that there exists an element x 2 G H of order p. We know that x 62 Z.G/ and so Œa; x ¤ 1 or Œb; x ¤ 1. Obviously, ha; b; xi D G. First suppose Œa; x ¤ 1. Since hai E G, ha; xi Š MpmC1 is minimal nonabelian of order p mC1 . But jGj D p mCnC1 and n 2 which implies that M D ˆ.G/ha; xi D ha; xi hb p i is a maximal subgroup of G which is neither abelian nor minimal nonabelian. Assume at the moment that also Œx; b ¤ 1. In that case, we get hbihzi E G and so hb; xi is nonmetacyclic minimal nonabelian of order p nC2 . Then it follows that hb; xi must be a maximal subgroup of G with jGj D p nC3 (and so m D 2). But this case will be studied in part (ii) of this proof (where G possesses a nonmetacyclic minimal nonabelian maximal subgroup). Hence we may assume Œx; b D 1. This gives Œx; ab D Œx; a D z r , r 6 0 .mod p/ and so hx; abi is minimal nonabelian which forces that hx; abi must be a maximal subgroup of G. Now, habi covers H=hai Š Cpn n n n n and so o.ab/ p n , where n 2. We have .ab/p D ap b p D ap . If n m, then o.ab/ D p n and habi \ hai D ¹1º. Since hab; zi E G, we see that hab; xi is a nonmetacyclic minimal nonabelian subgroup of order p nC2 . In that case, hab; xi must be a maximal subgroup of G (with m D 2) and again this will be studied in part (ii) of the proof. It follows that we may assume n < m. In that case, o.ab/ D p m and habi hzi so that habi E G. Hence hab; xi is metacyclic minimal nonabelian of order p mC1 and so hab; xi must be a maximal subgroup of G. From jGj D p mCnC1 follows n D 1, contrary to our assumption.

102

p-groups G with p > 2, d.G/ > 2, exactly one exceptional maximal subgroup 79

We may assume that Œa; x D 1 and then Œb; x ¤ 1. Since hbi hzi E G, hb; xi is nonmetacyclic minimal nonabelian of order p nC2 . If hb; xi is a maximal subgroup of G, then this case will be treated in part (ii) of this proof. Thus we may assume that hb; xi is not a maximal subgroup of G and so M D ˆ.G/hb; xi with m > 2. It follows that hab; xi, which is minimal nonabelian, must be a maximal subgroup of G. n n Since habi covers H=hai, we obtain that o.ab/ p n , n 2, and .ab/p D ap . If n m, then o.ab/ D p n and habi\hai D ¹1º and so hab; xi is nonmetacyclic minimal nonabelian of order p nC2 . In that case, hab; xi must be a maximal subgroup of G with m D 2, a contradiction. We may assume that n < m and then o.ab/ D p m with habi hzi and so hab; xi is metacyclic minimal nonabelian of order p mC1 . But then jGj D p mCnC1 implies n D 1, contrary to our assumption. Therefore we may assume that there are no elements of order p in G H . We know that for an element c 1 2 Z.G/ H , c p 2 Z.H / ˆ.Z.H // and so c p is not a p-th power of any element in Z.H /. But Ã1 .H / D Z.H / H 0 D hzi and so H is a powerful group. Then c p D hp for some element h 2 H Z.H / (see Proposition 26.10). It follows hc 2 G H and .hc/p D hp c p D 1 and so hc is of order p, a contradiction. (ii) It remains to consider the case where H is nonmetacyclic minimal nonabelian. We may set m

n

ha; b j ap D b p D 1; Œa; b D z; z p D Œa; z D Œb; z D 1i; where we may assume m 2, n 1 since jGj p 5 . Here we have H 0 D hzi, jH j D p mCnC1 , and jGj D p mCnC2 . Also, hzi is a maximal cyclic subgroup in H , Z.H / D hap i hb p i hzi D ˆ.H / D ˆ.G/ and for each x 2 Z.G/ H we have x p 2 Z.H / ˆ.Z.H //. (ii1) First assume n D 1 so that Z.H / D hap ihzi and for an element c 2 Z.G/H we have c p D apr z s , where both integers r; s are not divisible by p. Suppose that r 0 .mod p/ and then s 6 0 .mod p/. Replacing c with another suitable gener2 0 ator of hci, we may assume that c p D ap r z for some integer r 0 . Take the element 0 2 0 c 0 D apr c 2 Z.G/ H ; then we get .c 0 /p D ap r c p D z. Thus G D H hc 0 i with .c 0 /p D z and hzi D H 0 and so we have obtained a p-group from Proposition 71.1(ii) which is an A2 -group, a contradiction. We have proved that c p D apr z s with r 6 0 .mod p/. Set a0 D ar so that o.a0 / D p m ;

Œa0 ; b D z r

and

.ca0 /p D c p .a0 /p D z s ;

where ha0 ; bi D H . Consider the subgroup U D hca0 ; bi. Since Œca0 ; b D z r and o.b/ D p, we have in case s 6 0 .mod p/ that U Š Mp3 and in case s 0 .mod p/ that U Š S.p 3 /. However, c p D .a0 /p z s , G D ha0 ; b; ci, o.c/ D p m , m > 1, and hci \ U D ¹1º and so we get G D U hci which are the groups stated in part (b) of our theorem. Here M D U hc p i and all p C 1 maximal subgroups containing hci are abelian.

80

Groups of prime power order

We have to show that all subgroups hc i a0 ; c j bi are minimal nonabelian maximal subgroups of G (not containing hci) for all integers i; j (mod p) unless i 1 .mod p/ and j 0 .mod p/ holds. Indeed, Œc i a0 ; c j b D Œa0 ; b D z r and ˆ.hc i a0 ; c j bi/ D hc pi .a0 /p D apri z si apr D apr.i1/ z si ; c pj D aprj z sj ; z r i D ˆ.G/ D ˆ.H / if either i 6 1 .mod p/ or j 6 0 .mod p/. (ii2) It remains to consider the case n 2. In this case, for an element c 2 Z.G/H we have c p D api b pj z k , where at least one of the integers i; j; k is 6 0 .mod p/. First suppose that i 0 .mod p/ and j 0 .mod p/ so that k 6 0 .mod p/. 2 0 2 0 We may deﬁne c p D ap i b p j z k for some integers i 0 ; j 0 . Considering the element 0 0 c 0 D api b pj c 2 Z.G/ H , we get .c 0 /p D ap

2i 0

b p

2j 0

cp D zk ;

k 6 0

.mod p/;

and so G D H hc 0 i with hc 0 i \ H D hzi D H 0 and this is an A2 -group from Proposition 71.1(ii), a contradiction. Now assume that one and only one of the integers i; j is 0 .mod p/. Because of the symmetry, we may assume i 6 0 (mod p) and j 0 .mod p/. We have ˆ.hai b j c; bi/ D hapi b pj c p D z k ; b p ; Œai b j c; b D z i ¤ 1i < ˆ.G/ and ˆ.hab; ar ci/ D hap b p ; apr c p D ap.iCr/ b pj z k ; Œab; ar c D z r ¤ 1i < ˆ.G/ for some suitable r 6 0 .mod p/ such that i C r 0 .mod p/. Hence we deduce that ˆ.G/hai b j c; bi and ˆ.G/hab; ar ci are two distinct maximal subgroups of G which are neither abelian nor minimal nonabelian, a contradiction. Finally, we consider the case where both i and j are 6 0 .mod p/. We have ˆ.hai b j c; bi/ D hapi b pj c p D z k ; b p ; Œai b j c; b D z i ¤ 1i < ˆ.G/ and ˆ.ha; b r ci/ D hap ; b pr c p D api b p.j Cr/ z k ; Œa; b r c D z r ¤ 1i < ˆ.G/ for some suitable r 6 0 .mod p/ such that j C r 0 .mod p/. Hence we deduce that ˆ.G/hai b j c; bi and ˆ.G/ha; b r ci are two distinct maximal subgroups of G which are neither abelian nor minimal nonabelian, a contradiction. Our theorem is proved. Lemma 102.2. Let G be a p-group, p > 2, which possesses exactly one maximal subgroup M which is neither abelian nor minimal nonabelian. Suppose that d.G/ D 3 and G has at most one abelian maximal subgroup. Then ˆ.G/ D Z.G/ D ˆ.H / for each minimal nonabelian maximal subgroup H of G. Also, jG 0 j > p, jM 0 j D p and d.M / 3 which implies jGj p 5 .

102

p-groups G with p > 2, d.G/ > 2, exactly one exceptional maximal subgroup 81

Proof. From the ﬁrst two paragraphs of the proof of Theorem 102.1 we know that jG W Z.G/j p 3 and jG 0 j > p. Let H be a maximal subgroup of G which is minimal nonabelian. Then ˆ.H / D Z.H / ˆ.G/ and jH W ˆ.H /j D p 2 which gives that ˆ.H / D ˆ.G/. Let K ¤ H be another maximal subgroup of G which is minimal nonabelian. Then Z.K/ D ˆ.K/ D ˆ.G/ which implies CG .ˆ.G// hH; Ki D G and so ˆ.G/ D Z.G/. Since jM W ˆ.G/j D p 2 and ˆ.G/ D Z.G/, we have M D S ˆ.G/, where S is minimal nonabelian and S \ ˆ.G/ D ˆ.S/ < ˆ.G/ D Z.M /. This implies M 0 D S 0 Š Cp and d.M / 3. Theorem 102.3. Let G be a p-group, p > 2, satisfying d.G/ D 3 which has exactly one maximal subgroup M which is neither abelian nor minimal nonabelian. Suppose that G has exactly one abelian maximal subgroup A. Then we have ˆ.G/ D Z.G/, G 0 Š Ep2 , jM 0 j D p, d.M / 3, 1 .G/ D E Š Ep3 , and E 6 ˆ.G/. If E 6 A, then we have the following possibilities: mC1

2

m

D b p D t p D 1; Œb; t D z; ap D z n ; b p D u; Œt; a D u; (a) G D ha; b; t j ap Œu; a D Œu; t D Œa; b D Œz; t D 1i, where m 2 and n 6 0 .mod p/. We have jGj D p mC4 ;

G 0 D hu; zi Š Ep2 ;

ˆ.G/ D Z.G/ D hap ; ui Š Cpm Cp ;

1 .G/ D E D hu; z; t i Š Ep3 ;

A D ha; bi is abelian of type .p mC1 ; p 2 /, M D hb; ti hap i, where hb; ti is a nonmetacyclic minimal nonabelian group of order p 4 , and all other p 2 C p 1 maximal subgroups of G are minimal nonabelian. 2

2

(b) G D ha; b; t j ap D b p D t p D 1; ap D z; u D Œt; a; Œb; t D ul z; b p D un z s ; Œu; a D Œu; t D Œa; b D Œz; t D 1i; where l; n; s are integers mod p, n 6 0 .mod p/, n is a square (in GF.p/) and .l C s/2 4n .mod p/. The group G is a special group of order p 5 with G 0 D hu; zi Š Ep2 ;

1 .G/ D E D hu; z; t i Š Ep3 ;

A D ha; bi Š Cp2 Cp2 , M D haj b; tiG 0 , where j .1=2/.l s/ .mod p/, and all other maximal subgroups of G are minimal nonabelian. If E A, then we have: ˛

2

˛1

(c) G D ha; b; d j ap D b p D d p D 1; ap D z; c D Œa; b; z D Œd; a; d p D n r s p c z ; Œd; b D c ; c D Œc; d D Œc; a D Œc; b D 1i, where ˛ 2, n; r; s are integers mod p, n 6 0 .mod p/. We have jGj D p ˛C3 ;

G 0 D hz; ci Š Ep2 ;

ˆ.G/ D Z.G/ D hap ; ci Š Cp˛1 Cp

and A D ˆ.G/hb; as d i. If ˛ 3, then M D ˆ.G/hb; d i with s 6 0 .mod p/. If ˛ D 2, then M D ˆ.G/hb; a r d i with r 6 s .mod p/. All other maximal subgroups of G (distinct from A and M ) are minimal nonabelian.

82

Groups of prime power order

Proof. From Lemma 102.2 we conclude that ˆ.G/ D Z.G/ (and so G is of class 2), jM 0 j D p, d.M / 3, jG 0 j > p and jGj p 5 . By Exercise 1.69(a), we get that jG 0 W .A0 H 0 /j D jG 0 W H 0 j p, where H is a minimal nonabelian maximal subgroup of G. Hence jG 0 j p 2 and so jG 0 j D p 2 . If G 0 D hvi Š Cp2 , then all p 2 C p C 1 maximal subgroups of the nonabelian group G=hv 2 i are abelian, a contradiction (see Exercise 1.6(a)). Hence G 0 Š Ep2 . We see also that each of the p C 1 subgroups of order p in G 0 is the commutator group of exactly p nonabelian maximal subgroups of G. p For each x; y 2 G we have .xy/p D x p y p Œy; x. 2 / D x p y p and so G is regular. By Theorem 7.2, 1 .G/ is of exponent p and j1 .G/j D jG W Ã1 .G/j. Suppose that j1 .G/j p 5 . Let H be a minimal nonabelian maximal subgroup of G. Then we obtain that jH \ 1 .G/j p 4 , contrary to the structure of H . We have proved that j1 .G/j p 4 . Suppose that G has no normal elementary abelian subgroup of order p 3 . Then we get j1 .A/j p 2 and so A is metacyclic. If H is any minimal nonabelian maximal subgroup of G, then jH j p 4 and the fact that 1 .H / Š Ep3 is not possible imply that H is metacyclic. In that case, M (with d.M / 3) is the only maximal subgroup of G which is not metacyclic. By Proposition A.40.12 of Berkovich, G is an L3 -group. In this case, 1 .G/ Š S.p 3 / (the nonabelian group of order p 3 and exponent p) and G=1 .G/ is cyclic of order p 2 . But then Ep2 Š G 0 1 .G/, contrary to the fact that G 0 Z.G/. We have proved that G possesses a normal elementary abelian subgroup E Š Ep3 of order p 3 . Suppose that G has an elementary abelian subgroup F of order p 4 . Since G 0 Š Ep2 and G 0 Z.G/, we have G 0 F and so F E G. Also, j1 .G/j p 4 implies that F D 1 .G/. If G=F is noncyclic, then there is a minimal nonabelian maximal subgroup K of G containing F , contrary to the structure of K. Hence G=F is cyclic of order p. Let a 2 G F be such that hai covers G=F . Then o.a/ D p m , m 2, which implies that 1 .hai/ D hzi ˆ.G/ D Z.G/ and so z 2 F . On the other hand, G=G 0 is abelian of rank 3 which forces z 2 G 0 . Set F D G 0 hf1 ; f2 i for some f1 ; f2 2 F G 0 so that ha; f1 ; f2 i D G and ˆ.G/ D G 0 hap i. As ŒF; hai D G 0 , we may choose f1 2 F G 0 so that Œf1 ; a D z and thus hf1 ; ai Š MpmC1 . This gives p p hf1 ; aiG 0 Š Cp MpmC1 . Since Œf1 ; f2 a D z and .f2 a/p D f2 ap Œa; f2 . 2 / D ap , it follows that 1 .hf2 ai/ D 1 .hai/ D hzi and hf1 ; f2 ai Š MpmC1 . Hence we conclude that hf1 ; f2 aiG 0 Š Cp MpmC1 is another maximal subgroup of G (distinct from hf1 ; aiG 0 ) which is neither abelian nor minimal nonabelian, a contradiction. We have proved that G does not possess an elementary abelian subgroup of order p 4 . Assume that 1 .G/ D S is a nonabelian subgroup of order p 4 and exponent p. Then we have Z.S/ D G 0 . If G=S is noncyclic, then there is a minimal nonabelian maximal subgroup H of G containing S, which contradicts the structure of H . Hence G=S is cyclic of order p. Let t; t 0 2 S G 0 so that S D G 0 ht; t 0 i and then we have 1 ¤ Œt 0 ; t D z 2 G 0 , ht; t 0 i Š S.p 3 / and S Š S.p 3 / Cp . Let M be the unique maximal subgroup of G containing S so that M is neither abelian nor minimal

102

p-groups G with p > 2, d.G/ > 2, exactly one exceptional maximal subgroup 83

nonabelian and let A be the unique abelian maximal subgroup of G. Then A > G 0 and A \ S Š Ep3 so that we may assume that A \ S D G 0 ht 0 i. Also, A covers G=S and so if c 2 A M , then hci covers A=.A \ S / Š G=S and o.c/ p 2 which implies 1 .hci/ ˆ.G/ \ S D Z.G/ \ S D Z.S/ D G 0 : We have ˆ.G/ D G 0 hc p i, and so G D hc; t; t 0 i and Œc; t 0 D 1. If Œc; t 2 hzi, then G=hzi is abelian, a contradiction. Hence Œc; t D u 2 G 0 hzi so that G 0 D hu; zi. The other p 2 C p 1 maximal subgroups of G (which are distinct from A and M ) are of the form T ˆ.G/, where T is one of the following minimal nonabelian subgroups: hc; .t 0 /i ti and hc j t 0 ; c k ti, where i; j; k are integers mod p and both j and k cannot be congruent 0 .mod p/ (see Exercise P9 and P11). Here T ˆ.G/ must be minimal nonabelian and this will be the case if and only if ˆ.T / ˆ.G/ D G 0 hc p i. It is enough to consider hc; ti and hc j t 0 ; cti for any integer j mod p. We have ˆ.hc; ti/ D hc p ; Œc; t D ui D ˆ.G/ if and only if 1 .hci/ ¤ hui which gives 1 .hci/ D hul zi for some integer l mod p. Further, we get ˆ.hc j t 0 ; cti/ D hc pj .t 0 /p D c pj ; .ct/p D c p t p D c p ; Œc j t 0 ; ct D uj zi D hc p ; uj zi: But 1 .hci/ D 1 .hc p i/ D hul zi and so for j D l, ˆ.hc l t 0 ; cti/ D hc p i < ˆ.G/ which shows that hc l t 0 ; ctiˆ.G/ is another maximal subgroup of G (distinct from M ) which is neither abelian nor minimal nonabelian, a contradiction. We have proved that 1 .G/ D E Š Ep3 . (i) Assume that Ep3 Š E D 1 .G/ 6 Z.G/ D ˆ.G/ and E 6 A, where A is the unique abelian maximal subgroup of G. Then we see that A \ E D G 0 , A covers G=E and A is metacyclic. Since G=G 0 is abelian of rank 3, we have d.G=E/ D 2 and so there exists at least one maximal subgroup H of G which is minimal nonabelian. If H=E is noncyclic, then E ˆ.H / D ˆ.G/ D Z.G/, a contradiction. Hence we have H=E Š .H \ A/=G 0 Š Cpm , m 1. Since d.G=E/ D 2, G=E Š A=G 0 is abelian of type .p m ; p/. Let a 2 H \ A be such that hai covers .H \ A/=G 0 . Noting m that 1 .G/ D E, we have o.a/ D p mC1 and 1 ¤ z D ap 2 G 0 . Let t 2 E G 0 so that Œt; a D u 2 G 0 hzi and so G 0 D hu; zi since H is nonmetacyclic and so H 0 D hui is a maximal cyclic subgroup in H . Since A=G 0 Š Cpm Cp , there is an element b 2 A H such that 1 ¤ b p 2 G 0 , b p 2 G 0 hzi and Œa; b D 1. Indeed, if b p 2 hzi, then the subgroup hb; 2 .hai/i contains elements of order p in A H , a contradiction. We have ˆ.H / D ˆ.G/ D hap ; ui, G D ha; b; t i and A D hai hbi is abelian of type .p mC1 ; p 2 /, m 1. If Œb; t 2 hui, then G=hui would be abelian, a contradiction. Hence Œb; t 2 G 0 hui and so replacing b with b 0 D b s (with some s 6 0 .mod p/), we may assume from the start that Œb; t D ul z for some integer l mod p.

84

Groups of prime power order

We have A D ha; bi and so we have to check the p 2 Cp other maximal subgroups of G which are of the form T ˆ.G/, where T is one of the minimal nonabelian subgroups: ha; b i t i, haj b; ak ti, where i; j; k are any integers mod p (see Exercise P11). (i1) First assume m 2 so that ˆ.G/ > G 0 . Consider T D hb; ti, where ˆ.hb; ti/ D hb p ; Œb; ti G 0 < ˆ.G/ and so M D hb; tiˆ.G/ must be our unique maximal subgroup of G which is neither abelian nor minimal nonabelian. All other maximal subgroups of G (which are distinct from A and M ) must be minimal nonabelian. Indeed, ˆ.ha; b i ti/ D hap ; b ip t p D b ip ; Œa; b i t D u1 i D ˆ.G/: Further, for j 6 0 .mod p/ we have ˆ.haj b; ti/ D hapj b p ; Œaj b; t D uj ul zi: Here hapj b p i covers ˆ.G/=G 0 and 1 .hapj b p i/ D hzi. Hence if l 6 0 .mod p/, then ˆ.hal b; ti/ D hapl b p i < ˆ.G/, a contradiction (since hal b; tiˆ.G/ will be another maximal subgroup which is neither abelian nor minimal nonabelian). Therefore we conclude that l 0 .mod p/ and so Œb; t D z. Finally, if both j and k are not congruent 0 .mod p/, then we have ˆ.haj b; ak ti/ D hapj b p ; apk ;

Œaj b; ak t D uj zi D ˆ.G/:

We have b p 2 G 0 hzi and so b p D un z s , where n and s are some integers mod p with n 6 0 .mod p/. Then we replace a with a0 D an b s and u with u0 D un z s and get m .a0 /p D z n ; b p D u0 ; Œt; a0 D Œt; an b s D un z s D u0 : Writing again a instead a0 and u instead u0 , we see that we have obtained the relations stated in part (a) of our theorem. (i2) Now assume m D 1 so that jGj D p 5 and G 0 D Z.G/ D ˆ.G/ D hu; zi and therefore G is a special p-group. In this case, we have ap D z;

Œt; a D u;

Œa; b D 1;

Œb; t D ul z;

b p D un z s ;

where l; n; s are some integers mod p with n 6 0 .mod p/. Also, A is abelian of type .p 2 ; p 2 /. For all integers i mod p we have ˆ.ha; b i ti/ hap D z; Œa; b i t D u1 i D G 0 D ˆ.G/ and so all subgroups ha; b i ti are minimal nonabelian maximal subgroups of G. Hence among the rest of the p 2 nonabelian maximal subgroups haj b; ak tiG 0 (j; k are any integers mod p) there is exactly one which is not minimal nonabelian. We have ˆ.haj b; ak t i/ D hun z sCj ; z k ; ulj zi;

102

p-groups G with p > 2, d.G/ > 2, exactly one exceptional maximal subgroup 85

and so if k 6 0 .mod p/, then ˆ.haj b; ak t i/ D hu; zi D G 0 . It remains to examine the case k 0 .mod p/, where we must have ˆ.haj b; ti/ D hun z sCj ; ulj zi ¤ G 0 D hu; zi for exactly one j . It follows that the quadratic congruence ˇ ˇ ˇ n ˇ s C j ˇ ˇ D j 2 C j.s l/ C .n sl/ 0 ˇl j 1 ˇ

.mod p/

must have exactly one solution in j . This occurs if and only if .s l/2 4.n sl/ 0

.mod p/

or, equivalently, .s C l/2 4n .mod p/. In this case, j .1=2/.l s/ .mod p/ and for that integer j the maximal subgroup M D haj b; tiG 0 is the only one which is neither abelian nor minimal nonabelian. We have obtained the groups from part (b) of our theorem. (ii) Assume that Ep3 Š E D 1 .G/ 6 Z.G/ D ˆ.G/ and E A, where A is the unique abelian maximal subgroup of G. Since d.G=E/ D 2, there exists (at least one) maximal subgroup H containing E which is minimal nonabelian. Then H is nonmetacyclic, H=E ¤ ¹1º is cyclic and Z.H / \ E D G 0 . Indeed, if H=E is noncyclic, then we have E ˆ.H / D ˆ.G/ D Z.G/, contrary to our assumption. Taking an element a 2 H E such that hai covers H=E and an element b 2 EG 0 , we get 1 .hai/ G 0 and ˛ H D ha; b j ap D b p D 1; c D Œa; b; c p D Œa; c D Œb; c D 1i; where ˛ 2, H 0 D hci, Z.H / D ˆ.H / D hap i hci, and jGj D p ˛C3 . Setting ˛1 ap D z, we have G 0 D hc; zi, E D hbi G 0 Š Ep3 because hci is a maximal cyclic subgroup in H and therefore hci ¤ hzi. Now, ha; ci is an abelian normal subgroup of G of type .p ˛ ; p/ which possesses exactly p cyclic subgroups hac i i (i any integer mod p) of order p ˛ . But Œa; b i D c i and so we get NH .hai/ D ha; ci and all subgroups hac i i are conjugate in H . Therefore N D NG .hai/ covers G=H and so G D NH with N \H D ha; ci. As N=G 0 Š G=E is abelian of rank 2, it follows that N=G 0 is abelian of type .p ˛ ; p/. Hence there exists an element d 2 N H such that 1 ¤ d p 2 G 0 and hd i normalizes hai. But N is a maximal subgroup of our group G which does not contain E and so N is nonabelian (noting that in our case E A). This gives 1 ¤ Œd; a 2 hzi and so replacing d with a suitable power d j (j 6 0 .mod p/), we may assume from the start that Œd; a D z. If d p 2 hzi, then we get hd; ai Š Mp˛C1 in which case there are elements of order p in hd; aiH , a contradiction. We have proved that d p 2 G 0 hzi so that hd; ai is a metacyclic minimal nonabelian maximal subgroup of G D ha; b; d i. Now, H D ha; bi is a minimal nonabelian maximal subgroup of G containing E and the other p maximal subgroups of G containing E are hb; ai d iˆ.G/, where i is any integer mod p and

86

Groups of prime power order

ˆ.G/ D G 0 hap i. For exactly one i , hb; ai d iˆ.G/ is the unique abelian maximal subgroup A of G, i.e., Œai d; b D 1 and then c i Œd; b D 1 and so we may set Œd; b D c s for some integer s mod p. Since d p 2 G 0 hzi, we may set d p D c n z r for some integers n; r mod p with n 6 0 .mod p/. All p 2 maximal subgroups of G which do not contain E are hb j a; b k d iˆ.G/, where j; k are any integers mod p. They are all metacyclic minimal nonabelian since ˆ.hb j a; b k d i/ D hap ; d p D c n z r ; Œb j a; b k d D c sj Ck z 1 ¤ 1i D ˆ.G/; where we have used the facts hap i hzi and n 6 0 .mod p/. We obtain that A D hb; as d iˆ.G/ is the unique abelian maximal subgroup of G and in the set of the p 1 nonabelian maximal subgroups hb; ai d iˆ.G/ for i 6 s .mod p/ there is exactly one which is not minimal nonabelian. We compute for all i 6 s .mod p/ ˆ.hb; ai d i/ D hb p D 1; .ai d /p D api c n z r ; Œb; ai d D c i s ¤ 1i D hapi z r ; ci: If ˛ 3, then hz r i hap i and so in this case ˆ.hb; ai d i/ ¤ ˆ.G/ if and only if i 0 .mod p/. Then we have M D hb; d iˆ.G/ and in this case s 6 0 .mod p/. If ˛ D 2, then ˆ.hb; ai d i/ D hz i Cr ; ci since ap D z. Hence in this case we have ˆ.hb; ai d i/ ¤ ˆ.G/ if and only if i r .mod p/. Then M D hb; ar d iˆ.G/ and in this case we must have r 6 s .mod p/. We have obtained the groups stated in part (c) of our theorem. (iii) It remains to consider the case Ep3 Š E D 1 .G/ Z.G/ D ˆ.G/. We shall show that this difﬁcult case cannot occur. If Hi is any minimal nonabelian maximal subgroup of G, then ˆ.Hi / D ˆ.G/ E Š Ep3 and so jHi j p 5 which implies jGj p 6 . By the ﬁrst paragraph of this proof, we know that there are exactly p nonabelian maximal subgroups of G whose commutator subgroup is equal to M 0 D hmi. Obviously, G=M 0 is an A2 -group of order p 5 with .G=M 0 /0 D G 0 =M 0 Š Cp and G=M 0 has exactly pC1 abelian maximal subgroups. By Proposition 71.1, there is a minimal nonabelian maximal subgroup H=M 0 of G=M 0 and an element d 2 G H with 1 ¤ d p 2 G 0 and Œhd i; G D M 0 . We have H 0 D hci with c 2 G 0 M 0 and G 0 D hc; mi which implies that H is a minimal nonabelian maximal subgroup of G (noting that H ¤ M since H 0 ¤ M 0 ). There are exactly p C 1 maximal subgroups Xi of G (i D 1; 2; : : : ; p C 1) which contain hd i. Then Xi \ H is an abelian maximal subgroup of H . Further, we obtain that Xi0 D Œhd i; .Xi \ H / M 0 and so Xi is either abelian or Xi0 D hmi. It follows that ¹X1 ; : : : ; XpC1 º D ¹A; M; H1 ; : : : ; Hp1 º, where Hj (j D 1; 2; : : : ; p 1) are 0 0 minimal nonabelian with .Hj / D hmi D M . But Hj (containing E) is nonmetacyclic and so M 0 is a maximal cyclic subgroup in Hj which implies d p 2 G 0 M 0 and we may set d p D c s mt , where s; t are some integers (mod p) with s 6 0 .mod p/.

102

p-groups G with p > 2, d.G/ > 2, exactly one exceptional maximal subgroup 87

Consider H D H1 so that H is a minimal nonabelian maximal subgroup of G containing hd i and .H /0 D hmi. Choose an element a 2 .H \ H / ˆ.G/ so that H D hd; a i and Œd; a D m. By Exercise P9, ˆ.H / D hd p ; Œd; a D mih.a /p i D G 0 h.a /p i; and we know that ˆ.H / D ˆ.G/. If 1 .ha i/ G 0 , then E 6 ˆ.G/, a contradiction. Hence we have 1 .ha i/ 6 G 0 which implies ˆ.H / D ˆ.G/ D h.a /p i G 0 and this is an abelian group of type .p; p; p 1 /, where o.a / D p , 2 and H \ H D ha i G 0 . It follows that .H \ H /=G 0 Š Cp and since H=G 0 is noncyclic abelian, there exists b 2 H .H \ H / such that 1 ¤ .b /p 2 G 0 and ha ; b i D H . We may set Œa ; b D c, where hci D H 0 . Also, hci is a maximal cyclic subgroup in H which gives .b /p 2 G 0 hci and we have G D ha ; b ; d i, jGj D p C4 , 2. We may set .b /p D c v mw , where v; w are some integers (mod p) with w 6 0 .mod p/. Suppose that Œd; b D 1 so that A D hd; b iˆ.G/. Also, hd; a i D H and so we investigate other maximal subgroups hd; .a /i b iˆ.G/ containing hd i, where i 6 0 .mod p/. We get ˆ.hd; .a /i b i/ D hd p 2 G 0 hmi; Œd; .a /i b D mi ; .a /pi .b /p i D ˆ.G/: But then there is no subgroup M in the set of maximal subgroups of G containing hd i, a contradiction. Hence we have Œd; b D mr with r 6 0 .mod p/. Since the maximal subgroups A and M are contained in the set of the p C 1 maximal subgroups of G which contain hd i, it follows that all p 2 maximal subgroups hd i a ; d j b iˆ.G/ of G (i; j are any integers mod p) which do not contain hd i must be minimal nonabelian. For each integer i mod p we must have ˆ.hd i a ; b i/ D hd pi .a /p ; .b /p D c v mw ; Œd i a ; b D cmri i D ˆ.G/ and this will be the case if and only if hc v mw ; cmri i D G 0 D hc; mi or, equivalently, ˇ ˇ ˇv w ˇ ˇ ˇ ˇ1 ri ˇ D vri w 6 0 .mod p/: But r 6 0 .mod p/ and so if v 6 0 .mod p/, then the congruence vri w 0 .mod p/ would have a solution in i , a contradiction. Hence v 0 .mod p/ and so .b /p D mw . For each integer i mod p we must have ˆ.hd i a ; db i/ D hd pi .a /p ; d p .b /p D c s mt Cw ; Œd i a ; db D cmri 1 i D ˆ.G/

88

Groups of prime power order

and this will be the case if and only if hc s mt Cw ; cmri1 i D G 0 D hc; mi or, equivalently, ˇ ˇ ˇs t C w ˇ ˇ ˇ ˇ1 ri 1ˇ D .sr/i s t w 6 0 .mod p/: But sr 6 0 .mod p/ and so the congruence .sr/i s t w 0 .mod p/ would have a solution in i , a ﬁnal contradiction. Our theorem is proved. Lemma 102.4. Let P be a p-group with jP 0 j D p. If P possesses a minimal nonabelian maximal subgroup H , then P has exactly p C 1 abelian maximal subgroups. Proof. By Exercise P.10, P D H CP .H / and in our case CP .H / is abelian so that CP .H / D Z.P /. But then all p C 1 maximal subgroups of P containing Z.P / are abelian and so we are done (see Exercise 1.6(a)). Theorem 102.5. Let G be a p-group, p > 2, with d.G/ D 3 which has exactly one maximal subgroup M which is neither abelian nor minimal nonabelian. Suppose that G has no abelian maximal subgroups. Then ˆ.G/ D Z.G/, 1 .G/ D G 0 Š Ep3 , jM 0 j D p, d.M / 3 and we have one of the following possibilities: 2

2

2

(a) G D ha; b; c j ap D b p D c p D 1; ap D z; b p D y; c p D x; Œa; b D z; Œa; c D yz ˇ ; Œb; c D x 1 y ı z ; Œx; b D Œx; a D Œy; a D Œy; c D Œz; b D Œz; c D 1i, where ˇ; ı; are integers mod p, 6 0 .mod p/ and .ˇ ı/2 C 4 is not a square in GF.p/. We have jGj D p 6 , 1 .G/ D G 0 D ˆ.G/ D Z.G/ D hx; y; zi Š Ep3 and so G is a special p-group. Also, M D ha; biG 0 and all other maximal subgroups of G are nonmetacyclic minimal nonabelian. 2

2

n

n1

D z; Œa; b D z; (b) G D ha; b; c j ap D b p D c p D 1; ap D x; b p D y; c p Œc; a D x ˛ y z ˇ ; Œc; b D x y z ı ; Œx; b D Œx; c D Œy; a D Œy; c D Œz; a D Œz; b D 1i, where n 3, ˛; ˇ; ; ı; ; are integers mod p, 6 0 .mod p/, 6 0 .mod p/, and .˛/2 C4 is not a square in GF.p/. In this case, we have jGj D p nC4 , 1 .G/ D G 0 D hx; y; zi Š Ep3 and ˆ.G/ D Z.G/ D hc p ; x; yi is abelian of type .p n1 ; p; p/. Also, M D ha; biˆ.G/ and all other maximal subgroups of G are nonmetacyclic minimal nonabelian. Proof. Let 1 D ¹H1 ; H2 ; : : : ; Hp2 Cp ; M º be the set of maximal subgroups of G, where Hi (i D 1; : : : ; p 2 C p) are minimal nonabelian and M is neither abelian nor minimal nonabelian. From Lemma 102.2 we get ˆ.G/ D Z.G/ D Z.Hi / D Z.M /, d.M / 3, jM 0 j D p and jG 0 j > p. By a result of A. Mann (see Exercise 1.69(a)), jG W .H10 H20 /j p and so p 2 jG 0 j p 3 . Suppose that for some Hi ¤ Hj we have Hi0 D Hj0 or Hi0 D M 0 . Then, by Exercise 1.69(a), jG 0 j p 2 and so jG 0 j D p 2 . If G 0 Š Cp2 , then the nonabelian group G=1 .G 0 / has p 2 C p C 1 abelian maximal subgroups, which is a contradiction by Exercise 1.6(a). Hence G 0 Š Ep2 . Let X be any ﬁxed subgroup of order p in G 0 . Then there is a minimal nonabelian maximal subgroup Hi (i 2 ¹1; : : : ; p 2 C pº) such that Hi0 ¤ X so that Hi =X is minimal nonabelian. By Lemma 102.4, G=X has exactly

102

p-groups G with p > 2, d.G/ > 2, exactly one exceptional maximal subgroup 89

p C 1 abelian maximal subgroups. Hence there are exactly p C 1 maximal subgroups of G whose commutator subgroup is equal X. But G 0 has p C 1 subgroups of order p and so G must have .p C 1/2 D p 2 C 2p C 1 maximal subgroups, a contradiction. We have proved that ¹H10 ; H20 ; : : : ; Hp0 2 Cp ; M 0 º is the set of p 2 C p C 1 pairwise distinct subgroups of order p in G 0 and so, in particular, G 0 Š Ep3 . Assume that there is an element t 2 G G 0 of order p. Since G=G 0 is abelian of rank 3, G=.G 0 ht i/ is noncyclic. But then there is a maximal subgroup Y of G containing G 0 ht i which is minimal nonabelian, a contradiction (with the structure of Y ). We have proved that 1 .G/ D G 0 Š Ep3 and all minimal nonabelian maximal subgroups of G are nonmetacyclic. Set T =G 0 D 1 .G=G 0 / Š Ep3 . If G=T is noncyclic, then there is a maximal subgroup K of G containing T which is minimal nonabelian. Since K > T , T is abelian of type .p 2 ; p 2 ; p 2 /. But K 0 < G 0 and so K 0 is not a maximal cyclic subgroup in K, contrary to the fact that K is nonmetacyclic minimal nonabelian. We have proved that G=T is cyclic. (i) First assume that T D G, i.e., G=G 0 Š Ep3 and so in this case G is a special group of order p 6 with G 0 D 1 .G/ Š Ep3 . We determine the structure of M . We have M D G 0 S , where S D ha; bi is minimal nonabelian and G 0 \ S D ˆ.S/ < G 0 since d.M / 3. Set S 0 D M 0 D hzi Š Cp . If ˆ.S/ D hzi, then hzi is a unique subgroup of order p in S which implies that S would be cyclic, a contradiction. Hence ˆ.S/ D 1 .S/ Š Ep2 and so S is metacyclic of order p 4 and exponent p 2 . We may choose a; b 2 S G 0 so that ap D z, b p D y 2 ˆ.S/ hzi and Œa; b D z. Since Ã1 .M / D Ã1 .S/ D hy; zi M 0 D hzi, it follows that M is a powerful group. By Proposition 26.10, each element in hy; zi is a p-th power of an element in M . Let c be an element in G M . Suppose that c p 2 hy; zi. Then there is m 2 M such that mp D c p . We get p .mc/p D mp c p Œc; m. 2 / D 1;

contrary to the fact that 1 .G/ D G 0 . Hence we have c p D x 2 G 0 hy; zi and so G 0 D hx; y; zi and G D ha; b; ci. All p 2 C p maximal subgroups of G which are distinct from M must be minimal nonabelian. We obtain ˆ.ha; ci/ D hz; x; Œa; ci D G 0 and so Œa; c D y r y 0 , where r 6 0 0 .mod p/ and y 0 2 hx; zi. Replacing c with c 0 D c r , where r 0 6 0 .mod p/ is such 0 that rr 1 .mod p/, we get 0

0

0

0

0

Œa; c 0 D Œa; c r D Œa; cr D y rr .y 0 /r D y.y 0 /r ; 0

0

0

where c 0 2 G M , .y 0 /r 2 hx; zi and .c 0 /p D .c p /r D x r D x 0 2 G hy; zi. Writing again c and x instead of c 0 and x 0 , respectively, we see that we may assume from the start Œa; c D yy 00 and c p D x, where y 00 2 hx; zi and so we may set Œa; c D x ˛ yz ˇ for some integers ˛; ˇ mod p. From ˆ.hb; ci/ D hy; x; Œb; ci D G 0 follows that Œb; c D x y ı z , where ; ı; are some integers mod p with 6 0 .mod p/.

90

Groups of prime power order

Maximal subgroups of G containing hci are ha; ciG 0 and hai b; ciG 0 . Therefore we must have for all integers i mod p ˆ.hai b; ci/ D hyz i ; x; Œai b; c D x ˛i C y i Cı z ˇ i C i D G 0 which is equivalent to ˇ ˇ ˇ 0 1 i ˇˇ ˇ ˇ 1 0 0 ˇˇ D i 2 C .ı ˇ/i 6 0 ˇ ˇ˛i C i C ı ˇi C ˇ

.mod p/:

Hence the quadratic congruence i 2 C .ı ˇ/i 0 .mod p/ should not have any solution in i which is equivalent to the requirement that .ˇ ı/2 C 4 is not a square in GF(p). We have to examine the p 2 maximal subgroups hc j a; c k biG 0 of G (j; k are integers mod p) which do not contain hci. For all k 6 0 .mod p/ we get ˆ.ha; c k bi/ D hz; x k y; Œa; c k b D x k˛ y k z kˇ C1 i D G 0 which is equivalent to ˇ ˇ ˇ0 0 1 ˇˇ ˇ ˇk 1 0 ˇˇ m D k 2 k˛ D k.k ˛/ 6 0 ˇ ˇk˛ k kˇ C 1ˇ

.mod p/:

Hence we must have ˛ 0 .mod p/ and so Œa; c D yz ˇ . For all j 6 0 .mod p/ we obtain ˆ.hc j a; bi/ D hx j z; y; Œc j a; b D x j y ıj z j C1 i D G 0 which is equivalent to ˇ ˇ j 0 ˇ ˇ 0 1 ˇ ˇj ıj

ˇ ˇ 1 ˇ ˇ D j.j C C 1/ 6 0 0 ˇ j C 1ˇ

.mod p/:

Since 6 0 .mod p/, we must have 1 .mod p/ and so Œb; c D x 1 y ı z . We have obtained the groups of order p 6 stated in part (a) of our theorem. It remains to check that all maximal subgroups hc j a; c k biG 0 are minimal nonabelian unless j k 0 .mod p/ in which case ha; biG 0 D M . Indeed, ˆ.hc j a; c k bi/ D hx j z; x k y; Œc j a; c k b D x j y ıj Ck z j Cˇ kC1 i < G 0 if and only if ˇ ˇ ˇj ˇ 0 1 ˇ ˇ ˇk ˇ D k 2 C k.ˇj ıj / j 2 0 1 0 ˇ ˇ ˇj ıj C k j C ˇk C 1ˇ

.mod p/:

102

p-groups G with p > 2, d.G/ > 2, exactly one exceptional maximal subgroup 91

The quadratic congruence in k k 2 C kj.ˇ ı/ j 2 0

.mod p/

(where j 2 GF.p/ is ﬁxed) has a solution in k if and only if the discriminant ()

j 2 .ˇ ı/2 C 4j 2 D ..ˇ ı/2 C 4/j 2

is a square in GF(p). But we know that .ˇ ı/2 C 4 is not a square in GF(p) and so we must have j 0 .mod p/. From () we then get k 2 0 .mod p/ and so also k 0 .mod p/ and we are done. (ii) Now assume that T < G, where T =G 0 D 1 .G=G 0 / Š Ep3 and T =G 0 is cyclic. Hence G=G 0 is abelian of type .p m ; p; p/, m 2, and the unique maximal subgroup of G containing T is obviously equal to M . There are normal subgroups U and V of G such that G D U V , U \ V D G 0 , U=G 0 Š Ep2 and V =G 0 Š Cpm , m 2. Let c be an element in V G 0 such that hci covers V =G 0 . We have o.c/ D p n , n 3, where n1 D z, where z 2 G 0 . Then M D n D m C 1 (noting that 1 .G/ D G 0 ). Set c p p 0 p hc iU , ˆ.G/ D Z.G/ D G hc i is abelian of type .p n1 ; p; p/ and jGj D p nC4 . Let a; b 2 U G 0 be such that U D G 0 ha; bi, where ap ; b p 2 G 0 and G D ha; b; ci. Since each minimal nonabelian maximal subgroup Hi of G (i D 1; : : : ; p 2 C p) is nonmetacyclic and contains ˆ.G/ and hzi is not a maximal cyclic subgroup in ˆ.G/, it follows that Hi0 ¤ hzi (for all i ) and so M 0 D hzi. Therefore 1 ¤ Œa; b 2 hzi and so we may assume Œa; b D z. Now, G=hzi has the unique abelian maximal subgroup M=hzi and all other maximal subgroups of the factor group G=hzi are minimal nonabelian. Hence G=hzi is an A2 -group with d.G=hzi/ D 3 and of order p nC3 > p 4 (since n 3), G 0 =hzi Š Ep2 , G 0 =hzi Z.G=hzi/ and G=hzi has a normal elementary abelian subgroup of order p 3 , namely hG 0 ; 2 .hci/i=hzi. Therefore G=hzi is an A2 -group from Proposition 71.4(b) which implies 1 .G=hzi/ D hG 0 ; 2 .hci/i=hzi. Set ap D x and b p D y and consider the abelian group M=hzi. If the abelian subgroup U=hzi of order p 4 and exponent p 2 has rank > 2, then 1 .U=hzi/ > G 0 =hzi, which contradicts the above fact. Hence we get U=hzi Š Cp2 Cp2 which implies G 0 D hx; y; zi. Since ˆ.hc; ai/ D hc p ; x; Œc; ai D ˆ.G/ and ˆ.hc; bi/ D hc p ; y; Œc; bi D ˆ.G/, we must have Œc; a D x ˛ y z ˇ ; Œc; b D x y z ı for some integers ˛; ˇ; ; ı; ; mod p with 6 0 .mod p/ and 6 0 .mod p/. The maximal subgroups of G containing hci are ha; ciˆ.G/ and hai b; ciˆ.G/. Therefore we must have for all integers i mod p ˆ.hc; ai bi/ D hc p ; x i y; Œc; ai b D x ˛i C y i C z ˇ iCı i D ˆ.G/ which is equivalent to ˇ ˇ 0 0 ˇ ˇ i 1 ˇ ˇ˛i C i C

ˇ 1 ˇˇ 0 ˇˇ D i 2 C . ˛/i 6 0 ˇi C ı ˇ

.mod p/:

92

Groups of prime power order

Hence the quadratic congruence i 2 C . ˛/i 0 .mod p/ should not have any solution in i which is equivalent to the requirement that . ˛/2 C 4 is not a square in GF(p). We have obtained the groups stated in part (b) of our theorem. It remains to check that all maximal subgroups hc j a; c k biˆ.G/ are minimal nonabelian unless j k 0 .mod p/ in which case ha; biˆ.G/ D M . Note that ˆ.G/ D hc p i hxi hyi and

2

ˆ.ˆ.G// D hc p i hzi;

where ˆ.G/=ˆ.ˆ.G// Š Ep3 . We have ˆ.hc j a; c k bi/ D hc pj x; c pk y; Œc j a; c k b D x j ˛k y j k z ıj ˇ kC1 i < ˆ.G/ if and only if ˇ ˇ ˇj 1 0 ˇˇ ˇ ˇk 0 1 ˇˇ D k 2 C .˛ /j k j 2 0 ˇ ˇ 0 j ˛k j k ˇ

.mod p/:

The quadratic congruence in k ()

k 2 C .˛ /j k j 2 0

.mod p/

(where j 2 GF.p/ is ﬁxed) has a solution in k if and only if the discriminant j 2 .˛ /2 C 4j 2 D ..˛ /2 C 4/j 2 is a square in GF(p). But we know that .˛ /2 C 4 is not a square in GF(p) and so we must have j 0 .mod p/. From () we then get k 2 0 .mod p/ and so (noting that 6 0 .mod p/) k 0 .mod p/ and we are done.

101

p-groups G with p > 2 and d.G / D 2 having exactly one maximal subgroup which is neither abelian nor minimal nonabelian

Here we determine up to isomorphism the title groups (Theorems 101.1 and 101.2). As a result we get a complete classiﬁcation of p-groups (which are not A2 -groups) from Lemma 76.5 (Theorems 100.1 and 101.1). Also, p-groups of Lemma 76.13 are completely determined (Theorems 100.2, 100.3 and 101.2). Theorem 101.1. Let G be a two-generator p-group, p > 2, with exactly one maximal subgroup M which is neither abelian nor minimal nonabelian. If G has an abelian maximal subgroup A, then we have G D hh; k j Œh; k D v; Œv; k D z; Œv; h D z ; v p D z p D Œz; h D Œz; k D 1; hp D z ; k p

nC1

D z i;

where n 1 and ; ; are integers mod p with 6 0 .mod p/. We have G 0 D hv; zi Š Ep2 ;

jGj D p nC4 ;

Z.G/ D hk p ; zi;

G 0 \ Z.G/ D hzi Š Cp ;

ˆ.G/ D Z.G/G 0 ;

ŒG 0 ; G D hzi

and so G is of class 3. Also, S D hv; hi Š S.p 3 / (if 0 .mod p/) or S Š Mp3 (if 6 0 .mod p/), S is normal in G, G D Shki;

S \ hki hzi; M D Shk p i;

A D CG .G 0 /;

G=S Š CpnC1 ; d.M / D 3;

G=Z.G/ Š S.p 3 /;

M 0 D hzi;

1 D ¹A; M; M1 ; : : : ; Mp1 º;

0 D hzi. Finally, G is an where all Mi are minimal nonabelian with M10 D D Mp1 L3 -group if and only if 6 0 .mod p/ and in that case 1 .G/ Š S.p 3 /, G=1 .G/ is cyclic of order p nC1 (n 1) and Z.G/ D hk p i Š CpnC1 is cyclic.

Proof. Obviously, A is a unique abelian maximal subgroup of G (otherwise, by Exercise 1.6(a), all p C 1 maximal subgroups of G would be abelian). By Exercise 1.69(a),

101

p-groups G with p > 2, d.G/ D 2, exactly one exceptional maximal subgroup 67

jG 0 W .A0 M10 /j p, where M1 is a minimal nonabelian maximal subgroup of G and so jG 0 j p 2 . But if jG 0 j D p, then this fact together with d.G/ D 2 would imply (Exercise P9) that G is minimal nonabelian, a contradiction. Hence jG 0 j D p 2 . Then it follows from jGj D pjG 0 jjZ.G/j (Lemma 1.1) that jG W Z.G/j D p 3 . Set 1 D ¹A; M; M1 ; : : : ; Mp1 º, where all Mi .i D 1; : : : ; p 1/ are minimal nonabelian. We have Z.G/ Mi (otherwise d.G/ D 3) and so Z.G/ D Z.Mi / D ˆ.Mi / for all i D 1; : : : ; p 1. Also, ˆ.Mi / < ˆ.G/ < Mi and so ˆ.G/ is abelian. For each x 2 G A, CA .x/ D Z.G/ and so x p 2 Z.G/. Hence G=Z.G/ is generated by its elements of order p and so G=Z.G/ Š S.p 3 / because d.G/ D 2 and so G=Z.G/ cannot be elementary abelian. This implies G 0 \ Z.G/ Š Cp ;

ˆ.G/ D Z.G/G 0 ;

cl.G/ D 3:

Also, Mi0 D M 0 D G 0 \ Z.G/ for all i D 1; : : : ; p 1. If d.M / D 2, then M 0 Š Cp would imply that M is minimal nonabelian, a contradiction. Hence we get d.M / 3. In particular, jM j p 4 and so jGj p 5 . (i) First assume that G 0 D hvi Š Cp2 is cyclic. Since hv p i D Mi0 is not a maximal cyclic subgroup in Mi > ˆ.G/ D Z.G/G 0 , it follows that all Mi (i D 1; : : : ; p 1) are metacyclic. In particular, j1 .ˆ.G//j p 2 . Suppose that A is also metacyclic so that M (with d.M / 3) is the only nonmetacyclic maximal subgroup of G. By a result of Y. Berkovich (see Proposition A.40.12), G is an L3 -group. But then we have G 0 1 .G/, where 1 .G/ is of order p 3 and exponent p and so G 0 Š Ep2 , a contradiction. It follows that A must be nonmetacyclic in which case 1 .A/ 6 ˆ.G/. Let a be an element of order p in Aˆ.G/ and let k 2 GA be such that hˆ.G/; ki D M1 . Since Œk; v ¤ 1, we may replace k with another generator of hki so that we may assume that Œk; v D v p . As hk; vi0 D hv p i Z.G/, it follows that hk; vi is minimal nonabelian and so hk; vi D M1 . By Exercise P9, we have Z.G/ D ˆ.M1 / D hk p ; v p ; Œk; v D v p i D hk p ; v p i: All maximal subgroups of G distinct from A D ˆ.G/hai are ˆ.G/hai ki D Z.G/hai k; vi; where i is any integer mod p. Since Œai k; v D Œai ; vk Œk; v D v p 2 Z.G/, it follows that hai k; vi is minimal nonabelian. Again, by Exercise P9, ˆ.hai k; vi/ D h.ai k/p ; v p ; Œai k; v D v p i D h.ai k/p ; v p i: The factor group G=hv p i is minimal nonabelian (since we have d.G=hv p i/ D 2 and .G=hv p i/0 Š Cp ) and so computing in G=hv p i, we get p

.ai k/p D aip k p Œk; ai . 2 / x;

where x 2 hv p i:

68

Groups of prime power order p

But aip D 1 and Œk; ai 2 hvi so that Œk; ai . 2 / 2 hv p i which gives .ai k/p D k p y for some y 2 hv p i. By the above, ˆ.hai k; vi/ D hk p y; v p i D hk p ; v p i D Z.G/ and so we conclude that ˆ.G/hai ki D Z.G/hai k; vi D hai k; vi is minimal nonabelian for all i D 1; : : : ; p 1. It follows that G is an A2 -group, a contradiction. (ii) We have proved that G Š Ep2 . Since ŒG; G 0 D G 0 \ Z.G/ Š Cp , we get by the Hall–Petrescu formula (Appendix 1) for any x; y 2 G, .xy/p D x p y p l for some l 2 G 0 \ Z.G/. We have A D CG .G 0 / and we take an element k 2 G A such that ˆ.G/hki D M1 is a minimal nonabelian maximal subgroup of G. Then for any element v 2 G 0 Z.G/, we get Œv; k D z, where hzi D G 0 \ Z.G/. Since hk; vi0 D hzi, hk; vi is minimal nonabelian and so hk; vi D M1 . In particular, ˆ.hk; vi/ D hk p ; v p ; Œv; ki D hk p ; zi D ˆ.M1 / D Z.G/: Thus hk p i covers Z.G/=hzi, where jZ.G/j p 2 . Set Z.G/=hzi Š ˆ.G/=G 0 Š Cpn with n 1 so that jGj D p nC4 . Consider the abelian group G=G 0 of rank 2. Since .G 0 hki/=G 0 Š CpnC1 , there is a subgroup S=G 0 of order p such that G D S hki and S \ hki hzi. Let h 2 S G 0 so that hp 2 hzi since hp 2 Z.G/ \ G 0 D hzi. Assume that S A in which case h 2 A ˆ.G/ and G D hh; ki. We may assume that Œh; k D v and examine all maximal subgroups ˆ.G/hhi ki of G (i is any integer mod p) which are distinct from A. We have Œv; hi k D Œv; kŒv; hi k D Œv; k D z and so hv; hi ki is minimal nonabelian. On the other hand, ˆ.hv; hi ki/ D hv p D 1; .hi k/p D hip k p l; Œv; hi k D zi D hk p ; zi D Z.G/ (where l 2 hzi) since .hi k/p D k p l 0 for some l 0 2 hzi. This means that ˆ.G/hhi ki D hv; hi ki and so all these p maximal subgroups of G are minimal nonabelian. But then G is an A2 -group, a contradiction. We have proved that S 6 A D CG .G 0 / and so 1 ¤ Œv; h 2 hzi. Since G D hh; ki, we may set Œh; k D v and Œv; k D z, where v 2 G 0 Z.G/ and hzi D G 0 \ Z.G/. nC1 Also, Œv; h D z , hp D z , and k p D z , where ; ; are integers mod p with 6 0 .mod p/. Here S D hv; hi Š S.p 3 / or Mp3 , S is normal in G, G D S hki with hki \ S hzi and M D S Z.G/ D S hk p i with d.M / D 3. It remains to examine all p maximal subgroups ˆ.G/hhi ki (i D 0; 1; : : : ; p 1) of G which are distinct from M D ˆ.G/hhi D S hk p i. We compute Œv; hi k D Œv; kŒv; hi D zz i D z i C1 ; where the congruence i C 1 0 .mod p/) has exactly one solution i 0 for i (noting 0 that 6 0 .mod p/). Hence A D ˆ.G/hhi ki is an abelian maximal subgroup of G

101

p-groups G with p > 2, d.G/ D 2, exactly one exceptional maximal subgroup 69

and for all other i 6 i 0 .mod p/ we see that hv; hi ki is minimal nonabelian and moreover, ˆ.hv; hi ki/ D hv p D 1; .hi k/p D hip k p l; Œv; hi k D z i C1 ¤ 1i for some l 2 hzi. Hence ˆ.hv; ai ki/ D hk p ; zi D Z.G/ and so ˆ.G/hhi ki D hv; hi ki is a minimal nonabelian maximal subgroup of G. Our theorem is proved. Theorem 101.2. Let G be a two-generator p-group, p > 2, with exactly one maximal subgroup H which is neither abelian nor minimal nonabelian. If G has no abelian maximal subgroup, then 1 D ¹H; H1 ; : : : ; Hp º, where Hi (i D 1; : : : ; p) are nonmetacyclic minimal nonabelian, G 0 Š Ep3 , W D ŒG; G 0 Š Ep2 , W Z.G/ (and so G is of class 3) and CG .G 0 / D ˆ.G/ is abelian. Moreover, ¹H 0 ; H10 ; : : : ; Hp0 º is the set of p C 1 subgroups of order p in W and the following holds. (a) If jGj p 6 , then we have G D hh; x j hp

mC1

m

D 1; Œh; x D v; hp D u; Œv; h D z;

Œv; x D u˛ ; v p D z p D Œu; x D Œz; h D Œz; x D 1; x p 2 hu; zii; where m 2 and ˛ is an integer mod p with ˛ 6 0 .mod p/. Here jGj D p mC4 ;

G 0 D hu; z; vi Š Ep3 ;

Z.G/ D hhp i hzi Š Cpm Cp ;

W D ŒG; G 0 D hu; zi Z.G/; ˆ.G/ D Z.G/ hvi:

Finally, H D ˆ.G/hxi, where in case x p 2 W hui we have d.H / D 3 and in case x p 2 hui we have d.H / D 4 and Hi D hv; x i hi (i D 1; : : : ; p) is the set of p nonmetacyclic minimal nonabelian maximal subgroups of G. (b) If jGj D p 5 , then 2

G D hh; x j hp D 1; Œh; x D v; hp D u˛ ; Œv; h D z; Œv; x D u; v p D z p D Œu; x D Œz; h D Œz; x D 1; hp D uˇ z i; where ˛; ˇ; are integers mod p with ˇ 6 0 .mod p/. We have ˆ.G/ D G 0 D hu; z; vi Š Ep3 and W D ŒG; G 0 D hu; zi D Z.G/ Š Ep2 . If p 5, then we have ˛ .mod p/. In that case, ˛ 0 .mod p/ implies Ã1 .G/ D hui and 1 .G/ D H Š S.p 3 / Cp and ˛ 6 0 .mod p/ implies Ã1 .G/ D W , 1 .G/ D G 0 and H Š Mp3 Cp . Also, all p maximal subgroups Hi D G 0 hx i hi (i integer mod p) are nonmetacyclic minimal nonabelian. If p D 3, then either ˇ D 1 and 6 ˛ .mod 3// or ˇ D 1 and ˛ .mod 3/. In that case, H D G 0 hxi Š S.27/ C3 or H Š M27 C3 and all three maximal subgroups Hi D G 0 hx i hi (i integer mod p) are nonmetacyclic minimal nonabelian.

70

Groups of prime power order

Proof. We set 1 D ¹H; H1 ; : : : ; Hp º, where Hi (i D 1; : : : ; p) are minimal nonabelian. Since H is neither abelian nor minimal nonabelian, we have jH j p 4 and so jGj p 5 . First suppose that two distinct minimal nonabelian maximal subgroups of G have the same commutator subgroup, say, H10 D H20 . Then considering G=H10 (see Exercise 1.6(a)), we see that all maximal subgroups of G=H10 are abelian and so we get H 0 D H10 D D Hp0 D hzi Š Cp : By Exercise 1.69(a), we have jG 0 W .H10 H20 /j D jG 0 W H10 j p, and so jG 0 j p 2 . But if jG 0 j D p, then this fact together with d.G/ D 2 implies (see Exercise P9) that G is minimal nonabelian, a contradiction. Hence jG 0 j D p 2 . Also, d.H / 3 and so H is nonmetacyclic. Indeed, if d.H / D 2, then (noting that jH 0 j D p) H would be minimal nonabelian, a contradiction. Suppose for a moment that G 0 D hvi Š Cp2 is cyclic. Then H10 D D Hp0 D hv 2 i and G 0 D hvi Hi so that Hi0 is not a maximal cyclic subgroup in Hi and therefore Hi is metacyclic for all i D 1; : : : ; p. By Proposition A.40.12, G is an L3 -group. But in that case, G 0 is of exponent p, a contradiction. We have proved that G 0 Š Ep2 . We notice that ˆ.G/ D H1 \ H2 is a maximal normal abelian subgroup of G. Taking elements h1 2 H1 ˆ.G/ and h2 2 H2 ˆ.G/, we have hh1 ; h2 i D G and so s D Œh1 ; h2 2 G 0 hzi and s 62 Z.G/. Indeed, if s 2 Z.G/, then G=hsi would be abelian, a contradiction. In particular, ŒG; G 0 D hzi D H10 and so G is of class 3. Since s 62 Z.G/, we have s 62 Z.H1 / or s 62 Z.H2 / and we may assume without loss of generality that s 62 Z.H1 /. Suppose that there is an element x 2 H1 ˆ.G/ such that x p 2 hzi. Then G 0 hxi is minimal nonabelian of order p 3 and so G 0 hxi D H1 , contrary to jGj p 5 . Assume that hzi D H10 is not a maximal cyclic subgroup in H1 . Then there is v 2 ˆ.G/ such that v 2 D z. This implies that all Hi , i D 1; : : : ; p, are metacyclic. Again by Proposition A.40.12, G is an L3 -group. This means that the subgroup U D 1 .G/ is of order p 3 and exponent p and G=U is cyclic of order p 2 . We have G 0 U and so if U is nonabelian, then CG .G 0 / covers G=U and CG .G 0 / is an abelian maximal subgroup of G, a contradiction. If U is elementary abelian, then jG W CG .U /j D p since a Sylow p-subgroup of GL3 .p/ is isomorphic to S.p 3 / and so is of exponent p. But in that case, CG .U / is an abelian maximal subgroup of G, a contradiction. Hence hzi D H10 is a maximal cyclic subgroup in H1 which yields that H1 is nonmetacyclic. Noting that jH1 j p 4 , we get E D 1 .H1 / D 1 .ˆ.G// Š Ep3 which also implies that all Hi are nonmetacyclic. By the previous paragraph, 1 .hh1 i/ D hui E and u 2 E G 0 . Further, we have H1 D Ehh1 i and so H1 is a splitting extension of G 0 by hh1 i, where o.h1 / D p n , n 2. Since G=G 0 is abelian of rank 2, we get G D H1 F with H1 \ F D G 0 and jF W G 0 j D p. We have G=F Š H1 =G 0 Š Cpn . If F is nonabelian, then CG .G 0 /

101

p-groups G with p > 2, d.G/ D 2, exactly one exceptional maximal subgroup 71

covers G=F and so CG .G 0 / is an abelian maximal subgroup of G, a contradiction. Hence F is abelian. Assume that Ã1 .F / 6 hzi. Then we obtain that jÃ1 .F /j D p and G 0 D hzi Ã1 .F / Z.G/, a contradiction. Hence Ã1 .F / hzi and so for an element x 2 F G 0 we have x p 2 hzi. Since G satisﬁes G D hh1 ; xi, we may set Œx; h1 D s 2 G 0 hzi and Œs; h1 D z, i where H10 D hzi Z.G/. Then x h1 D xs, s h1 D sz and s h1 D sz i for all i 1. We get 2 x h1 D .xs/h1 D .xs/.sz/ D xs 2 z i

i

and claim that we have x h1 D xs i z .2/ for all i 2. Indeed, by induction on i, iC1

x h1

i

i

i

i

i C1 2

D .x h1 /h1 D .xs/h1 D .xs i z .2/ /.sz i / D xs iC1 z .2/Ci D xs i C1 z .

/:

p p p Our formula gives x h1 D xs p z . 2 / D x and so F hh1 i is an abelian maximal subgroup of G, a contradiction. We have proved that H10 D hz1 i, H20 D hz2 i; : : : ; Hp0 D hzp i are pairwise distinct subgroups of order p in G 0 \ Z.G/. By Exercise 1.69(a), we get jG 0 W .H10 H20 /j p and so jG 0 j p 3 . Set W D hz1 ; : : : ; zp i so that W is an elementary abelian subgroup of order p 2 contained in G 0 \ Z.G/ which implies that G 0 is abelian of exponent p 2 . We have G D hx; yi for some x; y 2 G. If Œx; y 2 Z.G/, then G=hŒx; yi is abelian which implies that G 0 D hŒx; yi is cyclic, contrary to the fact that W G 0 . Thus Œx; y 2 G 0 W which gives

jG 0 j D p 3 ;

W Š Ep2 ;

¹1º ¤ ŒG; G 0 W Z.G/

and so G is of class 3. Let hzpC1 i be the subgroup of order p in W with hzpC1 i ¤ hzi i for all i D 1; : : : ; p. For any ﬁxed i 2 ¹1; : : : ; pº we consider G=hzi i, where Hi =hzi i is abelian (and two-generated) and Hj =hzi i is minimal nonabelian for all j ¤ i, j 2 ¹1; : : : ; pº. This implies that H=hzi i must be nonabelian (Exercise 1.6(a)). If G=hzi i is metacyclic, then Theorem 36.1 gives that G is also metacyclic, contrary to Ep2 Š W G 0 . Hence G=hzi i is nonmetacyclic. Suppose that H=hzi i is minimal nonabelian. Then G=hzi i is a nonmetacyclic A2 -group. If jG=hzi ij > p 4 , then Theorem 65.7(a) implies that G 0 =hzi i Š Ep2 . Suppose that jG=hzi ij D p 4 and G 0 =hzi i Š Cp2 . In that case, each maximal subgroup of G=hzi i is metacyclic and so G=hzi i is minimal nonmetacyclic. By Theorem 69.1, G=hzi i is a group of maximal class and order 34 . But in that case, G 0 =hzi i cannot be cyclic (see Exercise 9.1(c)). We have proved that in any case G 0 =hzi i Š Ep2 . Assume now that H=hzi i is not minimal nonabelian and we know already that H=hzi i is nonabelian. By Theorem 101.1, we have again G 0 =hzi i Š Ep2 . As a consequence we get Œx; yp 2 hzi i for each i D 1; : : : ; p which implies that Œx; yp D 1 and so G 0 Š Ep3 is elementary abelian. By Lemma 36.5(b), G 0 =ŒG; G 0 is cyclic and so ŒG; G 0 D W (since ŒG; G 0 W ). Since .G=W /0 Š Cp and d.G=W / D 2, G=W is minimal nonabelian (Exercise P9)

72

Groups of prime power order

and so H=W is abelian which implies ¹1º ¤ H 0 W . Suppose that H 0 D hzj i for some j 2 ¹1; : : : ; pº. Then G=hzj i is nonabelian with at least two distinct abelian maximal subgroups Hj =hzj i and H=hzj i. But then .G=hzj i/0 Š Cp (Exercise P1), a contradiction. We have proved that H 0 D hzpC1 i or H 0 D W . Suppose that H 0 D W Z.G/. In that case, d.H / 3. Indeed, if d.H / D 2, then H is a two-generator group of class 2 in which case H 0 must be cyclic (Proposition 36.5(a)), a contradiction. Consider G=hzpC1 i with d.G=hzpC1 i/ D 2 and having minimal nonabelian maximal subgroups Hi =hzpC1 i for all i D 1; : : : ; p. The remaining maximal subgroup H=hzpC1 i is neither abelian nor minimal nonabelian since d.H=hzpC1 i/ 3. But .Hi =hzpC1 i/0 D W=hzpC1 i for all i D 1; : : : ; p, contrary to the ﬁrst part of this proof. Hence we must have H 0 D hzpC1 i. We have proved that H 0 ; H10 ; : : : ; Hp0 are p C 1 pairwise distinct subgroups of order p in W . Since H is not minimal nonabelian, we have d.H / 3 and so jH j p 4 and jGj p 5 . Also, 1 .Hi / D G 0 ˆ.G/ for all i D 1; : : : ; p, where ˆ.G/ is abelian. Therefore we have either CG .G 0 / D ˆ.G/ or CG .G 0 / is a maximal subgroup of G. In any case, there exist two minimal nonabelian maximal subgroups of G, say H1 and H2 , such that G 0 6 Z.H1 / and G 0 6 Z.H2 /. Then H1 \ H2 D ˆ.G/ and taking some elements h1 2 H1 ˆ.G/ and h2 2 H2 ˆ.G/, we have hh1 ; h2 i D G and so v D Œh1 ; h2 2 G 0 W . Indeed, if v 2 W , then G=hvi is abelian, a contradiction. We may set Œv; h1 D z1 and Œv; h2 D z2 so that H10 D hz1 i, H20 D hz2 i and W D hz1 ihz2 i. All maximal subgroups of G are H1 D ˆ.G/hh1 i and ˆ.G/hhi1 h2 i, where i is any integer mod p. We compute Œv; hi1 h2 D Œv; h2 Œv; hi1 h2 D z2 .z1i /h2 D z1i z2 ¤ 1 which shows CG .v/ D ˆ.G/ and so CG .G 0 / D ˆ.G/. As hv; h1 i (with Œv; h1 D z1 ) is minimal nonabelian, we have hv; h1 i D H1 D G 0 hh1 i. Hence H1 =G 0 Š Cpm is pm m cyclic of order p , m 1, and h1 2 W hz1 i. The abelian group G=G 0 is of rank 2 and so G=G 0 is of type .p m ; p/ and jGj D p mC4 . Finally, ˆ.G/ D G 0 hh1 i D hh1 i hvi hz1 i p

p

is abelian of type .p m ; p; p/. pm

(i) First suppose that m 2. Set u D h1 so that u 2 W H10 , o.h1 / D p mC1 p 3 and ˆ.G/=G 0 is cyclic of order p m1 p since H1 =G 0 Š Cpm . Consider any Hi for 2 i p so that Hi \ H1 D ˆ.G/. Let hi 2 Hi ˆ.G/ and v 2 G 0 W so that 1 ¤ Œhi ; v D zi and Hi0 D hzi i. Since hhi ; vi is minimal nonabelian, we have hhi ; vi D Hi D G 0 hhi i and so Hi =G 0 is also cyclic of order p m . Further, we have p p p ıp hi 2 ˆ.G/ G 0 and hhi i covers ˆ.G/=G 0 . It follows hi D h1 k for some k 2 G 0 and ı 6 0 .mod p/. Then p2

hi

ıp 2

D h1

and so

pm

hhi i D hui;

pm

where u D h1 ;

101

p-groups G with p > 2, d.G/ D 2, exactly one exceptional maximal subgroup 73

and this implies that u 2 W Hi0 . We have proved that u 62 Hi0 for all i D 1; : : : ; p which forces hui D H 0 . Since G=G 0 is abelian of type .p m ; p/, m 2, we get G D H1 F with H1 \F D G 0 and jF W G 0 j D p. For the maximal subgroup ˆ.G/F of G we have .ˆ.G/F /=G 0 D .ˆ.G/=G 0 / .F=G 0 / Š Cpm1 Cp and so .ˆ.G/F /=G 0 is not cyclic which implies that ˆ.G/F D H . Taking the elements h1 D h 2 H1 ˆ.G/ and x 2 F ˆ.G/, we have o.x/ p 2 , G D hh; xi and m so v D Œh; x 2 G 0 W . We set again u D hp and we know that H 0 D hui. Also j set Œv; h D z1 D z, where H10 D hzi, W D hui hzi, v h D vz and v h D vz j for j 1. Since CG .G 0 / D CG .v/ D ˆ.G/, we have x p 2 W D hu; zi and Œv; x D u˛ with ˛ 6 0 .mod p/. We have obtained all relations stated in part (a) of our theorem. From Œh; x D v we get Œh2 ; x D Œh; xh Œh; x D v h v D .vz/v D v 2 z: j

We prove by induction on j 2 that Œhj ; x D v j z . 2 / . Indeed, j

j

Œhj C1 ; x D Œhhj ; x D Œh; xh Œhj ; x D v h .v j z . 2 / / D .vz j /.v j z . 2 / / j

j

j C1 2

j

D v j C1 z j C. 2 / D v j C1 z .

/:

p

In particular, Œhp ; x D v p z . 2 / D 1 and so hp 2 Z.G/ since G D hh; xi. We get Z.G/ D hhp i hzi Š Cpm Cp

and ˆ.G/ D Z.G/ hvi:

Further, we have H D F hhp i with F \ hhp i D hui. If x p 2 W hui, then F is minimal nonabelian and so d.H / D 3. If x p 2 hui, then F D hv; xi hzi, where hui D Z.hv; xi/ and hv; xi Š S.p 3 / or Mp3 and so d.H / D 4. Finally, we have to check all p maximal subgroups Hi D ˆ.G/hx i hi of G which are distinct from H and we have to show that they are minimal nonabelian. We have Œv; x i h D Œv; hŒv; x i h D z.u˛i /h D zu˛i ; where hzu˛i i are pairwise distinct subgroups of order p in W for i D 1; : : : ; p since ˛ 6 0 .mod p/). Thus hv; x i hi is minimal nonabelian with hv; x i hi0 D hzu˛i i ¤ hui. By the Hall–Petrescu formula (Appendix 1), we get for all r; s 2 G, m

m

p m m m . / .p / .rs/p D r p s p c2 2 c3 3 ;

where c2 2 G 0 and c3 2 W D ŒG; G 0 . But m 2 and so .rs/p D r p s p . We get m

m

m

m

m

m

m

.x i h/p D .x i /p hp D hp D u and so o.x i h/ D p mC1 which together with hv; x i hi D G 0 hx i hi and G 0 \hx i hi D hui implies that jhv; x i hij D p mC3 . But jGj D p mC4 and so hv; x i hi D Hi is minimal nonabelian and we are done.

74

Groups of prime power order

(ii) Suppose that m D 1. Then jGj D p 5 and G 0 D ˆ.G/ with CG .G 0 / D G 0 . We have H1 \ H D G 0 and since 1 .H1 / D G 0 , we get 1 ¤ hp 2 W D ŒG; G 0 D Z.G/ Š Ep2 for an element h 2 H1 G 0 . Also, we have Ã1 .G/ W . Take an element x 2 H G 0 so that x p 2 W , G D hh; xi and v D Œh; x D G 0 W . Set Œv; x D u so that 1 ¤ u 2 W and hui D H 0 . Then Œv; h D z 62 hui from which we conclude that H10 D hzi and W D huihzi. Since hv; xi is minimal nonabelian, we have hv; xi ¤ H which implies x p D u˛ (for some integer ˛ mod p) and H D hv; xi hzi, where hv; xi Š S.p 3 / or Mp3 . Since H1 is nonmetacyclic minimal nonabelian, hzi is a maximal cyclic subgroup in H1 which implies hp D uˇ z with ˇ 6 0 .mod p/. All p maximal subgroups Hi D G 0 hx i hi (i is any integer mod p) of G which are distinct from H must be minimal nonabelian. Since Œv; x i h D Œv; hŒv; x i h D z.ui /h D zui ¤ 1 and hv; x i hi is minimal nonabelian with hv; x i hi0 D hui zi and so we must have that Hi D hv; x i hi, we get h.x i h/p i ¤ hui zi or, equivalently, ()

h.x i h/p ; ui zi D hu; zi

for all integers i .mod p/. (ii1) First we assume p 5 in which case G is regular. By the Hall–Petrescu formula (Appendix 1), we have in our case for all r; s 2 G, .p/ .p/ .rs/p D r p s p c2 2 c3 3 ; where c2 2 G 0 and c3 2 W D ŒG; G 0 and so .rs/p D r p s p . Hence .x i h/p D x pi hp D u˛i .uˇ z / D u˛iCˇ z : Our condition () is equivalent to ˇ ˇ ˇ˛i C ˇ ˇ ˇ ˇ D .˛ /i C ˇ 6 0 ˇ i 1ˇ

.mod p/

for all integers i .mod p/, where we know that ˇ 6 0 .mod p/. This is equivalent to ˛ 0 .mod p/ and so ˛ .mod p/. If ˛ 0 .mod p/, then Ã1 .G/ D hui and

1 .G/ D H Š S.p 3 / Cp :

In case ˛ 6 0 .mod p/ we get Ã1 .G/ D W and 1 .G/ D G 0 so that H Š Mp3 Cp . (ii2) Finally, we suppose p D 3. In that case, the Hall–Petrescu formula gives for all r; s 2 G, .rs/3 D r 3 s 3 c3 , where c3 2 W . This is not a sufﬁcient information because

101

p-groups G with p > 2, d.G/ D 2, exactly one exceptional maximal subgroup 75

we have to know exactly the element c3 . Using the usual commutator identities (see 7, p. 98) together with xy D yxŒx; y, we compute exactly .rs/3 . We get .rs/2 D r.sr/s D r.rsŒs; r/s D r 2 s.sŒs; rŒs; r; s/ D r 2 s 2 Œs; rŒs; r; s; .rs/3 D r 2 s 2 Œs; rŒs; r; s rs D r 2 s 2 rŒs; rsŒs; r; rŒs; r; s D r 2 .s 2 r/sŒs; rŒs; r; sŒs; r; rŒs; r; s: But we have r 2 .s 2 r/sŒs; r D r 2 .rs 2 Œs 2 ; r/sŒs; r D r 3 s 3 Œs 2 ; rŒs 2 ; r; sŒs; r D r 3 s 3 Œs; rŒs; rs Œs 2 ; r; sŒs; r D r 3 s 3 Œs; r.Œs; rŒs; r; s/Œs 2 ; r; sŒs; r D r 3 s 3 Œs; r; sŒs 2 ; r; s and so .rs/3 D r 3 s 3 Œs; r; sŒs 2 ; r; sŒs; r; sŒs; r; rŒs; r; s D r 3 s 3 Œs 2 ; r; sŒs; r; r:

(1)

Also, we get Œs 2 ; r D Œs; rs Œs; r D Œs; rŒs; r; sŒs; r D Œs; r2 Œs; r; s and then Œs 2 ; r; s D ŒŒs; r2 Œs; r; s; s D ŒŒs; r2 ; s D Œs; r; sŒs;r Œs; r; s D Œs; r; s2 which together with (1) yields .rs/3 D r 3 s 3 Œs; r; s2 Œs; r; r D r 3 s 3 Œs; r; s1 Œs; r; r D r 3 s 3 Œs; Œs; rŒŒs; r; r: Thus we have obtained the formula ()

.rs/3 D r 3 s 3 Œs; Œs; rŒŒs; r; r:

Since hhp ; zi D huˇ z ; zi D hu; zi (noting that ˇ 6 0 .mod 3/),we have to use our condition () only for i D 1; 2. By (), .xh/3 D x 3 h3 Œh; Œh; xŒŒh; x; x D u˛ uˇ z Œh; vŒv; x D u˛Cˇ C1 z 1 ; and so from hu˛Cˇ C1 z 1 ; uzi D hu; zi we get ˇ ˇ ˇ˛ C ˇ C 1 1ˇ ˇ ˇ D ˛ C ˇ 1 6 0 (2) ˇ 1 1 ˇ

.mod 3/:

We compute Œh; x 2 D Œh; xŒh; xx D vv x D v.vu/ D v 2 u and so by (), .x 2 h/3 D x 6 h3 Œh; v 2 uŒv 2 u; x 2 D u2˛ .uˇ z /z 2 u D u˛Cˇ C1 z

C1

:

76

Groups of prime power order

From our condition () for i D 2 we get hu˛Cˇ C1 z C1 ; u2 zi D hu; zi or, equivalently, ˇ ˇ ˇ˛ C ˇ C 1 C 1ˇ ˇ ˇ D ˛ C ˇ C 1 6 0 .mod 3/: (3) ˇ 1 1 ˇ

Now, (2) and (3) hold if and only if .˛ C ˇ 1/.˛ C ˇ C 1/ 6 0

.mod 3/:

This is equivalent to ..ˇ 1/ C .˛ //..ˇ 1/ .˛ // 6 0

.mod 3/

.ˇ 1/2 .˛ /2 6 0

.mod 3/:

or

Hence if ˇ D 1, then 6 ˛ .mod 3/ and if ˇ D 1, then ˛ .mod 3/. We have obtained the groups stated in part (b) of our theorem which is now completely proved.

103

Some results of Jonah and Konvisser

In this section we will present another approach to counting theorems due to Jonah and Konvisser [KonJ, JKon]. Deﬁnition 1. Let G be a p-group. The line M1 M2 determined by two distinct maximal subgroups M1 and M2 of G is the set of all maximal subgroups of G containing M1 \ M2 . Obviously, jM1 M2 j D p C 1, i.e., every line contains exactly p C 1 points. Proposition 103.1. Let … be a ﬁnite projective space in which each line contains p C1 points and let f be a function from the points of … to the set of integers. Suppose that there is a point m0 2 … with the following property: If m1 and m2 are two points different from m0 on the same line through m0 , then f .m1 / f .m2 / .mod p/. P Then m2… f .m/ f .m0 / .mod p/. Proof. We have X

f .m/ D f .m0 / C

m2…

XX

f .m/ ;

L0 … m2L0

where L runs over all lines of … through m0 and L0 D L ¹m0 º. Furthermore, X f .m/ jL0 jf .m/ 0 .mod p/ m2L0

since f is constant (mod p) on the points of L0 and jL0 j D p by hypothesis. The two displayed formulas yield the result. Applying these ideas to p-groups, we get the following Theorem 103.2. Let G be a p-group, C be a nonempty set of proper subgroups of G and ˛.H / be the number of elements of C contained in H G. Suppose that: (i) For each proper normal subgroup M of G which contains an element of C, one has ˛.M / 1 .mod p/.

94

Groups of prime power order

(ii) There is a maximal subgroup M0 of G with the property: if a maximal subgroup M1 ¤ M0 contains an element of C, then so do all maximal subgroups on the line M0 M1 . Then ˛.G/ 1 .mod p/ or, what is the same, jCj 1 .mod p/ since ˛.G/ D jC j. Proof. By Hall’s enumeration principle and Proposition 103.1, we have X ˛.M / ˛.M0 /j1 j 1 .mod p/ ˛.G/

M 21

since, by hypothesis, if M 2 1 ¹M0 º, then ˛.M / ˛.M0 / 1 .mod p/ and j1 j 1 .mod p/. At this point it is appropriate to give the following Deﬁnition 2. Let C be a nonempty set of proper subgroups of a p-group G. (a) A maximal subgroup M0 of G which satisﬁes Lemma 103.2(ii) is called an origin (for C). (b) A proper subgroup L of G is called a local origin if for each M 2 1 the intersection L \ M contains an element of C . (c) A pair E1 ; E2 of distinct elements of the set C is said to support good lines if each maximal subgroup M < G containing E1 \ E2 also contains an element of C . Proposition 103.3 (Method of locating origins). Let C be a nonempty set, closed under conjugation, of proper subgroups of a p-group G satisfying ˛.M / D 0 or ˛.M / 1 .mod p/ for each M 2 1 . Then G contains an origin M0 if any at least one of the following three conditions is satisﬁed: (a) G contains a local origin L, in which case any maximal subgroup M L is an origin. (b) Each pair of distinct G-invariant elements E1 ; E2 2 C supports good lines, in which case any maximal subgroup M of G containing an element of C is an origin. (c) A maximal subgroup M of G contains a family E1 ; : : : ; Em of G-invariant elements of the set C satisfying: if E is a G-invariant element of the set C , then for some i 2 ¹1; : : : ; mº, the pair E, Ei supports good lines, in which case M is an origin. Proof. Statement (a) follows from the deﬁnitions. Indeed, let L M0 2 1 and let M 2 1 ¹M0 º. Then ˛.M0 / 1 .mod p/ and ˛.M / 1 .mod p/ since L is a local origin. It follows that M0 is an origin. (b) Let M0 2 1 contain an element of the set C . Then M0 contains a G-invariant element E0 2 C since ˛.M0 / 1 .mod p/ in view of ˛.M0 / > 0. For the same

103

Some results of Jonah and Konvisser

95

reasons, each M1 2 1 contains a G-invariant element E1 of the set C whenever M1 contains an element of the set C . Thus each M on the line M0 M1 contains an element of the set C for M > M0 \ M1 E0 \ E1 , which in turn implies that M contains a G-invariant element of C because the G-invariant pair E0 ; E1 supports good lines by hypothesis. Statement (c) follows as (b). Exercise 1 coincides with the case n D 2 of Theorem 1.10(b). Exercise 1. Let G be a noncyclic p-group, p > 2. Then the number of noncyclic subgroups of order p 2 in G is congruent to 1 .mod p/. Solution. We use induction on jGj. Let C be the set of all abelian subgroups of type .p; p/ in G; then C is nonempty (Lemma 1.4) and G-invariant. By induction, if Ep2 Š E M1 2 1 , then ˛.M1 / 1 .mod p/ so that M1 contains a G-invariant abelian subgroup E1 of type .p; p/. One may assume that ˛.G/ > 1. Let E2 ¤ E1 be a G-invariant element of the set C (if E2 does not exist, the result is obvious). Then the product E1 E2 is a group of order p 4 and exponent p since p > 2. Since .E1 E2 /\M contains an abelian subgroup of type .p; p/ for all M 2 1 , it follows that E1 E2 is a local origin; then M is an origin. By Lemma 103.2, ˛.G/ 1 .mod p/. Exercise 2 (Sylow). Let G be a group of order p m and n < m. Then the number of subgroups of order p n in G is congruent to 1 .mod p/. Hint. Let C be the set of all subgroups of order p n in G. As above, C contains a G-invariant element E1 . If E2 ¤ E1 is another G-invariant element of the set C (if E2 does not exist, the result is obvious), then, obviously, E1 E2 is a local origin (indeed, M \ E1 E2 contains an element from C for M 2 1 ). Now the result follows from Lemma 103.2. In the previous two exercises one can take as C the set of all G-invariant subgroups of the set C deﬁned in these exercises. The following result is also known (see Theorem 10.4). Proposition 103.4. Let G be a p-group, p > 2. If G contains an elementary abelian subgroup of order p 3 , then the number of such subgroups in G is congruent to 1 .mod p/. Proof. We use induction on jGj. Let C be the set of all elementary abelian subgroups of order p 3 in G. Then, as above, there exists a G-invariant element E1 2 C . Let E2 ¤ E1 be a G-invariant element in C (see Exercises 1 and 2). Set H D E1 E2 ; then exp.H / D p since p > 2 and cl.H / 2 (see Theorems 7.1(b) and 7.2(b)). Then one of the following assertions holds: (i) jH j D p 4 . Then E1 \ E2 Z.H /. Let a maximal subgroup M of G contain E1 \ E2 . If H M , then H and so M contains an element of C. If H 6 M , then

96

Groups of prime power order

H \M is an element of C since jH \M j D p 3 and E1 \E2 < H \M < H so H \M is elementary abelian of order p 3 since E1 \ E2 Z.H / and exp.H \ M / D p. We see that the pair E1 ; E2 supports good lines. (ii) jH j p 5 . If M 2 1 , then jM \ H j p 4 so M \ H contains an abelian subgroup of order p 3 which is an element of the set C hence the pair E1 ; E2 supports good lines. Now the result follows from Proposition 103.3 and Lemma 103.2. Exercise 3. Suppose that a p-group G has order > p 3 , p > 2. Then the number of abelian subgroups of order p 3 in G is congruent to 1 .mod p/. Using the above approach, the following results are established in [KonJ]. Theorem 103.5. Let G be a p-group, p > 2, and let C be any one of the following classes of abelian subgroups of G: (i) elementary abelian subgroups of order p k for some ﬁxed k 5. (ii) abelian subgroups of order p k for some ﬁxed k 5. (iii) abelian subgroups of ﬁxed index p or p 2 . Suppose that the set C is nonempty. Then jCj 1 .mod p/ except in the index p 2 case (case (iii)) where the number can also be exactly 2. Corollary 103.6 ([Alp2]). If G is a p-group, p > 2, having an abelian subgroup of index p 3 , then there is a normal abelian subgroup of the same index in G. Proof. Indeed, let A be an abelian subgroup of index p 3 in G, and let A < M 2 1 ; then jM W Aj D p 2 . Then, by Theorem 103.5, the number of subgroups of index p 2 in M is not divisible by p since p > 2, and so one of these subgroups is normal in G. Corollary 103.7. Suppose that a p-group G possesses an elementary abelian subgroup Ek of order p k , p > 2, k < 6. Then the normal closure EkG of Ek in G contains a G-invariant elementary abelian subgroup of order p k . Indeed, the number of elementary abelian subgroups of order p k in EkG is not divisible by p so one of them is G-invariant. Exercise 4. Suppose that a 2-group G contains an abelian subgroup E of type .2; 2/. Then one of the following holds: (a) E G is dihedral or semidihedral. (b) E G contains a G-invariant abelian subgroup of type .2; 2/. (Hint. Use Theorem 1.17(b).)

104

Degrees of irreducible characters of p-groups associated with ﬁnite algebras

All results of this section are due to Isaacs [Isa11] In what follows q is a power of a prime p, F D GF.q/ (the ﬁeld of cardinality q, where q is a power of a prime), R a ﬁnite dimensional F-algebra (as always, with the identity element). Let J D J.R/ be the Jacobson radical of R (= the maximal nilpotent two-sided ideal of R; see [Lang, p. 658]); then 1 C J D ¹1 C x j x 2 J º is a p-subgroup of R , the group of units in R (Theorem 34.6). We refer to the group 1 C J arising in this way from an F-algebra R as an F-algebra group G. Let x 2 G; then x D 1 C a for some a 2 J . Observe that jGj D jJ j so G is a p-group. The upper unitriangular group UT.n; q/ is a Sylow p-subgroup of GL.n; q/ and it has the form 1 C J , where J is the space of strictly n n upper-triangular matrices over F; then J is the radical of the F-algebra F 1 C J . Thompson conjectured that the degrees of complex irreducible characters of the group UT.n; q/ are degrees of q. This conjecture is a corollary of the following general Isaacs’ theorem which is the main result of this section: Theorem 104.1 ([Isa11, Theorem A]). Let G be an F-algebra group. Then the degrees of all irreducible complex characters of G are powers of q. The theorem is a consequence of the following six lemmas. Lemma 104.2. Let N be a normal subgroup of a group G and suppose that G=N is abelian. Let 2 Irr.G/ and suppose that N D 2 Irr.N /. Then every irreducible constituent of G has the form for some 2 Lin.G=N /.1 Proof. Note that .1N /G D G=N , the regular character of G=N , and so the irreducible constituents of .1N /G are exactly the characters as in the statement of the lemma. Since (see [BZ, Theorem 5.1.2]) Irr. G / Irr.. N /G / and . N /G D .1N N /G D .1N /G ; the result follows. particular, all irreducible constituents of G are extensions of to G. The lemma is a particular case of a known result of Gallagher. 1 In

98

Groups of prime power order

Lemma 104.3. Let G; N and G=N be such as in Lemma 104:2. Let 2 Irr.N / be G-invariant and suppose that 2 Irr. G /. Then there exists a subgroup M of G with N M G and 2 Irr.M / such that N D and M D e, where e 2 D jG W M j. Furthermore, vanishes on the set G M . Proof. Take M to be the unique smallest subgroup of G containing N such that vanishes on G M and let 2 Irr. M / be arbitrary. To see that M D e for some integer e > 0, let 2 Irr. M / be arbitrary. Then ; 2 Irr. M / and so, by Lemma 104.2, D for some 2 Lin.M=N / since M=N is abelian. Let K D ker./; then N K M . Since G=N is abelian, extends to a character of G=N and 2 Irr../G / D Irr. G / (consider . /M ). Now, is an irreducible constituent of M D . /M , and Lemma 104.2 then tells us that D for some 2 Lin.G=M /. For the elements x 2 M K we see that .x/ D .x/ ¤ 1 D .x/ and thus .x/ D 0. By the deﬁnition of M , then K D M and hence D 1M . We conclude that D , i.e., M D e, as claimed. Now e 2 D he; ei D h M ; M i D jG W M j; where the third equality holds since vanishes on G M . It remains to show that N is irreducible. If this is false, it follows from the fact that M=N is abelian that is induced from a character of some subgroup T with N T < M [BZ, Theorem 7.2.2]. Next, T is normal in M , and thus vanishes on M T . It follows that vanishes on G T , a contradiction since T < M . For an arbitrary positive integer q we shall say that G is a q-power-degree group if all of its irreducible characters have degrees that are powers of q. Lemma 104.4. Fix a positive integer q and let G be a ﬁnite group. Let M G G and suppose that H is a collection of subgroups of G. Assume the following: (1) M is the intersection of any two distinct members of the set H . S S (2) G D H 2H H (i.e., G=M D H 2H H=M is a nontrivial partition). (3) jG W H j D q for each H 2 H . (4) G=M is abelian of order q 2 . (5) M and all members of the set H are q-power-degree groups. Then G is a q-power-degree group. 1 Proof. By (1)–(4), jH j D qq1 D q C 1 and if H and K are any distinct members of the set H , we have HK D G by the product formula. To show that G is a q-power-degree group, let 2 Irr.G/ be arbitrary. We choose 2 Irr. M / and let T D IG ./, the inertia subgroup of in G. Let H 2 H and ˛ 2 Irr. H /. Then 2

.1/ ˛.1/ H .1/ D jH W M j.1/ D q.1/:

104

Degrees of irreducible characters of p-groups associated with ﬁnite algebras

99

Since M and H are q-power-degree groups, however, and so .1/ j ˛.1/, it follows that ˛.1/ 2 ¹.1/; q.1/º: In the ﬁrst case, ˛M D and thus H T . Otherwise, we get H D ˛ and this forces H \ T D M. If M < T , then since T is partitioned by subgroups T \ H (H 2 H ), we must have T \ H > M for some H 2 H . Therefore, by the result of the previous paragraph, H T . If, in fact, H < T , then T \K > M for some other member K 2 H different from H , and in this case T HK D G. We see, therefore, that the only possibilities are T D M , T D G or T 2 H . Suppose that T < G. By the Clifford Correspondence Theorem 7.2.2, we know that D ˛ G for some ˛ 2 Irr.T /. Since either T D M or T 2 H , it follows that T is a q-power-degree group, and so ˛.1/ is a q-power. As .1/ D jG W T j˛.1/ and jG W T j 2 ¹q; q 2 º; the result follows in this case. We are left with the situation where T D G so is G-invariant. In this case, Lemma 104.3 yields the existence of a subgroup Q with M Q G and a character 2 Irr.Q/ such that Q D e and M D . Since .1/ D .1/ is a q-power and .1/ D e.1/, it sufﬁces to show that e is either 1 or q. To this end, we shall use the additional information (from Lemma 104.3) that e 2 D jG W Qj and that vanishes on the set G Q. Let H 2 H and consider ˛ 2 Irr. H /. Since is H -invariant and both and ˛ have q-power degrees, we have seen that ˛.1/ < jH W M j.1/ D q.1/ so we get ˛.1/ D .1/. We know that .1/ D e.1/ and it follows that H is a sum of exactly e irreducible constituents and so h H ; H i e. Write S D H \ Q and note that H vanishes on H S since it vanishes on H Q. Also, S is irreducible since M D 2 Irr.M / and M S. We thus have ejH W S j h H ; H ijH W Sj D h S ; S i D heS ; eS i D e 2 : It follows that e jH W S j and so jS=M j D q=jH W Sj q=e. Now Q=M is covered by subgroups .Q \ H /=M as H runs over H , and we have just seen that each of these subgroups has order at least qe . Since the pairwise intersections of these subgroups are trivial and jG=Qj D e 2 , we have jQ=M j D

q2 q2 jQjq 2 D D 2 jGj jG W Qj e

and so X q2 1 D jQ=M j 1 D .j.Q \ H /=M j 1/ 2 e H 2H q q 1 D .q C 1/ 1 : jH j e e

100

Groups of prime power order

If e < q, we can divide by qe 1 to get qe C 1 q C 1 and this yields e D 1. We have now shown that either e D 1 or e D q, and the proof is complete. Note that G=M in Lemma 104.4 has a prime exponent since this quotient group is equally partitioned (see 68). The following lemma resembles the known Nakayama’s lemma. Lemma 104.5. Let J D J.R/ and suppose that U J is a multiplicatively closed subspace. If J D U C J 2 , then U D J . Proof. We show by induction on m that J D U CJ m for any positive integer m. Since J is nilpotent, the result will follow. We can assume that m > 2 and J D U C J m1 . Then, since U J is a multiplicatively closed subspace, we get J 2 D J.U C J m1 / J U C J m ; J U D .U C J m1 /U U C J m and so J 2 U C J m . It follows that J D U C J 2 U C J m U C J m1 D J and we conclude that J D U C J m . Lemma 104.6. Let J D J.R/ and suppose that U J is an ideal of R. Set G D 1CJ and N D 1 C U . Then N is a normal algebra subgroup of G and G=N is naturally isomorphic to the algebra group corresponding to R=U . Proof. Write RN to denote the F-algebra R=U and, as usual, let rN 2 RN denote the image of r 2 R. Observe that GN D 1 C JN and so GN is an F-algebra group and the image of G under the group homomorphism g 7! g. N The result now follows from the observation that N D 1 C U is the kernel of this homomorphism. Before stating the next lemma, we introduce a bit more notation. If G D 1 C J is an F-algebra group, it is clear that the collection of the F-algebra subgroups of G is closed under intersection. It follows that for each element x 2 G there is a unique smallest F-algebra subgroup containing x, and we write A.x/ to denote this subgroup. In fact, it is easy to construct A.x/ explicitly. Write x D 1 C u and observe that the smallest multiplicatively closed F-subspace of J containing u is uFŒu. It follows that A.x/ D 1 C uFŒu. The key observation here is that A.x/ is abelian for all x 2 G. This follows since the multiplication in uFŒu is clearly commutative. Lemma 104.7. Let G be an F-algebra group for R that is not of the form A.x/ for any element x 2 G. (In particular, this holds if G is nonabelian.) Then there exists an F-algebra subgroup M G G and a family H of F-algebra subgroups of G satisfying properties (1)–(4) of Lemma 104:4.

104

Degrees of irreducible characters of p-groups associated with ﬁnite algebras 101

Proof. We have G D 1CJ , where J D J.R/ for some ﬁnite dimensional F-algebra R, and we can assume that R D F 1 C J . If J 2 has codimension 1 in J , we can write J D J 2 C Fu for some element u 2 J . It follows that J D J 2 C uFŒu and hence, by Lemma 104.6, we have J D uFŒu and A.u/ D A.1 C u/, a contradiction. We conclude that the codimension of J 2 in J is at least 2. Now let U be an F-subspace of J having codimension 2 and containing J 2 . Note that U and all F-subspaces of J containing U are ideals of R and that the multiplication in J.R=U / is trivial. Let M D 1 C U and let H be the collection of the subgroups 1 C V , where V runs over all hyperplanes of J that contain U . By Lemma 104.6, M is normal in G and G=M is isomorphic to 1 C J.R=U /, which is abelian. All of the assertions are now clear. We mention that Lemma 104.7 essentially continues to hold even if the ﬁeld F is inﬁnite. Obviously, we need to delete the assertion about the indices of the members H 2 H and M , but everything else goes through without change. Proof of Theorem 104.1. We are given an F-algebra group G, where jFj D q (q being a power of a prime), and we show by induction on jGj that G is a q-power-degree group. We can certainly assume that G is nonabelian and thus, by Lemma 104.7, we get a normal algebra subgroup M and a collection of algebra subgroups H satisfying certain properties. In particular, all of these subgroups are proper in G and thus each is a q-power-degree group by the inductive hypothesis. It follows from Lemma 104.4 that G is a q-power-degree group, as required. It follows from Theorem 104.1 that UT.n; q/ is a q-power-degree group. The important paper [Isa11] contains a number of related results.

105

On some special p-groups

This section was inspired by a letter of Isaacs in which he disproved one problem posed by the ﬁrst author on character degrees (see below). Most main results of this section are also due to Isaacs. 1o . Here we repeat some considerations from 46. Let s > 2 be a positive integer, F D GF.p s /, F0 be the prime subﬁeld of F and S D Gal.F=F0 / D hi; the set of all automorphisms of F. It is known that o./ D s and if .x/ D x for some x 2 F, then x 2 F0 . We will construct a special p-group P D Ap .s; / (see 46) and study its structure. For x; y 2 F, write .x; y/ to denote the matrix .aij / with ai i D 1; i D 1; 2; 3; a12 D x; a13 D y; a23 D .x/; a21 D a31 D a32 D 0: The set of matrices .x; y/ (x; y 2 F) is a group (denote it by P D Ap .s; /) with respect to matrix multiplication; obviously, jP j D jFj2 D p2s . The identity element of P is .0; 0/. It is easy to check that (i) We have (1) .x; y/.x1 ; y1 / D .x C x1 ; y C y1 C x.x1 //;

.x; y/1 D .x; y C x.x//

and the commutator (2)

Œ.x; y/; .x1 ; y1 / D .0; x.x1 / x1 .x//:

If .x1 ; y1 /.x; y/ D .x; y/.x1 ; y1 /, then by (2) we get x.x1 / D x1 .x/ or, if x1 ¤ 0, x11 x D .x11 x/. In that case, x D x1 x0 , where x0 2 F0 . Thus (ii) If x ¤ 0, then (3) CP ..x; y// D ¹.xx0 ; y 0 /; x0 2 F0 ; y 0 2 Fº;

jCP ..x; y//j D jF0 jjFj D p sC1 :

Obviously, Z.P / D ¹.0; y/; y 2 Fº, and so jZ.P /j D jFj D p s and Z.P / Š FC , the additive group of the ﬁeld F. In particular, Z.P / Š Eps . It follows that G is a group with small abelian subgroups (see 20).

105

On some special p-groups

103

By induction on n, we get (4)

1 .x; y/n D nx; ny C n.n 1/x.x/ : 2

Setting in the above formula n D p, we obtain (5)

.x; y/p D .0; p.p 1/=2 x.x//:

If p > 2, then .x; y/p D .0; 0/ is the identity element of P . In case p D 2, we get .x; y/2 D .0; x.x//. If p D 2 and .x; y/ is an involution, then we obtain x D 0 so .x; y/ 2 Z.P /. Thus (iii) exp.P / D p if p > 2 and exp.P / D 4 if p D 2; in the last case, we have Z.P / D 1 .P /. It follows from (3) and (iii) that if .x; y/ 2 P Z.P /, then CP ..x; y// D h.x; y/; Z.P /i is isomorphic to Cp2 Eps1 if p D 2 and is isomorphic to EpsC1 if p > 2. Therefore, sC1 2s if p D 2 and o..x; y// D 4, then .x; y/2 is contained in exactly 22.21/ D 2s1 cyclic subgroups of order 4. It follows from (iii) that if p D 2, then P has exactly 22s 2s D 2s1 .2s 1/ 2.2 1/ cyclic subgroups of order 22 and 2s 1 subgroups of order 2. Therefore (iv) If p D 2, then Ã1 .P / D Z.P /. In particular, ˆ.P / D Ã1 .P /P 0 D Z.P / (by (i), P 0 Z.P /). It follows from (ii) that jP 0 j p s1 (since jP 0 j is at least the size of a conjugacy class). We will prove that jP 0 j D p s . Let x1 ; : : : ; xs be a normal basis of the extension F=F0 consisting of primitive elements of F.1 Then, by (ii), the commutators ci D Œ.xi ; 0/; .1; 0/ D .0; xi .xi /;

i D 1; : : : ; s:

We claim that c1 ; : : : ; cs are linearly independent. Indeed, assume c1a1 : : : csas D .0; 0/ for some a1 ; : : : ; as 2 ¹0; : : : ; p 1º. Set ui D xi .xi /, i D 1; : : : ; s. Then a2 u2 C C as us D 0 or, what is the same, .a1 x1 C C as xs / D a1 x1 C C as us : 1 The elements x ; : : : ; x constitute a normal basis of the extension F=F if .x / 2 ¹x ; : : : ; x º s s 1 0 1 i for all i and 2 Gal.F=F0 /. A normal basis consisting of primitive elements does always exists; see [Ward, Kapitel 8].

104

Groups of prime power order

This implies a1 x1 C C as xs 2 F0 and so a1 D D as D 0, proving our claim. Therefore jhc1 ; : : : ; cs ij D p s and so P 0 D Z.P /. Thus (v) P is special, P 0 D Z.P / D ˆ.P / is elementary abelian of order p s . By (3) and (v), we obtain (vi) The class number of P is k.P / D jZ.P /j C

jP j jZ.P /j D p s C p sC1 p: p s1

Next, jLin.P /j D jP W P 0 j D p s

and

jIrr1 .P /j D k.G/ jLin.P /j D p sC1 p;

where Irr1 .P / is the set of nonlinear characters of P . Let s be odd. If 2 Irr.P /, then .1/2 jP W Z.P /j D p s and so .1/2 p s1 . In this case, X .1/2 p s1 .p sC1 p/ D p 2s p s p 2s p s D jP j jP W P 0 j D 2Irr1 .P /

and so (vii) If s is odd, then the degree of any nonlinear irreducible character of P is equal 1 1 to p 2 .s1/ , i.e., cd.P / D ¹1; p 2 .s1/ º. It is easy to check that if s is even, then jcd.P /j > 2. According to 46, in this case, jcd.G/j D 3. An automorphism of a group G is said to be central or normal, if it centralizes Inn.G/, the group of inner automorphisms of G. In this case, for any x; g 2 G one has .x/1 .g/.x/ D .x 1 gx/ D x 1 .g/x and so

.g/..x/x 1 / D ..x/x 1 /.g/

hence, since im./ D G, we get .x/ D xzx , where zx 2 Z.G/ for all x 2 G, and we conclude that induces the identity automorphism on G=Z.G/. Conversely, if 2 Aut.G/ induces the identity automorphism of G=Z.G/, it is central. The set Autc .G/ of central automorphisms of a group G is a normal subgroup of Aut.G/.2 (viii) (L. Kazarin) Let p D 2 and let s > 1 be odd, P D A2 .s; /, where is a generator of Gal.F=F0 /, 2 Autc .P /. Then .a; b/ D .a; b C .a// for all .a; b/ 2 P , where D is a linear operator of the s-dimensional vector space GF.2s /=GF.2/. 2 Indeed,

if ˛ 2 Aut.G/, 2 Autc .G/ and g 2 G, then .˛ 1 ˛/.g/ D ˛ 1 ..˛.g// D ˛ 1 .˛.g/z/ D g˛.z/

for some z 2 Z.G/, and we conclude that ˛ 1 ˛ 2 Autc .G/.

105

On some special p-groups

105

2

We claim that jAutc .P /j D 2s . Indeed, .a; b/ D .a; 0/.0; b/. Let us show that ¹a.a/ j a 2 GF.2s /º D GF.2s /: In fact, if a.a/ D a1 .a1 / (a; a1 2 GF.2s //, then .a1 a1 / D .a1 a1 /1 so that a D a1 in view of o./ D s is odd. It follows that for each b 2 GF.2s / there exists a unique a 2 GF.2s / such that b D a.a/, and so .0; b/ D .a; 0/2 . Therefore is deﬁned uniquely by its action on elements of the form .a; 0/. Let .a; 0/ D .a; .a//, where W GF.2s / ! GF.2s /. Since .a; 0/2 D .0; a.a//, then for b D a.a/ we have .0; b/ D ..a; 0/2 / D ..a; 0/ /2 D .a; .a//2 D .0; a.a// D .0; b/; i.e., ﬁxes all elements of Z.P /. In this case, .a; b/ D ..a; 0/.0; b// D .a; .a//.0; b/ D .a; b C .a//: If .a; b/ D .0; 0/ and D is linear, then a D 0 and b D .a/ D .0/ D 0, and so is an injection so a surjection. We claim that is linear. We have .a; 0/ .c; 0/ D .a; .a/.c; .c// D .a C c; .a/ C .c/ C a.c//: On the other hand, .a; 0/.c; 0/ D .a C c; a.c// and ..a; 0/.c; 0// D .a C c; .a C c/ C a.c//: Since is an automorphism, we get .a C c/ D .a/ C .c/. It is easy to check that the mapping W .a; b/ 7! .a; b C .a// is an automorphism of P for every linear map W GF.2s /=GF.2/ ! GF.2s /=GF.2/: Thus, .a; b/ D .a; b C .a// D .a; b/.0; .a//; and 2 Autc .P / since .0; .a// 2 Z.P /. It is easy to see that for every there is a unique central automorphism of P corresponding to . Since the number of ’s is equal to the number of s s matrices over GF.2/ (every such matrix contains s 2 entries equal 0 or 1), the claim (viii) follows. 2o . Let P , F, and p s be as in the previous subsection. Set bD

ps 1 D 1 C p C C p s1 : p1

The multiplicative cyclic group F of order p s 1 has a unique subgroup C of order b since b j p s 1. We deﬁne an action of C on P as follows: .x; y/c D .xc; yc pC1 / .c 2 C /:

106

Groups of prime power order

Note that .c/ D c p for all c 2 F ( is a generator of Aut.F/). We have .x; y/c .u; v/c D .xc; yc 1Cp /.uc; vc 1Cp / D ..x C u/c; .y C v C x.u//c 1Cp / D ..x; y/.u; v//c ; and we conclude that .x; y/ 7! .xc; yc pC1 / is an action. If .x; y/c D .x; y/, then c D 1, i.e., every element of C induces a ﬁxed-point-free automorphism of P . Therefore CP D .C; P / is a Frobenius group with kernel P and cyclic complement C of order b. We will construct a p-solvable group G such that b.Q/ does not divide .1/ for all 2 Irr.G/, where Q 2 Sylp .G/ and b.Q/ D max¹.1/; 2 Irr.Q/º. This example is taken from Isaacs’ unpublished note. We retain the above notation. In what follows we will assume that s D p > 2. Then o./ D p. We deﬁne an action of S D hi on P D Ap .p; / as follows: .x; y/ D ..x/; .y//: Set G D .C S/ P . If .x; y/ D .x; y/, then x; y 2 F0 , and so jCP ./j D jF0 j2 D p 2 ;

jCZ.P / ./j D p:

Thus ﬁxes exactly p classes of Z.P / and p 2 classes of P and p classes of P =P 0 D P =Z.P /. By Brauer’s permutation Lemma 10.3.1(c), ﬁxes exactly p linear characters of both groups Z.P / and P . If ¤ 1Z.P / is one of them in Lin.Z.P // and H D ker./, then ﬁxes all linear characters of Z.P / with kernel H . It follows that ﬁxes exactly one maximal subgroup of Z.P /. We choose so that .c/ D c p for all c 2 F. If c 2 C and .c/ D c, then we have c D 1 since .jC j; p/ D 1. Therefore S C D .S; C / is a Frobenius group with kernel C and complement S D h i. The group .S; C / has exactly jC j D b subgroups of order jSj D s D p and all of them are conjugate in SC (Sylow). If T is a subgroup of order p in .S; C / D .T; C /, then T ﬁxes exactly one maximal subgroup HT of Z.P /. Since b is equal to the number of maximal subgroups of Z.P / and to the number of subgroups of order p in .S; C /, there is a one-to-one correspondence T $ HT , where T and HT are as above. By the result of the previous paragraph, S ﬁxes exactly p 2 p D p.p 1/ nonlinear irreducible characters of P (these characters are irreducible constituents of ˛ P , where ˛ runs over all p 1 nonprincipal linear characters of Z.P /=HS and HS is the unique maximal subgroup of Z.P / ﬁxed by S). It follows that ˛ SP is the sum of p 2 irreducible characters of SP of degree p .p1/=2 . In particular, the inertia subgroup IG .˛/ D SP , and so all irreducible constituents of ˛ G have degree bp .p1/=2 . If T is a subgroup of order p in SC , then it is conjugate to S in SC . Therefore, if HT is the unique maximal subgroup of Z.P / ﬁxed by T and ˇ is a nonprincipal linear character of Z.P /=HT , then IG .ˇ/ D TP . If is a nonlinear irreducible character of G=Z.P /, then .1/ divides jG W P j D bp. Therefore, the p-part of the degree of every irreducible character of G does not exceed p .p1/=2 .

105

On some special p-groups

107

The Sylow p-subgroup SP of G has an irreducible character of degree p .pC1/=2 , however. This is because S deﬁnitely does not stabilize all nonlinear irreducible characters of P . (It stabilizes only those that have a particular ﬁxed maximal subgroup of Z.P / as kernel.) This proves that b.SP / does not divide .1/ for all 2 Irr.G/. The same question is open for p D 2. Isaacs [Isa4] proved that for every set S of powers of a prime p containing p 0 D 1 there exists a p-group P such that cd.P / D S (see Theorem 105.2 below). Next, he showed that for every p-group P there exists a group G with cyclic Sylow p-subgroup C such that cd.G/ D cd.P /. We will prove this result. More exactly, we will show that for every set S of powers of a prime p containing 1 D p 0 and with maximum member p a there exists a group G having a cyclic Sylow p-subgroup C of order p a such that cd.G/ D S. For each member p e 2 S, let Ve be an elementary abelian q-group on which a cyclic group of order p e acts faithfully and irreducibly. Let C act on Ve with kernel of order p ae (in that case, C acts on Ve irreducibly), and let W be the direct product of the groups Ve for all p e 2 S so that C acts on W . Let G D C W be the semidirect product of W with C . We claim that cd.G/ D S. Obviously, it is enough to prove that p a 2 cd.G/ (by Ito’s theorem (see Introduction, Theorem 17), the set cd.D/ contains only powers of p not exceeding jC j D p a ). Let W D Ve1 Ves ;

where S D ¹1; p e1 ; : : : ; p es D p a º:

Let xi 2 Ve#i , i D 1; : : : ; s, x D x1 : : : xs . Then the normal closure of hxi in G is equal to W , the socle of G. Therefore, by [BZ, Corollary 9.6] (Gasch¨utz’ theorem), Irr.G/ has a faithful character . We claim that .1/ D p es D p a . Note that C Va is a Frobenius group. As Va 6 ker. /, it follows that Va has a nonlinear constituent, and so .1/ p a in view of cd.C Va / D ¹1; p a º. Since .1/ divides p a by Ito’s theorem (see Introduction, Theorem 17), we get .1/ D p a , as desired. Of course, if cd.G/ D S, where S is as in the previous paragraph, then G has an abelian normal p-complement. Problem 1. Let P be a Sylow p-subgroup of a p-solvable group G. Study cd.P / in the case where p 2 does not divide .1/ for all 2 Irr.G/. Problem 2. Let P D Ap .s; /, where is a generator of Aut.GF.p s //, s > 2 is even. Find cd.P /. Exercise. Let G be a group of class two of maximal order generated by n > 1 elements of order p (for the matrix construction of this group, see [Macd3]). Prove that (a) exp.G/ D p if p > 2 and exp.G/ D 4 if p D 2, jGj D p n.nC1/=2 . (b) G is special and jZ.G/j D p n.n1/=2 . (c) If x 2 G Z.G/, then jG W CG .x/j D p n1 , i.e., G is a p-group with small abelian subgroups; see 20. (d) k.G/ D p n.n1/=2 C p n.nC1/=2.n1/ p n.n1/=2.n1/ .

