North-Holland Mathematical Library Board of Advisory Editors:
M. Artin, H. Bass, J. EelIs, W. Feit, P. J. Freyd, F. W. Gehring, H. Halberstam, L. V. Hiirmander, M. Kac, J. H. B. Kemperman, H. A. Lauwerier, W. A. J. Luxemburg, F. P. Peterson, I. M. Singer and A. C. Zaanen
VOLUME 25
NORTH-HOLLAND PUBLISHING COMPANY AMSTERDAM • NEW YORK • OXFORD
The Representation Theory of
Finite Groups Walter FEIT Yale University New Haven, CT 06520 U.S.A.
1982
NORTH-HOLLAND PUBLISHING COMPANY AMSTERDAM • NEW YORK • OXFORD
C) 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
Published by:
North-Holland Publishing Company — Amsterdam • New York • Oxford
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 library) Bibliography: p. Includes index. 1. Modular representations of groups. groups. I. Title. QA171.F36 512' .22 ISBN 0-444-86155-6
PRINTED IN THE NETHERLANDS
2. Finite
80-29622
To 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 representations 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 representations 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 [1935], 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 he had established the basic properties of projective modules he avoided modules and representations 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. Thompson [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 I 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 1969. 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
ix
PREFACE
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. (IV.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 0 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 [19424 and may be found in the bibliography. W. Feit
September, 1980
CONTENTS
vii
PREFACE
1
CHAPTER I
1. Preliminaries 2. Module constructions 3. Finiteness conditions 4. Projective and relatively projective modules 5. Complete reducibility 6. The radical 7. Idempotents and blocks 8. Rings of endomorphisms 9. Completeness 10. Local rings 11. Unique decompositions 12. Criteria for lifting idempotents 13. Principal indecomposable modules 14. Duality in algebras 15. Relatively injective modules for algebras 16. Algebras over fields 17. Algebras over complete local domains 18. Extensions of domains 19. Representations and traces
CHAPTER
II
1 4 5 8 15 17 21 23 28 33 36 38 42 46 50 53 63 69 74
79
1. Group algebras 2. Modules over group algebras 3. Relative traces 4. The representation algebra of R[G] 5. Algebraic modules 6. Projective resolutions
79 79 87 92 93 95
xi
xii
CONTENTS
CHAPTER III
1. Basic assumptions and notation 2. F[G] modules
3. Group rings over complete local domains 4. Vertices and sources 5. The Green correspondence 6. Defect groups 7. Brauer homomorphisms 8. R[G X G] modules 9. The Brauer correspondence
CHAPTER IV
1. Characters 2. Brauer characters 3. Orthogonality relations 4. Characters in blocks 5. Some open problems 6. Higher decomposition numbers 7. Central idempotents and characters 8. Some natural mappings 9. Schur indices over Q, 10. The ring A(R[G]) 11. Self dual modules in characteristic 2
CHAPTER
V
1. Some elementary results 2. Inertia groups 3. Blocks and normal subgroups 4. Blocks and quotient groups 5. Properties of the Brauer correspondence 6. Blocks and their germs 7. Isometries
8. 7r-heights 9. Subsections 10. Lower defect groups 11. Groups with a given deficiency class
CHAPTER VI
1. Blocks and extensions of R 2. Radicals and normal subgroups 3. Serial modules and normal subgroups 4. The radical of i[G] 5. The radical of REG] 6. p -radical groups
97 97 98 104 111 115 123 128 131 136
140 140 142 144 149 165 171 178 181 185 186 188
192 192 195 198 202 206 209 215 226 230 240 245
248 248 249 253 256 262 265
CONTENTS
CHAPTER
VII
269
1. Blocks with a cyclic defect group 2. Statements of results 3. Some preliminary results 4. Proofs of (2.1)-(2.10) 5. Proofs of (2.11)-(2.17) in case K = 6. The Brauer tree 7. Proofs of (2.11)-(2.19) 8. Proofs of (2.20)-(2.25) 9. Some properties of the Brauer tree 10. Some consequences 11. Some examples 12. The indecomposable
[G] modules in B
13. Schur indices of irreducible characters in Ê 14. The Brauer tree and field extensions 15. Irreducible modules with a cyclic vertex
340
CHAPTER VIII
1. Groups with a Sylow group of prime order 2. Tensor products of /[1'si] modules 3. Groups of type I-2(P) 4. A characterization of some groups 5. Some consequences of (4.1) 6. Permutations groups of prime degree 7. Characters of degree less than p 1 8. Proof of (7.1) 9. Proof of (7.2) 10. Proof of (7.3) 11. Some properties of permutation groups 12. Permutation groups of degree 2p 13. Characters of degree p -
CHAPTER IX 1. 2. 3. 4.
