The Representation Theory of Finite Groups (North-Holland Mathematical Library)

North-Holland Mathematical Library Board of Advisory Editors : Artin, H. Bass, 1. Eells, W. Feit, P. 1. Freyd, F. W. Gehring, H. Halberstam, L. V. Hormander, M. Kac, 1. H. B . Kemperman, H. A. Lauwerier, W. A. 1. Luxemburg, F. P. Peterson, I. M. Singer and A. C. Zaanen

M.

The Representation Theory of Finite Groups Walter FElT

Yale University New Haven, cr 06520 U. S.A.

VOLUME 25

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1982

NORTH-HOLLAND PUBLISHING COMPANY AMSTERDAM· NEW YORK· OXFORD

NORTH-HOLLAND PUBLISHING COMPANY AMSTERDAM· NEW YORK· OXFORD

© NORTH-HOLLAND PUBLISHING COMPANY

-

1982

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the copyright owner.

ISBN: 0 444 86155 6

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To

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Sole distributors for the U.S.A. and Canada:

Elsevier Science Publishing Company, Inc. 52 Vanderbilt Avenue New York , N.Y. 10017

Library of Congress Cataloging in Publication Data Feit,

Walter,

1930-

The representation theory of finite groups. (North-Holland mathematical l ibrary) Bibliography;

p.

Includes index.

1.

Modular representations of groups.

groups.

I.

Title.

QAl71.F36 512'.22 ISBN 0-444-86155-6

PRINTED IN THE NETHERLANDS

2.

80-29622

Finite

SIDNIE

PREFACE

Ordinary characters of finite groups were first defined by Frobenius in 1896. In the next 15 years the theory of characters and complex representa­ tions was developed by Frobenius, Schur and Burnside. During this time L. E. Dickson [1902] , [1907a] , [1907b] considered representations of groups with coefficients in a finite field. He called these modular represen­ tations and showed that if the characteristic p of the coefficient field does not divide the order of the group G then the methods used for complex representations of G can be used without any essential changes . If however p does divide the order of G Dickson showed that the theory is quite different. He proved that the multiplicity of any irreducible representation as a constituent of the regular representation of G is divisible by the order of a Sylow p -group. Apparently no one considered modular representations after Dickson until Speiser [1923] studied the connnection between ordinary and modular representations. He showed that if the characteristic p of the finite field does not divide the order of the group G then the modular representations of G can be derived from the ordinary representations by reduction modulo a prime divisor of p in the ring of integers in a suitable number field. The subject was then dormant until Brauer [ 1 935], at the suggestion of Schur, showed that the number of absolutely irreducible representations of a finite group over a field of characteristic p is equal to the number of p '-classes of G. Shortly thereafter it was realized that before one could get at the deeper properties of modular representations of finite groups, it was necessary to study ring theoretic properties of group algebras. This work was begun by Brauer and Nesbitt [1939a] , [1939b], and independently by Nakayama [1938] . During the next 10 to 15 years the theory of modular representations was developed by Brauer, Osima and others. Amongst

viii

PREFACE

other things Brauer [1941c] gave a complete description of the ordinary and modular characters in a block of defect 1 . It was characteristic of Brauer's work that once h e had established the basic properties of proj ective modules he avoided modules and representa­ tions whenever possible and tried to deal only with characters. However in his work on blocks of defect 1 he found it necessary to use some delicate arguments concerning representations. Using a result of D .G. Higman [1954] as his starting point, Green [1959a], [1962b] introduced a new point of view into the subject which emphasized the study of modules. Thomp­ son [1967b] showed how this point of view could lead to a generalization of Brauer's work which would handle blocks with a cyclic defect group. This was then done in full generality by Dade [1966] . I gave a course on modular representations during the academic year 1968-9. At that time no book on the subject existed and I wrote some lecture notes which constitute roughly the first five chapters of this book. A second set of notes, covering roughly Chapters VI-XI, appeared almost a decade later. The delay was at least partly due to the fact that the material on blocks with a cyclic defect group was in a form that did not lend itself easily to exposition. During the intervening time some simplifications and generalizations have been found and this material is presented in Chapter VII. This book is meant to give a picture of the general theory of modular representations as it exists at present. It does not include material concerning modular representations of specific classes of groups such as symmetric groups or groups of Lie type. The first six chapters should be read more or less in order, though some results are not needed until much later in the book. The last six chapters are essentially independent of each other except that the material in Chapter VIII is based on the results of Chapter VII. Lectures on this material and circulation of these notes have elicited valuable advice from colleagues and students. I wish to express my thanks to M. Benard, E. Cline, E.C. Dade, L. Dornhoff, D. Fendel, R. Gordon, M. Isaacs, W. Knapp, M . Schacher, P. Schmid, L.L. Scott, G. Seligman, R. Steinberg, T. Tamagawa, Y . Tsushima, A. Watanabe and W. Willems for suggesting many improvements and corrections. Above all l owe a great debt to H. Blau, D. Burry, J.H. Lindsey II and D. Passman. Passman read an early version of the notes which appeared in 1 969. Blau and Burry read the first seven chapters of the manuscript and Lindsey read Chapters VII-XI. They have all made innumerable valuable and pertinent comments and their critical scrutiny has brought many errors

PREFACE

ix

and obscurities to light. The impact of their suggestions can be seen throughout the book. Most especially is this the case in Chapter VII which has benefited greatly from many improvements suggested by Blau, Burry and Lindsey, and Sections 6 and 7 of Chapter V which are based to a large extent on Blau's suggestions. Finally it gives me great pleasure to thank Ms. Donna Belli for her unfailing patience while transforming my handwriting into a superbly typed manuscript. A remark about notation. (lV.a.b) for instance denotes the assertion in Chapter IV, Section a, designated as Lemma, Theorem, Corollary or equation a.b. However references to this statement in Chapter IV itself will simply by denoted by (a.b). The symbol D always indicates the fact that a proof is complete. All references in the text are given by author, date and possibly a letter, such as Brauer [ 1 942a], and may be found in the bibliography. W. Feit September, 1980

CONTENTS

Vll

PREFACE

1

CHAPTER I 1. Preliminaries

2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.

14. 15; 16.

17.

1 8. 19.

Module constructions Finiteness conditions Projective and relatively projective modules Complete reducibility The radical Idempotents and blocks Rings of endomorphisms Completeness Local rings Unique decompositions Criteria for lifting idempotents Principal indecomposable modules Duality in algebras Relatively injective modules for algebras Algebras over fields Algebras over complete local domains Extensions of domains Representations and traces

CHAPTER II 1. 2. 3. 4. 5. 6.

4 5 8 15 17 21 23 28 33 36 38 42 46 50 53 63 69 74

79

Group algebras Modules over group algebras Relative traces The representation algebra of R [ G] Algebraic modules Projective resolutions

79 79 87 92 93 95 xi

xii

CONTENTS

CHAPTER III 1. 2. 3. 4. 5. 6. 7. 8. 9.

Basic assumptions and notation F[ GJ modules Group rings over complete local domains Vertices and sources The Green correspondence Defect groups Brauer homomorphisms R [ G x G] modules The Brauer correspondence

CHAPTER IV 1. 2. 3. 4. 5. 6. 7. 8.

9. 10. 11.

Characters Brauer characters Orthogonality relations Characters in blocks Some open problems Higher decomposition numbers Central idempotents and characters Some natural mappings Schur indices over Qp The ring A�(1�[ G]) Self dual modules in characteristic 2

CHAPTER V 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

Some elementary results Inertia groups Blocks and normal subgroups Blocks and quotient groups Properties of the Brauer correspondence Blocks and their germs Isometries 7T-heights Subsections Lower defect groups Groups with a given deficiency class

CHAPTER ·VI 1. 2. 3. 4. 5. 6.

Blocks and extensions of R Radicals and normal subgroups Serial modules and normal subgroups The radical of R [ G] The radical of R [ G] p -radical groups

CONTENTS

97 97 98 104 111 115 123 1 28 131 136

140 140 142 144 149 165 171 178 181 185 1 86 188

192 1 92 195 198 202 206 209 215 226 230 240 245

269

CHAPTER VII 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.

Blocks with a cyclic defect group Statements of results Some preliminary results Proofs of (2. 1)-(2 . 1 0) Proofs of (2. 1 1)-(2. 1 7) in case K K The Brauer tree Proofs of (2. 1 1)-(2 . 1 9) Proofs of (2.20)-(2.25) Some properties of the Brauer tree Some consequences Some examples The indecomposable R[ G] modules in B Schur indices of irreducible characters in 13 The Brauer tree and field extensions Irreducible modules with a cyclic vertex =

Groups with a Sylow group of prime order Tensor products of R[N] modules Groups of type Lz(P) A characterization of some groups Some consequences of (4.1) . Permutations groups of prime degree Characters of degree less than p 1 Proof of (7. 1) Proof of (7.2) Proof of (7.3) Some properties of permutation groups Permutation groups of degree 2p Characters of degree p -

CHAPTER IX 1. 2. 3. 4.

The structure of A (G) A (G) in case a Sp -group of G is cyclic and R is a field Permutation modules Endo-permutation modules for p -groups

248 248 249 253 256 262 265

CHAPTER X 1. 2. 3. 4. 5.

269 273 281 289 290 299 301 302 305 308 317 322 333 336 337

340

CHAPTER VIII 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 1 1. 12. 13.

xiii

Groups with a normal p '-subgroup Brauer characters of p -solvable groups Principal indecomposable characters of p-solvable groups Blocks of p -solvable groups Principal series modules for p -solvable groups

340 34 1 347 351 358 361 365 371 380 381 384 387 391

396 397 402 405 407

411 411 419 421 424 427

xiv

CONTENTS 6. The problems of Chapter IV, section 5 for p -solvable groups 7. Irreducible modules for p -solvable groups 8. Isomorphic blocks

CHAPTER XI 1. An analogue of Jordan's theorem

CHAPTER XII 1. Types of blocks

2. 3. 4. 5. 6.

Some properties of the principal block Involutions and blocks Some computations with columns Groups with an abelian S2-groUP of type (2m, 2m ) Blocks with special defect groups 7. Groups with a quaternion S2-groUP 8. The Z*-theorem

428 433 440

444 444

CHAPTER I

450 450 455 456 459 463 465 467 471

BIBLIOGRAPHY

476

SUBJECT INDEX

500

1. Preliminaries

The purpose of this chapter is to provide the background from ring theory which is needed for the study of representations of finite groups. No attempt will be made to prove the most general results about rings. Most of this material can be found in one or more of the books listed at the end of this section . We assume that the reader is familiar with basic properties of rings and modules. In this section we wilJ introduce some terminology and conven­ tions and state without proof some of the basic results that are needed . Throughout these notes the term ring will mean ring with a unity element. All modules are assumed to be unital. In other words if V is an A module for a ring A then v 1 = v for all v E V. Let A be a ring. An element e in A is an idempotent if e Z = e I O. Two idempotents e I, e 2 are orthogonal if e I ez = e 2e 1 = O . An idempotent e in A is primitive if it is not the sum of two orthogonal idempotents in A. An idempotent e in A is centrally primitive if e is in the center of A and e is not the sum of two orthogonal idempotents which are in the center of A. An A module will always be a right A module . It will sometimes be necessary to refer to left A modules or two sided A modules. In these latter cases the appropriate adjective will always be present. Results will generally be stated in terms of modules. It is evident that there always exist analogous results for left modules. Let V be an A module. A subset {Vi } of V generates V if V LViA. V is finitely generated if a finite subset generates V. {Vi } is a basis of V if .

=

2

CHAPTER I

[I

(i) {Vi } generates V. (ii) If LV i ai = 0 for ai E A then ai 0 for all i. If V has a basis then V is a free module. The term A -basis and A -free will be used in case it is not clear from context which ring is involved. The regular A module AA is defined to be the additive group of A made into an A module by multiplication on the right. The left regular A module AA is defined similarly. Right ideals or left ideals of A will frequently be identified with submodules of AA or left submodules of AA. We state without proof the following elementary but important results. =

LEMMA

1.1. An A module is free if and only if it is a direct sum of

submodules, each of which is isomorphic to AA. LEMMA

1.2. A finitely generated free A module has a finite basis.

LEMMA 1.3. An A module V is the homomorphic image of a free A module. If V is finitely generated it is the homomorphic image of a finitely generated

free A module.

At times it is convenient to use the terminology and notation of exact sequences. In this notation Lemma 1.3 would for instance be stated as follows.

1]

PRELIMINARIES

3

If (0) � V1 � V2 � V are A modules then V2/ VI is a constituent of V. If V2/ VI is irreducible it is an irreducible constituent of V. A normal series of V is an ordered finite set of submodules (0) = Va � . . . � Vn = V. The factors of the normal series (0) = Va � . . . � Vn = V are the modules �+I/�. The normal series (0) Wo � . . . � Wm = V is a refinement of the normal series 0 = Va � . . . � Vn = V if there exists a set of indices 0 � j 1 < . . . < jn � m such that Vi = lVJi. The refinement is proper if m > n. A normal series without repetition is a normal series in which (0) is not a factor. A composition series is a normal series without repetition such that every proper refinement is a normal series with repetition. Two normal series of V are equivalent if there is a one to one correspondence between the factors of each such that corresponding factors are isomorphic. The following result is evident. =

LEMMA

1.5. A normal series is a composition series if and only if each of its

factors is irreducible.

The next two results are the basic facts concerning normal series. They are stated here without proof.

LEMMA 1 .4. If V is an A module then there exists an exact sequence

THEOREM

with U a free A module. If V is finitely generated then there exists such an exact sequence with U finitely generated.

THEOREM

If V = VI E9 V2 for submodules VI, V2 then VI or V2 is a component of V. If U is isomorphic to a component of V we will write U I V. V is indecomposable if (0) and V are the only components of V. V is deco'fnposable if it is not indecomposable. Tfmis a nonnegative integer then m V denotes the direct sum of m modules each of which is isomorphic to V. Thus (1.1) and (1.2) assert that a finitely generated A module is free if and only if it is isomorphic to mAA for some nonnegative integer m. Let V -I- (0). V is irreducible if (0) and V are the only submodules of V. V is reducible if it is not irreducible. Note that (0) is neither reducible nor i'rre, (ii). This is a direct consequence of (7. 1). (ii) => (i). Let e be a primitive idempotent. Thus AA = (1 - e )A EB eA. If eA = VI EB V2 with VI, V2 nonzero then by (7. 1) e is not primitive contrary to assumption. An analogous argument shows that (ii) is equivalent to (iii). 0

+ ... +

LEMMA 7.3. Let A = II EB . . . EB In where each � is a nonzero ideal of A. en with ei E L. Then {ed is a set of pairwise orthogonal Let 1 = e l central idempotents and Ii = eiA is a ring with unity element ei for each i.

CHAPTER I

22

8]

[7

Conversely if {ei } is a set of pairwise orthogonal central idempotents then (Lei )A = EB eiA where eiA = Aei is an ideal of A.

RINGS OF ENDOMORPHISMS

23

LEMMA 7.7. Assume that A satisfies A.C.C. Then A = EB�=1 Aei where

{ei} is the set of centrally primitive idempotents in A. For each i, Aei is not the direct sum of two nonzero ideals of A. If A = EB�= Ai where each Ai is a nonzero ideal which is not the direct sum of two nonzero ideals (of A or Ai ) then m = n and A i = Aei after a suitable rearrangement. ]

PROOF. If a E A then

a = 1 · a = a . 1 = e]a + . . . + en a = ae 1 + . . . + aen · Hence eia = aei and eiej = Oijei for all i, j. Since eiA C Ii and A = EB eiA it follows that eiA = Ii and eia = a = aei for ail a E 1;. Conversely by (7. 1) (Lei )A = EB eiA and eiA = Aei since each ei is central. 0

PROOF. This follows directly from (7.3), (7.4) and (7.6).

0

COROLLARY 7.4. Let e be a central idempotent in A. Then e is centrally primitive if and only if eA is not the direct sum of two nonzero ideals.

Let e be a centrally primitive idempotent in A. The block B = B (e) corresponding to e is the category of all finitely generated A modules V with Ve = V. If Ve = V then V is said to belong to the block B = B (e ). We will simply write V E B.

PROOF. Immediate by (7.3).

THEOREM 7.8. Assume that A satisfies A.C.C. Let V be a finitely generated

0

indecomposable nonzero A module. There exists a unique block B with V E B. If V E B then every submodule and homomorphic image of V is in B. Furthermore V E B = B (e) if and only if ve = v for all v E V.

LEMMA 7.5. If A satisfies A.C.C. then 1 is a sum of pairwise orthogonal

primitive idempotents.

PROOF. Let {ei } be the set of centrally primitive idempotents in A. By (7.6) 1 = Lei . Thus if V is any A module then V = EB Vei . Hence if V is indecomposable and V I- (0) then V = Vei for a unique value of i. If W is a submodule of V then clearly Wei = W, ( V/ W)ei = V/ W and Wej = (0) = (V/W)ej for i l- j. Clearly ve = v for all v E V = Ve. 0

PROOF. By (3.8) AA is a direct sum of a finite number of indecomposable . modules. Thus (7. 1) and (7.2) imply the result. 0 LEMMA 7.6. Assume that A satisfies A.C.C. Then (i) A contains only finitely many central idempotents. (ii) Two centrally primitive idempotents are either equal or orthogonal. (iii) l = L�=)ei where {e), . . . , en } is the set of all centrally primitive

idempotents in A.

PROOF. (i) By (7.5) 1 = L fi where {fi } is a set of pairwise orthogonal primitive idempotents. Let e be a central idempotent. Then e = Lefi and fi = efi + (1 - e) f;. Thus efi = 0 or efi = fi since fi is primitive. Hence e = L Ii where j ranges over those summands with eli = Ii· Thus e is the sum of elements in a subset of {fi}. This proves (i). (ii) If e), e2 are centrally primitive idempotents then e) = e1e2 + e)(l - e2)' Thus e) = e]e2 or e 1 e 2 = O. Similarly e 2 = e)e2 or e1e 2 = O. Thus either e1e 2 = 0 or e) = e]e 2 = e 2· (iii) Let {e), . . . , en } be the set of all centrally primitive idempotents in A. Then e = L�=1 ei is a central idempotent by (ii). If e I- 1 then 1- e is a central idempotent and so by (i) is a sum of centrally primitive idempo­ tents. Since ei (1- e ) = 0 for i = 1, . . . , n, this is impossible. 0

.

8. Rings of endomorphisms

Let A be a ring. LEMMA 8. 1 (Schur). Let V, W be irreducible A modules. HomA ( V, W) = (0) if V;I< W and EA ( V) is a division ring.

Then

, PROOF. Let f E Hom A ( V, W). Then either f is an isomorphism or fe y ) = (0) since V and W are irreducible. Thus HornA ( V, W) = (0) if V;I< W, and every nonzero element in EA ( V) is an automorphism of V and so has an inverse. Hence EA ( V) is a division ring. 0 LEMMA 8.2. Let e be an idempotent in A and let V be an A module. Define f: Ve � HomA (eA, V) by f( v )ea = vea. Then f is a group isomorphism. If

24

[8

CHAPTER I

8]

RINGS OF ENDOMORPHISMS

25

A is an R algebra then Ve and HomA (eA, V) are R modules and f is an R -isomorphism. If V = eA then f : eAe � EA (eA ) is a ring isomorphism. Thus in particular EA (AA ) = A = EA (AA ).

coefficients in A. In other words A is an A -free A module with basis n {eii. / i, j = 1 , . . . , n } where

PROOF. By (2 .4) and (2.5) Ve and Hom (eA, V) are R modules if A is an R -algebra. In this case f is clearly an R -homomorphism. In any case f is a group homomorphism and if V = eA, f is a ring homomorphism. If f(v ) = 0 then v = ve = 0 and f is a monomorphism. If h E HomA (eA, V) then f(h (e )) = h. Thus f is an epimorphism. The last statement follows by setting e = 1 . D

This definition is equiva lent to the assertio n that A = A Q9z Z where Z n n denotes the ring of rational integers. It follows easily that for i = 1 , . . .,n e � = eii and eiiAneii = A. Furthermore A n is also a left A module with aeij = eija for a E A. Thus A n is a two-sid ed A module .

LEMMA 8.3. Let e be an idempotent in A. If N is a right ideal in eAe then

N = NA n eAe. The map sending N to NA sets up a one to one correspon ­ dence between the right ideals of eAe and a set of right ideals of A . Thus if A satisfies either A. C. C. or D . C. C. so does eAe.

PROOF. Let N be a right ideal of eAe. Clearly NA n eAe is a right ideal of eAe and N � NA n eAe. Since N = Ne and NA n eAe � eAe it follows that

NA n eAe = NAe n eAe

=

NeAe

n

eAe � N.

The remaining statements follow. D LEMMA 8.4. Let e be an idempotent in A. Then J (eAe ) = eJ (A )e. PROOF. Suppose that a E J(A ). Then eae E J(A ) and (6.4) (i) and (6. 5) imply that eae + b = eaeb for some b E A. Multiply by e on both sides to get that eae + ebe = eae ebe and so by (6.4) (i) eae is right quasi-regular in eAe. Since eJ(A )e is an ideal of eAe (6.4) (ii) and (6.5) impiy that

eJ(A )e � J(eAe ). Let I be a primitive ideal in A and let V be a faithful irreducible A il module. Then Ve is an eAe module. If Ve = (0) then J(eAe) �;;;>e/1e k I. Suppose that Ve "l (0). Let (0) C W = We � Ve where W is 'eA� module. Thus V = WA since V is irreducible. Hence Ve = WeAe � W and so Ve = W. Thus Ve is an irreducible eAe module and so VJ(eAe ) = VeJ(eAe ) = (0). Hence J(eAe ) � 1. Since I was an arbitrary primitive ideal in A this yields that J (eAe ) � J (A ) and so J (eAe ) = eJ (eAe )e � eJ(A )e. D an

For any integer n > 0 let An be the ring of all n by

n

matrices with

LEMMA 8.5. If A satisfies A .C.C. or D .C.C. so does An . PROOF. Since A n is a finitely generated A module the result follows from (3.5). D LEMMA 8.6. Let V be an A module. Assume that V = VI EB · · · EB Vn with = Vi for i, j = 1, . . . , n. Then EA ( V) = EA ( VI) n . V; PROOF. For i 1 , . . , n let ei l be an isomor phism from VI to V; where e l l is the identit y map. Let e l i be the inverse map sendin g V; to VI . Let eij = ei le lj. For i, j = 1 , . . . , n define eij E EA ( V) by eijVs = eij (Oj vs ) for =

S Vs E Vs, s = 1 , . . . , n. If x E EA ( VI) define x E EA ( V) by xv" = es l xe lsvs for s = 1, . . . , n. (8.7) The map sendin g x to x is clearly a monom orphis m of EA ( VI) into EA ( V). Let E denote the image of EA ( VI ) under this map. Let F be the ring generated by E and all eij, i, j = 1 , . . . , n. It follows from (8.7) that for all i, j and all x E E

Hence if x "l 0

Xeij = eijX = 0 for all i, j. The .ring F is an E modul e genera ted by e ·· i ,. = 1 , then (8.8) yields that for s, t = 1 , . . . Xij E E. , n,

o = ess 2: i,j eijXij el/

=

estXst .

'J"

(8.8) •

•

•

, n.

If � "e··IJxIJ·· = 0

CHAPTER I

26

[8

Thus by (8 . 8) Xst = 0 for all s, t. Hence F is an E -free E module. This implies that F = En . It remains to show that EA ( V) � F. Let y E EA ( V). Since 2:eii = 1 it follows that y = 2: i.j eiiyejj . Thus it suffices to show that eiiyejj E F for all i, j. There exists z E EA ( Vi ) such that eiiyejj = eijZ. However z = ej lxe 'j for some x E EA ( VI). Thus by (8.7)

eiiyejj

=

eijXejj E F.

0

THEOREM 8.9. Assume that V is a completely reducible nonzero A module with a composition series. Let V = EB;n= 1 EB;'� I V;j where each Vij is ir­ reducible and Vij = V,t if and only if i = s. Let Di = EA ( Vi I ) and let U; = EB;� 1 Vij for all i. Then the following hold. (i) Di is a division ring for i = 1, . . . , m. (ii) EA ( Vi ) = (Di ) ni is a simple ring for i = 1 , . . . , m. (iii) EA ( V) = EB;: 1 EA ( U; ). Every ideal in EA ( V) is of the form

EB EA ( � ) where j ranges over a subset of {I, . . . , m }. (iv) EA ( V) is a semi-simple ring which satisfies D.C.C. and left D .C.C.

PROOF . (i) This follows from (8. 1). (ii) EA ( Vi ) = ( Di ) ni by (8.6). It is easily verified that' (Di ) ni is a simple ring. (iii) Let ei be the proj ection of V onto Vi for i = 1 , . . . , m. Thus ei I:- 0 and eiej = Oijei for all i, j. If x E EA ( V) then by (5. 1 1) xei = eiXej for all i. Since 2:ej = 1 it follows that x = 2:j xej . Hence

ejx

=

L j ejxej = ejxej = xej .

Thus ej is in the center of EA ( V) for all i. Therefore by (7.3 ) EA ( V) = EB�= l eiEA ( V). Clearly ejEA ( V) = EA ( U; ). If I is an ideal of EA ( V) then I = EB;'� l eJ and eJ is an ideal of ejEA ( V). Since ejEA ( V) is simple eJ = (0 ) or ejEA ( V). Hence I = EB ejEA ( V) where j ranges over all i with eJ I:- (0) . (iv) By (iii) every ideal in EA ( V) contains an idempotent. Thus J(EA ( V» = (0 ) and EA ( V) is semi-simple. Since EB�= l Di contains only finitely many right or left ideals it satisfies D .C.C. and left D .C.C. By (ii) and (iii) EA ( V) is finitely generated as a module or left module over EB;: I Dj • Thu's EA ( V) satisfies D .C.C. and left D .C.C. 0 THEOREM 8.10 (Artin-Wedderburn). Let A be a semi-simple ring with D .C.C. Then A = EB�= I Aj where each Ai is an ideal of A which is a simple ring with D .C.C. Furthermore the Ai are uniquely determined.

RINGS OF ENDOMORPHISMS

27

PROOF. By (6. 10) AA is completely reducible. By (8.2) A = EA (AA ). The result follows from ( 8.9) . 0

THEOREM 8. 1 1 (Artin-Wedde rburn). Let A be a simple ring with D.C.C.

Then A = Dn for some division ring D and some integer n > O. Furthermore n is unique, D is uniquely determined up to isomorphism and any two irreducible A modules are isomorphic. If V is an irreducible A module then

EA ( V) = D.

PROOF. By (8.2 ) A = EA (AA ), and AA is completely reducible by ( 6. 10) . Since A is simple ( 8.9) implies that A = Dn for some division ring D and some integer n. Furthermore any two irreducible submodules of AA are isomorphic. Thus by (6. 1 ) any two irreducible A modules are isomorphic. Hence if e is a primitive idempotent in A (8.2) yields that D = eAe and so D is uniqu ely determined up to isomorphism. By (8.9) AA is a sum of n irreducible submodules. Thus by (1 . 7) Dm = Dn if and only if m = n. 0 LEMMA 8. 12. Suppose that A is semi-simple and satisfies D .C.C. Then

every nonzero ideal contains a central idempotent and every nonzero right ideal contains an idempotent.

+

PROOF. The first statement follows from (8. 10). Let V be a right ideal of A. By (6. 10 ) AA = V EB V' for some right ideal V' of A. Let 1 = e e' with e E V, e ' E V'. Then by (7. 1 ) e is an idempotent and V = eA. 0 LEMMA 8. 13. Let V be a finitely generated free A module. Then EA ( V) is a finitely generated free A module. PROOF. V = mAA for some integer m. Hence by (8.2) and ( 8.6) EA ( V) = Am•

0

8. 14. Let R be a commutative ring which satisfies A.C.C. Let V be a finitely generated R module. Then ER ( V) is a finitely generated R -algebra and satisfies A.C.C. and left A.C.C.

LEMMA

PROOF. By ( 1 .3 ) there exists a finitely generated free R module F with F/ W = V for some submodule W. Let E I be the subring of ER (F) consisting of all endomorph isms of F which send W into itself. Since R is commutative V, F, W are two sided R modules. Hence by (2.4) and (2.5)

28

CHAPTER I

[9

ER (F) and E I are R modules. By (8. 13) ER (F) satisfies A.C.C. and so E I is a finitely generated R module and satisfies A.C.C. If f E E I then f induces an endomorphism ! of F / W = V. The map sending f to ! is a homomorphism of E I into ER ( V). By (4. 1) F is projective since it is free. Thus for any g E ER ( V) there exists a commuta­ tive diagram

F

Y 181 � F V�O where t is the natural proj ection of F onto V. If w E W then tfw = gtw = 0 and so fw E W. Hence f E El and ! = g. Thus the map sending f to ! is an epimorphism of E I onto ER ( V). Thus ER ( V) is a finitely generated R -algebra and satisfies A.C.C. Since R is commutative ER ( V) also satisfies left A.C.C. 0 LEMMA 8.15. Let R be a commutative ring which satisfies A.C.C. and let A

be a finitely generated R - algebra. Then J (R )A � J (A ). If V is a finitely generated A module then EA ( V) is a finitely generated R -algebra.

PROOF. If V is a finitely generated A module then VR is a finitely generated R module. Hence by (8. 14) ER ( V) is a finitely generated R -algebra and satisfies A.C.C. Since EA ( V) is a submodule of the R module ER ( V) it follows that EA ( V) is a finitely generated R -algebra. Let V be an irreducible A module. Thus EA ( V) is a finitely generated R -algebra. By (8. 1) EA ( V) is a division ring. Hence the center of EA ( V) is a finitely generated R -algebra which is a field and so the image of R in EA ( V) is a field. Thus J(R )EA ( V) = EA ( V)J(R ) = (0). Hence J(R )A annihilates every irreducible A module and so J(R )A � J(A ). 0 9. Completeness

Let A be a ring and let V be an A module. Suppose that I is an ideal in A such that n �=o VI i = (0) where 1 0 = A. Then for any real number c with o < c < 1 define the norm II II� on V as follows II O II� = 0, II v ll� = C

i

if v E VI', v E VI i + 1 •

9]

29

COMPLETENESS

Define d � by

d Hv, w ) = Il v - w II� · It is easily verified that for u, v, w E V dHu, v ) � max {dHu, w ), d Hv, w )} and . that d � is a metric on V. Two such metrics dL dr are equivalent, written as d � d � : if a sequence in V is a Cauchy sequence with respect to d � if and only if it is a Cauchy sequence with respect to d � : . It is easily seen that d � d f for 0 < c, c ' < 1 . From now o n a fixed c with 0 < c < 1 i s chosen and we write dl d � . If I is an ideal of A and n 7= 0 VI' (0) we will say that dr is defined on V. V is complete with respect to d, if dr is defined on V and every Cauchy sequence in V converges with respect to the metric d, . In the special case V AA these definitions will be applied to the ring A. Analogous definitions can of course be made for left A modules. �

�

=

=

=

LEMMA 9. 1 . Let V be an A module and suppose that dl is defined on V. Let

{vd be a sequence of elements in V. Then (i) {Vi } is a Cauchy sequence if and only if for every integer m > 0 there exists an integer n > 0 such that Vi - Vj E VI m for i, j > n. (ii) limi---->ex> Vi v if and only if for every integer m > 0 there exists an integer n > .0 such that Vi - v E VI m for i > n. =

PROOF. Immediate from the definition. 0 LEMMA 9.2. Let V be an A module. Suppose that dr is defined on V and on

A. Then (i) Addition is continuous from V X V to V. (ii) Multiplication is continuous from V x A to V. (iii) Addition and multiplication are continuous on A. (iv) If dl is defined on the A module W and f E HOIDA ( V, W) then f is continuous.

PROOF. (i) It follows directly from (9. 1 ) (ii) that lim( Vi ± Wi ) = lim Vi ± lim Wi · (ii) Let lim Vi = v, lim a i = a where Vi E V, ai E A for all i. By (9. 1) (ii) for every integer m > 0 there exists an integer n > 0 such that Vi - V E vrn and ai - a E rn for i > n. Thus

30

CHAPTER I

31

COMPLETENESS

Viai - va = Viai - Via + via - va

(6.7) I(A /I) = I/l. Thus by (6. 9) I n c;;;, I for some integer n > O . i n c;;;, 1 m c;;;, . Hence n�=o VI i c;;;, n�=o VI (0) and dJ is defined on V. Since I m dJ • 0 J m for every integer m it follows from (9. 1) that d/ PROOF. By

=

Hence lim Vi ai = va. (iii) Immediate by (i) and (ii). (iv) Suppose that {v; } is a sequence in V with lim Vi = O. Then (9. 1) (ii) implies that for every integer m > 0 there exists an integer n > 0 such that Vi = Uiai for some Ui E V, ai E 1 m and all i > n. Hence fVi (fUi )ai E WI m for i > n. Thus lim fVi = O. If lim Wi = W this implies that =

(lim fWi ) - fw = lim(fwi - fw ) Hence f is continuous.

=

lim f (Wi - W ) = O.

0

THEOREM 9.3. Let V be a finitely generated A module. Assume that A is

complete with respect to d[ and dl is defined on V. Then V is complete with respect to d[.

PROOF. Let {V I , . . . , v. } be a set of generators of V. Let {wJ be a Cauchy sequence in V. There exists a sequence {m (n )} of nonnegative integers such that Wi - Wj E VI m ( n ) for all i, j 3 n, m (n ) � m (n + 1) and limn�oo m (n ) = 00. Let W I = 2:;= 1 vtb l t with bi t E A and let Wn + 1 - Wn = 2::= 1 Vt ant with ant E I m ( n ) . Define n-I bnt = bit + j=L1 ajt for all n, t. Then Wn = 2:; = 1 vtbnt and n +k - I b" +k,, - bnt jL aj, E Im ( n ). =n =

Thus for each t { bn, } is a Cauchy sequence in A. Let bt = limn->oo bn, and let W = 2:;= 1 vtbt . Then lim(w - wn ) =

� vt {lim( bt - bnt )} = O.

,=1

Thus W = lim Wn and so V i s complete. 0 LEMMA 9.4. Let V be an A module and suppose that d[ is defined on V. Let 1 = I(A ). If I c;;;, I and A /I satisfies D. C.C. then dJ is defined on V and

dl :-- dJ•

�

LEMMA 9.5. Suppose that A is complete with respect to d/ . Then I c;;;, I(A ). PROOF. Let a E l. Let bi = l + a + · · · + a i for i = O , l, . . . . Then bi - bj E 1 m for i, j � m. Thus {b; } is a Cauchy sequence in A. Let b = lim bi . Th�n (1 - a )b = lim( l - a )bi = lim(l - a i + l ) = 1 . Hence a i s right quasi-regular. Since a was arbitrary in I (6.4) and (6.5) imply that I c;;;, 1. 0 Let I = I (A ). A is complete if dJ is defined on A and A is complete with respect to dJ• The A module V is complete if dJ is defined on V and V is complete with respect to dJ• A is complete on modules if every finitely generated A module is complete. LEMMA 9.6. Suppose that dJ is defined on every finitely generated A module. If V is a finitely generated A module then every submodule of V is closed. If

furthermore A is complete then A is complete on modules.

PROOF. Let W be a submodule of V. Since dJ is defined on V / w, (0) is closed in V/ W. Let f be the natural projection of V onto V / W. Thus 1 W = f- (0) and so W is closed since f is continuous by (9.2) (iv). If A is complete then by (9.3) A is complete on modules. 0 In the rest of this section we will be concerned with finding criteria which ensure that a ring A is complete on modules. LEMMA 9.7. If I(A ) is nilpotent then A is complete on modules. In particular if A satisfies D .C.C. then A is complete on modules. PROOF. If I(A ) is nilpotent and V is an A module then any Cauchy sequence on V is ultimately constant and thus converges. 0 LEMMA 9.8 (Nakayama). Let W be a finitely generated A module. Assume

that WI(A ) = W. Then W = (0).

CHAPTER I

32

[9

PROOF. Suppose that WI (0). Choose a set of generators W I , . . • , Wn with n minimal. Since WJ(A ) = W, Wn = L7� 1 Wi ai with a i E J(A ). Thus Wn (1 - an ) = L�:i Wi ai . Since an E J (A ), 1 - an has an inverse in A. Hence Wn is in the module generated by W I , . . . , Wn - l and so W is generated by W I , . . . , Wn -l contrary to the fact that n was minimal. Thus W = (0). D The proof of the next result is due to 1. N. Herstein. LEMMA 9.9. Suppose that R is commutative and satisfies A.C.C. Let V be a finitely generated R module and let I be an ideal of R. If W = n��o Vr then WI = W. PROOF. Clearly WI � W. By A.C.C. choose U maximal among all sub­ modules of V such that U n W = WI. Let a E I. We will first show that Va m � U for some integer m. For each s let Vs = {v I va s E U}. Since R is commutative Vs is a submodule of V. Clearly Vs � VS + l for each s. Hence by A.C.C. there exists an m with Vm = U;� o VS . Clearly WI � ( Va m + U) n W. Suppose that W E ( Va m + U) n W. Thus W = va m + u for some v E V, u E U. Since wa E WI � U and ua E U l this implies that va m + E U and v E Vm + l = Vm • Hence va m E U and so W = va m + u E U. Thus W E U n W = WI. Therefore ( Va m + U) n w = WI. The maximality of U now implies that Ya m � U as required. Let a 1, , an be a set of generators for the ideal I. Choose m such that Va 7' � U for i = 1, . . . , n. Since R is commutative I m n � 10 where 10 is the ideal of R generated by a 'In , . . . , a �'. Thus VIm n � U and so W � VIn", � U. Hence W = U n W = WI. D • • •

THEOREM 9. 10. Suppose that R is commutative and satisfies A.C.C. Let J = J(R ). Then dJ is defined on every finitely generated R module. In particular n ��o r = (0). PROOF. Let V be a finitely generated R module. Let W = n �� o VJ i . By (9.9) WJ = W. Thus W = (0) by (9.8). D THEOREM 9 . 1 1 . Let A be a finitely generated algebra over the commutative ring R. Assume that R satisfies A.C.C., R /J(R ) satisfies D .C.C. and R is

complete. Then (i) A satisfies A.C.C., A /J(A ) satisfies D .C.C. and A is complete on modules.

LOCAL RINGS

33

(ii) If V is a finitely generated A module then EA ( V) satisfies A.C.C., EA (V)/J(EA ( V» satisfies D .C.C. and EA ( V) is complete on modules. PROOF. By (8.15), (ii) follows from (i). Clearly A satisfies A.C.C. Since A /J(R )A is a finitely generated R /J(R ) module it satisfies D .C.C. Thus by (8. 15) A /J(A ) satisfies D . C.C. It remains to show that .A is complete on modules. AJ(R ) = J(R )A is an ideal of A and (AJ(R )Y = AJ(R Y for all i. Thus dAJ ( R ) is equal to the metric dJ ( R ) defined on AR as an R module. Hence by (9.6) and (9. 1 0) A is complete with respect to dAJ ( R ) . If V is a finitely generated A module then

n V(AJ(R )Y = n VAJ(R )i = n VJ(R Y = (0) i �O i �O i �O by (9. 10). Thus by (9.3) V is complete with respect to dAJ ( R ) . Since A /AJ(R ) is a finitely generated R /J(R ) module it satisfies D .C.C. Thus by (9.4) V is complete with respect to dJ ( A ) . Hence A is complete on modules. D LEMMA 9. 12. Let R be commutative. Assume that R satisfies A.C.C., R /J ( R ) satisfies D .C.C. and R is complete. Let B � A be finitely generated R -algebras with 1 E B. Then B n J(A ) � J(B ). PROOF. Let x E B n J(A ). By (9. 11) A is complete. Therefore L� X i converges in A. Since B is an R submodule of the R module A it is closed i by (9.6). As 1 E B this implies that L� X i E B. Since (1 - X ) L� X = 1 it follows that X is right quasi-regular in B. As B n J(A ) is an ideal of B, (6.5) implies that B n J(A ) � J(B ). 0 10. Local rings

ring A is a local ring if the set of all nonunits in A form an ideal. A local ring is also said to be completely primary. A

LEMMA 10.1 . Let A be a ring. The following are equivalent. (i) A is a local ring. (ii) J(A ) is the unique maximal ideal of A and contains all the nonunits

in A.

(iii) J (A ) contains all the nonunits in A. (iv) A /J (A ) is a division ring.

34

[10

CHAPTER I

PROOF. (i) :? (ii). Let I be the ideal in A consisting of all nonunits in A. Since every right ideal of A consists of nonunits it follows that I contains every right ideal of A and so I is the unique maximal right ideal of A. Thus J (A ) = I since J (A ) is the intersection of all the maximal right ideals of A. (ii) :? (iii). Clear. (iii) :? (iv). Let ii denote the image of a in A = A /J(A ). If ii � 0 then a is a unit in A and so ab = ba = 1 for some b E A. Thus iib = bii = 1 and A /J (A ) is a division ring. (iv) :? (i). Suppose that a � J (A ) then there exists b E A such that ab = 1 - c for some c E J(A ). Thus ab is a unit and so a has a right inverse in A. Similarly a has a left inverse in A. Thus a is a unit in A. 0 LEMMA 10.2. Let A be a commutative ring. Then A is a local ring if and

only if A has a unique maximal ideal.

PROOF. If A is local then by (10.1) (ii) A has a unique maximal ideal. If A has a unique maximal ideal I then A /I is a commutative ring with no nontrivial ideals. Hence A /I is a field and so A is local by ( 10.1) (iv). 0 LEMMA 10.3 (Fitting). Let A be a ring and let V be an A module. Assume that V satisfies D .C.C. and A.C.C. Let f E EA ( V). Then there exists an

integer n and subm.odules U, W such that V = U EB W where U is the kernel of r +i for all j = 0, 1, . . . and W = r +i V for all j = 0, 1, . . . .

PROOF. Let � be the kernel of Ji and let "'1 = ji V. Then � � Ui+ 1 and "'1 + 1 � "'1 for all j. Let U = U:=o Un, W = n:=o Wn• By A.C.C. and D . C.C. there exists an integer n with Un = U and Wn = W. It remains to show that V = U EB W. Suppose that u E U n W. Then r u = 0 and u = rv for some v E V. Thus F n v = 0 and so v E U. Thus u = r v = O. Hence U n W = (0) and U + W = U EB W. If v E V then rv E W = W2n and so rv r (v - rw) = 0 and v - rw E U. Then Thus V = U EB W. 0

=

F"w for some W E V. Hence v = (v - rw ) + rw E U t W.

THEOREM 10.4. Suppose that A is a ring which satisfies A.C.C. Assume that A is complete on modules and A /J(A ) satisfies D .C.C. Let V be a finitely

generated indecomposable A module. Then EA ( V) is a local ring.

PROOF. Let J = J(A ). Since v/ vpn is a finitely generated A /J m module

10]

LOCAL RINGS

35

for every integer m > 0, (6. 1 1) and (6. 13) imply that V / VJ m satisfies A.C.C. and D .C.C. Let f E EA ( V). For every integer m ? 0 fVJ m � VJ m . Thus f induces an endomorphism on V/ VI m . (10.3) applied to V/ VI m asserts the existence of an integer n (m ) and submodules Um , Wm of V such that Urn =

and

{u I f n( m)+ u E i

VI m }

Wm = f n (m )+i V + VI m

for all j ? 0,

for all j ? 0

(10.5) (10.6) (10.7)

It

may be assumed that n (m + I ) ? n (m ) for all m. Then

f n( m + l) v + VJ m + l � f n ( m) v + VJ m = Wm . If v E Um +1 then f n ( rn + l ) v E VI m+l � VI m and so f n( rn )+jv E VJ rn for suit­ able j ? O. Thus v E Urn. Therefore Um + 1 � Urn, Wm + l � Wm and V = Urn + Wrn for all m. Let U = n : =o Urn, W = n: =o Wrn. We will show that V = U EB W Suppose that u E u n W. Thus for any m, u E Um n Wm and so by . (10.7) u E VJ m . Hence u E n : =o VJ m = (0). Therefore u n W = (0) and U + W = U EB W. Let v E V. By (10.7) v = Um + Wm for every integer m with Urn E Urn , Wm E Wm. Thus for any integers i, j Wm + 1

=

o = v - v = Ui - Uj + Wi - Wj . Hence for i, j > m Ui - Uj = Wi - Wi E Urn n Wrn � VI m . Therefore {Ui }, {w; } are Cauchy sequences in V. Let u = lim Ui, W = lim Wi . By (9.6) Urn and Wrn are closed in V. Since Ui E Urn for i > m and Wi E Wm for i > m this implies that u E Um and W E w'n for any integer m. Hence u E U and W E W. Furthermore v

= lim v = lim Ui + lim Wi =

u + w.

Therefore V = U EB W. Suppose that U = (0). Then V = W = Wrn for all m and so by (l0.7) Urn = VJ m for all m. If v E V with fv = 0 then v E U and so v = O. Thus f is a monomorphism. Let v E V. Then by (10.6) v = fVm + Zrn for all m, with Zm E VI m . Hence lim Z m = 0 and so {fvm } is a convergent sequence. If fu E VJ m for some m then u E Urn = VJ m and so {vm } is a Cauchy

36

CHAPTER I

[1 1

UNIQUE DECOMPOSITIONS

sequence. Since f is continuous by (9.2) this implies that v = lim fvm = f(lim vm ). Hence f is an epimorphism and so f is an automorphism. For each f E EA ( V) let U = Uf , W = Wf be defined as above. Thus either V = Uf or V = Wf . If V = Wf then f is a unit in EA ( V). Further­ more if V = Uf then V = WI- f . Hence every element in EA ( V) is either a unit or is quasi-regular. Suppose that EA ( V) is not a local ring. Then there exist two non units whose sum is a unit. Thus ther � exist f, g nonunits with f + g = 1. Hence g = 1 - f is a non unit contrary to the fact that f is quasi-regular. Hence . EA ( V) is a local ring. 0

i =2

o

Vi = Wi for

A ring A has the unique decomposition property if for any finitely generated A module V the following hold: (i) V is a direct sum of a finite number of indecomposable A modules. (ii) If V = EB �= l Vi = EB7= 1 V\tj where each V; , V\tj is indecomposable and nonzero then m = n and V; = W; after a suitable rearrangement.

COROLLARY 1 1 .2. Let A be a ring such that every finitely generated A

module is a direct sum of a finite number of indecomposable modules. .Assume further that if V is a finitely generated indecomposable A module then EA ( V) is a local ring. Then A has the unique decomposition property.

LEMMA 1 1 .1. Let A be a ring and let V be an A module. Suppose that V = EB�� I Vi and V = "L7= 1 V\tj where each Vi, V\tj is a nonzero indecomposable A module. Assume that EA ( Vi ) and EA ( V\tj ) are local rings for all i, j. Then m = n and after a suitable rearrangement Vi = Wi for i. = 1, . . . , m.

v = ft e t w + (v - ft e t w ) E fl V + ;=2 � Vi . I11 Consequently V = fl VI EB EB i=2 Vi . Since fl Vt � WI this implies that WI = fl VI EB { WI n EB�=2 V; }. As Wt is indecomposable and fl VI i (0) we see that WI = fl VI is isomorphic to VI and V = WI $ EB;'�2 Vi . Therefore

=2

Hence by induction m = n and after a suitable rearrangement

1 1 . Unique decompositions

PROOF. Induction on the minimum of m and n. If m = 1 or n = 1 the result is clear since V is indecomposable. It may be assumed that V = EB;: I Vi = EB7= 1 V\tj . Let ei , t be the proj ection of V onto Vi , V\tj respectively . Thus e l and el te l are all in EA ( VI). Since el = "L7=1 e l tel and EA ( VI) is local it follows that e I fje I is a unit in EA ( VI) for some j. Hence by changing notation it may be assumed that e I /l e l is an automorphism of VI = el V. Thus fl VI � WI and the kernel of fl on VI is (0). We next show that V = fl VI EB EB : 2 Vi . If U E f l VI n EEr= 2 Vi then e l u = 0 and u = fl e l v for some v E V. Thus e l fl el v = e l u = 0 and e l v = 0 as e I /l e l is an automorphism on VI . Hence u = fl e I v = O. Therefore fl VI n EEr�2 Vi = (0). Suppose that v E V. Then e l v E el V = e l fl e t V. Hence e t V = e I / l e l w for some w E W. Thus e t(v - fl e t w ) = O and v - ft e t w E EB �= 2 Vi . Hence

j

37

PROOF. Clear by ( 1 1 . 1). 0 THEOREM 1 1 .3. Suppose that A is a ring such that A satisfies A.C.C. and A /J(A ) satisfies D .C.C. Assume that A is complete on modules. Then A

has the unique decomposition property.

, PROOF. Immediate by (l0.4) and (11.2).

0

THEOREM 1 1 .4 (Krull-Schmidt). Suppose that A satisfies D.C.C. Then A

has the unique decomposition property.

_

PROOF. By (6.1 1) A satisfies A.C.C. Furthermore A is complete on modules. The result follows from (11 .3). 0 The next result was independently discovered by Borevich and Faddeev R is a complete discrete valuation domain.

[1959], Reiner [1961] and Swan [1960] in case

THEOREM 1 1.5. Suppose that R is a complete commutative ring which satisfies A.C.C. and R /J(R ) satisfies D .C.C. Let A be a finitely generated R -algebra.

Then A has the unique decomposition property.

PROOF. Immediate by (9. 1 1) and (11 .3).

0

38

CHAPTER I

1 2 . Criteria for lifting idempotents

Let A be a ring and let Z be the ring of rational integers. I am indebted to C. Huneke and N. Jacobson for the proofs of (12.1 }-(12.3) below. LEMMA 12.1. Let I be a nil ideal of A. For x E A let i denote the image of x

in A = A /I. If ii is an idempotent then there exists f(t) E Z[t] with no constant term such that f(a ) is an idempotent in A and f(a ) = ii. PROOF. Since a (1 - a ) E I and so is nilpotent, there exists a positive integer n with a n (l - a )" = O. By raising {a + (1 - a )} = 1 to the (2n - l) st power we see that

a n g(a ) + (l - a ) " h (a ) = 1, where g et), h (t) E Z[t]. Let f(t) = t n g(t). Then f(a ) = a ng (a ) = a n g(a ) {a n g (a ) + (1 - a ) " h (a )} = f(a f + a n ' (l - a )" g(a)h (a ) = f(a f Hence f(a ) is an idempotent or zero. Thus it suffices to show that f(a ) = ii l Since a (a " g(a ) + (l - a ) " h (a » = a it follows that a n + g(a ) = ii. Since a ,, + 1 = a n as ii is an idempotent this yields that a ng (a ) = ii. 0 LEMMA 12.2. Let I be a nil ideal of A. For x E A let i denote the image of x

in A = A /I. If el and e2 are commuting idempotents in A with el = e2, then

PROOF. Since (1 - e I )e2, (1 - e2)eI E I, they are nilpotent. However {(1 - ei )ej }2 = (1 - ei )ej for all i, j. Hence (1 - eI)e2 = 0 = (1 - e2)e I . There­ fore e l = e2eI = e l e2 = e2. 0 THEOREM 12.3. Assume that A is complete on modules. Let I be an ideal of

A with I � leA ). For x E A let i denote the image ofx in A = A /I. If ii is an idempotent in A then there exists a sequence of polynomials Ii (t) E Z[ t] with no constant term such that it (a)} converges in A and if e = lim Ii (a ) then e is an idempotent with e = ii. PROOF. Since l(A )/l(A ) " is nilpotent in A /l(A ) " , (12. 1) may be applied. 2 Thus there exists fn (t) E Z[t] with fn (a ) - fn (a ) E l(A ) " and fn (a ) = ii.

39

CRITERIA FOR LIFTING IDEMPOTENTS

[12

t'urtherrno:re fn (a ) and fn + ,(a) commute and map onto idempotents in A /l(A )" with fn+ I (a ) = ii = fn (a ). Thus fn (a ) - fn + I (a ) E l(A ) " by (12.2). Consequently {fn (a )} is a Cauchy sequence and so converges as A is c()mplete. This implies the required statement . 0 THEOREM 12.4 (Brauer, Nakayama). Assume that A is complete on mod­

ules. Let I be an ideal of A with I � l(A ). For a E A let ii denote the image of a in A = A /I. If eJ, . . . , en is a set of pairwise orthogonal idempotents in A then eJ , . . . , en is a set of pairwise orthogonal idempotents in A. If x . . . , Xn is a set of pairwise orthogonal idempotents in A then there exists a set of pairwise orthogonal idempotents e I , . . . , en in A such that ei = Xi for i = 1, . . , n. I,

.

PROOF. Since I � l(A ) it follows that n�=o I i = (0). Thus if e is an idempotent in A with e = 0 then e = e i E I i for all i and so e E n�=o r = (0). Hence e I: 0 and so e is an idempotent. The first statement follows. The second statement is proved by induction on n. If n = 1 it follows from (12.3). Suppose that n > 1 . By induction there exists a set of pairwise orthogonal idempotents e, e3, . . . , en such that e = x I X2 and ei = Xi for i = 3, . . . , n. Let x I = b and let a = ebe. Thus ii = Xl and ea = ae = a. By (12.3) there exists an idempotent eo which is a limit of polynomials in a with eo = ii. As e commutes with a it follows that eeo = e o e Hence e I = ee o is an idempotent such that el = X I and eel = el e = el . Let e2 = e - e l . Thus e2 = X2, e2e I = 0 = e]e2 and e� = e2. Therefore if i = 1, 2, j = 3, . . . , n then eiej = eieej = 0 and ejei = ejeei = O. Hence e I, . . . , en is a set of pairwise orthogonal idempotents with ei = Xi for i = 1, . . . , n. 0

+

.

COROLLARY 12.5. Assume that A is complete on modules. Assume further that A /l(A ) satisfies D .C.C. Let N be a right ideal in A. Then either N � 1 (A ) or N contains an idempotent. PROOF. Assume that N g: 1 (A ). Let ii denote the image of a in A A /l(A ). Then N is a nonzero right ideal of A and A is semi-simple and satisfies D .C.C. Thus by (8. 12) there exists an idempotent x E N. Since l(A ) is the inverse image of N in A it may be assumed that x = ii for a E N. If f(t) E Z[t] with no constant term then f(a ) E N. Hence by (12.3) there exists a sequence of elements {aj } in N such that lim aj = e I S an idempotent. By (9.6) N is closed in A and thus e E N. 0 =

N+

40

CHAPTER I

[12

CRITERIA FOR LIFTING IDEMPOTENTS

COROLLARY 12.6. Assume that A satisfies D .C.C. Let N be a right ideal in A. Then either N is nilpotent or N contains an idempotent. PROOF. If N does not contain an idempotent then N � J(A ) by (12.5). Hence N is nilpotent by (6.9). 0 COROLLARY 12.7. Assume that A is complete on modules. Assume further that A /J(A ) satisfies D .C.C. Then A is a local ring if and only if 1 is the

unique idempotent in A.

PROOF. Suppose that A is a local ring. Let e be an idempotent in A, e "l 1. Then (1 - e)e = e (1 - e ) = 0 and so e and 1 - e are both nonunits. Thus 1 = e + (1 - e ) lies in J(A ) which is impossible. Suppose that 1 is the only idempotent in A. Then by (12.5) every maximal right ideal in A is contained in J(A ). Thus A /J(A ) has no proper right ideals and so A /J(A ) is a division ring. Thus A is local by (10.1). 0 As a consequence of (12.7) one can obtain an alternative proof of (1 1 .5) as follows: By (1 1 .2) it suffices to show that EA ( V) is a local ring for every indecomposable A module. Since 1 is clearly the only idempotent in EA ( V) for V indecomposable only the hypotheses of (12.7) need to be verified for EA ( V). These follow from (8. 15) (ii) and (9. 1 1). This is the proof given by Swan [1960]. These results can be used to prove the following version of Hensel's Lemma.

LEMMA 12.8. Let R be an integral domain which is complete with respect to dI • Assume that R satisfies A.C.C., R /I satisfies D .C.C. and R is integrally closed in its quotient field K. For x E R let i denote the image of x in R = R /J(R ). Then R is a local ring. If f.(t) E R [t] with f(t) monic and f(t) = go(t )ho(t) with (go(t), ho(t» = 1 then there exist g (t), h (t) E R [t] such that g (t) = go(t), h (t) = ho(t) and f(t) = g(t)h (t). PROOF. By (9.4) and (9.5) I c: J(R ) and R is complete. Since R contains no idempotent other than 1 (12.5) implies that every ideal of R is contained in J(R ) and so by (10.2) R is a local ring and R is a field. Let p (t) E R [t], p (t) monic such that p (t) is irreducible in K[t]. Let L = K(a ) where a is a root of p (t) and let S = R (a ). Then S is an integral domain which is a finitely generated R -algebra. Thus 1 is the only idempotent in S. By (6. 13) and (9. 1 1) S satisfies A.C.C., S/SJ(R ) satisfies

.

41

and S is complete. Thus by (12.4) S = S/SJ(R ) contains no idempotent other than 1. Since R is a field and S = R [t]/(p (t) the Chinese remainder theorem implies that p (t) is a power of an irreducible polyno­ mial in R [ t ] . Now let f(t), go(t), ho(t) be as in the statement of the result . It may be assumed that go(t), ho(t) are monic. Since R is integrally closed in K, f(t) = fI�= l Pi (t) "; where each Pi (t) is irreducible in K and is monic with Pi (t) E R [t]. Hence by the previous paragraph for each i, Pi (t) I go(t) o r Pi (t) I ho(t). Let g(t) = fIg (t) "i where j ranges over all values of i with Pi (t) I go(t). Let h (t) = f(t)/g (t). Then it follows that g(t), h (t) have the desired properties. 0 It follows from Gauss' Lemma that if R is a unique factorization domain, and in particular, if R is a principal ideal domain then R is integrally closed in its quotient field. Thus (12.8) applies in these situations. The proof of the next result is essentially due to Dade [1973] and was brought to my attention by D. Burry.

THEOREM 12.9. Let A be a finitely generated algebra over the commutative ring R. Assume that R satisfies A.C.C., R /J(R ) satisfies D.C.C. and R is

complete. For x E A let i denote the image of x in A = A /J(R )A. Then the . mapping x � i defines a one to one correspondence between the set of all central idempotents in A and the set of all central idempotents in A. PROOF. By (8. 15) J(R )A c: J(A ). Hence in particular J(R )A contains no idempotents. Thus if e is a central idempotent in A then e is a central . idempotent in A. Suppose conversely that ii is a central idempotent in A. By (9. 1 1) A is complete on modules. Hence (12.4) implies that ii = e for some idempo­ tent e. We next show that e is central.

eAe EB eA (1 - e ) EB (1 - e )Ae EB (1 - e )A (1 - e). This i s the Peirce decomposition. Since e is central in A it follows that eA (1 - e) = e(l - e)A = 0, (1 - e)Ae = (1 - e)eA = O. Thus eA (l - e) C: J(R )A = AJ(R ) and so eA (l - e ) C: eA (I - e )J(R ). Since eA (1 - e) is an R module, Nakayama's Lemma, (9.8) implies that eA (1 - e) = O. Similarly (1 - e )Ae = O. Consequently A = eAe EB (1 - e )A (1 - e ) and so every element in A is of the form eae + (1 - e )b (1 - e ). Therefore e is in the center of A. A

=

CHAPTER I

42

Suppose that er = e2 for central idempotents e r and e2. Then ei - ere2 E J(A ) and (ei - ere2)2 = ei - e re2 for i = 1, 2. Thus ei - er e 2 = 0 and so

er = ere 2 = e2.

PRINCIPAL INDECOMPOSABLE MODULES

[ 13

0

13. Principal indecomposable modules

Let A be a ring satisfying A.C.C. An A module V is a principal indecomposable. module if V is indecom­ posable, V � (0) and V I AA .

LEMMA 13. 1 . Suppose that A has the unique decomposition property. Then every principal indecomposable A module is a finitely generated projective module. Conversely every finitely generated projective A module is a direct sum of principal indecomposable modules.

PROOF. If V I AA then V is a homomorphic image of AA and so is finitely generated. In view of the unique decomposition property the last statement will follow if every indecomposable proj ective module is a principal indecomposable module. Let V be a finitely generated indecomposable proj ective module. Then V I mAA for some integer m. Let AA = EB � where each Vi is indecom­ posable. Thus mAA = EB m � and so V = � for some i by the unique decomposition property. 0 In the rest of this section we will be concerned with rings satisfying the following conditions

43

13.4. Suppose that A satisfies (13.2). Let e be an idempotent in A.

Then e is primitive if and only if eAe is a local ring.

PROOF. If e is primitive then eA is indecomposable by (7.2) and so eAe is a local ring by (8.2) and (13.2) (iii). If e is not primitive then eAe contains two idempotents whose sum is e. Thus eAe is not a local ring. 0 THEOREM 13.5. Suppose that A satisfies (13 .2). Let e be a primitive idempo ­

tent in A. Then eJ (A ) is the unique maximal submodule of eA eA / eJ(A ) is irreducible. If e r is another primitive idempotent in A eA = erA if and only if eA /eJ(A ) = erA /erJ(A ).

and then

PROOF. Let N be a right ideal of A with N c;;, eA, N � eA. If N � J (A ) then by (13.2) (ii) there exists an idempotent f E N. Thus ef = f, e � f. Hence

(efe )2 = efefe = efe. Thus efe = 0 or efe is an idempotent in eAe. By (13.4) eAe is a local ring and so is the unique idempotent in eAe. Thus fe efe = e or fe = efe O . If fe e then eA c;;, fA c;;, N which is not the case, hence fe = O. Therefore (e - fY = e and so e commutes with f. Thus f ef = fe = 0 . which contradicts the fact that f is an idempotent. Hence N c;;, J(A ) and so N = eN � eJ(A ). If eA = eJ(A ) then e E eJ(A ) c;;, J(A ) and so 1 - e is a unit contrary to e (1 - e ) = O. The first statement is proved. If eA = elA then since eJ(A ), erJ(A ) is the unique maximal submodule of eA, erA respectively it follows that eA /eJ(A ) = erA/erJ(A ). Suppose that eA /eJ(A ) = erA /e r1(A ). Since eA, e J A are proj ective , modules there exists a map f : eA � e r A with trf t where t, tl are the natural projections of eA, e l A onto eA /eJ(A ) = e r A /eIJ(A ). Since trer J (A ) = (0) the first statement above implies that f is an epimorphism. Let V be the kernel of f. Then eA / V = erA and so eA = V EO elA since erA is proj ective. As eA is indecomposable this implies that V (0) and eA = erA as required. 0 e

.