108

Groups of prime power order

(e) For any positive integer n such that 2r n, Irr.G/ has a character of degree p r . P (f) Find T.G/ D 2Irr.G/ .1/. (g) (Mann) If p > 2, then we have Aut.G/=Autc .G/ Š Aut.G=G 0 / Š GL.n; p/, jAutc .G/j D jG 0 jn and Autc .G/ is elementary abelian. (h) (Mann) If p D 2, then Aut.G/=Autc .G/ Š Sn . (i) If n > 5, then G 0 has non-commutators. (In fact, this is true in the case n > 3. Moreover, if n D 4, then G has an epimorphic image H of order p 8 such that not all elements of H 0 are commutators [Macd3]. Numerous examples of groups, in which not all elements of G 0 are commutators, have been published; see, for example, [Fit], [Isa6] and [Macd3].) 3o . Let X be a set of powers of p such that 1 2 X. We will prove that there exists a p-group G such that cd.G/ D X . As we know (see the exercise, this is the case if X D ¹1; p; : : : ; p e º. Isaacs [Isa4] has shown that such a G exists for any X; moreover, G may be taken to be of class 2. Let U be an abelian group which acts on an abelian group A. Set b A D Lin.A/, the group of linear characters of A. As we know (see [BZ, 1.9]), there exists a natural isomorphism of b A onto A. If ˛ 2 b A, u 2 U and a 2 A, then, setting ˛ u .a/ D ˛.uau1 /; we can deﬁne an action of U on b A. The following lemma is of independent interest. Lemma 105.1 ([Isa4]). Let U be an abelian group which acts on an abelian group A and S be the set of the sizes of the orbits in this action. Write G D U b A, where the b semidirect product is constructed with respect to the action of U on A, induced by the given action of U on A. Then cd.G/ D S. Furthermore, if ŒA; U; U D ¹1º, then G has the nilpotence class 2. Proof. Let 2 Lin.b A/ and let T be the stabilizer of in U . Since the cyclic group b A= ker./ is centralized by an abelian subgroup T , it follows that b AT = ker./ is abelAT / is linear. Obviously, b ian and thus every 2 Irr.b AT D IG ./, the inertia subgroup of in G, and thus every 2 Irr.G / has degree jG W b AT j D jU W T j. Therefore, cd.G/ D ¹jU W T j j T is a stabilizer in U of some 2 Lin.b A/º: However, there is a natural correspondence between Lin.b A/ and A, and this deﬁnes a permutation isomorphism of the actions of U on A and Lin.b A/, respectively, and we conclude that cd.G/ D S, as required. Now suppose that ŒA; U; U D ¹1º. If ˛ 2 b A, we have for u 2 U and a 2 A that Œ˛; u.a/ D .˛ 1 ˛ u /.a/ D ˛.a1 /˛.uau1 / D ˛.Œa; u1 /:

105

On some special p-groups

109

Therefore, if v 2 U , we get by the above displayed formula, Œ˛; u; v.a/ D Œ˛; u.Œa; v 1 / D ˛.Œa; v 1 ; u1 / D ˛.1/ D 1 and so Œb A; U; U D ¹1º. Since b A is abelian and Œb A; U b A, this yields CG .Œb A; U / b b b b b U A D G, i.e., ŒA; U Z.G/. As G=ŒA; U D U A=ŒA; U is abelian, it follows that cl.G/ 2, completing the proof. Now we are ready to prove the main result of this section. Theorem 105.2 ([Isa4]). Let p be a prime and let 0 D e0 < e1 < < em be integers. Then there exists a p-group that is generated by elements of order p and has nilpotence class 2 such that cd.G/ D ¹p ei j 1 i mº. Proof. Suppose that U is an elementary abelian p-group of rank em with generators u1 ; : : : ; uem . Let A be an elementary abelian p-group with basis a1 ; z1;1 ; : : : ; z1;e1 ; : : : ; am ; zm;1 ; : : : ; zm;em (m blocks of sizes e1 ; : : : ; em , respectively) and deﬁne an action of U on A as follows. Put .zi; /u D zi; for all i; ; and .ai /u D ai if > ei and .ai /u D ai zi; if

ei . Since the automorphisms of A deﬁned this way all have order p and commute pairwise, this does deﬁne an action of U on A. Let us compute the sizes of the orbits of this action. Write Z D hzi; j ei i A and let a 2 A. Obviously, Z Z.G/, where G D U b A is deﬁned as in Lemma 105.1. Assume that a 2 A Z. Then there exists a unique subscript i such that we can write a D bcz, where b 2 haj j j < ii;

1 ¤ c 2 hai i;

z 2 Z:

Suppose that u 2 U . If u involves the generator u with ei , the exponents of zi; in a and in au will not be equal, and u does not centralize a. Thus we obtain that CU .a/ hu j > ei i. Since the reverse inclusion is obvious, we ﬁnally obtain CG .a/ D hu j > ei i. We now have jU W CU .a/j D p ei and we see that the orbit sizes of the action of U on A are precisely the numbers p ei for 0 i m. Since ŒA; U Z Z.G/, we have ŒA; U; U D ¹1º and the result follows by Lemma 105.1.

106

On maximal subgroups of two-generator 2-groups

To facilitate the proof of Theorem 106.3, we ﬁrst prove Lemmas 106.1 and 106.2. The lemma below follows from Propositions 71.3, 71.4 and 71.5, but we prefer to give an independent proof. Lemma 106.1. If a nonmetacyclic A2 -group G is two-generator, then p > 2.1 Proof (Berkovich). Assume that this is false, i.e., there exists a nonmetacyclic twogenerator 2-group G which is an A2 -group. Since G is not metacyclic, it has a maximal subgroup, say A, such that d.A/ > 2 (Corollary 36.6); then d.A/ D 3 (Schreier; see Appendix 25). From Lemma 65.1 we deduce that A is not an A1 -group so it is abelian. Let 1 D ¹M; N; Aº. If one of the subgroups M; N , say M , is abelian, then A \ M D Z.G/ so Z.G/ D ˆ.G/ since G=Z.G/ Š E4 Š G=ˆ.G/, and we conclude that G is an A1 -group, contrary to the hypothesis. Thus M and N are A1 -subgroups. We have Z.M / D ˆ.M / < ˆ.G/ < A so CG .Z.M // AM D G, and we conclude that Z.M / D Z.G/ since Z.G/ < ˆ.G/ (otherwise, G is an A1 -group). Similarly, Z.N / D Z.G/. Since the two-generator group G=Z.G/ of order 8 has two noncyclic subgroups M=Z.G/ and N=Z.G/ of order 4, it is generated by involutions. It follows that G=Z.G/ Š D8 . Suppose that R G G satisﬁes G=R Š D8 . We claim that then R D Z.G/. Indeed, it follows from d.G/ D 2 that R < ˆ.G/. Since G=R has exactly one cyclic subgroup of order 4, one may assume, without loss of generality, that M=R Š E4 holds so R D ˆ.M / D Z.M /. In that case, CG .R/ AM D G so that R D Z.G/ (compare indices). Thus there is in G only one normal subgroup, namely Z.G/, such that the corresponding quotient group is isomorphic to D8 . N D Set GN D G=Ã1 .A/. Then GN is nonabelian of order 16 since AN Š E8 and d.G/ N D 4, GN is not of maximal class. Then, by Proposition 10.17, d.G/ D 2. As exp.G/ GN has no nonabelian subgroup of order 8 (otherwise, d.G/ > 2) so it is an A1 -group. N of order 2 such that N Š E4 . In this case, there exists RN < Z.G/ It follows that Z.G/ 0 N RN Š D8 since a nonabelian group G= N RN has a subgroup A= N RN Š E4 . RN ¤ GN ; then G= By the ﬁrst paragraph, R D Z.G/. Let B=Z.G/ < G=Z.G/ be cyclic of order 4; then a 2-group G which is a two-generator A2 -group is metacyclic; this will be the case if and only if jG 0 j D p 2 by Corollary 65.3. 1 Thus

106

On maximal subgroups of two-generator 2-groups

111

B is abelian. As B ¤ A (indeed, A=Z.G/ Š E4 ), we get a contradiction since, by the above, there is in G only one abelian maximal subgroup. Lemma 106.2 ([BJ3, Lemma 2.1]). Let G be a nonmetacyclic p-group and let R < G 0 be G-invariant of order p. If G=R is minimal nonabelian and all nonabelian maximal subgroups of G are two-generator, then G is an A2 -group and p > 2. Proof. By hypothesis, jG 0 j > p so G is not an A1 -group (Lemma 65.1). We also have d.G/ D d.G=R/ D 2 since R < G 0 ˆ.G/. Let M 2 1 be nonabelian. Then d.M / D 2 by hypothesis, and M=R, as a maximal subgroup of the A1 -group G=R, is abelian which implies in view of jRj D p that M 0 D R. Hence M is an A1 -group (Lemma 65.2(a)) and so G is an A2 -group. By Lemma 106.1, p > 2. Note that if R < G 0 is G-invariant of order p and G=R is a metacyclic A1 -group, then G is a metacyclic A2 -group. Indeed, G is metacyclic by Theorem 36.1. Since jG 0 j D p 2 , G is an A2 -group by Corollary 65.3. Now we are ready to prove our main result. Theorem 106.3 ([BJ3, Theorem 2.2]). If a 2-group G and all its nonabelian maximal subgroups are two-generator, then G is either metacyclic or minimal nonabelian. Proof. Assume that G is a counterexample of minimal order. Then G is nonabelian (otherwise, G is metacyclic since, by hypothesis, d.G/ 2). Assume that jG 0 j D 2. Then, since we have d.G=G 0 / D 2, G is an A1 -group by Lemma 65.2(a), contrary to the assumption. It follows that jG 0 j > 2. Let R < G 0 be G-invariant of order 2; then GN D G=R is nonabelian. Since GN satN < jGj, it is either metacyclic or minimal nonabelian by isﬁes the hypothesis and jGj induction. If GN is metacyclic, then G is also metacyclic by Theorem 36.1, contrary to the assumption. Thus GN D G=R is a nonmetacyclic A1 -group. By hypothesis, G and all its nonabelian maximal subgroups are two-generator. Then, by Lemma 106.2, G is an A2 -group and p > 2, a ﬁnal contradiction. Let a nonabelian 2-group be neither metacyclic nor minimal nonabelian. If all nonabelian maximal subgroups of G are two-generator, then d.G/ D 3 (Theorem 106.3). If G is a 2-group with d.G/ > 2 and all maximal subgroups of G are two-generator, then the quotient group G=K4 .G/, where Kn .G/ is the n-th member of the lower central series of G, is described up to isomorphism in 71 (the order of such quotient group does not exceed 29 ). However, see 113 which shows that there exist such groups of arbitrary class. Remark. Let G be a two-generator p-group, p > 2, possessing an abelian maximal subgroup, say A, and suppose that all nonabelian maximal subgroups of G are two-generator. Suppose that jG 0 j > p (if jG 0 j D p, then G is minimal nonabelian by Lem-

112

Groups of prime power order

ma 65.2(a)). Let R < G 0 be a G-invariant subgroup of order p and let G=R be of maximal class. We claim that then G is also of maximal class. Indeed, since R < ˆ.G/, we get p 2 D j.G=R/ W .G 0 =R/j D jG W G 0 j D pjZ.G/j by Lemma 1.1, i.e., Z.G/ is of order p so coincides with R. It follows that G is of maximal class, as was to be shown. Conversely, if a p-group G of maximal class and order > p 3 has an abelian maximal subgroup, say A, then all nonabelian subgroups of G are of maximal class so two-generator. Indeed, if H 2 1 is nonabelian, then H is of maximal class by Fitting’s lemma. Next, if F < G is nonabelian, then F H 2 1 , and since H \ A is maximal abelian in H , we conclude that F is of maximal class by induction in H , completing the proof. If all nonabelian maximal subgroups of a p-group G are twogenerator and jG=Ã1 .G/j p 4 , then G has an abelian maximal subgroup. Indeed, G=Ã1 .G/ has a maximal subgroup, say A=Ã1 .G/, which is not generated by two elements (Theorem 5.8(b)); then d.A/ > 2 so A is abelian by hypothesis.

107

Ranks of maximal subgroups of nonmetacyclic two-generator 2-groups

Let G be a nonmetacyclic two-generator 2-group. By Theorem 44.5, there is in G a maximal subgroup C such that d.C / > 2 (in fact, by Schreier’s theorem on the number of generators of subgroups, d.C / D 3; see Appendix 25). Let A; B; C be all maximal subgroups of G. It is important to know all possibilities for d.A/ and d.B/ and also on the structure of A and B. As Theorem 107.1 asserts, there are exactly two possibilities: d.A/ D d.B/ D 2 and d.A/ D d.B/ D 3. Theorem 107.1 ([BJ3, Theorem 5.1]). Suppose that G is a nonmetacyclic two-generator 2-group. Then the number of two-generator maximal subgroups of G is even. Proof. By hypothesis, G is nonabelian. If A 2 1 , then exp.G=Ã1 .A// 4 so that Ã2 .G/ Ã1 .A/ D ˆ.A/ and hence d.A=Ã2 .G// D d.A/. Therefore, without loss of generality, one may assume that Ã2 .G/ D ¹1º; then exp.G/ D 4. It follows that G=G 0 is abelian either of type .4; 2/ or .4; 4/. Since G is not metacyclic, at least one maximal subgroup of G is not two-generator by Corollary 36.6. Let 1 D ¹A; B; C º and d.C / D 3 (see the paragraph preceding the theorem). Assume that the statement of the theorem is false. Then one may assume d.A/ D 2 and d.B/ D 3.D d.C //. Let R be a G-invariant subgroup of index 2 in G 0 . Then G=R is minimal nonabelian and nonmetacyclic (Theorem 36.1). By what has just been said, jG=Rj 25 . If jG=Rj D 25 , then all maximal subgroups of G=R are not generated by two elements (see Lemma 65.1), contrary to the assumption. Thus jG=Rj D 24 . In this case, G=G 0 is abelian of type .4; 2/ and j1 .G=R/j D 23 (Lemma 65.1). All maximal subgroups of G=R which are different from C =R D 1 .G=R/ are abelian of type .4; 2/. Hence B=R is abelian of type .4; 2/ and so it is two-generator. By assumption, we have on the one hand d.B/ D 3 implying that R ¤ ˆ.B/ (note that jRj D jˆ.B/j), and on the other hand that ˆ.B/ is a G-invariant subgroup of index 23 in B. Write GN D G=ˆ.B/; then GN (of order 24 ) is not of maximal class since it contains a subgroup BN Š E8 . It follows that jGN W GN 0 j D 8 D jG=G 0 j so ˆ.B/ < G 0 (this also follows from Taussky’s theorem), and we conclude that jG 0 W ˆ.B/j D 2. Since G 0 has only one G-invariant subgroup of index 2 (Lemma 36.5), we conclude that ˆ.B/ D R. This is a contradiction since, by the above, R ¤ ˆ.B/. Thus d.B/ D 2 so that the number of two-generator maximal subgroups of G is even, as was to be

114

Groups of prime power order

shown. (There exists an A2 -group G of order 26 and d.G/ D 3 all of whose seven maximal subgroups are A1 -groups so two-generator (see 71).) Thus, if the number of two-generator maximal subgroups of a 2-group G is odd, then either G is metacyclic or d.G/ D 3. Exercise. Let G be a p-group such that jG=G 0 j D p 2 . Then G has no maximal subgroup of rank p C 1. If A 2 1 with d.A/ D p, then G=ˆ.A/ Š †p2 , a Sylow p-subgroup of the symmetric group of degree p 2 . Solution. Let A 2 1 . Then G=ˆ.A/ is of maximal class (to prove, use Lemma 1.1 and induction) so that d.A/ < p C 1 (Theorem 9.5). Now let A 2 1 with d.A/ D p. Then G=ˆ.A/ Š †p2 (Exercise 9.13). Theorem 107.2 ([BJ3, Theorem 5.2]). Suppose that G is a nonmetacyclic two-generator 2-group. Then one of the following holds: (a) Every maximal subgroup of G is not generated by two elements if and only if G=G 0 has no cyclic subgroup of index 2. (b) Exactly one maximal subgroup of G is not generated by two elements if and only if G=G 0 has a cyclic subgroup of index 2. Proof. (a) Suppose that all maximal subgroups of a given nonmetacyclic two-generator 2-group G are not two-generator (then their ranks are equal to 3 by Appendix 25). We claim that, in this case, G=G 0 has no cyclic subgroup of index 2. Assume that this is false. As in the proof of Theorem 107.1, one may assume that Ã2 .G/ D ¹1º hence exp.G/ D 4 and G is nonabelian. Let R be a G-invariant subgroup of index 2 in G 0 . Then G=R is a nonmetacyclic A1 -group (Theorem 36.1 and Lemma 65.2(a)). As in the proof of Theorem 107.1, 23 < jG=Rj 25 and R is the unique G-invariant subgroup of index 2 in G 0 . Assume that jG=Rj D 24 ; then G=G 0 is abelian of type .4; 2/ and the maximal subgroups of G=R are abelian of types .4; 2/, .4; 2/ and .2; 2; 2/. Let B=R < G=R be abelian of type .4; 2/. By hypothesis, we have B=ˆ.B/ Š E8 . Since G=ˆ.B/ is not of maximal class (indeed, it contains a subgroup Š E8 , namely B=ˆ.B/), it follows that jG=ˆ.B/ W .G=ˆ.B//0 j D 8 (Taussky’s theorem) and so G 0 \ ˆ.B/ D R which yields ˆ.B/ > R, and this is a contradiction (recall that d.B=R/ D 2). As in the proof of Theorem 107.1, we get R D ˆ.B/, and this is a contradiction. Thus jG=Rj D 25 so G=G 0 is abelian of type .4; 4/. It follows that, in the case under consideration, G=G 0 has no cyclic subgroup of index 2. Now suppose that G=G 0 has no cyclic subgroup of index 2. Let R be (the unique) G-invariant subgroup of index 2 in G 0 . Write GN D G=R. Then GN is a nonmetacyclic N Š E8 (Lemma 65.1). Since A1 -group by Theorem 36.1 and Lemma 65.2(a) so 1 .G/ G=G 0 has no cyclic subgroup of index 2, it is abelian of type .2m ; 2n /, m; n > 1, m1 ; p n1 / hence noncyclic. N N therefore it follows that G= 1 .G/ is abelian of type .p N N N N 1 .G/ N is maximal Therefore 1 .G/ ˆ.G/ since d.G/ D d.G/ D 2. Thus, if A=

107

Ranks of maximal subgroups of nonmetacyclic two-generator 2-groups

115

N N N in G= 1 .G/, then d.A=R/ D 3 since A=R is abelian and contains 1 .G/ Š E8 , and we conclude that d.A/ D 3. Since A 2 1 is arbitrary, the proof of (a) is complete. Statement (b) follows from (a) and Theorem 107.1. Suppose that a nonmetacyclic two-generator p-group G contains a metacyclic abelian maximal subgroup A. If G has a normal elementary abelian subgroup E of order p 3 , then A has a cyclic subgroup of index p. Indeed, G D AE and G=.A \ E/ Š .A=.A \ E// .E=.A \ E//. Since d.G/ D 2, we conclude that A=.A \ E/ is cyclic, and our claim follows in view of A \ E D 1 .A/. We have G 0 A \ E. In the following two theorems we shall consider the nonmetacyclic two-generator 2-groups G containing exactly one non-two-generator maximal subgroup. Theorem 107.3 ([BJ3, Theorem 5.3]). Suppose that G is a nonmetacyclic two-generator 2-group which possesses exactly one maximal subgroup K which is not two-generator. If 1 D ¹M; N; Kº is the set of maximal subgroups of G, then jG 0 W K 0 j D 2, G=K 0 is nonmetacyclic minimal nonabelian, both M and N are either metacyclic or nonmetacyclic and G=G 0 has a cyclic subgroup of index 2. Proof. By Theorem 107.2, G=G 0 has a cyclic subgroup of index 2. We know that d.M / D 2 D d.N / and d.K/ D 3. If K 0 D ¹1º, then Theorem 106.3 implies that G is minimal nonabelian in which case both M and N are metacyclic since, in the case under consideration, 1 .M / Š 1 .N / Š E4 and M; N are abelian. Therefore one may assume that G is not minimal nonabelian and hence jG 0 j > 2 (Lemma 65.2(a)); then K 0 > ¹1º. We have K=ˆ.K/ Š E8 so that d.G=ˆ.K// D 2 implies that G=ˆ.K/ is nonmetacyclic minimal nonabelian of order 24 (this follows from the previous paragraph; this also follows from Proposition 10.17). Hence we get K 0 G 0 \ ˆ.K/ and jG 0 W .G 0 \ ˆ.K//j D 2. Assume that K 0 < G 0 \ ˆ.K/. In this N where G, N MN , NN are two-generator groups and KN is abelian case, consider G=K 0 D G, N with d.K/ D 3. By Theorem 106.3, GN must be minimal nonabelian. This is a contraN 0 is of order > 2. We have proved that K 0 D G 0 \ ˆ.K/, and diction since GN 0 D .G/ 0 so, by the above, jG W K 0 j D 2 and therefore GN D G=K 0 is nonmetacyclic minimal N D 3). nonabelian (recall that d.K/ Now assume that M is nonmetacyclic but N is metacyclic. Then M 0 ¤ ¹1º since d.M / D 2, and let R < M 0 be a G-invariant subgroup with jM 0 W Rj D 2. In this case, we consider GN D G=R so that MN is nonmetacyclic minimal nonabelian (Theorem 36.1 and Lemma 65.2(a)). Note that NN is metacyclic and we set EN D 1 .MN / Š E8 (LemN EN is noncyclic, then EN ˆ.G/ N since d.G/ N D 2, and so EN Š E8 ma 65.1). If G= N N EN is cyclic of oris contained in metacyclic subgroup N , a contradiction. Hence G= N N N N der 4 since M > E in view of d.M / D 2 < 3 D d.E/. Let aN 2 GN MN be such N EN and hNzi D hai that hai N is cyclic of order 8, hai N covers G= N \ EN Š C2 so that, in N since ha; N D GN (recall that GN is not split over E). N We have particular, hNz i 2 Z.G/ N Ei 2 0 2 N N N N N M D haN iE and M D ŒaN ; E D hui, N where uN 2 E hzN i since uN is not a square N (recall that MN is minimal nonabelian). in MN by Lemma 65.1, and so hzN ; ui N Z.G/

116

Groups of prime power order

N But then Let vN 2 EN hNz ; ui N so that vN aN D vN N with N 2 hz; N ui N hence N 2 Z.G/. 2 N aN D .vN / N N aN D vN N 2 D vN vN aN D .vN /

N D ¹1º, a contradiction since MN D haN 2 ; Ei N is nonabelian. The proof is and so ŒaN 2 ; E complete. It follows from Theorem 107.3 and Corollary 36.6 that the number of metacyclic maximal subgroups in any nonmetacyclic two-generator 2-group is even. Let us show that provided X is a nonmetacyclic two-generator 2-group of minimal possible order, then it is an A1 -group of order 16 which is uniquely determined. Since X=Ã2 .X/ is nonmetacyclic two-generator (Corollary 44.9), we get Ã2 .X/ D ¹1º so that exp.X/ D 4. By Taussky’s theorem and Theorem 36.1, X=X 0 is abelian of type .4; 2/. If R < X 0 is X-invariant of index 2, then X=R is nonmetacyclic and minimal nonabelian (Theorem 36.1) so that R D ¹1º. It follows that X is nonmetacyclic minimal nonabelian of order 16 so it is determined uniquely (Lemma 65.1). Theorem 107.4 ([BJ3, Theorem 5.4]). Let G be a smallest group from Theorem 107.3, where maximal subgroups M and N of G are both nonmetacyclic. Then G is uniquely determined as follows: G D ha; b j a2 D b 4 D 1; Œa; b D c; Œb; c D m; c 2 D m2 D Œm; a D Œm; b D Œa; c D 1i; where jGj D 25 ;

G 0 D hc; mi Š E4 ;

K3 .G/ D Z.G/ D hmi

and so cl.G/ D 3. The three maximal subgroups of G are M D hG 0 ; bi;

N D hG 0 ; abi

(which are both nonmetacyclic minimal nonabelian of order 24 ) and K D hG 0 ; a; b 2 i Š C2 D8 :1 Our group G exists as a subgroup of the alternating group A8 . Proof. By the paragraph preceding the theorem, a smallest possibility for two-generator nonmetacyclic subgroups M and N is the nonmetacyclic minimal nonabelian group H16 D hx; y j x 2 D y 4 D 1; Œx; y D z; z 2 D Œz; x D Œz; y D 1i of order 24 . Therefore we try to construct our two-generator group G under the assumption that jGj D 25 and the set of maximal subgroups ¹M; N; Kº is such that 1 Thus

G is an A3 -group.

107

Ranks of maximal subgroups of nonmetacyclic two-generator 2-groups

117

M Š N Š H16 and K is a nonabelian group of order 24 with d.K/ D 3 (indeed, K must be nonabelian by Theorem 106.3). Since G is neither of maximal class (and so jG 0 j < 23 by Taussky’s theorem) nor minimal nonabelian (and so jG 0 j > 2 by Lemma 65.2(a)), we have jG 0 j D 4. As d.G/ D 2, we conclude that G=G 0 is abelian of type .4; 2/ so G D AB, where A \ B D G 0 , A=G 0 Š C2 and B=G 0 Š C4 . Let a 2 A G 0 and b 2 B G 0 be such that A D hG 0 ; ai and B D hG 0 ; bi. We have ˆ.G/ D G 0 hb 2 i so that G D ha; b; G 0 i D ha; bi, and we set c D Œa; b ¤ 1. By Proposition 10.17, since K is neither of maximal class nor minimal nonabelian, we get K D U Z.K/, where U is nonabelian of order 8. Note that K=ˆ.K/ Š E8 . We obtain G DM [N [K

and

exp.M / D exp.N / D exp.K/ D 4

so

exp.G/ D 4

since G D M [N [K. It follows from jˆ.G/j D 8 that ˆ.G/ is abelian. Assume that ˆ.G/ is abelian of type .4; 2/; then every set of generators of ˆ.G/ contains an element of order 4 so, since ˆ.G/ D Ã1 .G/, we get exp.G/ D 8, which is a contradiction. Thus ˆ.G/ Š E8 so G 0 Š E4 and K Š D8 C2 since Z.K/ Š E4 . It follows that jG 0 W K 0 j D 2 so G=K 0 is nonmetacyclic A1 -group (of order 24 and hence we get G=K 0 Š H16 by Lemmas 65.1 and 65.2(a)); in particular, M 0 D N 0 D K 0 since M=K 0 and N=K 0 are abelian. Since jK 0 j D 2, we have K 0 D hmi Z.G/ and so G 0 D hc; mi.Š E4 /, where o.c/ D 2. As we have shown, G=K 0 is a nonmetacyclic A1 -group. Therefore, by Lemma 65.1, the generator of G 0 =K 0 is not a square in G=K 0 , and we conclude that a2 2 hmi. Since K=G 0 Š E4 and G=G 0 is abelian of type .4; 2/, we get K=G 0 D 1 .G=G 0 / so that M=G 0 Š C4 Š N=G 0 . It follows from ˆ.G/ Š E8 that ˆ.G/ D hb 2 i G 0 so that Ahb 2 i=hmi Š E8 and so we must have K D Ahb 2 i with ˆ.K/ D hmi. The other two maximal subgroups are G 0 hbi and G 0 habi so that we may set M D B D G 0 hbi and N D G 0 habi. We have M D hb; c; mi D hb; ci since m 2 ˆ.M /, and similarly N D hab; ci, so that Œb; c D Œab; c D m since M 0 D N 0 D K 0 D hmi. We compute m D Œab; c D Œa; cb Œb; c D Œa; cb m which gives Œa; c D 1 and therefore A D ha; c; mi is abelian of order 8. Further, Œa; b 2 D Œa; bŒa; bb D cc b D c.cm/ D m: If a2 D m, then ha; b 2 i Š D8 , and then a0 D ab 2 is an involution. Replacing a with a0 (if necessary), we may assume from the start that a is an involution, and then A D ha; c; mi Š E8 (see the above two displayed formulas; recall that A is abelian since Œa; c D 1 D Œa; m D Œc; m). We get K D hci ha; b 2 i Š C2 D8 and the structure of G is uniquely determined as given in the statement of the theorem. It is easy to see that Z.G/ D hmi D K3 .G/ so cl.G/ D 3. The existence of G as a subgroup of A8 is established if we set a D .2; 7/.3; 8/

and b D .1; 2; 3; 4/.5; 6; 7; 8/:

118

Groups of prime power order

It is enough to check that the permutations a and b satisfy all the above relations. Since we get m D .1; 6/.2; 7/.3; 8/.4; 5/ ¤ 1G and hmi D Z.G/, our permutation representation is faithful. Theorem 107.5 ([BJ3, Theorem 5.5]). Suppose that G is a nonmetacyclic two-generator 2-group and let M; N; K be all its maximal subgroups such that M; N are nonmetacyclic two-generator and d.K/ D 3 (see Theorem 107.3). Then there is R G G such that G=R is isomorphic to the group from Theorem 107.4. Proof. We may assume that jGj > 25 (otherwise, in view of Theorem 107.4, there is nothing to prove). Since two-generator abelian 2-groups are metacyclic, the (nonmetacyclic two-generator) subgroups M and N are nonabelian. It follows that G is neither abelian nor minimal nonabelian; in particular, jG 0 j > 2 (Lemma 65(a)). In view of Theorem 106.3, K is nonabelian. Let L1 be a G-invariant subgroup of index 2 in M 0 (since d.M / D 2, the quotient group M 0 =K3 .M / is cyclic so L1 is uniquely determined). By Theorem 36.1, M=L1 is not metacyclic and neither is G=L1 . We also have L1 < ˆ.M / < ˆ.G/ D N \ K. Since d.N=L1 / D 2 D d.G=L1 /, it follows that d.K=L1 / D 3 (Corollary 36.6). By Theorem 107.3, N=L1 is nonmetacyclic so nonabelian. Thus the group G=L1 satisﬁes the hypothesis. Therefore, without loss of generality, one may assume that L1 D ¹1º. Let L2 be a G-invariant subgroup of index 2 in N 0 . As above, N=L2 is not metacyclic so is M=L2 , and d.K=L2 / D 3, so again G=L2 satisﬁes the hypothesis, and we may assume that L2 D ¹1º. By Lemma 65.2(a), M and N are A1 -groups since, by construction, jM 0 j D 2 D jN 0 j. Now set L3 D Ã2 .M /. Then, by Supplement to Corollary 36.6, M=L3 is not metacyclic and, as above, G=L3 satisﬁes the hypothesis so, arguing as above, one may assume that L3 D ¹1º. Similarly, one may assume that Ã2 .N / D ¹1º. In that case, we get exp.M / D exp.N / D 4. Since M and N are nonmetacyclic two-generator, they are nonabelian so A1 -groups as epimorphic images of A1 -groups. From Lemma 65.1 we deduce jM j 25 (indeed, an A1 -group of exponent 2n has order 22nC1 ). By assumption, we must have jGj > 25 so jM j D 25 and M=M 0 is abelian of type .4; 4/, jGj D 26 . In that case, by Lemma 65.1, Z.M / D ˆ.M / D 1 .M / Š E8 Š 1 .N / D Z.N / D ˆ.N /: Since Z.M / D ˆ.M / < ˆ.G/ < N

and

Z.N / < ˆ.G/ < M;

it follows that Z.M / D Z.N / D Z.G/.Š E8 / (indeed, we have Z.G/ < M in view of d.G/ D 2). The quotient group G=Z.G/ of order 8 is not elementary abelian since d.G/ D 2. It follows that G=Z.G/ contains a cyclic subgroup, say C =Z.G/, of order 4. In that case, C is abelian so d.C / D 3. Since K is a unique maximal subgroup of G not generated by two elements, we get C D K so that K is abelian. This is a ﬁnal contradiction: by the above, G has no abelian maximal subgroups.

107

Ranks of maximal subgroups of nonmetacyclic two-generator 2-groups

119

Let G be a two-generator 2-group all of whose maximal subgroups are not twogenerator (in that case all those subgroups are three-generator by Appendix 25); then G is nonabelian. Denote by H32 the unique A1 -group of order 32 and exponent 4 (see Lemma 65.1; all maximal subgroups of H32 are abelian of type .4; 2; 2/). Supplement to Theorem 107.5 ([BJ3]). If all maximal subgroups of a two-generator 2-group G are not two-generator, then there is R G G such that G=R Š H32 . Proof. Indeed, if M < G is maximal, then ˆ.M / D Ã1 .M / Ã2 .G/ so that, as we have noticed, d.M=Ã2 .G// D d.M / D 3. Thus G=Ã2 .G/ satisﬁes the hypothesis so one may assume that Ã2 .G/ D ¹1º holds. By Theorem 107.2(a), G=G 0 is abelian of type .4; 4/. If R is G-invariant of index 2 in G 0 , then G=R Š H32 (Lemmas 65.2(a) and 65.1), as desired. It follows that if a group G from the supplement has minimal possible order, then G Š H32 . Remark 1. Suppose that a nonabelian two-generator p-group G, p > 2, has an abelian subgroup A of index p. (i) If 1 .G/ D G, then exp.G=G 0 / D p so jG=G 0 j D p 2 and G is of maximal class (to prove this, we use Lemma 1:1 and induction). (ii) If 1 .G/ 6 A, then jG W G 0 Ã1 .A/j D jG W ˆ.G/j D p 2 (indeed, A1 .G/ D G, G 0 Ã1 .A/ G 0 Ã1 .G/ D ˆ.G/ and G=G 0 Ã1 .A/ is elementary abelian since it is abelian of rank 2 and generated by elements of order p as an epimorphic image of G=Ã1 .A/ D A1 .G/=Ã1 .A/). Thus jG=Ã1 .A/ W .G=Ã1 .A//0 j D p 2 . It follows from (i) that G=Ã1 .A/ is of maximal class; then d.A/ D d.A=Ã1 .A/ p by Exercise 9.13. Remark 2. Let p > 2 and let G be a two-generator p-group containing a maximal subgroup M such that d.M / D p C 1 (see Appendix 25). Let us study the structure of G=ˆ.M /. To simplify our task, one may assume that ˆ.M / D ¹1º; then M Š EppC1 so G is not of maximal class (Theorems 9:5 and 9:6). It follows that jZ.G/j > p (if jZ.G/j D p, then jG W G 0 j D pjZ.G/j D p 2 by Lemma 1.1, so G is of maximal class contrary to what has just been said). Thus we have jG=G 0 j D pjZ.G/j p 3 . If G=G 0 j > p 3 , then, in view of M Š EppC1 , p C 1 4, M=G 0 < G=G 0 and d.M=G 0 / > 2 D d.G=G 0 / D d.G/, we get a contradiction. Thus jG=G 0 j D p 3 hence exp.G=G 0 / D p 2 since d.G=G 0 / D d.G// D 2, and we conclude that G=G 0 is abelian of type .p 2 ; p/. Write H=G 0 D 1 .G=G 0 /; then M 1 .G/ H so that M D H D 1 .G/ since M 2 1 . Let x 2 G M ; then G D hx; M i, o.x/ D p 2 and N D G, N the group x p 2 Z.G/ since M is abelian. Write GN D G=hx p i. Since 1 .G/ N G is of maximal class (Remark 1(i)), isomorphic to †p2 , a Sylow p-subgroup of the symmetric group of degree p 2 by Exercise 9.13. If KN < GN is maximal and nonabelian, then KN is of maximal class (Fitting’s lemma). However, K is not of maximal class since d.G/ D 2 (here we use Theorem 12:12(a)), so jZ.K/j > p. We have Z.K/ < M (otherwise, G D M Z.K/ so cl.G/ D 2, a contradiction). Thus CG .Z.K// MK D G so Z.K/ D Z.G/ since Z.G/ < ˆ.G/ < K in view of d.G/ D 2.

108

p-groups with few conjugate classes of minimal nonabelian subgroups

Let 1 .G/ be the number of conjugate classes of A1 -subgroups (= minimal nonabelian subgroups) of a p-group G (see 76). If 1 .G/ 1, then G is either abelian or an A1 -group (Theorem 10.28). Therefore only the case where 1 .G/ > 1 is interesting. It follows from Remark 76.1 that if a p-group G is neither abelian nor minimal nonabelian, then 1 .G/ p. Proposition 108.1. Suppose that a p-group G contains a normal nonabelian subgroup N of index p n such that G=N is noncyclic. Then we have 1 .G/ n C p. If, in addition, 1 .G/ D n C p, then all proper G-invariant subgroups of N are abelian. Proof. By hypothesis, n 2. Let N D N0 < N1 < < Nn D G be a chain of G-invariant subgroups such that G=Nn2 is noncyclic; then all indices of this chain are equal to p. Since a nonabelian p-group is generated by its minimal nonabelian subgroups (Theorem 10.28), there is a minimal nonabelian subgroup Si < Ni such Si 6 Ni 1 , i D 1; : : : ; n 2. We have G=Nn2 Š Ep2 . Let S0 N0 D N be minimal nonabelian and Nn1 D Nn1;1 ; : : : ; Nn1;pC1 all (nonabelian) maximal subgroups of G containing Nn2 . Let Sn1;i < Nn1;i be minimal nonabelian not contained in Nn2 , i D 1; 2; : : : ; p C 1. Since .n 2/ C 1 C .p C 1/ D n C p minimal nonabelian subgroups S0 ; S1 ; : : : ; Sn2 ; Sn1;1 ; : : : ; Sn1;pC1 are not conjugate in pairs (indeed, normal closures of these subgroups in G are pairwise distinct), we get 1 .G/ n C p. Now suppose that 1 .G/ D n C p. Assume that there is in N a proper nonabelian G-invariant subgroup R; then G=R is noncyclic of order p nC1 . By the previous paragraph, 1 .G/ .n C 1/ C p > n C p, a contradiction. Thus all proper G-invariant subgroups of N are abelian. Proposition 108.2. Suppose that a p-group G of order p m > p 3 is of maximal class with abelian A 2 1 . Then 1 .G/ D p. Proof. It is well known that all nonabelian subgroups of G are of maximal class so all minimal nonabelian subgroups of G have the same order p 3 , and the number of minimal nonabelian subgroups in G equals p m3 . Indeed, 1 D ¹H1 ; : : : ; Hp ; Aº, where H1 ; : : : ; Hp are of maximal class (Fitting’s lemma). Then, by induction, Hi contains

108

p-groups with few conjugate classes of minimal nonabelian subgroups

121

p m13 D p m4 minimal nonabelian subgroups, and Hi \ Hj D ˆ.G/.< A/ is abelian for all i ¤ j , so the number of minimal nonabelian subgroups in G is equal to p p m13 D p m3 , as asserted. If Si Hi is minimal nonabelian, i D 1; : : : ; p, then S1 ; : : : ; Sp are pairwise nonconjugate since SiG D Hi for i D 1; : : : ; p (indeed, if a G-invariant subgroup has index > p in G, then it is contained in ˆ.G/; see Exercise 9.1(b)). We have jG W NG .Si /j D p m4 for all i D 1; : : : ; p. Indeed, NG .Si / is nonabelian hence of maximal class. Since all normal subgroups of NG .Si / of index > p are abelian, we get jNG .Si /j D pjSi j D p 4 , proving our claim. Thus all minimal nonabelian subgroups of Hi are conjugate in G for i D 1; : : : ; p and do not lie in Gj (j ¤ i ) hence 1 .G/ D p. Proposition 108.3. Suppose that G is a group of exponent p. Then 1 .G/ D p if and only if G is of maximal class and order > p 3 with an abelian subgroup of index p. In this case, jGj p p , and so we must have p > 3. Proof. By hypothesis, jGj > p 3 . If G is of maximal class with an abelian subgroup of index p, then 1 .G/ D p (Proposition 108.2). Now let 1 .G/ D p. Then, by Theorem 76.17, d.G/ D 2 and G has an abelian maximal subgroup. As ˆ.G/ D G 0 , we get jG W G 0 j D p 2 . It follows that G is of maximal class so jGj p p (Theorem 9.5). Since G is not of maximal class, we get p > 3. Let .G/ be the number of conjugate classes of maximal cyclic subgroups of a p-group G. Then .G/ k D 1 C p C C p d.G/1 . Indeed, if x 2 G ˆ.G/, then hxi is a maximal cyclic subgroup of G. The group G=ˆ.G/ has exactly k subgroups of order p, say M1 =ˆ.G/; : : : ; Mk =ˆ.G/. Let Ci Mi be cyclic such that Ci 6 ˆ.G/ for all i . Then C1 ; : : : ; Ck are maximal cyclic subgroups in G. Since we have CiG Ci ˆ.G/ D Mi for all i and Mi \ Mj D ˆ.G/ (i ¤ j ), it follows that C1 ; : : : ; Ck are pairwise nonconjugate in G so that .G/ k. It is interesting to classify the p-groups G satisfying .G/ D k (all 2-groups of maximal class satisfy this condition). Exercise. If G is a nonabelian p-group, then 1 .G/ d.G/1. Improve this estimate in the case d.G/ > 3. (Hint. If A is minimal nonabelian, then the subgroup Aˆ.G/ is nonabelian of index p d.G/2 in G and G-invariant. Apply Proposition 108.1.)

Problems Problem 1. Estimate the maximal possible number of conjugate classes of minimal nonabelian subgroups in a p-group of maximal class and order p m . Problem 2. Estimate the maximal possible number of conjugate classes of minimal nonabelian subgroups in representation groups of elementary abelian group Epn . Problem 3. Estimate the maximal possible number of conjugate classes of minimal nonabelian subgroups in metacyclic groups of order p m .

109

On p-groups with metacyclic maximal subgroup without cyclic subgroup of index p

According to [MS, Lemma 4.9], if p > 3 is a prime and if a is an automorphism of order p of the abelian group L of type .p 2 ; p 2 /, then a centralizes 1 .L/ (the proof of this result is also reproduced in [AS, Lemma A.1.30]). We offer a generalization of this result. Our proof is based on entirely other ideas. Theorem 109.1. Suppose that p > 2 and L is a metacyclic p-group without cyclic subgroup of index p. An element a 2 Aut.L/ of order p does not centralize 1 .L/ if and only if p D 3, the partial holomorph G D hai L is a 3-group of maximal class and one of the following holds: (a) L is abelian of type .3m ; 3n /, m > 1; n > 1 and jm nj 1. m

n

m1

(b) L D ha; b j a3 D b 3 D 1; ab D a1C3

i is minimal nonabelian, jmnj 1.

Proof. Suppose that G D hx; G1 i is a 3-group of maximal class and order > 34 with regular subgroup G1 of index 3 such that o.x/ D 3. Then G1 , the fundamental subgroup of G, is metacyclic so d.G1 / D 2 (see Theorems 9.6(b) and 9.11). If G1 is abelian of type .3m ; 3n /, then jm nj 1 since G has no normal cyclic subgroup of order 32 (indeed, G has a normal abelian subgroup of type .3; 3/ and it is a unique G-invariant subgroup of order 9 by Exercise 9,1; if k D min¹m; nº, then Ãk .G1 / is cyclic of order p jmnj and characteristic in L so normal in G). Next, we obtain that CG .1 .G1 // D G1 so x does not centralize 1 .G1 /. Now suppose that G1 is nonabelian. Then jG10 j D 3 since G10 is G-invariant cyclic so G1 is minimal nonabelian (Lemma 65.2(a)). In this case, by Lemma 65.1 (see also Exercise 1.8(a)), G1 has the same deﬁning relations as L in part (b) of our theorem. Since Ãjmnj .G1 / is cyclic, we get again jm nj 1. As, by hypothesis, G1 =1 .G1 / is noncyclic, we also have 1 .G1 / ˆ.G1 / D Z.G1 / since d.G1 / D 2. Since again CG .1 .G1 // D G1 , the element x does not centralize 1 .G1 /. Next we assume that G is not a 3-group of maximal class. Now suppose that L and a are as in the statement of our theorem. We have to describe the structure of L. Put G D ha; Li. Since L is regular, the subgroup 1 .L/ is abelian of type .p; p/; clearly, 1 .L/ G G. As j1 .G/j p 3 , G is nonmetacyclic. Since L has no cyclic subgroup of index p, the quotient group L=1 .L/ is noncyclic.

109

Metacyclic maximal subgroup

123

Suppose that a does not centralize 1 .L/. It follows that H D ha; 1 .L/i is nonabelian of order p 3 and exponent p since p > 2. We have G D LH since H 6 L and L 2 1 . Assume that G is regular. Then exp.1 .G// D p (Theorem 7.2(b)) so, considering the intersection L \ 1 .G/, we conclude that j1 .G/j D pj1 .L/j D p 3 D jH j whence 1 .G/ D H (H is deﬁned in the previous paragraph). It follows that G has no elementary abelian subgroups of order p 3 . Since G is neither metacyclic nor a 3-group of maximal class (Theorems 9.5 and 9.6), we can conclude from Theorem 13.7 that G=1 .G/ is cyclic, a contradiction since G=1 .G/ Š L1 .G/=1 .G/ Š L.L \ 1 .G// D L=1 .L/; and the last quotient group, as we have noticed, is noncyclic. Now let G be irregular (by assumption in the ﬁrst paragraph of the proof, G is not a 3-group of maximal class). It follows from Theorems 9.5 and 9.6 that G is not of maximal class for p > 3 as well. Therefore, by Theorem 12.1(b), G D L1 .G/, where 1 .G/ is of order p p and exponent p. Since L \ 1 .G/ D 1 .L/ is elementary abelian of order p 2 , we get p D 3. It follows that 1 .G/ D H D ha; 1 .L/i is nonabelian of order 33 and exponent 3 so G has no elementary abelian subgroups of order 33 . In this case, since G=H is noncyclic, the group G is of maximal class (Theorem 13.7), a ﬁnal contradiction. As we have noticed, the regular maximal subgroup of a 3-group of maximal class and order > 34 is such as in conclusion of Theorem 109.1. Corollary 109.2. Suppose that p > 2 and L is an abelian group of type .p m ; p n /, m > 1, n > 1. An element a 2 Aut.G/ of order p does not centralize 1 .L/ if and only if p D 3, G D ha; Li is a 3-group of maximal class and jm nj 1. Remark. Now we consider a similar but more complicated situation for p D 2. Suppose that a metacyclic 2-group L without cyclic subgroup of index 2 is maximal in a 2-group G and 1 .L/ Z.L/; then 1 .L/ is a four-subgroup. In this case, G is not of maximal class. Suppose that there is an involution a 2 GL which does not centralize 1 .L/. We claim that G is not metacyclic. Assume that this is false. Then the nonabelian subgroup hx; 1 .L/ Š D8 so G is of maximal class (Proposition 10.19), contrary to what has just been said. By hypothesis, CG .1 .L// D L. Now we use Theorem 66.1 classifying minimal nonmetacyclic 2-groups. Let H be a minimal nonmetacyclic subgroup of G. If H Š E8 , then L \ H D 1 .L/ so CG .1 .L// HL D G, a contradiction. In case H Š Q8 C2 , we have Z.H / D 1 .H / D 1 .L/ < H \ L so we get CG .1 .L// HL D G, a contradiction. Now assume that jH j D 25 (this H is special with center of order 4). As exp.H / D 4, it follows that 1 .L/ D 1 .H /, and since 1 .H / D Z.H /, we get CG .1 .L// HL D G, a contradiction. Thus H Š D8 C4 is of order 16. Similarly, if F < G is minimal nonabelian, then either jF j D 8 or F Š M2n (Lemma 65.1). As we know (see Theorem 10.28), it is possible to choose F so that F 6 L.

124

Groups of prime power order

If a noncyclic p-group L which contains a cyclic subgroup Z of index p is not a 2-group of maximal class (then 1 .L/ is abelian of type .p; p/), is maximal in a p-group G D ha; Li and an element a 2 G L of order p does not centralize 1 .L/, then H D ha; 1 .L/i is nonabelian of order p 3 and G D HZ with H \ Z D Z.H / and 1 .H / D H . If, in addition, p D 2, then all minimal nonmetacyclic subgroups of G have order 24 and are isomorphic to D8 C4 (Theorem 66.1). Let L be a maximal subgroup of a p-group G D hx; Li, where p > 2 and 1 .L/ is abelian of type .p; p/ (then L is either metacyclic or an irregular 3-group of maximal class by Theorem 13.7) and o.x/ D p. Suppose that x does not centralize 1 .L/. Then H D hx; 1 .L/i is nonabelian of order p 3 and exponent p. Clearly, G has no subgroup of order p 4 and exponent p (if such a subgroup, say F , exists, then we get j1 .F \ L/ D p 3 > p 2 D j1 .L/j, a contradiction). Then either G D H C , where H D 1 .G/ and C is cyclic, or G is a 3-group of maximal class (Theorem 12.1(a)). In the ﬁrst case, jL=Ã1 .L/j D j1 .L/j D p 2 (Theorem 7.2(d)) so L is metacyclic (Theorem 9.11), and this case we have already considered. Now assume that G and L are 3-groups of maximal class and jGj > 34 . Then CG .1 .L// D G1 is metacyclic and x does not centralize 1 .G1 /. Such a G1 , by Theorem 109.1, is either abelian or minimal nonabelian so G1 ¤ L; we see that L \ G1 D ˆ.G/ is abelian of index 3 in L. Exercise 1. Let G be a p-group, p > 2, and suppose that a noncyclic M 2 1 is metacyclic. If CG .1 .M // D M , then either M contains a cyclic subgroup of index p or G is a 3-group of maximal class. (Here we do not assume that G splits over M .) Exercise 2. Let G be a 2-group and suppose that M 2 1 is metacyclic but neither cyclic nor of maximal class. Suppose that CG .1 .M // D M . Classify the minimal nonabelian subgroups of G which are not contained in M . Answer. Q8 , D8 and M2n .

110

Equilibrated p-groups

We begin with the following Deﬁnition (compare with [BDM]). A p-group G is said to be weakly equilibrated (in short WE-group) if it follows from G D AB with A; B < G that one of the factors A; B is G-invariant. A group G is said to be equilibrated (brieﬂy: E-group) if every of its subgroup is weakly equilibrated. If, in addition, G is a p-group, then it is called a WEp -group and an Ep -group, respectively. Obviously, the properties WE, E, WEp and Ep are inherited by quotient groups. It is not true that the property WEp is inherited by subgroups (for example, if the fundamental subgroup G1 of a 3-group G of maximal class and order 35 is nonabelian, then G1 is not an E3 -group which follows from Exercise 5 below). However, as Lemma 110.1 shows, this group G is a WE3 -group. A minimal nonabelian WEp -group is an Ep -group since abelian p-groups are Ep -groups. Next, we follow [BDM]. We consider p-groups only. Note that in [BDM] the nonnilpotent groups are also studied. In what follows G is a non-Dedekindian WEp -group of order p m (Dedekindian p-groups are Ep -groups). Since groups of order p 3 are Ep -groups, one may assume that m > 3. As all subgroups of Mpm of order > p are normal, Mpm is an Ep -group. Moreover, all groups from Theorem 1.25 are Ep -groups. Exercise 1. (a) A nonmetacyclic minimal nonabelian p-group G of order p m > p 3 is not a WEp -group. (b) All two-generator WE2 -groups are metacyclic. (c) If G is a nonmetacyclic two-generator WEp -group, p > 2, then jG W G 0 j D p 2 . (d) If a metacyclic p-group G is a non-WEp -group with jG 0 j D p and G D AB, where A; B are not G-invariant, then A and B are cyclic. Solution. (a) We ﬁrst assume that m D 4. In this case, by Lemma 65.1, 2

G D ha; b j ap D b p D 1; c D Œa; b; c p D Œa; c D Œb; c D 1i: Set A D hai and B D hap c; bi. Then jAj D jBj D p 2 and A \ B D ¹1º so we have G D AB by the product formula. Since G 0 D hci 6 A and G 0 6 B, neither A nor B

126

Groups of prime power order

are G-invariant, and our claim follows. Now let m > 4. In this case, r

s

G D ha; b j ap D b p D 1; c D Œa; b; c p D 1; Œa; c D Œb; c D 1i: 2

We also assume that r s; then r > 1. Set T D hap ; b p i. Then the quotient group G=T is a nonmetacyclic minimal nonabelian p-group of order p 4 so, by the above, it is not a WEp -group. In this case, G is also not a WEp -group since the property WEp is inherited by epimorphic images. (b) Suppose that a two-generator WE2 -group G is nonmetacyclic; then it is nonabelian and jG W G 0 j > 4 (Taussky’s theorem). By Theorem 36.1, if R < G 0 is G-invariant of index 2, then G=R is nonmetacyclic and minimal nonabelian. Therefore, by part (a), G=R is not a WE2 -group since jG=Rj > 23 , and our claim follows. (c) Assume, by way of contradiction, that jG W G 0 j > p 2 . If R < G 0 of G-invariant of index p, then G=R is nonmetacyclic and minimal nonabelian (Theorem 36.1) so it is not a WEp -group by (a) since jG=Rj > p 3 ; then G is not a WEp -group as well. (d) A noncyclic subgroup of G contains 1 .G/ > G 0 and a subgroup containing 1 .G/ is normal. If C < G is cyclic such that G=C is cyclic, then A \ C D ¹1º D B \ C , and the claim follows. (In the case under consideration, G is minimal nonabelian by Lemma 65.2(a).) Exercise 1. A 2-group G of maximal class is an E2 -group. Solution. Assume that G D AB, where A; B < G, and jAj jBj, A and B are not normal in G; then jAj 4, jBj 4. Let C < G be cyclic of index 2. Then, since jB \ C j jA \ C j, we get B \ C A \ B. As B \ C G G, one may assume that B \ C D ¹1º. In this case, jG W Aj jBj D 2 so A G G, contrary to the assumption. It follows that G is also an E2 -group since all its proper subgroups are either cyclic or of maximal class. Exercise 2. Suppose that G is a two-generator nonmetacyclic p-group of order p 4 , p > 2. Then G is a WEp -group if and only if G is of maximal class and has no subgroup isomorphic to Ep3 . Solution. Let G be a WEp -group. We have to prove that G has no subgroup Š Ep3 . By Exercise 1(a), G is not minimal nonabelian. Let B < G be minimal nonabelian. Since d.G/ D 2, we get CG .B/ < B so G is of maximal class (Proposition 10.17). Assume that G has a subgroup U Š Ep3 . Then U D Z.G/ A, where A Š Ep2 , so AG D ¹1º, i.e., A is not G-invariant. Let V < G be of order p 2 such that V 6 U ; then V is not normal in G (otherwise, V D ˆ.G/ < U ).1 In this case, CG .V \ U / D U V D G so U \ V D Z.G/, and we get V \ A D ¹1º. Then G D AB by the product formula, and we conclude that G is not a WEp -group, contrary to the assumption. Now suppose that G has no subgroup Š Ep3 . We have to prove that then G is a WEp -group. Assume that G D AB, where A and B are not normal in G; then jAj D p 2 D jBj. If A does not exist, then U is generated by all subgroups of G of order p 2 so G has a cyclic subgroup of index p, a contradiction (Theorem 1.2). 1 If V

110

127

Equilibrated p-groups

is noncyclic, then Z.G/ < A (otherwise, Z.G/A Š Ep3 , contrary to the assumption). If A is cyclic, then Z.G/ D Ã1 .G/ < A since G=Ã1 .G/ is of order p 3 (Theorem 9.11) and exponent p. Thus, in any case, Z.G/ is contained in A and B so, by the product formula, jABj < p 4 D jGj, a contradiction. Now it is clear that G is an Ep -group if it has no subgroup Š Ep3 . Exercise 3. Suppose that G is a p-group of maximal class and order > p 4 , p > 3. Then G is not a WEp -group. Solution. By Theorems 9.5 and 9.6, jG=Ã1 .G/j p 4 . Let Ã1 .G/ N G G be such that jG=N j D p 4 ; then N < ˆ.G/, exp.G=N / D p and G=N (of maximal class and exponent p) has a subgroup Š Ep3 . By Exercise 3, G=N is not a WEp -group so is G. 2

2

Exercise 4. If p > 2 and G D ha; b j ap D b p D 1; ab D a1Cp i is nonabelian metacyclic of order p 4 and exponent p 2 , then G is not a WEp -group. Solution. We have Z.G/ D hap i hb p i D 1 .G/ D Ã1 .G/ and G 0 D hap i. Set B D hbi; then B is not G-invariant. The quotient group G=hap b p i is nonabelian so it has a nonnormal subgroup A=hap b p i of order p. It follows that A is cyclic (indeed, all noncyclic subgroups of G are normal). In this case, A \ B D ¹1º, and so we get G D AB by the product formula. Thus G is not a WEp -group. n

n

n1

Remark. Moreover, if p > 2, then G D ha; b j ap D b p D 1; ab D a1Cp i is not a WEp -group. The group G is metacyclic minimal nonabelian (Lemma 65:2(a)). We have Ãn1 .G/ D 1 .G/ Z.G/. If L is of order p and L 62 ¹1 .hbi/; G 0 º, then n1 there is x 2 G such that L D hx p i (Theorem 7:2(c)). Then, by the product formula, G D U V , where U D hbi, V D hxi are not G-invariant. Indeed, V \ G 0 D ¹1º, and if V G G, then G D hai V is abelian, a contradiction. From this it also follows that m n n1 G D ha; b j ap D b p D 1; ab D a1Cp i, where m < n, is not a WEp -group nm (consider the quotient group G=hb p i). It is easy to show that a nonabelian metacyclic group G of order 24 and exponent 4 is a WE2 -group. To prove this, it sufﬁces to notice that all subgroups of order 2 of G are characteristic and exactly one of them is maximal cyclic. Indeed, let G D AB, where A; B < G are not G-invariant. We have G Š H2 D hx; y j x 4 D y 4 D 1; x y D x 3 i: Since G=hx 2 y 2 i Š Q8 , we get AG ¤ hx 2 y 2 i ¤ BG . Then one of subgroups AG ; BG contains G 0 D ha2 i so is G-invariant. Exercise 5. If G is a two-generator WEp -group, p > 2, of order > p 3 that satisﬁes jG=G 0 j > p 2 , then G is metacyclic. (Hint. Use Exercise 1(c). This also follows from Exercise 1(a) (take in G 0 a G-invariant subgroup of index p).) Note that if G is a 3-group of maximal class not isomorphic to †32 2 Syl3 .S32 /, then all subgroups of G are two-generator (see Exercise 9.13).

128

Groups of prime power order

Lemma 110.1 (compare with [BDM, 4, Example 4]). A 3-group G of maximal class and order > 34 is a WE3 -group. Proof. We use induction on jGj. Write Z D Z.G/. Assume that G D HK, where H; K < G are not G-invariant. By the paragraph preceding the lemma, Exercises 9.13, 3 and induction, in the factorization G=Z D .HZ=Z/.KZ=Z/ one of the factors, say HZ=Z, is G-invariant; then HZ G G. By assumption, Z 6 H so that HZ D H Z. Since H 6 ˆ.G/, we get H Z 2 1 (every normal subgroup of index > p in a p-group X of maximal class is contained in ˆ.X/). All members of the set 1 , apart from the fundamental subgroup G1 , are of maximal class. It follows that H Z D G1 . Since G1 is metacyclic, the subgroup H is cyclic. Then ˆ.H / D ˆ.G1 /, a cyclic subgroup of order > 3, is G-invariant, a ﬁnal contradiction (see Exercise 9.1(c)).2 It follows that if G is a 3-group of maximal class and order 3m > 34 , then G is an E3 -group provided its fundamental subgroup G1 is either abelian or m is even (see the remark after Exercise 5).3 Indeed, our group G is an E3 -group if and only if G1 is a WE3 -group since if G1 is a WE3 -group, then every proper subgroup of G is also a WE3 -group so G is an E3 -group (if H 2 1 ¹G1 º, then H is of maximal class and H \ G1 is abelian). Now we are ready to prove the following Theorem 110.2 ([BDM, Theorem 4.1]). Suppose that G is a nonabelian two-generator Ep -group of order > p 3 and p > 2. Then one of the following holds: (a) G is metacyclic (see Exercise 5). (b) G is a 3-group of maximal class (all such groups are described in Lemma 110.3 below). (c) G is of maximal class and order p 4 without subgroup Š Ep3 ; see Exercise 3. Proof. Suppose that G is neither metacyclic nor of maximal class. Then jGj > p 4 (Exercise 3). Next, by Exercise 3 and Theorem 9.11, jG=Ã1 .G/j D p 3 . By Exercise 6, jG=G 0 j D p 2 . From Theorem 36.16 and Corollary 44.9 it follows that G=Ã2 .G/ is neither metacyclic nor of maximal class, where Ã2 .G/ D Ã1 .Ã1 .G//. Therefore one may assume that Ã2 .G/ D ¹1º; then exp.G/ D p 2 . Let R Ã1 .G/ be G-invariant of order p. Then, by induction, either G=R is of maximal class and order p 4 or a 3-group of maximal class and order 35 . In both cases, GN D G=R has no subgroup Š Ep3 so all its maximal subgroups are two-generator. (i) Let GN be of maximal class and order p 4 . Then GN has an abelian subgroup T =R of type .p 2 ; p/ (Exercise 3). By Lemma 65.2(a), Exercises 1(a) and 5, T is abelian of type .p 2 ; p 2 / or .p 2 ; p; p/ since exp.G/ D p 2 . Since jG=G 0 j D p 2 , we get jZ.G/j D 2 Thus, if a p-group of maximal class and order > p 4 is a WE -group, then p 2 ¹2; 3º; see Exerp cises 2, 4 and Lemma 110.3(b), below. 3 However, we do not assert that if m is even, then G is an E -group; see Lemma 110.3. 3

110

Equilibrated p-groups

129

1 0 p jG=G j

D p so that Z.G/ D R, and we conclude that G is of maximal class. In the case under consideration, since jG=Ã1 .G/j D p 3 , we get p D 3 (Theorem 9.5). (ii) Let GN be a 3-group of maximal class and order 35 (recall that exp.G0 / D 32 ). By Theorem 13.7, there is in G a normal subgroup E Š E27 . As G=R is also of maximal class, we conclude that E D Ã1 .D/, where D=R is the fundamental subgroup of the quotient group G=R. Let H=E D Z.G=E/; then H G G is of order 34 . Since jH j D 34 > 33 D jEj D j1 .D/j, it follows that exp.H / D 9 so that K D Ã1 .H / is G-invariant of order 3. By induction, G=K is of maximal class and of order 35 with normal subgroup E=K of order 33 and exponent 3, contrary to Theorem 9.6(c). Lemma 110.3 (see [BDM, Example 4]). A 3-group G of maximal class and order 3m > 34 is an E3 -group if and only if its fundamental subgroup G1 is either abelian or sC1 s s G1 D ha; b j a3 D b 3 D 1; ab D a1C3 i: Proof. All subgroups of G are two-generator (see Exercise 9.13). By Lemma 110.1, G is a WE3 -group. If the fundamental subgroup G1 of G is abelian, it is a WE3 -group. Then all proper subgroups of G which have (metacyclic) abelian fundamental subgroups are WE3 -groups. Thus G is an E3 -group if G1 is a WE3 -group. Assume that G1 is nonabelian. Then G1 , as we know, is minimal nonabelian (this follows from Exercise 9.1(b) and Lemma 65.2(a)). As we know, G is an E3 -group of and only if G1 is a WE3 -group (so an E3 -group). Next, by Exercise 5 and the remark following it, G1 is not a WE3 -group if m 1 is even; then m is odd. Now let m D 2k. By Lemma 65.1 and Theorems 9.5 and 9.6, we have r

s

G1 D ha; b j a3 D b 3 D 1; ab D a1C3

r1

i;

jr sj D 1; r Cs D m1 D 2k 1:

Since E9 Š 1 .G1 / < G1 , all noncyclic subgroups of G1 are G1 -invariant. Let s D r C 1. Then G1 =Ãr .G1 / is not a WE3 -group (Exercise 5 and the remark following it) so G1 and G are not E3 -groups. Now let r D s C 1. We claim that then G is a WE3 -group. Assume that this is false. Then G D AB, where A and B are nonnormal in G so A and B do not contain G 0 hence they have trivial intersections with hai. It follows that jAj p s , jBj p s so that jGj D jABj p 2s < p rCs D jGj, a contradiction. Thus, in the case under consideration, G is a WE3 -group. Exercise 6. Let G be an extraspecial group of order p 2mC1 , m > 1. (a) Prove that G is a WEp -group. (b) If, in addition, exp.G/ D p, then G is not an Ep -group. (c) If m > 2 and p > 2, then G is not an Ep -group. Solution. (a) Assume that G D AB, where A; B < G are not G-invariant. Then we have G 0 6 A; B so A; B are abelian of orders p m (see 4). By the product formula, jABj < p 2mC1 D jGj, a contradiction.

130

Groups of prime power order

(b) It sufﬁces to assume that m D 2. We have G D D D1 , where D and D1 are nonabelian of order p 3 . Let L < D1 be of order p such that L ¤ G 0 . Set H D D L. It sufﬁces to show that H is not an E3 -group. Let L1 < D of order p be such that L1 ¤ G 0 and set A D L L1 . Then A Š Ep2 is not normal in H since D \ A D L is not normal in D. Let L2 < Z.H / of order p be such that L2 62 ¹L; G 0 º and let L3 < D of order p be such that L3 62 ¹L1 ; G 0 º. Set B D L2 L3 . Then B is not normal in H since B \ D D L2 is not normal in D. Since A \ B D ¹1º, we get H D AB is not a WEp -group so G is not an Ep -group. (c) Let G D D1 Dm1 Dm , where D1 ; : : : ; Dm1 are nonabelian of order p 3 and exponent p. Then D1 D2 is not an Ep -group by (b) so that G is not an Ep -group. m

n

Exercise 7. Prove that the metacyclic p-group G D ha; b j ap D b p D 1; ab D m1 a1Cp ; n mi, p > 2, is not a WEp -group. (Hint. See the last paragraph of the proof of Lemma 110.3.) r

s

Exercise 8. Let G D ha; b j ap D b p D 1; ab D a1Cp Then G is a WEp -group if and only if s < r.

r1

i, p > 2, r C s > 4.

Solution. Using the argument in the proof of Theorem 110.3, which is also applied to minimal nonabelian metacyclic p-groups with p > 3, we have to prove that if s < r, then G is a WEp -group. Assume that this is false. Then G D AB, where A and B are not G-invariant. Since G 0 < 1 .G/, we get j1 .A/j D j1 .B/j D p so A and B are cyclic and A \ hai D ¹1º D B \ hai. It follows that jAj p s , jBj p s so that jGj jAjjBj p 2s < p rCs D jGj, a contradiction. Thus G is a WEp -group if and only if s < r. Exercise 9. Let a WE-group G be nonnilpotent and solvable. Then we have G D D P , a semidirect product, where P 2 Sylp .G/ for some prime p is G-invariant and D is Dedekindian. Solution. Let H < G be a maximal nonnormal subgroup of G; then jG W H j D p a for some prime p. Let P 2 Sylp .G/. Since G D HP and H is not G-invariant, we get P G G. Let F=P < G=P . Since G D HP HF , we obtain G D HF so F G G. It follows that G=P is Dedekindian. It remains to apply Hall’s Theorem A.28.4 on solvable groups. Exercise 10. Let a nonabelian WEp -group G D AB be such that jG=G 0 j D p 2 and A G G, B < G. Prove that jG W Aj D p. Solution. If jG W Aj > p, then A G 0 D ˆ.G/ so G D AB D B, a contradiction. Exercise 11. Let a nonabelian two-generator WEp -group G D AB, where A G G and B < G. Prove that G=A is cyclic. (Hint. If G=A is noncyclic, then A ˆ.G/.) Exercise 12. Let a nonabelian two-generator non-WEp -group G D AB, where A and B are maximal nonnormal subgroups of G. Then G=AG and G=B G are cyclic. (Hint. If G=AG is noncyclic, then A < AG ˆ.G/.)

110

Equilibrated p-groups

131

Exercise 13. Let G be a 2-group such that G=1 .G/ Š Q8 . Is it true that G is not a WE2 -group? (Hint. See Lemma 42.1(c) about the structure of G. By that lemma, G is metacyclic.) Exercise 14. Suppose that G is a metacyclic p-group. Then one of the following holds: (a) jGj D p 3 . (b) G is a 2-group of maximal class. (c) G has a subgroup isomorphic to Mpn , n > 3. (d) 1 .G/ Z.G/. Solution. Suppose that (a), (b) and (c) do not hold; then G is non-Dedekindian and 1 .G/ Š Ep2 . Let A G be minimal nonabelian. If jAj D p 3 , then either p D 2 and G is of maximal class or G D A (Proposition 10.19), contrary to the assumption. It follows that 1 .A/ Z.A/ (Lemma 65.1). Since all such A generate G (Theorem 10.28) and 1 .A/ D 1 .G/, we conclude that 1 .G/ Z.G/, i.e., (d) holds. Lemma 110.4 ([BDM, Theorem 4.2]). Let G be a metacyclic p-group, p > 2, let hai D M G G and G=M D hbM i. Then G D hbihbai. Proof. One has G D ha; bi. Suppose that ab D ak ; then we get k 1 .mod p/ since o.b/ is a power of p and p j .k o.b/ 1/. Let us show that hbihbai D hbaihbi. We have ˇ1 ˇ1 b ˛ .ba/ˇ D b ˛Cˇ ab : : : ab a D b ˛Cˇ a1CkCCk : Thus we must show that every integer can be expressed modulo p n D o.a/ in the form 1 C k C C k ˇ 1 . If this is not so, there exist integers ˇ; with 0 ˇ < < p n such that 1 C k C C k ˇ 1 D 1 C k C C k 1 .mod p n /: Hence, since p does not divide k, we get .1/

1 C k C C k ˇ 1 0

.mod p n /;

and 0 < ˇ < p n . It follows that k 6 1 .mod p n /. But k 1 .mod p/; thus, if k D 1 C p m s with GCD.s; p/ D 1, then 1 m < n. It follows from (1) that k ˇ 1 k ˇ 1 D

0 k1 sp m

.mod p n /

so k ˇ 1 .mod p mCn /. We have .2/

k ˇ D .1 C sp m / ˇ D 1 C . ˇ/sp m C Cp mCn ;

where C is a positive integer. By (2), we obtain that p n j . ˇ/ so ˇ p n , a contradiction.

132

Groups of prime power order

Theorem 110.5 ([BDM, Theorem 4.2]). Let G be a metacyclic WEp -group, p > 2. Then cl.G/ 2. Proof. We retain the same meaning for M, a and b as in the statement of Lemma 110.4. Since G D hbihbai is a WEp -group, either hbi G G or hbai G G; hence Œb; a, a generator of G 0 , is contained either in hbi or hbai. But also Œb; a 2 M D hai. It follows that Œb; a commutes with b and a hence Œb; a commutes with hb; ai D G, and we conclude that G 0 D hŒb; ai Z.G/. Exercise 15 ([BDM, Lemma 4.3(a)]). Suppose that a nonabelian Ep -group G has exponent p. Then jGj D p 3 . Solution. By hypothesis, p > 2. Assume that jGj D p 4 . Then G contains an abelian subgroup A of index p so, by Exercise 3, cl.G/ D 2. By Lemma 65.1, G is not minimal nonabelian. Therefore it follows from Proposition 10.17 that G D D L, where L Z.G/ is of order p. As in the solution of Exercise 6(b), G is not a WEp -group. Now let jGj > p 4 . Since G is not minimal nonabelian (Lemma 65.1), it contains a nonabelian subgroup H of order p 4 . By what has just been proved, the subgroup H is not a WEp -group so G is not an Ep -group. Thus jGj D p 3 . Exercise 16. Let G be an Ep -group of order > p 4 , p > 2, containing a nonabelian subgroup A of order p 3 and exponent p. Prove that one of the following holds: (a) 1 .G/ D A and G D AC , where C is cyclic. (b) G is a 3-group of maximal class. Solution. Assume that G has a normal subgroup U Š Ep3 . Let U1 < U be a G-invariant subgroup which is as small as possible and such that U1 6 A. Set H D AU1 . If A \ U1 D ¹1º, then jU1 j D p and H D A U1 is not a WEp -group (see solution of Exercise 6(b)). Thus A \ U1 > ¹1º. Moreover, all minimal normal subgroups of G contained in U are contained in A. It follows that jH j D p 4 . Assume that U1 D U . Then H is not of maximal class (Exercise 3) so, by Proposition 10.17, CH .A/ 6 A. Since 1 .H / D H , we get H D A C for some C of order p, and then H is not a WEp -group. Thus jU1 j D p 2 and U1 6 Z.H / (otherwise, A is a direct factor of H ). In this case, we have CH .U1 / Š Ep3 . Then, by Exercise 3, H is not of maximal class so cl.H / D 2, and we conclude that exp.H / D p since p > 2. Then H D A C again, a contradiction. Thus U does not exist and the result now follows from Theorem 13.7. Theorem 110.6 ([BDM, Lemma 4.3(c)]). If an Ep -group G contains a proper subgroup H of maximal class and order > p 3 , p > 2, then p D 3 and G is of maximal class. Proof. Assume that G is not of maximal class. By Theorems 9.5 and 9.6, one may assume that jH j D p 4 (indeed, H contains a subgroup of maximal class and order p 4 ); then exp.H / D p 2 and H has no subgroup Š Ep3 (Exercise 3). Theorem 13.7 yields

110

Equilibrated p-groups

133

that either H is a 3-group of maximal class or H D C 1 .H /, where C is cyclic of order p 2 and 1 .H / is nonabelian of order p 3 and exponent p. By Exercise 16, the second case is impossible since a group from Exercise 16(a) has no such a subgroup as H (it is important that jGj > p 4 ). Thus H is of maximal class and order 34 . Again H has no subgroups of order 33 and exponent 3. Indeed, if H has a nonabelian subgroup A of order 33 and exponent 3, then, by Exercise 16, G D AC , where A D 1 .G/ and C is cyclic of order > 32 . Such a G, however, has no subgroup of maximal class and order 34 . By Exercise 13.10(a), we have H < M G, where jM W H j D p and M is not of maximal class. One may assume without loss of generality that G D M ; then G=G 0 Š E33 (Theorem 12.12(a)). By Theorem 12.12(b), GN D G=K3 .G/ is of order 34 and exponent 3 since jGj > 34 . As above, GN D SN CN , where SN is nonabelian of order 33 and jCN j D 3. As we know, such a GN is not an E3 -group, a contradiction. Thus G is a 3-group of maximal class. For further details on Ep -groups, see [BDM] and [Sil]. In view of Theorem 110.2, the classiﬁcation of Ep -groups, p > 2, will follow from the solution of the following Problem 1. Classify the p-groups all of whose nonabelian two-generator subgroups are either metacyclic or of maximal class. Problem 2. Classify the metacyclic WEp -groups.

111

Characterization of abelian and minimal nonabelian groups

If G is abelian, then G 0 D ˆ.G/0 and it appears that this property characterizes abelian groups (Theorem 111.1). If G is either abelian or minimal nonabelian, then H 0 D ˆ.G/0 for all H 2 1 , where 1 is the set of all maximal subgroups of G, and Theorem 111.3 shows that this property is characteristic for groups all of whose maximal subgroups are abelian. In this section we consider also nonnilpotent groups. In Lemma J we collect some known results cited in what follows. Lemma J. Let G be a ﬁnite group. (a) (R. Baer [Bae6]) If P 2 Syl.G/ is G-invariant, then ˆ.P / D P \ ˆ.G/. (b) (Redei; see Exercise 1.8(a)) If G is a nilpotent minimal nonabelian group, then G is a p-group, jG 0 j D p, Z.G/ D ˆ.G/ is of index p 2 in G and one of the following holds: (i) p D 2 and G is the ordinary quaternion group, m

n

m1

(ii) G D ha; b j ap D b p D 1; ab D a1Cp ; m > 1i is metacyclic of order p mCn , m n (iii) G D ha; b j ap D b p D c p D 1; c D Œa; b; Œa; c D Œb; c D 1i is nonmetacyclic of order p mCnC1 . In particular, if exp.G/ D p, then p > 2 and jGj D p 3 . (c) (Miller–Moreno; see Lemma 10.8) If G is a nonnilpotent minimal nonabelian group, then we have jGj D p a q b , G D P Q, where P 2 Sylp .G/ is cyclic, Q D G 0 2 Sylq .G/ is a minimal normal subgroup of G, Z.G/ D ˆ.G/ has index p in P . (d) (Wielandt) A group G is nilpotent if and only if G 0 ˆ.G/ (or, what is the same, G=ˆ.G/ is nilpotent). (e) If G is nilpotent and noncyclic, then G=G 0 is also noncyclic. (f) [Gas1] If a Sylow p-subgroup is normal in G=ˆ.G/, then a Sylow p-subgroup is normal in G. If H is normal in G, then ˆ.H / ˆ.G/. (g) .G=ˆ.G// D .G/, where .G/ is the set of prime divisors of jGj.

111

Characterization of abelian and minimal nonabelian groups

135

(h) [BG] If NF .F \ H / ¤ F \ H ¤ NH .F \ H / for any two distinct F; H 2 1 , then either G is nilpotent or G=ˆ.G/ is (nonnilpotent) minimal nonabelian. The converse assertion is also true. (i) If G is a nonabelian p-group such that G 0 Z.G/ has exponent p, then we have ˆ.G/ Z.G/. (j) (Fitting; see Corollary 6.5) Let Q be a normal abelian Sylow q-subgroup of a group G. Then Q \ Z.G/ is a direct factor of G. Let us prove Lemma J(a). Since ˆ.P / D D ˆ.G/ \ P (Lemma J(f)), it sufﬁces to prove the reverse containment. To this end, one may assume that ˆ.P / D ¹1º; then P is elementary abelian. We have to show that p 62 .ˆ.G//. Let H be a p 0 -Hall subgroup of G. Then P D D L, where L is H -admissible (Maschke). In this case, G D HP D .HL/ D, a semidirect product with kernel D, so D D ¹1º since D ˆ.G/. Thus, in general case, ˆ.P / D D. Let us prove Lemma J(i). Take x; y 2 G. Since cl.G/ D 2, we have 1 D Œx; yp D Œx; y p so that Ã1 .G/ Z.G/, and we obtain ˆ.G/ D G 0 Ã1 .G/ Z.G/. Let 1 (1n ) be the set of maximal (nonnormal maximal) subgroups of a group G. If G is not nilpotent, then the set 1n is not empty (Lemma J(d)). Theorem 111.1. The following conditions for a group G are equivalent: (a) G 0 D ˆ.G/0 . (b) G is abelian. Proof. As we have noticed, (b) ) (a) so it remains to prove the reverse implication. We have G 0 D ˆ.G/0 ˆ.G/ so G is nilpotent (Lemma J(d)); then G D P1 Pk , where P1 ; : : : ; Pk are Sylow subgroups of G. We have (1) (2) (3)

ˆ.G/ D ˆ.P1 / ˆ.Pk /; G 0 D P10 Pk0 ; ˆ.G/0 D ˆ.P1 /0 ˆ.Pk /0 :

Equality (1) follows from Lemma J(a), and (2), (3) are known properties of direct products. Since ˆ.G/0 D G 0 , it follows from (2) and (3) that ˆ.Pi /0 D Pi0 so Pi satisﬁes the hypothesis for i D 1; : : : ; k. To complete the proof, it sufﬁces to show that Pi is abelian for i D 1; : : : ; k so one may assume that k D 1, i.e., G is a p-group. Assume that G is nonabelian of the least possible order. We get ˆ.G/0 D G 0 > ¹1º so that ˆ.G/ is nonabelian. Let R < ˆ.G/0 be G-invariant of index p. Since R < ˆ.G/0 D G 0 < ˆ.G/, we get ˆ.G=R/0 D .ˆ.G/=R/0 D ˆ.G/0 =R D G 0 =R D .G=R/0 whence the nonabelian group G=R satisﬁes the hypothesis so we must have R D ¹1º by induction. In that case, we get jˆ.G/0 j D jG 0 j D p so G 0 Z.G/. By Lemma J(i), ˆ.G/ Z.G/ so that ˆ.G/ is abelian, contrary to what has been said above.

136

Groups of prime power order

Lemma 111.2. Let G be a noncyclic p-group and D D hˆ.H / j H 2 1 i. Then D ˆ.G/ and jˆ.G/ W Dj p with equality only for p > 2. Proof. Since all members of the set 1 are G-invariant, we have D ˆ.G/ (Lemma J(f)). As all maximal subgroups of the quotient group G=D are elementary abelian, G=D is either elementary abelian or nonabelian of order p 3 and exponent p (Lemma J(d)). In the ﬁrst case, D D ˆ.G/; in the second case, p > 2 and jˆ.G/ W Dj D jˆ.G=D/j D p. Thus, in both cases, jˆ.G/ W Dj p, as desired. If H 2 1 , then ˆ.G/ < H so ˆ.G/0 H 0 . Below we consider the extreme case when ˆ.G/0 D H 0 for all H 2 1 . Theorem 111.3. The following assertions for a group G are equivalent: (a) ˆ.G/0 D H 0 for all H 2 1 . (b) G is either abelian or minimal nonabelian. Proof. Obviously, groups all of whose maximal subgroups are abelian satisfy condition (a) hence implication (b) ) (a) is true. Therefore it remains to prove the reverse implication. In what follows one may assume that G is nonabelian. (i) Suppose that G is a nonabelian p-group satisfying condition (a). Note that ˆ.G/, ˆ.G/0 and H 0 are G-invariant for all H 2 1 . If ˆ.G/ is abelian, then all maximal subgroups of G are also abelian by hypothesis, and G is minimal nonabelian so (b) holds. Next we assume that ˆ.G/0 > ¹1º. Let R < ˆ.G/0 be G-invariant of index p N is nonabelian and jˆ.G/ N 0 j D p. Take H 2 1 . We have and set GN D G=R; then ˆ.G/ .4/

N 0 D ˆ.G=R/0 D .ˆ.G/=R/0 D ˆ.G/0 =R D H 0 =R D HN 0 ˆ.G/

N since is of order p so, by Lemma J(i), ˆ.HN / Z.HN / and hence ˆ.HN / Z.ˆ.G// N N N N N N N ˆ.H / ˆ.G/ H . Setting D D hˆ.H / j H 2 1 i, we conclude D Z.ˆ.G// N is abelian since jˆ.G/ N W Dj N p by Lemma 111.2, contrary to what has so ˆ.G/ already been said. Thus, if G is a p-group, then (a) ) (b). In what follows we assume that G is not a prime-power group. Write T D ˆ.G/0 ; then T D H 0 for all H 2 1 by hypothesis. Therefore all maximal subgroups of G=T are abelian so the quotient group G=T is either abelian or minimal nonabelian. As in (i), one may assume that ˆ.G/ is nonabelian. (ii) Suppose that G is nilpotent; then G D P1 Pk , where Pi 2 Sylpi .G/, i D 1; : : : ; k, k > 1. It follows from (2) and (3) which are true for any nilpotent group that ˆ.Pi /0 D Hi0 for all maximal subgroups Hi of Pi so Pi is either abelian or minimal nonabelian by (i). Then ˆ.Pi / is abelian for all i so ˆ.G/ is abelian in view of (1), contrary to the assumption. Thus the theorem is true provided G is nilpotent. Next we assume that G is not nilpotent. In this case, by Lemma J(d), G=T is nonnilpotent (and minimal nonabelian). (iii) It remains to consider the case where G=T is minimal nonabelian and nonnilpotent. It follows from the structure of G=T (Lemma J(c)) that ˆ.G=T / D ˆ.G/=T is

111

Characterization of abelian and minimal nonabelian groups

137

cyclic. Remembering that ˆ.G/ is nilpotent and T D ˆ.G/0 , we conclude that ˆ.G/ is cyclic (Lemma J(e)) so abelian, a ﬁnal contradiction. Corollary 111.4. Suppose that a group G is neither abelian nor minimal nonabelian. Then there exists a nonabelian H 2 1 such that ˆ.G/0 < H 0 . Proof. We have ˆ.G/0 H 0 for all H 2 1 . Assume that ˆ.G/0 D H 0 for all nonabelian H 2 1 ; then ˆ.G/ is nonabelian (take a nonabelian H 2 1 ) so the set 1 has no abelian members. In that case, ˆ.G/0 D H 0 for all H 2 1 by hypothesis, so that G is minimal nonabelian (Theorem 111.3), contrary to the hypothesis. We claim that the following conditions for a group G are equivalent: (a) H 0 ˆ.G/ for all H 2 1 , (b) either G is nilpotent or G=ˆ.G/ is nonnilpotent minimal nonabelian, (c) NF .F \ H / ¤ F \ H ¤ NH .H \ F / for all distinct F; H 2 1 . Obviously, (a) , (b) and (b) ) (c). Next, (b) follows from (c) by Lemma J(h). Below we use some known results on Frobenius groups (see [BZ, 10.2]). Recall that G is said to be a Frobenius group if there is a subgroup H < G, H > ¹1º, such that H \H x D ¹1º for all x 2 G H . In that case, G D H N is a semidirect product with kernel N , Sylow subgroups of H are either cyclic or generalized quaternion. Lemma 111.5. The following conditions for a nonnilpotent group G are equivalent: (a) All members of the set 1n are abelian. (b) G=Z.G/ D .U=Z.G// .Q1 =Z.G// is a Frobenius group with elementary abelian kernel Q1 =Z.G/ D .G=Z.G//0 and a cyclic complement U=Z.G/, U 2 1 . Proof. If U; V are two distinct abelian maximal subgroups of a nonabelian group G, then U \ V D Z.G/. As we have noticed, the set 1n is nonempty. T Suppose that G satisﬁes condition (a). Given H 2 1n , set HG D x2G H x ; then HG D Z.G/ by the previous paragraph. Set GN D G=Z.G/; then GN D HN QN 1 is a Frobenius group with kernel QN 1 and complement HN . Since HN is abelian, it is cyclic. There is in QN 1 an HN -invariant Sylow subgroup by Sylow’s theorem. Therefore, since N the subgroup QN 1 is a p-group; moreover, QN 1 is a minimal normal HN is maximal in G, subgroup of GN so it is elementary abelian. All nonnormal maximal subgroups of GN are conjugate with HN (Schur–Zassenhaus) so all members of the set 1n are conjugate in G (indeed, all members of the set 1n , being abelian, contain Z.G/). Thus (a) ) (b). Conversely, every group as in (b) satisﬁes condition (a). In fact, if H 2 1n , then Z.G/ < H . By (b), H=Z.G/ is cyclic so H is abelian. The proof of Lemma 111.5 shows that if the set 1n has at least one abelian member, then the group G has the same structure as in Lemma 111.5(b). Corollary 111.6. A nonnilpotent group G satisﬁes ˆ.G/0 D H 0 for all H 2 1n if and only if GN D G=ˆ.G/0 is as in Lemma 111.5(b).

138

Groups of prime power order

Proof. Suppose GN D G=ˆ.G/0 is as in Lemma 111.5(b) and HN is a nonnormal maxiN then HN is cyclic by hypothesis, so H 0 ˆ.G/0 . As ˆ.G/ < H , mal subgroup of G; 0 we get ˆ.G/ H 0 so H 0 D ˆ.G/0 and whence G satisﬁes the hypothesis. Now suppose that ˆ.G/0 D H 0 for all H 2 1n . Then all nonnormal maximal subgroups of G=ˆ.G/0 are abelian so G=ˆ.G/0 is as in Lemma 111.5(b). Proposition 111.7. Suppose that a nonnilpotent group G is not minimal nonabelian and such that M 0 D ˆ.G/ for all nonabelian M 2 1 . Then (a) G D P Q is a semidirect product with kernel Q D G 0 2 Sylq .G/, P 2 Sylp .G/ is of order p, ˆ.G/ D ˆ.Q/ > ¹1º. Set H D P ˆ.G/. The set 1 consists of two conjugacy classes of subgroups with representatives H and Q. (b) P G D G, where P G is the normal closure of P in G. (c) G=ˆ.G/ is minimal nonabelian so that Q=ˆ.G/ is a minimal normal subgroup of G=ˆ.G/. (d) H 0 < ˆ.G/ if and only if G is minimal nonnilpotent. (e) If ˆ.G/ is abelian, then G is either minimal nonnilpotent or a Frobenius group. Proof. By hypothesis, all maximal subgroups of G=ˆ.G/ are abelian so it is nonnilpotent and minimal nonabelian by Lemma J(d,c). We obtain that ˆ.G=ˆ.G// D ¹1º so that jG=ˆ.G/j D pq b , where a subgroup of order p is not normal in G=ˆ.G/ (Lemma J(b,c)). By hypothesis, the set 1 has a nonabelian member so ˆ.G/ > ¹1º. Lemma J(g) yields .G/ D .G=ˆ.G// D ¹p; qº. Due to Lemma J(f), Q 2 Sylq .G/ is normal in G. We have G D P Q, where P 2 Sylp .G/. From Lemma J(a) we deduce ˆ.Q/ D Q \ ˆ.G/. By the above, ˆ.G/ D P1 ˆ.Q/, where P1 is a maximal subgroup of P . The subgroup H D P ˆ.Q/ 2 1 and all nonnormal maximal subgroups of G are conjugate with H . Since P1 Q=Q is the unique normal maximal subgroup of G=Q, it follows that G=Q Š P is cyclic. Let, as above, P1 < P be maximal; then, as we have noticed, P1 ˆ.G/. The subgroup P1 Q D P1 Q D M 2 1 . Assume that P1 > ¹1º and M is nonabelian. We have P1 6 M 0 D ˆ.G/, a contradiction. Thus, if P1 > ¹1º, then M is abelian. Similarly, P1 6 .P ˆ.G//0 D H 0 so that H is also abelian. It follows that G is minimal nonabelian, contrary to the hypothesis. Thus we get P1 D ¹1º so that jP j D p and Q 2 1 . This completes the proof of (a). By Lemma J(a), ˆ.G/ D ˆ.Q/. Assume that there is L G G of index q. Since L 2 1 , we get ˆ.G/ < L, and this is a contradiction since G=ˆ.G/ has no normal subgroup of index q. It follows that P G D G, and the proof of (b) is complete. Suppose that H 0 < ˆ.G/. Then H is abelian by hypothesis. In this case, we have CG .ˆ.G// H G P G D G so that ˆ.G/ D Z.G/ since Z.G=ˆ.G// D ¹1º, and we conclude that G is minimal nonnilpotent since G=ˆ.G/ is minimal nonabelian. The proof of (d) is complete. Assume that ˆ.Q/.D ˆ.G// is abelian and G is not minimal nonnilpotent. Then, by (d), H D P ˆ.Q/ is nonabelian so H 0 D ˆ.Q/ by hypothesis. In this case, due to

111

Characterization of abelian and minimal nonabelian groups

139

Lemma J(j), NH .P / D P so that H is a Frobenius group as jP j D p. Since G=ˆ.Q/ is a Frobenius group, it follows that G is also a Frobenius group (in the case under consideration, p > 2, by Burnside). This completes the proof of (e) and, thereby, of the proposition. Let a nonabelian group G be such that H 0 D ˆ.H / for all nonabelian H G. Then G has no minimal nonnilpotent subgroups by Lemma J(d). It follows that the group G is nilpotent. Let A G be minimal nonabelian. Then A is a p-subgroup for some prime p and A0 D ˆ.A/ has order p and index p 2 in A (Lemma J(b)) so that jAj D jA0 jp 2 D p 3 . Thus all minimal nonabelian subgroups of G have order equal to the cube of a prime. Suppose that P 2 Syl2 .G/ is nonabelian. Then P is among 2-groups described in 90. Now suppose that G D Q H , where Q 2 Sylq .G/ is nonabelian and H > ¹1º. Assume that Z H is cyclic of order p 2 . Then K D QZ does not satisfy the hypothesis since ˆ.K/ 6 K 0 . Thus either G is a prime power or G D Q H , where Q 2 Sylq .G/ is nonabelian and exp.H / > 1 is square free. It follows that if H is also nonabelian, then exp.Q/ D q and we conclude that q > 2.

Problems Problem 1. Study the nonabelian p-groups G of exponent > p all of whose subgroups H G satisfy Ã1 .ˆ.H // D Ã1 .H /. (Obviously, we must have p > 2 since for any 2-group X we have ˆ.X/ D Ã1 .X/.) Problem 2. Classify the p-groups G satisfying H 0 D ˆ.G/0 for all those H G that are neither abelian nor minimal nonabelian. Problem 3. Study the p-groups G which satisfy H 0 D ˆ.G/0 for all H 2 i , where i 2 ¹1; : : : ; d.G/ 1º is ﬁxed. (For i D 1, see Theorem 111.3.)

112

Non-Dedekindian p-groups all of whose nonnormal subgroups have the same order

In this section we classify the groups G with the title property. If a subgroup H < G is nonnormal, it is cyclic since it is not generated by its (G-invariant) maximal subgroups. The p-groups all of whose nonnormal subgroups are cyclic are classiﬁed in Theorem 16.2. Therefore Theorems 112.3 and 112.4 are partial cases of Theorem 16.2. However, our proofs of the theorems are essentially shorter and not so involved. Moreover, our approach is entirely different. Therefore we decided to present the proofs of Theorems 112.3 and 112.4. We also classify the nonnilpotent groups with the title property. Theorem 112.3 was proved by D. Passman [Pas], and Theorem 112.4 was also proved by G. Zappa [Zap] (his proof is essentially longer). If G is a non-Dedekindian minimal nonabelian p-group, then (see Exercise 1.8(a) or Lemma 65.1) G D ha; bi, where m

n

m

n

(MA1) ap D b p D c p D 1, Œa; b D c, Œa; c D Œb; c D 1, jGj D p mCnC1 , G D hbi .hai hci/ D hai .hbi hci/ is nonmetacyclic. (MA2) ap D b p D 1, ab D a1Cp cyclic.

m1

, jGj D p mCn and G D hbi hai is meta-

If a minimal nonabelian p-group is Dedekindian, then it is isomorphic to Q8 . If G is a nonnilpotent minimal nonabelian group, then (Miller–Moreno [MM]) (MNA) G D P Q, where P 2 Sylp .G/ is cyclic, G 0 D Q 2 Sylq .G/ is an elementary abelian minimal normal subgroups (p; q are distinct primes), Z.G/ D Ã1 .P /, jQj D q b , where b is a minimal positive integer such that p j q b 1 hence b is the order of q modulo p. The group from (MNA) is said to be an A.p; q/-group. Below we freely use the following fact. The diagonal subgroup X of the direct product G D A B of two isomorphic nonabelian groups A and B is not G-invariant. Indeed, assume that this is false; then A \ X D B \ X D ¹1º so CG .X/ AB D G and X 2 Z.G/ is abelian, a contradiction since X Š A and A is nonabelian. The following lemma may be useful in proofs of such results as Theorems 112.2 and 112.4. For its proof, see Lemma 96.1.

112

Nonnormal subgroups have the same order

141

Lemma 112.1 (= Lemma 96.1). Let G D ha; bi be a minimal nonabelian p-group as in (MA1) or (MA2). (a) All nonnormal cyclic subgroups of G have the same order if and only if one of the following holds: (a1) G is as in (MA2) with m n. (a2) G is as in (MA1) with m D n. (b) All nonnormal subgroups of G have the same order if and only if one of the following holds: (b1) G 2 ¹D8 ; S.p 3 /º. (b2) G is as in (MA2) with m n. (c) All nonnormal cyclic subgroups of G of same order are conjugate if and only if G Š Mpt . (d) All nonnormal subgroups of G are conjugate if and only if G Š Mpt . Let nc.G/ denote the number of conjugate classes of nonnormal subgroups of a group G. If G is nilpotent and nc.G/ D 1, then G is primary. Therefore, in Theorem 112.2, we conﬁne to p-groups. Theorem 112.2 (O. Schmidt [Sch3]). Let G be a p-group. If nc.G/ D 1, then we have G Š MpnC1 . Proof. See Lemma 96.4. Theorem 112.3 (Passman [Pas, Proposition 2.4] = Theorem 1.25). If all nonnormal subgroups of a non-Dedekindian p-group G of order > p 3 have the same order p, then one and only one of the following holds: (a) G Š Mpn . (b) G D Z.G/1 .G/, where Z.G/ is cyclic and 1 .G/ is nonabelian of order p 3 so 1 .G/ 2 ¹S.p 3 /; D8 º (note, that D8 Z.G/ Š Q8 Z.G/). (c) The group G D Q D is extraspecial of order 25 , where Q Š Q8 and D Š D8 . Proof. It is easy to check that the groups (a)–(c) satisfy the hypothesis. By Proposition 1.23 (see also Lemma 96.2), jG 0 j D p (otherwise, there is in G 0 a G-invariant subgroup R of index p such that G=R contains a nonnormal subgroup K=R; then K is not G-invariant of order > p, contrary to the hypothesis). If G is minimal nonabelian, then either jGj D p 3 or G Š Mpn , n > 3, by Lemma 112.1(a). Next we assume that G is not minimal nonabelian. Let U < G be minimal nonabelian; then U 0 D G 0 so U G G. In view of Lemma 4.3, G D U C , where C D CG .U /. Lemma 112.1(b) yields U 2 ¹Q8 ; D8 ; MpnC1 ; S.p 3 /º since, if U is non-Dedekindian, it satisﬁes the hypothesis of that lemma. By assumption, C 6 U . Let K < G be nonnormal of order p; then K is contained in exactly one subgroup of G of order p 2 (otherwise, K G G). Considering K1 .Z.G//, we con-

142

Groups of prime power order

clude that Z.G/ is cyclic and so G has no subgroup Š Ep3 (indeed, if such a subgroup exists, it must lie in Z.G/). Next we consider all four possibilities for U . (i) Let U Š D8 . Assume that C has a subgroup, say V , of order 2 that satisﬁes V ¤ Z.U /. If A < U is nonnormal, then A V Š E4 is not G-invariant (indeed, .A V / \ U D A is not U -invariant), a contradiction. Thus C is either cyclic or Š Q8 (recall that jG 0 j D 2) so G D U C is a group from (b) or (c). In what follows we assume that G has no subgroups Š D8 . (ii) Let U Š Q8 . Since G is not Dedekindian, we get exp.C / > 2 by Theorem 1.20. Let L C be cyclic of order 4; write H D U L. Assume that U \ L D ¹1º; then H D U L. Since 1 .H / < Z.H /, exp.H / D 4 and H is non-Dedekindian (Theorem 1.20), it has a nonnormal cyclic subgroup of order 4, a contradiction. Thus jH j D 16. Then H D U L has a subgroup Š D8 (see Appendix 16), contrary to the assumption. Next we assume that G has no subgroups Š Q8 . (iii) Let U Š S.p 3 /, p > 2. As in (i), C is cyclic (of order > p with U \C D Z.U / since jGj > p 3 ), and we get a group from (b) since then U D 1 .G/ by the regularity of G. In what follows we assume that G has no subgroups Š S.p 3 /. (iv) Suppose that U Š MpnC1 . Assume that there is K < G of order p such that K 6 U . Put H D 1 .U /K; then H of order p 3 is generated by elements of order p. It follows from (i) and (iii) that H Š Ep3 . In this case, H has a non-G-invariant subgroup of order p 2 (otherwise, H Z.G/, which is a contradiction). Thus we obtain 1 .G/ D 1 .U / Š Ep2 and C is cyclic by (ii). There is S C such that S 6 U but Ã1 .S/ < U . Let T < U be cyclic of index p. Then U \ S Z.U / D ˆ.U / < T , jT j jS j so (the abelian group) T S D T L, where L of order p, and L is not contained in U (otherwise, S < U ), a ﬁnal contradiction. Let G be a non-Dedekindian p-group such that jH=HG j p for all H < G. If a subgroup H < G is nonnormal of maximal order, then G=HG is one of groups of Theorem 112.3. Let us show how to prove Theorem 112.2 making use of Theorem 112.3. Suppose that H < G is nonnormal; then H is cyclic. Set jH j D p s . If s D 1, then G Š MpnC1 (Theorem 112.3). If s > 1, then G=1 .H / Š Mpm by induction, and the proof is ﬁnished as in Theorem 96.4. D. Passman [Pas, Theorem 2.9] considered the p-groups without nonnormal subgroups U < V such that jV W U j D p. Suppose that, as in [Pas, Theorem 2.9], a non-Dedekindian p-group G has no two nonnormal subgroups U < V with jV W U j D p. Let H < G be nonnormal of maximal order. Then H is cyclic since all its maximal subgroups are G-invariant, and G=HG is as in Theorem 112.2.1 1 As

Passman [Pas] has noticed, the converse is also true, i.e., if all nonnormal subgroups of G are cyclic, then it has no chain U < V with nonnormal U; V such that jV W U j D p. Therefore Theorem 2.9

112

Nonnormal subgroups have the same order

143

We consider an important partial case of [Pas, Theorem 2.9]. Note that the statement of Theorem 112.4 is neither stated in [Pas] nor follows from [Pas, Theorem 2.9] (since in the last theorem groups of order < 28 were not considered). For a complete classiﬁcation of groups from [Pas, Theorem 2.9], see Theorem 16.2. Now we are ready to prove the main result of this section. Theorem 112.4. Suppose that all nonnormal subgroups of a p-group G have the same order p > p; then 1 .G/ D 1 .Z.G//, j1 .G/j p 2 . If j1 .G/j D p, then G Š Q16 , D 2. Now we let 1 .G/ Š Ep2 ; then G=1 .G/ is abelian. Let H < G be nonnormal; then H is cyclic, G=HG is as in Theorem 112.3 and one and only one of the hollowing holds: m

n

(a) G D ha; b j ap D b p D 1; ab D a1Cp m n D > 1.

m1

i is metacyclic minimal nonabelian,

(b) G D Q L, where Q Š Q8 , L Š C4 , and D 2. (c) G is minimal nonmetacyclic and special of order 25 , D 2.2 (d) G is special of order 26 with Z.G/ D G 0 D ˆ.G/ D 1 .G/ Š E4 , D 2.3 Proof. By hypothesis, jGj > p 3 . The hypothesis is inherited by non-Dedekindian subgroups. Note that all nonnormal subgroups of any proper non-Dedekindian epimorphic image of G are also of same order (which, however, is < p ). We have 1 .G/ Z.G/ since > 1. Let H < G be nonnormal; then all maximal subgroups of H are G-invariant so they do not generate H , and we infer that H is cyclic. If U=HG < G=HG is not normal, then we have jU j D jH j so jU=HG j D p, and hence G=HG is as in Theorem 112.3. Next, H is contained in exactly one (G-invariant) subgroup of G of order p C1 so NG .H /=H has exactly one subgroup of order p. We assume that j1 .G/j > p (if j1 .G/j D p, then G Š Q16 ). It follows that H 1 .Z.G//=H is cyclic so j1 .Z.G//j D p 2 since H is cyclic and j1 .G/j > p. If G is minimal nonabelian, it is metacyclic as in Lemma 112.3(b2) with m n > 1. Next we assume that G is not minimal nonabelian. If G is of maximal class, then j1 .G/j D p since 1 .G/ Z.G/; then, as we know, G Š Q16 . Next we assume that G is not of maximal class; then 1 .G/ D 1 .Z.G// Š Ep2 and so G=1 .G/ is Dedekindian since all subgroups of G containing 1 .G/ are noncyclic so G-invariant. from [Pas] yields an almost full classiﬁcation of p-groups all of whose nonnormal subgroups are cyclic. See 16. 2 Deﬁning relations of G, namely, G D ha; b; c j a4 D b 4 D Œa; b D 1; c 2 D a2 ; ac D ab 2 ; b c D ba2 i; are given in Theorem 66.1. 3 Deﬁning relations of G, namely, G D ha; b; c; d j a4 D b 4 D Œa; b D 1; c 2 D a2 b 2 ; ac D a1 ; b c D a2 b 1 ; d 2 D a2 ; ad D a1 b 2 ; b d D b 1 ; Œc; d D 1i; are given in Theorem 16.2(ix).

144

Groups of prime power order

Assume that G=1 .G/ is nonabelian. By the previous paragraph, there exists Q8 Š Q=1 .G/ G=1 .G/ for some Q G and all maximal subgroups of Q containing 1 .G/ are abelian since 1 .G/ Z.G/; then jZ.Q/j D 8 and jQ0 j D 2 (Lemma 1.4). In this case, Q0 < 1 .G/, a contradiction since Q=1 .G/ Š Q8 is nonabelian. Thus G=1 .G/ is abelian so G 0 1 .G/. We conclude that G 0 is of order p 2 and exponent p. (a) By the previous paragraph, G 0 1 .G/ Š Ep2 , and since 1 .G/ Z.G/, we get cl.G/ D 2. If d.G/ D 2, then G is minimal nonabelian (Lemma 65.2(a)), contrary to the assumption. Thus we have d.G/ > 2 so G is not metacyclic. If p > 2, then, since .Ep2 Š/1 .G/ Z.G/, G has no minimal nonmetacyclic subgroups (Lemma 69.1(a,b)) so it is metacyclic, d.G/ D 2, contrary to what has just been said. Next we assume that p D 2 and d.G/ > 2. (b) Suppose that Q < G is nonabelian of order 8; then Q Š Q8 as 1 .Q/ Z.G/, and QGG since Q is noncyclic. Write C D CG .Q/; then G=C is isomorphic to a noncyclic subgroup of D8 , a Sylow 2-subgroup of Aut.Q/ Š S4 , the symmetric group of degree 4. As G 0 Z.G/ C , we get G=C Š E4 so G D Q C is a central product by the product formula since Q\C D Z.Q/ has index 4 in Q. Since G is not Dedekindian, there is a cyclic L C of order > 2. If Q \ L > ¹1º, then Q L has a nonnormal subgroup of order 2 < 2 , a contradiction. Thus Q \ L D ¹1º. Let elements u; v 2 Q generate different cyclic subgroups of order 4 and let x be a generator of L. If o.x/ > 4, then huxi and hux 2 i of different orders are not v-invariant, a contradiction. Thus exp.C / D 4 and so D 2. Suppose that C contains a minimal nonabelian subgroup, say D; then we obtain j1 .D/j j1 .G/j D 4 so D is metacyclic by Lemma 65.1; then jDj 16 since exp.D/ exp.C / D 4. If jDj D 8, then D Š Q8 and QD D Q D since Q \ L D ¹1º for a cyclic L < C of order 4. Then the diagonal subgroup of Q D (of order 8 > 4) is not normal in Q D, a contradiction. Thus D Š H2;2 D ha; b j a4 D b 4 ; ab D a3 i, the unique nonabelian metacyclic group of order 16 and exponent 4. Then Q \ D D Z.Q/ D hzi is a maximal cyclic subgroup of D of order 2 since j1 .G/j D 4; we have z D a2 b 2 . To obtain a contradiction, one may assume that G D Q D. Let u and v be generators of distinct cyclic subgroups of Q of order 4. Put H D hub; ai. We have .ub/2 D u2 b 2 D zb 2 D a2 b 2 b 2 D a2 ;

Œub; a D Œb; a D a2

so H Š Q8 . Since Œub; v D Œu; v D z ¤ a2 (indeed, Q \ hai D ¹1º), we get z 62 H , so H \ Q D ¹1º. As Q; H G G, we get F D QH D Q H . Then the diagonal subgroup X of F is not normal in F , a contradiction since jXj D 8. Thus D does not exist so C is abelian and d.C / 2 as 1 .C / 1 .Z.G// Š E4 . If C is abelian of type .4; 4/, there is in C a cyclic subgroup L of order 4 containing Q \ C D Z.Q/, contrary to what has already been proved. Thus C must be abelian

112

Nonnormal subgroups have the same order

145

of type .4; 2/ so, if C4 Š L1 < C , then L1 \ Q D ¹1º and G D Q L1 (indeed, by the product formula, jGj D 25 D jQ L1 j). In what follows we assume that G has no nonabelian subgroups of order 8. Let us show that D 2. Indeed, by (a), d.G/ > 2. Then G=G 0 contains a subgroup E=G 0 Š E8 . Since d.E/ > 2, E is neither abelian (since j1 .G/j D 4) nor minimal nonabelian. Let T < E be minimal nonabelian. We claim that T is metacyclic of order 16 and exponent 4. Indeed, 8 < jT j < jEj D jG 0 jjE=G 0 j 25

and 1 .T / 1 .G/ Š E4

so T is metacyclic (see (MA1) and (MA2)) and jT j D 16 and hence T Š H2;2 since exp.T / D 4. Since T has a nonnormal (cyclic) subgroup of order 22 , we get D 2. Then H Š C4 , jHG j D 2 and C2 Š HG ˆ.G/. By the above, GN D G=HG is a group from Theorem 112.3. (c) Let GN D G=HG D QN CN , where QN Š Q8 , QN \ CN D 1 .CN / and CN Š C2n , n > 1. Suppose that M < G is minimal nonmetacyclic. Since M has no nonabelian subgroup of order 8 and 1 .G/ Š E4 , we have jM j D 25 (Theorem 66.1). In this case, N D QN 2 .CN / has order 16 so 2 .G/ L and jLj D jHG jjLj N D 25 . LN D 2 .G/ 5 Since 2 .M / D M has order 2 (Theorem 66.1), we get L D M so 2 .G/ D M . Suppose that .2 .G/ D/M < G. In this case, there is in G a cyclic subgroup V of order 8. Then V G G (since D 2 by (b)) and Z D V \ M is cyclic of order 4 and G-invariant. It follows from the hypothesis and G 0 D M 0 Š E4 that G=Z is nonabelian and Dedekindian since D 2 and E4 Š G 0 6 Z; next, by assumption, we have jG=Zj > 8 (indeed, jGj > jM j D 25 ). Therefore M=Z Š Q8 so there are exactly four maximal subgroups of M that does not contain Z (indeed, M contains seven maximal subgroups and only three of them contain Z). Let K be one of such subgroups that is nonabelian (K exists since M has at most one abelian maximal subgroup in view of jM 0 j D 4). Then K \ V D 1 .V / as Z < V . Since M=1 .V / is nonabelian, it contains at most three abelian maximal subgroups by Exercise 1.6(a). Therefore one may assume that K is chosen so that K=1 .V / is nonabelian. We have K V =1 .V / D .K=1 .V // .V =1 .V //, and this group is non-Dedekindian by Theorem 1.20 since jV =1 .V /j D 4. Therefore since K=1 .V / 2 ¹D8 ; Q8 º, the group K V =1 .V / (of exponent 4) possesses a nonnormal subgroup, say T =1 .V /, of order 4.4 Since jT j D 8 > 4, we get a contradiction. Thus G D M is minimal nonmetacyclic as in part (c) of the theorem. (d) Suppose that G=HG D GN D DN QN is extraspecial of order 25 , where DN Š D8 and QN Š Q8 . Since HG < Z.G/ and GN is a monolith, we get G=Z.G/ Š E16 so jˆ.G/j 4 and exp.G/ D 4. Suppose that M < G is minimal nonmetacyclic. As in 4 This is clear if K= .V / Š Q (in that case, all subgroups of order 2 are normal there). In case 1 8 K=1 .V / Š D8 and R=1 .V / < K=1 .V / is noncentral so of order 2, R=1 .V / 1 .V =1 .V // of order 4 is not normal in K=1 .V /.

146

Groups of prime power order

part (c), we obtain jM j D 25 and M is special. Then Z.M / D M 0 D ˆ.M / Š E4 . It follows that ˆ.G/ D ˆ.M / so Z.G/ D ˆ.G/ D G 0 D M 0 and hence G is special of order 26 . If G is as in (c) or (d), then every subgroup of G of order > 4 contains 1 .G/ D G 0 so is G-invariant. This means that this G satisﬁes the hypothesis.5 Note that the central product G D U C , where U D ha; b j a4 D b 4 D 1; ab D a3 i Š H2;2

and C D hci

such that a2 D c 2 , has nonnormal subgroups of order 4 and 2 (for example, hbi and haci), however G=ha2 b 2 i Š D8 C4 . For completeness, we prove the following Proposition 112.5. Let G be a nonnilpotent group and a positive integer n be ﬁxed. Suppose that .H / D n for all nonnormal H < G.6 Then G D PQ is minimal nonabelian as in (MNA), where P 2 Sylp .G/ is cyclic, G 0 D Q 2 Sylq .G/, and one of the following holds: (a) jP j D p n , jQj D q. (b) n D 1, jP j D p, Q Š Eq 2 . Proof. Groups (a) and (b) satisfy the hypothesis. Now let M G be minimal nonnilpotent. Then M D P Q, where P 2 Sylp .M / is cyclic and Q D G 0 2 Sylq .M / (see Lemma 10.8). Since P is not normal in M so neither in G and .M / ¤ .P /, we get jP j D p n and so M G G. Then Q G G since Q is characteristic in M . Since a maximal subgroup of M containing P is not normal in M , it follows that P is maximal in M so Q is a minimal normal subgroup of M and M is minimal nonabelian. By Frattini’s lemma, G D NG .P /M D NG .P / Q is a semidirect product with kernel Q. As NG .P / is not normal in G, we get NG .P / D P so G D NG .P /Q D PQ D M . Thus G is a minimal nonabelian group. Let n > 1. Then jQj D q (otherwise, G also has a nonnormal subgroup of order q and .q/ D 1 < n) so jGj D p n q. Let n D 1 and jQj > q. Then Q has no proper subgroup of order q 2 so jQj D q 2 . Also, G of order pq 2 satisﬁes the hypothesis (this group is as in (b)). Proposition 112.5 is a generalization of the nonnilpotent part of the following. Corollary 112.6 (O. Schmidt [Sch3]). If G is a nonnilpotent group with nc.G/ D 1, then G is minimal nonabelian as in Proposition 112.5(a). 5 It is noticed in Theorem 16.2(ix) that all maximal subgroups of the group G from part (d) are minimal nonmetacyclic. Let us prove this. If S 2 1 , then S (of order 25 ) is nonmetacyclic since exp.G/ D 4. If M S is minimal nonmetacyclic, then, as in the ﬁrst paragraph of (d), we get jM j D 25 D jS j so M D S. 6 Recall that .H / is the number of prime factors of n, i.e., ƒ.H / D a C C a provided that 1 k Q jH j D kiD1 p ai is the standard decomposition.