The structure of A (G) A (G) in case a Se -group of G is cyclic and R is a field Permutation modules Endo-permutation modules for p-groups
CHAPTER
269 273 281 289 290 299 301 302 305 308 317 322 333 336 337
X
1. Groups with a normal p'-subgroup 2. Brauer characters of p-solvable groups 3. Principal indecomposable characters of p -solvable groups 4. Blocks of p -solvable groups 5. Principal series modules for p-solvable groups
340 341 347 351 358 361 365 371 380 381 384 387 391
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
1. 2. 3. 4. 5. 6. 7. 8.
XII
Types of blocks Some properties of the principal block Involutions and blocks Some computations with columns Groups with an abelian S2-group of type (2 - ,2"') Blocks with special defect groups Groups with a quaternion S,-group The Z*-theorem
428 433 440
444 444
450 450 455 456 459 463 465 467 471
BIBLIOGRAPHY
476
INDEX
500
SUBJECT
CHAPTER I
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 will introduce some terminology and conventions 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 y • 1 = y for all y E V. Let A be a ring. An element e in A is an idempotent if e 2 = e/ O. Two idempotents e e2 are orthogonal if e e2 = e 2e = 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 { v,} of V generates V if V = E v,A. V is finitely generated if a finite subset generates V. { y,} is a basis of V if
2
CHAPTER I
El
(i) {v,} generates V. Ey,a, =0 for a, E A then a, =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 A, is defined to be the additive group of A made into an A module by multiplication on the right. The left regular A module A A is defined similarly. Right ideals or left ideals of A will frequently be identified with submodules of AA or left submodules of ,A. We state without proof the following elementary but important results.
(ii) If
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 A,. LEMMA 1.2. A finitely generated free A module has a finite basis. LEMMA 1.3. An A module Vis 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. LEMMA 1.4. If V is an A module then there exists an exact sequence 0—>141 —>u—>V—>0 with U a free A module. If V is finitely generated then there exists such an exact sequence with U finitely generated.
If V = V, el V, for submodules V,, V, then V, or V, is a component of V. If U is isomorphic to a component of V we will write U IV. V is indecomposable if (0) and V are the only components of V. V is decomposable if it is not indecomposable. If m is a nonnegative integer then mV 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 mA A for some nonnegative integer m. Let V/ (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
1]
PRELIMINARIES
3
If (0) C V, C V, C V are A modules then V,/ V, is a constituent of V. If V2/11, is irreducible it is an irreducible constituent of V. A normal series of V is an ordered finite set of submodules (0) = V, C • • • C V„ = V. The factors of the normal series (0) = V0 C • C V„ = V are the modules V,. The normal series (0) = W, C • • C W„, = V is a refinement of the normal series 0 = V, C • • C V„ = V if there exists a set of indices 0 fi < • < j„ m such that 14 = W0 . 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.
THEOREM 1.6 (Schreier). Two normal series of V have equivalent refinements.
1.7 (Jordan—Holder). If V has a composition series then any two composition series are equivalent.