=

=

=

=

=

HYPOTHESIS 13.2. (i) A satisfies A.C.C. (ii) If N is a right ideal of A then either N � J(A ) or N contains an

idempotent. (iii) If V is a finitely generated indecomposable A module then EA ( V) is a local ring.

LEMMA 13.3. Suppose that A satisfies A.C.C., A /J(A ) satisfies D .C.C. and A is complete on modules. Then A satisfies (13.2). In particular if A satisfies D.C.C. then A satisfies (13.2). PROOF. (13.2) (i) is clear. (13.2) (ii) follows from (12.5). (13.2) (iii) follows from (10.4). 0

=

COROLLARY 13.6. Suppose that A satisfies (13 .2). For any A module V let Rad( V) be the radical of V. The map sending V to V/Rad( V) sets up a one

to one correspondence between isomorphism classes of principal indecompos­ able modules and isomorphism classes of irreducible modules.

PROOF. This follows from (6. 1) and (13.5). 0

44

CHAPTER I

[ 13

THEOREM 13.7. Assume thatA is complete on modules and satisfies A.C.C. Assume further that A /J(A ) satisfies D .C.C. Let I be an ideal of A with I � J(A ). For any A module V and v E V let i5 denote the image of v in V = V/ VI. Then (i) If W is a finitely generated projective A module then W = P for some finitely generated projective A module P. (ii) If P is a finitely generated projective A module then P is a finitely generated projective A module. If P is a principal indecomposable A module then P is a principal indecomposable A module. (iii) The map sending P to P sets up a one to one correspondence between isomorphism classes of finitely generated projective A modules and isomorphism classes of finitely generated projective A modules. P�OOF. By ( 1 1 .3) A and A have the unique decomposition property. (i) It suffices to consider the case that W is indecomposable. Thus by (13.1) W = xA with x an idempotent in A. By (12.4) x = e for some idempotent in A. Thus W = eA. (ii) It may be assumed that P = eA for some primitive idempotent e in A. Thus P = eA is proj ective. By (13.5) P has a unique maximal submodule and so is indecomposable. (iii) By (i) and (ii) it suffices to show that if P I , P2 are principal indecomposable modules then PI = P2 if and only if PI = P2• Since Pi /Rad Pi = Pi /Rad Pi as A /J (A ) = A /J (A ) as A modules the result follows from (13.6). 0 LEMMA 13.8. Suppose that A satisfies (13.2). Let e be a primitive idempotent

in A. Then EA (eA /eJ(A )) = eAe /J(eAe) = eAe /eJ(A )e. If A is a finitely generated R -algebra for some commutative ring R then these isomorphisms are R - isomorphisms.

PROOF. The second equality follows from (8.4). If f E EA (eA ) then since f is an A -endomorphism f(eJ(A )) � eJ(A ). Thus f induces an endomorphism ! of eA /eJ(A ). The map sending f to ! is clearly a ring homomorphism and an R -homomorphism in case A is an R -algebra. If h E EA (eA /eJ(A )) then h can be viewed as a map h : eA � eA /eJ(A ). Since eA is projective this implies the existence of f E EA (eA ) with tf = h where t : eA � eA /eJ(A ) is the natural proj ec­ tion. Then ! = h. Thus the map sending f to ! is an epimorphism of EA (eA ) onto EA (eA /eJ(A )). Since EA (eA /eJ(A )) is a division ring by Schur's Lemma, (8. 1) and EA (eA ) = eAe is a local ring by (8.2) and (13.4) the result follows. 0

PRINCIPAL INDECOMPOSABLE MODULES

45

13.9. Suppose that A satisfies (13.2). Let V be a finitely generated

module with a composition series and let e be a primitive idempotent in A. The following are equivalent. (i) HOffiA (eA, V) -I (0). (ii) Ve I (0). (iii) V has a composition factor isomorphic to eA /eJ(A ).

(iv). If EA ( V) I: R then there exists a monic irreducible polyno­ mial f(t) E R [t] which has a root in EA ( V) R. Let fo(t) E R [t] such that fo(t) is monic and fo(t) = f(t). Let S = R [a] where a is a root of fo(t). Since S /( 7T)S is a field S is a finite unramified extension of R and f (t) has a root in S. Thus EA ( V) 0 R S is not a division ring. Hence by (18.4) EAs ( Vs ) is not a division ring and so by (8. 1) Vs is reducible. (iv) => (v). Clear from the definition of a splitting field. (v) => (i). Let U1 , , Uk be a complete set of representatives of the isomorphism classes of principal indecomposable A modules. By (16.4) there exists j with Ir�. ( � , V) = 1 . Let F be an extension field of R. Thus by (18.4) IF (( � )F, VF) = 1 . Let U;, . . . , U:n be a complete set of representa­ tives of the principal indecomposable AF modules. Since ( �)F I AF this implies that there exists i with IF ( U:, VF) = 1. Hence by (16.4) VF is irreducible. D -

•

•

•

·

·

An A module V is absolutely indecomposable if for every finite extension S of R, Vs is an indecomposable As module.

A module and let E = EA ( V). If E/l(E) = Ii then V is absolutely indecomposable. If R is algebraically closed then every finitely generated indecomposable A module is absolutely indecomposable.

LEMMA 18.6. Let V be a finitely generated indecomposable

�ROOF. Since E /1 (E) is a division ring which is a finitely generated

R -algebra the second statement follows from the first. Suppose that E/l(E) = R. Let S be a finite extension of R. By assumption l(R )S � l(S) � l(Es ). Thus l(R )Es � l(Es ). Since l(E)s /l(R )Es is a nilpotent ideal in Es/l(R )Es this implies that l(E)s � l(Es). Since Es/l(E)s = S/l(R )S it follows that Es /l(Es) = S/l(S) has only one idempotent. Thus by (12.3) Es has only one idempotent and so by (18.4) EAs ( Vs) has only one idempotent. Hence Vs is indecomposable. D

A module. There exists a finite unramified extension S of R such that Vs = E9�=1 Vi where each Vi is an absolutely indecomposable S module and EAs ( Vi )/l(EAs ( Vi )) = S.

LEMMA 18.7. Let V be a finitely generated indecomposable

PROOF. If S is a finite unramified extension of R then Vs is the direct sum of n (S) nonzero indecomposable As modules for some integer n (S) with

·

·

73

n'( S) < dimR ( V). Choose S so that n (S) is maximum. Clearly every component of Vs remains indecomposable when tensored with any finite unramified extension of S. Replacing R by S it may be assumed that Vs is indecomposable for any finite unramified extension S of R. For any finite unramified extension S of R let D (S ) = EAs ( Vs )/l(EAs ( Vs )). Thus D (S ) is a finite dimensional S-algebra which is a division ring. If T is a finite unramified extension of S then l(S) T � l(T) � l(EA T ( VT)) by assumption and so l(S)EAT ( VT) � l(EA T ( VT))' Thus since l(EA s ( Vs ))r/l(S)EA T ( VT) is nilpotent, D ( T) is a homomorphic image of D (S) 0s T. Choose a finite unramified extension S of R such that dimsD (S) is minimum. If D (S ) I: S there exists a monic irreducible polynomial f(t) E S[t] of degree at least two which has a root in D (S). Choose fo(t) E S [t] such that fo(t) is monic and faCt) = f(t). Let T S[a] where a is a root of f�(t). Then it is easily seen that T is a finite unramified extension of S. Since f(t) has a root in T, D (S ) 0s T is not a division ring and so dimr D ( T) < dimsD (S) contrary to the choice of S. Hence D (S ) = S and Vs is absolutely indecomposable by (18.6). 0 =

The converse of the first statement in (18.6) is false . It is however almost true, the difficulty only occurs if R has inseparable extensions. This is discussed in Huppert [1975] and a counterexample due to Green is given there. It should be observed that Huppert's definition of absolute indecomposability differs from that given in the text. In any case (18.7) is important for some applications. LEMMA 18.8. Let U be a principal indecomposable A module and let L = U/Rad( U) where Rad( U) is the radical of U. Then L is absolutely

ir.reducible if and only if EA ( U)/l(EA ( U)) = R.

(13.8) EA (L) = EA ( U)/l(EA ( U)). The result follows from COROLLARY 18.9. There exists a finite unramified extension S of R such that

is splitting field of As. a

By (18.7) there exists a finite unramified extension S of R such that = E9 U; where EA s ( U; )/l(EAs ( U; )) = S for each i. By (18.8) S is a " ... I ,H-, r. nfield of As. 0

74

19. Representations and traces

Let R be a commutative ring and let A be a finitely generated R -free R -algebra. An R -representation of A or simply a representation of A is an algebra homomorphism f : A � ER ( V) where V is a finitely generated R -free R module and f(1) = 1 . If f : A � ER ( V) is a representation of A define va = vf (a ) for v E V, a E A. In this way V becomes an A module . If conversely V is an R -free A module define f : A � ER ( V) by vf(a ) = va for v E V, a E A. Then f is an R -representation of A. Thus there is a natural one to one correspon­ dence between representations of A and finitely generated R -free A modules. The module corresponding to a representation is the underlying module of that representation. All adj ectives such as irreducible etc. will be applied to the representation f if they apply to the underlying module of f. Two representations fl ' f2 with underlying modules VI , V2 are equivalent if VI is isomorphic to V2 • It is easily seen that fl is equivalent to h if and only if there exists an R -isomorphism g : VI � V2 such that 1-2 = g�lflg. Let V be a finitely generated R -free R module with a basis consisting of n elements. By (8.6) ER ( V) = R n . Thus a representation of A defines an algebra homomorphism of A into R n . Assume that R is an integral domain and let f be an R -representation of A with underlying module V. Define the function tv : A � R by tv (a ) is the trace of f(a ) where f(a ) E R n . It is clear that tv is independent of the choice of isomorphism mapping ER ( V) onto R n . The function tv is the trace function afforded by V or the trace function afforded by f. It is easily seen that if V = W then tv = two Furthermore if K is the quotient field of R then tVK (a ) = tv (a ) for a E A and V a finitely generated R -free A module. Suppose that R satisfies the same hypotheses as in section 17. Let S be an extension of R. Then for any finitely generated R -free A module V we have tvs (a ) = tv (a ) for a E A. Throughout the remainder of this section F is a field and A is a finitely

generated F-algebra. All modules are assumed to be finitely generated.

Under these circumstances trace functions are sometimes called charac­ ters. See for instance Curtis and Reiner [1962] . However we prefer reserve the term character for a different, though closely related, concept which will be introduced later. THEOREM 19. 1 . Let V, W be absolutely irreducible A modules. Then V =

if and only if tv = two

REPRESENTATIONS AND TRACES

CHAPTER I

75

. If V = W then tv = tw o Suppose that V� W. Let Iv, Iw be the annihilator of V, W respectively. Let I = Iv n Iw . It suffices to prove the result for the algebra A /1. Thus by changing notation it may be assumed that 1 = (0). Hence J(A ) = (0). Since A has two nonisomorphic irreducible inodules the Artin-Wedderburn theorems (8. 1 0) and (8. 1 1) imply that A = Av EEl Aw where A v and Aw are simple rings and V, W is a faithful absolutely irreducible Av, Aw module respectively. Since V and W are absolutely irreducible it follows from (8. 1 1 ) that A v = Fm and Aw = Fn for some m, n and the isomorphisms sending Av to F,n and Aw to F" are equivalent to representations with underlying modules V, W respectively. Choose a E Av such that a corresponds to the matrix

Then tv (a ) = l and tw (a ) = O. Hence tv l- tw.

0

Suppose that K is a finite Galois extension of F and u is an automor­ phism of K over F. Then (T defines an automorphism of AK by (a 0 x t = a 0 x for a E A, x E K. This automorphism will also be denoted by (T. If V is an AK module let v a = {v" I v E V} where cr

Vcr + WIT = (v + W )'T

v'Ta 'T = (va )"

for v, W E V,

for v E V, a E AK •

Clearly vcr is an AK module and V O. Let G be the Galois group of L over F. Then A has a unique irreducible module W up to isomorphism and the following hold. (i) AL = EijaEGAa, with Acr = Ln for all u E G. (ii) There is an irreducible AL module M such that WL = Eiju E G MeT, {MU} is. a

complete set of representatives of the isomorphism classes of irre ­ ducible AL modules and each MeT is absolutely irreducible. Furthermore if tMer (a) = tMT (a ) for all a E A then u = T.

76

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CHAPTER I

REPRESENT AT IONS AND TRACES

PROOF. Let W be an irreducible A module. By (8. 1 1) every irreducible A module is isomorphic to W. It is easily seen that dimF W = n [L : F] . If a- E G then 0- defines an automorphi sm of A = LI1 in a natural way. This will also be denoted by a-. Let L = F(c ) and let

' qbsolutely irreducible A module W such that WK = V and tv (a) = tw (a ) for all a E A.

b=

(�

o

o

Let U be an n -dimensiona l L space. For 0- E G define fa : A L � EL ( U) by u (fcra ) ua cr. Thus each fcr is an absolutely irreducible representatio n of AL. Let u" be the underlying AL module and let ta = tu". Then tcr tT. Hence if a- -I: T then ta (b) = c " -I: c T tT (b). Thus Ucr � UT for a- -I: T by =

=

=

(19. 1).

Let I be the annihilator of E9 Ua• By (8. 10) and (8. 1 1) AL II = EBa EG Acr , where Ua is a faithful irreducible Acr module . Hence Aa = Ln for all a- E G and dimL AL

=

(

)

n 2(L : F) = dimL E9 Aa a EG

= dimL (AL I I).

Thus I = (0) and (i) is proved . Furthermore every irreducible AL module is isomorphic to some Ua and each Ucr is absolutely irreducible. For some 0- E G, Ua I WL. As wr = WL the unique decomposition property ( 1 1 .4) implies that E9aEG U,T I WL• Thus WL = E9cr EG Ua as

M

=

(

)

n [L : F] = dimL E9 Ucr . (TEG UI • By (19. 1 ) M a -' = Ucr for E G. 0

dimL WL = dimF W

Define

W be an irreducible A module such that V is a composition factor of WK ' By factoring out the annihilator of W and changing notation it may be assumed that W is a faithful A module. Thus J(A ) = (0) and by the Artin-Wedderburn theorems (8. 10) and (8. 1 1) and the fact that finite division rings are fields it follows that A = Ln for some finite extension field L of F. It may be assumed that A = Ln . By replacing K by KL and V by VKL it may be assumed that L � K. Let M and G be defined as in (19.2). Then V = M'f!

W : @ Xi

for suitable w ; E W. Thus ws = w I for all j and all y E H. Hence w I E InvH ( V) and v = Li W I @ Xi = Tr � ( w l @ l) . 0

(i) By (1.4.8) V I ( VH ) G . Hence by (3.4) HO( G, H, V) ' HO (G, H, ( VH )G ) = (0).

LEMMA 3.5. Let H be a subgroup of G. Let 2l be a nonempty set of subgroups of H and let � = {H n A x I A E 2l , X E G}. Let W be an R [H] module. Then v E LA T r� {Inv A ( WO )} if and only if v = Tr�(w @ l) for some w E LBE'8 Tr;;{InvB ( W)}, Furthermore HO(G, 91, W G ) = HO(H, � , W). Thus in particular HO(G, ( I ), W G ) = HO(H, ( 1 ), W).

(ii) Immediate by (i) and (2.7).

E 21

0

LEMMA 3.7. Let H be a subgroup of G. The following hold. (i) Let U, V, W be R [G] modules. If f E HomR [H]( U, V) and g E HomR[ O ) ( V, W) then Tr�(gf) = g Tr�(f ). If f E HomR[ G ]( U, V) and g E HomR[H) ( V, W) then Tr�(gf) = Tr�(g)f. (ii) Let A be a ring which is an R [G] module and suppose that (ab)x = (ax ) (bx ) for a, b E A and X E G. Then Tr�(ab ) = a Tr�(b ) and Tr�(ba ) = Tr�(b )a for a E Inv G (A ) and b E Inv H (A ). Furthermore Tr�(lnvH (A » is an ideal in the ring Invo (A ). (iii) If V is an R [G] module then Tr�(ER[H]( V») is an ideal in ER[o ] ( V).

PROOF. Suppose that v = Tr�(w @ 1) with w = Tr;;( u ) where B = H n A with A E 91 and u E InvB ( W). Thus v = Tr�(u @ 1 ) . Hence by (3.2) u @ x - I E Invw-rnA ( W G ) and v = Tr � nAX( u @ l ) = Trik r n A( u @ x - I ) = Tr�{Tr�x-r n A( U @ x - I )}. Thus v E LA E'll Tr�{InvA ( W G )}. Suppose conversely that v = Tr � ( u ) for some A E 91 and u E Inv A ( W G ). Let {Xi } be a complete set of representatives of the (H, A ) double cosets in G. By the Mackey decomposition (2.9) and (3.4) u = L Ui where Ui = Tr�Xi ( Wi @ Xi ) for some Wi E W. Therefore by (3.2) x

PROOF. Immediate by definition and (2.4).

0

The following fundament al result is essentially due to D. G. Higman [1954].

nA

,

THEOREM 3.8. Let H be a subgroup of G and let V be a finitely generated [G] module. The following are equivalent. (i) V is R [H]-projective. (ii) V I ( VH ) G .

.R

90

(iii) V I w G for some finitely generated R [H] module W. (iv) HomR ( V, V) is R [H]-projective. (v) HO(G, H, HomR ( V, V)) = (0) . (vi) There exists f E HomR[H] ( V, V) = InvH {HomR ( V, V)} such that Tri;(f ) = 1 . (vii) V is R [H]-injective. PROOF. By (1.4.8) (i), (ii) and (iii) are equivalent. (iii) => (iv). Clear by (2.7).

(iv) => (v). Clear by (3.6). (v) => (vi). Immediate by definition and (2.4). (vi) => (vii). Suppose that W is an R [G] module with V � W such that VH I WH o Thus there exists a projection e of W onto V which is an R [H]-homomorphism. Hence Tri;(ef ) E HomR[ G ] ( W, W). Let {x; } be a cross section of H in G. If w E W then Tri;( ef) = Li {(wx ;-')ef}xi E Li ( Wef)xi � Li ( Vf)Xi � V and if v E V then w

= v Tri;(f) = V. Hence Tri;(ef) is a projection of W onto V and so V I W as required. (vii) => (ii). Let {Xi } be a cross section of H in. G with XI = 1. Define g : V � ( VHf by gV = L i vx ;-' @ xi ' Thus g = Tri;(h ) where h : V� V @ 1 with hv = v @ 1 . If gv = 0 then vx I ' @ X = V @ 1 = O. Hence g is an R [ G]-monomorphism. Let W = {Li ", , Vi @ Xi }. Then W is an R [H] module. Clearly g( V) n W = (0). If LVi @ Xi E V G then I

= g(v,) + L (Vi - v, X ;-' ) @ Xi E g ( V) + W. i", 1 Hence {( VH ) G h-l = g ( V)H E9 W. Therefore VH 1 { ( VH f }H and so V I ( VH ) G since V is R [H]-injective. D COROLLARY 3.9. Let H be a subgroup of G. Let V be a finitely generated R -free R [ G] module. Then V is R [H]-projective if and only if H()( G, H, V* @ V) = (0).

91

RELATIVE TRACES

[3

CHAPTER II

Clear by (2.8) and (3.8).

D

COROLLARY 3.1 0 . Suppose that 1 G : H I has an inverse in R for some

subgroup H of G. Then every finitely generated R [ G] module is R [H]­ projective. PROOF. Let f = (1/ 1 G : H I )1 E HomRIHJ( V, V). Then Tr�;(f) =

sult follows from (3.8).

D

1.

The re­

COROLLARY 3 . 1 1 . Suppose that 1 G 1 has an inverse in R. Then every finitely

generated R [G] module is R -projective and every finitely generated R -free [ G] module is projective. If furthermore V is an indecomposable R [ G ] module and W is a submodule of V with WR I VR then W = (0) or W = V.

R

PROOF. Clear by (3. 10) and (1.4.6).

D

COROLLARY 3. 1 2 (Maschke). Let F be a field whose characteristic does not

divide 1 G I . Then every finitely generated F[ G] module is projective. F[ G ] is semi -simple and every finitely generated F[ G] module is completely reducible. PROOF. By (3. 1 1) every finitely generated F[ G] module is projective. Thus every finitely generated F[ G] module is completely reducible. Since F[ G]F[ G ] is completely reducible it follows that F[ G] is semi-simple. D

The following result in case H = (1) is due to Green [1974a]. LEMMA 3.13. Let H be a subgroup of g. Let V, W be finitely generated R -free R [G] modules. Let g E HomR[ G ]( V, W). The following are equiva ­

lent. (i) g E Tri;(HomR[H]( V, W)). (ii) Suppose that U is a finitely generated R [H]-projective R [G] module such that V � W � 0 is an exact sequence and V � W � 0 is split. Then there exists h E HomR[ G ]( V, V) with g = fh. (iii) Suppose that V is a finitely generated R [H]-injective R [G] module such that O � V � V is an exact sequence and O � Vn � Vn is split. Then there exists h E HomR[ G ]( V, W) with g = hf. H

H

PROOF. (i) => (ii). There exists t E HomR[H]( W, V) with ft = 1. Let g =

92

CHAPTER II

[4

Tr�( go) with go E HomR [H] ( V, W). Let h Tr2(tgo) E HomR[ G ] ( V, U). Then by (3.7) fh Tr2(ftgo) = Tr2(go) g. (ii) � (i). By Higman's Theorem (3.8) there exists go E HomRlH] ( U, U) with Tr�(go) 1 . Thus by (3.7) Tr� (fgo h ) = f Tr�( go) h fh g. The dual of the argument above shows that (i) is equivalent to (iii). 0

5]

ALGEBRAIC MODULES

=

=

=

= =

=

93

Clear by definition. 0 In view of (4.2) the Grothendieck algebra is not too interesting in case R is a field. However even in that case the representation algebra may be infinite dimensional as a C module. If .0 is a nonempty set of subgroups of G let Ac� (R [OD be the C submodule of Ac (R [ G D generated by all ( V) where V is R [H]-projective for some H E S) . 4.3. Let .0 be a nonempty set of subgroups of G. Then Acs;,(R [G D is an ideal in Ac (R [ G D. If furthermore R [ G] has the unique decomposition property then Ac�(R [GD is free as a C module with basis ( Vd where Vi ranges over a complete set of representatives of the isomorphism classes of finitely generated indecomposable R [G] modules which are R [H]-projective for some H E S). LEMMA

4. The representation algebra of

R [ G]

For any finitely generated R [G] module V let ( V) denote the class of all R [ G] modules which are isomorphic to V. For any commutative ring C the representation algebra or Green Algebra Ac (R [GD is defined as follows. Ac (R [GD is the C module generated by the set of all isomorphism classes ( V) of finitely generated R -free R [G] modules subject to the relations ( VI EB V2) ( VI) + ( V2) ' Multiplication is defined by ( VI) ( V2) ( VI @ V2) ' It is easily seen that Ac (R [GD is a commutative ring with 1 = (R ) where R InvR[ G ](R ). The Grothendieck algebra A �(R [GD equals A c (R [ G])/I where I is the ideal of Ac (R [GD generated by all ( U) - ( V) + ( W) where U, V, W are finitely generated R [ G] modules such that there exists an exact sequence O � U � V � W � O. =

=

=

LEMMA 4. 1 . If R [ G] has the unique decomposition property then Ac (R [ G D is free as a C module with basis ( Vi ) where Vi ranges over a complete set of representatives of the isomorphism classes of finitely generated indecomposable R [ G] modules.

PROOF. Clear.

0

4.2. Assume that R is a field. Let C be an integral domain whose quotient field has characteristic 0 and let ( vt denote the image of ( V) in A �(R [ G ]). Let L I , . . . , Ln be a complete set of representatives of the isomorphism classes of irreducible R [ G] modules. Then ( V)O = 2:� I Ci (Li ) if and only if Li occurs as a composition factor of V with multiplicity Ci . In particular ( VI)O ( V2)0 if and only if VI V2. LEMMA

=

=

�

0

PROOF. Clear by (2.3). 0

Let Z be the ring of rational integers. A finitely generated R [G] module is (R, S))-projective or simply .0-projective if V E A z,� (R [ G D. If R [ G] has the unique decomposition property then clearly V is ,\) -proj ective if and only if V EB Vi where for each i there exists Hi E .0 such that Vi is R [Hi ]-projective. For future reference we introduce the following notation. If V and W are finitely generated R [G] modules then V W(.0) means that (V) - ( W) E Az,s;,(R [GD. In particular if V is .0-projective then V 0(.0) and conversely. V

=

==

. 5.

==

Algebraic . modules

Throughout this section C is the field of complex numbers. The Green ring Ac (R [GD is generally very large. This section contains the definitions and basic properties of some interesting subrings. These ideas are due to Alperin [1976c], [1976e] . r----"'-.. If V is an R [ G ] module let V" V @ . . . @ V. An element E Ac (R [G]) is algebraic if it is the root of a nonzero polynomial with integer coefficients. In other words, if there exist integers ao, . . . , ak , not all 0, with ao + . . . + ak k O. An R [G] module V is algebraic if V is R -free and ( V) is an algebraic element of Ac (R [GD. n

x

=

x

=

[5

CHAPTER II

94

LEMMA 5 . 1 . Suppose that R [ G] has the unique decomposition property. Let

be an R -free R [G] module. The following are equivalent. (i) V is algebraic. (ii) There exist a finite number of indecomposable R -free R [ G] modules WI , . . . , Wm such that if W is indecomposable and W I vn for any n, then W = Wi for some i.

V

PROOF. (i) � (ii). There exist nonnegative integers ao, . . . , ak , bo, . . . , bj with j < k such that ak � 0 and 2:�=o ai ( vy = 2:�=o bi ( VY. Thus every indecomposable component of Vk is a component of V i for some i < k by the unique decomposition property. Hence if s � k then a routine induc­

tion argument shows that every indecomposable component of component of V i for some i < k. Thus (ii) holds. (ii) � (i). There exist integers aij such that ( V ) = a l l ( W I ) + . . + a m ( Wm ) , ( Vr a2 l ( WI ) + . . + a2m ( Wm ), .

=

VS

is a

i

.

The linear dependence of these equations shows that ( V) must satisfy a nonzero polynomial of degree at most m + 1 with integer coefficients. 0 LEMMA 5.2. Suppose that R [G] has the unique decomposition property. Let V, VI , V:� be R -free R [ G] modules. (i) If V is algebraic and W I V then W is algebraic: (ii) If VI and V2 are algebraic then VI E9 V2 and VI @ V2

95

PROJECTIVE RESOLUTIONS

(i) Clear since ( vn )H = ( VH t . (ii) Induction on 1 H I · By (2.3) and the Mackey theorems (2.9) and (2. 10) ( UG t + 1 = ( U G t 0 uG = { « UG t )H 0 U}G

= { ( UG )Ht 0 U} G =

{( � ( U �F n H)Hr 0 uf'

Thus ( UGt+1 is a direct sum of terms of the form {QSr:i ( U�Xi n Hy-{} G . If x E G then UX is an algebraic R [HX ] module. Thus by (i) U ;r n H is algebraic and so by induction ( U�X nH) H is algebraic. There are at most 1 G H I pairwise nonisomorphic R [H] modules of the form ( U�IXnH) H . Thus by (5. 1) and (5.2) there exist a finite number of R [H] modules WI , . . . , Wm such that every indecomposable component of H U ( 0�:i �Xi nH) is isomorphic to some Wi for all n � O. Hence there exist a finite number of indecomposable R [G] modules VI , . . . , Vk such that every indecomposable component of W;- is isomorphic to some Vj. By (5.1) UG is algebraic. The result follows from (5.1). 0 :

Suppose that R = F is an algebraically closed field. An indecomposable F[ G] module V is irreducibly generated if there exist irreducible F[ G] modules L I , . . , Ls with V I L I 0 . 0 Ls . An F[ G] module is irreducibly generated if every indecomposable component is irreducibly generated. Let R = F be an algebraically closed field and let G be a group. In view of (5. 1) every irreducibly generated F[G] module is algebraic if and only if every irreducible F[ G] module is algebraic. It is natural to ask when this condition is satisfied. If char F = 2 and G SL2(2n ) then Alperin [1976c] has shown that this is the case. If G is solvable this condition also holds. See (X.7. 1). By (5.2) and (5.3) a projective F[ G] module is algebraic. It will be proved in (111.2. 18) that if char F P then every projective F[G] module is irreducibly generated if and only if Op (G) = (1). . .

.

=

are algebraic.

PROOF. (i) Immediate by (5. 1).

(ii) By (5. 1) there exist indecomposable R [G] modules Wl l , • • • , W l m, W2I , • • • , W2m such that for all s, t > 0 Vf @ V� is the direct sum of modules of the form Wl i 0 W2j • Thus (5. 1) and the unique decomposition property imply that VI 0 V2 and VI E9 V2 are both algebraic. 0 LEMMA 5.3. Suppose that R [K] has the unique decomposition property for

every subgroup K of G. Let H be a subgroup of G. (i) If V is an algebraic R [G] module then VH is an algebraic R [H] module. (ii) If U is an algebraic R [H] module and V I UG then V is an algebraic R [G] module.

=

6. Projective resolutions

Let V be an R [ G] module. A projective resolution of sequence

V

is an exact (6. 1)

. where each clearly

Pi

is a projective R [G] module. If

M n

=

dn (Pn ) c;;;, PII _ I then

96

[6

CHAPTER II

(6.2) is an exact sequence. LEMMA 6.3. Let V be an R [G] module. Let

CHAPTER III

be exact sequences with Pi , P; projective for all i. Then there exist projective R [G] modules Q, Q ' with M" EB Q = M� EB Q'. PROOF. An immediate consequence of Schanuel ' s Lemma (1.4.3).

0

An R [G] module is periodic if V = M" in (6.2) for some n � O. The smallest integer n � 0 with V M" is the period of V. Let Vo(G) = R as R modules with Vo(G) = Invc ( Vo(G» . =

LEMMA 6.4. Let F be a field. Suppose that Vo( G) is a periodic F[ G] module with period n. If n > 0 then every F[ G] module is periodic and its period

divides n. If n = 0 then every F[ G] module is projective.

PROOF. Since V V (9 Vo(G) the result follows from (2.7) and (6.3) by tensoring every term in (6.2) with V. 0 =

Projective resolutions of F[ G] modules and perio�ic F[ G] modules have been studied by various authors. See for instance Alperin [1973], [1976b], [1977b], Carlson [1977], [1978], [1979] . We will not pursue this subject in this book except in a special case in Chapter VII, section 10.

1.

Basic assumptions and notation

Throughout the remainder of this book the following notation will be used. G is a group. p is a fixed rational prime. R is an integral domain satisfying the following conditions: (i) R satisfies A.C.C. and R /J(R ) satisfies D .C.C. (ii) R is complete. (iii) J(R ) = ( ) is a principal ideal. (iv) p = 1 + . . . + 1 E ( ) 17'

'-r---I

17'

.

P

K is the quotient field of R. All modules are assumed to be finitely generated. If V is an R [G] module and v E V then v denotes the image of v in V = V/(1T) V. Similarly R = R /(1T) and R [G] = R [ G]/(1T)R [G] = R [G]. In case = 0 R = R = K is simply a field of characteristic p. Otherwise R is a complete discrete valuation ring. In view of these assumptions the results of Chapter I sections 17, 18, 19 and the results of Chapter II apply directly to R [ G]. Furthermore many results from Chapter I will frequently be used when applicable. An exposition of much of the material in this chapter can be found in Green [1974b] or Michler [1972a] . The main purpose of this chapter, and indeed of the whole book, is to . study R [G] modules. However before doing so it is necessary to investi­ gate F[ G] modules where F is a field. This is done in the next section. 17'

07

98

2.

F[ G ]

F(G) MODULES

[2

CHAPTER III

99

By (II.2.6) and (II.2. 10)

modules

Throughout this section F is a field and G is a group. Let Vo(G) denote the F[G] module consisting of invariant elements with dimF Vo( G ) l . Observe that all F[ G ] modules are F -free. =

LEMMA 2. 1 . Let V, W be F[G] modules. Then dimF (Inv G ( V * @ W» =

I ( V, W).

( V G )* @ W G = EB ( V;';X nA @ W w nA) G . x

The result follows from (2.1) and (2.3). 0 THEOREM 2.5 (Frobenius Reciprocity). Let H be a subgroup of G. Let V be

an F[G] module and let W be an F[H] module. Then I( W G, V) I( W, VJ-I ) and I( V, W G ) = I( VJ-I , W).

=

PROOF. Immediate by (II.2.4) and (II.2.8). 0

PROOF. Clear by (2.4).

LEMMA 2.2. Suppose that char F = P > O. Let V be an F[ G] module.

THEOREM 2.6 (Nakayama Relations). Let H be a subgroup of G. Assume

(i) If P I dimF V. Then the multiplicity of Vo( G ) in V * @ V is at least 2. If V is absolutely irreducible then V * @ V is not completely reducible. (ii) If P � dimF V then Vo(G ) I V* @ V.

PROOF. By (II.2.8) V* @ V = HomF ( V, V) = Fn where n = dimF V. The

group G acts on Fn by conjugation. Thus if Wo is the set of scalars in Fn and WI is the set of matrices of trace 0 both Wo and WI are F[ G] modules. Clearly Wo = Vo( G). If P � n then V * @ V = Wo EB WI and (ii) is proved. Suppose that p i n. It follows that Wo � WI . If w E Fn and E G then ( w - w ) E WI . Thus F / WI = Vo(G). Hence the multiplicity of Vo(G) in V* @ V is at least 2. If V is absolutely irreducible then by (1.8. 1) and (2.1) dimF (Inv G ( V * @ V» = 1 . If V * @ V is completely reducible then the first statement implies that dimF (Inv G ( V* @ V» ;:?: 2. Thus V * @ V is not completely reducible. 0 x

x

n

LEMMA 2.3. Let H be a subgroup of G and let W be an F[H] module. Then

dimF {InvJ-I ( W)} = dimF {Inv G ( W G )}. PROOF. Immediate by (II.3.4). 0 THEOREM 2.4 (Mackey [1951]). Let H, A be subgroups of G. Let V be an

F[ H] module and let W be an F[A ] module. Then I ( V G , W G ) = 2: I( V�xnA ' W w nA) x

where in G.

x

ranges over a complete set of (H, A ) double coset representatives

0

that F is a splitting field of F[ H] and of F[ G]. Let { U }, { Vj } be a complete system of representatives of the isomorphism classes of principal indecompos­ able F[ G], F[ H] modules respectively. Let L;, � be the irreducible F[ G], F[H] module corresponding to U, Vj respectively. Then (i) (L; )J-I � EB a;jMj for each i if and only if V? = EB a;j V; for all j. (ii) ( V; )J-I = EB b;j Vj for each i if and only if M? � EB bijLi for each j.

PROOF. (i) Let ( Li )J-I � EB aij� and let V ? = EB a:j U. By (1. 16.4) a;j = ' I( Vj, (L; )J-I). By (1. 16.9) a[j = I( V?, L; ). The result follows from (2.5). (ii) Let ( U )J-I = EB b;j Vj and let M? � L bijL;. By (1. 16.4) bi� = I( V;, M?). By (1.16.9) b;j = I« V; )J-I, Mj ). The result follows from (2.5). D

LEMMA 2.7. Let T 0. By (3. 14) P is the vertex of every component of VP• Hence P is contained in a vertex of V by (4. 1). 0

, COROLLARY 4. 13. Suppose that P <J G, P a p-group. Let V be an irreducible R [G] module with vertex Po. Then P � Po and PolP is the vertex of V as an R [G IP] module. PROOF. Immediate by (2. 13), (4. 12) and definition.

0

The proof of the next result is due to D. Burry. See Green [1971], Proposition 3.41, Erdmann [1977b], Lemma 4.6. LEMMA 4. 14. Let V be an indecomposable R [G] module and let S be a finite extension of R. Let A be a vertex of V. Then A is a vertex of every component of Vs.

PROOF. If H is a subgroup of G and W is an S [H] module, let W denote W as an R [H] module. The definition implies that WG = Wo. Similarly

= (Ms )G for M an R [H] module. . Let W be a nonzero indecomposable S [ G] module such that W I Vs . As V I ( VA )G it follows that Vs I {( VS )A }G and so W is S [A ]-projective. If A is not a vertex of W then there exists a subgroup B of A with I A : B I = P such that W is S [B ]-projective. As W I Vs, W I ( V;) = [S : R ] V. Hence W nV for some integer n > 0 and so V I ( WB )G as (W;t = ( Ws )G contrary to the fact that A is a vertex of V. 0 (MG )s

=

5.

The Green correspondence

For most 6f this section we follow Green [1964] . The following notation be used throughout this section.

will

[5

CHAPTER III

1 16

P is a p -subgroup of 0 and Ii is a subgroup of 0 with No (P) C H. If � is a set of subgroups of 0 write A E o � if A X E � for some x E O. We define the following sets of subgroups. X' = X' (P, H) = {A I A C P n px for some x E O - H}, V = V(P, H) = {A I A C H n px for some x E 0 H}, 21 = 21(P, H) = {A I A C P and A � X' } . Observe that since No (P) C H, X' consists of proper subgroups of P. Thus P E 21. Clearly .r C V · The aim of this section is to define a one to one correspondence between indecomposable R [0] modules with vertex in 21 and indecomposable R [H] modules with vertex in 21 which preserves many properties of modules. Such a correspondence can frequently be used to study various properties of an R [ 0] module by considering the corresponding R [H] module. The next four results are necessary preliminaries. -

°

LEMMA 5 . 1 . Suppose that A is a subgroup of P. The following are

equivalent. (i) A E o X'. (ii) A E X'. (iii) A E V . (iv) A E H V .

THE GREEN CORRESPONDENCE

117

H W EB WI E9 W' EB W� == ( UO )H == U == W EB WI (V) . W' EB W; == O(V).

Hence

( WO )H == W(V).

0

LEMMA 5.3. Let V be an indecomposable R [0] module with vertex A � P. Then there exists an indecomposable R [H] module W with vertex A such that V I WO

O(g) .

and

VH == W(V).

Furthermore

V == O(X')

if and only if

W ==

PROOF. Let U be an indecomposable R [A ] module with V I U o . Let w i U H such that V I W e . Hence W has vertex A. Thus VH I ( WO )H and

by (5.2) ( WO )H == W(V). Hence VH == O(V) or VH == W(V). If V == O(X') then A E o X' and so A E V by (5. 1). Hence VH == W == O(V). If V� O(X') then A � o X' and so A gH V by (5. 1). Thus by (4.6) (ii) VH � O(V) and so VH == W� O(V). 0 LEMMA 5.4. Let W be an indecomposable R [H] module with vertex A C P. Then there exists an indecomposable R [ 0] module V with vertex A such

that w i

VH

V == O(X').

and

WO == V(X').

Furthermore

W == O(X')

if and only if

PROOF. Let WO = E9 V; where each V; is indecomposable. By the

PROOF. (i) � (ii). A Z C P n px for some z E 0, x E 0 H. Thus A P n p z- 1 n pxz - l . Either z - I � H or xz - � H. Thus A E X'. (ii) � (iii). Clear since .r C V . (iii) � (iv). Immediate. (iv) � (i). A Y e H n px for some y E H, x E 0 H. Thus A � P n H y -l n pxy - 1 • Hence A E X' since xy - � H. 0 -

)

�

W EB W',

Let V = E9 V; where each Vi is an indecomposable R [ 0] module with vertex in P. Define the R [H] module f( V) as follows: f ( V) = E9 f( V; ), f( V; ) = (0) if V; == O(X') , (5.5) f( V; ) = W if V; � O(.t') where W is defined as in (5.3). By (5.3) f is well defined on isomorphism classes of indecomposable R [0] modules. By combining (5.2), (5.3), and (5.4) one immediately gets

Hence ( UO )H = it follows from (11.2.9)

THEOREM 5.6 (Green [1964]). Let f be defined by (5.5). Then f defines a

-

)

LEMMA 5.2. Let W be an R [P]-projective R [H] module. Then ( WO )H == W(V).

PROOF. There exists an R [P] module U such that W I U H. Let U I-I W EB WI . By the Mackey decomposition ( W ? )H = WI EB W; for some W', W EB WI EB W' EB W�. Since H n px E V for that ( UO )H == UH (V). Thus

(II.2.9)

x

W�. gH

Mackey decomposition (II.2.9) w i ( WO )H and so w i ( Vi )H for some i. Say w i ( VI )H. If W == O(X' ) then WO == O(X') and so VI == 0 == WO (X'). Suppose that W� O(X'). Then W� O(V) by (5. 1 ) and so ( VI)H � O(V). Since ( WO )H == W(V) by (5.2) it follows that ( Vi )H == O(V) for i > 1 . Thus by (5.3) . VI � O(X') and V; == O(X') for i > 1 . Hence WO == VI � O(X'). 0

( WO )H

=

=

one . to one correspondence from the set of all isomorphism classes of

CHAPTER III

1 18

[5

indecomposable R [G] modules with vertex in 2l onto the set of all isomorphism classes of indecomposable R [H] modules with vertex in 2l . The mapping f has the following properties. (i) If V is an R [P]-projective R [G] module then f( V) == VH (�) and f( V) G V(.f). (ii) Let V be an indecomposable R [ G] module with vertex in 2l . Then V and f ( V) have a common vertex and source. If W is an indecomposable R [H] module with vertex in �l then f( V) = W if and only if V I w G or equivalently W I VI-I. ==

THE GREEN CORRESPONDENCE

119

(iv) Since V * * = V it follows from (11.2.7) that V and V* have a common vertex. By (11.2.6) f( V*) = f( V)*. 0 LEMMA 5.8. Let f be the Green correspondence with respect to (G, P, H).

Assume that .1' = {(1)}. Suppose that V is an R -free R [G] module which is R [P]-projective and that V is indecomposable. Then f( V) = f( V).

The mapping sending V to f( V) defined in (5.5) is called the Green correspondence with respect to (G, P, H). In case G, P and H are specified by the context it is simply called the Green correspondence. The remainder of this section is concerned with investigating further properties of this mapping.

PROOF. If V is projective then f( V) = f( V) = (0). Suppose that V is not projective. By (3. 17) V is R [P]-projective . Thus by (5.6) f( V) == f( V) (g) . Hence f( V) = f( V) EB EB Wi where each W; is an indecomposabl e R [H] module which is R [P]- projective such that W; == O(gJ ). Let A be a vertex of � with A � P. Since W; == O(g) , there exists E H with A Z � H n px for some x E G - H. Thus A � P n p xZ- 1 E .1'. Consequently A = (1) and W; is R -projective. Therefore f( V) = f( V) EB U for some projective module U. By (1. 17. 1 1), U = (O). 0

LEMMA 5.7. Let V, VI , V1 be indecomposable R [G] modules with vertices

LEMMA 5.9 (Green). Let C be a commutative ring. Let Y) be a set of

in P. Let f be the Green correspondence with respect to (G, P, H). Then the following hold. (i) f( VI ) 0 f( V1) == f( VI @ V2) (.f). (ii) f(HomR ( VI, V2» == HomR (f( V I ), f( V2») (.1' ). (iii) If is an automorphism of R [G] with R a = R and G The result follows from (6.7). 0 The following notation will be used throughout the rest of this section. e is a centrally primitive idempotent in R [ G ] . D i s a defect group o f e . N = N o (D ). s is the Brauer mapping with respect to (G, D, N). so(e ) is defined as in (7.2). f is the Green correspondence with respect to (G x G, D!\ N x N). The main results of this section are the following theorems. THEOREM 8.8 (Brauer, Rosenberg). so(e ) is a centrally primitive idempotent

in N with defect group D.

THEOREM 8.9 (Green). D Ll is a vertex of the R [ G x G] module R [G]e and

f(R [G]e) R [N]so(e). =

Before proving (8.8) and (8.9) some preliminary lemmas are needed. LEMMA 8 . 10 . The R [G x G] module R [G]e is R [D x D] projective. PROOF. In (8.7) let A = B = G, e

=

e l = e2 and P = Q = D. 0

G] MODULES

135

Hence fe W) and Ue are R [N x N] modules and fe W) = R [N]so(e ). 0 PROOF OF (8.8) AND (8.9). By (8.5) R [ G ] e has a vertex D g for some p -group

By (8. 10) Dg c;;, o x o D x D. Thus Do c;;, o D. By (4.6), (8. 12) and (8. 13) c;;, o xo Dg. Thus Do = o D. Hence D Ll is a vertex of R [G]e. Let 21 = 21(DLl, N x N) and x X(DLl, N x N) be defined as in section 5. Since every group contained in x has order strictly less than I D I it follows that DLl E 21. Thus by (5.6) a unique indecomposable component of R [G]eN X N has D Ll as a vertex. Thus by (8. 12) and (8.13) the R [N x N] module R [N] soC e ) is indecomposable. Therefore soC e ) is a centrally primitive idempotent in R [N] and by (5.6) f(R [ G ] e ) R [N]so(e ). This completes the proof of (8.8) and (8.9). 0 Do. D Ll

=

=

. THEOREM 8. 14 (Green). Let P be a Sp -subgroup of G with D c;;, P. Then D = P n pz for some z E Co (D ). PROOF. By (8.6) and (8.9) R [G ] e has vertex D Ll. As R [G]e I R [G], every

component of (R [G]e )pxp has vertex (P, P, y ) for some y E G by (8.2) and (8.3). Hence D Ll = pX P (P, P, y ). Therefore by (8. 1) D Ll = (P, P, ) for some E G. Hence {( a, a ) I a E D} = D Ll = (P, P, z ) {( a, a Z ) I a E P n zpz - I } . Consequently D P n zpz - I and z E Co (D ). Thus D = DZ = z

=

R [G]e.

PROOF. By (8. 10) and (8.3) (D, D, z ) is a vertex of R [G]e for some E G. Since D <J G it follows from (8.1) (i) that D Ll is a vertex of R [G ] e. 0 z

D

Ll is a vertex of every indecomposable component of the R [N x N] module R [N] so(e ).

PROOF. Let so(e ) = L ei where {ei } is a set of centrally primitive idempo­

tents in R [N] . By (6.9) and (7.4) D is a defect group of every ei. Hence by (8. 1 1) DLl is a vertex of R [N]ei for all i as required. 0 LEMMA 8.13. R [N]so(e ) I R [G]eN X N.

PROOF. Let H N and W = R [N]so(e ). Let f, g and U be defined as in (7.6). Thus by (7.6) R [G]eN = fe W) EB Ue. Clearly f and g are R [N x N]=

x

z

LEMMA 8. 1 1 . If D <J G then D Ll is a vertex of the R [G x G] module

LEMMA 8. 12 .

R [G

p n pz.

=

0

An alternative proof of (8. 14) can be found in Thompson [1967a] . Refinements can be found in Lam [1976], O'Reilly [1975] . Green [1968] has observed that (8. 14) c�n be used to prove the following result. _

THEOREM 8.15 (Alperin [1967]). There exist Sp -subgroups P, Q of G such

that D = P n Q and Np (D ), No (D ) are both Sp -subgroups of No (D).

PROOF. Let A be a Sp-subgroup of N o (D) and let P be a Sp -subgroup of G with A c;;, P. By (8.14) there exists z E Co (D ) c;;, N o (D ) such that D = P n

pz

Q where Q P Z . Thus N p (D ) = P n N o (D ) A and N o (D) = No (Dy A Z are both Sp -subgroups of N o (D ). 0

n

=

=

=

1 36

-

9. The Brauer correspondence

The results in this chapter are primarily due to Brauer [1956] , [1959] . See also Rosenberg [1961]. Alternative treatments can be found in Hamernik and Michler [1972] ' Michler [1972b], Reynolds [1966b]. Throughout this section S is a finite extension of R such that 5 is a splitting field of S[H] for every subgroup H of G. Thus by (1. 16.1) S is a splitting field of the center of S[ H] for every subgroup H of G and by (1. 16.2) there is a natural one to one correspondence between blocks of S [H] and central characters of 5[H] for every subgroup H of G. Let 0 be a subgroup of G. If h is an 5-Iinear map from Z(S ; G : G) to S define h G : Z(S ; G : G ) � S as follows. If C is a conjugate class of G then h G ( (; ) = L h (ti ) where Ci ranges over all conjugate classes of G with C � C. Clearly h G is an S-linear map. Furthermore if 0 � A � G for some group A then ( h A )G = h G.

Let B be a block of S [0] and let X be the central character of S[ G ] corresponding to B. The 5-linear map X G need not be a central character of 5[G]. If however X G is a central character of S[G] we say that B G is defined and let B G be the block of S [GJ corresponding to X G. In case B G is defined the map sending B to B G is called the Brauer correspondence. LEMMA 9.2. Suppose that 0 � A � G for groups 0 and A. Let B be a block of S [ 0]. If B A is defined and either one of B G or (B A )G is defined then so is the other and (B A )G = B G. PROOF. Clear by (9. 1).

0

LEMMA 9.3. Let be an automorphism of S [G] such that sa = S and = for all E G. Let B be a block of S [ 0] . Then B G is defined if and only if (B O" )G is defined. If B G is defined then (B CT )G = (B G t. x

o-

x

x

a

PROOF. Immediate from the definitions.

0

THEOREM 9.4. Let P be a p -subgroup of G and let H be a subgroup of G with PC G (P) � H � N G (P). If B is a block of S [H] then B G is defined. Let e, e be the centrally primitive idempotents of S [H], S [ G] corresponding to B, B G respectively and let X be the central character of S[ H] corresponding to E. Then i G = X . s and so( e )e = e where s is the Brauer mapping with respect to

(G, P, H).

137

THE BRAUER CORRESPONDENCE

CHAPTER III

-:::;-

-

� ; .; . . If C is a conjugate class of G th �n A G ( C) = A ( C n H) by ; definition. Thus by (6.1 1) A

_

---.:.=... -

____ -__

-

A

X G (C) = A ( C n CP (H» = A {s( C)}.

Thus X G = X . s. Since s is a ring homomorphism it follows that X is a ring homomorphism. Since X G (1) = 1, X G I- 0 and so X G is a central character of S[G]. Thus B G is defined. Since l = X G (e) = X (s(e» it follows that s(e)e = e and so so( e )e = e. 0 C

LEMMA 9.5. Let D � 0 be subgroups of G with C G (D ) � O. Let B be a block of 0 with defect group D. Then B = B #0 for some block B # of N o (D ) and B # G = B G is defined. PROOF. Let H = No (D ). Let s be the Brauer mapping with respect to (G, 15, H). Let e be the centrally primitive idempotent in S [ G ] corre­ sponding to B. By (8.8) so(e) is a centrally primitive idempotent in S [H] . B # = B (so(e» . Since D C (D ) = D C a (D ) � H � N (D ) G G

o it follows from (9.4) that B # G is defined and B # = B is defined. Thus by (9.2) B G is defined. 0 LEMMA 9.6. Let 0 be a subgroup of G. Let B be a block of S [ G ] with defect group D. Suppose that B G is defined. Let D be a defect group of E G. Then 15 � c D.

PROOF. Let X be the central character of 5[ 0] corresponding to B. Thus A = X c is the central character of S'[ G] corresponding to B G. By (6. 10) there exists a conjugate class C of G with defect group D such that A (C) I- O. Thus X (L ) I- 0 for some conjugate class L of q with L � C. Let Do be a defect group of L. Clearly Do � G D. By (6.10) D � G Do. Thus 15 � G D.

0

THEOREM 9.7 (First Main Theorem on Blocks ) (Brauer). Let D be a p-:subgroup of G. Let 6 be a subgroup of G with NG (D ) � G. The map sending B to B G defines a one to one mapping from the set of all blocks of S [ G ] with defect group D to the set of all blocks of S [ G] with defect group D. Before proving (9.7) it is necessary to prove the following purely group theoretic lemma.

1 38

1 39

THE BRAUER CORRESPONDENCE

CHAPTER III

LEMMA 9.8. Let D be a p -subgroup of G and let N = No (D). If C is conjugate class of G with defect group D then C n Co (D) is a class of N with defect group D. Conversely if L is a conjugate class of N defect group D then L = C n Co (D) for some conjugate class C of G defect group D.

'-' Vlt/ U,Ii (1 ) = O. This contradiction establishes result. 0

.'.'1Ul.�H1Ul

3. 16. Let a be a character or a Brauer character of G and let P be a Sp -group of G. Let Cj be the conjugate class of G with Xj E Cj. Let 1]0 be defined as in (1.2) . Then

LEMMA

� (XS, 1]« ) -- ill I G 1 Xs (1 ) � a (xj ) ws ( Cj ). -1

A

If a = Xt then

� (X" 1]0 ) -- ill G 1 Xt (1) � Xs ( Xj ) wt (Cj ) . -I

I

A

PROOF. By definition

combination of irreducible characters of G which vanishes on all p -singular elements. Then (J is a T-linear combination of the cJ>i.

PROOF. By (2.5) and ( 3.4) () = � CicJ>i for complex numbers Ci = «(), ({Ji ) ' . The result follows from (3. 12). 0

Ci. By (3.3)

LEMMA 3.14. Let B be a block. If () = L-s bsXs with bs E K define () B = �Xs EB bsXs. (i) If (J = �s bsXs vanishes on all p -singular elements of G then so does () B • . (ii) �XsEB XS (x )Xs (y ) = 0 for all x, y with x a p '-element in G and y a p-singular element.

() = � CicJ>i for Ci E K. Thus (} B = �i. (ii) Let x be a p '-element in G. Define () = �s Xs (x)Xs. Then by the second orthogonality relation (} ( y ) = 0 for all p-singular elements y in G. The result follows from (i). 0 PROOF. (i) By (2.5) and (3.4),

LEMMA 3.15. Let {3 be a faithful Brauer character of G. Suppose that {3 takes

on exactly n distinct values. Then each ({Ji is an irreducible constituent of one of the Brauer characters {3 0 , f3 I , {3 2 , , f3 n - 1 . •

•

.

f3 and let Aj be the set of p '-elements in G with (3 (x ) = aj. Choose Zj E Aj. If ({Ji is not a constituent of f3 s for s = 0, . . . , n - 1 then by (3.3) �j f3 S (Zj ) �ZEAj cJ>i (Z -I) = O for s = O, . . . , n - l. ' The nonvanishing of the Vandermonde determinant implies that �ZEAj cJ>i (Z -I) = 0 for all j. As f3 is PROOF. Let al, . . . , an be the n distinct values assumed by z

This proves the first formula. If

(XS, 1]0 ) -- ill IG1

�

� '

X

G

a

= XI then

Xs (X )XI (X - I ) _- ill G1

I

�

� '

X

G

Xs (x )Xt (X ) -1

is symmetric in s and t. Thus the second formula follows from the first. 4.

Characters in blocks

The following notation will be used in this section and the next. Cj is the conjugate class of G containing Xj. ( 1 G I ) = a. For each s, 1]s = 1]0 is defined as in (1.2) with a = Xs. 1]8 = �t astXI. v

aSI = als for all s, t. (ii) ast = 0 if xs, Xt are in distinct blocks. (iii) �s astasu = p a atu for all t, u.

LEMMA 4.1. (i)

PROOF. (i) Clear by (3. 16). (ii) By (3.7) (iii)

ast = ( 1]s, XI ) = P a

(� dsi({Ji, 2: dtj({Jj )' = p a .?

for XS, XI in distinct blocks.

•

J

.,J

dsidl/Yij

=0

0

I SO

[4

CHAPTER IV

LEMMA 4.2. Let B be a block and let A be the central character of R [0] corresponding to B. If XS E B then Ws A. Thus XS and XI are in the same block if and only if Ws = Wt. =

PROOF. Let e be the central idempotent in R [0] corresponding to B. Then Ws (e ) = 1 and so Ws (e) r!= O. Since Ws is a central character of R [OJ this implies that Ws = A. 0 LEMMA 4.3. Suppose that B is a block of defect d. Let XS E B. Then Ws (x ) = 0 if x is in a class of defect less than d, and Ws (x ) r!= 0 for some x in a class Co of defect d such that a defect group of Co is a defect group of B. PROOF. Clear by (III.6. 10) and (4.2).

0

p a - d I Xs (1) .

LEMMA 4.4. Let B be a block of defect d and let Xs E B. Then If for some j, XS (x j ' ) � 0 (mod ) then � has defect at most d. If XS (x j ' )ws (Cj ) � 0 (mod ) then Cj has defect d and v (X, (1» = a - d. 17

17

PROOF. By (III.6.S) and (2.3) p a I Cj I xs (x t ) - Ws (Xj ) Xs (l)

_

--

,

-

d Xs (l). I

Since

is an algebraic integer it follows that v ( I � I ) ?= v (Xs (1» ?= a - d if (mod ) Thus Cj has defect at most d. If also Ws ( � ) � 0 (mod ) then by (4.3) a - d � v (Xs ( l» � v ( 1 � I ) � a - d. Thus � has defect d and v (Xs ( l» = a - d. 0

XS (x j ' ) � 0 17

17

.

THEOREM 4.5. Let B be a block of defect d. Then (i) d is the smallest integer such that p , XS (1) for all Xs E B. (ii) d is the smallest integer such that p a - d , 1 . Let n = km. By induction there exists v E Hm,do (C) with do � 4m 2 , v I- 0 such that the one dimensional space spanned by v is fixed by H. If { Xi } is a cross section of H in G and w = TIi (VXi ) E H mk,dok ( C ) then G preserves the one dimensional space spanned by w. The result follows by induction. Thus it may be assumed that V is not induced by any C [ H] module for any subgroup H of G with HI- G. This implies that any normal abelian subgroup of G is central. By Bertrand's postulate there exists a prime p with 2n + 1 < P < 2(2n + 1). Thus p � 4n + 1 and so p - 1 � 4n. Let P be a Sp -group of G. Then P is abelian and P <J G. See Feit and Thompson [1961]. Thus P � Z(G) and so G = P x Go by Burnside's transfer theorem. Clearly . d (G) = d ( Go). By (4.29) d(Go) � (p - 1)n � 4n 2 • 0

&s (1) = Lu cLsXu ( 1 ) � Lu duiXu (1) = cJ>i (1),

SOME OPEN PROBLEMS

0.

�

O

COROLLARY 4.33. Let 0 be a subgroup of G. Let B be a block of G. Suppose that (Xu )o is irreducible for every Xu in B. Then ('Pi )O is an irreducible Brauer character for every 'Pi in B. PROOF. Clear by (4.31). 0 As a consequence of (4.33) we will prove the following results of Isaacs and Smith . For various refinements see Isaacs and Smith [1976], Pahlings

[1977].

. COROLLARY 4.34. Let P be a Sp -subgroup of G and let N = No (P). Let B be the principal block of G. The following are equivalent. (i) G has p- length 1 . (ii) If Xu is in B then (Xu )N is irreducible. (iii) If 'Pi is in B then ('Pi )N is irreducible. PROOF. (i) =? (ii). Since G = Op ',p,p' (G) the Frattini argument implies that G = Op , (G) N. If Xu E B then 0r, (G) is in the kernel of Xu by (4. 12) and so (Xu )N . is irreducible. (ii) =? (iii). This follows from (4.33) . . (iii) =? (i). P = Op (N) is in the kernel of irreducible Brauer characters of N and so P is in the kernel of 'Pi for every 'Pi in B. Thus P � Op',p (G) by (4.12). 0 5. Some open problems

In this section we list some open problems in the theory . A discussion of these and related questions can be found in Brauer [1963]. Some nontrivial examples of modular character tables and Cartan invariants can be found in James [1973], [1978] . These can be used to illustrate the problems below.

1 66

CHAPTER IV

167

SOME OPEN PROBLEMS

(I) If I C I 'Pj (xi )1 'Pj (1) E R is it true that I C I 'Pj (xi )1 'Pj (1) = A ( C ), where is the central character of R [ G] corresponding to the block B ? In answer to an earlier question, Willems [ 1981] has pointed out that

I C I 'Pj (xdl'Pj (1) need not be in R. As an example let p = 2 and let G = II , the smallest lanka group. If Xi is an element of order 3 and 'Pj is chosen suitably then 'Pj (l) = 56 and 'Pj (Xi ) = - 1 . Thus I C I 'Pj (xi )I'Pj (l) - 209/2 � R. See Fang [1974] .

=

(II) Is I G I /'Pj (l) E R for all j ? Willems [1981] has shown that the answer to (II) is yes if and only if I e , and let H = Cc (P) = Cc (y ). Let e be the centrally primitive idempotent in R [H] with B = B (e). By (III.9.4) B O is defined. Let XI E B = B (e) where e is a centrally primitive idempotent in R [ G] and let V be an R -free R [ G ] module which affords XI. Let s be the Brauer mapping with respect to ( G, P, H) and let 0 be the character afforded by V s ( e ). By (IIL4. 1 1) and Nagao's theorem (IIL7.5), XI (xy ) = O (xy ) for all p '-elements x in H. Since d i; I: 0 this implies that 'P r is afforded by an irreducible constituent of V s ( e ) . Hence s ( e ) e = e. Therefore B C = B by (IIL9.4). The last statement follows from (2.4). D

HO

HO

1 73

HIGHER DECOMPOSITION NUMBERS

CHAPTER IV

1 72

o

For any p -element y the p-section containing y or simply the section containing y is the set of all elements in G whose p -part is conjugate to y. Clearly the p -section containing y = 1 is the set of all p '-elements in G. Let

be a complete set of representa tives of the p '-classes in Cc (y). It is easily seen that {yx n is a complete set of representat ives of the conjugate classes in G which lie in the section containing y. Let CY = (c n denote the Cart an matrix of C c (y ). 6.2. Let B be a block of G. Let y, z be p-elements in G. (i) If Y is not conjugate to z in G then for ali i, j

LEMMA

Ls (d;i )* d :j XLsEB ( d ;i ) * d :j = O. (ii) Ls ( d ;i ) * d ;j = c� for all i, j. =

(iii) LxsEB (d ;;) * d ;j = 0 unless 'P r and 'P J lie in the same block B of Cc (y) and B C = B. In that case LxsEB (d ;; ) * d ;j = c � . By definition

('P r(x J )) * ' (d ;; )* ' (d :; ) ('P f (x j )) = (Xs (yx J )) * ' (Xs (zx j )) =

( � Xs (yx r)*Xs (zx j )).