112

Nonnormal subgroups have the same order

147

The following proposition is also a particular case of Theorem 16.2. Proposition 112.7. Let G be a non-Dedekindian two-generator p-group of order > p 3 with cyclic derived subgroup G 0 . If all nonnormal subgroups of G are cyclic, then one of the following holds: (a) G is minimal nonabelian and metacyclic as in (MA2). (b) p D 2 and either G Š Q16 or G D ha; b j a8 D 1; a4 D b 4 ; ab D a3 i is of order 25 . Conversely, all nonnormal subgroups of groups from (a) and (b) are cyclic. Proof. First we check the last assertion. Let G be minimal nonabelian metacyclic as in (MA2). Then G 0 < 1 .G/ Š Ep2 so all nonnormal subgroups of G do not contain 1 .G/. If H < G is nonnormal, then j1 .H /j D p hence H , being abelian, is cyclic (Proposition 1.3) so the groups from (a) satisfy the hypothesis. Now assume that G is as in (b). Clearly, Q16 satisﬁes the hypothesis. Now let G be the second group in (b). If H < G is nonnormal and noncyclic, H 6 2 .G/ as 1 .G/ D R G G is the unique proper noncyclic subgroup of 2 .G/ Š C4 C2 . Thus exp.H / D 8.D exp.G// so H 2 1 , a contradiction. Next we ﬁnd the groups satisfying the hypothesis. Suppose that G is not a 2-group of maximal class (otherwise, G Š Q16 ). In this case, there is in G a normal subgroup R Š Ep2 ; then all subgroups of G=R are normal so G=R is Dedekindian. If p > 2, then G=R is abelian. Since G 0 R and G 0 is cyclic, we get jG 0 j D p so G is minimal nonabelian (Lemma 65.2(a)). In this case, G is metacyclic (otherwise, G has a nonnormal abelian subgroup of type .p; p/ which is noncyclic). Next we assume that p D 2 and G is not minimal nonabelian (see the ﬁrst paragraph); then we have jG 0 j > 2 (Lemma 65.2(a)). It remains to show that then G is as the second group in (b). We have G 0 6 R so G=R is nonabelian Dedekindian, and since d.G=R/ D 2, we get G=R Š Q8 , jGj D 25 and R < ˆ.G/. Assume that G has a subgroup E Š E8 ; then E Z.G/ since all subgroups of E of order 4 are G-invariant. In this case, G is minimal nonabelian, a contradiction. Thus G has no subgroup Š E8 so j2 .G/j D 8 and G (of order 25 ) is as in Lemma 42.1(c) and hence has deﬁning relations given in (b). Remark. Let p > 2 and let G be a nonabelian p-group all of whose nonnormal subgroups are cyclic. As in the proofs of Theorems 112.3 and 112.4, G has no subgroup Š Ep3 so G is a group from Theorem 13.7. If R G G is abelian of type .p; p/, then G=R is abelian by Theorem 1.20. If G is metacyclic, then, being cyclic, jG 0 j D p and G is minimal nonabelian by Lemma 65.2(a). If G is a 3-group of maximal class and jGj > 33 , then jGj D 34 , 1 .G/ Š E9 . If G D EC , where E D 1 .G/ Š S.p 3 / and C is cyclic, then all subgroups of G of type .p; p/ are normal so jG 0 j D p. Then, by Lemma 4.3, G D E CG .E/, where CG .E/ is cyclic. Problem. Classify the p-groups that have no three nonnormal subgroups of pairwise distinct orders.

113

The class of 2-groups in 70 is not bounded

Suppose that all maximal subgroups of a nonabelian p-group G are generated by two elements but d.G/ D 3. By Theorem 70.2, if such a group G is of class > 2, then p D 2, jGj 27 , and G=K3 .G/ is isomorphic to the Suzuki group of order 26 given in Theorem 70.1(d). If G is of class 3, then these groups are completely determined in 70, where in particular for p > 2, jGj p 5 and G is of class 2 and for p D 2, jGj 28 . Theorem 113.1 (E. Crestani). Let G be a 2-group all of whose maximal subgroups are generated by two elements but d.G/ D 3. Then the nilpotence class of G is not bounded. Proof. Eleonora Crestani from University of Padova has invented the following ingenious construction of a 2-group Gn of order 24nC4 for each ﬁxed n 2 such that d.Gn / D 3, each maximal subgroup of Gn is generated by two elements and the class of Gn is at least n. Therefore the nilpotence class of such groups is not bounded. In fact, E. Crestani has also proved that the nilpotence class of Gn is exactly 2n, but that would require further tedious computations. In order to construct our 2-group Gn for each ﬁxed n 2, we start with the homocyclic group H D hg4 ; g5 ; g6 ; g7 i Š C2n C2n C2n C2n of order 24n . Let K be the splitting extension of H by the cyclic group hg3 i Š C4 of order 4, where g3 induces the automorphism of order 4 on H given by (1) g34 D 1;

g4g3 D g41 g62 ;

g5g3 D g51 g72 ;

g6g3 D g6 g41 ;

g7g3 D g7 g51 :

g2

We compute gi 3 D gi1 for i D 4; 5; 6; 7 and so g32 inverts H which implies that all elements in Hg32 are involutions. Let L D Khg2 i be the cyclic extension of order 2 of K, where (2)

g22 D g6 g72 ;

g4g2 D g41 g62 g74 ; g5g2 D g51 g62 g74 ;

g6g2 D g6 g42 g52 ; g7g2 D g7 g4 g5 ;

g3g2 D g31 g7 :

Here we have to show that g2 induces an automorphism on K D hg3 ; g4 ; g5 ; g6 ; g7 i. It is enough to see that gig2 , i D 3; 4; : : : ; 7, generate K (which is obvious) and then

113

149

The class of 2-groups in 70 is not bounded

we have to show that the relations (1) for K remain invariant when we replace gi with gig2 , i D 3; : : : ; 7. For example, .g3g2 /2 D .g31 g7 /2 2 Hg32 (since K=H Š C4 ) and we know that all elements in Hg32 are involutions and so o.g31 g7 / D 4. Then we g2 verify .g4g2 /g3 D .g4g2 /1 .g6g2 /2 . Indeed, g2

1

.g4g2 /g3 D .g41 g62 g74 /g3

g7

2

D .g41 g62 g74 /.g3 g3 g7 /

D .g4 g62 g74 /g3 g7 D g41 g62 g62 g42 g74 g54 D g43 g54 g64 g74 ; and for the right-hand side we get the same result: .g4g2 /1 .g6g2 /2 D g4 g62 g74 g62 g44 g54 D g43 g54 g64 g74 : In the same way we show that the last three relations in (1) remain invariant under the above replacement. Then we have to show that g22 acts the same way on K as the inner automorphism of K induced by g6 g72 (since g22 D g6 g72 ). First we compute that g22 ﬁxes each of the elements g4 ; g5 ; g6 ; g7 in H and then we show that g2

g g72

g 3 2 D g3 6

:

Finally, Gn D Lhg1 i, where (3)

g12 D g32 g61 g73 ; g4g1 D g41 g62 g76 ; g5g1 D g51 g62 g74 ; g6g1 D g6 g42 g53 ; g7g1 D g7 g4 g5 ;

g2g1 D g2 g32 ;

g3g1 D g31 g6 :

We proceed in exactly the same way as in the previous paragraph. First we show that g1 induces an automorphism on L D hg2 ; g3 ; g4 ; g5 ; g6 ; g7 i, where obviously gig1 , i D 2; : : : ; 7, generate L. We only check that the relations (1) and (2) for L remain invariant when we replace all gi with gig1 , i D 2; 3; : : : ; 7. Then we check that g12 acts the same way on the generators g2 ; g3 ; : : : ; g7 of L as the inner automorphism of L induced by g32 g61 g73 (since g12 D g32 g61 g73 ). The group Gn is constructed. Our group Gn exists because it is obtained with three consecutive cyclic extensions (of orders 4, 2, 2) of the homocyclic group H Š C2n C2n C2n C2n , n 2. From our relations (1), (2) and (3) we see that the subgroups H; K; L are normal in Gn . It is easy to see that the nilpotence class of Gn is at least n. Indeed, the involution g32 inverts the cyclic subgroup hg4 i of order 2n and so hg4 ; g32 i Š D2nC1 is a dihedral group of order 2nC1 and class n. Hence Gn is of class at least n. We show that Gn0 D hg32 ; g4 ; g5 ; g6 ; g7 i D H hg32 i and since G=Gn0 Š E8 , we also have ˆ.Gn / D Gn0 , d.Gn / D 3 and Gn D hg1 ; g2 ; g3 i. Indeed, H D hg4 ; g5 ; g6 ; g7 i and K D H hg3 i are normal in Gn and so H hg32 i is also normal in G as K=H Š C4 . From our relations (1), (2), (3) we get g12 ; g22 ; g32 2 H hg32 i and Œg2 ; g1 D g32 ; Œg3 ; g1 D g32 g6 ; Œg3 ; g2 D g32 g7 ; Œg3 ; g6 D g4 ; Œg3 ; g7 D g5 ; which shows that Gn =.H hg32 i/ Š E8 and Gn0 D H hg32 i D ˆ.G/. Thus d.Gn / D 3 and Gn D hg1 ; g2 ; g3 i.

150

Groups of prime power order

It remains to prove that each maximal subgroup of Gn is generated by two elements. Our seven maximal subgroups of Gn are Hi D Ti Gn0 , where Ti (i D 1; 2; : : : ; 7) is one of the subgroups: hg1 ; g2 i; hg1 ; g3 i; hg2 ; g3 i; hg1 g2 ; g3 i; hg1 ; g2 g3 i; hg2 ; g1 g3 i; hg1 g2 ; g1 g3 i: Here we shall freely use the relations (1), (2), (3) (without quoting). (i) H1 D hg1 ; g2 iGn0 D hg1 ; g2 i since hg1 ; g2 i Gn0 . Indeed, we have Œg1 ; g2 D g32 ;

g12 D g32 g61 g73 ;

g22 D g6 g72

and so Œg1 ; g2 g12 g22 D g71 which shows that g32 ; g6 ; g7 2 hg1 ; g2 i. Furthermore, we have g6g1 .g72 /g1 D g6 g72 g51 and so g5 2 hg1 ; g2 i. Finally, g7g1 D g7 g5 g4 and so g4 2 hg1 ; g2 i and we are done. (ii) H2 D hg1 ; g3 iGn0 D hg1 ; g3 i since hg1 ; g3 i Gn0 which shows the following relations (in this order): Œg3 ; g1 D g32 g6 ;

g12 D g32 g61 g73 ;

g6g1 .g72 /g1 D g6 g72 g51 ;

g7g1 D g7 g5 g4 :

(iii) H3 D hg2 ; g3 iGn0 D hg2 ; g3 i since hg2 ; g3 i Gn0 which show the relations Œg3 ; g2 D g32 g7 ;

g7g3 D g7 g51 ;

g7g2 D g7 g4 g5 ;

g22 D g6 g72 :

(iv) H4 D hg1 g2 ; g3 iGn0 D hg1 g2 ; g3 i since hg1 g2 ; g3 i Gn0 . Indeed, we have .g1 g2 /2 D g5 g7 , .g5 g7 /g3 D g52 g73 and so g32 ; g5 ; g7 2 hg1 g2 ; g3 i. Further, we get Œg3 ; g1 g2 D g42 g6 .g51 g71 / and so g42 g6 2 hg1 g2 ; g3 i. Finally, we conclude .g42 g6 /g3 D g4 g63 2 hg1 g2 ; g3 i and so g4 ; g6 2 hg1 g2 ; g3 i. (v) H5 D hg1 ; g2 g3 iGn0 D hg1 ; g2 g3 i D T5 as T5 Gn0 . Indeed, Œg2 g3 ; g1 D g6 which implies that g6 2 T5 . Further, g12 D g32 g61 g73 implies that g32 g73 2 T5 , .g2 g3 /2 D g73 g51 g6 gives g73 g51 2 T5 and then also g32 g51 2 T5 . From the relations g6g1 D g6 g42 g53 and g6g2 g3 D g4 g52 g74 g63 it follows g42 g53 2 T5 , g4 g52 g74 2 T5 and then also g41 g51 g74 2 T5 . In view of .g4 g5 g74 /g1 D g43 g53 g72 2 T5 , we obtain that .g41 g51 g74 /3 .g43 g53 g72 / D g710 and so g72 2 T5 which also forces g4 g5 2 T5 . Then by the above, the relation .g4 g5 /2 .g42 g53 / D g51 gives that g5 2 T5 . But from the above facts it also follows that g4 ; g32 ; g7 2 T5 and we are done. (vi) H6 D hg2 ; g1 g3 iGn0 D hg2 ; g1 g3 i D T6 as T6 Gn0 . Indeed, Œg2 ; g1 g3 D g71 and so g7 2 T6 . From g22 D g6 g72 it follows that g6 2 T6 and .g1 g3 /2 D g32 g41 g73 yields g32 g41 2 T6 . Further, the relation g7g2 D g7 g4 g5 implies g4 g5 2 T6 . From g6g1 g3 D .g4 g53 /g63 g76 it follows g4 g53 2 T6 which together with a previous result gives g52 2 T6 . Finally, g7g1 g3 D g41 g52 g62 g73 implies with the previous results that g4 ; g5 ; g32 2 T6 and we are done. (vii) H7 D hg1 g2 ; g1 g3 iGn0 D hg1 g2 ; g1 g3 i D T7 since T7 Gn0 . Indeed, from .g1 g2 /2 D g5 g7 and .g1 g3 /2 D g32 g41 g73 we conclude that t1 D g5 g7 2 T7 and

113

The class of 2-groups in 70 is not bounded

151

t2 D g32 g41 g73 2 T7 . We compute Œg1 g2 ; g1 g3 D g32 g42 g51 g6 g71 and this gives t3 D g32 g42 g51 g6 g71 2 T7 . Further, .g5 g7 /g1 g3 D g43 g55 g64 g75 and so t4 D g43 g55 g64 g75 2 T7 . Finally, we obtain .g32 g41 g73 /g1 g2 D g32 g53 g62 g75 which implies that t5 D g32 g53 g62 g75 2 T7 . Now, the ﬁve elements t1 ; t2 ; t3 ; t4 ; t5 also lie in Gn0 and note that we have ˆ.Gn0 / D 2 2 2 2 hg4 ; g5 ; g6 ; g7 i so that Gn0 =ˆ.Gn0 / Š E25 . We claim that ht1 ; t2 ; t3 ; t4 ; t5 i D Gn0 and so Gn0 T7 , as required. Indeed, considering the cosets ti ˆ.Gn0 / (i D 1; : : : ; 5), we see that t1 ˆ.Gn0 / contains g5 g7 , t2 ˆ.Gn0 / contains g32 g4 g7 , t3 ˆ.Gn0 / contains g32 g5 g6 g7 , t4 ˆ.Gn0 / contains g4 g5 g7 and t5 ˆ.Gn0 / contains g32 g5 g7 and these new ﬁve elements are obviously linearly independent mod ˆ.Gn0 /.

114

Further counting theorems

In what follows let e.X/ be the number of solutions of x p D 1 in a p-group X and let 1 .X/ be the set of all maximal subgroups of a p-group X. In this section we closely follow [HupB, VIII.5]. Theorem 114.1 ([Bla10] = [HupB, Theorem VIII.5.1]). Let G be a p-group and let ¹1º < N 1 .Z.G//. Then X e.G=M / D .j1 .N /j 1/e.G=N / C e.G/: M 21 .N /

Proof. Let k D j1 .N /j; then jN j D .p 1/k C 1 since N is elementary abelian. If Z N is of order p, then j1 .N=Z/j D

jN j p .p 1/k C 1 p k1 jN W Zj 1 D D D : p1 p.p 1/ p.p 1/ p

Let Y D ¹.x; M / j x 2 G N; M 2 1 .N /; x p 2 M º: For x 2 G N and x p 2 N , let Yx D ¹M 2 1 .N / j x p 2 M º: Clearly, jYx j is the number of maximal subgroups of the quotient group N=hx p i. Thus p jYx j D k if x p D 1 and jYx j D k1 p if x ¤ 1 (see the ﬁrst displayed formula in the proof). Hence X X X X k1 X k1 k1 D ; jYx j D kC jYj D k C p p p p p p x x x D1

x ¤1

x D1

where x runs through the set of all elements of G N for which x p 2 N , i.e., all x with .xN /p D N . Thus the number of x 2 G N for which x p 2 N is equal to .e.G=N / 1/jN j. Hence k1 k1 jYj D .e.G/ jN j/ k C .e.G=N / 1/jN j p p (1) k1 k1 kjN j: D e.G/ k C e.G=N /jN j p p

114

153

Further counting theorems

On the other hand, suppose that M 2 1 .N /. The number of elements x 2 G N for which x p 2 M is .e.G=M / p/jM j, so X

jYj D

.2/

.e.G=M / p/

M 21 .N /

jN j : p

Comparing the two expressions for jYj in (1) and (2), we obtain e.G/

.p 1/k C 1 k1 C e.G=N /jN j kjN j D p p

X

.e.G=M / p/

M 21 .N /

jN j ; p

or, since jN j D .p 1/k C 1, k D j1 .N /j, we get e.G/

k1 jN j C e.G=N /jN j kjN j D p p

X

e.G=M / kjN j;

M 21 .N /

and we conclude that e.G/ C e.G=N /.j1 .N /j 1/ D

X

e.G=M /;

M 21 .N /

as was to be shown. Thus e.G/ D

X

e.G=M / .j1 .N /j 1/e.G=N /:

M 21 .N /

Lemma 114.2 ([HupB, Lemma VIII.5.2]). Suppose that a p-group G D AB (central product). Let D D A \ B, jDj D p and exp.A=D/ D exp.B=D/ D p. Then 1 .jAj e.A//.jBj e.B// e.G/ D e.A/e.B/ C : p p1 Proof. We have G Š .A B/=h.x; x 1 /i for a generator x of D. Thus pe.G/ D j¹.a; b/ 2 A B j ap D x i ; b p D x i for some i 2 ¹0:1; : : : ; p 1ººj: If i D 0, the number of solutions of ap D x i .D 1/ is e.A/; otherwise, it is equal to jAje.A/ since exp.A=D/ D p. Similar statements hold for B. Hence p1 jAj e.A/ jBj e.B/ p1 p1 .jAj e.A//.jBj e.B// ; D e.A/e.B/ C p1

pe.G/ D e.A/e.B/ C .p 1/

completing the proof.

154

Groups of prime power order

We have, by Lemma 114.2, 1 e.Q8 Q8 / D .2 2 C .8 2/2 / D 20; 2 1 e.D8 D8 / D .62 C .8 6/2 / D 20; 2 1 e.Q8 D8 / D .2 6 C 6 2/ D 12: 2 Theorem 114.3 ([HupB, Lemma VIII.5.3]). Suppose that G is a 2-group, d.G/ D d and jˆ.G/j D 2. Then e.G/ 2 ¹2d ; 2d ˙2d r º, where r is a positive integer satisfying 1 2r d . Further, if e.G/ D 2d ˙ 2d 2 d , then G is extraspecial. Proof. Clearly, exp.G/ D 4. If G is abelian, then e.G/ D 2d . Now let G be nonabelian. Then, by Lemma 4.2, G D EZ.G/, where E is extraspecial. Let E D ES.m; 2/ and d.Z.G// D s; then d D 2m C s , where D 0 if exp.Z.G// D 4 and D 1 if exp.Z.G// D 2. We have cd.E/ D ¹ .1/ j 2 Irr.G/º D ¹1; 2m º: By the Frobenius–Schur formula (see [BZ, 4.6]), we get e.E/ D 22m ˙ 2m (indeed, Irr1 .E/ D ¹ º, where .1/ D 2m ). Suppose that exp.Z.G// D 2. Then G D E T , where T Š E2s1 , and so (3)

e.G/ D e.E/e.T / D .22m ˙ 2m /2s1 D 22mCs1 ˙ 2mCs1 :

In this case, d D 2m C s 1;

m C s 1 D .2m C s 1/ m D d m;

2m d:

Now suppose that exp.Z.G// D 4. Then Z.G/ D C4 E2s1 , d.Z.G// D s and e.Z.G// D 2s . Let us apply Lemma 114.2 with A D E and B D Z.G/. We have (4)

2e.G/ D 2s Œ22m ˙ 2m C 22mC1 .22m ˙ 2m / D 22mCsC1

so e.G/ D 22mCs D 2d , and we obtain again the desired result. 1 Now assume that e.G/ D 2d ˙ 2d 2 d . Then, by (3) and (4), exp.Z.G// D 2 and d D 2m C s 1;

1 d DmCs1 2

(by (3))

so that 2m C s 1 D 2.m C s 1/. It follows that s D 1 so G D EZ.G/ D E is extraspecial, completing the proof. Theorem 114.4 ([HupB, Theorem VIII.5.4]). Suppose that G is a 2-group of class at most 2 and exponent at most 4. Suppose that ¹x 2 G j x 2 D 1º D N is a subgroup of G. Then jGj jN j3 ; indeed, jGj < jN j3 if G is not special.

114

Further counting theorems

155

Proof. Since the theorem obviously holds if G is abelian (in this case, jGj jN j2 j), we will assume, in what follows, that G is nonabelian. We prove the theorem by induction on jN j. If jN j D 2, then G is isomorphic to Q8 (Proposition 1.3), and the assertion is clear. Suppose then that jN j > 2. Clearly, ˆ.G/ N as exp.G=N / < exp.G/ D 4. First suppose that G 0 < N . We have N L where L=G 0 D 1 .G=G 0 /. Since L=G 0 is elementary abelian, we get L=G 0 D .N=G 0 / .K=G 0 / for some K L. As G 0 < N , we get K < L so induction may be applied to K. Thus jKj j1 .K/j3 D jN \ Kj3 D jG 0 j3 so jLj D jN jjKj=jG 0 j jN jjG 0 j2 . But jL W G 0 j D j1 .G=G 0 /j D jG=G 0 W Ã1 .G=G 0 /j D jG W ˆ.G/j so jG W Lj D jˆ.G/ W G 0 j. Hence jGj D jG W LjjLj D jˆ.G/ W G 0 jjLj jˆ.G/ W G 0 jjN jjG 0 j2 D jˆ.G/jjN jjG 0 j jN jjN jjG 0 j D jN j2 jG 0 j < jN j3 since G 0 < N by assumption. Suppose that G 0 D N ; then ˆ.G/ D N also in view of G 0 ˆ.G/ N D G 0 . Since G is of class 2, we obtain that N Z.G/. Suppose next that N < Z.G/. Choose z 2 Z.G/ N . Let M be a maximal subgroup of G not containing z (M exists since z 62 ˆ.G/). Thus G D M hzi; M \ hzi D Z; where Z D hz 2 i ¤ ¹1º (indeed, 1 .G/ D N < Z.G/ so o.z/ > 2). We apply induction to M=Z. If x 2 M and x 2 2 Z, we have x 2 D .z 2 /m D z 2m for some m and xz m 2 N since .xz m /2 D x 2 z 2m D x 2 x 2 D 1). Hence z m 2 M \ hzi D Z j3 and xZ 2 N=Z. Thus jM=Zj jN=Zj3 , and this implies jM j jN ; then jZj2 jGj D 2jM j

2jN j3 1 D jN j3 < jN j3 2 jZj 2

since jZj D 2 (recall that exp.G/ D 4). Thus one may assume that G 0 D ˆ.G/ D Z.G/ D N ; then G is special. We have e.G/ D jN j D 2n , and e.G=N / D 2d , say. Next, j1 .N /j D 2n 1. Hence, by Theorem 114.1, X (5) e.G=M / D .j1 .N / 1/e.G=N / C e.G/ D .2n 2/2d C 2n : M 21 .N /

By Theorem 114.3, e.G=M / 2 ¹2d ; 2d ˙ 2d r º, where the positive integer r is such that 2r d . Substituting in (5), we obtain an equation of the form X 2nCd 2d C1 C 2n D .˙2si /; i

where si 12 d . Suppose that n < 12 d . Then si > n for is divisible by 2nC1 , which is a contradiction since 2nC1 does not divide 2n . Hence n 12 d and jGj D 2d Cn

all i, so the right-hand side j .2nCd 2d C1 /, but 2nC1 3n 2 D jN j3 .

156

Groups of prime power order

Theorem 114.5 ([HupB, Theorem VIII.5.5]). Let G be a nonidentity 2-group and suppose that Aut.G/ permutes the set Inv.G/ of involutions of G transitively. Further, let j1 .Z.G//j D q. Then (a) 1 .Z.G// D 1 .G/. (b) (Gross [Gro]) jGj D q m for some positive integer m. (c) (Shaw) If G is nonabelian and of exponent 4, then G is special and jGj 2 ¹q 2 ; q 3 º. Proof. Statement (a) is obvious. k1

(b) Given an involution t 2 G, write mk D j¹y 2 G j y 2 D tºj (k 1). Since Aut.G/ permutes the set of involutions of G transitively, mk is independent of t , and the number of elements of order 2k in G is .q 1/mk (k 1). Hence jGj 1 D .q 1/.m1 C m2 C / and so q 1 divides jGj 1. Let jGj D q m 2s , where 2s < q. Since 2s q m 1 2s .q m 1/ 2s 1 jGj 1 D D C q1 q1 q1 q1 is an integer, we get s D 0 so jGj D q m , proving (b). (c) Since exp.G/ D 4, we get exp.G=1 .Z.G/// D 2 hence G=Z.G/ is elementary abelian. Thus G 0 1 .Z.G//. As G is nonabelian, we have G 0 > ¹1º. By hypothesis, 1 .Z.G// is a minimal characteristic subgroup of G so G 0 D 1 .Z.G//. Therefore G=G 0 is elementary abelian so G 0 D ˆ.G/. If G 0 < Z.G/, then Z.G/ contains an element z of order 4. Now, if y 2 G Z.G/, y 2 is an involution. Since z 2 is also an involution, we obtain y 2 D .z 2 /˛ for some ˛ 2 Aut.G/. But then y 2 D .z ˛ /2 and y.z ˛ /1 2 Z.G/ so y 2 Z.G/, which is a contradiction. Hence G 0 D Z.G/ D 1 .G/. By (b), jGj D q m for some positive integer m, and by Theorem 114.4, m 3. Since G is nonabelian, m > 1 so m D 2 or 3.1

1 Both

values of m are possible.

115

Finite p-groups all of whose maximal subgroups except one are extraspecial

In a letter Z. Janko reported that he proved that if all maximal subgroups of a 2-group G except one are extraspecial, then G is of maximal class and order 24 . Below we consider the similar problem for all p. Recall that all maximal abelian subgroups of any extraspecial group of order p 2mC1 have the same order p mC1 . In particular, if this group possesses an abelian subgroup of index p, then m D 1. Theorem 115.1. All nonabelian maximal subgroups of a nonabelian p-group G except one are extraspecial if and only if G is either minimal nonabelian or of order p 4 . Proof. All nonabelian maximal subgroups of a nonabelian group G of order p 4 are extraspecial unless G is minimal nonabelian. In what follows we assume that G is not minimal nonabelian and jGj > p 4 (clearly, jGj D p 2m for some m > 2 by Theorem 4.7(a)). Then there is a nonabelian E 2 1 . Since, by Exercise 1.6(a), the number of nonabelian members of the set 1 is at least p > 1, one may assume that E is extraspecial. Assume that there is an abelian A 2 1 . Since A \ E is an abelian maximal subgroup of the extraspecial group E, we get jEj D p 3 (Theorem 4.7(d)); then jGj D pjEj D p 4 ; contrary to the assumption. Thus the set 1 has no abelian members. Since j1 j p C 1 3, there is an extraspecial F 2 1 ¹Eº. Note that E 0 G G and E 0 D ˆ.E/ ˆ.G/ < F so E 0 D F 0 since F 0 is a unique normal subgroup of order p in F . Thus all extraspecial members of the set 1 have the same derived subgroup E 0 . By what has been said in the previous paragraph, all members of the set 1 except one are extraspecial. Since G has a maximal subgroup whose center is of order > p and this subgroup is neither abelian nor extraspecial, it follows that the number of extraspecial members in the set 1 is divisible by p (indeed, we have j1 j D 1 C p C C p d.G/1 ). Set GN D G=E 0 . We consider the following two cases. (i) Suppose that GN is nonabelian. Since GN has at most one nonabelian member and the number of abelian maximal subgroups of GN is at least p, we conclude that G=E 0

158

Groups of prime power order

is minimal nonabelian (see the previous paragraph and Exercise 1.6(a)). As EN and FN N D GN so GN 2 ¹D8 ; S.p 3 /º, where S.p 3 / is are elementary abelian, we get 1 .G/ 3 N D p 4 , contrary nonabelian of order p and exponent p. In this case, jGj D jE 0 jjGj to the assumption. (ii) Thus GN is abelian. As GN D EN FN , it follows that GN is elementary abelian. Since N D p 2mC1 so GN Š Ep2mC1 and jGj D 22mC2 N jEj D p 2m for some m > 1, we get jGj 0 0 with G D E . Therefore G is not extraspecial, and we conclude that jZ.G/j > p. Since G=Z.G/ is noncyclic, it contains two distinct maximal subgroups M=Z.G/ and N=Z.G/. As, by assumption, M and N are nonabelian and, obviously, M and N are not extraspecial (indeed, jZ.G/j > p and Z.G/ M \ N ), we get a contradiction. Exercise 1. Classify the p-groups all of whose maximal subgroups except one are elementary abelian. Answer. The dihedral group D8 is the unique group with that property. Corollary 115.2. If all maximal subgroups of a nonabelian p-group G except one are extraspecial, then G is of maximal class and order p 4 . A p-group G is said to be a CZ-group if G 0 is its unique minimal normal subgroup. In this case, Z.G/ is cyclic. If x; y 2 G, then we infer that 1 D Œx; yp D Œx; y p so Ã1 .G/ Z.G/ and hence ˆ.G/ D G 0 Ã1 .G/ Z.G/, and we conclude that ˆ.G/ is also cyclic (see 4 on such G). A minimal nonabelian p-group X is a CZ-group if and only if its center is cyclic (in this case, X is either of order p 3 or X Š Mpn ). Obviously, extraspecial p-groups are CZ-groups. Let G be a CZ-group. By Lemma 4.2, G D .A1 An /Z.G/, where A1 ; : : : ; An are minimal nonabelian. In our case, Z.G/ is cyclic so Z.Ai / is cyclic for all i. Then it follows that Ai 2 ¹D8 ; Q8 ; S.p 3 /; Mpk º. At least one member of the set 1 is not a CZ-group. Theorem 115.3. Suppose that all maximal subgroups of a nonabelian p-group G of order p m > p 4 except one are CZ-groups and let F 2 1 be an arbitrary CZ-subgroup. Then Z.G/ is cyclic. (a) If G=F 0 is abelian, then G is a CZ-group.1 Next we assume that G=M 0 is nonabelian for all CZ-members M 2 1 . Then we have F 0 D M 0 for all such M since Z.G/ is cyclic. (b) If the quotient group G=F 0 is minimal nonabelian, then d.G/ D 2 and jG 0 j D p 2 . Also, G=Z.F / is abelian of type .p 2 ; p/ and Z.F / < ˆ.G/. (c) Suppose that G=Z.F / is abelian (or, what is the same, G 0 Z.F /). Then G=Z.F / is abelian of type .p 2 ; p/ and d.F / 3. If M 2 1 ¹F º is a CZ-subgroup, then Z.M / ¤ Z.F / and M=Z.F / is cyclic; moreover, M Š Mpm1 . 1 But

we do not assert that all CZ-groups satisfy the hypothesis.

115

Finite p-groups all of whose maximal subgroups except one are extraspecial 159

Proof. Assume that Z.G/ is noncyclic. Then all members of the set 1 that contain Z.G/ are not CZ-groups. Since G=Z.G/ is noncyclic, the set 1 has at least p C 1 > 1 members which are not CZ-groups, and this is a contradiction. Thus (i) Z.G/ is cyclic. By hypothesis, the set 1 has two distinct members F and H that are CZ-groups. By (i), F 0 D H 0 since F 0 ; H 0 Z.G/. Then, by Exercise 1.6(a), the set 1 .G=F 0 / contains at least p C 1 abelian members. Assume that the set 1 .G=F 0 / has a nonabelian member M=F 0 . Then M is not a CZ-group (otherwise, M 0 D F 0 so M=F 0 is abelian), and we conclude that d.G/ 3. Therefore, by Exercise 1.6(a), the set 1 .G=F 0 / has at least p 2 nonabelian members so the set 1 has at least p 2 members that are not CZ-groups, and this is a contradiction. Thus (ii) G=F 0 is either abelian or minimal nonabelian for any CZ-subgroup F 2 1 , D F 0 for all CZ-subgroups H 2 1 ¹F º.

H0

We consider two cases, mentioned in (ii), separately. A. Suppose that G=F 0 is abelian; then G 0 D F 0 . Since Z.G/ is cyclic by (i), G 0 is a unique minimal normal subgroup of G so G is a CZ-group. B. Suppose that G=F 0 is minimal nonabelian. Since F 0 ˆ.F / ˆ.G/, we get d.G/ D d.G=F 0 / D 2. By Lemma 65.1, d.F / D d.F=F 0 / 3, and this is true for any CZ-member of the set 1 . Due to Lemma 4.2 (or Lemma 4.3), F is a product of minimal nonabelian subgroup and Z.F / so F=Z.F / Š Ep2 hence jG=Z.F /j D p 3 . Since G=Z.F / is noncyclic and d.G/ D 2, we obtain Z.F / < ˆ.G/. Suppose that the quotient group G=Z.F / is abelian; then G=Z.F / (of order p 3 ) is abelian of type .p 2 ; p/ since d.G/ D 2. Let H 2 1 ¹F º be a CZ-subgroup. We have Z.H / ¤ Z.F / (otherwise, Z.F / D Z.G/ and so G is minimal nonabelian, a contradiction). Then the quotient group H=Z.F / (recall that Z.F / ˆ.G/ < H ) is cyclic of order p 2 (indeed, F=Z.F / is the unique noncyclic maximal subgroup of G=Z.F / which is abelian of type .p 2 ; p/) so H is metacyclic since Z.F / is cyclic (indeed, F is a CZ-group). In this case, H is minimal nonabelian (Lemma 65.2(a)). Since Z.H / is cyclic and jH j > p 3 , we get H Š Mpm1 . Remark 1. Suppose that a p-group G is neither abelian nor minimal nonabelian. Let 10 be the set of nonabelian members of the set 1 . It follows from Exercise 1.6(a) that j10 j p. We suppose that the set 10 contains p 1 non-extraspecial members. One may assume that jGj D p m > p 4 (groups of order p 4 satisfy the hypothesis). Then there is an extraspecial E 2 1 . Since jEj 25 , it follows that E has no abelian subgroup of index p, and we conclude, as above, that 10 D 1 so the set 1 contains p 1 non-extraspecial members. Because of j1 j p C 1, there is an extraspecial F 2 1 ¹Eº. As in the proof of Theorem 115.1, Z.G/ is cyclic. Since E 0 ; F 0 Z.G/, N of maximal subgroups of GN conwe get E 0 D F 0 . Set GN D G=E 0 . Then the set 1 .G/ tains two distinct (elementary) abelian members, therefore, by Exercise 1.6(a), it conN D EN FN D GN hence, if GN is abelian, tains at least p C 1 abelian members. Next, 1 .G/

160

Groups of prime power order

it has exponent p. Assume that GN is neither abelian nor minimal nonabelian. Then the N contains at least p nonabelian members whose inverse images in G are not set 1 .G/ extraspecial, contrary to the hypothesis. Thus, if GN is nonabelian, it is minimal nonN D p 3 (Lemma 65.1). (i) Assume that GN is elementary abelian; then abelian and so jGj G is not extraspecial since jGj D pjEj D p 2.kC1/ , where jEj D p 2kC1 . It follows that jZ.G/j > p. Since G=Z.G/ is noncyclic, all p C 1 members of the set 1 containing Z.G/, are not extraspecial, contrary to the hypothesis (recall that the set 1 has no abelian members). (ii) Now let GN is minimal nonabelian of order p 3 . Then we get 0 j D p 4 , contrary to the assumption. N jGj D jGjjE Recall that a p-group G is said to be an An -group if it has a nonabelian subgroup of index p n1 but all its subgroups of index p n are abelian. The p-groups which are A2 -groups are classiﬁed in 71. Remark 2. Let a p-group G be neither abelian nor minimal nonabelian. Suppose that all nonabelian members of the set 1 are either extraspecial or minimal nonabelian. Then G is an A2 -group. Indeed, by hypothesis, jGj > p 3 . If jGj D p 4 , it satisﬁes the hypothesis. In what follows we assume that jGj > p 4 . If all nonabelian maximal subgroups are minimal nonabelian, then G is an A2 -group. In the following we assume that there is an extraspecial E 2 1 . If A 2 1 is either abelian or minimal nonabelian, then E \ A is abelian of index p in E so jEj D p 3 . Then jGj D p 4 , contrary to the assumption. Thus all maximal subgroups of G are extraspecial. This is impossible since there is in G a maximal subgroup with center of order > p. Theorem 115.4. Let a p-group G be neither abelian nor minimal nonabelian. Suppose that all nonabelian members of the set 1 are either minimal nonabelian or of maximal class. Then one and only of the following holds: (a) G is an A2 -group. (b) G is a p-group of maximal class with abelian subgroup of index p. (c) G is a 3-group of maximal class with nonabelian fundamental subgroup G1 (in this case, G1 is minimal nonabelian). Proof. Suppose that G is not an A2 -group; then jGj > p 4 . In this case, there is in the set 1 a member M of maximal class. Then, by Theorem 12.12(a), G is either of maximal class or d.G/ D 3; so, in the second case, the set 1 contains exactly p C 1 members that are either abelian or minimal nonabelian (Theorem 13.6). (i) Suppose that G is of maximal class. One may assume that p > 3 as all p-groups of maximal class and order > p 4 , p 3, satisfy the hypothesis (see 9). If the set 1 has an abelian member, then all nonabelian members of the set 1 are of maximal class (Fitting’s lemma) so G satisﬁes the hypothesis. Next we assume that 1 has no abelian members. If jGj > p pC1 , then the fundamental subgroup G1 of G must be minimal nonabelian since it is not of maximal class. In this case, as p p1 D j1 .G1 /j p 3 (Lemma 65.1), we get p D 3, contrary to the assumption.

115

Finite p-groups all of whose maximal subgroups except one are extraspecial 161

Now let jGj D p m p pC1 ; then p > 3 since m > 4. In this case, the set 1 has at most two distinct members that are not of maximal class. Let A 2 1 be not of maximal class. Then A is minimal nonabelian. We have A0 D Z.G/. Since cl.G/ > 3, we conclude that A is a unique member of the set 1 which is not of maximal class (otherwise, cl.G=Z.G// D 2). As j1 .A/j p 3 , it follows that G has no subgroup of order p 5 and exponent p. Since m > 4, we get exp.A/ > p. Next, the metacyclic abelian subgroup A=A0 is the fundamental subgroup of G=A0 (recall that A0 D Z.G/). It follows that jG=A0 j D p 4 so jGj D p 5 and A=A0 is abelian of type .p 2 ; p/. Since exp.G=A0 / D p, we get a contradiction. (ii) Suppose that 1 contains exactly p C 1 members, say A1 ; : : : ; ApC1 , that are either abelian or minimal nonabelian. Then all intersections M \Ai (i D 1; : : : ; pC1) are abelian and have index p in M (here M 2 1 is of maximal class). Since m > 4, M contains at most one abelian subgroup of index p. It follows that M \ A1 D D M \ ApC1 so G=.M \ A1 / contains p C 2 distinct subgroups M=.M \ A1 /; A1 =.M \ A1 /; : : : ; ApC1 =.M \ ApC1 / of order p, a contradiction since the abelian group G=.M \A1 / of type .p; p/ contains exactly p C 1 subgroups of order p. Exercise 2. Classify the p-groups all of whose nonabelian maximal subgroups are either of maximal class or have derived subgroup of order p. (Hint. Use 137.) Exercise 3. Classify the p-groups all of whose nonabelian maximal subgroups are either absolutely regular or of maximal class. (Hint. Use Theorem 12.1(b).) Exercise 4. Classify the p-groups all of whose nonabelian maximal subgroups are either metacyclic or of maximal class. Exercise 5. Classify the p-groups all of whose nonabelian members of the set 2 are either minimal nonabelian or of maximal class. Exercise 6. Classify the p-groups containing p extraspecial maximal subgroups.

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Groups covered by few proper subgroups

1o Introduction. We say that a group G is covered by proper subgroups A1 ; : : : ; An if (1)

G D A1 [ [ An :

We have, in (1), G > ¹1º and n > 1. A group is covered by its proper subgroups if and only if it is noncyclic (check!). Every noncyclic group is covered by (proper) cyclic subgroups. A group is not covered by two proper subgroups. Covering (1) is said to be irredundant if every proper subset of the set ¹A1 ; : : : ; An º does not cover G. In what follows we assume that (1) is an irredundant covering of G by proper subgroups. Remark 1. If, in (1), jA1 j jAn j, then jGj jA1 jn .n 1/ < njA1 j and hence jG W A1 j < n. Let M be a maximal subset (with respect to inclusion) of pairwise noncommuting elements of a nonabelian group G. We denote the set of all such subsets by ƒ.G/. Write .G/ D max¹jMj j M 2 ƒ.G/º: For any abelian group G we set .G/ D 1. If H is a subgroup of G, then we have .H / .G/. Recall that two groups G and G1 are lattice isomorphic if there is a bijective mapping of the set of subgroups of G onto the set of subgroups of G1 such that, provided F; H G, we have .F \ H / D F \ H and hF; H i D hF ; H i. If groups G and G1 are lattice isomorphic, then the inequality .G/ ¤ .G1 / is possible owing to the fact that some nonabelian groups are lattice isomorphic to abelian groups (indeed, n the group MpnC1 is lattice isomorphic to the abelian group S of type .p ; p/). As Lemma 116.2(a) shows, if M 2 ƒ.G/, then G D x2M CG .x/ and this covering is irredundant. Every nonabelian group contains two noncommuting elements. If a; b 2 G satisfy ab ¤ ba, then the elements a; b; ab are pairwise noncommuting, i.e., .G/ 3. For p-groups one can prove a stronger result. Lemma 116.1. Let G be a nonabelian p-group. Then (a) If G is minimal nonabelian, then .G/ D p C 1. (b) .G/ p C 1.

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Groups covered by few proper subgroups

163

Proof. (a) Since all maximal subgroups of G are abelian, any two noncommuting elements of G are contained in distinct maximal subgroups of G. Thus .G/ p C 1. If M1 ; : : : ; MpC1 are all the maximal subgroups of G and xi 2 Mi ˆ.G/ for all i , then, for i ¤ j , hxi ; xj i D G is nonabelian, so xi xj ¤ xj xi . Hence x1 ; : : : ; xpC1 are pairwise noncommuting elements so .G/ p C 1, completing the proof. (b) Let H be a minimal nonabelian subgroup of G. Then .H / D p C 1 by (a), and so .G/ .H / D p C 1. The following lemma establishes a connection between members of the set ƒ.G/ with some irredundant coverings of a nonabelian group G. Part (b) of this lemma also shows that the members of ƒ.G/ of cardinality .G/ have a special property. If G is a minimal nonabelian p-group of order p n , then jƒ.G/j D .p n1 p n2 /pC1 ; but G has only one irredundant covering (see Lemma 116.3(b) and Theorem 116.5). Lemma 116.2. Let G be a nonabelian group and M 2 ƒ.G/. Then (a) We have

[

(2) If N M1 2 ƒ.G/ and an irredundant covering.

S

CG .x/ D G:

x2M

x2N

CG .x/ D G, then N D M1 ; in particular, (2) is

S (b) Suppose, in addition, that jMj D .G/. If x 2 M, then hG y2M¹xº CG .y/i is abelian. S Proof. (a) Assume that there is g 2 G x2M CG .x/. Since M M [ ¹gº, it follows from the maximality of M that S gx D xg for some x 2 M; then g 2 CG .x/, contrary to the choice of g. Thus x2M CG .x/S D G. Now assume that there is u 2 M such that x2M¹uº CG .x/ D G. Then there exists v 2 M ¹uº such that u 2 CG .v/, so that u; S v are distinct commuting members of the set M, a contradiction. Thus the covering x2M CG .x/ D G is irredundant. S (b) Given x 2 M, set D D G y2M¹xº CG .y/. The set D is nonempty. Assume that there are noncommuting u; v 2 D. By the choice, every element of the set M¹xº does not commute with u and v. It follows that .M ¹xº/ [ ¹u; vº M2 2 ƒ.G/, a contradiction since jM2 j > jMj D .G/. Thus all elements of the set D commute so the subgroup hDi is abelian. It follows from Lemma 116.2(a) that if H < G is such that .H / D .G/ and if we have M D ¹x1 ; : : : ; x .G/ º 2 ƒ.H /, then there are i ¤ j with CG .xi / 6 H and CG .xj / 6 H . Indeed, there is an index i such that CG .xi / 6 H since .G/ [ kD1

CG ..xk / D G > H

(Lemma 116.2(a)).

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Groups of prime power order

If for all j ¤ i we have CG .xj / H , then H [ CG .xi / D G, which is impossible since H and CG .xj / are nonincident. 2o p-groups. In this section we study the p-groups containing a maximal subset (with respect to inclusion) of pairwise noncommuting elements of cardinality p C 1. Some related results are also established and discussed. Next, we study the p-groups which are covered by 2p proper subgroups. It is proved that if a p-group G admits an irredundant covering by p C 2 subgroups, then p D 2. A noncyclic p-group G admits an irredundant covering by pC1 maximal subgroups (indeed, if T G G is such that G=T is abelian of type .p; p/, then p C 1 maximal subgroups of G containing T cover G). Moreover, Lemma 2.1 showsTthat if a p-group pC1 G is covered by p C 1 proper subgroups A1 ; : : : ; ApC1 , then jG W i D1 Ai j D p 2 , i.e., all Ai are maximal in G. Lemma 116.3 is known; however, we prove it to make our exposition self-contained. Lemma 116.3. Suppose that a noncyclic p-group G of order p m is covered by n proper subgroups A1 ; : : : ; An as in (1). Then (a) n p C 1. (b) If n D p C 1, then the covering (1) is irredundant and jG W In particular, the Ai are maximal in G.

TpC1 i D1

Ai j D p 2 .

Proof. (a) If n p, then ˇ ˇ n ˇX # ˇ m1 ˇ ˇ A 1/ D p m p < jG # j; i ˇ p.p ˇ iD1

a contradiction. (b) Now let n D p C 1. Then the covering (1) is irredundant by (a). First assume that A1 ; : : : ; ApC1 are maximal in G; then jAi \ Aj j D p m2 for i ¤ j . We have ! p [ .Ai ApC1 / : (3) G D ApC1 [ iD1

Since Ai Aj D Ai .Ai \ Aj / for i ¤ j , the right-hand side of (3) contains at most p m1 C p.p m1 p m2 / D p m D jGj elements so (3) is a partition of G. Then it follows that Ai \ ApC1 D Aj \ ApC1 and .Ai ApC1 / \ .Aj ApC1 / D ¿ for all distinctTi; j < p C 1 (indeed, one pC1 can take in (3) Ak instead of ApC1 ). We conclude that i D1 Ai D A1 \ ApC1 has 2 index p in G. It follows from the above computation (see the displayed formula after (3)) that, in fact, all subgroups A1 ; : : : ; ApC1 must be maximal in G (otherwise, we obtain that SpC1 j i D1 Ai j < jGj).

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165

It follows from Lemmas 116.2 and 116.3 that if G is a nonabelian p-group, then we have .G/ pC1. In Theorem 116.5(b), the p-groups G which satisfy .G/ D pC1 are classiﬁed. Lemma 116.4. Let H be a minimal nonabelian subgroup of a p-group G. Then the intersection ƒ.H / \ ƒ.G/ is not empty if and only if G D H CG .H / (central product); in this case, ƒ.H / ƒ.G/. Proof. (i) Let M 2 ƒ.H / \ ƒ.G/; then we have S jMj D p C 1 (Lemma 116.1(a)). By hypothesis and Lemma 116.2(a), we get G D x2M CG .x/ so, by Lemma 116.3(b), T T jG W x2M CG .x/j D p 2 . Since x2M CG .x/ D CG .M/ and hMi D H , we get CG .H / D CG .M). As CG .H / \ H D Z.H / has index p 2 D jG W CG .M/j in H , we obtain G D H CG .H / by the product formula. (ii) Now suppose that we are given an (arbitrary) p-group G D H CG .H /, where H is minimal nonabelian, S and let M D ¹x1 ; : : : ; xpC1 º 2 ƒ.H /. Then we conclude G D H CG .H / x2M CG .x/, so M 2 ƒ.G/ by Lemmas 116.2(a) and 116.3(a). Thus ƒ.H / ƒ.G/ since M is arbitrary.1 Theorem 116.5. Let G be a nonabelian p-group. (a) If M 2 ƒ.G/ has cardinality p C 1, then jG W CG .x/j D p for all x 2 M and jG W CG .M/j D p 2 . (b) .G/ D p C 1 if and only if G D H Z.G/, where H is an arbitrary minimal nonabelian subgroup of G, H \ Z.G/ D Z.H /. If, in addition, G is of exponent p, then G D H E, where H is nonabelian of order p 3 and E is abelian. Proof. Statement (a) follows from Lemma 116.3(b). (b) Suppose that .G/ D p C 1. Let H G be minimal nonabelian. Then we see that M 2 ƒ.H / has cardinality p C 1 (Lemma 116.1(a)) so that M 2 ƒ.G/ by hypothesis. By Lemma 116.4, G D H CG .H /. We claim that CG .H / D Z.G/. It sufﬁces to show that CG .H / is abelian (indeed, then CG .CG .H // H CG .H / D G). Assume that this is false. Then CG .H / contains two noncommuting elements b; b1 . Let M D ¹a1 ; : : : ; apC1 º 2 ƒ.H /. Take a 2 L¹Z.H /[¹a1 ºº, where L is an (abelian) maximal subgroup of H containing a1 (such an a exists since jL Z.H /j > 1). Then, as Œab; ai D Œa; ai ¤ 1 for i > 1 (indeed, for i > 1 the subgroup ha; ai i D H is nonabelian), we see that the p C 1 elements ab; a2 ; : : : ; apC1 are pairwise noncommuting. Note that Œab; ab1 D Œb; b1 ¤ 1 and for i > 1 we have Œab1 ; ai D Œa; ai ¤ 1. It follows that the p C 2.> .G// elements ab; ab1 ; a2 ; : : : ; apC1 are pairwise noncommuting, which is a contradiction. Thus CG .H / is abelian so coincides with Z.G/. Let us show that for our group G D H Z.G/ we have .G/ < p C 2 (we have ƒ.H / ƒ.G/, but our assertion is stronger). Indeed, assume that g1 ; : : : ; gpC2 2 G do not assert that, in the case under consideration, .G/ D .H / (however, this equality holds by Theorem 116.5). 1 We

166

Groups of prime power order

are pairwise noncommuting. Then we have gi D hi zi , where hi 2 H and zi 2 Z.G/ (i D 1; 2; : : : ; p C 2). Let i ¤ j . Then Œhi ; hj D Œhi zi ; hj zj D Œgi ; gj ¤ 1 so the minimal nonabelian p-group H contains the p C 2 pairwise noncommuting elements h1 ; : : : ; hpC2 , contrary to Lemma 116.1(a). Now suppose that G D H Z.G/ is of exponent p (here H is of order p 3 as minimal nonabelian group of exponent p by Lemma 65.1, and Z.G/ is elementary abelian). In this case, H \ Z.G/ D Z.H / is of order p so Z.G/ D Z.H / E, where E is elementary abelian. Then G D H E, and this completes the proof of (b). Theorem 116.5(b), in particular, classiﬁes the nonabelian p-groups possessing exactly p C 1 distinct centralizers of noncentral elements.2 Proposition 116.6. The following assertions for a nonabelian p-group G are equivalent: (a) If H G is minimal nonabelian, then ƒ.H / ƒ.G/. (b) G D .B1 Bk /Z.G/, where B1 ; : : : ; Bk are minimal nonabelian. Proof. (a) ) (b). We proceed by induction on jGj. Let B1 G be minimal nonabelian. Then G D B1 CG .B1 / by Lemma 116.4. If CG .B1 / is abelian, we are done. If CG .B1 / is nonabelian, the result follows by induction applied to CG .B1 /. (b) ) (a). Let G be as in (b) and H G be minimal nonabelian. Since jG 0 j D p, then, by [B1, Lemma 4.3(a)], we obtain G D H CG .H / so ƒ.H / ƒ.G/ by Lemma 116.4. Remark 2. The argument in part (ii) of the proof of Lemma 116.4 shows that if H is a nonabelian subgroup of an arbitrary group G D H CG .H /, then ƒ.H / ƒ.G/. 3o Nonnilpotent groups. In this subsection G is a nonnilpotent group. We consider coverings of nonnilpotent groups by a few proper subgroups. Minimal nonabelian and minimal nonnilpotent groups play a crucial role in what follows. Let p be a prime divisor of jGj such that G has no normal p-complement. Then there is in G a minimal nonnilpotent subgroup H D Q P , where P D H 0 2 Sylp .H / and Q 2 Sylq .H / is cyclic (this follows from Frobenius’ normal p-complement theorem; see, for example, [Isa5, Theorem 9.18]). We have jP j D p bCc , where b is the order of p modulo q and p c D jP \ Z.H /j (see [BZ, Lemma 11.2]). In this case, there are in H exactly p b Sylow q-subgroups, say Q1 D hx1 i; : : : ; Qpb D hxpb i: Then x1 ; : : : ; xpb are pairwise noncommuting elements (indeed, if i ¤ j, then hxi ; xj i is nonnilpotent since it has two distinct Sylow q-subgroups Qi and Qj so this twogenerator subgroup coincides with H ). If ¹y1 ; : : : ; ys º is a maximal subset of pairwise 2 Note

that [P] yields an estimate of the index jG W Z.G/j in terms of .G/.

116

Groups covered by few proper subgroups

167

noncommuting elements of P , then ¹y1 ; : : : ; ys ; x1 ; : : : ; xpb º is a maximal subset (with respect to inclusion) of pairwise noncommuting elements of H of cardinality p b C s p b C 1 (note that s D 1 if and only if P is abelian). If P is nonabelian, then s pC1 (Lemma 116.1(b)). Thus .G/ .H / D p b Cs p b C1. Theorem 116.7. Let G be a nonabelian group and p be a prime divisor of jGj. (a) If G has no normal p-complement, then .G/ p C 1. If, in addition, p is the minimal prime divisor of jGj, then .G/ p 2 C 1. (b) Suppose that G has a normal p-complement however a Sylow p-subgroup is not a direct factor of G. Then .G/ p C 2. If, in addition, .G/ D p C 2, then either p D 2 and q D 3 or p is a Mersenne prime. (c) If G D P A, where P is nonabelian, A is abelian and .G/ < p C 2, then P is such as in Theorem 116.5(b) and A is abelian. Proof. Statement (a) was proved in the paragraph preceding the theorem.3 (b) Now assume that G has a normal p-complement H but P 2 Sylp .G/ is not a direct factor of G. In this case, the p-solvable group G contains a nonnilpotent subgroup PQ, where Q 2 Sylq .H /; then we get Q D PQ \ H G PQ. Therefore PQ contains a minimal nonnilpotent subgroup F D P1 Q1 , where P1 2 Sylp .F / is cyclic and Q1 D F 0 2 Sylq .F /. Then jQ1 j D q bCc , where b is the order of q modulo p and q c D jQ1 \ Z.F /j. As above, there is M 2 ƒ.F / of cardinality q b C 1. Since q b p C 1, we get jMj p C 2. Now assume that jMj D p C 2; then q b D p C 1 so either p D 2 and q D 3 or q D 2 and p is a Mersenne prime.4 Statement (c) now follows from Remark 2 and Theorem 116.5(b). Proposition order of a group G and let SpC1 116.8. Let p be a minimal prime divisor ofTthe pC1 G D i D1 Ai be an irredundant covering. Then jG W iD1 Ai j D p 2 . In particular, jG W Ai j D p for i D 1; : : : ; p C 1. Proof. It follows from Remark 1 that if p is the minimal prime divisor of jGj, then it is not covered by p proper subgroups (otherwise, there is in G a proper subgroup of index < p, contrary to Lagrange’s theorem). One may assume that jA1 j jApC1 j. Then, by Remark 1, jG W A1 j < p C 1 so that jG W A1 j D p and hence A1 G G. First assume that all Ai are maximal in G. Set jGj D g and jG W Ai j D ki , where i D 2; : : : ; p C 1. Note that ki p for all i . We have ! pC1 [ .Ai A1 / : (4) G D A1 [ iD2 3 If 4 If

G Š A4 , the alternating group of degree 4, p D 2, then .G/ D 22 C 1. G Š A4 and p D 3, then .G/ D 5 D 3 C 2.

168

Groups of prime power order

Further, G D A1 Ai for i > 1. Since Ai A1 D Ai .Ai \A1 / and jAi W .Ai \A1 /j D jG W A1 j D p so that jG W .Ai \ A1 /j D jG W Ai jjAi W .Ai \ A1 /j D pki we obtain jAi A1 j D jAi .Ai \ A1 /j D

g g g D ki pki ki

for i > 1; 1

1 : p

The right-hand side of formula (4) contains v elements, where pC1 pC1 1 X g 1 X g g g g 1 g D C C 1 1 p D g: v C 1 p p ki p p p p p p iD2

i D2

SinceTv D g, it follows that (4) is a partition of G and ki D p for all i . In this case, pC1 jG W i D1 Ai j D p 2 . SpC1 Now let Ai Bi < G, where Bi are maximal in G for all i . Then G D i D1 Bi is an irredundant covering of G by the ﬁrst sentence of the proof, and so jG W Bi j D p for all i by the previous paragraph. If Ai < Bi for some i , then, taking in the above displayed formula SpC1 Aj D Bj2 for j ¤ i , we get a contradiction. Thus Bi D Ai for all i and so jG W i D1 Ai j D p by the previous paragraph. Lemma 116.3(b) is a partial case of Proposition 116.8. Let G be a non-p-nilpotent group. Then, using Theorem 116.7, one can show the following results: (a) If p D 2, then .G/ 5. (b) If p > 2, then .G/ p C 1. (c) If p > 2 is a minimal prime divisor of jGj, then .G/ p 2 C 1. 4o On the number of maximal subgroups appearing in some coverings of p-groups. In this subsection we consider irredundant coverings of a p-group by k proper subgroups, where p C 1 < k 2p. It is impossible to avoid some repetition in computations (otherwise, the proofs would be unreadable). Remark 3. We claim that if a p-group G of order p 3 is neither cyclic nor Q8 , it admits an irredundant covering by 2p subgroups. Indeed, let T G G be such that G=T is abelian ofStype .p; p/. Let A1 =T; : : : ; ApC1 =T be all subgroups of order p in G=T . pC1 Then G D i D1 Ai is an irredundant covering. Since G is neither cyclic nor isomorphic to Q8 , one may assume that A1 is noncyclic (here we use Theorem 1.2 which implies that if a p-group contains > p cyclic subgroups of index p, it is isomorphic to Q8 ). In this case, there exists in T an A1 -invariant subgroup T0 such that A1 =T0

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Groups covered by few proper subgroups

169

is abelian of type .p; p/. Let T D T1 ; T2 ; : : : ; TpC1 be all maximal subgroups of A1 containing T0S . Then G is covered by 2p subgroups A2 ; : : : ; ApC1 ; T2 ; : : : ; TpC1 since pC1 A1 A2 [ . i D2 Ti /, and this covering is irredundant. Lemma 116.9. If a p-group G admits an irredundant covering by p C 2 subgroups A1 ; : : : ; ApC2 , then (a) If p > 2, then at least p C 1 of the Ai ’s are maximal in G. (b) If p D 2, then at least two of the Ai ’s are maximal in G. Proof. Let jA1 j jApC2 j and jGj D p n . By Remark 1.1, we get jG W A1 j D p. Assume that jG W ApC1 j > p. Then ˇpC2 ˇ p pC2 X X ˇ[ ˇ ˇ ˇ Ai ˇ jA1 j C jAi A1 j C jAi A1 j p Dˇ n

i D1

Dp

n1

i D2

C .p 1/.p

n1

i DpC1

p

n2

/ C 2.p n2 p n3 /

D p n p n3 .p 2 3p C 2/ D p n p n3 .p 1/.p 2/: If p > 2, then p n p n p n3 .p 1/.p 2/ < p n , which is a contradiction. Thus, if p > 2, then at least p C 1 subgroups Ai are maximal in G, completing this case. Now let p D 2 and assume that jA2 j < 2n1 . Then ˇX ˇ 4 X ˇ 4 ˇ ˇ Ai ˇˇ jA1 j C jAi A1 j D 2n1 C 3.2n2 2n3 / 2 Dˇ n

i D1

D2

n1

i D2

n3

C3 2

D 7 2n3 < 2n ;

a contradiction. Thus, if p D 2, then at least two Ai are maximal in G. S Let G D 4iD1 Ai be an irredundant covering of a 2-group G of order 2n > 23 , jA1 j jA2 j jA3 j jA4 j. We claim that if jG W A3 j D 2, then jG W A4 j D 2 so all Ai are maximal in G. Then we have (5)

G D A1 [ .A2 A1 / [ .A3 A1 A2 / [ .A4 A1 A2 A3 /:

Assume that jG W A4 j > 2. We have jG W .A1 \ A2 \ A3 /j D 23 (Lemma 116.3) hence jA3 A1 A2 j D 2n3 . Further, A3 A1 A2 D .A3 A1 / .A3 A1 / \ A2 D A3 .A3 A1 / .A3 \ A2 / .A1 \ A2 / D A3 .A3 A1 / .A3 \ A3 / .A3 \ A1 \ A2 / :

170

Groups of prime power order

It follows that jA3 A1 A2 j D 2n1 2n2 .2n2 2n3 / D 2n3 : Now it is clear that jA4 A1 A2 A3 j < 2n3 . Therefore the right-hand side of formula (5) contains v elements, where v < 2n1 C 2n2 C 2n3 C 2n3 D 2n ; which is a contradiction. Thus, in the case under consideration, either exactly two or four of the Ai ’s are maximal in G. Let G be a 2-group which is not generated by two elements. We claim that then G S admits an irredundant covering G D 4iD1 Ai , where Ai 2 1 for all i . Without loss of generality one may assume that ˆ.G/ D ¹1º. Let A1 ; A2 2 1 be distinct, and set T D A1 \ A2 . Let T < A3 2 1 ¹A1 ; A2 º and let S < T be of index 2; then A3 =S Š E4 . Let T =S; T1 =S; T2 =S < A3 =S be of index 2. As G=Ti Š E4 , there are distinct B1 ; B2 2 1 ¹A3 º such that Bi \ A3 D Ti , i D 1; 2. Since A3 is a subset of the set A1 [B1 [B2 (indeed, A3 D T [T1 [T2 is a subset of A1 [B1 [B2 ), it follows that G D A1 [ A2 [ B1 [ B2 is a covering (indeed, by the above, G D A1 [ A2 [ A3 is a subset of A1 [ A2 [ B1 [ B2 ). Due to the fact the intersection of any three distinct elements of the set ¹A1 ; A2 ; B1 ; B2 º has index p 3 in G, our covering is irredundant (Lemma 116.3(b)). Similarly, if p > 2 and a p-group G is not generated by two elements, then it admits an irredundant covering by 2p maximal subgroups. Lemma 116.10. Suppose that A; B; C; D are pairwise distinct maximal subgroups of a p-group G of order p n such that jG W .A \ B \ C /j D p 3 . Then (6)

jA [ B [ C j D 3p n1 3p n2 C p n3 ;

(7)

jD .A [ B [ C /j p n1 3p n2 C 3p n3 p n4 :

Proof. Note that if distinct U; V < G are maximal, then jG W .U \ V /j D p 2 . By the inclusion-exclusion identity, jA [ B [ C j D .jAj C jBj C C j/ .jA \ Bj C jB \ C j C jC \ Aj/ C jA \ B \ C j D 3p n1 3p n2 C p n3 ; proving (6). By hypothesis, A \ B ¤ B \ C ¤ C \ A (if, for example, A \ B D A \ C , then A \ B D .A \ B/ \ .A \ C / D A \ B \ C , a contradiction since jA \ Bj D p n2 > p n3 D jA \ B \ C j). We have (8)

D .A [ B [ C / D D .D \ .A [ B [ C //;

116

and so

Groups covered by few proper subgroups

171

D .A [ B [ C / D D .D \ A/ [ .D \ B/ [ .D \ C / :

By the inclusion-exclusion identity, j.D \ A/ [ .D \ B/ [ .D \ C /j D jD \ Aj C jD \ Bj C jD \ C j jD \ A \ Bj C jD \ A \ C j C jD \ B \ C j C jD \ A \ B \ C j: If A \ B D, then D \ A \ B \ C D D \ C has order p n2 , a contradiction since A \ B \ C D \ A \ B \ C has order p n3 by hypothesis. Thus A \ B 6 D, and the same is true for A \ C and B \ C . Then, by the product formula and the previous displayed formula, we have jD \ A \ Bj D jD \ A \ C j D jD \ B \ C j D p n3 ; jD \ .A [ B [ C /j D j.D \ A/ [ .D \ B/ [ .D \ C /j D 3p n2 3p n3 C jD \ A \ B \ C j: Since jD \ A \ B \ C j 2 ¹p n3 ; p n4 º, we obtain jD \ .A [ B [ C /j 3p n2 3p n3 C p n4 : Now (7) follows from (8). Theorem 116.11. If a group of order p n admits an irredundant covering by p C 2 subgroups, then p D 2. Proof. Assume that a group G of order p n admits an irredundant covering by p C 2 proper subgroups A1 ; : : : ; ApC2 and p > 2. By Lemma 116.9(a), oneT may assume that SpC1 pC1 A1 ; : : : ; ApC1 are maximal in G. Since G ¤ iD1 Ai , we have jG W i D1 Ai j p 3 . One may assume, without loss of generality, that jG W .A1 \ A2 \ A3 /j D p 3 . We may also assume that jG W ApC2 j D p. Then, by (6), we have jA1 [ A2 [ A3 j D 3p n1 3p n2 C p n3

(9) and, for i > 3, (10)

jAi .A1 [ A2 [ A3 /j < p n1 3p n2 C 3p n3 D p n3 .p 2 3p C 3/

by (7). Set A1 [ A2 [ A3 D U . We have (11)

GDU [

pC2 [ iD4

! .Ai U / :

172

Groups of prime power order

Therefore, taking into account (9) and (10), we obtain jGj D p n .3p n1 3p n2 C p n3 / C .p 1/p n3 .p 2 3p C 3/ D p n p n3 .p 2 3p C 2/ D p n p n3 .p 1/.p 2/ < p n since p > 2, a ﬁnal contradiction. Thus we must have p D 2. Proposition 116.12. If a p-group G is covered by at most k 2p proper subgroups A1 ; : : : ; Ak (we do not assume that this covering is irredundant), then at least p of these subgroups are maximal in G. If p > 3 and p C 2 < k < 2p, then at least p C 1 summands in our covering are maximal in G. Proof. (i) In view of Lemma 116.9, one may assume that p > 2. Let jA1 j jAk j. Then jG W A1 j D p as p > 2 (Remark 1). Since we do not assume that our covering is irredundant, one can add new summands to obtain k D 2p. We also may assume, by way of contradiction, that A1 ; : : : ; Ap1 are maximal in G and Ap ; : : : ; A2p have index p 2 in G. Indeed, if, for example, jG W Ai j > p 2 (i > p 1), one can replace Ai by a subgroup that contains Ai and has index p 2 in G. If, for example, jG W Ai j > p (i < p), one can replace Ai by a maximal subgroup of G that contains Ai . We have (12)

G D Ap1 [

p2 [

! .Ai Ap1 / [

i D1

2p [

! .Ai Ap1 / :

iDp

The right-hand side of formula (12) contains v elements, where v p n1 C .p 2/.p n1 p n2 / C .p C 1/.p n2 p n3 / D p n p n3 .p 1/2 < p n D jGj; a contradiction since v D jGj D p n . Thus at least p subgroups Ai (i 2p) are maximal in G. (ii) To prove the last assertion, one may assume, by way of contradiction, that A1 ; : : : ; Ap are maximal in G and jG W Ai j D p 2 for i > p (see (i)). (Here p > 3 since 3 C 2 D 2 3 1.) We may also assume that k D 2p 1 (if k < 2p 1, one can add to our union 2p 1 k new summands of order p n2 ). Then, as above, we obtain jGj D p n p n1 C .p 1/.p n1 p n2 / C .p 1/.p n2 p n3 / D p n1 C .p 1/.p n1 p n3 / D p n p n3 .p 1/ < p n ; a contradiction.

116

Groups covered by few proper subgroups

173

Proposition 116.13. Suppose that a p-group G of order p n p 4 , p > 2, is covered by k proper subgroups, say A1 ; : : : ; Ak , where p C 2 < k 2p. Let, in addition, any p C 2 subgroups Ai do not cover G, jG W Ai j D p for i p and jG W Ai j > p for i > p. Then (a) k D 2p and our covering is irredundant. Tp (b) j i D1 Ai j D p n2 . (c) jAi j D p n2 for i > p. T2p (d) j i DpC1 Ai j D p n3 . Proof. By the second assertion of Proposition 116.12, we get k D 2p so (a) is true. We have ! ! p1 2p [ [ .Ai Ap / [ .Ai Ap / : (13) G D Ap [ i D1

i DpC1

(c) Assume that jApC1 j jA2p j and jA2p j < p n2 . Then the right-hand side of (13) contains v elements, where v p n1 C .p 1/.p n1 p n2 / C .p 1/.p n2 p n3 / C .p n3 p n4 / D p n1 C .p 1/.p n1 p n3 / C .p n3 p n4 / D p n p n4 .p 1/2 < p n D jGj; which is a contradiction. This proves (c). (b, d) We have jG W Ai j D p 2 for i > p by (c). In this case, the right-hand side of (13) contains v elements, where v p n1 C .p 1/.p n1 p n2 / C p.p n2 p n3 / D p n D jGj; so (13) is a partition. This implies (b) and (d). A similar argument shows that if a p-group G is covered by p 2 proper subgroups, then at least two of these subgroups are maximal in G. Indeed, if only one summand of our covering is maximal in G (see Remark 1), we obtain p n p n1 C .p 2 1/.p n2 p n3 / D p n p n3 .p 1/ < p n ; a contradiction. Corollary 116.14. If a nonabelian p-group G has at most 2p pairwise noncommuting elements, then the centralizers of at least p of these elements are maximal in G. Remark 4. Let G be a group of maximal class and order p nC2 , n 2, with abelian subgroup A of index p. In this case, every x 2 G A satisﬁes jCG .x/j D p 2 (indeed,

174

Groups of prime power order

CA .y/ D Z.G/ is of order p and y p 2 Z.G/ for all y 2 G A), and the number of maximal abelian subgroups of order p 2 not contained in A is equal to jG Aj D pn p.p 1/ (indeed, if B is such a subgroup, then jB Aj D p.p 1/). These p n subgroups together with A cover G and this covering is irredundant. The so-obtained set of cardinality p n C 1 is contained in ƒ.G/. It is easy to see that .G/ D p n C 1. Remark 5. Let G be a nonabelian p-group. If x 2 G and Ax is a maximal abelian subgroup of CG .x/, then Ax is also a maximal abelian subgroup of G. Let M 2 ƒ.G/. Take in CG .x/ a maximal abelian subgroup Ax for every x 2 M (indeed, if B > Ax is abelian, then, by choice, B 6 CG .x/, a contradiction). Then j¹Ax j x 2 Mºj D jMj. It follows that G has at least .G/ maximal abelian subgroups. If the group G has exactly p C 1 maximal abelian subgroups, say A1 ; : : : ; ApC1 , they cover G. In this case, A1 ; : : : ; ApC1 are maximal in G (Lemma 116.2(b)) and G has the structure described in Theorem 116.6(b). Remark 6. Let B1 ; : : : ; Bn beSall maximal abelian subgroups of a nonabelian group G. Then .G/ n since G D niD1 Bi (it is possible that this covering may be redundant). If Bi \ Bj D Z.G/ for all i ¤ j (in this case, the considered covering is irredundant), then .G/ D n. Indeed, take xi 2 Bi Z.G/ for all i . We claim that ¹x1 ; : : : ; xn º 2 ƒ.G/ (indeed, if ¹x; x1 ; : : : ; xn º G, then x 2 Bi for some i so xxi D xi x). For example, let G be a Sylow 2-subgroup of the simple Suzuki group Sz.q/, where q D 22mC1 . Then Z.G/ D ˆ.G/ has index 22mC1 in G. If A < G is maximal abelian, then jA W Z.G/j D 2. It follows that there is M 2 ƒ.G/ of cardinality 22mC1 1; moreover, all members of the set ƒ.G/ have the same cardinality. Remark 7. Let G D AB be a central product of nonabelian groups A and B. Further, let M D ¹a1 ; : : : ; am º 2 ƒ.A/ and N D ¹b1 ; : : : ; bn º 2 ƒ.B/. Then m 1 C n elements of the set M1 D ¹a2 ; : : : ; am ; a1 b1 ; a1 b2 ; : : : ; a1 bn º are pairwise noncommuting. We claim that M1 2 ƒ.G/. Assume that there exists x 2 G M1 such that all elements of the set M1 [ ¹xº are pairwise noncommuting. We have x D ab, where S a 2 A and b 2 B. For i > 1 we have 1 ¤ Œab; ai D Œa; ai so that a 2 D D A niD2 CA .ai /. By Lemma 116.1(b), the subgroup hDi is abelian so Œa; a1 D 1 since a1 2 D. For i D 1; : : : ; n we have 1 ¤ Œab; a1 bi D Œab; bi D Œb; bi . We conclude that n C 1 > .B/ elements b; b1 ; : : : ; bn 2 B are pairwise noncommuting, which is a contradiction. It is easy to deduce from this by induction that if G is an extraspecial group of order p 2mC1 , then .G/ mp C 1. It follows from Remark 7 and Lemma 116.1(a) that if G is a nonabelian p-group such that .G/ 2p, then CG .H / is abelian for all minimal nonabelian H < G.

116

Groups covered by few proper subgroups

175

Problems Below we state some related problems. Problem 1. Classify the 2-groups which do not contain ﬁve pairwise noncommuting elements. (See Lemma 1.3 and Theorem 116.11.) Problem 2. Does there exist a p-group G admitting an irredundant covering by n subgroups, where p C 1 < n < 2p? If ‘yes’, classify such the groups. Problem 3. Describe the set of positive integers n such that there is an elementary abelian p-group admitting an irredundant covering by n maximal subgroups. Problem 4. Let M; N be groups and .M / D m, .N / D n. Then .M N / D mn. (i) Estimate .M N / in terms of M , N and M \ N . Consider the case where M; N are p-groups of maximal class. (ii) Find .G/, where G is an extraspecial group of order p 2mC1 . Problem 5. Classify the pairs of groups N G G such that .G=N / D .G/. (The set of such pairs is inﬁnite: let G be a minimal nonabelian p-group with noncyclic center and N < G with G 0 6 N .) Problem 6. Find .†pn /, where †pn is a Sylow p-subgroup of the symmetric group Spn of degree p n . Deal with the same problem for UT.m; p n /, a Sylow p-subgroup of the general linear group GL.m; p n /. Problem 7. Study the nonabelian p-groups G such that CG .H / is abelian for all minimal nonabelian H G. (See the paragraph following Remark 7.) Problem 8. Study the nonnilpotent groups G such that ƒ.H / ƒ.G/ for all minimal nonnilpotent H G. (Compare this with Lemma 2.2(b).) Problem 9. Study the groups that are covered by (i) minimal nonnilpotent subgroups, (ii) minimal nonabelian subgroups, (iii) Frobenius subgroups. Problem 10. Classify the p-groups that are covered by subgroups of maximal class. Problem 11. Let H be a proper subgroup of maximal class of a p-group G such that ƒ.H / ƒ.G/. Study the structure of G. Problem 12. Find .G/, where G 2 ¹An ; Sn º (for example, .A5 / D 21; see also [Br].)

117

2-groups all of whose nonnormal subgroups are either cyclic or of maximal class

The 2-groups all of whose nonnormal subgroups are cyclic are completely determined in 16. The ﬁrst step in classifying the title groups was done by the ﬁrst author in Theorem A.40.25. We ﬁnish the classiﬁcation of the title groups by proving the following result which solves Problem 1978. Theorem 117.1. Let G be a 2-group all of whose nonnormal subgroups are either cyclic or of maximal class. Assume, in addition, that G has a nonnormal subgroup of maximal class. Then we have one of the following possibilities: (a) G is generalized quaternion Q2n , n 5. (b) G D hy; b j y 8 D b 4 D 1; b y D bu; y 2 D ua; uy D uz; ay D a1 ; a2 D b 2 D Œa; b D z; Œu; a D Œu; b D 1i, where G is of order 25 and class 3 with 2 .G/ D hui ha; bi Š C2 Q8 , Z.G/ D hzi Š C2 , G 0 D hu; zi Š E4 and ˆ.G/ D hu; ai Š C2 C4 . (c) G is any 2-group of class 2 with 1 .G/ D G 0 Š E4 having a nonnormal quaternion subgroup. Conversely, each group in (a), (b) and (c) satisﬁes the assumptions of our theorem. Proof. Let H be any nonnormal subgroup of maximal class. Suppose that H is not generalized quaternion. Let V be a four-subgroup in H . Then V E G and so H Š D8 . But H has another four-subgroup W ¤ V and W is normal in G. It then follows that H D hV; W i E G, a contradiction. Therefore H is generalized quaternion. If G is of maximal class, then G Š Q2n , n 5 (part (a) of our theorem), and so we assume in the sequel that G is not of maximal class. This implies that G possesses a normal four-subgroup U and G=U is Dedekindian. Let Q be a nonnormal subgroup of maximal possible order in G such that Q Š Q2m , m 3. Set N D NG .Q/ and hzi D Z.Q/ so that Q < N < G, N E G and z 2 Z.N /. Then each subgroup X of N with X > Q is normal in G and so N=Q is Dedekindian. If X1 =Q and X2 =Q are two distinct subgroups of order 2 in N=Q, then X1 E G, X2 E G and X1 \X2 D Q, a contradiction. We have proved that N=Q ¤ ¹1º is either cyclic or N=Q Š Q8 . Let t be any involution in G Q. Since U D hz; ti E G and U \Q D hzi, we have j.UQ/ W Qj D 2 which implies t 2 N . On the other hand, Q=hzi Š D2m1 (if m 4)

117 2-groups with nonnormal subgroups being cyclic or of maximal class

177

or Q=hzi Š E4 (if m D 3) and G=U is Dedekindian. Hence m D 3 and Q Š Q8 . Set L D QU so that L E G and L=Q D 1 .N=Q/. Each involution in G is contained in N and so each involution in G is also contained in L. If CL .Q/ Q, then L would be of maximal class (and order 24 ), contrary to the fact that U is a normal four-subgroup in G. Hence L D V Q with V \ Q D hzi. If V Š C4 , then Q is a unique quaternion subgroup in L. In that case, Q E G, a contradiction. Hence V Š E4 , L D hti Q and 1 .G/ D U D hz; ti. Set C D CG .Q/ so that C \ L D ht; zi and C E G. If C does not cover N=Q, then N=C Š D8 (since Aut.Q8 / Š S4 ), contrary to the fact that G=U is Dedekindian. Thus we get N D Q C with Q \ C D hzi and C =hzi Š N=Q is either cyclic or quaternion. Set Q D hv; wi, where hvi, hwi, hvwi are three cyclic subgroups of index 2 in Q. Now, hv; ti, hw; ti and hvw; ti are normal in G and so hzi Z.G/, jG W NG .hvi/j 2, jG W NG .hwi/j 2 and NG .hvi/ \ NG .hwi/ D N . Hence either jG W N j D 2 or jG W N j D 4 and in the second case G=N Š E4 , where NG .hvi/=N , NG .hwi/=N and NG .hvwi/=N are three distinct subgroups of order 2 in G=N . In any case, there is an element v of order 4 in Q and an element x 2 G N with x 2 2 N so that v x D v t, where 2 ¹1; 1º. Hence Œv; x D t or zt and so, in particular, hz; ti G 0 . Clearly, z is not a square in C . Indeed, if there is y 2 C with y 2 D z, then o.vy/ D 2, where v 2 Q hzi and y 2 C L and so vy 2 N L, contrary to 1 .G/ D ht; zi. (i) First assume C > ht; zi. Then C =hzi is either cyclic of order 4 or C =hzi Š Q8 . In the ﬁrst case, there is c 2 C such that hci covers C =hzi and then (since z is not a square in C ) C D hci hzi, Z.N / D C and 1 .hci/ is characteristic in C . It follows that 1 .hci/ Z.G/ and U D ht; zi Z.G/. Suppose that C =hzi Š Q8 . Then we get ¹1º ¤ C 0 U D ht; zi. If C 0 D U , then, by a result of O. Taussky (Lemma 1.6), C is of maximal class and order 24 , a contradiction. Hence jC 0 j D 2 and since C 0 covers U=hzi, we have U D C 0 hzi. Then it follows that C 0 Z.G/ and so again U D ht; zi Z.G/. We know that G=U is Dedekindian. Assume that G=U is nonabelian, i.e., G=U is Hamiltonian. In that case, G=U has a quaternion subgroup V =U Š Q8 . Set W=U D Z.V =U / D .V =U /0 so that W is abelian of type .4; 2/ and therefore all elements in W U are of order 4. Let S1 , S2 , S3 be three maximal subgroups of V which contain U . Then Si =U Š C4 and since U Z.G/, each Si is an abelian maximal subgroup of V , i D 1; 2; 3. By Exercise P1, jV 0 j D 2. On the other hand, V 0 covers W =U and so there are involutions in W U , a contradiction. Hence G=U is abelian and so G 0 D U D 1 .G/ Z.G/. We have obtained the groups from part (c) of our theorem. (ii) Now assume C D ht; zi D U so that N D L D ht i Q Š C2 Q8 and jGj D 25 or 26 . Suppose, in addition, that 2 .G/ D N . In that case, Theorem 52.1(d) implies that G is a unique group of order 25 with Z.G/ D hzi, G 0 D ht; zi and ˆ.G/ Š C4 C2 which appears in part (b) of our theorem. From now on we may assume that there is an element x 2 G N of order 4 and so 1 ¤ x 2 2 U . In this case,

178

Groups of prime power order

we obtain j.G=U / W 1 .G=U /j 2 which implies that the Dedekindian group G=U cannot be Hamiltonian and so G=U is abelian which gives G 0 D U . We have also hzi Z.G/ U . Since 1 .G/ D U , the case U hxi Š D8 is not possible and so x centralizes U . Because x normalizes N D L and does not normalize Q, there is an element v of order 4 in Q with v 2 D z such that x does not normalize hvi. On the other hand, hv; ti E G and so we may set v x D vt for a suitable element t 2 U hzi which gives Œv; x D t. But U D hz; ti Z.N hxi/ and so hv; xi is minimal nonabelian with .hv; xi/0 D hŒv; xi D ht i. Also, hv; xi ht; zi D U and so hv; xi E G. This implies that .hv; xi/0 D ht i Z.G/ and so Z.G/ D ht; zi. We have again obtained some groups from part (c) of our theorem.

118

Review of characterizations of p-groups with various minimal nonabelian subgroups

Recall that a group G is said to be minimal nonabelian if it is nonabelian but all its maximal subgroups are abelian. A p-group H is minimal nonabelian if and only if H D ha; bi is two-generated and jH 0 j D p, i.e., Œa; bp D Œa; b; a D Œa; b; b D 1. In that case, ˆ.H / D Z.H / so that jH W Z.H /j D p 2 . Exercise 1.8(a) ([Red] and Lemma 65.1). Let G be a minimal nonabelian p-group. Then jG 0 j D p and G=G 0 is abelian of rank two and G is one of the following groups: m

n

(a) G D ha; b j ap D b p D 1; ab D a1Cp (G is metacyclic). m

m1

i, m 2, n 1, jGj D p mCn

n

(b) G D ha; b j ap D b p D c p D 1; Œa; b D c; Œa; c D Œb; c D 1i is nonmetacyclic of order p mCnC1 and if p D 2, then m C n > 2. Next, G 0 is a maximal cyclic subgroup of G. (c) G Š Q23 . Our group G is nonmetacyclic if and only if G 0 is a maximal cyclic subgroup of G. Proposition 10.19. Let G be a metacyclic p-group containing a nonabelian subgroup B of order p 3 . Then (a) if p D 2, then G is of maximal class, (b) if p > 2, then jGj D p 3 , i.e., G D B. Proposition 10.28. A nonabelian p-group is generated by its minimal nonabelian subgroups. Theorem 10.33. All minimal nonabelian subgroups of a 2-group are dihedral if and only if G D hxi A, where ax D a1 for all a 2 A. Theorem 10.34(a). A p-group G, p > 2, is generated by minimal nonabelian subgroups of exponent p unless it contains a subgroup Š †p2 2 Sylp .Sp2 /. The following result generalizes Theorem 10.28 and plays an important role in what follows. Lemma 57.1. Let G be a nonabelian p-group and let A be any maximal abelian normal subgroup of G. Then for each x 2 G A there is a 2 A such that ha; xi is minimal nonabelian.

180

Groups of prime power order

It follows that minimal nonabelian subgroups of G cover the set G A and so they generate G (see also Theorem 10.28). We have the following open problem. Classify ﬁnite p-groups which are covered by its minimal nonabelian subgroups. We have solved only the following two special cases of this problem. Lemma 65.2(a). Let G be a nonabelian p-group such that G 0 1 .Z.G// and d.G/ D 2. Then G is minimal nonabelian. Theorem 76.A. If G is a nonabelian p-group containing p 2 C p C 1 minimal nonabelian subgroups, then all its subgroups of index p 3 are abelian. Theorem 92.1. Let G be a nonabelian p-group such that for each minimal nonabelian subgroup H of G and each x 2 H Z.G/ we have CG .x/ H . Then G is one of the following groups: (a) G is minimal nonabelian. (b) p D 2, d.G/ D 3 and G D ha; b; c j a4 D b 4 D c 4 D 1; Œa; b D c 2 ; Œa; c D b 2 c 2 ; Œb; c D a2 b 2 ; Œa2 ; b D Œa2 ; c D Œb 2 ; a D Œb 2 ; c D Œc 2 ; a D Œc 2 ; b D 1i; where G is a special 2-group of order 26 with ha2 ; b 2 ; c 2 i D G 0 D Z.G/ D ˆ.G/ D 1 .G/ Š E8 and G is isomorphic to an S2 -subgroup of the simple group Sz.8/. (c) p > 2, d.G/ D 2, G is of order p 5 and 2

2

G D ha; x j ap D x p D 1; Œa; x D b; Œa; b D y1 ; Œx; b D y2 ; p

p

b p D y1 D y2 D Œa; y1 D Œx; y1 D Œa; y2 D Œx; y2 D 1; ˇ

ap D y1˛ y2 ; x p D y1 y2ı i; where in case p > 3, 4ˇ C .ı ˛/2 is a quadratic non-residue (mod p). Here ˆ.G/ D G 0 D hb; y1 ; y2 i D 1 .G/ Š Ep3 ; Z.G/ D K3 .G/ D Ã1 .G/ D hy1 ; y2 i Š Ep2 : Conversely, all the above groups satisfy the assumptions of the theorem. Theorem 92.2. Let G be a nonabelian p-group such that whenever A is a maximal subgroup of any minimal nonabelian subgroup H in G, then A is also a maximal abelian subgroup of G. Then each abelian subgroup of G is contained in a minimal nonabelian subgroup and one of the following holds: (a) G is minimal nonabelian. (b) G is metacyclic. (c) G is isomorphic to the group of order 26 deﬁned in Theorem 92.1(b). (d) G is isomorphic to a group of order p 5 deﬁned in Theorem 92.1(c).

118

Review of p-groups with various minimal nonabelian subgroups

181

Theorem 92.2 is in fact a proper generalization of Theorem 92.1 since the assumption of Theorem 92.1 implies the assumption of Theorem 92.2 but not vice versa. The assumptions of both Theorems 92.1 and 92.2 imply that each abelian subgroup of G is contained in a minimal nonabelian subgroup and so in both cases G is indeed covered by its minimal nonabelian subgroups. If we know something about the structure of minimal nonabelian subgroups of a 2-group G, then we are able in some cases to determine the structure of G. In studying 2-groups all of whose minimal nonabelian subgroups are of exponent 4 (which, in general, is still an unsolved problem) the following lemma is crucial. Lemma 57.2. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are of exponent 4 and let A be a maximal normal abelian subgroup of G. Then all elements in G A are of order 4 and so either exp.A/ D 2 or exp.A/ D exp.G/. If x 2 G A with x 2 2 A, then x inverts each element in Ã1 .A/ and in A=1 .A/. If exp.G/ > 4, then either G=A is cyclic of order 4 or G=A Š Q8 . Theorem 90.1. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to D8 or Q8 . Then G is one of the following groups: (a) G is generalized dihedral (i.e., jG W H2 .G/j D 2). (b) G D H Z.G/, where H is of maximal class and Ã1 .Z.G// Z.H /. (c) G D H Z.G/, where H is extraspecial and Ã1 .Z.G// Z.H /. Theorem 57.3. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to H2 D ha; b j a4 D b 4 D 1; ab D a1 i. Then the following holds: (a) If G is of exponent 8, then G has a unique abelian maximal subgroup A. We have exp.A/ 8 and E D 1 .A/ D 1 .G/ D Z.G/ is of order 4. All elements in G A are of order 4 and if v is one of them, then CA .v/ D E and v inverts ˆ.A/ and on A=E. (b) If G is of exponent 4, then G D K V , where exp.V / 2 and for the group K we have one of the following possibilities: (b1) K Š H2 is of order 24 . (b2) K is the minimal nonmetacyclic group of order 25 (see Theorem 66.1(d)). (b3) K is a unique special group of order 26 with Z.K/ Š E4 in which every maximal subgroup is minimal nonmetacyclic of order 25 (from (b2)): K D ha; b; c; d j a4 D b 4 D 1; c 2 D a2 b 2 ; Œa; b D 1; ac D a1 ; b c D a2 b 1 ; d 2 D a2 ; ad D a1 b 2 ; b d D b 1 ; Œc; d D 1i: (b4) K is a splitting extension of B D B1 Bm , m 2, with a cyclic group hbi of order 4, where Bi Š C4 , i D 1; 2; : : : ; m, and b inverts each element of B (and b 2 centralizes B).

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Theorem 57.4. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to H16 D ha; t j a4 D t 2 D 1; Œa; t D z; z 2 D Œa; z D Œt; z D 1i. Then 1 .G/ is a self-centralizing elementary abelian subgroup and G is of exponent 4. Moreover, the centralizer of any element of order 4 is abelian of type .4; 2; : : : ; 2/. Theorem 57.5. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to H32 D ha; b j a4 D b 4 D 1; Œa; b D z; z 2 D Œa; z D Œb; z D 1i. Then 1 .G/ Z.G/ and G is of exponent 4 and class 2. Theorem 57.6. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to M16 D ha; t j a8 D t 2 D 1; at D a5 i. Then E D 1 .G/ is elementary abelian and if A is a maximal normal abelian subgroup of G containing E, then A is of type .2s ; 2; : : : ; 2/, s 2, jG W Aj 4, each element x in G A is of order 8, jE W CE .x/j D 2, x inverts each element in A=E and in Ã1 .A/ and we have the following possibilities: (a) s > 2 in which case jG W Aj D 2 and G=E Š Q2s is generalized quaternion of order 2s . (b) s D 2 in which case either G Š M16 V with exp.V / 2 or G=E Š Q8 and jG W Aj D 4. In Theorem 92.6 we classify nonabelian 2-groups all of whose minimal nonabelian subgroups are isomorphic to Q8 or H2 D ha; b j a4 D b 4 D 1; ab D a1 i. This theorem generalizes a result of N. Blackburn concerning p-groups G which possess nonnormal subgroups and such that the intersection of all nonnormal subgroups is nontrivial (Corollary 92.7). Namely, it is easy to see that in such p-groups G we must have p D 2 and each minimal nonabelian subgroup of G is isomorphic to Q8 or H2 . Theorem 92.6. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to Q8 or H2 D ha; b j a4 D b 4 D 1; ab D a1 i. Then the following holds: (a) If G is of exponent > 4, then G has a unique abelian maximal subgroup A, jG 0 j > 2 and all elements in G A are of order 4. We have 1 .A/ D 1 .G/ Z.G/ and if x 2 G A, then x inverts each element of A=1 .A/. (b) If G is of exponent 4, then G D K V , where exp.V / 2 and for the group K we have one of the following possibilities: (b1) K Š Q8 or K Š H2 . (b2) K D ha; b; c j a4 D b 4 D c 4 D Œa; b D 1; c 2 D a2 ; Œa; c D b 2 ; Œb; c D a2 i is the minimal nonmetacyclic group of order 25 . (b3) K is a unique special group of order 26 with Z.K/ Š E4 given in Theorem 57.3(b3) in which every maximal subgroup is isomorphic to the minimal nonmetacyclic group of order 25 (from (b2)). (b4) K Š Q8 C4 .

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183

(b5) K Š Q8 Q8 . (b6) G D K V has an abelian maximal subgroup B of exponent 4 and an element v 2 G B of order 4 which inverts each element of B. (b7) K D Q C is a central product of Q D ha; bi Š Q8 and C D hc; d j c 4 D d 4 D 1; c d D c 1 i Š H2 with Q \ C D hc 2 d 2 i D Z.Q/, where K is special of order 26 and Z.K/ D 1 .K/ Š E4 . Conversely, in each of the above groups in part (b) every minimal nonabelian subgroup is isomorphic to Q8 or H2 . Corollary 92.7 ([Bla7]). Let G be a ﬁnite p-group which possesses nonnormal subgroups and let R.G/ be the intersection of all nonnormal subgroups. If R.G/ > ¹1º, then p D 2, jR.G/j D 2 and G is one of the following groups: (a) G Š Q8 C4 E2s , s 0. (b) G Š Q8 Q8 E2s , s 0. (c) G has an abelian maximal subgroup A of exponent > 2 and an element x 2 G A of order 4 which inverts each element in A. Theorem 93.1. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are metacyclic and have exponent 4 and assume that D8 and H2 D ha; b j a4 D b 4 D 1; ab D a1 i (and possibly Q8 ) actually appear as subgroups of G. Then G has a unique abelian maximal subgroup A of exponent > 4 and an involution t 2 G A such that t inverts each element in A=1 .A/ and in 2 .A/ and there is an element a 2 A of order > 4 such that at D a1 with 1 ¤ 2 1 .A/, where in case that Ã1 .A/ is cyclic we have 62 Ã1 .A/. Conversely, all these 2-groups satisfy the assumptions of the theorem. Theorem 94.1. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are nonmetacyclic and have exponent 4. Then exp.G/ D 4, 1 .G/ is elementary abelian and if A is a maximal normal abelian subgroup of G containing 1 .G/, then ˆ.G/ 1 .A/ is elementary abelian and Ã1 .A/ Z.G/. Theorem 98.1. Let G be a nonabelian 2-group all of whose minimal nonabelian subgroups are isomorphic to M2nC1 , n 3 ﬁxed. Then E D 1 .G/ is elementary abelian. Let A be a maximal normal abelian subgroup of G containing E. Then A D CG .E/, all elements in G A are of order 2n , G=A is elementary abelian of order 4, and if x 2 G A, then jE W CE .x/j D 2, A0 D CA .x/ D CE .x/hx 2 i, A=A0 is cyclic, hA; xi0 is cyclic of order jA=A0 j, hA; xi0 \ A0 D 1 .hxi/, x inverts A=A0 and Ãn2 .A/. We have two possibilities: (a) exp.A/ D 2m 2n in which case jG W Aj D 2 and A is of type .2m; 2n2; 2; : : : ; 2/. (b) exp.A/ 2n1 in which case A is of type .2n1; 2r ; 2; : : : ; 2/ with 1 r n 2 and either jG W Aj D 2 or jG W Aj D 4, where in the last case G=n2 .A/ Š Q8 . Conversely, if G is a 2-group of type (a) or (b), then each minimal nonabelian subgroup of G is isomorphic to M2nC1 , n 3 ﬁxed.

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Finally, we consider the following general problem. Theorem 95.1. Let G be a nonabelian 2-group of exponent 2e (e 3) which does not possess a minimal nonabelian subgroup of exponent 2e . Then G has a unique maximal normal abelian subgroup A. The factor group G=A is either cyclic or generalized quaternion and each element in G A is of order < 2e . Also, G possesses at least one metacyclic minimal nonabelian subgroup and jCG .1 .A// W Aj 2. Proposition (see commentary of Mann to Problem 115). If p > 2 and all minimal nonabelian subgroups of a p-group G are of exponent p and exp.G/ > p, then we have jG W Hp .G/j D p. Exercise A40.79. Given a positive integer n and a p-group G of order p n , put Dn .G/ D hH 0 j H < G; jG W H j D p n i. Obviously, Dn .G/ is a characteristic subgroup of G. If G is nonabelian, then D1 .G/ < G if and only if G=D1 .G/ is minimal nonabelian. Exercise 124.7. Let A be a proper minimal nonabelian subgroup of a p-group G. Suppose that all A-containing subgroups of G of order pjAj are two-generator. Then A=A0 is a maximal abelian subgroup of NG .A/=A0 . Theorem 111.3. The following assertions for a group G are equivalent: (a) ˆ.G/0 D H 0 for all H 2 1 . (b) G is either abelian or minimal nonabelian. Theorem 124.30. Let H Š H2;p be a proper subgroup of a metacyclic p-group G. Here H2;p is the unique nonabelian metacyclic group of order p 4 and exponent p 2 . Then one of the following holds: (a) If p > 2, then jG W H j D p, jG 0 j D p 2 and either 3

2

3

2

2

G D hz; u j z p D up D 1; z u D z 1Cp i or G D hz; u j z p D up D 1; z u D z 1Cp i: (b) If p D 2, H G G, then jG W H j D 2. (c) If p D 2 and H is not normal in G, then 1 .G/ D 1 .H / and G=1 .G/ is either dihedral or semidihedral. Proposition 124.31. Suppose that H Š Mpn , n > 3, is a proper subgroup of a metacyclic p-group G. Then NG .H /=H is cyclic. Next assume that n D 4. (a) If p > 2, then jNG .H /=H j D p. (b) If p D 2, then jNG .H /=H j 4.

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In this section we review characterizations of p-groups of maximal class obtained in the book. We hope that this section will help the readers. Recall that a group G of order p m is called a p-group of maximal class if m > 2 and cl.G/ D m 1. Thus p-groups G of maximal class are nonabelian. Proposition 1.3. Suppose that a p-group G of order p m has only one subgroup of order p n , 0 < n < m. Then either G is cyclic or p D 2, n D 1 and G Š Q2m . Lemma 1.4. Let N be a normal subgroup of a p-group G. If N has no G-invariant abelian subgroup of type .p; p/, then N is either cyclic or a 2-group of maximal class. In particular, if N and G are as in Lemma 1.4 and Z2 .G/ \ N is cyclic, then N is either cyclic or a 2-group of maximal class. For N D G we get a known result of P. Roquette [Roq]: A p-group without normal abelian subgroup of type .p; p/ is either cyclic or 2-group of maximal class. Proposition 1.6 (O. Taussky). If G is a nonabelian 2-group such that jG W G 0 j D 4, then G 2 ¹D2m ; Q2m ; SD2m º. It follows from Proposition 1.6 that the groups D2m , Q2m and SD2m exhaust all groups of maximal class and order 2m (m > 2). Therefore some results stated below are trivial for p D 2. Proposition 1.8 (Suzuki). Let G be a nonabelian p-group and let A < G be of order p 2 . If CG .A/ D A, then G is of maximal class. In particular, if x 2 G # is such that jCG .x/j D p 2 , then G is of maximal class. By Theorem 9.6(f), if G is a p-group of maximal class, there is x 2 G such that jCG .x/j D p 2 . Exercise 9.1. Let G be a p-group of maximal class and order p m . Then (a) jG W G 0 j D p 2 , jZ.G/j D p, nonabelian epimorphic images of G are of maximal class. (b) If 1 i < m 1, then G has only one normal subgroup of order p i . (c) If p > 2 and m > 3, then G has no cyclic normal subgroups of order p 2 .

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Note that Exercise 9.1(b) and Lemma 1.4 imply Exercise 9.1(c). Lemma 9.1. Let G be a noncyclic subgroup of order p m , m > 2. If G contains, for any i D 1; : : : ; m 2, only one normal subgroup of order p i , then G is of maximal class. The converse is trivial. A number of most important properties of p-groups of maximal class is listed in the following two theorems. Theorem 9.5. Let G be a group of maximal class and order p m , m p C 1. Then ˆ.G/ and G=Z.G/ have exponent p. If m D p C 1, then the group G is irregular and jÃ1 .G/j D p. It follows from Theorem 9.5 that a p-group of maximal class and order > p pC1 is irregular. Theorem 9.6. Let G be a group of maximal class and order p m , p > 2, m > p C 1. Then (a) jG W Ã1 .G/j D p p . In particular, Ã1 .G/ D Kp .G/. (b) There is G1 2 1 such that jG1 W Ã1 .G1 /j D p p1 . This subgroup G1 is absolutely regular. (c) G has no normal subgroups of order p p and exponent p (this follows from (b) and Exercise 9.1(b); here the condition m > p C 1 is essential). Moreover, if N G G and jG W N j > p, then N is absolutely regular since N < G1 . Next, if N G G of order p p1 , then exp.N / D p, N D 1 .G1 / D 1 .G 0 /. In particular, G has no normal cyclic subgroups of order p 2 (see Exercise 9.1(c)). (d) Let Z2 D Z2 .G/ be a normal subgroup of order p 2 in G, G0 D CG .Z2 /. Then G0 is regular such that j1 .G0 /j D p p1 (see (b)). (e) Let 1 D ¹M1 D G0 ; M2 ; : : : ; MpC1 º, where G0 is deﬁned in (d). Then the subgroups M2 ; : : : ; MpC1 are of maximal class (and so irregular; see Theorem 9.5). Thus the subgroups G1 from (b) and G0 from (d) coincide. The subgroup G1 is called the fundamental subgroup of G (in general, G1 is deﬁned for all m 4). (f) In this part, m 3 (i.e., we do not assume as in other parts that m > p C 1). The group G has an element a such that jCG .a/j D p 2 , i.e., CG .a/ D ha; Z.G/i (this is trivial for m 2 ¹3; 4º). (g) ŒKi .G/; Kj .G/ Ki Cj C1 .G/ for all i; j 2 ¹1; : : : ; m 1º. Here K1 .G/ D G1 .1 By Exercise 9.1(b) for jGj > p 3 , G > G1 D K1 .G/ > K2 .G/ D G 0 > > Km1 .G/ D Z.G/ > Km .G/ D ¹1º is a chain of characteristic subgroups; all indices of this chain are equal to p. 1 In

what follows we do not use this property.

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Let G1 be the fundamental subgroup of a p-group G of maximal class, jGj > p pC1 , and exp.G1 / D p nC1 . If k n, then jk .G/j D p k.p1/ . If p D 3, then G1 is metacyclic: G1 is either abelian or minimal nonabelian (this follows from Theorem 9.11, Exercise 9.1(c) and Lemma 65.2(a)). Let G be a p-group of maximal class and order p m > p pC1 and p C 1 n < m. Then the number of subgroups of maximal class and order p n in G is equal to p mn . This follows from Theorem 9.6(e). Indeed, by that theorem, our claim follows in case n D m 1. To prove this for remaining values n, one should apply induction. Theorem 9.7. If a p-group G is such that G=KpC1 .G/ is of maximal class, then G is also of maximal class.2 Theorem 9.8. Let G be a p-group, p > 2. (a) (1st Hall regularity criterion) If G is absolutely regular, it is regular.3 (b) (Hall) If G has no normal subgroups of order p p1 and exponent p, it is regular.4 (c) (2nd Hall regularity criterion) If jG 0 W Ã1 .G 0 /j < p p1 , then G is regular. In particular, if G 0 is cyclic, then G is regular. (d) If G is irregular, then G 0 has a characteristic subgroup of exponent p and order p p1 . Theorem 9.11 ([Bla2, Theorem 2.6(i)]; Huppert; see also Corollary 36.7). If p > 2, then G is metacyclic if and only if jG W Ã1 .G/j p 2 . Exercise 9.13. If a p-group of maximal class, p > 2, possesses a subgroup H such that d.H / > p 1, then G Š †p2 , a Sylow p-subgroup of the symmetric group of degree p 2 .5 In particular, d.H / p for all H G (this is also true for p D 2). Remark 10.5. If A is a subgroup of a p-group G such that NG .A/ is of maximal class, so is G. Proposition 10.17. Let B be a nonabelian subgroup of order p 3 of a p-group G. If B satisﬁes CG .B/ < B, then G is of maximal class. It follows that if a 2-group G has at most two proper subgroups isomorphic to D8 , then G 2 ¹SD16 ; D16 ; SD32 º: Note that we do not know all 2-groups which contain only one subgroup isomorphic to Q8 . 2 For

a more general result, see Theorem 12.9. is also true for p D 2 since absolutely regular 2-groups are cyclic. 4 This is a partial case of Theorem 12.1(a). 5 This is not true for p D 2 as the dihedral group D 16 shows. 3 This

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Proposition 10.19. If a metacyclic p-group G contains a nonabelian subgroup of order p 3 , then either jGj D p 3 or G is a 2-group of maximal class.6 Proposition 10.24. Let R be a subgroup of order p of a nonabelian p-group G. If there is only one maximal chain connecting R with G, then either CG .R/ Š Ep2 (in this case, G is of maximal class by Proposition 1.8) or G Š Mpn . Two very important characterizations of p-groups of maximal class are contained in the following theorem which is due to Blackburn. These results are most frequently cited in our book. Theorem 12.1. (a) If a p-group G has no normal subgroup of order p p and exponent p, it is either absolutely regular or irregular of maximal class. (b) Suppose that a p-group G is not absolutely regular. If G contains an absolutely regular subgroup H of index p, then either G is of maximal class or G D H 1 .G/, where 1 .G/ is of order p p and exponent p. In particular, if all regular subgroups of an irregular p-group G are absolutely regular, then G is of maximal class and j1 .G/j D p p1 . The most important assertion of Lemma 1.4 follows from Theorem 12.1(a). Theorem 1.2 follows from Theorem 12.1(b) and Proposition 1.6. It follows from Theorem 12.1(a) that if N is a normal subgroup of a p-group G and N \ Zp .G/ is absolutely regular, then N is either absolutely regular or of maximal class. Recall that ck .G/ is the number of cyclic subgroup of order p k of a p-group G. Lemma 12.3. Let a p-group G be of maximal class and order p m > p pC1 , p > 2. Then (a) c1 .G/ 1 C p C C p p2 .mod p p /. (b) c2 .G/ p p2 .mod p p1 /. (c) If n > 2, then p p1 j cn .G/ D cn .G1 /. (d) The number of subgroups of order p p1 and exponent p in G is 1 .mod p mp /. Condition m > p C 1 is important in Theorem 12.1. Note that the classiﬁcation of groups of order p pC1 is one of the most important problems in p-group theory. Proposition 12.6. Suppose that a group G of order p m is not of maximal class and m > n > p C 1. Let N be the set of normal subgroups D of G such that G=D is of maximal class and order p n . Then jN j 0 .mod p/. The following theorem generalizes Theorem 9.7. Theorem 12.9. Suppose that a p-group G has only one normal subgroup L of index p pC1 . If G=L is of maximal class, then G is also of maximal class.7 6 This

follows from Proposition 10.17. is easily seen that this theorem follows from Proposition 12.6 (this was not noticed in 12 so that theorem was proved independently). 7 It

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Theorem 12.12. (a) If a two-generator p-group G contains a subgroup of maximal class and index p, then G itself is of maximal class. If d.G/ D 3, then G 0 D ˆ.G/, i.e., G=G 0 Š Ep3 . (b) Suppose that a group G of order p m , m > 3, contains a subgroup H of maximal class and index p. If G is not of maximal class, then exactly p 2 maximal subgroups of G are of maximal class. If, in addition, p > 2 and m > 4, then the remaining p C 1 maximal subgroups of G have no two generators and their intersection .G/ has index p 2 in G. If, in addition, H is irregular, then G=KpC1 .G/ is of order p pC1 and exponent p. Proposition 12.13. Let G be a p-group. If A 2 1 is absolutely regular and M < G is irregular of maximal class, then G is also of maximal class. Theorem 13.2. If G is a group of order p m which is neither absolutely regular nor of maximal class, then (a) c1 .G/ 1 C p C C p p1 .mod p p /. In particular, the number of solutions of x p D 1 in G is divisible by p p . (b) If k > 1, then p p1 divides ck .G/. Theorem 13.2 implies Theorem 10.17. Recall that ek .G/ is the number of subgroups of order p k and exponent p in a p-group G. The following theorem is the most important counting result. Theorem 13.5 and 13.6. If a p-group G is neither absolutely regular nor of maximal class, then (a) ep .G/ 1 .mod p/. (b) Let jGj D p m > p n p pC1 . Then the number of subgroups of maximal class and order p n in G is divisible by p 2 . It follows that if N is a normal subgroup of a p-group G and N has no G-invariant subgroup of order p p and exponent p, then it is either absolutely regular or of maximal class. Next, if Zp .G/ \ N is absolutely regular, then N is either absolutely regular or of maximal class. Proposition 13.14. (a) If 1 .G/ is either absolutely regular or irregular of maximal class, so is the p-group G. (b) A p-group G is of maximal class if and only if 2 .G/ is of maximal class. If an irregular p-group G is of maximal class, then one of the following holds: (i) j1 .G/j D p p1 , (ii) jG W 1 .G/j p. If G is of maximal class, then 2 .G/ D G.

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Groups of prime power order

Proposition 13.16. Suppose that a nonabelian p-group G has a cyclic subgroup U of order p 2 such that CG .U / is cyclic. Then G is of maximal class. It is important in the hypothesis of Proposition 13.16 that jU j D p 2 ; for jU j > p 2 the assertion does not hold. For a proof, see Appendix 40. Remark 13.4. Let G be of order p m > p k p pC1 , and M 2 1 . If H < G is of maximal class and order p k such that H 6 M and all such H are of maximal class, then G is also of maximal class. Remark 13.6. Let a p-group G satisfy the following conditions: (i) G contains a proper abelian subgroup A of order p 3 . (ii) Whenever A < H G and jH W Aj D p, then jZ.H /j D p. Then: (a) G is of maximal class, (b) jG W Aj D p. To justify (b), it sufﬁces to take A in CG .R/, where R G G is of order p 2 . Exercise 13.10. (a) Let H < G, where G is a p-group. If every subgroup of G of order pjH j containing H is of maximal class, then G is also of maximal class. (b) Let A be a proper absolutely regular subgroup of a p-group G, p > 2, exp.A/ > p, such that whenever A < B G with jB W Aj D p, then 1 .B/ D B. Then G is of maximal class. If, in addition, jAj > p p , then jG W Aj D p. Proposition 13.18. Let M < G be of maximal class, G is a p-group. (a) Set D D ˆ.M /, N D NG .M / and C D CN .M=D/. Let t be the number of subgroups K G of maximal class such that M < K and jK W M j D p. Then t D c1 .N=M / c1 .C =M /. If G is not of maximal class, then t 0 .mod p/. (b) Suppose, in addition, that M is irregular and G is not of maximal class and a positive integer k is ﬁxed. Then the number t of subgroups L < G of maximal class and order p k jM j such that M < L is a multiple of p. Theorem 13.19. Let G be a group of maximal class of order p m and exponent p e , m > p C 1. Then (a) If 3 k e, then k .G/ G1 , where k .G/ D hx 2 G j o.x/ D p k i, i.e., every element in G G1 has order p 2 . If m > p C 1, then exp.G1 / D p e . (b) If x 2 G G1 , then x p 2 Z.G/ so Hp .G=Z.G// D G1 =Z.G/. Here Hp .G/ D hx 2 G j o.x/ > pi is the Hughes subgroup of G. (c) If H < G, H 6 G1 and jH j > p p , then H is of maximal class. (d) If H < G is irregular, then 1 .G1 / < H . (e) If H < G is of order p p and H 6 G1 , then 1 .G1 / < H . (f) If p > 2 and L < G is of order p p1 , then either L < G1 or Z.G/ < L. (g) If L < G is of order p p1 and L 6 G1 , then 1 .G/ normalizes L. It follows from Theorems 9.5 and 13.19(b) that if G is of maximal class, then we have 2 .G/ D G.

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Theorem 36.12. Let G be a p-group of exponent p e > p 2 and 1 < k < e. Let U be a maximal member of the set of subgroups of G with exponent p k . (a) If U is absolutely regular, then G is also absolutely regular, U D k .G/ and the subgroup U is not of maximal class. (b) If U is irregular of maximal class, then G is also of maximal class. Lemma 36.13. Let G be a p-group of order > p pC1 with j2 .G/j D p pC1 . Then one of the following holds: (a) G is absolutely regular. (b) G is an Lp -group. n

(c) p D 2 and G D ha; b j a2 D 1; a2

n1

D b 4 ; ab D a1C2

n2

i.

Theorem 36.14. Let G be a p-group and suppose that H < G is a maximal member of the set of subgroups of G of exponent p 2 . Suppose, in addition, that jH j D p pC1 . Then one of the following holds: (a) H D 2 .G/ so G is as in Lemma 36.13. (b) G is a 2-group of maximal class (in this case, H is also of maximal class). Recall that Ã2 .G/ D Ã1 .Ã1 .G//. We have Ã2 .G/ Ã2 .G/ (strict inequality is possible). Theorem 36.16. Suppose that a p-group G is such that G=Ã2 .G/ is of maximal class. Then G is also of maximal class. Proposition 25.8(c). If G is an irregular p-group of maximal class and W G ! G0 is a lattice isomorphism, then G is also of maximal class.