THEOREM
Let R be a commutative ring. A is an R- algebra if (i) A is an R module and A is a ring. (ii) (ab)r = a (br) = (ar)b for all a, b E A, r E R. A is a finitely generated R-algebra if A is an R -algebra and is finitely generated as an R module. A is a free R -algebra if A is an R -algebra and is free as an R module. It follows directly from these definitions that if A is an R -algebra then br = (1 r)b for r E R, b E A. If A is a free R -algebra then the map sending r to 1 • r is an isomorphism of R into the center of A. In this case we will generally identify R with a subring of the center of A. If R is a field then clearly every R -algebra is a free R -algebra.
4
[2
CHAPTER I
Let R be a commutative ring and let G be a finite group. Let
R[G] =rcx x E
where
E r„x ± E s„. = E (r.„ r„x)(
,x)=
E x.yEG
rx syxy =
E(E zEG
rxy -isy ) z.
yEG
Then R[G] is the group algebra of G over R. It is clear that R[G] is a free R -algebra.
Books E. Artin, C. Nesbitt and R. M. Thrall, Rings with Minimum Condition, University of Michigan, 1944. N. Bourbaki, Algèbre, Hermann, Paris. H. Cartan and S. Eilenberg, Homological Algebra, Princeton University, 1956. C. W. Curtis and I. Reiner, Representation Theory of Finite Groups and Associative Algebras, Interscience, 1962. N. Jacobson, Structure of Rings, A.M.S. Colloquium Publications, vol. 37, 1956. O. Zariski and P. Samuel, Commutative Algebra, Van Nostrand, New York, 1958.
2. Module constructions Let A be a ring and let V, W be A modules. HomA ( V, W) is the additive abelian group of all homomorphisms from V to W. The same notation will be used in case V, W are left A modules. Under suitable hypotheses Hom A ( V, W) can be made into an A module or a left A module. The same notation will be used to denote this module. The context should prevent any possible confusion from arising. If f E HomA ( V, W) then the image of y under f will be denoted by fv. If V, W are left A modules this image will be denoted by vf. Thus if y E V, a E A then (fv)a = f(va), a(vf)= (av)f if V is a module, left module
respectively. EA (V) denotes the ring of endomorphisms of V. Thus V is a left EA (V) module. Similarly if V is a left A module then V is an EA (V) module. The ring EA (V) is sometimes called the inverse endomorphism ring of V. Let VA , AV denote the fact that V is an A module, left A module
3)
FINITENESS CONDITIONS
5
respectively. If B is another ring let A Vs denote the fact that V is a left A module and a B module such that (av)b = a(vb) for a E A, b E B, y E V. Under these circumstances V = A VB is a two sided (A, B) module. If A = B then V = A VA is a two sided A module. An ideal of A always means a two-sided ideal of A. It is frequently convenient to identify ideals of A with two-sided submodules of A A A . Let A, B be rings. Let a E A, b E B. Let f EHom A (V, W) and let y E V, w E W then the following hold. LEMMA 2.1.
VA OA A
WB is a B module with (v w)b = v wb.
LEMMA 2.2. B VA OA W is a left B module with b(v w)= by Øw. LEMMA 2.3. Hom A (A VB A W) is a left B module with v(bf)— (vb)f. LEMMA 2.4. Hom A ( B VA, W4 ) is a B module with (fb)v = f(bv). LEMMA 2.5. Hom A ( VA, B WA ) is a left B module with (bf)v = b(fv). LEMMA 2.6. HomA (A V, A Ws ) is a B module with v(f b)= (vf)b.
In particular (2.5) implies LEMMA 2.7. HomA ( VA.AAA ) is a left A module with (af)v = a(fv).
If A is a R -algebra where R is a commutative ring then any A module is both a left and right R module. Thus (2.3) implies LEMMA 2.8. If R is a commutative ring and A is an R-algebra then for any A module V, Hom R(V, R) is a left A module with v(af)= (va)f.
3. Finiteness conditions Let A be a ring and let V be an A module. V satisfies the ascending chain condition or V satisfies A.C.C. if every ascending chain of submodules contains only finitely many distinct ones. V satisfies the descending chain condition or V satisfies D.C.C. if every descending chain of submodules contains only finitely many distinct ones. The next two results are well known.