If y i� . not conjugate to z then Ls xs (yx r)* Xs (zx j ) = 0 for all i, j. By (3.6) ('P r(x J )) is nonsingular. This implies one of the equations in (i). Since

Ls Xs (yx r)*Xs (yx J )

=

I Cc (YX r ) I Oij = I C Cdy) (x r) l oij

it follows from (3.8) that

( � Xs (yx r)*Xs (YXn ) = ('P r(x n) * '( c �) ('P i (x J )).

Hence the non singularity of (cp i(x J )) implies (ii). Fix i, j. Let B" B 2 respectiv ely be the block of Cc (y ), Cc (z ) respectively with 'P r E BI and 'P J E B 2 • If ( d ;i ) * d :j I: 0 for some s then by (6. 1) B � B ? and XS E B �. Thus

=

L

X,EB

(d�i) * d:j = L (d;i)*d:j =0

otherwise.

This completes the proof of (i), and by (ii)

L

XsEB

(d;i)*d ;j = c � if B � = B ? = B =0

otherwise.

CHAPTER IV

1 74

Since c � = 0 if B I -I B 2 , (iii) follows:

k

0

The next result is a refinement of (3. 14). See Osima [1960b] . LEMMA 6.3. Let B be a block. If () = Ls bsXs with bs E K define (} B = Lxs E B bsXs. (i) If () vanishes on a p -section of G then (} B vanishes on that p-section. (ii) If Z I and Z 2 are in different p-sections of G then

L XS (ZI) * Xs (Z 2) = o.

Xs E B

PROOF. (i) Suppose that () vanishes on the p -section of G which contains the p -element y. Then LS,i bsd ;i 'P r (x ) = (} (xy ) = 0 for all p '-elements x in C G (y ). Thus Ls bsd �i = 0 for all i. Hence by (6. 1) L bsd �i = 0 for all i and Li bsd ;i 'P r (x ) = 0 for every p I-element x in C G (y ) as so () (xy ) = L B

xs E

B

xs E B

required. (ii) Let (} = Ls Xs (Z I)*Xs. Then (} (z ) = O for every element Z in the p -section containing Z 2 . The result follows from (i). 0 (6.3) can be used to generalize (4.23) (ii) as follows. LEMMA 6.4. Let p and q be distinct primes. Suppose that y is a p -element in G and x is a q -element in G such that no conjugate of x commutes with any

conjugate of y. (i) Let B (p ) be a fixed p-block of G and let B (q ) be a fixed q -block of G. Then L Xs (x )Xs (y ) = 0 where Xs ranges over all the irreducible characters which lie in B (p) and also in B (q ). (ii) There exists a nonprincipal irreducible character which is in the principal p-block and also in the principal q -block. Xs E B (q ) Xs (x )Xs. By assumption no element of G is in the p -section containing y and also in the q -section containing X - I . Thus by (6.3) (ii), (} (z ) = 0 for every element Z in the p -section containing y. Hence B by (6.3) (i) (} (p )(y ) = O. (ii) Immediate by (i). 0

PROOF. (i) Let () = L

The following notation will be used in the rest of this section . B I , B 2 , are all the blocks of R [ G ] . {Yi } i s a complete set of representatives o f the conjugate classes i n G consisting of p -elements. Thus {YiX ;i} is a complete set of representatives of the conjugate classes in G. •

• •

1 75

HIGHER DECOMPOSITION NUMBERS

[6

is the number of conjugate classes in G.

k(Bm ) is the number of irreducible characters in Bm. I (Yi ) is the number of conjugate classes in C G (Yi ) consisting of p '_

elements.

I (y;, Bm ) is the number of irreducible Brauer characters of C G (Yi ) which lie in blocks B of C G (Yi ) with B G = Bm. X = (Xs (YiX r)) is a character tab Ie of G. If y is a p -element in G then 'PY = ('P i(xI)) is a table of irreducible �rauer characters of C G (y ). 'P is the direct sum of the matrices 'PYi• D = (d�;) where s is the row index and (i, j) is the column index. Dm is the submatrix of D where s ranges over all values with Xs E Bm and for each i, j ranges over values for which 'P;i is in a block B of C G (Yi ) with = Bm. Thus D m is a k (Bm ) x ( L; l(y;, Bm )) matrix. If T is a ring of algebraic numbers in K and S is a subset of G define

fJ G

{ a I a = � asXs, as E T, (z ) = 0 for Z S } . ChT (S, Bm ) = { a I a = L asXs, as E T, a (z ) = 0 for Z s}.

, ChT (S) =

a

g

g

Xs E Bm

In case T is the ring of rational integers write Ch(S ) = ChT (S) and Ch(S, Bm ) = ChT (S, Bm ). LEMMA 6.5. (i) I(y; ) = Lm I(y;, Bm ) for all i. (ii) k = Lm k (Bm ) = L; l(y; ). (iii) X , 'P, D are all k x k matrices and X = D'P if the rows and columns are

suitably arranged.

PROOF. (1II.2.8) and (6. 1) irnply (i). (ii) and (iii) are clear by definition. 0 LEMMA 6.6. (i) D is the direct sum of the matrices Dm after a suitable

arrangement of rows and columns. (ii) Let T be a field of algebraic numbers in K. Then for all m dimT (ChT ( G, Bm )) = k (Bm ) =

L l(y;, Bm ) i

(iii) Dm is a nonsingular k (Bm ) x k (Bm ) matrix and ± (det Dm ? is a power of p for each m. (iv) Let y be a p-element. There exists a block B of C G (y ) with B G = Bm if and only if y is conjugate to an element of the defect group of Bm.

1 76

CHAPTER IV

PROOF. (i) Immediate by (6. 1). By definition dimT (ChT ( G, Bm )) k (Bm ). By (6.5) D is a square matrix. As D is nonsingular each Dm is a square matrix. Thus in view of (i) both (ii) and (iii) will follow once it is shown that ± (det Dm )2 is a power of p. The mapping which sends every element of G into its inverse permutes the rows and columns of D and sends Dm to D!. Thus (det Dm )2 ± det(D!'Dm ). By (6.2) D!' Dm is the direct sum of Cartan matrices. Hence ± (det Dm ? is a power of p by (3.9). (iv) Suppose that Y is not conjugate to any element of the defect group P of Bm. By (2.4) d �i = 0 for all XS E Bm. Since Dm is nonsingular, this implies that Bm -I jj 0 for any block jj of Co (y). Suppose that y is conjugate to an element of P. It may be assumed that y E P. By (III. 6. 10) and (4.2) there exists a p '-element x E Co (P) such that for XS E Bm, � -I 0, where x is in the conjugate class C. Choose XS of height 0 in Bm. Thus XS (x ) -1 0. Since x E Co (P) it follows that x E Co (y) and XS (xy ) -1 0, and so Xs (xy ) -1 0. Hence d �i -1 0 for some i. The result follows from the second main theorem on blocks (6. 1). 0

'rationals, where h ranges over all the integers with 1 � h � I G I and (h, I G I ) = 1 and ah is the image of a under the automorphism which sends , Zn be a complete set of a I G I th root of unity onto its h th power. Let z representatives of the conjugate classes in S. For i = 1, . . . , n define (}i = Ls asiXs where asi = Lh ahXs (z �h). Thus each asi is a rational integer. If Z E G and Z is not conjugate to z ;n for any m with (m, I G I ) = 1 then (}i (Z ) = Lh ah Ls Xs (Z � h )Xs (Z ) = O and so Oi E Ch(S). Suppose that Lj bjOj = 0 for rational integers bj• Then for each i

==

1

1,

=

•

LEMMA 6.9. Let T be a subfield of K consisting of algebraic numbers. Let S be a union of p -sections in G. Assume that at least one of the following assumptions holds. (i) T contains a primitive ( I G Ip )th root of unity. (ii) If Z E S then Z h E S for all integers h with (h, I G I ) = 1. Then dimT (ChT (S, Bm )) = L Y; ES I( Yi, Bm ) for all m. PROOF. In view of (6.8) dimT (ChT (S)) = L Y;ES I (y; ) under either assump­ tion. Thus if the result if false then for some m ', dimT (ChT (S, Bm, )) < 2: I( Yi, Bm, ) y;ES by (6.7) (ii). Hence by (6.7)

l(y;, Bm ) dimT (ChT (S)) = 2: m dimT (ChT (S, Bm )) < 2: m 2:

0

PROOF. Clearly rank(Ch(S)) � n. Thus it suffices to exhibit n linearly independent elements in Ch(S). Let {ah } be a normal basis of the field of I G Ith roots of unity over the

.

The following refinement of (6.6) (ii) is due to W. Wong [1966] .

PROOF. (i) Since G - S is a union of p -sections this is clear by (6.3) (i). (ii) Suppose that Ls asXs E ChT (S, Bm ). Then Ls,j asd �j 'P J(x ) = 0 for all Y = Yi g S and all p '-elements x E Co (y ). Hence Ls asd �j = 0 for all Y = Yi g S and all j such that 'P ] is in a block jj of C o (y ) with jj O = Bm. Thus the vector (as ) satisfies L Y;ES I ( Yi' Bm ) homogeneous linear equations. By (6.6) (iii) these equations are linearly independent. Therefore by (6.6) (ii)

LEMMA 6.8 (Suzuki [1959]). Let S be a union of n conjugate classes in G. Assume that if Z E S then z h E S for all integers h such that (h, I G I ) = 1 . Then rank(Ch(S)) = n.

•

where in the last sum U, h ) ranges over all pairs with Zi conjugate to Z 7Since {ah } is a basis of the field of I G I th roots of unity over the rationals this implies that bi = O. Thus {(}i } is a linearly independent set. 0

LEMMA 6.7. Let T be a subring of K which consists of algebraic numbers. Let S be a union of p-sections in G. Then (i) ChT (S) = Eft ChT (S, Bm ). (ii) If T is a field then dimT (ChT (S, Bm )) � LY;ES I ( Yi, Bm ) for all m.

dimT (ChT (S, Bm )) � k (Bm ) - 2: I( Yi, Bm ) = 2: / ( Yi, Bm ). y;ES y;ES

177

HIGHER DECOMPOSITION NUMBERS

y;ES

= 2: l(y; ) = dim; (ChT (S)). y;ES -

This contradiction establishes the result.

0

Xs is a p-conjugate of XI if XI = X � for some automorphism of the field of I G Ith roots of unity over the rationals which leaves the p'-th roots of unity fixed. Two p-conjugate characters have the same irreducible Brauer (J"

1 78

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CENTRAL IDEMPOTENTS AND CHARACTERS

characters as constituents and so lie in the same block. XS is p-rational if it has no p-conjugate other than itself. LEMMA 6. 10. Let M be a group of automorphisms of the field of I G I th roots of unity over the rationals such that the p ' th roots of unity are in the fixed field of M. Distribute the irreducible characters in Bm into families of conjugates under the action of �\1. Let ni be the number of irreducible characters in the ith family. Then the following hold. (i) The number of orbits of the columns of Dm under the action of M is equal to the number of families in Bm. (ii) If P � 2 or if the 4th roots of unity are in the fixed field of M then after suitable relabeling the ith orbit of columns of Dm consists of ni columns.

a

PROOF. If P � 2 or if the 4th roots of unity are in the fixed field of M then M is cyclic. Since Dm is nonsingular by (6.6) (iii) the result follows from a combinatorial lemma of Brauer. See Brauer [1941c], Lemma 2.1 or Feit [1967c], (12. 1). 0 COROLLARY 6. 1 1 . Suppose that p � 2. The number of p-rational irreducible characters in Bm is at least as great as the number 1(1, Bm ) of irreducible Brauer characters in Bm.

1 79

Xs � ( Z ) Z . es - I G(l)I Z� G XS LEMMA 7.2 (Osima). Let B be a block of R [G ] and let e be the centrally primitive idempotent in R [ G] corresponding to B. Then -1

_

PROOF. Xs E B if and only if ws (e ) � O. Thus by (7. 1) e=

X�B es = I � I Z�G { X�B Xs (l )Xs ( ) } Z-

l

Z.

Thus the first equation follows from

In particular this implies by (2.5) that the coefficient of any p-singular element in e is zero. If is a p '-element then z

PROOF. Since the ordinary decomposition numbers are rational Dm has at least 1(1, Bm ) rational columns. The result follows from (6. 10). 0 Some results concerning the fields generated by the entries of X, 'P and D can be found in Reynolds [1974] . 7. Central idempotents and characters

Some of the formulas in this section have been used by Osima as the starting point for an alternative development of much of the theory of blocks. For this approach see Osima [1952b], [1955], [1960a], [1964], Iizuka [1956] , [1960a], [1960b], and Curtis and Reiner [1962] . The following result is well known and easily verified. LEMMA 7. 1 . Let es be the centrally primitive idempotent of K[ G] corre ­ sponding to Ws . Then

This proves the second equation. D LEMMA 7.3. Let e be a centrally primitive idempotent in R [G]. Let B be the block corresponding to e and let A be the central character of R [ G ] corresponding to B. For each j let G denote the conjugate class of G with Xj · E G. Then e = Lj aj¢ with aj E R. If iijA ( G ) � 0 then B and G have a common defect group.

PROOF. By (7.2) e = Lj aj¢ with aj E R. Let D be a defect group of B and let � be a defect group of G. If iij � 0 then Pj � G D by (III.6.5) since e E ZD (G ; G). If A (G) � 0 then D � G Pj by (III. 6. 10). D The next result is due to Osima [1966] in case character.

a

is the principal Brauer

[7

CHAPTER IV

1 80

THEOREM 7.4. Let e be a centrally primitive idempotent in R [ G ] and let B be the block corresponding to e. Let Xr be a character of height 0 in B. Let a be a rational integral linear combination of {'Pi I 'Pi E B } with v(a (l)) v(Xr (l)). Let 11« be defined as in (1.2) and let 11a = Ls bsX,,· Define =

1 &ill " 11« ( Z ) z. br z�O Then fa E R [ G] and e - la E J(Z(G : G)). f J«

-1

- IG1

_ _

8]

SOME NATURAL MAPPINGS

181

. where X; E C, Xj E �, Xs E Cs. If j G lp ;:= P Q then 1 C I I � I 0 (mod p Q ) but I G 1 X (l) � 0 (mod p2Q ). Hence C�e = O. Therefore {M(l - eo)}2 = O. Since MeM(l - e) = 0 this implies that M 2 = (Meot Since ei c�rresponds to a block Bi of defect 0, Bi contains a unique irreducible R [ G ] module and Ze; = Rei. Hence Me; = Zei = Rei and so Mea = E9�= Rei. Therefore ==

==

1

R e- 2 = ffi R-e-·,. 0 M2 = ffi W ; =1 i=1 , W n

n

PROOF. By (4.7) fa E R [ G ] . By (4.1) and (7. 1) Jfa

8. Some natural mappings

- &.ill " � &D1 " s ( Z - I ) Z z� x - b, x�B XS (1) 1 G 1 O

� LB &ill br Xs ( 1 ) es X;E where es is the centrally primitive idempotent in K[ G] corresponding to, OJs• Thus if XS � B then OJs (e - f:, ) = OJ" (e ) - OJs (fa ) = O. If XS E B then =

� OJs ( e Jfa ) -- OJs ( e ) OJs If V a ) -- 1 X'b(1) , XS (1) ' Hence by (4.7) OJ" (e - fa ) 0 (mod ) Thus Ws (e - fa ) 0 (mod 1T) for all s and so A (e - 1) = 0 for every central character of R [ G ] . 0 _

_

_

==

1T .

==

The next result is due to Iizuka and Watanabe [1972], who also prove further results of a similar nature concerning blocks of positive defect. This result generalized earlier results of Brauer [1946a], [1956] . THEOREM 7.5. Let Z be the center of R [ G ] and let M be the subspace of Z spanned by all C such that C is a conjugate class of defect O. Then M is an ideal of Z and the number of blocks of defect 0 is equal to dimR M2 . PROOF. By (111.6.2) M is an ideal of Z. Let eI, . . . , en be all the centrally primitive idempotents of R [G] corresponding to blocks of defect O. By (7.3) e; E M for i = 1, . . . , n. Let eo = L�= I ei. Then M = Meo EB M(l - eo). We next show that M2 = (Meot Let C, � be classes which have defect O. Let e be a centrally primitive idempotent of R [ G] which corresponds to a block B of positive defect. Then by (7.1) Xi ) (Xj)X (xs) c" C C".C.e = L L 1 I I � I X ( X G1 ( ) S

XE B

1

X l

To a large extent this section consists of results which are reformulations of some results proved earlier in this chapter . We will follow Broli(� [1976b], [1978b] quite closely. See also Serre [1977]. The notation introduced in section 6 after (6.4) will be used with the following modifications. T will denote an arbitrary subring of K. Instead of ChT (S, B ) we will write ChT (G : S, B ) to emphasize the dependence on G. We will also write ChT (G : G) = ChT (G). Greg denotes the set of all p i-elements in G. The Grothendieck algebra A �(K[ G D is isomorphic to Ch( G) in the obvious way. By (3. 12) Ch(G : Greg ) is isomorphic to the Grothendieck algebra A � (R [ G D. Let Ch T,proj (G) denote the ring of all T-linear combi­ nations of { 4\ }. Let ChT,proj (G : B ) = ChT,proj (G) n ChT (G : B ). As usual the subscript T will be omitted in case T = Z, the ring of rational integers. Define the linear map b = b O, c = c o, d = d O as follows: where b ( CPi ) = Lu du;Xu, b : ChT,proj (G) � ChT (G), c : ChT,proj ( G) � ChT (G : Greg ), where C ( CPi ) = Lj Cii'Ph where d (Xu ) = Li dui'Pi. d : ChT (G) � ChT ( G : Greg ), By (3. 1) the following diagram is commutative. Ch T,proj (G)

Y

� d

ChT ( G) � ChT (G : Greg )

It is clear from the definitions that b, c, d respect blocks. In other words

. (d(O)) B = d ( O B ) for 0 E ChT (G). Similarly (b (O)) B = b ( O B ) and (c (O)t = c( 0 B ) for 0 E Ch T,proj ( Q). Furthermore if X E Ch T ( G) and cP E Ch T,proj ( G)

then by (3.4)

[8

CHAPTER IV

1 82

(8. 1) ( d (X ), Ith root of 1 , (y ) such that ?;( yx ) = (y )?;(x ) for x E Cc (y ). Define the linear map Wy of ChT (C c (y )) by Wy (?;) = ( y )?; for any irreducible character ?; of C c (y ). Then for any ?; E ChT (C o (y )) and any E Cc (y ) it follows that (wy (?;)) (x ) = ?;( yx ). For any subgroup H of G let Res� : ChT (G) � ChT (H) be defined by Res�(e ) = eH• Let Ind� : ChT (H) � ChT (G) be defined by Ind�(e) = e C. Let c Y' c denote the map c = c CG( Y ) for the group C c (y ). Define Wi:

Wi:

Wi:

x

LEMMA 8.2. (i) The following diagram is commutative. ChT,proj (Cc (y ))

bY:/

ChT (G)

--�)

�

ChT (Cc (y) : C c ( Y ) reg)

(ii) If Y/ E Ch1�proj (Cc (y)) then b y,C ( Y/ ) vanishes outside the p -section which contains y and it is a class function such that (b Y' c ( y/ )) (yx ) = y/ ( x ) for x E C c ( Y )rcg . (iii) If e E ChT (G) then d Y' c ( e ) is an R -valued function on Cc (y ) defined by (d y, C (e )) (x ) = e (yx ) for x E Cc ( Y ) rcg . PROOF. Statements (ii) and (iii) are direct consequences of the definitions. Then (i) follows from (ii) and (iii). D LEMMA 8.3. If e E ChT (G) and Y/ E ChT,proj (Cc (y )), then ( d Y C (e ), Y/ )cG( Y ) = (e, b Y' c (y )) c . '

PROOF. By (8. 1) and the Frobenius reciprocity theorem (III.2.5) (d Y' C (0 ), Y/ )cdY ) = (Resgo (Y ) ( 0 ) , (wy f l o b Co( Y ) (y ))Cd Y ) = ( 0, b Y'C (y/ )) c . D

SOME NATURAL MAPPINGS

1 83

Let B be a block of G and let 13 1 , 132, • • • be all the blocks of C c (y ) with B F = B. If ?; is a class function defined on Cc (y ) let ?;B = Li ?;B;. The next result is equivalent to the second main theorem on blocks (6. 1). THEOREM 8.4. Let 0 E ChT (G) and let Y/ lowing hold. (i) b Y' o ( 1 satisfies (i), (ii), (iii) of (10.1) with Ui = 1 for all i. PROOF. Clearly l o / p E -A �l) (G/P) and satisfies (i), (ii), (iii) of (10. 1). The result follows from (4.26). D A much more important class of examples arises from the work of Steinberg [1968]. He showed that if G is a simple group of Lie type then the Steinberg character of G satisfies condition (i) of (10.1) with Ui = ± 1 for all i. This motivated Ballard [1976] and Lusztig [1976], who indepen­ dently showed that the Steinberg character satisfies condition (iii) of (10.1). Their work led to (10.1). It is not always true that A �l) (G) is a principal ideal of A O( G). It can be shown by explicit computation that for p = 5 and G = PSL (9) Ao there is no YJ in A �l)(G) which satisfies the conditions in (10.1). Also if G = II , the first Janko group, then it is not difficult to verify that no 1] in A �l ) (G) satisfies the conditions of (10.1) for any prime p 1 1 11 1 . The proof of (10. 1) i s based o n the following elementary result. 2

A O ( G) = A � (R [ G]) is the ring of all integral linear combinations of generalized Brauer characters of G. The ideal A �1) (G) of A O( G) can be identified with the ring of integral linear combinations of characters afforded by projective R [G] modules. The main result of this section gives a criterion for an element to be an ideal generator of A �1) (G), and hence in particular a criterion for A �I)(G) to be a principal ideal. The argument is from Feit [1976]. See also Alperin [1976e] . THEOREM 10. 1 . Let X I , . . . , Xm be a complete set of representatives of all the conjugate classes of G consisting of p '-elements. Let v( I C G (xi ) l ) = ai for 1 � i � m. Let YJ E A �o< G). The following are equivalent. (i) For 1 � i � m, YJ (Xi ) = P aiUi, for some unit Ui in some algebraic number field. (ii) rr �= I YJ (Xi ) = ± rr�= l p ai. (iii) For all E A �l) (G) there exists 'Ya E A O ( G) with YJ'Ya = Furthermore if YJ satisfies (i), (ii), (iii) then 'Ya is uniquely determined by ll'

THE RING A�(1� [ G])

10]

ll' .

ll'.

:=

LEMMA 10.4. Let 0 be a complex valued class function on G. Let y , Ym be a set of pairwise nonconjugate elements of G such that O(Yi ) -I 0 for 1 � i � m. Let X be the complex vector space consisting of all complex valued class functions f on G such that f(y ) = 0 if Y is not conjugate to any Yi. Let L ( 0 ) be the linear transformation on X defined by L ( O)f = Of. Then dim X = m, the characteristic roots of L (O) are O (YI), . . . , O(Ym ) and det L (O ) = �;: 1 O (Yi). 1,

•

•

•

PROOF. Define fi E X by f; (Yi ) = Oii. Then {fi } is a basis of X and so dim X = m. Furthermore L (O)f; = O(Yi )f; for all i. Thus L (O) has the required characteristic values and so the required determinant. D

1 88

CHAPTER IV

[11

PROOF OF (10 . 1) . Let Pi be a Sp -group of C G (Yi ). Since yt E A 71) (G) it E A 71) « Yi > X Pi ). Thus yt (Yi ) = p Q;Ui for some algebraic follows that yt integer Ui . Therefore n�= 1 yt (Yi ) = n�= 1 Ui n�= l pQ;. Clearly n;: 1 yt (Yi ) is rational valued and so n�= 1 Ui is a rational integer. The equivalence of (i) and (ii) is now clear. Condition (iii) holds if and only if ytA O( G) = A 71) (G). Since ytA o( G) � A 71) ( G), this holds if and only if I A O( G) : ytA O( G) I = I A O( G) : A 71) ( G) I. By (10.4) yt (Yi ) = ± I1 Ui IT p Q;. I A O(G) : ytA O(G) I = I det L (y ) I = ± I1 i=1 i=1 =1 By definition I A O( G) : A 71) (G) I = det C, where C is the Cart an matrix of G. By (3. 1 1) det C = n�= 1 p Q;. Thus condition (iii) holds if and only if n�= 1 Ui = ± 1 , which is equivalent to condition (i). If yt satisfies (ii) then yt (Yi ) -I- 0 for all i. Thus Ya (Yi ) = a ( Yi )yt (Yi t l and so Yo is uniquely determined by a. D (y; ) x P;

i

1 1 . Self dual modules in characteristic 2

Throughout this section F is a field of characteristic 2 and V is an absolutely irreducible F[ G] module which affords a real valued Brauer character. Equivalently V = V*. V is of quadratic type if there is a nondegenerate G-invariant quadratic form on V. V is of symplectic type if there is a nondegenerate G-invariant alternat­ ing form on V. We will show that V is essentially always of symplectic type and then give some criteria for V to be of quadratic type. The results in this section are primarily due to Willems [1977] . However the first two results are older. See e.g. Fang [1974] Lemma 1 or James [1976] Theorem 1 . 1 . We will first introduce some notation. A is the representation of G with underlying module V. E is the space of all matrices on V where X � A (x )' XA (x ) for x E G where the prime denotes transpose. Thus E is an F[ G] module with E = V Q9 V = V @ V* by (II.2.8). S = {X E E l X = X'}, T = {X + X' I X E E}. Clearly T � S and S, T are both F[ G] modules. Vo(G) is the F[G] module with dimF Vo(G) = 1 and InvG ( Vo (G» = Vo (G).

SELF D UAL MODULES IN CHARACTERISTIC 2

11]

1 89

THEOREM 1 1 . 1. If V;;6 Vo ( G) then V is of symplectic type. PROOF. Since V = V* there exists M E E with M- 1A (x )M = A (x - I ), for all x E G. Thus M'A (x )'(M't l = A (x - 1 ) and so A (X - I), = (M't I A (x )M' for all x E G. Hence M'M- I A (x ) = A (x )M'M - 1 for all E G. Thus by Schur's lemma M' = cM for some c E F. Thus M = cM' = c 2 M. Hence c 2 = 1 and so c = 1 . Therefore M' = M. Furthermore A (x )'MA (x ) = M for all x E G. Thus M defines a G-invariant symmetric bilinear form on V. Let W be the subspace consisting of all isotropic vectors in V. Thus W is a G-invariant subspace and so W = (0) or W = V. If W = (0) then dimF V = 1 and V = Vo (G). If W = V then M defines a nondegenerate symplectic form on V. 0 x

COROLLARY 1 1 .2. -Let 'P be a real valued irreducible Brauer character of G in characteristic 2. Then either 'P = 1 G or 'P (1) is even. PROOF. Let U be an F[ G] module afforded by 'P. If U is of symplectic type then 'P (1) = dimF U is even. The result follows from ( 1 1 . 1) with U = V. 0 For i < j let X;j denote the matrix in E with 1 in the (i, j) and U, i ) entries and 0 elsewhere. Let Xii denote the matrix with 1 in the (i, i ) entry and 0 elsewhere. Then {Xij I i � j } is an F-basis of S and {Xij I i < j } is an F-basis of T. By (1.19.3) there exists a finite subfield Fo of F and an Fo[G] module Vo such that V = ( VO)F. Let V�2) denote the Fo[ G] module derived from V by applying the automorphism of Fo which sends a to a 2. Let V(2) = ( V�2 ) )F. LEMMA 1 1 .3. (i) There exists a submodule Eo of E with EIEo = Vo(G). (ii) EIS = T. (iii) S IT = V(2) .

PROOF. (i) Let E 1 denote the space of all matrices on V where X � A (X - l )XA (x ) for x E G. Then as F[G] modules EI V Q9 V* = V Q9 V = E. Let E2 be the set of all scalar matrices in E. Then E2 = Vo( G) is a submodule of E 1 • The result follows as El = E f. (ii) Define f : E � T by f(X) = X + X'. Then f is an F[ G]­ homomorphism of E onto T with kernel S. (iii) Without loss of generality F may be replaced by its algebraic closure. �

1'

I' I i.

190

[11

CHAPTER IV

Thus the map sending a to a 2 is art automorphism of F. Let x E G and let A (x ) = ( aij ) . Then A (x )' XssA (x ) = ( aSiasj ) = L a ;i Xii + X 0

Let M = ( mij ) be an upper triangular matrix in E, i.e. Define the quadratic form OM on V by

OM (v) = L mijXiXj where i,j

v =

mij =

0 for i > j.

(xt, . . . , Xn ) .

Define fM E HomF (S, F) by

fM « Xij )) = L mijXij. i,j

LEMMA 1 1 .4. Let M be an upper triangular matrix in E. Then OM is G-invariant if and only if fM E HomF[ G ] (S, F) where F = Vo(G).

PROOF. Let v = (X l , . . . , Xn ) E V and let A

= ( aiJ E E.

Then

OM (vA ) = L mijaisajtXsxt. i,j,s,t Thus OM (vA ) = OM (v ) if and only if the following two equations are satisfied for all s < t. L mijaisajs = msS,

(1 1 .5)

i,j

L mij ( aiS ajt + ajS ait ) = i,j

SELF DUAL MODULES IN CHARACTERISTIC 2

191

THEOREM 1 1 .7. V is of quadratic type if and only if there exists a submodule So of S with S/So = Vo(G). PROOF. Every quadratic form on V is of the form OM for some upper triangular II).atrix M. Every map in HomF (S, F) is of the form fM for some upper triangular matrix M. Hence the result follows from (11 .4). 0

i

for some X E T.

1 1J

COROLLARY 1 1 .8. If V is not of quadratic type then there exists a nonsplit exact sequence O � Vo (G) � W � V � O. PROOF. By (11 .7) Vo(G) is not a homomorphic image of S. Thus by (11 .3) (i) and (ii) there exists an exact sequence O � To � T � Vo ( G ) � O. Thus by (1 1 .3) (iii) there exists an exact sequence O � Vo (G)� S/To � V (2) � 0. As Vo (G) is not homomorphic image of S it follows that the sequence is not split. Hence there also exists a nonsplit sequence of the required form. 0 a

COROLLARY 1 1 .9. If V is not in the principal 2-block then V is of quadratic type. PROOF. Clear by (1 1 .8). 0 COROLLARY 1 1 . 10. If G is solvable then V is of quadratic type.

mst .

(11 .6)

For all s < t

fM (A I XssA ) = L mijaisajs, i,j

fM (A ' XstA ) = L mij ( aiS ajt + aitajs ) . i,j Thus fM (A ' XA ) = fM (X) for all X E S if and only if (1 1 .5) and (11.6) are satisfied. Consequently fM (A IXA ) = fM (X) for all X E S if and only if OM is A -invariant. 0

PROOF. Induction on ' G ' . If ' G ' = 1 the result is clear. Suppose that ' G ' > 1 . If V is not in the principal 2-block the result follows from (11 .9). If V is in the principal 2-block then O2',2 (G) is in the kernel of V by (4. 12). As O2',2 (G) I (1) the result follows by induction. 0

1]

193

SOME ELEMENTARY RESULTS

PROOF. Let {L; } be the set of all conjugate classes of H and let E L;. Let nx ) = 0 if x E G - H. If C is a conjugate class of G with E C the definition of induced characters implies that Z;

z

CHAPTER V

Thus

=

The notation and assumptions introduced at the beginning of Chapter IV will be used throughout this chapter. Also the following notation will be used. If B is a block of G then AB is the central character of R [ G ] corresponding to B. v is the exponential valuation defined on K with v (p ) = 1 . Xl, X2, .. are all the irreducible characters of G. 'P I, 'P2, . . . are all the irreducible Brauer characters of G. WS is the central character of K[ G ] corresponding to XS. If H is a subgroup of G and h : Z(K; H : H�K is linear, define G h : Z(K ; G : G�K by � h G (C) = h (C n H) for any conjugate class C of G. Some of the results in this chapter can be proved without the assumption that K and R are splitting fields for every subgroup of G. See for instance Brow§ [1972] , [1973] ; Hubbart [ 1972] ; Reynolds [197 1 ] . A

1 . Some elementary results

For the results in this section and the next see Brauer [1967], Fong [1961 ] and Reynolds [1963]. LEMMA 1 . 1 . Let H be a subgroup of G and let � be an irreducible character of H. Let � G = �s asXs and let w be the central character of K[H] corresponding to �. Then

L asXs (l)ws s

= �G (l) w G .

I G : H I � ( I ) L� c w (L; ) = � G (I) w G ( C). . �

0

LEMMA 1 .2. Let H be a subgroup of G and let Bo be a block of H. Assume that Bo contains an irreducible character � such that �G is irreducible. Then B� is defined and �G E B � .

PROOF. Let �G = XS and let w be the central character of K[H] corre­ sponding to �. By (1 . 1) Ws = w G . Thus w G = Ws is a central character of R [ G] . 0 LEMMA 1 .3. Let Bo be a block of the subgroup H of G for which B� is defined. Let � be an irreducible character in Bo and let �G = �s as Xs . If B is a

block of G then

v( � a X (1) ) > ( � a X (1)) X

v

X

B

s s

V (� G (1»

B

s s

= v ( �G (1»

if B -I- B�, if B

=

Bf

PROOF. Let w be the central character of K[H] corresponding to �. Let e be the central idempotent in R [ G ] corresponding to B. Then Ws (e ) = 1 if XS E B, and Ws (e ) = 0 if XS g B. Also w G (e) 0 if B -I- B � and w G (e) = 1 if B = B �. The result now follows from ( 1 . 1). 0

=

COROLLARY 1 .4. Let Bo be a block of the subgroup H of G for which B � is defined. Assume that Bo and B � have the same defect. If � is an irreducible character of height O in Bo then some character XS of height 0 in B � occurs as a constituent of �G with multiplicity as ¢ 0 (mod p ). PROOF. Clear by (1.3) .

0

1 94

CHAPTER V

[1

THEQREM 1 .5. Let B be a block of G with defect group D. Suppose that A B (C) -I 0 for a conjugate class C of G. Then there exists z E C with z E CG (D ) such that the p -factor y of z is in Z(D). PROOF. By (III.9.7) there exists a block Bo 2f NG (D) with defect group D such that HR = B. Let Ao = A o Since A R( C}f 0 there exists a conjugate class Co of NG (D) with Co � C and Ao(Co) -1 0. Choose z E Co. By (III.6.10) z E CG (D). If ( is an irreducible character in Bo then ( z ) -I O. Hence by (IV.2.4) y is conjugate in NG (D) to an element of D and so y E D. Since D � CG (z ) � CG (y ) it follows that y E Z(D). D B

PROOF. (i) Let Xr be a character of height 0 in B R . By (III.6.10) and (IV.4.8) there exists a p '-element x such that D is a Sp -group of CG (x ) and X' (x )wr (x - I ) 1'= 0 (mod ) If y E Z(D) then D is a Sp -group of CG (xy ) and Xr (xy ) Xr (x ) 1'= 0 (mod ) Thus v (wr (xy » = v (wr (x » = O. Hence Wr (xy ) 1'= 0 (mod ) for all y E Z(D ). Let ( be an irreducible character in Bo and let W be the central character of K[H] corresponding to (. Then G wr = W . Hence if y E Z(D) there exists a conjugate z of xy in H with W (z ) -I o. By (1.5) the p -factor of z is conjugate to an element of Z(Do). (ii) and (iii) are immediate by (i). (iv) If H = (I) then B R has defect 0 by (iii) and so contains exactly one irreducible character Xt . Hence if ( is the unique character in Bo then G ( = LXs (l)Xs and so by (1.3) G v (I G I) = v «( (1» = V (Xt (1)2) = 2 v (I G I). Thus v (I G I) = 0 contrary to assumption. D 7T .

7T

7T .

LEMMA 1 .7. Let B be a block of G and let be an automorphism of G such that X � = XS for all Xs E B. If y is an element in a defect group of B then y O- is conjugate to y in G. (J'

PROOF. By (IV.4.8) there exists a p '-element x and a character Xr of height

195

in B such that Xr (x )wr (x - I ) 1'= O (mod 7T) and a Sp -group D of CG (x ) is a defect group of B. Replacing y by a conjugate it may be assumed that y E D. Hence Xr (yx ) Xr(X ) 1'= 0 (mod ) If y and y O- are not conjugate in G then by (IV.6.3) ==

L

Xs E B

'

THEOREM 1 .6. Let Bo be a block of the subgroup H of G for which B R is defined. Let Do be a defect group of Bo and let D be a defect group of B R . Then (i) Every element in Z(D) is conjugate to an element in Z(Do). (ii) The exponent of Z(D) is at most equal to the exponent of Z(Do). (iii) If B� has defect 0 then B R has defect O. ' (iv) If p i I G I then H-I (I) .

==

INERTIA GROUPS

I Xs ( YX ) 1 2 =

7T

L

Xs E B

.

xs « yx )" )Xs (yx ) = 0

Since I Xs (yx )I � 0 for all s this implies that Xs (yx ) = 0 for all XS E B. Hence in particular X' (yx ) = 0 contrary to the previous paragraph. D 2.

Inertia groups

Throughout this section G is a fixed normal subgroup of G. In general a , tilde sign will be attached to the quantities associated with G. For instance X\, X 2 , . . . are all the irreducible characters of G. WS is the central character of K[ G ] corresponding to Xs . Let A be a subset of G which is a union of conjugate classes of G. Let 0 be a complex valued function defined on A which is constant on the conjugate classes of G which lie in A. If z E G define OZ by OZ (x ) = O(x z-) for all x E A . Then O Z is defined on A and is constant on the conjugate classes of G which lie in A. The inertia group T( 0) of 0 in G is defined by T(8) = {z I z E G, OZ = O} . Clearly G � T(O) . If W is an irreducible R [ G ] module which affords cPi then it is easily seen that T ( W) = T (cPi )' Similarly if V is an R -free R [ G ] module which affords Xs then it is easily seen that T( VK ) = T(Xs ). However T( V) need not be equal to T( VK )' Let B be a block of G. The inertia group T(B ) of 13 in G is defined by

I

T(13 ) = {z z E G, 13 Z = B} .

LEMMA 2. 1 . Let 13 be a block of G. If Xs, cPi E 13 then T(cPi ) � T(13) and T (Xs ) � T(13 ) .

PROOF. Clear by (IV.4.9).

D

THEOREM 2.2. For any group H with G � H � G let �s (H) be the set of all irreducible characters of H whose restriction to G have Xs as a constituent. Let S)l?(H) be the set of all irreducible Brauer characters of H whose restriction to G have 'Pi as a constituent. (i) If T(cPi ) � H then the map sending 8 to 8 G defines a one to one correspondence between �?(H) and S)l?(G) .

1 96

[2

CHAPTER V

2]

INERTIA GROUPS

197

(ii) If V is an irreducible R [G] module which affords a Brauer character in 2I? (G) then a vertex of V is contained in T (iPd. (iii) If T(Xs ) � H then the map sending () to () G defines a one to one correspondence between 2Is (H) and 2I s (G). (iv) If Xt E 2Is (G) then there exists an R -free R [G] module which affords Xt and whose vertex is contained in T(Xs ) .

PROOF. If such a chain exists then by (2.3) consecutive characters belong to blocks which cover a fixed block of G. Suppose conversely that XS and Xt are in blocks which cover a fixed block of G. Let Xm be an irreducible constituent of (Xs )o . By (2.3) there exists XT in the same block as Xt such that Xm is a constituent of (XT )O . The chain xs, XT, Xt satisfies the required conditions. 0

PROOF. (i) It suffices to prove the result i n case T(iPi ) = H. Let 1/1 E 91?(H). By (11.2.9) (I/I G ) H = 1/1 + () where iPi is not an irreducible constituent of (}o . Let YJ be an irreducible constituent of I/I G with YJH = 1/1 + (} 1 . By (111.2.12) 1/10 = e iP for some integer e > O and YJ (I) = I G : H I I/I(1) = I/I G (I). Hence I/I G = YJ E 2I?( G). Furthermore if 1/11, 1/12 E 2I?(H) and I/I ? = I/Ii' then ( I/I ?)H (1/Ii')H and so 1/1 1 1/12 ' Therefore the map sending () to (} G is one to one. It remains to show that it is onto. Let 'Pj E 2I?( G). Let V be an R [ G] module which affords 'Pj and let � be an R [0] module which affords iPi . By (111.2.12) Va is completely reducible and so I( Vo, W; ) � O. Thus by (111.2.5) I( VH' W�) � 0 and so I ( VH W) � O for some irreducible constituent W of W�. Since ( W:-I)o = IH : 0 I W; it follows that W; is a constituent of W Thus if 1/1 is the Brauer character afforded by W then 1/1 E 2I? (H). Since I ( V, W G ) = I( VH' W) � 0 the irreducibility of V and WG implies that V = W G and hence 'Pj = 1/1 G as required. (ii) Clear by (i). (iii) If p is replaced by a prime not dividing I G I this follows from (i). (iv) Clear by (iii). 0

THEOREM 2.5 (Fang [1961], Reynolds [1963]). Let 13 be a block of O. The map sending B to B G defines a one to one correspondence between the set of all blocks of T(13 ) which cover 13 and the set of all blocks of G which cover 13. Furthermore if B is a block of T (13 ) which covers 13 the following hold. (i) The map sending () to () G defines a one to one correspondence between the sets of all irre_ducible characters, irreducible Brauer characters respec tively, in BA and BA O . (ii) With respect to the correspondence defined in (i) B and B 0 have the same decomposition matrix and the same Cartan matrix. (iii) B and B G have a defect group in common. (iv) 13 is the unique block of G covered by B.

i

=

=

'

H.

block B of G covers a block 13 of G if there exists XS E B and Xt E B such that Xt is a constituent of (Xs )o . A

LEMMA 2.3. The blocks of G covered· by the block B of G form a family of blocks conjugate in G. If 13 is covered by B and XS E B then some constituent of (Xs )o belongs to 13. If Xt E 13 there exists XS E B such that Xt is a constituent of (Xs )O . PROOF. This is a reformulation of (IVA . I 0).

0

LEMMA 204. XS and Xt belong to blocks which cover the same block of 0 if and only if there exists a chain Xh Xs, Xh , . . . , Xj" = Xt such that for any m either Xjm and Xim +' are in the same block of G or (Xim )O and (Xim -l-J G have a common irreducible constituent. =

PROOF. Let B be a block of T(13 ) which covers 13 and let � be an irreducible character in B. By (1.2), (2.1) and (2.2) (iii) B O is defined, ( 0 E B 0 and B 0 covers 13. Suppose that B is a block of G which covers 13. Let Xt E B. By (2.3) (Xt )o has an irreducible constituent Xs E B. By (2.1) and (2.2) (iii) Xt = � G for some irreducible character � of T(B ) such that Xs is an irreducible constituent of (a . Let B be the block of T(13 ) with � E B. Then B covers B and B 0 = B by (1.2). Let 1/11 , 1/12 be irreducible Brauer characters of T (B ) in blocks which cover 13. If iP E 13 and iP is an irreducible constituent of (1/1 ?)o then iP is a constituent of (l/Ii )O for i = 1 , 2. Thus if I/I ? = I/I? then (1/11)0 and (1/12)0 have a common irreducible constituent which is in 13, and so 1/11 = 1/12 by (2.2). Therefore the map sending () to () G defines a one to one correspondence between the sets of all irreducible characters, irreducible Brauer characters respectively, of T(13 ) and G which lie in blocks that cover 13. This map clearly preserves decomposition numbers and hence also Cartan invariants. Since the decomposition matrix of a block is indecomposable it follows that a character or a Brauer character () of T (B) is in B if and only if () 0 E B o . Thus the map sendIng B to B O is one to one as required. Furthermore (i) and (ii) are proved. (iii) Let D be a defect group of B. By (111.9.6) D � D for some defect group D of B o . By (i) B and B 0 have the same defect. Thus D = D.

[3

CHAPTER V

198

(iv) Since R Z

=

R for all

z

E T(R) this is clear by (2.3).

PROOF. By (2.5) B = fj G for some block fj of T(R) which covers R. The result follows from (2.5) (iii). 0

�E� 3. 1 . _Let a � Z (R ; G : G ) n R [O]. If A is a central character of R [ G] then A (a ) A G (a ) . =

0

LEMMA 3.2. Let {Rim } be the set of all blocks of 0 where Rim = Bjn for some

E G if and only if i = j. Let eim be the centrally primitive idempotent of R [ 0] with Rim B (eim ). Let ei Lm eim . Then {ei } is the set of all primitive idempotents in Z(R ; G : G) n R [0] . =

PROOF. Any idempotent in Z (R ; G : G) n R [ 0] is a sum of centrally primitive idempotents in R [0] and is invariant under conjugation by , elements of G. The result follows. 0

LEMMA 3.3 (Fang [1961]). Let Rim and eim be defined as in (3.2). For each i let Bil , Bi2 ' . . . be all the blocks of G which cover some Rim and let eim be the centrally primitive idempotent in R [ G ] with Bim B ( eim ). Then ei Lm eim = L m eim .

for all

Aa (a ) = AB (a )

PROOF. By (3.2) and (3.3) B covers R if and only if Aa (e) = AB (e) for every idempotent in Z(R ; G : G) n R [0]. By (1.16.1) R is a splitting field of Z(R ; G : G) n R [0]. The result follows since two central characters of a commutative algebra are equal if they agree on all idempotents. 0 unique block B of G which covers R.

The notation of the previous section will be used in this section. For the results in this section see Brauer [1967a], [1968] ; Fang [1961] ; Passman [1969].

=

Then B covers R if and only if E Z(R ; G : G) n R [O] .

LEMMA 3.5. Suppose that GIO is a p -group. If R is a block of 0 there is a

3. Blocks and normal subgroups

z

a

199

BLOCKS AND NORMAL S UBGROUPS

O.

0

COROLLARY � .6. Let B be a block of G and let R be a block of 0 covered by B. Then T(B ) contains a defect group of B.

PROOF. Clear by definition.

3]

=

=

PROOF. By (3.2) {ei } is a s�t of pairwise orthogonal central idempotents in R [G]. Thus it suffices to show that eimei � 0 for all i, m. Let V be a nonzero R [G] module in Bim . By (2.3) ( VG )ei � O. Hence Vei = Veim ei � O and so eim ei � O. 0

LEMMA 3.4 (Passman [1969]). Let B be a block of G and let R be a block of

PROOF. Suppose that BI and B2 are blocks of G which cover 13. Let Ai = AB. for i 1 , 2. If C is a conjugate class of G consisting of p '-elements the� C E Z(R ; G : G)n R [0]. Thus by (3.4) A 1 ( C ) = A13 (C) A ( C ) Hence by (lV.4.2), (IV.4.3) and (IV.4.8) (iii) B1 = B2 • 0 =

=

2

,

A block B of G is regular with respect to 0 if AB (C) = 0 for all conjugate classes C of G which are not contained in O. B is weakly regular with resp'?..c t to 0 if there exists a conjugate class C of G with C .� 0 such that AB ( C) � 0 and the defect of B is equal to the defect of C. In case 0 is determined by the context the phrase "with respect to 0 " will be omit�ed. By (111.6.10) a regular block is weakly regular. Further­ more if AB (C) � 0 and the defect of B is equal to the defect of C then B and C have a common defect group.

LEMMA 3.6. Let B be a block of G. The following are equivalent. (i) B is regular. (ii) B = R G for every block R of 0 which is covered by B. (iii) B = 13 G for some block R of O.

PROOF. (i) =? (ii). Let 13 be a block of 0 which is covere (ii). There exists a conjugate class C of G with C � G and AB (C) � 0 su�h that B_ and C have a common defect group D. By (3.4) AB; (C) = A8 (C) = AB (C) � 0 and so by (III.6.10) a defect group of Bi is a

PROOF. If E G is defined then E G is regular by (3.6) and E G covers E by (3.7). The converse follows from (3.7). The last sentence is clear. D

. subgroup of some defect group of B. (ii) � (iii). Trivial. _ (iii) :::;> (iv). Since AB (e) = 1 it follows that iicAB ( C) � 0 for s2me C. By (IY.7.2) C consists of p '-elements. By (3.3) C � G. If ei = Lai,cC for each i then for some i ii i, C � O. Hence by (IV.7.3) the defect of C is equalJ:o the defect of Bi and so is at most equal to the defect of B. Since A (C) � 0, (III.6.10) now implies that B and C have the same defect. (iv) :::;> (i). Trivial. D

LEMMA 3.9. Let B be a block of G with defect group D. If C G (D) � G then B is regular. PROOF. If AB (C) � 0 for some conjugate class of G then by (III.6.10) D � C G (z ) for some E C. Hence z E C G (D) � G and so C � G. 0 z

LEMMA 3. 10. Suppose that P is a p -group with P <J G and C G (P) � G. Then every block B of G is regular with respect to G. For every block E of G, the block E G is defined and is the unique block covering E.

PROOF. Let B be a block of G and let D be the defect group of B. By (III.6.9) P � D. Thus C G (D) � C G (P) � G �nd B is regular by (3.9). Let B be a block of G. By (3.8) E G is defined and is the unique block covering B. D COROLLARY 3.11 . Suppose that P is a p-group with C G (P) � P <J G. Then the principal block is the unique block of G. PROOF. By (3. 10) with G = P, E G is the unique block of G where E is the unique block of G. Hence G has exactly one block which must necessarily be the principal block. D

THEOREM 3.12 (Fang [1961]). Let E be a block of G and let BJ, B 2 , be all the blocks of G which cover E and let B = Bj for some j. For each i let ei be the centrally primitive idempotent of R [ G] with Bi = B ( ei ) and let e = Lei . The following are equivalent. (i) B is weakly regular. (ii) A defect group of any block Bi is conjugate to a subgroup of a defect group of B. (iii) The defect of any Bi is at most equal to the defect of B. • • •

COROLLARY 3. 13. Let E be a block of G. The following hold. (i) There exists a weakly regular block B of G which covers B. (ii) Let 13 be a block of T(E) which covers E. Then 13 is weakly regular if and only if 13 G is weakly regular. PROOF. (i) Choose a block B of maximum defect among those which cover B. By (3.12) B is weakly regular. (ii) By (2.5) 13 and 13 G have the same defect. Thus E has maximum defect among all blocks 13i of T(B ) which cover E if and only if E G has maximum defect among all blocks E f'. The result follows from (3.12). D THEOREM 3.14 (Fang [1961]). Let E be a block of G and let B be a block of G which covers B and is weakly regular. There exists a defect group D of B with D � T(E). For any defect group D of B with D � T(E) I T(B ) : DG I � O (mod p ) and D n G is a defect group of B. PROOF. By (2.6) T(B ) contains a defect group D of B. Thus by (2.5) and (3. 13) (ii) it may be assumed that G = T(E). Let e be the central idempotent o f R [G] with E = B (e). Since G = T(B ), e = LacC with ac E R and C ranging over the conjugate classes of G. By (3.3) and (3. 12) (iv) there exists a conjugate class C of G with C � G, iic � 0, AB ( C) � 0 such that D is a defect group of C. Let E C with D � C G (z ) . Let L be the conjugate class of G with E L. Thus C = U L where x ranges over a cross section of N G (L) in G. By (3.4) z

x

x

z

202

CHAPTER V

A B (C) = AB (C) =

2:X AB (L X ) = 2:X A�

Since T(13 ) = G this implies that

BLOCKS AND QUOTIENT GROUPS

-1

(L) .

be an R [ G ] module in a natural way. Similarly every character or Brauer character of GO is a character or Brauer character of G.

to

I T(13 ) : NG (L ) I AB (L ) = AB ( C ) � O .

Thus I T(13 ) : NG (L )I � O (mod p ) and AB (i ) � o. Since NG (L ) = CG (z)G and D is a Sp group of CG (z ) it follows that I T(13 ) : Da I � 0 (mod p ). . GeAB (L Slllce ) 0 it follows from (IV.7.3) that 13 and L have a common defect group. Clearly D n a is a Sp -group of Co (z ) and so is a defect group of L. D -

A

=

LEMMA 3.15. Let B be a weakly regular block of G and let 13 be a block of G covered by B. Let XS be of height 0 in B and let Xt be a constituent of (Xs )a with Xt E B._ Then the r�mification index of Xs is not divisible by p, Xt has height 0 in B_ and I T(B ) : T(Xt )1 � O (mod p ) . PROOF. By (2.5) and (3. 13) it may be assumed that G = T(13 ). Let D b� a defect group o! B. Let d = v (I D /) and let d = v (I D n a l). By (3. 14) d . the defect of B. By (III.2.12) (Xs )6 = e LzX : where { z } is a cross sect�on O. By (5.4) (ii) there exists a block pair (D *, B *) in G which properly extends (Do, Bo). Let H = D *Co (Do). Suppose that D * C D Z . Let Ho be the kernel of Bo. By (IV.4. 1 1) Ho is a p i_group. Furthermore Ho is the kernel of B f/ = B * H and 071', (Co (D Z » C Ho . It follows from (6.4) that 07T, (Co (D Z » C Ho n D * Co (D *) and so is in the kernel of B * . The result follows by induction. (ii) Since the kernel of Bo contains all 'TT '-elements in Co (Do) the

THEOREM 6.5. Let 'TT be a set of primes with p E 'TT. Let B be a block of G

-

with defect group D. Let N = No (D) and let B l be the germ of B. Assume that 0 ,(N ) is in the kernel of BI . Suppose that Go is a subgroup of G, Bo is a block of Go with defect group Do, Co (Do) C Go and B � = B. Then 071" ( Go) is in the kernel of Bo .

-

7T

PROOF. By (111.9.6) it may be assumed that Do C D. By (5.4) Bo = B i'o for

some block Bz of Do Co (Do). Let d = v(I D I), do = v(I Do I). The proof is by induction on d - do . Suppose that d - do = O. By (111.9.2) B i' = B � = B and B i' = (B�) G. Thus B � = BI by (111.9.7). By (3. 10) Bl covers Bz. Therefore 0 ,(N) = 071',(D Co (D » is in the kernel of Bz• Thus 07T,(GO) n D Co (D ) is in the kernel of Bz and so 071',(GO) is in the kernel of Bo by (6.4). Suppose that d do > O. By (5.4) (ii) there exists a block pair (D *, B *) in G which properly extends (Do, Bz). Let H = D *Co (Do). By induction 071', (H) is in the kernel of B � = B * H. Since B � covers Bz it follows that 0 7T ,(H) = 07T,(DoCo (Do» is in the kernel of Bz• As 07T,(GO) n DoCo (Do) C 071',(DoCo (Do» , (6.4) implies that 07T'( Go) is in the kernel of Bo = B fio . 0

assumptions of (i) are satisfied. Thus the result follows from (i).

71'

The following example shows that the assumptions in (6.7) (ii) that

-

with defect group D. Assume that 07T,(No (D » is in the kernel of the germ of B. Suppose that y is a p-element in G such that Co (y ) has a normal S71',-subgroup. If X = Xi E B then X (yx ) = x (y ) for every 'TT ' -element x in Co (y ) . 'P I ranges over blocks of Co (y ) which have 07T (Co (y » in their kernel. Thus 'P Hx ) = 'P Hi ) for every 'TT I -element x in Co (y ). 0

071" (Co (Do» is a S7T'-group of Co (Do) cannot be dropped even in case 'TT = {p } . Let (y ) be a cyclic group of order 4. Let S = S5 be the symmetric group on 5 letters and let A = A5 �e the alternating group. Let G = (y ) S where (y ) <J G and y Z = Y for E A, Y Z = Y - I for E S - A. Let p = 2. Let t be the irreducible character of degree 4 of S which is a constituent of the permutation representation of S on 5 letters. Then tA is irreducible. Thus tA is in a block of defect O. By (3.5) the block of S containing t is the only block of S covering that of tAo Thus it is weakly regular. Hence by (3 .14) there exists an involution x E S - A such that (x ) is a defect group of the block of S containing t. Therefore x is necessarily a transposition and Cs (x ) = (x ) W where W is the symmetric group on the three letters fixed by x. Consider t as a character of G which has (y ) in its kernel and let B be the block of G which contains { By (4.5) D = (x, y ) is a defect group of B. Thus in particular B is not the principal block of G. Let Do = (y ) . Then DoCo (Do) = (y ) A and by (3. 10) B = B � for some block Bo of Do Co (Do). Oz, (Do Co (Do» = (l) is in the kernel of Bo . lt is easily verified that No (D ) = ( x y ) W Thus W' = Oz, (No (D» . Hence by (4.3) the principal block of No (D ) is the only block which has W' in its kernel. Thus by (6 2) Oz' (No (D» is not in the kernel of the germ of B. z

COROLLARY 6.6. Let 'TT be a set of primes with p E 'TT. Let B be a block of G

PROOF. By (lV.6. 1) and (6.5) X (yx ) = Lj d � 'P Hx ) where

_

z

,

THEOREM 6 . 7 . Let 'TT be a set of primes with p E 'TT. Let B be a block of G

with defect group D. Let Do be a p -group contained in G and let Bo be a block of DoCo (Do) such that B � = B. (i) Suppose that whenever Do C D z for some E G, 071" (Co (D Z » is in the kernel of Bo. Then 07T, (No (D» is in the kernel of the germ of B. (ii) Suppose that 07T ' (C O (Do» is a S7T ' -groUP of Co (Do) and 07T ' (C O (Do» is in the kernel of Bo. Then 07T, (No (D» is in the kernel of the germ of B. z

-

0

.

.

THEOREM 6.8. Let L, M, HI, Hz be subgoups of G with LHI C MH2 and

H1 , H2 p i_groupS. Assume that [M, H2] C H2, [L, Hd C HI, L C M and H2 n HI L C HI . Let B be a block of L with defect group D such that CM (D) C L. Assume that HI n L is in the kernel of B. Let B be the block of LHI which has HI in its kernel and corresponds to B by the natural

CHAPTER V

214

[6

isomorphism from L /L n HI to LHdHI . Assume furthermore that B MH2 is defined. Then B M is defined and H2 n M is in the kernel of B M. If B M is the block of MH2 corresponding to B M by the natural isomorphism of M/M n H2 onto MH2/ H2 , then 13 MH2 = B M. The following diagram illustrates the statement. M ----

B

'

L

---

PROOF. By (111.9.5) B M is defined. Since M n H2 <J M and M n H2 n L � L n HI it follows from (6.4) (i) that M n H2 is in the kernel of B M. Since [L, H2] � H2 and L n H2 � L n HI is in the kernel of B, there exists a block B 2 of LH2 with H2 in its kernel such that B 2 corresponds to B by the isomorphism L /L n H2 = LH2/H2. By (4.3) B 2 has D as a defect group. The Frattini argument implies . that C MH21 H2 (DH2/H2) = C M (D )H2/H2 and so C MH2 (D ) � C M (D )H2 � LH2 . Thus by (I1I.9.5) B lj is defined for LH2 � H � MH2 . We will next show that (6.9) Let ? E B and let t E 13 2 with tL = r Let {Xs } be the set of all irreducible characters in B rH2 and let as = (XS, t MH2)MH2 • By (1 . 3 ) v «(MH2( 1)) =

(LasXs (1)). As ( is the unique irreducible character of LHI with ? � tL and HI is its kernel and since HI is in the kernel of each X" it follows from Frobenius 1J

reciprocity (I1I.2.5) that

as

= « XS ) L H J , () U-l J = «XS ) L , ?) L = «XS ) M , ? M )M .

Since H2 is in the kernel of each Xs by (6.4) (i), each (Xs ) M is irreducible. Thus there exists a block B ' of M such that {(XS )M } is the set of all irreducible characters in B '. Hence 13 ' = 13 MH2 . However

v ( L as (Xs ) M ( l) ) = v «(MH2 (1)) = v (I MH2 : LHl l x ( 1 )) = v (I M : L I ? (1)) = V (? M (1)) . Hence B ' = B M by (1 .3), and (6.9) is proved. Since H2 n HI L � HI it follows that HI H2 n LHI = HI . Consider the natural isomorphisms

7]

ISOMETRIES

215

L /L n HI = LHJHI = LHI/LHI n HI H2 = LHI H2/ HI H2 = LH2/ LH2 n HI H2 . Let B3 be the block of LHI H2 \vith HI H2 in its kernel which corresponds to B. Let B � be the block of LH2 with LH2 n HI H2 in its kernel which corresponds to B3 • Let ? be an irreducible character in B, let t be the unique irreducible character in B 2 with (L = ? and H2 in its kernel. Let () be the unique irreducible character of LH, H2 with HI H2 in its kernel such that ()L = r Then () E 133 by definition. Furthermore ()LH2 E B � . Since « ()LHJL = ? = (L it follows that B � = 13 2 . Thus 13 2 corresponds to 133• Since B iHJH2 is defined, (6.4) implies that B i H J H2 = B3 and so by (6.9) � B H2 = 13 rH2 = B M. By (6.4) B �H2 = 13 MH2 . Therefore 13 MH2 = B M . 0 In (6.8) it is essential to assume that B MH2 is defined. This does not follow from the other conditions. For instance let M = L = HI = (1) . Let l/J be the central character of (I) , let C� {I} be a conjugacy class of H2 and let C' = {x - I I x E C} . Then l/J H2 (C)l/J H2 (C' ) = 0 but l/J H2 (CC' ) = 1 C I � o .

7. lsometries

Let X be a set of characters of the group N and let A be a union of conjugate classes of N. The following notation will be used in this section. M(N) is the ring of all generalized characters of N. V(N, A : X) is the vector space of all complex linear combinations of characters in X which vanish on N - A . If X is the set of all irreducible characters in a union 5 of blocks of N let V(N, A : 5) = V(N, A : X). If X is the set of all irreducible characters of N let V(N, A) =

V(N, A : X). Let V(N, N : X) = V(N : X) and let V(N, N) = V(N). M(N, A : 5 ) = Let M(N, A : X) = V(N, A : X) n M(N). Let V(N, A : 5 ) n M(N) where. 5 is a union of blocks of N. Let M(N, A ) = V(N, A ) n M(N) and let M(N : X) = V(N : X) n M(N). The inner product of characters defines a natural metric on M(N, A : X) and on V(N, A : X) . If Y is a set of irreducible characters of G and a = LasXs with complex as let a Y = LXs E yaSXS ' If Y is the set of all irreducible characters in a union 5 of blocks of G let a s = a Y. It is evident that if a is a generalized

character of G then so is a Y. ' 'TT is a set of primes and 'TT is the complementary set of primes.