120

Nonabelian 2-groups such that any two distinct minimal nonabelian subgroups have cyclic intersection

Our technique with minimal nonabelian subgroups of p-groups gives also the following rather deep result. This solves a problem stated by the ﬁrst author for p D 2. Theorem 120.1. Let G be a nonabelian 2-group such that any two distinct minimal nonabelian subgroups of G have cyclic intersection. Then we have one of the following possibilities: (a) G is minimal nonabelian. (b) G is a 2-group of maximal class. (c) G D Q V , where Q Š Q2n , n 3, is generalized quaternion and V ¤ ¹1º is elementary abelian. (d) G D ha; h j a8 D h4 D 1; Œa; h D v; v 4 D 1; Œv; h D v 2 ; a2 D vh2 i, where jGj D 25 ;

G 0 D hvi Š C4 ;

Z.G/ D 1 .G/ D hh2 ; v 2 i Š E4 ;

ˆ.G/ D hh2 ; vi Š C2 C4 ;

cl.G/ D 3;

and maximal subgroups of G are hh; vi Š H2 (minimal nonabelian), hai hh2 i (abelian of type .8; 2/) and hah; vi hh2 i Š Q8 C2 . Conversely, each of the above groups satisﬁes the assumption of our theorem. Proof. Since minimal nonabelian 2-groups and 2-groups of maximal class satisfy the assumption of our theorem, we may assume that G is neither minimal nonabelian nor a 2-group of maximal class. Suppose that G possesses noncommuting involutions. Since any two noncommuting involutions generate a dihedral subgroup, it follows that D D hz; u; t j z 2 D u2 D t 2 D 1; Œu; z D Œt; z D 1; Œu; t D zi Š D8 is a subgroup of G, where Z.D/ D D 0 D hzi and hu; zi and ht; zi are two four-subgroups of D. If CG .D/ D, then G is of maximal class (Proposition 10.17), a contradiction. Hence CG .D/ 6 D and let v 2 CG .D/ D be such that v 2 2 hzi. Then

120

2-groups with cyclic intersections of minimal nonabelian subgroups

193

we get D1 D hz; u; vt i Š D8 with D \ D1 D hz; ui Š E4 , contrary to our assumption. We have proved that 1 .G/ is elementary abelian of order 4 (since G is not generalized quaternion). Assume that E D 1 .G/ 6 Z.G/. Let t 2 E be a noncentral involution and set H D CG .t/ so that E H and H < G. Let S be a subgroup of G such that S > H 2 and jS W H j D 2 and take an element x 2 S H . Then t 0 D t x ¤ t and .t 0 /x D t x D t which gives W D ht; t 0 i Š E4 , CG .W / D H and S D NG .W /. Set z D t t 0 so that z x D z, Œx; t D z and so M D hx; ti D W hxi is minimal nonabelian with M 0 D hzi. We have either 1 .hxi/ D hzi (in which case M Š M2n , n 4, since M Š D8 is not possible because there are no noncommuting involutions) or 1 .hxi/ D hui with u 2 E W (in which case M is nonmetacyclic minimal nonabelian with 1 .M / D hui W Š E8 ). Suppose that M D hx; ti ¤ S . In this case, M \ H < H and take an element h 2 H M so that y D hx 2 S H and y 62 M . We get t y D t hx D t x D t 0 ;

.t 0 /y D t y D t 2

and

z y D .t t 0 /y D z:

Thus M1 D hy; ti is minimal nonabelian with M \ M1 ht; t 0 i Š E4 , contrary to our assumption. We have proved that S D M D hx; ti D W hxi is minimal nonabelian. If, in addition, 1 .hxi/ D hzi, then E D W E G. But we know that NG .W / D S and so S D G is minimal nonabelian, a contradiction. Hence E D 1 .hxi/ W D 1 .G/ Š E8 and CG .E/ D H so that G=H is isomorphic to a subgroup of D8 and S < G. Since H is a maximal normal abelian subgroup of G, we may use Lemma 57.1. For any element g 2 G S , there is h 2 H such that hg; hi is minimal nonabelian. By our assumption, hg; hi \ 1 .S/ is cyclic and so 1 .hg; hi/ Š C2 which forces hg; hi Š Q8 . In particular, o.g/ D 4 and g 2 2 E so that g induces an involutory automorphism on E (noting that CG .E/ D H < S). Let e 2 E be such that e 0 D e g ¤ e which together with 2 .e 0 /g D e g D e shows that .ee 0 /g D ee 0 ¤ 1. Hence hg; ei is minimal nonabelian with hg; ei0 D hee 0 i. But hg; ei \ S he; e 0 i Š E4 ; contrary to our assumption. We have proved that E D 1 .G/ Z.G/. Suppose that all minimal nonabelian subgroups of G are quaternion. In that case, Theorem 90.1 implies that G D Q V , where Q Š Q2n , n 3, is generalized quaternion and V ¤ ¹1º is elementary abelian, and so we have obtained the groups stated in part (c) of our theorem. In what follows we assume that the group G possesses a minimal nonabelian subgroup H D ha; bi which is not quaternion. Then j1 .H /j 4 and so 1 .H / Š E4 or 1 .H / Š E8 . Since 1 .H / Z.G/, it follows by the assumption of our theorem that H E G and 1 .H / Z.H / D ˆ.H /. Also (Exercise P9), ˆ.H / D ha2 ; b 2 ; Œa; bi. Suppose that there is an involution t 2 GH . As t 2 Z.G/, we get Œat; b D Œa; b and H1 D hat; bi is minimal nonabelian with ˆ.H1 / D h.at/2 D a2 ; b 2 ; Œa; bi D ˆ.H /

194

Groups of prime power order

so that H \ H1 1 .H / is noncyclic, a contradiction. Hence we have proved that 1 .H / D 1 .G/ D E and so 4 jEj 8. Let K be any minimal nonabelian subgroup of G which is distinct from H . If K is not quaternion, then j1 .K/j 4 and 1 .K/ 1 .H / D 1 .G/ so that H \ K is noncyclic, a contradiction. We have proved that each minimal nonabelian subgroup of G which is distinct from H is quaternion. Suppose that G=H is not elementary abelian. Then there is a subgroup L > H such that L=H Š C4 . Let L0 be a unique subgroup of index 2 in L which contains H . If x 2 L L0 , then hxi covers L=H and so o.x/ 8 since there are no involutions in L0 H (noting that 1 .H / D 1 .G/). On the other hand, Lemma 57.1 implies that L is generated by its minimal nonabelian subgroups which are equal to H and quaternion subgroups of L. This shows that L possesses a quaternion subgroup which is not contained in L0 . But this contradicts the above fact that all elements in L L0 are of order 8. We have proved that G=H ¤ ¹1º is elementary abelian. In what follows we shall determine the structure of any subgroup T > H with jT =H j D 2. (i) First we show that T possesses at least one abelian maximal subgroup A. Assume that this is false. Let ¹X; Y; Zº be the set of maximal subgroups of H so that X; Y and Z are abelian and at least one of them, say X, is normal in T . In that case, X is a maximal normal abelian subgroup of T . For each x 2 T H , there is a 2 X such that hx; ai is minimal nonabelian (Lemma 57.1) and so hx; ai Š Q8 . In particular, we obtain o.x/ D 4 and 1 ¤ x 2 2 1 .H / ˆ.H / and x 2 2 Z.T /. It follows that ˆ.T / D ˆ.H / which implies that Y and Z are also maximal normal abelian subgroups of T . Let x0 be a ﬁxed element in T H so that X hx0 i is nonabelian and each minimal nonabelian subgroup of Xhx0 i is isomorphic to Q8 . By Theorem 90.1, X hx0 i D R V , where R Š Q2n , n 3, and exp.V / 2. Since x0 62 ˆ.X hx0 i/ and o.x0 / D 4 with x02 2 Z.T /, it follows that x0 inverts X. Considering in the same way the nonabelian subgroups Y hx0 i and Zhx0 i, we conclude that x0 also inverts Y and Z. But X [ Y [ Z D H and so x0 inverts on H . This implies that H is abelian, a contradiction. Hence we have proved that T possesses at least one abelian maximal subgroup A. (ii) We have 1 .H / D ˆ.H / D Z.T / so that exp.H / D 4 and either jH j D 24 if 1 .H / Š E4 or jH j D 25 if 1 .H / Š E8 . Indeed, let x 2 T .A [ H /. By Lemma 57.1, there is a 2 A such that hx; ai is minimal nonabelian. Since hx; ai ¤ H , we have hx; ai Š Q8 . In particular, we obtain o.x/ D o.a/ D 4, x 2 D a2 D z 2 Z.T / and x inverts a. Suppose that x centralizes an element b of order 4 in A. If b 2 D z, then ab is an involution and .ab/x D a1 b D .ab/z, contrary to the fact that 1 .G/ Z.G/. Hence b 2 ¤ z and we consider the subgroup hab; xi. We have Œab; x D Œa; xb Œb; x D z b D z

and so hab; xi0 D hzi

which implies that hab; xi is minimal nonabelian. But .ab/2 D a2 b 2 D zb 2 ¤ z and x 2 D z so that hab; xi 6Š Q8 , contrary to the fact that hab; xi ¤ H (since, by choice,

120

2-groups with cyclic intersections of minimal nonabelian subgroups

195

x 2 T .A [ H /). We have proved that x does not centralize any element of order 4 in A and so CA .x/ D 1 .H / D 1 .T / which implies that 1 .H / D Z.T /. On the other hand, we conclude that ˆ.H / D Z.H / and ˆ.H / A so that ˆ.H / Z.T /. This forces ˆ.H / D 1 .H /, exp.H / D 4, jH j D 24 if 1 .H / Š E4 and jH j D 25 if 1 .H / Š E8 . (iii) We have jT 0 j D 4 and so A is the unique abelian maximal subgroup of T . Each maximal subgroup X of T which is distinct from the subgroups A and H is of the form X D R1 .H / with R Š Q8 and R \ 1 .H / D Z.R/. We have jH j D 4j1 .H /j and so jT j D 8j1 .H /j as 1 .H / D ˆ.H /. In view of Lemma 1.1, we get jT j D 2jZ.T /jjT 0 j, where Z.T / D 1 .H / so that jT 0 j D 4. By Exercise P1, T has exactly one abelian maximal subgroup. Let X be any maximal subgroup of T distinct from A and H . Then X is nonabelian and each minimal nonabelian subgroup of X is isomorphic to Q8 . By Theorem 90.1, we have X D R V , where R Š Q2n , n 3, and exp.V / 2. On the other hand, 1 .H / D Z.T / D ˆ.H / ˆ.T / and so 1 .H / X which gives Z.R/ V D 1 .X/ D 1 .H /: Since jX j D jH j D 4j1 .H /j, it follows that R Š Q8 . (iv) We have d.T / D 2. Suppose that d.T / > 2 in which case we have ˆ.T / D ˆ.H / D 1 .H / D Z.T /, exp.T / D 4, T 0 Z.T / and d.T / D 3. (iv1) Assume, in addition, that 1 .H / Š E4 so that H D ha; b j a4 D b 4 D 1; Œa; b D z; z 2 ¹a2 ; b 2 ºi Š H2 ; where a 2 A and A Š C4 C4 . Let c 2 A H , where 1 ¤ c 2 2 1 .H / D ha2 ; b 2 i, c 2 ¤ a2 and T D ha; b; ci. If Œc; b D 1, then T =hzi is abelian, contrary to jT 0 j D 4. Hence Œc; b ¤ 1 and hc; bi is minimal nonabelian so that hc; bi Š Q8 (as hc; bi ¤ H ). It follows Œc; b D b 2 and c 2 D b 2 . Consider the subgroup hcb; ai with Œcb; a D z so that hcb; ai Š Q8 since hcb; ai ¤ H . We get z D a2

and

.cb/2 D c 2 b 2 Œb; c D c 2 b 2 b 2 D c 2 D b 2 ¤ a2 ;

a contradiction. (iv2) It remains to consider the case 1 .H / Š E8 so that H D ha; b j a4 D b 4 D 1; Œa; b D d; d 2 D Œd; a D Œd; b D 1i; where a 2 A and 1 .H / D ha2 ; b 2 ; d i Š E8 . Let c 2 AH , where 1 ¤ c 2 2 1 .H / and T D ha; b; ci. If Œc; b D 1, then T =hd i is abelian, contrary to jT 0 j D 4. Hence we get Œc; b ¤ 1 and hc; bi is minimal nonabelian so that hc; bi Š Q8 (since hc; bi ¤ H ). It follows that c 2 D b 2 D Œc; b. We have Œc; ab D Œc; b D b 2 so that hc; abi Š Q8 since hc; abi is minimal nonabelian distinct from H . But .ab/2 D a2 b 2 d ¤ c 2 D b 2 , a contradiction.

196

Groups of prime power order

(v) Now we determine the structure of T . As d.T / D 2, we have A \ H D ˆ.T /;

ˆ.H / D 1 .H / D Z.T / < ˆ.T /;

1 .H / D 1 .T /:

Since jˆ.T / W 1 .H /j D 2, it follows that ˆ.T / is abelian of type .4; 2/ or .4; 2; 2/ (depending on the possibilities 1 .H / Š E4 or 1 .H / Š E8 ). Take some elements h 2 H A and a 2 A H so that ha; hi D T . If Œa; h 2 1 .H / D Z.T /, then T =hŒa; hi is abelian and so T 0 D hŒa; hi, contrary to jT 0 j D 4. Hence we obtain that Œa; h D v 2 ˆ.T / 1 .H / which implies o.v/ D 4, T 0 D hvi Š C4 and T is of class 3. Set v 2 D z so that H 0 D hzi. Since H 0 is not a maximal cyclic subgroup in H , it follows that H is metacyclic, 1 .H / Š E4 and so jT j D 25 . We have H D hh; v j h4 D v 4 D 1; Œh; v D v 2 D zi Š H2 ; where 1 .H / D hh2 ; zi Š E4 . For any element x 2 T .A [ H / there is an element k 2 A such that hx; ki is minimal nonabelian (Lemma 57.1) and so hx; ki Š Q8 which implies that o.x/ D 4 and x 2 2 1 .H /. Also, ˆ.H / D 1 .H / and so all elements in T A are of order 4. If any element in A H is of order 4, then Ã1 .T / D 1 .H /, contrary to A\H D ˆ.T /. Hence o.a/ D 8 and a2 2 .A\H /1 .H /. All elements in A H are of order 8 and so exp.T / D 8. By (iii), we get hahiˆ.T / Š Q8 C2 . Since Œah; v D Œa; vh Œh; v D Œh; v D z, we must have hah; vi Š Q8 and therefore v 2 D z D .ah/2 D ahah D a2 a1 h1 h2 ah D a2 h2 Œa; h D a2 h2 v and so a2 D vh2 . The structure of T is uniquely determined: T D ha; h j a8 D h4 D 1; Œa; h D v; v 4 D 1; Œv; h D v 2 ; a2 D vh2 i; where all elements in AH are of order 8 and A D haihh2 i is abelian of type .8; 2/. It is now easy to determine the structure of G. We know that G=H ¤ ¹1º is elementary abelian. Also, all minimal nonabelian subgroups of G are equal to H Š H2 and some quaternion subgroups. Suppose that jG=H j > 2. Then there exists a subgroup G0 > H such that G0 =H Š E4 . If T =H is a subgroup of order 2 in G0 =H , then the structure of T is uniquely determined in (v) and for any other subgroup L=H of order 2 in G0 =H , we have L Š T . In particular, exp.G0 / D exp.T / D 8. By Theorem 92.6, G0 has a unique abelian maximal subgroup A0 and A0 \ T D A is the unique abelian maximal subgroup of T and is of type .8; 2/. All elements in A0 H are of order 8 and 2 .A0 / D A0 \ H D A \ H which is abelian of type .4; 2/. There is no such abelian group A0 of order 25 . We have proved that G D T and so we have obtained the group of order 25 stated in part (d) of our theorem.

121

p-groups of breadth 2

In this section we prove a result of Parmeggiani and Stellmacher [PS1] on p-groups of breadth 2. We recall that the breadth bG .x/ of an element x in a p-group G is deﬁned as p bG .x/ D jG W CG .x/j and the breadth b D b.G/ is max¹bG .x/ j x 2 Gº. We know that for a p-group G we have b.G/ D 1 if and only if jG 0 j D p (see Exercise 2.7 and Proposition 121.9). This result is also proved in [Kno], where we additionally ﬁnd the result that jG 0 j p 3 for p-groups with b.G/ D 2. Theorem 121.1. Let G be a p-group. Then b.G/ D 2 if and only if one of the following holds: (a) jG 0 j D p 2 , or (b) jG 0 j D p 3 and jG W Z.G/j D p 3 . Moreover, if b.G/ D 2 and jG 0 j D p 3 , then jA=Z.G/j D p for every maximal normal abelian subgroup A of G. We shall prove Theorem 121.1 in a series of lemmas. For that purpose we need the following deﬁnitions. Let A be a normal abelian subgroup in a p-group G and x 2 G. Then p bA .x/ D jA W CA .x/j;

bA D bA .G/ D max¹bA .x/ j x 2 Gº;

BA D ¹x 2 G j bA .x/ D bA .G/º; Mx D ACG .x/;

Gx D hMax

TA D ¹x 2 G j bA .x/ D 1º; \ j a 2 Ai; Dx D Max : a2A

Lemma 121.2. For every x 2 G, ŒA; x D ŒA; hxi D ¹Œa; x j a 2 Aº. In particular, jŒA; xj D p bA .x/ D jA W CA .x/j. Proof. The map a ! Œa; x 2 A (a 2 A) is a homomorphism from A into A since 0 for any a; a0 2 A we have Œaa0 ; x D Œa; xa Œa0 ; x D Œa; xŒa0 ; x. The kernel of this homomorphism is CA .x/ and so A=CA .x/ Š ¹Œa; x j a 2 Aº D A0 , which is a subgroup of A. Suppose that Œa; x i 2 A0 for some i 1. Then Œa; x i C1 D Œa; x i x D Œa; xŒa; x i x D Œa; xŒa; x i ŒŒa; x i ; x 2 A0 and so ŒA; x D ŒA; hxi which gives jŒA; hxij D p bA .x/ .

198

Groups of prime power order

Lemma 121.3. Let u; v 2 G be such that ŒA; u \ ŒA; v D ¹1º. Then CA .u/ CA .v/ or bA .u/ < bA .uv/. Proof. Suppose that CA .u/ 6 CA .v/. Then there is a 2 CA .u/ CA .v/. Thus we get auv D av ¤ a and 1 ¤ Œa; uv D Œa; vŒa; uv D Œa; v so that ŒA; uv \ ŒA; v ¤ 1. We have ()

ŒA; u ŒA; v D ŒA; uvŒA; v:

Indeed, for any a 2 A, we obtain Œa; uv D Œa; vŒa; uv D Œa; vŒa; uŒŒa; u; v 2 ŒA; u ŒA; v; and Œa; u D Œa; .uv/v 1 D Œa; v 1 Œa; .uv/v

1

D Œa; v 1 Œa; .uv/ŒŒa; .uv/; v 1 2 ŒA; uvŒA; v; which proves (). From () and ŒA; uv\ŒA; v ¤ ¹1º it then follows (using also Lemma 121.2) that p bA .u/CbA .v/ < p bA .uv/CbA .v/ which gives bA .u/ < bA .uv/ and we are done. Lemma 121.4. Let x; y 2 TA , T D hx; yi and TN D .T CG .A//=CG .A/. Then one of the following holds: (a) T CG .A/ TA and ŒA; x D ŒA; T ; moreover, CA .x/ ¤ CA .y/ if TN is not cyclic. (b) T CG .A/ TA and CA .x/ D CA .T /; moreover, ŒA; x ¤ ŒA; y if TN is not cyclic and in this case ŒA; T D ŒA; x ŒA; y. (c) CA .T / D CA .x/ \ CA .y/ and ŒA; T D ŒA; xy D ŒA; x ŒA; y. Proof. If TN is cyclic, then jT =TG .A/j D p and so CA .x/ D CA .y/ D CA .T /, and (b) holds. Indeed, we conclude that j.Ahxi/ W .CA .x/hxi/j D p so that if a 2 A CA .x/, then Œa; x 2 A \ .CA .x/hxi/ D CA .x/. Hence 1 ¤ Œa; x 2 Z.ha; xi/ and therefore ha; xi0 D hŒa; xi is of order p. This implies that ha; xi is minimal nonabelian and so Œa; x p D 1 and x p 2 CG .A/. Similarly, we get y p 2 CG .A/ and therefore if TN is cyclic, then jT =CT .A/j D p. In what follows we may assume that the quotient group TN is not cyclic. Assume that ŒA; x D ŒA; y D hsi (where o.s/ D p) and CA .x/ D CA .y/. If a 2 A CA .x/, then we may set ax D as and ay D as j with j 6 0 .mod p/. We have seen in the previj ous paragraph that s 2 Z.Ahxi/ and s 2 Z.Ahyi/. Hence we have ax D as j and j so ax y D as j Cj D a and x j y 2 CG .A/, contrary to our assumption that TN is not cyclic. We have proved that both ŒA; x D ŒA; y and CA .x/ D CA .y/ cannot hold. Assume ﬁrst that xy 2 TA . If, in addition, ŒA; x D ŒA; y D ŒA; T , then by the above CA .x/ ¤ CA .y/ and (a) holds since for each v 2 T CG .A/, ŒA; v D ŒA; T

121 p-groups of breadth 2

199

is of order p and so v 2 TA . If ŒA; x ¤ ŒA; y, then we obtain ŒA; x \ ŒA; y D ¹1º and Lemma 121.3 implies that CA .x/ D CA .y/ D CA .T / and (b) holds since for each v 2 T CG .A/ we conclude that CA .v/ D CA .T / is of index p in A and so v 2 TA . We have jA W CA .T /j D p, CA .T / Z.AT / and A0 D ŒA; x ŒA; y CA .T /. Considering .AT /=A0 , we see that both x and y centralize A=A0 and so T D hx; yi centralizes A=A0 which gives ŒA; T A0 and ŒA; T D ŒA; x ŒA; y. Assume now that xy 62 TA . Since xy 62 CG .A/ (since TN is noncyclic), we have CA .xy/ D CA .x/ \ CA .y/ and so bA .xy/ D 2 and jŒA; xyj D p 2 . On the other hand, for each a 2 A Œa; xy D Œa; yŒa; xy D Œa; yŒa; xŒŒa; x; y 2 ŒA; xŒA; y: Thus ŒA; xy D ŒA; x ŒA; y D ŒA; T and (c) holds. Lemma 121.5. Let x; u; v 2 G be such that uv 1 62 CG .A/ and bA .x/ > 1. Then at least one of the elements u; v; uv; ux; vx; uvx is not in TA . Proof. Assume that D ¹u; v; uv; ux; vx; uvxº TA : Consider

T1 D huv 1 ; ui;

T2 D huv 1 ; vxi;

T3 D hu; vxi:

Then case (a) or (b) of Lemma 121.4 applies to each of the subgroups T1 ; T2 ; T3 . Indeed, considering T3 D hu; vxi, we have u; vx; u.vx/ 2 TA and so (a) or (b) of Lemma 121.4 holds. Consider T1 D huv 1 ; ui, where uv 1 62 CG .A/. We have v 2 T1 and so T1 D huv 1 ; ui D hu; vi, where u; v; uv 2 TA and so (a) or (b) of Lemma 121.4 holds for T1 . In particular, T1 CG .A/ TA and so uv 1 2 TA . Finally, consider T2 D huv 1 ; vxi. Since uv 1 ; vx; uv 1 vx D ux 2 TA , again (a) or (b) of Lemma 121.4 holds for T3 . By Lemma 121.4, we have jŒA; Ti j D p or p 2 for each i 2 ¹1; 2; 3º. Then ﬁx i; j 2 ¹1; 2; 3º such that i ¤ j and jŒA; Ti j D jŒA; Tj j. Since i ¤ j , we have hTi ; Tj i D hi and

Ti \ Tj 6 CG .A/:

Indeed, each Tk hi, k D 1; 2; 3. Also, hT1 ; T2 i D huv 1 ; u; vxi D hu; v; xi D hi; hT1 ; T3 i D huv 1 ; u; vxi D hT1 ; T2 i D hi; hT2 ; T3 i D huv 1 ; u; vxi D hT1 ; T2 i D hi: Finally, uv 1 2 T1 \ T2

and uv 1 62 CG .A/;

u 2 T 1 \ T3

and u 2 TA ;

vx 2 T2 \ T3

and vx 2 TA :

Hence Ti \ Tj 6 CG .A/.

200

Groups of prime power order

Assume that jŒA; Ti j D jŒA; Tj j D p 2 . Then case (b) of Lemma 121.4 applies to Ti and Tj , and every element in hi CG .A/ has the same centralizer on A. Since x 2 hi, it follows that x 2 TA , a contradiction. Assume now that jŒA; Ti j D jŒA; Tj j D p. Then, due to Lemma 121.4(a) and (b), every element in hiCG .A/ has the same commutator with A. But x 2 hiCG .A/ and so jŒA; xj D p and x 2 TA , a contradiction. Lemma 121.6. Let x 2 G and a 2 A. Then the following hold: (a) Œa; Max \ Mx ŒA; x. (b) ŒA; Dx ŒA; x D ŒDx ; x. (c) ŒGx ; x ŒA; G. Proof. Let y 2 Max D CG .ax/A so that y D cb for some c 2 CG .ax/ and b 2 A. It follows ay x y D .ax/y D .ax/cb D .ax/b D ax b and this gives Œa; y D a1 ay D a1 ax b .x 1 /y D x b .x 1 /y D .x b x 1 /.x.x 1 /y / D Œb; x 1 Œx 1 ; y: In particular, Œx 1 ; y 2 ŒA; G and so (noting that ŒA; G E G) Œy; x 2 ŒA; G which gives ŒGx ; x ŒA; G, and this is (c). Assume that y 2 Mx and so y D a0 d , where a0 2 A and d 2 CG .x/. We get Œx 1 ; y D Œx 1 ; a0 d D Œx 1 ; d Œx 1 ; a0 d D Œx 1 ; .a0 /d 2 ŒA; x 1 D ŒA; x: If, in addition, y 2 Max , then we know from the previous paragraph that Œa; y can be written as Œa; y D Œb; x 1 Œx 1 ; y, where b 2 A and so Œa; y 2 ŒA; x, which is (a). We note that ŒMx ; x D ŒA; x. Indeed, if y 2 Mx , then y D a0 d with a0 2 A, d 2 CG .x/ and so Œa0 d; x D Œa0 ; xd Œd; x D Œ.a0 /d ; x 2 ŒA; x: This result together with (a) yields Œa; Dx Œa; Max \ Mx ŒA; x ŒDx ; x ŒMx ; x D ŒA; x and thus ŒA; Dx ŒA; x D ŒDx ; x and this is (b). Lemma 121.7. Let bA > 1. Suppose that ŒG; z ŒA; G for all z 2 G satisfying () Then G 0 D ŒCG .A/; G.

bA .z/ > 1 and

2bA .z/ bA :

121 p-groups of breadth 2

201

Proof. Let x 2 BA and y 2 G. Then we have ŒA; x ŒA; yŒA; yx. Indeed, we obtain that ŒA; y D ¹Œa; y j a 2 Aº and for each a 2 A, Œa; yx D Œa; xŒa; yx D Œa; xŒa; yŒŒa; y; x D Œa; yŒaŒa; y; x D Œa; yŒay ; x which gives ŒA; yx D ¹Œa; yŒay ; x j a 2 Aº. Thus, Œay ; x 2 ŒA; yŒA; yx for all a 2 A and ¹Œay ; x j a 2 Aº D ŒA; x which proves our assertion. It follows that we have one of the following two cases (considering the subgroup ŒA; yŒA; yx which contains ŒA; x, where jŒA; xj D p bA ): (1) y or yx satisfy () for all y 2 G, or (2) y and yx 2 TA for some y 2 G, and bA D 2. Assume that we are in case (1). Then we have ŒG; y ŒA; G or ŒG; yx ŒA; G for all y 2 G. Since x satisﬁes (), we also obtain ŒG; x ŒA; G. Thus, if for an element y 2 G, ŒG; yx ŒA; G, we have for all g 2 G, Œg; yx D Œg; xŒg; yx and so Œg; yx 2 ŒA; G which implies Œg; y 2 ŒA; G since ŒA; G E G and therefore ŒG; y ŒA; G for all g 2 G. Thus G 0 D ŒA; G D ŒCG .A/; G. Suppose that we are in case (2). Then ŒG; z ŒA; G for all z 2 G .TA [ CG .A// and thus ŒG; z ŒCG .A/; G for every z 2 G TA . Assume that there exist u; v 2 G such that Œu; v 62 ŒCG .A/; G, i.e., ŒG; u 6 ŒCG .A/; G and ŒG; v 6 ŒCG .A/; G. Then also Œuv; v; Œux; v; Œvx; u; Œuvx; v 62 ŒCG .A/; G. Indeed, Œuv; v D Œu; vv 62 ŒCG .A/; G; Œux; v D Œu; vx Œx; v 62 ŒCG .A/; G since Œx; v 2 Œx; G ŒA; G, Œu; v 62 ŒCG .A/; G, and Œvx; u D Œv; ux Œx; u 62 ŒCG .A/; G; Œuvx; v D Œuv; vx Œx; v 62 ŒCG .A/; G since Œx; v 2 ŒA; G and Œuv; vx 62 ŒCG .A/; G. Hence u; v; uv; ux; vx; uvx 2 TA . Moreover, Œu; vv

1

1

D Œuv ; v D Œv.uv 1 /; v/ D Œuv 1 ; v 62 ŒCG .A/; G

and so uv 1 62 CG .A/. But then Lemma 121.5 gives a contradiction. Lemma 121.8. Suppose bA D 1. Then either (i) jA W .A \ Z.G//j D p and CA .x/ D CA .y/ for every x; y 2 G CG .A/, or (ii) jŒA; Gj D p. If, in addition, A is a maximal normal abelian subgroup of G, then in case (i) jG W Z.G//j p 1Cb and in case (ii) b.G=ŒA; G/ < b.G/.

202

Groups of prime power order

Proof. If G=CG .A/ is cyclic, then there is g 2 G such that G D CG .A/hgi. It follows that jA=CA .g/j D p, CA .g/ D Z.G/ \ A and ŒA; g D ŒA; G is of order p so (i) and (ii) hold. Hence we may assume that there exist x; y 2 G such that for T D hx; yi the quotient group T CG .A/=CG .A/ is not cyclic. Then cases (a) or (b) of Lemma 121.4 hold for T . Assume that (b) holds for T , i.e., CA .x/ D CA .y/ and ŒA; x ¤ ŒA; y. Then for every z 2 G CG .A/ either ŒA; z ¤ ŒA; x or ŒA; z ¤ ŒA; y. Hence Lemma 121.4 applied to hz; xi or hz; yi gives CA .x/ D CA .y/ D CA .z/ and thus CA .x/ D A \ Z.G/. This is (i). Assume that (a) holds for T , i.e., CA .x/ ¤ CA .y/ and ŒA; x D ŒA; y. Thus for every z 2 G CG .A/ either CA .z/ ¤ CA .x/ or CA .z/ ¤ CA .y/. Hence another application of Lemma 121.4 gives ŒA; x D ŒA; y D ŒA; z and thus ŒA; G D ŒA; x is of order p. This is (ii). Let A be a maximal normal abelian subgroup of G. Suppose that (i) holds and let a 2 A Z.G/. Then CG .a/ D A

and jG W Z.G/j D jG W AjjA W Z.G/j D jG W CG .a/jp p bC1 :

Assume that (ii) holds. Then D D ŒA; G is a normal subgroup of G of order p such that b.G=D/ < b D b.G/. Proposition 121.9 (Knoche). Suppose that b D b.G/ D 1. Then jG 0 j D p. Proof. We suppose, in addition, that A is a maximal normal abelian subgroup of G so that CG .A/ D A. For every x 2 G A we have G D CG .x/A D Dx D Gx . Hence Lemma 121.6(b) gives jŒG; Aj D p and Lemma 121.6(c) yields ŒG; x ŒG; A and so G 0 D ŒG; A is of order p. Proof of Theorem 121.1. Let A be a maximal normal abelian subgroup of G and suppose that b D b.G/ D 2. If bA D 2 for some maximal normal abelian subgroup A of G, then for each x 2 G such that bA .x/ D 2 we have Mx D CG .x/A D G so that Gx D G and Lemma 121.6(c) gives ŒG; x ŒA; G. Hence the condition () of Lemma 121.7 is satisﬁed and so G 0 D ŒA; G. Since G D Max D CG .ax/A for every a 2 A, we have G D Dx and so Lemma 121.6(b) gives ŒA; G ŒA; x. Thus ŒA; G D ŒA; x is of order p 2 and so jG 0 j D jŒA; xj D p 2 . Hence we may assume that bA D 1 for every maximal normal abelian subgroup A of G. We discuss the two cases of Lemma 121.8. Assume that case (i) of Lemma 121.8 holds. For a 2 A Z.G/ we get CG .a/ D A and thus jG=Z.G/j p 3 . Now, b D 2 gives jG=Z.G/j D p 3 and jŒA; Gj D p 2 since a has exactly p 2 conjugates in G. Consider GN D G=ŒA; G so that GN is either N G= N AN Š Ep2 and so GN has more than one abelian maximal abelian or AN D Z.G/, subgroup which implies jGN 0 j D p. Thus jG 0 =ŒA; Gj p and so we get (a) or (b) of our theorem.

121 p-groups of breadth 2

203

Assume that case (ii) of Lemma 121.8 holds, i.e., jŒA; Gj D p. Set GN D G=ŒA; G. N and b.G/ N 1. Hence by Proposition 121.9, jGN 0 j p. As jŒA; Gj D p Then AN Z.G/ and b D 2, we conclude (by Proposition 121.9) that jG 0 j D p 2 and we get case (a) of our theorem. Suppose that G satisﬁes (a) or (b) of our theorem. Then clearly b 2 and Proposition 121.9 gives b D 2 since jG 0 j > p. Theorem 121.1 is proved.

122

p-groups all of whose subgroups have normalizers of index at most p

This section is written by the second author. Here we give a proof of the following result which was proved for p > 2 and predicted for p D 2 by J. P. Bohanon in [Boh]. Theorem 122.1. Let G be a p-group such that for each subgroup H of G we have jG W NG .H /j p. Then the following holds: (a) If p D 2, then jG W Z.G/j 24 . (b) If p > 2, then jG W Z.G/j p 3 . Our proof is elementary but very involved. In particular, we do not use any computer results for 2-groups of orders 28 which have the title property and also we do not use the representation theory for such groups in case p > 2. We recall that the breadth bG .x/ of an element x in a p-group G is deﬁned as p bG .x/ D jG W CG .x/j; and the (element) breadth b.G/ of G is max¹bG .x/ j x 2 Gº: Also, for an abelian normal subgroup A in a p-group G and an element x 2 G we set p bA .x/ D jA W CA .x/j and deﬁne bA D bA .G/ D max¹bA .x/ j x 2 Gº; BA D ¹x 2 G j bA .x/ D bA .G/º; TA D ¹x 2 G j bA .x/ D 1º: We deﬁne the subgroup breadth sb.G/ of a p-group G by p sb.G/ D max¹jG W NG .H /j j H Gº: Note that the title groups are either Dedekindian or have the subgroup breadth 1.

122

p-groups all of whose subgroups have normalizers of index at most p

205

We shall ﬁrst prove some preliminary results (Propositions 122.2 to 122.18 ) about p-groups with the subgroup breadth 1. Some of these results are of independent interest since they give additional information about such groups which is not contained in Theorem 122.1. In particular, Propositions 122.2, 122.7, 122.8, 122.13 and 122.16 give important properties of p-groups with the subgroup breadth 1. Proposition 122.2. Let G be a p-group with sb.G/ D 1. Then ˆ.G/ is abelian. Proof. Obviously, ˆ.G/ is Dedekindian and so either ˆ.G/ is abelian or p D 2 and ˆ.G/ is Hamiltonian, i.e., ˆ.G/ D Q V , where Q Š Q8 and exp.V / 1. Suppose that we are in the second case, where V ¤ ¹1º (by a classical result of Burnside). Set U D Z.Q/ V so that U D Z.ˆ.G// is normal in G and jU W V j D 2. If V E G, then ˆ.G=V / D ˆ.G/=V Š Q8 , contrary to a classical result of Burnside. Thus V is not normal in G and set T D NG .V / so that ˆ.G/ < T and jG W T j D 2. Take an element g 2 G T . We have V g ¤ V and V g < U . It follows that V0 D V \ V g is normal in G and jV W V0 j D 2. Hence ˆ.G=V0 / D ˆ.G/=V0 Š Q8 C2 and such groups G=V0 are classiﬁed in Theorem 85.1. In particular, G=V0 contains a subgroup H of order 26 given by H D hx; y j x 8 D y 8 D 1; x 4 D y 4 D z; x 2 D a; y 2 D b; Œa; b D z; ay D at; t 2 D Œt; a D Œt; b D 1; t x D t y D t z; b x D btz; .xy/2 D tz ; D 0; 1i; where ha; bi Š Q8 ;

ˆ.H / D ha; bi ht i Š Q8 C2 ;

Z.H / D Z.ha; bi/ D hzi:

We have ay D at;

b x D bt z D b 1 t;

ayx D .at/x D at x D atz D a1 t

and so NH .ha; bi/ D ˆ.H / is of index 4 in H , a contradiction. Proposition 122.3. Let G be a p-group with sb.G/ D 1. If A G is a maximal normal abelian containing ˆ.G/ (noting that ˆ.G/ is abelian by Proposition 122.2), then 1 bA .G/ 2, i.e., for each x 2 G A we have p jA W CA .x/j p 2 . Proof. We infer that that G=A ¤ ¹1º is elementary abelian. Let x 2 G A so that x p 2 A and j.Ahxi/ W Aj D p. Set M D NAhxi .hxi/ and A0 D M \ A so that j.Ahxi/ W M j p and jA W A0 j p. We have M 0 A0 \ hxi D hx p i Z.Ahxi/; and so M is of class 2. Suppose that M is of class 2. For any a 2 A0 we have

206

Groups of prime power order

Œa; x 2 hx p i ¤ ¹1º since M 0 hx p i and M 0 ¤ ¹1º. We get Œa; xp D Œa; x p D 1 and so ŒA0 ; hxi D 1 .hxi/ D M 0 is of order p (see Lemma 121.2) which implies jA0 W CA0 .x/j D p. Since CG .A/ D A, we obtain p jA W CA .x/j p 2 , and we are done. Proposition 122.4. If G D Q1 Q2 , where ŒQ1 ; Q2 D ¹1º and Q1 Š Q2 Š Q8 , then we have sb.G/ > 1. Proof. First suppose that Q1 \ Q2 D Z.Q1 / D Z.Q2 / D G 0 D hzi so that G is extraspecial of order 25 and “type +”. In this case, G possesses a foursubgroup S such that S \ hzi D ¹1º. We have ŒNG .S/; S S \ hzi D ¹1º and so NG .S/ D CG .S / D S hzi which gives jG W NG .S/j D 4. Assume now that G D Q1 Q2 , and set Q1 D ha1 ; b1 i with Q10 D hz1 i and Q2 D ha2 ; b2 i with Q20 D hz2 i. Considering the subgroup U D ha1 a2 ; b1 b2 i with U 0 D hz1 z2 i, we have G D Q1 U with Q1 \ U D ¹1º and so NG .U / D U NQ1 .U /, where NQ1 .U / D CQ1 .U /. But .b1 b2 /a1 D b1 b2 z1 ;

.a1 a2 /b1 D a1 a2 z1 ;

.b1 b2 /b1 a1 D b1 b2 z1

and so CQ1 .U / D hz1 i which gives NG .U / D U hz1 i and we are done. Proposition 122.5 (J. Shareshian). Let G be a 2-group with sb.G/ D 1 and we assume that G possesses two involutions which do not commute. Then jG W Z.G/j 24 and jG 0 j 4 so that b.G/ 2. Proof. Let s; t be involutions in G which do not commute. Then hs; ti Š D2n is a dihedral subgroup of order 2n , n 3. If n > 3, then sb.D2n / > 1 and so D D hs; ti Š D8 . Set Z.D/ D hzi and note that ¹s; szº and ¹t; tzº are already complete conjugate classes in G of s and t, respectively. This implies hs; zi E G, ht; zi E G, D E G, G D D C , where C D CG .D/ and D \ C D hzi. If C is Dedekindian, then the fact that Z.C / D Z.G/ implies jG W Z.G/j 24 and jG 0 j 4. Suppose that C has a nonnormal subgroup H such that hzi 6 H and let K ¤ H be a conjugate of H in C . Then the four subgroups ht; H i D hti hH i;

ht; Ki;

htz; H i;

htz; Ki

are pairwise distinct and conjugate subgroups in G, a contradiction. Hence, assuming that C is not Dedekindian, each nonnormal subgroup of C contains hzi. Such groups C are completely determined by N. Blackburn (see Corollary 92.7) and so we have one of the following three possibilities: (1) C Š Q8 C4 E2s ; (2) C Š Q8 Q8 E2s (s 0); (3) C has an abelian normal subgroup A of exponent > 2 with jC W Aj D 2 and an element g of order 4 in C A such that g 2 D z and g inverts each element in A.

122

p-groups all of whose subgroups have normalizers of index at most p

207

In case (1), jG W Z.G/j D 24 and jG 0 j 4. In case (2), sb.Q8 Q8 / > 1 (by Proposition 122.4), a contradiction. Assume that we are in case (3). Note that .st/2 D z so that stg is an involution that inverts on A. As sb.D16 / > 1, we must have exp.A/ D 4. Suppose that there exists b 2 A such that o.b/ D 4 and b 2 ¤ z D g 2 . We obtain that hstg; bi Š D8 so that .stg/b D stgb 2 . Also, .stg/s D stgz and .stgb 2 /s D stgzb 2 . It follows that the four involutions stg, stgz, stgb 2 , stgzb 2 are pairwise distinct and conjugate, a contradiction. Therefore we have an element h 2 A such that h2 D z, hg D h1 , hh; gi Š Q8 and C Š Q8 E2s . But then C is Dedekindian, contrary to our assumption. Proposition 122.6 (J. Shareshian). Let G be a 2-group with sb.G/ D 1, Z.G/ is cyclic and all involutions of G commute with each other. Then G contains at most three involutions. Proof. Suppose that our proposition is false. Since 1 .G/ is elementary abelian of order 8, G possesses a normal elementary abelian subgroup E of order 8. Let z be the unique involution in E which lies in Z.G/, and let t be any involution in E hzi. Since jG W CG .t/j D 2, there are exactly two conjugates ¹t; t 0 º of t in G and so the four-subgroup ht; t 0 i is normal in G which implies z 2 ht; t 0 i and so t 0 D tz. This gives ŒG; E D hzi. Set E D hz; s; t i and consider the four-subgroup S D hs; ti. We have ŒNG .S/; S S \ hzi D ¹1º and so NG .S/ D CG .S/ D CG .s/ \ CG .t/, where CG .s/ and CG .t/ are subgroups of index 2 in G. Since jG W NG .S/j D 2, we get CG .s/ D CG .t/ D CG .E/ with jG W CG .E/j D 2. Let g 2 G CG .E/ so that s g D sz and t g D t z. But then .st/g D st and so st 2 Z.G/, a contradiction. Proposition 122.7. Let G be a metacyclic p-group with jG 0 j D p 2 . If sb.G/ D 1, then p D 2 and either G Š Q16 or G D ha; b j a8 D 1; b 2 D a4 ; ab D a1C4 i; n

where n 2 and D 0; 1. Proof. First suppose jGj D p 4 . Then G has a cyclic subgroup H of index p. If p > 2, then G Š Mp4 and jG 0 j D p, a contradiction. We have p D 2 and G is of maximal class. If G is dihedral or semidihedral, then there exists an involution i 2 G H and jG W CG .i/j D 4, a contradiction. Hence we obtain in this case that G Š Q16 . Now suppose that jGj > p 4 . In view of Corollary 65.3, a metacyclic p-group G is an A2 -group if and only if jG 0 j D p 2 . According to Proposition 71.2, we have two possibilities for the structure of G: (a)

m

n

G D ha; b j ap D 1; b p D ap

m1

; ab D a1Cp

m2

i;

where m 3, n 2, D 0; 1, and in case p D 2, m 4; (b)

p D 2;

G D ha; b j a8 D 1; ; b 2 D a4 ; ab D a1C4 i;

where n 2, ; D 0; 1.

n

208

Groups of prime power order

Suppose that G is a 2-group in (b). If D 0, then jG W NG .hbi/j D 2 implies that b centralizes ha2 i, a contradiction. Hence D 1 and we have obtained all 2-groups stated in our proposition. It remains to be shown that all groups in (a) are not of subgroup breadth 1. Suppose that G is a p-group from (a). If D 0, then jG W NG .hbi/j D p gives that b centralizes ap , a contradiction. Hence D 1 and so we may set G D ha; b j ap D b p D z; z p D 1; ap m

n

m1

D v; ab D avi;

where m 2, n 2, and if p D 2, then m 3. We have jGj D p mCnC1 , Œa; b D v, 2 Œap ; b D z, Œap ; b D 1, G 0 D hvi and G is of class 2 if and only if m 3 since in 2 that case hvi hap i Z.G/. (i) First assume m 3 so that G is of class 2. In that case, we also have Œa; b p D z, 2 Œa; b p D 1. (i1) Suppose m D n. Since hbai covers G=hai Š Cpn , it follows that o.ba/ p n . On the other hand, n

p n n n .ba/p D b p ap Œa; b. 2 / D z 1 z D 1

and so G D haihbai with hai \ hbai. But ap normalizes hbai and so we conclude that hap ihbai D hap i hbai which gives CG .ap / D G, a contradiction. n (i2) Suppose m > n. Then there is a0 2 hai such that ha0 i < hai and .a0 /p D z. Here again hba0 i covers G=hai Š Cpn and so o.ba0 / p n . On the other hand, n

p n n n .ba0 /p D b p .a0 /p Œa0 ; b. 2 / D z 1 z D 1

since a0 2 hap i and so Œa0 ; b 2 hzi. Thus G is a splitting extension of hai by hba0 i. It follows that ap normalizes hba0 i and so ap centralizes hba0 i, ap 2 Z.G/, a contradiction. m (i3) Suppose n > m. Then there is b 0 2 hbi such that hb 0 i < hbi and .b 0 /p D z 1 . Since b 0 2 hb p i, it follows that Œa; b 0 2 hzi. We get pm 2

.b 0 a/p D .b 0 /p ap Œa; b 0 . m

m

m

and .b 0 a/p

m1

D .b 0 /p

m1

ap

m1

/ D z 1 z D 1

p m1 2

Œa; b 0 .

/ D .b 0 /pm1 v ¤ 1;

and so o.b 0 a/ D p m and hbi \ hb 0 ai D ¹1º since .b 0 /p v is an element of order p m1 which is not contained in hzi (noting that v 2 hai and .b 0 /p 62 hai). Now, we get 0 0 nC1 0 m G D hb; b ai D hbihb ai since o.b/ D p , o.b a/ D p , hbi \ hb 0 ai D ¹1º, and jGj D p mCnC1 . But b p normalizes hb 0 ai and so Œb 0 a; b p 2 hvi \ hb 0 ai D ¹1º which gives b p 2 Z.G/, contrary to Œa; b p D z. m1

122

p-groups all of whose subgroups have normalizers of index at most p

209

(ii) Assume that m D 2 so that G is of class 3, G 0 D hvi, ŒG; G 0 D hzi and p > 2. (ii1) Suppose, in addition, that m D n D 2 so that jGj D p 5 . Since hbai covers G=hai Š Cp2 , we have o.ba/ p 2 . By the Hall–Petrescu formula (Appendix A.1), 2

2

p 2 2 2 . / .p / .ba/p D b p ap c2 2 c3 3 ;

where c2 2 hvi, c3 2 hzi and so .ba/p D z 1 z D 1 which gives o.ba/ D p 2 . Hence G D haihbai with hai \ hbai D ¹1º. On the other hand, ap D v normalizes hbai and so v centralizes ba. We get CG .v/ ha; bai D G, contrary to v b D vz. 2 (ii2) Suppose n > 2. Then there is b 0 2 hbi such that hb 0 i < hbi and .b 0 /p D z 1 . Note that ha; b 0 i is a proper subgroup of the A2 -group G and so ha; b 0 i is abelian or minimal nonabelian which gives Œa; b 0 2 hzi. We get 2

p2

.b 0 a/p D .b 0 /p ap Œa; b 0 . 2 / D z 1 z D 1 2

2

2

and so o.b 0 a/ p 2 . Also, we obtain p

.b 0 a/p D .b 0 /p ap Œa; b 0 . 2 / D .b 0 /p v ¤ 1 since .b 0 /p v is an element of order p which is not contained in hzi (indeed, hvi hai and .b 0 /p 62 hai, and so .b 0 /p v 62 hai ). Hence G D hbihb 0 ai with hbi \ hb 0 ai D ¹1º. Then b p normalizes hb 0 ai and so Œb p ; hb 0 ai hb 0 ai \ hvi D ¹1º. Thus we obtain that CG .b p / hb; b 0 ai D G and so b p 2 Z.G/. But then hbi induces on hai an automorphism of order p which implies that jG 0 j D p, a contradiction. Our proposition is proved. Proposition 122.8. Let G be a p-group with sb.G/ D 1. Then b.G/ 2 if p > 2 and b.G/ 3 if p D 2. Proof. Let A be a maximal normal abelian subgroup of G which contains ˆ.G/ (noting that ˆ.G/ is abelian by Proposition 122.2). By Proposition 122.3, 1 bA .G/ 2. Let x 2 A so that NG .hxi/ A and jG W NG .hxi/j p. We have CG .x/ A and NG .hxi/=CG .x/ is elementary abelian acting faithfully on hxi. Hence, if p > 2, then jNG .hxi/=CG .x/j p and if p D 2, then jNG .hxi/=CG .x/j 4. In the ﬁrst case bG .x/ 2 and in the second case bG .x/ 3. Let x 2 G A and assume that jA W CA .x/j D p. First suppose, in addition, that the subgroup A does not normalize hxi in which case NG .hxi/ covers G=A and we have NG .hxi/ \ A D CA .x/. Since NG .hxi/=CG .x/ is elementary abelian of order p in case p > 2 and elementary abelian of order 4 if p D 2, we get again bG .x/ 2 for p > 2 and bG .x/ 3 for p D 2. Now suppose that A normalizes hxi so that NG .hxi/ A and jG W NG .hxi/j p. Assume that p > 2 and that CG .x/ does not cover NG .hxi/=A. Then jNG .hxi/=.ACG .x//j D p and NG .hxi/=CG .x/ Š Cp2 . Let y 2 NG .hxi/CG .x/ be such that hyi covers NG .hxi/=CG .x/ so that y p 2 ACA .x/ 2 and y p 2 CA .x/. In that case, we have o.x/ p 3 and x y D xv, where v 2 hx p i and o.v/ D p 2 . Consider the metacyclic subgroup H D hxihyi with H 0 D hvi Š Cp2 .

210

Groups of prime power order

Since sb.H / D 1, Proposition 122.7 gives a contradiction. Hence if p > 2, CG .x/ must cover NG .hxi/=A and so in this case bG .x/ 2. Now assume that p D 2 and also NG .hxi/=.ACG .x// Š E4 which implies that NG .hxi/=CG .x/ Š C4 C2 . Then there exists again an element y 2 NG .hxi/ such that y 2 2 A CA .x/ and x y D xv, v 2 hx 2 i, o.v/ D 4 and o.x/ 24 . Considering the metacyclic subgroup K D hxihyi with K 0 D hvi Š C4 , Proposition 122,7 gives again a contradiction. Hence we obtain that jNG .hxi/=.ACG .x//j 2 which implies in this case bG .x/ 3. We consider now an element g 2 GA such that jA W CA .g/j D p 2 . Set P D Ahgi, where g p 2 CA .g/. By Lemma 121.2, jŒA; gj D p 2 . If hgi E P , then P 0 A \ hgi D hg p i Z.P /;

ŒA; g hg p i;

and so P is of class 2 and for each a 2 A, Œa; gp D Œa; g p D 1 and this implies ŒA; g D 1 .hgi/, contrary to jŒA; gj D p 2 . Hence jP W NP .hgi/j D p which yields that NG .hgi/ covers G=A. Set A1 D NA .hgi/ so that A1 E G. Suppose that in case p > 2, jNG .hgi/ W .A1 CG .g//j D p and in case p D 2, jNG .hgi/ W .A1 CG .g//j D 4. Then in both cases there is y 2 NG .hgi/ .A1 CG .g// such that y p 2 A1 CA .g/ and g y D gv, where v 2 hg p i, o.v/ D p 2 , o.g/ p 3 and in case p D 2, o.g/ 24 . Considering the metacyclic subgroup L D hgihyi with L0 D hvi Š Cp2 , Proposition 122.7 gives a contradiction. Hence in case p > 2, CG .g/ covers NG .hgi/=A1 and so bG .g/ D 2, and in case p D 2, jNG .hgi/=.A1 CG .g//j 2 so that in this case bG .g/ 3 and we are done. Proposition 122.9. Let G be a p-group with sb.G/ D 1. If G is a minimal counterexample to Theorem 122.1, then 1 .Z.G// G 0 . Proof. Let G be a minimal counterexample to Theorem 122.1 and suppose that there is i 2 1 .Z.G// with i 62 G 0 . Set Z=hi i D Z.G=hi i/ so that for p > 2, jG=Zj p 3 and for p D 2, jG=Zj 24 . Since ŒG; Z hii and i 62 G 0 , we get ŒG; Z D ¹1º and so Z D Z.G/, contrary to the fact that G was a minimal counterexample to Theorem 122.1. Proposition 122.10. Let G be a p-group with sb.G/ D 1. Let A be a maximal normal abelian subgroup of G containing ˆ.G/ such that bA .G/ D 1 (i.e., for each x 2 GA, jA W CA .x/j D p). Then Theorem 122.1 holds. Proof. Let G be a minimal counterexample to this proposition. Using Lemma 121.8, we have either jA W Z.G/j D p or jŒG; Aj D p. (i) First assume jA W Z.G/j D p. Let a 2 A Z.G/ so that CG .a/ D A. In case jG=Aj p 2 , we get jG=Z.G/j p 3 , which contradicts the fact that G is a minimal counterexample. Hence jG=Aj p 3 and then Proposition 122.8 forces p D 2 and bG .a/ D 3 so that jG=Aj D 23 . But then jG=Z.G/j D 24 , a contradiction. (ii) Now assume ŒG; A D hzi is of order p. We may also assume jA=Z.G/j > p which forces jG=Aj > p. For any x; y 2 G A such that hx; yiA=A Š Ep2 we have, by Lemma 121.4, CA .x/ ¤ CA .y/. We show that ˆ.A/ Z.G/. Indeed, if a 2 A

122

p-groups all of whose subgroups have normalizers of index at most p

211

and g 2 G are such that Œa; g ¤ 1, then Œa; g 2 hzi Z.G/ so that ha; gi0 D hzi and therefore ha; gi is minimal nonabelian. Then ap 2 ˆ.ha; gi/ D Z.ha; gi/ and so Œap ; g D 1 which gives ap 2 Z.G/. Due to Proposition 73.8, A possesses a basis ¹a1 ; a2 ; : : : ; an º, n 2, (noting that Ã1 .A/ Z.G/) such that z 2 han i. Set A0 D ha1 ; : : : ; an1 i and G0 D NG .A0 / so p that jG W G0 j p. We get ŒG0 ; A0 A0 \hzi D 1 so that A D A0 han i D Z.G0 /, where jA W A j D p. If jG0 W Aj > p, then there are elements x; y 2 G0 A such that hx; yiA=A Š Ep2 and CA .x/ D CA .y/ D A , contrary to the above remark. Hence jG0 =Aj p which together with jG=Aj > p gives G=A Š Ep2 and jG W G0 j D p. Let g1 ; g2 2 G A so that hg1 ; g2 i covers G=A. Then CA .g1 / ¤ CA .g2 / (by the above remark) yields CA .g1 /\CA .g2 / D Z.G/, A=Z.G/ Š Ep2 and for each x 2 AZ.G/, jG W CG .x/j D p. If p D 2, then jG=Z.G/j D 24 , a contradiction. We must have p > 2 and then Proposition 122.8 implies b.G/ 2. By Proposition 73.8, Z.G/ has a basis ¹z1 ; z2 ; : : : ; zr º, r 1, such that z 2 hz1 i. In case r > 1, we consider the subgroup S D hz2 ; : : : ; zr i ¤ ¹1º and the factor group G=S. Let g 2 G A and assume that Œg; A S. But then Œg; A S \ hzi D 1 and so g centralizes A, a contradiction. Hence A=S is a maximal normal abelian subgroup of G=S . Let a 2 A Z.G/ and assume ŒG; a S . Then ŒG; a S \ hzi D 1 and so a 2 Z.G/, a contradiction. Hence Z.G/=S D Z.G=S/ and so jG=S W Z.G=S/j D p 4 . This contradicts the fact that G was a minimal counterexample to our proposition. It follows that S D ¹1º and so Z.G/ D hz1 i is cyclic. Suppose that z1 is not a p-th power of any element in the subgroup A. Then for each p x 2 A Z.G/ we have x p 2 hz1 i and so hz1 i is a cyclic subgroup of the maximal possible order in A. Then there is a complement U Š Ep2 of hz1 i in A. We obtain that jNG .U / W Aj p and ŒNG .U /; U U \ hzi D ¹1º and so NG .U / centralizes A, a contradiction. Hence we may set A D hai hbi, where ap D z1 , hap i D Z.G/, z 2 hap i, and o.b/ D p. We shall prove that there are no elements of order p in G A. Assume that there is an element c of order p in G A. First suppose Œb; c ¤ 1 so that hŒb; ci D hzi and T D hb; ci Š S.p 3 / is the nonabelian group of order p 3 and exponent p with Z.T / D hzi. All p conjugates of b in G are contained in hb; zi and all p conjugates of c in G are contained in hc; zi which implies hb; zi E G, hc; zi E G, hT i E G, and G D T H , where H D CG .T / with T \ H D hzi. We have Z.H / D Z.G/ D hz1 i and H=Z.H / Š Ep2 . Since all p C 1 maximal subgroups of H containing hz1 i are abelian, we obtain that jH 0 j D p, H 0 hz1 i and so H 0 D hzi and G 0 D hzi. Due to the fact that ˆ.G/ D ˆ.T /ˆ.H / hz1 i and ˆ.A/ D hz1 i, we get ˆ.H / D hz1 i and so H is minimal nonabelian. It follows that either H Š S.p 3 / or H Š Mpn , n 3. In any case, there is an element d of order p in H Z.H /. We have ŒNG .hb; d i/; hb; d i hb; d i \ hzi D ¹1º; and so NG .hb; d i/ D CG .hb; d i/ D CG .b/ \ CG .d / D .H hbi/ \ .T hz1 ; d i/ D hz1 ihb; d i

212

Groups of prime power order

which is of index p 2 in G, a contradiction. Thus we have proved that Œb; c D 1. Since .Ahci/0 D hzi, we get NAhci .hb; ci/ D CAhci .hb; ci/ D CAhci .c/ which is of index p in Ahci. Due to jG W NG .hb; ci/j D p, it follows that NG .hb; ci/ covers G=A. But A E G and hb; ci \ A D hbi and so NG .hbi/ D CG .hbi/ also covers G=A and so b 2 Z.G/, a contradiction. We have proved 1 .G/ D 1 .A/ D hb; zi Š Ep2 : Since jG=Z.G/j D p 4 , we have jGj p 5 . If b.G/ D 1, then Proposition 121.9 (Knoche) implies jG 0 j D p. In case b.G/ D 2, Theorem 121.1 yields that jG 0 j D p 2 . By Theorem 13.7, G is metacyclic (since G cannot be a 3-group of maximal class because jG 0 j p 2 and jGj p 5 ). If jG 0 j D p, then G is minimal nonabelian in which case jG=Z.G/j D p 2 , a contradiction. Finally, if jG 0 j D p 2 , then Proposition 122.7 gives a contradiction. Proposition 122.11. Let G be a 2-group with jG 0 j D 2 such that G=Z.G/ Š E16 . Suppose that G possesses an abelian subgroup S which satisﬁes G 0 \ S D ¹1º and S=.S \ Z.G// Š E4 . Then jG W NG .S/j D 4 and so sb.G/ > 1. Proof. We have ŒNG .S/; S S \ G 0 D ¹1º and so NG .S/ D CG .S/ Z.G/S, where jG W .Z.G/S/j D 4. If CG .S/ > Z.G/S , then G has an abelian maximal subgroup which implies (Lemma 1.1) jGj D 2jZ.G/jjG 0 j and so jG=Z.G/j D 4, a contradiction. Hence CG .S/ D Z.G/S and we are done. Proposition 122.12. Let G be a 2-group of order 25 possessing a maximal subgroup M Š E16 . If there is an element x 2 G M such that CM .x/ Š E4 , then sb.G/ > 1. Proof. By Lemma 99.2, there is an involution i 2 G M so that CM .i/ D CM .x/. But then jG W CG .i/j D 4 and we are done. Proposition 122.13. Let G be a p-group with sb.G/ D 1. If jG 0 j D p, then we have G=Z.G/ Š Ep2 unless p D 2, G=Z.G/ Š E16 and G Š .Q8 D/ E2n , n 0, Q8 \ D D Z.Q8 / D D 0 , where D Š D8 or D Š P is the nonmetacyclic minimal nonabelian group of order 24 . Proof. Since jG 0 j D p, Lemma 4.2 gives G D A1 A2 Ak Z.G/ with k 1, where A1 ; : : : ; Ak are minimal nonabelian and A01 D A02 D D A0k D G 0 D hzi being of order p. If k D 1, then G=Z.G/ Š Ep2 and we are done. We may assume k > 1 and we set G D .A1 A2 /C , where C D CG .A1 A2 / and .A1 A2 / \ C D Z.A1 A2 / D Z.A1 /Z.A2 / so that jG=C j D p 4 and Z.G/ C . Note that jG 0 j D p implies b.G/ D 1 so that we may use Proposition 122.10. It follows that C D Z.G/ must be abelian and p D 2. Thus it remains to study a 2-group G D .A1 A2 /Z.G/, where A1 and A2 are minimal nonabelian with A01 D A02 D G 0 D hzi. We have G 0 D hzi A1 \ A2 Z.A1 A2 / D Z.A1 /Z.A2 / and G=Z.G/ Š E16 :

122

p-groups all of whose subgroups have normalizers of index at most p

213

(i) First assume that there is an involution t 2 G Z.G/. Let b 2 G Z.G/ be such that Œb; t D z and then B1 D hb; ti is minimal nonabelian. Set C D CG .B1 / so that G D B1 C with B1 \ C D Z.B1 /. Since G=Z.G/ Š E16 , C is nonabelian. Let B2 be a minimal nonabelian subgroup of C and then C D B2 Z.G/. We set G1 D B1 B2 so that G D G1 Z.G/. Suppose that B2 possesses a nonnormal cyclic subgroup hb 0 i. Set hui D 1 .hb 0 i/ and S D ht i hb 0 i so that u ¤ z, where hzi D G 0 D B20 D B10 . Then 1 .S/ D ht; ui does not contain z. But then Proposition 122.11 gives a contradiction. Hence B2 is Dedekindian which implies that B2 Š Q8 , B1 \ B2 D hzi and Z.B1 / D Z.G1 /. Set B2 D hr; si. (i1) Suppose that B1 is metacyclic. Then either B1 Š D8 or B1 Š M2m , m 4. In the second case, Z.B1 / is cyclic of order 4 so that there is v 2 Z.B1 / such that v 2 D z and then i D vr is a noncentral involution in G1 .B1 [ B2 /. We obtain S D hi; ti Š E4 , S \ Z.G/ D ¹1º and z 62 S so that Proposition 122.11 gives a contradiction. Hence in this case we must have G D .Q8 D8 /Z.G/. (i2) Now assume that B1 is nonmetacyclic so that we may set m

B1 D ht; b j t 2 D b 2 D 1; Œt; b D z; z 2 D Œz; t D Œz; b D 1i; where m 2 and Z.B1 / D hb 2 i hzi. Assume that m 3 in which case there is an element w of order 4 in hb 2 i Z.B1 / so that w 2 ¤ z and .wr/2 D w 2 z. Considering the abelian subgroup S D hti hwri, we have 1 .S/ D ht; w 2 zi 6 hzi, and again Proposition 122.11 gives a contradiction. Hence in this case we have m D 2 and so G D .Q8 P/Z.G/, where P is the nonmetacyclic minimal nonabelian group of order 24 . (ii) Assume that there exist no involutions in G Z.G/. We have seen above that G D .A1 A2 /Z.G/, where we may assume that A1 6Š Q8 (noting that Q8 Q8 possesses noncentral involutions). There is an element d 2 A1 Z.A1 / such that z 62 hd i and 1 ¤ d 2 2 Z.G/, where hzi D G 0 . Let b 2 G be given such that Œd; b D z and B1 D hd; bi is minimal nonabelian, where b; d 2 B1 Z.B1 /. Note that hd i=hd 2 i is a noncentral subgroup of order 2 in G=hd 2 i. We have G D B1 C , where C D CG .B1 / and B1 \C D Z.B1 /. If C =hd 2 i is abelian, then we obtain that C 0 hd 2 i\hzi D ¹1º and C D Z.G/ is abelian, contrary to G=Z.G/ Š E16 . Let B2 =hd 2 i be a minimal nonabelian subgroup of C =hd 2 i, where B2 hd 2 ihzi and B20 D hzi. If B2 =hd 2 i 6Š Q8 , then there is b 0 2 B2 Z.B2 / such that hb 0 i \ hd 2 ; zi hd 2 i. Working in the factor group .B1 B2 /=hd 2 i, we get (as in (i)) that sb..B1 B2 /=hd 2 i/ > 1, a contradiction. Hence B2 =hd 2 i Š Q8 so that B1 \ B2 D hd 2 i hzi. By our results in (i) (applied to .B1 B2 /=hd 2 i), we get B1 =hd 2 i Š D8 or P. Set 1 .hd 2 i/ D hui and since B2 =hd 2 i Š Q8 , we have 1 .B2 / D hu; zi Š E4 . Since Z.B2 / D hd 2 ; zi, any minimal nonabelian subgroup X of B2 covers B2 =hd 2 ; zi and z 2 X. As j1 .X/j 4, X is metacyclic and z is a square in X so that there is an element x 2 X hd 2 ; zi such that x 2 D z, hzi D X 0 , hxi E G and hxi is a maximal cyclic subgroup of G. If there is y 2 B1 such that y 2 D z, then xy is a noncentral in-

214

Groups of prime power order

volution in G, a contradiction. Hence z is not a square in B1 and therefore B1 is nonmetacyclic minimal nonabelian. If B1 =hd 2 i Š D8 , then we have Z.B1 / D hd 2 i hzi and there are no involutions in B1 hd 2 ; zi so that 1 .B1 / D hu; zi, contrary to the fact that B1 is nonmetacyclic. We have proved that B1 =hd 2 i Š P. Set E D 1 .B1 / so that E Š E8 , Z.B1 / D hd 2 iE and E \ hd 2 ; zi D hu; zi. By the structure of B1 (being nonmetacyclic minimal nonabelian), there exists e 2 B1 Z.B1 / such that e 2 D l is an involution in E, where E D hu; z; li and B1 D hd; ei. Let y 2 B2 hd 2 ; zi be such that hx; yi covers the factor group B2 =hd 2 ; zi. Then hx; yi is minimal nonabelian with y 2 2 hd 2 ; zihd 2 i. Assume that o.d 2 / 4 and let d 0 2 hd 2 i be an element of order 4 so that .d 0 /2 D u. Consider the subgroup S D he; xd 0 i Š C4 C4 with 1 .S/ D hl; uzi Š E4 and z 62 S. By Proposition 122.11, we get a contradiction (noting that S \Z.G/ D 1 .S/). Hence o.d 2 / D 2 and so d 2 D u. If y 2 D z, then hx; yi Š Q8 and B2 D huihx; yi. In case y 2 D uz, we conclude that B2 D hx; yi Š H2 is minimal nonabelian, where H2 D hx; y j x 4 D y 4 D 1; x y D x 1 i, x 2 D z, y 2 D uz and u is not a square in H2 . We have proved that B1 D hd; e j d 4 D e 4 D 1; Œd; e D z; z 2 D Œz; d D Œz; e D 1i; where d 2 D u, e 2 D l, 1 .B1 / D hu; l; zi, and B1 B2 Š B1 Q8 or B1 B2 Š B1 H2 . If B1 B2 Š B1 Q8 with B1 \ Q8 D hzi, then consider B1 =huz; lzi Š Q8 so that .B1 Q8 /=huz; lzi Š Q8 Q8 , which is a contradiction by Proposition 122.4. Hence we must have B2 Š H2 D hx; y j x 4 D y 4 D 1; x y D x 1 i; where x 2 D z, y 2 D uz and u is not a square in B2 . In this case, we consider the subgroup S D hy; dei Š C4 C4 . Since S \ Z.G/ D 1 .S / D huz; ulzi 6 hzi; Proposition 122.11 gives a contradiction. (iii) We have proved that G D .D8 Q8 /Z.G/ or G D .P Q8 /Z.G/. It remains to be proved that Z.G/ is elementary abelian. We may set G D .A1 A2 /Z.G/, where A1 D hb; t j b 4 D t 2 D 1; b 2 D z; b t D bzi Š D8

(a) or

A1 D hb; t j b 4 D t 2 D 1; b 2 D u; Œb; t D z; z 2 D Œz; b D Œz; t D 1i Š P

(b) and

A2 D hx; y j x 2 D y 2 D z; z 2 D 1; Œx; y D zi Š Q8

with A1 \ A2 D hzi:

122

p-groups all of whose subgroups have normalizers of index at most p

215

First assume that Z.G/ has an element v of order 4 such that v 2 D s 62 A1 A2 . In both cases (a) and (b), we consider the abelian subgroup S D hvx; ti of G which satisﬁes 1 .S/ D ht; szi 6 hzi so that Proposition 122.11 gives a contradiction. Now assume that Z.G/ has an element v of order 4 such that v 2 2 Z.A1 A2 /. In case (a), we must have v 2 D z and then we consider the subgroup S D hvx; ti Š E4 , where vx is an involution in G .A1 A2 / commuting with t and so z 62 S and Proposition 122.11 gives a contradiction. In case (b), we have either v 2 D z or v 2 D u since there is an outer automorphism ˛ of A1 induced by t ˛ D t and b ˛ D bt, where .bt/2 D uz and so u˛ D uz. If v 2 D z, we consider the subgroup S D hvx; ti Š E4 with z 62 S. In case v 2 D u, we consider the subgroup S D hvx; ti Š C4 C2

with 1 .S/ D ht; uzi 6 hzi:

In both cases, we get a contradiction with the help of Proposition 122.11. Our proposition is proved. Proposition 122.14. Let G be a 2-group with sb.G/ D 1. Then b.G/ 3 and let a be an element in G with eight conjugates. Then we have o.a/ D 8, jG W NG .hai/j D 2 and NG .hai/=CG .a/ Š E4 . Proof. We have b.G/ 3 by Proposition 122.8. Let a be an element in G with eight conjugates. Set C D CG .a/ so that jG W C j D 8. Also set H D NG .hai/ so that jG W H j 2. Since jH=C j 4, we have o.a/ 8. Suppose that o.a/ D 2n with n 4. Assume that there is an element x 2 H C such that ax D a1 z , where n1 z D a2 and D 0; 1. This implies hxi \ hai hzi. On the other hand, a2 normalizes hxi and so Œa2 ; x hxi \ hai hzi. But .a2 /x D .a1 z /2 D a2 and so Œa2 ; x D a4 with o.a4 / 4, a contradiction. It follows that H=C contains only the involutory automorphism induced by a ! az. Hence H=C is cyclic. Assume that H C contains an element y which induces on hai an automorphism of order 4 given n2 by ay D av, where v D a2 . Then ha; yi is metacyclic with ha; yi0 D hvi Š C4 . But then Proposition 122.7 gives a contradiction. We get H=C Š C4 , jG W H j D 2, and if y 2 H C is such that hyi covers H=C , then ay D a1 v, where v is an element of order 4 in hai and v 2 D z with 1 .hai/ D hzi. Note that .a4 /y D .a1 v/4 D a4 and so v y D v 1 and ay D .a1 v/y D .a1 v/1 v y D az 2

which implies hyi \ hai hzi. On the other hand, a2 normalizes hyi and so Œa2 ; y hyi \ hai hzi. But .a2 /y D .a1 v/2 D a2 z and so Œa2 ; y D a4 z which is of order 4, a contradiction. We have proved that o.a/ D 8 and then jG W H j D 2 and H=C Š E4 . Proposition 122.15. Let G be a 2-group with sb.G/ D 1. Let a be an element in G with eight conjugates. By Proposition 122.14, we have o.a/ D 8, jG W NG .hai/j D 2, and NG .hai/=CG .a/ Š E4 . If b; c 2 NG .hai/ CG .a/ are such that ab D a1 and

216

Groups of prime power order

ac D a1 z with z D a4 , then o.b/ D o.c/ D 8 and 1 .hbi/ D 1 .hci/ D hbi \ hai D hci \ hai D hzi: We have

L D hb; a2 i Š M16 , where b 4 D a4 D z, z 2 D 1, Œa2 ; b D z, L E G, t D a2 b 2 is an involution and 1 .L/ D ht; zi.

b a D ba2 , .a2 /a D a2 and so a induces on L an outer automorphism of order 4 2 as b a D ba4 D bz, where a2 induces on L an inner automorphism of order 2.

Aut.L/ D D h˛i, where D Š D8 and ˛ is an outer involutory automorphism of L induced by b ˛ D b 1 and .a2 /˛ D a2 .

Proof. Let a be an element in G with eight conjugates so that o.a/ D 8 and then set C D CG .a/, H D NG .hai/, where jG W H j D 2 and H=C Š E4 . Let b; c 2 H C be such that hb; ci covers H=C and ab D a1 , ac D a1 z with z D a4 . This implies that hbi\hai hzi and hci\hai hzi. On the other hand, a2 normalizes hbi and hci. If hbi \ hai D ¹1º, then a2 centralizes hbi, a contradiction. In case hci \ hai D ¹1º, a2 centralizes hci, a contradiction. Hence we have hbi \ hai D hci \ hai D hzi; m

l

and so we may set b 2 D z and c 2 D z, where m 1 and l 2. Indeed, if c 2 D z, then ha; ci Š SD16 , and so ha; ci possesses a noncentral involution i with Cha;ci .i/ D hi; zi Š E4 , a contradiction. We shall study the structure of the subgroup M D ha; b; ci, where M 0 ha2 i since Œa; b D a2 , Œa; c D a2 . If b has only four conjugates in M , then b c D ba2i (for some integer i ) since b has already four conjugates in hb; ai with hb; ai0 D ha2 i and this gives Œb; c 2 ha2 i and M 0 D ha2 i so that b.M / D 2. Similarly, if c has only four conjugates in M , then c b D ca2j (for some integer j ) and this gives Œc; b 2 ha2 i and M 0 D ha2 i so that b.M / D 2. Hence b has only four conjugates in M if and only if c has only four conjugates in M and in that case M 0 D ha2 i and b.M / D 2. Conversely, if M 0 D ha2 i, then b.M / D 2 and so both b and c have exactly four conjugates in M . We aim to show that o.b/ D o.c/ D 8. First assume that b or c has eight conjugates in M . By the previous paragraph, both b and c have exactly eight conjugates in M . From Proposition 122.14 we then conclude that o.b/ D o.c/ D 8. In the sequel we assume that both b and c have exactly four conjugates in M and so M 0 D ha2 i. Set Œb; c D a2i (i integer) and compute Œb 2 ; c D Œb; cb Œb; c D .a2i /b a2i D a2i a2i D 1; Œb; c 2 D Œb; cŒb; cc D a2i .a2i /c D a2i a2i D 1; and so hb 2 ; c 2 i Z.M /.

122

p-groups all of whose subgroups have normalizers of index at most p

217

Assume m 3, where b 2 D z and set b 0 D b 2 so that o.b 0 / D 8, .b 0 /4 D z 0 0 0 and b 2 Z.M /. It follows that hb ; ai D hb i hai with hb 0 i \ hai D hzi. Because of .b 0 a/2 D .b 0 /2 a2 ¤ 1 and .b 0 a/4 D .b 0 /4 a4 D zz D 1, we obtain that o.b 0 a/ D 4. Set S D hb 0 ai and we claim that S has more than two conjugates in M . Indeed, we have S b D hb 0 a1 i and S c D hb 0 a1 zi. If S D S b , then either b 0 a D b 0 a1 (and then a2 D 1) or b 0 a D .b 0 a1 /1 D .b 0 /1 a (and then .b 0 /2 D 1), a contradiction. In case S D S c , either b 0 a D b 0 a1 z (and then a2 D z) or b 0 a D .b 0 /1 az (and then .b 0 /2 D z), a contradiction. If S b D S c , then either b 0 a1 D b 0 a1 z (and then z D 1) or b 0 a1 D .b 0 /1 az (and then .b 0 /2 D a2 z), a contradiction. Thus m 2. l l2 so that o.c 0 / D 8, .c 0 /4 D z and Assume l 3, where c 2 D z and set c 0 D c 2 c 0 2 Z.M /. It follows that hc 0 ; ai D hc 0 i hai with hc 0 i \ hai D hzi. Then we have T D hc 0 ai Š C4 and we claim that T has more than two conjugates in M . Indeed, we obtain T b D hc 0 a1 i and T c D hc 0 a1 zi. If T D T b , then either c 0 a D c 0 a1 or c 0 a D .c 0 /1 a, a contradiction. In case T D T c , we conclude either c 0 a D c 0 a1 z or c 0 a D .c 0 /1 az, a contradiction. Finally, if T b D T c , then either c 0 a1 D c 0 a1 z or c 0 a1 D .c 0 /1 az, a contradiction. Thus l 2 and so l D 2 and o.c/ D 8. Suppose o.b/ D 4 so that b 2 D z, ha; bi Š Q16 since ab D a1 . Then m

Q D ha2 ; bi Š Q8 ;

m2

c 2 2 Z.M /;

o.c 2 / D 4;

and so c 2 62 Q. But we have c 4 D z, where hzi D Z.Q/ and so hQ; c 2 i D Q hc 2 i with Q \ hc 2 i D hzi and therefore hQ; c 2 i contains a subgroup isomorphic to D8 . By Proposition 122.5, jG 0 j 4 and so b.G/ 2, contrary to the fact that a has eight conjugates in G. Hence o.b/ D 8. We have proved that in any case o.b/ D o.c/ D 8, as required. We have L D hb; a2 i D hb; a2 b 2 D t j b 8 D t 2 D 1; b t D bz; z D b 4 D a4 i Š M16 : Since b a D ba2 , it follows that the only two distinct cyclic subgroups hbi and hba2 i of order 8 in L are conjugate under a and so hb; b a D ba2 i D L is normal in G. From the fact that hb 2 i D Z.L/ is normal in G and 2 .L/ D hb 2 ; a2 i Š C4 C2 , we infer that ha2 i is also normal in G (noting that ha2 i and hb 2 i are the only two cyclic subgroups of order 4 in 2 .L/). Each automorphism ˛ 2 Aut.L/ sends b onto one of the eight elements of order 8 in the set .hbi hb 2 i/ [ .hba2 i hb 2 i/ and ˛ sends a2 onto one of the two elements in ¹a2 ; a2 D a2 zº and so jAut.L/j D 8 2 D 16. Since Inn.L/ Š L=Z.L/ Š E4 , we have jAut.L/=Inn.L/j D 4. Also, Inn.L/ D hia2 ; ib i, where ix is the inner automorphism of L induced by conjugation with the element x 2 L Z.L/. We deﬁne the outer automorphisms ˇa (induced by conjugation with the element a 2 G) and ˛ by b ˇa D b a D ba2 ;

.a2 /ˇa D .a2 /a D a2

and b ˛ D b 1 ;

.a2 /˛ D a2 ;

where .ˇa /2 D ia2 and ˛ 2 D 1. We check that Inn.L/hˇa i Š D8 and that ˛ com-

218

Groups of prime power order

mutes with ˇa and ib and so Aut.L/ D .Inn.L/hˇa i/ h˛i Š D8 C2 : Our proposition is completely proved. The following proposition is the key result of the whole section. Proposition 122.16. Let G be a 2-group with sb.G/ D 1. Then b.G/ 2, i.e., the element breadth of G is at most 2. Proof. Let G be a minimal counterexample to this proposition. Note that in this case Proposition 122.8 implies that b.G/ 3. Then G possesses an element a which has exactly eight conjugates in G. We use Proposition 122.15 together with the notation and all details stated in that proposition. Set C D CG .L/ so that L \ C D Z.L/ D hb 2 i, a 2 G .LC / and ¹1º ¤ G=.LC / is elementary abelian of order at most 4. We aim to show that C is abelian. Suppose that C is nonabelian. Note that C cannot be Hamiltonian since C has a central subgroup hb 2 i Š C4 . Let hxi be a nonnormal cyclic subgroup of C such that z 62 hxi and 0 0 let c 0 2 C be such that hx c i ¤ hxi and then z 62 hx c i. Now, t D a2 b 2 is a noncentral involution in L and t ¤ t b D t z. Set S D hx; ti D hxi hti and we claim that S D hx; ti;

0

S1 D hx c ; ti;

0

S2 D hx; tzi and S3 D hx c ; tzi

are pairwise distinct conjugates of S. If S D S1 , then 0

S \ C D hxi D S1 \ C D hx c i; a contradiction. In case S D S2 , we get S \ L D hti D S2 \ L D htzi; a contradiction. If S1 D S2 , then 0

S1 \ C D hx c i D S2 \ C D hxi; a contradiction. Thus we have proved that each nonnormal subgroup of C contains hzi. By a result of N. Blackburn (see Corollary 92.7), we have three possibilities for the structure of C . (1) C Š Q8 Q8 E2s (s 0) which is not possible since C has a central subgroup hb 2 i which is cyclic of order 4. (2) C Š Q8 C4 E2s (s 0) and let Q Š Q8 be a quaternion subgroup of C . If z 2 Q, then the subgroup hb 2 ; Qi contains a subgroup isomorphic to D8 . In that case, Proposition 122.5 implies jG 0 j 4 which gives b.G/ 2, a contradiction. Hence Q D hx; yi with Z.Q/ D hz 0 i ¤ hzi. But then hb 2 xi is nonnormal in C and z 62 hb 2 xi, a contradiction.

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p-groups all of whose subgroups have normalizers of index at most p

219

(3) C possesses an abelian maximal subgroup A with exp.A/ > 2 and an element v 2 C A of order 4 such that x v D x 1 for all x 2 A. But C has a central subgroup hb 2 i Š C4 , a contradiction. We have proved that C is abelian. (i) First assume that bG .b/ D 3, i.e., b has exactly eight conjugates in G. By Proposition 122.14, G possesses an element d such that b d D b 1 and so, by Proposition 122.15, our group G induces on L Š M16 the full automorphism group of L (which is isomorphic to the group D8 C2 ) and so we may choose our element d so that b d D b 1 , .a2 /d D a2 , d 2 2 C , o.d / D 8, d 4 D z and G D .LC /ha; d i. Also, the automorphism of L induced by d commutes with all automorphisms of L and so f D Œd; a 2 C . Consider the subgroup H D hb; d j b 8 D d 8 D 1; b 4 D d 4 D z; b d D b 1 i; where Z.H / D hd 2 i. Since .bd /2 D bdbd D bd 2 b d D bd 2 b 1 D d 2 ;

Œb; bd D Œb; d D b 2 ;

it follows that the cyclic subgroup S D hbd i of order 8 is not normal in H . Thus two conjugates of S in G are already contained in H . On the other hand, .bd /a D b a d a D ba2 df D .bd /a2 f; which forces a2 f 2 H . Since a2 f 2 LC and H \ .LC / D hb; d 2 i, we deduce that 2 a2 f must be contained in the abelian subgroup hb; d 2 i. But b a f D .bz/f D bz ¤ b, a contradiction. (ii) Now assume that bG .b/ D 2, i.e., there is no element in G which inverts hbi. We get G D .LC /hai, where C D CG .L/ is abelian and jG W .LC /j D 2. Since b inverts hai, b 2 centralizes hai and so CG .hb 2 i/ hL; C; ai D G so that b 2 2 Z.G/. We have NG .hai/ D haiNLC .hai/ and noting that L D ha2 ; bi normalizes hai, we obtain that NLC .hai/ D LC0 , where C0 D NC .hai/, C0 hb 2 i and jC W C0 j D 2 (Proposition 122.14). Let x 2 C C0 so that C E G implies Œa; x D f0 2 C and ax D af0 . Hence we have f0 62 hai and so f0 2 C hzi since hai \ C D hzi. Consider the subgroup K D ha; b j a8 D b 8 D 1; a4 D b 4 D z; ab D a1 i; where Z.K/ D hb 2 i. Since .ab/2 D abab D ab 2 ab D ab 2 a1 D b 2

and Œab; a D Œb; a D a2 ;

the cyclic subgroup habi of order 8 is not normal in K and so two conjugates of habi in G are already contained in K. On the other hand, .ab/x D af0 b D .ab/f0 which

220

Groups of prime power order

implies f0 2 K. But K \ .LC / D ha2 ; bi D L and so f0 2 .L \ C / hzi. Thus f0 D .b 2 /˙1 and therefore replacing b with b 1 (if necessary), we may set f0 D b 2 and so ax D ab 2 . Then we get 2

ax D .ab 2 /x D .ab 2 /b 2 D ab 4 D az and so x 2 2 C0 but x 2 62 C1 , where C1 D CC .a/ hb 2 i. On the other hand, C centralizes ha2 i and so in C0 C1 we cannot have elements which invert hai. This gives jC0 W C1 j D 2;

x 2 2 C0 C1 ;

C1 D Z.G/:

For any y 2 C0 C1 we have ay D az and ayb D a1 z. By Proposition 122.15, we have o.yb/ D 8 and .yb/4 D z. Then we get y 4 b 4 D z and y 4 D 1 so that C0 is of exponent 4. Suppose that Z.G/ D C1 D hb 2 i is cyclic. If, in addition, G possesses noncommuting involutions, then Proposition 122.5 implies that jG 0 j 4 which gives b.G/ 2, a contradiction. By Proposition 122.6, G has at most three involutions and so 1 .G/ D 1 .L/ D hz; ti Š E4 : But exp.C0 / D 4 and jC0 W hb 2 ij D 2 and so there exist involutions in C0 hb 2 i, a contradiction. Hence we get C1 D Z.G/ > hb 2 i so that there is a central involution u 2 C1 hb 2 i. Considering the factor group GN D G=hui, then the fact that G was a N 2. On the other hand, from minimal counterexample to our proposition gives b.G/ 2 b 1 x 2 x a D a , a D ab , a D az we deduce that aN has in GN eight conjugates, a ﬁnal contradiction. Our proposition is proved. Proposition 122.17. Let G be a 2-group with sb.G/ D 1 which is a minimal counterexample to Theorem 122.1. Then jG 0 j D 4, and let z 2 G 0 be a central involution in G. If R=hzi D Z.G=hzi/, then G=R Š E16 . Proof. Let G be a 2-group with sb.G/ D 1 which is a minimal counterexample to Theorem 122.1. If jG 0 j D 2, then Proposition 122.13 implies jG=Z.G/j 24 , a contradiction. Hence jG 0 j > 2 and so b.G/ > 1 by Proposition 121.9 (Knoche). On the other hand, Proposition 122.16 gives b.G/ 2 and so b.G/ D 2. Since jG=Z.G/j 25 , Theorem 121.1 implies jG 0 j D 4. N 1 and Let z 2 G 0 be a central involution in G and set GN D G=hzi so that sb.G/ 0 N N Š E4 or G=Z. N N Š E16 . Assume that jGN j D 2. By Proposition 122.13, G=Z. G/ G/ N N Š E4 and so we have only to rule out this case. Set Z.G/ N D R=hzi, where G=Z. G/ G=R Š E4 and r 2 R if and only if ŒG; r hzi. We have ŒG; R D hzi. Indeed, if ŒG; R D ¹1º, then we get R D Z.G/, a contradiction. Also, ŒG; r hzi implies that jG W CG .r/j 2. Let A be a maximal normal abelian subgroup of G which contains ˆ.G/ so that ¹1º ¤ G=A is elementary abelian. In view of Propositions 122.3 and 122.10, there is g 2 G A such that jA W CA .g/j D 4 and then Lemma 121.2 implies jŒA; gj D 4 so

122

p-groups all of whose subgroups have normalizers of index at most p

221

that ŒA; G D G 0 . If jG W Aj D 2, then we have jG=Z.G/j D 8, a contradiction. Hence jG W Aj 4. Since ŒG; G 0 hzi, we have G 0 R. As ŒA; g 6 hzi, it follows that A 6 R. For each x 2 .AR/ A, x 2 TA , i.e., jA W CA .x/j D 2. Therefore R does not cover G=A because g 62 TA . Hence jA W .A\R/j D 2 and jG W .AR/j D 2 which gives G D .AR/hgi. We have R0 hzi and ŒA; R D hzi (noting that jR W .A \ R/j 2) and so .AR/0 D hzi. Since jA W CA .g/j D 4, g does not centralize A\R and so we get ŒA \ R; g D hzi which gives jA \ R W CA\R .g/j D 2 and therefore CA .g/ is a subgroup of index 2 in A \ R and so g 2 2 CA\R .g/. Also, R0 hzi and ŒR; g D hzi imply .Rhgi/0 D hzi. Since G is a minimal counterexample to Theorem 122.1, we have jG=Z.G/j 25 and so Propositions 122.5 and 122.9 imply that 1 .G/ is elementary abelian and 1 .Z.G// G 0 . In particular, D8 is not a subgroup of G. First suppose jG=Aj D 4. If R is abelian, then CG .CA .g// hR; g; Ai D G, CA .g/ Z.G/ and jG W CA .g/j D 24 , a contradiction. Hence R is nonabelian with R0 D hzi and A \ R is an abelian maximal subgroup of R. Then Lemma 1.1 gives jRj D 2jZ.R/jjR0 j and so Z D Z.R/ is a subgroup of index 2 in A\R. If CA .g/ D Z, we get CG .Z/ hR; g; Ai D G and jG W Zj D 24 , a contradiction. It follows that Z0 D CA .g/ \ Z is of index 4 in A \ R and Z0 D Z.Rhgi/ with j.Rhgi/=Z0 j D 24 . Since .Rhgi/0 D hzi and sb.Rhgi/ D 1, we may apply Proposition 122.13 and get (noting that D8 is not a subgroup of G) Rhgi Š .P Q8 / E2s , s 0, where P is the nonmetacyclic minimal nonabelian subgroup of order 24 and P \ Q8 D hzi. Now assume jG=Aj > 4 so that j.AR/=Aj 4. Note that we have .AR/0 D hzi and Z.AR/ < A which gives j.AR/=Z.AR/j 23 . Applying Proposition 122.13, we get AR Š .P Q8 / E2s , s 0. We have shown that in any case G possesses a maximal subgroup M D .P Q/S, where P D hb; t j b 4 D t 2 D 1; Œb; t D z; z 2 D Œb; z D Œt; z D 1i; Q D hr; si Š Q8 ;

P \ Q D hzi and S Š E2s ; s 0:

Set b 2 D u so that Z.P / D hu; zi Š E4

and 1 .P / D hu; z; t i Š E8 :

Since hbi and hbzi are two distinct conjugates in G of the cyclic subgroup hbi Š C4 , it follows that hb; bzi D hbi hzi is normal in G and so Ã1 .hb; bzi/ D hui is also normal in G. Thus u 2 Z.G/ and since 1 .Z.G// G 0 , we get G 0 D Z.P / Z.G/ which shows that G is of class 2. Suppose that S ¤ ¹1º and let x be an involution in S. We have ht; xi Š E4 and ht; xi \ G 0 D ¹1º. Then ŒNG .ht; xi/; ht; xi ht; xi \ G 0 D ¹1º; and so NG .ht; xi/ D CG .ht; xi/ D CG .x/ \ CG .t/:

222

Groups of prime power order

But 1 .Z.G// D G 0 D hu; zi whence x 62 Z.G/ and so CG .x/ D M . Because of jM W CM .t/j D 2, we have jG W .CG .x/\CG .t//j D 4, a contradiction. Hence we get S D ¹1º and so M D P Q, Z.M / D Z.P / D G 0 Z.G/. Since jG=Z.G/j 25 , we have Z.P / D G 0 D Z.G/. We shall show that there are no involutions in G M . Indeed, let i be an involution in G M and note that ht; ii Š E4 and ht; ii \ Z.G/ D ¹1º so that E D hz; u; t; ii is isomorphic to E16 . Also, b 2 D u and E E G and therefore jEhbij D 25 . We have ŒNG .ht; ii/; ht; ii ht; ii \ G 0 D ¹1º

and so

NG .ht; ii/ D CG .ht; ii/

which together with jG W NG .ht; ii/j D 2 gives CG .t/ D CG .i/ D CG .ti/ D EQ, where b 62 EQ and jG W .EQ/j D 2. In particular, CE .b/ D hu; zi Š E4 and so Proposition 122.12 implies sb.Ehbi/ > 1, a contradiction. We have proved that there are no involutions in G M and so 1 .G/ D 1 .M / D 1 .P / D hz; u; t i Š E8 : Note that P =hui Š D8 and so M=hui Š D8 Q8 and therefore, applying Proposition 122.13 on the factor group G=hui, we see that there is an element v 2 G M such that v 2 D u and Œv; M hui. If b v D b, then v 2 D b 2 D u gives that vb is an involution in G M , a contradiction. Hence b v D bu D b 1 and so hb; vi Š Q8 . Since t and t b D t z are the only conjugates of t in G, we cannot have t v D tu and so Œt; v D 1. If v centralizes Q D hr; si, then hhb; vi; Qi D hb; vi Q Š Q8 Q8 ; and then Proposition 122.4 gives a contradiction. In case r v D ru and s v D su, we have .rs/v D rs. Hence we may choose generators r; s of Q so that Œr; v D 1 and Œs; v D u. The structure of G (of order 27 ) is uniquely determined. We compute .bt s/2 D .bt/2 s 2 D uz z D u and

Œv; bts D u u D 1

and so v bts is an involution in G M , a ﬁnal contradiction. Our proposition is proved. Proposition 122.18. Let G be a 2-group with sb.G/ D 1 which is a minimal counterexample to Theorem 122.1. Then jG 0 j D 4 and let z 2 G 0 be a central involution in G. We have G D AB, where A and B are normal subgroups of G, A \ B D hzi;

A Š Q8 ; C4 ; E4 or C2

and if A Š Q8 , then ŒA; B D 1, and G 0 D B 0 , B=hzi Š Q8 D8 with Q8 \ D8 D Z.Q8 / or B=hzi Š Q8 P with Q8 \ P D Z.Q8 / D P0 , where P is the nonmetacyclic minimal nonabelian group of order 24 .

122

p-groups all of whose subgroups have normalizers of index at most p

223

Proof. By Proposition 122.17, we have jG 0 j D 4 and if z 2 G 0 is a central involution in G, then G=hzi is isomorphic to one of the groups stated in the second half of Proposition 122.13. More precisely, we have G D AB, where A and B are normal subgroups of G, A \ B D hzi, A=hzi Š E2n , n 0, B=hzi Š Q8 D with Q8 \ D D Z.Q8 / D D 0 , D Š D8 or P, where P is the nonmetacyclic minimal nonabelian group of order 24 . Further, G 0 B and set T =hzi D Z.B=hzi/ so that T D G 0 (in case B=hzi Š Q8 D8 ) or T =hzi Š E4 (in case B=hzi Š Q8 P ) and in both cases B=T Š E16 and Z.B/ T . If B < G, then the fact that G is a minimal counterexample to Theorem 122.1 gives jB=Z.B/j 24 and so in this case Z.B/ D T . Suppose that A is nonabelian so that in this case A0 D hzi and A=A0 is elementary abelian. We may also assume that A is not Hamiltonian and so sb.A/ D 1. By Proposition 122.5, D8 is not a subgroup of A. Also, P cannot be a subgroup of A since P=P0 is not elementary abelian. Then, by Proposition 122.13, we get A=Z.A/ Š E4 . Let A0 be a minimal nonabelian subgroup of A so that A D A0 Z.A/. But we have A00 D hzi D ˆ.A0 / D ˆ.A/ and so A0 Š Q8 and A0 \ Z.A/ D hzi with Z.A/=hzi being elementary abelian. If there is v 2 Z.A/ such that v 2 D z, then A0 hvi contains a subgroup isomorphic to D8 , a contradiction (Proposition 122.5). It follows that Z.A/ is elementary abelian and so A is Hamiltonian after all. We have proved that A is either abelian with Ã1 .A/ hzi or A is Hamiltonian with A0 D hzi. Suppose that A possesses a central involution t satisfying t ¤ z and A > ht; zi. Set G0 D Bht i so that G0 < G. As B < G, we get T D Z.B/ and T =hzi D Z.B=hzi/. We have ŒB; t hzi and G0 =hzi D .ht; zi=hzi/ .B=hzi/ Š C2 .B=hzi/: Hence Z.G0 =hzi/ D .ht iT /=hzi and so Z.G0 / ht iT . On the other hand, we have G0 < G and so (by the minimality of G) Z.G0 / D ht iT . In particular, t centralizes B and so CG .t/ hA; Bi D G. Hence t is a central involution in G which is not contained in G 0 (since G 0 B), contrary to Proposition 122.9. We have proved that A Š Q8 , C4 , E4 or C2 . Assume that A Š Q8 and let hvi be any cyclic subgroup of order 4 in A. As v 2 D z and Œv; B hzi, we have hvi E G. Set again G0 D hviB so that G0 < G which gives T D Z.B/ and T =hzi D Z.B=hzi/. Due to the fact that G0 =hzi Š C2 .B=hzi/, we have Z.G0 =hzi/ D .hviT /=hzi and so Z.G0 / hviT . But jG0 =.hviT /j D 24 and G0 < G and so Z.G0 / D hviT . In particular, v centralizes B. But hvi was any cyclic subgroup of order 4 in A Š Q8 and so ŒA; B D 1. It remains to prove that G 0 D B 0 . Assume jB 0 j D 2 so that G 0 D hzi B 0 Z.G/ and therefore G is of class 2. Also, B < G and so A > hzi which gives (because of the minimality of G) T D Z.B/ and B=Z.B/ Š E16 . By Proposition 122.13, we obtain that B D .Q8 P/ E2s , s 0, with B 0 D Z.Q8 / D hz 0 i D P0 D Q8 \ P

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Groups of prime power order

since D8 is not a subgroup of G. If z 2 Q8 P, then z 2 Z.P/ hz 0 i and B=hzi Š .Q8 D8 / E2s and so s D 0. In case z 62 Q8 P, we get B=hzi Š .Q8 P/ E2s1 and so s D 1. Thus it follows that we have either B D hzi .Q8 P/ or B D Q8 P. If A Š Q8 , then ŒA; B D 1 and so hA; Q8 i D AQ8 , which is a contradiction (by Proposition 122.4). Hence A Š E4 or A Š C4 and ŒA; B D hzi since G 0 D hz; z 0 i. First assume B D hzi .Q8 P/, where we set P D hb; t j b 4 D t 2 D 1; b 2 D u; Œb; t D z 0 ; .z 0 /2 D Œb; z 0 D Œt; z 0 D 1i and Q8 D hr; s j r 2 D s 2 D z 0 ; Œr; s D z 0 i. Since ¹t; tz 0 º are the only two conjugates of t in G and ¹hbi; hbz 0 iº are the only two conjugates of hbi Š C4 in G, we have ht; tz 0 i E G and hb; bz 0 i E G and so P E G. This gives ŒA; P A \ P D ¹1º and so A centralizes P. Since A E G and ŒA; B D hzi, we have jG W CG .A/j D 2 and so we may set ŒA; r D 1 and ŒA; s D z. Consider the cyclic subgroup hsbi Š C4 with .sb/2 D z 0 u and .sb/t D .sb/z 0 so that ¹hsbi; hsbz 0 iº are the only two conjugates of hsbi in G (and they are contained in Q8 P). For a 2 A hzi, .sb/a D szb D .sb/z and .sb/z 62 Q8 P, a contradiction. Finally, suppose that B D Q8 P, where we may set P D hb; t j b 4 D t 2 D 1; b 2 D z; Œb; t D z 0 ; .z 0 /2 D Œb; z 0 D Œt; z 0 D 1i (noting that both z and zz 0 are squares in P) and Q8 D hr; s j r 2 D s 2 D z 0 ; Œr; s D z 0 i. Also, ŒA; B D hzi and since ¹t; t z 0 º are the only two conjugates of t in G, we have ŒA; t D 1 (as ŒA; t D z is not possible) and so A centralizes t . Assume that A Š E4 and let i be an involution in A hzi. We have ŒNG .hi; ti/; hi; ti hi; ti \ G 0 D ¹1º since G 0 D hz; z 0 i. Hence NG .hi; ti/ D CG .hi; ti/ and since jG W NG .hi; ti/j D 2 and hi; ti \ Z.G/ D ¹1º (because 1 .Z.G// G 0 by Proposition 122.9), we obtain CG .t/ D CG .i/ D CG .it/ D hi; t; zi Q8 : As ŒA; B D hzi, we have Œi; b D z and so b i D bz D b 1 which gives hb; ii Š D8 , a contradiction. We have proved that A D hvi with v 2 D z and Œv; t D 1 (because A centralizes t ). Since A E G, we get jG W CG .v/j D 2 and so jB W CB .v/j D 2. If v centralizes Q8 , then Œv; b D z and so hv; bi Š Q8 centralizes Q8 D hr; si, which is a contradiction (by Proposition 122,4). Hence we may assume Œv; r D 1 and Œv; s D z. Suppose that v centralizes b. Then i 0 D vb is an involution and so Œi 0 ; t D 1 by Proposition 122.5. But v centralizes t and so also b centralizes t , a contradiction. Hence Œv; b D z and the structure of G is uniquely determined.

122

p-groups all of whose subgroups have normalizers of index at most p

225

We see that vt brs is an involution in G B. Indeed, .vt brs/2 D .vt/2 .brs/2 Œbrs; vt D 1 and then we must have Œt; vtbrs D 1. On the other hand, computing in G we get Œt; vtbrs D Œt; vŒt; tŒt; bŒt; rŒt; s D z 0 which is a ﬁnal contradiction. Our proposition is proved. Proof of Theorem 122.1 for p D 2. Let G be a 2-group with sb.G/ D 1 (i.e., the subgroup breadth of G is equal to 1) and assume that G is a minimal counterexample to Theorem 122.1, i.e., jG=Z.G/j 25 but for each proper subgroup or factor group X of G, we have jX=Z.X/j 24 . Then Proposition 122.5 gives that 1 .G/ is elementary abelian and Proposition 122.9 yields 1 .Z.G// G 0 . The starting point in our proof is Proposition 122.18 which gives the following facts. We have jG 0 j D 4, and let z be ﬁxed involution in G 0 \ Z.G/. Then G D AB, where A and B are normal subgroups of G with A \ B D hzi, A Š Q8 , C4 , E4 or C2 and if A Š Q8 , then ŒA; B D ¹1º. Further, B D CD with C \ D D G 0 D B 0;

ŒC; D hzi;

C =hzi Š Q8

and D=hzi Š D8 or P;

where P is the nonmetacyclic minimal nonabelian subgroup of order 24 . Hence C and D are also normal subgroups of G. At this point it is possible to determine the structure of C . For each x 2 C G 0 , 2 x 2 G 0 hzi and so hz; xi D G 0 hxi is abelian. It follows that all three maximal subgroups of C which contain G 0 are abelian and so jC 0 j D 2 (Exercise P1). But C 0 covers G 0 =hzi and so G 0 D hziC 0 Š E4 and since C 0 Z.G/, we get G 0 Z.G/ and so G is of class 2 whose commutator subgroup is a four-subgroup. We have 1 .C / D G 0 and z is not a square in C . Set hc 0 i D C 0 so that G 0 D B 0 D hz; c 0 i. If ˆ.C / D hc 0 i, then there is a maximal subgroup Q of C which does not contain hzi in which case C D hziQ with Q Š Q8 and Q0 D hc 0 i. In case ˆ.C / D G 0 , we infer that C is minimal nonabelian of order 24 and exponent 4. Since 1 .C / D G 0 Š E4 , it follows that C is metacyclic and so c 0 is a square in C . There are r; s 2 C G 0 such that r 2 D c 0 , s 2 D c 0 z and Œr; s D c 0 . In this case, C D hr; s j r 4 D s 4 D 1; r 2 D c 0 ; s 2 D zc 0 ; Œr; s D c 0 i Š H2 ; where H2 is deﬁned by H2 D hx; y j x 4 D y 4 D 1; x y D x 1 i. In any case, there are elements r; s 2 C G 0 of order 4 such that r 2 D c 0 , s 2 D c 0 z , D 0; 1 and Œr; s D c 0 . We see that .rs/2 D r 2 s 2 Œs; r D c 0 c 0 z c 0 D c 0 z . Fix this notation in the rest of the proof which we split into two parts: Part 1, where D=hzi Š D8 , and Part 2, where D=hzi Š P with P being the nonmetacyclic minimal nonabelian group of order 24 . It is interesting to note that Part 1 will be more difﬁcult than Part 2.

226

Groups of prime power order

Part 1, where D=hzi Š D8 . Each of the three maximal subgroups of D containing G 0 are abelian and so D 0 D hd 0 i is of order 2 and d 0 2 G 0 hzi and so we may set d 0 D c 0 z , D 0; 1. If ˆ.D/ < G 0 , then ˆ.D/ D D 0 D hd 0 i and D D hzi D0 , where D0 Š D8 , a contradiction (since dihedral groups cannot be subgroups in G). Hence we have ˆ.D/ D G 0 and so D is minimal nonabelian of order 24 and exponent 4. If 1 .D/ D G 0 , then D Š H2 and if 1 .D/ Š E8 , then D Š P. In case D Š H2 , the fact that D=hzi Š D8 shows that z is a square in D. If D Š P, then d 0 is the only involution in Z.D/ which is not a square in D. In any case, there is y 2 D such that y 2 D z. Also, G 0 D B 0 D Z.B/ Z.G/ and jB=Z.B/j D 24 so that we must have G > B which implies A > hzi. If A Š Q8 , then we know that ŒA; B D ¹1º and so A hyi (with A \ hyi D hzi) contains a subgroup isomorphic to D8 , a contradiction. It follows that A Š E4 ; or C4 . In view of jG=Z.G/j 25 , we have A 6 Z.G/ and so ŒA; B D hzi and G 0 D B 0 D Z.B/ D Z.G/ D ˆ.G/. (i) First assume D D hb; t j b 4 D 1; b 2 D z; t 2 D 1; Œb; t D d 0 ; .d 0 /2 D Œd 0 ; b D Œd 0 ; t D 1i Š P; where ¹t; t b D t d 0 º are the only two conjugates of t in G. On the other hand, we obtain ŒC; t ŒC; D hzi and so we must have ŒC; t D ¹1º which gives CB .t/ D hti C . Since ŒC; D hzi, we may set Œb; r D z 1 and Œb; s D z 2 , where 1 ; 2 2 ¹0; 1º. We compute Œt; tbr D d 0 , Œt; tbs D d 0 and Œt; tbrs D d 0 , which shows that no one of the elements ¹t br; t bs; t brsº could be an involution (since 1 .G/ is elementary abelian). On the other hand, ..tb/r/2 D .tb/2 r 2 Œr; tb D zd 0 c 0 z 1 D zz z 1 D z 1CC1 which gives C 1 0 .mod 2/, ..tb/s/2 D zd 0 c 0 z z 2 D zz z z 2 D z 1CCC2 which gives C C 2 0 .mod 2/, ..tb/.rs//2 D zd 0 c 0 z z 1 C2 D z 1CCC1 C2 which gives C C 1 C 2 0 .mod 2/. From the above relations we get 1 D 0 and D 0. But then we have d 0 D c 0 and so D 0 D C 0 which gives ŒC; D D hzi since B 0 D G 0 . This implies 2 D 1 and then the above relations also give D 1 so that C Š H2 . The structure of B is uniquely determined. Consider the abelian subgroup S D ht; si D ht i hsi, where s 2 D zc 0 . Due to the fact that ŒNB .S/; S S \ G 0 D hzc 0 i, an element x 2 B lies in NB .S/ if and only if Œt; x 2 hzc 0 i and Œs; x 2 hzc 0 i. We have CB .t/ D ht i C and t b D tc 0 (where c 0 ¤ zc 0 ) and so NB .S/ ht i C . Also, r 2 CB .t/ but Œs; r D c 0 ¤ zc 0 and so NB .S/ is a proper subgroup of CB .t/ which shows jB W NB .S/j 4, a contradiction. (ii) Now assume D D hb; f j b 4 D 1; f 4 D 1; b 2 D z; f 2 D d 0 ; Œb; f D d 0 i Š H2 ;

122

p-groups all of whose subgroups have normalizers of index at most p

227

where 1 .D/ D 1 .C / D G 0 D B 0 D hz; c 0 i D hz; d 0 i D Z.B/ D Z.G/ D ˆ.G/: Here we have two possibilities for c 0 d 0 D z , namely, D 0 or 1. (ii1) First assume D 0, i.e., d 0 D c 0 which implies ŒC; D D hzi since B 0 D G 0 . We set Œb; r D z 1 ; Œb; s D z 2 ; Œf; r D z 3 ; Œf; s D z 4 ; where 1 ; 2 ; 3 ; 4 2 ¹0; 1º, and compute .f r/2 D c 0 c 0 z 3 D z 3 ; .f s/2 D c 0 c 0 z z 4 D z C4 ; .f .rs//2 D c 0 c 0 z z 3 C4 D z C3 C4 : Assume 3 D 0 so that Œf; r D 1 and i D f r is an involution. Then i b D f b r b D f c 0 rz 1 D ic 0 z 1

and i s D f s r s D f z 4 rc 0 D ic 0 z 4

so that we have 4 D 1 ;

i bs D i

and

CB .i/ D G 0 hf; r; bsi:

Suppose, in addition, that 1 D 1 and consider the abelian subgroup S D hi i hf i Š C2 C4 with f 2 D c 0 . Since ŒNB .S/; S S \ G 0 D hc 0 i, an element x 2 B lies in NB .S/ if and only if Œi; x 2 hc 0 i and Œf; x 2 hc 0 i. As i b D i.c 0 z/, we conclude that NB .S/ CB .i/. But bs 2 CB .i/ and f bs D .f c 0 /s D f z 4 c 0 D f .zc 0 / and so NB .S/ is a proper subgroup of CB .i/ and so jB W NB .S/j 4, a contradiction. Thus we have proved that 1 D 4 D 0 and so Œb; r D Œf; s D 1. But ŒC; D D hzi and so we must have 2 D 1 and Œb; s D z. If D 0, then hr; si Š Q8 and hr; si hf i (with hr; si \ hf i D hc 0 i) contains a subgroup isomorphic to D8 , a contradiction. Hence D 1 and so s 2 D .rs/2 D c 0 z and therefore C Š H2 so that the structure of B is in this case uniquely determined. Let us consider the abelian subgroup S D hi i hbsi Š C2 C4 with .bs/2 D c 0 z. Since ŒNB .S /; S S \ G 0 D hc 0 zi, an element x 2 B lies in NB .S / if and only if Œi; x 2 hc 0 zi and Œbs; x 2 hc 0 zi. As i b D ic 0 , we have NB .S / CB .i/. But r 2 CB .i/ and Œbs; r D c 0 ¤ c 0 z and so NB .S / is a proper subgroup of CB .i/ and therefore jB W NB .S /j 4, a contradiction. We have proved that 3 D 1 and so Œf; r D z. If Œf; s D z, then Œf; rs D 1 and so interchanging s and rs (if necessary) and noting that s 2 D .rs/2 D c 0 z , we may assume from the start that Œf; s D 1 and so 4 D 0. Assume that D 1. Then j D f rs is an involution in B G 0 , j r D f z r sc 0 D jzc 0 , j s D jc 0 and ¹j; jzc 0 ; jc 0 º are three pairwise distinct conjugate involutions in B, a contradiction. Hence we obtain D 0 so that r 2 D s 2 D .rs/2 D c 0 , hr; si Š Q8 and u D f s is an involution in

228

Groups of prime power order

B G 0 . We have ur D f z sc 0 D uzc 0 and so ¹u; uzc 0 º are the only two conjugates of u in G. Also, ub D f c 0 sz 2 D uc 0 z 2 which gives 2 D 1, Œb; s D z and ubr D u so that CB .u/ D G 0 hf; s; bri. Let us consider the abelian subgroup S0 D hui hf i Š C2 C4 with f 2 D c 0 . Since ŒNB .S0 /; S0 S0 \ G 0 D hc 0 i, an element x 2 B lies in NB .S0 / if and only if Œu; x 2 hc 0 i and Œf; x 2 hc 0 i. As ub D uc 0 z, we infer that NB .S0 / CB .u/. But br 2 CB .u/ and Œf; br D c 0 z ¤ c 0 and so NB .S0 / is a proper subgroup of CB .u/ and so jB W NB .S0 /j 4, a contradiction. (ii2) We have proved that D 1, i.e., d 0 D c 0 z. Since ŒD; C hzi, we may set again Œb; r D z 1 ; Œb; s D z 2 ; Œf; r D z 3 ; Œf; s D z 4 ; where 1 ; 2 ; 3 ; 4 2 ¹0; 1º, and we look for involutions in B .C [ D/. We compute .f r/2 D c 0 z c 0 z 3 D z 1C3 ; .f s/2 D c 0 z c 0 z z 4 D z 1CC4 ; .f .rs//2 D c 0 z c 0 z z 3 C4 D z 1CC3 C4 : If 3 D 1, then .f r/2 D 1, .f s/2 D z 1C.C4 / , .f .rs//2 D z C4 and so f r is an involution and either f s or f rs is an involution. But in that case, we conclude that Œf r; f s D c 0 z 1C4 ¤ 1 and Œf r; f rs D c 0 z 4 ¤ 1, contrary to the fact that 1 .G/ must be elementary abelian. Hence 3 D 0 and then .f r/2 D z;

.f s/2 D z 1C.C4 / ;

.f rs/2 D z 1C.C4 / :

If C 4 1 .mod 2/, then both f s and f rs are involutions which is not possible since Œf s; f rs D c 0 ¤ 1. Hence C 4 0 .mod 2/ and so 4 D . We have proved that Œf; r D 1 and Œf; s D z which gives .f r/2 D .f s/2 D .f rs/2 D z. Also note that b 2 D .bf /2 D z and so if there is an involution i 2 A hzi, then i commutes with b (otherwise, Œi; b D z and then hi; bi Š D8 ), bf , f r, f s, f rs. But we have hb; bf; f r; f s; f rsi D B and so ŒA; B D 1, contrary to the fact that we know that ŒA; B D hzi. Hence A D hai Š C4 with a2 D z. Suppose that there are no involutions in G B. Then Œa; b D Œa; bf D Œa; f r D Œa; f s D Œa; f rs D z: From Œa; b D Œa; bf D z follows Œa; f D 1. But then Œa; r D Œa; s D Œa; rs D z, a contradiction. Hence it follows that a commutes with at least one element x in the set L D ¹b; bf; f r; f s; f rsº in which case ax is an involution in G B. We check that no two distinct elements in the set L commute. If a would commute with two distinct elements x; y 2 L, then ax and ay are involutions and then 1 D Œax; ay D Œx; y, a contradiction. We have proved that a must commute with exactly one of the ﬁve elements in L. But we may interchange s and rs and so it is enough to consider the following four cases: (a) Œa; b D 1;

(b) Œa; bf D 1;

(c) Œa; f r D 1;

(d) Œa; f s D 1:

122

p-groups all of whose subgroups have normalizers of index at most p

229

(ii2a) We assume Œa; b D 1 so that i D ab is an involution in G B and Œa; bf D Œa; f r D Œa; f s D Œa; f rs D z: This implies Œa; f D z, Œa; r D Œa; s D 1. Since i f D .ab/f D az bc 0 z D ic 0 , ¹i; ic 0 º are the only two conjugates of i in G. Then i r D .ab/r D a bz 1 D iz 1 gives 1 D 0 and i s D .ab/s D a bz 2 D iz 2 gives 2 D 0. We have CB .i/ D hbihr; si D hbiC and i f D ic 0 . Consider the abelian subgroup S D hi i hbri Š C2 C4 with .br/2 D zc 0 . Since ŒNG .S/; S S \ G 0 D hzc 0 i, an element x 2 G lies in NG .S/ if and only if Œi; x 2 hzc 0 i and Œbr; x 2 hzc 0 i. As i f D i c 0 , we have NG .S/ CG .i/. But s 2 CG .i/ and .br/s D br s D .br/c 0 and so NG .S/ is a proper subgroup of CG .i/ and therefore jG W NG .S/j 4, a contradiction. (ii2b) We assume that Œa; bf D 1 so that i D abf is an involution in G B and Œa; b D Œa; f r D Œa; f s D Œa; f rs D z. This gives Œa; f D z, Œa; r D Œa; s D 1. Since i f D .abf /f D az bc 0 z f D ic 0 ; i r D .abf /r D a bz 1 f D iz 1 ; i s D .abf /s D a bz 2 f z D iz C2 ; we get 1 D 0 and 2 D . If D 0, then we consider the abelian subgroup S1 D hi i hasi Š C2 C4 with .as/2 D zc 0 . As ŒNG .S1 /; S1 S1 \ G 0 D hzc 0 i, an element x 2 G lies in NG .S1 / if and only if Œi; x 2 hzc 0 i and Œas; x 2 hzc 0 i. Since i f D ic 0 and jG W CG .i/j D 2 (noting that 1 .Z.G// G 0 ), we obtain NG .S1 / CG .i/. But we have r 2 CG .i/ and .as/r D .as/c 0 and so NG .S1 / is a proper subgroup of CG .i/ and therefore we get jG W NG .S1 /j 4, a contradiction. If D 1, then we consider the abelian subgroup S2 D hi i hsi Š C2 C4 with s 2 D zc 0 . Since ŒNG .S2 /; S2 S2 \ G 0 D hzc 0 i, an element x 2 G lies in NG .S2 / if and only if Œi; x 2 hzc 0 i and Œs; x 2 hzc 0 i. Because i f D ic 0 and jG W CG .i/j D 2, we have NG .S2 / CG .i/. But r 2 CG .i/ and s r D sc 0 and so NG .S2 / is a proper subgroup of CG .i/ and therefore jG W NG .S2 /j 4, a contradiction. (ii2c) We assume Œa; f r D 1 so that i D af r is an involution in G B and Œa; b D Œa; bf D Œa; f s D Œa; f rs D z: This gives Œa; f D 1, Œa; r D 1, Œa; s D z. We have i b D .af r/b D az f c 0 z rz 1 D ic 0 z 1 ; i s D .af r/s D az f z rc 0 D ic 0 z C1 ; and so 1 D C1. As i r D .af r/r D i and Œi; sb D 1, we have CG .i/ D ha; f; r; sbi.