6
CHAPTER I
[3
3.1. Let V be an A module. The following statements are equivalent. (i) V satisfies A.C.C. (ii) Every nonempty set of submodules of V contains a maximal element. (iii) Every submodule of V is finitely generated.
LEMMA
3.2. Let V be an A module. The following statements are equivalent. (i) V satisfies D.C.C. (ii) Every nonempty set of submodules of V contains a minimal element.
LEMMA
LEMMA
3.3. Let V,, V,, W be submodules of V. If V, + W C V 2 + W and V2,e V.
v1 n wc v2 n w then
Suppose that V2 C V. Choose y E V , v0 V2 . Thus y E V, + W C V 2 + W and so v = v2 + w with v2 E V2, w E W. Hence
PROOF.
— v2 = w E V, n Thus v E
V2
w C V2 n wc V2-
contrary to the choice of v.
E
3.4. Let W be a submodule of V. If W and V I W both satisfy A.C.C. or D.C.C. then so does V and conversely. LEMMA
The converse is clear. Suppose that W and V/ W satisfy A.C.C. (D.C.C.). Let v,} be an ascending (descending) chain of submodules of V. Then { V, n WI and I( V, + W)/ WI are ascending (descending) chains of modules. Hence there are only finitely many distinct modules in fv, n WI and fy+ WI. If V,n1,V=ynWandV, + W= V, + W then by (3.3) V, = V, since {V,} is a chain. Thus {y} contains only finitely many distinct modules. 0 PROOF.
{
The ring A satisfies A.C.C. or A is Noetherian if AA satisfies A.C.C. The ring A satisfies D.C.C. or A is Artinian if AA satisfies D.C.C. Left Noetherian and left Artinian are defined analogously. 3.5. Let V be a finitely generated A module. If A satisfies A.C.C. or D.C.C. so does V. LEMMA
Since mA,,, /A 4 (m — 1)A,,, it follows by induction and (3.4) that a finitely generated free A module satisfies A.C.C. or D.C.C. if A does. The result follows from (1.3) and (3.4). E PROOF.
FINITENESS CONDITIONS
7
Although on occasion it may not be necessary to invoke either chain condition, the main object of this chapter is the study of finitely generated modules over rings which satisfy either A.C.C. or D.C.C. and (3.5) will be used continually. COROLLARY 3.6. Let R be a commutative ring and let A be a finitely generated R -algebra. If R satisfies A.C.C. (D.C.C.) then A satisfies A.C.C. and left A.C.C. (D.C.C. and left D.C.C.). PROOF. Since R is commutative this follows from (3.5). E LEMMA
3.7. V has a composition series if and only if V satisfies both A.C.C.
and D.C.C.
PROOF. Suppose that V satisfies both A.C.C. and D.C.C. The set of submodules which have a composition series contains (0) and so is nonempty. Thus there exists a submodule W maximal among submodules which have a composition series. If W/ V choose U minimal with the property that W C U. Thus U/ W is irreducible and U has a composition series contrary to the choice of W. Thus W = V as required. Suppose that V has a composition series with n factors. By Schreier's Theorem (1.6) any chain of submodules contains at most n distinct ones. Thus V satisfies A.C.C. and D.C.C. E LEMMA 3.8. If V satisfies either A.C.C. or D.C.C. then Vis a direct sum of a finite number of indecomposable modules.