216

[7

CHAPTER V

If p E 7T then Sp ( 7T, N) is the union of all p -blocks B of N such that if D is a defect group of B then 07 T , (NN (D)) is in the kernel of the germ of B. If Z E G then Z-rr denotes the 7T -part of z. The 7T-section of G containing Z is the set of all elements in G whose 7T-part is conjugate to Z7r ' V7r (N, A : X) = {a I a E V(N, A : X), a (yx ) = a (y ) for any element y E A and any element x E 07r , (CN (y ))} .

M7r (N, A : X) = V7r (N, A : X) n M(N) . Similarly the subscript 7T applied to any of the sets defined above means that it is intersected with V7r (N, A : X). In this section we will be concerned with the following two hypotheses.

7]

21 7

ISOMETRIES

PROOF.

Let

C = CG (y). Since y, X-rr have relatively prime orders,

C c (X7r ) = CG (YX7r ) = CG « yx )7r ) � C . By (7.3)

(xxoY = xu for some Z E C,

u E 07r ' ( C) n C c (X7r ) = 07T' ( C) n CG « yx) 7r ) � 07T, (CG (yx ) ,, ) .

Therefore

{3 (yxxo) = (3 « yxx oY ) = {3 (yxu ) = {3 (yx )

by the definition of

V7T ( G).

0

7.5. (i) a E V7r (N, A : X) if a E V(N, A : X) and for y E A, is a linear combination of irreducible characters of CN (Y7T ) all of which have 07T (CN ( Y )) in their kernel. (ii) Suppose that (7.2) is satisfied. Then a E V7r (N, A : X) if and only if a E V(N, A X) and a is constant on 7T-sections in N. (iii) Suppose that (7. 1) is satisfied. If Z E G and Z7r E A then Z7r' = z [ Z2 with Zl E CN (Z7T ) and Z2 E 07r, (CG (Z7r )). If furthermore z : E A for some x E G and z :, = z � z � with z ; E CN (z :) and z � E O-rr, (CG (z :)) then a (z7T zl) = a (z :z O for all a E V7T (N, A ).

LEMMA a C N ( Y7T )

7T

HYPOTHESIS 7. 1 . (i) N is a subgroup of G. A is a subset of N which is a union of 7T-sections of N such that any two 7T-elements in A which are conjugate in G are conjugate in N. (ii) If y is a 7T-element in A then CG (y ) = CN (y )07r, (CG (y)). HYPOTHESIS (ii)

7.2. (i) (7. 1)

is statisfied.

07r, (CG (y)) is a S7T'- groUP of CG (y) for all 7T-elements y in A.

We first prove some elementary results.

LEMMA 7.3. Let H <J G, H a 7T'-subgroup. Let x E H, Y E G. Then there exists u E CG (Y7r ) n H and Z E H such that yx = (yu y . PROOF.

:

PROOF.

(i) and (ii) are immediate by definition. (iii) The first statement follows directly from (7. 1). If z : E A then by (7. 1) (i) there exists y E N such that z ; = Z7r ' Then z ? z ? = z ;' = (z ;Y (z �Y Since xy E CG (Z7r ), z ? and (z �Y are in 07T, (CG (Z7T ))' Let xy = uv with u E N and v E 07r' (CG (Z7T )). Then

Define T

= {(Y7r y t l z ranges over a cross section of CH ( Y7r ) in H, t E C H (Y7r Y } '

T � y7rH and ' T ' = ' H ' . Thus T = Y7T H. Since H <J G it follows that (yx )7r E Y7rH and so (yx ) 7r = (Y7r Y t for some Z E H and some 7T '-element t E CG (y �). Thus t = 1 and (yx )7r = y :. As H <J G, (yx )7r' = Y7r ' XO for some Xo E H. Hence (yx ):�I = (Y7T, xoy-1 = Y7r' u for some u E H. Since Y7T' u centralizes (yx ) :-- 1 = Y7r it follows that u E CG (Y7r) n H. Therefore (yx y-I = Y7T « yx ):�) = Y7r Y7r' U = yu. 0 LEMMA 7.4. Suppose that (3 E V7r ( G ). Let y be a 7T-element in G. Let x E CG (y ) such that x7r, y have relatively prime orders and let Xo E 07r ' (CG (y )). Then (3 (yx ) = (3 (yxxo) .

Then

.

==

Since

u, y, z : z ;

and

(Z7TZ IYv

==

(z7rz [Y

(mod

07T, (CG (Z7T ))) '

Z7T Zl are all in N it follows that

(z : z ;Y = (Z7TZ1Y ZO for some Zo E 07T, (CG (Z7T )) n N � 07T , (CN (Z7T )). Therefore

a (z7T zJ) = a «z7TzlYZO) == a «z :z ;Y ) = a (z : z O. 0 Suppose that (7. 1) is satisfied. If a E V7T (N, A ) define a T E V7T ( G ) as follows. If Z7r is not conjugate to an element of A let a T (z ) = O. If z : E A for some x E G then ZX E CG (z :) and so z :, = Zl Z2 where zl E CN (z :) and z2 E 07r, (CG (z :)). Let a T (z ) = a (z :z J). By (7.S) (iii) a T is well defined. Thus T is a linear mapping from V7T (N, A ) to V7T ( G ).

CHAPTER V

218

[7

Suppose that (7.1) is satisfied and A is a union of p -sections for some p E If y is a p-element in A, Xl a p'-element in CN (y ), X2 E 07T ' (C O (y » and X = XI X2 then a T (yx ) = a (YXI)'

LEMMA 7.6.

77.

PROOF. By (7.4) aT (yx ) = aT implies that a T (yx 1 ) = a (yx 1 ) '

(YXI)' Since yXI E A the definition of 0

LEMMA 7.7. Suppose that (7. 1 ) is satisfied. (i) If a E V7T (N, A ) and {3 E V7T (G) then (aT, {3 )o (ii) If a I , a2 E V7T (N, A ) then (aJ , a2)N = (a � , a ;)o .

T

= (a, {3N)N.

PROOF. (i) If Y is a 17 -element in G let {y};', {y }� respectively, be the 17-section of G, N respectively, containing y. Let ')' = a T (3 * . Then ,), (z ) = 0

if z E {y };' and y is a 17 -element of G which is not conjugate to any element of A. If y is a 17-element in A then Co (y ) = CN (y )07T' (Co (y » . Thus there exists a cross section {Xi } of CN (y ) � Co (y) with {xJ � 07T' (C O (y » such that every 17 '-element in Co (y) is of the form ZXi for some i and some 17 '-element Z E Co (y). Since ,), E V7T (G) this implies that 1

�

(z ) L.J ,), I G I zE{y}� -

=

1

1

�'

�

I C o (y ) I x ECG (y) ')' (yx ) = I CN (y ) I x ECN (y) ')' (yx ) '

L..J

L.J

= � L ')' (z ) I N I z E{y};;:

where 2:' means that X ranges only over 17 '-elements. Thus if y ranges over 17-elements in A which form a complete set of representatives of all 17-sections in G which meet A then

(a T, {3 )o = (ii) Let

I

b i Ly T

L ')' (Z ) =

z

E {y}�

I

�I L

{3 = a; in (i). Then

I I Z�N al (z )a2 (z )*

(a �, a D o = (aJ , (a ;)N )N =

�

Y

L ')' (Z ) = (a, {3N )N .

Z

E {y};;'

aI (z )a ; (z )*

� I z� = I�I Z�N aI (z )a2 (z )* = (aJ , a2)N .

=

I

0

The existence of T is a very useful tool for the investigation of certain questions concerning the structure of finite groups provided M7T (N, A Y �

7]

ISOMETRIES

219

M(G) or equivalently T maps generalized characters in V7T (N, A ) onto generalized characters of G. A discussion of some of these applications can for instance be found in Feit [1967c] or Smith [1974] , [1976a] . In case 17 is the set of all primes it is easily seen that a T = a 0 for all a E V7T (N, A ) and so M7T (N, A Y � M(G). This case, which gave rise to the whole method, was first studied by Brauer and Suzuki . who were the first to realize the importance of results of this type. It is not known whether M7T (N, A Y � M(G) whenever hypothesis (7. 1) is satisfied. However Reynolds [1963] and Leonard and McKelvey [ 1 967] have shown that M7T (N, A Y � M(G) provided hypothesis (7.2) is satisfied. This result and its proof is a generalization of an earlier theorem of D ade [1964] who proved the same conclusion under the stronger hypothesis that (7.2) holds and CN (y) is a 17:-group for every 17-element y in A. Dade's result in turn generalized a result where the conclusion was proved under still stronger hypotheses, see Feit and Thompson [ 1 963], section 9. All these results are of course generalizations of the case that 17 is the set of all primes. In this section we will primarily be concerned with showing that under suitable additional hypotheses T is related to the Brauer correspondence. The basic approach in this section is due to Brauer. Some of the material can be found in Brauer [ 1 964a] , Gorenstein and Walter [1962] , section 1 1 , Niccolai [1974] , Reynolds [1967] , [1968], [1970], Walter [1966] and Wong [1966] . Glauberman [1968] Remark 3.2, first pointed out that Brauer's approach yields Dade's result in case 17 = {p} for some prime p. In this case a proof that M7T (N, A Y � M(G) will be given below.

THEOREM 7.8. Let p E 17. Assume that (7. 1) holds and A is a union of p-sections of N. Let Bo be a p -block in Sp ( 17, N) such that V(N, A Bo) 1= (0). Let D be a defect group of Bo and let Bo be the germ of Bo with respect to D. Then (i) NN (D)C o (D) = NN (D) 07T ' (CO (D » . (ii) 01T, (NN (D» is in the kernel of Eo and there exists a unique block Eo of NN (D ) Co (D) with 07T' (C O (D» in its kernel such that Eo and Eo coincide as blocks of NN (D) NN (D) Co (D) 07T. (NN (D»" 07T' (C O (D» N (iii) Let y E A n D and let B (y ) be a block of CN (y) with B (y ) = Bo . Then 07T' (CN (y» is in the kernel of B (y ). If E (y ) is a block of Co (y) with 07T' (C O (y» in its kernel such that B (y ) and E (y ) coincide as blocks of :

[7

CHAPTER V

220

CN (y) �� 07T , (CN (y » - 0 7T , (CG (y» G then B (y ) = B f? -

PROOF. Let y E A n D and let B (y ) b e a block o f CN (y ) with B (y t = Bo . Let Dl be a defect group of B (y ). By (5.4) there exist block pairs (Di, Bi ) in N for i = 1 , . . . , n such that (D i+l , B i+1) properly extends (Di ' Bi ) for i = I , . . . , n - l, B 7N (y) = B (y ) and Dn is a defect group of Bo . The following assertion will be proved by induction. and Bi is in For i = 1 , . . . , n CG (Di ) = CN (Di ) 07T, (CG (Di »

Sp (7T, Di CN (Di » .

Assume that the statement has been proved for i < m. If m = 1 let P = (y ) . If m > 1 let P = D m - l . Hence by (7 . 1 ) or induction CG (P) = CN (P) 07T' (CG (P» , Let Z E CG (Dm ) � CG (P), Thus Z = ZI Z2 with ZI E CN (P) and Z2 E 07T, (CG (P» . Since Dm � NN (P) it follows that Dm {07T' (CG (P» n N} is a group and [Dm' Z rl = [Dm, Z ;-1] is contained in 07T , (CG (P» n N. As Dm is a Sp-group of Dm {07T, (CG (P» n N} there exists l Z3 E 07T. (CG (p» n N with ZI Z ;- E NN (Dm ). Thus

Z3 Z2 E NG (Dm ) n 07T, (CG (P»

�

CG (Dm ) n 07T, (CG (P»

C 07T, (CG (Dm »

.

I

Since Z = (Z I Z ;- ) (Z3 Z2) this implies that CG (Dm ) = CN (Dm ) 07T, (CG (Dm » as the opposite inclusion is trivial. By (6.5) Bi is in Sp (7T, Di CN (Di » . The statement follows by induction. (i) Since V ( N, A : Bo) I:- (0), (IV.6.2) implies the existence of y E A n D and a block B (y) of CN (y ) with B (y t = Bo. Since Dn is conjugate to D in N the result follows from the previous paragraph. (ii) Immediate by (i). (iii) Let Bi be a block of Di CG (Di ) such that 07T' (Di CG (Di » is in the kernel of Bi and Bi coincides with Bi as blocks of

Then Di is a defect group of Bi and so (Di Bi ) is a block pair in G. Suppose that B f" = Bo for i = 1 , . . . , n. It will first be proved by induction on n - i that B f' = B f? Suppose that n - m = O. Replacing Dm b y a conjugate i n N i t may be assumed that D = Dm . By the first main theorem on blocks (1II . 9 . 7 ) '

7]

ISOMETRIES

221

B �N(D) = 13o . Thus B �N (D)CG(D) = Bo by (6. 8) with M = NN (D ), L = D CN (D ), HI = H2 = OTr, (CG (D» , and so B � = B f? Suppose that n - m > o and the result has been proved for m < i � n. Thus B �+1 = B f? Since (D m+l , B m+l) extends (Dm, Bm ) it follows from (6. 8) that (D m+l , B m+l) extends «Dm , Bm ). Therefore B � = B �+1 = B f? This proves the statement in the previous paragraph. Since B 7N (y) = B (y ) it follows from (6.8) applied to CN (y ) and CG (y ) y G that B 7° ( ) = B (y ) . Thus B (y ) = B 7 = B f? D

E 7T. Assume that (7.1) holds and A is a union of p-sections of N. Let {A }� be the set of all elements in G whose 7T-part is conjugate to an element of A. For any p -block Bo of N such that V(N, A : Bo) � (0) and Bo � Sp (7T, N) let Bo be defined as in (7.8) (ii). Let B be a p -block of G with B � Sp (7T, G) and let BO l , B02 , . . . be all the p -blocks of N such that V (N, A : BOi ) � (0), BOi � Sp ( 7T, N) and B &; = B. Then the following hold. (i) If a E V7T (N, A : UiBoi ) then a T E V ( G, {A }� : B ) . (ii) V7T (N, A : U i Boi r � VTr ( G, {A }� : B ). G (iii) If a E VTr (N, A : U i BOi ) then a T = (a )B . (iv) If B � Sp ( 7T, G ) then MTr (N, A : U i Boi r � M7T (G, {A }� : B ) . THEOREM 7.9. Let p

Tr

PROOF. If Y is a p -element in G and (3 is a function defined on CG (y ) let (3y be the function on the set of all p '-elements x in CG (y ) defined by {3y (x ) = (3 (yx ). (i) Let aT = (a T )B + B . It suffices to show that By = 0 for all p -elements y in G. Let y be a p -element in G. Let SI be the union of all p -blocks 13 of CG (y) with B G = B and let S2 be the union of the remaining p -blocks of CG (y ). For t = 1 , 2 let V'(CG (y) : S/ ) be the space of all complex linear combinations of irreducible Brauer characters in St . By (IV.6.2) (a T ): E V' (CG (y ) : SI) and By E V' (CG (y ) : S2)' it suffices to show that a ; E V' (CG (y ) : Sl) because then

By = a ; - (a T ): E V' (CG (y ) : SI) n V' (CG (y ) : S2) = (0) . If y is not conjugate to an element of A then a ; = 0 by definition. Thus it may be assumed that y E A. Let a = L ai where ai is a linear

combination of irreducible characters in BOi. If Y is not conjugate to an .element of a defect group of BOi then (ai )Y =10 by (IV.2.4). If x is a p '-element in CG (y ) then X = Xl X2 where Xl E CN (y ) and X2 E 07T, (CG (y» . By (7.6) a ;(x ) = ay (Xl ) ' Thus a ; = ay is a linear combination of irreducible

222

CHAPTER V

(7

Brauer characters of Co (y )!07T' (C O (y » C N (y )/07T, (CN (y » . Hence (4.3) and (7.8) (iii) imply that a � E V' (Co (y ) : SI) as was to be shown. (ii) Clear by (i). (iii) Let y be a p -element in G. If y is not conjugate to an element of A then both a T and a O vanish on the p -section of G which contains y. Hence by (IV.6.3) (a O )B also vanishes on the p -section which contains y. Thus it suffices to show that if y is a p -element in A then a T and (a O )B agree on the p -section of G which contains y. Let C = Co (y ). By the Mackey decomposition (11.2.9) a 0 (yx ) = 2: {a �z n c f (yx ) for every p I-element x in C, where NzC ranges over all the (N, C) double cosets of G. If yx E c A Z then y E c A Z and so by (7. 1) (i) there exists Z l E N with Z I Z E C. Thus NzC = NC. Hence { a ° } Y = { (a Nnct}y . Therefore by (IV.6.1) { ( a O )B } y = ({ (a Nnct} S ) y where S is the union of all blocks Bi of C with B ? = B . B y (7.4) ( a Nn c) y i s a linear combination o f Brauer characters o f N n C all of which have 07T, (N n C) in their kernel. Since C/07T, ( C) = N n C/07T{N n C) there exists a linear combination f3 of Brauer charac­ ters of C all of which have 07T'( C) in their kernel such that (a Nne) y and f3 agree as functions on the set of p i-elements in C/07T,(C). Since y E Z(C) it follows that { ( a Nne)C}y = { ( a Nnc)y } c. Thus if cP is the character afforded by an indecomposable proj ective module of C which has 07T'( C) in its kernel then =

z

({ a Nnct}y, CP)� = « a Nne) , cP Nn c)J.,nc = ( f3 Nn C, cP Nnc)J.,nc = ( f3, CP )� where the prime indicates that the summation in the inner product ranges over p i-elements. Thus by (IV.3.3) { ( a Nnct}y f3 is a linear combination of irreducible Brauer characters of C none of which have 07T' ( C) in their s. c kernel. Hence by (6.5) ({ (a Nn c) y ) y f3 By the second main theorem on blocks (IV.6.1) (a N nc) y is a linear combination of Brauer characters in blocks B ( y ) of N n C with B (y t � UiBoi . Thus f3 is the corresponding linear combination of Brauer charac­ ters in blocks B'(Y ) of C. Since 13 g = B it follows from (7.8) (iii) that f3 S = f3. It remains to show that f3 = (aT ) Y ' Let x be a p i-element in C. Then x = X I X2 with X I a p i-element in N n C and X2 E 07T' ( C). Therefore since B is in Sp ( Tr, G) it follows from (7.6) that -

=

Thus f3 = (a T )y as required. (iv) Clear by (iii). 0

223

ISOMETRIES

7]

LEMMA 7 . 10 Let p E Tr. Assume that (7.2) holds and A is a union of p-sections of N. Let Bo be a p -block of N such that YeN, A : Bo) ;;6 (0) and Bo � Sp ( Tr, N). Let 130 be defined as in (7.8) (ii). Then 13 � � Sp ( Tr, G ). .

PROOF. There exists an irreducible character X in Bo such that X does not vanish on the p -section which contains some p -element y E A. Let D be a defect group of Bo . Thus D is a defect group of 130 • By (IV.2.4) it may be assumed that y E D. Since 07T' (Co (y » is a S7T' - group of Co (y) it follows that 07T, (D Co (D » is a S7T'-group of D Co (D ). There exists a block pair (D, B2) in G with B �N(D)CG(D) = 130• Since 130 covers B2 it follows that 07T, (D Co (D » is in the kernel of B2• By (6.7) (ii) B � = B ? � Sp ( Tr, G). 0 THEOREM 7 . 1 1 . Let p E Tr. Assume that (7.2) holds and A is a union of p -sections of N. Let {A };' be the set of all elements in G whose Tr -part is conjugate to an element of A. Let B 1 , • • • , Bn be all the p -blocks of G in Sp ( Tr, G). For each i let Bi l , Bi2 , . . . be all the p -blocks of N in Sp ( Tr, N) such that V ( N, A : Bij ) ;;6 (0) and 13 g = Bi where 13 ij is defined as in (7. 8) (ii). For each i let Si = U jBij . Then the following hold. (i) V(N, A : B ij) = V7T (N, A : B ij) for all i, j. (ii) V(G, {A };' : Bi ) = V7T ( G, {A };' : Bi ) � r all i. (iii) V7T (N, A ) = V7T (N, A : lJ �= 1 Si ) = EB�=1 V7T (N, A : Si ). (iv) V7T ( G, {A };') = V7T ( G, {A };' : U �=I Bi ) = EB�= l V7T (G, {A };' : Bi ) . (v) V7T (N, A : Si r = V7T (G, {A };' : Bi ) for i = 1 , . . . , n. (vi) If {A };' n N = A then M7T (N, A : Si r M7T (G, {A };' : B; ) for i = 1 , . . . , n. (vii) Suppose that for every p -element y in A and every integer h with (h, I G i) 1, y h E A. The for i = 1 , . . . , n =

=

rank M7T (N, A : Si r = rank M7T ( G, {A };' : Bi ) = dim V7T (G, {A };' : Bi ) . PROOF. (i) and (ii). Since 07T'(C O (Z7T » � 07T'(C O (zp » for all z E {A };' these results follow from (6.5). (iii) The second equality follows from (IV.6.3) and (i). Clearly

V7T (N, A : .CJ = 1 Si ) � V7T (N, A ) .

Suppose that a E V7T (N, A ). For y a p -element in A , let B b e a block of N with (a B ) y ;;6 O. If B 1 ;;6 B2 then by the second main theorem (IV.6. 1) no irreducible Brauer character of C N (y) is a component of both (a Bt) y and (a B2) y . By (7.4) (a B ) y is a linear combination of irreducible Brauer

224

[7

CHAPTER V

characters of eN (y ) with O rr , ( CN (y» in their kernels. Hence B � Sp (71", N) by (6.7) and (7.2). Thus 13 0 � Sp (71", G) by (7. 10) and (iii) follows. (iv) and (v). The second equality in (iv) follows from (IV.6.3) and (ii). Let m = dim V7T (N, A ). Then m = dim V� (G, {A };'). Since V7T (N, A : Si r � V7T ( G, {A };' : Bi ) by (7.9) (i), both (iv) and (v) follow from (iii). (vi) By (7.9) (i v )

7]

a E M7T (G, {A };' : Bi ). Let {3 = aN. Since {A };' n N = A it follows that {3 E V7T (N, A ) and a = {3 T. By (iii) {3 = L{3j with {3j E V7T (N, A : Sj ). Thus by (iv) and (v) {3j = 0 for i � j. Hence {3 E V7T (N, A : Si ). Since (3 E M(N) the result follows. (vii) Since A is a union of p -sections it follows that if z E {A };' and (h, 1 G I) = 1 then Z h E {A }�. Thus by (IV.6.9) rank M7T (N, A : S; ) =

dim V7T (N, A : Si ) and rank M7T (G, {A };' : Bi ) = dim V7T (G, {A };' : Bi ). The result follows from (v). 0

COROLLARY

7.12. Suppose that (7.2) holds and

then a T is a generalized character of G.

PROOF.

Clear by (7.9) (iii) and (7. 1 1) (iii).

71"

= {p}. If a E M7T (N, A )

0

Suppose that (7.2) is satisfied. Let X be a set of characters of N/07T, (N) with M(N, A : X) � M7T (N, A ). The set X is coherent if T can be extended to a linear isometry from M(N : X) into M( G). If � is an irreducible character in X and T is coherent theri " C W = 1 . Thus ± C is an irreducible character of G. The next result yields some information concerning the values assumed by C on certain elements of G. This result was stated without proof in Feit [1967c] , p. 175 . Unfortunately the fact that A has to satisfy assumption (i) of (7. 1 3) below was omitted in the statement.

THEOREM 7.13. Assume that (7.2) is satisfied and 07T, (N) is a S7T'-groUP of N. Let X be a set of irreducible characters of N/07T, (N). Suppose that the

following assumptions are satisfied. (i) If P E 71" then the set of p -singular elements in A is a union of p -sections of N. (ii) X is coherent. (iii) V7T (N, A : X) = V7T (N, N 0 7T ' (N) : X). Then the following conclusions hold. (i) If � E X then (C, a T)o = ( � ;. , a )N for all a E V7T (N, A : X). -

225

(ii) Let X = {�s } and let g = L£'s (l)£,s . For each t there exists YI E M(N) such that (YI, £,s ) = 0 for all s and an integer al such that

Furthermore there exists Y E V (N) and a rational number

a

such that

£, ; (z ) = £,r ( z ) + £'t (l) {ag (z ) + y (z )} for all z E A.

M7T (N, A : S r � M7T (G, {A };' : Bi ).

Let

ISOMETRIES

(i) Let 71"1 be the set of all primes p E 71" such that if a E V7T (N, A : X) then a vanishes on all p -singular elements in N. Let 71"z = 71" - 71"1 ' For each p E 71"z let S (P ) be the union of Sp ( 71", G) and all p -blocks of G of defect O. We will first show that if � E X and p E 71"z then e£,T E S (P ) for e = 1 or - 1. Choose p E 71"z . Let £' E X and let e = 1 or - 1 such that e£,T is a character of G. Let Ap be the set of all p -singular elements in A. By assumption (i) Ap is a union of p -sections and (7.2) is satisfied if A is replaced by Ap . Furthermore V7T (N, Ap ) � Vrr (N, A ) and T defined on V7T (N, Ap ) is the restriction to V7T (N, Ap ) of T defined on V7T (N, A ). Let X = {£,s } with £' = £'1 . For each s let as = £,s (I)£, - £,(l)£,s . By assump­ tion (iii) {as } is a basis of Vrr (N, A : X). Since p E 71"z there exists t such that at does not vanish on Ap . Let at = {31 + {3z where {31 vanishes on N Ap and {3z vanishes on Ap . Thus {31 � 0 and {3 � = ({3 fY)S (P) by (7.9) (iii) and (7. 1 1) (iii) applied to (31 E V7T (N, Ap ) . Therefore by (IV.3.4) a ; = () + 11 where () E V (G : S (p » and 11 is a complex linear combination of charac­ ters afforded by projective R [GJ modules not in S (P ). Furthermore () � 0 and a ; = £'t (l)£,T - £,(1)£, ; . Thus either e£,T E S (P ) or �t ( l )C vanishes on all p -singular elements of G. In the latter case C vanishes on all p -singular elements of G and so e£,T is in a p -block of defect O. Hence in any case e£,T E S (P ). Let £' E X. Define 1/1 as follows :

PROOF.

-

1/1 (z ) = £,T (z ) =0

if z is p -singular for some p E 71"z and the 71"-section of z in G meets A, otherwise.

Since e£,T E U p E 7T2 S (P ) it follows from (6.5) that C is constant on p -singular 71"-sections which meet A for all p E 71"z . Thus 1/1 E V7T (G). Hence if a E V7T (N, A : X) (7.7) implies that

(£,T, a T)o = ( l/I, a T )o = ( I/IN, a )N = (£';. , a )N '

226

CHAPTER V

[8

(ii) Let ( ? ;)N = ?r + '22 ari ?i + ')' where (?h ')'1 ) = 0 for all j and implies that for all i, j, t ,

t.

£&2 ati - (1) a l ?, l

for all

.

]

and

t.

at/ .

{ ��l) g (z ) ?,

+ ')'t

(z )

}

= ?, ( 1)

{(

s g ( z ) + ')'s (z ) ?s l)

s

}

for all Z E A. Thus the second equation in the statement of (ii) follows by setting and

8. 7r-heights

In this section we will use the notation of Chapter IV, section 8. Let

v ( I G I) = a .

Let e be a centrally primitive idempotent in R [ G ] . Let B = B (e ) be the block corresponding to e. L et D = D (B ) be the defect group of B and let d = d (B ) denote the defect. Thus v (1 G : D (B ) i) = a - d (B ). In this section we will primarily be concerned with ChR (G, B ) and ZR (G )e. Observe that ChK ( G, B ) is the dual space of ZK (G )e. See BroU(� [1976b] , [1978b], [1979] , [1980] for the results in this section and further related results.

LEMMA

If a E ZR (G)e and e E ChR (G, B) then v ( e, (a)) � v ( 1 G : D (B ) I ) .

81 .

.

By (111.6.6) e = Tr ( ) for some E R [ G ] . Since a E ZR (G) it follows that ae = Let { Xi } be a cross section of D in G. Since e is a class function on G it follows that

PROOF.

gc

Trg (ac).

c

acxi) = I G : D I e (ac ).

0

For a E ZR (G)e let s (a ) be the largest integer such that 1Ts(a ) 1 O (a ) in R for all e E ChR (G, B ) . Define the 1T -height of a, hIT (a ) by ( 1T h,, (a » = (1T·s (a ) I G : D I - I) ,

The first equation in the statement of (ii) follows by setting at = lf Z E A then ? ( l) C (Z ) - ?I (l) ? ;(S) = ?s (l) ?I (Z ) - ?, ( l)?S (Z ) for all s and t. Thus the first equation in the statement of (ii) implies that ?s (1)

O(a ) = e (ae ) = e ('22x � I This implies the result.

(l )a'i = ?i (l)ati for all i, j and t. Hence in particular

227

1T-HEIGHTS

Then (i)

?i (l)a'i - ?i (l)a'i = «C)N' ?i (l) ?i - ?i (l) ?i )N - ( ?,' bi (l) ?i - ?i (l) ?i )N = ( C' ?i ( l) C - ?i (l) C) o - {?i ( 1)8it - ?i (1)8jl } = O .

Therefore ?i

8]

where the parentheses denote a fractional ideal of K. For e E ChR (G, B ) let t ( e ) be the largest integer such that e I (Ii) I e (a ) in R for all a E ZR (G )e. Define the 1T height of e, h7T ( e ) by -

(1T h " (Ii) = ( 1T ' (IJ) I G : D I- I ) . o

By (8. 1)

h7T (a ) and h7T ( 0 ) are nonnegative integers for a E ZR (G)e

E ChR (G, B ).

and

Xu E B. By (IV.7. 1) v (Xu (1 » = v (Xu (e ) . For a E ZR (G) XU (a )JXu ( 1) = Wu (a ) E R. Therefore (Xu ( 1» = (Xu (e » � (1T h,, (xu ) I G : D i) � (Xu (1 » . Thus if (1T m ) = (p) it follows that h7T (Xu)lm is the height of Xu. In particular Xu has height 0 if and only if it has 1T-height O. This argument also shows Suppose that

that h7T (e ) = O. Since e is a primitive idempotent in ZR (G) it follows that ZR (G)e is a local ring. Thus ZR (G)eJJ (ZR (G)e) = R as K was chosen large enough. Let A be the central character of ZR (G )e. By (IV.4.2) A = for any

Wu

Xu E B.

LEMMA 8.2. Let a E ZR (G)e. The following are equivalent. (i) For every e E Ch (G, B ) with h7T ( e ) = o, v ( e (a » = v ( I G : D I ). (ii) There exists e E ChR (G, B) with v ( e (a » = v(1 G : D I). (iii) h7T (a ) = O. (iv) a � J (ZR (G)e). R

(i) � (ii) and (ii) � (iii) are immediate. (iii) � (iv). Suppose that a E J (ZR ( G ) e ) . Then (a (1) = (a ) = o for all E B. Hence e (a ) == o (mod 1Tp v{/ O : O i l) for all e E ChR (G, B ) and so h7T (a ) rf O. (iv) � (i). Let e E ChR (G, B) with v ( e (a » > p v, where v = v (1 G : D I). Let 0 = Then

PROOF.

Xu )JXu

Xu

'22 cuXu.

O (� ) =

p

L Cu � p = L cu (� p ) wu ( a).

Wu

228

CHAPTER V

[8

By assumption ( 8 (a )jp V ) = O. Since a � J (ZR (G)e ) it follows that Wu (a ) = A (a ) I: 0 for all u. Therefore

I cu

(X�P» )

==

0

(mod 1T )

corresponding to B. L et X = Xu be it character of height 0 in B. L et Co be the conjugate class of G containing xy. Since D is a Sp-group of Cc (xy ) and X (xy ) == X (x ) � O (mod 1T ) it follows that v (x (xy » = v (X (x » = O and so Therefore A (Co e ) = A (Co) = Wu (Co) I: O . Thus Co e � J(ZR (G)e ) but Co e E ZR (G : S )e. 0

Thus h-rr ( 8 ) I: O.

8.6. Define a on G by a (x ) = 1 if x is a p -element and a (x ) = 0 otherwise. Then a B E ChR (G, B ) and h 7T ( a B ) = O .

LEMMA

0

8.3. Let 8 E ChR (G, B ). The following are equivalent. (i) For every a E ZR (G)e with h-rr (a ) = 0, v ( 8 ( a » = v (1 G : D /). (ii) There exists a E ZR (G)e with v ( 8 (a » = v (1 G : D /) . (iii) h7T ( 8 ) = O.

LEMMA

(i) � (ii) and (ii) � (iii) are immediate. (iii) � (i). This follows from the fact that (8.2) (iii) implies (8.2) (i).

PROOF.

0

Let 8 E ChR (G, B ). Then h7T ( 8 ) = 0 if and only if v ( 8 (1» =

If v ( 8 (1» = v(1 G : D I) then clearly h7T (8) = O. Suppose that h7T ( 8 ) = O. Since h7T (e ) = 0 and v ( 8 ( e » = v ( 8 (1» , the result follows from (8.3). 0

PROOF.

8.5. Let y be a p-element in G and let S be the p -section which contains y. (i) If y is not conjugate to an element of D then ZR (G : S )e = (0). (ii) If Y is not conjugate to an element of Z(D ) then ZR (G : S ) e � J(ZR (G)e ). (iii) If Y is conjugate to an element of Z(D ) then ZR (G : S ) e g;, J(ZR (G)e ).

LEMMA

(i) By (IV.2.4) 8 (x ) = 0 for all x E S and all 8 E ChR (G,B ). Since ChK (G, B) is the dual space of ZK (G)e this implies the result. (ii) This follows from ( 1 .5). (iii) It' may be assumed that y E Z(D ). By (111.6. 101 there exists a p ' -element x such that D is a Sp -group of Cc (x ) and A ( C) I: 0 if C is the conjugate class containing x and A is the central character of R [GJ

PROOF.

22 9

7T'-HEIGHTS

v (Wu ( Co» = v (1 G : D 1) - v (X (1» = v(wu ( C» = O .

and so

LEMMA 8.4 . v (1 G : D /).

8]

PROOF. Let {qi } be the set of all primes distinct from p which divide 1 G I · Thus q � I E R for all i. Hence by (IV . 1 .3) a E ChR(G) and so a B E ChR (G, B ). Let 8 E ChR (G, B ) with h-rr ( 8 ) = O. Let a = L 8 (x -l)x. Let S = Greg and let 8s be defined as in Chapter IV, section 8. Then 8s (a )e E ZR (G)e and by (IV.8.S) a B (8s (a )e ) = a ( 8s (a )e ) = a ( 8s (ae » = a ( 8s (a » = 8 (1) .

Thus h7T ( a B ) = 0 by (8.3).

0

8.7 (Broue [1978b]). Let B be a block of G with defect group D. Let e be a centrally primitive idempotent corresponding to B. Let y E Z(D ) and let S be the p -section which contains y. Let a = ae = LXES ax x. Then v(ay ) ;?: v (1 Cc (y ) : D I ). Furthermore v ( ay ) = v(1 C c (y ) : D I) if and only if a � J(ZR (G)e ).

THEOREM

Let a be defined as in (8.6). By (8. 1) v(a B (a » ;?: v ( 1 G : D I). By (8.2) equality holds if and only if a = ae � J (ZR (G)e ). Since

PROOF.

a B (a ) = a (ae ) = a (a ) = ay 1 G : Cc (y ) I , equality holds if and only if

COROLLARY Then

v (ay ) = v(1 Cc (y ) : D I).

0

8.8 (Brauer [ 1 976a]). Let B be a block with defect group D.

where Xu, 'Pi ranges over all irreducible, irreducible Brauer characters respec ­ tively in B.

[9

CHAPTER V

230

PROOF. Let e be the centrally primitive idempotent of R [ G ] correspond­ ing to B. By (IV.7. 1) and ( IV . 7.2 ) e E ZR (G : Greg) and e = 2": ax x with I G I a l = 2": Xu (1)2 = 2": 'Pi (1) 0 then nB < pd - h .

LEMMA 9. 9.

PROOF. Let Xu be an irreducible character in B of height h. Let S (y, B ) be a major subsection associated to B. By (9.2) m �� B ) '" O. By (9.4) (i) (iv) p d - h m �� B ) is a totally positive algebraic integer. By (9.3) �p d- h m �� B ) :<s pd - h , where the sum ranges over all major subsections associated to B. Thus the arithmetic-geometric inequality implies that

Thus nB :<S p d - h . If equality holds then m �� B ) p -( d - h) for all major B subsections and in particular m �l�B ) = P -( d - h) and so v(m �1� » = h - d. By (9.4) (iv) h = O . 0 =

Let B be a block with defect group D. Let 1 D 1 p d and let n be the largest integer such that p n :<S e, the inertial index of B. Let h be the height of an irreducible character in B. Then h :<s n + d v(I Z (D ) I) .

COROLLARY 9. 10.

=

-

Thus

p h I Z(D ) I :<s p de
O.

T(D)/D C G (D ) is isomorphic to a p '-subgroup of GL, (P ).

; =1

The result follows from (9. 10) . . D

Suppose that B has an abelian defect group of rank r > O. Then the height of an irreducible character in B is less than �r(r + 1) .

COROLLARY 9.12.

PROOF. Clear by (9. 1 1).

D

The estimate in (9. 1 1) is extremely crude and obvious improvements can be made in (9. 1 1) and (9. 1 2). However these methods do not appear strong enough to answer Question (IX) in Chapter IV.

Let B be a block of defect d with an abelian defect group D. Let k k (B) denote the number of irreducible characters in B. Suppose that the inertial index of B is 1 . Then B contains a unique irreducible Brauer character, k (B ) = p d, every irreducible character in B has height 0, every decomposition number is 1 and the unique Cartan invariant is p d.

THEOREM 9 . 1 3 =

PROOF. By (9.8) there are at least pd subsections associated to B. Thus by (IV.6.S) k (B ) � p d. Hence by (lVA.21) it suffices to show that B contains exactly one irreducible Brauer character. This will be done by induction on

IGI·

If D � Z(G) the result follows from (4.6). Suppose that D g Z(G). Choose y E D, y � Z(G). Let S (y, B ) be a subsection associated to B. Since D is abelian it is a major subsection. Let {Xu } be the set of all ir:educible characters in B. As D is a defect group of 13, the inertial index of B is 1 . As y � Z( G), induction may be applied to C G (y) '" G. Thus 13 �as a unique irreducible Brauer character 'P r . Hence the Cart an matrix of B is (p d ). By (9.5) d � ; '" 0 for all u. The arithmetic-geometric inequality implies that

238

CHAPTER V

[9

Therefore

Since there are p d subsections associated to B, (IV.6.5) implies that E has a unique irreducible Brauer character for each such subsection S (y, E ). This is the case in particular for S (1, B ). 0 If D is not assumed to be abelian in (9.13) the result is easily seen to be false . Consider for instance the case G = D. A suitable generalization of (9. 13) has been proved by Broue and Puig [1980] . It is a good deal more complicated than (9. 1 3). We next prove a technical preliminary result. LEMMA 9. 1 4 .

Let (gij ) be a positive definite symmetric matrix with integral entries. Let Q be the corresponding hermitian form. Let q be the minimum of Q (e, e ) as e ranges over all nonzero vectors with integral coordinates. Let s be a primitive p nth root of 1 for some n and let tr denote the trace from Q( s ) to Q. If a is a nonzero vector whose entries are algebraic integers in Q(s ) then tr(Q (a, a )) � [Q(s ) : Q] q = p n - I (p - 1)q. i

PROOF. Let t = p n-l (p - 1). Let 'L::� ei s , where each ei is a vector with i integer coordinates. Then Q ( a, a ) = 'L:,j:o Q (ei' ej ) . Since tr(1) = t, tr(s ) = _ p n - 1 if i == 0 (mod p n -1) but i � O (mod p n ) and tr(s i ) = O otherwise, it follows that tr( Q ( a, a )) = p n - I

1-1

2: As ,

s =o

(9. 15)

where

i,j with i,j ranging over all values congruent to s mod p n - I and 0 � i,j < can be rewritten as As

with

i,j

= 2: i<j Q (ei - ej, ei - ej ) + 2:i Q (ei' ei )

as above.

t. This (9. 16)

9]

23 9

SUBSECTIONS

For a fixed s let No = No be the number of i with O :!S i < t, i == s (mod p n- I ) and ei = O. Let Nl = N� be the number of i with 0 � i < t, i == s (mod p n - I) and ei nonzero. Then No + NJ = P - 1. If No = p - 1 then As = O. If No = 0 then the second term in (9. 16) is at least (p - 1)q and so As � (p - 1)q. If 0 < No < p - 1 then the first term in (9. 16) is at least No NI q � No q and the second term is at least NJ q and so As � (p - 1)q. Since a -1 0, some ei is nonzero . Thus As � 0 for all s and As � (p - 1)q for at least one value of s. The result follows from (9. 1 5). 0

Let . B be a block of defect d. Let S (y, E) be a major subsection asspciated to B. Let k (B) denote the number of irreducible characters in B and let I (E) denote the number of irreducible Brauer characters in E. (i) Let (; be the Cartan matrix of E and let 6 be the quadratic form corresponding to p d {;- I . Let q (E ) be the minimum value assumed by 6 on the set of all nonzero vectors with integral coordinates. Then q (E)k (B ) � p d I (E) . (ii) If B contains no irreducible characters of positive height then k (B )2 � P 2d I (E ) .

THEOREM 9.17.

PROOF. Let {Xu } be the set of all irreducible characters in B. Choose n so that if s is a primitive p n th root of 1 then m �� B ) E Q( s ) for all u, v. (i) By (9.5) m ��13 ) -1 0 if XW has height O. Thus by (9.4) (ii)

m �� 13 ) = 2: I m �� 13 ) 1 2 > 0 v

for all

u.

By (9.4) (i) pdm �� 13) = 6 ( a , a ) , where a = ( d �i) ' Hence p n - l (p - 1)q (E) � tr (p d m �� 13 ») by (9. 14). If this is summed over all u then (9. 14) (iii) implies that p n - l (p _ 1)q (E)k (B) � tr(p d I (E )) = p n - 1 (p - 1) p d I (B ) . The result follows. (ii) By (9.5) pd m �� B ) is a nonzero algebraic integer for all u, v. By (9.4) (ii) 'Lu l m � 13) 1 2 = m � 13 ) . If k = k (B ) and t = p n - l (p - 1) the arithmetic-geometric inequality implies that

= kt1 2: p 2d m ( y, B )a _

a

vv

,

240

CHAPTER V

[10

where (T ranges over the Galois group of Q(£ ) over over v, (9.4) (iii) implies that

Q. If this is summed

k � k1t L L p2d m 0 and Co (y) has deficiency class r for every element y of order p in G then G has deficiency class r. PROOF. Let y be a p -element, let Bo be a nonprincipal block of Co (y ) and let d be the defect of Bo . By the third main theorem on blocks (5.4), B&' = B is defined and the defect of B is at least d. By (6.2) B is not the principal block of G. Thus if G has deficiency class r then d < r. Hence Co (y) has deficiency class r. Conversely suppose that G is not of deficiency class r with r > O. Let B be a nonprincipal block of defect d � r and let D be the defect group of B. Since r > 0 there exists an element y of order p in Z(D ). By (6.2) and (9.2)ti) there exists a nonprincipal block of Co (y ) of defect d and so Co (y) is not of deficiency class r. 0 An old result of Landau [1903] implies that for a given integer k, there are only finitely many finite groups which have at most k classes. Thus (IV.4. 1 8) implies the next result.

11]

GROUPS WITH A GIVEN DEFICIENCY CLASS

THEOREM 1 1 .6.

247

There exist only finitely many groups G of deficiency class whose Sp -group has a given order.

In case ( 1 1 .6).

p = 2, Brauer [ 1 97 4c]

0

has proved the following generalization of

Let G be a group of deficiency class r � 0 for the prime 2. Suppose that a Sz-group P of G contains an elementary abelian subgroup of order 2'+1 . Let I P 1 = 2 a • Then there exists a bound f(a ) depending only on a such that I G I � f(a ). THEOREM 1 1 .7.

The proof of ( 1 1 .7) is not very difficult but depen ds on specia l prope rties of involu tions. It is not known wheth er an analog ous result is true for odd primes . . If y is a p -elem ent in G it is possib le to apply (1 1 .4) to Co (y). Thus ( 1 1 .5) can be used to give an induct ive characteriza tion of group s of positiv e deficiency class. Such a characterization does not seem to be possib le for group s of deficiency class 0 since the secon d statem ent of ( 1 1 .5) is false in this case. For instan ce let q be a prime with q == 1 (mod p) and let G be a Frobe nius group of order pq. Then G does not have deficiency class o for p but I Co (y) I = p for every eleme nt y of order p in G and so Co (y ) has deficie ncy class 0 for every eleme nt y of o�der p in G. This construction shows the existe nce of infinit ely many groups of deficie ncy class 1 with a Sp -group of order p. It follows from ( 1 1 . 6) that an analog ous situati on canno t occur for deficie ncy class O. It has been shown that a simple group of Lie type in charac teristic p has deficiency class 1, Dagge r [ 1971], 1. E. Humphreys [ 1 97 1 ] . This also shows the existen ce of infinit ely many group s with deficiency class 1 . Amon g these there are only finitel y many with a given Sp -group . A related result of a slightly different sort can be found in Tsushima [1977] .

irreducible R [G] module up to isomorphism. Then irreducible R [ G ] module up to isomorphism. 2]

RADICALS AND NORMAL SUBGROUPS

PROOF. Clear by (1. 19.4).

249

B

contains a unique

0

LEMMA 1 . 3 . Let H <J G and let V be a projective indecomposable R [GJ module. Then VH = E9 U; where each U; is a projective indecomposable R [H] module and for each i there exists x E G with Vi = V�.

CHAPTER VI

PROOF. Immediate by (1. 13 .6) and Clifford' s theorem (111.2. 12). 1 . Blocks and extensions o f

R

will be used The notatio n introd uced at the beginning of Chapt er III on will also throug hout this chapte r. The follow ing assum ptions and notati be used. K is a finite extension of Qp , the field of p -adic numb ers. A R is the ring of integers in K. i n K of K By (111.9. 10) and (1. 19.4) there exists aAfinite unramified. extens � . condItIons are such that if R is the ring of intege rs in K then the follow mg

A satisfie d. K is a G. of H up subgro every for [H] R of field g splittin a is R (I) splittin g field for every p '-subgr oup H of G. (u) be th� (II) Let B be a block of R [G] with defect group D.A Let such that If [G] R of B block a Galois group of K over K. Then there exists ' V is an irreducible R [ G] module in B then _

I

V Q9 R R

=

r-1

L yO"j

j =O

with Y E 13.

13 i If furthermore 13 U) = u is the block of is a defect group of B U) by (111.9. 10).

R [ G]

which contai ns v ui then

R [G] module in B is R [D]-projective and there exists an irreducible R [G] module in B with vertex D.

LEMMA 1 .2. Let B

0

be a block of R [G]. Suppose that B contains a unique 248

R [G] covers

the block

b of R [H]

LEMMA 1 .4. Suppose that H <J G and the block B of R [ G] covers the block

. b of R [H]. Let D be a defect group of B. Suppose that D CG (D) � H. Then the following hold. (i) B = b G is the unique block of R [GJ which covers b. (ii) D is a defect group of b. PROOF. (i) Choose the notation so that B covers G, where G is defined as in (II). by (V.3.9) B is regular and so by (V.3.7) 13 = G G is the unique block which covers G. Then 13 U) is the unique block which covers G U). This implies that B is the unique block which covers b. The definition of the Brauer homomorphism implies that B = b G . (ii) By (V.3. 14) D is a defect group of G and so by (II) D is a defect group of b. 0

D

Then !!. very LEMMA 1 . 1 . Let B be a block of R [ G] with defect group D.

PROOF. Clear by (111.4. 14).

S uppose th at H <J G. The block B of A A if B covers b U) for some j.

0

2.

Radicals and normal subgroups

Throughout this section H is a normal subgroup of G. The results in this section up to (2.7) are mostly due to Knorr [ 1 976] . Some related results have been proved by Ward [1968], Huppert and Willems [ 1 975], Willems [ 1 975] .

LEMMA 2. 1 . Let V be an R [H]-projective R [ G] module and let M = V/ VJ(R [G]). Then M is R [H]-projective if and only if VJ(R [G]) = VJ(R [H]).

250

[2

CHAPTER VI

PROOF. Let I = leR [ G D and let 10 =J(R [HD. As loR [ G ] = R [G]lo, Vlo

is an R [ G ] module. Suppose that VI = Vlo. Let A = R [ G ] , B = R [H] and S (l.4. 1 1) implies that M is R [H]-projective. Suppose conversely that M is R [H]-proj ective. Then _

=

10•

Then

O � ( Vl/ Vlo)H � ( V/ Vlo)H � MH � O is a split exact sequence as ( V / Vlo)H is completely reducible. Thus O � Vl/ Vlo � V/ Vlo � M � O is a split exact sequence as M is R [H]-proj ective. Hence V / Vlo = M E9 Vl/ Vlo and so VI/ V10 = ( V / VJo)l Ml + ( VI / V10)1 = ( VI / V10)1. Thus Vl/ Vlo = (0) by Nakayama's lemma (I.9.8) and so VI = Vlo. D =

COROLLARY 2.2. Let V be a projective indecomp�sable R [ G ] module and let M = V/ VI ( R [ G D. Then H contains a vertex of M if and only if Vl ( R [ GD = Vl( R [ il D· PROOF. Clear by (2. 1).

D

2J

As U;1 (R [H]) � U;1 ( R [ G D for all i the result follows. (iv) � (i). By ( 1 . 1) there exists an irreducible R [ G ] module M in B whose vertex is a defect group D of B. As M = V / Vl(R [ G D for some projective indecomposable module V, the result follows from (2.2). D

COROLLARY 2.4. Let V be an irreducible R [H] module. Assume that every component of vG lies in a block with a defect group which is contained in H. Then VG is completely reducible. PROOF. By assumption there exist centrally primitive idempotents {ei } such that a defect group of each ei lies in H and such that VGe = VG, where e = L ei . By (2.3)

(

be the block corresponding to e. The following are equivalent. (i) H contains a defect group of B. (ii) Vl( R [ G D n = Vl( R [HD n for all R [ G ] modules V in B and all integers n ;?; O. (iii) l(R [ G De = l( R [HD R [ G ] e. (iv) Vl(R [ G D = Vl(R [HD for all projective indecomposable R [ G ] modules V i n B.

)

VGJ(R [GD = V Gl(R [ G De = V _® R [ G ] l( R [ G De R[ H ) = V ® l (R [ G D e R[ H ) = V ® l ( R [HD R [ G ] e = VJ (R [ H] ) ® R [ G ] e R[H) R[H) = (0).

Hence VG is completely reducible.

THEOREM 2.3. Let e be a centrally primitive idempotent in R [ G] and let B

25 1

RADICALS AND NORMAL SUBGROUPS

D

COROLLARY 2.5. Suppose that p ,{ 1 G : H I . Let V be an irreducible R [H] module. Then VG is completely reducible.

_

PROOF. (i) � (ii). By ( 1 . 1) every module in B is R [H]-proj ective. Thus (ii) follows from (2. 1) by induction on n. (ii) � (iii). This is clear 'as eR [ G ] = R [ G] e is a module in B. (iii) � (iv). Let R [ G ] e = EB Vi , where each Vi is indecomposable. Each proj ective indecomposable R [ G ] module in B is isomorphic to some Vi . By the hypothesis EB VJ( R [ G D = R [ G ] el( R [ G D = R [ G ] el(R [HD = EB VJ( R [HD·

PROOF. Since H contains a defect group of every block the result follows from (2.4). D The next result is quite old. See e.g. Green and Stonehewer [1969] , Villamayor [ 1 959] .

COROLLARY 2.6. l(R [G]) = J(R [HD R [ G ] if and only if p

,{

I G : HI .

PROOF. If l(R [ G]) = l(R [HD R [ G ] then R [G]/l(R [GD is R [H]­

projective by (2. 1). Thus every irreducible R [ G ] module is R [H]­ projective and so p ,{ I G : H I by ( 1 . 1). If P ,{ 1 G : HI then every R [ G ] module is R [H]-proj ective. Thus (2. 1) implies the result . D If W is an indecomposable R [ G ] module let w ( W) = WH ( W) denote

[2

CHAPTER VI

252

the integer so that WH is a direct sum of w ( W) nonzero indecomposable R [H] modules. The integer w ( W) is called the H-width of W or simply the width of W. Observe that if U is a proj ective R [ G ] module then w ( U) = w ( U / UJ( R [ G ]) = w ( U / UJ( R [H])

by Clifford's theorem (III.2. 1 2) and (I. 1 3 .7). THEOREM 2.7. Let B be a block of R [ G] which covers the block b of R [H] .

3]

SERIAL MODULES AND NORMAL SUBGROUPS

253

UJ ( R [ G ] )" / UJ( R [ G ])n+ l � (0). Let UH = ��= 1 Pi , where each Pi is a projective indecomposable R [ H ] module. By (V.2.3) and (1 .3) each Pi is

serial. By (2.3)

VH = ( UJ( R [ H] )" / UJ( R [ H] ) n + l )H

w

=

P;l ( R [H])" / PiJ ( R [ H] )n + 1 E9 i= l

and so w ( V) ::s; w = w ( U ) . The minimality of w now implies that V is irreducible and w ( V) = w. The result follows as the Cart an matrix is indecomposable. 0

Assume that H contains a defect group of B. Then the following statements are equivalent. (i) Every indecomposable projective R [ G ] module in B is serial. (ii) Every indecomposable projective R [H] module in b is serial. If furthermore (i) or (ii) is satisfied then all irreducible modules in B have the same width.

It should be mentioned that in general not all irreducible R [ G ] modules in V will have the same width. For instance suppose that p = 2, H = SL2(8), G is the semi direct product of H with a cyclic group of order 3 and R contains the field of 8 elements. Then the principal block of G contains an irreducible R [ G] module of degree 6 and width 3, and of course it also contains the one dimensional module of width 1 .

PROOF. Suppose that (i) is satisfied. Let U be a proj�ctive indecomposable R [ G ] module in B of maximum width. Let UH = E9 �= 1 Pi , where each Pi is a projective indecomposable R [H] module. Then w = w ( U ) =

COROLLARY 2.8. Let B be a block of R [ G ] which covers the block b of

w ( U / UJ( R [H]) .

Let n be a positive integer such that PJ ( R [H] )" � (0). By (1 .3) PiJ( R [H] )" � (0) for i = 1, . . . , w. Let V = UJ( R [H])" / UJ( R [H] )"+I . Then w

VH = E9 P;l(R [H] )" / PiJ( R [H] ) ,, +

1

R [H] . Assume that H contains a defect group of B. Then the following statements are equivalent. (i) Every indecomposable projective R [ G ] module in B is serial. (ii) Every indecomposable R [ G ] module in 13 is serial. (iii) Every indecomposable projective R [H] module in b is serial. (iv) Every indecomposable R [ H] module in B is serial.

PROOF. Clear by (I. 16. 14) and (2.7).

i=1

and s o w ( V) � w . By (2.3) V i s irreducible since U i s serial. Let Uo b e the projective indecomposable R [ G ] module which corresponds to V. Then w ( Uo) = w ( V) � w ( U) . Hence the choice of U implies that w ( Uo) = w. The indecomposability of the Cartan matrix (I. 16.7) now implies that every projective indecomposable R [ G ] module in B has width w and so every irreducible R [ G ] module in B has width w. Thus (i) implies the last statement. Furthermore PiJ( R [H] )" / PiJ( R [H] ) ,, +1 is irreducible as w ( V) = w. Hence each Pi is serial. Since w is the width of every projective indecomposable R [ G ] module in B, the previous paragraph implies that if U is any projective indecom­ posable R [ G ] module in B then UH is a direct sum of serial modules. Hence (ii) holds. It remains to show that (i) follows from (ii). Let U be of minimum width among the proj ective indecomposable R [ G] modules in B. Let n be a positive integer such that V = _

0

3. Serial modules and normal subgroups

Throughout this section the following hypothesis and notation will be assumed to hold. HYPOTHESIS 3. 1 . (i) H I (y ) = I G I ( c1>" 'P l)= I G I .

Y

y

Therefore in particular c (1) = lPl (1) = P n is the order of a Sp -group P of G. Hence P contains all the p -elements in G and so P i = Xu + Xv' (iii) Let X = XO if I A I = 1 and let X = XA for some A E A if I A I I- 1. Then

ooX (x ) + u! ouXu (x ) = O =1 for every p '-element x in G. (iv) Suppose that G = Na - 1 . If I A I I- 1 then 01 = . . . = De = 1 and D O = - 1 . If I A I = 1 then the notation can be chosen so that 01 = . . . = De = 1 and DO = - 1. THEOREM 2. 16. Suppose that K = K. Then every irreducible character in B is of height O. Furthermore Xu (1) Dut/l (1) (mod p a ) for u = 0, . . . , e. ==

XA (1)

==

- ooet/l(l)

(mod p a ) for A E A.

The congruences in (2. 16) are very strong if D is a Sp -group of G and get weaker as the power of p in I G D I gets larger. If for instance I D 1 2 / 1 G I they contain no information. Suppose that K = K. For 0 � k � a - I let t/lk be the unique irreducible Brauer character in bk . Let {z } be a cross section of T (bk ) in Nk. Then {b %} is the collection of blocks of Ck covered by Bk • Thus {t/I %} is the collection of irreducible Brauer characters in blocks covered by Bk • If x E Dk - Dk+ 1 then CG (x ) = Ck . Let d (u, x, t/I %) or d CA, x, t/lt) denote the corresponding higher decomposition number for Xu or XA respectively where u = 1, . . . , e and A E A or possibly u = 0 in case I A 1 = 1. :

THEOREM 2.17. Suppose that K = K. For 0 � k � a - 1 there exist E k = such that for x E Dk - Dk + 1 the following hold. d (u, x, t/I %) = EkOu for 1 � u � e or for u = 0 if I A I = 1 .

- EkOO d (A, x, t/l k) = 1 Ck I z

� L.J

w E T(bd

±1

'WZ (x ) for A E A,

where , is an irreducible constituent of the restriction to D of an irreducible constituent of ACK in bk. Furthermore if 0 � j � a - I and E :), . . . , E j are the signs for the group G then there exists y = ± 1 such that E ; = yEi for 0 � i � j. If G = G for some j with O � j � a - 1 then Ej = . . . = Ea - l = 1 unless

2]

STATEMENTS OF RESULTS

279

= 2 and j = 0, in which case the notation may be chosen so that Ei = 1 for O � i � a - 1.

p

It follows from (2.17) that if I A I I- 1 and A E A then XA has a higher decomposition number not equal to ± 1 unless p = 2, e = 1 and A 2 = 1 . In the latter case if XI is the nonexceptional character in B and is the irreducible character of D with kernel equal to DI, we can define X � = XAa for A E A U {I} and note that {X a satisfies (2. 1 1}-(2. 17) with Do, ot, E I , . . . , Ea - I unchanged but Eo replaced by - Eo. Then Xa = X ; is the nonexceptional character. If A runs over A in (2.17) then it is easily seen that , runs over a complete set of representatives of the T(bo)-conjugate classes of nonprinci­ pal irreducible characters of D. ll'

THEOREM 2.18. Let r be the number of blocks of R [G] which are algebrai­ cally conjugate to B. Then the Schur index mo of XA over K is dOi for some i and is independent of A E A, and II XA W = m� r is independent of A E A. By (2.9) m = It�. (Li' Li ) is independent of i. Let m o , r be defined as in (2. 18). THEOREM 2. 19. (i) If 1 � u, i � e then Qp (Xu ) = Qp (4\ ), mQp (Xu ) = 1 and IIXu W = m. (ii) Given i with 1 � i � e then either there exists u, v with 1 � u, v � e such that cJ>i = Xu + Xv or there exists u with 1 � u � e and cJ>i = Xu + (e/emo)Xo. If A = 1 and e I- e then replacing K by a purely ramified\ extension it may be assumed that = 1 and so Xo can be recognized from the decomposition of the cJ>i . In this case it is called the exceptional character in B. Thus B contains an exceptional character unless I D 1 - 1 = e. Let aui = dUi for u I- 0 and let a Oi = e /(ema) if d Oi I- O. Thus if 1 � i � e, cJ>i = auiXu + aviXv for suitable u, v. It follows from (2.19) that I (e/e ). Since e /e = [ Qp (Xo, i = 11 + 11 ' for some i where 11 ' = 0 or 11 ' is a character. Let eJ, . . . , ep a be all the irreducible characters of Ca - l in ba - l . Then I (11ca - " es ) - (11ca - " et ) I � 1 for all s, t. If 11 rf 0 and 11 rf cJ>i then (11ca - " es ) rf (11ca - p et ) for some s, t. PROOF. By (1. 17. 12) there exists an R -free R [G] module Y such that YK affords 11 and Y is indecomposable. If Y is projective then 11 = cJ> and the

result is trivial since YCa -1 is projective. Suppose that Y is not projective. By (1 .5) (i) Yo = Y EB A I EB A2 where A l is projective and A2 is a sum of modules in blocks other than B. By (111.5.8) and (1.4) (i) Y = Y is indecomposable. If 11 1 and 112 are the characters afforded by (A 1)K and (A2)K respectively then «112)ca _ p eS ) = 0 for all s and «11 1)ca _l, es ) is independent of s. Hence it may be assumed that G = 0 and Y = Y. It follows from (3.5) that Y is serial and YCa -1 = EB Ux, where U is an indecomposable R [Ca - d module in ba - 1 and x runs over a cross section of T(ba -1) in Na-I• Thus YCa _ 1 = Yo EB A, where Yo = U and A is a sum of modules in blocks other than ba - 1 • Hence it may be assumed that G = Ca - 1 • Now (5.5) implies that (11ca - p es ) � 1 for all s. This yields the result. 0 LEMMA 5.7. All decomposition numbers in B are

0

or 1 .