230

Groups of prime power order

If D 0, then we consider the abelian subgroup S1 D hi i hri Š C2 C4 with r 2 D c 0 . Since ŒNG .S1 /; S1 S1 \ G 0 D hc 0 i, an element x 2 G lies in NG .S1 / if and only if Œi; x 2 hc 0 i and Œr; x 2 hc 0 i. Because i b D ic 0 z and jG W CG .i/j D 2, we have NG .S1 / CG .i / < G. But sb 2 CG .i/ and Œr; sb D zc 0 and so NG .S1 / is a proper subgroup of CG .i/ and therefore jG W NG .S1 /j 4, a contradiction. If D 1, then we consider the abelian subgroup S2 D hi i hari Š C2 C4 with .ar/2 D zc 0 . Since ŒNG .S2 /; S2 S2 \ G 0 D hzc 0 i, an element x 2 G lies in NG .S2 / if and only if Œi; x 2 hzc 0 i and Œar; x 2 hzc 0 i. Because i b D .af r/b D ic 0 and jG W CG .i/j D 2, we get NG .S2 / CG .i/ < G. But sb 2 CG .i/ and Œar; sb D c 0 and so NG .S2 / is a proper subgroup of CG .i/ and therefore jG W NG .S2 /j 4, a contradiction. (ii2d) We assume Œa; f s D 1 so that i D af s is an involution in G B and Œa; b D Œa; bf D Œa; f r D Œa; f rs D z: This gives Œa; f D 1, Œa; r D z, Œa; s D 1. We have i b D .af s/b D az f c 0 z sz 2 D ic 0 z 2 ; and so D 0. Also,

i s D .af s/s D a f z s D iz ;

i r D .af s/r D az f sc 0 D izc 0

which gives 2 D 1 and i f D i , i br D i and therefore CG .i/ D ha; f; s; bri. We consider the abelian subgroup S D hi i hsi Š C2 C4 with s 2 D c 0 . Since ŒNG .S/; S S \G 0 D hc 0 i, an element x 2 G lies in NG .S/ if and only if Œi; x 2 hc 0 i and Œs; x 2 hc 0 i. As i b D ic 0 z and jG W CG .i/j D 2, we have NG .S/ CG .i/ < G. But br 2 CG .i/ and Œs; br D zc 0 and so NG .S/ is a proper subgroup of CG .i/ and therefore jG W NG .S/j 4, a contradiction. Part 2, where D=hzi Š P. Here P is the nonmetacyclic minimal nonabelian subgroup of order 24 . We set W=hzi D Z.D=hzi/ D ˆ.D=hzi/ so that W > G 0;

jW W G 0 j D 2;

W=hzi Š E4 ;

W=hzi D Z.B=hzi/

and hence Z.B/ W . There is y 2 D such that y 2 2 W G 0 (noting that ˆ.D/ covers W=hzi). For each g 2 G, Œy 2 ; g D Œy; g2 D 1 and so y 2 2 Z.G/ and therefore W Z.G/, W D Z.B/, and B=Z.B/ Š E24 . Since ŒA; B hzi and A=hzi is elementary abelian, we get W D ˆ.G/. Each of three maximal subgroups of D containing W are abelian and so D 0 D hd 0 i is of order 2, where d 0 2 G 0 hzi. First suppose that z 62 ˆ.D/ and then D D hzi P0 , where P0 Š P so that we may set P0 D hb; t j b 4 D t 2 D 1; Œb; t D d 0 ; .d 0 /2 D Œd 0 ; t D Œd 0 ; b D 1i:

122

p-groups all of whose subgroups have normalizers of index at most p

231

Since ¹t; t b D td 0 º are the only two conjugates of t in G and ŒC; D hzi, we must have Œt; r D Œt; s D 1. Similarly, hbi and hb t i D hbd 0 i are the only two conjugates of hbi in G and so Œb; r D Œb; s D 1. We have proved that ŒC; D D ¹1º. If D 0 D C 0 , then B 0 D C 0 D D 0 is of order 2, contrary to B 0 D G 0 . Hence we have d 0 ¤ c 0 and so d 0 D c 0 z. We claim that hbri Š C4 , h.br/t i D hbr c 0 zi and h.br/s i D hbr c 0 i are three distinct cyclic subgroups of order 4 and this gives a contradiction. Indeed, the elements of order 4 in hbri are br and .br/.br/2 D br b 2 c 0 , the elements of order 4 in hbr c 0 zi are br c 0 z and br b 2 z, and the elements of order 4 in hbr c 0 i are br c 0 and br b 2 . We see that the above six elements of order 4 are pairwise distinct and we are done. We have proved that z 2 ˆ.D/ and so W D ˆ.D/ with D=W Š E4 which together with jD 0 j D 2 implies that D is minimal nonabelian of order 25 . Since D=hzi Š P is nonmetacyclic of exponent 4, it follows that D is nonmetacyclic of exponent 4 or 8. If D is of exponent 4, then D D hb; f j b 4 D f 4 D 1; Œb; f D d 0 ; .d 0 /2 D Œd 0 ; b D Œd 0 ; f D 1i; where

1 .D/ D W D ˆ.D/ D Z.D/ D hb 2 ; f 2 ; d 0 i Š E8 :

In case that z is not a square in D, we get 1 .D=hzi/ D W=hzi Š E4 which implies that D=hzi is metacyclic minimal nonabelian, contrary to D=hzi Š P. Hence z is a square in D so that we may choose the generators b and f of D so that b 2 D z. If D is of exponent 8, then D D hb; t j b 8 D t 2 D 1; Œb; t D d 0 ; .d 0 /2 D Œd 0 ; b D Œd 0 ; t D 1i; where b 4 D z since D=hzi Š P is of exponent 4. We have proved that in any case z is a square in D. Suppose that A Š Q8 . Then we know that ŒA; B D ¹1º and let b 0 2 D be such that .b 0 /2 D z. But then the subgroup A hb 0 i with A \ hb 0 i D hzi possesses a subgroup isomorphic to D8 , a contradiction. If A Z.G/, then jG=Z.G/j D 24 , a contradiction. Hence we have A Š C4 or E4 and ŒA; B D hzi. (i) First suppose that D is of exponent 4, i.e., D D hb; f j b 4 D f 4 D 1; Œb; f D d 0 ; .d 0 /2 D Œd 0 ; b D Œd 0 ; f D 1i; where b 2 D z and d 0 2 G 0 hzi. Now, hf i (where f and ff 2 are its elements of order 4) and hf b i D hf d 0 i (where f d 0 and ff 2 d 0 are its elements of order 4) are the only conjugates of hf i in G. Hence if Œf; C D hzi, then hf i is conjugate under C to hf zi (where f z and ff 2 z are its elements of order 4) and so hf zi is distinct from hf i and hf d 0 i, a contradiction. Similarly, hf bi (where f b and f bf 2 zd 0 are its elements of order 4) and h.f b/b i D hf bd 0 i (where f bd 0 and f bf 2 z are its elements of order 4) are the only conjugates of hf bi in G. Hence if Œf b; C D hzi, then hf bi

232

Groups of prime power order

is conjugate under C to hf bzi (where f bz and f bf 2 d 0 are its elements of order 4) and so hf bzi is distinct from hf bi and hf bd 0 i, a contradiction. Since D D hf; f bi, we get ŒC; D D ¹1º which implies D 0 ¤ C 0 (since B 0 D G 0 ) and so d 0 D c 0 z. We claim that hf ri Š C4 has more than two conjugates in G which gives a contradiction. Indeed, hf ri (where f r and f rf 2 c 0 are its elements of order 4), h.f r/b i D hf rc 0 zi (where f rc 0 z and f rf 2 z are its elements of order 4), and h.f r/s i D hf rc 0 i (where f rc 0 and f rf 2 are its elements of order 4) are three pairwise distinct conjugates of hf ri. (ii) Now assume that D is of exponent 8, i.e., D D hb; t j b 8 D t 2 D 1; Œb; t D d 0 ; .d 0 /2 D Œd 0 ; b D Œd 0 ; t D 1i; where b 4 D z and d 0 2 G 0 hzi. We have here W D hb 2 i hc 0 i D hb 2 i hd 0 i Š C4 C2 and W D Z.B/ D Z.G/ D ˆ.G/. Since ¹t; t b D td 0 º is the complete conjugate class in G, ŒC; t D hzi is not possible and so we get Œt; r D Œt; s D 1. We have r 2 D c 0 and s 2 D .rs/2 D c 0 z , D 0; 1. Assume at the moment that D 1 so that C Š H2 . Then replace s with s 0 D sb 2 , where o.b 2 / D 4 and b 2 2 Z.B/. We obtain .s 0 /2 D .sb 2 /2 D s 2 b 4 D c 0 z z D c 0 and Œs 0 ; r D Œs; r D c 0 so that hr; s 0 i Š Q8 . Replacing C D hr; si with C D hzihr; s 0 i Š C2 Q8 , we may assume (writing again s instead of s 0 and C instead of C ) from the start that D 0, i.e., s 2 D .rs/2 D c 0 and C D hzi hr; si, where hr; si Š Q8 . Since Œb; hr; si hzi, we may choose the generators r; s of hr; si so that Œb; r D 1. Assume that d 0 D c 0 in which case we must have Œb; s D z since B 0 D G 0 . Consider the abelian subgroup S D ht i hb 2 si Š C2 C4 , where .b 2 s/2 D zc 0 . Note that ŒNB .S/; S S \ B 0 D hzc 0 i. We conclude that jB W CB .t/j D 2 and t b D tc 0 so that NB .S/ CB .t/. But r centralizes t and .b 2 s/r D .b 2 s/c 0 and so r 62 NB .S/ and therefore jB W NB .S/j 4, a contradiction. We have proved that d 0 D c 0 z and so we have the following two possibilities: Œb; s D 1 or Œb; s D z. But we shall show that these two possibilities give isomorphic groups. Indeed, if Œb; s D z, then replace b with b D br and replace s with s D ts. We have .b /4 D .br/4 D z; Œb ; t D Œbr; t D zc 0 ; where hc 0 i D hs ; ri0 and Q D hs ; ri Š Q8 since .s /2 D s 2 D c 0 , r 2 D c 0 , and Œs ; r D Œts; r D c 0 . In addition, Œhb ; ti; Q D ¹1º. Indeed, t commutes with Q , Œb ; r D Œbr; r D Œb; r D 1 and Œb ; s D Œbr; ts D Œb; tŒb; sŒr; tŒr; s D zc 0 z 1 c 0 D 1: Thus we may assume from the start that Œb; s D 1. The structure of B is uniquely determined.

122

p-groups all of whose subgroups have normalizers of index at most p

233

It is easy to see that there are involutions in G B. Indeed, if A Š E4 , then there is an involution i 2 A hzi D A B. Suppose that A D hvi Š C4 with v 2 D z. Since b 2 2 Z.G/ and b 4 D z, i D vb 2 is an involution in G B. Let i be an involution in G B from the previous paragraph. Then we deduce that hi; ti Š E4 , hi; ti \ G 0 D ¹1º and hi; ti \ Z.G/ D ¹1º. From ŒNG .hi; ti/; hi; ti hi; ti \ G 0 D ¹1º it follows that NG .hi; ti/ D CG .hi; ti/ D CG .i/ \ CG .t/: But jG W NG .hi; ti/j D 2 and so we must have CG .i/ D CG .t/ D hi; ti hb 2 ; r; si, which is a subgroup of index 2 in G. As Œt; b D c 0 z ¤ 1, we have also Œi; b ¤ 1. Note that ŒA; b hzi and so if i 2 A, then Œi; b D z. If i D vb 2 (in case A D hvi Š C4 ), then Œi; b D Œvb 2 ; b D Œv; b D z. Hence in any case Œi; b D z. Consider the abelian subgroup S D hit i hb 2 si Š C2 C4 , where .b 2 s/2 D b 4 s 2 D zc 0 . Note that ŒNG .S/; S S \ G 0 D hzc 0 i. In view of .it /b D iz tc 0 z D it c 0 , we must have NG .S/ CG .it / D CG .t/. On the other hand, r centralizes it but .b 2 s/r D b 2 s c 0 and so r 62 NG .S/. We obtain that jG W NG .S/j 4, a contradiction. Theorem 122.1 is completely proved for p D 2. Proof of Theorem 122.1 for p > 2. Let G be a p-group, p > 2, with sb.G/ D 1 which is a minimal counterexample to Theorem 122.1 and so jG=Z.G/j > p 3 . By Proposition 122.9, 1 .Z.G// G 0 . Due to Proposition 122.8, b.G/ 2, i.e., the element breadth of G is at most 2. If b.G/ D 1, then Proposition 121.9 implies that jG 0 j D p. But then Proposition 122.13 gives jG=Z.G/j D p 2 , a contradiction. Hence we obtain b.G/ D 2. In that case, Theorem 121.1 implies jG 0 j D p 2 (because jG 0 j D p 3 would imply jG=Z.G/j D p 3 ). Also, we note that G cannot possess an abelian maximal subgroup because in that case Lemma 1.1 gives jGj D pjZ.G/jjG 0 j and then we conclude jG=Z.G/j D p 3 , a contradiction. N D 1 and sb.G/ N D 1. Let 1 ¤ z 2 G 0 \ 1 .Z.G// so that for GN D G=hzi, b.G/ N Set R=hzi D Z.G/ and then Proposition 122.13 gives G=R Š Ep2 . We have r 2 R if and only if ŒG; r hzi in which case jG W CG .r/j p. Also, ŒG; R D hzi because if ŒG; R D 1, then jG=Z.G/j D p 2 , a contradiction. As ŒG; G 0 hzi, we have G 0 R. Let A be any maximal normal abelian subgroup in G which contains ˆ.G/, where ˆ.G/ is abelian (Proposition 122.2). By Proposition 122.10, we have sbA .G/ D 2, and we ﬁx an element g 2 G such that jA W CA .g/j D p 2 and then jŒA; gj D p 2 (Proposition 121.2) so that ŒA; g D G 0 . This implies that A 6 R and R does not cover G=A so that jA W .A \ R/j D p and jG W .AR/j D p, where g 2 G .AR/ so that G D .AR/hgi. For each x 2 .AR/ A, x 2 TA , i.e., jA W CA .x/j D p (noting that for each r 2 R we have jG W CG .r/j p). Since jA W CA .g/j D p 2 , g does not centralize R \ A and so ŒR \ A; g D hzi which implies j.R \ A/ W CR\A .g/j D p and therefore CR\A .g/ D CA .g/ so that g p 2 CR\A .g/. As ŒR; R hzi and Œg; R D hzi, we have .Rhgi/0 D hzi. Also, R0 hzi and ŒA; R D hzi (noting that R 6 A since G has no abelian maximal subgroup) and so .AR/0 D hzi.

234

Groups of prime power order

Assume that jG=Aj > p 2 so that j.AR/=Aj p 2 . We have Z.AR/ < A and so j.AR/ W Z.AR/j p 3 , which contradicts Proposition 122.13 applied to AR. Hence we get G=A Š Ep2 and then jR W .R \ A/j D p. Since .Rhgi/0 D hzi, we may apply Proposition 122.3 to Rhgi and we get j.Rhgi/ W Z.Rhgi/j D p 2 . Suppose that R is abelian. Then CG .CA .g// hA; R; gi D G which gives Z.G/ D CA .g/ and jG W Z.G/j D p 4 . Now assume that R is nonabelian so that R0 D hzi. In that case, R \ A is an abelian maximal subgroup of R and then we get from jRj D pjZ.R/jjR0 j (Lemma 1.1) that jR W Z.R/j D p 2 , where Z D Z.R/ is a subgroup of index p in R \ A. But we infer that j.Rhgi/ W Z.Rhgi/j D p 2 , and so there is an element h 2 .Rhgi/ R such that h 2 Z.Rhgi/. We get CG .Z/ hA; R; hi D G and so again Z D CA .g/ D Z.G/ and jG W Z.G/j D p 4 . Thus we have proved that for any maximal normal abelian subgroup A of G containing ˆ.G/, G=A Š Ep2 and CA .g/ D Z.G/ is of index p 4 in G, where g 2 G A is such that jA W CA .g/j D p 2 . We shall ﬁx this notation for A, R and g in the rest of the proof. (i) We prove that 1 .G/ is elementary abelian. Suppose that x; y are some elements of order p in the group G such that Œx; y ¤ 1. Then C D CG .x/ D NG .hxi/ implies that jG W C j D p and y 2 G C so that the set p1 ¹x; x y ; : : : ; x y º contains p conjugates of x in G and they are all central elements p1 in C which implies that N D hx; x y ; : : : ; x y i is an elementary abelian normal subgroup in G, N C , and jN j p 2 . Set N0 D CN .y/ so that jN W N0 j D p and x 2 N N0 since jG W CG .y/j D p. We have N0 D Z.hx; yi/ and hx; yi=N0 Š Ep2 so that 1 ¤ Œx; y D z 0 2 N0 and hx; yi0 D hz 0 i is of order p which implies that hx; yi is minimal nonabelian. Hence S D hx; yi Š S.p 3 / which is the minimal nonabelian group of order p 3 and exponent p and so N0 D hz 0 i and N Š Ep2 . All p conjugates of x in S are contained in hx; z 0 i and they are all conjugates of x in G and so hx; z 0 i E G. Similarly, all p conjugates of y in S are contained in hy; z 0 i and they are all conjugates of y in G and so hy; z 0 i E G. This fact implies that S D hx; yi E G, G D S P , where P D CG .S/ and S \ P D hz 0 i D Z.S/. If P is abelian, then G 0 D S 0 D hz 0 i is of order p, a contradiction. Hence P is nonabelian and so P possesses nonnormal subgroups (Lemma 1.18). If all nonnormal subgroups of P contain hz 0 i, then Corollary 92.7 (N. Blackburn) implies p D 2, a contradiction. Let P1 be a nonnormal subgroup of P such that z 0 62 P1 and set U D hP1 ; xi D P1 hxi. We have U \ P D P1 and U \ S D hxi. If h 2 G normalizes U , then h normalizes U \ S D hxi and so h centralizes x. But h also normalizes U \ P D P1 and so we get NG .U / D CG .x/ \ NG .P1 /. On the other hand, we have CG .x/ D P hxi and jG W CG .x/j D p. But P1 is not normal in P hxi (since P1 is a nonnormal subgroup of P1 ) and so NG .U / is a proper subgroup of P hxi which gives jG W NG .U /j p 2 , a contradiction.

122

p-groups all of whose subgroups have normalizers of index at most p

235

(ii) 1 .G/ is elementary abelian of order p 3 . First suppose that G possesses a normal elementary abelian subgroup F of order p 4 . Set F0 D F \ G 0 so that jF0 j p 2 . Let x; y be any elements in F F0 such that hx; yi Š Ep2 and hx; yi \ G 0 D ¹1º. Then ŒNG .hx; yi/; hx; yi hx; yi \ G 0 D ¹1º and so NG .hx; yi/ D CG .hx; yi/ D CG .x/ \ CG .y/: Note that for each x 2 F F0 we have jG W CG .x/j D p since 1 .Z.G// G 0 . As jG W NG .hx; yi/j D p, we have C D CG .x/ D CG .y/ and so C D CG .F / is a subgroup of index p in G. In particular, ˆ.G/ centralizes F and so we may choose a maximal normal abelian subgroup A of G so that A F ˆ.G/. Recalling our results about A and g 2 G A with jA W CA .g/j D p 2 (from the beginning of our proof), we know that G=A Š Ep2 and CA .g/ D Z.G/. Since CF .g/ 1 .Z.G// G 0 and jG 0 j D p 2 , we see that jCF .g/j p 2 and so F covers A=CA .g/. But C D CG .F / is a subgroup of index p in G and so C > A and C centralizes hZ.G/; F i D A, contrary to CG .A/ D A. Now assume that 1 .G/ Š Ep2 . In view of Theorem 13.7, G is either metacyclic (with G 0 Š Cp2 ) or a 3-group of maximal class. In case that G is metacyclic, Proposition 122.7 implies that sb.G/ > 1, a contradiction. If G is a 3-group of maximal class, then jG=G 0 j D 32 and so jGj D 34 . But then jG W Z.G/j 33 , a contradiction. Thus we have proved that 1 .G/ Š Ep3 . (iii) We have Ep3 Š O D 1 .G/ ˆ.G/. Suppose that this is false. Let N be a maximal subgroup of G which does not contain O so that O \ N D 1 .N / Š Ep2 . Also we know that N is nonabelian. By Theorem 13.7, N is either metacyclic or a 3-group of maximal class. Suppose that N is metacyclic. If N 0 Š Cp2 , then Proposition 122.7 implies that sb.N / > 1, a contradiction. Hence we have jN 0 j D p which gives that N is minimal nonabelian and then ˆ.N / D Z.N / and jN W Z.N /j D p 2 . Let y 2 O N so that jN W CN .y/j D p and thus y centralizes ˆ.N / D Z.N / which gives Z.N / D Z.G/ and jG W Z.G/j p 3 , a contradiction. We have proved that N is a 3-group of maximal class. In that case, jN W N 0 j D 32 . If jN 0 j D 3, then jN j D 33 , jGj D 34 and jG W Z.G/j 33 , a contradiction. Hence jN 0 j D 32 and jGj D 35 . We have G 0 N \ O and so G 0 D N 0 D N \ O Š E9 : Since N=N 0 Š E9 and O=N 0 Š C3 , we get ˆ.G/ D ˆ.N / D N \ O D G 0 . As G has no abelian maximal subgroup, we have CG .O/ D O so that O D A is a maximal normal abelian subgroup of G containing ˆ.G/. We recall our results about A, R and g from the beginning of the proof. Since R G 0 , we obtain R \ A D G 0 , g 3 2 G 0 , and N D Rhgi is another maximal subgroup of G which does not contain O since N \A D G 0 D 1 .N / Š E9 and j.N /0 j D 3. By Theorem 13.7, N is metacyclic and so N is minimal nonabelian with Z.N / D ˆ.N / and jN W Z.N /j D 32 .

236

Groups of prime power order

Let v 2 A G 0 so that v centralizes a maximal subgroup of N and so v centralizes ˆ.N / D Z.N / which forces Z.N / Z.G/. But then jG W Z.G/j 33 , a contradiction. It is now easy to complete the proof of Theorem 122.1 for p > 2. To do this, we recall again our results about A ˆ.G/, R, g and z from the beginning of the proof. We have ˆ.G/ .Rhgi/ \ A D R \ A, where G=A Š Ep2 , jA W .R \ A/j D p, j.R \ A/ W CR\A .g/j D p, and CA .g/ D CR\A .g/ D Z.G/. By (iii), we conclude that Ep3 Š O D 1 .G/ ˆ.G/ and so O R \ A. Since 1 .Z.G// G 0 , we have O 6 Z.G/ and so Ep2 Š G 0 D O \ Z.G/ and O covers .R \ A/=Z.G/. In particular, the group G is of class 2 with an elementary abelian commutator subgroup of order p 2 . For each x; y 2 G we have Œx p ; y D Œx; yp D 1 and so x p 2 Z.G/ which implies that ˆ.G/ D Ã1 .G/G 0 Z.G/. But then O ˆ.G/ Z.G/, a ﬁnal contradiction. Theorem 122.1 is completely proved. It follows from Theorem 122.1 that if G is a metacyclic group of order p 2mC1 , where m > 1 if p > 2 and n > 2 if p D 2, then there is in G a subgroup H such that jG W NG .H /j > p.

123

Subgroups of ﬁnite groups generated by all elements in two shortest conjugacy classes

Here we present some results of I. M. Isaacs (Subgroups generated by small classes in ﬁnite groups, Proc. Amer. Math. Soc. 136 (2008), 2299–2301) which generalize the results of A. Mann about subgroups of p-groups which are generated by all elements of the ﬁrst two class sizes. Let M.G/ be the subgroup of a ﬁnite group G generated by all elements that lie in conjugacy classes of the two smallest sizes. The following simple result is the key for all following arguments. Lemma 123.1. Let A be an abelian normal subgroup of a group G. Let x be a noncentral element of G and a 2 A. Then jCG .Œa; x/j > jCG .x/j, and so the G-class of Œa; x is smaller than that of x. Proof. Consider the map u ! Œu; x (u 2 A) from A into A. Since for any u; v 2 A we have Œuv; x D Œu; xv Œv; x D Œu; xŒv; x, it follows that our map is a homomorphism onto ŒA; x D ¹Œu; x j u 2 Aº with kernel CA .x/. Thus ŒA; x is a subgroup of A and jŒA; xj D jA=CA .x/j. Set H D ACG .x/ so that jH j D .jAjjCG .x/j/=jCA .x/j and jAj=jCA .x/j D jH j=jCG .x/j: Obviously, ŒA; x E H . Indeed, if c 2 CG .x/ and Œu; x 2 ŒA; x with u 2 A, then Œu; cc D Œuc ; x c D Œuc ; x 2 ŒA; x. We may assume that Œa; x ¤ 1 because x is noncentral (and so in case Œa; x D 1, our result is trivial). Hence the conjugacy class of Œa; x in H has jH j=jCH .Œa; x/j elements so that jH j=jCH .Œa; x/j < jŒA; xj D jAj=jCA .x/j D jH j=jCG .x/j which gives jCH .Œa; x/j > jCG .x/j and we are done. Lemma 123.2. For an abelian normal subgroup A of a group G, ŒA; M.G/ Z.G/. Proof. If x 2 M.G/ is contained in a conjugacy class of G of size m, where m is the smallest class size exceeding 1, then, by Lemma 123.1, we have Œa; x 2 Z.G/ for each a 2 A. Hence all generators of M.G/ centralize A=.A \ Z.G//. Thus M.G/ centralizes A=.A \ Z.G// and so ŒA; M.G/ Z.G/.

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Groups of prime power order

Theorem 123.3. Let G be a group that contains a self-centralizing abelian normal subgroup A. Then M.G/ is nilpotent of class at most 3. Proof. Set M D M.G/ so that Lemma 123.2 yields ŒA; M Z.G/ and therefore ŒA; M; M D ¹1º. Since ŒM; A; M D ¹1º, the three-subgroups lemma implies that ŒM; M; A D ¹1º and then M 0 CG .A/ D A. But in that case we get ŒM 0 ; M; M ŒA; M; M D ¹1º, and so M is nilpotent of class at most 3. A maximal normal abelian subgroup in a supersolvable group G is self-centralizing and so we get the following result. Corollary 123.4. Let G be a supersolvable group. Then M.G/ is nilpotent of class at most 3. In particular, if G is a p-group, then M.G/ is of class at most 3. Theorem 123.5. Let G be a group and assume that M.G/ has a nonidentity solvable normal subgroup. Then Z.M.G// is nontrivial. Proof. Write M D M.G/. By our assumption, the Fitting subgroup F.M / of M is nontrivial. Let Z D Z.F.M // so that Z is a nontrivial characteristic abelian subgroup of M , and therefore Z E G. Due to Lemma 123.2, ŒZ; M Z.G/. If ŒZ; M ¤ ¹1º, then Z.G/ is nontrivial. By the deﬁnition of M D M.G/, Z.G/ M and so Z.M / is nontrivial and we are done in this case. If ŒZ; M D ¹1º, then ¹1º ¤ Z Z.M /. In any case, Z.M / is nontrivial. Finally, we prove a result about arbitrary ﬁnite group G (without any assumptions). Theorem 123.6. Let G be any ﬁnite group. Then the Fitting subgroup of M.G/ has the nilpotence class at most 4. Proof. Write again M D M.G/ and set F D F.M /. Let n be the nilpotence class of F so that Kn .F / ¤ ¹1º and KnC1 .F / D ¹1º. We want to show that n 4 and so we may assume that n 3 so that Kn2 .F / is deﬁned. If this normal subgroup is abelian, Lemma 123.2 yields ŒKn2 .F /; M Z.G/ and then Kn .F / D ŒKn2 .F /; M; M D ¹1º, a contradiction. Hence Kn2 .F / is nonabelian and then ¹1º ¤ ŒKn2 .F /; Kn2 .F / K2n4 .F /: Since KnC1 .F / D ¹1º, we must have 2n 4 < n C 1 and so n < 5, as required.

124

The number of subgroups of given order in a metacyclic p-group

In this section we compute the number of subgroups of given order in a metacyclic p-group and also prove some structure results. In the following lemma we gather some known results which we use in what follows. Lemma J. Let G be a nonabelian metacyclic p-group. (a) Let G be of order p 4 and exponent p 2 ; then G is minimal nonabelian. (i) If p D 2, then G Š H2 D ha; b j a4 D b 4 D 1; ab D a3 i, all subgroups of order 2 are characteristic in G, exactly one maximal cyclic subgroup of G has order 2. 2 2 (ii) If p > 2, then G D ha; b j ap D b p D 1; ab D a1Cp i, all maximal cyclic subgroups of G have order p 2 . (b) (Proposition 10.19) If G has a nonabelian subgroup of order p 3 , then it is of maximal class. In particular, if p > 2, then jGj D p 3 . (c) If p D 2 and L G G is such that G=L is nonabelian of order 8, then L is characteristic in G. (d) (Theorem 1.2 and Lemma 1.6) There are four and only four series of nonabelian 2-groups with cyclic subgroup of index 2, namely D2n , Q2n , SD2n , M2n . (e) j1 .G/j ¤ p 2 if and only if G is a 2-group of maximal class. Let us prove Lemma J(a). There is in G a normal cyclic subgroup A D hai such that G=A is cyclic; then we have jAj D p 2 . If B < G is cyclic and such that AB D G, then jBj D p 2 ; let B D hbi. Since jG W CG .A/j D p and the group of automorphisms of A is generated by the automorphism a ! a1Cp of order p, one can write ab D a1Cp hence the group G has the same deﬁning relations as in (a). Therefore, if p > 2, then G is regular, and (ii) follows easily. Now let p D 2. The subgroup G 0 D ha2 i is characteristic in G. Next, G has exactly three subgroups of order 2: G 0 D ha2 i, U D hb 2 i and V D ha2 b 2 i, and all these subgroups are central. Since G=V Š Q8 and V is the unique subgroup of order 2 not contained in a cyclic subgroup of order 4, it follows that V is characteristic in G, and hence U is also characteristic in G.

240

Groups of prime power order

Let us prove Lemma J(c). We have Ã2 .G/ L. If Ã2 .G/ D L, then it is nothing to prove. Now let Ã2 .G/ < L. Then jG W Ã2 .G/j D 24 , so G=Ã2 .G/ is nonabelian metacyclic of order 24 and exponent 4, and now the result follows from Lemma J(a). In what follows we freely use the following fact: If G is a metacyclic p-group, then 1 .G/ Š Ep2 unless G is either cyclic or a 2-group of maximal class. 1o Mann’s theorem on the number of subgroups of given order in certain metacyclic p-groups. Let G be a metacyclic group of order p n and exponent p e and let f D ne; clearly, f e. We also write e D e.G/ and f D f .G/. This notation is applicable to subgroups and epimorphic images of G as well. We begin with the following Deﬁnition. A p-group G is said to be quasi-regular if it is metacyclic and (QR1) if p D 2, then G has no nonabelian sections of order 8. (QR2) if H is a section of G, then Ã1 .H / D ¹x p j x 2 H º, i.e., all elements of Ã1 .H / are p-th powers. A more general condition than (QR2) was considered by Mann (see 11) for all p-groups, where p is an arbitrary prime. All metacyclic p-groups, p > 2, are quasi-regular (Theorem 7.2(c)). If G is a quasiregular p-group, then exp.m .G// p m for all positive integers m (this is proved by induction on jGj). The group H2 (see Lemma J(a)) is not quasi-regular since it has two nonabelian sections of order 8. The groups M2n are quasi-regular. Exercise. Metacyclic quasi-regular p-groups G are powerful (see 26). Solution. This is obvious for p > 2 (indeed, G 0 ˆ.G/ D Ã1 .G/). Now let p D 2 and write GN D G=Ã2 .G/. We have to prove that GN is abelian. Assume that this is N 2 ¹23 ; 24 º. Then jGj N ¤ 23 by (QR1). By Lemma J(a), there exists false. We have jGj only one nonabelian metacyclic group of order 24 and exponent 4, namely H2 . For this 2 i Š D so G N N does not satisfy condition (QR1), which is a congroup we have G=hb 8 tradiction. Let sm .G/ (s0m .G/, cm .G/) be the number of subgroups (noncyclic subgroups, cyclic subgroups) of order p m in a p-group G. In what follows, c; e; f; m; n; t are positive integers. The next theorem was inspired by Mann’s letter to the ﬁrst author at June 28, 2009. Theorem 124.1 (Mann [M33] for p > 2; see also 5o ). Let G be a quasi-regular metacyclic group of order p n and exponent p e < p n , m n, and set f .G/ D f D n e. Then one of the following holds: (a) If m f , then sm .G/ D

p mC1 1 p1 .

(b) If f < m e, then sm .G/ D (c) If m > e, then sm .G/ D

p f C1 1 p1

p nmC1 1 p1

.

so that sm .G/ is independent of m.

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The number of subgroups of given order in a metacyclic p-group

241

As we know, this is the unique result presenting exact values for sm .G/ for such a general class of p-groups. Of course, Theorem 124.1 is not true for general metacyclic 2-groups as 2-groups of maximal class show. We need the following Lemma 124.2. Suppose that G is a quasi-regular metacyclic group of order p n and exponent p e . Given a cyclic A G of order p e , there is a cyclic B G such that G D AB and A \ B D ¹1º. Proof. We use induction on jGj. One may assume that the group G is nonabelian and f D n e > 1 (otherwise, the result is either trivial for n D e or follows from Introduction, Exercise 4, and Theorem 1.2). Then the subgroup Ã1 .G/ is noncyclic of exponent p e1 by (QR2)). We have jA \ Ã1 .G/j D p e1 . Therefore, by induction, Ã1 .G/ D U V , where U D A \ Ã1 .G/, V is cyclic and U \ V D ¹1º. It follows that A \ V D ¹1º. Let V D hvi. Then, by (QR2), there is b 2 G such that v D b p . Set B D hbi. Then A \ B D ¹1º and G D AB. (Clearly, jBj D p f .) Note that for general abelian p-groups cyclic subgroups of maximal order are complemented (Introduction, Exercise 4). As Lemma 124.2 shows, this property also holds for quasi-regular metacyclic p-groups so for all metacyclic p-groups, p > 2. Generalized quaternion groups show that this is not true, in general, for p D 2. However, it is not true that if p > 2 and A G G is cyclic and such that G=A is cyclic, then A is complemented in G. Example: G is abelian of type .p; p n /, n > 2, and A is a cyclic subgroup of order p 2 not contained in ˆ.G/. Indeed, then G=A is cyclic, but A is not complemented in G (otherwise, we have j2 .G/j D p 4 , a contradiction since, in fact, j2 .G/j D p 3 ). Proof of Theorem 124.1. One may assume that 1 < m < n. If f D 1, then G is either abelian of type .p e ; p/ or G Š Mpn (see Theorem 1.2). In both these cases, we get sm .G/ D p C 1, and the same result is obtained from (a–c). In what follows we assume that f > 1; then G has no cyclic subgroup of index p. By Lemma 124.2, G D AB, where A; B < G are cyclic of orders p e , p f , respectively. Taking into account that G is quasi-regular and f e, we get ()

jk .G/j D p 2k .k f / and

G=f .G/ Š Cpef D Cpn2f :

We have (1)

sm .G/ D s0m .G/ C cm .G/:

If H G is noncyclic, then 1 .G/ H since j1 .G/j D p 2 D j1 .H /j (Lemma J(e)), and so (2)

s0m .G/ D sm2 .G=1 .G//:

One can rewrite (1) as follows: (3)

sm .G/ D sm2 .G=1 .G// C cm .G/:

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Groups of prime power order

(A) We ﬁrst compute cm .G/ for m e (if m > e, then cm .G/ D 0). Since G is quasi-regular and p m1 .p 1/ D '.p m / is the number of elements of order p m in Cpm , it follows that cm .G/ D

(4)

jm .G/ m1 .G/j : p m1 .p 1/

(i) Suppose that m f . Then we get jm .G/j D p 2m , jm1 .G/j D p 2m2 so that, by (4), cm .G/ D

(5)

p 2m p 2m2 D p m1 .p C 1/: p m1 .p 1/

(ii) Suppose that f < m e. By (), we have jm .G/j D p 2f Cmf D p mCf

and jm1 .G/j D p 2f Cm1f D p mCf 1 :

Therefore, by (4), (6)

cm .G/ D

p mCf p mCf 1 D pf : p m1 .p 1/

(B) In view of (3), it remains to compute sm2 .G=1 .G//. Here we use induction on m. We have p n1 D jG=1 .G/j D p n2 ; e1 D e.G=1 .G// D e 1;

(7)

f1 D f .G=1 .G// D f 1 (for example, f1 D n1 e1 D .n 2/ .e 1/ D n e 1 D f 1). (j) Let m f ; then m 2 f 2 < f 1 D f1 D f .G=1 .G//. By induction (see part (a) of the statement), sm2 .G=1 .G// D

p m1 1 p m2C1 1 D p1 p1

so that, by (3) and (5), we get sm .G/ D

p m1 1 p mC1 1 C p m1 .p C 1/ D ; p1 p1

completing this case. (jj) Let f < m e; then f1 D f .G=1 .G// D f 1 m 2 e 2 < e 1 D e.G=1 .G// D e1 : If m 2 D f 1 or, what is the same, m 1 D f , then, by induction (see part (a)), sm2 .G=1 .G// D

pf 1 p m2C1 1 D p1 p1

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243

so, by (3) and (6), we obtain sm .G/ D p f C

p f C1 1 pf 1 D ; p1 p1

as in (b), and we are done in this case. Now let m 2 > f 1 or, what is the same, m > f C 1. Then, by induction (see part (b)), pf 1 p f 1C1 1 sm2 .G=1 .G// D D p1 p1 so, by (3) and (6), we have sm .G/ D p f C

p f C1 1 pf 1 D ; p1 p1

completing case (jj). (jjj) Now let m e C 1 or, what is the same, m 2 e 1 D e.G=1 .G// D e1 . Then sm .G/ D sm2 .G=1 .G// since all subgroups of G of order p m are noncyclic hence contain 1 .G/. Let m 2 D e 1 D e1 , i.e., e C 1 D m; then, by induction (see part (b)), sm2 .G=1 .G// D

pf 1 p ne 1 p nmC1 1 p .f 1/C1 1 D D D ; p1 p1 p1 p1

and this coincides with the formula in (c). Now let m 2 > e 1, i.e., m > e C 1. Then, by induction (see part (c)), sm2 .G=1 .G// D

p nmC1 1 p .n2/.m2/C1 1 D ; p1 p1

and this coincides with the formula in (c). Since all possibilities for m are considered, the proof is complete. Corollary 124.3 (Mann [M33] for p > 2). If G and m; n; f; e are as in Theorem 124.1 (i.e., G is quasi-regular), then sm .G/ D snm .G/. Proof. One may assume that G is noncyclic and 1 < m < n. (i) Let m f ; then n m n f D e. If n m D e, then m D n e D f . We have, by Theorem 124.1(b,a), snm .G/ D se .G/ D

p mC1 1 p f C1 1 D D sm .G/: p1 p1

If n m > e, then m < n e D f . We have, by Theorem 124.1(c,a), snm .G/ D

p mC1 1 p n.nm/C1 1 D D sm .G/: p1 p1

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Groups of prime power order

(ii) Let f < m e; then f D n e n m < n f D e. If n m D f , then m D n f D e. We have, by Theorem 124.1(a,b), snm D sf .G/ D

p f C1 1 D se .G/ D sm .G/: p1

If f < n m < e, then, by Theorem 124.1(b), snm .G/ D

p f C1 1 D sm .G/: p1

(iii) Let m > e, then n m < n e D f so, by Theorem 124.1(a,c), snm .G/ D

p nmC1 1 D sm .G/; p1

and the proof is complete. The group G D H2 with n D 4, e D f D 2 is not quasi-regular, however, sm .G/ is such as in Theorem 124.1. 2o The number of cyclic subgroups of given order in a metacyclic 2-group. In this subsection we shall ﬁnd the number of cyclic subgroups of given order in a metacyclic 2-group. In this and the following subsections we suppose that G is a metacyclic group of order 2n . Set w D w.G/ D max¹i j ji .G/j D 22i º but jwC1 .G/j ¤ 22.wC2/ : In this case, we write R.G/ D w .G/. Then we get jR.G/j D 22w and G=R.G/ is either cyclic or a 2-group of maximal class (Lemma J(e)). We shall retain this notation in what follows. If w D 0, then either G is cyclic or of maximal class (Lemma J(e)). Part A of the following known theorem completes the case w D 0. Theorem 124.4. Suppose that G is a 2-group of maximal class and n > 3. A. If n > m > 2, then cm .G/ D 1. (a) If G Š D2n , then c1 .G/ D 2n1 C 1 and c2 .G/ D 1. (b) If G Š Q2n , then c1 .G/ D 1, c2 .G/ D 2n2 C 1. (c) If G Š SD2n , then c1 .G/ D 2n2 C 1 and c2 .G/ D 2n3 C 1. B. We have sm .G/ D 2nm C 1 for 2 m < n. If G is a group from Theorem 124.4, then its maximal subgroups are known (for example, if G D SD2nC1 , then 1 D ¹C; D; Qº, where C is cyclic, D is dihedral and Q is generalized quaternion). This allows us, using induction and Hall’s enumeration principle, to prove that theorem. In what follows w D w.G/ is a positive integer.

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Theorem 124.5. Suppose that G satisﬁes R.G/ D G; then jGj D 2n D 22w . In that case, we have G D AB, where A G G and B are cyclic of the same order 2w and cm .G/ D 3 2m1 for all m 2 ¹1; : : : ; wº. Proof. Since G is metacyclic, there is a cyclic AGG such that G=A is cyclic. It follows from exp.G/ D 2w and jAj 2w that jG W Aj 2w D exp.G/ whence jAj D 2w . If B < G is cyclic such that AB D G, then, since exp.G/ D 2w , we get jBj D 2w , A \ B D ¹1º, and the ﬁrst assertion of the theorem is proven. Since j t .G/j D 22t for all nonnegative integers t w, we get jm .G/ m1 .G/j 22m 22m2 D D 3 2m1 ; 2m1 2m1

cm .G/ D completing the proof.

In view of Theorem 124.5, we suppose in what follows that .w .G/ D/R.G/ < G. By Theorem 124.5, if m w, then cm .G/ D 3 2m1 . Therefore in the following two theorems we assume that m > w and R.G/ < G. Theorem 124.6. Suppose that G=R.G/ is nonidentity cyclic and m > w. Then we have cm .G/ D 2w . Proof. We obtain jm .G/j D 22wC.mw/ D 2wCm ; jm1 .G/j D 22wC.m1w/ D 2wCm1 : Therefore cm .G/ D

jm .G/ m1 .G/j 2wCm 2wCm1 D D 2w ; 2m1 2m1

and the proof is complete. In what follows we assume that the quotient group G=R.G/ is of maximal class. If L=R.G/ < G=R.G/ is cyclic of order 2s , s > 0, then we infer that cm .L/ D 2w by Theorem 124.6. In case that L1 =R.G/ < G=R.G/ is another cyclic subgroup of order 2s , we conclude that L \ L1 has no cyclic subgroups of order 2wCs . If L < G is cyclic of order 2m > 2w , then, by the product formula, LR.G/=R.G/ is cyclic of order 2mw since jL \ R.G/j D 2w . Therefore we get Theorem 124.7. Suppose that G=R.G/ is a 2-group of maximal class and m > w. Then cm .G/ D 2w cmw .G=R.G//. Since cmw .G=R.G// is known (Theorem 124.4.A), we have completed the computation of cm .G/ if G=R.G/ is of maximal class. Thus the number cm .G/ is computed for all metacyclic 2-groups.

246

Groups of prime power order

3o The number of subgroups of given order in a metacyclic 2-group. In this subsection we compute the number sm .G/ of subgroups of order 2m in G, where G > ¹1º is a metacyclic 2-group. We ﬁrst consider a metacyclic 2-group G D R.G/.D w .G// (see the second paragraph of 2o ) for which this problem is solved without difﬁculty (however, this is a key result as we shall see in the sequel). In what follows G is a metacyclic 2-group of order 2n with R.G/ > ¹1º or, what is the same, w 1 (see Theorem 124.4). Theorem 124.8. If G is a metacyclic 2-group such that G D R.G/, w 1 (in this case, jGj D 22w ) and 1 m 2w, then one of the following holds: (a) If m w, then sm .G/ D 2mC1 1. (b) If 1 t w, then swCt .G/ D 2wtC1 1. Proof. It is easily checked that the theorem is true for w D 1 and t D w. Next we assume that w > 1 and t < w. As in the proof of Theorem 124.1, formulas (1), (2) and (3) hold. Therefore, in view of Theorem 124.5, it remains to ﬁnd the number of noncyclic subgroups of order 2m . If H < G is noncyclic, then 1 .G/ H so the number of noncyclic subgroups of order 2m in G is equal to sm2 .G=1 .G//. We proceed by induction on w. Obviously, R.G= t .G// D G= t .G/ for t w. (a) Suppose that m w. In our case, we obtain that w 0 D w.G=1 .G// D w 1, jG=1 .G/j D 22w2 . We have m 2 < w 1 D w 0 . Let H < G be of order 2m . If H is noncyclic, then the number of such H is G is equal to sm2 .G=1 .G// D 2.m2/C1 1 D 2m1 1 by induction (see the previous paragraph). If H < G is cyclic, then the number of such H is equal to cm .m .G// D 3 22m1 (Theorem 124.5) so, by (3), we obtain sm .G/ D 3 2m1 C .2m1 1/ D 2mC1 1; completing the proof of (a). (b) In what follows we assume that m D wCt , where 1 t < m. As exp.G/ D 2w , we obtain that exp.G= t .G// D 2wt . If H G is of order 2wCt , then t .G/ < H and jH= t .G/j D 2wt , and so swCt .G/ D swt .G= t .G//. As we know, we have R.G= t .G// D G= t .G/; therefore, using the formula from the previous sentence and (a) for G= t .G/, we get swCt .G/ D swt .G= t .G// D 2wtC1 1 by induction, and the proof is complete. It is easy to show that the groups from Theorem 124.8 coincide with metacyclic 2-groups G of order 22w and exponent 2w . Indeed, if G is such a group, then 1 .G/ is abelian of type .2; 2/ by Lemma J(e), and now, by induction, ji .G=1 .G//j D 22i for all i m 1, and our claim follows.

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The number of subgroups of given order in a metacyclic p-group

247

It follows that the number sm .G/ from Theorem 124.8 is computed by the same formulas as in Theorem 124.1. The group H2 is a group from Theorem 124.8 but not from Theorem 124.1. Note that a group G D AB from Theorem 124.8 has a noncyclic center. This is the case if one of the subgroups A; B, say A, is normal in G since 1 .A/ Z.G/ and, in view of jAut.A/j D 2m1 , 1 .B/ centralizes A so contained in Z.G/. Now suppose that A and B are not normal in G. Then there exists x 2 GNG .A/ such that Ax ¤ A. By Ore (see Lemma A.28.8), AAx ¤ G soT A \ Ax 1 .A/ > ¹1º for all x 2 G by the product formula. It follows that AG D x2G Ax > ¹1º so Z.G/ 1 .A/ > ¹1º. Similarly, Z.G/ 1 .B/ > ¹1º, and our claim follows since 1 .A/1 .B/ Z.G/ is noncyclic. Suppose that a metacyclic 2-group G is such that G=R.G/ is nonidentity cyclic and R.G/ > ¹1º. We claim that in this case G D AB, where A; B < G are cyclic and A \ B D ¹1º. Indeed, suppose that exp.G/ D 2e and, as above, exp.R.G// D p w ; then w < e. Let A < G be cyclic of order 2e ; then A \ R.G/ is cyclic of order 2w . By Theorem 124.5, R.G/ D .A \ R.G//B, where B < R.G/ is cyclic of order 2w and .A \ R.G// \ B D ¹1º. It follows that A \ B D ¹1º, so AB D G by the product formula. Let 1 D ¹A; B; C º be the set of all maximal subgroups of a noncyclic metacyclic 2-group G. Then, supposing 2m < jGj, we have, by Hall’s enumeration principle, (8)

sm .G/ D sm .A/ C sm .B/ C sm .C / 2sm .ˆ.G//:

We may assume that G has no cyclic subgroup of index 2 (see Theorem 124.4). In case 2mC2 D jGj D 2n , we get sm .G/ D sn2 .G/ D 3 3 2 1 D 7 D 22 C 3 by (8). We shall freely use this fact in what follows. Next we retain the notation introduced in this paragraph. Now we compute sm .G/ for the case where G=R.G/ is nonidentity cyclic. The following lemmas will help us to state the inductive hypothesis. If m w, then we have sm .G/ D sm .R.G//, and this number is computed in Theorem 124.8(a). Therefore in what follows we consider case m > w only. Lemma 124.9. Suppose that G is a metacyclic 2-group that satisﬁes jG=R.G/j D 2. If 1 t w C 1, then swCt .G/ D 2wtC2 1. Proof. If t D w C 1, then w C t D 2w C 1 D n, where 2n D jGj; then 1 D swCt .G/ D sn .G/ D 1 D 2w.wC1/C2 1; as in the statement. Next we assume that t < w C 1. We have exp.G/ D 2wC1 . If w D 1, then t D 1, G is abelian of type .4; 2/, and so swCt .G/ D s2 .G/ D 3 D 22 1 D 211C2 1, as in statement. Therefore one may also assume that w > 1.

248

Groups of prime power order

In view of Theorem 124.8(b), swCt .R.G// D 2wtC1 1 so it sufﬁces to ﬁnd the number of subgroups H < G of order 2wCt such that H 6 R.G/. Let H be such a subgroup; then exp.H / D 2wC1 D exp.G/. If H is cyclic, then m D w C 1 so t D 1, hence the number of such H is equal to cwC1 .G/ D 2w (Theorem 124.6), and we get swC1 .G/ D swC1 .R.G// C cwC1 .G/ D .2w1C1 1/ C 2w D 2wC1 1; completing this case. Next we assume that jH j D 2.wC1/Ct , where 0 < t < w. In this case, H is noncyclic and t .G/ < H . Suppose that H 6 R.G/. Since exp.H / D 2wC1 D exp.G/, it follows that H= t .G/ is of order 2wCtC12t D 2wtC1 and exponent 2wtC1 , and we conclude that H= t .G/ is cyclic. We have w.G= t .G// D w t. Therefore the number of such H is equal to cwtC1 .G= t .G// D 2wt (Theorem 124.6). Since swC1Ct .R.G// D 2w.tC1/C1 1 D 2wt 1 (Theorem 124.8), we obtain swCtC1 .G/ D 2wt C .2wt 1/ D 2wtC1 1: Replacing t C1 by t in the last equality, we get swCt .G/ D 2wtC2 1, as required. Lemma 124.10. Suppose that a metacyclic 2-group G is such that G=R.G/ is cyclic of order 4. (a) We have swC1 .G/ D 2wC1 1. (b) If 2 t m, then swCt .G/ D 2wtC3 1. Proof. (a) Since jwC1 .G/=R.G/j D 2, we have, by Lemma 124.9, swC1 .G/ D swC1 .wC1 .G// D 2w1C2 1 D 2wC1 1; and (a) is proven. (b) If t D w C 2, then 2wCt D jGj so swC.wC2/ .G/ D 1 D 2w.wC2/C3 1; as in the statement. In what follows we assume that t < w C 2. If w D 1, then jGj D 24 has a cyclic subgroup of index 2, t D 2 (by hypothesis), s1Ct .G/ D s3 .G/ D 3 D 22 1 D 212C3 1; and this coincides with the required result. In the sequel we assume that w > 1. Set U D wC1 .G/; then jU W R.G/j D 2 D jG W U j. As swCt .U / D 2wtC2 1 (Lemma 124.9), it sufﬁces to compute the number of H < G of order 2wCt such that H 6 U . Let H be such a subgroup. Then exp.H / D 2wC2 D exp.G/. If H is cyclic, then jH j D 2wC2 (this holds if and only if t D 2), and so the number of such H in G

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249

is equal to cwC2 .G/ D 2w (Theorem 124.6), and we get swC2 .G/ D cwC2 .G/ C swC2 .U / D 2w C .2w2C2 1/ D 2wC1 1 D 2w2C3 1; and this coincides with the required result for t D 2. (Note that if t D 2 and H 6 U , then H is cyclic.) Now we suppose that jH j D 2.wC2/Ct , where 0 < t < w. In this case, H is noncyclic of exponent 2wC2 D exp.G/, t .G/ < H and H= t .G/ is cyclic of order 2wCtC22t D 2wtC2 (cyclic since jH= t .G/j D 2wtC2 D exp.H= t .G/). As exp.R.G= t .G// D 2wt , the number of such H equals cwt C2 .G= t .G// D 2wt (Theorem 124.6). Since swCtC2 .U / D 2w.tC2/C2 1 D 2wt 1 (Lemma 124.9), we get swCt C2 .G/ D 2wt C .2wt 1/ D 2wtC1 1: Replacing in the last equality t C 2 by t , we get swCt .G/ D 2wtC3 1, completing the proof. Lemma 124.11. Suppose that G is a metacyclic 2-group such that G=R.G/ is cyclic of order 23 , w > 0. (a) If t D 1, then swC1 .G/ D 2wC1 1. (b) If t D 2, then swC2 .G/ D 2wC1 1. (c) If 3 t < w C 3, then swCt .G/ D 2wtC4 1. Proof. Statements (a) and (b) follow from Lemmas 124.9 and 124.10, respectively. Indeed, let t D 1. Then swC1 .G/ D swC1 .wC1 .G// D 2w1C2 1 D 2wC1 1

(Lemma 124.9):

Now let t D 2. Then swC2 .G/ D swC2 .wC2 .G// D 2w2C3 1 D 2wC1 1

(Lemma 124.10):

(c) One may assume that t < w C 3. If w D 1, then t D 3, by hypothesis, G has a cyclic subgroup of index 2 so s1C3 .G/ D 3 D 213C4 1. Next we assume that w > 1. Set U D t C2 .G/. Let H < G be of order 2wC3 . Since swC3 .U / D 2w3C3 1 D 2w 1 (Lemma 124.10), it sufﬁces to count the number of those subgroups H that are not contained in U . In this case, we get exp.H / D 2wC3 so H is cyclic. Then the number of such H is equal to cwC3 .G/ D 2w (Theorem 124.6). We get swC3 .G/ D 2w C .2w 1/ D 2wC1 1 D 2w3C4 1; and this coincides with the required result for t D 3.

250

Groups of prime power order

Suppose that jH j D 2wC3Ct , where 0 < t < w, and H < G, H 6 U . Then we conclude that t .G/ < H , H= t .G/ is cyclic of order 2wtC3 . The number of such H is equal to cwtC3 .G= t .G// D 2wt since exp.R.G= t .G/// D 2wt (Theorem 124.6). All such H are not contained in U as exp.H / D 2wC3 > 2wC2 D exp.U /. Since swC3Ct .U / D 2w.tC3/C3 1 D 2wt 1 (Lemma 124.10), we get swC3Ct .G/ D 2wt C .2wt 1/ D 2wtC1 1: Replacing in the last equality t C 3 by t , we get swCt .G/ D 2wtC4 1, completing the proof. The proofs of the previous three lemmas are similar. Our goal there was to state inductive hypothesis and show as to attain this. In any case, Lemma 124.9 must be proved (it is the basis of induction). Now we are ready to prove the following Theorem 124.12. Suppose that G is a metacyclic 2-group of order 2n such that w > 1 and G=R.G/ is cyclic of order 2c (in this case, n D 2w C c/.1 (a) If 1 t < c, then swCt .G/ D 2wC1 1. (b) If c t w C c, then swCt .G/ D 2wtCcC1 1. Proof. We proceed by induction on c. As in previous three lemmas, one may assume that t < w C c. (i) First we consider case t c. Since w > 0, G is noncyclic. Set U D wCc1 .G/; then exp.U / D 2wCc1 D 1 c1 > 1. 2 exp.G/, jG W U j D 2 and U=R.U / D UR.G/=R.G/ is cyclic of order 2 The theorem holds for c D 1; 2; 3 (Lemmas 124.9–124.11) so one may assume that c > 3. We have exp.G/ D 2wCc . By induction, swCt .U / D 2wtC.c1/C1 1 D 2wtCc 1 since t > c 1 D log2 .jUR.G/=R.G/j/. Therefore it sufﬁces to ﬁnd the number of those H < G of order 2wCt that are not contained in U . All such subgroups have the same exponent 2wCc D exp.G/. If such an H is cyclic, then jH j D 2wCc (this holds if and only if t D c), and the number of such H in G is equal to cwCc .G/ D 2w (Theorem 124.6). Therefore we obtain swCc .G/ D cwCc .G/ C swCc .U / D 2w C .2wcCc 1/ D 2wC1 1; and this coincides with the required result for t D c. Next we assume that jH j D 2wCcCt , where 0 < t < w. Then t .G/ < H and H= t .G/ is of order 2wCcCt2t D 2wtCc D exp.H= t .G// whence H= t .G/ is cyclic. We have swCcCt .U / D 2w.cCt /C.c1/C1 1 D 2wt 1 by induction. w D 1, then the number of proper subgroups of given order in G is equal to 22 1 since G has a cyclic subgroup of index 2 and cl.G/ 2. 1 If

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The number of subgroups of given order in a metacyclic p-group

251

Since R.G= t .G// D R.G/= t .G/ and exp.R.G= t .G/// D 2wt , the number of such H equals cwtCc .G= t .G// D 2wt (Theorem 124.6). Therefore swCcCt .G/ D 2wt C .2wt 1/ D 2wtC1 1: Replacing in the displayed formula t C c by t , we get swCt D 2wtCcC1 1, and the proof of (b) is complete. (ii) It remains to prove (a); then t < c. In this case, H < V D wCt .G/. The subgroup V =R.G/ is cyclic of order 2t . By part (b), applied to V , we obtain that swCt .V / D 2wtCt C1 1 D 2wC1 1. Since swCt .G/ D swCt .V /, the proof of (a) is complete. To complete the computation of the value sm .G/, it remains to consider the case where G=R.G/ is a 2-group of maximal class and order 2c . Note that the factor group G=R.G/ contains a cyclic subgroup of index 2 (Theorem 1.2). In the notation of (8) (namely, 1 D ¹A; B; C º is the set of maximal subgroups of G) we assume in the sequel that C =R.G/ is cyclic. These three cases will be covered in Theorems 124.16, 124.18 and 124.19 and in the supplements to them. In those theorems we consider the case w > 1, t c 1. In the supplements we consider the cases w D 1 and t D 1 separately. In the case under consideration, R.G/ < ˆ.G/ as d.G=R.G// D 2 D d.G/. The following two lemmas are the bases of induction for Theorems 124.16 and 124.18, respectively. Lemma 124.13. Suppose that G is a metacyclic 2-group such that G=R.G/ Š Q8 and w C t > w > 1, t w C 3. (a) If t D 1, then swC1 .G/ D 2wC1 1. (b) If 2 t w C 3, then swCt .G/ D 2wtC4 1. Proof. If t D w C 3, then we have w C t D n, where 2n D jGj, so we obtain that swCt .G/ D 1 D 2w.wC3/C4 1, as in (b). If t D w C 2, then w C t D n 1 so swCt .G/ D 3 D 22 1 D 2w.wC2/C4 1; as in (b). In what follows we assume that t < w C 2. We have R.A/ D R.B/ D R.C / D R.ˆ.G// D R.G/ and the quotient groups A=R.G/, B=R.G/ and C =R.G/ are cyclic of order 4, jˆ.G/=R.ˆ.G/j D 2. If t D 1, then, by Lemma 124.9, swC1 .G/ D s.wC1 .G// D 2w1C2 1 D 2wC1 1; and the proof of (a) is complete. Now let t > 1. Then swCt .A/ D swCt .B/ D swCt .C / D 2wtC3 1 (Lemma 124.10); swCt .ˆ.G// D 2wtC2 1

(Lemma 124.9)

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Groups of prime power order

so, by (8), swCt .G/ D 3 .2wtC3 1/ 2 .2wtC2 1/ D 2wtC4 1; and the proof is complete. Lemma 124.14. Suppose that G is a metacyclic 2-group such that G=R.G/ Š D8 and w C t > w > 1, 1 t w C 3. (a) If t D 1, then swC1 .G/ D 2wC3 2wC1 1 D 3 2wC1 1. (b) If t > 1, then swCt .G/ D 2wtC4 1. Proof. In the notation of identity (8), we have R.A/ D A, R.B/ D B so we obtain w.A/ D w.B/ D w C 1, R.C / D R.ˆ.G// D R.G/ so w.C / D w.ˆ.G// D w. In case t D w C 3, we get w C t D n, and, as in the previous lemma, we infer that swCt .G/ D 1 D 2w.wC3/C4 1, as in (b). If t D w C 2, then swCt .G/ D 3 D 22 1 D 2w.wC2/C4 1; as in (b). Next we assume that t < w C 2. For t D 1, we have swC1 .A/ D swC1 .B/ D 2.wC1/C1 1 D 2wC2 1 (Theorem 124.8(a)); swC1 .C / D swC1 .wC1 .C // D swC1 .ˆ.G// D 2w1C2 1 D 2wC1 1

(Lemma 124.9)

so that, by (8), swC1 .G/ D 2 .2wC2 1/ C .2wC1 1/ 2.2wC1 1/ D 2wC3 2wC1 1 D 3 2wC1 1; as in (a). If t > 1, then w C t D .w C 1/ C .t 1/ so that swCt .A/ D swCt .B/ D 2.wC1/.t1/C1 1 D 2wtC3 1 swCt .C / D 2wtC3 1 swCt .ˆ.G// D 2wtC2 1

(Theorem 124.8(b)); (Lemma 124.10); (Lemma 124.9)

so that, by (8), swCt .G/ D 3.2wtC3 1/ 2.2wtC2 1/ D 2wtC4 1; completing the proof.

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Lemma 124.15. Let G be a metacyclic 2-group of order 2n such that G=R.G/ Š Q24 , w > 1 and 1 t w C 4. Then (a) swC1 .G/ D 2wC1 1. (b) If 2 t w C 4, then swCt .G/ D 2wtC5 1. Proof. If t D w C 4, then swC.wC4/ .G/ D sn .G/ D 1 D 2w.wC4/C5 1; as in (b). In case t D w C 3, we get swC.wC3/ .G/ D 3 D 22 1 D 2w.wC3/C5 1; as in (b). If t D 1, then, by Lemma 124.9, swC1 .G/ D swC1 .wC1 .G// D 2w1C2 1 D 2wC1 1; as in (a). We have w.M / D w for M 2 ¹A; B; C; ˆ.G/º. Now let t D 2. Then swC2 .A/ D swC2 .B/ D 2w2C4 1 D 2wC2 1 (Lemma 124.13); swC2 .C / D 2w2C4 1 D 2wC2 1 swC2 .ˆ.G// D 2

w2C3

1D2

wC1

1

(Lemma 124.11); (Lemma 124.10) :

Therefore, by (8), we get swC2 .G/ D 3.2wC2 1/ 2.2wC1 1/ D 2wC3 1 D 2w2C5 1; as in (b). Next we assume that 2 < t < w C 3. By Lemma 124.13, Theorem 124.12 and identity (8), swCt .G/ D 2.2wtC4 1/ C .2wtC4 1/ 2.2wtC3 1/ D 2wtC5 1; and the proof is complete. Now we are ready to prove the following Theorem 124.16. Suppose that G is a metacyclic 2-group such that G=R.G/ Š Q2c and w C t > w > 1, c 1 t w C c. Then swCt .G/ D 2wtCcC1 1. Proof. We proceed by induction on c. The statement of the theorem is true for c D 3; 4 (Lemmas 124.13, 124.15) and t 2 ¹w C c 1; w C cº (direct checking; see Lemma 124.15). Next assume that c > 4 and t < w C c 1. As in Lemma 124.15, we

254

Groups of prime power order

have w.M / D w for M 2 ¹A; B; C; ˆ.G/º. We obtain A=R.G/ Š Q2c1 Š B=R.G/;

C =R.G/ Š C2c1 ;

ˆ.G/=R.G/ Š C2c2 :

Then, by induction, swCt .A/ D swCt .B/ D 2wtCc 1: By Theorem 124.12, swCt .C / D 2wt Cc 1;

swCt .ˆ.G// D 2wtCc1 1:

Therefore, by (8), we obtain swCt .G/ D 3.2wt Cc 1/ 2.2wtCc1 1/ D 2wtCcC1 1; and the proof is complete. There is no problem to compute swCt .G/ for t < c 1. Our method works also in this case. Now suppose that w D 1 and G=R.G/ is of maximal class and order 2c . Taking into account that C and ˆ.G/ have cyclic subgroups of index 2, cl.C / 2 and ˆ.G/ is abelian (indeed, CG .R.G// ˆ.G/), we obtain s1Ct .C / D s1Ct .ˆ.G// D 3. Therefore, by (8), we get (9)

s1Ct .G/ D s1Ct .A/ C s1Ct .B/ C 3 2 3 D s1Ct .A/ C s1Ct .B/ 3:

Supplement 1 to Theorem 124.16. Let G=R.G/ Š Q2c , w D 1, jGj D 2n D 22Cc , 1 t < c. Then (a) s2 .G/ D 3 (in that case, t D 1). (b) If 1 < t < c, then s1Ct .G/ D 21tCc C 3. Proof. One may assume that 1 C t < n 2.D c/ (we have sn2 .G/ D 22 C 3 D 21tCc C 3, where t D n 3 and c D n 2; see the displayed formula following (8)). If t D 1, then s1C1 .G/ D s2 .2 .G// D s2 .ˆ.G// D 3 since 2 .G/. ˆ.G// is abelian of type .4; 2/ and ˆ.G/ has a cyclic subgroup of index 2. Next we also assume that t > 1. By the same displayed formula following (8), if c D 3 and t D 2, then we get s3 .G/ D 22 C 3 D 212C3 C 3. If c D 4, then t D 2 (by assumption, 1 < t < n 3 D c 1) so, by (9) and the previous paragraph, s1C2 .G/ D 2.22 C 3/ 3 D 23 C 3 D 212C4 C 3; as in (b). In case c D 5, we get t 2 ¹2; 3º so, by (9), s1C2 .G/ D 2.23 C 3/ 3 D 24 C 3 D 212C5 C 3

(by the previous paragraph),

s1C3 .G/ D 2 .22 C 3/ 3 D 23 C 3 D 213C5 C 3 (by the ﬁrst paragraph).

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The number of subgroups of given order in a metacyclic p-group

255

If c D 6, then t 2 ¹2; 3; 4º. We have, by (9) and the previous paragraphs, s1C2 .G/ D 2.24 C 3/ 3 D 25 C 3 D 212C6 C 3; s1C3 .G/ D 2.23 C 3/ 3 D 24 C 3 D 213C6 C 3; s1C4 .G/ D 2 .22 C 3/ 3 D 23 C 3 D 214C6 C 3: Now let jG=R.G/j D 2c , where t 2 ¹2; : : : ; c1º, c > 3. We claim that s1Ct .G/ D We prove this by induction on c. By the above, this is true for c D 3; 4; 5; 6 so one may assume that c > 6. Applying induction and (9), we obtain 21tCc C3.

s1Ct .G/ D 2.21tCc1 C 3/ 3 D 21tCc C 3; and we are done. We see that the cases w D 1 and w > 1 are essentially different. Supplement 2 to Theorem 124.16. Let G=R.G/ Š Q2c , w > 1 and t D 1. Then we have swC1 .G/ D 2wC1 1. Proof. For arbitrary c 3 we have, by Lemma 124.9, swC1 .G/ D swC1 .wC1 .G// D 2w1C2 1 D 2wC1 1; as required. Lemma 124.17. Suppose that G is a metacyclic 2-group such that G=R.G/ Š D24 , w C t > w > 1 and 1 < t < w C 3. Then swCt .G/ D 2wtC5 1 D 2wtC.4C1/ 1. Proof. It follows from the structure of the maximal subgroups of G that w.M / D w for M 2 ¹A; B; C; ˆ.G/º. By Lemmas 124.14, 124.12 and (8), we have swCt .G/ D 2.2wtC4 1/ C .2wtC4 1/ 2.2wtC3 1/ D 2wtC5 1; and we are done. Now we are ready to prove the following Theorem 124.18. Suppose that G is a metacyclic 2-group such that G=R.G/ Š D2c , w C t > w > 1, c 1 t < w C c 1. Then swCt .G/ D 2wt CcC1 1. Proof. We proceed by induction on c. The theorem holds for c D 3; 4 (Lemmas 124.14 and 124.17) so one may assume that c > 4. As above, we have w.M / D w for each M 2 ¹A; B; C; ˆ.G/º. By induction, Theorem 124.12 and (8), we obtain swCt .G/ D 2.2wtCc 1/ C .2wtCc 1/ 2.2wtCc1 1/ D 2wtCcC1 1; completing the proof.

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Now we consider the case G=R.G/ Š D2c , w D 1. As above, jGj D 2n . Supplement 1 to Theorem 124.18. Suppose that G=R.G/ Š D2c , w D 1 (in this case, n D c C 2) and 1 t < c. Then we have s1Ct .G/ D 21tCc C 3. (If t D c, then s1Cc .G/ D j1 j D 3.) Proof. By the paragraph containing (8), sn2 .G/ D 22 C 3 D 21.n3/Cn2 C 3 (here t D n 3 and c D n 2), as in the statement. Therefore one may assume in the sequel that t < n 3 D .c C 2/ 3 D c 1. Let t D 1. If H < G is of order 4 and H ¤ R.G/, then HR.G/ of order 8 contains exactly two ¤ R.G/ subgroups of order 4 since HR.G/ is abelian of type .22 ; 2/ by Lemma J(b). Since s1 .G=R.G// D 2c1 C 1, we get s1C1 .G/ D 1 C 2.2c1 C 1/ D 2c C 3 D 211Cc C 3; as in the statement. In what follows we assume that t > 1; then c > 3. Let c D 4. Then t D 2. Using (9), we obtain s1C2 .G/ D 2 .22 C 3/ 3 D 23 C 3 D 212C4 C 3 by the displayed formula following (8). Let c D 5. Then t 2 ¹2; 3º and s1C2 .G/ D 2.23 C 3/ 3 D 24 C 3 D 212C5 C 3; s1C3 .G/ D 2 .22 C 3/ 3 D 23 C 3 D 213C5 C 3 (the ﬁrst equality follows from the previous paragraph and the second one follows from the displayed formula following (8)). Now we will prove by induction on c that s1Ct .G/ D 21tCc C 3. This is true for c D 3; 4; 5 and t D 1. By induction, Theorem 124.12 and (9), we get s1Ct .G/ D 2.21tCc1 C 3/ 3 D 21tCc C 3; as required. Supplement 2 to Theorem 124.18. Suppose that G=R.G/ Š D2c , w > 1. Then we have swC1 .G/ D 2wCc1 C 2wC1 1. Proof. Let X1 =R.G/; : : : ; X2c1 =R.G/; X2c1 C1 =R.G/ D Z.G=R.G// be all subgroups of order 2 in G=R.G/. Then R.Xi / D R.G/;

swC1 .Xi / D 2w1C2 1 D 2wC1 1 for all i

(Lemma 124.9)

and swC1 .R.G// D 2w1C1 1 D 2w 1

(Theorem 124.8(b)).

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257

We get Xi \ Xj D R.G/ for i ¤ j and, given H < G of order 2wC1 not contained in R.G/, there is exactly one i 2c1 C 1 such that H < Xi .D HR.G//. Therefore swC1 .G/ D

2c1 XC1

swC1 .Xi / 2c1 swC1 .R.G//

i D1

D .2c1 C 1/.2wC1 1/ 2c1 .2w 1/ D 2wCc1 C 2wC1 1; and the proof is complete. (It is easy to check that, for t D 1 and c D 3, the obtained result coincides with Lemma 124.14(a).) In the following supplement we consider the groups G such that G=R.G/ Š D2c , w > 1 and t D 2. Note that the cases t D 3; : : : ; c 2 are more difﬁcult and, to treat them, one has to use the enumeration principle and induction on c for each value of t . Supplement 3 to Theorem 124.18. Suppose that G is a metacyclic 2-group such that G=R.G/ Š D2c , w > 1 and t D 2. Then swC2 .G/ D 2wCc2 C 2wC1 1. Proof. Suppose that X1 =R.G/; : : : ; X2c2 =R.G/ < G=R.G/ are abelian of type .2; 2/ and X2c2 C1 =R.G/ D Y =R.G/ < G=R.G/ cyclic of order 4. For i ¤ j we have Xi \ Xj D wC1 .Y /. Given H < G of order 2wC2 , there is i 2 ¹1; : : : ; 2c2 C 1º such that H < Xi since jH \ R.G/j 2w . Therefore (10)

swC2 .G/ D

2c2 XC1

swC2 .Xi / 2c2 swC2 .wC1 .Y //:

i D1

We have R.Xi / D Xi for i 2c2 and R.Y / D R.G/. By Theorem 124.8(b), swC2 .Xi / D s.wC1/C1 .Xi / D 2.wC1/1C1 1 D 2wC1 1;

i 2c2 ;

and, by Lemmas 124.10, 124.9, respectively (recall that Y D X2c2 C1 ), swC2 .Y / D 2w2C3 1 D 2wC1 1;

swC2 .wC1 .Y // D 2w2C2 1 D 2w 1:

Therefore we get, by (10), swC2 .G/ D 2c2 .2wC1 1/ C .2wC1 1/ 2c2 .2w 1/ D 2wCc2 C 2wC1 1; and the proof is complete. The cases G=R.G/ 2 ¹Q2c ; SD2c º, where w > 1 and t D 2, are considered similarly. Theorem 124.19. Suppose that G is a metacyclic 2-group such that G=R.G/ Š SD2c and w C t > w > 1, c 1 t w C c. Then swCt .G/ D 2wtCcC1 1.

258

Groups of prime power order

Proof. It is easily seen that the theorem holds for t 2 ¹w C c 1; w C cº. Next we assume that t < w C c 1. By Theorems 124.12, 124.16, 124.18 and (8), we have swCt .G/ D 3.2wtCc 1/ 2.2wtCc1 1/ D 2wt CcC1 1; and the proof is complete. Supplement 1 to Theorem 124.19. Let G=R.G/ Š SD2c , w D 1 and t 1. Then (a) If t D 1, then s1C1 .G/ D s2 .G/ D 2c1 C 3. (b) If 1 < t c 1, then s1Ct .G/ D 21tCc C 3. Proof. In case t D c 1, the result follows from the paragraph containing (8). Now let t < c 1. By Supplements 1 to Theorems 124.16 and 124.18 and (9), s1C1 .G/ D s2 .G/ D 3 C .211Cc1 C 3/ 3 D 2c1 C 3; s1Ct .G/ D 2.21tCc1 C 3/ 3 D 21tCc C 3

for 1 < t < c;

completing the proof. Supplement 2 to Theorem 124.19. Suppose that G=R.G/ Š SD2c , w > 1. Then we have swC1 .G/ D 2wCc2 C 2wC1 1. Proof. Let X1 =R.G/; : : : ; X2c2 =R.G/; X2c2 C1 =R.G/ D Z.G=R.G// be all subgroups of order 2 in G=R.G/. Then swC1 .Xi / D 2w1C2 1 D 2wC1 1 for all i (Lemma 124.9), swC1 .R.G// D 2w1C1 1 D 2w 1 (Theorem 124.8) and Xi \ Xj D R.G/ for all i ¤ j . We obtain (see Supplement 2 to Theorem 124.18) swC1 .G/ D .2c2 C 1/.2wC1 1/ 2c2 .2w 1/ D 2wCc2 C 2wC1 1; and we are done. The lower restrictions for t in Theorems 124.16, 124.18 and 124.19 are made for technical reasons (otherwise, the statements of these theorems would be more complicated). However, if necessary, it is not difﬁcult, using the above approach, consider the c 3 cases for which t 2 ¹2; : : : ; c 2º. If G is a noncyclic p-group of order p n with sk .G/ D 1 for some k 2 ¹1; : : : ; n1º, then we have k D 1, p D 2 and G is a generalized quaternion group (Proposition 1.3). By Sylow’s theorem, sk .G/ 1 .mod p/ for the same k. It follows that if G is neither cyclic nor generalized quaternion and k < n, then sk .G/ 1 C p. Therefore it is natural to classify the noncyclic p-groups G of order p n > p 3 satisfying sk .G/ D 1 C p for some ﬁxed k 2 ¹2; : : : ; n 2º (sn1 .G/ D 1 C p if and only if d.G/ D 2).

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The number of subgroups of given order in a metacyclic p-group

259

Proposition 124.20. Let G be a group of order p n > p 3 satisfying sk .G/ D 1 C p for some ﬁxed k 2 ¹2; : : : ; n 2º. Then one of the following holds: (a) G is abelian of type .p n1 ; p/. (b) G Š Mpn . n2

(c) p D 2, k D 2 and G D ha; b j a2 is metacyclic as in Lemma 42.1(c).

n3

D 1; n > 4; a2

D b 2 D 1; ab D a1 i

Proof. It is easy to see that G is not a 2-group of maximal class (otherwise, by Theorem 124.4, sk .G/ 1 .mod 4/ so sk .G/ ¤ 1 C 2). In that case, G has a normal abelian subgroup R of type .p; p/ (Lemma 1.4). (i) Suppose that k D 2. Assume that G has a subgroup E of order p 3 and exponent p; then E is nonabelian (otherwise, we have s2 .E/ D 1 C p C p 2 > 1 C p), s2 .E/ D 1 C p D s2 .G/. It follows that all subgroups of G of order p 2 are contained in E so exp.G/ D p, s2 .G/ 1Cp C2p 2 .mod p 3 / (Theorem 5.9), a contradiction. Thus E does not exist. Assume that G=R has two distinct subgroups X=R and Y =R of order p. Then X and Y contain together at least s2 .X/ C s2 .Y / 1 D 2p C 1 > p C 1 distinct subgroups of order p 2 , a contradiction. Thus G=R is either cyclic or generalized quaternion. In the ﬁrst case, G is one of groups (a, b). If G=R is generalized quaternion, then we obtain that j2 .G/j D 23 so G is as in (c) by Lemma 42.1(c). (ii) Suppose that k > 2. Then, as in part (i), G=R contains exactly one (proper since k < n 1) subgroup of order p k1 . In that case, G=R is cyclic. The subgroup CG .R/ is abelian with cyclic subgroup of index p and jG W CG .R/j p. Then 1 .G/ D R so G is either as in (a) or in (b). Now let s1 .G/ D 1 C p. If p > 2, then G is either metacyclic or a 3-group of maximal class (Theorem 13.7). Such G are also described very well for p D 2 in 82. Proposition 124.21. Let M be a metacyclic p-group and let G be a p-group of order p n > p 3 , p > 2. If s2 .G/ D s2 .M /, then one of the following holds: (a) G is metacyclic. (b) G D EC , where E D 1 .G/ is nonabelian of order p 3 and exponent p and C is cyclic of order > p. (c) G is of maximal class and order p 4, all maximal subgroups of G are two-generator. Proof. One may assume that the p-group M is noncyclic; then G is also noncyclic. We have s2 .G/ D s2 .M / D ¹1 C p; 1 C p C p 2 º (Theorem 124.1). Assume that G is nonmetacyclic. Then s2 .G/ D 1 C p C p 2 by Theorem 124.20, and so 2 .G/ is nonmetacyclic by Remark 41.2. (i) Assume that G has a subgroup E of order p 4 and exponent p. Then we obtain that s2 .E/ D 1 C p C 2p 2 C sp 3 for some nonnegative integer (Theorem 5.9) so s2 .E/ > 1 C p C p 2 D s2 .G/, a contradiction.

260

Groups of prime power order

(ii) Assume that G has an elementary abelian subgroup E of order p 3 . Then we get s2 .E/ D 1 C p C p 2 , so, by the ﬁrst paragraph, all subgroups of G of order p 2 are contained in E. It follows that exp.G/ D p, and so, by (i), G D E is of order p 3 , contrary to the hypothesis. (iii) It follows from (ii) that either G is a 3-group of maximal class or G D EC , where E D 1 .G/ is nonabelian of order p 3 and exponent p and C is cyclic (Theorem 13.7). The group G D EC satisﬁes the hypothesis (note that one of such groups is of maximal class; then its order is equal to p 4 ). Suppose that G is a 3-group of maximal class and order > 34 . Let G1 be the fundamental subgroup of G (see 9); then G1 is metacyclic without cyclic subgroup of index 3 (Theorems 9.6 and 9.11) so that s2 .G1 / D 1 C 3 C 32 D s2 .M / D s2 .G/: It follows that all subgroups of G of order 32 are contained in G1 so that 2 .G/ G1 . However (see Proposition 13.14(b)), 2 .G/ D G > G1 , and this is a contradiction. Thus, if G is a 3-group of maximal class, then n D 4. By (ii), G has no subgroup of type .3; 3; 3/, and this holds if and only if G is not isomorphic to a Sylow 3-subgroup of the symmetric group of degree 9. 4o A property of metacyclic 2-groups with nonabelian section of order 8. In this subsection G is a metacyclic 2-group containing subgroups L G M such that M=L is nonabelian of order 8 (M=L is said to be a section of G). In what follows we freely use the following known fact: There is only one nonabelian metacyclic group of order 16 and exponent 4. Before we denoted this group by H2 . If G D R.G/, then R.Ã1 .G// D Ã1 .G/. We use this obvious fact in the proof of the following Theorem 124.22. Suppose that a metacyclic 2-group G has a nonabelian section of order 8. If G is not of maximal class, then the following hold: (a) There is R G G such that G=R Š Q8 and G=Ã2 .G/ Š H2 . (b) Ã1 .G/ has no nonabelian section of order 8. Proof. (a) We use induction on jGj. Let L G M G be such that M=L is nonabelian of order 8. One may assume that M < G (otherwise, there is nothing to prove). Let M H 2 1 . Then, by induction, there is S G H such that H=S is nonabelian of order 8. By Lemma J(c), S is characteristic in H so normal in G. Due to Lemma J(b), G=S is of maximal class. Let R=S D Z.G=S/; then G=R is nonabelian of order 8, as was to be shown. (b) Assume that Ã1 .G/ D ˆ.G/ has a nonabelian section of order 8. Then, by (a), there is S G Ã1 .G/ such that Ã1 .G/=S is nonabelian of order 8. Due to Lemma J(c), S is characteristic in Ã1 .G/ so normal in G. By Lemma J(b), the quotient group G=S is of maximal class (of order 25 ). Thus ˆ.G=S / D Ã1 .G=S/ is nonabelian of order 8, contrary to Lemma 1.4.

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The number of subgroups of given order in a metacyclic p-group

261

5o Mann’s proof of Theorem 124.1 for p > 2. The content of this subsection is taken from [Man33] (only in one place we made a change using Lemma 124.2). We retain the notation introduced in 1o . Proof of Theorem 124.1. Below G is a metacyclic p-group, p > 2. We use the same notation as in 1o . By Lemma 124.2, G is the product of two cyclic subgroups with trivial intersection, say G D hxihyi D AB; A\B D ¹1º. By regularity, p e is the maximum of the orders of x and y, and thus we may assume that jAj D p e , and then jBj D p f . Hence e f . Since G 0 Ã1 .G/, we have Ãe1 .G 0 / D ¹1º. In a regular group, by Theorem 7.2(f), ŒG; Ãk .G/ D Ãk .G 0 /. Taking k D e 1, we get ŒG; Ãe1 .G/ D Ãe1 .G 0 / D ¹1º, e1 and hence Ãe1 .G/ Z.G/, and in particular N D hx p i is a central subgroup of order p. Write zm .G/ for the number of subgroups of G of order p m that do not contain N . It is clear that (11)

sm .G/ D sm1 .G=N / C zm .G/:

Note that in the quotient group G=N the invariants corresponding to e and f are e 1 and f (arranged in order). We now evaluate zm .G/. Let H be any subgroup of order p m not containing N . Since 1 .G/ D N Ãf 1 .B/ has order p 2 , we see that j1 .H /j D jH \ 1 .G/j D p, and thus H is cyclic since p > 2. If m > f , then we have jH j > jG W Aj, therefore H \ A ¤ ¹1º, and H N so that zm .G/ D 0. Now let m f . Then jm .G/j D jm .A/m .B/j D p 2m ;

jm1 .G/j D p 2.m1/ ;

and G contains p 2m p 2.m1/ elements of order p m and so p 2m p 2.m1/ D p m1 .p C 1/ .p 1/p m1 cyclic subgroups of order p m . Let z be a generator of N . If a cyclic subgroup H conm1 tains N , then H is generated by some element t such that t p D z. Then the elem1 powers are equal to z are the elements of the coset t m1 .G/, ments of G whose p their number is p 2.m1/ , and H contains p m1 of them. It follows that there are p m1 cyclic subgroups of order p m containing N , and zm .G/ D p m . Now the ﬁrst formula of the theorem follows by induction, noting that m 1 f .G=N /. Next, let f m e. If m D f , then the result follows from the previous case, so we assume that m > f . Then we get sm .G/ D sm1 .G=N /. Here e > f , implying e.G=N / D e 1 and f .G=N / D f , and thus e.G=N / m 1 f .G=N /, and we apply induction. Finally, assume that m e. Again sm .G/ D sm1 .G=N /, and once more the case of equality m D e is covered by the previous case, therefore we may assume that m > e, implying m 1 e.G=N /, and we apply induction.

262

Groups of prime power order

6o The p-groups with the same number of subgroups of some order as in a metacyclic p-group. Mann reported that, given a metacyclic p-group G, p > 2, he classiﬁed the p-groups H such that s1 .H / D s1 .G/ and s2 .H / D s2 .G/ (see also [Man33, Proposition 2] and Theorem 124.31). First we offer another proof of this result. Then we shall prove a related result. Proposition 124.23. Suppose that G and H are p-groups of exponent > p, p > 2, and G is metacyclic. Write Lpk .G/ D ¹x 2 G j o.x/ p k º. If jLp2 .H /j D jLp2 .G/j, then one of the following holds: (a) H is metacyclic. (b) p D 3 and H is of maximal class and order 34 ; moreover, H is minimal nonmetacyclic. (c) j1 .H /j D p 3 and H=1 .H / > ¹1º is cyclic. In particular (Mann), if s1 .H / D s1 .G/ and s2 .H / D s2 .G/, then H is as in parts (a–c) with nonabelian 1 .H / of order p 3 . Proof. One may assume that H is nonmetacyclic. Then jLp2 .H /j D jLp2 .G/j 2 ¹p 3 ; p 4 º: Write R D 1 .G/; then R is abelian of type .p; p/. Let M=R D 1 .G=R/. Then we get M D 2 .G/ D Lp2 . If jM=Rj D p, then j2 .H /j D p 3 and so H has a cyclic subgroup of index p so it is as in (a). Next we assume that jM j D p 4 . Then we obtain j2 .H /j D p 4 .

If j1 .H /j D p 2 , then H is a 3-group of maximal class. Since 2 .H / D H , we get jH j D 34 so H is as in (b).

If j1 .H /j D p 3 , then H is as in (c).

Now let s1 .H / D s1 .G/.D 1 C p/ and s2 .H / D s2 .G/ 2 ¹1 C p; 1 C p C p 2 º. In this case, jLp2 .H /j D j2 .H /j D p 4 so that, by the above, H is as in (a–c) (with 1 .H / is nonabelian of order p 3 in (c)). The analogous problem for p D 2 is surprisingly complicated. Denote by sŒk .G/ the number of subgroups of index p k in a p-group G. Proposition 124.24. Suppose that G and H be p-groups of exponent > p and assume that G is metacyclic. If sŒ1 .H / D sŒ1 .G/ and sŒ2 .H / D sŒ2 .G/, then one of the following holds: (a) H is metacyclic. (b) p > 2 and K3 .H / D Ã1 .H / has index p 3 in G. Proof. One may assume that the group G is noncyclic. Then we have sŒ1 .G/ D 1Cp, i.e., 2 D d.G/ D d.H /. Using Hall’s enumeration principle, we further conclude that sŒ2 .G/ 2 ¹1 C p; 1 C p C p 2 º. If sŒ2 .G/ D 1 C p, then G has a cyclic subgroup of

124

The number of subgroups of given order in a metacyclic p-group

263

index p. Therefore, if sŒ2 .H / D pC1, then, by what has just been said, H has a cyclic subgroup of index p so metacyclic. Now let sŒ2 .H / D p 2 C p C 1. In this case, using Hall’s enumeration principle, we conclude that the ranks of all maximal subgroups of H are equal to 2. Then, by Proposition 36.3, if p > 2, then we get case (b), if p D 2, then H is metacyclic. Corollary 124.25. Suppose that G and H are p-groups and assume that G is metacyclic. If sk .H / D sk .G/ holds for all positive integers k, then H is either metacyclic or minimal nonmetacyclic of order 34 . Proposition 124.26. Let G be an absolutely regular p-group of exponent > p, p > 2. If H is such a p-group that c1 .H / D c1 .G/ and c2 .H / D c2 .G/, then one of the following holds: (a) H is absolutely regular. (b) H is irregular of maximal class and order p 2.p1/ and j1 .H /j D p p1 . Proof. Let j1 .G/j D p r and j2 .G/j D p rCs ; then, since G is absolutely regular, we have r; s p 1; c2 .G/ D

c1 .G/ D 1 C p C C p r1 D c1 .H /;

p rCs p r D p r1 .1 C p C C p s1 / D c2 .H /: p.p 1/

By Theorems 13.2(a) and 9.5, H is either absolutely regular or irregular of maximal class. Assume that the group H is irregular of maximal class. Then we obtain r D p 1, j1 .H /j D j1 .G/j D p p1 . Let jH j D p n and let F be an absolutely regular maximal subgroup of H ; then n p C 1 (see Theorems 9.5 and 9.6). By Theorem 13.19, all elements of the set H F have the same order p 2 . Therefore there are in H exactly p n p n1 D p n2 p.p 1/ cyclic subgroups of order p 2 not contained in F . Let j2 .F /j D p p1Ct ; then we get t p 1 since F is absolutely regular, and we have c2 .F / D

p p1Ct p p1 D p p2 .1 C p C C p t 1 / p.p 1/

so that c2 .H / D p p2 .1 C p C C p t 1 / C p n2 D p r1 .1 C p C C p s1 / D p p2 .1 C p C C p s1 /: As n p C 1, we get n D 2 C .r 1/ C .s 1/ D r C s .p 1/ C .s 1/ D 2.p 1/:

264

Groups of prime power order

7o On subgroups of metacyclic p-groups. We begin with the following exercises. Exercise 1. Let G be a nonabelian metacyclic p-group that is not generalized quaternion. Then the following conditions are equivalent: (a) All minimal nonabelian subgroups of G have no cyclic subgroups of index p. (b) 1 .G/ Z.G/. Solution. (a) ) (b). If a subgroup H G is minimal nonabelian, then we obtain that 1 .G/ D 1 .H / Z.H /. Since G is generated by its minimal nonabelian subgroups (Theorem 10.28), we get 1 .G/ Z.G/. (b) ) (a). Assume that a minimal nonabelian F G has a cyclic subgroup of index p. Then we obtain 1 .G/ 6 F (otherwise, F would be abelian). It follows that p D 2 so F Š Q8 (see Exercise 1.8(a) or Lemma 65.1). In this case, G is of maximal class (Lemma J(b)) so G is generalized quaternion, contrary to the hypothesis. Exercise 2. Let a metacyclic p-group G be neither abelian nor minimal nonabelian. If 1 .G/ 6 Z.G/, then G contains at least k p 1 minimal nonabelian subgroups with a cyclic subgroup of index p. If k D p 1, then jG 0 j D p 2 so all subgroups of index p 2 in G are abelian. Solution. One may assume that G has no cyclic subgroup of index p; then G has no nonabelian subgroup of order p 3 . If p D 2, then k 1 (Exercise 1). Now let p > 2. Then CG .1 .G// D C is maximal in G and the minimal nonabelian subgroups of C have no cyclic subgroups of index p. If A < G is minimal nonabelian without cyclic subgroup of index p, then A 6 C and we obtain that 1 .G/ D 1 .A/ Z.A/ so that CG .1 .G// CA D G, contrary to the hypothesis. Thus every minimal nonabelian subgroup of G not contained in C has a cyclic subgroup of index p. It follows from 76 that k p 1 in all cases. Now suppose that k D p 1. Then, by Lemma 76.5(a), 1 D ¹C; A1 ; : : : ; Ap1 ; Aº, where A1 ; : : : ; Ap1 are minimal nonabelian and A is abelian. In view of Lemma 64.1(u), we get jG 0 j pjA01 A0 j D p 2 . If jG 0 j D p, then G is minimal nonabelian (Lemma 65.2(a)), contrary to the hypothesis. Thus jG 0 j D p 2 . In this case, all subgroups of G of index p 2 are abelian (Corollary 65.3), as asserted. Exercise 3. Let G and k be such as in Exercise 2. Study the structure of G provided k D p. (Hint. See Lemma 76.13.) Lemma 124.27. Let G > ¹1º be a metacyclic p-group, A < B G. Then A0 < B 0 unless B is abelian. Proof. Assume that B is nonabelian and A0 D B 0 . Take in B 0 a subgroup L of index p. Then B=L is minimal nonabelian (Lemma 65.2(a)) so A=L is abelian. In this case, we get A0 L < B 0 , as was to be shown. Theorem 124.28. Suppose that G is a metacyclic p-group with derived subgroup of order p n > p and let t be the number of maximal subgroups of G having derived subgroup of order p n1 . Then t 2 ¹p; p C 1º.

124

The number of subgroups of given order in a metacyclic p-group

265

Proof. By Lemma 64.1(u), there exists H 2 1 such that jH 0 j D p n1 so that t > 0. Assume that there are distinct A; B 2 1 such that jA0 j jB 0 j < p n1 . Since G 0 is cyclic, we get A0 B 0 . Then, by Lemma 64.1(u), we have jG 0 j pjA0 B 0 j D pjB 0 j < p p n1 D p n ; contrary to the hypothesis. Thus there is in 1 at most one member whose derived subgroup has order < p n1 . Since G is noncyclic so j1 j D p C 1 and G has no maximal subgroup whose derived subgroup has order p n (Lemma 124.27), we conclude that t p. We are done since t p C 1. Using induction and Hall’s enumeration principle, one obtains Corollary 124.29. Suppose that G is a metacyclic p-group with derived subgroup of order p n > p 2 . If t denotes the number of subgroups H of G with jH 0 j D p n2 and jG W H j D p 2 , then p 2 p t .p C 1/2 . Proof. Let M1 ; : : : ; Ms be all members of the set 1 whose derived subgroups have order p n1 ; then s 2 ¹p; p C 1º (Theorem 124.28). Let t.Mi / be the number of those maximal subgroups of Mi whose derived subgroups have order p n2 and let D 0 if jˆ.G/0 j < p n2 and D 1 if jˆ.G/0 j D p n2 ; then we have t.Mi / 2 ¹p; p C 1º (Theorem 124.28). By Hall’s enumeration principle, tD

s X

t .Mi / p :

i D1

It follows that

p2

p t .p C 1/2 .

Exercise 4. Suppose that G is a metacyclic 2-group of order > 22k , k > 1. If all minimal nonabelian subgroups of G are isomorphic to k

k

k1

Hk D ha; b j a2 D b 2 D 1; ab D a1C2

i;

then k1 .G/ D Z.G/ is abelian of type .2k1 ; 2k1 / and G=Z.G/ is dihedral. Solution. By Proposition 10.28, the group G is generated by minimal nonabelian subgroups. It follows that all minimal nonabelian subgroups are not normal in G (otherwise, they would be coincide). If A < G is minimal nonabelian, then we obtain that k1 .G/ D k1 .A/ D Z.A/ and thus we get k1 .G/ Z.G/ since such subgroups A generate G (Theorem 10.28). Moreover, since A \ Z.G/ D Z.A/, we get k1 .G/ D Z.G/. Therefore the nonabelian metacyclic quotient group G=k1 .G/ is generated by abelian subgroups of type .2; 2/, and we conclude that it is dihedral since jG=1 .G/j 8 (see Theorem 1.2). Exercise 5. If A metacyclic 2-group G contains two distinct minimal nonabelian subgroups A and B of order 22k and exponent 2k , then G=k1 .G/ is either dihedral or semidihedral. If G=k1 .G/ is dihedral, then k1 .G/ D Z.G/.

266

Groups of prime power order

Solution. We have k1 .G/ D k1 .A/ D k1 .B/. As A=k1 .G/, B=k1 .G/ are nonnormal four-subgroups of G=k1 .G/, it follows that G=k1 .G/ is of maximal class (Lemma J(e)); moreover, G=k1 .G/ is either dihedral or semidihedral. Suppose that G=k1 .G/ is dihedral. If H=k1 .G/ is abelian of type .2; 2/, then k k k1 either H is abelian or H Š ha; b j a2 D b 2 D 1; ab D a1C2 i. In both cases, we obtain k1 .G/ D Z.H /. As such subgroups H generate G, we get k1 .G/ Z.G/: Now the equality of these two subgroups follows since Z.G/ \ H D k1 .G/. Exercise 6. Suppose that G is a metacyclic p-group and B < G minimal nonabelian. Then B=B 0 is a maximal abelian subgroup of G=B 0 . Solution. Assume that B=B 0 < A=B 0 , where A=B 0 is abelian. Then A0 D B 0 , contrary to Lemma 124.27. Exercise 7. Let A be a proper minimal nonabelian subgroup of a p-group G. Suppose that all subgroups of G containing A of order pjAj are two-generator. Then A=A0 is a maximal abelian subgroup of NG .A/=A0 . Hint. Let A < T NG .A/ be such that jT W Aj D p and T =A0 is abelian. Since d.T / D 2 and T 0 D A0 is of order p, it follows from Lemma 65.2(a) that T is minimal nonabelian, which is a contradiction. Exercise 8 (= Proposition 10.17). Let B be a nonabelian subgroup of order p 3 of a p-group G. If CG .B/ < G, then G is of maximal class. Solution. Set N D NG .B/. Then, by N=C -Theorem, jN j D p 4 so N is a unique subgroup of order p 4 in G containing B. We claim that d.N / D 2 (otherwise, N contains at least two distinct abelian subgroups of order p 3 so jZ.N /j D p 2 ; then CG .B/ 6 B, a contradiction). Clearly, Z.G/ D Z.B/ D B 0 . Then, by Exercise 7, B=B 0 is a maximal abelian subgroup of G=B 0 so G=B 0 is of maximal class (Proposition 1.8). Since Z.G/ D B 0 is of order p, it follows that G is of maximal class as well. Exercise 9. Let H < G, where G is a metacyclic p-group. If H > R.H /, then we have R.H / G G. (Hint. We have R.H / D e .G/, where exp.R.H // D p e .) Exercise 10. Let G be a metacyclic 2-group and let hn .G/ be the number of subgroups H < G such that H D R.H / is of order 22n . Then one of the following holds: (a) hn .G/ D 1. (b) hn .G/ is even. In that case, if hn .G/ > 0, then G=n1 .G/ is either dihedral or semidihedral. Solution. Let H < G be such that H D R.H / is of order 22n . If H G G, then we have H D n .G/ and hn .G/ D 1. Now let H be nonnormal in G; then all such H

124

The number of subgroups of given order in a metacyclic p-group

267

are nonnormal in G. In this case, hn .G/ is even. Then n1 .H / D n1 .G/ and H=n1 .H / Š E4 is non-G-invariant. By Lemma J(e), It follows that G=n1 .G/ is either dihedral or semidihedral. Exercise 11. Prove that there exists a metacyclic p-group G such that G contains a maximal normal abelian subgroup A with nonabelian G=A. Solution (Mann). Let G be a semidirect product of a cyclic group A of order p n by a k cyclic group B whose generator induces on A the automorphism x ! x 1Cp . The order of B is the order of the automorphism. Choose n D 2k C 2. Then G 0 is the subgroup Ãk .A/ and Z.G/ D Ãnk .A/. Let L be the unique subgroup between G 0 and Z.G/, and let D D CB .L/. Then LD is a maximal normal abelian subgroup with nonabelian factor group. Many more examples can be produced by choosing other values of n and k to make the distance between G 0 and Z.G/ bigger. 2

2

Let H2;p D ha; b j ap D b p D 1; ab D a1Cp i. Then H2;p is the unique nonabelian metacyclic group of order p 4 and exponent p 2 . For p D 2 we have H2;2 D H2 . Theorem 124.30. Let H Š H2;p be a proper subgroup of a metacyclic p-group G. Then one of the following holds: (a) If p > 2, then jG W H j D p, jG 0 j D p 2 , G=G 0 is abelian of type .p 2 ; p/ and G=Z is cyclic of order p 2 for some cyclic Z G G. (b) If p D 2 and H G G, then jG W H j D 2, jG 0 j D 4, G=G 0 is abelian of type .4; 2/ and G=Z is cyclic of order 4 for some cyclic Z G G. (c) If p D 2 and the subgroup H is not normal in G, then we have 1 .G/ D 1 .H / and jG W CG .1 .G//j 2. If, in addition, jGj > 25 , then G=1 .G/ is either dihedral or semidihedral. Proof. Let G have a proper normal nonabelian subgroup H of order p 4 and exponent p 2 . Then we have R D 1 .H / D 1 .G/ G G. Set C D CG .R/. Let L < R be of order p, L ¤ H 0 . Then H=L C =L is nonabelian of order p 3 so H D C if p > 2 and C =L is of maximal class if p D 2 (Lemma J(b)). In the case under consideration, jG W C j p. Let p > 2; then H D C . In this case, G 0 < C so jG 0 j D p 2 since G 0 is cyclic and G is not minimal nonabelian (see Lemma 65.2(a). There is a cyclic Z G G such that G=Z is cyclic. Since G 0 < Z, it follows that G=G 0 is abelian of type .p 2 ; p/. Now let p D 2 and H G G; then L is characteristic in H (Lemma J(a)). Then C D G since L G G so G=L is of maximal class (Lemma J(b)). Since L 6 G 0 , it follows that jG W G 0 j D 8 so G=G 0 is abelian of type .4; 2/. As above, there is a cyclic Z G G such that G=Z is cyclic of order 4. Let H be not normal in G and C as above; then p D 2. Suppose that jGj > 25 . Take L < R.D 1 .G// of order 2, L ¤ H 0 . Then H=L C =L is nonabelian of order 8 so C =L is of maximal class. In this case, C =1 .G/ G=1 .G/ is of maximal class with nonnormal abelian subgroup H=1 .G/ of type .2; 2/, and we conclude that G=1 .G/ is either dihedral or semidihedral.

268

Groups of prime power order

Proposition 124.31. Suppose that H Š Mpn , n > 3, is a proper subgroup of a metacyclic p-group G. Then NG .H /=H is cyclic. Next, assume that n D 4. (a) If p > 2, then jNG .H /=H j D p. (b) If p D 2, then jNG .H /=H j 4. Proof. Assume that NG .H /=H is noncyclic. Write N D NG .H /. In this case, we have H 6 ˆ.N /. Indeed, otherwise 1 .H / D 1 .N / Z.N / (see the proof of Lemma 1.4). Then 1 .H / Z.H / so H is abelian, a contradiction. Thus NG .H /=H is cyclic. Next we assume that n D 4. By Exercise 8, H=H 0 is a maximal abelian subgroup of N=H 0 . If p > 2, then a Sylow p-subgroup of Aut.H=H 0 / is nonabelian of order p 3 and exponent p so N=H , being cyclic, has order p. Now let p D 2. Then Aut.H=H 0 / Š D8 so N=H is a cyclic subgroup of D8 , and we conclude that N=H is cyclic of order 4.

Problems Below we state some related problems. Problem 1. Study the metacyclic p-groups, p > 2, possessing a nonabelian section of order p 3 . (See Theorem 124.22.) Problem 2. Find the number of subgroups of given order in a nonmetacyclic minimal nonabelian p-group. Problem 3. Let G be an abelian group of type .p a1 ; : : : ; p ak /. Find the types of all maximal subgroups of G counting multiplicities. Problem 4. Find the number of subgroups of given order in an abelian p-group of rank 3. Problem 5. Let G be a homocyclic p-group of rank d . (i) Find the number of subgroups of given order in G. (ii) Find the number of subgroups of rank 2 and given order in G. Problem 6. Given a positive integer k > 1, classify the p-groups G such that there exists a metacyclic p-group M satisfying sk .G/ D sk .M /. Problem 7. Study the p-groups G such that sk .G/ D sk .A/ for all positive integers k and some abelian p-group A. Problem 8. Given p > 2 and n > p, does there exist an absolutely regular p-group A (see 9) and an irregular p-group G of the same order p n satisfying sk .G/ D sk .A/ for all k 2 ¹2; : : : ; n 1º?

125

p-groups G containing a maximal subgroup H all of whose subgroups are G -invariant

Here we solve a problem stated by the ﬁrst author. Theorem 125.1. Let G be a nonabelian p-group containing a maximal subgroup H such that all subgroups of H are G-invariant. Then there is an element g 2 G H such that one of the following holds: (i) p D 2, H is Hamiltonian, i.e., H D Q V , where Q Š Q8 , exp.V / 1, and g 2 Z.G/, o.g/ 4. (ii) p D 2, H is abelian of exponent 2e , e 2, and either g inverts each element e1 in H or e 3 and g satisﬁes hg D h1C2 for all h 2 H . In both cases, Z.G/ D CH .g/ D 1 .H / is elementary abelian and o.g/ 4. e1

for all h 2 H ,

e1

for all h 2 H ,

(iii) p D 2, H is abelian of exponent 2e , e 3, and hg D h1C2 where Z.G/ D CH .g/ D e1 .H /.

(iv) p > 2, H is abelian of exponent p e , e 2, and hg D h1Cp where Z.G/ D CH .g/ D e1 .H /.

Proof. (i) First assume that H is nonabelian. Then p D 2 and H D Q V , where Q Š Q8 and exp.V / 1. Set U D Z.Q/ V D 1 .H / so that U Z.G/ since each subgroup of order 2 in U is normal (and so central) in G. But each subgroup of Q is also normal in G and so G D Q C , where C D CG .Q/, Q \ C D Z.Q/ and H \ C D U . Thus C is abelian and so C D Z.G/. We take an element g 2 C H so that g 2 Z.G/, g 2 2 U and o.g/ 4. (ii) Now assume that H is abelian, p D 2, and 2 .H / 6 Z.G/. Let g be an element in G H . We have CH .g/ D Z.G/ and g 2 2 Z.G/. Let v be an element of order 4 in H Z.G/. Then v g D v 1 D vz, where z D v 2 2 Z.G/. Suppose that Z.G/ possesses an element w of order 4. If w 2 D z, then vw is an involution in H Z.G/ and vw is central in G, a contradiction. Hence we get w 2 D u ¤ z and hv; wi Š C4 C4 . But then we obtain .vw/g D .vz/w D .vw/z and z 62 hvwi as .vw/2 D zu which implies that hvwi is not normal in the group G, a contradiction. We have proved that Z.G/ D 1 .H / D CH .g/ is elementary abelian. For any x 2 H Z.G/ we obtain that o.x/ 4 and Chxi .g/ D 1 .hxi/ and so either x g D x 1 or o.x/ > 4 and x g D x 1 x 0 , where hx 0 i D 1 .x/. Also, g 2 2 Z.G/ and so o.g/ 4.

270

Groups of prime power order

Suppose that g does not invert each element in H . Then exp.H / D 2e with e 3. If g inverts each element of order 2e in H , then g inverts each element in H since elements of order 2e generate H , a contradiction. Hence there is an element h1 2 H of e1 order 2e such that g does not invert hh1 i and so hg1 D h1C2 . There exists a basis 1 ¹h1 ; : : : ; hr º, r 1, of H which contains h1 (since h1 is of the maximal possible order in H ). Assume that there is hi 2 ¹h1 ; : : : ; hr º such that i > 1, o.hi / D 2e and hgi D h1 i . But then .h1 hi /g D h1C2 1 e1

e1

e1

1 2 h1 i D .h1 hi / h1

e1

e1

¤ h21 which shows that hh1 hi i is not normal in G, and 1 .hh1 hi i/ D h21 h2i e1 a contradiction. Hence we must have hgi D h1C2 . Suppose that there is an elei 0 ment hj 2 ¹h1 ; : : : ; hr º such that j > 1, o.hj / D 2e , e 0 < e, but hjg ¤ hj1

e 0 1

e 0 3 and hjg D hj1C2

and so

:

0

Let h01 2 hh1 i be such that o.h01 / D 2e which implies that .h01 /g D .h01 /1 . But then .h01 hj /g D .h01 /1 hj1C2

e 0 1

and

1 .hh01 hj i/ D .h01 /2

e 0 1

hj2

e 0 1

¤ hj2

e 0 1

which shows that hh01 hj i is not normal in G, a contradiction. Hence we must have hjg D hj1 D hj1C2

e1

since o.hj / < 2e . We have proved that hgi D h1C2 for all 1 i r and so for i e1 all h 2 H , hg D h1C2 and we are done. (iii) Assume that H is abelian, p D 2, and 2 .H / Z.G/. Then H is of exponent 2e , e 3, and let g 2 G H so that g centralizes 2 .H /. If x 2 H Z.G/, then o.x/ 8, Chxi .g/ D hx 2 i and so x g D xx 0 , where hx 0 i D 1 .hxi/. Since all e1 elements of order 2e generate H , there is h1 2 H of order 2e such that hg1 D h11C2 . Let ¹h1 ; : : : ; hr º, r 1, be a basis of H containing h1 . Suppose that there is an element hi 2 ¹h1 ; : : : ; hr º such that i > 1, o.hi / D 2e and hgi D hi . But then e1

.h1 hi /g D .h1 hi /h21 e1

a contradiction. Hence hgi D h1C2 i 0 o.hj / D 2e , e 0 3, e 0 < e, and

e1

e1

and h21

62 hh1 hi i;

. Assume that there is hj 2 ¹h1 ; : : : ; hr º, j > 1, e 0 1

hjg D hj1C2

:

0

Let h01 2 hh1 i with o.h01 / D 2e which gives .h01 /g D h01 . But then e 0 1

.h01 hj /g D .h01 hj /hj2

125

p-groups G with a maximal subgroup having only G-invariant subgroups e 0 1

so that hj2

271

62 hh01 hj i, a contradiction. Hence hjg D hj D hj1C2 e1

It follows that for each h 2 H , hg D h1C2

e1

:

, Z.G/ D e1 .H /, and we are done.

(iv) Finally, assume p > 2 so that H is abelian of exponent p e , e 2. We have Z.G/ D CH .g/, where g is an element in G H and g satisﬁes g p 2 Z.G/. For each x 2 H Z.G/, g induces on hxi an automorphism of order p and so x g D xx 0 , where 1 ¤ x 0 2 1 .hxi/. Since all elements of order p e generate H , there is h1 2 H of order p e such that, replacing g with g l for some l 6 0 .mod p/ if necessary, we may set 1Cp e1 hg1 D h1 : Let ¹h1 ; : : : ; hr º, r 1, be a basis of H containing h1 . Let hi 2 ¹h1 ; : : : ; hr º be such that i > 1, o.hi / D p e but hgi D hi . Then p e1

.h1 hi /g D .h1 hi /h1

;

p e1

1Cp e1 0 hi ,

62 hh1 hi i, a contradiction. Hence we may set hgi D hi where h1 h0i 2 1 .hhi i/. Then 1Cp e1 1Cp e1 0 hi hi

.h1 hi /g D h1 and .h1 hi /p

e1

p e1

h0i D h1

p e1 0 hi /

.hi

D .h1 hi /1Cp

e1

where

h0i

p e1 p e1 hi i

2 1 .hh1 hi i/ D hh1

1Cp e1

which gives h0i D 1. Thus hgi D hi . Let hj 2 ¹h1 ; : : : ; hr º be such that j > 1, 0 g 0 e 0 0 o.hj / D p , e 2, e < e but hj D hj hj , where 1 ¤ hj0 2 1 .hhj i/. Let h01 2 hh1 i 0 be such that o.h01 / D p e which gives that .h01 /g D h01 . But then .h01 hj /g D .h01 hj /hj0 ; where hj0 62 hh01 hj i, a contradiction. Hence 1Cp e1

hjg D hj D hj and so for each h 2 H , hg D h1Cp

e1

, e1 D Z.G/, and we are done.