PROOF. Suppose that V satisfies A.C.C. Let { V,} be the set of submodules of V such that V/ V, is not the direct sum of a finite number of indecomposable modules. If the result is false (0) is in this set and hence it is nonempty. Let W be maximal in this set. Thus V/W is decomposable. Hence there exist submodules WI , W2 of V with W C W,, W C W2 and V/ W W,/ W W2/ W. The maximality of W implies that each of WI/ W — V/ W2 and W2/ W — V/ W, is the direct sum of a finite number of indecomposable modules. Thus so is V/ W contrary to the choice of W. Suppose that V satisfies D.C.C. If the result is false choose W minimal among those submodules of V which are not the direct sum of a finite number of indecomposable modules. Then W = W, FED W2 for some / (0), W2 / (0). Hence each of W, and W2 is the direct sum of a finite number of indecomposable modules. Therefore so is W contrary to the choice of W. E
8
CHAPTER I
[4
3.9. Suppose that A satisfies either A.C.C. or D.C.C. Let V be a finitely generated A module. Then V is the direct sum of a finite number of indecomposable modules. COROLLARY
PROOF. Clear by (3.5) and (3.8). E
4. Projective and relatively projective modules Throughout this section A is a ring which satisfies A.C.C. All the results in this section will be stated in terms of finitely generated modules. This is all that will be required in the sequel. However it is well-known that analogous results hold for arbitrary modules even if A does not satisfy any chain condition. A finitely generated A module P is projective if every exact sequence 0—> W—> V—>P—>0 with V, W finitely generated A modules is a split exact sequence. We state without proof the following basic result. THEOREM 4.1. Let P be a finitely generated A module. The following are equivalent. (i) P is projective. (ii) P V for some finitely generated free A module. (iii) Every diagram
U—* V —>C1 with U, V finitely generated A modules in which the row is exact can be completed to a commutative diagram
/ U—* V ---> O. COROLLARY 4.2. If V, V 1 , V2 are finitely generated A modules with V = V, Eel V2 then V is projective if and only if V, and V2 are projective. PROOF. Clear by (4.1) (ii). LI
4]
9
PROJECTIVE AND RELATIVELY PROJECTIVE MODULES
LEMMA 4.3 (Schanuel). Suppose that P, and
P2 are finitely generated projective A modules and the following sequences are exact
0 -->
0 W
f,
--->0 P, -24V —> 0
with W W 2, V finitely generated A modules. Then P i e
W2 --"=. P2 e
W1.
PROOF. Let U be the submodule of P, ED P2 defined by U = ul, u2) (I u = (2u2}.
Define g: U —> P, by g( u,, u) = u,. Thus g is an epimorphism with kernel u2)1 u2 e f2,( W2 ) W2. Hence U P,ED W2 since P, is projective. Similarly U P2 e wi. E j
LEMMA 4.4. Let P be a finitely generated projective A module. Suppose that U, V, W are finitely generated A modules and 0-“J j-> V >W—>0 is exact. Then there is an exact sequence of abelian groups
0—> HomA (P, U)—>Hom A (P, V)—> HomA (P, W)—> O. If A is an R -algebra for some commutative ring R then this is an exact sequence of R modules.
PROOF. For h E HomA (P, V) let s(h) --- th E HomA (P, W). Then s is a group homomorphism and s is an R -homomorphism in case A is an R -algebra. The kernel of s is clearly Hom A (P, f( U)) Hom A (P, U). Since P is projective (4.1) (iii) implies that s is an epimorphism. Let B be a sub ring of A with 1 E B. For any A module V let VB denote the restriction of V to B. Write A B = (A A )s . If V = VB is any B module then by (2.1) VB OS BAA is an A module.
LEMMA 4.5. Let B be a subring of A such that B A is a finitely generated free left B module and A B is a finitely generated free B module. Let V be a finitely generated A module. Then there exists an exact sequence
10
[4
CHAPTER I
0 —>W —It>VB0BAA
-L >V —>0
such that 0-3 WB —> h (VB 1 3:>)BAA) B
VB —> 0
is a split exact sequence. If furthermore VB is a projective B module then is a projective B module. PROOF.
Let {y,,
, y} be a basis of B A. Then
VB Ø BAA = {2
r=1
and E:'=,
v
WB
VI
0 y, v, E
0y 1 = 0 if and only if v, =0 for i = 1, . , n. Define
W={2vi0y,
v iy ; = 0} .