PROOF. Suppose this is not the case. Then there exists an irreducible character X with cJ>i = 2X + 11 ' for some i, where 11 ' = 0 or 11 ' is a character. If e is any irreducible character of Ca - I then (2Xca_p e) is even contrary to

(5.6).

0

LEMMA 5.8. Let X be an R -free R [ G] module in B such that XK is irreducible and X is indecomposable. Then one of the following holds. (i) XK is irreducible. (ii) There exists a principal indecomposable R [0] module U and an exact sequence

5]

PROOFS OF (2. 1 1 )-(2 . 1 7) IN CASE K

K

293

PROOF. IK (XK, XK) = 1 . Since X is not projective. this implies that

dimR HO( G, (I), HOffiR (X, X» = 1 . Thus b y (1.5) (iii) dimR HO( G, (I), HomR (X, X » = 1 . Since X = X by (111.5.8) and (1.4) (i), it is serial. Thus. there exists a principal indecomposable R [ 0] . module U and exact sequences

O � Y � U � X � o. o � v � U � X � O, As X is R -free, Y is a pure submodule of U and so Y = V. By (2.6) I ( U) = p a . Hence I (X) � p a /2 or I ( Y) � p a /2. Let_ Yo = X if I (X) � p a /2 and let Yo = Y if I ( Y) � p a /2. Thus in any case I ( Yo) � p a /2. By (111.5. 12) HO (G, (1), HomR ( Yo, Yo» is a nonzero cyclic R module. By (3 . 10) 0 = Tr&>(HomR ( Yo, Yo» = Tr&>(HomR ·( Yo, Yo» . Thus by (111.5.15) HomR[ G ] ( Yo, Yo) = HO(G, (I), HomR ( Yo, Yo» = R. Hence rankR HomR[ G ] ( Yo, Yo) = 1 and so IK « Yo)K, ( YO)K ) = 1. Therefore Yo is irreducible. 0 LEMMA 5.9. Let 0 � k � a - 1. Suppose that G = Cko Let G O = Ck /Da - I • Define � for 0 � j � a - 1 by Da - 1 � � and � /Da - l = CGo(DdDa - l). Then G p. Thus by induction all the results are true for Ck-1 and also for G O = GIDa-l• Let A be the subset of A consisting of all those characters in A which have Da-1 in their kernel. By (VA.S) there exists a unique block B O of G O with B O � B. By induction there exist irreducible characters Xl , XI., A E i1 ° in B and these are precisely all the irreducible characters in B which have Da -1 in their kernel. Let () J , ()A, A E A be the irreducible characters of Ck-1 in bk- l • For any element x in G let Xp, Xp' denote the p -part, p '-part of x respectively. Let co, . . . , Ck be the signs defined in (2. 17) for the group G O . Let c b, . . . , d-l be the signs defined in (2. 17) for the group Ck-1• By induction and (2.1S) the values of ou, U = 0, . . . , e are determined in all smaller cases. Let � be defined as in (S.9). By induction there exists 1" = ± 1 such that the signs for b� multiplied by 1" yield the signs for (b:�\lt Thus if a is the nonexceptional character of Hk-dDa-1 in (b :.".\ 1 )0 = (b � _l) Hk - I /Dk - l , then the generalized decomposition number of X l at xDa-1/Da-1 for x E D - Dk is 1" 0 times the generalized decomposition number of a, where 0 = 01 for b� . If x E D then (S.9) (iv) implies that the sign c which equals the generalized decomposition number of a in (b :�\It at xDa-11Da- 1 must equal the generalized decomposition number of a in b :.".-;1 at x. By (S.9) (ii) H a = {3 k -l , for {3 E bk-1 and so by (2. 17) applied to Ck-I, the decomposition number of {3 at x is equal to c. By the observation following (2. 17) we may assume that {3 = () Hence there . exists l' = ± 1 such that c j = YCj for 0 � j � k - 1 . If A E A U {I} then induction applied t o Ck-1 yields that if � i s an u, v

°

l.

5]

PROOFS OF (2. 1 1 )-(2. 1 7) IN CASE K

=

K

295

irreducible constitutent of the restriction to D of an irreducible constituent of ACk _1 in bk-1 then o

()A (X )

=

�(Xp )t{!k -l(XP)

/�; / ZEN�Ck_1

C (xp )t{!f(xp)

if Xp E Di - Di+1, 0 � i � k - 2.

As CG (x ) � Ck-1 for x E D - Dk this implies that

if Xp EGD,

If xp E Dj - Dj+1 for some j � k then by induction and (S.9) (iv) XA (x ) = - CjOo�(xp )t{!k (xp ) , where DO is defined by induction on G O . If k < a - I then since G O = CGo(Dk /Da-1), induction implies that DO = 1 and Ci = 1 for k � i < a - 1. If Xp E Da-1 then Xp is in the kernel of XA and so XA (x ) = XA (xp ) = �(xp )t{!dxp). Thus for A E A a U {I} o if Xp EGD, �(xp )t{!k (xp ) (S.10) XA (x ) = - ci oa '" yz (xp ) 'f' i ( Xp')

� Z�i �

./. z

If p a = 4 let DO = - 1 , 0 1 = 1 . In general 0 1 can be defined by induction and 01 + DO = O. Since c ; = CiY for 0 � i � k - 1, comparison of the last two equations yields that for A E A 0 U {I} �(Xp ) [t{!d Xp ) - Y0 1 t{! f- 1 (Xp)] if Xp EGDk, G XA (x ) - 1'0 1 () A (x ) o otherwise. _

For

fL

{

{

E A define 71>,-

(x ) =

g (xp ) [X1 (Xp) - 1'0 1 () f'(xp)] o

otherwise, where g is an irreducible constituent of the restriction to D of an

[5

CHAPTER VII

2 96

irreducible constituent of j.LCk - 1 in bk- t � The previous equation implies that if A E A 0 U {I} then XA = 71A + ),01 () ? For j.L E A define XIL = 711L + ),01 () �. It follows from the characterization of characters (IV . 1 . 1) that 711-'- is a generalized character. Thus also XIL is a generalized character. Direct computation shows that XIL satisfies equation (5. 10) with A replaced by j.L and ( replaced by g. In particular XIL (1) > O. We next compute Il xlL W . Let y S E Di - Di+1 and let A be the set of all elements in G whose p-part is y S. If i � k then LA I XIL (X ) 12 = LA I X 1 (X ) 12 by (5. 10). Suppose that i < k then

� 1 x. (x ) 1' =


0, B has index of inertia equal to 1 . Thus every irreduc­ ible character in B is irreducible as a Brauer character. Hence (11.2) (i) and (11.2) (ii) also hold in this case. The next two results are needed for the proof of (1 1 .5) which is concerned with the construction of examples that show that (1 1 .2) (i) is false in all other cases. (1 1 .5) below is also of interest because it shows that there appears to be no converse to (1. 18.2). For the rest of this section K is a finite unramified extension of Qp and R is the ring of integers in K. Let X be the R [D ] module which has {(1 - y Y 1 1 � i � p a - I} as an R -basis. Let Y be the unique submodule of X of index p. Then {zi l l � i � p a _ l} is an R -basis of Y where z l = p (l - y ), zi = (l - y Y for 2 � i � p a 1 . Y = VI EB V2 where VI i s the vector space over R spanned by ZI and V2 is the vector space over R spanned by {Zi / 2 � i � p a - I}. -

LEMMA 1 1 .3. If a > 1 and p ;l 2 then VI and V2 are R [D ] modules. PROOF.

By definition

for 2 � i � p a - 2.

Thus it suffices to show that Zp - l ( 1 - y ) E V2• Let YI be the R module spanned by Z and let Y2 be the R module spanned by {Zi 1 2 � i � p a - I}. Thus Yj = Vi for j = 1, 2. zi (I -' y ) = zi (l - y ) = Zi + I E V2 a

I

"

zp" - .(l - y) =

"

� (� ) ( - l)Y = p�. (� ) ( - l)Y P�I a =

(� ) ( - IY [(y

- 1) + lY.

[11

CHAPTER VII

Thus

320

z,"- l (l - y )

:�1 (�") < - lY[s (y - 1) + 1] (y - 1) ',�' (�") < - 1) '5 /�1 (�" ) ( - 1)' I (y 1) == P� (�Q) ( - I ts + (1 - lya - (1 - 1) == (y - 1) :�l (�Q ) ( - I ts (mod Y2).

�

�

Let

Thus

f(x ) = (1 - x ya =

� (�Q) ( - ltx s.

Hence /,(1) = 0 and zp a_ I (I - y ) == (y - 1) [f'(I) ± p Q ] == ± p Q (y - 1) (mod Y2). Since a > 1 this implies that zp a_ I (I - y ) E pYI + Y2 and so zp a- l (1 - y ) E V2• 0

LEMMA 1 1 .4. (i) V2 is an indecomposable R [D] module. (ii) Suppose that p r£ 2 and a > 1. Let N = DE be a dihedral group with E = <x ) of order 2. Then yN = ZI EB Z2 where = WI EB W2 , WI is the trivial R [N] module, dimR W2 = p Q - 2 and W2 has a one dimensional socle which is the R [N] module whose kernel is D.

2

1

PROOF. (i) By definition X/pX is a serial indecomposable R [D] mod­ ule. There is a natural map of X/pY onto X/pX. This map necessarily sends Y/pY onto an indecomposable submodule A of X/pX of codimension 1. Hence dimR A = P Q - 2. Thus the minimum polynomial of y on A has degree p a - 2 and so the minimum polynomial of y on Y/p Y has degree at least p Q - 2. This implies that V2 is indecomposable. (ii) Let fl = HI + x ), h = Hl - x ) in R [N]. Then fi yN � yN for i = 1, 2 as Y is the unique submodule of X with dimR (XI Y) = 1 and X = J(R [D]). Let Zi = t y N. Then Zi r£ 0 for i = 1, 2. As ( yN)D = Y EB Y it follows that (Zi ) D = Y for i = 1, 2. Hence ( yN )D =

llJ

VI EB VI EB V2 EB V2 where each summand is indecomposable by (i). Furthermore 21 EB 22 = yN = Vf"EB Vf". As dimR 2i = p Q - 1, dimR Vf" = 2 and dimR Vf" = - 2) it follows that 2i = W; I EB W;2 with dimR W; I = 1, dimR W; 2 = (p Q - 1) and W; 2 is indecomposable. Since V� = <x )(x ) exactly one of W; say WI is the trivial R [N] module. Let "'l = Wl j for j = 1, 2. It remains to show that S ( W2) is not the trivial R [N] module. Let 'P be the Brauer character afforded by W2 • No irreducible con­ stituent of ( yN)K has D in its kernel hence no irreducible constituent of (ZI)K has D in its kernel. Thus 1 + 'P (x ) = O. As 'P (x ) = - 1 it follows from (2.8) that S ( W2) is not the trivial R [N] module. 0 SOME EXAMPLES

321

2(P Q

J,

I,

THEOREM 1 1.5. Suppose that p r£ 2 and a > 1 . Let q be a prime with q + 1 == p Q (mod p Q +I ). Let G = PSL2(q). Then a Sp -group of G is cyclic of order p Q. G has an irreducible character () of degree q (the Steinberg character), which is afforded by a Q[ G] module. Furthermore there ex­ ists an R -free R [ G] module x such that XK affords () and X = MI EB M2 where MI is the trivial R [G] module and M2 is an irreducible R [G] module with dimR M2 = q - 1.

PROOF. The existence of () and the structure of the Sp -group of G are well known facts. Let D be a Sp -group of G. Then Co (D) = D x H for some p '-group H and I No (D) : Co (D) I = 2. In particular No (D)/H = N where N is defined in (11.4). Furthermore D n D Z = (1) for z � No (D). Thus the Green correspondence between G and No (D) is defined for modules with nontrivial vertex in D. Let X be the R [No (D)] module with kernel H such that X as an R [N] module is isomorp!Iic to ZI. Let X be the R [G] module which corresponds to X. Thus X = WI EB W2. Hence by (111.5.8) X = M, EB M2 where M = W; for i = 1, 2. Let B be the principal p -block of G. Then the index of inertia of B is 2. Since () (z ) = - 1 for z E Co (D ) - {1} it follows that () is in B, this implies that the Brauer tree for B is 0-------0--

1 q q-l Let 'P I , 'P2 be the irreducible Brauer characters in B where 'P I is the principal Brauer character. Thus M, affords 'P I . The principal block of

R [No (D)] has a unique indecomposable module, namely W2 , which has no invariants and whose dimension over R is congruent to - 2 (mod p a ). It follows from (111.5.10) and (111.5.14) that a module which affords 'P 2 corresponds to W2 . Thus M2 affords 'P 2 . Hence in particular rankR X = q. Since X has no invariants, neither does X. Thus the character afforded by XK is sum of exceptional characters and possibly e. As rankR X = e (1) and the degree of every exceptional character in B is q - 1 it follows that XK affords e. 0 322

CHAPTER VII

[12

The next result which is due to Benard and L. Scott, Benard [1976] will be used in section 1 3. In some sense it explains why it is necessary to assume that a > 1 in ( 1 1 .3).

LEMMA 1 1 .6. Let R be the ring of integers in an unramified extension of Qp. Let X be an R -free R [D] module such that XK is irreducible. Then X is indecomposable. If furthermore Y is an R -free R [D ] module with YK = XK then Y = X.

PROOF. If rankR X = 1 the result is obvious. Suppose that rankR X > 1 . Consider the map f : X X defined b y f(v) = v ( 1 - y ). Then f is an R [D ]-monomorphism and ' X : f(X) , = ± det(1 - y ). Since XK is ir­ reducible, the characteristic values of y are all the primitive p th roots of unity for some n. As rankR X > 1, n � O. Hence det(1 - y) = II(1 - �) = p, where � ranges over all primitive p th roots of 1 . Therefore ' X : f ( X) , = ' R , . This implies i n particular that X i s inde­ composable. Furthermore f(X) is the unique maximal submodule of X and so f (X) is the unique maximal submodule of X. Thus X isomorphic to its unique maximal submodule. Suppose that YK = XK• It may be assumed that YK = XK• If Y is replaced by a multiple it may be assumed that Y � X. Since ' X : Y , is finite there exists a chain of submodules Y = Xm � Xm-1 � � Xo = X such that Xi is maximal in Xi - I for 1 :::;:; i :::;:; m. By the previous paragraph Xi = Xi-1 for 1 :::;:; i :::;:; m and so Y = X. 0 �

11

11

IS

• • •

12. The indecomposable

R [ G]

modules in

B

This section contains a description of all the nonzero indecomposable R [ G] modules in B in terms of the Brauer tree of B and of its exceptional

THE INDECOMPOSABLE R [G] MODULES IN B

vertex. The results in this section �ere all proved by Janusz [1969b] and Kupisch [1968] [1969] in case R = R. Given the results of section 2 for K in general, Janusz's arguments go through without any change. The structure of the principal indecomposable R [ G] modules in B is described by (2.20). In this section ' D ' e - 2e pairwise nonisomorphic nonzero indecomposable R [G] modules in B are described. These modules are neither irreducible nor projective. Thus (2. 1) and (2.2) imply that up to isomorphism all the indecomposable R [ G] in B are accounted for. The fact that the number of modules described is equal to , D ' e - 2e involves a counting argument due to Dade. In his paper Janusz also shows that given any tree whatsoever, there exists a split symmetric algebra having only finite number of indecompos­ able modules up to isomorphism such that these indecomposable modules can be described in terms of the given tree. In particular this result indicates that the question of what trees can occur as Brauer trees may be very difficult. If some tree does not occur it must be possible to decide that some symmetric algebra with only finitely many indecomposable modules up to isomorphism is not group algebra. (See the remark on p. 306.) In this section 'T = 'TB denotes the Brauer tree. Vertices of 'T will be denoted by P, Pi or Q. The exceptional vertex, if it exists, is denoted by Pex • E or Ei will denote either an edge of 'T or O. If Li corresponds to the edge E, write Li E. If E = 0 let (0) E. Set t = ( ' D , - 1)/ e. By (2. 13) and (2.19) t is the multiplicity of any L; in XOi for dO i � 0, do; � O. If du; � 0 let Xu; be defined as before (2.20). If dUi � 0 then the statement dual to (2.20) implies that T(Xu;) = L for some j with dUj � O. Let P be a vertex of incident to the edge E. Let L E and let Xu correspond to P. Define V(E, P) = Xu; where T(Xu; ) = L. By (2.25) V(E, P) is determined up to isomorphism by E and P. By (2.20) V(E, P) is serial. Let E be an edge of and let PI and P2 be the vertices incident to E. If Ei is incident to Pi we will define modules V(E" E, E2 : ) for suitable integers We will allow the case that Ei 0 or Ei = E for exactly one of i = 1 or 2 if Pi = Pp . Several cases will be considered. We begin with some preliminary definitions. Let Vi (E) = Rad( V(E, Pi » for i = 1 , 2 . Let U be the indecomposable projective R [G] module corresponding to L where L E and let Mo(E) = U/S C U). Then the dual of (2.20) implies that Rad(Mo(E» = V1(E) EB V2 (E). Suppose that E, E1, E 2 are distinct with E � 0 such that 12]

323

a

�

�

'T

�

'T

n.

n

=

�

324

[12

CHAPTER VII

Pz

PI

-- 0 -- 0 --

is contained in Define modules Mi (n) for i = 1, 2 for suitable integers n as follows. If Ei = 0, M (1) = V; (E). If Ei -1 0 and Pi -I Pex let M (1) be the submodule of V; (E) such that S ( Vi (E)/M (1 » � Ei . Since V; (E) is serial Mi (1) is uniquely determined by (2. 13). Suppose that Ei -1 0 and Pi = Pex • Let 1 � n � t. Let M (n ) be the submodule of Vi (E) such that S ( Vi (E)/Mi (n » Li where Li � Ei and L occurs with multiplicity n as a constitutent of V; (E)/M (n). Since V; (E) is serial Mi (n) is uniquely determined by (2. 13) and (2.19). Define V(Ej, E, Ez : 1) = Mo(E)/(Ml (1) EB Mz(1» . If Pi = Pex and 1 � n � t define V(Er, E, Ez : n ) = Mo(E)/(M (n) EB � (1» , where {i, j} = {1, 2}. Suppose that El -I E = Ez with E -1 0 such that Pz = Pex and El

Ez

E

T.

12]

THE INDECOMPOSABLE R [ G ] MODULES IN B

325

phic composition factors in common unless they are Li and Lj, and E and E I have a common vertex.

PROOF. Suppose that Ls -l Lt are both composition factors of V and V'. Then Cis, Cit, Cjs, Cjt are all not zero. If i, j, s, t are pairwise distinct then there exist m, n, v, w such that dmidms -I 0, dnidnt -I 0, dvjdvs -I 0 and dwjdwt -I O. This implies that contains the following subgraph. T

=

PI

E

where the notation is the obvious one. This contradicts the fact that is a tree. Suppose that {i, j, s, t} is a set of 3 distinct elements, say i, j, t with s = i. Then an argument similar to that in the previous paragraph implies that has the subgraph T

T

w

D

Pex

-- 0 --0 -

El

E2

is contained in If El = 0 the corresponding edge is missing. Define MI (1) as above. If L � E then L occurs with multiplicity t - 1 as a constitutent of Vz(E). If 1 � n � t - 1 let Mz(n ) be the submodule of Vz(E) with S ( Vz(E)/Mz(n» = L such that L occurs with multiplicity n as a constitu­ ent of Vz(E)/Mz(n ). Define V(Er, E, Ez : n ) = V(Ez, E, E I : n ) = Mo(E)/(Ml (1) EB Mz(n » . In all cases V = V(Er, E, Ez : n ) V(Ez, E, E I : n ) is indecomposable as T( V) is irreducible. If furthermore Ei = 0 then V is serial as Mo(E)/M (1) = V (E, Pj ) for i -I j and so is serial. T.

=

LEMMA 12.1 . Let V = V(E I , E, Ez : n ) and V' = V(E � , E I, E� : n ') be de ­ fined as above where E -I 0, E ' -I 0, E I -I Ez and E � -I E �. Let Li � E and Lj � E '. Suppose that Li f:. Lj . Then V and V' cannot have two nonisomor-

n

v

contrary to the fact that is a tree. If {i, j} = {s, t} then clearly T

is a subgraph of T.

E

E'

� D

It will be necessary to consider the following two types of subgraphs of Let k be any integer with k ;?; 1 . (I)

T.

[ 12

CHAPTER VII

326

12]

Then

THE INDECOMPOSABLE R [ 0 ] MODULES IN B

X is represented by the direct sum of cones. L

oLi-1/o�Li +1

(II)

0

In type (I) Eo, . . . , Ek are distinct edges. In type (II) Eh+i and Eh+2s+1-i correspond to the same edge for i = 1, . . . , s. Possibly one or both of Po, Pk+l may equal Q. Given a graph of type (I) or (II) let .1 denote either the set of all even integers i with O ::':S i ::':S k or the set of all odd integers i with O ::':S i ::':S k. Given a graph of type (I) or (II) let Vi (ni ) = V(Ei - l, Ei, Ei+1 : ni ) where ni � 1 and E-l = Ek+1 = O. Let Li � Ei. Thus in case of type (II) Lh+j = Lh +2s+l-j for O ::':S j ::':S s. Observe that Li � Lj if i I- j but i == j (mod 2). The definition of Vi (ni) implies that S(Vo(no» = L 1 , S(Vdnk » = Lk-I and for o < i < k, S ( V; (ni » = Li -I EB Li + I . Choose .1 . Define X = EBi E .1 V; (ni). Then

m 2L2j-1 EB j=1 ( � 2L' - ) EB L'rn + ' S(X) = I Lo EB ( � 2L2j) EB L2m i

'

Lo EB C� 2 L ) 2i

If 2 Li

I SeX) there exist submodules

if 0 E .1, k

= 2m,

if 0 E ,1, k

=

if 0 � Ll, k

= 2m,

2m + 1 ,

=

if 0 � .1, k = 2m + 1. �

1,

interest in this section. The module V(Ei - J, Ei, Ei + l : represented by the cone

L,.,

ni )

with

Ei - I I- 0

Li

/�o L; _ '

and

Ei+ 1 1- 0

o oLi +1/ �Li+3 Li+2

0

Intuitively M is constructed from X by identifying isomorphic modules at the feet of adj acent cones and may be visualized as the following connected graph, where the dotted lines may not exist.

The next three results are concerned with showing that this picture is indeed accurate. Let XO = M and let ° denote the image of any element or submodule of X in XO = M. We will also write V; (ni t = V�)(ni )' Since V; (ni ) n Y = (0) for all i, V?(n ) = Vi (ni). If furthermore r is a subset of .1 such that whenever i E r then i + 2 � r then {EBiEr V; (ni )}O = EBi Er V�) (ni) ' LEMMA 12.2. (i) T(M) � EBiE.1 Li. (ii) SCM) = EBi E{O" ,. ,k l-.1 Li •

Lil C S( V;-I(n-I» and Li 2 C S(Vi+l(ni+I» with Li l Li2 Li . Let gi : Lil Li2 be an isomorphism. Let Y be the submodule of X consisting of all elements of the form Li (Vi gi (Vi » where i ranges over integers with 2Li I SeX) and Vi E Lil• Define M = X/ Y. In effect M is obtained from X by identifying the submodules Li Li 2 whenever 2Li I S (X). The module M is the primary object of =

327

may be

PROOF. (i) By definition YC

Therefore

SeX) C EB i E.1 S ( V; (ni » C EB i E.1 Rad( V; (ni »

C Rad X.

T(M) = reX) = EB i E.1 T(V; (ni» = EB i E.1 Li• (ii) Clearly S(X)o = EBi E{ , ... ,kl- .1 Li and S (X)O C S(XO) = S(M). O Thus it suffices to show that any irreducible submodule of M is in S(X)o . Suppose the result is false . There exists an indecomposable submdodule Z of S 2(X) which is not irreducible such that S (Z) C Y and ZO i.s irreducible . Let r be the set of all i such that ZO is isomorphic to a

[12

CHAPTER VII

328

composition factor of Vi (ni ). Z may be replaced by Z n {EBiEr V; ( ni )} ' Thus it may be assumed that Z C; EBiEr V; (ni ). Since Z is indecomposable S (Z) C; Rad(Z). Thus S (Z) Rad(Z) as S 2 (Z) Z. The dual of (2.20) implies that I (Z) 2 or 3. Let E be the edge corresponding Zo o If ZO is isomorphic to a constituent of S (Z) then (2.23) implies that E is incident with Pex and furthermore E is the only edge of T which is incident with Pex • Assume that j, j + 2 E r for some j. Then Lj + 1 is a composition factor of S ( V} (nj )) and S ( V} +2(nj+2)) . Thus ZO and Lj+1 are both isomorphic to composition factors of V} (nj ) and Vh+2 (nj +2 ). Suppose first that Ei and Ei +2 do not have a common vertex for any i, i + 2 E r. By (12. 1) Lj+ 1 = Z o o Thus r {j, j + 2} or the following configu­ ration occurs in T with s > 2 =

=

=

=

�

o

/ �

0 --- 0

o

E; +2S - 1

E; + 2S

0

,

where possibly Ej = Ej +2s+l • In this case r C; {j, j + 2, j + 2s}. Therefore in any case Y n EBiEr V; (ni ) C; Lj+ 1 = Z o o Since Y n Z I- (0) it follows that Y n Z = ZO and so S (Z) = ZO which is not the case since both vertices incident with Ej + 1 are incident with other edges in T. Suppose next that Ej and Ej +2 have a common vertex. By (12.1) ZO = Lj or Lj +l • Up to symmetry the following configuration must occur in 'T. E;

0

E; + I

Pex

0

E;+2

Y n EB

iEr

V; (ni ) C; Lj

+

I

If ZO = Lj + 1 then r = {j, j + 2 } . Let Yo Y n ( V} (I) EB V} + 2 (nj +2)). Then o Yo = Lj + 1 = Z o o Thus I (Z) � I (ZO) + I ( YO) 2 and so S (Z) = Z o There­ fore Yo, Z C; z, EB Z2 where ZI C; V} (1), Z2 C; V} +2 (nj +2 ). Both ZI and Z2 are indecomposable, all composition factors of ZI EB Z2 are isomorphic to ZO and I (ZI) 1, I (Z2 ) 2. Therefore Rad(Z) C; Rad(ZI EB Z2 ) = Rad(Z2) C; V} +2 (nj +2). Thus Yo n Rad(Z) = (0) and so Y n Z (0). Hence Z = ZO contrary to the reducibility of Z. If ZO = Lj then r C; {j, j + 2, j + 4} and if j + 4 E r then Lj + 3 = Lj other­ wise ZO = Lj is not a constituent of V} +4(nj + 4). Hence =

EB Lj +3 = Lj + EB ZO. I

Since Pex is not incident with Ej it follows that ZO ,{ S (Z). As S (Z) n Y I- (0) it follows that S (Z) = Lj + l • Thus Z n V} +4(nj + 4) = (0) and so Z is isomorphic to submodule Z' of V} (1) EB V} + lnj +2) with S (Z') n Y = Lj + l • Since Lj occurs with multiplicity 1 in V} +2 (nj +2) and Lj C; S ( V} +2 (nj +2)) it follows that Z' C; V} (1) EB ZJ, where Z, C; V} +2 (nj +2) and ZI = Lj + l • Thus Rad(Z') C; V} (1). As Rad(Z') n Y = S (Z') n YI- (0) this contradicts the fact that V} (1) n Y = (0). o Therefore if i E r then i + 2 g r. Consequently {EBiEr V; ( ni )} = EBi Er V?{nd · Hence Z = ZO contrary to the fact that Z is reducible. 0 LEMMA 12.3. Let i E ,1. (i) If Ei is not incident with Pex then Li occurs as a composition factor of M with multiplicity at most 2. If the multiplicity is 2 then Li I S (M). (ii) If Ei is incident with Pex then the multiplicity of Li as a composition

PROOF. If k � 2 then ,1 = {i} and M V; (ni ). The result is clear. Suppose that k > 2. (i) As Ei is not incident with Pex , Li occurs with multiplicity 1 in V; (ni ). If j I- i and Li is a constituent of V} (nj ) then the following subgraph of T must occur (up to symmetry) with P I- Pex, where Ej +2 may not be there. =

Ei

P

0 ----- 0 ------ 0 ------ 0

In any case this implies that Li I S (X) and so Li I S (M) and the multiplicity of Li " in S (M) is. l . (ii) As Ei i s incident with Pex , one of the following subgraphs o f T must occur (up to symmetry) Ei

=

=

329

factor in M is equal to the multiplicity of Li as a composition factor of V; (ni ). This is ni in case (I) and ni + 1 in case (II).

E; +2s + 1

0

THE INDECOMPOSABLE R [ O J MODULES IN B

12]

0 ------- 0 --- 0

=

=

Ei

Ei + 1

0 -- 0 --0

Pex

Pex

In the first case Li occurs with multiplicity ni + 1 in � (ni ) and multiplicity 1 in Vi + 2(ni+2) and in the socle of both. Thus Li occurs with multiplicity ni + 1 in M. In the second case L occurs with multiplicity ni in Vi (ni ) and does not occur in any V; (nj ) with j � i, j E .1 . 0 CHAPTER VII

330

[12

LEMMA 12.4. M is indecomposable.

Suppose that M = MI EB M2• For s = 1, 2 Iet /' be the projection of M onto Ms. For any i E .1 , V7(ni ) C V7(ni )fl EB V7(ni )f2 . We will first show that V7(ni ) V7(ni )fs for s = 1 or 2. Suppose this is not the case. Hence V7( n )fs � (0) for s = 1, 2. Let Ws = V7(ni )fs for s = 1, 2. Then 2Li I T( WI EB W2) and so Li occurs with multiplicity at least 2 as a constituent of M. Suppose that Ei is not incident with Pex • Then Li I S CM) by (12.3) and so Ws L for s = 1 or 2. Since Li g S ( V7(ni » it follows that V7(ni )fi V7(ni ) for s � t. Suppose that Ei is incident with Pex • Let n be the multiplicity of Li as a composition factor of M. By (12.3) (ii) the multiplicity of Li as a composi­ tion factor of Vi (nd is n - 1 . Let k be the smallest integer such that Li occurs with multiplicity n - 1 as a constituent of S k (Rad( � (ni » ): Since Rad( WI EB W2) = {Rad( Vi (ni » } fl EB {Rad( � (n » } f2 and this is a sum of at most 4 serial modules it follows that for s = 1 or 2 S k (Rad( Ws » � (0) and L; occurs with multiplicity n - 1 as a constituent of Rad( Ws ). Hence Li occurs with multiplicity n as a constituent of W,. Therefore L; does not occur as a constituent of Wt for s � t by (12.3) (ii) and so Wt = O. Hence W, � (ni ) for s = 1 or 2 in all cases. Suppose that i + 1 E .1 and Li I S (Ml ). Then L .{' S (M2) by (12.2) (ii) and so V7+ I (ni +l)h � V7+ I (ni + I ). Thus V7+ 1 (n-rl)fl V7+ 1 (ni +l ) and Li+2 1 S (MI). By changing notation if necessary it may be assumed that either 1 E .1 and L o I S (Ml ) or 0 E .1 and L I I S (MI). Hence by iterating the result of the previous paragraph we see that Li I S (M1) for all i E {1, . . . , k } - .1 . Conse­ quently (12.2) (ii) implies that S (M) C MI . Thus M2 = (0). 0 PROOF.

=

=

=

=

=

The definition of M shows that M depends on whether the graph is of type I or II, on the integers ni, the set .1 and the functions gj . It will later turn out that M is independent of the choice of functions gi .

LEMMA 12.5. M determines the type of graph and the ordered set Eo, . . . , Ek

12]

THE INDECOMPOSABLE R [ G J MODULES IN B

33 1

up to a reversal of order. Once the order is fixed, the set .1 is also determined by M except in case of a graph of type II with Po = Pk + = Q. Furthermore the integers nj for i = 1 , . . . , k are uniquely determined. I

PROOF. By (12.2) the graph is of type I if and only if S (M) and T(M) have no common composition factors. Furthermore {Eo, . . . , Ek } is determined by S (M) and T(M). If the graph is of type I the ordering Eo, . . . , Ek is determined up to a reversal by S (M) and T(M) since is a tree. By (12.2) .1 is determined by T(M). If the graph is of type II, (12.2) and (12.3) imply that {Eh+l, • • • , Eh+s} is determined by the common composition factors of S (M) and T(M). Thus Q is determined as Q � Pex • Eo, . . . , Ek are determined up to a reversal of order by S (M), T(M) and the fact that is a tree by (12.2). Thus {Po, Pk+I} is also determined. If Po � Q then .1 is determined by the condition that Lo I S (M) or L o I T(M) but not both. Similarly .1 is determined if Pk+1 � Q. If Po = Pk+1 = Q then the definition of M shows that .1 may be replaced by {1, . . . , k } - .1 . Since there is at · most one i E .1 with Pex incident to Ej it follows that nj = 1 for all j except possibly for j equal to such an i. In which case ni is determined by /(M) as is known. 0 T

T

T

Let 'Y be a graph of type I or II. Let n ( 'Y) be the number of modules M which can be associated to 'Y depending on .1 and the integers nj . In view of (12.4) and (12.5) there are at least 22n ('Y) pairwise nonisomorphic, nonpro­ jective, reducible, indecomposable, nonzero R [G] modules in B, where 'Y ranges over all subgraphs of of type I or II. We will now compute n ('Y) and then count the number of possibilities for 'Y . This will turn out to be I D i e - 2e. Once this is done it follows from (2.1) and (2.2) that every nonzero indecomposable R [G] module in B is projective, irreducible or isomorphic to one of the modules M. Case (i): 'Y is of type I. Pj � Pex for 1 � i � k. Then n ('Y) = 2 depending on the choice of .1 . Case (ii): 'Y is of type I. Pj = Pex for some i with 1 � i � k. Then n ( 'Y ) = 2 t depending on the choice of .1 and ni . Case (iii): 'Y is of type II. Po � Pk+I. Then 1 � nj � t - 1. Thus n ('Y) = 2( t - 1) depending on the choice of .1 . Case (iv): 'Y is of type II. Po = Pk+l = Q. Then 1 � nj < n - 1 . Thus n ( 'Y ) = (t - 1) as M does not depend on the choice of .1 . T

332

[12

CHAPTER VII

LEMMA 12.6. If T has no exceptional vertex then Ln (y) = I D i e - 2e. PROOF. Only Case (i) can occur. Given any pair of distinct edges in T there is a unique graph of type I with these as extreme edges since T is a tree. There are G) ways of choosing such a pair of edges. Hence

2: n (y) = 2

(; ) = e 2 - e = I D i e - 2 e

since 1 D 1 - 1 = e. 0 PROOF. In view of (12.6) it may be assumed that T has an exceptional vertex. Let PI, P2, . . . be all the vertices in T such that Pi "I Pex and Pi, Pex bound a common edge. Consider the subgraphs Tt, T2 , . . . of T such that Ti consists of Pex and all vertices and edges with the property that a path from any vertex in Ti - {PeJ to Pex goes ghrough Pi . Since T is a tree, Tl, T2, . . . yields a partition of the set of edges in T. Let ei be the number of edges in Ti . Thus Li ei = e. In Case (i) {Po, Pk+I} is any set of vertices in Ti for any i such that Po and Pk+1 do not bound an edge. The number of possible pairs of vertices in Ti is . � ei (ei + 1). The number of pairs which bound an edge is the number of edges ei . Hence there are � ei (ei - 1) possible sets {Po, Pk+1} in Ti . Therefore the number of possibilities for y is � Li (e; - ei ). Thus if y ranges over all possible graphs in Case (i), Ly n (y) = Li (e; - ei ). In Case (ii) Po may be chosen in any Ti - {Pex} and Pk+1 in any Tj - {PeJ with i "l j. Thus the number of possibilities for y is Li<j eiej ' Hence if y ranges over all possible graphs in Case (ii), Ly n ( y ) 2 t Li <j eiej . In Case (iii) 'Y lies in Ti for some i. There are �ei (ei - 1) ways of choosing {Po, Pk+1} in Ti . Thus there are � Li ei ( ei - 1) ways of choosing y in T. Thus if y ranges over all possible graphs in Case (iii), Ly n ('Y ) = (t - l)L(e ; - ei ). In Case (iv) Q may be chosen arbitrarily in T - {Pex}. Thus there are e choices for Q. Hence if y ranges over all possible graphs in Case (iv), Ly n (y) (t - l ) e. Consequently if y ranges over all possible graphs of type I or II in T =

=

2: y n (y) = 2:i (e T - ei ) + 2 t 2: i<j eiej + (t - 1) 2:i (e i - ei ) + (t - l)e

(� eiY - t � ei + (t - l)e

333

By the remarks above, ( 12.7) completes the description of all the indecomposable R [G] modules in B up to isomorphism. PROOF OF (2.26). If V is an irreducible or a principal indecomposable R [ G] module in B then S ( V) = T( V). Thus the result follows from (12.2) and ( 12.3). 0 13. Schur indices of irreducible characters in iJ

THEOREM 12.7. L n (y) = I D I e - 2e.

=t

SCHUR INDICES OF IRREDUCIBLE CHARACfERS IN B

13 ]

= te 2 - e = e (et + 1) - 2e = 1 D i e - 2e. 0

Throughout this section K K. Thus B = iJ, Xu = Xu , cPi 'Pi, XA = X" for O :!S u :!S e, 1 :!S i :!S e, A E A. Let {3 be an irreducible Brauer character in Bk for some k with O :!S k :!S a. Let M = Qp ((3 ) and let S be the ring of integers in M. By (1. 19.3) and (2.9) S is a splitting field for any irreducible S[N ] module in Bi for =

=

O :!S i :!S a.

PROOF. Clear by (2.9) and (2. 1 9). 0 If I A 1 "1 1 then XA - XI-< vanishes on all p '-elements of G for A, JL E A . Thus the maximal unramified subfield of Qp (X,, ) is the same for all A E A. Denote it by F. By (2. 1 5) F c;;;, M. This yields LEMMA 1 3.2. Suppose that I A 1 "1 1 . Let F be the maximal unramified subfield of Qp (XA ) for some A E A . Then F is the maximal unramified subfield of Qp (XI-< ) for all JL E A. Furthermore F c;;;, M and [M(XA ) : Qp (XA )] = [M : F] for all A E A. Let m (X ) = mQp (X ) denote the Schur index of the irreducible character X over Qp . The object of this section is to prove THEOREM 13.3. m (Xu ) = 1 for 1 :!S u :!S e and one of the following holds. (i) 1 A 1 = 1 and m (Xo) = [M(Xo) : Qp (Xo)] . (ii) I A 1 "1 1 . For A E A, m (X" ) = [M : F ] . It is easily seen that (13. 1}-(13.3) imply

[13

THEOREM 13.4. If X is an irreducible character in B and

of order 9, where x acts as an outer automorphism of order 3 on Suz(8) and x 3 is in the center of G. Let L ' = Qn and let L" be the cubic unramified extension of Qn. A nonprincipal 1 3-block of L "[G] has no exceptional characters . The tree T ' of a faithful 1 3-block B' of L '[ G] is as follows . 0 42

f.L � A, f.L � A. By (2. 17) (X - XIL )No = ) ± (A - f.L . Hence « X - XIL )No, A )No = ± 1 . Therefore either (XNo, A )No or « XIL )No, A )No is relatively prime to m (A ). As X and XIL are algebraically conjugate it follows that m (X) = m (XA ). Thus by changing notation if necessary it may be assumed that (XNo' A )No is relatively prime to m (A ).' By (13.9) m (X )XNo is afforded by a Qp (A ) [No] module. Thus by (IV.9. 1 ) m (A ) I m (X ) (XNo, A )No• Therefore m (A ) I m (x). Hence [M : F] I m (x ) by (13.8) and (13.9). 0

where the numbers at the vertices are the degrees of the corresponding characters . The tree T" is the triple unfolding of T ' hinged at Po and is the same as the tree of the principal 1 3-block given in section 9.

14. The Brauer tree and field extensions

15. Irreducible modules with a cyclic vertex

PROOF OF (13.7) IN CASE (ii). Choose

Let K = K and let X = ,{\ be an exceptional character in 13 if I A I � 1 and let X = XO if I A I = 1 . Let M = Qp (Xu ) for 1 � u � e. Thus M is defined as in the previous section. Suppose that L ' and L " are fields with Qp (X) � L ' � L " � M(X ) � K.

Let B" be a block of L "[ G] which splits into algebraically conjugate blocks, one of which is 13 if L " is extended to K = K. Let B ' be a block of L '[G] which splits into algebraically conjugate blocks, one o f which is B" i f L ' i s extended t o L ". Let T ' , T" b e the Brauer tree o f B ', B" respectively. According to (2.9) and (2. 1 9) there exists an integer m and a fixed vertex Po on T ' , which is the exceptional vertex if there is an exceptional vertex, such that T" is an "unfolding" of T' in the following sense. Each edge of T ' is replaced b y m edges i n T " and each vertex other than Po i s replaced b y m vertices. The vertex Po remains fixed. According to (13.3) the vertex Po in T ' corresponds to a character with Schur index m (X ), while the vertex Po in T" corresponds to a character with Schur index m (x)/m. Thus if, as in section 9, the vertices of T', T" are associated to irreducible L '[ G], L "[ G] modules respectively rather than characters then each vertex of T' is replaced by m vertices in T". It so

0

192

3

105 0 Po

0 42

This section contains a proof of the following result.

THEOREM 15. 1 . Let V be an irreducible R [G] module with a cyclic vertex P = (x >. Then P is a defect group of the block which contains V. This result is independent of the other material in this chapter. It is due to Erdmann [ 1977a] and we will give her proof here. In case G is p -solvable it had been proved earlier by Cliff [1977] . We will first prove a preliminary result .

LEMMA 15 .2. Let (1) � P = (x > <J G and let V be an indecomposable R [G] module with vertex P. (i) There exists an integer k with 0 � k � I P I and p l' k such that Vp is a projective R [P]/R [P] (1 - x t module. (ii) V(1 - x) = {v l v E V, v (l - x t - 1 = O}. (iii) If h E Tr� p)(HomR [(xp)l ( V, V» then h ( V) � V(l - x ). (iv) V/ V(l - x ) is an R [G] module which is R [P]-projective and which has P in its kernel. Furthermore V/ V(l - x ) is a projective R [G /P] module. (v) Let .x be a nonempty collection of proper subgroups of P. If HO ( G, .x, V) R then V / V(l - x ) is irreducible. =

338

CHAPTER VII

[ 15

PROOF. (i), (ii). For 0 � k � I p i , let Vk denote an indecomposable R [P] module of R -dimension k. Ry (II.2.11) ( Vf)p = I G : p i Vk• Since V is R [P]-proj ective it follows that Vp = m Vk for some integer m > O. As P is a vertex of V, p {' k. Now (i) and (ii) follow directly. (iii) In view of (ii) it suffices to show that h ( V(l - x t - I ) = (0). Let w = v (l - X ) k - I E V(l - x t - I . By (i) wx = w. By (ii) k = np + s for 0 < s < p and some integer n ?!= O. By assumption h = Tr� p)(g) for some g E HomR (xP)( V, V). Since (1 - x Y = 1 - x P it follows that

g(w ) = g(v (l - x Y (l - x )"P ) = g (v (l - x Y ) (l - x P ) ". Since 1 + x + . . . + x p - I = (1 - X y - l and np + p - 1 ?!= k this implies that {Tr�: � ) (g )} (w ) = g (w ) (l + x + . . . + x P - I ) = g(v (l - x Y ) (l - x P ) " (1 - x y - I = O. Thus

h (w ) = (Tr(G {Tr x P ) (g)} ) ( w ) = 0 x)

(x ) (

as required. (iv) V(l - x ) = (v(l - z ) I v E V, z E P) is an R [ G ] module as P <J G. Clearly P is in the kernel of V / V (1 - x ). Let ( Vp ) G = EB w; with W1 = V. By (i) ( W'; )p = m (R [P]/R [P] (l - x t ) for all i and some integer m depending on i. Thus ( Vp) G ( 1 - x ) = EB W; (1 - x ). Observe that ( Vp) G (1 - x ) = ( V(1 - x )p ) G since they have the same R -dimension and ( Vp ) G (1 - x ) � ( V(l - x )p ) G . Consequently

V/ V(l - x ) = WI / WI (l - x ) I ( Vp) G /( Vp) G (1 - x ) = ( Vp) G /( V(l - x )p) G = « V/ V(1 - x ))p t. Since ( V/ V(l - x ))p = Invp ( V/ V( l - x )) the previous paragraph implies that « V/ V(l - x ))p ) G is a direct sum of copies of R [ G / P ] R [ G / P ) . Thus V/ V(1 - x ) is a proj ective R [G / P] module. (v) Since every proper subgroup of P is contained in (x P ) the hypothesis implies that HO ( G, ( x P ) , V) = R. As EndR [ G ) ( V) is a local ring and Tr� p)(HomR(xp)( V, V)) is an ideal of EndR [ G )( V) by (II.3.7), it follows that if h E HomR [ G ) ( V, V) but h is not an isomorphism then h E Tr� p)(HomR[xp] ( V, V)). Thus by (iii) h ( V) � V(1 - x ). Suppose that the result is false. By (iv) V/ V(l - x ) is a projective R [G/P] module . Thus by (1.16.8) there exists an irreducible submodule M � V/ V(1 - x ) and an epimorphism f : V/ V(I - x ) � M. Let t be the natural proj ection of V onto V / V(l - x ). Thus the row in the following diagram is exact.

1 5]

IRREDUCIBLE MODULES WITH A CYCLIC VERTEX

339

V

If

v � V/ V(1 - x ) � O. Since Vp and ( V/ V(l - x ))p are R [P]/R [P] (l - x t modules and Vp is projective by (i) there exists an R [P]-homomorphism Ii such that the following diagram is commutative.

As V is R [P]-projective there exists an R [ G ]-homomorphism h such that the following diagram is commutative.

V

1' / v � V/ V(l - x ) � O. Since the result is assumed to be false f = t h is not an epimorphism. Thus h is not an epimorphism and so h ( V) � V (1 - x ). Hence f = t h = 0 contrary to assumption. 0 0

0

PROOF OF (15. 1). By (III.4. 14) it may be assumed that R is a splitting field of G and of all its subgroups. Let N = N G (P). Since P is cylic there exists G with CG (P) � G <J N such that p {' I N : G I and P centralizes all p '-elements in G. If P is the defect group of any block of G then P � P and so CN CP) � G. Thus by (V.3.10) every block of N is regular with respect to G. Let f be the Green correspondenc e with respect to ( G, P, N). Apply (15.2) to N with f( V) in place of V. By (111.5. 10) the hypothesis of (v) is satisfied. By (III.7.8) it suffices to show that if B is the block of N which contains f( V) then P is a defect group of B. Let W = f( V)/f( V) (l - x ). Then W is in B. By (15.2) W is in a block of R [G / P] of defect O. Since p {' I N : G I Clifford's theorem (111.2. 12), implies that Wo is a direct sum of irreducible modules in blocks of defect 0 of R [G/P]. By (V.4.5) each of these is in a block of R [G] with defect group P. Hence by (V.3. 14) B has defect group P. 0

TENSOR PRODUCTS OF R [NJ MODULES

By (VII.2.S) the composition factors of V('P, 1) in ascending order afford . the Brauer characters 'Pa - 1 , . . . , 'Pa . . Th e next result IS. an Immediate consequence of this fact.

2)

m 'T"

CHAPTER VIII

s of Let p be a prime . The object of this chapter is to apply the result s result the of Some p. order of Chapter VII to study groups with a Sp-group cyclic a has G that proved in this chapter can be generalized to the case ver this Sp-group and satisfies some other subsidiary conditions. Howe more general case will not be considered here. used The notation introduced at the beginning of Chapter VII will be ing follow the ion addit throughout sections 1, 2 and 3 of this chapter. In assumptions and notation will be used in these sections. K = K. P is a Sp-group of G. P = (y ). I P 1 = p. C = Co (P). N = No (P). H. By Burnside's transfer theorem C = P x H for some p i_group of D in If B is a p -block of positi ve defect of G then P plays the rolethan that Chapter VII. However the situation is a good deal simplerNa-I = G. If described in Chapter VII since C = Co = Ca-1 and N = No = le which R [N] modu M is a nonp rojec tive R [G] module let AI be the corresponds to M in the Green correspondence. e deno tes the inertial index of the principal p -bloc k of G. By (VII.2 .3) I N : C I = e. nonzero The group N has no p-blo ck of defect O. Let V be a 2.6) that (VII. and indecomposable R [N] modu le. It follows from (VII. 2.4) and V) ( S V is serial , I ( V) � P and V is determined up to isomorphism by let V ('P, 1) I ( V). If 'P is an irreducible Brauer character of N and 1 � I � p I» = l. In be an R [G] modu le such that S ( V('P, 1» affords 'P and I ( V('P,s 'P. Defin e particular V('P, 1) is an irreducible R [G] modu le which afford V('P, 0) = (0).

(I - I )

LEMMA 1 . 1. Let 'P be an irreducible Brauer character of N and let 1 � I � P. Then the following hold. (i) V('P, 1) * = V('P * a 1-\ I). (ii) For x in N let detv( p. By (2.8) LN @ L :' has (p - 2s)m projective direct summands and for 0 � i � s - 1, LN @ L :' has m direct summands M with [(M) = 2i + 1 . Let n be the number of direct summands M of LN @ L f,; with 1 < [(M) < d. As 3s - 2 > p it follows that p - s < 2(s - 1) + 1 . Thus n = � (p - s - 3)m if s is even and n = � (p - s - 2)m if s is odd. Hence in any case n 3 � (p - S - 3)m.

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4]

A CHARACTERIZATION OF SOME GROUPS

35 1

Let !VI be a direct summand of. LN ® L 'tv with 1 < I (M) < d. Since 1 < I (M), P is not in the kernel of M. Let M be the R [G] module corresponding to M. The minimality of d implies that MN f:; M. Hence MN has a nonzero projective summand. This implies that LN (29 L 'tv has at least n nonzero projective summands. Consequently (p - 2s )m � ! (p - s - 3)m. Therefore p � 3s - 3. As 3 ,f p, P � 3s - 2. D

that 1 + 2(s - 3) � P - 2s and so 4s - 5 � p � 3s - 1. Thus s � 4 and so p � 3s - 1 < 13 contrary to assumption. Therefore 3s - 1 < p. By (3.7) and (3.8) mj � 1 for 1 � i � s - 1 and mj � 2 for at least s - l values of i with 1 � i � s - 1 . Hence (3.6) implies that 1 + 2(s - 2) � P - 2s. Thus 4s - 3 � P and so 4d � 3p - 3 as required. D

PROOF OF (3.3). In view of (3. 1) 'P (1) = 1. Thus m = 1 and d = do. Suppose that e � i (p - 1). By (3.1) d > e and so by (VU.2.7) d � p - e > � (p - 1) contrary to assumption. Therefore e � �(p - 1) and d = p - s with s � e. By (2.8)

4. A characterization of some groups

s -- I

p - I -s

LN (29 L :' = E9 V(a \ 2i + 1) EB E9 V(a \ p). i =O i =s Since d < � (p - 1) it follows that s > i (p + 3). D

with 1 � i � s - 1. Then

PROOF. Since P is not in the kernel of Mi, the minimality of d implies that d � 2i + 1 in case mj = O. If i rf s - 1 then d � 2(s - 2) + 1 = 2p - 2d - 3. Thus 3d � 2p - 3 < 2p - 2 contrary to (3. 1). If i = s - 1 then p - s = d � 2(s - 1) + 1 and so p � 3s - 1 . D LEMMA 3.8. There is at most one value of i with mi

=

THEOREM 4.1 . Suppose that G = G ' and a Sp -group P of G has order p. Let 1 + np be the number of Sp -groups in G. Let H = Op ( G ). Then one of the following holds. (i) p > 3, GIH = PSL2 (p). (ii) P = 2a + 1 > 3, GIH = SL2 (p - 1). (iii) There exist positive integers h, u such that n -_ hup +uu2++1 u + h This result was first proved by Brauer [1943] under the additional assumption that CG (P) = P. The proof of (4.1) given here is quite similar to that given in the above mentioned papers except that the results of section 3 are used to simplify a portion of the argument. The proof of (4. 1) requires the following result of Zassenhaus [1936] which will here be stated without proof. ,

For 1 � i � s - 1 let M; be an indecomposable R [ G] module with Mi = V ( a i, 2i + 1). Let dimR Mi = 2i + 1 + mip. We need two subsidiary results. LEMMA 3.7. Suppose that mi = 0 for some i = s - 1 and 3s - 1 � p.

The purpose of this section is to provide a proof of the following result due to Brauer and Reynolds [1958] .

1.

PROOF. Suppose that mi = 1. Since M; !Vi i t follows that M; = M; . Thus (M; )N = Ai EB V(a i, p) for some j with V(a j, p) = V(a j, p)*. Hence a 2j = 1 by (1. 1). Since by (3.6) s � j � p - s - 1 and s > i (p + 3) it follows that j = Hp - 1) in case e = p - 1 or Hp - 1). Thus the result is proved unless e = � (p - 1). If e = Hp - 1) then j = Hp - 1), Hp - 1) or Hp - 1). Since G = G', (1 . 1 ) (ii) implies that a jPa - p (p - I )/2 = 1 . Thus a i-(p - I )/2 = 1 and so j !(p - 1) (mod e ). As e = �(p - 1) it follows that j rf Mp - 1), � (p - 1). Thus j = Hp - 1) and the result is proved also in this case. D �

==

Suppose that 3s - 1 � p. By (3.7) and (3.8) mj � 1 for 1 � i � s - 2 and mi � 2 for at least s - 3 values of i with 1 � i � s - 2. Hence (3.6) Implies

THEOREM 4.2. Ld G be a triply transitive permutation group on an odd number of letters m. Assume that only the identity fixes at least 3 letters. Then m = 2 a + 1 for some a � 1 and G = SL2 (2a ). Define the function 2 F(p, u, h ) = hup + U + u + h . u+1 We first prove some arithmetical lemmas. LEMMA 4.3. Let n � 1 be an integer. Suppose that u is a nonnegative integer such that (l + up) I (p - 1)(1 + np). Then there exists an integer h � 0 such that n = F(p, u, h ) and (u + 1) I h (p - 1).

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[4

PROOF. Let 1 + up = m , m 2 with m , I (p - 1) and m 2 1 (1 + np). Then p 1, up + 1 0 (mod m , ) . Hence u + 1 == 0 (mod m , ) . Also up + 1 == 0, 1 + np == 0 This implies that (n - u )p 0 (mod m 2) and so n - u 0 (mod m 2) . Thus up + 1 = m , m 2 divides (u + 1)(n - u ). Define the integer h by (u + 1)(n - u ) = h (up + 1). Hence h (up + 1) 0 (mod (u + 1)). Since u - 1 (mod (u + 1)) this implies that (u + 1) 1 h (p - 1). Furthermore n = F(p, u, h ). Suppose that h = - h ' < O. Then (u + 1) (u - n ) = h ' (up + l» O and so u 2 = U (h P + n - 1) + h ' + n > uh p. Thus u > h 'p. This contradicts the fact that (u + 1) 1 h '(p - 1). Hence h � O. 0 ==

==

==

==

==

==

I

I

LEMMA 4.4. Let n � 1 be an integer. Suppose that v is a positive integer such that (vp - 1) I (p - 1) (1 + np ). Then there exists an integer h � 0 such that u = (n - h )/v � O, n = F(p, u, h ) and (u + 1) l h (p - l). If u = 0 then v = pn - n + 1 . PROOF. Let vp - 1 = m , m2 with m , 1 (p - 1) and m 2 1 (1 + np ). Then p 1, vp - 1 == 0 (mod m ) Hence v - I == 0 (mod m I). Also vp - 1 == 0, np + 1 == 0 This implies that (n + v )p 0 (mod m2) and so n + v == 0 (mod m2) . Thus vp - 1 = m , m 2 divides (v - 1) (n + v ). Define the integer h by (4.5) (v - 1)( n + v) = h (vp - 1). Thus h � O. For fixed n, p, h let f(X) = (X - 1) (X + n) - h (Xp - 1). Thus f(v ) = O. Hence f(v ') = O where v ' = (h - n )/v and so u = - v ' is an integer. Since f(l) < 0 and v � 1 it follows that v ' < 1 and so v ' � 0 as v ' is an integer. Therefore u � O. As f( - u) = 0 it follows that (u + l) (n - u) = h (up + 1) and so n = F(p, u, h ). Furthermore h (p - l) == - h (up + l) == O (mod (u + l)). If u = 0 then n = h and (4.5) implies that v = pn - n + 1 . 0 ==

I

==

.

4]

A CHARACTERIZATION OF SOME GROUPS

353

The proof of (4. 1) will now be given. Suppose the result is false. Let G be a counter example of minimum orde!. We will show in a series of lemmas that the assumed existence of G leads to a contradiction. The notation introduced at the beginning of section 1 will be used for the remainder of this section with the following modifications. t = (p - 1)/e. B is the principal p-block of G. If t l- 1 , tl , . . . , tt are the exceptional characters in B. Xo, . . . , Xe and 0o, . . . , Oe are defined as in (VII.2.12) for the principal p -block of G where XI is the principal character of G. If t = 1 let t, = Xo. LEMMA 4.6. (i) If OJ = 1 for Xj an irreducible character then Xj (l) = 1 or 1 + np. (ii) If OJ = - 1 for Xj an irreducible character then Xj (l) = P - 1 or p (pn - n + 1) - 1. (iii) If tl- l and 00 = 1 then tti (1) 1 + np for 1 � i � t. (iv) If t I- 1 and 00 = - 1 then tti (1) = p - 1 or p (pn - n + 1) - 1 . =

PROOF. Let X be an irreducible character of G. Then 1 G c (y ) I X y ) ( X (1) I C is an algebraic integer. Since I G Cc (y ) 1 = e (1 + np ) = (p - 1)(1 + np )/t this implies that (1 + np)(p - l)X (Y ) tX (l) is an algebraic integer. Suppose that X = Xj is a nonexceptional character in the principal p-block. By (VII.2. 17) Xj (Y ) = OJ and Xj (l) = OJ + mp for some integer m. Thus x (1) I (p - 1) (1 + np). Suppose that t I- 1 and X = ti. Then by (VII.2. 17) L:=, ti (y ) = 00 and L: = I ti (1) = 00 + mp for some integer m . Thus tX (l) I (p - 1) (1 + np ). (i), (iii). Let X = Xj with OJ = 1 or let X = 2:: = 1 ti where t I- 1 and 00 = 1 . B y (4.3) n = F(p, u, h ) where x (1) = 1 + up. Since G i s a counterexample either h = 0 or u = O. If u = 0 then X(l) = 1 and so in particular t = 1 . If h = 0 then n = F(p, u, 0) = u and x (1) = 1 + np. (ii), (iv). Let X = Xj with OJ = - 1 or let X = 2:: = 1 ti where t I- 1 and 00 = - 1 . By (4.4) n = F(p, u, h ) where X (1) = vp - 1 with u = (n - h )/v. Since G is a counterexample either h = 0 or u = O. If h = 0 then :

:

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n = F(p, u, O) = and u = n /v. Thus v X (l) = p (pn - n + 1) - 1 by (4.4). q u

=

1 and x (1) = p - 1. If u

=0

then

The next result is a refinement of (4.6). LEMMA 4.7. (i) If 0i = 1 for Xi irreducible, j � 1 then Xi (1) = 1 + np. (ii) If 0i = - 1 for Xi irreducible then Xi (1) = p - 1 . (iii) If t � 1 and D O = 1 then t(; (l) = 1 + np for 1 ::s; i ::s; t. (iv) If t� 1 and DO = - 1 then t(j ( l ) = p - 1. PROOF. (i) Since G = G', Xj (l) > 1 for j � 1 . The result follows from (4.6) (i). (ii), (iii), (iv). There are at most e values of j with OJ = 1 since the Brauer tree is connected. By (4.6) Xi (1) ::S; 1 + np if 0i = 1 and X I (l) = 1 . Let co = 1 if Do = 1 and co = 0 if 00 = - 1 . Let c 1 = 1 - co. Then (VII.2.1S) (iii) implies that < e (1 + np ) = � (1 + np)

t 1 = t {p (pn - n + 1) - I}.

4]

A CHARACTERIZATION OF SOME GROUPS

355

Let G O = G/Op , (G). Every irreducible character in B has 01' (G) in its kernel and so may be identified with a character of G O . Suppose that there exists an irreducible nonprincipal Brauer character 'P in B with 'P (I) ::s; Hp - 1). As GO is simple, (3.1) implies that G O = PSL2 (p ) contrary to assumption. (i) Suppose that t = 1 . Let x be an element in N which maps onto an element of order p - 1 in G O . By (4.7) there exists j with Xi (l ) = P - 1. Thus (Xj )N is irreducible and so (Xi ) is the character afforded by the regular representation of (x ) . Hence the linear transformation corresponding to x in the representation which affords Xi has determinant - 1, contrary to the fact that G = G'. Thus t� 1 . If 0 0 = - 1 then by (4.7) (i (I) � Hp - 1) which i s impossible. Thus 00 = 1 . B y (4.7), t( 1 ( 1 ) = 1 + np. Suppose that ou 1 for some u with 2 ::s; u ::s; e. For any j let Pi denote the vertex on the Brauer tree corresponding to Xi' The path from Po to Pu contains a vertex Ps with Os = - 1 . Since p] is an end point of the tree, p] does not occur on this path. Thus XS has at least two nonprincipal Brauer constituents. By (4.7) xs (1) = p - 1 . Hence there exists an irreducible nonprincipal Brauer character 'P with 'P (1) ::S; Hp - 1) which is not the case. Consequently 0i = - 1 for 2 ::S; j ::s; e. The result follows from (4.7). (ii) This is a direct consequence of (i) and (VII.2.IS) (iii). (iii) Since GO is simple it follows from (3.2) that a Sp -group of GO is self centralizing. Hence (x)

C:

Thus Xi (1) � p (pn - n + 1) - 1 and t(i (1) � p (pn - n + 1) - 1. The result follows from (4.6). 0 LEMMA 4.8. (i) t� 1 , t( l ( l ) = 1 + np, XI (1) = 1 and Xi (l ) = P - 1 for 2 ::S; j ::S; e. (ii) 1 + (1 + np )/t = (p - l ) ( e - 1). (iii) G is simple and CG (y ) = (y ).

where G O has 1 + nop Sp-groups. Thus 1 + nop ::s; 1 + np and (1 + np )/t = (1(1) is an inte:ger. By (ii) p - 1 and (1 + np )/t are relatively prime. It follows that (1 + np)lt 1 (1 + nop )/t. Thus 1 + np ::s; 1 + nop and so n = no. The minimality of G now implies that G = G O . 0

PROOF. Let G] be the subgroup of G generated by all elements of order p in G. Thus G] <J G and G] has 1 + np Sp -groups. Suppose that G I � G. By induction p > 3 and G]/O p ' (G 1 ) is either isomorphic to PSL2 (p ) or p = 2Q - 1 amd G]/Op , (G]) = SL2 (p - 1). Hence G /Op' (G) has a normal subgroup A such that either A = PSL2 (p ) or p = 2Q - 1 and A SL2 (p - 1). Thus G/Op , (G) is isomorphic to a sub­ group of the automorphism group of A. Since G = G ' this implies that G /Op ,(G) = A contrary to assumption. Hence G = G]. This in particular implies that G/Op , (G) is simple.