126

The existence of p-groups G1 < G such that Aut.G1/ Š Aut.G /

Benjamin Sambale has constructed a 2-group G of order 32 which possesses a maximal subgroup G1 such that Aut.G1 / Š Aut.G/. We give here a simple proof of this fact which solves Problem 211 (for p D 2). Theorem 126.1. There exist 2-groups G1 < G with jGj D 32 and Aut.G1 / Š Aut.G/. Proof. We construct our 2-group G as a splitting and faithful extension of its abelian normal subgroup H D ha; b j a4 D b 2 D Œa; b D 1i of type .4; 2/ with a cyclic subgroup hci Š C4 of order 4, where ac D a1 b and b c D ba2 . Then ha2 i D Z.G/ and c 2 inverts each element in H and so all elements in Hc 2 are involutions. Thus all elements in G .H hc 2 i/ D Hc [Hc 1 are of order 4 and so G1 D H hc 2 i D 1 .G/ is a characteristic subgroup of G. Since H D 2 .G1 /;

ha2 ; bi D 1 .H /;

ha2 i D Ã1 .H / D Z.G/;

we conclude that H , ha2 ; bi Š E4 and ha2 i are also characteristic subgroups of G. As b D aac , we have G D ha; ci and so G is given in terms of generators and relations as follows: ()

G D ha; c j a4 D c 4 D .aac /2 D Œa; aac D 1; ac D a1 i: 2

Note that hai and habi are the only two cyclic subgroups of order 4 in H and since haic D hac i D ha1 bi D habi, we have NG .hai/ D G1 . Hence for each x 2 G G1 and y 2 H 1 .H /, hyi and hy x i are two distinct cyclic subgroups of order 4 in H , which gives that hx; yi D G. Each automorphism of G sends a to one of ¹a; a1 ; ab; a1 bº and c is sent to one of 16 elements in G G1 which shows that jAut.G/j 4 16 D 64. Conversely, take a 2 ¹a; a1 ; ab; a1 bº (4 possibilities) and c 2 G G1 D Hc [ Hc 1 (16 possibilities). We know from the previous paragraph that ha ; c i D G and we verify that a and c satisfy all relations in () (in place of a and c, respectively) and so a ! a , c ! c induces an automorphism of G, hence jAut.G/j D 4 16 D 64. Indeed, a and c are elements of order 4. Also, a and .a /c are two elements in H 1 .H / so that a .a /c 2 1 .H / and therefore .a .a /c /2 D 1. Obviously, a and .a /c (lying in the abelian subgroup H ) commute. Finally, .c /2 2 G1 H

126 The existence of p-groups G1 < G such that Aut.G1 / Š Aut.G/

273

and so .c /2 is an involution which inverts each element in H and therefore we obtain 2 .a /.c / D .a /1 . It is easy to show that jAut.G1 /j D 64. We have G1 D ha; b; d D c 2 j a4 D b 2 D Œa; b D d 2 D 1; ad D a1 ; b d D bi;

()

and for each 2 Aut.G1 / we conclude that a 2 ¹a; a1 ; ab; a1 bº (4 possibilities), b 2 ¹b; ba2 º (2 possibilities) and d 2 G1 H (8 possibilities) which implies that jAut.G1 /j 4 2 8 D 64. For the converse, if we consider a 2 ¹a; a1 ; ab; a1 bº, b 2 ¹b; ba2 º and d 2 G1 H , then obviously ha ; b ; d i D G1 and a ; b ; d satisfy all relations in () since each element in G1 H is an involution which inverts each element in H . We have proved that jAut.G1 /j D 64. It remains to show Aut.G1 / Š Aut.G/. First we determine the structure of Aut.G/. The restriction of Aut.G/ to H provides a homomorphism f W Aut.G/ ! Aut.H /. If 2 ker.f /, then a D a;

bDb

D .ac 1 ac/ D a.c /1 ac

which gives ac D a1 b D ac and so c 2 Hc. We may set D h with a h D a and c h D hc (h 2 H ) so that ker.f / D ¹ h j h 2 H º. The map h ! h (h 2 H ) is an isomorphism from ker.f / onto H and so ker.f / Š C4 C2 and Aut.G/=ker.f / Š Aut.H / Š D8 and therefore f is a homomorphism from Aut.G/ onto Aut.H /. Now we construct a complement of ker.f / in Aut.G/. Deﬁne L D h˛; ˇ 2 Aut.G/; where a˛ D ab; c ˛ D c and aˇ D a; c ˇ D c 1 i: Then a˛ D .ab/˛ D .a ac 1 ac/˛ D .a2 c 1 ac/˛ 2

D a2 .c 1 ab c/ D a2 a1 b ba2 D a1 2

and c ˛ D c so that o.˛/ D 4 and obviously o.ˇ/ D 2. Then we show that .˛ˇ/2 D 1. Indeed, a˛ˇ D .ab/ˇ D ab ˇ D a.ac 1 ac/ˇ D a acac 1 D a2 cac 1 and c ˛ˇ D c 1 . This gives (noting that a2 2 Z.G/) a.˛ˇ / D .a2 cac 1 /˛ˇ D .a2 cac 1 /2 c 1 .a2 cac 1 /c 2

D a2 cac 1 a2 cac 1 c 1 a2 ca D a2 ca2 c 1 a D a 2

and c .˛ˇ / D c which shows .˛ˇ/2 D 1. We have proved that L D h˛; ˇ j ˛ 4 D ˇ 2 D .˛ˇ/2 D 1i Š D8 . Since Z.L/ D h˛ 2 i and ˛ 2 62 ker.f /, we get ker.f / \ L D ¹1º. Thus we have shown so far that Aut.G/ is a semidirect product of ker.f / Š H and

274

Groups of prime power order

L Š Aut.H / Š D8 . It is easy to justify that CAut.G/ .ker.f // D ker.f /. Indeed, if we have CAut.G/ .ker.f // > ker.f /, then ˛ 2 centralizes ker.f /. In particular, we obtain 2 2 2 that ˛ 2 a D a ˛ 2 . But c ˛ a D c a D ac and c a ˛ D .ac/˛ D a1 c, a contradiction. We have proved that Aut.G/ is isomorphic to the holomorph of H Š C4 C2 . Now we determine the structure of Aut.G1 /, where G1 is given in (). The restriction of Aut.G1 / to H gives rise to a homomorphism f0 W Aut.G1 / ! Aut.H / with ker.f0 / D ¹h j h 2 H º; where ah D a; b h D b; d h D hd: The map h ! h .h 2 H / gives an isomorphism ker.f0 / Š H . Thus we conclude Aut.G1 /=ker.f0 / Š Aut.H / Š D8 . The automorphisms 2 Aut.G1 / with d D d form a complement L0 Š D8 of ker.f0 / in Aut.G1 /. Let 2 L0 with a D a1 b, b D ba2 so that 2 inverts each element in H and therefore h 2 i D Z.L0 /. In case CAut.G1 / .ker.f0 // > ker.f0 /, we infer that 2 centralizes ker.f0 /. In particular, we 2 2 2 have a 2 D 2 a . But then d a D .ad / D a1 d and d a D d a D ad , a contradiction. Hence CAut.G1 / .ker.f0 // D ker.f0 / and so Aut.G1 / is also isomorphic to the holomorph of H Š C4 C2 and we are done. A similar example was presented in [Li4].

127

On 2-groups containing a maximal elementary abelian subgroup of order 4

In the following theorem Problem 1591 is solved. Theorem 127.1 (Janko). Let G be a 2-group which contains a maximal elementary abelian subgroup A of order 4. If G contains also an elementary abelian subgroup E of order 16, then G is isomorphic to one of the groups of Theorem 49.3.1 Proof. By Exercise 51 in Appendix 40, G has no normal elementary abelian subgroup of order 8. Due to Exercise 52 in Appendix 40, every subgroup of G is generated by four elements. In particular, G has no elementary abelian subgroup of order > 16. Obviously, G is neither abelian nor a 2-group of maximal class. Then G possesses a normal four-subgroup U . If A Z.G/, then AE is elementary abelian of order > 16, contrary to what has just been said. Therefore, since A is a maximal elementary abelian subgroup of G, it follows that 1 .Z.G// < A and so, in particular, Z.G/ is cyclic. Set hzi D 1 .Z.G//; then we have hzi A \ U and jG W CG .U /j D 2. Note that jE \ CG .U /j 8 and so A ¤ U which gives A \ U D hzi. But then A 6 CG .U / and so A \ CG .U / D hzi. Let t 2 A CG .U / so that CG .t/ D ht i CCG .U / .t/ by the modular law, and therefore z is the only involution in L D CCG .U / .t/ (indeed, if z 2 F < CCG .U / .t/, where F is abelian of type .2; 2/, then A < ht; F i Š E8 , contrary to the hypothesis). Hence L is either cyclic of order 2m , m 2, or L is generalized quaternion. Indeed, if L is cyclic of order 2, then G would be of maximal class (Proposition 1.8), a contradiction. Suppose that L Š C2m , m 2. By Theorem 48.1, in that case G does not possess an elementary abelian subgroup of order 16. Hence we have L Š Q2m , m 3, and then G is isomorphic to one of the groups from Theorem 49.3 and we are done. Thus the groups from Theorem 127.1 are determined up to isomorphism. 1 In Theorem 49.3 the 2-groups G containing an involution t such that C .t/ D hti Q, where Q G is a generalized quaternion group, are classiﬁed if G also possesses an elementary abelian subgroup of order 16.

276

Groups of prime power order

Remark. Let t be an element of order p of a p-group G such that CG .t/ D hti Q, where j1 .Q/j D p is a generalized quaternion group. Then 1 .CG .t// Š Ep2 is a maximal elementary abelian subgroup of G. In that case, G has no normal subgroup of order p pC1 and exponent p (see Exercise 51 in Appendix 40). Now, if p D 2, by the proof of the theorem, G has no elementary abelian subgroup of order 25 by Exercise 52 in Appendix 40. G. Glauberman and N. Mazza have proved that if p > 2 and a p-group G possesses a maximal elementary abelian subgroup of order p 2 , then it has no elementary abelian subgroup of order p pC1 . As the symmetric group Sp2 , p > 2, shows, this estimate is best possible. Exercise. Suppose that a p-group G, p > 2, possesses a maximal elementary abelian subgroup A of order p 2 . If there is in G an elementary abelian subgroup E of order p 3 , then there is a 2 A# such that CG .a/ D hai Q, where Q has exactly one subgroup of order p. Next, G has no normal subgroup of order p pC1 and exponent p (compare with Glauberman–Mazza’s result from the paragraph preceding the exercise). In view of Theorem 13.7, one may assume that E GG. In that case, AE is of maximal class and order p 4 . Assume, in addition, that p > 3; then exp.AE/ D p. Let AE H G, where H is as large as possible among all subgroups of exponent p. As CH .A/ D A, it follows that H is of maximal class (Proposition 1.8), so jH j p p (Theorem 9.5).

128

The commutator subgroup of p-groups with the subgroup breadth 1

Recall that the subgroup breadth sb.G/ of a p-group G is deﬁned by p sb.G/ D max¹jG W NG .H /j j H Gº: This section is a continuation of 122 and here we prove some results which are crucial on the way of classifying completely p-groups G with sb.G/ D 1. Theorem 128.1. If a p-group G has the subgroup breadth 1, then jG 0 j p 2 . Proof. By Propositions 122.8 and 122.16, the element breadth b.G/ of G is at most 2. Recall that the element breadth b.G/ of a p-group G is deﬁned by p b.G/ D max¹jG W CG .x/j j x 2 Gº: If b.G/ D 1, then by Proposition 121.9 (Knoche), jG 0 j D p. We may assume in the sequel that b.G/ D 2. Then we use Theorem 121.1 (Parmeggiani–Stellmacher) and conclude that we must be in part (b) of that theorem (otherwise, jG 0 j D p 2 and we are ﬁnished) which implies that jG 0 j D p 3 and jG W Z.G/j D p 3 . By Proposition 122.2, ˆ.G/ is abelian. Let A be a maximal normal abelian subgroup of G containing ˆ.G/ so that G=A ¤ ¹1º is elementary abelian. If jG=Aj D p, then Lemma 1.1 gives jGj D pjZ.G/jjG 0 j, which contradicts jG 0 j D jG W Z.G/j D p 3 . Hence jG=Aj p 2 . But Z.G/ < A and so G=A Š Ep2 and jA W Z.G/j D p. For each x 2 G A, x p 2 Z.G/ and so G=Z.G/ is of order p 3 and exponent p. Also, for each x 2 G A, CA .x/ D Z.G/ and so jŒA; xj D p (Lemma 121.2) and ŒA; x Z.G/ since A=Z.G/ Z.G=Z.G//. (i) First assume that G=Z.G/ is nonabelian. Then p > 2 and G=Z.G/ Š S.p 3 / (the nonabelian group of order p 3 and exponent p). Since G 0 covers A=Z.G/, we get jG 0 \ Z.G/j D p 2 . For any x; y 2 G A such that Ahx; yi D G we have ŒA; x ¤ ŒA; y. Indeed, if ŒA; x D ŒA; y, then we obtain A=ŒA; x D Z.G=ŒA; x/ which implies that G=ŒA; x has (at least) two distinct abelian maximal subgroups. But then j.G=ŒA; x/0 j D p and then jG 0 j D p 2 , a contradiction. Hence if Gi =A, i D 1; 2; : : : ; p C 1, are p C 1 pairwise distinct subgroups of order p in G=A, then ŒA; Gi D ŒG 0 ; Gi (i D 1; 2; : : : ; p C 1) are p C 1 pairwise distinct subgroups of order p in G 0 \ Z.G/ so that G 0 \ Z.G/ Š Ep2 . On the other hand,

278

Groups of prime power order

for each x 2 A Z.G/ we have CG .x/ D A which shows that there are no elements of order p in AZ.G/. In particular, all elements in the set G 0 Z.G/ are of order p 2 . Let a 2 G 0 Z.G/ and then set z D ap so that Ã1 .G 0 / D hzi and G 0 is abelian of type .p 2 ; p/. Let M be any maximal subgroup of G which contains Z.G/. Since G 0 M , we must have M > A. This simple remark will be used often in the following arguments. Let g 2 G A be such that Œa; g D z. Then g p 2 Z.G/ and CG .g/ D Z.G/hgi. Indeed, we have CA .g/ D Z.G/ and so if CG .g/ > Z.G/hgi, then CG .g/ covers G=A and so CG .g/ would be a maximal subgroup of G containing Z.G/ but not containing A, contrary to our simple remark from the previous paragraph. On the other hand, jNG .hgi/ W CG .g/j D p and so o.g/ p 2 and NG .hgi/ D Ahgi. Indeed, if hgi E G, then A=Z.G/ and .Z.G/hgi/=Z.G/ are two distinct normal subgroups of order p in the factor group G=Z.G/, contrary to G=Z.G/ Š S.p 3 /. Also, if NG .hgi/ ¤ Ahgi, then NG .hgi/ is a maximal subgroup of G which contains Z.G/ but does not contain A, contrary to our simple remark. We have proved that hai normalizes hgi. Since Œa; g D z, we have 1 .hgi/ D hzi. If o.g/ p 3 , then there is an element h 2 hg p i Z.G/ such that o.h/ D p 2 and hp D z 1 . But then ah 2 AZ.G/ and .ah/p D ap hp D zz 1 D 1, a contradiction. Hence o.g/ D p 2 and 1 ¤ g p 2 hzi. We get ha; gi D U Š Mp3 with U \ A D hai. Since 1 .U / Š Ep2 , there is an element g 0 of order p in U A. But then we must have CA .g 0 / D Z.G/ and jG W CG .g 0 /j D p which shows that CG .g 0 / covers G=A. Hence CG .g 0 / is a maximal subgroup of G containing Z.G/ but not containing A, contrary to our simple remark. (ii) Assume that G=Z.G/ is abelian. In that case, we get G=Z.G/ Š Ep3 and so ˆ.G/ Z.G/. For any x; y 2 G we have Œx; yp D Œx p ; y D 1 and so G 0 Š Ep3 . If there is an element x 2 G Z.G/ of order p, then we infer that jG W CG .x/j D p and so CG .x/ is an abelian maximal subgroup of G, a contradiction. Hence 1 .G/ Z.G/. If x; y; z 2 G Z.G/ are such that G D Z.G/hx; y; zi, then G 0 D hŒx; y; Œx; z; Œy; zi. Let h be any element in G Z.G/. Then o.h/ p 2 and Œhhi; G Š Ep2 . We have CG .h/ D Z.G/hhi with jG W CG .h/j D p 2 because G has no abelian maximal subgroup. If hhi E G, then Œhhi; G hhi and so Œhhi; G is cyclic of order p 2 , a contradiction. It follows that jG W NG .hhi/j D p. Then we get Œhhi; NG .hhi/ hhi \ Œhhi; G and so 1 .hhi/ Œhhi \ G: We have proved the following fact which will be used often in the following arguments: () For each h 2 G Z.G/ we have 1 .hhi/ Œhhi; G G 0 .

128

The commutator subgroup of p-groups with the subgroup breadth 1

279

(ii1) Suppose that p > 2. Amongst all elements in G Z.G/ let a be one of smallest possible order p m , m1 m 2. Then we set t1 D ap so that (by ./) t1 2 G 0 . Further, let b be any element in G .Z.G/hai/ such that Œa; b D t1 . From () we conclude that such an element b exists. Let c 2 G .Z.G/ha; bi/ so that setting t2 D Œa; c and t3 D Œb; c, we obtain ht1 ; t2 ; t3 i D G 0 , where Z.G/ha; b; ci D G. Since Œb; a D t11 and Œb; c D t3 , we get Œhbi; G D ht1 ; t3 i. As Œab; a D t11 and Œab; c D t2 t3 , we have Œhabi; G D ht1 ; t2 t3 i: First suppose that o.b/ > o.a/ D p m . Then, using (), we get ˇ

1 .hbi/ D 1 .habi/ D ht1˛ t3 i D ht1 .t2 t3 /ı i D ht1 t2ı t3ı i; where not both ˛ and ˇ are 0 .mod p/ and also not both and ı are 0 .mod p/. We get ı 0 .mod p/ and then ˇ 0

˛ 6 0

.mod p/;

6 0

.mod p/;

.mod p/:

But then b p 2 Z.G/;

o.b p / p m ;

hb p i ht1 i

which implies that there is an element y 2 hb p i such that y p .ay/p

m1

D ap

m1

yp

m1

m1

D t11 . We obtain

D t1 t11 D 1

and so o.ay/ p m1 and ay 2 G Z.G/, which is a contradiction. We have proved that for each b 2 G Z.G/ with Œa; b D t1 we have o.b/ D p m . It follows that for each x 2 Z.G/, o.xb/ D p m and so o.x/ p m which gives that exp.Z.G// p m . Hence all elements in .Z.G/hai/ Z.G/ are also of order p m . m1 ˇ Since Œhbi; G D ht1 ; t3 i, the fact stated in () implies b p D t1˛ t3 , where not both ˛ and ˇ are 0 .mod p/. Then .ab/p

m1

D ap

m1

bp

m1

D t1 t1˛ t3 D t1˛C1 t3 2 G 0 ˇ

ˇ

ˇ

and so the minimality of o.b/ D p m yields that o.ab/ D p m and t1˛C1 t3 ¤ 1. On the other hand, the fact that Œhabi; G D ht1 ; t2 t3 i gives together with () that .ab/p

m1

ˇ

D t1˛C1 t3 D t1 .t2 t3 /ı D t1 t2ı t3ı ;

where not both and ı are 0 .mod p/. We conclude that ı 0 .mod p/ and then ˇ 0 .mod p/ and ˛C1

.mod p/

with

˛ 6 0

.mod p/

and

˛ C 1 6 0

.mod p/:

280

Groups of prime power order

Hence b p compute

m1

D t1˛ and so, replacing b with b 0 D b , where ˛ 1 .mod p/, we

.ab 0 /p

m1

D ap

m1

.b 0 /p

m1

D t1 .b p

m1

/ D t1 t1 D t1 t11 D 1: ˛

Since ab 0 62 Z.G/ and o.ab 0 / p m1 , we get a contradiction to the minimality of o.b/ D p m . (ii2) Assume that p D 2. Amongst all elements in G Z.G/ let a be one of maximal possible order 2m , m1 m 2. Then set t1 D a2 so that () implies t1 2 Œa; G. Let b be any element in G .Z.G/hai/ such that Œa; b D t1 . Take an element c 2 G .Z.G/ha; bi/ and set t2 D Œa; c and t3 D Œb; c, where G 0 D ht1 ; t2 ; t3 i. Since Œb; c D t3 and Œb; a D t1 , ˇ () implies 1 .hbi/ ht1 ; t3 i and therefore 1 .hbi/ D ht1˛ t3 i, where ˛; ˇ 2 ¹0; 1º and not both ˛ and ˇ are equal to 0. As Œc; b D t3 and Œc; a D t2 , it follows from () that 1 .hci/ ht2 ; t3 i and therefore 1 .hci/ D ht2 t3 i, where ; 2 ¹0; 1º and not both and are equal to 0. First assume m D 2 so that all elements in G Z.G/ are of order 4. We have ˇ

ˇ

.ab/2 D a2 b 2 Œb; a D t1 t1˛ t3 t1 D t1˛ t3 ;

Œab; a D t1 ;

Œab; c D t2 t3 ;

ˇ

and so Œab; G D ht1 ; t2 t3 i. Using (), we get t1˛ t3 2 ht1 ; t2 t3 i and so ˇ D 0, ˛ D 1, .ab/2 D t1 , and b 2 D t1 . We have

.ac/2 D t1 t2 t3 t2 D t1 t2C1 t3 ;

Œac; a D t2 ;

Œac; b D t1 t3 ;

and so Œac; G D ht2 ; t1 t3 i. Again using (), we get t1 t2C1 t3 2 ht2 ; t1 t3 i and so D 1. We have .bc/2 D t1 t2 t3 t3 D t1 t2 ;

Œbc; b D t3 ;

Œbc; a D t1 t2

and so Œbc; G D ht3 ; t1 t2 i. Applying (), we obtain that t1 t2 2 ht3 ; t1 t2 i, and so D 1, c 2 D t2 t3 . Finally, we have .ab c/2 D .ab/2 c 2 Œc; ab D t1 t2 t3 t2 t3 D t1 ;

Œabc; a D t1 t2 ;

Œabc; b D t1 t3

and so Œabc; G D ht1 t2 ; t1 t3 i. Using (), we get t1 2 ht1 t2 ; t1 t3 i, a contradiction. We have proved that m 3, where a is an element of maximal possible order 2m in G Z.G/. Now suppose that for each b 2 G .Z.G/hai/ such that Œa; b D t1 we m1 ˇ have always o.b/ D 2m . We get b 2 D t1˛ t3 , where not both ˛ and ˇ are equal to 0. m1 Since Œab; a D t1 , Œab; c D t2 t3 , we obtain (using ()) .ab/2 D t1 .t2 t3 /ı , where not both and ı are equal 0. We note that Œa; ab D t1 and so, by our assumption, we have o.ab/ D 2m . On the other hand, .ab/2

m1

ˇ

ˇ

D t1 t1˛ t3 D t1˛C1 t3 :

128

The commutator subgroup of p-groups with the subgroup breadth 1 ˇ

ˇ

281

Hence t1˛C1 t3 D t1 t2ı t3ı and this implies ı D 0 and then t1˛C1 t3 D t1 gives ˇ D 0 and ˛ C 1 .mod 2/. Since ı D 0, we must have D 1 and then ˛ D 0. But then both ˛ and ˇ are equal to 0, a contradiction. By the previous paragraph, we may take an element b in G .Z.G/hai/ such that m1 m1 Œa; b D t1 and o.b/ < 2m . In that case, we have .ab/2 D a2 D t1 . It is easy to see that exp.Z.G// D 2m1 . Indeed, in any case, exp.Z.G// 2m since all elements in G Z.G/ are of order at most 2m . Suppose that there is z 2 Z.G/ with o.z/ D 2m . Then b 0 D bz is of order 2m and since Œb 0 ; G D Œb; G D ht1 ; t3 i, we have (by ()) 1 .hb 0 i/ 2 ht1 ; t3 i and so .b 0 /2

m1

D .b/2

m1

m1

z2

m1

D z2

m1

2 ht1 ; t3 i:

m1

2 ¹t3 ; t1 t3 º, then .az/2 2 ¹t3 ; t1 t3 º, a contradiction as Œaz; G D ht1 ; t2 i. If z 2 m1 Therefore z 2 D t1 . However, Œc; G D Œcz; G D ht2 ; t3 i and so 1 .hci/ D t2 t3 , where ; 2 ¹0; 1º and not both and are equal to 0 and 1 .hcz/ ht2 ; t3 i. In case m1 o.c/ < 2m , we get .cz/2 D t1 , a contradiction. If o.c/ D 2m , then c2

m1

D t2 t3

m1

and .cz/2

D t2 t3 t1 ;

a contradiction. Hence exp.Z.G// D 2m1 and so all elements in Z.G/a and Z.G/ab are of order 2m and all elements in Z.G/b are of order < 2m . Consider ac, where Œac; b D t1 t3 , Œac; a D t2 so that Œac; G D ht2 ; t1 t3 i and therefore (by ()) 1 .haci/ ht2 ; t1 t3 i: On the other hand, .ac/2 2m1

m1

m1

D t1 c 2

:

2m1

D 1, then .ac/ D t1 , a contradiction. Hence we have o.c/ D 2m and so If c m1 m1 2 c D t2 t3 , where not both and are equal 0. Then .ac/2 D t1 t2 t3 2 ht2 ; t1 t3 i m1 2 and so D 1 and c D t 2 t3 . Consider .ab/c, where Œabc; a D t1 t2 , Œabc; c D t2 t3 so that, by (), 1 .habci/ ht1 t2 ; t2 t3 i: On the other hand, ..ab/c/2

m1

m1

D t1 t2 t3 2 ht1 t2 ; t2 t3 i; m1

D t3 and .abc/2 D t1 t3 . and so D 0 and we get c 2 Finally, we consider bc, where Œbc; b D t3 , Œbc; a D t1 t2 so that (by ()) 1 .hbci/ ht1 t2 ; t3 i: On the other hand, .bc/2

m1

D t3 .

282

Groups of prime power order

We see that hbi normalizes the subgroups hai;

habi;

hci;

haci;

hbci;

habci;

and all elements in the cosets Z.G/a;

Z.G/ab;

Z.G/c;

Z.G/ac;

Z.G/bc;

Z.G/abc

are of order 2m and they are also normalized by hbi. Let b0 be an element of smallest possible order 2s in Z.G/b, where 2 s m 1, s1 ˇ m 3. Then b02 D t1˛ t3 , where neither ˛ nor ˇ are equal to 0. Since a2

m1

D t1 ;

c2

m1

D t3 ;

m1

.ac/2

D t1 t3 ;

there is an element x of order 2s in one of the subgroups ha2 i; s1

so that x 2 Z.G/ and x 2

hc 2 i;

h.ac/2 i

ˇ

D t1˛ t3 . Then we get s1

.xb0 /2

ˇ

ˇ

D t1˛ t3 t1˛ t3 D 1

and so o.xb0 / 2s1 and xb0 2 Z.G/b, a ﬁnal contradiction. Thus our theorem is proved. Theorem 128.2. Let G be a p-group with the subgroup breadth 1 and let p > 2. Then G possesses an abelian maximal subgroup and G 0 (of order p 2 ) is elementary abelian. Proof. By Theorem 122.1(b), jG W Z.G/j p 3 . Due to Theorem 128.1, jG 0 j p 2 . If jG W Z.G/j D p 2 , then G=Z.G/ Š Ep2 and this implies jG 0 j D p. In case jG 0 j D p, Proposition 122.13 yields G=Z.G/ Š Ep2 . In what follows we may assume that jG=Z.G/j D p 3 and jG 0 j D p 2 . Also, we assume that G does not possess any abelian maximal subgroup. Let A be a maximal normal abelian subgroup of G containing ˆ.G/ (which is abelian) so that G=A is elementary abelian of order p 2 . As Z.G/ < A and jG=Z.G/j D p 3 , we have G=A Š Ep2 and jA W Z.G/j D p. For any g 2 G A, g p 2 Z.G/ and so G=Z.G/ is of order p 3 and exponent p. (i) First assume that G=Z.G/ is nonabelian and so G=Z.G/ Š S.p 3 / (the nonabelian group of order p 3 and exponent p). Since G 0 covers A=Z.G/, we infer that G 0 \ Z.G/ Š Cp . For any x 2 A Z.G/, CG .x/ D A which implies o.x/ p 2 . Suppose that there is an element y 2 GA of order p. As CA .y/ D Z.G/, it follows that that CG .y/ must cover G=A, which contradicts the structure of G=Z.G/ Š S.p 3 /. We have proved that 1 .G/ Z.G/. In particular, G 0 D hai Š Cp2 and set ap D z. Let g 2 GA so that o.g/ p 2 and 1 ¤ g p 2 Z.G/. As Œg; a ¤ 1 and Œg; a 2 Z.G/, we have hŒg; ai D hzi, hg; ai0 D hzi and hg; ai is minimal nonabelian. Then replacing g with g i for some i 6 0 .mod p/ (if necessary), we may assume that Œg; a D z.

128

The commutator subgroup of p-groups with the subgroup breadth 1

283

If hgi E G, then A=Z.G/ and .hgiZ.G//=Z.G/ would be two distinct normal subgroups of order p in G=Z.G/ Š S.p 3 /, a contradiction. Hence jG W NG .hgi/j D p. If NA .hgi/ D Z.G/, then NG .hgi/ must cover G=A and NG .hgi/ \ A D Z.G/, contrary to the structure of G=Z.G/. It follows that NG .hgi/ D Ahgi so that a normalizes hgi and so Œg; a D z implies that 1 .hgi/ D hzi. If o.g/ p 3 , then there is an element l 2 hg p i Z.G/ such that o.l/ D p 2 and l p D z 1 . But then .al/p D ap l p D zz 1 D 1 and so o.al/ D p and al 2 A Z.G/, a contradiction. Hence we have o.g/ D p 2 and ha; gi Š Mp3 with ha; gi \ A D hai. On the other hand, 1 .ha; gi/ Š Ep2 and so there are elements of order p in ha; gi A, a contradiction. (ii) Now assume that G=Z.G/ is abelian and so G=Z.G/ Š Ep3 . Hence ˆ.G/ Z.G/ and so for any x; y 2 G we have Œx; yp D Œx p ; y D 1 which implies that G 0 Š Ep2 . If x; y 2 G Z are such that .hx; yiZ.G//=Z.G/ Š Ep2 , then Œx; y ¤ 1 since G has no abelian maximal subgroups. Let a; b 2 G Z.G/ be such that .ha; biZ.G//=Z.G/ Š Ep2 which implies that Œa; b D t1 ¤ 1. Let c be an element in G .ha; biZ.G// so that ha; b; ciZ.G/ D G. Set Œa; c D t2 ¤ 1 and we claim that ht1 ; t2 i D G 0 . Indeed, suppose t2 D Œa; c D t1i for some i 6 0 .mod p/. Then Œa; b i c D Œa; bi Œa; c D t1i t1i D 1 and .ha; b i ciZ.G//=Z.G/ Š Ep2 , a contradiction. j Assume that Œb; c D t1 for some j 6 0 .mod p/. Then we get j j

Œb; aj c D Œb; aj Œb; c D t1 t1 D 1 which together with .hb; aj ciZ.G//=Z.G/ Š Ep2 gives a contradiction. Thus we have 1 ˇ proved that Œb; c D t1˛ t2 , where ˇ 6 0 .mod p/. As haˇ b; b ˛ˇ c; biZ.G/ D G, we have 1 .haˇ b; b ˛ˇ ciZ.G//=Z.G/ Š Ep2 : On the other hand, Œaˇ b; b ˛ˇ

1

c D Œa; bˇ ˛ˇ

1

ˇ

Œa; cˇ Œb; c D t1˛ t2

ˇ

t1˛ t2 D 1;

a contradiction. We have proved that G possesses an abelian maximal subgroup A. Then Z.G/ < A and jA=Z.G/j D p 2 . For each x 2 G A, CA .x/ D Z.G/ and so x p 2 Z.G/. Hence G=Z.G/ is generated by its elements of order p and therefore exp.G=Z.G// D p. In particular, A=Z.G/ Š Ep2 . For a ﬁxed element g 2 G A the map a ! Œa; g

.a 2 A/

284

Groups of prime power order

is a homomorphism from A into A with the kernel CA .g/ D Z.G/ and so A=Z.G/ Š ¹Œa; g j a 2 Aº D G 0 which implies that G 0 Š Ep2 . Our theorem is proved. Theorem 128.3. Let G be a 2-group with the subgroup breadth 1. Then G possesses a normal abelian subgroup of index 4. Proof. Suppose that this result is false. Then we get jG=Z.G/j 24 and so, by Theorem 122.1(a), jG=Z.G/j D 24 . Let A be a maximal normal abelian subgroup of G so that G=A is of order 23 . Since Z.G/ < A and jG=Z.G/j D 24 , we have jG=Aj D 23 and jA=Z.G/j D 2. By Proposition 122.16, the element breadth of G is at most 2. But if a 2 A Z.G/, then CG .a/ D A and so jG W CG .a/j D 23 , a contradiction.

129

On two-generator 2-groups with exactly one maximal subgroup which is not two-generator

By Theorem 71.6, a nonmetacyclic two-generator 2-group G contains an even number of two-generator maximal subgroups. We begin a study of the structure of that G provided it contains exactly two two-generator maximal subgroups (Problem 1969). Theorem 129.1 (Janko). Let G be a two-generator 2-group with exactly one maximal subgroup which is not two-generator. Then the other two maximal subgroups are either both metacyclic (and such groups G are completely determined in 87) or both are nonmetacyclic. Proof. Set 1 D ¹A; B; C º, where d.C / 3. By Schreier’s inequality (Appendix 25), we have d.C / D 3. Since G is nonmetacyclic, G is nonabelian and d.A/ D d.B/ D 2. If both A and B are metacyclic, then such groups G are completely determined in 87. Therefore we may assume that A is nonmetacyclic and so A is nonabelian. In what follows we assume (by way of contradiction) that B is metacyclic. Let H 2 ¹A; B; C º. Then we have ˆ.H / D Ã1 .H / Ã2 .G/. On the other hand, we infer that Ã2 .A/ Ã2 .G/ and Ã2 .A/ E G so that A=Ã2 .A/ is two-generator and nonmetacyclic (Corollary 44.9). Also, B=Ã2 .A/ is metacyclic (but noncyclic) and d.C =Ã2 .A// D 3. Finally, G=Ã2 .G/ is nonmetacyclic (Corollary 44.9) and therefore G=Ã2 .A/ is nonmetacyclic with d.G=Ã2 .A// D 2. It follows that we may assume that Ã2 .A/ D ¹1º. Hence exp.A/ D 4 and so exp.G/ 8. Since B is metacyclic of exponent 8, we get jBj 26 . Suppose that jBj D 26 . Then B \ A is a metacyclic subgroup of exponent 4 and order 25 , a contradiction. Hence B is metacyclic of order 25 and so jGj 26 . If jGj D 24 , then jAj D 23 which together with d.A/ D 2 implies that A is metacyclic, a contradiction. It follows that jGj D 25 or jGj D 26 . (i) First assume that jGj D 25 . If jA0 j D 4, then, by a result of O. Taussky (Lemma 1.6), A is of maximal class and so A is metacyclic, a contradiction. Hence jA0 j D 2 which together with d.A/ D 2 implies that A is the nonmetacyclic minimal nonabelian group of order 24 . Then we get E D 1 .A/ Š E8 and E E G. If G=E Š E4 , then E D ˆ.G/ (since d.G/ D 2) and so E B, a contradiction (since B is metacyclic). Hence G=E Š C4 and let g 2 G A be such that hgi covers G=E. In that case, g 2 2 A E is of order 4 and so o.g/ D 8 and hgi \ E Š C2 . Since CA .E/ D E and

286

Groups of prime power order

G=E Š C4 , it follows that CG .E/ D E and 1 .G/ D E so that G is a faithful and nonsplitting extension of E Š E8 by C4 . As A D Ehg 2 i is a unique maximal subgroup of G containing E, it follows that C does not contain E. Hence C \ E D U Š E4 and C covers G=E so that C =U Š C4 . But 1 .C / D U and so if h 2 C U is such that hhi covers C =U , then o.h/ D 8. It follows that C has a cyclic subgroup of index 2, contrary to d.C / D 3. (ii) It remains to consider the case jGj D 26 . In this case, jˆ.G/j D 24 and ˆ.G/ is a maximal subgroup of A. On the other hand, ˆ.G/ < B and B is metacyclic and so ˆ.G/ is a metacyclic group of exponent 4. Hence we have either ˆ.G/ Š C4 C4 or ˆ.G/ Š H2 D hx; y j x 4 D y 4 D 1; x y D x 1 i. In any case, 1 .ˆ.G// Š E4 and Ã1 .ˆ.G// D 1 .ˆ.G//. Since A does not possess elements of order 8, we have for each x 2 A ˆ.G/, x 2 2 1 .ˆ.G// which gives ˆ.A/ D Ã1 .A/ D 1 .ˆ.G//, contrary to d.A/ D 2. Our theorem is proved.

130

Soft subgroups of p-groups

We present here some work of L. Hethelyi on soft subgroups of p-groups. A subgroup A of a nonabelian p-group G is called a soft subgroup if it satisﬁes CG .A/ D A and jNG .A/ W Aj D p. Note that a nonabelian p-group P is of maximal class if and only if P possesses a soft subgroup of order p 2 . Our induction arguments will be based on the following: Lemma 130.1. Let A be a soft subgroup in a nonabelian p-group G, N D NG .A/, N and let K be the core of A in G, GN D G=K. Then NN is soft in GN and NN D AN Z.G/. 0 Moreover, jN j D p if jG W Aj > p. N AZ. N G/. N Now we see Proof. We may assume that G > N . Obviously, NN D NGN .A/ N is normal in G, N hence trivial. Then jNN W Aj N D p implies the second asthat AN \ Z.G/ N We infer that C N .NN / C N .A/ N N N .A/ N D NN , i.e., NN is sertion: NN D AN Z.G/. G G G N N self-centralizing. Let us denote N2 D NG .N /, so NGN .N / D N2 . For x 2 N2 N , Ax ¤ A is a maximal subgroup in N , hence Ax \ A D Z.N /, jN W Z.N /j D p 2 , jN 0 j D p. Counting the conjugate subgroups Ax , x 2 N2 , we obtain jN2 W N j D p N (noting that N has at most p C 1 abelian maximal subgroups), thus NN is soft in G. Lemma 130.2. Suppose that A is a soft subgroup in a nonabelian p-group G and set jG W Aj D p n , where n 1. Then the subgroups of G containing A form a chain A D N0 < N1 < < Nn1 D M < Nn D G; where NG .Ni1 / D Ni and jNi W Ni1 j D p, i D 1; : : : ; n. Proof (see also Appendix 40, Exercise 72). By induction, Lemma 130.1 yields at once jNi W Ni 1 j D p. Now if A X G, then let Nj be the largest subgroup of our chain contained in X . Suppose Nj < X. Since Nj C1 D NG .Nj / D NX .Nj /, we have Nj C1 X , a contradiction. Hence Nj D X. In what follows we shall use the notation introduced in Lemma 130.2. Moreover, for X Y , core.X W Y / denotes the largest normal subgroup of Y contained in X and X Y stands for the smallest normal subgroup of Y containing X. Corollary 130.3. For any v 2 G M we have hA; Av i D M , where M is the unique maximal subgroup of G containing a soft subgroup A of G.

288

Groups of prime power order

Proof. By Lemma 130.2, we have hA; Av i D Nj for some j n 1. Since A Nj 1 and A Njv , it follows that v normalizes Nj . But from NG .Nj / D Nj C1 we infer v 2 Nj C1 and so Nj D M . Corollary 130.4. There exists a 2 A such that CG .a/ D A, where A is a soft subgroup of G. T Proof. As A D CG .A/ D a2A CG .a/, the assertion follows from Lemma 130.2. Proposition 130.5. Let M be the unique maximal subgroup of a p-group G containing a soft subgroup A of G. For the upper central series of M we have: (i) Zi .M / D core.Ni 1 W G/ D Ni 1 \ Niv1 for any v 2 G M , i D 1; : : : ; n. (ii) Zi C1 .M /=Zi .M / is elementary abelian of order p 2 , i D 1; : : : ; n 1. (iii) Zi .G/ Zi .M /, i D 1; : : : ; n. Proof. (i) For i D 1 we have N0 D A and let v 2 G M be ﬁxed. By Corollary 130.3, hA; Av i D M . Now, Z.M / D CM .A/ \ CM .Av / D A \ Av . Finally, Lemma 130.1 proves (i) by induction. v (ii) We have Zi .M / D Ni 1 \ Ni1 Ni 1 \ Niv Ni \ Niv D Zi C1 .M /. The index in both steps is 1 or p, so jZi C1 .M /=Zi .M /j p 2 : If the index is p 2 , inserting also Ni \ Niv1 in the middle, we see that the factor is elementary abelian. Indeed, .Ni 1 \ Niv /=Zi .M / and .Ni \ Niv1 /=Zi .M / are two distinct subgroups of order p in Zi C1 .M /=Zi .M /. (iii) For i D 1 we have Z1 .G/ core.A W G/ D Z1 .M /. Lemma 130.1 is again applicable and gives our result. Now, as a corollary to Proposition 130.5 (i) and (iii), we obtain: Theorem 130.6. Let A be a soft subgroup of a p-group G with jG W Aj D p n , where n 1. Then there is a unique maximal subgroup M of G containing A. The nilpotence class of M is n and the nilpotence class of G is at least nC1. Since A is also soft in Ni , i D 1; : : : ; n 1, every above statement is also true for Ni . Corollary 130.7. The nilpotence class of Ni is i C 1 for i D 0; : : : ; n 1. Corollary 130.8. We have jA W core.A W G/j p n1 . Proof. Making use of Proposition 130.5 (i) and (ii), we obtain jA W core.A W G/j D p nC1 jM W core.A W G/j D p nC1 jZn .M / W Z1 .M /j p nC1 p 2.n1/ D p n1 :

130

Soft subgroups of p-groups

289

Proposition 130.9. Let M be the unique maximal subgroup of a p-group G containing a soft subgroup A of G. For the lower central series of M we have: 0 0 0v W G/ D hNnC1i ; NnC1i i for any v 2 G M , where (i) Ki .M / D n:cl:.NnC1i i D 2; : : : ; n;

(ii) Ki .M /=Ki C1 .M / is elementary abelian of order p 2 , i D 2; : : : ; n. Proof. (i) Let i D n and we may assume that n > 1. Since Kn .M / is normal in G, Kn .M / Z1 .M / A and cl.M=Kn .M // D n1, Lemma 130.1 implies that the factor group N1 =Kn .M / is soft in G=Kn .M /. Indeed, if N1 =Kn .M / is nonabelian, then AKn .M / would be soft in G=Kn .M / and then cl.M=Kn .M // D n. Hence we obtain that N10 Kn .M /. Conversely, for T D hN10 ; N10v i we have Z2 .M /=T D .N1 \ N1v /=T Z.hN1 ; N1v i=T / D Z.M=T /; hence cl.M=T / < cl.M /, so T Kn .M / also holds and therefore Kn .M / D hN10 ; N10v i D n:cl:.N10 W G/: Now, for i < n, (i) follows by backwards induction. (ii) Suppose that i D n. Since we may assume that n > 1, Lemma 130.1 implies jKn .M /j D jhN10 ; N10v ij jN10 j2 p 2 : The result then follows by induction. Proposition 130.10. Suppose that M is the unique maximal subgroup of a p-group G containing a soft subgroup A of G. For the series of commutator subgroups Ni0 we have Ni0 A D Ni1 , i D 1; : : : ; n and so N10 < N20 < < Nn0 D G 0 and so, in particular, G 0A D M . Proof. Clearly, we have Ni0 A Ni 1 and Ni0 A E Ni . Now Lemma 130.2 yields that Ni0 A D Ni 1 . Theorem 130.11. Suppose that A and B are soft subgroups of a p-group G contained in the same maximal subgroup M of G. Then B is conjugate to A. Proof. Suppose that jG W Aj D p 2 . Then jG=Z.M /j D p 3 , A=Z.M / 6 Z.G=Z.M //, B=Z.M / 6 Z.G=Z.M //, and jA=Z.M /j D jB=Z.M /j D p. However, we obtain that jM=Z.M /j D p 2 and so A=Z.M / is conjugate to B=Z.M /. Thus, for jG W Aj D p 2 , A is conjugate to B. Suppose that jG W Aj > p 2 . Set N1 D NG .A/ and GN D G=Z.M / with bar conN D NG .B/ are soft subgroups of GN contained in MN (by vention. Then NN1 and NGN .B/ N is conjugate to NN1 and Proposition 130.5(i) and Lemma 130.1). By induction, NGN .B/ so NG .B/ is conjugate to N1 . Thus we can assume B < N1 . However, B is conjugate to A in N2 D NG .N1 / by the previous paragraph.

290

Groups of prime power order

Theorem 130.12. Suppose that A is a maximal normal abelian subgroup of a nonabelian p-group G. Assume that G=A is cyclic and that G possesses a soft subgroup B distinct from A. Then G D AB. Proof. Suppose that this theorem is false. Since A ¤ B, neither A nor B is of index p in G. As AB < G and AB E G, we get jG W .AB/j D p and so M D AB is the unique maximal subgroup of G containing B. We have A\B D Z.M / E G. Now set N1 D NG .B/ so that jN1 W Bj D p and N1 M . Let x 2 G N1 be such that x normalizes N1 . Then B\B x D Z.N1 / Z.M / and jB W Z.N1 /j D p. Set R D AZ.N1 /; then R E G and G=R Š Cp2 . Since CG .B/ D B, we have for each b 2 B Z.N1 /, CG .b/ D B. Let jBj D p k and jM j D p n so that jB Z.N1 /j D p k p k1 D p k1 .p 1/ and jM Rj D p n p n1 D p n1 .p 1/: Note that all conjugates in G of the subset B Z.N1 / lie in M R and there are exactly jG W NG .B/j D jG W N1 j D p nC1 =p kC1 D p nk such distinct conjugates. For any g 2 G N1 we have .B Z.N1 // \ .B Z.N1 //g D ¿ since for each b 2 B Z.N1 /, CG .b/ D B. Hence the number of elements contained in the union of all these conjugates is p nk p k1 .p 1/ D p n1 .p 1/, which is the number of all elements contained in M R. It follows that each element y 2 M R is conjugate in G to an element b 2 B Z.N1 / and so CG .y/ M . On the other hand, G=R Š Cp2 and so if v 2 G M , then v p 2 M R but CG .v p / 6 M , a contradiction. Our theorem is proved. Theorem 130.13. Suppose that A is a maximal normal abelian subgroup of a nonabelian p-group G. Assume that G=A is cyclic and that G possesses a soft subgroup B distinct from A. Then we have: (i) d.G=Z.G// D 2, and if jG W Aj D p ˛ , then G=.G 0 Z.G// is of type .p ˛ ; p/. (ii) G is a CF-group, i.e., the index of any term of the lower central series of G beyond G 0 in its predecessor is at most p. Proof. (i) By Theorem 130.12, we have G D AB and so Z.G/ D A \ B. Let M be the unique maximal subgroup of G containing B. We have G 0 A \ M . On the other hand, G 0 B is a normal subgroup of G containing B and so G 0 B D M which implies A \ M D G 0 Z.G/. Since G=.G 0 Z.G// D A=.G 0 Z.G// M=.G 0 Z.G//; we infer that G=.G 0 Z.G// is of type .p ˛ ; p/. Take an element b 2 B Z.G/ so that hbi covers B=Z.G/ and take an element a 2 A M . Then hbi covers G=A and G D ha; bi.A \ M / D ha; biG 0 Z.G/ D ha; biZ.G/ which implies d.G=Z.G// D 2.

130

Soft subgroups of p-groups

291

(ii) Since ap 2 G 0 Z.G/, we have Œap ; b 2 K3 .G/. Now Œap ; b D ap .b 1 ap b/ D ap .ap /b D .a1 ab /p D Œa; bp ; so Œa; bp 2 K3 .G/. Since G 0 D hŒa; b; K3 .G/i, jG 0 W K3 .G/j D p. It follows by an easy induction that jKi .G/ W KiC1 .G/j D p for Ki .G/ ¤ ¹1º: in fact, if i 3 and Ki 1 .G/ D hu; Ki .G/i with up 2 Ki .G/, then Ki .G/ D hŒa; u; Œb; u; Ki C1 .G/i and Œa; u D 1 as u 2 G 0 A and a 2 A. Also, we obtain that Œb; up 2 Ki C1 .G/ since Œb; up D .ub /p up D Œb; up 2 Ki C1 .G/. Indeed, Œb; up D .b 1 up b/up D .b 1 u1 b/p up D ..u1 /b u/p D Œb; up since u 2 G 0 ; .u1 /b 2 G 0 and G 0 A. Thus G is a CF-group. Theorem 130.14. Suppose that G is a p-group of class greater than 2 and A is an abelian subgroup of G of index p. Then the following are equivalent: (a) Every maximal abelian subgroup of G is soft. (b) G has a soft subgroup distinct from A. (c) G=Z.G/ is a p-group of maximal class. (d) jZ2 .G/ W Z.G/j D p. Proof. The implication (a) ) (b) is trivial. (b) ) (c). By Theorem 130.13, G is a CF-group and G=.G 0 Z.G// is of type .p; p/. N GN 0 is of type .p; p/ and GN is a CF-group, so GN is of maximal If GN D G=Z.G/, then G= class. The implication (c) ) (d) is trivial. (d) ) (a). Let B ¤ A be an abelian maximal subgroup of G so that AB D G and A \ B D Z.G/. Set N D NG .B/ so that N D .N \ A/B. We have ŒN \ A; B B \ A D Z.G/ so N \ A Z2 .G/. Hence jN W Bj D j.N \ A/ W Z.G/j D p and B is soft. Theorem 130.15. Suppose that G is a nonabelian p-group which possesses a maximal normal abelian subgroup A such that G=A is cyclic. If jZ.G/ \ G 0 j D p, then G has a soft subgroup distinct from A. Proof. Note that G D Ahbi for some element b 2 G A so that CA .b/ D Z.G/ and CG .b/ D Z.G/hbi D B. We prove that B is soft in G. First of all, B is abelian and self-centralizing and set N D NG .B/. Further, we have jN W Bj D j.N \ A/ W Z.G/j and j.N \ A/ W Z.G/j D jŒN \ A; hbij D p by Lemma 121.2.

131

p-groups with a 2-uniserial subgroup of order p

Let us say that a subgroup H of a p-group G is n-uniserial (n 1) if for each index i D 1; 2; : : : ; n there is a unique subgroup Ki such that H Ki and jKi W H j D p i . In case the subgroups of G containing H form a chain, we say that H is uniserially embedded in G. We shall classify here all p-groups G which possess a subgroup of order p which is 2-uniserial and so extend some results for p > 2 of N. Blackburn and L. Hethelyi to the case of 2-groups. Theorem 131.1 (Janko). Let G be a p-group and let L be a 2-uniserial subgroup of order p. Then L is uniserially embedded in G (and so G is one of the groups in Exercise A40.77 of Appendix 40) or p D 2 and m

G D ha; t; x j a2 D t 2 D Œa; t D 1; u D a2 vDa

2m2

m1

;

; x 2 D tv; ax D a1 ; t x D tui;

where m 3;

jGj D 2mC2 ;

Z.G/ D htvi Š C4 ;

ˆ.G/ D hti ha2 i Š C2 C2m1 ;

G 0 D ha2 i Š C2m1

and here L D hti Š C2 is 2-uniserial but not 3-uniserial in G (and so ht i is not uniserially embedded in G). Proof. Let L be a 2-uniserial subgroup of order p in G. (i) First assume p > 2. Set N D NG .L/ D CG .L/. If jN=Lj D p, then N is a self-centralizing abelian subgroup of order p 2 and jNG .N / W N j D p. This means that N is a soft subgroup in G and so Lemma 130.2 implies that all subgroups of G containing N form a chain. Hence in this case L is uniserially embedded in G. We may assume jN=Lj > p. Since there is only one subgroup of order p in N=L, it follows that N=L is cyclic. If N D G, then L is uniserially embedded in G and we are done. Thus we may assume N < G. In this case, L is not characteristic in N and so N is not cyclic. Set R D 1 .N / Š Ep2 and N1 D NG .R/ so that N < N1 since R is characteristic in N . As there is only one subgroup of order p in N1 =R, we see that N1 =R is cyclic. Since R < N , we infer that R D 1 .N1 / is characteristic in N1 , and we conclude

131 p-groups with a 2-uniserial subgroup of order p

293

that N1 D G. In that case, 1 .G/ D R Š Ep2 and G possesses a cyclic subgroup of index p which does not contain L. Thus L is uniserially embedded in G and we are done in case p > 2. (ii) Assume now p D 2 and set L D hti. Since the involution t is contained in exactly one subgroup of order 4 in G, Theorem A.14.1 implies that we have one of the following possibilities: (a) G is cyclic. (b) CG .t/ D ht i C2m , m 1. (c) CG .t/ D ht i Q2m , m 3. (d) G D ha; bi with deﬁning relations m

a2 D b 4 D 1; t 2 D 1;

m 2;

u D a2

ab D a1 t ;

m1

;

b 2 D ut;

D 0; 1:

Here Z.G/ D ht; ui Š E4 , G=hti Š Q2mC1 , G 0 D ha2 t i, and G is metacyclic if and only if D 0. In case (a), ht i is uniserially embedded in G. In cases (c) and (d), CG .t/=ht i is a generalized quaternion group and so there are two distinct cyclic subgroups S=hti and T =hti of order 4 contained in CG .t/=ht i. But then S and T are two distinct subgroups of order 8 which contain hti and so ht i is not 2-uniserial in G, a contradiction. It remains to consider the above case (b), where CG .t/ D hti C with C D hai and o.a/ D 2m , m 1. If CG .t/ D G, then hti is uniserially embedded in G. From now on we assume CG .t / < G and so G is nonabelian. We may apply Theorem 48.1 in which such groups G are classiﬁed. In cases (a) and (b) of that theorem, G is either of maximal class or G Š M2n , n 4, and in these cases hti is uniserially embedded in G. In cases (d–h) of that theorem, we have m 2 and t is contained in a subgroup D Š D8 . On the other hand, t is also contained in ht i 2 .hai/ which is an abelian subgroup of type .2; 4/. Hence in all these cases ht i is not 2-uniserial in G, a contradiction. We have proved that we must be in case (c) of Theorem 48.1. From here we know that T D CG .t/ D hti hai is of index 2 in G, o.a/ D 2m with m 3 and t 2 ˆ.G/ so that ˆ.G/ D ht i ha2 i, where ha2 i D ˆ.T /. Hence there is an element m1 x 2 G T such that x 2 2 ht; a2 i ha2 i. Set u D a2 and 2 .hai/ D hvi Š C4 2 so that v D u, u 2 Z.G/ and 1 .T / D ht; ui Š E4 . Also, we have t x D tu and so x 2 62 ht; ui and ˆ.G/ D ht i ha2 i D hx 2 ; a2 i. Obviously, x does not centralize a2 . Indeed, if Œx; a2 D 1, then x centralizes ˆ.G/ D hx 2 ; a2 i and so x would centralize t, a contradiction. Thus we have proved that x induces on T an involutory automorphism such that x t D tu and x does not centralize ha2 i. In Theorem 34.8(d) the structure of Aut.T / is determined. In particular, we get ax D a1 u , where D 0; 1. If D 1, then we

294

Groups of prime power order

replace a with a0 D at and obtain .a0 /x D .at/x D ax t x D .a1 u/ .tu/ D a1 t D .at/1 D .a0 /1 : Hence writing again a instead of a0 , we may assume from the start that ax D a1 . We see that CT .x/ D ht vi D Z.G/ Š C4 since .tv/2 D u. Also, ha2 i Z.G/ and o.x 2 / 4 and so x 2 D .tv/˙1 D t v ˙1 . Replacing v with v 1 if necessary, we may assume that x 2 D t v and so the structure of G is uniquely determined. Let xt r as (r; s are integers) be any element in G T . Then we compute .xt r as /2 D xt r as xt r as D x 2 .t r as /x t r as D t v .tu/r as t r as D tvur D .tv/˙1 ; and see that all elements in G T are of order 8. It follows that ht ihvi is the only subgroup of order 8 containing ht i which shows that ht i is 2-uniserial in G. But G is not m3 3-uniserial since ht i ha2 i and ht; xi Š M24 are two distinct subgroups of or4 der 2 containing ht i, and the theorem is proved.

132

On centralizers of elements in p-groups

We present here some results of N. Blackburn [Bla12] about the centralizers of elements in p-groups. Suppose that K is a p-group and s is an element of order p in the center of K. We are concerned with the question: What are the p-groups G for which K G and K D CG .s/? In particular, is there a bound on the order or class of G? In fact, we show by Theorem 132.1 that these two last questions are the same. Theorem 132.1. Let G be a p-group of class k. Suppose that s 2 G and jCG .s/j D p r . Then jGj p rk . Proof. Let G D K1 .G/ > K2 .G/ > > Kk .G/ > KkC1 .G/ D ¹1º be the lower central series of G. For 1 i k, set H D hs; Ki .G/i. Every conjugate of s in H is of the form s h D sŒs; h (h 2 H ) so the number of conjugates of s in H is at most jH 0 j. Thus jH 0 j jH W CH .s/j jH j=p r jKi .G/j=p r : But we see that Ki .G/=KiC1 .G/ Z.G=Ki C1 .G// and H=Ki .G/ is cyclic and so H=Ki C1 .G/ is abelian. Then we get H 0 Ki C1 .G/ and so, by the above inequality, jKi C1 .G/j jKi .G/j=p r which implies jKi .G/ W Ki C1 .G/j p r for i D 1; : : : ; k. Hence jGj p rk . If s is an element of order p in a p-group G, then set C D CG .s/. Then, as a rule, we have inﬁnitely many p-groups G with such a ﬁxed centralizer C . For example, if C Š Ep2 , then there are inﬁnitely many p-groups of maximal class which all have such a small centralizer of s. However, there are examples for the structure of C which is small in a rather strong sense, where there exist only ﬁnitely many p-groups G with such a ﬁxed centralizer C . This will be seen in the next two theorems. Theorem 132.2. Let s be an element of order p in a p-group G and set C D CG .s/. m1 If there exists a positive integer m such that s D x p for some x 2 C and if C has mC1 no elementary abelian subgroup of order p , then jG W C j < p m . Proof. If this theorem is false, then there is a subgroup H of G such that H C and jH W C j D p m . Thus s has p m conjugates s1 ; s2 ; : : : ; spm in H . If si D s ui (ui 2 H ),

296

Groups of prime power order p m1

put xi D x ui ; thus x1 ; : : : ; xpm are distinct conjugates of x in H since xi D si . If M is a maximal subgroup of H containing C , then xi 2 M as x 2 M and M is normal in H . Since jM W C j D p m1 , it follows that p m1

si D xi

2 C;

i D 1; : : : ; p m :

Indeed, if si 62 C , then C \ hxi i D ¹1º and so, by the product formula, we obtain jM j jC jjhxi ij D jC jp m , a contradiction (because jH W M j D p and jH j D jC jp m ). But then si ; sj commute for any i; j 2 ¹1; 2; : : : ; p m º. Hence hs1 ; : : : ; spm i is elementary abelian of order p mC1 , a contradiction. Theorem 132.3. Let s be an element of order p in a p-group G and set C D CG .s/. Suppose that s is the p-th power of some element x 2 C and jC j D p l , where l 3 and 2l p C 4. Then jG W C j < p l2 . Proof. If l D 3, then, by the structure of groups of order p 3 , hsi D 1 .ˆ.C // and so hsi is characteristic in C . But then NG .C / centralizes hsi which implies G D C and we are done. Now we assume l > 3 so that p is odd and 2l p C 3. If our assertion is false, there exists a subgroup H of G such that H C and jH W C j D p l2 . Let M be a maximal subgroup of H such that M C . Thus jM j D p 2l3 p p and so M is regular (Theorem 7.1(b)). It follows that the elements of M of order at most p form a characteristic subgroup E of M (Theorem 7.2(b)). We have M E H and s 2 E. As jH W C j D p l2 , s has p l2 conjugates s1 ; : : : ; spl2 in H . If si D s ui (ui 2 H ), p p put xi D x ui . Since x 2 C , we have xi 2 M . If i ¤ j , we get xi ¤ xj . By Theo1 l2 rem 7.2(a), we infer that xi xj 62 E. Thus jM W Ej > p and so jM W Ej p l1 . Similarly, jEj > p l2 since si 2 E. Hence jEj p l1 which implies jM j p 2l2 , a contradiction. Our theorem is proved. In his paper [Bla12], N. Blackburn has studied 2-groups G which possess an involution s such that CG .s/ is the nonmetacyclic minimal nonabelian group of order 16. We shall describe these groups G more closely by using the results from 51 and 130. Theorem 132.4. Suppose that G is a 2-group of order 25 which possesses an involution s such that C D CG .s/ is the nonmetacyclic minimal nonabelian group of order 24 . Then E D 1 .C / is a self-centralizing elementary abelian group of order 8. Also, E D ˆ.G/ and so G is isomorphic to a unique group of order 25 given in Theorem 51.3. Proof. We may set C D CG .s/ D hr; t j r 4 D t 2 D 1; Œr; t D z; z 2 D Œz; r D Œz; t D 1i; where

hzi D C 0 ;

Z.C / D ˆ.C / D hr 2 ; zi

and E D 1 .C / D hr 2 ; z; t i Š E8

with CG .E/ D E:

132

On centralizers of elements in p-groups

297

Since G centralizes hzi D C 0 , it follows that s 2 ¹r 2 ; r 2 zº and G (normalizing Z.C /) fuses r 2 and r 2 z. We may set s D r 2 and for any v 2 G C , s v D sz. Note that for any t 0 2 E Z.C / we have .rt 0 /2 D r 2 .t 0 /2 Œt 0 ; r D sz and so we may put r v D rt 0 with some t 0 2 E Z.C /. It follows that t 0 2 G 0 and so E ˆ.G/ gives E D ˆ.G/. Then Theorem 51.3 applies and such a group G is uniquely determined. Theorem 132.5. Let G be a 2-group of order > 25 which possesses an involution s such that C D CG .s/ is the nonmetacyclic minimal nonabelian group of order 24 . Then we have jGj D 2nC3 , n 3, jG 0 j D 2n , Z.G/ D hzi D C 0 is of order 2, G is of class n C 1 (coclass 2) and G is a CF-group (i.e., the index of any term of the lower central series of G beyond G 0 D K2 .G/ in its predecessor is at most 2). Moreover, Z.C / D hz; si Š E4 , G 0 \C D W ¤ Z.C / is a normal four-subgroup of G, K3 .G/ is abelian of order 2n1 and rank 2, s inverts K3 .G/, L D hsiK3 .G/ is normal in G, G=L Š D8 , s 62 G 0 , ˆ.G/ D hsiG 0 , and so d.G/ D 2. Finally, E D 1 .C / is a selfcentralizing elementary abelian subgroup of order 8, E < ˆ.G/, and we have one of the following possibilities: (a) n D 3, E is normal in G and G is isomorphic to one of the four distinct groups of order 26 given in Theorem 51.4. (b) n > 3, E is not normal in G, G has no normal elementary abelian subgroup of order 8, W is the unique normal four-subgroup in G, and G is one of the groups described in (c) and (d) of Theorem 51.6. Proof. We may set C D CG .s/ D hr; t j r 4 D t 2 D 1; Œr; t D z; z 2 D Œz; r D Œz; t D 1i; where hzi D C 0 ;

Z.C / D ˆ.C / D hr 2 ; zi

and E D 1 .C / D hr 2 ; z; t i Š E8

with CG .E/ D E:

C 0 , it follows that s

Since NG .C / > C centralizes hzi D 2 ¹r 2 ; r 2 zº and NG .C / (normalizing Z.C /) fuses r 2 and r 2 z. This gives jNG .C / W C j D 2 and we may assume s D r 2 (noting that .rt/2 D r 2 t 2 Œt; r D r 2 z). Also, Z.G/ Z.C / and so Z.G/ D hzi is of order 2. Set A D hr; zi D hrihzi Š C4 C2 so that CG .r/ C and therefore CG .r/ D A, which implies that A is a self-centralizing abelian subgroup of type .4; 2/. As NG .A/ centralizes hsi D Ã1 .A/, it follows that NG .A/ D C which together with jC W Aj D 2 gives that A is a soft subgroup in G. We may apply all results from 130 about soft subgroups. Set jG W Aj D 2n so that jGj D 2nC3 with n 3. By Lemma 130.2, all subgroups of G containing A form a chain. As CG .r/ D A, the conjugacy class of r has 2n elements and so jG 0 j 2n . However, if jG 0 j > 2n , then jG 0 j D 2nC1 and so jG=G 0 j D 4.

298

Groups of prime power order

By a result of O. Taussky (Lemma 1.6), G would be of maximal class, a contradiction (since G has an abelian subgroup A of type .4; 2/). We have proved that jG 0 j D 2n . Let M be a unique maximal subgroup of G containing the soft subgroup A. By Theorem 130.6, the nilpotence class of M is equal to n and the class of G is at least n C 1. However, if the class of G is greater than n C 1, then the class of G is n C 2 and so G would be of maximal class, a contradiction. We have proved that G is of class n C 1. But jG=G 0 j D 23 and so the index of any term of the lower central series of G beyond G 0 in its predecessor is at most 2 which implies that G is a CF-group. By Proposition 130.10, we have G 0 A D M . On the other hand, we have hzi D C 0 and so hzi G 0 \ A. But jM j D 2nC2 and so, by the product formula, G 0 \ A D hzi. In particular, s D r 2 62 G 0 and so hsiG 0 D ˆ.G/, jG W ˆ.G/j D 4, and d.G/ D 2. The subgroup A is also a soft subgroup in M and let K ( C ) be a unique maximal subgroup of M containing A. By Proposition 130.10 (applied to M ), K D AM 0 with A \ M 0 D hzi, M 0 is normal in G, and jG 0 W M 0 j D 2. On the other hand, we have K3 .G/ D ŒG; G 0 M 0 and so the fact that G is a CF-group gives K3 .G/ D M 0 . Since s D r 2 2 ˆ.M / and M 0 ˆ.M /, we get ˆ.M / D hsiM 0 ;

where jˆ.M / W M 0 j D 2:

Note that ˆ.M / is normal in G and so ˆ.M /M 0 is a normal subset of G with exactly jM 0 j D 2n1 elements. Hence the involution s and all its 2n1 D jG W C j conjugates must be contained in ˆ.M / M 0 . It follows that all elements in sM 0 D sK3 .G/ are involutions. This implies that K3 .G/ is abelian. Also, L D hsiK3 .G/ D ˆ.M / and so L is normal in the group G. We have jLj D 2n so that G=L is nonabelian of order 8 since G 0 6 K > L. But M=ˆ.M / D M=L Š E4 and so G=L Š D8 . As AM 0 D K with A\M 0 D hzi and A < C K with jC W Aj D 2, it follows that C \ M 0 D W is of order 4. We know that s inverts W . If W Š C4 , then W hsi Š D8 , contrary to the fact that C is minimal nonabelian. Hence we infer that W Š E4 and so E D hsi W D 1 .C / Š E8 . Since CM 0 .s/ D W and s inverts M 0 , we obtain W D 1 .M 0 /. It follows that M 0 is abelian of rank 2 and W is a normal four-subgroup in G with W ¤ Z.C / D hz; si since s 62 G 0 . Also, s 2 ˆ.G/ and W M 0 ˆ.G/ gives E D hsi W D 1 .C / ˆ.G/. But G 0 > M 0 and so G 0 6 E and therefore E < ˆ.G/. Also, E is self-centralizing in G and so we are in a position to apply the results from 51. If n D 3, then M 0 D K3 .G/ is of order 4 and so K3 .G/ D W . All four conjugates of s in G lie in sW and so E D hsi W is normal in G. Theorem 51.4 is applicable and we get exactly four groups of order 64. If n > 3, then M 0 D K3 .G/ is of order > 4 and so all 2n1 conjugates of s in G (forming the set sK3 .G/) do not lie in E. This gives that E is not normal in G. In this case, Theorem 51.6 is applicable. Our group G has no normal elementary abelian subgroup of order 8 and W is the unique normal four-subgroup in G. The group G is then described in parts (c) and (d) of Theorem 51.6.

132

On centralizers of elements in p-groups

299

Exercise. Suppose that x is a noncentral element of a p-group G, o.x/ D p, such that C D CG .x/ is minimal nonabelian such that C =1 .G/ is nonidentity cyclic. Prove that (a) 1 .C / Š Ep3 . (b) jZ.G/j D p. (c) If x 2 R < 1 .C / is of order p 2 , then we have jNG .R/ W C j D p. In particular, jNG .C / W C j D p so that NG .C / D NG .R/. Study the structure of NG .C / in detail.

133

Class and breadth of a p-group

Let G be a p-group and let x be an element of G. We recall that the breadth of x is the nonnegative integer b.x/ such that p b.x/ D jG W CG .x/j: The breadth of the group G, denoted by b.G/ or just b, is the maximum of the breadths of its elements. Thus p b is the size of the largest of the conjugacy classes of G. It was shown by C. R. Leedham-Green, P. M. Neumann, and J. Wiegold in [LGNW] that there is a close relationship between the breadth b of a p-group G and its nilpotence class c D cl.G/. In particular, we present here a (technically somewhat simpler) proof of their result: Theorem 133.1. Let G be a p-group of the breadth b and class c. Then c<

p b C 1: p1

Since p=.p 1/ D 1 C 1=.p 1/ 2, we get a bound c < 2b C 1 or c 2b which is independent of p. A slightly stronger result was proved by M. Cartwright (Bull. London Math. Soc. 19 (1987), 425–430) in case of 2-groups, where he has shown that c 53 b C 1. The Class-Breadth Conjecture. For any p-group G we have c b C 1. This bound is attained by any group G of order p n (n 3) and maximal class n 1 since such a group always possesses an element x such that jCG .x/j D p 2 and so we have b D b.G/ D b.x/ D n 2. However, the class-breadth conjecture was shown in [ENOB] to be false at least for p D 2. But the conjecture remains open for p-groups of odd order. The class-breadth conjecture has been established under various conditions. For example, if b < p, then c bC1 (Theorem 133.5) and if G is metabelian, then c bC1 (Theorem 133.6). In order to prove Theorem 133.1, we have to prove ﬁrst the following two technical lemmas. If G D K1 .G/ > K2 .G/ > > Kc .G/ > KcC1 .G/ D ¹1º

133

Class and breadth of a p-group

301

is the lower central series of a p-group G of class c, then we deﬁne the two-step centralizer Ci (i D 1; 2; : : : ; c 1) by Ci D CG .Ki .G/=Ki C2 .G// D ¹g 2 G j Œg; x 2 Ki C2 .G/ for all x 2 Ki .G/º: It is clear that Ci is normal in G, Ci Ki C2 .G/ and Ci is a proper subgroup of G for all i D 1; 2; : : : ; c 1. Lemma 133.2. Let G be a p-group and x 2 G. Then b.x/ s.x/, where s.x/ denotes the number of indices i (i D 1; 2; : : : ; c 1) for which x 62 Ci . Proof. We use induction on the length c of the lower central series of G. The lemma is certainly correct for p-groups with the lower central series of length 1 (in that case, G is abelian), for then b.x/ and s.x/ are both 0 whatever x may be. Let us suppose that the lemma is known to be right for p-groups with the lower central series of length c 1. We set S.x/ D ¹i j 1 i c 1 and x 62 Ci º so that s.x/ D jS.x/j. Put GN D G=Kc .G/ and embellish other symbols with a bar to N or simply to denote the cordenote images under canonic epimorphism of G onto G, N responding notion in G. Notice that if 1 i c 2, then Ci D CGN .Ki .G/=.Ki C2 .G// D Ci =Kc .G/ so that our convention is unambiguous; and if 1 i c 2, then xN 2 Ci if and only N x/ if x 2 Ci . An instance for our inductive hypothesis is that b. N sN .x/. N There are two cases to consider. Suppose ﬁrst that x 2 Cc1 which means that x centralizes Kc1 .G/. Then we have N x/, N x/ S.x/ D S. N so that s.x/ D sN .x/; N and since certainly b.x/ b. N the inductive hypothesis gives N x/ b.x/ b. N sN .x/ N D s.x/; as required. The other possibility is x 62 Cc1 , i.e., x does not centralize Kc1 .G/. In this case, N x/[¹c1º, N S.x/ D S. N and so s.x/ D sN .x/C1. N On the other hand, as Kc1 .G/ Z.G/ and therefore Kc1 .G/ is centralized by x, N we have (noting that Kc1 .G/ CGN .x/ N and CG .x/ CGN .x/) N N jGN W .CG .x/Kc1 .G//j D jG W .CG .x/Kc1 .G//j < jG W CG .x/j; jGN W CGN .x/j the last inequality is strict because Kc1 .G/ 6 CG .x/ (by our assumption). This gives N x/ N x/ b. N < b.x/ and so b. N b.x/ 1, and therefore by the above, N x/ b.x/ b. N C 1 sN .x/ N C 1 D s.x/: This case also yields the desired inequality and induction completes the proof.

302

Groups of prime power order

Lemma 133.3. Let G be a p-group of class c 2. Then there exists an element g 2 G which lies in t.g/ two-step centralizers Ci (1 i c 1) and t.g/ < p1 .c 1/. Proof.PFor x 2 G, let ti .x/ D 1 if x 2 Ci and ti .x/ D 0 if x 62 Ci , 1 i c 1. Thus c1 i D1 ti .x/ D t .x/ is the number of two-step centralizers Ci in which x lies. We estimate the average of ti .x/ over G: 1 1 1 X jCi j ; ti .x/ D jGj jGj p x2G

the latter inequality arises from the fact that Ci is a proper subgroup of G and hence has index at least p. It follows that c1 1 XX 1 X c1 : ti .x/ D t.x/ jGj jGj p i D1 x2G

x2G

In this inequality the summands t .x/ are nonnegative, and one of them, t.1/ D c 1, c1 exceeds c1 p . Therefore at least one summand t.g/ for some g 2 G is less than p , and this is what the lemma states. Proof of Theorem 133.1. We combine both Lemmas 133.2 and 133.3 together with the notation introduced there. Since s.g/ C t .g/ D c 1, we get b b.g/ s.g/ D .c 1/ t.g/ > .c 1/ 1 p1 : D .c 1/ 1 D .c 1/ p p This gives c <

p p1 b

1 .c 1/ p

C 1 and we are done.

Lemma 133.4. If Ci ,S i D 1; 2; : : : ; c 1, are the two-step centralizers in a p-group G of class c 2 and if c1 iD1 Ci ¤ G, then c bC1 and so the class-breadth conjecture holds. S Proof. If x 2 G, x 62 Ci , then we have s.x/ D c 1 and, using Lemma 133.2, we get c 1 b.x/ b and c b C 1. Theorem 133.5. Let G be a p-group with b D b.G/ < p. Then c D cl.G/ b C 1 and so the class-breadth conjecture holds. Proof. From Theorem 133.1 follows 1 b p bC1D 1C bC1DbC1C c< p1 p1 p1 p1 bC1C D b C 2; p1 and since both c and b are integers, we get c b C 1.

133

Class and breadth of a p-group

303

Theorem 133.6. Let G be a p-group such that Kp .G/ is contained in Z.G 0 /. Then the class-breadth conjecture holds. In particular, if G is metabelian, then the class-breadth conjecture holds. Proof. Let Ci , 1 i c 1, be the two-step centralizers of the lower central series of length c of G. That is, Ci D CG .Ki .G/=Ki C2 .G//. We note that G cannot be covered by less than p C1 proper subgroups. Indeed, considering a normal subgroup G0 of G such that G=G0 Š Ep2 , we see that Ep2 is partitioned by exactly p C 1 subgroups of order p. We may assume that c p C 2. Indeed, if c p C 1, then the two-step centralizers C1 ; C2 ; : : : ; Cc1 contain at most p proper subgroups of G and so they cannot cover G. But then, by Lemma 133.4, the class-breadth conjecture holds. We ﬁrst use the fact that Kp .G/ Z.K2 .G// to show Ci CiC1 whenever we have p i < c 1. To do this, suppose that g 2 Ci . Now Ki C1 .G/ is generated by the set of commutators ¹Œx; a j x 2 Ki .G/; a 2 Gº, and so we need only to prove that ŒŒx; a; g 2 KiC3 .G/ whenever x 2 Ki .G/ and a 2 G. From the Hall–Witt identity (Introduction, Exercise 13) Œx; y 1 ; zy Œy; z 1 ; xz Œz; x 1 ; yx D 1 we get by setting y D a1 and z D g 1

Œx; a; ga Œa1 ; g 1 ; xg Œg; x 1 ; a1 x D 1: As x 2 Ki .G/ Kp .G/ while Œa1 ; g 1 2 K2 .G/, we have Œa1 ; g 1 ; x D 1 by our assumption. Hence 1

Œx; a; ga Œg; x 1 ; a1 x D 1: Since g 2 Ci and x 1 2 Ki .G/, it follows that Œg; x 1 2 Ki C2 .G/ and so we obtain Œg; x 1 ; a1 2 KiC3 .G/ which implies that Œg; x 1 ; a1 x 2 Ki C3 .G/. Thus we get 1 Œx; a; ga 2 Ki C3 .G/ and so Œx; a; g 2 Ki C3 .G/ and g 2 CiC1 . We have shown that Cp CpC1 Cc1 : Therefore

c1 [ i D1

Ci D Cc1 [

p1 [ i D1

Ci :

S This exhibits ic1 D1 Ci as a union of p subgroups of index at most p in G; it therefore cannot be the whole of G (see 116). Lemma 133.4 then completes the proof.

134

On p-groups with maximal elementary abelian subgroup of order p 2

1o . Glauberman and Mazza [GM] have proved that if a p-group G, p > 2, possesses a maximal elementary abelian subgroup R of order p 2 (i.e., R is not contained in an elementary abelian subgroup of G of order p 3 ; in general, E is a maximal elementary abelian subgroup of a p-group G provided 1 .CG .E// D E), then G has no elementary abelian subgroup of order p pC1 . The proof of this deep result is not elementary. The result above is not true for p D 2 (see 127). Some information on the structure of the groups with title property is contained in Theorem 134.6. In particular, if we have exp.RG / D p, then G has no elementary abelian subgroup of order p pC1 . The proof of this result is fairly involved. However, we cannot present an elementary proof of the theorem of Glauberman–Mazza. In this section we continue to study the structure of groups from [GM] clearing the structure of CG .R/ also in case p D 2. In that case, there is G containing an elementary abelian subgroup of order 24 , and all such G are determined in Theorem 127.1. If a 2-group G has a maximal elementary abelian subgroup R of order 4, then every subgroup of G is generated by four elements so G has no elementary abelian subgroup of order 25 . Indeed, in view of a theorem of MacWilliams (see 50), it sufﬁces to show that G has no normal elementary abelian subgroup of order 8. Assume that E Š E8 is a normal subgroup of G. Then, since RE is not of maximal class, we get CRE .R/ > R (Proposition 1.8), and CRE .R/ is elementary abelian by the modular law, and this is a contradiction. The same MacWilliams’ result shows that every subgroup of G is generated by four elements. Note that the wreath product G D Q2n wr C2 has a maximal elementary abelian subgroup of order 4 and a maximal subgroup B (the base of this wreath product) with d.B/ D 4. We begin with the following auxiliary result. Proposition 134.1. Suppose that a p-group G contains a maximal elementary abelian subgroup R of order p 2 and assume that R is not normal in G. If x 2 R Z.G/, then CG .x/ D CG .R/ has no metacyclic subgroup of order p 4 and exponent p 2 . Proof. By hypothesis, G is nonabelian and 1 .Z.G// < R. Write C D CG .R/ and N D NG .R/; then jN W C j D p by N/C-Theorem (see Proposition 12 in Introduction). Obviously, 1 .C / D R. Assume, by way of contradiction, that L C is meta-

134

On p-groups with maximal elementary abelian subgroup of order p 2

305

cyclic of order p 4 and exponent p 2 . Clearly, Z.G/ \ R D U has order p so Z.G/ is cyclic. If y 2 RU , then CG .y/ D C.D CG .R// since R D hyiU and U Z.G/. Since exp.L/ D p 2 and L is metacyclic, we have L D AB, where A and B are cyclic of order p 2 such that A\B D ¹1º. At least one of the subgroups 1 .A/, 1 .B/ is different from U.D 1 .Z.G//; denote that subgroup by hxi; then, as we have noticed, CG .x/ D C . Let, for deﬁniteness, x 2 A. We have hxi U D R. If G has no normal abelian subgroup of type .p; p/, it is a 2-group of maximal class (Lemma 1.4), and such a group G has no subgroup isomorphic to L, a contradiction. Let E G G be abelian of type .p; p/; then we have R ¤ E by hypothesis, and U < E. Write F D hx; Ei, where x is chosen in the previous paragraph. Clearly, we get x 62 E (otherwise, E D hxi U D R). Since R < F and 1 .F / D F , it follows that F is nonabelian of order p 3 . Recall that A < L is the cyclic subgroup of order p 2 containing x. Write W D A E (the natural semidirect product with kernel E); then F < W and F=E is a unique subgroup of order p in the cyclic group W=E of order p 2 . Since the centralizer CW .E/ has index p in W , it contains F so F is abelian, contrary to what has just been proved. Thus L does not exist. Theorem 134.2. Suppose that a nonabelian p-group G, p > 2, possesses a maximal elementary abelian subgroup R of order p 2 . Set C D CG .R/. Then one of the following holds: (a) G is metacyclic; then R D 1 .G/. (b) G is a p-group of maximal class. If, in addition, R G G, then p D 3.1 (c) C has a cyclic subgroup of index p so it is abelian of type .p n ; p/, n > 1. Proof. If R Z.G/, then C D G has no elementary abelian subgroup of order p 3 so it is metacyclic (Theorem 13.7). In what follows we assume that R 6 Z.G/; then Z.G/ is cyclic. It follows that, in any case, C is metacyclic. Suppose that R G G. If G contains an elementary abelian subgroup A of order p 3 , then CRA .R/ > R is elementary abelian, contrary to the hypothesis. Then, by Theorem 13.7, one of the following holds: (i) G is metacyclic. (ii) G D 1 .G/Z, where 1 .G/ is nonabelian of order p 3 and exponent p and Z is cyclic. (iii) G is a 3-group of maximal class. In case (i), there exist no further restrictions on the structure of G. In case (ii), CG .R/ has a cyclic subgroup of index p. Finally, in case (iii), if jGj > 34 , we have R D 1 .ˆ.G// and CG .R/ D C has no cyclic subgroups of index 3 by Exercise 9.1(c); in this case, C is either abelian or minimal nonabelian. In what follows we assume that R is not G-invariant; then G is not metacyclic. We also assume that G is not of maximal class. Then jC j > p 2 (Proposition 1.8). Then, by Proposition 134.1, the (metacyclic) centralizer C has no metacyclic subgroup of order p 4 and exponent p 2 . We claim that in the case under consideration C 1 If

a 3-group G of maximal class is not isomorphic to a Sylow 3-subgroup of the symmetric group of degree 32 , then all maximal elementary abelian subgroups of G have order 32 (Exercise 9.13). If p > 3, then there is a p-group G of maximal class and order > p 4 that has no such subgroup as R (this is the case if 1 .G/ ˆ.G/).

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has a cyclic subgroup of index p. One may assume that jC j > p 4 . Since C is regular (Theorem 7.1(c)), we get j2 .C /j p 4 and exp.2 .C // D p 2 so that j2 .C /j D p 3 by what has just been said. It follows that C =R has only one subgroup of order p, and hence it is cyclic. If Z < C is maximal such that R 6 Z (Z exists since R 6 ˆ.C /), then Z is cyclic of index p in C . Since R Z.C /, the subgroup C is abelian of type .p n ; p/ as in (c). Note that the p-groups G, p > 2, such that CG .x/ is abelian of type .p n ; p/ for some x 2 G of order p were studied in great detail in the rarely cited important Blackburn’s result (see Theorem A.38.10); that paper yields essential additional information on groups in part (c) of Theorem 134.2. A subgroup A of a p-group G is said to be soft in G, if it satisﬁes CG .A/ D A and jNG .A/ W Aj D p (see 130). Thus soft subgroups are abelian. A subgroup C from Theorem 134.2(c) is soft. Moreover, if a nonnormal R < G is of order p 2 , then we get jNG .R/ W CG .R/j D p, and if, in addition, CG .R/ is abelian, then it is soft in G. Soft subgroups have a number of remarkable properties (see [Het4] and further papers of L. Hethelyi listed in MathSciNet; see also 130). One of such properties is proved in the following remark (note that this proof is distinct from the original one due to L. Hethelyi [Het4]). Remark 1 (the result of this remark coincides with Lemma 130.2). Let A be a nonnormal maximal abelian subgroup of a p-group G, jNG .A/ W Aj D p. Let us prove that if A < H < G, then jNG .H / W H j D p (it follows that there is only one maximal chain connecting A with G). Set N0 D NG .A/; then jN0 W Aj D p and N0 is nonabelian. Set N1 D NG .N0 /. Then N0 contains jN1 W N0 j > 1 conjugates of A under N1 . Since N0 is nonabelian, the number of abelian subgroups of index p in N0 is equal to p C 1 (see Exercise 1.6(a)), therefore we get jN1 W N0 j D p. The intersection of all abelian subgroups of index p in N0 coincides with Z.N0 / D Z G N1 . The quotient group N1 =Z is nonabelian since its subgroup A=Z (of index p 2 ) is not normal. Since CG .A/ D A, we obtain Z.G/ Z < A. Let R Z.G/ be of order p. Then either A=R or N0 =R is a maximal abelian subgroup of G=R since N1 =R, having a nonabelian epimorphic image of N1 =Z.G/, is nonabelian.2 Clearly, the pair K=R < G=R, where K=R is the above chosen maximal abelian subgroup of G=R containing A=R, satisﬁes the identity jNG=R .K=R/ W K=Rj D p by the above as K 2 ¹A; N0 º. Thus K=R is soft in G=R. By induction, there is only one maximal chain connecting K=R and G=R so there is only one maximal chain connection A and G. Indeed, there is nothing to prove in case K D A. If K > A, then K D N0 so the result also holds since N0 is a unique subgroup of G of order pjAj containing A. Similarly, by induction, we obtain the second assertion on indices. Remark 2. If R and G are as in Theorem 134.2, then every subgroup H G such that R < H and exp.H / D p has order p p . Indeed, we have jCH .R/j D p 2 so H is 2 In

fact (see Lemma 130.1), N0 =R is abelian, but we do not use this fact.

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of maximal class (Proposition 1.8), and now the claim follows from Blackburn’s theory of p-groups of maximal class (see Theorems 9.5, 9.6). Assume that there is in G a normal subgroup L of order p pC1 and exponent p. Then CRL .R/ D R by the modular law, so RL is of maximal class (Proposition 1.8). This is a contradiction since a p-group of maximal class cannot contain such a subgroup as L (Theorem 9.6). The case p D 2 is considered in the following theorem. If a 2-group G contains a maximal elementary abelian subgroup R G G of order 4, then G has no normal elementary abelian subgroup of order 23 ; the structure of such G is in some detail described in 50. Note that a minimal nonmetacyclic group X of order 25 satisﬁes j1 .X/j D 4 and d.X/ D 3 (the group X is special). Theorem 134.3. Suppose that a nonabelian 2-group G contains a maximal elementary abelian subgroup R of order 4 and assume that R is not normal in G. Then one of the following holds: (a) The centralizer CG .R/ has a cyclic subgroup of index 2 (so it is abelian). (b) The centralizer CG .R/ D Q Z, where Q is a generalized quaternion group and jZj D 2.3 Proof. Set C D CG .R/; then 1 .C / D R. As R is not normal in G, the subgroup C has no metacyclic subgroup of order 16 and exponent 4 by Proposition 134.1, and this allows us to describe the structure of C . If C is abelian, we get case (a) since C has no abelian subgroup of type .4; 4/ and so j2 .C /j 23 (see the proof of Theorem 134.2). Suppose that C is nonabelian. Then C contains a minimal nonabelian subgroup A. As 1 .A/ 1 .C / D R, it follows that A is metacyclic (Lemma 65.1). Assume that jAj > 8. Then R < A since 1 .A/ Š E4 Š R D 1 .C /, so R D 1 .A/ Z.A/, and we infer that A has no cyclic subgroup of index 2 (otherwise, A will be abelian). Since, by Proposition 134.1, A has no metacyclic subgroup of order 16 and exponent 4, we get a contradiction. Therefore jAj D 8. As A 6Š D8 , we obtain A Š Q8 . Thus all minimal nonabelian subgroups of C are isomorphic to Q8 . It follows that C D Q Z, where Q is a generalized quaternion group and jZj D 2 (see Corollary A.17.3), and the proof is complete. Proposition 134.4 ([GM, Lemma 2.5] for p > 2). Suppose that a p-group G that is not a 2-group of maximal class contains a non-G-invariant maximal elementary abelian subgroup R of order p 2 . Then G has only one normal elementary abelian subgroup of order p 2 unless p D 2, and G contains a proper subgroup of order 24 that is isomorphic to the group D8 C4 of order 16. 3 In this case, Z.G/ D .Q/ has order 2 since this subgroup is characteristic in C and Z.G/ is cyclic. 1 It follows that CG .Z/ D C . The 2-groups G containing an involution x such that CG .x/ D Q hxi, where Q is either cyclic or a generalized quaternion group, are described in 48, 49.

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Proof. Assume that E and F are distinct G-invariant abelian subgroups of type .p; p/ in G. Since Z.G/ is cyclic, we get E \F D U , where U D 1 .Z.G// so jE \F j D p and the subgroup H D EF has order p 3 by the product formula. For the subgroups E=U and F=U we get E=U; F=U Z.G=U /. If H is abelian, it is elementary, and so R 6 H . In case that H is nonabelian, it is either of exponent p > 2 or isomorphic to D8 (this follows from the description of groups of order p 3 ). In that case, all noncyclic subgroups of index p in H are normal in G since H=U Z.G/ and U D ˆ.H /. It follows that R 6 H , again. Write D D HR; then 1 .D/ D D and, as U D H \ R has order p, we get jDj D p 4 by the product formula. Since E=U and F=U are central subgroups of G=U , it follows that D=U Š Ep3 so that d.D/ D 3 and cl.D/ D 2. Suppose that H is abelian. Then CD .R/ is of exponent p so it coincides with R by hypothesis, and it follows from Proposition 1.8 that cl.D/ D 3 > 2, contrary to the last sentence of the previous paragraph. Now let H be nonabelian. By Proposition 10.17, we have CD .H / 6 H since D is not of maximal class, and so Z.D/ has order p 2 . It follows that Z.D/ is cyclic (otherwise, we obtain R < RZ.D/ Š Ep3 , contrary to the hypothesis). In that case, we have p D 2 (if p > 2, then D D 1 .D/ is of exponent p, a contradiction). It follows that D Š D8 C4 has order 24 (note that D8 C4 Š Q8 C4 ). In particular, if, in Proposition 134.4, p > 2, then G has only one normal abelian subgroup of type .p; p/, as asserted in [GM, Lemma 2.5]. Deﬁnition. A proper subgroup A of a p-group G is said to be generalized soft if, whenever A H < G, then jNG .H / W H j D p. (In that case, there is only one maximal chain connecting A and G, but the converse is not true.) In the following proposition we consider the p-groups containing a subgroup of order p that is, as a rule, generalized soft. Proposition 134.5. Suppose that a p-group G contains a subgroup L of order p such that there is only one maximal chain connecting L and G. Then one of the following holds: (a) G is abelian with cyclic subgroup of index p. n

(b) G D ha; b j ap D b p D 1; b a D a1Cp (c) G is a p-group of maximal

n1

i Š MpnC1 .

class.4

Proof. Write N D NG .L/; then N=L is cyclic. If N D G, we have case (a). Next we assume that N < G. If jN=Lj D p, then G is of maximal class by Proposition 1.8. Now assume that jN=Lj > p. Since L is not G-invariant, it is not characteristic in N so N is not cyclic. Let R D 1 .N / and N1 D NG .R/. Since R is characteristic in N , we get N < N1 . By hypothesis, N1 =R is cyclic. Since R < N < N1 , it follows that 4 Not

all p-groups of maximal class contain such a subgroup as L (for example, a p-group G of maximal class, p > 3, such that j1 .G/j D p p1 has no such subgroup).

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R D 1 .N1 / is characteristic in N1 , and we conclude that N1 D G. In that case, G=R is cyclic so G possesses a cyclic subgroup of index p hence G Š MpnC1 by Theorem 1.2. Remark 3. Below we describe the pairs L < G of 2-groups such that L Š E4 , L is not G-invariant and there is only one maximal chain connecting L with G. We write C D CG .L/; then C < G. One may assume that L < C (otherwise, G is of maximal class by Proposition 1.8). In that case, C =L > ¹1º is cyclic so C is a maximal abelian subgroup of G of rank 2 or 3. If jC =Lj D 2, then we obtain that C 2 ¹E8 ; C4 C2 º. Such G are described in 50, 77. Next assume that jC =Lj > 2. Let d.C / D 3. Then T D 1 .C / Š E8 is a proper characteristic subgroup in C . In that case, we see that NG .T /=T > C =T > ¹1º is cyclic by hypothesis, and so T D 1 .NG .T // is characteristic in NG .T /, and we conclude that NG .T / D G hence 1 .G/ D T . It follows that G=1 .G/ is cyclic and 1 .G/ Š E8 . Then G has a cyclic subgroup of index 4 (such G are described in 74). Now let C be abelian of rank 2; then L D 1 .C / so C has a cyclic subgroup of index 2 by Proposition 134.1. In that case, NG .L/=L is cyclic by hypothesis. Therefore it follows from L < C < NG .L/ that L D 1 .NG .L// is characteristic in NG .L/ so NG .L/ D G, i.e., L G G, contrary to the hypothesis. 2o . In this subsection we prove two theorems. We suppose that the p-groups R, E and G appearing in these theorems satisfy the following Hypothesis. (1) p > 2, R is a nonnormal maximal elementary abelian subgroup of order p 2 of a p-group G. (2) E is a maximal elementary abelian subgroup of G, E ¤ R. (3) G is not of maximal class. Let us prove that if R and G satisfy the Hypothesis, then Z.G/ is cyclic. Indeed, Z.G/ < CG .R/ and 1 .CG .R// D R so that 1 .Z.G// < R, and our claim follows. Theorem 134.6. Let p; G; R; E be as in the Hypothesis. Suppose that exp.RG / D p and let RG M G G, where M is as large as possible subjecting exp.M / D p. Then one of the following holds: (a) The subgroups RG and M are of maximal class, jM j p p . (b) If G is irregular, then jM j D p p . (c) If j1 .G/j D p p , then G=1 .G/ is cyclic. (d) If jM j D p p , then ME is of maximal class and order p pC1 so that jEj p p . Theorem 134.7. Let p; G; R; E be as in the Hypothesis. Suppose that exp.RG / > p. Then the following hold: (a) There is M G G of order p p and exponent p. Write H D MR and F D M C , where C D CG .R/; then H is of maximal class and order p pC1 , H < F .

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(b) Let H < T G, where jT W H j D p; then the subgroup T is not of maximal class and T D M2 .C / is the unique subgroup of G of order pjH j containing H so that NG .H /=H is cyclic. (c) F=M is abelian of type .p n1 ; p/, H G F and F=H is cyclic of order p n1 . (d) If R < B < H with jB W Rj D p, then B is nonabelian and CG .B/ is cyclic. (e) The subgroup H is not normal in G. (f) 1 .Z.G// Ã1 .C /. (g) If jME W M j > p, then R 6 ME. (h) If A G G is of order p k < p p and exponent p, then A < M . Our proofs of Theorems 134.6 and 134.7 are essentially based on Blackburn’s theory of p-groups of maximal class (see 9) and Hall’s results on regular p-groups (see 7). In both theorems G is nonabelian. Remark 4. If R is as in the Hypothesis, then it is not contained in ˆ.G/. Assume, by way of contradiction, that R ˆ.G/; then ˆ.G/ is noncyclic. In this case, there is in ˆ.G/ a G-invariant abelian subgroup L of type .p; p/ (Lemma 1.4). As L Z.ˆ.G// (consider CG .L/!), the subgroup RL is elementary abelian properly containing R, a contradiction. It follows from Z.G/ < CG .R/ that Z.G/ is cyclic (see the paragraph following the Hypothesis. Note that if G satisﬁes the Hypothesis, it has no normal subgroup of order p pC1 and exponent p. Indeed, assume that B G G is of order p pC1 and exponent p. Write H D BR. By Theorem 9.6(c), H is not of maximal class as B H so that CH .R/ > R (Proposition 1.8), and we conclude that CB .R/ 6 R. If L < CB .R/ is of order p is such that L 6 R, then RL D R L is an elementary abelian p-group properly containing R, which is a contradiction. Thus B does not exist. Proof of Theorem 134.6. By Proposition 1.8, if R < K G and exp.K/ D p, then K is of maximal class and order p p . Therefore, by assumption, exp.G/ > p. Obviously, G is nonmetacyclic. In this theorem exp.RG / D p. Since R is not normal in G, we get R < RG . Let M be a maximal G-invariant subgroup of exponent p containing RG . Then, by Proposition 1.8, the subgroups RG and M are of maximal class as CM .R/ D R D CRG .R/ (see part (1) of the Hypothesis). By Proposition 1.8 and Theorem 9.5, we obtain that jM j p p . Set Q D M C , where C D CG .R/. Since, by Theorem 134.2(c), C is abelian of type .p n ; p/, n > 1, and R D M \ C , then Q=M Š C =.C \ M / D C =R is cyclic of order p n1 . (i) By Theorem 7.2(b), if G is regular, then M D 1 .G/. There is no further restriction on the structure of G in this case. We claim that if G is irregular, then jM j D p p . Indeed, by Theorem 12.1(a), there is N G G of order p p and exponent p since, by assumption, G is not of maximal class.

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As we know, M p p . One may assume that M 6 N (otherwise, there is nothing to prove). Let N1 be a G-invariant subgroup of N satisfying N1 6 M and such that N1 is as small as possible. Then jMN1 j D pjM j p pC1 hence exp.MN1 / D p 2 by the choice of M . Therefore MN1 is irregular hence jMN1 j p pC1 (Theorem 7.1(b)), and we conclude that MN1 is of maximal class and order p pC1 (Theorem 7.2(b)). Since M 6 ˆ.MN1 / (otherwise, we have MN1 D N1 , contrary to the proved equality jMN1 j D p pC1 ), we infer that jMN1 W M j D p (Exercise 9.1(b)); then jM j D p p , as asserted. It is important in what follows that R < M . (ii) Let T =M G=M be of order p such that T 6 Q.D M C / (see the second paragraph of the proof; such a T exists if and only if G=M is noncyclic). Since CT .R/ D T \ C D R; the subgroup T is of maximal class (Proposition 1.8). (iii) In what follows we suppose that jM j D p p (this is the case provided G is irregular by (i), however, we do not assume here that G is irregular). Set W D ME, where E < G is as in part (2) of the Hypothesis. We claim that W is of maximal class. This is the case if E < M . Now let E 6 M ; then M < W . To prove our claim, take M < U ME such that jU W M j D p. By the modular law, we get U D M.U \ E/ so that 1 .U / D U . Assume that exp.U / D p. Then U is not of maximal class since jU j D p pC1 (Theorem 9.5) so that CU .R/ > R (Proposition 1.8), which is a contradiction. Thus exp.U / > p so that U is irregular (Theorem 7.2(b)). As jU j D p pC1 , we conclude that U is of maximal class (Theorem 7.1(b)). Thus all subgroups of order p pC1 lying between M and W D ME are of maximal class. It follows from Exercise 10.10 that W is also of maximal class. By Exercise 9.1(b), jW j D pjM j D p pC1 since M 6 ˆ.W /, and we conclude that jEj < jW j D p pC1 . (iv) Suppose that j1 .G/j D p p ; then M D 1 .G/ (see the deﬁnition of M in the second paragraph of the proof). Assume that G=M is noncyclic. Assume, in addition, that p D 3. Then G has no elementary abelian subgroup of order p 3 . In this case, however, G=M is cyclic by Theorem 13.7, contrary to the assumption. Thus p > 3. Since Q=M D M C =M is cyclic, there is T =M < G=M of order p such that T =M 6 Q=M (Proposition 1.3); then T 6 Q. It follows that T \Q D M so that CT .R/ D R, and Proposition 1.8 implies that T is of maximal class. Since jT j D p pC1 , we infer that T is irregular (Theorem 9.5). Let T < L G be such that jL W T j D p. The subgroup L is not of maximal class since M G L, jLj > p pC1 and M is of order p p and exponent p (see Theorem 9.6(c)). Therefore, by Theorem 12.12(b), L=Kp .L/ is of order p pC1 and exponent p. It follows that L has no absolutely regular subgroup of index p (otherwise, jL=Ã1 .L/ p p ). Let D < M be G-invariant of order p 2 and set V D CG .D/; then V is maximal in G as D 6 Z.M / in view of jZ.M /j D p < jDj. The intersection M \ V D 1 .V / has order p p1 hence V is absolutely regular since it is not of maximal class in view of D Z.V / (Theorem 12.1(b)). Then L \ V , as

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a subgroup of the absolutely regular group V , is absolutely regular of index p in L, contrary to what has been said above. Thus G=M is cyclic, as was to be shown. The proof of our theorem is completed. Proof of Theorem 134.7. By hypothesis, in this case exp.RG / > p hence RG is irregular in view of 1 .RG / D RG (Theorem 7.2(b)). Since G is not of maximal class, there is M G G of order p p and exponent p (Theorem 12.1(a)). We have R 6 M since exp.RG / > p D exp.M /. (i) We claim that H D MR is of maximal class and order p pC1 . Indeed, we have jM \ Rj D p since Z.G/ is cyclic, and so jMRj D p pC1 by the product formula. Next, CH .R/ D H \ CG .R/ D R since CH .R/ D R CM .R/ D R by the modular law, and our claim follows from Proposition 1.8 (in the case under consideration, exp.CH .R// D p by Theorems 7.1(b) and 7.2(b) since cl.CH .R// < p). (ii) Let R < S < H , where jH W S j D p; then we get S D R.S \M / by the modular law, so exp.S/ D p since 1 .S/ D S and jSj D p p (Theorems 7.1(b) and 7.2(b)). If H G G, then, considering the action of G on maximal subgroups of H via conjugation, we obtain S G G (indeed, since exp.H / D p 2 and d.H / D 2, H has at most p maximal subgroups of exponent p and one of them, namely M , is normal in G), and this is a contradiction since RG S and exp.RG / > exp.S/. Thus H is not normal in G. (iii) Write F D M C ; then we obtain MR D H < F since jF j p 2 jM j > jH j (recall that R 6 M and exp.C / > p; see Theorem 134.2(c)). Let H < T G, where jT W H j D p; then T is not of maximal class since M G T (Theorem 9.6(c)). Then we have CT .R/ > R (Proposition 1.8) so that 2 .C / < T . By the product formula, we obtain T D M2 .C / M C D F so that T is the unique subgroup of NG .H / of order pjH j containing H , and we infer that NG .H /=H is cyclic since p > 2 (Proposition 1.3). By Theorem 12.12(b), T =Kp .T / is of order p pC1 and exponent p so that Ã1 .T / D Kp .T / D Kp .H / D Z.M / D Ã1 .H / D Ã1 .2 .C // D R \ Z.G/ D 1 .Z.G//: It follows that F=M is abelian of type .p n1 ; p/ so that H GF and hence F=H is cyclic of order p n1 . If R < B < H with jB W Rj D p, then B is nonabelian of order p 3 and exponent p, and CT .B/ D 2 .C /; moreover, CG .B/ is cyclic since R < B. (iv) Let jM j D p p and jME W M j > p. We have to prove that R 6 ME. Assume that this is false. Since R \ M D Z.M / is of order p, we have jMRj D p pC1 by the product formula. Next, CMR .R/, being of exponent p, coincides with R, and we conclude that MR is of maximal class (Proposition 1.8). Let M < U < ME be such that jU W M j D p and U ¤ MR. As U D M.U \ E/ by the modular law, we obtain that 1 .U / D U . Assume that exp.U / D p. Write H D UR (note that U G ME). Since H is not of maximal class (Theorem 9.6(c)), we get CH .R/ > R. Again by the modular law, we conclude that CH .R/ D M.U \ CH .R// so that exp.CH .R// D p,

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313

contrary to Theorem 134.2(c) since H is neither metacyclic nor of maximal class (note that MR is also of maximal class). Hence we obtain that exp.U / > p so U is irregular (Theorem 7.2(b)). Since jU j D pjM j D p pC1 , it follows that U is of maximal class (Theorem 7.1(b)). Thus all subgroups of G of order pjM j lying between M and ME are of maximal class. Therefore ME is also of maximal class (Exercise 10.10). Since M 6 ˆ.ME/, we get jME W M j D p (Exercise 9.1(b)), contrary to the assumption. (v) Now let A G G be of order p k < p p and exponent p. We have to prove that A satisﬁes A < M , where M is as in part (a) of the statement of our theorem. This is true for k D 1 since Z.G/ is cyclic. Now assume that we have proved that any normal subgroup of G of order < p k and exponent p is contained in M . Set J D AM . Then, by assumption, jA \ M j D p k1 . By the product formula, we infer that jJ j D p pC1 . Due to Exercise 9.1(b), J is not of maximal class since it has two nonincident proper normal subgroups of different orders. It follows that cl.J / < p, and we conclude that J is regular (Theorem 7.1(b)). Since 1 .J / D J , it follows from Theorem 7.2(b) that exp.J / D p. This is a contradiction since, as we have noticed (see the paragraph preceding the proof of Theorem 134.6), G has no normal subgroup of order p pC1 and exponent p. Thus we get A < M . Exercise 1. Theorem 134.7(h) also holds under the condition of Theorem 134.6. (Hint. Argue as in part (v) of the proof of Theorem 134.7.) Exercise 2. Suppose that a pair of p-groups R < G satisfy the Hypothesis. If M and N are two distinct G-invariant subgroups of order p p and exponent p, then MN is of maximal class and order p pC1 . Solution. Let A < M be a G invariant subgroup of index p. By Theorem 134.7(h) and Exercise 1, we get A < N so that MN is of order p pC1 . As we know, exp.MN / D p 2 so MN is irregular. It follows that MN is of maximal class. Exercise 3. Suppose that R is a nonnormal maximal elementary abelian subgroup of a p-group G, p > 2. Then C D CG .R/ is abelian with cyclic subgroup of index p unless NG .R/ is either metacyclic or a 3-group of maximal class. Solution. By Theorem 13.7, C is metacyclic. Set N D NG .R/; then jN W C j D p. Since N has no normal elementary abelian subgroup of order p 3 , the result follows from Theorem 13.7. Remark 5. Suppose that G is a p-group and M G G is of order p p and exponent p such that every subgroup of order p pC1 between M and G has an exponent p 2 . Then G has no elementary abelian subgroup of order p pC1 . Indeed, let E G be elementary abelian and let E 6 M (otherwise, there is nothing to prove). Write H D ME and let H < U G, where jU W H j D p. Then, by hypothesis, exp.U / D p 2 . By the modular law, U D M.U \ E/ so that 1 .U / D U , and hence U is of maximal class (Theorems 7.1(b) and 7.2(b)) and order p pC1 . Since U is arbitrary, we conclude as

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above that the group G is of maximal class and order p pC1 (Exercise 10.10). It follows that jEj p p , as asserted. (This argument is the same as in part (iv) of the proof of Theorem 134.7.)

Problems Problem 1. Suppose that G, p > 2, possesses a maximal elementary abelian subgroup of order p 2 and H G. (i) Is it true that d.H / p? (ii) Is it true that jH j < p pC1 provided exp.H / D p? Problem 2. Suppose that G, p > 2, possesses a maximal elementary abelian subgroup n1 of order p n . Is it true that G has no elementary abelian subgroup of order p 1Cp ? Problem 3. Study the p-groups all of whose minimal nonabelian (so all nonabelian) subgroups are generalized soft. Problem 4. Study the p-groups containing a cyclic generalized soft subgroup of order p n . (The problem is nontrivial even for n D 2.) In the following three problems R is a nonnormal maximal elementary abelian subgroup of order p 2 in a p-group G. Problem 5. Study the structure of RG , where exp.R/ > p. Consider in detail the case when RG is of maximal class. Problem 6. Study the structure of G provided RG is of maximal class. Problem 7. Study the structure of G provided CG .R/ is metacyclic. Problem 8. Study the p-groups G containing a nonabelian subgroup B of order p 3 such that CG .B/ is cyclic. (See the statement of Theorem 134.7.)

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Finite p-groups generated by certain minimal nonabelian subgroups

Recall that a group is said to be minimal nonabelian if it is nonabelian but all its proper subgroups are abelian. A nonabelian group contains a minimal nonabelian subgroup. Therefore it is natural to believe that minimal nonabelian groups essentially impact on the structure of a nonabelian group. As many places of our book show, knowledge of all minimal nonabelian subgroups of a nonabelian p-group G allows us to do strong conclusions on its structure. This is analogous to impact of minimal nonnilpotent subgroups on the structure of a nonnilpotent group (this is true in view of Frobenius’ normal p-complement theorem). To justify our thought, we present the following two examples (other examples are given after the proof of Lemma 135.5): (i) If a p-group G possesses at most p 2 C p C 1 minimal nonabelian subgroups, then all its subgroups of index p 3 are abelian (see 76). (ii) If all minimal nonabelian subgroups of a 2-group G have the same order 8, then either the Hughes subgroup has index 2 in G or G D H Z.G/, where H is either of maximal class or extraspecial and Ã1 .Z.G// Z.H / (see Theorem 90.1). In what follows all considered groups have prime power order. Let MA.G/ be the set of all minimal nonabelian subgroups of a p-group G. If G is nonabelian, we have, by Lemma 135.7 below, G D hMA.G/i, i.e., G is generated by all members of the set MA.G/ (this result is important in the proof of assertion (i) above). Since minimal nonabelian subgroups are small, that result shows that in a complicated p-group G the set MA.G/ is large. However, in some cases, certain welldescribed subsets of that set also generate G, and in what follows we present some of such subsets. There is, for p > 2, only one, up to isomorphism, minimal nonabelian p-group generated by elements of order p and its order is p 3 (Exercise 1.8(a)). We denote this group by S.p 3 /. The dihedral group D8 of order 8 is the only minimal nonabelian 2-group generated by involutions. Let k be a positive integer and let MAk .G/ be the set of minimal nonabelian subgroups A of a p-group G such that A D k .A/. Since for any p-group X > ¹1º we

316

Groups of prime power order

have exp.X=1 .X// < exp.X/, it follows that if A 6Š D8 is as above and k > 1, then ()

exp.A/ exp.1 .A// exp.A=1 .A// p p k1 D p k ;

since, in view of jA0 j D p, the quotient group A=1 .A/ is abelian of exponent p k1 and exp.1 .A// D p. In the following paragraph we shall use the following fact. Let G be a nonabelian p-group. Further, assume that all containing ˆ.G/ subgroups of G of order pjˆ.G/j are abelian. Then CG .ˆ.G// possesses all subgroups T of G of order pjˆ.G/j such that ˆ.G/ < T . Since all such T generate G, it follows that ˆ.G/ Z.G/. Let G be a nonabelian p-group. Suppose that ˆ.G/ 6 Z.G/. In this case, there is a sequence ˆ.G/ D T0 < T1 < < Td.G/ D G of subgroups and A1 ; : : : ; Ad.G/ of minimal nonabelian subgroups such that T1 is nonabelian and, for i D 1; : : : ; d.G/, one has Ai Ti but Ai 6 Ti1 (Lemma 135.7 below); then jTi W Ti1 j D p and Ai Ti 1 D Ti . It follows that G D Td.G/ D hA1 ˆ.G/; A2 ; : : : ; Ad.G/ i D hA1 ; A2 ; : : : ; Ad.G/ i: Thus, in the case under consideration, G is generated by d.G/ minimal nonabelian subgroups. If, however, ˆ.G/ Z.G/, the same argument shows that G is generated by d.G/ 1 minimal nonabelian subgroups (note that there is a containing ˆ.G/ subgroup of G of order pjˆ.G/j that is not contained in Z.G/). The obtained estimate is not best possible (for example, an extraspecial group G of order p 1C2m is generated by m < 2m D d.G/ minimal nonabelian subgroups since it is a central product of m nonabelian subgroups of order p 3 ). Let k D ı.G/ denote the minimal number of minimal nonabelian subgroups, say A1 ; : : : ; Ak , generating a nonabelian p-group G. Set T0 D ˆ.G/;

T1 D ˆ.G/A1 ;

T2 D T1 A2 ;

:::;

Tk D Tk1 Ak D G:

Since, for all i , ˆ.Ai / ˆ.G/ has index p 2 in Ai , we get jTi W Ti 1 j p 2 . It follows that p d.G/ D jG W ˆ.G/j D jTk W Tk1 j : : : jT1 W T0 j p 2k D p 2ı.G/ ; and we obtain ı.G/ d.G/ 2ı.G/: As we have noticed, the lower estimate is attained for extraspecial p-groups. This estimate can be slightly improved in the case ˆ.G/ Z.G/. In this section, however, we study the generation of a nonabelian p-group G by minimal nonabelian subgroups satisfying certain additional properties. Write Dk .G/ D hMAk .G/i;

Dk .G/ D hA j A 2 MAk .G/; exp.A/ D p k i:

Obviously, Dk .G/; Dk .G/ k .G/.

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317

In what follows we establish criteria guaranteeing the equality Dk .G/ D G and also study the p-groups G satisfying Dk .G/ < G for a positive integer k (see Theorem 135.1(b) and Remark 2). Let us prove that if G is a nonabelian regular group of exponent p e , then we have G D De .G/. Assume that G is a counterexample of minimal order. Then we conclude that e > 1 (Lemma 135.7) and G is not minimal nonabelian. Setting H D e1 .G/, we have exp.H / D p e1 (Theorem 7.2(b)). By Lemma 135.7, there exists a minimal nonabelian subgroup A < G such that A 6 H ; then exp.A/ D p e . If M 2 1 is nonabelian of exponent p e , then we get De .M / D M by induction (such M exist: take M containing A). By Theorem 7.2(b), there exists in G at most one maximal subgroup of exponent p e1 (otherwise, H D G). Therefore all other members of the set 1 have exponent p e . Since the number of abelian maximal subgroups in G is 0, 1 or p C1 and j1 j 1 .mod p/, there are at least p nonabelian members of the set 1 . It follows that there are in 1 at least p 1 nonabelian members of exponent p e . Since G is nonabelian regular, we have p > 2 so p 1 > 1. It follows that there is a nonabelian N 2 1 ¹M º of exponent p e . By induction, we see that De .N / D N . Therefore De .G/ De .M /De .N / D MN D G, contrary to the assumption. It is proved in Lemma 30.3 that if a nonabelian 2-group G D 1 .G/, then we have G D D1 .G/. To clear up our path, we offer another proof of this equality. Since G is nonabelian, it contains two noncommuting involutions a and b. Then ha; bi is dihedral so it contains a subgroup A Š D8 . Set H D D1 .G/. Let x 2 G A be an involution. If x 2 CG .A/, we get U D Ahxi D A hxi. There are in U exactly four maximal subgroups not containing x, and these subgroups generate U and are isomorphic to D8 , and we conclude that x 2 U H . Now assume that x does not centralize A. In this case, there is in A D 1 .A/ an involution y such that xy ¤ yx. Then hx; yi is dihedral so it is generated by subgroups isomorphic to D8 , and again x 2 hx; yi H . Thus all involutions are contained in H so H 1 .G/ D G. Considering the group S.p 3 / as an analog of D8 , we note that a result similar to the result of the previous paragraph does not hold for p > 2 as a Sylow p-subgroup †p2 of the symmetric group of degree p 2 shows. Indeed, †p2 has exactly two maximal subgroups, say E and M , both of exponent p, E is abelian and M is nonabelian,1 and so D1 .G/ D M < G (Lemma 135.7) since M is nonabelian in view of cl.†p2 / D p > 2, and any L 2 1 ¹M:Eº satisﬁes D1 .L/ D L \ E hence MA1 .L/ D ¿. As Theorem 30.1(d) shows, if a nonabelian p-group G D 1 .G/, p > 2, has no subgroup Š †p2 , then G is generated by subgroups isomorphic to S.p 3 /. However, the proof of this result was omitted there. Here (see Theorem 135.1(a)) we offer its 1 The group G D † p 2 is the wreath product of two groups of order p. Let E be the base of G; then E is elementary abelian of order p p and G D hxi E is a semidirect product, o.x/ D p. Next, we have CG .x/ D hxi Z.G/, where jZ.G/j D p, so jCG .x/j D p 2 , and hence G is of maximal class (Proposition 1.8). Set M D hxi ˆ.G/; then M D 1 .M / is regular so exp.M / D p. The set E [ M contains exactly 2p p p p1 1 elements of order p, and this number coincides with the number of elements of order p in G. Thus G has exactly two maximal subgroups E and M of exponent p.