If VB is a projective B module then since AB is free it will follow by (4.2) and the first part of this result that WB is projective. Define f by f (E r;_, v, y,)= viy,. Let h be the identity map on W. If a EA then yia = E,"=, b,,y, for i = 1, . . . , n, 6,1 E B. Thus n
f(E Vi
n
n
ri
= E viyia = E E v,b,iyi = f (E E =
v,
=
v,
yi)
ya) .
Hence f is an A -homomorphism. Clearly W is the kernel of f. Let 1 = co); with c, E B. Define g : VB —> (VB BAA)B by gv = vv, y,. If b E B then (gv)b = (E vc, ( )y,) b = v® co,,b = y ØbE ,=1
= E vbc, y, = g(vb). Thus g is a B-homomorphism. Since fgv = v for v E V it follows that f is an epimorphism, g is a monomorphism and g(VB )n WB = (0). It only remains to show that (VB OB BAA)B = WB ± g(VB). This follows from the fact that if w E (VB OB BAA)B then f(w — gfw)= 0 and so w — gfw E WB while w = (w — gfw)+ gfw.
Let B be a subring of A with 1
E B such that B satisfies A.C.C.
and
BA
PROJECTIVE AND RELATIVELY PROJECIIVE MODULES
4]
11
and AB are finitely generated as left and right B modules respectively. A finitely generated A module P is projective relative to B or simply B-projective if for every pair of finitely generated A modules W, V the exact sequence 0—>W —* h V>P—>0
splits provided 0—> WB
VE
PB —> 0
is a split exact sequence. LEMMA 4.6. Let B be a subring of A with 1 E B such that B satisfies A.C.C., AB is a finitely generated free B module and B A is a finitely generated left B module. Let V be a finitely generated A module. Then V is projective if and only if VB is a projective B module and Vis B -projective.
PROOF. Clear by definition and (4.1). 0 COROLLARY 4.7. Let A be a finitely generated algebra over a field F. Let V be a finitely generated A module. Then V is projective if and only if V is F-projective.
PROOF. Clear by (4.6). E THEOREM 4.8. Let B be a subring of A with 1 E B such that B satisfies A.C.C., AB is a finitely generated free B module and B A is a finitely generated free left B module. Let P be a finitely generated A module. The following are equivalent. (i) P is B-projective. (ii) PIP@ OB BAA • module W. (iii) P W OB BAA for some finitely generated B (iv) Every diagram
g
L
V >
with LI, V finitely generated A modules in which the row is exact can be completed to a commutative diagram
12
[4
CHAPTER I
ig
u
v,o
provided there exists a commutative diagram PB
g
v.
v„
O.
PROOF. (i) » (ii). Clear by (4.5). (iii). Trivial. (ii) (iii) (iv). Suppose that U, V are finitely generated A modules such that the row in the following diagram is exact
g
U -Lo V —0
and
V Ig u,
vB
o
is commutative. Let W OB BAA = P P' and define go , ho by go(u + u') = gu, h o(u + u')= hu for u E P, u'E P'. Let y i, y2, , y„ be a B -basis of B A. Define f: W OBBAA U by f(Y:=, w, ho(w, 01)y, where w, E W. We show that f is an A -homomorphism. Let a E A with y,a = Eç=i1),,y) , b, E B. Then YI
/1
®Yia)-=
i=1 ; =i
PI
wibii
Yi)
n
E E ho(wibii 0 1 ))); i=1;=-1
E ho(w, Ø i)yia E E ho(w, 01)bifyi = i=1
i=11=1
= f(E ®Yi)a•
41
PROJECTIVE AND RELATIVELY PROJECTIVE MODULES
13
Now by definition tf
wi
yi) =
tho (w, 01)y,
=E go(wi 01* = go (E wi ® Y i). Hence if f also denotes the restriction of f to P the diagram
is commutative as required. (iv) (i). Suppose that V, W are finitely generated A modules such that W 2-). V—L> P-->0 is exact and —> W8 ±› V8 .±›
--*0
is a split exact sequence. Hence by assumption there exists a commutative diagram
A where fgw = w for w E P. Thus g is a monomorphism and g(P)nh(W)= (0). If v E V then fv = fgfv. Hence (v — gfv)+ gfv E h(W)+ g(P). Thus V = h(W)(1) g (P).