Let 8 1 , 82, , be all the irreducible characters of G which are not in B. Then each 8; is in a block of defect O. Let 8; (1) = pai. Let q be a prime with q I e. Let x be an element of order q in N. •

•

LEMMA 4.9. (i) t = P - 2 - n. (ii) 2: a ; = n.

=

PROOF. (i) Since te = p - I this follows from (4.8) (ii). (ii) By (4.8) (i)

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357

particular this implies that (cI>, IN ) � 1 and (cI>, IN ) = 1 if and only if cI>( z ) = 1 for all z E N - P. This last condition holds if and only if cI> = cI> the principal indecomposable character which corresponds to the principal Brauer character. Hence «O; )N, IN ) � a; for all i. Thus (4.9) (ii) and (4.12) imply that «0; )N, IN ) = a; for all i. Therefore (O; )N = a;cI>o and so O; (x ) = a;. D 0,

:

= 1 + ( np 1 t + (p - 1 )2 ( e -,- 1) + P 2 L a T . Direct computation using (i) yields the result.

4]

D

LEMMA 4.10. Let z be a q -singular element in G. Then the following hold. (i) Xi (z ) = 0 for 2 � i � e. (ii) ti (Z ) = - 1 for 1 � i � t.

PROOF. (i) By (4.8) (ii) (p - 1) is relatively prime to (1 + np )/t. As I G 1 = p(p - 1)(1 + np )It, this implies that Xi is in a block of defect 0 for q, where 2 � i � e. The result follows. (ii) This follows from (i) and (VII.2. 1S) (iii). 0

LEMMA 4.13. e = 2. P = 2a + 1 for some a > 1 . Every element of even order in G is conjugate to x. A Sr group of G has order p - 1 .

PROOF. Let {Y/i } b e the set of all irreducible characters o f G. If z is· a q -singular element then (4. 10) yields that o = L y/; (I)y/; (z ) = 1 - (l + np) + p L a;O; (z ).

Thus 2: aiO; (z ) = n. Suppose that z is not conjugate to x. Then (4. 10) and (4.1 1) imply that -]

LEMMA 4.1 1 . Oi (x ) = ai for all i.

PROOF. Let T = ' + Xl - 2:;= 2 Xj . By (VII.2.1S) (iii) T(z ) = 0 for z E N, z i- y i with 1 � i � P - 1. By (VII.2. 17) T(y ; ) = e - A (Y ; ) for all i, where A is a faithful irreducible character of N. Thus I

p

L T(z ) = ep - L A (y i ) = ep. zE N Hence (TN, IN ) = 1 . Since (4.10) (i) holds for a n arbitrary prime divisor q of e i t follows that if 2 � j � e then ;=1

p-I

L X (z ) = L Xj (Y ; ) = O. z EN j Thus «Xj )N, IN ) = O. Hence also «'; )N, IN ) = 0 for 1 � i � t by the previous paragraph. Let p be the character afforded by the regular representation of G. Then (pN, IN ) = 1 + np. Hence ( 2: 0; (I) (O; )N, IN ) = np and so ; =0

(4.12) If cI> is a principal indecomposable character of N then cI> = 'P + 2: A where 'P is an irreducible character of N/P and A ranges over all the faithful irreducible characters of N. For each i, (O; )N is a sum of a; principal indecomposable characters of N. Thus cI>( z ) = 'P (z ) for z E N - P. In

0 = L Y/i (X )y/; (z ) = 1 + t + L Oi (X )Oi (z ) = 1 + t + L a;O; (z ) = 1 + t + n > O. This contradiction shows that every q -singular element in G is conjugate to x. Thus every q -singular element in G has order q. Furthermore Co (x ) is a Sq -group of G. By (4. 10) and (4. 12)

I Co (x ) 1 = L 1 Y/i (x ) 1 2 = 1 + t + L a �· = 1 + t + n.

Hence by (4.9) (i) I Co ( x ) 1 = P - 1 is even. Thus q = 2 and p = 2a + 1. If a = 1 then a Srgroup of G has order 2 contrary to the simplicity of G. Since N /P is cyclic of order e it follows that e = 2. D LEMMA 4. 14. Let S be a Srgroup of G. Then No (S) is a Frobenius group of order (p - l) (p - 2).

PROOF. By (4. 13) every element of S - (1) has order 2. Thus S is abelian. Since no element of odd order commutes with an involution, No (S) is a Frobenius group with Frobenius kernel equal to S. By (4.13) 1 S 1 = p - 1 and any two involutions in G are conjugate. Hence by a theorem of Burnside any two involutions in No (S) are conjugate. D

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PROOF OF (4. 1). By (4.13) t = 4 (p - 1 ) Thus by (4.9) (i) n = 4 (p - 3). Hence I G 1 = p(p - l) (p - 2). By (4. 14) G has a permutation repres� ntation on p letters in which No (S) is the subgroup leaving a letter fixed. Smce No (S) I. S a Frobenius group, any faithful permutation representation of No (S) on G has a triply transitive permutation representation on p letters. As I G I = p (p - 1) (p - 2) the assumptions of (4.2) are satisfied and so by (4.2) G = SL2 (p - 1). 0 .

5. Some consequences of (4.1)

The next two results are due to Brauer and Reynolds [1958] . The first of these was originally proved by Brauer under the additional assumption that Co (P) = P. Nagai [1952], [1953], [1956], [1959] has proved several results related to Brauer's original result in which analogous conclusions are reached under various assumptions about n. A slightly different sort of related result can be found in Hung [1973] . Herzog [1969], [1970], [1971] and more recently Brauer [1976b], [1979], have proved related results in case the Sp -group of G is assumed to be cyclic but not necessarily of prime order. THEOREM 5.1. Suppose that G = G' and a Sp -group P of G has order p. Let 1 + np be the number of Sp -groups in G. Let H = Op,(G). Then one of the following holds. (i) p > 3, G /H = PSL2 (P ). (ii) p = 2a + 1 > 3, G/H = SL2 (p - 1). (iii) n ?; ! (p + 3). PROOF. Suppose that neither (i) nor (ii) holds. By (4.1) n = F (p, u, h) for some positive integers h and u where 2 F(p, u, h) -_ hup +uu++1 u + h . It is easily seen that for fixed positive h, F (p, u, h ) is an increasing function of u and for fixed positive u, F(p, u, h ) is an increasing function of h. Thus n ?; F (p, 1 , 1) = ! (p + 3). 0

COROLLARY 5.2. Suppose that p is a prime factor of I G I and p] > I G I· Assume further that the following conditions are satisfied. (i) G = G'. (ii) A Sp -group of G is not normal in G.

5]

SOME CONSEQUENCES OF (4. 1 )

359

(iii) G has no normal subgroup of order 2. Then p > 3 and either G = PSL2 (p ) or p - 1 = 2" and G = SL2 (p - 1). PROOF. Let P be a Sp -group of G. If I p i ?; p2 then I G P 1 < p and Sylow'S theorem implies that P <J G contrary to assumption. Thus I P I = p. If (5. 1) (i) or (5.1) (ii) holds then I G : H I > 1p3 and so H = (1) and the result is proved. If (5. I) (iii) holds then I G I ?; 2P (1 + np ) ?; 2P (1 + � (p + 3)P ) = P (p + 1) (p + 2) > P 3 contrary to assumption. 0 :

The next result shows that in a very special case, groups which have a character that satisfies the condition of (IV. IO. I) can be classified. COROLLARY 5.3. Assume that a Sp -group of G has order p and G has an irreducible character YJ such that YJ vanishes on p -singular elements and YJ (x ) = ± I Co (x ) lp for every p '-element x of G. Then G is of type L2 (p). PROOF. Let H be the last term in the descending commutator series of G. If p I I G : H I then G is p -solvable and the result is proved. If p ,f I G : H I then by Clifford ' s theorem YJH is irreducible as YJ (1) = p. Thus by changing notation it may be assumed that G = G'. . Let x be a p '-element in G and let Z = Z(G). If x does not commute with a p-element other than 1 then YJ (x ) = ± 1 . If x commutes with a p-element other than 1 then YJ (x ) = ± P = ± YJ (1) and so x E Z. Thus if G has exactly 1 + np S,, -groups then G has exactly (l + np ) (p - l) I Z I p­ singular elefuents. Hence I G I = I G i ll YJ 1 12 = I G 1 - (1 + np ) (p - 1) I Z I - I Z I + I z i p 2 . Therefore I Z I (p2 - 1) = I Z I (1 + np ) (p - 1) and so 1 + np = 1 + p. Hence n = 1. If P = 2a + 1 > 5 then the number of Sp -groups is 1 + 1 (p - 3)p > l + p. Since SL2 (4) = PSL2 (5) the result follows from (5. 1). 0 The proof of the next result in case p = 5 or 7 depends on the classification of all finite groups which have a faithful complex representa­ tion of degree at most 4. This classification can be found in Blichfeldt [1917] . THEOREM 5.4 (Brauer and Tuan [1945]). Let G be a simple group with I G I = pq bm, where p, q are primes, b, m are positive integers and m < p - 1.

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Then p > 3, q = 2 and one of the following holds. . (i) p = 2a ± 1 and G = PSL2 (p), . (ii) p = 2a + 1 and G = SL2(p - 1). Conversely the groups listed in the conclusion satisfy the assumptions. PROOF. The converse follows easily from the fact that in case (i) I G 1 = !p (p - l ) (p + 1) = 2ap (p ± 1)/2 and in case (ii) I G 1 = 2 ap (p - 2). In proving the result it may clearly be assumed that q ,( m. If p = q then Sylow ' s theorem implies that a Sp -group is normal in G contrary to assumption. Thus p f: q. Hence if P is a Sp-group of G then I P 1 = p. If P = 2 or 3 then G is solvable by Burnside ' s theorem. Thus p � 5. By (VII.2.15) (iii) there exists a character ( f: Ie in the principal p -block B of G such that (d, q ) = 1 where d = ((1). Thus d I m and so d < p - 1 . Consequently ( is an exceptional character in B by (VII.2.16). Suppose that d :S ! (p + 1). If P = 5 or 7 the result follows by inspection of finite linear complex groups in dimension at most 4. If P � 1 1 then G = PSL2 (p) by (3.1). Thus (p + l)(p - 1) = � p I G 1 = 2q bm. Since (p + 1 , p - 1) = 2 it follows that q b divides either p + 1 or p - 1 . Hence either p - 1 = q b and p + 1 = 2 m or p + 1 = q b and p - 1 = 2 m . In either case q = 2 and statement (i) holds. Suppose that d > ! (p + 1). Since d < P it follows from (3.2) that Ce (P) = P. Let e = I Ne (P) : P I and let t = (p - l)le. By (VII.2. 16) d ·= p - e. Let e = q Ch where h I m. Thus d = p - q Ch and so (d, h ) = 1. Thus dh 1 m. Since ! < d this implies that h = 1 . Hence e = q C and d = p - q C. Since d < P - 1 it follows that 'c > O. These results imply that (5.5) 1 + d = 1 + P - q C = 2 + te - q C = 2 + (t - l)q c. m

If q f: 2 then (VII.2. 15) (iii) implies the existence of a nonprincipal nonex­ ceptional character X in B with (X (1), q) = 1. Hence X (1) I m and so X (1) < P - 1 which is impossible as X (1) ± 1 (mod p) by (VII.2.16) and X (l) > 1. Therefore q = 2. Suppose that c > 1. Then (5.5) and (VII.2.15) (iii) imply that there exists an irreducible character X in B with X(l) ¢ 0 (mod 4), X f: I e, ( Thus X(l) = n or 2n with n I m. Since X (l) :S 2 n < 2(p - 1) and x (1) = ± 1 (mod p ) it follows that x(1) = p ± 1 . Hence n = ! (p ± 1). As n I m and m < 2 n this implies that m = Hp ± 1). Consequently d :S Hp + 1) contrary ==

i.

6]

PERMUTATION GROUPS OF PRIME D EGREE

361

to assumption. Thus c = 1. Therefore d = p - 2 and so m = p - 2. There­ fore e = 2 and I G 1 = p 2b (p - 2). Since e = 2 there exists a unique irreducible character X in B with X f: I e, (i. By (VII.2.15) (iii) X (l) = P - 1. Hence (x(1), m ) = 1 and so p - 1 = r. By (IVA.23) X is not in a 2-block of G of full defect b. Since p - 2 is odd, X is the only irreducible character in B which is not in a 2-block of G of defect b. Since C e (P) = P there are no elements in G of order 2p. Thus by (IVA.24) X (l) 0 (mod 2b). Consequently X (l) = 2b and so I G 1 = p (p - l) (p - 2). Since e = 2 this implies that G contains exactly (1 + !(P - 3)p) Sp-groups. Statement (ii) now follows from (5.1). 0 ==

6. Permutation groups of prime degree

Let G be a transitive permutation group of prime degree p. Then a G has order p. Thus the material of Chapter VI is applicable to the study of G. See for instance Ito [1960a], [1960b], [1962b], [1963a], [1963b], [1963c], [1964], [1965a], [1965b], Michler [1976a], Neumann [1972a], [1972b], [1973], [1975], [1976] . Similarly the methods of Chapter VI can be applied to transitive permutation groups of degree 2p and 3p if it is assumed that p 2 does not divide the order of such a group. See for instance Ito [1962c], [1962d], L. Scott [1969], [1972]. In this section we will only prove a classical result of Burnside and a result of Neumann. See also Klemm [1975], [1977] Mortimer [1980] for related results. Sp -group of

THEOREM. 6. 1. Let 'Po be the principal Brauer character of G and let 5. 1 < e < p - 1 . There exists a cyclic subgroup E of order e in N such that N = Ee and E n e = (I). (iii) e = P x Z, Z is cyclic and ( I Z I, e ) = 1. (iv) G is not of type L2 (p). There exists a faithful irreducible character � of G with �(1) = p - e. =

CHARACfERS OF DEGREE LESS THAN P

7]

-1

367

LEMMA 7.5. It suffices to prove (7. 1), (7.2) and (7.3) for groups G which satisfy ( ) * .

PROOF. Suppose that G is a counterexample to (7. 1), (7.2) or (7.3) of minimum order. We will show that G satisfies ( ) Since G is a counterexample, �(1) < (p - 1). Let Go be the subgroup of G generated by all elements of order p in G. Theorem B of Hall-Higman (VII. I0.2) implies that Op ' (Go) is in the center of Go. Since P 11 Go it follows that Go/Go n Z is simple and Go = Gb. If G "I Go then the minimality of 1 G 1 implies that Gol Go n Z is isomorphic to PSL2 (P ) for p > 7 or to SL2 (2a ) for 2a = p - 1 > 4. The Schur multipliers of these groups are well known. Since Go is generated by elements of order p, this yields that Go is isomorphic to one of PSL2 (p ), SL2 (P ) for p � 7 or SL2 (2a ) for 2a = p - 1 > 4. None of these groups admit an outer automorphism which stabilizes a character of degree less than p - 1. Hence G = GoZ and G is not a counterexample. Consequently G = Go. Thus G I Z is simple, G = G' and (i) is satisfied. Since G is a counterexample G is not of type L 2 (p). Thus by (3. 1) �(1) > ! (P - 1). By (3.2) e = P x Z. Hence by (VII.2.16) «1) = p - e and (iv) is satisfied. As G has a faithful irreducible character, Z is cyclic. Furthermore I Z I I �(1) = p - e. Hence ( I Z I, e ) = 1 and (iii) is satisfied. Since G = G' 1 < �( l ) < p - 1. Thus 1 < e < p - 1 by (iv). Nle is a group of automorphisms of P and so is cyclic. As NIP is abelian this implies the existence of a cyclic subgroup E of NI Z of order e. Thus (ii) is satisfied since (e, 1 Z 1 ) = 1. 0 * .

For the rem'ainder of this section assume that ( ) is satisfied. *

Since � is faithful there exists a faithful irreducible character 71 of Z such that �(z ) = 71 (z )�(1) for z E Z. There is a one to one correspondence between p-blocks of G of defect 1 and characters 71 u, 0 � u � I Z 1 - 1. Hence by the first main theorem on blocks (111.9.7), G has exactly 1 Z i p -blocks of defect 1. Let Bo, B1, denote all the p -blocks of defect 1 where the notation is chosen so that for a character 8 in Bu , 8z = 8 (1)71 u. Thus Bo is the principal p-block. Since NIP is abelian, the index of inertia of Bu is equal to e for O � u � I Z I - 1. Let A I, . . . , At denote all the irreducible characters of NI Z which do not have P in their kernel. For 0 � u � 1 Z 1 - 1 let ��ul, . . . , ��u ) denote the exceptional characters in •

•

•

368

[7

CHAPTER Vln

Define 8 (u ) = ± 1 by L: = I ?jU)(I) == 8(u ) (mod p ) . Thus 8 (� ) = 8o for the block Bu as defined in section 2 of Chapter VII. By (VII .2 . 17) the notation can be chosen so that for 0 � u � I z 1 - 1 S ?j u ) (y Sz ) = - 8 ( U )Ai (y )71 U ( z ) (7.6) for 1 � s � P - 1 , z E Z, 1 � i � t. Bu .

7.7. Let X be an R -free R [G] module which affords the character g. Suppose that gz = g (I)71 u . Let g = {3 + Y + P where (3 = L; =l hj?t >, y is a character in Bu which is orthogonal to every ?Ju ) and is orthogonal to every character in Bu. Let h = L; = I hj. Assume that p > 7, e � 3 and t � 3. Then the following hold. (I) Suppose that X is indecomposable and g (l) == e (mod p ) or g (l) == e + 1 (mod p). Then one of the following occurs. (i) 8 ( u ) = 1 ; h � t - 1 . (ii) 8 (u ) = - 1 ; h � 1 . (II) Suppose that X = WI EB W2 , where each � is indecomposable and g (l) == 2e (mod p ). Then one of the following occurs. (i) 8 (u ) = 1 ; h � t - 2. (ii) 8 (u ) = - 1; h � 2 .

LEMMA

PROOF. Let be a faithful irreducible character of P. For 0 � i � P - 1 and for any R -free R [ G] module Y let as ( Y) be the multiplicity of S as a constituent of the character afforded by YP• If Y affords the character 8 let as (8) = as ( Y). If 8 is an irreducible constituent of p then 8z = 8 (1)71 u and 8 is not in Bu . Hence 8 is in a block of defect O. Thus as (8) = ao (8) for O ::s; s � p - 1. Therefore as ( g ) - ak ( g ) = as ( {3 ) - ak ((3 ) + as ( y ) - ak ( y ) (7.8) for 0 � s, k � P - 1 . If 8 is an irreducible constituent of y then by (VII.2.17) as (8) = ad 8) for l � s, k � p - l and ao (8) - as (8 ) == 8 (1) (mod p ) for l ::s; s ::S; p - 1. Therefore a o ( y ) - al (-y ) == y (l) (mod p ) . Thus (7.6) implies g (I) == - 8(u )eh + ao( y ) - as ( Y ) (mod p ) for l � s � p - 1. (7.9) By (7.6) E

E

{ (Z hjAj ) - ak (Z hjA )}

as ( {3 ) - ak ( {3 ) = - 8 ( U ) as

j

for O ::s; s, k � P - 1 .

(7.10)

CHARACTERS OF D EGREE LESS THAN P

7]

-1

369

In proving the result it may be assumed that h � t - 1 otherwise there is nothing to prove. Hence by (7. 10) there exists m with 1 � m ::s; p - 1 such that ao ( {3 ) - am ((3 ) = O. Thus by (7.8) ao ( g ) - am ( g ) = ao ( Y ) - am ( y ). Hence (7.9) yields that (7. 1 1) g (l) == - 8 (u )eh + ao( g ) - am ( g ) (mod p ). (I) Since X is indecomposable it follows from (VII.I0.9) that l ao( g) - am ( g ) I � 1 . Let g (l) = e + c (mod p) where c = 0 or 1 . B y (7.1 1) e + c == - 8 (u )eh + c ' (mod p) where c ' = 0 , ± 1 . Thus {1 + 8 (u )h }e + c - c ' == 0 (mod p). If 1 {1 + 8 (u )h }e + c - c ' l � p then h � (t - 1) as I c - c ' I ::s; 2 < e and the result is proved. Thus it may be assumed that { I + 8 (u)h }e = c ' - c. Hence c ' - c == 0 (mod e ) and so c ' - c = O. Thus 8 ( u )h = - 1 and so 8 (u ) = - 1, h = 1 as required. ( II ) By (V II .I0.9) l a () ( g ) - a", (g) I ::S; 2. By (7. 1 1) 2e == - 8(u )eh + c (mod p) with c = 0, ± I , ± 2. Thus {2 + 8 ( u )h }e - c == 0 (mod p ). If 1 {2 + 8(u)h }e - c l � p then h � t - 2 as 1 c 1 � 2 and the result is proved. Thus it may be assumed that {2 + 8 (u )h}e = c. Hence c 0 (mod ) and so c = 0. Therefore 2 + 8 ( u )h = 0 and so 8 ( u ) = - 1, h = 2. 0 ==

e

7. 12. Suppose that the assumptions of (7.7) are satisfied. Assume furthermore that 8 (u ) = 1 and ?\u ) ( l) > (p - 1). Then the following hold. ( I ) If g (l) e (mod p) then g ( l ) � (t - 1) ?\U) (1) + p - 1 . If g ( l ) e + 1 (mod p) then g (l) � (t - 1)? \ u ) (l). ( II ) If g (1) == 2e ( mod p ) then g (l) � (t - 2) ?\U ) (1) + p - 1. LEMMA

==

==

PROOF. If h � r, all the statements are trivial. Suppose that h � t - 1. If g (l) == e + 1 (mod p), then h = t - 1 by (7.7) and the result is clear. Suppose that g(l) e (mod p ) . By (7.7) h = t - 1. Therefore e == g (l) == - e (t - 1) + y (I) == e + 1 + y(1) (mo d p ). Hence y (l) == - 1 (mod p) and so y (l) � P - 1 . The result is proved in this case. Suppose that g(l) == 2e (mod p). If h � t - 1 the result is clear. Suppose that h < t - 1 . Thus by (7.7) h = t - 2. Therefore 2e == g(l) == - (t - 2)e + y (l) == 1 + 2e + y (l) (mod p). Hence )' (1) - 1 (mod p ) and so )' (1) � P - 1 . This implies the result. 0 ==

==

7. 13. Let f.L be an irreducible character of N/C For 1 � i � P - 1 and 0 � u � I Z 1 - 1 let W(i, u ) denote the indecomposable R [GJ module

LEMMA

CHAPTER VIII

370

[7

------such that W(i, u ) = V( T/ u /Lo< 2i + 1). If K is replaced by a suitable finite extension field then the following hold. (i) For 0 � i < He - 1) there exists an R -free R [G] module M(i, u ) such that M(i, u ) = W(i, u ) EB W(e - 1 - i, u ). (ii) There exists an R -free R [ G] module M (u ) such that M (u ) = WO (e - 1), u ) if e is odd and M(u ) = W(�e, u ) if e is even. PROOF. By (I.17. 12) there exists an R -free R [N] module Y(u ) for O � u � I z l - 1 such that Y(u ) = V( T/ U/La (e- l l/2, e ) if e i s odd and Y(u ) = V( T/ u /La e/2, e + 1) if e is even. Let M(u ) be defined by M(u ) = Y(u ). By (I1I.S.8) M(u ) has the required properties. By (1.17.12) there exists an R -free R [N] module X(i, u ) for O � u � I z l - 1, O � i � He - 1) such that X(i, u ) = V( T/ u /La i, 2e ). By (VI.2.8) there exists an exact sequence o � V (T/ U /La \ 2i + 1) � V ( T/ U /La \ 2e ) � V( T/ U p,a - i - 1 , 2(e - 1 - i ) + l) � O. As a e = 1 it follows from (1 . 1 8 .2) that if K is replaced by a suitable 'finite extension field then there exists an R -free R [N] module Y (i, u ) such that

Y(i, u ) = V( T/ u /La \ 2i + l) EB V( T/ u/La e- i -1, 2(e - 1 - i ) + 1). Let M(i, u ) be defined by M(i, u ) = Y(i, u ). The result follows from (III.S.8). 0 �

The next result is implicit in Brauer [1966c]. LEMMA 7. 14. Suppose that 0 � u, v, w � 1 z 1 - 1 with w = u + v (mod 1 z 1 ). Assume that the following conditions are satisfied. (i) o ( u ) = o (v) = o ( w ) = l. (ii) � \m ) (l) < p - 1 for m = u, v, w. Let �iU) �)v ) L� =l hijd �W ) + r, where r is orthogonal to each � �w ) . Then L� =l h;jk � t. =

PROOF. Let 'P h . " ., 'Pe be all the irreducible characters of E where 'Pj = 'P L Then lPj = 'P [E for j = 1, . . "' e are all the principal indecomposable charac­ ters of PE. By abuse of notation let (A'i )PE = A i . For fixed m, Uint ) } is a set of irreducible characters, any two of which agree on p i-elements. Thus for m = u, v or w (�i m ) pE = lP[( m ) - Ai, where f(m ) does not depend on i. Hence

8]

PROOF OF (7, 1)

371

(7. 15) Let PE, PEP be the character afforded by the regular representation of E, EP respectively. Then AilPs = { ( Ai )E'Ps yE ( P tE = PP . E E Similarly =

lP[(U ) lP[(v)

of

= { ( lPf( U )

E {(t + E 'Pf (v)Y = PE 'Pf(u ) 'P f ( v ) YE = ( tPE tE + ('P f(u )+f(v ) tE = tppF. + lPf (U )+f(v ) .

For each i, (Ai ) p is a sum of e distinct nonprincipal irreducible characters P. Thus

Hence A iAj = ()l + () 2 where ()l is a sum of at most e principal indecompos­ able characters of PE and ()2 is a linear combination of the ( Ai ) PE. Since A iAj vanishes on E - { l } it follows that () l = 0 or () l PPF.. Therefore (7. 15) becomes =

(7. 16) where 0 = 0 or 1. The set { lPs } U {Ai } is a basis for the additive group of integral linear combinations of irreducible characters of PE. Furthermore if a character of PE is expressed in terms of this basis then the coefficient of lPs is nonnegative for ,all s. By (7. 16) the coefficient of any lPs in (� :u ) �jv) )PE is at most t. Since (� kW ) ) PE = lPf( w ) - Ak the result follows. 0 8. Proof of (7. 1)

This section contains a proof of the following result which implies (7. 1). THEOREM 8 . 1 . Suppose that condition ( ) of section 7 is satisfied with p > 7 and � a faithful irreducible character. Assume that there exists u, v, w with , 0 � u, v, w � 1 z I � 1 such that w == u + v (mod 1 z I ) and the following conditions are satisfied. (i) o (u ) = o (v) = o ( w ) = l . (ii) � = � \ u ) , �\U ) (l) < p - 1 and �\v ) (l) < p - l . Then e = 2. *

CHAPTER VIII

372

[8

8]

PROOF OF (7. 1 )

373

LEMMA 8.2. Statement (7. 1) follows from (8.1).

Furthermore

PROOF. Suppose that (8. 1) has been proved. In proving (7. 1) it may be assumed that ( ) is satisfied by (7 .S). Let u = Then YJ 2 u has the same order as YJ U since I z I is odd. Hence there is a field automorphism which sends YJ U to YJ 2 u. Thus the exceptional character in B 2 u is algebraically conjugate to ( and so 8 (2u ) = 8 ( u ) = 1. Now (7. 1) follows from (8. 1). 0

p 2 - e (2p - e ) = (p - e )2 = dimR XjU ) 0 Xjv) � 2: dimR W(k, w ). k= (J Suppose that e is odd. Then e� p 2 _ e (2p - e ) � k�= 2 dimR { W(k, w ) EB W(e - k - 1, w )} 0 + dimR W(H e - 1), w ) � H e - l) (t - 2) (2p - e ) + He - l)(p - 1) + (t - 1) (2p - e ) + (p - 1). Hence p 2 � (2p - e ){e + ! et - � t - e + 1 + t - l} + He + 1) (P - 1) = (2p - e )d (e + 1) + H e + l) (p - 1) = He + 1){2pt - te + p - I} = pt(e + 1). Thus p � t (e + 1) = te + t = P - 1 + t al)d so t � 1 contrary to the fact that e < p - 1 . Stlppose that e is even. Let f = dimR W O e - 1, w). Then

*

v.

For the rest of this section assume that the hypotheses of (8. 1) are satisfied. Furthermore assume that e > 2. A contradiction will be derived from this situation. Let xjm ) be an R -free R [GJ module which affords ( jm) for m = u, 1 � i � t such that Xjm ) is indecomposable. Let Aj = A � . Thus ( jU ) (y ) = ( J V ) ( y - l ). Furthermore there exist irreducible characters v, v ' of N/C such that v,

LEMMA 8.3 . (\w ) (l) = p - e. PROOF. By (2.7) e-l

V(T] wvv'a k , 2k + l) EB A, k =(J where A is a projective R [N] module. Let f-t = vv' and let W(k, w ) be defined as in (7.13). Thus (Xju ) 0 Xjv ) )N = E9

e-l

(

e- l

Xju ) 0 Xjv) = E9

W(k, w ) EB A ' k =(J for some projective R [G] module A '. Suppose that (\w ) (l) I: p - e. Thus (\w ) (l) � 2p - e. By (7. 12) and (7. 13) dimR W(k, w ) + dimR W(e - 1 - k, w ) � (t - 2)(\w ) (l) + p - 1 � (t - 2) (2p - e ) + p - 1 for 0 � k � � (e - 1). If e is even then dimR W(e /2, w ) � (t - l )(\w ) (l) � (t - 1) (2p - e ). If e is odd then dimR W(He - 1), w ) � (t - l)(\w ) (l) + p - 1 ;?; (t - 1) (2p - e ) + p - 1 .

Hence

!

p 2 - e (2p - e ) � (��-2 k =(J dimR { W(k, w ) EB W(e - 1 - k, w )} + dimR WO e, w ) + f � (! e - l) (t - 2) (2p - e ) + (� e - l) (p - 1) + (t - 1) (2p - e ) + f = (2p - e ){! et - e - t + 2 + t - I } + (� e - 1) (p - 1) + f.

pet + p = P 2 � (2p - e ) (� et + 1) + G e - 1 )et + f = pet - !e2t + �e2t - et + f + 2p - e. Therefore o � - et + p - e + f = 1 - e + f.

Hence f � e - 1 < � (p - 1). Furthermore f e - 1 (mod p) and so f > 1 as e > 2. Thus P is not in the kernel of WO e - 1 , w ). Hence by (3.1) G is of type L 2 (p) contrary to assumption. 0 ==

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374

LEMMA 8.4. There is an irreducible Brauer character 'P such that (\W) = 'P as a Brauer character. PROOF. Since 8 (w ) = 1, the principal Brauer character does not occur as a constituent of (\w ) . If the result is false then some nonprincipal irreducible Brauer character has degree at most � (p - e ) < � (p - 1). Thus by (3. 1) G is of type L 2 (p) contrary to assumption. 0 LEMMA 8.5. Let 'P be defined as in (8.4). Let 8 be the unique nonexceptional irreducible character in Bw which has 'P as a constituent. Then 8 (1) ;:-; 2p - 1. PROOF. If the result is false then 8 (l) = P - 1 as 8(1) - 1 (mod p). Thus 8 = 'P + 'P I as a Brauer character and 'P l(l) = e - 1. The principal Brauer character can occur as a constituent of 8 with multiplicity at most 1. Since e > 2, 'P 1 (1) > 1 and so there exists a non principal irreducible Brauer character in Bw of degree at most e - 1 < Hp - 1). Thus by (3. 1) G is of type L 2 (p) contrary to assumption. 0 Let h = � hijk , (�U ) (lv ) = a8 + i:= l hijk ( �w ) + r; k k where r i s orthogonal t o 8 and to all ( ); ) . ==

=1

LEMMA 8.6. (i) h ::::; t. (ii) If e is odd, a + h ;:-; He - l) (t - 2) + t - 1. (iii) If e is even, a + h ;:-; � e (t - 2) + 1. PROOF. (i) This follows from (7. 14) and (8.3). Let 'P be defined as in (8.4) . For any R -free R [ G] module Y let n ( Y) be the multiplicity with which 'P occurs as a constituent of the Brauer character afforded by Y. Thus a + h = n (Xj u ) 0 Xjv) ). Let W(k, w ) be defined as in (7.13). Thus by (2.7) e-I

XjU) 0 Xjv) = EB W(k, w ) EB A k =0 for some projective R [G] module A. (ii) By (7.7) and (7.13) 2 a + h ;:-; (e f0/ {n ( W(k, w )) + n ( W(e - k - 1, w ))} k + n ( WO (e - 1), ) ) ;:-; He - 1) (t - 2) + (t - 1). =

w

8]

PROOF OF (7. 1)

375

(iii) By (7.7) and (7. 13) 2-2 a + h ;:-; e/2: {n ( W (k, w )) + n ( W (e - k - 1, ))} + n ( WO e, w )) k =O ;:-; O e - l) (t - 2) + (t - 1) = O e ) (t - 2) + 1. 0 w

LEMMA 8.7. If 1 ::::; k ::::; p - 1 then (p - l)r(y k ) = a (p - 1) - he + ep - e 2 • PROOF. By (7.6) (jU) (y )(jV) (y ) = As (y )As (y -l) for suitable s. Thus (jU ) (y )(jv ) (y ) = e + g (y ), where g (y ) is a sum of e 2 - e primitive p th roots of 1. Let Tr denote the trace from the field of p th roots of 1 to the rationals. Since rz = r(1)YJ w and r is orthogonal to all (�w ) it follows that r(y ) is rational. As 8 (y S ) = - 1 it follows from (7.6) that e (p - 1) - e 2 + e = Tr{( lU ) (y )(jV ) (y )} = - a (p - 1) + he + (p - 1)r(y ). This implies the result. , LEMMA 8.8. h + e

==

0

1 (mod t).

PROOF. Divide the equation in (8.7) by e and read modulo t.

0

LEMMA 8.9. r(l) > (p + l){a + e - 1 - (e + h - l)/t} ;:-; (p + l)a. PROOF. Since rz = r(l)YJ w and r is orthogonal to 8 and all ( �w ) it follows that where the first sum is over characters Xk in Bw with Xk (1) 1 (mod p), the second sum is over characters Xk in Bw with Xk (1) - 1 (mod p) and ro is a sum of characters in blocks of defect O. Therefore if 1 ::::; s ::::; p - 1 r (y S ) = L' Ck - L" Ck ::::; L c £ . Thus by (8.7) ==

L' Ck

i

;:-; r(y ) = a + e + (1 - e - h ).

==

CHAPTER VIn

376

[8

As the principal character of G occurs in �l u ) �jv ) with multiplicity at most 1 this implies that r (l) � 2:' CkXk (1) � { ( L ' Ck ) - 1 } (p + 1) + 1 > (p + l){a + e - 1 + t1 (1 - e - h )}. This proves the first inequality. If the second inequality is false then a > a + e - 1 + (1/t) (1 -:- e - h ). Hence by (8.6) (i) (e - 1) < t1 (e + h - 1) = t1 (e - 1) + th � t1 (e - 1) + 1 . Thus ( t - l)(e - 1 ) < t. Since e � 3 this implies that t < 2 and so t = 1 contrary to the fact that e < p - 1 . 0 LEMMA 8. 10.

P

� 13 and t � 4

or

p = 13 and t = 3.

PROOF. If P = 11 then e = 5. Thus �\U ) (I) = 6 < � (p - 1) contrary to (3. 1). Thus P � 13. Hence by (3.3) P - e � � (p - 1) and so p + 3 � 4e. Thus et = p - 1 � 4e - 4 and so t � 4 - 4/ e. Thus t � 4 if e > 4. If e � 4 and t � 3 then p = 13 and t = 3. 0 LEMMA 8. 1 1 . e is even. PROOF. Suppose that e is odd. Thus t is even. Let (8.6) (ii)

C

= ! et - ! t - e. By

a � C + t - h. By (8.9)

(p - e )2 = � l u ) (1 ) � jv) (1) = a () (1) + h (p - e ) + r (1 ) > a (2p - 1) + h (p - e ) + a (p + 1) = 3pa + h (p - e ).

Therefore (p - e )2 > 3P (c + t - h ) + h (p - e ) = 3pc + t (p - e ) + ( t - h ) (2p + e ). Hence by (8.6) (i) (p - e ) (p - e - t) > 3pc + (t - h )(2p + e ) > 3(p - e )c.

8]

PROOF OF (7. 1)

377

Therefore Thus

et - e - t + 1 = P - e - t > 3c = � et - � t - 3e. O > ! et - ! t - 2e - 1 = ! t(e - 1) - (2e + 1).

This implies that + 1) _6_ t < 2(2e e - l = 4+ e - l · Hence either t = 4 or e = 3 and t = 6. If e = 3, t = 6 then p = 19, c = 3. Thus (S. 12) implies that 160 > 171. Hence t = 4. Now c = e - 2 and (8. 12) implies that (3e + 1) (3e - 3) > 3(4e + l) (e - 2). Thus 0 > e 2 - 5e - 1 and so e � 5. Since 21 = 4.5 + 1 is not a prime e = 3. Thus by (8.8) h 2 (mod 4) and so h = 2 by (8.6) (i). Now (S. 12) implies that 60 > 39 + 58. This contradiction establishes the result. 0 ==

Define the integer b by b = ! e (t - 2) + 1 - t = ! et - e - t + 1 .

(8. 13)

B y (8.6) (iii) and (8. 1 1) a � b + t - h. LEMMA 8. 15. e > t. PROOF. Suppose that e � t. By (S.6) (i) and (8.8) h = t + 1 - e. Thus by (8.9), r(l) > (p + l) (a + e - 2) and so (p - e y > a (2p - 1) + h (p - e ) + (a + e - 2) (p + 1) = 3pa + (t + 1 - e ) (p - e ) + (e - 2) (p + 1). Therefore by (8. 14) (p - e ) (p - t - 1) > 3pb + 3P (e - 1) + (e - 2) (p + 1) > (p - e ) {3b + 3(e - 1) + (e - 2)}. Hence et - t = P - t - 1 > � et - 3e - 3t + 3 + 4e - 5 = � et + e - 3t - 2. Therefore

(8. 12)

O > ! et + e - 2t - 2 = t O e - 2) + e - 2.

[8

CHAPTER VIII

378

Hence (e /2 - 2) < 0 contrary to the fact that e ;?: 4 as e is even. LEMMA S.16. t = 3, e

=

8]

0

4 and p = 13.

=

Thus / Z / / 9. If S is a subset of G let SO denote its image in GO = G /Z. There exists an element x E N of order e = 4 such that XO also has order 4 since ( / Z /, 4) = 1. Suppose that Bu + v is the principal block. Thus YJ u+v = 10 and GO has an irreducible character ��u+V) of degree 9. Let (X�u+v) )NO = V(a j, 9). By (1. 1) a (x Y = a (x )9j = detv (ai,9) (x ) = 1 . Thus a j = 1No and s o InvNo (X)u+v)) i- (0). Hence

> a (2p - 1) + h (p - e ) + a (p + 1) = 3pa + h (p - e ) (S. 17) ;?: 3pb + t(p - e ) + (t - h ) (2p + e ).

Hence

(p - e ) (p - e - t) > 3pb + (t - h ) (2p + e ) > 3(p - e )b.

379

The proof of (S. l) will be completed by showing that the case in (S. 16) cannot occur. For the rest of this section it will be assumed on the contrary that e = 4, t 3 and p = 13. By (3. 1) (�U )(l), (�v) (l) and (�u+V) (l) are all at least S. Hence (�u) (l) = ��V) (l) = ��u+v) (l) = P --C. e = 9.

PROOF. Suppose that t ;?: 4. By (S.9) and (S. 14) (p - e )2 ;?: a (2P - 1) + h (p - e ) + r (1 )

Thus

PROOF OF (7. 1 )

(S. lS)

et - e - t + 1 = P - e - t > 3b = � et - 3e - 3t + 3. Therefore O > ! et - 2e - 2t + 2 = to e - 2) - 2(e - 1). By (S. 10) and (S.15) e i- 4. Therefore 12 - 4 = 4 + -(S.19) -t < 4e e -4· 4 e -Since e > t this implies that t � 6. Thus by (S. 10) t = 4, 5 or 6. By (S. 1 1) e is even. If t = 6 then (S.19) implies that e < 10 and so e = S contrary to the fact that te + 1 is a prime. If t = 4 then b = e - 3. Hence (S. lS) implies that (3e + 1) (3e - 3) > 3(4e + l)(e - 3). Thus 0 > e 2 - ge - 2 and so e < 10. Hence e = 6 or S contrary to the fact that te + 1 is a prime. If t = 5. Then b = �e - 4 and (S.lS) implies that (4e + 1)(4e - 4) > 3(5e + l)(�e - 4). Therefore 0 > 13e 2 - S7e - 16 and so e < S. Thus e = 6. Hence p = 31, b = 5. Now (S.lS) implies that 35 > (5 - h )6S. Hence h ;?: 5 and so h = 5 as h � t. By (S. 14) a ;?: 5. Hence (S.9) and (S. 17) imply that 252 ;?: 61a + 125 + 32(a + 3) ;?: (61 + 32)5 + 125 + 96. Hence 125 ;?: 93 + 25 + 19 which is not the case. By (S. 10) the result is established. 0

Thus Invoo (X�u+V)) i- (0) contrary to (S.4). Therefore B is not the u+v principal block. By (1.1) �

Thus

v (x ) = detxju)(x ) = 1,

(X� U ) )N Hence by F .7)

=

Vi

= detx\v) (x ) = 1.

V( YJ u, p - e ),

(X�u) 0 X}v))N 3 12 (S.20) = EB V( YJ u+va k , 2k + l) EB EB V(YJ " +va \ p). k =8 k =O Therefore 3 X(u) ' 0. Suppose that n is even. Then by ( 10.5) 1 + np I (p - 2)2. Let d = (1 + np, (p - 2» . Thus 2n + 1 0 (mod d ). Hence 1 + np I (2n + 1 )2 . If 1 + np = (2n + 1 )2 then p = 4(n + 1) which is not the case. Thus 1 + np � ! (2n + lY. This implies that p � 2n + 2 contrary to ( 10.3 ) . Suppose that n is odd. Hence by ( 10.3) and (10.5) {(n + l)p - 1} = ((1) I (p - 2)2 . If d = (p - 2, (n + l )p - 1) then 2n + 1 0 (mod d ) and so {(n + l)p - 1} 1 (2n + 1Y. If (n + l )p - 1 = (2n + lY then p = 4n2 n+ +4n1 + 2 = 4n + _2_ n+l Hence = 1 as (n + 1) 1 2 and so p = 5 contrary to assumption. Therefore {(n + l )p - 1} ::oS H2n + 1)2 and so p � 2n + 1 contrary to ( 10.3). 0 *

Iz

I>

==

==

n

384

CHAPTER VIn

[11

1 1. Some properties o f permutation groups The material in this section, except for (1 1 .8), is independent of the previous results of this chapter and is included h ere as it will be needed in the next section. We will state several results without proof. If G is a permutation group on a set D, let ax denote the image of a under the action of x for a E D, x E G. If .1 is a subset of G let G", denote the subgroup of all elements in G which leave every element of .1 fixed. In other words GLl = {x I x E G, ax = a for all a E L1 }. If .1 = {a} let Ga = G{ a } . For a E D let a G = {ax I x E G}. A proof of the following result can be found in Wielandt [ 1 964] (13.1).

1 1. 1 (Jordan). Let G be a primitive permutation group on D. Let .1 be a subset of D with 1 � 1 .1 1 < i D 1 - 1 . If Gil is transitive on D - .1 then G is doubly transitive on D.

THEOREM

If G is a permutation group on D then Wielandt has defined the group G (2) to consist of all permutations on D which preserve all the orbits of G in D x D. Clearly G � G (2) . Suppose that G is a permutation group on D. Let .1 be a subset of D and let H be a subgroup of G such that ax E L1 for a E .1, x E H. For x E H define the permutation x il as follows ax il = ax, If a E .1, if

a E D - .1,

ax il

=

[1969] (6.5).

THEOREM 1 1 .2 (Dissection Theorem). Let G be a permutation group on D. Let .1 be a subset of D and let H be a subgroup of G such that ax E .1 for a E .1, x E H. Assume that for all a E .1, b E D - .1 , H = HaHb. Then

H il

X

H fl -Ll � H( 2) .

PROOF. We will first show that if x E H then X il E H(2 ). It suffices to prove that if (a, b ) E D x D then there exists z E H such that (a, b )z = (a, b )X Ll. There are three cases . If a, b E .1 let z = x. If a, b E D - .1 let z = 1 . Suppose that a E .1, b E D - .1 . B, y assumption x = XaXb with Xa E Ha, Xb E Hb. Let z = Xb. Then

385

(a, b )x il = (ax, b ) = ( aXaXb, b ) = (axb, b ) = (a, b )Xb. This proves that x Ll E H (2). Let x E H then x = x Llx n - il . Since X Ll E H(2) it follows that x n-Ll E H (2) . Thus HilH f2 - Ll = H Ll X H f2- Ll � H (2 ) . 0 We will next state some results without proof about rank 3 permutation groups. These are due to D.G. Higman [1964] though special cases had previously been considered by Wielandt [1956]. We will restrict our attention only to those results which are necessary for the considerations of the next section. Let G be a rank 3 permutation group on D. Assume that I D I is even. Let I G X + f) be the character afforded by the permutation representation of G on D where X, f) are irreducible characters of G. Since I D I is even, X (I) /: 8 (1) and so X, 8 are rational valued. For a E D let {a }, .1 (a ), r(a ) be the orbits of Ga where .1 (ax ) = .1 (a )x and r(ax ) = r(a )x for x E G. Let k = I .1 (a ) l , 1 = l r(a ) l. Since I G I is even, k /: and so .1 and r are self paired in the sense that

+

I

.1 (a )

=

{ax I ax - 1 E .1 (a )},

Define the integers

fL,

A by

I d (a ) n d (b ) 1

r (a ) = {ax I ax - 1 E r (a )}.

C . {I + +A + + (A -fL)(k + I) + k + I �

if

b E .1 (a),

if

b E r(a ).

(1 1 .3)

Then A, fL are independent of the particular choice of

a.

Let H il = {x il I x E H}. The following result is due to Wielandt

SOME PROPERTIES OF PERMUTATION GROUPS

1 1]

I r(a ) n r(b)

1=

- k fL - 1

1-k

1

if

b E r(a ),

if

b E .1 (a ).

a, b.

Furthermore

(11 .4)

The following conditions are satisfied. 2 Let d = (A - fL ) + 4(k - fL ) . Then d is a perfect square and

}

8 (1) 2k = + X (I)

2 Yd

2

.

(11 .5)

If furthermore s, t are the characteristic values of the incidence matrix associated to the orbit .1 then

}

s = (X - fL ) ± Yd (11.6) t 2 · Let p be a prime. Let G be a primitive permutation group on 2p letters . In case p = 5, examples of this situation are provided by A s and Ss acting on the set of 2-element subsets of {1 , . . . , 5}. All other known examples of

386

[1 1

CHAPTER VIn

such groups G are doubly transitive groups. The main object of this and the next section is to investigate such groups G which are not doubly transitive. We begin by stating without proof the following result which is the starting point of all work done on the structure of such groups.

[1956], [1964]). Let p be a prime. Let G be a primitive permutation group on n where / n / = 2p. Assume that G is not doubly transitive. Then 2p = m 2 + 1 for some integer m. Furthermore G has rank 3 on n and the following conditions are satisfied, where the notation is that introduced above. (i) X (I) = p, 8 (1) = P - 1. (ii) k = � m (m - l), I = � m (m + l). (iii) s + t = - 1. THEOREM 1 1 .7 (Wielandt

THEOREM 1 1 .8 (Ito [1962cD. Let p, G, n be as in ( 1 1 . 7). Let P be a Sp -group of G. Assume that p > 3 and / N G (P) / = 2p. Then p = 5 and G = A s . PROOF. The Brauer tree of the principal p -block of G is as follows

1 p-l

0-----0----0

.

Thus there exists an irreducible character ( of G with ((1) = P - 2. Since Op' ( G ) = (I) it follows that ( is faithful. By (7.3) P = 2a + 1 for some a and G = SL2 (p - 1). Since G has a subgroup of index 2p the known properties of PSL2 (p - 1) imply that p = 5 and G = SL2 (4) = A s . 0 LEMMA

PROOF. Hence

11.9 (Ito [ 1 967a]). Let p, G, n be as in (11 .7). Then f.L = A + 1 = � (m - 1)2. A = H m + l) (m - 3), By (11 .6) and ( 1 1 . 7) (iii) A - f.L = - 1. By (11.5) and ( 1 1 . 7) '

± Vd = ± Vd(X ( I ) -

8(1)) = 2k - (k + I) 4( k - f.L ) = d - (A - f.L ? = m 2 - 1

=

- m.

PERMUTATION GROUPS OF DEGREE 2p

1 2]

387

LEMMA 1 1 . 10 (Ito [1967a]). Let p > 3, G, 12 be as ( 1 1 . 7). Then the following hold. (i) Ga is faithful in its action on r(a ). (ii) Let H be the kernel of the action of Ga on .1 (a ). Then H is an elementary abelian 2-group and every orbit of H on r(a ) has cardinality 1 or 2 . . m

PROOF. (i) Let H be the kernel of the action of Ga on r(a ). Let {cJ, . be an orbit of H in .1 (a). Since H � GT(a ) it follows that

. . , Cn }

r* = r(a ) n r(cJ) = . . , = r(a ) n r(cn ). By (11.4), ( 1 1 . 7) and (11.9), / r /* = m + A + 1 = m + f.L. Let i � j. Then r* � r(Ci ) n r(Cj ). By (11 .4), (1 1 . 7) and (11 .9) / r(Ci ) n r(Cj ) / :!S m + f.L. Hence r(Ci ) n r(Cj ) = r* for i � j. Thus {r(Ci ) - r*} is a collection of pairwise disjoint subsets of .1 (a ) and so n O m (m + 1) - m - H m - 1)2} = n O m (m + 1) - m - f.L } :!S � m (m - l). This implies that n :!S 2m/(m + 1) < 2 and so n = 1. Thus H acts trivially on Ll (a ) and so H = (I). (ii) Let {cJ , . . . , cn } be an orbit of H on r(a ). Since H � G4 (a) it follows that .1 * = .1 (a ) n .1 (c J ) = . , . = .1 (a ) n .1 (cn ). By (11 .3) and (11.9), 1 .1 /* = f.L = A + 1 . Let i � j. Then .1 * � Ll (Ci ) n .1 (Cj ). B y (1 1 .3) and (11 .9) / .1 (Ci ) n .1 (Cj ) / :!S f.L. Hence .1 (Ci ) n Ll (Cj ) = .1 * for i � j. Thus {.1 (Ci ) - Ll *} is a collection of pairwise disj oint subsets of r(a ) and so n O m (m - 1) - � (m - 1)2} = n O m (m - 1) - f.L } :!S ! m (m + 1). As p > 5, m > 3. Thus n :!S 2m/(m - 1) < 3. Thus n = 1 or 2. Hence every orbit of H on n has size 1 or 2 and so H is an elementary abelian 2-group.

0

and so by (1 1 .7) (ii) Thus

4f.L = 4k + 1 - m 2 = m 2 - 2m + 1 = (m - If 4A = 4f.L - 4 = m 2 - 2m - 3 = (m + l) (m - 3). 0

The next result is a weak form of a theorem of Ito but is sufficient for what is needed in section 12. Ito actually proved that H = (I) for p > 5 .

12. Permutation groups of degree

2p

Let p be a prime. Let G be a primitive permutation group on 12 where / 12 / = 2p. Assume that G is not doubly transitive on n. By ( 1 1 .7)

2p = m 2 + 1 for some integer following result.

m.

The object of this section is to prove the

THEOREM

prime.

[12

CHAPTER VIII

388

12.1

(L. Scott

[1969], [1970], [1972]). If m > 3 then m is not a

We will give Scott's proof as presented in [1970], [1972] which is a simplification of an earlier proof given in [1 96 9] This proof depends on the results in section 1 1 as well as on the results in Chapter VI. In particular (12. 1) implies that if p < 313 then p = 5, 41 or 1 13. The cases p = 41 or 1 13 have been shown to be impossible in Scott [1976] by special arguments. Thus it is known that if G exists with p > 5, then

12J

PERMUTATION GROUPS OF DEGREE 2p

389

Let A be the normal closure of x in M. Then A fixes all letters in D - 8. Since Mb is transitive on .1i n 8 and A <J M it follows that .1i n 8 � b A for i = 1, 2. Clearly b E b A . Hence E � b A • Thus by Jordan's theorem (11 .1), G is doubly transitive on D contrary to assumption. 0

.

p ? 313.

Throughout the remainder of this section it is assumed that (12.1) is false and G exists for some prime q = m > 3. A contradiction will be derived from this assumption. Let G( 2) be defined as in section 1 1 . Since G is not doubly transitive on D it follows that 0(2 ) is not doubly transitive on D. Thus G may be replaced by G( 2). Hence it may be assumed that G = G(2). Let 0 b e a Sq -group o f G.

LEMMA

12.2. I 0 I = q.

PROOF. Suppose that the result is false and 1 0 1 > q. Choose a E D with o � Ga . Since the orbits of Ga on D - {a } have cardinalities � q (q - 1) and � q (q + 1) by (1 1.7), it follows that a is the only point in D fixed by O. Furthermore every orbit of 0 on D - {a } has cardinality q since � q (q + 1) < q 2 . Let b E D - {a }. Then l O : H I = q, where H = Ob. Let 8 be the set of fixed points of H on D. Choose x E 0 - H. If c E D - 8 then c H = C O as I c H I = I c O l = q. Thus 8 is a union of orbits of O. For d E 8, C Od = C H = C O. This implies that QcOd = 0 for c E D - 8 and d E 8. Thus Wielandt's Dissection Theorem (11.2) implies that O E X O n-E � G. Thus it may be assumed that x fixes all letters in

D - 8. Let b E 8 - {a }. Let {b}, .1 1 = .1 (b), .1 2 = r(b) be the orbits of Gb on D. Thus x � Gb• Suppose that .1i � D - 8 for some i. Thus Gb � 3, B is not the principal q -block. 0

12.4. There exists a character I/J of A with 1/J (1) = � (q - 1) such that 8c = I/J + Po{3, Xc = (Po - 1 0 ) I/J + Po, where {3 is a character of A and Po is the character afforded by the regular representation of O.

LEMMA

PROOF. By

(VII.2. 17)

8c = ( apo ± 1 0 ) 1/J + Po {3 for some integ�r a. Since Ie is a constituent of 8c by the Frobenius reciprocity theorem and Ie is not a constituent of tf; it follows that (3 ;1 O. Furthermore I/J (l) ± 8(1) =+= � (mod q ). Thus 1/J(1) ? Hq - 1). If a ;l 0 ==

then

==

Hq2 - 1) = 8 (1) ? (q - l H (q - 1) + q = H q2 + 1). Thus a = 0 and so the + sign must occur. Hence 1/J (1) - � (mod q). As X i s i n the same block as 8 i t follows from (VII.2.17) that Xc = (a ' Po - 1)1/J + Po{3 ' since X (1) + 8 (1) 0 (mod q ). Since I e is a constituent of Xc it follows that {3 ' ;I O. Clearly a ' ;I O. Hence ==

==

390

CHAPTER VIII

Hq 2 + 1) = x(1) = (a ' q 1)I/I{1) + q/3 ' (1). As /3 ' (1) 3 1 and I/I (1) 3 Hq - 1) this is easily seen /3 ' (1) = 1 and 1/1(1) = Hq - 1). Hence /3 = 1A. 0

[ 12

-

to imply that

a' =

LEMMA 12.5. Let CPo be the principal indecomposable character of Ga corresponding to the trivial Brauer character. Then A is in the kernel of 3 . Thus

PROOF. By (13.7) (A I - JLk t

either a l (1)= O (mod p ) or {3 (I) = O (mod p ). By (13.6) and Frobenius reciprocity (111.2.5) (3 (1) � p. Hence a (1) = p and {3 (1) = P + 1 . Let () = a, Xk = {3. By (13.6) and Frobenius reciprocity (}N = A i". Direct computation yields «A l - JLk t )N = A l + A i" - (JL � + JL i:).

Henc� (XdN = 1 + (}N - « A 1 - JLk t )N is as required.

0

LEMMA 13 .9. t = O. PROOF. Suppose not. Let (), Xk be defined as in (13.8).

Assume first that e is odd. By (13.5) Xk = Xk . Thus Xk is on the real stem of the Brauer tree T of the principal block. As e is odd, T, and hence the real stem has an even number of vertices . Thus not both end points

394

CHAPTER VIII

[13

correspond to characters with degree d == 1 (mod p ). Since 1 G is an end point, Xk cannot be an end point and so Xk is reducible as a Brauer character. Let 'P be an irreducible Brauer character with 'P (1) � � Xk (1) = � (p + 1) which is a constituent of Xb Since Xk (1) == 1 (mod p) 'P is not the principal Brauer character. Hence 'P is faithful by (13.4). If P > 7 then � (p + l ) < Hp - 1) and so G is of type L 2 (P ) by (3. 1) contrary to assump­ tion. Since e is odd and e > 1 , p > 5. Suppose that p = 7. Then e = 3 and T , looks as follows , 1

a

8

+ 2t

e � a (1) � ! (p + 1) it follows that p � 7. Suppose that p = 7. 6 = e � a (1) � 4, which is not the case.

e I (p - 1) this implies that

Suppose finally that (A 2 - J.Lk )G = - Xk + a - {3. If a (1), {3 (1) � 0 (mod p ) then a (1), {3 ( 1 ) == ± 1 (mod p ) as (A 2 - J.Lk )G i s p -rational. Hence - 1 ± 1 ± 1 == 0 (mod p) and so p � 3 contrary to the fact that e � s + 2 t � 4. By (13.6) {3 (1) i' p. Hence a (1) = P and so (3 ( 1 ) = 1 contrary to (13 .4). 0 LEMMA 1 3 . 10.

e

=

2.

PROOF. By (13.9) t = O. Thus H is abelian. By' (13.4) and Burnside's transfer theorem H is a 2-group. Let x be the unique involution in the cyclic group E. If e i' 2 then (x ) is a characteristic subgroup of H. Hence N G (H) C N G « x ) = H. Thus H is a S 2-grOUP of G and Burnside's transfer theorem implies that G has a normal 2-complement contrary to (13 .4). D LEMMA 1 3 . 1 1 . There exist pairwise ters a, {3, y of G such that

PROOF OF (13. 1). Suppose first that one of a, {3, or y in . (13. 1 1) is an . ex�e � tlO�al char �cter in the principal block. Since (A I - A2)G is p -rational thIS ImplIes that 2 (P - 1) = (p - l )/e � 2 and so p = 5. The Brauer tree of the principal block is �o

=

Since

395

PROOF. Since E is a T.!. set it follows that II (A I - A2)G W = II (A I - A 2) H W = 4. The result follows as (A I - A2)G (1) = O. D

b

where the numbers denote the degrees of the corresponding characters. Thus either a = 3 or b = 3. The classification of all finite groups with a faithful complex representation of degree 3 implies that G = PSL2 (7) contrary to assumption. Hence e is even. Therefore s is even and so s � 2. By Frobenius reciprocity (J.L �, Xk ) = 1 and (A ?, Xk ) = 0 for all i. Hence (A2 - J.Lk )G = - Xk + a ± {3 for irreducible characters a, {3 of G . Suppose that (A2 - J.Ld G = - Xk + a + (3. Then a (l) + (3 (1) = P + 1 and so it may be assumed that a (1) � ! (p + 1). Since (l , (A 2 - J.LdG ) = 0 it G follows from (13.4) that a is faithful. Hence G is of type L 2 (p) by (3 . 1 ) if P > 7 contrary to assumption . Since 4�s

CHARACTERS OF DEGREE P

13]

distinct nonprincipal irreducible charac ­

1

a+1 a

Thus if a, {3, y are all in the principal block then 1 ± (a + 1) ± 2a = 0 and so a = 2 contrary to the fact that a + 1 == ± 1 (mod 5). Hence one of the degrees of a, {3, 'Y is 5. Thus the exceptional degree is 3 as 1 + 5 = 2.3 and so the classification of the finite groups with a faithful 3-dimensional complex representation yields that G = PSL2 (5) contrary to assumption . Thus none of a, {3, y is exceptional in the principal block. Since e = 2 by (13. 1 0) there are only 2 nonexceptional characters in the principal block unless p = 3 in which case there are 3 non exceptional characters. If p = 3 then one of a, (3, y has degree 3 and so G = PSL2 (5) contrary to assumption by inspection of the groups with a faithful complex representation of degree 3 . Suppose that p > 3 then a t least 2 o f the degrees a ( 1 ) , (3 (1), y (l) are p. Thus exactly two of them are p and the other degree is 1, 2p - 1 or 2 p + 1 . The case o f degree 1 i s impossibly b y (13.4). Hence the possible Brauer trees are as follows �

1 2p + 2

2p + 1

o-----cr----o

1 2p

-

1 2p - 2

If 8 is an irreducible character with 8 ( 1) = P then (8, (A I - A 2) G ) i' 0 by (13.6). Hence G has exactly 2 irreducible characters of degree p. Since (2p + 26) I I G I for 6 = 1 or - 1 this implies that 2 2 o == 1 G 1 = 2p + 1 + (2p + 6 f + � (p - 1) (2p + 26 ) ==

Hence

p

=

4

(mod p

5 and 6

=

+ 6 ).