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Groups of prime power order

proof independent of Theorem 30.1. Next, we study in Theorem 135.1(b) the p-groups G D 1 .G/, p > 2, such that D1 .G/ < G. Remark 1. Let G D 1 .G/, p > 2, be a nonabelian p-group. Suppose that whenever x; y 2 G are noncommuting elements of order p, then D1 .hx; yi/ D hx; yi. We claim that then G D D1 .G/. By hypothesis, G, being nonabelian, contains a subgroup L Š S.p 3 /. Let H D D1 .G/; then H > ¹1º. Assume that H < G. Then there is x 2 G H of order p. If x 2 CG .L/, then U D hx; Li D hxi L. Since U is generated by p 2 its maximal subgroups not containing x, and all these subgroups are isomorphic to L Š S.p 3 /, we get x 2 H , a contradiction. Now assume that x 62 CG .L/. Then there is y 2 L# (of order p) such that xy ¤ yx. In this case, by hypothesis, the nonabelian subgroup hx; yi D D1 .hx; yi/ hence x 2 H , a contradiction. Thus we have H D G. Theorem 135.1. Let G D 1 .G/ be a nonabelian p-group, p > 2. Then (a) (Theorem 30.1(d)) If G has no subgroup isomorphic to †p2 , then G D D1 .G/. (b) (Locating of elements of order p) Suppose that H D D1 .G/ < G and E is a maximal by inclusion normal elementary abelian subgroup of G. Then the set H [ E contains all elements of G of order p. Next, E is a unique maximal normal elementary abelian subgroup of G, jEj p p and G D HE; in particular, ˆ.H / H . Corollary 135.2. The result of Remark 1 follows from Theorem 135.1(a). Proof. Our group G has no subgroup isomorphic to †p2 since †p2 is generated by two noncommuting elements of order p and, as we have noticed, D1 .G/ < G. Therefore the result follows from Theorem 135.1(a). The next lemma is useful in proofs by induction (see the proofs of Lemma 135.8 and Theorem 135.1). Lemma 135.3 (= Lemma 30.2). Let k be a positive integer and A a proper subgroup of a p-group G D k .G/ such that k .A/ D A. Then we have A M 2 1 , where M D k .M /. In particular, there are two distinct U; V 2 1 such that k .U / D U and k .V / D V . Lemma 135.4. Suppose that a nonabelian p-group G has an abelian subgroup A of index p. If jG W G 0 j D p 2 , then G is of maximal class. Proof. We proceed by induction on jGj. One may assume that jGj > p 3 . By Lemma 1.1, jZ.G/j D p1 jG W G 0 j D p so Z.G/ is a unique minimal normal subgroup of G. Since Z.G/ < G 0 and jG=Z.G/ W G 0 =Z.G/j D jG W G 0 j D p 2 ; the quotient group G=Z.G/ is of maximal class by induction, and the result follows since jZ.G/j D p.

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319

Lemma 135.5. Suppose that A < G is a normal abelian subgroup of a p-group G such that A 6 Z.G/. Then for every x 2 G CG .A/ there is a 2 A such that hx; ai is a minimal nonabelian subgroup.2 Proof. Let x 2 G CG .A/. Bringing to mind our aim, one may assume, without loss of generality, that G D Ahxi; then CA .x/ D U DW A \ Z.G/. One may also assume that jA W U j D p (if not, let us consider instead of A an x-invariant subgroup A1 A of order pjU j such that U < A1 ). Since G is nonabelian, the quotient group G=U is noncyclic. It follows that G=U D .hx; U i=U / .A=U / is abelian of type .p k ; p/, where p k is the order of x modulo U . In this case, G=U contains two distinct cyclic subgroups M=U and N=U of order p k so maximal in G=U . Then M; N are abelian maximal subgroups of G, and we see that Z.G/ D M \N has index p 2 in G. By Lemma 1.1, jG 0 j D p1 jG W Z.G/j D p. Let a 2 A U.D A CG .x// and set H D hx; ai; then H is nonabelian, jH 0 j D jG 0 j D p and d.H / D 2. From Lemma 65.2(a) it follows that H is minimal nonabelian. Below we offer some applications of Lemma 135.5. Suppose that all minimal nonabelian subgroups of a nonabelian 2-group G have the same order 8. Let A be a maximal abelian normal subgroup of G. By Lemma 135.5, all elements of the set G A have order 2 or 4. Assume that there is x 2 G A of order 4. Then there is a 2 A such that H D ha; xi has order 8. Since ˆ.G/ H \ A, it follows that x 2 2 A, and we conclude that ˆ.G/ A. Such groups are classiﬁed in Theorem 90.1. Let p > 2 and A be a maximal abelian normal subgroup of a nonabelian p-group G of exponent > p. If all minimal nonabelian subgroups of G are isomorphic to S.p 3 /, then A is equal to the Hughes subgroup Hp .G/ of G. Indeed, by Lemma 135.5, all elements of the set G A have order p, and we conclude that Hp .G/ A as exp.A/ > p. Since A is generated by elements of maximal order, we get, in fact, Hp .G/ D A.3 Suppose that all minimal nonabelian subgroups M of a nonabelian p-group G satisfy k .M / Z.M /, where k is a ﬁxed positive integer. We claim that if A is a normal abelian subgroup of G such that exp.A/ p k and jAj is as large as possible, then k .G/ A. Assume that there is an element x 2 G A of order p k . Then hx; Ai is nonabelian (Corollary 10.2) so A 6 Z.G/. In this case, there is in A an element a such that H D ha; xi is minimal nonabelian. By hypothesis, x; y 2 Z.H / so H is abelian, a contradiction. Thus x does not exist, and we conclude that k .G/ A; moreover, k .G/ D A. Lemma 135.6. Suppose that G D k .G/ is a nonabelian p-group, p k > 2. Then the set MAk .G/ is not empty. 2 In particular, if A < G is a maximal normal abelian, then for each x 2 G A there is a 2 A such that the subgroup hx; ai is minimal nonabelian. This important result, which is due to Z. Janko, coincides with Lemma 57.1. 3 According to Mann’s commentary to item 115 in Research problems and themes I, one has the equality jG W Aj D p.

320

Groups of prime power order

Proof (compare with the proof of Lemma 30.4(a)). Let E < G be a G-invariant abelian subgroup of exponent p k of maximal order. Take x 2 G E of order p k . Then H D hx; Ei is nonabelian by Corollary 10.2. By Lemma 135.5, there is a 2 E such that the subgroup B D hx; ai is minimal nonabelian. Since o.x/ D o.a/ p k , we get k .B/ D B. By (), exp.B/ p k so that B 2 MAk .G/. Lemma 135.7 (= Theorem 10.28). A nonabelian p-group is generated by minimal nonabelian subgroups. We present an application of Lemma 135.7. Let G D D C , where D Š D8 and C Š C2n (n > 2). At ﬁrst sight, the set MA.G/ has no member of exponent 2n . But this is false. Indeed, we have n1 .G/ D Dn1 .C / < G. By Lemma 135.7, there is M 2 MA.G/ such that M 6 n1 .G/ hence exp.M / D 2n . Note that Lemma 135.7 follows from Lemma 135.5. Indeed, if, in Lemma 135.5, A is a maximal normal abelian subgroup of G, then x 2 G A is contained in some minimal nonabelian subgroup of G. Since hG Ai D G, our claim follows. The following lemma covers a partial case of Theorem 135.1(a) (but it does not follow from that theorem). Lemma 135.8. Suppose that a nonabelian p-group G D 1 .G/ has no subgroup isomorphic to †p2 . If G possesses an elementary abelian subgroup V of index p, then we have exp.G/ D p. Proof. Since G is nonabelian, we get p > 2. By hypothesis, there is x 2 G V of order p; then G D hxi V is a semidirect product with kernel V . Let x 2 F 2 1 ; then F D hxi .F \ V / by the modular law, so 1 .F / D F . By induction, exp.F / D p. Thus every maximal subgroup of G containing x has exponent p. If jGj p p , it is regular so of exponent p (Theorems 7.1(b), 7.2(b)). Next we assume that jGj p pC1 . If jG W G 0 j D p 2 , then G is of maximal class (Lemma 135.4) so we get G Š †p2 by Exercise 9.13, a contradiction. Thus jG W G 0 j > p 2 . Since 1 .G/ D G, it follows that G=G 0 is elementary abelian. Set H D hxi G 0 ; then G=H is elementary abelian of order > p. Let T1 =H; : : : ; Tn =H be all maximal subgroups of G=H ; thenSn p C 1. By the above, exp.Ti / D p for all i n. Therefore it follows from G D niD1 Ti that exp.G/ D p. Proof of Theorem 135.1. (a) Let G be a counterexample of minimal order; then we get jGj > p 3 and exp.G/ > p by Exercise 1.8(a) and Lemma 135.7 so G is irregular and jGj p pC1 . All proper nonabelian subgroups H D 1 .H / of G satisfy D1 .H / D H by induction. Due to Lemmas 135.6 and 135.3, there is a nonabelian U 2 1 such that 1 .U / D U ; then D1 .U / D U by induction. By the last assertion of Lemma 135.3, there exists V 2 1 ¹U º such that 1 .V / D V . One may assume that V ¤ D1 .V / (otherwise, D1 .G/ D D1 .U V / D1 .U /D1 .V / D U V D G). Then V is elementary abelian by induction. If x 2 GV is of order p, then G D hxi V , a semidirect product with kernel V , and now the result follows from Lemmas 135.8 and 135.7 since G has no subgroup isomorphic to †p2 .

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Finite p-groups generated by certain minimal nonabelian subgroups

321

(b) By hypothesis, we get H D D1 .G/ < G. Let E be a normal elementary abelian subgroup of G such that jEj is as large as possible. Assume that there is an element x 2 G .H [ E/ of order p. Set L D hx; Ei. By Corollary 10.2, L is nonabelian. If L is regular, then it follows from 1 .L/ D L that exp.L/ D p by Theorem 7.2(b), and so L D D1 .L/ by Exercise 1.8(a) and Lemma 135.7, a contradiction since L 6 H by the choice of x. Thus L is irregular so jEj p p (Theorem 7.1(b)). Assume that jZ.L/j D p. It follows from Lemma 135.4 that L is of maximal class (indeed, we have jG W G 0 j D pjZ.G/j D p 2 by Lemma 1.1) so L Š †p2 by Exercise 9.13). In that case, U D hx; ˆ.L/i is nonabelian (by Fitting’s lemma) of exponent p (Theorem 7.2(b)) so U D D1 .U /; then x 2 U H , contrary to the choice of x. Thus jZ.L/j > p. Note that Z.L/ D CE .x/. Let T =Z.L/ be an L-invariant subgroup of order p in E=Z.L/ so T is elementary abelian. Then V D hx; T i is nonabelian since x does not centralize T , and 1 .V / D V . We have cl.V / D 2 since jV =Z.L/j D p 2 . It follows from this and p > 2 that exp.V / D p (Theorems 7.1(b) and 7.2(b)). Then, by Lemma 135.7 and Exercise 1.8(a), we have V D D1 .V /, and we obtain x 2 V H , a contradiction. Thus the set G.H [E/ has no elements of order p, as was to be shown. Since G D 1 .G/ and the subgroup HE . H [ E/ contains all elements of G .D 1 .G// of order p, we conclude that G D HE. Because G=H Š E=.E \ H / is elementary abelian, we get ˆ.G/ H . In particular, E 6 H since, by assumption, H < G. Assume that E1 < G is another maximal G-invariant elementary abelian subgroup. Then EE1 is nonabelian, by Corollary 10.2, so it is of class 2 (Fitting’s lemma). It follows that exp.EE1 / D p (Theorems 7.1(b) and 7.2(b)) so D1 .EE1 / D EE1 by Exercise 1.8(a) and Lemma 135.7, and we conclude that E < EE1 H , contrary to the last sentence of the previous paragraph. The subgroup E from Theorem 135.1(b) has the following property: If x 2 G E is of order p, then the subgroup Hx D hx; Ei contains a subgroup isomorphic to †p2 . Indeed, since E 6 D1 .G/, it follows that D1 .Hx / < Hx , and the claim follows from Theorem 135.1(a). Theorem 135.1(a) can be restated as follows: If p > 2, then for a nonabelian p-group G one of the following holds: (i) 1 .G/ is elementary abelian. (ii) G contains a subgroup isomorphic to †p2 . (iii) D1 .G/ D 1 .G/. Assume that p > 2. The group †p2 satisﬁes the hypothesis of Theorem 135.1(b). Let G D †p2 L, where L > ¹1º is elementary abelian p-group. Let M; E < †p2 be maximal of exponent p, M be nonabelian and E be abelian. Write W D D1 .G/. We claim that W D M L. By Lemma 135.7 and Exercise 1.8(a), we get M L W . It remains to prove the reverse inclusion W M L. Let S.p 3 / Š A < G and set F D LA; then exp.F / D p, F is nonabelian so D1 .F / D F . We have to prove that A < M L. By the modular law, F D L.F \†p2 / so that T D F \†p2 is nonabel-

322

Groups of prime power order

ian of exponent p since exp.F / D p and L Z.F /. We get T Š S.p 3 / as jT j D p 3 . It follows that T D1 .†p2 / D M . Thus A < LT LM so that W D LM . Therefore the set of p-groups satisfying the hypothesis of Theorem 135.1(b) is inﬁnite. Remark 2. Let k > 1 and G D k .G/ be a nonabelian p-group. Suppose that G is such that Dk .G/ < G. Then G possesses only one maximal G-invariant abelian subgroup A of exponent p k . The set A [ Dk .G/ contains all elements of orders p k so that G D ADk .G/. In particular, A 6 Dk .G/ and G 0 Dk .G/. Indeed, we have MAk .G/ ¤ ¿ (Lemma 135.6) so Dk .G/ D H > ¹1º. By hypothesis, H < G. Let A be a G-invariant abelian subgroup of exponent p k such that jAj is as large as possible. We claim that the set H [A contains all elements of orders p k . Assume that this is false. Then there exists an element x 2 G .H [A/ of order p k . Set F D hx; Ai; then F is nonabelian by Corollary 10.2 since p k > 2. By Lemma 135.5, there is an element a 2 A such that L D hx; ai is minimal nonabelian. As o.x/; o.a/ p k , it follows that k .L/ D L, and we conclude that x 2 L 2 MAk .G/ D H by (), contrary to the choice of x. Thus x does not exist. Since HA . H [ A/ is normal in G and the set H [ A contains all elements of G of orders p k , we get G D k .G/ HA so that G D HA. In particular, A 6 H and G 0 H since G=H Š A=.H \ A/ is abelian. Assume that there is another maximal G-invariant abelian subgroup A1 of exponent p k . By the above, A1 6 H . Therefore, since A1 H [ A, it follows that A1 D .A1 \ H / [ .A1 \ A/ is a union of two nonincident subgroups, which is impossible. The groups from Theorem 135.1(b) and Remark 2 deserve further investigation. In particular, we do not even know if a group G from Remark 2 satisfying Dk .G/ > G does exist. Write k .G/ D jMAk .G/j. Clearly, k .G/ D k .Dk .G//. Proposition 135.9. Let G be a 2-group with nonabelian 1 .G/. Then (a) 1 .G/ D 1 if and only if G 2 ¹D8 ; SD16 º. (b) 1 .G/ D 2 if and only if G 2 ¹D16 ; SD32 º. (c) 1 .G/ D 3 if and only if 1 .G/ D D C , the central product, where D Š D8 , C is cyclic of order 4 and D \ C D Z.D/. Next, GN D G=CG .D/ 2 ¹E4 ; D8 º. If GN Š E4 , then G D D CG .D/, where CG .D/ is cyclic. If GN Š D8 , then we have G D DQ, where D \ Q D Z.D/ and Q is either cyclic or a generalized quaternion group, Q is not G-invariant, jQ W CG .D/j D 2. In the second case, G is as in Theorem 43.9 (in this case, G contains exactly seven involutions). Proof. Note that 1 .G/ D 1 .1 .G//. Since 1 .G/ is nonabelian, we conclude that 1 .G/ 1. As we know, all members of the set MA1 .G/ are isomorphic to D8 . Let D 2 MA1 .G/. If CG .D/ < D, then we see that G is of maximal class by Proposition 10.17. In this case, if jGj D 2n , then we get 1 .G/ D 2n3 if G Š D2n and 1 .G/ D 2n4 if G Š SD2n (clearly, 1 .Q2n / D 0). Therefore, if 1 .G/ D 1, then

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G 2 ¹D8 ; SD16 º, and if 1 .G/ D 2, then we have G 2 ¹D16 ; SD32 º. If CG .D/ 6 D, let L CG .D/ be cyclic of minimal order such that L 6 D; then we get jLj 4. In case jLj D 2, we obtain that 1 .DL/ D 1 .D L/ D 4 > 3, a contradiction. If jLj D 4, then DL D DL is the central product of order 16, and then 1 .DL/ D 3. The cases (a), (b) are completed. Now let 1 .G/ D 3; then G is not of maximal class. One may assume from the start that DGG. If L1 ¤ L is another cyclic subgroup of order 4 in CG .D/ (see the previous paragraph), then 1 .DL1 / D 3 and then 1 .G/ 5, a contradiction. Thus L is the unique cyclic subgroup of order 4 in CG .D/. Then we have 1 .CG .D// D 0. Indeed, if D1 2 MA1 .CG .D//, then, since 1 .CG .D// 2, CG .D/ is of maximal class by (a) and (b), so DD1 D D D1 is extraspecial of order 25 . If L0 < D1 is cyclic of order 4, then 1 .DL0 / D 3 by the above. As D1 6 DL0 , we get 1 .DD1 / > 3, a contradiction. It follows that CG .D/ is cyclic by Theorem 1.17(b). In this case, we have 1 .G/ D D L, where L is cyclic of order 4 since 1 .G/ D D1 .G/. The quotient group G=CG .D/ 2 ¹E4 ; D8 º is a subgroup of a Sylow 2-subgroup of Aut.D/ Š S4 containing D=Z.D/ Š E4 . If G=CG .D/ Š E4 , then G D D CG .D/, where CG .D/ is cyclic. Let GN D G=CG .D/ Š D8 . Let xN 2 GN DN be an involution and Q D hx; CG .D/i. Then Q has only one subgroup of order 2 coinciding with Z.D/ so Q is either cyclic or a generalized quaternion group. In that case, G is as in Theorem 43.9(b, c). Let G be a 2-group with 1 .G/ D 4. Take D8 Š D < G. If CG .D/ < D, then G is of maximal class so G 2 ¹D32 ; SD64 º. Next we assume that CG .D/ 6 D. Assume that CG .D/ D C > ¹1º is cyclic. Then 1 .DC / D 3. Let D8 Š D1 6 CD. Since G is not of maximal class, CG .D1 / 6 D1 . In this case, we obtain 1 .D1 CG .D1 // 3 so 1 .G/ 5 > 4, a contradiction. Thus CG .D/ is noncyclic. If G D D R, where R > ¹1º is either cyclic or generalized quaternion, then 1 .G/ D 4. It is easy to show that if a 2-group G contains exactly one subgroup Š Q8 , then we have G 2 ¹Q8 ; SD16 ; Q8 C2n º (the last group has order 2nC2 24 ). It is not easy to classify the 2-groups with exactly two or three subgroups isomorphic to Q8 . Proposition 135.10. Suppose that p > 2 and G is a p-group with nonabelian 1 .G/. If 1 .G/ D 1, then one of the following holds: (a) G D C 1 .G/, where C is cyclic and 1 .G/ Š S.p 3 /, jGj D p 2 jC j. (b) G is a group of maximal class and order 34 . If, in addition, G D 1 .G/, then we have G Š †32 . Proof. By hypothesis, we have D1 .G/ Š S.p 3 /. One may assume that D1 .G/ < G. If D1 .G/ D 1 .G/, then G has no normal elementary abelian subgroup of order p 3 so G is as in (a) or a 3-group of maximal class. In the second case, jGj D 34 so G is as in (b).4 4 According to a report of Mann, the set of groups X of maximal class and order p pC1 , p > 2, satisfying j1 .G/j D p p is nonempty.

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Groups of prime power order

Now let D1 .G/ < 1 .G/. Take x 2 1 .G/ D1 .G/ and write H D hx; D1 .G/i; then 1 .H / D H . If exp.H / D p, then we get 1 .G/ p > 1 by Theorem 5.8(b), a contradiction. Thus exp.H / > p so H is irregular, and we obtain p D 3 by Theorem 12.1(a). If H D 1 .G/, then G is of maximal class by Theorem 13.5 since H has at most two subgroups of order 33 and exponent 3. In this case, in view of D1 .G/ G G, we get jGj D 34 (Theorem 9.6(c)) so G D H . Thus we conclude that (the irregular 3-group) G D 1 .G/ > D1 .G/ has only one nonabelian subgroup D1 .G/ of order 33 and exponent 3 so G Š †32 by Theorem 30.6. Set ˛1 .G/ D jMA.G/j. It is known (see 76) that ˛1 .G/ p unless G is either abelian or minimal nonabelian. Recall that a p-group is said to be an A2 -group if all its subgroups of index p 2 are abelian but G has a nonabelian maximal subgroup (see 76). Given a noncyclic p-group G, set 2 D ¹H < G j ˆ.G/ < H; jG W H j D p 2 º. Proposition 135.11. If a p-group G possesses a proper A2 -subgroup, then there exist ˛1 .G/ .2p 3/ pairwise distinct minimal nonabelian subgroups generating G. Proof. Let H 2 1 be neither abelian nor minimal nonabelian (such an H exists by hypothesis). Then ˛1 .H / p by Remark 76.1. The subgroup H contains a member T of the set 2 ; then T ¤ Z.G/. Let H1 =T D H=T; H2 =T; : : : ; HpC1 =T be all subgroups of order p in G=T ; then we have H1 D H; H2 ; : : : ; HpC1 2 1 . One may assume that H D H1 ; : : : ; Hp are nonabelian. By Remark 76.1, H contains p 1 pairwise distinct minimal nonabelian subgroups, say S1 ; : : : ; Sp1 , that are not contained in T . For i 2 ¹2; : : : ; pº, let Mi be a minimal nonabelian subgroup of Hi that is not contained in T (if Hi is minimal nonabelian, then Mi D Hi ). By Lemma 135.7, we have G D hHp ; S1 i D hMA.Hp / [ ¹S1 ºi. Therefore G D hMA.G/ ¹S2 ; : : : ; Sp ; M2 ; : : : ; Mp1 ºi if p > 2; G D hMA.G/ ¹S2 ºi

if p D 2:

The proof is complete. Given a subgroup N of a nonabelian p-group X, denote by N .X/ the subgroup generated by all those minimal nonabelian subgroups of X which are not contained in N . If N is abelian, then N .X/ D X (Lemma 135.7), and N .X/ D ¹1º if and only if N D X is minimal nonabelian. Proposition 135.12. Suppose that N is a proper subgroup of a nonabelian p-group G. If N .G/ < G, then p D 2. Proof. By the paragraph preceding the proposition, N is neither abelian nor minimal nonabelian. Assume, by way of contradiction, that p > 2. We proceed by induction on jGj. First suppose that N 2 1 . Since j1 j 1 C p .mod p 2 / and, by Exercise 1.6(a), the set 1 contains 0, 1 or p C 1 abelian members, it follows that the set 1 contains

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325

at least p nonabelian members, say N D M1 ; M2 : : : ; ; Mp . Let i > 1. If A Mi is minimal nonabelian and such that A 6 N \ Mi , then A 6 N so that A N .G/. By induction, N \Mi .Mi / D Mi for i > 1. Therefore Mi N .G/ for i > 1. Since p 3, we get N .G/ M2 M3 D G, contrary to the hypothesis. Now suppose that a nonabelian N 62 1 . Then we have N < M1 2 1 , and M1 is nonabelian as well. By the previous paragraph, we see that M1 .G/ D G. However, G D M1 .G/ N .G/. Since, by hypothesis, N .G/ < G, this is a ﬁnal contradiction. Thus, if G is not minimal nonabelian, then p D 2. Let a nonabelian 2-group G be such that 1 D ¹N; M; Aº, where N; M are nonabelian and A is abelian. Then N .G/ M < G. Thus the set of nonabelian 2-groups satisfying the hypothesis of Proposition 135.12 is inﬁnite. However, some useful information on the pairs of 2-groups N < G satisfying N .G/ < G is contained in the following Proposition 135.13. Suppose that N is a proper subgroup of a nonabelian 2-group G. If N .G/ < G, then the following hold: (a) N 2 1 . (b) N .G/ 2 1 . (c) If N \ N .G/ D T , then there is an abelian A 2 1 such that T < A. (d) A is the unique abelian member of the set 1 . In what follows we assume that d.G/ > 2. (e) The set 1 has no minimal nonabelian members. (f) If a nonabelian K < G is such that K 6 N .G/ and K 6 N , then K \ N is nonabelian and N \K .K/ D K \ N .G/. (g) If a nonabelian L < G is such that L 6 N , L 6 N .G/ and L is minimal by inclusion subjecting the above conditions, then d.L/ D 2 and L\N .G/ .L/ D L \ N : Proof. As we have noticed, N is nonabelian. (i) First suppose that N 2 1 . Assume that the set 1 has no abelian members. Let M 2 1 ¹N º be arbitrary and let K 2 1 ¹N; M º be such that N \ M < K. Let S K be minimal nonabelian such that S 6 N \K.D N \M /. Then S 6 M . As S N .G/, we get N .G/ 6 M . Since M is arbitrary, it follows that N .G/ is not contained in any maximal subgroup of G, and we conclude that N .G/ D G, contrary to the hypothesis. Thus there is an abelian A 2 1 . Let N .G/ M 2 1 and let N \ M < L 2 1 ¹M; N º. We claim that L is abelian. Assume that this is false. Take S L such that S 6 N \L.D N \M /. Then

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S 6 M so that S 6 N .G/. Since S 6 N , we get a contradiction. Thus L is abelian, and so one may assume that L D A. (ii) Now we show that N; N .G/ 2 1 . Assume that N [ N .G/ [ A ¤ G. Take x 2 G .N [M [A/, where N .G/ M 2 1 . Then, by Lemma 135.5, there exists a 2 A such that the subgroup U D ha; xi is minimal nonabelian. By the choice of x, the subgroup U is not contained in N and N .G/, which is a contradiction. Thus we get G D N [ N .G/ [ A. Then it follows from Lemma 116.3 that N; N .G/ 2 1 , completing the proof of (a) and (b). (iii) Assume that there exists an abelian A1 2 1 ¹Aº. Set T D N \ A1 and let H; N; A1 be all maximal subgroups of G containing T . Clearly, T ¤ N \ A. Take x 2 H T . Then there is a1 2 A1 such that M D ha1 ; xi is minimal nonabelian. By the choice of x, we have M 6 N whence M N .G/. Such M cover the set H T . Since hH T i D H , it follows that H N .G/. Because H and N .G/ are distinct maximal subgroups of G, we get a contradiction. Thus A1 does not exist, completing the proof of (d). (iv) Supposing that d.G/ > 2, we will prove that the set 1 has no minimal nonabelian members. Assume that this is false and let H 2 1 be minimal nonabelian. Then, by hypothesis, either H D N or H D N .G/. (iv1) Assume that H D N .G/. Then H is the unique minimal nonabelian subgroup of G not contained in N , i.e., in the notation of 76, ˇ1 .G; N / D 1. Then, by Lemma 76.5, d.G/ D 2, contrary to the hypothesis. (iv2) Assume that H D N . Then N is the unique minimal nonabelian subgroup of G not contained in N .G/, and again, as in (iv1), ˇ1 .G; N .G// D 1 and d.G/ D 2, a contradiction. Thus H does not exist, so (e) is true. In what follows we assume that d.G/ > 2. (v) Let a nonabelian K < G be as in (f) (K exists since d.G/ > 2, for example, K 2 1 ¹N; N .G/; Aº). By hypothesis, K is not minimal nonabelian. Due to Lemma 135.7, K \ N .G/ is nonabelian (otherwise, all minimal nonabelian subgroups of K are contained in K \ N so K N by Lemma 135.7). Let S < K be minimal nonabelian such that S 6 K \ N .G/; then S 6 N .G/. By hypothesis, S < N so S K \ N . It follows that K\N .G/ .K/ D K \ N . Since one of the proper subgroups K \ N and ıK\N .K/ of K is nonabelian, we infer from Lemma 135.7 that the second is also nonabelian. This completes the proof of (f). (vi) Let L be such as in (g). Write T D N .G/ \ N \ A. Take x 2 N .G/ T , y 2 N T such that xy ¤ yx (since hN T i D N , such x; y exist). Let H D hx; yi; then H is nonabelian and H 6 N .G/, H 6 N and H < G as d.H / D 2 < d.G/. Then we have H \N .G/ .H / D H \ N by (f). If a nonabelian L H is minimal by inclusion such L 6 N .G/ and L 6 N , then the above argument shows that L is the desired subgroup.

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Remark 3. Suppose that p > 2 and G is a p-group such that 1 < 1 .G/ < p (see Proposition 11), say MA1 .G/ D ¹S1 ; : : : ; Sk º, 1 < k < p. Then all subgroups Si are normal in G. Set H D S1 S2 . If exp.H / D p, then, by Lemma 30.5(a), we have 1 .H / p > 1 .G/, a contradiction. Thus we get exp.H / > p so H is irregular (Theorem 7.2(b)), and we infer that jH j D p 4 , p D 3 by Fitting’s lemma and Theorem 7.1(b). By Theorems 30.7 and 9.6, the group G is of maximal class and order 34 . (In this case, G 6Š †32 .)

Problems Problem 1. Study the p-groups G D k .G/ containing exactly two distinct maximal subgroups U; V such that k .U / D U and k .V / D V . Problem 2. Study the nonabelian p-groups G D k .G/.D hx 2 G j o.x/ D p k i), k > 1, that are not generated by minimal nonabelian subgroups of exponent p k . (See Remark 2 and Theorem 135.9.) Problem 3. Let k > 1. Study the p-groups G > k .G/ such that H D k .H / for all H < G with exp.H / p k . Problem 4. Study the p-groups G such that 1 .ˆ.G// < ˆ.G/ however 1 .ˆ.H // D ˆ.H / .1 .G 0 / < G 0 however 1 .H 0 / D H 0 / for all H < G (two problems). Problem 5. Given k > 1, classify the nonabelian p-groups G D k .G/ containing exactly one minimal nonabelian subgroup of exponent p k . (Compare this with Proposition 135.10.) Problem 6. Classify the p-groups G, p > 2, with 1 .G/ p (see Remark 3). Problem 7. Classify the 2-groups G such that 1 .G/ Š D8 C2 .

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We study the p-groups all of whose nonabelian subgroups of a ﬁxed order are two-generator. A number of important results of this kind for abelian subgroups were proved by Blackburn and are used essentially in the Odd Order paper [FT]. Blackburn has classiﬁed the p-groups, p > 2, all of whose normal abelian subgroups are two-generator (see Theorem 13.7 and Theorem 69.4); for p D 2 this was done by the second author in 50. In contrast, in this section we study the p-groups, p > 2, in which certain nonabelian subgroups are two-generator. All our proofs are fairly short but involved. It is known a comparatively small number of structure results for groups of exponent p. One of such results is Theorem 136.1. Theorem 136.1. Suppose that G is a nonabelian group of order p m and exponent p and n is a ﬁxed integer such that 3 < n < m. Then the following assertions are equivalent: (a) All nonabelian subgroups of G of order p n are two-generator. (b) G is of maximal class with abelian subgroup of index p. In particular, m p. Since all nonabelian groups of order p 3 are two-generator, the hypothesis of Theorem 136.1 is fulﬁlled in case n D 3 for all nonabelian groups of exponent p. Therefore we assume there that n > 3. Note that p > 3 in this theorem (indeed, groups of exponent 3 are of class at most 2). It is impossible to state an analog of Theorem 136.1 for nonabelian p-groups of exponent > p since many such groups need not necessarily contain a nonabelian subgroup of order p n for a ﬁxed n > 2. For groups of exponent p the existence of such subgroups is due to the fact that minimal nonabelian subgroups of exponent p have order p 3 (see Lemma 136.2(ii)). We collect in Lemma 136.2 most results used in what follows. Some results are not stated in full generality. Lemma 136.2. Suppose that G is a nonabelian group of order p m . (i) (Theorem 5.8(b)) If G is of exponent p and n 2 ¹3; : : : ; m 1º, then the number of two-generator subgroups of order p n in G is divisible by p.

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p-groups in which certain proper nonabelian subgroups are two-generator 329

(ii) (Redei; see Lemma 65.1) Suppose that G is minimal nonabelian. Then d.G/ D 2;

jG 0 j D p;

jG W Z.G/j D p 2 :

If exp.G/ D p, then jGj D p 3 (in this case p > 2).1 Next, j1 .G/j p 3 and jG=Ã1 .G/j p 3 . (iii) (Exercise 13.10(a)) Let H < G. If all subgroups of G of order pjH j containing H are of maximal class, then G is also of maximal class. (iv) (Tuan; Lemma 1.1) If A < G is abelian of index p, then jGj D pjG 0 jjZ.G/j. (v) (Proposition 10.17) If B G is nonabelian of order p 3 such that CG .B/ < B, then G is of maximal class. (vi) (Exercise 1.6(a)) The number of abelian subgroups of index p in G is equal to one of the numbers 0, 1 or p C 1. (vii) (Theorem 12.12(a)) Suppose that G contains a subgroup H of maximal class and index p. If d.G/ D 2, then G is also of maximal class (otherwise, we have d.G/ D 3). (viii) (Blackburn; see Theorem 9.5) If G of maximal class is regular (in particular, of exponent p), then jGj p p . (ix) (Hall’s regularity criterion; see Theorem 9.8(a)) If G is irregular, then we have jG=Ã1 .G/j p p . In particular, absolutely regular p-groups are regular. (x) (Theorem 44.12) Suppose that N is a two-generator normal subgroup of G. If N ˆ.G/, then N is metacyclic. (xi) (Lemma 30.3(a, b)) Suppose that a p-group G D k .G/ contains a subgroup E D k .E/. Then we have E U 2 k with k .U / D U . There is also V 2 1 ¹U º with k .V / D V . (xii) (Blackburn; see Theorem 12.1(a)) If a group G is neither absolutely regular nor of maximal class, it has a normal subgroup of order p p and exponent p. (xiii) (Feit–Thompson [FT, Lemma 8.3]; see also Corollary 10.2) If E < G is a maximal normal elementary abelian subgroup of G, p > 2, then E satisﬁes 1 .CG .E// D E. (xiv) (Lemma 57.1) Let A < G be a maximal normal abelian subgroup. Then for any x 2 G A there is a 2 A such that the subgroup hx; ai is minimal nonabelian. (xv) (Theorem 10.28) Any nonabelian p-group G is generated by minimal nonabelian subgroups. (xvi) If G contains an abelian subgroup of index p and jG W G 0 j D p 2 , then G is of maximal class. 1 It

follows from this the fact important in what follows: a nonabelian group of exponent p contains a nonabelian subgroup of order p 3 .

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Groups of prime power order

(xvii) If G is of order p 4 and exponent p and d.G/ D 2, then G is of maximal class. (xviii) If G is a p-group of maximal class with abelian subgroup of index p, then all nonabelian subgroups of G are of maximal class so two-generator. Proof of Lemma 136.2(xvi). We proceed by induction on jGj. By Lemma 136.2(iv), jZ.G/j D p1 jG W G 0 j D p. Write GN D G=Z.G/. Then jGN W GN 0 j p 2 D jG W G 0 j so that jGN W GN 0 j D p 2 . Therefore, if jGj D p 3 , we are done. If jGj > p 3 , then GN is of maximal class by induction. Then G is also of maximal class since jZ.G/j D p. Proof of Lemma 136.2(xvii). Obviously, the group G has an abelian subgroup of index p. Since G 0 D ˆ.G/ and, by hypothesis, jG W G 0 j D p 2 , the result follows from Lemma 136.2(xvi). Proof of Lemma 136.2(xviii). One may assume that jGj > p 3 . We proceed by induction on jGj. Let H 2 1 be nonabelian and A 2 1 be abelian. As G D AH , we conclude that cl.H / D cl.G/ 1 (Fitting’s lemma) hence H is of maximal class. Now let U < G be nonabelian. Let U H 2 1 . By the above, H is of maximal class, and A \ H is abelian of index p in H . Therefore, by induction, U is of maximal class. Proof of Theorem 136.1. We ﬁrst have to prove that (a) ) (b). Assume that G is a counterexample of minimal order; then G is not of maximal class. Since exp.G/ D p, we have G 0 D ˆ.G/. (i) Suppose that n D m 1. Since j1 j 1 .mod p/, there is in G a non-two-generator subgroup, say A, of order p n (Lemma 136.2(i)). By hypothesis, A is abelian. Due to Lemma 136.2(ii), G is not minimal nonabelian since m > 4. Therefore there is a nonabelian M 2 1 . We have d.G/ d.M / C 1 D 3 since, by hypothesis, d.M / D 2. Next, M \ A is an abelian subgroup of index p in M . As jM W M 0 j D jM W ˆ.M /j D p 2 , the subgroup M is of maximal class (Lemma 136.2(xvi)). Then, by Lemma 136.2(vii), d.G/ D 3 since, by assumption, G is not of maximal class. Assume that m D 5 so n D 4. Let B < G be minimal nonabelian; then jBj D p 3 (Lemma 136.2(ii)). If CG .B/ < B, then G is of maximal class (Lemma 136.2(v)), contrary to the assumption. Thus there exists x 2 CG .B/ B. Then K D B hxi is nonabelian of order p 4 D p n and d.K/ D 3, contrary to the hypothesis. Thus m > 5. Now we are ready to complete the proof of our implication in case n D m 1. Take N G G such that N < G 0 and jG=N j D p 5 (recall that jG W G 0 j D p 3 ); then N > ¹1º, and G=N is neither abelian nor of maximal class. Therefore there is in G=N a nonabelian maximal subgroup M=N (Lemma 136.2(ii)); then M 2 1 is nonabelian. By hypothesis, d.M / D 2 hence d.M=N / D 2. Thus all nonabelian maximal sub-

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p-groups in which certain proper nonabelian subgroups are two-generator 331

groups of G=N are two-generator. By induction, G=N is of maximal class, which is a contradiction since d.G=N / D 3 > 2. Thus, if n D m 1, then G is of maximal class with and abelian subgroup of index p. (ii) Now let 3 < n < m 1. Let B < G be nonabelian of order p n (B exists by Lemma 136.2(ii)), and let B < H < G, where jH j D p nC1 . By hypothesis, all nonabelian maximal subgroups of H are two-generator so H is of maximal class by (i). Thus all B-containing subgroups of G of order pjBj D p nC1 are of maximal class. By Lemma 136.2(iii), the group G is of maximal class as well. Take L 2 1 such that d.L/ > 2 (L exists by Lemma 136.2(i)). Then either by hypothesis (provided m D 5) or by induction (provided m > 5), we conclude that the subgroup L is abelian. The implication (b) ) (a) follows from Lemma 136.2(xviii). Finally, the inequality m p follows from Lemma 136.2(viii). Remark 1. Suppose that a p-group G of order p m > p 4 is neither abelian nor minimal nonabelian. We claim that all proper nonabelian subgroups of G have centers of order p if and only if G is of maximal class with abelian subgroup of index p. Indeed, all minimal nonabelian subgroups have the same order p 3 (Lemma 136.2(ii)). If B < G is minimal nonabelian, then CG .B/ D Z.B/ < B is of order p. Therefore G is of maximal class (Lemma 136.2(v)). If R GG is of order p 2 , then CG .R/ has no minimal nonabelian subgroups by the above, so CG .R/ is abelian and maximal in G. Conversely, if G is of maximal class with abelian subgroup of index p, then all proper nonabelian subgroups of G have the same center of order p (Lemma 136.2(xviii)). The simple argument above was the starting point of some subsequent considerations. Theorem 136.3. Suppose that G is a nonabelian group of order p m > p 4 and assume that n 2 ¹4; : : : ; m1º is ﬁxed. If a p-group G has a nonabelian subgroup of order p n and all nonabelian subgroups of order p n have centers of the same order p, then G is of maximal class. Proof. By hypothesis, G is neither abelian nor minimal nonabelian. (i) Suppose that n D m 1. Let R G G be of order p 2 . Then jG W CG .R/j p. In this case, there is a maximal subgroup M of G such that R < M CG .R/. Since R Z.M / is of order p 2 > p, the subgroup M is abelian by hypothesis. Let H 2 1 be nonabelian; then jZ.H /j D p. In this case, H has an abelian subgroup H \ M of index p. By Lemma 136.2(iv), we have jH W H 0 j D pjZ.H /j D p 2 and so, by Lemma 136.2(xvi), H is of maximal class. In this case, all nonabelian members of the set 1 are of maximal class and d.G/ d.H / C 1 D 2 C 1 D 3. Assume, by way of contradiction, that G is not of maximal class. Then it follows from Theorem 12.12(c) that d.G/ D 3 and the number of subgroups of maximal class and index p in G is equal to p 2 . As we have noticed, all nonabelian members of the set 1 are of maximal class. It follows that G has exactly .p 2 C p C 1/ p 2 D p C 1 > 1 pairwise distinct abelian maximal subgroups so cl.G/ D 2. In this case, cl.G/ D 2

332

Groups of prime power order

hence cl.H / D 2, and we conclude that jH j D p 3 . Then jGj D pjH j D p 4 < p m , which is a contradiction. Thus, if n D m 1, then G is of maximal class. (ii) Now suppose that jGj D p m > p nC1 p 5 . Let H < G be nonabelian of order p n and H < F < G, where jF W H j D p. Then, by hypothesis, all nonabelian maximal subgroups of F have centers of the same order p so F is of maximal class by (i). Thus all subgroups of G containing H and having order pjH j are of maximal class. In this case, by Lemma 136.2(iii), the group G is also of maximal class, completing case (ii) and thereby the proof of our theorem. Theorem 106.3 asserts that if a 2-group G and all its nonabelian maximal subgroups are two-generator, then G is either metacyclic or minimal nonabelian. The following theorem extends this result to p-groups, p > 3. Note that in Theorem 136.4 we do not assume that d.G/ D 2 (we deduce this from another additional condition). Theorem 136.4. Suppose that all nonabelian maximal subgroups of a non-absolutely regular p-group G, p > 3, are two-generator. If, in addition, there is M 2 1 such that jM W M 0 j D p 2 , then G is of maximal class with abelian subgroup of index p. Proof. One may assume that exp.G/ > p (otherwise, the result follows from Theorem 136.1); then Ã1 .G/ > ¹1º. We have jG=Ã1 .G/j p p p 5 (the last inequality holds in view of p 5). Then, by Lemma 136.2(ii), there is in G=Ã1 .G/ a nonabelian maximal subgroup H=Ã1 .G/ hence d.H / D 2 by hypothesis, and we obtain that d.G/ d.H / C 1 D 2 C 1 D 3. Therefore, since Ã1 .G/ Ã1 .G/G 0 D ˆ.G/ and jG=Ã1 .G/j > p 3 , it follows that G=Ã1 .G/ is nonabelian. If K=Ã1 .G/ < G=Ã1 .G/ is nonabelian of index p (such a K=Ã1 .G/ exists since G=Ã1 .G/ is not minimal nonabelian in view of Lemma 136.2(ii)), then d.K/ D 2 by hypothesis, whence we get d.K=Ã1 .G// D 2. Thus any nonabelian maximal subgroup of the group G=Ã1 .G/ (of exponent p) is two-generator. It follows that G=Ã1 .G/ is of maximal class by Theorem 136.1. In particular, d.G/ D 2 since Ã1 .G/ ˆ.G/. By Lemma 136.2(i), there is A=Ã1 .G/ < G=Ã1 .G/ of index p such that d.A=Ã1 .G// > 2; then d.A/ > 2 hence A is abelian by hypothesis. Next, there exists M 2 1 such that jM W M 0 j D p 2 by hypothesis. In this case, M \ A is an abelian maximal subgroup of M , and it follows from Lemma 136.2(xvi) that M is of maximal class. As the two-generator p-group G contains a subgroup M of maximal class and index p, we infer that G is also of maximal class by Lemma 136.2(vii). Theorem 136.4 is an extension of Theorem 136.1. Supplement to Theorem 136.4. Let a p-group G of order > p 4 be neither abelian nor minimal nonabelian. If for all nonabelian M 2 1 we have jM W M 0 j D p 2 , then G is of maximal class. Proof. Let N G G be of index p 4 . Then G=N possesses an abelian subgroup A=N of index p. As jA W A0 j > p 2 , it follows by hypothesis that A 2 1 is abelian. Since G

136

p-groups in which certain proper nonabelian subgroups are two-generator 333

is not minimal nonabelian, there is a nonabelian M 2 1 . Then M \ A is abelian of index p in M so, by Lemma 136.2(xvi), M is of maximal class since jM W M 0 j D p 2 . Now, repeating word for word the last part of the proof of Theorem 136.3, we complete the proof. All maximal subgroups of a 3-group of maximal class and order > 34 are two-generator. The same is true for 2-groups of maximal class. However, there are inﬁnitely many 2-groups G with d.G/ D 3 all of whose maximal subgroups are two-generator; see 70, 113. In the proof of Theorem 136.7 we use the following two lemmas. Lemma 136.5. Let G be a p-group of order > p p , p > 2. If G contains a subgroup R of order p p and exponent p and if all such subgroups are (elementary) abelian, then 1 .G/ is also (elementary) abelian. Proof. Bringing to mind our aim, one may assume that G D 1 .G/. It remains to prove that G is abelian. We proceed by induction on jGj. We have R U 2 1 such that 1 .U / D U and V 2 1 ¹U º such that 1 .V / D V (Lemma 136.2(xi)). By induction, U and V are elementary abelian. Then we obtain that U V D G so cl.G/ 2 (Fitting’s lemma). Since G is not minimal nonabelian, we conclude that it is abelian (Theorem 136.1). Lemma 136.6. Let a non-absolutely regular nonabelian group G have order p pC1 , p > 3. Suppose that all nonabelian maximal subgroups of G are two-generator. Then one of the following holds: (a) 1 .G/ is elementary abelian of order p p and G=Ã1 .G/ is of maximal class so d.G/ D 2. Next, G is regular, jZ.G/j D p 2 and G=G 0 is abelian of type .p 2 ; p/. If B 2 1 ¹1 .G/º, then B=B 0 is abelian of type .p 2 ; p/. (b) G is of maximal class with abelian subgroup of index p. Proof. By Lemma 136.2(ii), G is not minimal nonabelian as jG=Ã1 .G/j p p > p 3 . Let a nonabelian K 2 1 ; then d.K/ D 2. By Theorem 136.1, G=Ã1 .G/ is nonabelian of maximal class with an abelian subgroup of index p. It follows that d.G/ D 2 since Ã1 .G/ ˆ.G/. If the quotient group A=Ã1 .G/ is maximal in G=Ã1 .G/ of rank > 2 (see Lemma 136.2(i)), then A is abelian. Suppose that G is not of maximal class; then jZ.G/j D p 2 so that cl.G/ D p 1 4: In this case, G is regular (Theorem 7.1(b)), and we obtain that 1 .G/ D M 2 1 (Theorem 7.2(d)) and exp.M / D p. Assume that M ¤ A. Then, since cl.G/ > 2, the subgroup M is nonabelian. The intersection A \ M is an abelian subgroup of index p in M . Since exp.M / D p and d.M / D 2, we get M 0 D ˆ.M / so that jM W M 0 j D p 2 , and Lemma 136.2(xvi)

334

Groups of prime power order

implies that M is of maximal class. As d.G/ D 2, it follows that G is of maximal class (Lemma 136.2(vii)), contrary to the assumption. Now let M D A; then M.D 1 .G// is elementary abelian of order p p . As we have noticed, jZ.G/j D p 2 . In this case, by Lemma 136.2(iv), jG W G 0 j D pjZ.G/j D p 3 so G=G 0 is abelian of type .p 2 ; p/ since d.G/ D 2. Let R < G 0 be G-invariant of index p. By Lemma 65.2(a), the quotient group G=R is minimal nonabelian. As R < M and M=R Š Ep3 , it follows that 1 .G=R/ Š Ep3 . Let B 2 1 ¹1 .G/º. Since B=R is abelian of type .p 2 ; p/, we get cl.B/ p 2. We conclude from BM D G that cl.B/ p 2 (Fitting’s lemma). Thus we obtain cl.B/ D p 2, and we infer that R D B 0 so that B=B 0 is abelian of type .p 2 ; p/. Theorem 136.7. Let a p-group G of order > p p , p > 3, be not absolutely regular, and let a ﬁxed natural number n be such that 3 < n < p. Suppose that all nonabelian subgroups of G of order p n and exponent p and all nonabelian subgroups of G of order p p and exponent p 2 are two-generator. Then one of the following holds: (a) 1 .G/ is elementary abelian. (b) G is of maximal class; in that case, 1 .ˆ.G// is (elementary) abelian. Proof. Suppose that 1 .G/ is nonabelian. By hypothesis, Theorems 9.5, 9.6, 9.8 and Lemma 136.2(xii), we get j1 .G/j p p1 with equality if and only if G is of maximal class. Suppose that G is of maximal class. Then G is irregular and j1 .ˆ.G//j D p p1 (Theorems 9.5, 9.6). Since, in view of p > 3, R D 1 .ˆ.G// is not metacyclic, we obtain d.R/ > 2 (Lemma 136.2(x)) so it is elementary abelian by Theorem 136.1 if n < p 1 and hypothesis if n D p 1. Next we assume that G is not of maximal class. In this case, there is in G a nonabelian subgroup R of order p p and exponent p (Lemma 136.5), and all such subgroups are of maximal class as n < p (Theorem 136.1). Take R < S G, where jS j D p pC1 . Assume that exp.S/ D p. Then all nonabelian subgroups of index p in S are twogenerator so S is of maximal class (Theorem 136.1), contrary to Lemma 136.2(viii). Thus we get exp.S/ D p 2 . By Lemma 136.6, S is of maximal class as 1 .S/. R/ is nonabelian. Since S runs over the set of all subgroups of G that contain R and have order pjRj, it follows, in view of Lemma 136.2(iii), that G is of maximal class. All p-groups of maximal class with an abelian subgroup of index p satisfy the hypothesis of Theorem 136.7; however, there is a group of maximal class and order p pC1 , p > 3, that does not satisfy this hypothesis [Bla3]. Theorem 136.8. Let G be a nonabelian p-group of order > p 4 and suppose that all its nonabelian subgroups of order p 4 are generated by two elements. Then one of the following holds: (a) G is of maximal class. (b) The subgroup 1 .G/ is abelian.

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p-groups in which certain proper nonabelian subgroups are two-generator 335

Proof. Assume that H D 1 .G/ is nonabelian. (i) We claim that if G contains a nonabelian subgroup B of order p 3 , then G is of maximal class. This is the case if CG .B/ < B by Lemma 136.2(v). Suppose that there is in CG .B/ B an element y of minimal possible order. Since y p 2 Z.B/, we get o.y/ p 2 and the subgroup K D hy; Bi has order p 4 and d.K/ D 3, contrary to the hypothesis since K is nonabelian. By Theorem 9.6(f), a p-group G of maximal class contains an element x such that CG .x/ is of order p 2 . Therefore, if CG .x/ < A < G with jA W CG .x/j D p, then A is nonabelian of order p 3 . Next we assume, in addition, that G is not of maximal class. Then, by the above, all subgroups of G of order p 3 are abelian. Since H D 1 .G/ is assumed to be nonabelian, we get jH j p 4 . Assume that G has no nonabelian subgroup of order p 4 . If G is regular, then we have exp.H / D p. Therefore, since H is nonabelian by assumption, we get jH j D p 3 (Lemma 136.2(ii)). But, by assumption, all subgroups of G of order p 4 which contain H are abelian, a contradiction. Now let G be irregular. As G is not of maximal class, it follows, by Lemma 136.2(xii), that there exists R G G of order p p and exponent p. Assume that p > 2. We claim that R is abelian. Indeed, if p D 3 and R < R1 < G, where jR1 j D 34 , then R1 is abelian by assumption; then R is also abelian. If p > 3, then R is abelian since it has no minimal nonabelian subgroup. Then, by Lemma 136.5, 1 .G/ is abelian so G is as in case (a). Now let p D 2. Then there are in the nonabelian subgroup H two noncommuting involutions x; y. In this case, hx; yi is dihedral so it contains a dihedral subgroup B of order 8. Then B is contained in a nonabelian subgroup of order 24 , contrary to the assumption. In what follows we assume that G has a nonabelian subgroup of order p 4 . By the above, all subgroups of G of order p 3 are abelian. It follows that all nonabelian subgroups of G of order p 4 are minimal nonabelian. Therefore, provided p D 2, G has a nonabelian (dihedral) subgroup of order 8 which is contained in some overgroup of order 16 which is not minimal nonabelian. Thus we must have p > 2. (ii) To obtain a contradiction, it remains to prove that the subgroup H.D 1 .G// is abelian. Therefore, bringing to mind our aim, one may assume, without loss of generality, that G D H.D 1 .G//. Let E < G be a G-invariant elementary abelian subgroup such that jEj is as large as possible. Let x 2 1 .G/ E be of order p. Then we infer that K D hx; Ei is nonabelian (Lemma 136.2(xiii)). If jEj < p 3 , then jEj D p 2 and K is nonabelian of order p 3 and exponent p, contrary to the assumption. Thus we get jEj > p 2 . In this case, by Lemma 136.2(xiv), there is a 2 E such that B D ha; xi is minimal nonabelian; then 1 .B/ D B. Therefore, by Lemma 136.2(ii), B is nonabelian of order p 3 and exponent p, a contradiction. Theorem 136.9. Let G be a nonabelian group of order p m > p 4 . (a) If exp.G/ D p and G possesses only one subgroup, say H , of order p 4 such that d.H / > 2, then G is of maximal class and m D 5.

336

Groups of prime power order

(b) Let p > 3 and let G be an irregular p-group. If G has only one subgroup, say H , of order p 4 and exponent p such that d.H / > 2, then one of the following holds: (b1) G is 5-group of maximal class, all subgroups of G of order 55 and exponent 5 are of maximal class. (b2) G D C 1 .G/, where 1 .G/ is of maximal class and of order 55 and exponent 5, C is absolutely regular. (b3) G is a group of maximal class, j1 .G/j D 54 . Proof. (a) Assume that G is of order p m > p 5 and exponent p. Let L G G be of order p 2 ; then C D CG .L/ has index p in G. If L < B < C is of order p 4 , then d.B/ > 2 (Lemma 136.2(xvii)). Since C of order > p 4 has only one non-two-generator subgroup of order p 4 , it follows that C is nonabelian. By Lemma 136.2(xv), there exists a nonabelian T < C of order p 3 with T 6 B. If U < L is of order p such that U 6 T , then D D T U ¤ B, jDj D p 4 , d.D/ > 2 and D ¤ B, contrary to the hypothesis. Thus m D 5. Due to Lemma 136.2(i), there is B 2 1 satisfying d.B/ > 2. By hypothesis, all members of the set 1 ¹Bº are two-generator so they are of maximal class (Lemma 136.2(xiii)). Assume that G is not of maximal class. Then the set 1 has exactly p 2 members of maximal class (Lemma 136.2(xxi)), and all other p C 1 > 1 members of 1 are not of maximal class so they are not two-generator (Lemma 136.2(xiii)), contrary to the hypothesis. Thus G is of maximal class and order p 5 , as was to be shown. (b) We have exp.G/ > p (Theorem 7.1(b)). If the subgroup 1 .G/ is abelian, then we see that j1 .G/j D p 4 by hypothesis. In this case, G is a 5-group of maximal class (Lemma 136.2(xii)) so G is as in (b3). Next we assume that 1 .G/ is nonabelian. Assume that G is not of maximal class. Then there is in G a subgroup R of order p p and exponent p (Lemma 136.2(xii)). Since p > 3, R is nonabelian (otherwise, R contains more than one subgroup isomorphic to Ep4 ). Then, by part (a), R is of maximal class and p D 5. Thus all subgroups of G of order 55 and exponent 5 have maximal class. Then, by Lemma 136.2(i) or Theorem 136.1, G has no subgroup of order 56 and exponent 5. Let L G G be abelian of type .5; 5/. It is easy to see that C D CG .L/ has index 5 in G (otherwise, LR is of order > 55 and exponent p). Assume that C has a G-invariant subgroup M of order 55 and exponent 5. Since exp.LM / D 5, we get L Z.M /, a contradiction since M , by the above, is of maximal class. Thus M does not exist. Therefore, by Lemma 136.2(xv), C is absolutely regular. Lemma 136.2(xi) then yields R D 1 .G/. Now G D CR is as in (b2). Now let G be of maximal class. Then K D 1 .ˆ.G// is of order p p1 and exponent p. Assume that p > 5; then K is nonabelian by hypothesis. Since K ˆ.G/, it follows that jZ.K/j > p (see Lemma 1.4). Then, by (a), K has two distinct nonabelian non-two-generator subgroups of order p 4 , contrary to the hypothesis. It follows that p D 5. If G has no subgroup of order 55 and exponent 5, it satisﬁes the hypothesis. Now assume that G has a subgroup R of order 55 and exponent 5. Then, by the above, R is of maximal class, completing the proof.

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p-groups in which certain proper nonabelian subgroups are two-generator 337

Remark 2. We present another application of Lemma 136.2(xiv). Let G D k .G/ be a p-group of exponent p k such that G is nonabelian but k .H / is abelian for all H 2 1 . We claim that then G is minimal nonabelian. In view of Lemma 136.2(xi), there is A 2 1 such that k .A/ D A. Then A is abelian by hypothesis. Let x 2 GA be of order p k . By Lemma 2(xiv), there is a 2 A such that H D ha; xi is minimal nonabelian. Since k .H / D H is nonabelian, it follows H D G, and we are done.

Problems Problem 1. Suppose that a group G is neither abelian nor minimal nonabelian with jGj D p m and p k D min¹jAj j A < G is minimal nonabelianº. Let a ﬁxed integer n be such that k < n < m. Study the structure of G provided all its nonabelian subgroups of order p n are two-generator. If p D 2 and n D m 1, then 70 shows that the set of 2-groups satisfying the condition of Problem 1 is inﬁnite. Problem 2. Study the p-groups of order > p 3 all of whose maximal subgroups except one have derived quotient groups of index p 2 . This problem is not solved even for groups of exponent p.2 Problem 3. Study the p-groups all of whose nonabelian subgroups of index p 2 are two-generator. (Note that the p-groups all of whose subgroups of index p 2 are abelian, are classiﬁed; see Lemma 65.1 and 71. All p-groups of maximal class with abelian subgroup of index p and 3-groups of maximal class satisfy this condition.) Problem 4. Study the p-groups all of whose nonabelian subgroups of order p 5 are two-generator. (See Theorem 136.8.)3 Problem 5. Study the p-groups of order p m and exponent p all of whose nonabelian subgroups of order p n , 2 < n < m 1 (n is ﬁxed), are nonnormal. (All p-groups of maximal class and exponent p > 5 with abelian subgroup of index p satisfy this condition.) Problem 6. Classify the p-groups all of whose nonabelian subgroups of class 2 are two-generator. Problem 7. Classify the p-groups G with exactly one nonabelian subgroup H of order p 4 such that d.H / D 3. (See Theorem 136.9.) Some close results are contained in 30, 135. p D 2, then G is of maximal class. Indeed, by Taussky’s theorem (Lemma 1.6), all nonabelian maximal subgroups of G are of maximal class (one may assume that jGj > 8). Assume that G is not of maximal class. Then, by Theorem 5.4, the number of members of maximal class in the set 1 is divisible by 4. Therefore the number of remaining members, say M , of the set 1 is equal to 3 since d.G/ D 3, and we get a contradiction since for these M we have jM W M 0 j > 4, again by Taussky’s theorem. 3 If G is a 2-group of order 25 and all its subgroups of orders 24 and 25 are two-generator, then G is metacyclic. This Blackburn’s result follows from Theorem 66.1. 2 If

137

p-groups all of whose proper subgroups have its derived subgroup of order at most p

Here we give a characterization of the title groups. In the proofs we partly use some ´ ideas of J. Q. Zhang, X. H. Li (Proposition 137.3) and V. Cepuli´ c, O. C. Piljavska (Proposition 137.5). To facilitate the proof of the main result (Theorem 137.7), we shall ﬁrst prove some auxiliary results. Proposition 137.1. Let G be a title group. Then for all x; y 2 G such that hx; yi < G we have o.Œx; y/ p and Œx; y 2 Z.G/. Proof. Suppose that Œx; y ¤ 1. Let X be a maximal subgroup of G containing hx; yi. Then X 0 D hŒx; yi E G with o.Œx; y/ D p and so Œx; y 2 Z.G/. Proposition 137.2. If G is a title group, then G 0 is abelian of order p 3 . Proof. We may assume that G is nonabelian. Let X ¤ Y be two maximal subgroups of G. Then jX 0 j p and jY 0 j p. By Exercise 1.69(a), we get jG 0 W .X 0 Y 0 /j p and so jG 0 j p 3 . By Burnside, G is abelian. Proposition 137.3 (Zhang and Li). If G is a title group and jG 0 j p 2 , then d.G/ 3. Proof. Since jG 0 j p 2 , G is not minimal nonabelian (Lemma 65.1) so there exists a maximal subgroup A with jA0 j D p; then A0 GG. Suppose that M 0 A0 for each maximal subgroup M of G. Then G=A0 is minimal nonabelian, and we get d.G=A0 / D 2 (Lemma 65.1), A0 ˆ.G/ and so d.G/ D 2, completing this case. We may assume that G has a maximal subgroup B such that B 0 6 A0 . In that case, we get jA0 j D jB 0 j D p and A0 \ B 0 D ¹1º. Let a1 ; a2 2 A and a3 ; a4 2 B be such that A0 D hŒa1 ; a2 i and B 0 D hŒa3 ; a4 i. Since jha1 ; a2 ; a3 ; a4 i0 j p 2 , we obtain ha1 ; a2 ; a3 ; a4 i D G by hypothesis, and so d.G/ 4. We assume, by way of contradiction, that d.G/ D 4. Then we have o.Œx; y/ p and Œx; y 2 Z.G/ for any x; y 2 G by hypothesis, and we conclude that G 0 is elementary abelian and G 0 Z.G/. In particular, cl.G/ D 2. For any k 2 ¹1; 2º and l 2 ¹3; 4º we have ha1 ; a2 ; al i < G and hak ; a3 ; a4 i < G hence ha1 ; a2 ; al i0 D hŒa1 ; a2 i and hak ; a3 ; a4 i0 D hŒa3 ; a4 i: It follows that

Œak ; al 2 hŒa1 ; a2 i \ hŒa3 ; a4 i D ¹1º:

137

p-groups G with jH 0 j p for all proper subgroups H of G

339

This implies Œa1 ; a2 a3 D Œa1 ; a2 Œa1 ; a3 D Œa1 ; a2 and Œa2 a3 ; a4 D Œa2 ; a4 Œa3 ; a4 D Œa3 ; a4 : But then ha1 ; a2 a3 ; a4 i is a proper subgroup of G and we have jha1 ; a2 a3 ; a4 i0 j p 2 , contrary to the hypothesis. The proof is complete. Proposition 137.4 (Berkovich). Let G be a nonabelian p-group. If d.G/ D 2, then we have H 0 < G 0 for each H < G. Proof. Let R < G 0 be a G-invariant subgroup of index p in G 0 . Then j.G=R/0 j D p and d.G=R/ D 2. This implies that G=R is minimal nonabelian. For each maximal subgroup H of G we have H 0 R < G 0 , and we are done. ´ Proposition 137.5 (Cepuli´ c and Piljavska). Let G be a title p-group with p > 2. Then for any a; b 2 G we have Œap ; b D Œa; b p D Œa; bp . Proof. Set g D Œa; b. If g commutes with a, then for each n 1 we prove by induction that Œan ; b D Œa; bn . Indeed, for n > 1, Œan ; b D Œaan1 ; b D Œa; ba

n1

Œan1 ; b D Œa; bŒan1 ; b

D Œa; bŒa; bn1 D Œa; bn : In particular, Œap ; b D Œa; bp . Now assume that Œg; a D z ¤ 1. Since hg; ai < G, Proposition 137.1 implies that o.z/ D p and z 2 Z.G/. We note that g a D a1 ga D g.g 1 a1 ga/ D gŒg; a D gz;

and so

i

g a D gz i for all i 1:

We have Œap ; b D Œa ap1 ; b D Œa; ba D Œa; ba

p1

Œa; ba

p2

p1

Œap1 ; b D Œa; ba

p1

Œa ap2 ; b

Œap2 ; b

and so continuing we ﬁnally get Œap ; b D Œa; ba D ga

p1

p1

ga

Œa; ba

p2

p2

: : : Œa; ba Œa; b

: : : g a g D .gz p1 /.gz p2 / : : : .gz/g p

D g p z .p1/C.p2/CC1 D g p z .p1/ 2 D g p D Œa; bp ; where we have used the fact that p > 2. Thus we have proved that in any case we get Œap ; b D Œa; bp . Now, Œa; b p D Œb p ; a1 D Œb; ap D Œa; bp ; and the proof is complete.

340

Groups of prime power order

Proposition 137.6. Suppose that G is a p-group which has one of the following properties: (a) jG 0 j p. (b) d.G/ D 2, jG 0 j D p 2 . (c) p > 2, d.G/ D 2, cl.G/ D 3, G 0 Š Ep3 , Ã1 .G/ Z.G/. (d) d.G/ D 3, cl.G/ D 2, G 0 Š Ep3 or Ep2 . Then G has the title property, i.e., jH 0 j p for each proper subgroup H of G. Proof. If G is a p-group with property (a), then obviously G has the title property. Suppose that G is a p-group in (b). By Proposition 137.4, for each H < G we have H 0 < G 0 and so G has the title property. Now assume that G is a p-group with property (c). Since cl.G/ D 3, it follows that ¹1º ¤ K3 .G/ Z.G/. But d.G/ D 2 and so ¹1º ¤ G 0 =K3 .G/ is cyclic and therefore K3 .G/ Š Ep2 . By assumption, Ã1 .G/ Z.G/ and so ˆ.G/ D Ã1 .G/G 0 is abelian and G=ˆ.G/ Š Ep2 . Also, Ã1 .G/K3 .G/ Z.G/ and in fact Ã1 .G/K3 .G/ D Z.G/. Indeed, if Ã1 .G/K3 .G/ < Z.G/, then G=Z.G/ Š Ep2 . But in that case, G has p C 1 abelian maximal subgroups and this implies (Exercise P1) jG 0 j D p, a contradiction. Let M be any maximal subgroup of G so that jM W ˆ.G/j D p. Then M is either abelian or satisﬁes Z.G/ D Z.M / and M=Z.M / Š Ep2 . In the second case, we may use Lemma 1.1 because ˆ.G/ is an abelian maximal subgroup of M . Then we conclude from jM j D pjZ.M /jjM 0 j that jM 0 j D p. We have proved that in this case G has the title property. Suppose that G is a p-group with property (d). For any x; y 2 G we have Œx p ; y D Œx; yp D 1 and so Ã1 .G/ Z.G/. It follows that ˆ.G/ D Ã1 .G/G 0 Z.G/ and G=ˆ.G/ Š Ep3 . Let M be any maximal subgroup of G satisfying M=ˆ.G/ Š Ep2 . Then p C 1 maximal subgroups of M containing ˆ.G/ are abelian. This implies that jM 0 j p, and we are done. Theorem 137.7. A p-group G has the property that each proper subgroup of G has its derived subgroup of order at most p if and only if one of the following holds: (a) jG 0 j p. (b) d.G/ D 2, jG 0 j D p 2 . (c) p > 2, d.G/ D 2, cl.G/ D 3, G 0 Š Ep3 , Ã1 .G/ Z.G/.1 (d) d.G/ D 3, cl.G/ D 2, G 0 Š Ep3 or Ep2 . Here we have ˆ.G/ D Z.G/. Proof. If G is a p-group in (a)–(d), then Proposition 137.6 implies jH 0 j p for each subgroup H < G. Suppose that G is a p-group all of whose proper subgroups have its derived subgroup of order p. If jG 0 j p, then we obtain the groups in part (a) of our theorem. 1 Such

p-groups G exist. See for example A2 -groups of order p 5 , p > 2, in Proposition 71.5(b).

137

p-groups G with jH 0 j p for all proper subgroups H of G

341

In what follows we assume that jG 0 j p 2 . By Proposition 137.2, G 0 is abelian of order p 2 or p 3 . Due to Proposition 137.3, we have d.G/ 3. (i) First assume that d.G/ D 2. If jG 0 j D p 2 , then we have obtained the groups in part (b) of our theorem. In the sequel we shall assume jG 0 j D p 3 . By Exercise 1.69(a), all p C 1 maximal subgroups Mi .i D 1; 2; : : : ; p C 1/ of G are nonabelian, jMi0 j D p and for any i ¤ j we have Mi0 \ Mj0 D ¹1º so that Mi0 Mj0 Š Ep2

and Mi0 Mj0 Z.G/:

If cl.G/ D 2, then d.G/ D 2 would imply that G 0 is cyclic, contrary to the existence of the subgroup Mi0 Mj0 Š Ep2 . Hence cl.G/ 3. But ¹1º ¤ K3 .G/ D ŒG; G 0 Mi0 Mj0 Z.G/ and so cl.G/ D 3. Set E D Mi0 Mj0 D G 0 \ Z.G/ Š Ep2 . Whenever a; b 2 G are such that ha; bi D G, then Œa; b 2 G 0 E. Indeed, if 1 ¤ Œa; b 2 E, then we have o.Œa; b/ D p and Œa; b 2 Z.G/. But then G=hŒa; bi is abelian and so G 0 D hŒa; bi is of order p, a contradiction. Let g D Œa; b, where ha; bi D G and g 2 G 0 E. For any x 2 G we have g x D ge with some e 2 E. Then i

g x D ge i

and so

p

g x D g:

It follows that Ã1 .G/ centralizes G 0 and so ˆ.G/ D Ã1 .G/G 0 centralizes G 0 . (i1) Now assume p > 2. Suppose, in addition, that G 0 is not elementary abelian. Then we get E D 1 .G 0 / and set ¹1º ¤ Ã1 .G 0 / D hsi < E so that G 0 =hsi Š Ep2 . If K3 .G/ D ŒG; G 0 D hsi, then cl.G=hsi/ D 2 so that d.G=hsi/ D 2 would imply that G 0 =hsi D .G=hsi/0 is cyclic, a contradiction. Hence there is an element c 2 G ˆ.G/ such that g c D gl with l D Œg; c 2 E hsi. Let d 2 G ˆ.G/ be such that hc; d i D G so that Œc; d 2 G 0 E. By Proposition 137.5, we get Œc; d p D Œc; d p D s j , where j 6 0 (mod p). Consider the maximal subgroup C D hˆ.G/; ci. Since g; c; d p 2 C , we have C 0 hŒg; c; Œc; d p i D hl; s j i D E Š Ep2 ; which is a contradiction. We have proved that G 0 Š Ep3 . For any x; y 2 G we get, by Proposition 137.5, Œx p ; y D Œx; yp D 1 and so Ã1 .G/ Z.G/. We have obtained the groups given in part (c) of our theorem. (i2) It remains to consider the case p D 2. Assume, in addition, that ¹1º ¤ K3 .G/ D ŒG; G 0 < E and set ŒG; G 0 D hui, where u is an involution in E Z.G/. Note that ˆ.G/ centralizes G 0 and for each x 2 G ˆ.G/ and y 2 G 0 E we have y x D yu0 with u0 2 hui. Set G0 D CG .G 0 / so that jG W G0 j D jG0 W ˆ.G/j D 2. Since G=hui is of class 2 and d.G=hui/ D 2, we conclude that G 0 =hui is cyclic. Hence if g 2 G 0 E, then g 2 D v

342

Groups of prime power order

is an involution in E hui and therefore E D 1 .G 0 / D hu; vi and Ã1 .G 0 / D hvi. Take some elements a 2 G0 ˆ.G/ and b 2 G G0 . Then ha; bi D G and therefore Œa; b D h 2 G 0 E with h2 D v, ha D h and hb D hu. Consider the maximal subgroup H D hˆ.G/; bi. Since Œa2 ; b D Œa; ba Œa; b D ha h D h2 D v

and

Œh; b D u;

we get H 0 hu; vi D E Š E4 , a contradiction. We have proved that K3 .G/ D ŒG; G 0 D E D G 0 \ Z.G/ Š E4 : Let a; b 2 G ˆ.G/ be such that ha; bi D G. Then g D Œa; b 2 G 0 E, Œg; a D c1 , Œg; b D c2 , where hc1 ; c2 i D E D K3 .G/. We set c3 D c1 c2 and get Œg; ab D Œg; bŒg; ab D Œg; bŒg; a D c2 c1 D c3 : We compute the commutator subgroups of our three nonabelian maximal subgroups X1 D hˆ.G/; ai;

X2 D hˆ.G/; bi

and X3 D hˆ.G/; abi;

where we note that we must have jXi0 j D 2 for i D 1; 2; 3. Since Œg; a D c1 and Œa; b 2 D Œa; bŒa; bb D gg b D g gc2 D g 2 c2 ; we infer that X10 D hc1 i and so we must have g 2 c2 2 hc1 i. This forces either g 2 D c2 or g 2 D c3 . As Œg; b D c2 and Œa2 ; b D Œa; ba Œa; b D g a g D gc1 g D g 2 c1 ; we obtain X20 D hc2 i and so we must have g 2 c1 2 hc2 i. This forces either g 2 D c1 or g 2 D c3 . With the above we get exactly g 2 D c3 . Since Œg; ab D c3 and Œa2 ; ab D Œa; aba Œa; ab D g a g D gc1 g D g 2 c1 (where we have used the fact that Œa; ab D Œa; b), we infer that X30 D hc3 i and so we must have g 2 c1 2 hc3 i. But we know that g 2 D c3 and so g 2 c1 D c3 c1 D c2 2 hc3 i, a contradiction. We have proved that such 2-groups do not exist! (ii) Finally, assume that d.G/ D 3. For any x; y 2 G we have hx; yi < G and so Proposition 137.1 implies that o.Œx; y/ p and Œx; y 2 Z.G/. But then G 0 is elementary abelian (of order p 2 or p 3 ) and G 0 Z.G/ and so we have obtained the groups from part (d) of our theorem. For any a; b 2 G we have Œap ; b D Œa; bp D 1 and so ˆ.G/ Z.G/. If Z.G/ 6 ˆ.G/, then there is a maximal subgroup M of G such that G D hM; xi, where x 2 Z.G/. But then G 0 D M 0 and so jG 0 j D 2, a contradiction. Hence ˆ.G/ D Z.G/. Theorem 137.7 is proved.

138

p-groups all of whose nonnormal subgroups have the smallest possible normalizer

1o . A p-group G is called an S1 -group if jNG .H / W H j D p for all nonnormal subgroups H of G. We determine here up to isomorphism all non-Dedekindian S1 -groups by using the results (but not the proofs) from [ZG]. This solves Problem 116(i) stated by the ﬁrst author. Lemma 138.1. Assume that G is a minimal nonabelian p-group of order > p 3 . If G is an S1 -group, then 2

2

G Š ha; b j ap D b p D 1; ab D a1Cp i D Mp .2; 2/: Proof. Suppose that G has an elementary abelian subgroup E of order p 3 . Then each subgroup of order p in E must be normal in G and so E Z.G/. Let X be a cyclic nonnormal subgroup in G. But then NG .X/ E and so jNG .X/ W X j p 2 , a contradiction. We have proved that G is metacyclic and so n

m

G Š ha; b j ap D b p D 1; ab D a1Cp

n1

i;

where n 2, m 1, n C m 4 and G 0 D hŒa; bi D hap i. Obviously, hbi is not normal in G and NG .hbi/ D hbi hap i which forces n D 2 and so m 2. Assume m2 m2 m2 m 3 so that b p 2 Z.G/, o.ab p / D p 2 and hab p i is not normal in G. m2 m2 m2 Indeed, if hab p i C G, then we have ŒG; hab p i hab p i \ G 0 D ¹1º and m2 p so ab 2 Z.G/ which forces a 2 Z.G/, a contradiction. But n1

NG .hab p and therefore jNG .hab p we are done.

m2

i/ W hab p

m2

m2

i/ D hai hb p i

ij p 2 , a contradiction. Hence m D 2 and

Theorem 138.2 ([ZG]). Assume that G is a nonabelian p-group, p > 2, and jGj > p 3 . Then G is an S1 -group if and only if 2

2

G Š ha; b j ap D b p D 1; ab D a1Cp i D Mp .2; 2/: Proof (Janko). Suppose that G is a nonabelian p-group, p > 2, jGj > p 3 and G is an S1 -group.

344

Groups of prime power order

First assume that G is nonabelian of order p 4 . Suppose, in addition, that G is not minimal nonabelian and let H be a minimal nonabelian subgroup of order p 3 in G. Then 1 .H / contains a G-invariant subgroup E Š Ep2 . Let E0 be a noncentral subgroup of order p of H with E0 < E. Then we must have NG .E0 / D E. On the other hand, CH .E/ D E and so CG .E/ 6 H , a contradiction. Hence G is minimal nonabelian and so Lemma 138.1 implies that G Š Mp .2; 2/. Now we use induction on jGj. Let G be a nonabelian S1 -group of order p 5 , and let N be a subgroup of order p in G 0 \ Z.G/. Then G=N is an S1 -group and so (by induction) G=N is either abelian or G=N Š Mp .2; 2/. (i) First assume that G=N is abelian. Then G 0 D N is of order p. Let K be a minimal nonabelian subgroup of G so that K 0 D G 0 D N and K satisﬁes either jKj D p 3 or K Š Mp .2; 2/ (Lemma 138.1). We get G D KCG .K/, where K \ CG .K/ D Z.K/ and CG .K/ 6 K (since jGj p 5 ). If Y is a nonnormal cyclic subgroup of K, then jNG .Y / W Y j p 2 , a contradiction. (ii) Now suppose that G=N Š Mp .2; 2/. Since G=N is metacyclic, it follows from Theorem 36.1 that G is also metacyclic. As G 0 Š Cp2 , we conclude that G is a metacyclic A2 -group of order p 5 . By Proposition 71.2(a), we have 3

2

2

G Š ha; b j ap D 1; b p D ap ; ab D a1Cp i; where D 0; 1. Consider the maximal subgroup L D haihb p i of G. Since hai is a cyclic subgroup of index p in L (and L is obviously noncyclic), there is an element x of order p in Lhai. But G=hai Š Cp2 acts faithfully on hai and so x does not centralize hai which implies that hxi is not normal in G. On the other hand, L is minimal nonabelian and so x centralizes hap i and therefore jNG .hxi/ W hxij p 2 , a contradiction. Since Mp .2; 2/ is obviously an S1 -group, our theorem is proved. Theorem 138.3. Let G be a non-Dedekindian 2-group. Then G is an S1 -group if and only if one of the following holds: (a) G is of maximal class distinct from Q8 . (b) G Š ha; b j a2 D b 4 D 1; ab D a1C2 D 1, then n 3. n

n1

i, where n 2, D 0; 1, and if

Proof (Janko). Suppose that G is a non-Dedekindian 2-group and G is an S1 -group. By Lemma 138.1, each minimal nonabelian subgroup H of G is isomorphic to D8 , Q8 or M2 .2; 2/ D H2 . (i) Suppose that G possesses a subgroup H Š D8 . Let t be a noncentral involution in H . Since jCH .t/ W htij D 2, we have CG .t/ < H and so CG .H / H . By Proposition 10.17, G is of maximal class. In what follows we may assume that D8 is not a subgroup of G. (ii) Now suppose that G possesses a subgroup H Š Q8 so that in this case we must have H < G. Set H D ha; bi and Z.H / D hzi. First assume that H has a cyclic subgroup C of order 4 such that C is not normal in G. Then NG .C / D H which implies

138

Smallest normalizers of nonnormal subgroups

345

that CG .H / H and so again (Proposition 10.17) G is of maximal class. In what follows we may assume that each cyclic subgroup of H is normal in G. This gives that H E G and G D HK, where K D CG .H / and H \ K D hzi. Suppose that K has a cyclic subgroup K1 D hki of order 4. If k 2 D z, then we have HK1 D H K1 with H \ K1 D hzi. But we know that in this case HK1 possesses a subgroup isomorphic to D8 , a contradiction. Hence we must have HK1 D H K1 , where we set k 2 D z 0 . In this case, we consider haki Š C4 , where .ak/2 D zz 0 and .ak/b D ak z, which implies that haki is not normal in HK1 . But ha; ki Š C4 C4 centralizes haki and so HK1 is not an S1 -group, a contradiction. We have proved that K is elementary abelian. But then G is Dedekindian, contrary to our assumption. (iii) In the sequel we assume that each minimal nonabelian subgroup of G is isomorphic to H2 . Then we use Theorem 57.3, where such 2-groups are classiﬁed. If G is a group from part (b) of Theorem 57.3, then G is of exponent 4 and by inspection we see that the only S1 -group there is the group H2 which appears in part (b1) of that theorem. We have obtained a group from part (b) of our theorem for the special case n D 2 and D 0. Indeed, in case (b2) of Theorem 57.3, G contains the unique minimal nonmetacyclic group H of order 25 given in Theorem 66.1(d) with H D ha; b; c j a4 D b 4 D Œa; b D 1; c 2 D a2 ; ac D ab 2 ; b c D ba2 i; and here hai is not normal in H , but ha; bi centralizes hai and so H is not an S1 -group. In case (b3) of Theorem 57.3 the groups there contain a minimal nonmetacyclic group of order 25 as a subgroup. Finally, in case (b4) of Theorem 57.3, we have a subgroup K D Bhbi, where o.b/ D 4, hbi \ B D ¹1º, b inverts each element of B, hbi is not normal in K, but 1 .B/ is elementary abelian of order 4 and 1 .B/ centralizes hbi and so such a group K is not an S1 -group. It remains to consider the groups G in Theorem 57.3(a) which are of exponent 8. Here G has a unique abelian maximal subgroup A. We have exp.A/ 8, E D 1 .A/ D 1 .G/ D Z.G/ is of order 4. All elements in G A are of order 4 and if b is one of them, then CA .b/ D E and b inverts each element of ˆ.A/ and of A=E. Since hbi is not normal in G and E centralizes hbi, we get E Š E4 and therefore A is of rank 2. If C D hci is any cyclic subgroup of order 4 in A, then C C G and so c b D c 1 . In case c 2 D b 2 , we infer that hb; ci Š Q8 , a contradiction. Thus we have proved that b 2 D u is not a n1 square in A and so A D hui hai with o.a/ D 2n , n 3, and we set a2 D z so that E D 1 .A/ D 1 .G/ D hu; zi. Since b inverts each element of A=E, we have ab D a1 with 2 E. If D u, then .ba/2 D ba ba D b 2 ab a D u a1 u a D 1; a contradiction. In case D uz, we obtain .ba/2 D ba ba D b 2 ab a D u a1 uz a D z:

346

Groups of prime power order n2

But ba inverts v D a2 (of order 4) and so hba; vi Š Q8 , a contradiction. We have proved that ab D a1 z , D 0; 1, and so we have obtained all groups stated in part (b) of our theorem. (iv) Conversely, if G is a 2-group of maximal class (distinct from Q8 ), then G is obviously a non-Dedekindian S1 -group. Assume that G is a 2-group from part (b) of our theorem. If n D 2, then D 0 and G Š H2 , which is an S1 -group. Suppose that n 3 n1 and set b 2 D u and a2 D z so that ab D a1 z and 1 .G/ D Z.G/ D hu; zi. i j Let x D b a with 2 1 .G/ (i; j are integers) be any element in G .hui hai/. Then we compute x 2 D .b i aj /2 D b i aj b i aj D b 2 .b 1 aj b/aj D u.ab /j aj D u .a1 z /j aj D uaj z j aj D uz j ; and we see that x 2 2 1 .G/ hzi. For any y 2 1 .G/ hzi we infer that G=hyi is dihedral or semidihedral and so G=hyi is an S1 -group. Let M be any subgroup of G. If M contains an element y 2 1 .G/hzi, then M=hyi is a subgroup of G=hyi and so either M E G or jNG .M / W M j D 2. Suppose that M does not contain any element in 1 .G/ hzi. Then, by the above, M hui hai and so M E G since every subgroup of huihai is normal in G. Hence G is an S1 -group. Our theorem is proved. 2o . Below we prove a more general result. Let G be a non-Dedekindian p-group and C < G be nonnormal cyclic of minimal possible order. Write jC j D p .G/ D p . Denote by C the set of all nonnormal cyclic subgroups of G of order p . Clearly, if H < G is non-Dedekindian, then we obtain that .H / .G/. If G satisﬁes the hypothesis of Theorem 138.4, then the members of the set C are maximal cyclic subgroups of G. Our main aim is to prove the following Theorem 138.4 (Berkovich). Let G be a non-Dedekindian p-group of order > p 3 such that any nonnormal cyclic subgroup of order p D p .G/ has index p in its normalizer. Then one and only one of the following holds: (a) If p > 2, then G is the unique nonabelian metacyclic group of order p 4 and exponent p 2 , D 2. (b) G is a 2-group of maximal class. In this case, D 1 unless G is generalized quaternion and D 2. (c) G D hx; y j x 2 D y 4 D 1; x y D x 1C2 i is metacyclic, where 2 ¹0; 1º, n 2, and if D 1, then n > 2. Here jGj D 2nC2 , G D hyi hxi (semidirect n1 product with kernel hxi), Z.G/ D hx 2 ; y 2 i D 1 .G/, ˆ.G/ D hx 2 ; y 2 i D hy 2 i G 0 is abelian of type .2n1 ; 2/, G=G 0 ia abelian of type .4; 2/, G=hy 2 i is of maximal class but not generalized quaternion. In this case, D 2. n

n1

Proof. If G is a 2-group of maximal class satisfying the hypothesis of the theorem, then all its abelian subgroups of order > 4 are normal cyclic and all its nonnormal sub-

138

Smallest normalizers of nonnormal subgroups

347

groups of order > 4 are of maximal class. If K < G is nonabelian, then a normal subgroup of K of index > 2 is contained in ˆ.K/ < ˆ.G/ and so are normal in G. Therefore, if U 2 C , then jU j 4 hence 2, and jNG .U / W U j D 2, and so G satisﬁes the hypothesis. This proves (b). Next we assume that our group G is not a 2-group of maximal class. (i) If we have U 2 C , then any abelian subgroup A of G containing U has order at most pjU j.1 Proof. Indeed, we have A NG .U / so that jA W U j jNG .U / W U j D p. (ii) G has no elementary abelian subgroup of order p 3 . Proof. Assume, by way of contradiction, that a subgroup E < G is elementary abelian of order p 3 . Then, by (i), all subgroups of order p contained in E are normal in G, and we conclude that E Z.G/. Let U 2 C . Then jE \ U j p hence, by the product formula, EU is abelian of order p 2 jU j, contrary to (i). (iii) All subgroups of order p are normal in G and j1 .G/j D p 2 (recall that we have assumed that G is not a 2-group of maximal class). In particular, > 1. Proof. Assume that a subgroup T < G is nonnormal of order p. Then D 1 hence jNG .T /j D pjT j D p 2 by hypothesis. Since NG .T / D CG .T /, it follows that G is of maximal class by Proposition 1.8 (then, by assumption, p > 2). This argument shows that any subgroup of G properly containing a nonnormal subgroup of order p is either of order p 2 or of maximal class. Then, since p > 2, G possesses a G-invariant abelian subgroup R of type .p; p/ (Lemma 1.4). As Z.G/ is cyclic, it follows that there is in R a non-G-invariant subgroup L of order p. Write E D CG .R/; then jEj D p1 jGj p 3 . Let B=R E=R be of order p. Then B is abelian of order p 3 D p 2 jLj (see (ii)), contrary to (i). Thus the group G has no nonnormal subgroups of order p, and we conclude that 1 .G/ Z.G/ so that j1 .G/j D p 2 by (ii). It follows from (iii) that G is not of maximal class also for p > 2. (iv) The group G possesses a minimal nonabelian subgroup H such that .H / D

.D .G//. Moreover, H is metacyclic of order p 4 and exponent p 2 and D 2. Proof. Let T 2 C and write N D NG .T /; then jN W T j D p so that jN j D pjT j D p C1 p 3 since > 1 by (iii). Assume that N Š Q8 . Then D 2 and CG .T / D T so G is of maximal class, a contradiction. Thus we have N 6Š Q8 so that N is abelian of type .p ; p/ by Theorem 1.2. Denote by N1 the normalizer of N in G; then N1 is nonabelian. The subgroup N GN1 has exactly c .N / D p cyclic subgroups of index p (here c .N / is the number of cyclic subgroups of order p in N ). Let N1 act via conjugation on cyclic subgroups of index p in N . Since the stabilizer of T in N1 coincides 1 By

hypothesis, if jNG .U / W U j > p, then U G G.

348

Groups of prime power order

with NN1 .T / D N , we get jN1 W N j D c .N / D p. Therefore the normalizer of any cyclic subgroup of N of order jT j in N1 (even in G by hypothesis) coincides with N . Assume that p D 2 and all minimal nonabelian subgroups of N1 are Š Q8 (G has no dihedral subgroup of order 8 by (iii)). Then, by Corollary A.17.3 and hypothesis, we get N1 D QE, where Q is generalized quaternion and exp.E/ 2. By construction, N1 ¤ Q (otherwise, all abelian subgroups of Q are cyclic but N < Q is noncyclic abelian) so that E > ¹1º. If jQj > 8, then there exists in Q a nonnormal cyclic subgroup M of order 4 (then D 2), and the subgroup NN1 .M / D NQ .M / E has order 8jEj > 2jM j, contrary to the hypothesis. Thus Q Š Q8 so that N1 is Dedekindian. In this case, T G N1 , a contradiction since NG .T / D N < N1 . Now let p be arbitrary. By the previous paragraph, there is in N1 a minimal nonabelian subgroup H 6Š Q8 ; then we have jH j > p 3 since 1 .H / Z.H / by (iii). By (ii) and Lemma 65.1, H is metacyclic. Assume that .H / > . Then jH j p .H /C2 > p C2 D jN1 j, which is a contradiction. Thus .H / D . Let L < H be nonnormal cyclic of order p . By hypothesis, jH j D p jNH .L/j D p C2 . Let Z G H be cyclic such that H=Z is cyclic; then H 0 < Z. It follows that L \ Z D ¹1º, and we conclude that H D L Z, a semidirect product with kernel Z, and jZj D jH W Lj D p 2 . Let 2 .L/ D hbi and Z D hzi. Then the cyclic subgroup U D hbzi of order p 2 is not normal in H so that .H / D 2 .G/ > 1 hence D 2 and jH j D jLjjZj D p 4 so H is metacyclic of exponent 4. As a by-product, we have also obtained N1 D H . (v) The theorem is true if p > 2. Proof. Indeed, by (ii), (iii) and Proposition 10.19, G is metacyclic. Take in G a minimal nonabelian subgroup H such that .H / D (H exists in view of (iv)). By (iv), H is of order p 4 and exponent p 2 . Let V be of order p such that V ¤ H 0 ; then we obtain V G G by (iii). In this case, H=V is a nonabelian subgroup of order p 3 in the metacyclic group G=V . It follows from Proposition 10.19 that H=V D G=V so G D H , as was to be shown. In what follows we assume that p D 2. Let H G be minimal nonabelian with

.H / D .D .G//. By (iv), H is metacyclic of order 16 and exponent 4 and D 2. (vi) The group G is as in (c). Proof. If G is minimal nonabelian, then it coincides with H so G is metacyclic of order 16 and exponent 4 by the paragraph preceding (vi). Next we assume that G is not minimal nonabelian; then we get H < G so jGj > 16. Set jGj D 2nC2 ; then n > 2. As we know, H D ha; b j a4 D b 4 D 1; ab D a3 i. Let L D hbi < H be nonnormal cyclic of order 4; then L is not normal in G so that

D 2 by (iii). In this case, by hypothesis, we obtain NH .L/ D ha2 ; bi D NG .L/ (the last equality holds by hypothesis). Therefore NG=Ã1 .L/ .L=Ã1 .L// D NG .L/=Ã1 .L/ is abelian of type .2; 2/. It follows from Proposition 1.8 that G=Ã1 .L/ is of maximal class. We claim that G is metacyclic. Assume that this is false. If Ã1 .L/ < G 0 , then

138

Smallest normalizers of nonnormal subgroups

349

jG W G 0 j D 4 so G is of maximal class by Proposition 1.6, contrary to the assumption. Thus Ã1 .L/ 6 G 0 so that G 0 \ L D ¹1º. In this case, G 0 Š .G=Ã1 .L//0 hence G 0 is cyclic of order 2n1 . As Ã1 .L/ D ˆ.L/ < ˆ.G/, we get d.G/ D d.G=Ã1 .L// D 2 and so G=G 0 is abelian of type .4; 2/. The subgroup F D L G 0 (semidirect product with kernel G 0 ) is a metacyclic maximal subgroup of G. Also, we have 1 .G/ < F as j1 .G/j D 4 D j1 .G 0 / 1 .L/j. We obtain G=G 0 D .F=G 0 / .K=G 0 /, where F=G 0 Š L is cyclic of order 4 and jK=G 0 j D 2. Since F \K D G 0 is cyclic, it follows that K has only one subgroup of order 2 since it does not contain Ã1 .L/, and we infer that K is either cyclic or generalized quaternion. We also have G D L K (semidirect product with kernel K) and jKj D 2n . Let M=G 0 < G=G 0 be cyclic of order 4 such that M ¤ F (note that G=G 0 contains exactly two cyclic subgroups of order 4 and F=G 0 is one of them); then M is metacyclic. Since d.G/ D 2, we have 1 D ¹F; M; K Ã1 .L/º; where F and M are metacyclic. Assume that KÃ1 .L/ is metacyclic. Then, by Theorem 66.1, we get jGj D 25 and exp.G/ D 4, a contradiction since, in this case, K is cyclic of order 8. Thus K Ã1 .L/ is nonmetacyclic so K is a generalized quaternion group of order 2n , where n > 2, by the above. If n > 3, then K possesses a nonnormal cyclic subgroup D of order 4 and NK .D/ Ã1 .L/ NG .D/ has order 16 D 4jDj > 2jDj, contrary to the hypothesis. Hence n D 3, K Š Q8 and b induces on K an involutory automorphism since b 2 centralizes K but b does not centralizes K. Let G 0 ; U; V be pairwise distinct cyclic subgroups of order 4 in K. Since G 0 G G, it follows that jG W NG .U /j 2 so U G G. Similarly, V G G. Then b inverts G 0 ; U; V so K, and we conclude that K is abelian (Burnside), a contradiction. The obtained contradiction shows that G is metacyclic. Let Z G G be cyclic such that G=Z is cyclic. Then we get G 0 < Z so jG=Zj < jG=G 0 j D 8. Since G has no cyclic subgroup of index 2 (Theorem 1.2), we obtain jG=Zj D 4. We have G D L Z (semidirect product since L \ G 0 D ¹1º). All the above allows us to write out deﬁning relations for G in the following form: G D ha4 D z 2 D 1; z a D z 1C2 n

n1

i;

where 2 ¹0; 1º (Theorem 1.2), and if D 1, then n > 2, so G is as in (c). (vii) It remains to check that the groups (a)–(c) listed in the conclusion of our theorem satisfy the hypothesis. The group from (a), obviously, satisﬁes the hypothesis. As we have noticed, 2-groups of maximal class satisfy the hypothesis (see the ﬁrst paragraph of the proof). Now let G be a group from (c) and jGj D 2nC2 > 24 . In this case, .G/ D D 2 since 1 .G/ D Z.G/ and the set C2 contains a nonnormal cyclic subgroup hai of

350

Groups of prime power order

order 4 (see the deﬁning relations for G). Let L < G be nonnormal cyclic of order 4 and Z D hzi. We have to prove that jNG .L/ W Lj D 2. Since cl.G/ > 2, CG .Z/ is the unique abelian maximal subgroup of G and LZ D G since L 6 CG .Z/, and so we obtain L\Z D ¹1º by the product formula. It follows that CG .L/ D LZ.G/ is abelian of type .4; 2/. Since NG .L/ D CG .L/, we ﬁnally get jNG .L/ W Lj D 2. The proof is complete. To prove that Theorem 138.3 follows from Theorem 138.4, it sufﬁces to show that all groups (a)–(c) from the conclusion of the latter theorem satisfy the hypothesis of Theorem 138.3. It sufﬁces to check only the groups from Theorem 138.4(c). We omit this routine checking. Note only that if H < G is nonnormal and contains 1 .G/, then jNG .H / W H j D 2 since G=1 .G/ is of maximal class; therefore it sufﬁces to assume that H is cyclic (see Proposition 10.19). Let p be a prime and n 4 be a natural number. Denote by tp .n/ the number of pairwise nonisomorphic groups of order p n satisfying the hypothesis of Theorem 138.4. If p > 2, then we have tp .n/ D 1 for n D 4 and tp .n/ D 0 for n > 4. Next, t2 .n/ D 4 for n D 4 and t2 .n/ D 5 for n > 4. This follows from Theorem 138.4. Corollary 138.5. If any nonnormal subgroup (cyclic subgroup) of a non-Dedekindian p-group G has index p in its normalizer, then G is one of the groups (a)–(c) from Theorem 138.4.

Problems Problem 1. Classify the non-Dedekindian p-groups G such that whenever C A < G, where C 2 C and A is abelian, then jA W C j p. Problem 2. Classify the non-Dedekindian p-groups such that whenever C < G is nonnormal cyclic, then jNG .C / W C j p 2 . Problem 3. Let H be a subgroup of a p-group G. Study the structure of G if for every maximal cyclic subgroup C of G not contained in H we have jNG .C / W C j D p. Problem 4. Let H be a maximal subgroup of a non-Dedekindian 2-group G. Describe the structure of G provided all minimal nonabelian subgroups of G not contained in H (see Theorem 10.28) are isomorphic to Q8 . Problem 5. Classify the non-Dedekindian p-groups G such that jNG .C / W C j D p for all nonnormal maximal cyclic C < G. Problem 6. Classify the non-Dedekindian p-groups G such that jNG .H / W H j D p for all maximal nonnormal H < G. Problem 7. Suppose that H < G is noncyclic. Study the structures of H and G provided all maximal abelian (cyclic) subgroups of H are maximal abelian (cyclic) in G.

138

Smallest normalizers of nonnormal subgroups

351

In conclusion we solve Problem 1 (see Theorem 138.7 and 138.8). Lemma 138.6. Let G be a non-Dedekindian metacyclic p-group of order > p 3 satisfying the hypothesis of Problem 1. Then G is one of the groups from Theorem 138.4. Proof. One may assume that G satisﬁes jGj > p 4 . Also assume that G is not a 2-group of maximal class. Then R D 1 .G/ is abelian of type .p; p/ (Lemma 1.4 and Proposition 10.19). Since CG .R/ contains an abelian subgroup A of type .p 2 ; p/, it follows that all subgroups of order p are normal in G whence R Z.G/. Let H G be minimal nonabelian. By what has just been said and Proposition 10.19, we get jH j > p 3 . As in the proof of Theorem 138.4, we deduce that jH j D p 4 and exp.H / D p 2 . Let K < 1 .G/ be of order p such that K ¤ H 0 . Then H=K is a nonabelian subgroup of order p 3 in G=K so, since jG=Kj > p 3 , it follows that G=K is a 2-group of maximal class (Proposition 10.19). By Taussky’s Proposition 1.6, we have jG W G 0 j > 4 so, by assumption, K 6 G 0 , and we conclude that G 0 is cyclic of index 8 in G hence G=G 0 is abelian of type .4; 2/. As above, there is a cyclic subgroup Z > G such that G=Z is cyclic of order 4. Let L be a cyclic subgroup of order 4 in H such that L \ H 0 D ¹1º. Then we have L 2 C ( D 2) and L \ Z D L \ G 0 D ¹1º. Hence G D LZ and CZ .L/ D 1 .Z/ shows that G is a group from Theorem 138.4(c). Theorem 138.7. If a non-Dedekindian p-group G of order > p 3 , p > 2, is a group of Problem 1, then G is metacyclic of order p 4 and exponent p 2 . Proof. In view of Lemma 138.6, it sufﬁces to show that G is metacyclic. Assume, by way of contradiction, that G is not metacyclic. As in the proof of Theorem 138.4, we have (i) 1 .G/ Z.G/ is of order p 2 . Then G is not a group of maximal class. (ii) If H G is minimal nonabelian, it is metacyclic of order p 4 and exponent p 2 . By assumption, H < G. Let R H be of order p such that R ¤ H 0 ; then we have R G G by (i). In this case, H=R is a nonabelian subgroup of order p 3 in G=R. Write N=R D NG=R .H=R/; then jN j > p 4 . By (i), N has no elementary abelian subgroup of order p 3 . Then N is a group from Theorem 13.7. Therefore, by (i), N is metacyclic of order > p 4 , contrary to Lemma 138.6. Not all arguments in the proof of Theorem 138.4 work under hypothesis of the following theorem (for example, one cannot assert that G=Ã1 .L/ is a 2-group of maximal class). Theorem 138.8. If G is a non-Dedekindian nonmetacyclic 2-group of Problem 1 and jGj > 25 , then D 2. Suppose that G 6Š Q2n C2 . Then (a) There is in G an abelian subgroup, say A, of index 2. Let G D hx; Ai. (b) 1 .G/ D Z.G/ is of order 4 and all elements of the set G A have order 4. Also, we have d.A/ D 2.

352

Groups of prime power order

(c) All subgroups of A of order 4 are normal in G. (d) Any cyclic subgroup of G of order 4 not contained in A is nonnormal in G. Conversely, any group G satisfying conditions (a)–(d) also satisﬁes the condition of Problem 1. Proof. Let RGG be abelian of type .2; 2/ (Lemma 1.4) and let B=R < CG .R/=R be of order 2; then B is abelian of order 8. By hypothesis, all subgroups of R of order 2 are normal in G so that R Z.G/. As in part (ii) of the proof of Theorem 138.4, we obtain j1 .G/j D 4 so 1 .G/ D R Z.G/. Arguing as in the proof of Theorem 138.4, one proves that any minimal nonabelian subgroup of G is either isomorphic to Q8 or metacyclic of order 16 and exponent 4. As, by hypothesis, G 6Š Q2m C2 , there is in G a metacyclic minimal nonabelian subgroup H of order 16 and exponent 4 (H exists in view of Corollary A.17.3 and what has been said in the previous paragraph). Since H has a nonnormal cyclic subgroup, say L, of order 4, we have D 2. By hypothesis, CH .L/ of order 8 is maximal abelian in G so that CG .L/ D CH .L/. As 1 .G/ Z.G/ < CG .L/, it follows that Z.G/ D 1 .G/. Let A be a maximal normal abelian subgroup of G; then 1 .G/ < A so that A is metacyclic but noncyclic since j1 .G/j D 4. For any x 2 G A there exists an element a 2 A such that U D ha; xi is minimal nonabelian (Lemma 57.1). Since exp.U / D 4, we get o.x/ D 4 so that all elements of the set G A have the same order 4. Because x 2 2 1 .G/ D Z.G/ centralizes A, it follows that x 2 2 A, and we conclude that G=A is elementary abelian. Therefore, since jGj > 25 , we get jAj > 8 (otherwise, G=A Š Aut.A/ Š D8 ). In this case, all cyclic subgroups of A of order 4 are normal in G by hypothesis. Let R < A be cyclic of order 4. Then jG W CG .R/j D 2 since R 6 Z.G/ in view of exp.Z.G// D 2. Assume that jG=Aj > 2. Then we infer that CG .R/ > A hence there is b 2 CG .R/ A of order 4 (recall that all elements of the set GA have order 4). Clearly, hb; Ai is nonabelian (indeed, A is maximal abelian in G) of order 2jAj and R < Z.hb; Ai/. By Corollary A.17.3, hb; Ai has a minimal nonabelian subgroup H1 6Š Q8 since hb; Ai 6Š Q2n E2s : the center of the last group has exponent 2 < 4 D exp.R/. By the above, H1 is metacyclic of order 16 and exponent 4. Since R centralizes H1 and R 6 H1 , there is in H1 R (of order 32) an abelian subgroup of order 16 containing a nonnormal cyclic subgroup of H1 of order 4 D 2 , contrary to the hypothesis. Thus jG W Aj D 2. Let D < G be cyclic of order 4 such that D 6 A. Then we have G D AD. Assume that D G G. Then G=.A \ D/ is abelian so A \ D D G 0 is of order 2. In this case, by Lemma 1.1, jGj D 2jG 0 jjZ.G/j D 16; contrary to the hypothesis. Thus all cyclic subgroups of G of order 4 not contained in A are nonnormal in G. It remains to show that our group G satisﬁes the condition of Problem 1 if it satisﬁes conditions (a)–(d) in the statement of the theorem. Note that the set C coincides with the set of all cyclic subgroups of G that are not contained in A.

138

Smallest normalizers of nonnormal subgroups

353

Let D 2 C .D C2 /. Assume that D < B < G, where B is abelian of order 16. As jB \ Aj D 8, the subgroup B \ A has a cyclic subgroup L of order 4. Then CG .L/ AD D G so that L < Z.G/, a contradiction since exp.Z.G// D 2 < 4 D exp.L/. We claim that if the 2-group G D hxiA of order 26 satisfying conditions (a)–(d) of Theorem 138.8 is metacyclic, then it is as in Theorem 138.4(c). Let x 2 G A and write X D hxi. We want to prove that then A D 1 .X/ C , where C is cyclic. If this is false, there is in A an element y of order 4 such that y 2 D x 2 (indeed, if all invariants of an abelian group, say B, are > p, then any element of B of order p is contained in a cyclic subgroup of B of order p 2 ; this follows from Theorem 6.1 with G D Q). Since hyi G G by (d), one has y x D x 1 (otherwise, y 2 Z.G/ D 1 .G/), and we conclude that hx; yi Š Q8 so that G is of maximal class (Proposition 10.19), which is a contradiction. Thus (e) A D 1 .X/ C , where C is cyclic (of index 2 in A). (f) We claim that the quotient group GN D G=1 .X/ is of maximal class. Indeed, GN has a cyclic subgroup AN Š C of index 2 by (e). Assume that GN is not of maximal class. Then, by Theorem 1.2, GN is either abelian of type .2n ; 2/ or isomorphic to M2nC1 . In the ﬁrst case, we get X G G, a contradiction. Thus GN Š M2nC1 , n 4, as jGj 26 . N then XN < B. N In this case, we get XN G BN Let BN be a noncyclic maximal subgroup of G; N D 2n 24 . It then follows that B is nonabelian by hypothsince BN is abelian, and jBj esis. The quotient group B=X is cyclic of order 2n1 23 . Then CB .X/ is abelian of order 2n 24 , contrary to the hypothesis. Thus GN is of maximal class. As in the proofs of Theorems 138.4 and 138.8, we get (recall that G is not of maximal class) (g) G 0 is cyclic and G=G 0 is abelian of type .4; 2/. (h) G is as in Theorem 138.4(c). Therefore, if A has no cyclic subgroup of index 2, then G is nonmetacyclic. Now we present a nonmetacyclic 2-group G satisfying conditions (a)–(d) of Theorem 138.8. Let m

n

G D hx; a; b j a2 D b 2 D 1; m; n 2; m C n 5; x 2 D a2

m1

; ab D ba; ax D a1 ; b x D b 1 i:

We have ha; xi Š Q2mC1 and hb; xi Š D2nC1 . Further, we have jGj D 2mCnC1 > 25 j and c D Œx; ai b j D a2i b 2 2 G 0 with c D 1 if and only if i m 1 and j n 1. m1 n1 It follows that Z.G/ D ha2 i hb 2 i is abelian of type .2; 2/. Clearly, we have 0 G D Œhxi; A. Since the noncyclic subgroup ha2 ihb 2 i of index 8 in G is contained in G 0 and jG W G 0 j > 4 (Proposition 1.6), it coincides with G 0 so that G 0 D ha2 i hb 2 i is abelian of type .2m1 ; 2n1 /, and we conclude that G is nonmetacyclic. If y 2 G A

354

Groups of prime power order

is arbitrary, then y inverts A since CG .y/ D hy; Z.G/i is abelian of type .4; 2/. It follows that G satisﬁes the condition of Problem 1. It is interesting to classify the nonnilpotent groups G such that whenever H < G is nonnormal, then either NG .H / D H or jNG .H / W H j is a prime. There is a nonsolvable group satisfying this condition, for example, PSL.2; 5/. Using Theorem 138.4, Frobenius’ normal p-complement theorem for p D 2 (see Lemma 10.8), the Odd Order Theorem and the classiﬁcation of simple groups with abelian Sylow 2-subgroup of type .2; 2/, it is easy to prove that PSL.2; 5/ is the unique nonsolvable group satisfying the above condition.

139

p-groups with a noncyclic commutator group all of whose proper subgroups have a cyclic commutator group

This section is written by the second author. Here we give a characterization of the title groups (Theorem 139.A). The surprising and deep result is that if G is a title group, then G 0 is elementary abelian of order p 2 or p 3 and for each H < G we have jH 0 j p. We use here a corresponding characterization result of A. Leone (see Theorem B in [Leo]), but most of our proofs are simpler. At the beginning we prove two very useful lemmas, and then the main result (Theorem 139.A) will be proved in a series of propositions concerning the title groups. Theorem 139.A. A p-group G has a noncyclic commutator group G 0 and each proper subgroup H of G has a cyclic commutator group H 0 if and only if for each H < G we have jH 0 j p and exactly one of the following holds: (a) d.G/ D 2, cl.G/ D 3, and G 0 Š Ep2 . (b) p > 2, d.G/ D 2, cl.G/ D 3, G 0 Š Ep3 and Ã1 .G/ Z.G/. (c) d.G/ D 3, cl.G/ D 2, G 0 Š Ep3 or Ep2 and ˆ.G/ D Z.G/. A p-group G is said be a KC-group if the commutator group G 0 is cyclic. Then the title groups are actually p-groups which are minimal non-KC-groups. Lemma 139.1. Let G D hx; yi be a nonabelian two-generator KC-group. If p > 2 or p D 2 and ŒG 0 ; G Ã2 .G 0 /, then Ã1 .G 0 / D hŒx; yp i D hŒx; y p i. Proof. By hypothesis, G 0 D hŒx; yi. If Ã1 .G 0 / D ¹1º, then jG 0 j D p and G is minimal nonabelian (Lemma 65.2(a)), and the result follows from Lemma 65.1. Next we assume that Ã1 .G 0 / > ¹1º and consider the quotient group GN D G=Ã2 .G 0 /. Then we N D 2 and GN 0 Š Cp2 . As G=Ã N 1 .GN 0 / is minimal nonabelian (Lemma 65.2(a)), get d.G/ for each H 2 1 we have jHN 0 j p. N (i) Assume that p > 2. In this case, we may apply Proposition 137.5 for the group G. p p 0 We get Œx; y D Œx; y , where 2 Ã2 .G /. But this gives hŒx; y p i D hŒx; yp i and we are done.

356

Groups of prime power order

(ii) Suppose that p D 2. Then we compute Œx; y 2 D Œx; yŒx; yy D Œx; yŒx; y.Œx; y1 Œx; yy / D Œx; y2 ŒŒx; y; y: By our assumption, we have ŒŒx; y; y 2 ŒG 0 ; G Ã2 .G 0 / and so we obtain again hŒx; y 2 i D hŒx; y2 i. Lemma 139.2. Let G D ha; b; ci be a p-group of Theorem 137.7(d), i.e., d.G/ D 3;

cl.G/ D 2;

G 0 Š Ep3 or Ep2

and

ˆ.G/ D Z.G/:

Then for each subgroup X of order p which is contained in G 0 there is a maximal subgroup L of G such that L0 D X. Proof. If G 0 Š Ep3 , then all p 2 C p C 1 maximal subgroups of G must have pairwise distinct commutator subgroups of order p. Indeed, if L ¤ K are two distinct maximal subgroups of G such that L0 D K 0 , then Exercise 1.69(a) would imply jG 0 j p 2 , a contradiction. Hence in this case each subgroup of order p in G 0 is a commutator subgroup of a maximal subgroup of G. Assume now that G 0 Š Ep2 . We may choose the generators a; b; c of G so that setting Œa; b D k and Œb; c D l we have G 0 D hk; li. But then for any i D 0; 1; : : : ; p 1 we obtain Œac i ; b D Œa; bŒc; bi D kl i and so again each subgroup of order p in G 0 is a commutator subgroup of one of the maximal subgroups ˆ.G/hb; ci and ˆ.G/hac i ; bi (i D 0; 1; : : : ; p 1) of G. Proposition 139.3. Let G be a p-group which is a minimal non-KC-group. Then we have: (i) Z.G/ ˆ.G/. (ii) G is metabelian. (iii) d.G/ 3. Proof. (i) Suppose Z.G/ 6 ˆ.G/. Then there is a maximal subgroup M of G such that G D hM; xi, where x 2 Z.G/. But then G 0 D M 0 is cyclic, a contradiction. (ii) Let M be a maximal subgroup of G so that M 0 is cyclic and M 0 E G. Since Aut.M 0 / is abelian, G=CG .M 0 / is abelian and so G 0 centralizes M 0 . Set H D hM 0 j M is any maximal subgroup in Gi so that H is characteristic in G and H Z.G 0 /. Each maximal subgroup M of G contains H and M=H is abelian. Thus G=H is either abelian or minimal nonabelian. This gives jG 0 =H j p and so G 0 is abelian. (iii) We consider G=ˆ.G 0 / so that G 0 =ˆ.G 0 / is elementary abelian of order p 2 . Each maximal subgroup M=ˆ.G 0 / of G=ˆ.G 0 / has its derived subgroup of order p and so, by Proposition 137.3, d.G=ˆ.G 0 // 3 and therefore d.G/ 3.