LI
COROLLARY 4.9. Suppose that A, B satisfy the assumptions of (4.8). Let V, V I V, be finitely generated A modules with V = V i e V,. Then V is ,
B-projective if and only if V, and V, are B-projective.
PROOF. Clear by (4.8) (iii). 0
14
[4
CHAPTER I
4.10. Suppose that A, B satisfy the assumptions of (4.8). Let P be a finitely generated A module such that PB is B-free. The following are equivalent. (i) P is B-projective. (ii) For every pair of finitely generated A modules W, V with W@, V@ finitely generated projective B modules the exact sequence COROLLARY
h 0->W-->
-f->P->0
splits provided
0-> W,
> ->
is a split exact sequence.
PROOF. (i) (ii). Clear. (ii) (i). Since (P CIO B--A A 1,B is a projective B module (4.5) implies that PI P@ 013 BAA • Thus by (4.8) P is B-projective. LI LEMMA 4.11. Suppose that A, B satisfy the assumptions of (4.8). Let S be a subset of B such that AS = SA. If P is an A module then PS is an A module. If, furthermore P is B-projective then PS and P /PS are B-projective. PROOF. By definition PSA = PAS = PS and so PS is an A module. Suppose that P is B-projective. By (4.8) -PB Ali A =P'EDV where P' P. Then P'S (i) VS = (P' (i) V)S = (PR
Thus PS
B AA) S = PS B 'A A .
is B-projective by (4.8). Furthermore P'/P'S V/VS (P' + V)/(P'+ V)S -
PB Ø BAA IPSB
Ø BAA =(PIPS)B BAA•
Hence P/PS = P'/P'S is B-projective by (4.8). E Observe that if R is a commutative ring which satisfies A.C.C., G is a finite group and H is a subgroup of G then A = R[G] and B = R[H] satisfy the hypotheses of (4.5)-(4.11).
5]
COMPLETE REDUCIBILITY
15
5. Complete reducibility
Let A be a ring and let V be an A module. V is completely reducible if every submodule of V is a component of V. 5.1. Let V be completely reducible. Then every homomorphic image of V is isomorphic to a submodule and every submodule of Vis isomorphic to a homomorphie image. LEMMA
PROOF.
If W C V let V = W W'. Then W---- V/ W' and V/ W
W'.LIl
LEMMA 5.2. If V is completely reducible then so is every submodule and every homomorphic image of V.
Let U = V/ W. Let U, C U. Then there exists a submodule V, of V with W C V, such that LI, = VI/W. Choose V2 with V = V, ED V2. Then U = U(i) (V2 ± W)/W. Thus U is completely reducible. The result follows from (5.1). 0
PROOF.
LEMMA 5.3. If V is completely reducible and V / (0) then V contains an irreducible submodule.
Let y E V, v# O. By Zorn's Lemma there exists a submodule W of V maximal with the property that y W. Let V= W W'. If W' is not irreducible then W' = W ED W2 for nonzero submodules WI , W2 of V. Since (WE = W it follows that 1) W43 W, for either i = 1 or i = 2. This contradicts the maximality of W. Thus W' is irreducible. E
PROOF.
w)n(w@w2)
THEOREM 5.4. The following statements are equivalent. (i) V is completely reducible.