- 1 as

p > 3.

Therefore

1 G 1 = 50 + 1 + 81 + 128 = 260. However 2 p + 2 6 8 and so (2p + 26 ) l' I G I .

lishes the result.

=

D

This contradiction estab­

THE STRUCTURE OF A ( G )

1J

CHAPTER IX

Throughout this chapter R is either a field of characteristic p or the ring of integers in a finite extension K of Qp . Let R denote the residue class field. Thus R = R in case R is a field. For any R [ G ] module V let ( V) denote the isomorphism class of R [ G ] modules which contains V. Let C be a field of characteristic O. The representation algebra Ac (R [ G ] ) was defined in Chapter II, section 4. We will frequently write Ac ( R [ G ]) = A ( G ) if no special reference to C or R is required. This chapter contains results related to the structure of A ( G ). Let Sj be a nonempty set of subgroups of G. Let A( l ) ( G ) = Ac� ( R [ G ]) be defined as in (IIA.3). If H is a subgroup of G let AH ( G ) = A{H}(G). By (IIA.3) A�(G) is an ideal of A ( G ) and if ( V) ranges over the isomorphism classes of indecomposable R -free R [H]-projective R [ G ] modules for H E f;) then {( V)} is a C-basis of A,\) ( G ). It is natural to ask for conditions under which Ac (R [ G ]) is semi-simple. In section 2 it is shown that this is the case if a Sp -group of G is cyclic and R = R is a field. (A complete proof is not given as a result of Lindsey is assumed.) In case p = 2 and 8 ,f I G I then Ac (R [ G ]) is also semi-simple . The reader is referred to Conlon [ 1 965] , [ 1 966] , [ 1 969] ; Donovan and Freislich [1976] , and earlier papers of Basev [ 1 961] and Heller and Reiner [ 1961] which contain a classification of R [P] modules in case p = 2 and P is a noncyclic group of order 4. For related results see Wallis [1969] , Hernaut [ 1 969] , Muller [ 1 974a] , [1974b], Erdmann [1979a] , Donovan and Freislich [1978] . In this connection we mention some other known results which will not be treated here. If R = Zp is the ring of p -adic integers and G has a cyclic Sp -group then Ac (Zp [ G] ) is semi-simple. See Reiner [ 1 966b]. If G has noncyclic Sp-groups then Ac (Zp [ G ]) contains nonzero nilpotent elements. See 396

397

Reiner [ 1 966a] , Gudivok, Goncarova and Rudko [1971] . This has been generalized to the case that R is a finite extension of the p -adic integers by Gudivok and Rudko [ 1 973] . If R = R is a field and p � 2 it has been shown by Zemanek [ 1 971] that if a Sp -group of G is not cyclic then Ae (R [ G ]) contains nilpotent elements. Zemanek [1973] has also investigated the case that G is of order 8. See also Yamauchi [1972] , Bondarenko [ 1 975] , Ringel [ 1 975] . Various types of representations and modules have been studied. See for instance Carlson [1974] , [ 1 976a] , [ 1 976b] or Janusz [ 1 970] , [1971], [1972] or Johnson [ 1 969a], [ 1 969b] . In sections 3 and 4 yet another type of module will be discussed.

1 . The structure of A ( G)

A (G) is a commutative algebra . Furthermore dime A ( G ) is finite if and only if G has a cyclic Sp -group. This section contains some results concerning the structure of A ( G ). Let P be a p -group in G. Let !S(P) be the set of all proper subgroups of P. Define Wp ( G ) = Ap ( G )IA 5( p ) ( G ). By (III.5.9) Wp ( G ) = Wp (N G (P)).

THEOREM 1 . 1 (Conlon [ 1 967] , [ 1 968]). A (G) = EB Wp (N G (P)), where P

ranges over a complete set of representatives of the conjugate classes of p -groups in G and the sum is a ring direct sum. The proof given here is a simplification of Conlon's proof. As an immediate consequence of ( 1 . 1) one gets

COROLLARY 1 . 2 (Green [ 1 964]). A ( G ) is semi-simple if and only if

. Wp (N G (P))

is semi-simple for every p -group P in

G.

The next result was proved by Lam [ 1 968] in case R is a field . Related results may be found in Conlon \[1967] , [ 1 968] ; Wallis [ 1 968] .

THEOREM 1 .3 . (i) Let P be a p-group in G. Then AI' (N G (P)) is isomorphic to

a subalgebra of EB Ap (H), where H ranges over all subgroups of NG (P) such that P � H and HIP is cyclic. (ii) A ( G ) contains a nonzero nilpotent element if and only if there exists a p-group P in G and a subgroup H of N G (P) with HIP a cyclic p i_group such that A (H) contains a nonzero nilpotent element.

[1

CHAPTER IX

398

Por future reference the following elementary result is included here. LEMMA 1 .4. Assume that R = R is an algebraically closed field. Let P be a

p-group and let H be a p '_group. Then A (P x H) = A (P) (9 A (H). PROOF. Clear by (IIL3 .7).

1]

is

THE STRUCTURE OF A (G)

split.

Since

U (9 VI I U (9 v.

U (9 VI

0

is

R [ T]-projective

399

it

follows

that

LEMMA 1 .7. Let T <J G. (i) Let x, y E A (R [GIT]) with {3 (x ) = {3 (y ). If u E A-r (R [0]) then

ux = uy. (ii) AT (R [G]) has an identity for multiplication.

0

The rest of this section is devoted to the proofs of ( 1 . 1) and (1 .3). Some preliminary results are proved first. Let A O(R [OJ) = A �(R [OJ) ' be the Grothendieck algebra. Then A O(R [GJ) may be identified with the algebra of C-linear combinations of Brauer characters afforded by R [G] modules. If V is an R [ G] module let f3v be the Brauer character afforded by V. Define the algebra homomorph­ ism f3 : A (R [GJ) � A O(R [GJ) by f3 « V)) = f3v for an R [O] module V.

PROOF. (i) Repeated application of (1 .6) in case R = R is a field yields the result. (ii) Let Vo be the irreducible R [ G] module which affords the principal Brauer character 'Po. Thus ( Vo) E A (R [ G I T]). By (1 .5) there exists x E A(1)(R [G/T]) � AT (R [O]) with f3 (x ) = 'Po. Hence by (i) ( V)x = ( V) ( Vo) = ( V) for all ( V) E AT (R [OJ). 0

LEMMA 1 .5 . A O (R [GJ) = A (1) (R [ GJ) = A (1) ( R [GJ).

LEMMA 1 .8. Let T <J G. Assum,e that R is not a field. Let U be an R -free

PROOF. The second isomorphism follows from (1. 13 .7). Let {'P; } be the set of Brauer characters afforded by the irreducible R [ G] modules. Let ctJi be the Brauer character afforded by the principal indecomposable R [G] module which corresponds t o 'Pi . Given i, then b y (1. 19.3) there exist irreducible Brauer characters 'Pij such that 'Pi = 2:j 'Pij . Thus ctJi = 2:j ctJij where ctJij corresponds to 'Pij . By (lV.3.3) this implies that ('Pi, ctJk )' = oi km where m -I- 0 depends on i. This implies that {ctJi } is a basis of A O(R [GJ). Hence the restriction of f3 to A(I)(R [ O J) is an isomorphism from A (1) (R [OJ) to A O(R [GJ). 0 The next result which generalizes (IIL2.7) first appeared [ 1 967] in case R is a field.

III

Conlon

LEMMA 1 .6. Let T <J O. Let U be an R -free R [ T]-projective R [O] module.

Let V be an R -free R [ G I T] module and let VI be a pure submodule of V. Then U (9 V = ( U (9 VI) EB ( U (9 VI VI).

PROOF. Since T is in the kernel of V and VI is a pure submodule, the sequence is a split exact sequence. Tensoring with UT shows that

R [ T]-projective R [ G] module. Let VI , V2 be R -free R [ G I T] modules such that VI and V_ 2 afford the same Brauer character. Then ( U (9 VI) = ( U (9 V2)' PROOF. By (lII.2.7) () (9 VI = () (9 V2 • Let f; be the natural projection from U (9 V; onto U (9 VI = U (9 V; . Since ( U (9 V; )y = (dim V; ) U there exists an R [ T]-homomorphism h such that the following diagram is commutative

�

( U (9 V2)y

1 12

( U (9 VI)y � ( U (9 VI)y � O . Since U (9 V2 is R [ T]-projective there exists an R [ G ]-homomorphism g : U (9 V2 � U (9 VI with f l o g = f2 . Let MJ be the kernel of f l . Thus MJ + g ( U (9 Y2 ) = U (9 VI and MI = 1T( U (9 VI) � Rad( U (9 VI)' Hence g is an epimorphism. Let M be the kernel of g. Since g (U (9 V2 ) is R -free, M is a pure submodule of U (9 V. Therefore rankR M + rankR ( U (9 VI ) = rankR M + rankR g ( U (9 V2) = rankR ( U (9 V2) = rankR ( U (9 VI)' Thus rankR M 0 and so M = (0). Hence g is an isomorphism of U (9 V2 onto U (9 VI . 0 =

CHAPTER IX

400

LEMMA 1 . 9 . Let t 1 . B y definition

AD (G) = LD SP (G) = SD (G) + LD) Sp (G). PE'13( ) PE'-/5 P� (D

Do E �(D), Do -I- D

then by induction LPE'B ( Do) Sp (G) = L PE 'B ( Do) A �(G) for suitable ideals A �(G) of A (G) with A �(G) = Wp (NG (P)). If DJ, D2 E �(D); D1, D2 -I- D then by definition If

and so A � c: E9 PE'-/5( D I ) n'B ( D2 ) SP (G) for P C: � ( D I ) n � ( D2) ' Furthermore AD (G) = SD (G) E A0( D)(G) and

A CS( D ) (G) = EB A �(G ) and AD (G)IA0 ( D)(G) = WD (NG (D)) . D PE'B P�D( )

1]

THE STRUcrURE OF A ( G )

401

By (1.9) Wp (N G (P)) is a ring with multiplicative identity for any P E �. Thus A0 ( D)( G) has a multiplicative identity. Therefore there exists an idempotent e E A (G) such that A8 ( D ) (G) = AD (G)e. Let A b(G) = AD (G) n A (G ) (l - e ). Then A b(G) is an ideal of A (G) and A b(G) n A0( D)(G) = (0). Since 1 - e maps onto 1 in A (G)IA0 ( D)(G) it follows that and

A b(G) = AD (G)IA0( D)(G) = WD (NG (D)) AD (G) = A b(G) E9 A8( D)(G).

0

PROOF OF (1 . 1). Let D be a Sp -group of G. Then AD (G) = A (G) and the result follows from (1. 10). 0 LEMMA 1 . 1 1 . (i) Suppose that R is not a field. Let e be the character afforded by an R -free R [ G] module. Then () = Lai1] F where each ai is rational and each 1]i is the character afforded by an R -free R [H] module for some cyclic subgroup H of G. (ii) Suppose that R = R is a field. Let 'P be the Brauer character afforded .

by an R [ G ] module. Then 'P = Lai t/J F where each ai is rational and t/Ji is the character afforded by an R [H] module for some cyclic p '-subgroup H of G.

PROOF. If R or K is a splitting field for G both results are clear as the dimension of the space of all Lai 1] F, Lai t/J F is the number of classes, p -singular classes respectively. If e T = () or 'P T = 'P for T ranging over a group M of field automorphisms then

PROOF OF (1 .3). (i) It may be assumed that G = N G (P), that is to say P 1 and let I P : PI I = p. If

LEMMA 2. 5. For 1 � s, t � / P / , V, @ V; is the direct sum of min(s, t)

2.

A ( G)

in case a

Sp- group

of

G

is cyclic and R is a field

Throughout this section A (G) = A (F[ G]) where F is a field of charac­ teristic p. The following result is due to O'Reilly [1964], [1965]. THEOREM 2. 1 . Let G be a group with a cyclic Sp -group. Then A (G) is

semi -simple.

MI and M2 are indecomposable F[ P] modules then MI @ M2 = Mo EB E9�= 1 Vj , where Mo is F[Pd-projective, each V; is indecomposable and dimF V; I- dimF Vi for i I- j. If I P I = p then (2.2) follows from (VIII.2.7). In the general case it is a good deal more complicated. For further results see Renaud [1979] . It will first be necessary to prove

THEOREM 2.3 (Green [1962b]). Let P be a cyclic p-group. Then A (P) is

semi -simple.

The proof of (2.3) given by Green depends on getting generators and relations for A (P) and is quite complicated. We will here give a short elegant proof due to Hannula, Ralley and Reiner [1967]. O'Reilly's proof of (2. 1) was simplified by Lam [1968] . In fact this simplification was the motivation for (1.3). However the proof still depended on Green's method and used results about generators and relations for A (P). Ultimately results concerning the tensor products of F[ G ] modules are of course needed but the proof given here uses only (2.2). It should be pointed out however that Green's method yields more

nonzero indecomposable F[ P] modules.

PROOF. Let Vs @ V; be the direct sum of n nonzero indecomposable modules. Since V'; = V; it follows that

n = dimF Invp ( V-. @ V ';) = dimF HomF[p] ( Vs , V; ).

The structure of min(s, t). D

V,·

and

Vr

shows that dimF HomF[p]( Vs , V; ) =

LEMMA 2 .6. Let m be an integer and let f(X) = L: = I min(i, j)X;Xj be a real quadratic form. Then f(X) is a positive definite form. PROOF. It " is easily seen that

f(X) = (XI + . . . + Xm Y + (X2 + · · · + Xm Y + · · · + X� . D PROOF OF (2.3). It may be assumed that C is the field of real numbers. It suffices to show that if u E A (P) with u 2 = 0 then u = O. Let u = L bs ( Vs ). Let ( V, ) ( V; ) = Lj aS1j ( Vi ). By (2.5) Lj astj = min( s, t). Thus o = U2

=

bs btastj ( Vi ) 2: s ,t

404

CHAPTER IX

[2

and so �S.I bsbt as1j = 0 for all j. By summing over j this implies that �S.I min(s, t)bsbt = O . Hence by (2.6) bs = 0 for all s and so u = O. 0

3]

I p : PI I = p· (i) For 1 ::::; 1 ::::; 1 P 1 and 'P an irreducible character of H there exists an indecomposable F[N] module V('P, I ), unique up to isomorphism, such that 'P is afforded by the socle of V and dimF V('P, I) = I. Every nonzero indecomposable F[N] module is isomorphic to some V ('P, I). V('P, I ) is F[PI]-projective if and only if I 0 (mod p ) and V('P, I) = V('P, 1)* if and only if 'P * a /-I = 'P where a is defined as in Chapter VII, section 1 . Furthermore V('P, l)p is indecomposable. (ii) If M I and M2 are indecomposable F[N] modules with M� = M for i = 1 , 2 then M I @ M2 = EB�= I Vi EB Mo where Mo is F[PI].:.projective and each V; is indecomposable with V � = Vi ' (iii) Suppose that 2 / 1 CH (P) I . For each I with 1 ::::; I ::::; 1 p i there exist exactly 2 irreducible characters 'P of H with 'P * a /-I = 'P. Each such 'P has Ho in its kernel where 1 CH (P) : Ho 1 = 2. ==

PROOF. (i) This is a special case of the results of Chapter VII, section 2. (ii) By (i) (M )p is indecomposable for i = 1, 2. By (2.2) M I @ M2 = EB �= I Vi EB Mo where Mo is F[PI]-proj ective and dimF Vi � dimF Vi for i � j. As (MI @ M2)* = MI @ M2 this implies that EB �= I Vi = ��= I V � . Thus V; = V� by the unique decomposition property and the fact that dimF V; � dimF Vj for i � j. 2 (iii) As H is cyclic, 'P * = 'P - I and so 'P *a /-1 = 'P if and only if 'P = a /- I . H Since a / :CH( P )/ = 1 there are exactly two choices for 'P. Clearly Ho is in the kernel of 'P. 0 PROOF OF (2. 1). It may be assumed that C is the field of complex numbers. Let FI be the algebraic closure of F. Then A (F[ G D is a subalgebra of A (FI[ GD. Thus it may be assumed that F = FI is algebraically closed as A (FI[ GD has finite C-dimension. Furthermore it s'uffices to show that A (F[ GD has no nonzero nilpotent elements. By (1 .3) it may be assumed that G = PH with P <J G where P is a cyclic p -group and H is a cyclic p '-group. Suppose that p = 2. Then G = P x H. Thus by (1 .4) A ( G) =

405

A (P) @ A (H).

Clearly A (H) is semi-simple. By (2.3) A (P) is semi-simple. Thus A (G) is semi-simple as required. Thus it may be assumed that p � 2. Let h be the order of (x ) = H and let x - I yx = y I . Define

The next result is required for the proof of (2. 1). LEMMA 2.7. Assume that F is algebraically closed. Let N = PH with P <J G where P is a cyclic p -group with 1 p i > 1 and H is a cyclic p '_group. Let

PERMUTATION MODULES

G,

= (x, y / y / P / = X 2h

= 1,

x - ' yx = y t ).

Then G is a homomorphic image of G I and so A (G) � A (G1). Hence it suffices to show that A ( G,) is semi-simple. Therefore by changing notation it may be assumed that 2 / 1 CH (P) I . The proof is by induction on 1 P I . If 1 P 1 = 1 then A (H) = C[ H] is semi-simple. Suppose that 1 p i > 1. Let 1 P : p, 1 = p. Let I = ApI (G). It suffices to show that I and A (G)/ I are semi-simple. The map sendi ng ( V) to ( l/p ) ( VG ) for any F[PIH] module V is easily seen to induce an isomorphism from A (PI G) onto 1. Thus by induction I is semi-simple. It remains to show that A (G)jI is semi-simple. Let S be the subspace of A (G) spanned by all ( V) with V indecompos­ able and V* = V. By (2.7) (ii) S + 1/1 is a subalgebra of A (G)/ 1. If 1 ::::; I ::::; 1 p i then (2.7) (ii) and (2.7) (iii) imply the existence of an indecom­ posable F[ G ] module V with dimF V = I and ( V) E S. Thus if V is an indecomposable F[ G ] module with ( V) E S and X is an indecomposable F[ H] module then the map sending ( V) @ (X) to ( V @ X) defines an algebra homomorphism of «S + 1)/1) @ A (H) onto A (G )/1. Since A (H) is semi-simple it suffices to show that (S + 1)/1 is semi-simple. Let Z = CH (P). By (2.7) (ii) and (iii) the map sending V to VCo (P) defines a one to one linear map of S onto A (P x Z). Since V is F[PI]-proj ective if and only if VCo (P) is F[Pd-projective, it follows that (S + 1)/1 = A (P x Z )/A pl (P x Z). Thus it suffices to show that A (P x Z) is semi-simple. This follows directly from ( 1 .4) and (2.3). 0 3.

Permutation modules

The set of all isomorphism classes of indecomposable R [G] modules is in general very large and complicated. This section and the next contains the definition of some special types of R [ G] modules. For H a subgroup · of G let Vo(H) denote the R -free R [0] module of R -rank 1 with Vo(H) = Inv H ( Vo(H)). Thus if R is a field then Vo(H) affords the principal Brauer character and if R is the ring of integers in the p-adic number field K then VO(H) @R K affords the trivial character. An R [ G ] module is a transitive permutation module if it is isomorphic to Vo(H)G for som� subgroup H of G. A direct sum of transitive permutation modules is called a permutation module.

406

[3

CHAPTER IX

LEMMA 3 . 1 . There are only finitely many isomorphism classes of transitive permutation modules. PROOF. Clear as G has only a finite number of subgroups. 0 LEMMA 3.2. Let V be a permutation module. Then the following hold. (i) V = V*. (ii) V is an algebraic module. (iii) If M is a subgroup of G then VM is a permutation module. (iv) If G is a subgroup of M then V M is a permutation module. PROOF. (i) Clear by (II.2.6). (ii) This follows from (II.5.3). (iii) Clear by the Mackey decomposition (II.2.9). (iv) Immediate from the definition . See (II.2. 1) (iv).

0

LEMMA 3.3. Let Vand W be permutation modules. Then the following hold. (i) V EB W is a permutation module. (ii) V (9 W is a permutation module. PROOF. (i) Immediate by definition. (ii) This follows from the Mackey tensor product theorem (II.2. 10). 0 LEMMA 3 .4. Suppose that G is a p-group. (i) A transitive permutation module is indecomposable. (ii) If V is a permutation module and W I V then W is a permutation

module.

PROOF. (i) This follows from (III.3. 8). (ii) This is a consequence of (i). 0 COROLLARY 3.5. Let Q be a subgroup of the p-group P and let V be an R [ Q] module. Then V is a permutation module if and only if v P is a

permutation module.

PROOF. If V is a permutation module then so is v P by definition. Suppose that vP is a permutation module. Then so is ( VP )o. By the Mackey decomposition (II.2. 10) V I ( V P ) o. Hence V is a permutation module by (3.4). 0

4J

ENDO-PERMUTATION MODULES FOR p-GROUPS

407

LEMMA 3.6. Let V be a transitive permutation module. Then the following

hold.

(i) rankR (Inv G ( V)) = 1 . (ii) V = WI EB Wz with WI indecomposable such that Inv G ( W2) = (0) and Inv G ( WI) "f= (0). PROOF. (i) This follows from (II.3.4). (ii) Immediate by (i). 0 The class of indecomposable modules defined in (3.4) are of interest and have been studied by Scott [1971], [1973] . Some related results can be found in Dress [1975] .

4. Endo-permutation modules for

p- groups

Throughout this section P is a p -group. Vo(H) is defined as in the previous section. An endo -permutation R [P ] module is an R -free R [P] module V such that V* (9 V = HomR ( V, V) is a permutation module. An endo -trivial R [P] module is an R -free R [P] module V such that V* (9 V = HomR ( V, V) = Vo(P) EB U, where U is a projective R [P] module. Since any proj ective R [P] module is a permutation module it follows that an endo-trivial R J P] module is an endo-permutation R [P] module . Clearly a permutation module for R [P] is an endo-permut ation R [P] module. These concepts were introduced by Dade [1978a] , [1978b], who also classified the endo-permut ation R [P] modules in case P is abelian. See also Carlson [1980a] . We will here only present basic elementary properties of endo-permut ation R [P] modules and compare some of these with the correspondin g properties for permutation modules. The next result · shows that endo-permutation R [P] modules arise naturally at least in case R = R is a field. THEOREM 4. 1. Let G

= HP where H = Op,(G) and P is a Sp -group of G. Let F = R be a splitting field of G and let V be an irreducible F[ H] module with G = T( V). Then V extends uniquely to an irreducible F[ G ] module W and Wp is an endo -permutation F[P ] module.

408

CHAPTER IX

[4

PROOF. The existence and uniqueness of W follows from (111.3. 16). Under the action x � y -1xy, F[H] becomes a permutation module M for F[ 0]. Let e be the centrally primitive idempotent of F[H] correspond­ ing to e. Then V* 09 V = eF[H] I MH • Since T( V) = 0, eF[H] = W 09 W* as F[ 0] modules and W* 09 W I M. Thus W� 09 Wp I Mp and so W� 09 Wp is a permutation module for F[P] by (3.4). 0 LEMMA 4.2. Let O � V � U � W � 0 be an exact sequence of R -free R [P] modules with U projective. Then V is an endo -permutation (endo ­

trivial) module if and only if W is an endo -permutation (endo -trivial) module. PROOF. Immediate by (III.S. 12). 0

It follows from (4.2) that for instance every factor in a proj ective resolution of Vo(P) is an endo-permutation module . In this way it can be shown that if P is neither cyclic nor a genera'lized quaternion group then there exist infinitely many endo-permutation R [P] modules. Also there exist endo-permutation R [P] modules V which are not algebraic and such that V� V*. This is in contrast to (3.1) and (3.2) (i), (ii).

4]

ENDO-PERMUTATION MODULES FOR p -GROUPS

409

LEMMA 4.6. Let V be an endo -permutation R [P] module and let 0 be a

subgroup of P. Then Vo is an endo -permutation R [0] module.

PROOF. Clear by definition. 0 In general the direct sum of endo-permutation modules need not be an endo-permutation module in contrast to (3.3) (i). Two en do-permutation R [P] modules V, W are compatible if V EB W is an en do-permutation R [P] module . LEMMA 4.7. Let V, W be endo -permutation R [P] m'odules. The following

are equivalent. (i) V and W are compatible. (ii) V* 09 W = HomR ( V, W) is a permutation module. (iii) W* 09 V = HomR ( W, V) is a permutation module.

PROOF. Since ( V* 09 W)* = W* 09 V, (ii) and (iii) are equivalent. Since

( V EB W) 09 ( V* EB W*) = ( V 09 V*) EB ( V 09 W*) EB ( W 09 V*) EB ( W 09 W*),

Do there exist only finitely many isomorphism classes of endo -permutation R [P] modules with V = V*?

the equivalence of (i) with (ii) and (iii) now follows from (3.3) (ii). 0

This is an open question. It follows from Dade's results that the answer is affirmative if P is abelian.

LEMMA 4.8. Let V be an indecomposable endo -permutation R [P] module

LEMMA 4.3. Let V be an endo -permutation R [P] module such that V" =

PROOF. Let .0 be the set of all subgroups Q of P with Q ;i P. By (111.4.9) HO ( P, .0, V* 09 V) ;i (0). Thus there exists an indecomposable component W of V* 09 V with HO (P, .0, W) ;i (0). Since V is an endo-permutation module, W = Vo(O ) for some subgroup Q of P. By (II.3.4) HO ( P, Q, W) = (0). Hence HO(P, .0, W) = (0) if Q E .0. Thus Q g .0 and so Q = p. 0

( vn )* for some positive integer n. Then V is algebraic.

PROOF. As V 2 n = vn 09 ( vn )* is a permutation module the result follows from (3.2) (ii). D LEMMA 4.4. If V, W are endo -permutation R [P] modules then so are V*,

V 09 W and HomR ( V, W) = V* 09 W.

PROOF. Clear by definition.

D

The next result indicates that compatibility of two endo-permutati on modules is a rare phenomenon. LEMMA 4.9. Let V, W be indecomposable endo -permutation R [P] modules

LEMMA 4.5. Let V be an endo -permutation R [P] module. If W I V then W

is an endo -permutation module.

PROOF. Since W* 09 W I V* 09 V the result follows from (3.4).

with vertex P. Then Vo(P) I V* 09 V.

0

with vertex P. Then V and W are compatible if and only if V = W.

PROOF. If V = W then V and W are compatible by (4.7). Suppose that V and W are compatible. Let S) be the set of all subgroups Q of P with Q ;i P. There exist nonnegative integers ao, bo for Q � P such that

CHAPTER IX

410

[4

V* ® W = apVo(P) EB EB ao ( Vo( O)r, O E.'Q

Thus

v ® V* = bp Vo(P) EB EB bo ( Vo( O )r· O E:i;;J

CHAPTER X

apV EB ffi ao ( Vo r = v ® V * ® W = bpW EB ffi bo ( Wo )G. oEe O E:i;;J Hence any indecomposable component of V ® V * ® W with vertex P is isomorphic to V. By (4.8) a p I- 0 and so V = W. 0 In general endo-permutation modules do not behave well with respect to induction. The next result is concerned with this situation .

Let 0 be a subgroup of P and let V be an endo -permutation R [ 0 ] module. The following are equivalent. (i) vP is an endo -permutation R [P] module. (ii) The endo -permutation R [ 0 x n o ] modules Voxn o and VOx no are

LEMMA 4. 10.

compatible for all x E P.

PROOF. The Mackey tensor product theorem (II.2. 10) implies that vP ® ( VP)* is a permutation module if and only if ( Vox no ® Vdn or is a permutation module for all x E P. Thus (i) is equivalent to (ii) by (3.5) and (4.7). 0

As might be expected the theory developed so far becomes somewhat simpler when applied to p -solvable groups. On the one hand certain results are true for p -solvable groups which are not true in general, on the other hand some questions which remain open in the general case can be settled for p -solvable groups. This chapter is primarily concerned with p -solvable groups though some of the results are proved in a more general context. The notation introduced at the beginning of Chapter IV will be used throughout this chapter.

1. Groups with a normal p ' -subgroup

The first three results in this section are slight refinements of results of Fong [ 1 960], [ 1 961],J1962] . The method used to prove (1 . 1) is adapted from Serre [1977] , it has its roots in the work of Schur. I am indebted to Watanabe [1979] who suggested that part of the proof of ( 1 . 1) which shows that Nx is nonempty for all x. This fills a gap in an earlier version of the result. See also Nobusato [1978] . For a related result see Tsushima [ 1 978c] .

Let H <J G. Let ? be an irreducible character of H. Assume that G = T(?), the inertia group of ? Let F be an algebraically closed field such that char F l' ' H ' . Let V be an irreducible F[ H] module which affords ? Then the following hold. (i) There exists a finite group {; and an exact sequence (1) '-;' Z � {; 4 G � (1),

LEMMA 1 . 1 .

41 1

412

[1

CHAPTER X

where Z is a cyclic group in the center of 0 and I Z I I I H 1 2 . Also 0 contains a normal subgroup H = H such that ZH = Z x H = f- I (H). The group 0 depends only on G and �, in particular it is independent of the choice of F. (ii) Let FI be the subfield of F generated by a primitive I H 1 2 root of unity. There exists an FI[ 0] module VI such that if V VI 0FJ F then f( Vff ) = V. Furthermore if W is an irreducible F [ G ] module with V a constituent of WH then W = V 0 W for some absolutely irreducible F [ 0 /H] module W. (iii) Let L1 (F) be the set of all Brauer characters afforded by irreducible F [ G ] modules W such that V is a constituent of WHo Let J (F) be the set of all Brauer characters afforded by F [ 0 /H] modules U such that Z is in the kernel of V 0 u. Then the map sending W to W defined by (ii) induces a one to one mapping from L1 (F) onto J (F). =

F' be a field with FI C F' C F. Let V' be an F' [ H] module which affords { Since char F' ,{' I H I, V' is irreducible. Let A be a representation with underlying module V'. Let SF' {det A (y ) l y E H}. Clearly S = SF' is independent of the choice of F'. For x E G let Nx be the set of all linear transformations z on V' such that z -I A ( y ) z A (x -I yx ) for all y E H and such that det z E S. Let � (1) = d. Let x E G. We will first show that there exists a linear transformation z E GLd (FI ) with det z = 1 and z l A (y )z = A (x - l yx ) for all y E H. It clearly suffices to prove this in case the order of x is q a for some prime q and some integer a � 1 . Let I H I = h . Let Fo b e the subfield o f FI generated b y a primitive h th root of 1. It may be assumed that A (y ) E GLd (Fo) for all y E H by a PROOF. Let

=

=

-

classical result of Brauer which follows directly from (IV. 1 . 1) (iii), see e.g. Feit [ 1 967c] (16.3). Since T(�) = G there exists Z o E GLd (Fo) such that z 0 1 A (y ) z o = d A (x - l yx ) for all y E H. Let a = (det zoyl and let z = azo. Then det z = 1 , I z - A (y)z = A (x -1yx ) for all y E H and z d = a dz g E GLd (Fo). Further­ more z q a = {31 is a scalar for some {3 E Fl. Thus {3 d = 1 and so

z q a E GLd (Fo).

If q ,{' d then z

GLd (Fo).

E GLd (Fo) C GLd (FI) as z q a and z d are both in

Suppose that q I d. Then (z, A (y ) l y E H) is a finite group whose exponent divides q ah. Thus it may be assumed by Brauer ' s theorem that z E GLd (Fo(E )), where E is a primitive q " th root of 1 for some n. As q I h, [Fo( E ) : Fo] is a power of q. Since a E Fo(E ) it follows that [ Fo( a ) : Fo] is a power of q. Assume first that either q -1 2 or q = 2 and 4 1 h. Then Fo(E ) is a cyclic

1]

413

GROUPS WITH A NORMAL p i-SUBGROUP

extension of Fo. As a d E Fo it follows that a = CEo for some c E Fo and some q " th root of 1, Eo such that Fo(Eo) = Fo ( a ) . Let q b = (d, q " ). Then [ Fo ( a ) : Fo] = [Fo(Eo) : Fo] I q C. Hence a E Fo(Eo) C FI and so z E GLd (FI). Suppose now that q = 2, 2 / h and 4 ,(' h. Then there exists Ho 2. Since p� 2 x p " == 1 + p k+n a (mod p n + k + l ) . =

Since x p " = 1 for some n � 0 this implies that a assumption. Thus x = 1 . 0

==

0 (mod p ) contrary to

PROOF OF (2.3). Induction on I G : Op' (G) I. If G = Op' (G) the result is clear. The existence of X follows from (2. 1). Let H = Op' ( G). Let � be an irreducible constituent of 'PH. By (V.2.5) and (1.2) it suffices to prove the result for the group G (�). By induction it may be assumed that I G : H I = I G (�) : Op,(G(�» I . Thus Op, (G (�» = Op' (Z), where Z is the center of G (�). Thus Op (G(�» � (I). Therefore it may be assumed that Op (G) � (I). Let XI, X2 be p -rational characters such that X l = X2 = 'P as Brauer characters. By (2.4) Op (G) is in the kernel of X l and X2 . Hence X l and X2 are characters of GlOp ( G ). Thus by induction X l = X2. 0 THEOREM 2.5 (Isaacs [1974]). Suppose that p � 2. Let G be a p-solvable group. The following statements are equivalent.

PRINCIPAL INDECOMPOSABLE CHARACTERS OF P -SOLVABLE GROUPS

3]

421

(i) Every irreducible p-rational character of G is irreducible as a Brauer

character. (ii) For each p -singular element x in G with x Xpxp, where Xp is the p-part of x there exists a power y of x such that I (xp ) I = I (y ) I but y is not conjugate to xp in CG (xp) =

PROOF. It follows from (IV.6. 10), or more directly from Brauer's com­ binatorial lemma quote9 in (IV.6. 10), that the number of irreducible p -rational characters of G is equal to the number of conjugate classes that contain an element z whose p -part x is conjugate to every power of x that has the same order a� x in C G (zp') where Zp' is the p '-part of z. Thus condition (ii) holds if and only if the number of p -rational irreducible characters of G is equal to the number of irreducible Brauer characters of G. By (2. 1) this is true if and only if condition (i) holds. 0 COROLLARY 2.6. Let G be a p -solvable group and let 'P be an irreducible Brauer character of G. Let be an automorphism of the field Q( 'P ). Then 'P T is an irreducible Brauer character. T

PROOF. Clear by (2. 1). 0 (2.6) is false for groups G in general. For instance SL2 (1l) has a unique irreducible Brauer character 'P for p = 11 of degree 2. However YS E Q('P ). Further results related to (2.3) and (2.5) can be found in Isaacs [1978] . Cliff [1977] has given an alternative treatment of the original Fong-Swan theorem and has obtained information about indecomposable modules of p-solvable groups. Gagola [1975] has generalized the Fong-Swan theorem and shown that if the vertex of an absolutely irreducible F[ G] module is contained in a normal p -solvable subgroup, then this module can be lifted to one in characteristic O.

3.

Principal indecomposable characters of p -solvable groups

The first three results in this section are due to Fong [1962].

P <J G with I p i = p n. Let 'P be an irreducible Brauer character of G /P. Let Qj, Qj o be the principal indecomposable character of G, G/P respectively, which corresponds to 'P. Then $ (1) =

LEMMA 3 . 1 . Suppose that

p n Qj O( l ) .

422

CHAPTER X

[3

PROOF. See (lV.4.26). 0 THEOREM 3.2. Let G be a p -solvable group. Let 'P be an irreducible Brauer character of G and let

(2 m ) is a power of 2. Hence in any case Co (P) contains a unique cyclic subgroup A of order q 2 1 which contains all nonzero scalars. Choose J E A with J2 = c1. If r = cf then x = J satisfies (i). Suppose that r � cf. Then g = cf r � O. Clearly g is a P-invariant alternating bilinear form. Since P acts irreducibly on V it follows that g is nondegen­ erate. By definition gl = - cg. There exists z E GL( V) with r = g. As Sylow groups are conjugate, z may be chosen so that pz = P. Let Jo = JZ. Then 10 E Co (P) and so Jo E A. Thus by changing notation it may be assumed that J E A, J2 = cI and r = cf. If - 1 = a 2 for some a E K then x = aJ satisfies (i). Suppose that - 1 � a 2 for all a E K. Hence q = 3 (mod 4) and c = - 1. Thus J 2 = - I and r = f. Hence j E Sp(f). Let x = y be defined as in (7.3). Then x satisfies (i). 0 =

-

-

-

LEMMA 7.5. Let q be the power of an odd prime and let K = Fq. Let V be a vector space over K and let f be a nondegenerate alternating bilinear form on V. Let V = VI EB V2 with V2 = vt and dim VI = dim V2 = 2n. Let f be the restriction of f to \1;. Let Pi be a S 2-groUP of Sp(fi ) Sp2 n (q). Assume that Pi acts absolutely irreducibly on \1;. Then P I may be identified with P2 . Let P = {(y, y ) y E P I = P2} Then there exists x E GL( V) such that r = - f, x =

l

.

CHAPTER X

436

[7

commutes with every element of P and some power of x is equal to - d 2 I for some d E K with - d 2 � 1 . PROOF. I f - 1 = a 2 for some a E K, the result follows from (7.4). Suppose that - 1 � a 2 for any a E K. Hence q == 3 (mod 4) and d 2 � 1 for all d E K. We may identify VI with V2 and fl with f2 . Define the linear transformation J on V = VI EB V2 by J : (v!, V 2) � ( - V2, V I ) . It is easily seen that J 2 = I and r = f. Furthermore J commutes with every element of P. If x = Y is defined as in (7.3 ) then x has the desired properties. 0 -

-

LEMMA 7.6. Let p be an odd prime. Let K = F2, let V be a vector space over K and let f be a nondegenerate quadratic form on V. Let P be a p-group with p � O(f). Assume that P acts irreducibly on V. If x E Z(P) {I} then vx � v -

for all v E V, v � O. PROOF. Clear. 0

7]

PROOF. (i) This is well known. (ii) The existence of a unique irreducible Brauer character 'PA afforded by an F[ 0 ] module such that A is a constituent of ('PA )Z is well known, as are all the other properties of 0 in the statement. Then G = T( 'PA ). The existence and uniqueness of XA now follow from (III.3.16). (iii) Straightforward verification. 0 = PO, V, Z, Z(c ) be defined as in (7.7). Let A, J.L be linear characters of Z with A � 1, J.L � 1 . Then one of the following occurs. (i) Let Af.L be the character of PZ defined by AJ.L (xz ) = AJ.L (z ) for x E P, Z E Z. Then

LEMMA 7.8. Let P, 0, G �

A

XA @ X�

=

and let F be an algebraically closed field of characteristic p. (i) Let f(x, y ) = [x, y ] . Then f defines nondegenerate alternating bilinear form from V to Z = F;. If q = 2 then f(x ) = x 2 defines a nondegenerate quadratic form on V. Let A ( 0) denote the group of all outer automorphisms of 0 and let Ao( 0) denote the subgroup consisting of all automorphisms which fix all the elements of z. Then Ao( 0) <J A (0) and A (0)/Ao( 0) = Aut(Z) is cyclic of order q - 1 . If q � 2 then Ao( 0) = Sp(f) = Sp2n (q ). If q = 2 then Ao( 0) = O(f) = 02 n (2) is an orthogonal group. In any case a subgroup H of 0 is abelian if and only if the image of H in V is isotropic. (ii) Let P be a p-group with P � Ao(O) and let G = OP be the semidirect product. If A is a linear character of Z with A � 1, then (up to isomorphism ) there exists a unique irreducible F[ G] module XA such that A is a constituent of the character afforded by (XA ) Furthermore (XA )o is irreducible and every irreducible F[ 0] module which does not have Z in its kernel is isomorphic to some (XA ) If H is a maximal abelian subgroup of 0 then H = Ho x z, / Ho l = q n and ,\" 0 = (XA )o where '\"(hz ) = A ( ) for h E Ho. (ii) For i = 1, 2 let Oi be extra special with Zi = Z( Oi ). Then ZI = Z2 = Z. Let c E KX and let A I , A 2 be linear characters of Z with A l = A �c. Let 0 = 0 1 X 02 and let Z(c ) = {(z, ) I z E Z} � Z(O). Then XA1 @ XA2 is an irreducible F[ 0 ] module with kernel Z (c ). Furthermore 0/ Z (c) is extra special. a

z.

0.

Z

Z

c

�

A (Af.L ) G.

(ii) P = 2. Let Oi = 0 for i = 1, 2. Let 00 = (0 1 X 02)!Z( - 1), let Po = {(x, x ) I x E P}, let Go = PoOo . Let XA20 be the J:reducible F[ Go] module such that (X A 20 ) = (XA ) @ (XA ) 2 • Let A be the character of PoZ(Oo)/Z( - 1) such that A « )) = A ( ) for x E P, E Z. Then 00

LEMMA 7.7. Let q be a prime and let 0 be an extra -special q -group with / 0 / = q 2n + l . Let Z = Z(O) and let V = O/Z. Let p be a prime distinct from q

437

IRREDUCIBLE MODULES OF p -SOLVABLE GROUPS

01

4

XZ I ' XZ 2

0

2

4

Z I Z2

Zi

PROOF. The proof is by induction on / 0 / . Without loss of generality it may be assumed that P is a Sp -group of Ao( 0). Suppose that V contains a nonisotropic proper P-invariant subspace. Let W be minimal among such spaces . Then W = W n W-L EB Wo for some P-invariant space Wo � (0) with Wo n W� = (0). The minimality of W implies that W = Woo Hence if W = WI then V = WI EB W2 , where each Wi is P-invariant, WI � (0) and wt- = W2 . Since P is a Sp-group of Ao(O) i t follows that P = P I X P2 , where Pi acts trivially o n "'f for i � j. There exist extra special groups with R /Zi = Wi for i = 1 , 2 where Zi = Z(R ), such that ( P I H1 x P2H2)/Zo = PO for some subgroup Zo of ZI x Z2. Thus XA = XA1 @ XA2 and X� = X�1 @ X� , where XAi , X�, are irreducible F[ PiR ] modules which do not have Zi in their kernels. Observe that if p = 2 and (i) is satisfied for a group then also (ii) is satisfied. Thus by induction it may pe assumed that either (i) or (ii) is satisfied for the groups PiR for) = 1, 2. Hence either XAi @ X�, = (�yiHi for i == 1 or 2 or XA�O @ XA10 = (A 4 ) G,o for i = 1 or 2. Since PHI Z2 n PH2 Z1 = PZ1 Z2 and POHl OZ(H20) n POH20Z(HlO) = POZ (HIO x H20) it follows from the tensor product theorem that (A;t YI H] @ (fit y2 H2 = (,Q )G and (fl)G[o @ (f!)G20 = (A4)G and the re­ sult is proved in either case. Hence it may be assumed that V has no proper nonisotropic P-invariant subspace.

438

[7

CHAPTER X

Suppose that (O)� V I � V with VI a P -invariant subspace. Then VI � v t . Thus vt = VI E9 Vo with Vo P-invariant. A s Va n v t = (0) and Vo is isotropic this yields that Vo = (0). Hence VI is a maximal isotropic subspace of V. Furthermore V = VI E9 V2 with V2 a maximal isotropic subspace which is P-invariant. Let M be the inverse image in 0 of \1;. Then M is abelian . Furthermore Mi = Mo x Z, where P normalizes Mia. Let A, Ii respectively be the character of PM! , PM2 respectively with PMlO, PM20 in its kernel such that A (xz ) = A (z ) for x E PMlO, Z E Z and Ii (xz ) = /L (z ) for x E PM20, Z E Z. Then A O = XA and Ii ° = Xw As PM, n PM2 = PZ, the Mackey tensor product theorem (II.2. 10) implies that XA

0 X�

= A°

0 Ii ° = (A;;, ) 0.

-c.

l

=

Y Q(O)P(O)Z(H)

-(' H ) Q(O)P(O)Z(H) _ - ( /\/L - ( /\/L -(' P(O) Z(H» _

Since

0 (0) P(O)Z(H)/Z (c ) = Go the result is proved.

IRREDUCIBLE MODULES OF p -SOLVABLE GROUPS

439

Let p � q be primes. Let 0 be an extra special q -group and let P be a p-group contained in Ao( 0). Let F be an algebraically closed field of characteristic p and let V be an irreducible F[ PO] module which does not have Z( 0) in its kernel. Then V is an algebraic module.

LEMMA 7.9.

PROOF. By (II.5.3) and (7. 8) either V 0 V or V 0 V 0 V 0 V is alge­ braic. Hence by definition

V is algebraic.

0

In effect the content of (7.8) is the assertion that if V is an irreducible F[ PO] module then the en do-permutation module Vp has the property that Vp = V� if P is odd and V� = ( V�)* if P = 2. This would be enough to

prove (7.9) which is the essential result needed for the proof of (7.2).

Thus (i) holds. Therefore it suffices to prove the result in case P acts irreducibly on V. Let A = f.-t Two cases will be considered. Case (I). There exists an automorphism (J of G such that x 0" = x for all x E P, 0 0" = 0, V 0" ;I v for v E V, v ;l 0 and z 0" = Z C for z E Z. Case (II). The conditions of Case (I) are not satisfied. Suppose that Case (II) holds. By (7.4) and (7.6) p = 2, c ;l a 2 for a E K and P acts absolutely irreducibly on V. Let c = - 1 . Let Po, 00 , Go be defined as in statement (ii). By (7.5) there exists an automorphism (J of Go such that XU = x for all x E Pa, O g = 00 , v O" ;I v for v E Vo, v ;l 0, where Vo = Oo/Z(Oo) and z O" = Z - I for z E Z. In Case (I) change the notation and let P = Po, 0 = 00, G = Go, V = Vo, XA = XA O, X� = X�o so that both cases can be handled simultane­ ously. 0 Let H = P(O)( 00 x 00), where prO) = {(x, x ) I x E Po}. Define 0 ( ) = {(y, y ) l y E Oo} and 0 (0") = {(y, y O" ) y E Oo}. Then prO) normalizes both 0 (0) and 0 (0"). Furthermore H/Z(c ) is extra special, Z(c ) � 0 0" and O ( O" )/Z(c) is abelian. Since H � PoOo x PoOo, there exists an irreducible F[H] module Y = (XA O 0 X�O) H with kernel Z (c ) . --Let A/L be the linear character of P(O)O (u)Z(H) with p(O) o (a) in its kernel H � dH � (z l , Z2) = A (Zl)/L (z2) = /L (z lCz2).0 Then ( � )QoxQo = Y QoxQo and so A/L = Y by (7.7) (ii). By definition 0 ( ) n Q ( O" ) � Z(H.) Thus the Mackey decomposition (II.2.9) implies that (XA O 0 X� o)Q(O)P(O)Z(H)

7]

0

Q(O)P (O)Z(H) •

PROOF OF (7.2). Let W be an irreducible F[ G] module and let 'P be the Brauer character afforded by W. It may be assumed that 'P is faithful and so Op (G) = (1). The proof is by induction on 'P (l) = dimF W. If 'P is induced by a character 'Po of a proper subgroup then 'Po is algebraic by induction and so 'P is algebraic by (II.5.3). Thus it may be assumed that 'P is not induced by a character of any subgroup. Let H be a minimal normal noncentral subgroup of G and let � be an irreducible constituent of 'PH. As G is p -solvable, H is a p '-group. Therefore G = T(�) is the inertia group of { Since H is noncentral � (1) > 1. Thus W = V 0 \tV by ( 1 . 1) �here ((1) = diriJp V. If dimF V < 'P ( l) then also dimF W < 'P (1) and so V and W are algebraic by induction. Hence W is algebraic by (II.5.2). Thus 'P (1) = �(1) and so VH is irreducible. Hence if P is a Sp -group of G, then VHP is irreducible . Furthermore V I ( VHP ) O and so by (II.5 .3) it suffices to show that VHP is algebraic. Thus by changing notation it may be assumed that G = HP. Suppose that P � Go <J G. It suffices to prove the result for Go by (II.5.3) since V I ( Voo f . We will now consider two cases depending on whether H is solvable or not. Suppose first that H is solvable. The minimality of H implies that H is an extra special q -group for some prime q ;l p. Thus W is algebraic by (7.9) . Suppose finally that H is not solvable. Let Z = Z( G) . The minimality of H implies that Z � H' and H/Z = MI x . . . X Mk, where M = M for all i and some simple group M. Then {M } is the set of all conjugate subgroups of M and P acts as a transitive permutation group on the set {M }. There exists a group Ho with Z(Ho) � Hb and Ho/Z(Ho) = M such 'that H is a homomorphic image of H, x . . . X Hk with R = Ho for all i. Thus G is a

440

.

[8

CHAPTER X

homomorphic image of G = P(HI X . , . X Hk ), and P permutes the set {H; } transitively. Therefore 'PH = I; II� I 1;;, where I;i is an irreducible character of H whose kernel contains all � for j � i. Hence 1 G : T(l;l ) 1 = k . By (111.3. 16) there exists an irreducible Brauer character 'P I of T(I;I ) with ( 'Pl) H = 1;1 . Hence 'P Y = 'P . Since 'P is primitive this implies that T(l;l) = G. Hence k = 1 and HI Z is simple. As HIZ is well behaved, a Sp -group of G is cyclic and so there are only a finite number of indecomposable F[ H] modules (up to isomorphism). Hence by (11.5. 1), W is algebraic. 0 =

=

ko( x ) X (x )ko(x ) == X (x ) k (x ) ko(x ) == == ko(x ) k (x ) k (x ) X ( l) k (x ) x (l)

Thus X H i s in the principal block o f R [ H ] .

Let X be an irreducible character in B with

Throughout this section the following notation will be used. G is a finite group. If x E G then k (x ) = 1 G : cG (x ) l . K i s a finite extension of Qp which i s a splitting field of G and all o f its subgroups. R is the ring of integers in K and 7r is a prime in R. We will be concerned with the following notation and assumptions.

=

HYPOTHESIS 8. 1 . (i) P is a Sp -group of G, p e G <J G and G G CG (P). (ii) B, B is the principal block of R [ G ] , R [ G ] respectively. (iii) A, A is the ideal of R [ G] , R [ G ] corresponding to B, B respectively. The object of this section is to prove the following result. THEOREM 8.2. Suppose that (8. 1) is satisfied. Then the algebras A and A are

=

Alperin [1976d] showed that A A in case GI G is solvable. In full generality (8.2) is due to Dade [1977] . His proof depends on his deep work on Clifford theory. See e.g. Dade [1971a] . For an alternative proof see Schmid [1980]. We will here adapt Alperin' s approach to the more general situation. We will first prove a series of lemmas.

Throughout the remainder of this section it will be assumed that (8. 1) is satisfied. LEMMA 8.3. Let H be a subgroup of G with G � H. Let X be an irreducible

character in B such that XH is irreducible. Then XH is in the principal block of

R [H].

PROOF. For x E H let ko(x ) = I H : CH (x ) l . A Sp -group of CG (x ) is con­ tained in G and hence in C H (x ). Thus k (x ) and ko(x ) are divisible by the same power of p and so ko(x )1 k (x ) � 0 (mod p). Since X is in B it follows from (IV 04.2) that (mod 7T).

0

LEMMA 804. Suppose that X is an irreducible character in B with T(X ) = G.

8. Isomorphic blocks

isomorphic.

441

ISOMORPHIC BLOCKS

8]

X

� XC . Then

Xc = X.

PROOF. Apply ( 1. 1) with char F = 0 and H = G. T!ms X = 8Yf where 8 is a character of G with 8c = X if G is identified with G <J G. Furthermore Yf is an irreducible character of G I G and (G I G )I Z GIG for Z in the center of G I G. Let y E G I G and let y be the image of y in GI G. By (8. 1 ) there exists x E CG (P) with y = .i where .i is the image of x in GI G. Thus X (x ) = 8 (Yo) Yf (y ) for some Yo E G. Since X E B, =

X (x ) k (x ) ==

k

)�O

(x x(l) Hence Yf (y ) � O. As Yf is irreducible and y is an arbitrary element of GIG this implies that Yf ( l ) = 1 . Hence x (l) = 8 (1) X (l). 0

(mod 7r).

=

LEMMA 8.5. Let S, S be the set of all irreducible characters in B, B respec ­ tively. For X E S let r(x) = Xc. Then r is a one to one map from S onto S. PROOF. Induction on 1 G : G I . If 1 G : G 1 = 1 the result is trivial. By induction it may be assumed that GI G is simple. We will first show that r maps S onto S. Suppose that 1 G : G 1 = q is a prime . Let X E S and let X be an G irreducible constituent of Xc . As q is a prime either Xc = X or X = X . Since X E B it follows from (IVA.2) that if x E CG (P) then x

:��S

(

x)

== k (X ) � O

(mod 7r).

G Thus X (x ) � O and so X� X as G = G CG (P). Hence Xc = X. By (8.3) Xc = X E S. If conversely X E S then by (V.2.3) X is a constituent of Xc for some X E S. Thus X = Xc and r maps S onto S in this case.

CHAPTER X

442

[8

Suppose that GIG is a noncyclic simple group. Let X E S. Let q be a prime and let Q be a Sq -group of G. By induction QG is the inertia group of X in QG and so Q C T(X). Since q was chosen arbitrarily this implies that G = T(X) · By (V.2.3) X C Xo for some X E S. Hence X = Xo by (8.4). If conversely X E S then by (V.2.3) there exists X E S with X C Xo. By (8.4) Xo = X· Thus r maps S anto S in any case. It remains to show that r is a one to one map. Let X E S with X = Xo. Let p be the character afforded by the regular representation of G I G. Then X O = xp· Hence if 0 E S with 00 = X = Xo then 0 C X o and so 0 = XA for some irreducible character A of GIG. Since 0 (1) = X ( 1) it follows that A (1) = 1. Thus if x E Co (P) then

k (X )X (X ) A (X ) k (x )O (x) k (x ) k (x ) X (x ) (mod ) X (l) X ( l) x (l) Hence A (x ) 1 (mod ) As G = G Co (P) this implies that A = 1 0 and so X = O. 0 ==

==

==

7T

==

7T

.

.

LEMMA 8.6. Let So, So be the set of all irreducible Brauer characters in B, B respectively. Let D, b be the decomposition matrix of B, B respectively and let C, C be the Cartan matrix of B, B respectively. For 'P E So let ro('P ) = 'Po. The

following hold. (i) ro is a one to one map of So onto So. (ii) Let r be defined as in (8.5). Let r (Xu ) = Xu for Xu E S and let ro( 'Pi ) = �i for 'Pi E So. Then D = b and C = C.

443

ISOMORPHIC BLOCKS

8]

Thus if 'P E So then (IV.3. 12) implies that 'P<x,O) is in the principal block of (x, G). Suppose that 'P h 'P E So with ('P I )O = 'Po. Since ('P l)(X,O), 'P(x,O) are in the principal block of (x, G) it follows by induction that ( 'P lh, o ) = 'P (x,O). Hence in particular 'Pl (X) = 'P (x ) for an arbitrary p '-element x E G. Hence 'P I = 'P and so ro is one to one in all cases. 0 LEMMA 8.7. Let V be a projective R [G] module in B. Let E = EndR[o] ( V) and let E = EndR[o] ( Vo ). Let f : E � E be defined by restriction. Then f is an algebra isomorphism of E onto E. PROOF. Clearly f is an algebra monomorphism. It remains to show that f is an epimorphism. Suppose that h E E and a E R with a - I h an R [G]­ endomorphism of V. Then a - 1h EndR ( V) n EndK[o]( VK ) = E. Thus f(E ) i s a pure submodule o f the R module E. Hence i t suffices t o show that rankR E = rankR E. Let U;, � be indecomposable projective R [GJ modules in B. By (8.5) rankR HomR[o] ( U;, �) = rankR HomR[o] « U; )o, (�)o ).

Since V is the direct sum of indecomposable projective R [G] modules in B it follows that rankR E = rankR E. 0 PROOF OF (8.2). Since A ' is an algebra with unity element it is anti­ isomorphic with the endomorphism ring of the F[ G] module A. Similarly A is anti-isomorphic with the endomorphism ring of the F[ GJ module A. By (8.7) it suffices to show that A o A as F[ GJ modules. Let {'Pi } be the set of all irreducible Brauer characters in B. Then by (8.6) {�i } is the set of all irreducible Brauer characters in B where �i = ('Pi )0. Let Vi be a proj ective indecomposable R [GJ module corresponding to 'Pi . Since C = C by (8.6) ( U; )o is the proj ective indecomposable R [GJ module corresponding to �i. Thus =

PROOF. In view of (8.5) it is clear that (ii) will follow as soon as (i) is proved. We will prove (i) by induction on 1 G : G I . If 1 G : G 1 = 1 the result is trivial. By induction it may be assumed that GIG is simple. If 'P E So then 'Po is an irreducible Brauer character by (8.5) and (IV.4.33). By (IV.4. 10) 'Po E So. If conversely � E So then there exists 'P E So with � C 'Po by (lV.4. 10). Thus � = 'Po by the previous sentence. Hence ro maps So onto So. It remains to show that ro is one to one. Suppose that 1 G : G 1 = q is a prime. Let 'P, 'P I E So with 'Po = ('PI)O . By (111.2. 14) there exists a linear character A of GIG with 'P I = 'PA. By (IV.3. 12) 'P (x ) = Lxu ES auXu (x) for all p '-elements x in G. Hence 'PA ( X ) = Lxu ES auXuA (x ) for all p '-elements x in G. Since Xo = XAo for all X E S it follows from (8.5) that XuA � B unless A = 1 0. Since 'PA E So this implies that A = 1 0 and 'PA = 'P. Thus ro is one to one in this case. Suppose that GIG is a noncylic simple group. Let x E G. If X E S then X<X,O) is irreducible and so X< X, O) is in the principal block of (x, G) by (8.3).

Ao

=

EB 'Pi (1) ( U; )o = EB �i ( 1) ( Vi )o

=

A.

0

1]

CHAPTER XI

1.

An analogue of Jordan's theorem

One of the oldest results in group theory is the following theorem. THEOREM 1 . 1 (Jordan). There exists an integer valued function l(n ) defined

on the set of positive integers with the following property. If the finite group G has a faithful representation of degree n over the complex numbers then G has a normal abelian subgroup A with 1 G : A / < l(n ).