139

p-groups G with noncyclic G 0 and cyclic H 0 for all subgroups H < G

357

Proposition 139.4. A p-group G is a minimal non-KC-group with H D hM 0 j M is any maximal subgroup in Gi being cyclic if and only if G 0 Š Ep2 and d.G/ D 2. Here we have cl.G/ D 3. Proof. Suppose that G is a p-group which is a minimal non-KC-group and H is cyclic. Then G 0 > H and each maximal subgroup of G=H is abelian. It follows that G=H is minimal nonabelian. This gives d.G=H / D 2 and so d.G/ D 2 and jG 0 W H j D p so that G 0 is abelian of type .p n ; p/, n 1, where jH j D p n and exp.G 0 / D p n . We set G D ha; bi, where Œa; b 2 G 0 H . The commutator group G 0 is generated by all conjugates of Œa; b in G and since G 0 is abelian, we have exp.G 0 / D p n D o.Œa; b/. Thus H and hŒa; bi are two distinct cyclic subgroups of order p n (and index p) in G 0 . Suppose, by way of contradiction, that n > 1. Then cn .G 0 / D p and since one of the cyclic subgroups, H , of order p n is normal in G, it follows that all of them are normal in G. But then hŒa; bi is normal in G and so G 0 D hŒa; bi is cyclic, a contradiction. It follows that n D 1, jH j D p and o.Œa; b/ D p so that G 0 Š Ep2 . Conversely, assume that G is a p-group with d.G/ D 2 and G 0 Š Ep2 . Let N < G 0 be a G-invariant subgroup of order p contained in G 0 . Then we infer that d.G=N / D 2 and .G=N /0 D G 0 =N is of order p and so G=N is minimal nonabelian. Hence, if M is any maximal subgroup of G, then M G 0 and M=N is abelian, which gives M 0 N . It is easy to see that such a p-group G is of class 3. Indeed, if G 0 Z.G/, then considering a subgroup K < G 0 of order p with K ¤ N , we conclude that G=K is minimal nonabelian and so for any maximal subgroup M of G we get also M 0 K and so M 0 N \ K D ¹1º. In this case, G would be minimal nonabelian and so jG 0 j D p, a contradiction. Hence we obtain that G 0 6 Z.G/ and so G is of class 3. Our proposition is proved. Proposition 139.5. Let G be a p-group which is a minimal non-KC-group with H D hM 0 j M is any maximal subgroup in Gi being noncyclic and d.G/ D 2. Then p > 2;

cl.G/ D 3;

G 0 Š Ep3 ;

and

Ã1 .G/ Z.G/:

Proof. Suppose that G is a p-group which is a minimal non-KC-group, H is noncyclic and d.G/ D 2. Then there exist maximal subgroups M and N such that H0 D M 0 N 0 is noncyclic. Then the factor group G=H0 has two distinct abelian maximal subgroups M=H0 and N=H0 and this gives jG 0 W H0 j p. Suppose for a moment that G 0 D H0 . Let X be a G-invariant subgroup such that M 0 X < H0 and jH0 W Xj D p. Since d.G=X/ D 2 and .G=X/0 D H0 =X is of order p, it follows that G=X is minimal nonabelian. But then N=X is abelian and so N 0 X , a contradiction. We have proved that jG 0 W H0 j D p and since d.G=H0 / D 2 and .G=H0 /0 D G 0 =H0 is of order p, it follows that G=H0 is minimal nonabelian. For each maximal subgroup L of G, L0 H0

358

Groups of prime power order

and so H0 D H and d.H / D 2. Also, for any two distinct maximal subgroups K and L of G we have K 0 L0 D H . Indeed, by Exercise 1.69(a), jG 0 W .K 0 L0 /j p and so K 0 L0 D H . Consider the factor group G=ˆ.H / so that H=ˆ.H / Š Ep2 and jG 0 =ˆ.H /j D p 3 . Using Theorem 137.7 for G=ˆ.H /, we see that G=ˆ.H / is a group from part (c) of that theorem. In particular, we have p > 2, cl.G=ˆ.H // D 3 and G 0 =ˆ.H / Š Ep3 so that d.G 0 / D 3. Let Mi (i D 1; 2; : : : ; p C 1) be the set of all maximal subgroups of G and so we infer that G 0 =Mi0 is noncyclic. For any ﬁxed i 2 ¹1; 2; : : : ; p C 1º we consider the factor group G=Mi0 . We see that d.G=Mi0 / D 2, .G=Mi0 /0 D G 0 =Mi0 is noncyclic, but for each maximal subgroup Mj =Mi0 (j D 1; 2; : : : ; p C 1) of G=Mi0 we have .Mj =Mi0 /0 .Mj0 Mi0 /=Mi0 H=Mi0 and so each maximal subgroup of G=Mi0 has its commutator group contained in the same cyclic subgroup H=Mi0 . We may apply Proposition 139.4 to conclude that G 0 =Mi0 Š Ep2 . Thus Mi0 (i D 1; 2; : : : ; p C 1) are pairwise distinct cyclic maximal subgroups of H . Set jMi0 j D p m , m 1, so that H is abelian of type .p m ; p/. If m > 1, then such a group H has exactly p cyclic maximal subgroups (of order p m ) and a noncyclic maximal subgroup (of type .p m1 ; p/). This is a contradiction (because we need in H exactly p C 1 pairwise distinct cyclic maximal subgroups) and so m D 1 and jG 0 j D p 3 . But we have d.G 0 / D 3 and so

ˆ.G 0 / D ˆ.H / D ¹1º:

Hence G is a group from part (c) of Theorem 137.7 and we are done. The following result is crucial for the whole investigation. Proposition 139.6. Let G be a p-group which is a minimal non-KC-group. Suppose that there exist nonabelian maximal subgroups M; N of G such that M 0 \ N 0 D ¹1º. If p > 2 or p D 2 and ŒL0 ; L Ã2 .L0 / for all L 2 1 , then H D hM 0 j M 2 1 i is elementary abelian and central in G. Proof. As M \N is abelian, there are m 2 M \N and n 2 N M with N 0 D hŒm; ni. Indeed, if N 0 D hŒn0 ; ni with n; n0 2 N M , then n0 D ni m for some m 2 M \ N and Œn0 ; n D Œni m; n D Œm; n. We apply Lemma 139.1 on the group hm; ni which is a nonabelian two-generator group with a cyclic commutator group hŒm; ni. It follows that hŒm; np i D hŒm; np i. But np 2 M and so Œm; np 2 N 0 \ M 0 D ¹1º and therefore Ã1 .N 0 / D ¹1º which gives jN 0 j D p. In an analogous way, we also get jM 0 j D p. Let L be any maximal subgroup of G with L0 ¤ ¹1º. Then either L0 \ M 0 D ¹1º or L0 \ N 0 D ¹1º since M 0 \ N 0 D ¹1º. This gives jL0 j D p and so L0 Z.G/. Hence H is elementary abelian and central in G. In what follows we may assume that d.G/ D 3. This implies that H D hM 0 j M is any maximal subgroup in Gi D G 0 ; i.e., G 0 is generated with commutator subgroups of all maximal subgroups of G.

139

p-groups G with noncyclic G 0 and cyclic H 0 for all subgroups H < G

359

Proposition 139.7. Let G be a p-group which is a minimal non-KC-group satisfying d.G/ D 3. Suppose that there exist maximal subgroups M; N of G such that M 0 ¤ ¹1º, N 0 ¤ ¹1º and M 0 \ N 0 D ¹1º. If p > 2 or p D 2 and ŒL0 ; L Ã2 .L0 / for all maximal subgroups L of G, then G is a p-group in Theorem 137.7(d), i.e., d.G/ D 3;

cl.G/ D 2;

G 0 Š Ep3 or Ep2 ;

and ˆ.G/ D Z.G/. Proof. By Proposition 139.6, G 0 is elementary abelian (noncyclic) and central in G. Due to Theorem 137.7, G is a p-group in Theorem 137.7 (d) and we are done. Proposition 139.8. Let G be a p-group which is a minimal non-KC-group satisfying d.G/ D 3. If p > 2 or p D 2 and ŒL0 ; L Ã2 .L0 / for all maximal subgroups L of G, then there exist maximal subgroups M; N of G such that M 0 ¤ ¹1º;

N 0 ¤ ¹1º and

M 0 \ N 0 D ¹1º:

Proof. Suppose, by way of contradiction, that L0 \ T 0 ¤ ¹1º for any maximal subgroups L; T of G with L0 ¤ ¹1º, T 0 ¤ ¹1º. Since G 0 is noncyclic, there exist maximal subgroups M and N of G such that hM 0 ; N 0 i is noncyclic and we have M 0 \N 0 ¤ ¹1º. Applying Proposition 139.6 on the factor group G=.M 0 \ N 0 /, we get jM 0 =.M 0 \ N 0 /j D p

and

jN 0 =.M 0 \ N 0 /j D p

so that jM 0 j D jN 0 j D p n , n > 1. We may set M 0 D hhi, N 0 D hki so that hp D k p . Then we have hk 1 ¤ 1 and .hk 1 /p D 1 so that hhk 1 i is a subgroup of order p in G 0 with hhk 1 i 6 M 0 and hhk 1 i 6 N 0 and we get M 0 \ N 0 D hhp i D hk p i. Let L be any maximal subgroup of G. Then we have either L0 M 0 \ N 0 or jL0 j D p n , L0 > M 0 \ N 0 and jL0 =.M 0 \ N 0 /j D p. We study the factor group G=.M 0 \ N 0 / which is a minimal non-KC-group (noting that hM 0 ; N 0 i=.M 0 \ N 0 / Š Ep2 ), where each maximal subgroup of G=.M 0 \ N 0 / has its commutator group of order at most p and d.G=.M 0 \ N 0 // D 3. It follows that G=.M 0 \ N 0 / is a group from Theorem 137.7(d) and so we may apply Lemma 139.2 on this group. Hence there is a maximal subgroup L of G with L0 D hhk 1 ; M 0 \ N 0 i. But then L0 is noncyclic, a contradiction. Lemma 139.9. Suppose that G is a 2-group which is a minimal non-KC-group such that exp.G 0 / D 4 and d.G/ D 3. Assume, in addition, that there is a maximal subgroup M of G such that M 0 Š C4 and ŒM 0 ; M ¤ ¹1º. If there exists another maximal subgroup N of G with N 0 ¤ ¹1º and M 0 \ N 0 D ¹1º, then the following statements hold: (i) M 0 \ L0 ¤ ¹1º for each maximal subgroup L of G with L ¤ N . (ii) N D CG .G 0 / D CG .M 0 /. (iii) jN 0 j D 2.

360

Groups of prime power order

Proof. For any a 2 M 0 and n 2 N we have Œa; n 2 M 0 \ N 0 D ¹1º

and so

CG .M 0 / D N :

Hence, if L is any maximal subgroup of G with L ¤ N , then L 6 CG .M 0 /. Therefore ¹1º ¤ ŒM 0 ; L M 0 \ L0 and this gives (i). If L is any maximal subgroup of G with L ¤ N , then M 0 \ L0 ¤ ¹1º together with M 0 \ N 0 D ¹1º implies that L0 \ N 0 D ¹1º. Hence ŒN 0 ; L N 0 \ L0 D ¹1º and so N 0 Z.G/. Also, ŒL0 ; N L0 \ N 0 D ¹1º and so CG .L0 / N which together with CG .M 0 / D N gives CG .G 0 / D N and (ii) is proved. We may choose m 2 M \ N and n 2 N M so that N 0 D hŒm; ni (noting that M \ N is abelian). As N 0 Z.G/, we may apply Lemma 139.1 on the group hm; ni. We get hŒm; n2 i D hŒm; n2 i D ¹1º, which gives jN 0 j D 2 and this is (iii). Proposition 139.10. Let G be a 2-group which is a minimal non-KC-group satisfying d.G/ D 3. Then ŒM 0 ; M Ã2 .M 0 / for all maximal subgroups M of G. Proof. Suppose, by way of contradiction, that there exists a maximal subgroup X of G such that ŒX 0 ; X 6 Ã2 .X 0 /. In particular, X 0 is cyclic of order 4 and so G 0 is of exponent 4. Note that the factor group G=Ã2 .G 0 / is a minimal non-KC-group such that d.G=Ã2 .G 0 // D 3 and .G=Ã2 .G 0 //0 is noncyclic of exponent 4. We want to show that such a 2-group G=Ã2 .G 0 / does not exist and so we may assume in the sequel that Ã2 .G 0 / D ¹1º. In other words, we may assume that G is 2-group which is a minimal non-KC-group with d.G/ D 3 and G 0 is noncyclic of exponent 4. First suppose that for each maximal subgroup L of G we have ŒL0 ; L Ã2 .L0 /. Then Proposition 139.8 implies that there exist maximal subgroups M; N of G such that M 0 ¤ ¹1º, N 0 ¤ ¹1º and M 0 \ N 0 D ¹1º. By Proposition 139.7, G 0 is elementary abelian of order 4 or 8, contrary to the fact that G 0 is noncyclic of exponent 4. Thus we have proved that there exists a maximal subgroup M of G that satisﬁes M 0 Š C4 and ŒM 0 ; M 6 Ã2 .M 0 /, i.e., ŒM 0 ; M ¤ ¹1º. (a) Assume that there is a maximal subgroup N of G such that N 0 ¤ ¹1º

and M 0 \ N 0 D ¹1º:

In this case, all assumptions of Lemma 139.9 are satisﬁed and so the statements (i), (ii) and (iii) of that lemma hold. Let L be any maximal subgroup of G with L ¤ N . Then L0 \ M 0 ¤ ¹1º, L0 \ N 0 D ¹1º, and either L0 Z.G/ or CG .L0 / D N . Assume that L0 Z.G/. Here N \L is abelian and so there are elements l 2 N \L and l 0 2 L N such that L0 D hŒl; l 0 i. By Lemma 139.1 applied to hl; l 0 i, we obtain that hŒl; l 0 2 i D hŒl; l 02 i D ¹1º (since .l 0 /2 2 N \ L) which implies jL0 j D 2 and so L0 D 1 .M 0 /. Suppose that CG .L0 / D N so that in this case L0 Š C4 and L0 \ M 0 1 .M 0 /. We see that G D hN; ti, where t 2 G N and t inverts L0 . Since CG .G 0 / D N and t inverts G 0 (noting that G 0 is generated by all X 0 , where X is any maximal subgroup of G), it follows that each subgroup of G 0 is normal in G. Set 1 .M 0 / D hzi so that

139

p-groups G with noncyclic G 0 and cyclic H 0 for all subgroups H < G

361

Ã1 .G 0 / D hzi. Then we obtain G 0 D 1 .G 0 /M 0 with 1 .G 0 / \ M 0 D hzi. We get G 0 D M 0 N 0 K with K 1 .G 0 /. Considering the factor group G=K, it follows that we may assume from the start that G 0 D M 0 N 0 is abelian of type .4; 2/ (since we want to show that G=K does not exist). Set N 0 D hui so that E D 1 .G 0 / D hu; zi Z.G/. We know that N D CG .G 0 /. Also set N D ˆ.G/ha; bi so that Œa; b 2 hui. Consider G=hzi so that G=hzi is a group from Theorem 137.7(d). This implies that ˆ.G=hzi/ D ˆ.G/=hzi D Z.G=hzi/. It follows that .ˆ.G//0 hzi \ hui D ¹1º and for any x 2 N ˆ.G/, Œˆ.G/; x hzi \ hui D ¹1º so that the three maximal subgroups of N containing ˆ.G/ are abelian. This also forces Œa; b D u. Take an element c 2 G N so that G D ha; b; ci. If Œa; c 2 E and Œb; c 2 E, then G=E would be abelian and so G 0 E, a contradiction. Hence (interchanging a and b if necessary), we may assume that Œa; c 2 G 0 E so that Œa; c2 D z. We may assume that Œb; c D z with D 0; 1. Indeed, suppose that Œb; c 62 hzi. Set L D ˆ.G/hb; ci so that L is a maximal subgroup of G which is distinct from N . By Lemma 139.9, L0 \ M 0 ¤ ¹1º and so L0 hzi. It follows that Œb; c 2 G 0 E. Then we have (since ˆ.G/hbi is abelian) Œab; c D Œa; cb Œb; c D Œa; cŒb; c 2 E: The maximal subgroup L1 D ˆ.G/hab; ci of G is distinct from N and so, in view of Lemma 139.9, L01 \ M 0 ¤ ¹1º and so L01 hzi and Œab; c 2 hzi. As G D ha; ab; ci and Œa; ab D Œa; b D u, we replace in this case b with ab so that we may assume from the start that Œb; c D z with D 0; 1. We consider the proper subgroup K D hac; b; a2 i of G (noting that a2 2 ˆ.G/). Then we have Œac; b D Œa; bc Œc; b D uc z D uz and Œa2 ; ac D Œa2 ; c D Œa; ca ŒŒa; c D Œa; c2 D z as a 2 N commutes with Œa; c (because N D CG .G 0 /). But then K 0 huz ; zi D E is noncyclic, a contradiction. (b) Assume that for all maximal subgroups L of G with L0 ¤ ¹1º, L0 \ M 0 ¤ ¹1º, where G 0 is noncyclic of exponent 4, M 0 Š C4 and ŒM 0 ; M ¤ ¹1º. For each maximal subgroup L of G we have either L0 M 0 or L0 Š C4 and L0 \ M 0 D hi i, where we set hi i D Ã1 .M 0 /. Thus Ã1 .G 0 / D hii and G 0 =hi i is elementary abelian of order 4 (since G 0 is noncyclic). It follows that G=hi i is a group from Theorem 137.7(d) and so G 0 =hii Š 1 .G 0 / Š E4 or E8 . Let F be a four-subgroup contained in 1 .G 0 / such that F > hii. Then Lemma 139.2 (applied on the group G=hi i) gives that G possesses a maximal subgroup L1 such that L01 covers F=hi i. But L01 is cyclic and so jL01 j D 2 and therefore F D L01 hi i, contrary to our assumption in (b). Our proposition is proved.

362

Groups of prime power order

We can now prove our main result. Proof of Theorem 139.A. If G is a p-group from (a), (b) or (c) of Theorem 139.A, then Proposition 137.6 implies that jH 0 j p for each subgroup H < G. Conversely, if G is a p-group with G 0 being noncyclic but for each H < G we have H 0 is cyclic, then Propositions 139.4, 139.5, 139.7, 139.8 and 139.10 show that jH 0 j p and we have one of the possibilities stated in parts (a), (b) or (c) of Theorem 139.A.

140

Power automorphisms and the norm of a p-group

An automorphism of a ﬁnite group G is called a power automorphism if it maps each subgroup of G onto itself. This will be the case if and only if maps each element x 2 G to some power x i of x (therefore such a is called a power automorphism). We denote by A.G/ the group of all power automorphisms of G and we shall prove here some elementary facts about A.G/ (Theorem 140.7). Then, in Corollary 140.8, we obtain a result of E. Schenkman [Sc] asserting that the norm of a p-group G (which is the intersection of the normalizers of all subgroups of G) is contained in the second center Z2 .G/ of G. Lemma 140.1. If 2 A.G/ and g 2 G, the inner automorphism induced by g 1 g is a power automorphism. Proof. Let H G and g 2 G. Then

H g D .H g / D .H /g D H g

whence (noting that g commutes with g ) H D Hg

g 1

D Hg

1 g

and so g 1 g induces a power automorphism of G. Lemma 140.2. If 2 A.G/ and g; h 2 G, then Œg 1 g ; hŒh1 h ; g D 1 and Œg 1 g ; h 2 hgi \ hhi: Proof. Since g h D .gh/ 2 hghi, it follows that g h commutes with gh. Hence (noting that Œg; g D 1 and Œh; h D 1) Œg 1 g ; hŒhh ; g 1 D gg h1 g 1 .g h /.gh/h g 1 D gg h1 g 1 .gh/.g h /h g 1 D 1: We have proved ()

Œg 1 g ; hŒhh ; g 1 D 1:

364

Groups of prime power order

By Lemma 140.1, g 1 g and h1 h induce inner power automorphisms on G and so Œg 1 g ; h 2 hhi and Œhh ; g 1 2 hgi. Hence for all g; h 2 G we get Œg 1 g ; h 2 hgi \ hhi: Interchanging g and h, we also have Œh1 h ; g 2 hgi \ hhi which shows that Œh1 h ; g D Œh1 h ; ghh

D Œh1 h ; h1 h ghh

D hh h h1 g 1 h h h1 h h1 h ghh

D g 1 .h1 h /g.hh / D g 1 g hh

D g hh

g 1 D Œhh ; g 1 :

Putting this into (), we ﬁnally get the required relation Œg 1 g ; hŒh1 h ; g D 1: Given an automorphism of a group G and g; h 2 G, we set Œg; D g 1 g and Œg; ; h D ŒŒg; ; h. Lemma 140.3. If 2 A.G/ and Œgi ; ; hj D 1 for i D 1; 2; : : : ; n, j D 1; 2; : : : ; m, then Œg1 g2 : : : gn ; ; h1 h2 : : : hm D 1: Proof. By Lemma 140.2 which asserts that Œg; ; hŒh; ; g D 1 (g; h 2 G), we may assume m n. We prove this lemma by induction on m. If m D 1, then n D 1 and Œg1 ; ; h1 D 1 by assumption. In case m > 1, we get Œg1 g2 : : : gn ; ; .h1 h2 : : : hm1 /hm D Œg1 g2 : : : gn ; ; hm Œg1 g2 : : : gn ; ; h1 h2 : : : hm1 hm D 1 by induction hypothesis. Lemma 140.4. If 2 A.hg; hi/ and Œg; ; h D 1, then is central in hg; hi, i.e., for each g1 2 hg; hi we have g1 D g1 z for some z 2 Z.hg; hi/. Proof. Set G D hg; hi and let g1 ; h1 2 G. Then g1 D a1 a2 : : : an ;

h1 D b1 b2 : : : bm ;

where each ai and bj is equal to g or h (noting that G is assumed to be ﬁnite). We claim that for each i; j we have Œai ; ; bj D 1. Indeed, Œg; ; g D Œg 1 g ; g D 1 since g 1 g is a power of g. Similarly, Œh; ; h D 1 and by our assumption Œg; ; h D 1. Finally, by Lemma 140.2, Œg; ; hŒh; ; g D 1 and so also Œh; ; g D 1. Using Lemma 140.3, we get Œg1 ; ; h1 D Œa1 a2 : : : an ; ; b1 b2 : : : bm D 1: If we set g11 g1 D z, then we get Œz; h1 D 1 and so z 2 Z.hg; hi/. Hence is central in hg; hi and we are done.

140

365

Power automorphisms and the norm of a p-group

Lemma 140.5. Let G be a p-group which is either abelian or minimal nonabelian with p > 2 and x; y 2 G such that o.x/ o.y/. Then for some integer u we have hxi \ hx u yi D ¹1º and

o.x u / D o.y/:

Proof. If hxi hyi, then there is an integer u with x u D y 1 so that o.x u / D o.y/ and x u y D 1 and we are done. We may assume that hxi 6 hyi. Since o.x/ o.y/, we have jhxi=.hxi \ hyi/j jhyi=.hxi \ hyi/j D p r with r 1. There exists an element x u D x 0 2 hxi hyi of order o.y/ such that r r .x 0 /p D y p , where jhx 0 i=.hxi \ hyi/j D p r . Since G is either abelian or minimal nonabelian with p > 2, we have for each a; b 2 G and s 1, s

s

1

s

.ab/p D ap b p Œb; a 2 p

s .p s 1/

Hence .x 0 y/p D 1 and so o.x 0 y/ p r . If .x 0 y/p r

.x 0 /p

r1

D yp

r1

r1

s

s

D ap b p : D 1, then

2 hxi \ hyi;

a contradiction. Hence o.x 0 y/ D p r and the element .x 0 y/p D .x 0 /p 0 0 of order p contained in hx yi. Suppose that hxi \ hx yi ¤ ¹1º. Then r1

.x 0 /p

r1

yp

r1

2 hxi

and so y p

r1

r1

yp

r1

is

2 hxi \ hyi;

contrary to jhyi=.hxi \ hyi/j D p r . Thus we have proved that hxi \ hx u yi D ¹1º and o.x u / D o.y/. Lemma 140.6. For a; b 2 G and 2 A.G/ we have Œab; D Œa; b Œb; . Proof. We have Œa; b Œb; D b 1 .a1 a / b .b 1 b / D b 1 a1 a b D .ab/1 .ab/ D Œab; : Theorem 140.7. Every power automorphism of a p-group G is central, i.e., for each g 2 G we have g D gz for some z 2 Z.G/. In particular, ﬁxes each element of G 0 . Proof. Let G be a p-group with a noncentral power automorphism . Then for some g; h 2 G we have Œg; ; h ¤ 1. By Lemma 140.2, d D Œg; ; h 2 hgi \ hhi Z.K/, where K D hg; hi. Since is noncentral in K=hd p i, we may assume d p D 1. We consider the subgroup H D hŒg; ; Œh; i of K. By Lemma 140.2, Œg; ; hŒh; ; g D 1

366

Groups of prime power order

and so we may assume that o.Œg; / o.Œh; /. Now, Œh; D hr for some integer r. Hence ŒŒg; ; Œh; D ŒŒg; ; hr D d r 2 hd i Z.H / since Œg; ; h D d 2 Z.K/ and so Œg; ; hr D Œg; ; hr . This shows that H is either abelian or minimal nonabelian with H 0 D hd i. In case p D 2, we can conclude that Œh; D h1 h D h1Ci D hr , where i is odd and so r D 1 C i is even, which gives d r D 1 and so in this case H is abelian. We have proved that the subgroup H D hŒg; ; Œh; i is either abelian or minimal nonabelian with p > 2, where H 0 D hd i and d D Œg; ; h. As o.Œg; / o.Œh; /, we may use Lemma 140.5 for the group H . It follows that there is an integer u such that hŒg; i \ hŒg; u Œh; i D ¹1º

()

and o.Œg; u / D o.Œh; /:

Now, d and Œg; are both contained in hgi and so one is a power of the other. However, d commutes with h while Œg; does not. Hence d is a power of Œg; , say d D Œg; t . Then setting v D u.1 t /, we get using Lemma 140.6 Œg v h; D Œg ut g u h; D Œg ut ; g

uh

Œg u h;

(since Œg ut ; is a power of g) D Œg ut ; h Œg u h; (and since Œg ut ; h D .g ut .g /ut /h D ...g 1 g /t /u /h D ..Œg; t /u /h D .d u /h D d u because d 2 Z.K/) D d u Œg u h; D d u Œg u ; h Œh; D d u Œg u ; Œg u ; ; hŒh; (since

Œg u ; D g u .g /u D .g 1 g /u D Œg; u

and Œg u ; ; h D ŒŒg u ; ; h D ŒŒg; u ; h D Œg; u .h1 Œg; h/u D Œg; u .Œg; h /u D .Œg; 1 Œg; h /u D Œg; ; hu D d u because Œg; commutes with d D Œg; 1 Œg; h 2 hgi and so Œg; also commutes with Œg; h ) D d u Œg; u d u Œh; D Œg; u Œh; :

140

Power automorphisms and the norm of a p-group

367

We have proved that Œg v h; D Œg; u Œh; : By (), we get hŒg; i \ hŒg v h; i D ¹1º. Since d D Œg; t is of order p, we have 1 ¤ Œg; 2 hgi. On the other hand, Œg v h; is a power of g v h. Hence we have either Œg v h; D 1 or hgi \ hg v hi D ¹1º. By Lemma 140.2, Œg v h; ; g D ŒŒg v h; ; g hg v hi \ hgi: Thus, in any case, Œg v h; ; g D 1. We get hg v h; gi D hg; hi and so, by Lemma 140.4, is central in hg; hi. But this implies that Œg; ; h D 1, a contradiction. Our theorem is proved. The norm N .G/ of a group G is the intersection of the normalizers of all subgroups of G. Corollary 140.8. The norm N .G/ of a p-group G is contained in the second center Z2 .G/ of G. Proof. Let a 2 N .G/. Then a induces in G an inner power automorphism. By Theorem 140.7, for all g 2 G we have g a D gz with some z 2 Z.G/. This implies that a 2 Z2 .G/ and we are done. Exercise 1. If M Š Mpn , then N .G/ is a noncyclic maximal subgroup of G. Exercise 2. If G is a nonabelian metacyclic group of order p 4 and exponent p 2 , then N .G/ D Z.G/. Exercise 3. If G 6Š Q8 is a 2-group of maximal class, then we have N .G/ D Z.G/ unless G Š Q2n , n > 3 (then N .G/ D Z2 .G/. Exercise 4. Find N .G/, where (a) G is minimal nonabelian p-group. (b) G is extraspecial. (c) G is special. (d) G is metacyclic. (e) jˆ.G/j D p. (f) jG 0 j D p. (g) G is a p-group of maximal class, p > 2. (h) G is a Sylow p-subgroup of the holomorph of a cyclic p-group. (i) G is an A2 -group. (j) G is minimal nonmetacyclic 2-group. Exercise 5. Find N .A B/ is terms of A and B.

141

Nonabelian p-groups having exactly one maximal subgroup with a noncyclic center

Theorem 141.1 solves a problem posed by the ﬁrst author. Note that if G is a nonabelian p-group all of whose maximal subgroups have a cyclic center, then p D 2 and G is a 2-group of maximal class. Theorem 141.1. A nonabelian p-group G has exactly one maximal subgroup with a noncyclic center if and only if Z.G/ is cyclic and G has exactly one normal abelian subgroup of type .p; p/. Proof. Suppose that G is a nonabelian p-group with exactly one maximal subgroup M such that Z.M / is noncyclic. If Z.G/ is noncyclic, then G=Z.G/ is cyclic since M is the only maximal subgroup of G containing Z.G/. But then G is abelian, a contradiction. Hence Z.G/ must be cyclic. Suppose that G has no normal abelian subgroup of type .p; p/. Then p D 2 and G is of maximal class (Lemma 1.4). But in that case, either each maximal subgroup of G has a cyclic center or G Š D8 . In the last case, G has two distinct maximal subgroups with a noncyclic center, a contradiction. Thus we have proved that G has a normal abelian subgroup U of type .p; p/. As jG W CG .U /j D p (because Z.G/ is cyclic), we get CG .U / D M . Also, 1 .Z.G// D hzi < U . Assume that G possesses another normal abelian subgroup V distinct from U of type .p; p/. Since CG .V / is a maximal subgroup of G with a noncyclic center (recall that Z.G/ is cyclic), we have CG .V / D M . In particular, we get V M , U \V D hzi and U V Š Ep3 with U V Z.M /. Take x 2 G M . We may choose u 2 U hzi and v 2 V hzi so that ux D uz and v x D vz 1 . Indeed, if ux D uz i and v x D vz j , where i; j are some integers with i 6 0 .mod p/ and j 6 0 .mod p/, we ﬁnd integers i 0 ; j 0 so that i i 0 1 .mod p/ and jj 0 1 .mod p/. Then, replacing u with 0 0 u0 D ui and v with v 0 D v j , we get (noting that i 0 6 0 .mod p/, j 0 6 0 .mod p/) 0

0

0

0

0

0

.u0 /x D .ui /x D .ux /i D .uz i /i D ui z i i D ui z D u0 z; and

0

0

0

0

0

0

.v 0 /x D .v j /x D .v x /j D .vz j /j D v j z jj D v j z 1 D v 0 z 1 :

Thus, writing again u instead of u0 and v instead of v 0 , we can assume from the start that ux D uz and v x D vz 1 . It follows that .uv/x D ux v x D uz vz 1 D uv:

141 p-groups with exactly one maximal subgroup having a noncyclic center

369

But then hz; uvi Š Ep2 and hz; uvi Z.G/, which contradicts the fact that Z.G/ is cyclic. We have proved that U is the unique normal elementary abelian subgroup of order p 2 in G. Suppose that G is a nonabelian p-group with a cyclic center Z.G/ which possesses a unique normal abelian subgroup U of type .p; p/. Since Z.G/ is cyclic, we conclude that jG W CG .U /j D p and so M D CG .U / is a maximal subgroup of G with a noncyclic center. Let N be any maximal subgroup of G with a noncyclic center Z.N /. Since Z.N / is normal in G, it follows that 1 .Z.N // contains an abelian subgroup V of type .p; p/ which is normal in G. Because Z.G/ is cyclic, we have CG .V / D N . By our assumption, V D U and so CG .U / D M D CG .V / D N , and we are done. Theorem 141.2. Let G be a nonabelian p-group all of whose nonabelian maximal subgroups have a cyclic center. Then one of the following holds: (a) G is minimal nonabelian. (b) G is a 2-group of maximal class. (c) G D LZ.G/, where L Š D8 ; Q8 or Mpn , n 3 (if p D 2, then n 4), and either Z.G/ is cyclic with Z.G/ > Z.L/ or G D Lht i, where t is an element of order p. (d) G has exactly one abelian maximal subgroup A of rank 2, Z.G/ is cyclic, and G possesses exactly one normal abelian subgroup of type .p; p/ (contained in A). Proof. We may assume that G is not minimal nonabelian. First assume that G contains more than one abelian maximal subgroup. Then we get G=Z.G/ Š Ep2 and jG 0 j D p. All p C 1 maximal subgroups of G containing Z.G/ are abelian and therefore each maximal subgroup M of the group G which does not contain Z.G/ is nonabelian. Set Z1 D M \ Z.G/ D Z.M / so that (by our assumption) Z1 is cyclic and so Z.G/ is of rank 2. Now let L be a minimal nonabelian subgroup of M so that M D LZ1 , G D LZ.G/ and L \ Z1 D Z.L/ ˆ.G/. It follows that L has a cyclic subgroup of index p and so L Š D8 ; Q8 or Mpn , n 3 (if p D 2, then n 4 ). Suppose that Z.G/ is not cyclic. Let t be an element of order p in Z.G/Z1 . If Lht i < G, then a maximal subgroup of G containing Lhti has a noncyclic center, a contradiction. Hence in this case G D L ht i. Assume that Z.G/ is cyclic. Then Z1 D ˆ.Z.G// D ˆ.G/ and so each maximal subgroup M1 of G which does not contain Z.G/ contains Z1 D ˆ.G/ and Z.M1 / D Z1 is cyclic. Now assume that G has exactly one abelian maximal subgroup A. If A is cyclic, then all maximal subgroups of G have cyclic center and then p D 2 and G is a 2-group of maximal class. We may assume that A is of rank 2 and so, by Theorem 141.1, Z.G/ is cyclic and G has exactly one normal abelian subgroup of type .p; p/ (contained in 1 .A/). Finally, assume that all maximal subgroups of G are nonabelian. Then all maximal subgroups of G have cyclic center. But then p D 2 and G is a 2-group of maximal class and such groups have a cyclic subgroup of index 2, a contradiction.

142

Nonabelian p-groups all of whose nonabelian maximal subgroups are either metacyclic or minimal nonabelian

We solve here Problem 1535(i) and prove the following result. Theorem 142.1. Let G be a nonabelian p-group all of whose nonabelian maximal subgroups are either metacyclic or minimal nonabelian. Then one of the following holds: (a) G is minimal nonabelian (= A1 -group). (b) G is an A2 -group, i.e., G is not minimal nonabelian but all nonabelian maximal subgroups of G are minimal nonabelian (and such groups are completely determined in 71). (c) G is metacyclic. Proof. Let G be a title group. If all maximal subgroups of G are abelian, then G is minimal nonabelian and we have obtained case (a) of our theorem. We may assume that G has nonabelian maximal subgroups and they are all either metacyclic or minimal nonabelian. It follows that d.G/ 3. First assume that G has more than one abelian maximal subgroup. Then G contains exactly p C 1 abelian maximal subgroups, d.G/ D 3 and jG 0 j D p. Let H be any nonabelian maximal subgroup of G. Then jH 0 j D p and H is either minimal nonabelian or H is metacyclic. But in the second case we have d.H / D 2 which together with jH 0 j D p implies that H is also minimal nonabelian. Hence in this case all nonabelian maximal subgroups of G are minimal nonabelian and so G is an A2 -group from case (b) of our theorem. From now on we may assume that G has at most one abelian maximal subgroup. If, in addition, all nonabelian maximal subgroups of G are minimal nonabelian, then again G is an A2 -group from part (b) of our theorem. Now suppose that all nonabelian maximal subgroups of G are metacyclic. We may also assume that G is nonmetacyclic (because of the possibility (c) of our theorem). If G is minimal nonmetacyclic, then, by the results of 66 and 69, G is either the minimal nonmetacyclic 2-group of order 25 or G is a 3-group of order 34 and maximal class. But these both groups are also A2 -groups and so they are included in part (b) of our theorem. Assume now that G has exactly one nonmetacyclic maximal subgroup M

142 p-groups with abelian, metacyclic or minimal nonabelian maximal subgroup 371

in which case M is the unique abelian maximal subgroup of G. If p D 2, then such groups are completely determined in 87. Considering all results of that section (Theorems 87.8, 87.10, 87.12, 87.14 to 87.17, and 87.19 to 87.21), we see in the case where G is not minimal nonabelian that the unique nonmetacyclic maximal subgroup of G is always nonabelian, a contradiction. Now suppose that p > 2. Then these groups are determined by Y. Berkovich in Proposition A.40.12. It follows that assuming jGj > p 4 (noting that in case jGj p 4 , G is an A2 -group ), G is an L3 -group, i.e., 1 .G/ D E is of order p 3 and exponent p and G=E is cyclic of order > p. Let M be the unique maximal subgroup of G containing E. Since M is nonmetacyclic, M is abelian and M D CG .E/. Let a 2 G M so that G D hE; ai and E \hai ¤ ¹1º. If CG .a/ > hai, then jCG .a/ W haij D p and CG .a/ is another abelian maximal subgroup of G (distinct from M ), a contradiction. Hence we get CG .a/ D hai. Let X be any maximal subgroup of G distinct from M . Then X covers G=E and X \ E Š Ep2 . Let a0 2 X be such that X D hX \ E; a0 i and ha0 i \ .X \ E/ ¤ ¹1º. Since ha0 i is a cyclic subgroup of index p in X and X is nonabelian, we get X Š Mpn , n 4, and so X is minimal nonabelian. Hence G is again an A2 -group from part (b) of our theorem. We have reduced our problem to a classiﬁcation of the title groups with the following properties: (i) G has at most one abelian maximal subgroup. (ii) G has a nonabelian maximal subgroup H which is minimal nonabelian but not metacyclic. (iii) G has a nonabelian maximal subgroup K which is metacyclic but not minimal nonabelian. In particular, K 0 is cyclic of order p 2 . We have E D 1 .H / Š Ep3 . If E ˆ.H /, then E ˆ.G/ and then each maximal subgroup of G would be nonmetacyclic, which contradicts our assumption (iii). Hence E D 1 .H / 6 ˆ.H / and we may set n

H D ha; b j ap D b p D 1; n 2; Œa; b D z; z p D Œz; a D Œz; b D 1i; where E D hap ; z; bi, ˆ.H / D Z.H / D hap ; zi and H 0 D hzi is a maximal cyclic subgroup of H . In particular, H=E is cyclic of order p n1 . If G=E is abelian, then G 0 E and so G 0 is elementary abelian. But this contradicts our assumption (iii), where K 0 is cyclic of order p 2 . It follows that G=E is nonabelian with a cyclic subgroup of index p. In particular, we have n 3 and so H=E is cyclic of order p 2 . All p C 1 maximal subgroups of G containing E are nonmetacyclic and so (by our assumption (iii)) d.G/ D 3 which forces ˆ.H / D ˆ.G/ since jG W ˆ.H /j D p 3 . Suppose that G possesses an abelian maximal subgroup A. By a result of A. Mann, we have jG 0 W .A0 H 0 /j p and so jG 0 W H 0 j p which implies that jG 0 j p 2 and so G 0 D K 0 Š Cp2 . On the other hand, G 0 ˆ.G/ D ˆ.H / and G 0 > H 0 , which contradicts the fact that H 0 D hzi is a maximal cyclic subgroup in H . n1

372

Groups of prime power order

Thus we have proved that all maximal subgroups of G are nonabelian. In particular, all p C 1 maximal subgroups of G containing E are nonabelian and nonmetacyclic and so they all must be nonmetacyclic minimal nonabelian. Let H1 ¤ H be another nonmetacyclic minimal nonabelian maximal subgroup of the group G containing E. Since H1 > ˆ.G/ D ˆ.H / and jH1 W ˆ.H1 /j D p 2 with ˆ.H1 / ˆ.G/, we have ˆ.H1 / D ˆ.H / D ˆ.G/ D Z.H / D Z.H1 /: This implies that ˆ.G/ Z.G/ and so G is of class 2. For any x; y 2 G we have Œx; yp D Œx p ; y D 1 and so G 0 is elementary abelian. But this contradicts our assumption (iii), where K 0 is cyclic of order p 2 . This is a ﬁnal contradiction and our theorem is proved.

143

Alternate proof of the Reinhold Baer theorem on 2-groups with nonabelian norm

Recall that the norm N .G/ of a group G is the intersection of the normalizers of all subgroups of G. Clearly, the subgroup N .G/ is characteristic in G and Dedekindian, i.e., N .G/ D Q E A, where Q 2 ¹¹1º; Q8 º, exp.E/ 2 and A is abelian of odd order. In this section we offer another proof of the following remarkable result due to Reinhold Baer. Theorem 143.1. If the norm N .G/ of a 2-group G is nonabelian, then N .G/ D G. Proof (Berkovich). Assume, by way of contradiction, that N .G/ < G. By Theorem 1.20, there is in N .G/ a subgroup Q Š Q8 . We have, by assumption, Q < G so that jGj 16. Let A D hai and B D hbi be distinct cyclic subgroups of Q of order 4. (i) If Q H G, then Q N .H /. This is obvious. (ii) If Q < H G and jH W Qj D 2, then H D Q C is Dedekindian. Assume that this is false. Then H has a nonnormal cyclic subgroup L of order 4 such that L 6 Q. As QL D H , it follows that Q does not normalize L, contrary to the hypothesis. This H is Dedekindian, and our claim follows. (iii) CG .Q/ has no cyclic subgroup of order 4. Assume, however, that the subgroup L D hxi CG .Q/ is cyclic of order 4. Set H D Q L; then we get 16 jH j 32. If jH j D 16, our claim follows from (ii). Now let jH j D 32; then H D Q L and haxi is not b-invariant, a contradiction. (iv) By (ii), jG W Qj > 2. (v) It follows from hypothesis and (iii) that if D D hd i < G is cyclic of order 4, then Q \ D > ¹1º. Indeed, assume that Q \ D D ¹1º. Then, by (iii), Q is not normal in F D QD so F=QF Š D8 . But the norm of D8 coincides with its center, and this is a contradiction. (vi) We claim that exp.G/ D 4. Assume that T < G is cyclic of order 8. Since Q normalizes T , we get H D QT G. Bringing to mind our aim, one may assume that G D QT . By (v), one has Q \ T > ¹1º so that 16 jGj 32. From (iv) we deduce jGj > 16. Let jGj D 32. Write H1 D AT and H2 D BT . Since, by (i), A N .H1 /,

374

Groups of prime power order

it follows that H1 is not of maximal class in view of A 6 ˆ.H1 /. Similarly, H2 is not of maximal class. Then, by Theorem 1.2, Ã1 .T / D ˆ.Hi / Z.Hi /, i D 1; 2. Thus Ã1 .T /, a cyclic subgroup of order 4, centralizes hA; Bi D Q, contrary to (iii). Now we are ready to complete the proof of our theorem. By (v) and (vi), Ã1 .G/ D Ã1 .Q/ so G=Ã1 .G/, being of exponent 2, is abelian, and we conclude that G 0 D Ã1 .G/ has order 2 since G is nonabelian. Then G=CG .Q/ is isomorphic to a subgroup of D8 2 Syl2 .Aut.Q//, and G=CG .Q/ contains a four-subgroup isomorphic to Q=Z.Q/. As Ã1 .Q/ CG .Q/, we get G=CG .Q/ Š Q=Z.Q/ since Q \ CG .Q/ D Z.Q/, and we infer that G D QCG .Q/. Since exp.CG .Q// D 2 by (iii), we deduce that CG .Q/ D Z.Q/ E, where E < CG .Q/. In that case, we obtain G D QCG .Q/ D Q.Z.Q/ E/ D QE; where Q \ E D ¹1º. Thus G D Q E so G is Dedekindian as exp.E/ D 2. Remark. In the proof of Theorem 143.3 we do not use the ﬁniteness of G. Thus that theorem is also true for inﬁnite 2-groups. The following theorem is a partial case of Schenkman’s result (see Corollary 140.8). Theorem 143.2. Suppose that G is an arbitrary ﬁnite group with nonabelian N .G/. Then P 2 Syl2 .N .G// centralizes all elements of G of odd order and P Z2 .G/. Proof. Let Q8 Š Q N .G/, let a 2 Q# and let x 2 G be of order p k , where p > 2 is a prime. Set H D ha; xi. Let y 2 hxi be of order p and F D ha; yi. Assume that F is nonabelian. In case o.a/ D 2, we see that F is a nonabelian group of order 2p so its norm equals ¹1º, a contradiction since a 2 N .G/. If o.a/ D 4, then F is minimal nonabelian with norm of order 2, a contradiction again. Thus F is abelian. Hence Q centralizes all subgroups of G of odd order. As, by Theorem 10.28, P 2 Syl2 .N .G// is generated by its minimal nonabelian subgroups all of which are isomorphic to Q8 , it follows that P centralizes all subgroups of G of odd order. Now let P P1 2 Syl2 .G/. By Theorem 143.1, the subgroup P1 is Dedekindian. Therefore Z.P / Z.P1 / (Theorem 1.20). Since, by the above, Z.P / centralizes all elements of G of odd order, we conclude that Z.P / Z.G/. As P =Z.P / P1 =Z.P /, P1 =Z.P / is abelian and P =Z.P / centralizes all elements of G=Z.P / of odd order, we get P =Z.P / Z.G=Z.P // so that P Z2 .G/, proving the last assertion. Exercise 1. Let N1 .G/ be the intersection of the normalizers of all cyclic subgroups of a 2-group G of order 8. If N1 .G/ contains a subgroup isomorphic to Q8 , then we have G D N1 .G/. (Hint. See the proof of Theorem 143.1.) Exercise 2. Let G 6Š SD16 be a 2-group, D < G and D Š D8 . Then there exists in G a subgroup L of order 2 such that L 6 D and D 6 NG .L/.

143

Alternate proof of Baer’s theorem on 2-groups with nonabelian norm

375

Exercise 3. Let M be a proper nonabelian subgroup of order p 3 of a p-group G, p > 2. Then, if 1 .G/ > 1 .M /, there is a subgroup L < G of order p such that L 6 M and M 6 NG .L/. Schenkman [Sc] has proved that N .G/ Z2 .G/ and ŒG 0 ; N .G/ D ¹1º for arbitrary groups G (ﬁnite and inﬁnite).

144

p-groups with small normal closures of all cyclic subgroups

A p-group G is called a BI.p k /-group (G 2 BI.p k / for short) with a ﬁxed integer k 0 if jhxiG W hxij p k for all x 2 G. Obviously, each p-group is a BI.p k /-group for some integer k and so we can only investigate such groups for some small k. Using to some extent the results of Heng Lv, Wei Zhou and Dapeng Yu [LZY], we shall establish here some remarkable properties of BI.p 2 /-groups for p > 2 (Theorems 144.6, 144.7 and 144.9). The corresponding problem for p D 2 is completely open. Look at similar problems stated in Problem 148. Lemma 144.1. Let G 2 BI.p k /. Then jG W NG .hai/j p k for all a 2 G. Proof. Let o.a/ D p n , n 1. Since G is a BI.p k /-group, we have jhaiG j p nCk . Hence there are at most .p nCk p n1 /=.p n p n1 / D .p kC1 1/=.p 1/ < p kC1 cyclic subgroups of order p n in haiG . This gives jG W NG .hai/j p k . Lemma 144.2. Suppose that G D ha; bi is a nonabelian p-group. If jhaiG W haij p and jhbiG W hbij p with hai \ hbi D ¹1º, then G 0 is of order p. Proof. If one of hai or hbi, say hai, is normal in G, then G is a semidirect product of hai and hbi. In that case, we see that hbi is not normal in G and so jhbiG W hbij D p and jhai \ hbiG j D p. But G 0 hai \ hbiG and so jG 0 j D p. Hence we may assume that jhaiG W haij D jhbiG W hbij D p: Set A D haiG and B D hbiG which yields jA W haij D p, G=A ¤ ¹1º and Ahbi D G. Similarly, jB W hbij D p, G=B ¤ ¹1º and Bhai D G. Hence we obtain G 0 A \ B. If hbi \ A D ¹1º or hai \ B D ¹1º, then jA \ Bj D p and so jG 0 j D p and we are done. It follows that we may assume hbi\A ¤ ¹1º and hai\B ¤ ¹1º. As hai\hbi D ¹1º, we see that hbi\A D hb1 i is of order p and A D haihb1 i and similarly, hai\B D ha1 i is of order p and so A \ B D ha1 i hb1 i Š Ep2 . In particular, o.a/ p 2 , o.b/ p 2 ,

144 p-groups with small normal closures of all cyclic subgroups

377

and G D haihbi is a product of the two cyclic subgroups hai and hbi with G 0 ha1 i hb1 i: For any x 2 G, haix \ hai ¤ ¹1º and so haiG ha1 i which implies ha1 i E G and so a1 2 Z.G/. Similarly, b1 2 Z.G/ and so G 0 ha1 i hb1 i Z.G/. In this case, hŒa; bi D G 0 is of order p and we are done. Lemma 144.3. If jGj D p mC2 and exp.G/ D p m with m 3 and p > 2, then we have jG 0 j p 2 . Proof. By Theorem 74.1, G is either metacyclic or an L3 -group. In the second case, G 0 < 1 .G/ and so jG 0 j p 2 . Let G be metacyclic, and let R be a normal elementary m1 abelian subgroup of order p 2 in G and a 2 G of order p m . Since ap centralizes R and G is metacyclic, we have jhai \ Rj D p and so .Rhai/=R is a cyclic subgroup of index p in G=R and G=R is noncyclic. Therefore there exists b 2 G .Rhai/ with b p 2 R so that o.b/ p 2 . Also, G=.Rhap i/ is elementary abelian of order p 2 which gives Rhap i D ˆ.G/ and so G D ha; bi. Since metacyclic p-groups for p > 2 are regular, we obtain that exp.hbiG / p 2 . Then G 0 hbiG gives exp.G 0 / p 2 . On the other hand, G 0 is cyclic and so jG 0 j p 2 and we are done. Lemma 144.4. Let G D hx; yi 2 BI.p 2 /, p > 2. If hxi \ hyi D ¹1º, then the following hold: (i) cl.G/ 4. In particular, cl.G/ 3 if o.x/ o.y/ p 3 . (ii) jG 0 j p 3 and exp.G 0 / p 2 . (iii) Ã2 .G/ Z.G/. Proof. We may assume that G is nonabelian. Let o.x/ D p m , o.y/ D p n . Since G is a BI.p 2 /-group, we have jhxiG W hxij p 2 . Then from jhxijjhx y ij=jhxi \ hx y ij D jhxihx y ij jhxiG j p mC2 2

we get jhxi \ hx y ij p m2 and x p 2 hxi \ hx y i. Hence 2

2

2

hx p i D h.x y /p i D h.x p /y i: 2

2

2

Let M D hx p ; yi. Then hx p i E M . By Lemma 144.1, we have x p 2 NG .hyi/ and 2 2 so hyi E M . But hxi\hyi D ¹1º and so Œx p ; y D 1 and x p 2 Z.G/. Similarly, we 2 get y p 2 Z.G/. Set H D hxiG and K D hyiG so that G 0 H \K. If jH \Kj p 3 , then we obtain cl.G/ 4. Hence if o.y/ p 2 , then jKj p 4 and thus H \ K < K which yields that cl.G/ 4. In this case, we also conclude that exp.G 0 / p 2 . Indeed, assume that H \ K D G 0 is cyclic of order p 3 . Then the group G is regular and K D hyiG is of

378

Groups of prime power order

exponent p 2 , a contradiction. We have a similar situation if o.x/ p 2 . So we need only consider the case where o.x/; o.y/ p 3 . Without loss of generality we may assume that m n 3. We shall prove that under this condition cl.G/ 3 which will complete the proof of (1). Since G is a BI.p 2 /-group and hxi \ hyi D ¹1º, we see that jH \ Kj p 4 . If jH \ Kj D p 3 , then xp m1

m1

yp

2 H \ K;

n1

2H \K

n1

and, by the above, x p ; yp 2 Z.G/. This implies that H \ K 2 Z2 .G/ and then 0 cl.G/ 3 and exp.G / p 2 . So we need to consider the case jH \ Kj D p 4 m2

where x p

m2

n2

; yp

n2

m1

2 H \K

n1

and so H \ K D hx p ihy p i. Set A D hx p ; y p i and so, by the above, we N get A Z.G/ since m; n 3. Now consider G D G=A and we use the “bar convention”. Then GN D hx; N yi N and we obtain N

N

jhxi N G W hxij N D jhyi N G W hyij N D p: By Lemma 144.2, jGN 0 j p and so jG 0 j p 3 . If jG 0 j p 2 , then cl.G/ 3. In case jG 0 j D p 3 , we have A G 0 . Since A Z.G/, we conclude that ŒG; G 0 A Z.G/ and so cl.G/ 3. Obviously, in any of the above cases, jG 0 j p 3 and exp.G 0 / p 2 . We have proved (1) and (2). Let g1 D x i y j 2 G. By the Hall–Petrescu formula (see Appendix 1), 2

2

2

.x i y j /p D .x i /p .y j /p a2l2 a3l3 a4l4 ; 2 where ai 2 Ki .G/, li D pi , i D 2; 3; 4. If p > 3, then p 2 j li for i D 2; 3; 4. Since exp.G 0 / p 2 , it follows that p2

2

2

2

g1 D .x i y j /p D .x i /p .y j /p 2 Z.G/: For any g D x k1 y m1 : : : x kr y mr 2 G, we see that 2

2

2

2

2

g p D .x k1 /p .y m1 /p : : : .x kr /p .y mr /p 2 Z.G/: Now we consider the case p D 3. We shall prove in the following four cases that exp.K3 .G// 3. (i) Suppose that o.y/ D 3. Then K D hyiG is of order 33 . Since K is generated by elements of order 3, it follows that exp.K/ 3. But G 0 K and so we are done. (ii) Suppose that o.x/ D o.y/ D 32 . Then jH j; jKj 34 and so jH \ Kj 33 . Obviously, exp.K3 .G// 3 if jH \ Kj 32 and so we need only consider the case jH \ Kj D 33 . Now we have jH j D jKj D 34 , jGj D 35 and jG 0 j jH \ Kj D 33 . If jG 0 j 32 , then exp.K3 .G// 3. It remains to consider the case that G 0 D H \ K

144 p-groups with small normal closures of all cyclic subgroups

379

is of order 33 . If G 0 is cyclic, then G is regular and so G D hx; yi is of exponent 32 , a contradiction. Therefore G 0 is noncyclic. By a result of Burnside, G 0 is abelian. Indeed, if G 0 were nonabelian, then Z.G 0 / is cyclic and so G 0 would also be cyclic, a contradiction. It follows that j1 .G 0 /j 32 , jG=1 .G 0 /j 33 and K3 .G/ 1 .G 0 /, and we are done. (iii) Suppose that o.x/ D 3m 33 and o.y/ D 32 . (We argue similarly in the case o.y/ D 3n 33 and o.x/ D 32 .) As above, we need only consider jH \ Kj D 33 and G 0 D H \ K. If G 0 is cyclic, then G is regular and so exp.K/ D 32 , a contradiction. Hence G 0 is noncyclic and then G 0 is noncyclic abelian with j1 .G 0 /j 32 . In that case, K3 .G/ 1 .G 0 /, and we are done. (iv) Suppose that o.x/; o.y/ 33 . By (2), we get jG 0 j 33 and we see by the above m1 n1 arguments that K3 .G/ hx 3 i y3 and so exp.K3 .G// 3. Since 32 j l2 ; l4 and 3 j l3 and exp.K3 .G// 3, it follows from the Hall–Petrescu formula that Ã2 .G/ Z.G/. Lemma 144.5. Suppose that G D hx; yi 2 BI.p 2 / is noncyclic, p > 2, o.x/ D p m and o.y/ D p n . If m n and m 3, then there exists y1 2 G such that G D hx; y1 i p and hxi \ hy1 i D ¹1º or hxi \ hy1 i D hy1 i with o.y1 / D p 2 , where p D 3. Proof. We proceed by induction on m C n. Let H D hxiG and K D hyiG . We ﬁrst prove our lemma in the special cases where n D 2 and m D n D 3, which illustrate the idea in the general case. (a) n D 2 and hxi \ hyi D hy p i. If p D 3, then y1 D y satisﬁes all the requirem2 ments. Hence we need only consider p > 3. Consider the subgroup M D hx p ; yi. m2 m2 Since jhx p iG j p 4 and y p 2 hx p i, we have jM j p 5 and so M is regular. m2 m1 p As y p D x dp for some integer d (by the assumption), we get .yx dp / D1 m1 by the regularity of M (Theorem 7.2(a)). Set y1 D yx dp . Then G D hx; y1 i and hxi \ hy1 i D ¹1º. (b) m D n D 3. We shall ﬁrst prove that here G satisﬁes cl.G/ 5, exp.G 0 / p 2 and exp.K3 .G// p in the following subcases. (b1) hxi \ hyi D hx p i. Then x p 2 Z.G/ and x p D .x p /y D .x y /p . Since G is a BI.p 2 /-group, we have jH j; jKj p 5 . By Theorem 74.1, either H is metacyclic (if jH j p 4 , then H is certainly metacyclic) or jH j D p 5 , j1 .H /j D p 3 and exp.1 .H // D p. If H is metacyclic, then H is regular. Since x; x y 2 H , we see that .x 1 x y /p D 1 (Theorem 7.2(a)) and exp.G 0 / p as G 0 1 .H /. Because H is metacyclic, j1 .H /j p 2 and so cl.G/ 3. So we need only consider the case that jH j D jKj D p 5

and j1 .H /j D j1 .K/j D p 3

with exp.1 .H // D exp.1 .K// D p. Set GN D G=1 .H / and then o.x/ N D p 2 and hxi N E GN so that GN D hxih N yi N is meta0 0 N N cyclic and so G < hxi. N Therefore jG j p and so K3 .G/ 1 .H /. We have proved that in this case cl.G/ 5, exp.G 0 / p 2 and exp.K3 .G// p.

380

Groups of prime power order 2

(b2) hxi \ hyi D hx p i. If H or K is an L3 -group, then the argument in the second paragraph of (b1) can be applied to see that G satisﬁes cl.G/ 5, exp.G 0 / p 2 and exp.K3 .G// p. So we need only consider the case where both H and K are metacyclic. (b2.1) H and K are metacyclic and H \ K is cyclic. Since G 0 H \ K , we obtain 2 2 2 G 0 D hŒx; yi. As x; x y 2 H and x p D .x p /y D .x y /p , we see that the regularity 2 2 of H implies .x 1 x y /p D 1 and so Œx; yp D 1. Hence jG 0 j p 2 , jK3 .G/j p and cl.G/ 3. (b2.2) H and K are metacyclic and H \ K is noncyclic. Since p is odd, we have j1 .H \ K/j p 2 . But H and K are metacyclic and so 1 .H / D 1 .H \ K/ D 1 .K/ is elementary abelian of order p 2 . Set GN D G=1 .H / so that GN D hx; N yi; N

hxi N \ hyi N D ¹1º;

N

jhxi N G W hxij N p;

N

jhyi N G W hyij N p:

By Lemma 144.2, we have jGN 0 j p and so K3 .G/ 1 .H /. Therefore cl.G/ 4, exp.G 0 / p 2 and exp.K3 .G// p. Now, 2

hxi \ hyi D hx p i or hxi \ hyi D hx p i; 2

2

and so x p D y kp , where 1 k p 1. Since cl.G/ 5, using the Hall–Petrescu formula we get 2 2 2 .xy k /p D x p y kp c2n2 c3n3 c4n4 c5n5 ; 2 where ci 2 Ki .G/, ni D pi , i D 1; : : : ; 5, and p 2 j n2 , p j n3 , p j n4 , p j n5 since 2 p > 2. Hence .xy k /p D 1. Let y0 D xy k so that G D hx; y0 i and o.y0 / p 2 . By (a), there exists y1 2 G such that hxi \ hy1 i D ¹1º or hxi \ hy1 i D hy13 i with o.y1 / D 9. (c) m D n 4. 2 2 3 (c1) hx p i hxi \ hyi. Hence we obtain x p 2 Z.G/ and hx p i E G. Consider 3 p 3 GN D G=hx i so that GN D hx; N yi N and o.x/ N D o.y/ N D p . By (b), there exists y1 2 G 2 N such that G D hx; N yN1 i and o.yN1 / p . Thus we get G D hx; y1 i and o.y1 / < o.x/. Let G0 D hx p ; y1 i. By induction, there exists y2 2 G0 such that G0 D hx p ; y2 i and p hx p i \ hy2 i D ¹1º or hx p i \ hy2 i D hy2 i with p D 3 and o.y2 / D 9. Therefore p G D hx; G0 i D hx; y2 i and hxi \ hy2 i D ¹1º or hxi \ hy2 i D hy2 i with o.y2 / D 9. (c2) hxi\hyi D hx p i. We shall prove in this part that cl.G/ 6, exp.G 0 / p 3 , exp.K3 .G// p 2 and exp.K4 .G// p. By Theorem 74.1, H is either metacyclic or H is an L3 -group and the same holds for K. (c2.1) First suppose that H (or K) is an L3 -group. Now consider the factor group N G D G=1 .H / D hx; N yi, N where hxi N \ hyi N D ¹1º. By Lemma 144.4, exp.GN 0 / p 2 . 0 2 N Then G is a cyclic group of order p as GN 0 HN D hxi. N Therefore exp.G 0 / p 3 , 2 0 exp.K3 .G// p and exp.K4 .G// p. Since jG j p 5 , we have cl.G/ 6. (c2.2) H and K are metacyclic and H \K is cyclic. Since G 0 H \K, we see that m1 0 G is cyclic and G 0 D hŒx; yi. Consider the factor group GN D G=hx p i D hx; N yi, N m1

144 p-groups with small normal closures of all cyclic subgroups

381

where hxi N \ hyi N D ¹1º. By Lemma 144.4, oŒx; N y N p 2 and so oŒx; y p 3 . Hence 0 3 2 exp.G / p , exp.K3 .G// p , exp.K4 .G// p and cl.G/ 4. (c2.3) H and K are metacyclic and H \ K is noncyclic. We have 1 .H / D 1 .K/ D 1 .H \ K/ Š Ep2 : N

N yi, N where hxi N \ hyi N D ¹1º and jhxi N G W hxij N p, Consider GN D G=1 .H / D hx; N 0 0 G jhyi N W hyij N p. By Lemma 144.2, we have jGN j p and so jG j p 3 which gives cl.G/ 4, exp.G 0 / p 3 , exp.K3 .G// p 2 and exp.K4 .G// p. m1 m1 m1 Since hxi \ hyi D hx p i, we have x p D y lp for some positive integer l p 1. By the Hall–Petrescu formula, we get m1

m1

m1

.xy l /p D xp y lp c2n2 c3n3 c4n4 c5n5 c6n6 ; m1 where ci 2 Ki .G/, ni D p i , i D 1; : : : ; 6, and p 3 j n2 , p 2 j nj , j D 3; : : : ; 6 m1 since p > 2. It follows that .xy l /p D 1 and so G D hx; xy l i with o.x/ > o.xy l /. By induction, there is an element y1 2 G such that G D hx; y1 i and hxi \ hy1 i D ¹1º p or hxi \ hy1 i D hy1 i with o.y1 / D p 2 , where p D 3. mk

(c3) hxi\hyi D hx p i for some k 2. In this case, we consider the factor group mkC1 N G D G=hx p i so that GN D hx; N yi. N Then we apply (c2) to ﬁnd g0 2 G such that GN D hx; N yN0 i with o.yN0 / < o.y/. N Hence we obtain G D hx; y0 i with o.y0 / < o.y/. By induction, there is y1 2 G satisfying all the requirements. mn (d) m > n 3. Let G1 D hx p ; yi. By induction, there exists an element y1 mn mn mn p p p such that G1 D hx ; y1 i and hx i \ hy1 i D ¹1º or hx p i \ hy1 i D hy1 i with o.y1 / D p 2 , where p D 3. Hence G D hx; G1 i D hx; y1 i satisﬁes all the requirements. Theorem 144.6. Let G 2 BI.p 2 / with p > 2. Then Ã2 .G/ Z.G/ and exp.G 0 / p 2 . Proof. We ﬁrst prove that Ã2 .G/ Z.G/. Let x 2 G be such that o.x/ D p m p 3 . For any y 2 G we consider G1 D hx; yi. 2 It sufﬁces to prove that Œx p ; y D 1. If hxi \ hyi D ¹1º, then Lemma 144.4 implies 2 2 that Ã2 .G1 / Z.G1 / and so Œx p ; y D Œx; y p D 1: (i) o.y/ p 2 . By the above argument, we need only consider the case o.y/ D p 2 2 2 and hxi \ hyi D hy p i. If m D 3, then x p 2 Z.G1 / and so Œx p ; y D 1. Let m > 3 p and x0 2 hxi be such that x0 D y p . By Lemma 144.1, jhxi W Nhxi .hyi/j p 2 and so hx0 i normalizes hyi and therefore Œx0 ; y 2 hy p i so that hx0 ; yi is of order p 3 and p p class 2. Now set y1 D x0 y so that y1 D x0 y p D 1 and G1 D hx; yi D hx; y1 i. 2 By Lemma 144.4, x p 2 Z.G1 / and we are done. (ii) o.y/ p 3 . Without loss of generality we may assume o.x/ o.y/. If there is y1 such that G1 D hx; yi D hx; y1 i and hxi \ hy1 i D ¹1º, then, by Lemma 144.4, we 2 have Ã2 .G1 / Z.G1 / which gives Œx p ; y D 1. Due to Lemma 144.5, there is only one exceptional case where p D 3, G1 D hx; yi D hx; y1 i, hxi \ hy1 i D hy13 i and 2 o.y1 / D 9. By (i) applied to G1 D hx; y1 i, we see that x 3 2 Z.G1 / and we are done.

382

Groups of prime power order

Now we prove exp.G 0 / p 2 . We claim that o.Œx; y/ p 2 for all x; x 2 G. If o.y/ p 2 , then jhyiG j p 4 . Suppose that exp.hyiG / p 3 . Then hyiG is metacyclic and so regular which implies that exp.hyiG / D exp.hyi/ p 2 , a contradiction. Therefore we get exp.hyiG / p 2 and so o.Œx; y/ p 2 since Œx; y 2 hyiG for each x 2 G. If o.x/ o.y/ p 3 , then consider G1 D hx; yi so that, by Lemma 144.5, there exists an element y1 such that G1 D hx; yi D hx; y1 i and hxi \ hy1 i D ¹1º or hxi \ hy1 i D hy13 i and o.y1 / D 9. If hxi \ hy1 i D ¹1º, then, by Lemma 144.4, o.Œx; y/ p 2 . In case hxi \ hy1 i ¤ ¹1º, we get o.y1 / D 32 and so we also see that o.Œx; y/ 32 . Indeed, by the above argument, we infer that jhy1 iG1 j 34 and exp.hy1 iG1 / 32 and since G10 hy1 iG1 , we get o.Œx; y/ 32 . It sufﬁces to show that exp.ha; bi/ p 2 for any a; b 2 G with o.a/; o.b/ p 2 . Set A D ha; bi, H D haiA and K D hbiA . Hence we get jH j p 4 , jKj p 4 and jAj p 6 . If jAj D p 6 , then jH \ Kj p 2 and then jA0 j p 2 , jK3 .A/j p and cl.A/ 3. By the Hall–Petrescu formula, exp.A/ p 2 . So it remains to consider the case jAj p 5 . Suppose that exp.A/ p 3 . By Theorem 74.1, A is either metacyclic or jAj D p 5 and A is an L3 -group so that j1 .A/j D p 3 and exp.1 .A// D p. If A is metacyclic, then A is regular, and therefore exp.A/ p 2 . In the second case, we get exp.2 .A// p 2 . But a; b 2 2 .A/, a contradiction. Hence exp.A/ p 2 . We have proved that exp.G 0 / p 2 . Theorem 144.7. Let G 2 BI.p 2 / with p > 2. Then cl.G/ 4. Proof. It sufﬁces to prove that Œx; y 2 Z3 .G/ for any x; y 2 G. We set o.x/ D p m , o.y/ D p n , H D hxiG and K D hyiG . If jH \ Kj p 3 , then Œx; y 2 Z3 .G/ since Œx; y 2 H \ K E G. (i) m 2 and n 2. Hence jH j; jKj p 4 and jH 0 j; jK 0 j p 2 . If y 2 H , then Œx; y 2 H 0 and so Œx; y 2 Z2 .G/ Z3 .G/. Similarly, Œx; y 2 Z3 .G/ if x 2 K. Now assume that x 62 K and y 62 H . Then jH \ Kj p 3 and then Œx; y 2 Z3 .G. (ii) m 2 and n 3 (and similarly for m 3 and n 2 ). Since jH j p 4 , we need only consider jH \ Kj D p 4 and then H K. By Lemma 144.3, jK 0 j p 2 and therefore Œx; y 2 Z3 .G/. (iii) m 3 and n 3. If hxi \ hyi D ¹1º, then jH \ Kj p 4 . As before, we need m1 n1 only consider jH \ Kj D p 4 . Then x p ; yp 2 H \ K. By Theorem 144.6, m1 n1 hx p ; y p i Z.G/. Hence we get H \ K Z3 .G/. We consider now the case hxi \ hyi ¤ ¹1º. Set G1 D hx; yi, where we can assume m n. By Lemma 144.5, p there exists y1 2 G1 such that G1 D hx; y1 i and hxi\hy1 i D ¹1º or hxi\hy1 i D hy1 i with o.y1 / D p 2 , where p D 3. If hxi \ hy1 i D ¹1º, we see that G10 Z3 .G/ as above and then Œx; y 2 Z3 .G/. In the second case, o.y1 / D 32 and so, by (ii) above, we also have Œx; y 2 G10 Z3 .G/. Lemma 144.8. Let G 2 BI.p 2 / with p > 2. Suppose that x; y 2 G, o.x/ D p m p 3 and o.y/ D p n p 3 . If hxi \ hyi D ¹1º, then Œx; y 2 Z2 .G/.

144 p-groups with small normal closures of all cyclic subgroups m1

383

n1

Proof. By Theorem 144.6, x p ; yp 2 Z.G/. Set H D hxiG and K D hyiG . 2 Since G is a BI.p /-group, we have jH \ Kj p 4 . If jH \ Kj p 2 , then we have m1 n1 ; yp 2H \K Œx; y 2 H \ K Z2 .G/. In case jH \ Kj D p 3 , we get x p m1 n1 p p and x ;y 2 Z.G/ and so H \ K Z2 .G/. We need only consider the case jH \ Kj D p 4 in which case H \ K D hx p

m2

ihy p

n2

i:

m2

n1

2 Z.G/. Since also y p 2 Z.G/, we get If m 4, then, by Theorem 144.6, x p Œx; y 2 H \ K Z2 .G/. Similarly, we have Œx; y 2 H \ K Z2 .G/ in case n 4. We may assume that jH \ Kj D p 4 and m D n D 3. Since hxi \ hyi D ¹1º, we have jH \ hyij D p 2 and so y p 2 H . Thus H D hx; y p i D hxihy p i is of order p 5 and HK E G is of order p 6 since HK D H hyi with y p 2 H . But jhxihyij D jhxijjhyij D p 6 and so HK D hxihyi is a product of two cyclic subgroups. By Corollary 36.8, H is metacyclic and so H is also regular. Thus HK is metacyclic of exponent p 3 which implies that .HK/0 is cyclic of order p 2 . Since Œx; y 2 .HK/0 and .HK/0 Z2 .G/, we are done. Theorem 144.9. Let G 2 BI.p 2 / with p > 2. Then we have cl.G/ 3 if and only if Œ2 .G/; G Z2 .G/. Proof. Obviously, we need only prove that our condition is sufﬁcient. Let x; y 2 G. If o.y/ p 2 (or o.x/ p 2 ), then Œx; y 2 Z2 .G/ by assumption. Hence we may assume that o.x/ o.y/ p 3 . Let G1 D hx; yi. By Lemma 144.5, there is y1 2 G1 such that G1 D hx; y1 i and hxi \ hy1 i D ¹1º or hxi \ hy1 i D hy13 i with o.y1 / D 9. If o.y1 / p 2 , we have Œx; y1 2 Z2 .G/ by assumption and Œx; y 2 G10 Z2 .G/. In case o.y1 / p 3 , we get hxi \ hy1 i D ¹1º and, by Lemma 144.8, Œx; y1 2 Z2 .G/ and Œx; y 2 G10 Z2 .G/. Therefore cl.G/ 3.

Appendix 27

Wreathed 2-groups

In this section we study the subgroup structure of wreath products of some small 2-groups with a group of order 2. k Let solk .G/ be the number of solutions of x 2 D 1 in a group G. Exercise 1. Let A be a ﬁnite group and G D A wr C , where C D hc j c 2 D 1i Š C2 . Then sol1 .G/ D jAj C sol1 .A/2 : (We have c1 .G/ D sol1 .G/ 1.) It follows that the set G .A Ac / contains exactly jAj involutions. Since CG .c/ D hci , where is the diagonal of the base A Ac , all involutions of the set G .AAc / are conjugate in G in view of jG W CG .c/j D jAj. Solution. Let B D A Ac be the base of G D C B (a semidirect product with kernel B). An element g 2 G B can uniquely be represented in the form g D xy c c for x; y 2 A. We have g 2 D xy c cxy c c D xcyxcy D x.yx/c y D .xy/.yx/c : If g 2 D 1, then xy D 1, i.e., y D x 1 . It follows that G B contains exactly jAj involutions and sol1 .G/ D jAj C sol1 .B/ D jAj C sol1 .A/2 : In particular, if A Š C2n , then sol1 .G/ D 2n C 4 so c1 .G/ D 2n C 3. Next, c1 .D8 wr C2 / D 8 C 62 1 D 43;

c1 .Q8 wr C2 / D 8 C 22 1 D 11:

1o . Let G D A wr C , where n

A D ha j a2 D 1i Š C2n ;

C D hc j c 2 D 1i Š C2 :

Then jGj D 22nC1 and n

n

G D ha; b; c j a2 D b 2 D c 2 D 1; Œa; b D 1; ac D bi: The group B D hai hac i is the base of the wreath product G. Every element of B can be written uniquely in the form ai .ac /j , where i; j D 1; 2; : : : ; 2n . If n D 1, then we have G Š D8 . So, in what follows, we assume that n > 1. The abelian subgroup B

A.27

Wreathed 2-groups

385

of type .2n ; 2n / is maximal in G so we get Z.G/ D CB .c/ < B. Let ai .aj /c 2 Z.G/, 0 i; j < p n . Then ai .aj /c D .ai .aj /c /c D aj .ai /c : It follows that ai D aj . It is easy to see that .aac /i D ai .ai /c 2 Z.G/ for all i . Thus we have proved assertion (a) of the following Lemma A.27.1. (a) Z.G/ D haac i Š C2n . (b) G is not metacyclic unless n D 1. Proof. It remains to prove (b). Due to (a), h1 .B/; ci is nonabelian so it is isomorphic to D8 . By Proposition 10.19, if a metacyclic 2-group contains a nonabelian subgroup of order 8, it is of maximal class. Since G is not of maximal class unless n D 1, it is not metacyclic for n > 1. Remark. Let G D A wr B be a p-group. Then d.G/ D d.A/ C d.B/. We claim that if ¹x1 ; : : : ; xr º and ¹y1 ; : : : ; ys º are minimal bases of the groups A and B, respectively, then B D ¹x1 ; : : : ; xr ; y1 ; : : : ; ys º is a minimal basis of G. Indeed, we have hBi D G, hB ¹xi ºi < G and hB ¹yj ºi < G for all i r and j s. In particular, jG W ˆ.G/j D jA W ˆ.A/jjB W ˆ.B/j D p rCs : It follows from Lemma 1.1 that 22nC1 D jGj D 2jZ.G/jjG 0 j D 2nC1 jG 0 j so that jG 0 j D 2n . As G D ha; ci, we have d.G/ D 2. Since ˆ.G/ is a subgroup of index 2 in an abelian group B of type .2n ; 2n /, we get the following Lemma A.27.2. (a) jG 0 j D 2n . (b) ˆ.G/ is abelian of type .2n ; 2n1 /. Lemma A.27.3. G=Z.G/ Š D2nC1 . In particular, cl.G/ D n C 1. Proof. We have .aZ.G//c D ac Z.G/ D a1 .aac Z.G// D a1 Z.G/ (Lemma A.27.1(a)). Since ha; Z.G/i D B and G D hB; ci, the result follows. Since G=ŒB; C is abelian, we get G 0 D ŒB; C . Lemma A.27.4. Let n > 1. (a) G 0 D ha1 ac i Š C2n and G=G 0 is abelian of type .2n ; 2/.1 (b) D D hc; G 0 i Š D2nC1 and G=D is cyclic; then j1 .G/j D 2jDj D 2nC2 . 1 This

agrees with the Ito–Ohara result; see Corollary 36.11 (indeed, as we shall prove, our nonmetacyclic group G is a product of two cyclic subgroups).

386

Groups of prime power order

(c) 1 .G/ D D1 .B/ is of order 2nC2 , D \ 1 .B/ D 1 .Z.G//. (d) D < 2 .G/, jG W n1 .G/j D 2. Proof. We have a1 ac D Œa; c 2 G 0 and o.a1 ac / D 2n D jG 0 j which proves (a) (Lemma A.27.2(a)). Since .a1 ac /c D .a1 ac /1 , we see that D D hc; G 0 i Š D2nC1 . As d.G/ D 2, D G G and Z.D/ is cyclic, it follows that D 6 ˆ.G/ (Lemma 1.4) so that the quotient group G=D is cyclic. By Exercise 1, c1 .G/ D 2n C 3 D c1 .D/ C 2. It follows from n > 1 that jD \ 1 .B/j D 2 so 1 .G/ D D1 .B/ (see Proposition 1.17(a)) is of order 2jDj D 2nC2 , completing the proof of (b) and (c). Note that D D 2 .D/ and 1 .B/ 6 D so we obtain 2 .G/ > D. It follows from the cyclicity of G=1 .G/ as an epimorphic image of G=D and j1 .G/j D 2nC2 that jn1 .G/j D 22n hence jG W n1 .G/j D 2, completing the proof of (d). Lemma A.27.5. The class number of G is equal to k.G/ D 22n1 C 2nC1 2n1 . Proof. We have cd.G/ D ¹1; 2º (Introduction, Theorem 17) so that jGj jG W G 0 j 22 2nC1 2 2nC1 D 2nC1 C D 22n1 C 2nC1 2n1 ; 22

k.G/ D jG W G 0 j C

and the proof is complete. Lemma A.27.6. We have o.ac/ D 2nC1 hence exp.G/ D 2nC1 . Proof. We obtain exp.G/ exp.B/ exp.C / D 2nC1 . The element .ac/2 D aac has order 2n hence o.ac/ D 2nC1 . Lemma A.27.7. The subgroup G 0 \ Z.G/ D 1 .G 0 / is of order 2 and ˆ.G/ D ha2 i haac i D ha2 i Z.G/ D G 0 Z.G/ is abelian of type .2n ; 2n1 /. Proof. Clearly, we see that G 0 \ Z.G/ D 1 .G 0 / since G 0 < D and D G G is of maximal class. Further, we get ˆ.G/ D Ã1 .G/ and so a2 ; aac 2 ˆ.G/ as aac D a2 Œa; c. Since ha2 i \ haac i D ¹1º, the ﬁrst equality in the displayed line follows. Because of G 0 \ ha2 i D ¹1º, B D G 0 hai and a2 2 ˆ.G/, we get ˆ.G/ D ha2 i G 0 (compare orders). Since G 0 Z.G/ ˆ.G/ and G 0 \ Z.G/ D 1 .G 0 /, we obtain G 0 Z.G/ D ˆ.G/ by the product formula. Let us compute ck .G/ for all positive integers k, k n C 1 D log2 .exp.G//. Since ck .B/ D

jk .B/ k1 .B/j 22k 22k2 D D 3 2k1 2k1 2k1

A.27

387

Wreathed 2-groups

for k n, it remains to compute the number solk .G B/ of elements of order 2k > 2 in the set G B; then ck .G/ D 3 2k1 C

solk .G B/ 2k1

(here 2k1 D '.2k /, where './ is Euler’s totient function). Let x D ai .ac /j c 2 G B, i; j D 0; 1; 2; : : : ; 2n 1. We have x 2 D ai Cj .ac /i Cj D .aac /i Cj 2 Z.G/: Suppose that x 2 D 1. Then ai Cj D 1 so i C j 0 .mod 2n /. We see that there are in G B exactly 2n elements of order 2, i.e., c1 .G/ D 2n C c1 .B/ D 2n C 3 (this coincides with the result of Exercise 1). Now suppose that x 4 D 1. In that case, we have 1 D x 4 D a2.i Cj / .ac /2.iCj / so 2.iCj / D 1. It follows that i C j 0 .mod 2n1 /. If i D 0, then j 2 ¹0; 2n1 º. a Similarly, for arbitrary i we have j 2 ¹2n1 i; 2n i º. Thus G B contains exactly 2 2n elements of order 4 so exactly 2 2n 2n D 2n elements of order 4. Now we prove for k 2 ¹2; : : : ; nº that G B contains exactly 2nC.k2/ elements of k order 2k . By the previous paragraph, this is true for k D 2. Let x 2 D 1 (x 2 G B). Then a.iCj /2

k1

.ac /.iCj /2

k1

D1

so that i C j 0

.mod 2n.k1/ /:

It follows that for arbitrary i we have j D s 2n.k1/ i;

where s D 1; 2; : : : ; 2k1 :

Thus for a ﬁxed i there are in G B exactly 2k1 elements ai .ac /j c of order 2k whence the set G B contains exactly 2k1 2n D 2nC.k1/ elements of order 2k , k < n C 1. By induction, the set G B has 2nCk2 elements of order 2k1 . Thus the set G B possesses exactly 2nCk1 2nCk2 D 2nCk2 elements of order 2k . Now let o.x/ D 2nC1 ; then o.x 2 / D 2n , where x 2 D ai Cj .ac /i Cj D .aac /i Cj . It follows that i C j is odd, and we conclude that G B has exactly 2n1 2n D 22n1 elements of order 2nC1 . Since 2n C 2n C 2nC1 C 2nC2 C C 22n1 D 22n D jG Bj; we have found all elements of the set G B. Let Nk .X/ be the number of elements of order 2k in a set X; then, with k D 3; : : : ; n, Nk .B/ D 3 22.k1/ ;

N1 .G B/ D N2 .G/ D 2n ;

Nk .G B/ D 2nCk2 :

We obtain the number ck .G/ from the following equality: Nk .G/ D Nk .G B/ C Nk .B/ D 2k1 ck .G/:

388

Groups of prime power order

Thus we have c1 .G/ D 2n C 3;

c2 .G/ D 3 2 C 2n1 ;

ck .G/ D 3 2k1 C 2n1 cn .G/ D 3 2

n1

C2

n1

for k D 3; : : : ; n; D 2nC1 ;

cnC1 .G/ D 22n1 =2n D 2n1 :

Since .ac/2 D aac 2 Z.G/ is of order 2n , we infer that Z.G/ has index 2 in haci so Z.G/ is not a maximal cyclic subgroup of G. It follows from the proof of Lemma 27.3 n1 n1 that Z.G=Z.G// D ha2 Z.G/i. However, the subgroup ha2 ; Z.G/i is abelian of type .2n ; 2/. Since G=Z.G/ Š D2nC1 has only one minimal normal subgroup, it follows that haci is not normal in G; moreover, it has exactly 2n1 conjugates in G. We claim that all cyclic subgroups of the group G that are not contained in ˆ.G/ are not G-invariant. Indeed, assume that X GG is cyclic and X 6 ˆ.G/. Since d.G/ D 2, it follows that G=X is cyclic, and hence G is metacyclic, a contradiction. Let cG 0 , uG 0 and cuG 0 be all involutions in the quotient group G=G 0 . Then we get c1 .hc; G 0 i/ D 2n C 1 (Lemma A.27.4(b)). Since c1 .G/ D 2n C 3, it follows that the subgroups L1 D hu; G 0 i and L2 D hcu; G 0 i contain together exactly four involutions (one of them is an element of G 0 ). Therefore one may assume that the subgroup L1 contains exactly one involution so L1 is cyclic or generalized quaternion. Since G 0 is contained in an even number of subgroups of maximal class and order 2jG 0 j D 2nC1 and D D hc; G 0 i Š D2nC1 is dihedral so of maximal class, we obtain L1 Š Q2nC1 . As CG .G 0 / > G 0 , it follows that L2 is abelian of type .2n ; 2/, and so G 0 is a maximal cyclic subgroup of G. Then 1 .G/ D DL1 D D1 .B/ (compare the number of involutions and use Theorem 1.17(a); DL1 is not of maximal class since 1 .B/ G DL1 ). Now we shall ﬁnd all three members of the set 1 . We have B 2 1 . Since G satisﬁes cl.G/ D n C 1 > 2, B is the unique abelian maximal subgroup of G and two remaining members of the set 1 have class at least n (Fitting’s lemma). The subgroup ˆ.G/ D ha2 i h.ac/2 i D ha2 i haac i D ha2 i Z.G/ is abelian of type .2n ; 2n1 /. Since c 62 B, the subgroup M D hc; ˆ.G/i D hc; a2 ; aac i is another maximal subgroup of G. It then follows from Lemma A.27.3 that M=Z.G/ is abelian of type .2; 2/ if n D 2 and isomorphic to D2n if n > 2. By the above, we have cl.M / D n; then Z.M / D Z.G/. Since hc; G 0 i \ Z.G/ D 1 .G 0 /, we obtain M D hc; G 0 iZ.G/ by the product formula so d.M / D 3. As .ac/2 D aac 2 Z.G/, we get G 0 \ haci D 1 .G 0 / so N D haciG 0 is the third (metacyclic) member of the set 1 . Indeed, we have N ¤ B since exp.N / D 2nC1 > 2n D exp.B/.2 2 This

agrees with the known result that a nonmetacyclic two-generator 2-group has an even number of metacyclic subgroups of index 2 (see Theorem 107.1).

A.27

389

Wreathed 2-groups

As haci \ hai D ¹1º, we get G D hacihai by the product formula. Since all members of the set 1 are pairwise nonisomorphic, they are characteristic in G. It follows that Aut.G/ is a 2-group.3 Let D ¹xx c j x 2 Aº be the diagonal subgroup of B. Then D Z.G/. Assume that G has an elementary abelian subgroup E of order 8. Then G D BE so E \B D 1 .B/ Š E4 , CG .E \B/ BE D G hence E \B Z.G/, a contradiction since Z.G/ D haac i is cyclic. Thus E does not exist. 2o . Let G D M wr C , where M is a 2-group of maximal class and order > 8 and C D hci Š C2 . Set B D M M c (B is the base of G). Since B 0 D M 0 .M 0 /c G 0 ˆ.G/ is G-invariant and 1 .G=B 0 / D G=B 0 , we deduce that G 0 D ˆ.G/. Therefore, since d.G/ D 3 (see the remark following Lemma A.27.1), we obtain G=G 0 Š E8 hence B 0 < G 0 (compare indices!). Next, D ¹xx c j x 2 M º is the diagonal subgroup of B which is isomorphic to M and < ˆ.G/ D G 0 so that M 0 ˆ.G/ D G 0 . Clearly, B 0 \ has index 4 in and jG W B 0 j D 25 . It follows that jG W B 0 j D 8 D jG W G 0 j whence B 0 D G 0 . Since M 0 \ D ¹1º, the subgroup M 0 is a semidirect product with kernel M 0 and has index 8 D jG W G 0 j in G. Thus G 0 D M 0 . Assume that G possesses a normal subgroup E Š E8 . Clearly, E 6 B since B has no normal elementary abelian subgroup of order 8 in view of jM j > 8. Then we have E \ B D Z.B/ so that CG .Z.B// BE D G, and hence Z.B/ D Z.G/ is noncyclic. Since CB .c/ D , it follows that Z.G/ < hence Z.G/ is cyclic, and this is a contradiction. Thus E does not exist. In the case under consideration, the subgroup Z Z c , where Z < M is cyclic of index 2, is a G-invariant metacyclic subgroup, and we have G=.Z Z c / Š D8 since Z Z c 6 G 0 and 1 .G=.Z Z c // D G=.Z Z c /. Thus the result of Theorem 50.1 is best possible. Also, the result of Theorem 50.3 is also best possible. 3o . In this subsection G D Q wr C , where n1

Q D ha; b j a2

D 1; a2

n2

D b 2 ; ab D a1 i Š Q2n ;

C D hc j c 2 i Š C2 :

We have jGj D 22nC1 ;

Z.G/ D hb 2 .b 2 /c i;

jZ.G/j D 2;

G D ha; b; ci

and d.G/ D 3 by the remark following Lemma A.27.1. The subgroup B D Q Qc is the base of the wreath product G. Let A D hai; then A1 D AAc is an abelian normal subgroup of G (indeed, A1 is B-invariant and c-invariant and G D hB; ci). It is easy to proof: If 2 Aut.G/ is of odd order, then D idG for n D 1; if n > 1, then stabilizes the chain G > Z.G/ > ¹1º so D idG again. 3 Another

390

Groups of prime power order

see that G=A1 Š D8 so A1 6 ˆ.G/. Since .ac/2 D aac is of order 2n1 D exp.B/, we have o.ac/ D 2n D exp.G/. We obtain that .bc/2 D bb c 62 A1 , .bb c /c D bb c so hbb c A1 i D Z.G=A1 /. Set T =A1 D hbb c A1 i. Then T G G and G=T is abelian of type .2; 2/. By Exercise 1, c1 .G/ D 2n C 3 and all involutions of the set G B are conjugate in G. By the previous paragraph, N1 .G B/ D 2n ;

N1 .G/ D 2n C 3;

sol1 .G/ D 2n C 4:

Every element of G B can be presented uniquely in the form z D xy c c for some x; y 2 Q. Note that xy c D y c x for all x; y 2 Q. We have .xy c c/2 D xy c cxy c c D xy c x c y D .xy/.yx/c ;

(1)

k

k1

.xy c c/2 D .xy/2

(1a)

..yx/c /2

k1

:

Note that sol1 .Q/ D 2

(2)

and, for k > 1;

solk .Q/ D 2n1 C 2k :

It follows from (2) that (3)

N2 .B/ D sol2 .Q/2 sol1 .Q/2 D .2n1 C 4/2 4 D 22.n1/ C 2nC2 C 12

and, for 2 < k < n, (4)

Nk .B/ D .2n1 C 2k /2 .2n1 C 2k1 /2 D .2n C 3 2k1 /2k1 :

Now let z D xy c c 2 G B, x; y 2 Q and z 4 D 1. Then, by (1), .xy/2 D 1 so x D y 1 u, where u2 D 1 and u 2 Q, and we infer that sol2 .G B/ D 2jQj D 2nC1 and hence N2 .G B/ D sol2 .G B/ sol1 .G B/ D 2n : It follows that N2 .G/ D 22.n1/ C 2nC2 C 2n C 12

(5) so that

1 c2 .G/ D N2 .G/ D 22n3 C 2nC1 C 2n1 C 6: 2

Let 2 < k < n, z 2 G B with z 2 D 1. Then we get x D y 1 u, where u 2 Q and o.u/ 2k1 (see (1a)). It follows that k

solk .G B/ D jQjsolk1 .Q/ D 2n .2n1 C 2k1 / (see (2)). Then (6)

Nk .G B/ D 2n .2n1 C 2k1 / 2n .2n1 C 2k2 / D 2nCk2

A.27

Wreathed 2-groups

391

and we obtain Nk .G/ D Nk .B/ C Nk .G B/ D .2n C 3 2k1 /2k1 C 2nCk2 (7)

D 3.2n1 C 2k1 /2k1 ; ck .G/ D 3.2n1 C 2k1 /

.2 < k n/:

Since exp.G/ D 2n , all elements of G of order 2n lie in the set G B in view of exp.B/ D 2n1 . If o.xy c c/ D 2n , then y D x 1 u, where u 2 Q with o.u/ D 2n1 . It follows that the number of elements of order 2n in G is equal to jQj 2n2 D 22n2 so that 22n2 D 2n1 : cn .G/ D '.2n / We state the obtained results in the following Proposition A.27.8. Suppose that G D Q wr C2 , where Q Š Q2n . Then G 0 D ˆ.G/ is of index 8 in G and (a) c1 .G/ D 2n C 3. (b) c2 .G/ D 22n3 C 2nC1 C 2n1 C 6. (c) If 2 < k < n, then ck .G/ D 3 .2n1 C 2k1 /. (d) cn .G/ D 2n1 . Exercise 2. Let G D D wr C2 be a 2-group, where D has a cyclic subgroup of index 2. Find ck .G/ and describe all maximal subgroups of G. Exercise 3. Let G D A wr C2 , where A is an arbitrary (ﬁnite) group, C2 D hci. Is it true that all involutions in the set G .A Ac / are conjugate in G? Exercise 4. Let G D M wr C2 be a 2-group, where M has a cyclic subgroup of index 2. Find cl.G/. Exercise 5. Consider the standard wreath product G D E1 wr E2 , where E1 ; E2 are elementary abelian groups of order 2n , 2m , respectively. Find c1 .G/. Exercise 6. Let A be a 2-group, G D A wr C2m . Express c1 .G/ in terms of A. Exercise 7. Let G D A wr Cp be a p-group. Express jG W G 0 j in terms of A. Exercise 8. Let G D M1 M2 , where M1 and M2 are 2-groups of maximal class. Find the number of subgroups of maximal class and given order in G. Exercise 9. Let G D M wr C2 , where M is a 2-group of maximal class. Find the number of nonabelian subgroups of order 8 in G.

392

Groups of prime power order

Exercise 10. Study the subgroup structure of the standard wreath product G D E2n wr M; where M is a 2-group of maximal class. Exercise 11. Study the subgroup structure of G D C2m wr C2n . Exercise 12. Study the subgroup structure of the standard wreath product G D C2n wr M; where M is a 2-group of maximal class. Exercise 13. Let G D A wr B be the standard wreath product. Is it true that G satisﬁes exp.G/ D exp.A/ exp.B/? Solution. Let a 2 A, b 2 B, o.a/ D m, o.b/ D n. Then .ab/n D aab

1

ab

2

ab

.n1/

is of order o.a/ D m. Exercise 14. Describe the structure of Aut(C2n wr C2 /. Exercise 15. Describe the structure of Aut.Q wr C2 /, where Q is a 2-group of maximal class. Exercise 16. Let G D Cpn wr Cp . Describe the structure of 1 .G/. Exercise 17. Prove an analog of Proposition A.27.8 for the groups G D D2n wr C2 and G D SD2n wr C2 .

Appendix 28

Nilpotent subgroups

1o Theorems of Sylow, Hall, Carter etc. We prove Sylow’s theorem in the following form: Theorem A.28.1 (Sylow). All maximal p-subgroups of a group G are conjugate and have order jGjp , and their number is 1 .mod p/. A subgroup P G is a Sylow p-subgroup of G provided jP j is a power of p and p does not divide jG W P j. According to Theorem A.28.1, the Sylow p-subgroups coincide with the maximal p-subgroups of G and are conjugate in G. Let Sylp .G/ denote the set of all Sylow p-subgroups of G and .G/ be the set of the prime divisors of jGj. Lemma A.28.2 (Cauchy). If p 2 .G/, then G contains a subgroup of order p. In particular, a maximal p-subgroup of G is > ¹1º. Proof. Suppose that G is a counterexample of minimal order. Then G has no proper subgroup C of order divisible by p and jGj ¤ p. If G has only one maximal subgroup, say M , then it is cyclic. Indeed, if x 2 G M , then hxi is not contained in M so we get hxi D G. Then hx o.x/=p i < G is of order p, a contradiction. If G is abelian and A ¤ B are maximal subgroups of G, then G D AB so jGj D

jAjjBj ; jA \ Bj

and p does not divide P jGj, a contradiction. In case that G is nonabelian, we obtain that jGj D jZ.G/j C kiD1 hi , where h1 ; : : : ; hk are sizes of noncentral G-classes. Since the hi are indices of proper subgroups in G, we see that p divides hi for all i . Then p divides jZ.G/j so Z.G/ contains an element of order p, and hence G is not a counterexample. The set S D ¹Ai ºniD1 of subgroups of a group G is said to be G-invariant if Axi 2 S for all i n and x 2 G. All members of a G-invariant set S are conjugate if and only if S has no nonempty proper G-invariant subset. Lemma A.28.3. Let S ¤ ¿ be a G-invariant set of subgroups of a group G. Suppose that whenever N ¤ ¿ is a G-invariant subset of S, then jN j 1 .mod p/. Then all members of the set S are conjugate in G.

394

Groups of prime power order

Proof. Assume that S has a proper G-invariant subset N ¤ ¿. Then S N ¤ ¿ is G-invariant so jSj D jN j C jS N j 1 C 1 6 1

.mod p/;

a contradiction. Thus S is a minimal G-invariant subset so all its members are conjugate in G. Let P be a maximal p-subgroup of a group G and P1 be a p-subgroup of G. In case PP1 G, we see that PP1 is a p-subgroup by the product formula, and we conclude that P1 P . If P1 is a maximal p-subgroup of G, then NP .P1 / D P \ P1 . Proof of Theorem A.28.1. One may assume that p divides jGj. Let S0 D ¹P D P0 ; P1 ; : : : ; Pr º be a G-invariant set of all maximal p-subgroups of G; then we have Pi > ¹1º for all i (Lemma A.28.2). Let r > 0 and let P act on the set S0 ¹P º via conjugation. Then the size of every P -orbit on the set S0 ¹P º is a power of p greater than 1 because the P -stabilizer of a ‘point’ Pi is equal to P \ Pi < P . Thus p divides r hence jS0 j D r C 1 1 .mod p/; and the ﬁrst assertion follows (Lemma A.28.3). Then p does not divide jG W NG .P /j. Since P is a maximal normal p-subgroup of NG .P / D N , the prime p does not divide jN=P j (Lemma A.28.2) whence p does not divide jG W N jjN W P j D jG W P j, i.e., jP j D jGjp . A standard statement of Sylow’s theorem is as follows: (a) A group G contains a subgroup of order jGjp . (b) The number of subgroups from (a) is 1 .mod p/. (c) All maximal p-subgroups of G have order jGjp . Remark 1. (1) (Frobenius) Let P be a p-subgroup of a group G and suppose that the set M D ¹P1 ; : : : ; Pr º Sylp .G/ is P -invariant and P 6 Pi for all i . Then we obtain that jMj 0 .mod p/ (let P act on M via conjugation) so, by Theorem A.28.1, the number of Sylow p-subgroups of G containing P is 1 .mod p/ (indeed, take as M the set of all Sylow p-subgroups not containing P ). Similarly, if P 2 Sylp .G/, then the number of p-subgroups of G of given order that are not contained in P is divisible by p (let P act on the set of the above p-subgroups!). (2) (Frobenius) Let M D ¹M1 ; : : : ; Ms º be the set of all subgroups of order p k in a group G of order p m , k < m. We claim that s D jMj 1 .mod p/. To prove this, we use induction on m. Let 1 D ¹G1 ; : : : ; Gr º be the set of all maximal subgroups of G. Because the number of subgroups of index p in the elementary abelian p-group

A.28

395

Nilpotent subgroups

G=ˆ.G/ of order, say p d , is equal to pd 1 D 1 C p C C p d 1 1 p1

.mod p/;

we get r D j1 j 1 .mod p/, so one may assume that k < m1. Let ˛i be the number of members of the set M contained in Gi and ˇj the number of members of the set 1 containing Mj for all i; j . Then, by double counting, ˛1 C C ˛r D ˇ1 C C ˇs ; By the above, r 1 .mod p/, and by induction, ˛i 1 .mod p/ for all i . Next, ˇj is the number of maximal subgroups in G=Mj ˆ.G/ so ˇj 1 .mod p/ for all j . By the displayed formula, jMj D s r 1 .mod p/. (3) (Frobenius; see also [Bur, Theorem 9.II]) It follows from (1) and (2) that the number of p-subgroups of order p k in a group G of order p k m is 1 .mod p/. (4) Suppose that the subsets S1 ; S2 Sylp .G/ are nonempty and disjoint. Further, let Pi 2 Si be such that the set Si is Pi -invariant, i D 1; 2; then jSi j 1 .mod p/, i D 1; 2 (see the proof of Theorem A.28.1). It follows that S2 is not P1 -invariant (otherwise, considering the action of P1 on S2 via conjugation, we get jS2 j 0 .mod p/ since P1 62 S2 ). (5) (Burnside) Let G be a non-p-closed group (i.e., we have Op .G/ 62 Sylp .G/) and P 2 Sylp .G/. Suppose that Q 2 Sylp .G/ ¹P º is such that D D P \ Q is a maximal, by inclusion, intersection of Sylow p-subgroups of G. Assume that D > ¹1º. We claim that N D NG .D/ is not p-closed. Assume that this is false. Then P1 2 Sylp .N / is nonnormal in N . It follows from the properties of p-groups that P \ P1 > D and Q \ P1 > D so P1 is not contained in P . Therefore, if P1 U 2 Sylp .G/, then we get U ¤ P . However, P \ Q D D < P \ P1 P \ U , a contradiction. The following assertion is easily checked. If R is an abelian minimal normal subgroup of a group G and G D HR, where H < G, then H is maximal in G and we have H \ R D ¹1º. Theorem A.28.4 (P. Hall [Hal1]). If is a set of primes, then all maximal -subgroups of a solvable group G are conjugate. By Theorems A.28.1 and A.28.4, a maximal -subgroup of a solvable group G is its -Hall subgroup, and this gives the standard form of Hall’s theorem. Let Hall .G/ be the set of all -Hall subgroups of G. Given H G, the subgroup \ HG D Hx x2G

is said to be the core of H in G.

396

Groups of prime power order

The proofs of Theorems A.28.4, A.28.6 and A.28.7 are based on the following Lemma A.28.5 ([Ore]). If distinct maximal subgroups F and H of a solvable group G > ¹1º have equal cores, then they are conjugate in G. Proof. One may assume that FG D ¹1º; then G is not nilpotent (otherwise, jGj is a prime and F D H D ¹1º). Let R be a minimal normal, say a p-subgroup, of G; then we have RF D G D RH , R \ F D ¹1º D R \ H . Let K=R be a minimal normal, say a q-subgroup, of G=R, q is a prime. In this case, K is nonnilpotent (otherwise, we get NG .K \ F / > F and so ¹1º < K \ F FG D ¹1º, contrary to the assumption) whence q ¤ p. Then, by the product formula, F \ K and H \ K are nonnormal Sylow q-subgroups of K hence they are conjugate in K (Theorem A.28.1) so are in G. It follows that then NG .F \ K/ D F and NG .H \ K/ D H are also conjugate in G, as was to be shown. Proof of Theorem A.28.4.1 We use induction on jGj. Let F and H be maximal -subgroups of G and R a minimal normal, say p-subgroup, of G. If p 2 , then R F and R H by the product formula, and F=R, H=R are maximal -subgroups of G=R so they are conjugate by induction; then F and H are also conjugate. Now assume that O .G/ D ¹1º; then p 2 0 . Let F1 =R, H1 =R be maximal -subgroups of G=R containing FR=R, HR=R, respectively; then we get F1x D H1 for some x 2 G and F1 =R; H1 =R 2 Hall .G=R/ by induction. Assume that F1 < G. Then, by induction, we obtain F 2 Hall .F1 / as a maximal -subgroup of F1 so F1 D F R. Similarly, H1 D H R. Therefore F x 2 Hall .H1 /. Then H D .F x /y D F xy for some y 2 H1 by induction, and we are done. Now let F1 D G; then G=R is a -group. Suppose that K=R is a minimal normal, say a q-subgroup, of G=R; then q 2 hence q ¤ p.2 0 /. Let Q 2 Sylq .K/; then we have K D Q R. By the Frattini argument, G D NG .Q/K D NG .Q/QR D NG .Q/R hence NG .Q/ 2 Hall .G/ is maximal in G. Assume FR < G; then F 62 Hall .G/. In this case, NG .Q/ \ FR, as a -Hall subgroup of FR (product formula!), is conjugate with F , contrary to the choice of F . Thus FR D HR D G and FG D ¹1º D HG by assumption; then F and H are maximal in G (see the paragraph preceding the statement of the theorem). In this case, by Lemma A.28.5, F and H are conjugate in G. Let us prove, for completeness, by induction on jGj that a group G is solvable if it has a p 0 -Hall subgroup for all p 2 .G/ (Hall–Chunikhin). Let Gp0 2 Hallp0 .G/, where p 0 D .G/ ¹pº, and let q 2 p 0 . If Gq 0 2 Hallq 0 .G/, then we conclude that Gp0 \Gq 0 2 Hallq 0 .Gp0 / by the product formula, so Gp0 is solvable by induction. Let R be a minimal normal, say an r-subgroup, of Gp0 ; then r 2 p 0 . In view of Burnside’s two-prime theorem, we may assume that j.G/j > 2, so there exists s 2 .G/¹p; rº. Let Gs 0 2 Halls 0 .G/; then we have G D Gp0 Gs 0 . By Theorem A.28.1, one may assume 1 This

proof is independent of the Schur–Zassenhaus theorem. See also Theorem A.43.5.

A.28

Nilpotent subgroups

397

that R < Gs 0 ; then we get R .Gs 0 /G . Next, .Gs 0 /G is solvable as a subgroup of Gs 0 . Thus G has a minimal normal, say an r-subgroup, which we denote by R again. In case R 2 Sylr .G/, we conclude that G=R Š Gr 0 is solvable so is G. If R 62 Sylr .G/, then Gr 0 R=R 2 Hallr 0 .G=R/. Let q 2 .G/¹rº; then R < Gq 0 so Gq 0 =R 2 Hallq 0 .G=R/. In this case, by induction, G=R is solvable, and the proof is complete. A subgroup K is said to be a Carter subgroup (= C-subgroup) of G if it is nilpotent and coincides with its normalizer in G. Theorem A.28.6 (Carter [Car]). A solvable group G possesses a C-subgroup and all C-subgroups of G are conjugate. Proof. We use induction on jGj. One may assume that G is not nilpotent. Let R be a minimal normal, say p-subgroup, of G. Existence. By induction, G=R contains a C-subgroup S=R so NG .S/ D S. Let T be a p 0 -Hall subgroup of S (Theorem A.28.4); then TR is normal in S since TR=R is a p 0 -Hall subgroup in the nilpotent subgroup S=R, and S D T P , where P 2 Sylp .S/. One may assume that T > ¹1º (otherwise, we see that S D P is a C-subgroup of G). Set K D NS .T /; then K D T NP .T / is nilpotent and KR D S (Theorem A.28.4 and the Frattini argument). If y 2 NS .K/, then y 2 NS .T / D K since T is characteristic in K, and so NS .K/ D K. If x 2 NG .K/, then x 2 NG .KR/ D NG .S/ D S so x 2 NS .K/ D K, and hence K is a C-subgroup of G. Conjugacy. Let K, L be C-subgroups of G; then KR=R, LR=R are C-subgroups of G=R so, by induction, LR D .KR/x for some x 2 G, and we have K x LR. One may assume that R is not contained in K (otherwise, K x D LR so R < L D NG .L/ since LR is nilpotent, and we obtain K x D L, as desired); then RK is nonnilpotent whence R is not contained in L. If LR < G, then .K x /y D L for y 2 LR by induction, and we are done. Now let G D LR; then G D KR (recall that .KR/x D LR) so G=R is nilpotent, and this is true for each choice of R. Thus R is the unique minimal normal subgroup of G so K and L are maximal in G and KG D LG . Then K and L are conjugate in G (Lemma A.28.5). Theorem A.28.7. Let G D N1 : : : Nk be a nonnilpotent group, where N1 ; : : : ; Nk are proper pairwise permutable and nilpotent subgroups. Then there exists an index i k such that the normal closure NiG < G. Lemma A.28.8 ([Ore]). If G D AB, where A; B < G, then A and B are not conjugate. Proof. Assume that B D Ag for g 2 G. Then g D ab, a 2 A and b 2 B, so we obtain 1 B D Aab D Ab and A D B b D B, G D A, a contradiction. Proof of Theorem A.28.7.2 Assume that G is a minimal counterexample. Then G is nonnilpotent. By [Wiel2] and [Keg1], G is solvable. Let R < G be a minimal normal 2 For

k D 2, Theorem A.28.7 was proved by Janko and Kegel [Keg1], independently.

398

Groups of prime power order

subgroup, say a p-subgroup; then we have G=R D .N1 R=R/ : : : .Nk R=R/, where all Ni R=R are nilpotent. If G=R is not nilpotent, we get .Ni R/G < G for some i by induction so NiG < G. Now let G=R be nilpotent. Then, by hypothesis, we get Ni R D G for all i so Ni is maximal in G and .Ni /G D ¹1º for all i. Therefore N1 N2 D G. By Lemma A.28.5, N1 and N2 are conjugate in G, contrary to Lemma A.28.8. Exercise 1. If K; L are distinct Carter subgroups of a solvable group G, then we have KL ¤ LK. Solution. Write M D hK; Li. Then K; L, being Carter subgroups of M , are conjugate in M (Theorem 28.6) so they are not permutable by Theorem 28.8. Exercise 2. (a) Let K; L < G, where G is an arbitrary group. If K and L are conjugate in hK; Li, then KL ¤ LK. (This is restatement of Lemma 28.8.) (b) Suppose that K H < G, where K is a Carter subgroup of a solvable group G and x 2 G H . Is it true that HH x ¤ H x H ? Exercise 3. Let G be solvable and H < G be such that all subgroups of G containing H coincide with their normalizers and x 2 G H . Is it true that HH x ¤ H x H ? Supplement 1 to Lemma A.28.5. If for arbitrary maximal subgroups F and H of a group G with equal cores we have .jG W F j/ D .jG W H j/, then G is solvable. Proof. Assume that G is a minimal counterexample. As the hypothesis is inherited by epimorphic images, G has only one minimal normal subgroup R, and G=R is solvable so R is nonsolvable (the equality R D G is possible). Let p 2 .R/, P 2 Sylp .R/ and NG .P / F < G, where F is maximal in G. By Frattini’s lemma, G D RF so jG W F j D jR W .F \ R/j and p 62 .jG W F j/. Let q 2 .jR W .F \ Rj/; then q ¤ p. Take Q 2 Sylq .R/ and let NG .Q/ H < G, where H is maximal in G. Then we get FR D G D HR so FG D ¹1º D HG . However, q 2 .jG W F j/ and q 62 .jG W H j/, a contradiction. A group G is said to be p-solvable, if every its composition factor is either a p- or p 0 -number. Supplement 2 to Lemma A.28.5. Suppose that G is a p-solvable group with minimal normal p-subgroup R. Assume that G possesses a maximal subgroup H that satisﬁes HG D ¹1º. Then all maximal subgroups of G with core ¹1º are conjugate. Proof. By hypothesis, we have j.G/j > 1. Let F be another maximal subgroup of G with FG D ¹1º; then G D FR D HR and F \ R D ¹1º D H \ R. If R 2 Sylp .G/, we are done (Schur–Zassenhaus). Now let p 2 .G=R/; then G=R is not simple. Let K=R be a minimal normal subgroup of G=R. Then, as in the proof of Lemma A.28.5, j.K/j > 1 so, taking into account that HG D ¹1º and G=R is p-solvable, we conclude that K=R is a p 0 -subgroup. In that case, F \K and H \K as p 0 -Hall subgroups

A.28

Nilpotent subgroups

399

of K are conjugate in K so in G (Schur–Zassenhaus) whence F D NG .F \ K/ and H D NG .H \ K/ are also conjugate in G. Supplement 3 to Lemma A.28.5 ([Gas4]). Let M be a minimal normal subgroup of a solvable group G. Suppose that K 1 and K 2 are complements of M in G such that K 1 \ CG .M / D K 2 \ CG .M /. Then K 1 and K 2 are conjugate in G. Proof (compare with [Weh, Theorem 3.7]). From K i M D G we conclude that K i is maximal in G, i D 1; 2. We have K i \ M D ¹1º so that .K i /G CG .M /, i D 1; 2. By the modular law, we get CG .M / D M CK i .M /; then K i \ CG .M / D CK i .M / so CK 1 .M / D CK 2 .M / by hypothesis, and hence CK i .M / D .K i /G , i D 1; 2. Then .K 1 /G D .K 2 /G so K 1 and K 2 are conjugate in G by Lemma A.28.5. 2o Groups with a cyclic Sylow p-subgroup. Here we prove two results on groups with a cyclic Sylow p-subgroup. Theorem A.28.9 (compare with [Wiel3]). Let P 2 Sylp .G/ be cyclic. If H G G and p divides GCD.jH j; jG W H j/, then H is p-nilpotent and G is p-solvable. Remark 2. Suppose that P 2 Sylp .G/ is cyclic. (1) Suppose, in addition, that p j jZ.G/j. Set N D NG .P /. Since N has no minimal nonnilpotent subgroup S satisfying S 0 2 Sylp .S/ (Lemma 10.8), it follows that N is p-nilpotent so P Z.N /. In this case, G is p-nilpotent (Burnside’s normal p-complement theorem). (2) Suppose that G, p, P and H are given as in Theorem A.28.9 and assume that the subgroup P1 D P \ H.2 Sylp .H // is normal in H so in G. Let T 2 Hallp0 .H / (Schur–Zassenhaus); then, by Schur–Zassenhaus and Frattini, H D P1 T;

G D H NG .T / D P1 T NG .T / D P1 NG .T /:

It follows from the last equality, the cyclicity of P and P1 < P that NG .T / D G. Thus T G G so H D P1 T is p-decomposable. (3) If ¹1º < H < G are arbitrary and R ˆ.H / is normal in G, then R ˆ.G/. Indeed, assuming that this is false, we get G D RM for some maximal subgroup M of G. Then, by the modular law, we get H D R.H \ M / so H D H \ M and R < H M , a contradiction. (4) If H G G are arbitrary and G D AH , where A is as small as possible, then we have A \ H ˆ.A/. Indeed, if this is false, then A D B.A \ H /, where B < A is maximal. Then G D AH D B.A \ H /H D BH , contrary to the choice of A. Proof of Theorem A.28.9. By the product formula, we have P1 D P \ H 2 Sylp .H /. Set N D NG .P1 /; then P N . Further, set N1 D NH .P1 / D N \ H . Then, by Remark 2(2) applied to the pair N1 < N , we get N1 D P1 T , where T 2 Hallp0 .N1 /, and so H is p-nilpotent (Burnside’s normal p-complement theorem).3 By Frattini’s 3 Since H \P

ˆ.P /, it follows, by the deep Tate’s theorem [Hup, Satz 4.4.7], that H is p-nilpotent.

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Groups of prime power order

lemma, G D HN so G=H Š N=.N \ H / D N=N1 . Therefore it remains to prove that N=N1 is p-solvable. To this end, we may assume that N D G; then P1 is normal in G. In this case, G=CG .P1 / as a p 0 -subgroup of Aut.P1 / is cyclic of order dividing p 1. By Remark 2(1), we see that CG .P1 / is p-nilpotent, and we conclude that G is p-solvable. Theorem A.28.10 ([Hal3, Theorem 4.61]). If P 2 Sylp .G/ is cyclic of order p m and k m, then ck .G/ 1 .mod p mkC1 /. Proof. Let C D ¹Z1 ; : : : ; Zr º be the set of all subgroups of order p k in G not contained in P . Let P act on the set C via conjugation. Then the P -stabilizer of Zi equals P \ Zi which is of order p k1 at most. Thus r D jC j 0 .mod p m.k1/ /. 3o Subgroup generated by some minimal nonabelian subgroups. Let p 2 .G/ and let Ap .G/ be the set of all minimal nonabelian subgroups A of an arbitrary (ﬁnite) group G such that A0 is a p-subgroup (in this case, A is either a p-group or p-closed and minimal nonnilpotent). Set Y Lp .G/ D hH j H 2 Ap .G/i and L.G/ D Lp .G/: p2.G/

It is known (Exercise 10.6 and Theorem 10.28) that L.G/ D G if G is a nonabelian p-group and G 0 L.G/ for arbitrary G. Theorem A.28.11. Given an arbitrary group G, the quotient group G=Lp .G/ is p-nilpotent. Moreover, if p j jG=Lp .G/j, then the Sylow p-subgroups of G are abelian. Proof. Take P 2 Sylp .G/. One may assume that P is not contained in Lp .G/. Then P is abelian (Theorem 10.28), proving the last assertion. Assume that G=Lp .G/ is not p-nilpotent. T