Several proofs of this theorem are known. See for instance Curtis and Reiner [1962] (36. 13) for a proof and references to other proofs . It is an immediate consequence of Jordan's theorem that the same conclusion holds if the field of complex numbers is replaced by any field whose characteristic does not divide 1 G / . However the result is false for fields whose characteristic divides 1 G I . For example let F be an algebrai­ cally closed field of characteristic p > 0 and let Gm = SL2(p m ). Each Gm has a faithful F-representation of degree 2 but a normal abelian subgroup of Gm has order at most 2 while / Gm I can be arbitrarily large. This section contains a proof of the following analogue of (1.1) which was first conjectured by O. H. Kegel. THEOREM 1.2 (Brauer and Feit [1966]). Let p be a prime. There exists an

integer valued function f(m, n) = It) (m, n) such that the following is satisfied. Let F be a field of characteristic p and let G be a finite group which has a faithful F-representation of degree n. Let p m be the order of a Sp -group of G. Then G has a normal abelian subgroup A with 1 G : A I < f(m, n ). Isaacs and Passman [1964] have used Jordan's theorem to show that if 444

AN ANALOGUE OF JORDAN'S THEOREM

445

the degree of every irreducible complex representation of the group G is bounded by some integer n then the conclusion of (1.1) holds (with a function different from l(n » . In a similar manner J. F. Humphreys [1972] has used (1 .2) to prove an analogous result in characteristic p. For a related result se�, J. F. Humphreys [1976] . The various proofs of Jordan's theorem (1 .1) yield different values for ' I (n ). None of these values seem to be anywhere near the best possible value. Since ( 1 . 1) is needed for the proof of (1.2) and several fairly crude estimates are also used in the proof of (1 .2) the value of f(m, n ) which can be derived from the given proof of (1 .2) is probably nowhere near the best possible result. Thus (1.1) and (1 .2) are both qualitative results which do not yield any useful bounds . The proof of (1 .2) will be given in a series of lemmas. Without loss of generality it may be assumed that the field F in (1 .2) is algebraically closed . For any group G let L J (G) denote the F[G] module which affords the principal Brauer character. Let B J (G) denote the principal p -block. If G is a finite group and P is a Sp -group of G, then the center Z(P) of P is a Sp -group of Cc (P) and Burnside ' s transfer theorem implies that Co (P) is the direct product a p -group and a p '_group. Thus PCc (P) = P x H for some p '-group H. Suppose that V is an indecomposable F[ G] module with dimF V¢ 0 (mod p). Th en P is a vertex of V. Let V denote the F[No (P)] module which corresponds to V in the Green correspondence. The crux of the proof of (1 .2) is contained in the next result. LEMMA 1 .3. Suppose that V is an irreducible F[ G] module with n = dimF V > 1 . Then at least one of the following holds. (i) G has a normal subgroup of index p. (ii) There exists an irreducible constituent L of V* ® V ® V* ® V �ith

L in BJ(G) and L � L J(G). (iii) Let P be a Sp -group of G. Let N = N o (P). There exists an irreducible constituent L of V* ® V with dimF L ¢ 0 (mod p) and L � L (G) such that if Ho is the kernel of LH then 1 H : Ho 1 < l(n ), where l(n) is defined by ( 1 . 1). J

PROOF. Assume that G satisfies neither (i) nor (ii). Let LJ = L J (G), L2, , Ls denote the distinct irreducible constituents of V* ® V. Let W be an F[ G] module, all of whose irreducible constituents are constituents of V* ® V ® V* ® V. We will show that the multiplicity of L J(G) in W is equal to dimF lnvo W. It clearly may be assumed that W is indecomposable. If W is not in B J (G) the result is trivial. Suppose that W •

•

•

446

CHAPTER XI

[1

1]

IS

III B I (G). Since (ii) is excluded, every composItIon factor of W is isomorphic to L I(G). If x is a p '-element in G then W(x ) is completely reducible . Thus x is in the kernel of W. Since G has no normal subgroup of index p it follows that G is in the kernel of W. Hence dimF W = 1 as W is indecomposable and the result is proved. This fact will be applied to several modules . Let W = Li @ Lj with 1 ::::; i, j ::::; s. Thus InvF W = HomFI G ] (Li , Lj ). For i = j, Schur's Lemma implies that dimF Inv G W = 1 . Thus by (III.2.2) and the previous paragraph dimF Li � 0 (mod p ) for 1 ::::; i ::::; s. For i -I- j, Schur's Lemma implies that InvF W = (0). Since (ii) is excluded it follows that no irreducible constituent of W is in BI(G). Hence by (IV.4. 14), Lr, . . . , L, lie in s distinct p-blocks. Let Y be an indecomposable direct summand of V* @ V. Since all the irreducible constituents of Y lie in one p -block it follows from the previous paragraph that all the irreducible constituents are isomorphic to Li for some i. Let b be the length of a composition series of Y. Thus L I ( G) occurs with mUltiplicity b in L ; @ Y. Since every irreducible constituent of L ; @ Y is a constituent of L ; @ Li and so of V* @ V @ V* @ V, it follows that b = dimF HomF[ G ] (L , Y). Let Yo be the socle of Y and let bo be the length of a composition series of Yo. The previous argument applied to Yo shows that bo = dimF HomF[ G ] (L , Yo). However HomF[ G ] (L , Y) = HomF[ G ] (L , Yo). Thus b bo and so Y = Yo. As Y is indecomposable this implies that Y = Li . Consequently V* @ V is completely reducible. Let =

s

V* @ V = EB aiLi . i=1

(1.4)

B y Schur's Lemma a I = 1 . Hence'& Y (III.2.2) dimF V � 0 (mod p ). Thus P is a vertex of V. Hence (III .5 .7), (V.6.2 ) and (1 .4) imply that s

V* @ V = EB a;lJi E9 S, i=1

(1 .5)

where each indecomposable direct summand of S has a vertex properly contained in P. By (III.7.7) L I = L I (N) and L is not in B I (N) for i � 2. By (III.3.7) VPXH = EB� = I ( Vi @ Xi ) , where XI , . . . , Xe are distinct ir­ reducible F[H] modules and VI , . . . , Ve are F [ P] modules which are conjugate under the action of N. Thus

e ( V* @ V)pxH = EB ( V i @ Uj ) @ (Xi @ Xj ) i.j= 1

447

AN ANALOGUE OF JORDAN'S THEOREM

where S is an F[ P x H] module which does not contain L I (P x H) as a constituent. By (1 .5) L I (N) I V* @ V. Thus L I (P x H) I ( V* @ V)PXH and so L I (P) I Vj @ Uj for some j. Since the modules Uj are conjugate under the action of N, this implies that L I (P) I Vj @ Uj for j = 1, . . . , e. Thus I

eL I (P x H) I ( V* @ V)PXH. Suppose that e > 1 . Since al

= 1 , (1 .5) implies that L I (P x H) I TpXH for some indecomposable direct summand T of S. Thus L I (P) I Tp. Since a vertex of T is properly contained in P this contradicts ( III .4 .6) . Thus e = 1 . Consequently

(1 .6) where X is an irreducible F[ H] module and V is an F[ P] module such that L I (P) I V * @ V. Let d dimF X. If d = 1 then X* @ X = L I (H) and H is in the kernel of V* @ V. Hence by ( 1 .5 ) H is in the kernel of Li for all i. Thus by (JV.4.14) Li is in BI(N) for all i and so Li is in B I (G) by (III.7.7) and (V. 6.2) . Thus s = 1. Hence by (1.4) n = 1 contrary to assumption . Thus d > 1 . Let M b e a n irreducible constituent o f V and let D b e the kernel o f M. By (III.2. 13) P � D. Thus N/D is a p '-group. Hence by the remark following ( 1 . 1) there exists a subgroup A o with D � Ao <J N such that A o/D is abelian and I N : Ao l < J(n ). Let A I = Ao n H and DI = D n H. Then A I <J N, I H : A I 1 < J(n ) and A I /D I is abelian. By (1 .6) MH = uX for some positive integer u. It follows that D I is the kernel of X. If A is an abelian group and X is an F[ A ] module then X* @ X contains L I (A ) with multiplicity at least dimF X, Apply this remark with A A JD I . Thus (X* @ X)A, contains L I (A I) with multiplicity at least d � 2. As X is irreducible, X* @ X contains L I (N) with multiplicity 1 . Thus there exists an irreducible constituent Z � L I (H) o f X* @ X such that L I (A I ) / ZA, . By Clifford 's theorem (III.2. 12), A I is in the kernel of Z. Since L I (P) @ Z I ( V* @ V)PXH by (1 .6) it follows from ( 1 .5 ) that LI(P) @ Z I TpXH for some direct summand T of EB aiL E9 S. By (III.4.6) P is a vertex of T. Thus by (1 .5) T = Lj is irreducible. Since Z;;6 L I (H), j -I- 1. Let Ho be the kernel of (Lj )H. Since A I belongs to the kernel of Z and A I <J N it follows from Clifford's theorem (III .2. 12) that A I � Ho. Thus I H : Ho i ::::; I H : A l l < J(n ). Consequently (iii) holds for L Lj • D =

I

=

=

LEMMA 1 .7. Let V be an irreducible F[ G] module with dimF V = n > 1 .

Suppose that G contains no normal subgroup of index p. Then there exists an irreducible constituent L of V* @ V @ V* @ V such that

448

CHAPTER XI

[1

1 < I G : Go I < I GLIPI 2 n4J(n)(p) I ,

where Go is the kernel of L. PROOF. Let L be an irreducible F[ G] module with dimF L = d. Let Go be the kernel of L and let iP be the trace function afforded by L. Suppose that iP has exactly e algebraic conjugates in F. Thus the subfield of F generated by all iP (x ) as x ranges over G has p C elements. By (1. 1 9.2) GIGo is isomorphic to a subgroup of GL d (p e ). Thus I G : Go I � I GL d (p e ) I � I GLde (p ) / . By assumption (1 .3) (i) does not hold. Thus by (1 .3) either ( 1 .3) (ii) or (1 .3) (iii) must hold . Suppose that (1 .3) (ii) holds. Choose L accordingly. G rf Go as L ;l: L , ( G ) . By (IV.4.9) and (IV.4. 1 8), iP has at most I p l 2 algebraic conjugates in F. Thus the result follows from the previous paragraph and the fact that d < n4. Suppose that (1 .3) (iii) holds . Choose L accordingly. A s above G rf Go. Let B" . . . , Ba be all the p -blocks of G which contain algebraic conjugates of iP. Since dimF L = d < n 4 the remarks above show that it suffices to prove that iP has at most I p 12J(n ) algebraic conjugates. Thus by (IV.4. 1 8) it suffices to show that a � J(n ). Let $ be the trace function afforded by i. By (III. 7.7) there are exactly a p -blocks of N G (P) which contain an algebraic conjugate of $. By (1 .3) (iii) the group N G (P)IHo has less than J(n ) p -blocks. Thus by (V.4.3) a < J(n ) as req uired. 0 Let G be a finite group and let V be an F[ G] module. Let 7( G, V) = (m, n, a ) where a Sp -group of G has order p m , dimF V = n and a is the multiplicity with which L , (G) occurs as a constituent of V* 0 V 0 V * 0 V. Define the partial ordering -< as follows : (m " n" a,) -< (m, n, a ) if one of the following is satisfied. (i) m, < m, n, � n. (ii) m , � m, n, < n. (iii) m , = n, n, = n, a, > a. If H is a subgroup of G, then clearly either 7 (H, VH ) = 7 (G, V) or 7(H, VH ) -< 7(G, V). Observe that if 7(G, V) = (m, n, a ) then a � n 4. Thus there are only finitely many triples 7( G" V, ) with 7( G" VI) -< 7( G, V).

LEMMA 1 . 8. Let P be a Sp -group of G and let / p I = p m . Let X be a faithful F[ G] module and let dimFX = n. Let g(m, n ) = / GLIPI2J( n ) n 4(p ) / , where

1]

AN ANALOGUE OF JORDAN'S THEOREM

449

J(n ) is defined by ( 1 . 1). Assume that G is not abelian. Then there exists Go <J G with I G : Go / < g(m, n ) such that 7(Go, XGo) -< 7(G, X). PROOF. It may be assumed that G has no normal subgroup of index p otherwise the result is trivial. Suppose first that every composition factor of X has F -dimension 1 . Let X, be a completely reducible F[ G J module with the same composition factors as X. Then P is the kernel of X, and G I P is abelian. Thus G is solvable and so G contains an abelian subgroup A with / G : A I = / p I . Let Go = nxEG x-'Ax. Then Go <J G and / G : Go l � I p i ! � g (m, n ). Since G is not abelian / P I � 1 . Thus 7( Go, XGo) -< 7( G, X). Suppose next that some irreducible constituent V of X has F -dimension at least 2. Let L and Go be defined as in (1 .7). If 7( G, X) = (m, n, a ) and 7(Go, XGo) = (m " n l , a l ), then clearly m l � m and n , = n. Since Go is the kernel of L and L ;I: L I ( G ), al > a. Thus 7(Go, XGo) -< 7(G, X). 0 LEMMA - 1 .9. There exists a function h (m, n, a ) such that if X is a faithful F[G] module, then I G : A I � h (7(G, X» for some normal abelian sub ­

group A of G.

PROOF. Let h (m, n, a ) = 1 if (m, n, a ) � 7( G, X) for any pair (G, X). If the result is false it is possible to choose a counterexample (G, X) so that h (m " n " a ) is defined for all (m l , n l , a l ) -< 7(G, X). Let g(m, n ) = I GLIPI2 n4J(n )(p ) I . D efine

and let

h (7( G, X» = g(m, n )ho( 7( G, X» 8( m.n) . If G is abelian let G = A. The result is clear in this case. Suppose that G is nonabelian. Let Go be defined by (1 .8). Thus Go contains a normal abelian subgroup Ao with I Go : Ao l � h (7(Go, XGo» ' Let A = nXEG x -IAox. Then I G A 1 < g(m, n )h (7(Go, XGo» 8( m.n). This implies that I G A 1 < h (7.( G, X» . 0 :

:

(1 .2) is now a direct consequence of (1 .9) if f(m, n ) is defined by

f (m, n ) = max h (m, n, a ). O::::::; a � n 4

1J

CHAPTER XII

I f a block B has a noncyclic defect group D then the situation i s vastly more complicated than in the case described in Chapter VII when D is cyclic. Such a block in particular contains an infinite number of indeco �­ posable modules, see for instance Hamernik [ 1 974a] , [ 1 975a] . �xcept I� very special cases when p = 2 and D contains a cyclic subgroup of mdex 2 It is hopeless to attempt to describe all the indecomposable modules in B. See Bondarenko [1975], Ringel [ 1 974], [ 1 975] . It is perhaps less obvious that there appears to be no method for constructing the irreducible Brauer characters in B. The purpose of this chapter is to study questions concerning the ordinary irreducible characters in a block B and to present some applications of this study. In this connection the concept of a basic set introduced in Chap ter . IV, section 3 plays an important role. After a general introductory sectIon most of the chapter deals with the case that p = 2 and D is a special type of 2-group. The material in this chapter originated with work of Brauer [1952] . The notation introduced at the beginning of Chapter IV is used throughout this chapter.

1. Types of blocks

The results in this section are due to Brauer [1961a], [ 1964a], [ 1 969b], [ 1 971a], [1971c], [1971 d] . See also Reynolds [1965] . For related results see Brauer [ 1 964c] . Let B be a block and let {Xu } be the set of irreducible characters in B. Let 'PB = {'Pi } be a basic set for B. Thus there exist rational integers dur such that Xu (x ) = 22i dui'Pi (x ) for all p i-elements x in G. As in Chapter IV, 450

45 1

TYPES OF BLOCKS

section 3 {dui } is called the set of decomposition numbers with respect to 'PB and {eij } is called the set of Cartan invariants with respect to 'PB where Cij = 22u duiduj . Let y be a p -element in G. Let {'P r} be the union of basic sets for all the blocks of CG (y ). Then for each Xu in B there exist algebraic integers d �i in the field of p n th roots of unity over Q for suitable n such that Xu (yx ) = 22i d �i'P r(x ) for all p I-elements x in CG (y ). The algebraic integers d �i are the higher decomposition numbers with respect to {'P r}. The set of columns d �i as 'P r ranges over a given basic set is a subsection with respect to {'P r}.

LEMMA 1 . 1 . Let B be a block of G. Let y, z be p-elements in G. Let {'P r}, {'P �} be the unions of basic sets for all blocks of CG (y ), CG (z ) respective/yo (i) If Xu is in B then d �i = 0 unless 'P r belongs to a basic set of a block B of CG (y ) with B G = B. (ii) If y is not conjugate to z in G then for all i, j L (d �i) * d �j = L (d �i)*d �j = o. U

� �B

(iii) If (c �) is the Cartan matrix of

i, j

CG (y) with respect to {'P r} then for all

Lu (d�i)*d �j = c �. PROOF. Immediate by (IV.6. 1) and (IV.6.2).

D

Given a basic set for the block B then the Cartan matrix (Cij ) of the block with respect to that basic set is the matrix of a positive definite integral quadratic form Q. If the basic set is replaced by another basic set then Q is replaced by a quadratic form which is equivalent to Q over Z. Thus to each block there is associated an equivalence class of integral quadratic forms.

LEMMA 1 .2. Suppose that p and d are given. There are only finitely many classes of integral quadratic forms which are associated to p-blocks of defect d of groups G. PROOF. Let Q be a quadratic form associated to a p -block of defect d. By (IV.4. 1 S) the dimension of Q is at most p 2d. By (IV.4. 16) every elementary divisor of the Cart an matrix is a power of p which is at most p d and so the discriminant of Q is bounded by a function of p and d. This implies the result by the reduction theory of quadratic forms. D

CHAPTER XII

452

[1

COROLLARY 1.3. There exists a bound f(P, d ) depending only on p and d

such that for each p-block of defect d of any group, a basic set can be chosen such that the Cartan invariants are at most equal to f(P, d ). PROOF. Immediate by (1 .2).

D

LEMMA 1 .4. Let B = Bo be the principal p -block of G and let % be the principal Brauer character of G. There exists a basic set for B containing 0/0 such that the Cartan invariants for this basic set lie below a bound fo(p, n ) depending only on p and n where a Sp -group of G has order p n. PROOF. Choose a basic set {'P; } for B as in (1 .3). Let % = 'Ldoi'Pi . Then d �i ::s; Cii ::s; f(p, n ). Furthermore the ideal of Z generated by all dOi is Z itself. Thus there exists a matrix of determinant ± 1 with entries in Z and first row equal to (do; ). The entries in this matrix are bounded by a function of the dO i and so by a function of p and n. D LEMMA 1 .5. Let B be a block of defect d, let y be a p -element in G and let B

be a block of C C (y) with B c = B. (i) Let {'P r} be a basic set for B and let m be the maximum of the Cartan invariants associated to {'P r}. Then the corresponding higher decomposition numbers d �i for Xu in B all belong to a finite set M(p, d, m ) depending only on p, d, and m. (ii) There exists a basic set for B such that the corresponding higher decomposition numbers d �i for Xu in B all belong to a finite set M(p, d) depending only on p and d.

1]

TYPES O F BLOCKS

453

over all the irreducible characters of B and the column index i is such that 'P ( Y, B ) = {'P r}.

Let T"b(B ) be the matrix of ordinary decomposition numbers of B with respect to the basic set 'P (y, B ). A block B of defect d is of a given type for a subsection (y, B ) if the pair of matrices Tl;( B ) , TY (B, B ) formed with respect to some basic set 'P (y, B ) of B is given. A block B is of a given type for an element y E Y if B is of a given type for all subsections (y, B ) corresponding to B. A block B is of a given type if it is of a given type for all elements y E Y. Let BO be a block of the group GO. Suppose that B and B O both have defect d. Define D O , yO, y O E yO , T"b°( B O) , T y o (B O , B O) for the group GO. Then B and B O are of the same type if the following conditions hold . (i) 1 Y I = 1 yO I and for a suitable ordering Y = { Yi }, yO = {y ?} with 1 B l (Cc ( Yi ), B ) 1 = 1 B l (Cco(y D, B O) I · (ii) After a suitable rearrangement BI(Cc ( Yi ), B ) = { Bij } and BI(Cco (y?), B O) = {B �} such that there exist basic sets for Bij and B ;� for all i, j with the property that T Yi (B, Bij ) = TIl (B O, B �) and T"bi(Bij ) = Ttl (B �) .

THEOREM 1 .6. For given p and d there exist only finitely many types of p -blocks of defect d (in the category of all finite groups). PROOF. If Y is defined a s above then from ( 1 .3 ) and ( 1 .5). D

1 Y 1 ::s; p d. Thus the result follows

If y is a p -element in G let BI(Cc (y ), B ) be the set of all blocks B of Cc (y ) with B c = B. Let D be a defect group of B. If y is not conjugate to an element of D then BI(Cc (y), B ) is empty by (IV.6.6) (iv). Let Y be a set of elements of D which consists of a complete set of

Suppose that the type of a block is given. It is natural to ask what additional information is required to compute the values of the irreducible characters in the block. Before considering this question a preliminary result will be proved. The following situation will be studied. Let D be a fixed subgroup of G with 1 D 1 = p d. Let y E D. Let K1 , , Kn denote the classes of p '-elements in Cc (y ). Assume that the following information is given. (i) I Ki 1 is known for i = 1 , . . . , n. (ii) It is known which conjugate class of G contains Ki • Denote this class by KF. (iii) It is known for which i = 1 , . . . , n the class KF has defect less than d and for which i the class KF has D as a defect group.

representatives of the conjugate classes of G which contain an element of D. For y E Y and B E BI(Cc (y ), B ) let 'P (y, B ) be a basic set for B. Define the matrix TY (B, B ) = (d �i) where the row index ranges over u as Xu ranges

LEMMA 1 .7. Let D be a subgroup of G with 1 D 1 = pd and let y E D. Suppose that (i), (ii), (iii) above are given. Suppose also that for each block B

PROOF. (i) If

(T

is an automorphism of the field generated by all d �i then (T. The result follows

« d �it)* = « d �;)*t· Thus by ( 1 . 1 ) 1 (d�it 1 2 ::s; m for all as each d � i is an algebraic integer. (ii) Clear by (i) and ( 1 .3).

D

• • •

454

[1

CHAPTER XII

of Co (y) of defect at most d there is a n irreducible character X in 13 such that X ( ) is known for E Ki , l � i � n. For any block b with defect group D and for E G let Xi

Xi

X

CUB

l x (x) ( ) _ / G : CoX (x) ( ) X

-

l

for some irreducible character X in B. Then it is possible to compute (x) for any p '-element in G. In particular it is possible to determine the number of blocks of G with defect group D.

2]

SOME PROPERTIES OF THE PRINCIPAL BLOCK

455

there exists a left inverse of T"bi(13ij ) which can be found explicitly. In other words the given basic set for 13ij can be expressed in terms of the irreducible characters in Bij .' Since TYi (B, 13ij ) is known this makes it possible to evaluate the irreducible characters Xu in B on the p -sections which contain elements of Yo. 0

CUB

X

PROOF. The last statement follows from the first by (IVA.3) and (IVA.8). It

remains to prove the first statement. Let X be a p '-element in G and let C be the conjugate class of G with X E C. Since y E D it follows from (IV.6.6) (iv) that if B is any block of G with defect group D then there exists a block 13 of Co (y ) with B 0 = B . By (III.9.6) 13 has defect at most d. If CU B is defined analogously to CUB it follows that CU B (x ) = Li CU B ( Xi ) , where Xi ranges over a complete set of representatives of the conjugate classes of Co (y ) which are contained in C n Co (y ). In particular since X is a p '-element, each such conjugate class is some Ki for 1 � i � n. If C i- K� for all i, the sum is empty and CUB (x ) = O. If C = K? for some i, then it is known by assumption which of the classes Ki lie in C and CUB ( Xi ) can be computed as X ( Xi ) is known by assumption. 0

THEOREM 1 .8. Let D be a subgroup of G of order p d. Let Y be a subset of D which is a complete set of representatives of conjugate classes of G which meet D and let Yo be a nonempty subset of Y. Suppose that for every y E Y, (i), (ii) (iii) are given and that for y E Yo the values on p '-elements in C o (y ) of the irreducible characters in blocks of defect at most d of Co (y ) are known. Let Yo be a fixed element in Yo, let 13 be a block of Co (Yo) for which 13 0 = B has defect group D and suppose that the type of B with respect to every element in Yo is known. Then it is possible to find the values of the irreducible characters in B on the p -sections which contain elements of Yo. y = yo. Thus it is possible to find

(x ) for p '-elements X in G. For each Yi E Yo it is possible to determine the set {13ij } = BI(Co (Yi ), B ). For each 13ij the values of the irreducible characters in Bij are known on p '-elements in Co (Yi ). By hypothesis the matrix T"hfBij ) of decomposition numbers with respect to a suitable basic set for Bij is known. Since all the elementary divisors of this matrix are 1 by (IV.3. 1 1)

PROOF. Apply

(1 .7) for

CUB

2. Some properties of the principal block

The results in this section will be useful for applications below. See Brauer [ 1 964b] . Let {Xu } be the set of all irreducible characters in the principal block Bo of G. For an element X E G define

A (G, x ) = L

Xu i n B o

Let

A (G) = A (G, 1).

(2. 1)

Thus

A (G) = L

Xu in B o

Clearly

I Xu (x ) 1 2.

2 Xu (1) .

(2.2)

A (G, x ) � 1 Co (x ) 1 ·

LEMMA 2.3. For x E G, A (G, x ) is a positive rational integer. Let i denote the image of x in G = G /Op (G ). Then A (G, x ) = A (G, i) � I CG (i ) l . In particular A (G) = A (G) � 1 G Op'( G) 1 and the equality sign holds if and only if G is of deficiency class 0 for p. ,

:

PROOF. By (IVA.9) A (G, x ) is a rational integer. Since Bo contains the principal character, A (G, x » O. By (IVA. 12) A (G, x ) = A (G, i). The in­ equalities are clear and equality holds if and only if Bo is the unique block of G. 0 LEMMA 204. Let y be a p-element in G and let x be a p '-element in Co (y ). Then A (G, yx ) = A (Co (y), x ). In particular A (G, y) = A (Co (y )). PROOF. By the second main theorem on blocks (IV.6. 1) and by (V.6.2) ( ) (f> r(x ) (f> J(x -I ), A (G, yx ) = L L i,j d �i d �j * Xu in B o

where (f> ; and (f> ] range over the irreducible Brauer characters in the principal block of Co (y ). By (IV.6.2) this implies that

456

CHAPTER XII

[3

3]

INVOLUTIONS AND BLOCKS

457

G which contains yx. By assumption the coefficient of C in C1 C2 is O. Thus a well known formula implies that

A ( G, yX ) = � c � 'P i(X )'P HX - I ) = A ( C o (Y ) , X ) , j,j

where 'P r , 'P J range over the irreducible Brauer characters in the principal block of Co (y ) . 0 LEMMA 2.5. If H <J G then A (H) � A (G) � I G : H I A (H).

PROOF. By (IVA. 10) every irreducible character in the principal block of H is a constituent of the restriction to H of some irreducible character in Bo. This implies the first inequality. If Bo, B " . . . are all the blocks which cover the principal block of H then I G : H I A (H) = 2:X (1)2 as X ranges over the irreducible characters in all Bj • 0 3. Involutions and blocks

Involutions (i.e. elements of order 2) play a special role in group theory. Brauer [ 1 957] was the first to realize their importance for the classification of simple groups. In that same paper he showed how the study of involutions can be connected with block theory. Most of the material in the rest of this chapter is based on these ideas. The results in this section are from Brauer [1964b], [1966a] . Related results can be found in Brauer [ 1 961b], [ 1 962a], [ 1966b], [ 1 974c] . THEOREM 3 . 1 . Let G be a group of even order and let p be a prime which

divides 1 G I . Let y be a p-element in G and let tl and t2 be involutions in G. Assume that y is never the p-part of 2 ,22 where 2j is conjugate to ti for i = 1, 2. Let B be a p-block of G and let {Xu } be all the irreducible characters in B. Then for all i

(3.2)

where the d �j are the higher decomposition numbers belonging to basic sets for the blocks of Co (y ). Furthermore (3.3)

PROOF. Let C denote the conjugate class of G which contains ti for i = 1, 2. Let x be a p '-element in Co (y ) and let C be the conjugate class of

where Xv ranges over all the irreducible characters of G and x is any p I-element in Co (y ). Let {'P n be the union of basic sets for all the p -blocks of Co (y). Thus Xv (yx ) = :2:i d �j 'Pi (x ) for every p I-element in Co (y ). Hence the linear independence of the set {'P n implies that

� d �i Xv (tl )Xv (t2) = 0 . v

Xv (1)

If 'P r is in the block fJ then d �i = 0 unless Xv is in fJ 0 by the second main theorem on blocks (IV.6. 1). This proves (3.2). Then (3.3) follows by multiplying (3.2) by 'P r(x ) and summing over all 'P r in basic sets for blocks o iJ with iJ = B. 0 If y E G we will say that x inverts y if x -I yx = y -l .

COROLLARY 3A. Let y be a p-element in G. If t l and t2 are involutions in G such that no conjugate of t, inverts y then the conclusion of (3. 1) holds. PROOF. Suppose that y is the p -factor of 2 1 22 where 2i is conjugate to ti for i = 1 , 2. Then 2 1 22 = yx for some p '-element x E Co (y ). Thus 1 -I )- 1 -I Z I (y x ) 2 1 = 2 1 2 1 22Z 1 = 22Z 1 = ( 2 1 22)- = ( yx .

As y is a power of yx it follows that 2 I inverts y contrary to assumption . The result follows from (3. 1). 0 It may happen that for fixed involutions tl and t2 there exist several nonconjugate elements y such that the conclusions of (3. 1) hold. Thus new equations can be obtained as linear combinations of the given ones. If it is possible to choose such linear combinations so that the Xu (y ) are replaced by rational integers au where 2:a;, is small, then it may be pos�ible to deduce properties of G. This idea will be used frequently in this chapter. The remaining results in this section are of importance for such applica­ tions. Let P be a p -group contained in G. Let {Xu } be the set of all irreducible characters contained in the principal block Bo. If () is a complex valued class function on P define

458

[3

CHAPTER XII

�

(3.5) au (0) = « Xu )p, O)p = I I y�p Xu (y )O (y - l ). Let a(O) denote the column (au (0)). If a = (au ) and b = (bu ) are two columns define (a , b) = Lu aub �, where * denotes complex conjugation. If 0 is a generalized character of P then au (0) is a rational integer by definition. Let A be defined as in (2.2). THEOREM 3.6. Let P be a p-subgroup of G and let N be a subgroup with P � N � Ne (P). Let U be a nonempty subset of P with N � Ne ( U). Assume

that the following conditions are satisfied. (i) If y 1 and Y2 are elements of U which are conugate in G then they are conjugate in N. (ii) w = A (Ce (y ))/ I CN (y ) I is independent of y in U. Let 0 and 11 be generalized characters of P such that 0 vanishes outside U. Then

I p I 2 S = L (X (Y ), X (z ))O(y - l )11 (Z ), y,z E P

where (X (y ), X (z )) is the inner product of the columns X (y ) = (Xu (y )) and X (� ) = (Xu (z )), and {Xu } is the set of all irreducible characters in Bo. By the second main theorem on blocks (IV.6. 1)

Lu Xu (y )Xu (z - 1 ) = L d �i(d �j)* cp ;(I) cp ] (1) = 0 if y is not conjugate to z in G. If y is conjugate to z then by (2.4) �0

(X(Y ), X(z )) = (X(Y ), X (y )) = A (G, y ) = A (Ce (y )).

Since 0 vanishes outside of U it follows that

I p I 2 S = L A (Ce (y ))O (y - l ) L 11 ( Z ), yEU

SOME COMPUTATIONS WITH COLUMNS

459

Since y can range over all of P this implies the result. 0 THEOREM 3.7. Let P be a p-subgroup of G. Let U be a subset of P. Assume that there exists an involution t in G such that no element in U is inverted by a G-conjugate of t. Let 0 be a generalized character of P which vanishes outside U. Let {Xu } be the set of all irreducible characters in Bo. Then the

following hold. au (O)Xu (t t = 0 . Lu Xu (1)

L au (O)Xu (1) = O. L au (O)Xu (t) = O.

(3. 8) (3.9) (3. 10)

If furthermore (lp, 0) I- 0 then either there exist at least two positive and two negative au (0) or G has a proper normal subgroup which contains (02,(G), t).

PROOF. Let (a( 0 ), a( 11 )) = s. By definition

(x (y ), X(z )) =

4]

PROOF. It follows from ( 3 . 4) that (3.3) holds for t = tl = t2 and all y E U. If (3.3) is multiplied by I p l - 1 0 (y - l ) and added over all y in U then the definition of au (0) shows that (3 .8) holds . By assumption 1 � U and no conjugate of t is in U. Thus by the second main theorem on blocks (IV.6. 1) the columns x (1) and X (t) for the blocks Bo are orthogonal to the column X (y ) for y E U. This proves (3.9) and (3. 10). Let Q be the diagonal quadratic form (au ( O )IXu (1)). If at most one of au (0) is positive or negative then a maximal isotropic subspace has dimension at most 1 . By (3.8), (3. 9) and (3. 1 0), (Xu (1)) and (Xu (t)) are vectors which lie in an isotropic subspace and so are proportional. If XO = Ie then Xo(l) = Xo(t) = 1 . It follows that Xu (1) = Xu (t) for all Xu in Bo with au (O) I- O. Choose Xu l- Xo with au ( O ) I- O. Then (02,(G), t) is in the kernel of Xu but G is not. Thus the kernel of Xu is the desired normal subgroup. 0

z

where z ranges over the G -conjugates of y in P. Since N � Ne ( U) it follows from (i) that z ranges over the N-conjugates of y in P. In other words z ranges over the distinct elements of the form y with x E N. Hence L X E N 11 (y X ) = I CN (y ) I L z 11 (z ). Now (ii) implies that x

4. Some computations with columns

This section contains results which indicate how computations with the columns a( 0 ) introduced in section 3 can be used to yield information

460

[4

CHAPTER XII

about irreducible characters of G. These methods are closely related to those introduced in Chapter V, section 7. Only elementary results will be discussed in this section. These are primarily from Brauer [1966a] . For (4. 9) (ii) see Feit [1974]. In the sequel the case p = 2 will be studied more closely and several applications will be presented. In case p � 2 several authors have used the "method of columns" and properties of isometries proved in Chapter V, section 7 to derive results about the structure of groups. See for instance G. Higman [1973], [1974] ; Smith and Tyrer [1973a] , [ 1973b] ; Smith [ 1 974] , (1976b] , [ 1976c], [1 977] . As an example we state here one such result without proof. THEOREM 4. 1 (Smith and Tyrer [1973b]). Let p � 2. Suppose that G has an abelian Sp -group P and 1 NG (P) : C G (p) 1 = 2. If G = G ' then P is cyclic. Consider the following hypothesis. HYPOTHESIS 4.2. (i) P is an abelian p -subgroup of G and N is a subgroup with P � N � N G (P). U is a nonempty subset of P. (ii) If y E U then C G (y) has a normal p -c;omplement. (iii) If y E U and y is conjugate in G to x E P then x E U and y is conjugate to x in N. (iv) If y E U there exists a Sp -group Py of C G (y ) such that CN (y) = Py CN (P) and P = Py n CN (P). The similarity of (4.2) and (V.7.1) is evident. LEMMA 4.3. Suppose that (4.2) is satisfied. Let () and characters of P such that () vanishes outside U. Then

77

be generalized

where x ranges over a cross section of CN (P) in N. PROOF. The assumptions of (3.6) are satisfied. If y E U then A (C G (y» = 1 Py 1 by (lV.4. 12), (2.4) and (4.2) (ii). If w is defined as in (3.6) then w - I = 1 CN (P) : p i by (4.2) (iv). The result now follows from (3.6). D The following hypothesis is also relevant. HYPOTHESIS 4.4. (i) P is an abelian Sp -group of G with 1 P 1 � 1 .

4]

461

SOME COMPUTATIONS WITH COLUMNS

(ii) N G (P)/C G (P) is cyclic of order m with 1 < m < 1 P 1 - 1 . (iii) If y E P - { I } then C G (y ) n N G (P) = C G (P). LEMMA 4.5. If (4.4) is satisfied then so is (4.2) with U = P - {I} and N = N G (P). PROOF. Suppose that (4.4) is satisfied. Then (4.2) (i) and (4.2) (iv) are immediate and (4.2) (iii) is a standard property of abelian Sylow groups. Burnside's transfer theorem shows that (4.4) (iii) implies (4.2) (ii). D If (4.4) is satisfied the following notation will be used. r = ( l p l - 1 )/m . ljio = h, ljil , . . . is the set of all irreducible characters of P. Thus each ljij for j > 0 has exactly m conjugates under the action of N = N G (P). The notation will be chosen so that {ljij / 1 ::;:;; j ::;:;; r} is a complete set of representatives of the orbits of N. By definition a ( ljij) = a( ljij ) for all j and x E N. THEOREM 4.6; Suppose that (4.4) is satisfied. There exists an integer s ::;:;; m such that the principal p-block Bo of G contains exactly r + s irreducible characters (1 , . . . , ( XI = 1 G, • • • , Xs . If Y is the p -part of z in G and

y E P - {I} then

r ,

(j (z ) = d + 8 L lji j ( y ) 1 � j ::;:;; r, x

Xu (z ) = au

1 � u ::;:;; s,

where x ranges over a cross section of C G (P) in N ; 8 = and au � O. Moreover

± 1;

d, a

(d - 8 f + (r - 1) d 2 + ! a ;, = m + 1 . u =1 If furthermore t is a p i-element then (j (t) = ( I (t) for all j and (rd - 8 )(j (t) + ! auXu (t) = O. 1

I,

•

•

•

, as E Z (4.7)

(4.8)

PROOF. Let bj = a(ljio - ljij ) = a (ljio) - a(ljij ). By (4.3) and (4.5) (bi , bj ) = m + 8ij • Let Ci = bi - bl . Then (Ci , cj ) = 1 + 8ij • The result now follows from a standard argument in character theory. See e.g. Feit [ 1 967c]. 0

CHAPTER XII

462

[5

Observe that in case P is cyclic (4.6) is a very special case of the results in Chapter VII.

COROLLARY 4.9.

Suppose that (4.4) is satisfied. (i) If m = 2 then the notation may be chosen so that in (4.6) d = 0, s = 2, a2 = 0 and �j ( 1) = X2(l) + D. (ii) If m = 3 then the notation may be chosen so that in (4.6) d = 0, s = 3 and au = ± 1 for all u. If furthermore G contains an involution w hich is in no proper normal subgroup then exactly two of 0, a2, a3 are equal to - 1 and �j (1)X2(1)X3(1) is the square of a rational integer.

PROOF. 1 d 1 � 1 by (4.7) as m � 3 . Hence either d = 0 or d = ° by (4.7). Suppose that d = D. If s = 1 then (4.8) implies that (r - 1 )0�j (1) = - 1 and so �j (1) = 1 which is not the case. Thus s ;?: 2. As m < 1 P 1 - 1 it follows that r ;?: 2. Hence (4.7) implies that 2 � r � m + 2 - s � m. Thus r = 2 if m = 2. If m = 3 then 2 � r � 3 and so 1 P 1 = 5 or 7. Thus 1 p i = 7 as m I I 'p 1 - 1 . Therefore r = 2 in any case. Hence i f j ' i- j then (4.6) implies that Thus the notation may be chosen so that d = ° in all cases. (i) By (4.7) s = 2 and a2 = ± 1 . Thus (4.8) implies that - O�j (1) + 1 + a2X2 ( 1 ) = 0. Hence �j (1) = oa2X2(1) +0 and so oa2 > 0. (ii) By (4.7) s = 3 and a u = ± 1 for all u . Suppose that G contains an involution which is in no proper normal subgroup. Let () = {(t/lo - t/llt } P. By (3 .7) exactly two of 0, a2, a3 are equal to - 1. Let V be a 4-dimensional vector space over Q with diagonal quadratic form

5]

GROUPS WITH AN ABELIAN 52-GROUP OF TYPE (2"', 2m )

Another example comes from are 6, 10, and 1 5 .

463

A7 for p = 7. The corresponding degrees

5. Groups with a n abelian Sr group o f type (21n , 2 m )

This section contains a proof of the following result. See Brauer [ 1 964b] Section VI. 5 . 1 . Let G be a group w hose S2-groUP P is abelian of type (2m , 2 m ) ith m ;?: 2. If 02, ( G ) = ( 1 ) then P <J G, Co (P) = P and 1 G : P / is 1 or 3. w

THEOREM

PROOF. The proof is by induction on / G / . Let G be a counterexample of minimum order. Then 02, ( G ) = (1). Let C Co (P), N = No (P). Let t l , t2, t3 be the involutions in P. If N = C then Burnside's transfer theorem implies the result. Since N/C acts faithfully on P by conjugation it follows that / N : C / = 3 and N/C permutes { tl , t2, t3} cyclically. If y E P - {1} then Co (y ) has a normal 2-complement by induction. Thus (IV.4. 12) implies that =

A (Co (y » = A (P) = 1 p i

=

22 m

for y E P - {1}.

(5 .2)

Let t/lo = 1 , t/l 1 , . . . be all the irreducible characters of P. Let r 2 1 ( 2 m - 1) ;?: 5. Then the notation may be chosen so that t/lo, . . . , t/lr is a complete set of representatives of the orbits of N on {t/li }. Let U = P - {1} in (3.6) then w = 3 / 1 N : P I . Hence (3.6) implies that

(a(t/li - t/lo), a(t/lj »

= Oij

(a(t/li - t/lo), a(t/lo»

= -3

=

for 1 � i, j � r, for 1 � i � r.

Define bi = a( t/li - t/lo) = a( t/I; ) - a( t/lo) for . 1 � i � r. Then

( bi , bj ) By (3 . 7) V has a 2-dimensional isotropic subspace and so V is the direct sum of two hyperbolic planes. Thus the discriminant of the form, which is equal to � 1 (1 )X2(I)X3(1 ) is a square in Q. 0 The situation described in (4. 1 0 ) occurs infinitely often. For instance, let G 1 ( q ) = PSL3(q ) and G - I(q ) = PSU3(q ). Then for E = ± 1 , G" (q ) contains a subgroup P x H which satisfies (4.4) of order (q 2 + Eq + 1 )/k, where 2 2 k = (q + Eq + 1 , 3). The degrees of the given characters are q\ q + Eq and 2 (q - 1) (q

-

E

).

Let Ci

=

=

3 + Oij

1 � i, j � r.

(5.3)

bi - br for 1 � i � r - 1. Then ( c i , Cj )

=

1 + Oij

for 1 � i, j � r - 1 .

A standard argument (see e.g. Feit [ 1 967c], §23) implies that the nonzero coefficients of the r - 1 columns C i appear in r rows and if the rows are suitably arranged, the matrix in the first (r - 1) of the rows is EI with E = ± 1 while all coefficients in the rth row are - E. By (5 .3) ( br, ci ) = - 1. Thus b r has the same coefficient b in each of the first r - 1 rows and the coefficient b + E in the r th row. There occur further

[5

CHAPTER XII

464

rows in which all the coefficients of (" . . . , Cr - l vanish. If the coefficients of br in these rows are 00, 0 " . . . then (5.3 ) implies that (5 .4) For the row corresponding to XO = 1 we have ao(t/Jo) = 1, aO(t/Ji ) = ° for i > 1. Hence the coefficient of br in this row is - 1, while the coefficient of each Cj vanishes. This shows that one OJ has the value - 1. Thus b = ° in (5.4) as r - 1 � 4. Thus if the rows are arranged suitably the nonzero coefficients of the columns b l , b2 , , br appear as follows 00, 00, . . . , 00 •

•

•

£ 0 ,...,0 0" £

°

°

£

°

(5.5)

where 00 = - 1 , XO = 1 and Xi denotes the irreducible character corre­ sponding to the (i + 1 )st row. In particular ai (t/Jj ) - ai (t/Jo) = OJ for i = 0, 1 , 2 and j = 1, . . . , r. Hence ai (t/J) - aj (t/Jo) = Oi for every nonprincipal irreducible character t/J of P and i = 0, 1, 2. It follows from (3.5), the definition of a-j (t/J), that if y E P then for ° � i � 2

Xi ( y ) = L ai (t/J )t/J( y ) . l'

Thus if

y E P - { I}, ° � i � 2 then Xi ( Y ) = L { ai (t/J) - ai (t/JO)}t/J( y ) = Oi L t/J( y ) = - Oi . �

���

(5 .6)

« Xi )p, 1) is an integer this implies that (5.7) Xi (1) - Oi (mod 22 m ) , i = 1 , 2. Choose t/J, as a nonprincip al character of P with t/J � = t/Jo . Then all elements of order less than 2 m belong to the kernel of t/J I and so

Since

==

I � I �p Xi (y ) (t/JI ( Y ) - t/Jo( Y )) = � I �S Xi ( y ) (t/J I ( y ) - t/JO ( Y )) ' I

ai (t/JI - t/lo) =

5]

BLOCKS WITH SPECIAL DEFECT GROUPS

465

where S is the set of all elements in P of order 2 m . As P is abelian no element in S is conjugate to its inverse in G and so in particular no element in S is inverted by a conjugate of tl • Hence (3 . 7) may be applied with U = S and () = t/JI - t/Jo . By (5.5) and (5.6) with y = t = t,

£X3(tt 02 01 XI(I) + X2 ( 1 ) + X3(1) - 0, - 1 + o I XI ( l ) + 02 X2 ( 1 ) + £X3(1) = 0, - 3 + £X3(t) = 0.

-1+

_

(5.8) (5. 9) (5. 10)

Suppose that XI ( l ) � 1 and X2 (1) � 1 . By (5.7) XI ( I ) � 15 and X2 (1) � 15. Hence (5 .8) and (5 . 10) imply that 9£ /X3(1) � 13/15. Thus £ = 1 and X3(1) � 10. By (5.9) X3(1) == 3 (mod 16) and so XlI ) = 3 contrary to (5.8). Hence Xi (1) = 1 for i = 1 o r 2. By (5.7) Oi = - 1 and s o P is in the kernel H of Xi by (5 .6). Thus H is a proper normal subgroup of G and 02,(H) � 02, (G) = (1 > and so by induction P <j H. Hence P <j G. The remaining statements are immediate as O2,( G) = (1). 0 6. Blocks with especial defect groups

Given a block B with a defect group D it is of interest to obtain information concerning the irreducible characters and irreducible Brauer characters in B, the decomposition matrix, the Cartan matrix and perhaps also the structure of certain modules and their sources. In case D is cyclic this situation has been discussed in Chapter VII. If D is noncyclic then virtually nothing is known for p odd, but much work has been done in case D is a special type of 2-group. In this sort of work the principal block is usually much easier to handle than a general block. The results of section 1 show, that given D, there are only a finite number of possibilities for various numbers attached to B. Unfortunately the number of possibilities can be very large even if D is not too complicated. Brauer [ 1 964b] was the first to make such investigations. He also used these to get information about the structure of the group G as in the previous section. The remainder of this chapter contains some further results of this sort . We will here only mention some of the literature on this subject. If B is the principal block and D is dihedral (including the 4-group as a special case) see Brauer [1964b], [ 1966a] ; Landrock [1976] ; Erdmann [ 1977b] . For the general case of D dihedral see Brauer [1971a] , [ 1 974b] ; Erdmann and Michler [ 1 977] ; Donovan [ 1 979] . For quasi-dihedral,

466

[6

CHAPTER XII

quaternion and similar 2-groups s_e e Brauer [1966a] ; Olsson [1975], [1 97 7] ; Erdmann [ 1 979b] ; Kiilshammer [ 1 980] . Olsson [1977] also has some results for p "1 2. By using the classification of simple groups with an abelian S2-grouP, questions about the principal block in case D is abelian are reduced to the study of groups of Ree type. For the study of these groups see Fong [1974] , Landrock and Michler [ 1 980a] , [1980b] . For the case of I D 1 = 8 and B arbitrary see Landrock [ 1 981 ] . In this section we will only prove some very elementary results which will be needed for the next section. See Brauer [ 1 964b] .

LEMMA 6.1. Let G have a S2-grouP P which is abelian of type (2, 2). Then either G has three or one conjugate class of involutions. In the former case G has a normal 2-complement and the irreducible characters in the principal 2-block Bo of G are the four characters of degree 1 of G /02, ( G ) = P. In the latter case the principal 2-block B o contains exactly four irreducible charac ­ ters Xo = 1, Xl , X2, XCI. For u = 1, 2, 3 there exists Cu = ± 1 such that Xu (y ) = Cu for every 2-singular element y in G. Furthermore Xu (1) == .su (mod 4) for u = 1, 2, 3.

PROOF. If No (P) Co (P) then G has a normal 2-complement by Burn­

1=

side's transfer theorem and the result follows. Suppose that No (P) "I Co (P). Then I No (P) : Co (P) 3 and G has only one conjugate class of involutions. Let t be an involution in G. Then Co (t) has a normal 2"'complement and so the principal Brauer character is the unique irreduc­ ible Brauer character in the principal block of Co (t). By (V.6.2) Xu ( y ) d �o for all u. By (IV.6.2) L;=o I d �o l 2 = c� = 4. Since Xu does not vanish on all 2-singular elements d �o "I 0 for all u. As d �o is rational for all u and d� = 1, it follows that Bo contains exactly 4 irreducible characters. Furthermore Cu = d �o = ± 1 for all u. As LyEP Xu (y ) == 0 (mod 4), (6.2) holds. Equation (6.3) follows from (IV.6.3) (ii). 0

=

=

COROLLARY 6.4. Suppose that G has exactly one class of involutions in (6.1). Then the notation can be chosen so that CI = 1, C3 - 1 . In that case 1 , XI , C2X2 is a basic set for Bo and the decomposition matrix and Carta'n matrix with respect to this basis are as follows.

GROUPS WITH A QUATERNION 52-GROUP

D =

(� n 0 1 0

c

1

=

1

c D 2 1

467

= (1 + &, ).

PROOF. Let x = 1 in (6.3) then the notation can be chosen so that C I = 1 and C3 = - 1 . By (6.3) 1 , Xl , C2X 2 is a basic set and D has the required form. Thus C = D 'D as required. 0 It is much more difficult to compute the decomposition matrix and Cartan matrix with respect to the basic set consisting of the irreducible Brauer characters. Landrock [1976] has done this and has shown that in case G has exactly one class of involutions in (6. 1) there are two possibilities as follows .

(� = (}

D =

(6.2) (6.3)

1 + c , X ,(x ) + C2X2 (X ) + C 3X 3(X ) = 0 for any 2'-element x in G.

=

7]

D

0 1 1 0 0 1

1

0

�), ,�)

c

=

1

c D· 2 1

(6.5)

2 C= 2 2 2 1

(6.6)

=

C D

Both of these cases occur infinitely often. A direct computation shows that if q is a prime power and G P SL2 (q ) then (6.5) occurs in case q == 3 (mod 8) and (6.6) occurs in case q == 5 (mod 8). If q == 3 c (mod 8) with C 1 the_n

=±

Xo(1) = 1, X I( 1 ) = q, X2 (1) = X 3 (1) = ! (q - .s ) .

If 'Po, 'P I , 'P 2 are the irreducible Brauer characters in Bo then 'Po(1) = 1 and

'P I (1) 'P2(1) = ! (q - 1). =

In either case Bo is the unique 2-block of G of defect 2 and so contains all the irreducible characters of odd degree. 7. Groups with a quaternion Sz- group

=

THEOREM 7. 1. Suppose that 02, ( G ) (1) and a Sz-group T of G is a (generalized) quaternion group. Then the center of G has order 2.

468

CHAPTER XII

[7

Brauer and Suzuki [1 959] first proved (7.1). Proofs of (7. 1) can also be found in Suzuki [1959] and Brauer [1964b]. All of these proofs depend on the theory of modular representations. It has been known for some time that if I T I > 8 then one can give an elementary proof using only the theory of ordinary characters. See e.g. Feit [ 1967c] Section 30. More recently Glauberman [ 1 974] has given a proof in case I T 1 = 8 which uses only ordinary characters. The proof given here is that in Brauer [ 1 964b] . It uses results proved earlier in this chapter. See also Dade [ 197 1 b] for the results in this section and the next. PROOF OF (7. 1). The proof is by induction on I G I . Let 1 T 1 = 2 n + ' with

T = (y, z I y 2n = 1, y 2n - 1 = Z 2 , z - ' yz = y - ' ) .

Let t = y 2n - 1 = Z 2 be the unique involution in T. Suppose that it has been shown that t is contained in a proper normal subgroup H of G. A Sz-group To of H contains only one involution and so is either cyclic or a quaternion group. Furthermore 0 2, (H) � 02 ,(G) = (I). If To is cyclic then H has a normal 2-complement and so To 1 by H = (u, x 1 u 2 = X n = 1, uxu = X - I ). Thus the noncyclic group of order 4 is the dihedral group of order 4 . LEMMA 8.2. Let u -I v be involutions in G. ' Let H = (u, v ). Then the following hold. (i) I H (uv ) I = 2 and H is a dihedral group. (ii) A Sr group of H either has order 2 or is a dihedral group. (iii) If 4 ,r I H I then u is conjugate to v in H. (iv) If 4 1 I H I then I H : H' I = 4 and u is not conjugate to v in H. (v) If 4 1 I H I then either I H I = 4 or I Z(H) I = 2 and u, v are both not in Z(H). :

PROOF. (i) This follows as u - I (uv )u = VU = (UV f l .

(ii) Let x = uv and let x have order mn where m is odd and n is a power of 2. Then (u, x m ) is a Srgroup of H. As ux m u = x -m it follows that either x m = 1 or (u, x m ) is a dihedral group. (iii) As 4 ,r I H I , (u ) and (v ) are Srgroups of H and so are conjugate in H. Since u, v is the unique element of ( u ), (v ) respectively of order 2 u is conjugate to v. (iv) (UV )2 = uvu - I V - I E H'. As 4 1 I H I it follows that I H : « uv )2 ) 1 = 4. Since « UV )2) <J H this implies that « UV )2 ) = H'. Hence I H : H' 1 = 4. Fur­ thermore uH' -I vH' and so v -I Y - I uy for any y E H. (v) Let x = uv. Then x has order 2m for some integer m > O. Since U - I X k U = x - k the only power of x in the center of H is the involution x l» . The result follows easily. 0 The next two results handle the simple implications in (8.1). LEMMA 8.3. Conditions (i) and (ii) of (8. 1) are equivalent. PROOF. (i) => (ii). Suppose that x - I tx E T then X - I txt E T and x - I txt has odd order. Thus x - I txt = 1 and so x - I tx = t - I = t. (ii) => (i). Let x E G and let s = x - I tx. Suppose that st = x - I txt has even order. Thus 4 1 I (s, t) I . By (8.2) (v) there exists an involution z -I s, t with z in the center of (s, t). Let D be a Srgroup of (s, t) which contains t. Then z E D and D is a dihedral group by (8.2) (ii). There exists y E (s, t) with y - I sy E D. By (8. 1) (iv) y - I sy -l t. Choose x E G with x - I Dx c T. Then x - I tx E T and x - I (y - I sy )x E T. Hence by assumption X� I tx = t = X - I (y - I sy )x and so t = Y - I sy contrary to what has been shown above. 0

8]

THE Z*-THEOREM

473

LEMMA 8.4. Condition (iii) of (8. 1) implies conditions (i) and (ii). PROOF. Clearly (8. 1) (iii) implies (8.1) (i). The result follows from (8.3). 0

In view of (8.3) and (8.4) the proof of (8. 1) will be complete once it is shown that (8. 1) (i) implies (8.1) (iii). This,will be done in a series of lemmas.

Throughout the rest of this section G is a minimal counterexample to the assertion that (8. 1 ) (i) implies (8. 1 ) (iii). Thus there exists an involution t E G with tE Z*(G) such that x - I txt has odd order for every x in G. Observe that every subgroup of G that contains t satisfies (8. 1) (i) and so does every factor group G /H where t E H. LEMMA 8.5. If H <J G then 02,(H) = (1). PROOF. The minimality of G implies that O2,( G) = (1). Hence Oz,(H) � O2,( G) = (1). 0

LEMMA 8.6. t is in the center of any 2-group which contains t. PROOF. Let D be a 2-group with t E D. If x E D then x - I txt E D and

X - I txt has odd order. Hence X - I txt = 1.

0

LEMMA 8.7. T contains an involution distinct from t. PROOF. If the result is false then T is either cyclic or a quaternion group. If

T is cyclic then G has a normal 2-complement. Thus G = T by (8.5) and the result is trivial. If T is a quaternion group then t E Z*( G) by (7. 1). 0 LEMMA 8.8. Let X be an irreducible character in the principal 2-block of G. Let s be an involution in G which is not conjugate to t. Then there exists a conjugate So of s with X (ts) = x (tso) and So E Co (t). PROOF. By (8.2) (iii) st has even order. Let z be the involution which is a power of st. Let To be a Srgroup of (s, t) with t E To. By (8.2) (ii) To is a dihedral group. By (8.6) z, t are both in the center of To. Thus I To 1 = 4 by (8.2) (v). Let So be a conjugate of s in (s, t) which lies in To . Then So -I z, t, So E Co (t) and z = tso.

474

CHAPTER XII

[8

Since st = yz = zy where y has odd order it follows from (V.6.3) applied to Co (x ) that

X (ts) = X (zy ) = X (z ) = X (tso). 0 LEMMA 8.9. Let s be an involution in G which is not conjugate to t. Let s ', t'

be conjugates of s, t respectively in G. Let X be an irreducible character in the principal 2-block of G. Then x (ts) = x (t's '). PROOF. If t' = x - I tx then t's ' = x - 1 (txs 'X - 1 )x. Thus it may be assumed that t' = t as X is a class function on G. By (8. 8) there exist conjugates so, sb of s in Co (t) such that

x(ts) = x (tSo) , X (ts ') = x (tsb). Let x - 'sbx = so. Then t, x - 1 tx are both in

(8. 10)

Co (so). By assumption x - 1 txt has odd order and so by (8.2) (iii) there exists y E (t, X - I tx ) � Co (so) with y -l ty = x - 1 tx. Thus yx - I E Co (t). Therefore

(yx - I r l(tso) (yx - I ) = t(yx - 1 r 1 so(yx - I ) = txsoX - I = tsf) . LEMMA 8. 1 1 . Let s be an involution in T with s "l t. Let X be an irreducible

character in the principal 2-block of G with X "I I. If X (s ) "I 0 then X (t) = - X (I). PROOF. By (8.3) s is not conjugate to t in G. Let Co, CI , . . . be all the conjugate classes of G where Co = {I}, t E C1, S E C2 • Let C = �X ECi X in the complex group algebra of G. Let C\ (;2 = �ai�j . Thus (8. 12) Let Xj E G for all j. If w is the central character corresponding to X then w ( G ) = / G / X (xj )IX(I). Thus

X (I)

I Cj I x(xj ) X (I) = Lj a . X (I) . ]

For x E G define a (x ) =·X (x )IX (I). By (8. 9) X (Xj ) = X (ts) whenever aj "l O. Hence

/ C 1 / a (t) / C2 / a (s ) = Lj aj / G / a (ts ) = a (ts ) Lj aj / G / ·

THE Z*-THEOREM

475

Thus (8. 12) implies that

a (t) a (s) = a (ts). Since t E Z(T) by (8.6), ts is an involution in T with ts "l t. The argument above applied to ts yields that a (t)a (ts ) = a (s ). Hence a (t)2 a (s ) = a (t)a (ts ) = a (s). Now suppose that X (s ) "l O and x "l 1 . Then a (s ) "l O and s o a (t? = 1. Hence a (t) = ± 1 . Thus X (t ) = ± X (I). Suppose that x (t) = X (I). Then t is i n the kernel H o f X. A s X "I I, H"I G. Thus the minimality o f G implies that t E Z*(H). By (8.5) Z*(H) = Z(H) and so t E Z(H). Let x E G. Then x - 1 tx is an involution in H. Therefore (x - 1 txt? = 1 . Thus x -1 txt = 1 as X - I txt has odd order. Consequently t E Z(G). 0 PROOF OF ( 8 . 1 ). By (8.7) T contains an involution s "l t. By (8.3) s is not conjugate to t in G. By (IV.6.3)

0 = L Xu (s)Xu (t) = L Xu (s )Xu (I) u u

Hence x (tso) = x (tsb) and the result follows from ( 8 . 10). 0

I C1 I x (t) I C2 I x( s )

8]

where Xu ranges over all the irreducible characters in the principal 2-block . of G. Thus (8. 1 1) implies that if Xo = 1 then

2 = Xo(s ) (Xo(t) + 1) = L Xu (s ) (Xu (t) + Xu (1» = L Xu (s )Xu (t) + L Xu (s )Xu (1) = O. u u This contradiction completes the proof. 0

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SUBJECT INDEX

S UBJECT INDEX

completely primary, 33 component, 2 composition series, 3 equivalent composition series, 3 constituent, 3 irreducible constituent, 3 cover, 1 69, 249 cross section, 79

Algebra, 3 finitely generated, 3 free, 3 Frobenius, 49 R-algebra, 3 serial, 58 sym metric, 49 uniserial, 58 Alperin-McKay conjectures, 171 annihilator, 17 Artin-Wedderburn Theorem , 26, 27 ascending chain condition, 5

decomposition matrix, 67 decomposition numbers, 67 for a basic set, 148 defect, 1 24, 1 26, 1 27 defect group, 124, 126, 1 27 deficiency class, 246 descending chain condition, 5 dual basis, 47

BO is defined, 1 36 basic set, 1 48 basis, 1 block, 23 block pair, 207 extend, 207 properly extend, 207 weakly extend, 207 Brauer character, 142 Brauer corresponden ce, 136 Brauer graph, 300 Brauer homomorphism, 1 29 Brauer mapping, 129 Brauer tree, 301, 305

elementary subgroup, 1 4 1 exceptional character, 277, 279 extension, 69 finite extension, 69 unramified extension, 69 First Main Theorem on blocks, 1 37 Fitting's Lemma, 34 Fong-Swan Theorem, 419 Frobenius reciprocity, 99

canonical character, 205 Cartan invariants, 55 for a basic set, 148 Cartan matrix, 55 central character, 54 character, 141 Clifford's Theorem, 101 coherent, 224

generalized character, 141 germ, 209 Green algebra, 92 Green correspondence, 1 1 3 Grothendieck algebra, 92 group algebra, 4 500

Hall-Higman Theorem B, 309 H-conjugate, 1 23 height, 1 5 1 Hensel's Lemma, 40 higher decomposition number, 1 72 with respect to a basic set, 45 1 Higman's Theorem , 89 idempotent, 1 centrally primitive, 1 primitive, 1 inertia group of a block, 195 of a characte r, 1 95 of a module, 86 inertial index, 235, 272 intertwining number, 53 invariant, 82 inverse endomorphism ring, 4 Jacobson radical, 1 8 Jordan-Holder Theorem, 3 kernel, 86 kernel of a block, 154 Krull-Schmidt Theorem , 37 linked idempotents, 45 local ring, 33 lower defect group, 241 associated to a section, 243 Mackey decomposition, 85 Mackey tensor product theorem, 85 major subsection, 230 Maschke's Theorem , 9 1 McKay conjecture, see Alperin-McKay metric complete, 3 1 complete o n modules, 31 defined on a module, 29 equivalent metrics, 29 module absolutely indecomposable, 72 absolutely irreducible, 7 1 A-faithful, 1 7 algebraic, 93 Artinian, 6 B-projective, 1 1 com patible endo-permutation, 409 complete, 29 completely reducible, 15

decomposable, 2 dual, 46 endo-permutation, 407 endo-trivial, 407 faithful, 86 finitely generated, 1 free, 2 .s)-projective, 93 indecomposable, 2 induced, 80 injective, 50 irreducible, 2 irreducibly generated, 95 left, 1 left Artinian, 6 left Noetherian, 6 left regular, 2 Noetherian, 6 of quadratic type, 188 of symplectic type, 1 88 periodic, 96 permutation, 405 principal indecomposable, 42 principal series, 427 projective, 8 reducible, 2 regular, 2 relatively injective, 50 relatively projective, 1 1 serial, 58 torsion free, 64, 65 transitive permutation, 405 two-sided, 5 uniserial, 58 multiplicity of a lower defect group, 243 Nakayama's Lemma, 31 Nakayama relations, 99 nil ideal, 19 nilpotent ideal, 19 nonsingular element in Hom R (A, R ), 49 normal series, 3 factors of, 3 proper refinement of, 3 refinement of, 3 without repetition, 3 orthogonal idempotents, 1 p-conjugate characters, 1 77 P-defective, 1 24 p-radical group, 266

501

502

SUBJECT INDEX

p-rational character, 1 78 p-section, 1 72 1T-height, 227 1T-section, 2 1 6 primitive ideal, 1 7 principal block, 1 54 Brauer character, 1 54 character, 154 projective resolution, 95 pure submodule, 64, 65 quasi-regular, 18 radical of a module, 1 6 o f a ring, 1 8 ramification index, 69 of a module, 101 real stem, 307 regular block, 1 99 representation, 74 equivalent representations, 74 representation algebra, 92 representation group of a character, 4 1 4 Reynold ' s ideal, 258 Schanuel's Lemma, 9 Schreier's Theorem, 3 Schur index, 1 85

Schur's Lem ma, 23 Second Main Theorem on blocks, 172 section, 1 72 semi-simple, 1 9 socle, 1 6 source, 1 13 splitting field of an algebra, 53 of a module, 53 subsection, 230 with respect to a basic set, 451 symmetric element in Hom R (A, R), 49 Third Main Theorem on blocks, 207 trace, 87 relative trace, 87 trace function, 74 type of block, 453 for an element, 453 for a subsection, 453 same, 453 Type L2(p), 347 underlying module, 74 unique decomposition property, 37 vertex, 1 12 weakly regular block, 1 99 width, 252