(ii) V is the direct sum of irreducible submodules. (iii) V is the sum of irreducible submodules. (ii). Consider the collection of sets of irreducible submodules of V whose sum is direct. By Zom's Lemma there is a maximal element {V,} in this collection. Let W = e V, and let V= W ED W'. If W' (0) then by (5.2) and (5.3) W' contains an irreducible submodule V'. Thus PROOF. (i)
16
CHAPTER I
[5
W + V' = V' ea e V, contrary to the maximality of { V, }. Hence W' = (0) and V = W as required. (ii) (iii). Clear. (iii) (i). Let W be a submodule of V. By Zorn's Lemma there exists a submodule W' of V maximal with the property that W n = (0). Thus W + W' = W e W'. Suppose that W ea W' V. Choose v E V, v W W'. By assumption v = v i + • • • + v„ where v, E y and y is irreducible for i = 1,..., n. Thus v, W W' for some j and so V, n (14, ciP)/ V,. Hence y n (cv cii= (0) as V, is irreducible. Thus W e W' + V, = W W' ED V, and so W n (tv y) = (0) contrary to the maximality of W'. Therefore V = W e W'. I=1
COROLLARY 5.5. Let V
be completely reducible.
The following
are
equivalent. (i) V satisfies A.C.C. (ii) V satisfies D.C.C. (iii) V has a composition series. (iv) V is the direct sum of a finite number of irreducible submodules.
By (3.7) and (5.4), (iv) is clearly equivalent to each of (i), (ii) and E
PROOF.
(iii).
The socle of V is the sum of all the irreducible submodules of V. In case V contains no irreducible submodules we set the socle of V equal to (0). The radical of V is the intersection of all maximal submodules of V. In case V contains no maximal submodule we set the radical of V equal to V. As immediate corollaries of (5.4) we get
COROLLARY 5.6. The socle of V is the maximal completely reducible submodule of V. COROLLARY 5.7. If V is completely reducible then its radical is
(0).
LEMMA 5.8. If AA is completely reducible then every A module is completely reducible. PROOF. Let V be an A module. Let v E V. The,map from AA to vA which sends a to va is a homomorphism. Thus vA is completely reducible by (5.2). Since v E vA this implies that V is in the socle of V. Thus V is equal to its socle and the result follows from (5.6). 0
61
17
THE RADICAL
5.9. Assume that V satisfies D.C.C. Then V is completely reducible if and only if the radical of V is (0).
LEMMA
PROOF. If V is completely reducible then its radical is (0) by (5.7). Assume that the radical of V is (0). By D.C.C. it is possible to choose a module which is minimal in the set of all submodules of V which are the intersection of a finite number of maximal submodules of V. This module is contained in every maximal submodule of V and hence in the radical. Thus it is (0). Consequently (0) = n :.=, W, where each W is maximal in V. Thus V/ W, is irreducible and so e,%, v/w, is completely reducible by (5.4). 1 v/W, be Let f : V —> V/ W, be the natural projection and let f: Vdefined by fy = .. f1y. Then f is a monomorphism. Hence V is isomorphic to a submodule of a completely reducible module. The result follows from (5.2). E LEMMA 5.10. Let V =
ED
V, where each V, is irreducible. Let W be an V, where j ranges over those i with irreducible submodule of V. Then W C W
PROOF. Let w E W, w 0. Then w = L,1 with y, E V1 . If v then the map sending wa into vi a for a E A is a nonzero homomorphism from W = wA to yiA = V,. Since W, V, are irreducible it is an isomorphism. Hence W V,. 0 .
COROLLARY 5.11. LetV=GV=631 W, where each
v, vv, is irreducible.
Let W be an irreducible module. Then EDV = W, where s ranges over all i with V, ---- W and t ranges over all j with W,-- --- W. Furthermore 631 V, is the submodule of V generated by all submodules of V which are isomorphic to W.
PROOF. Clear by (5.10). E
6. The radical Let A be a ring and let V be an A module. The annihilator of V is the set of all a E A such that Va = 0. Clearly the annihilator I of V is an ideal in A and V can be considered to be an A I I module. V is A-faithful if the annihilator of V is (0). An ideal / of A is primitive if the ring A II has an A// -faithful
18
[6
CHAPTER I
irreducible module. Clearly I is primitive if and only if I is the annihilator of an irreducible A module. The Jacobson radical of A or simply the radical of A is the intersection of all primitive ideals in A. This will be denoted by J(A). LEMMA 6.1. V is irreducible if and only if V submodule W of AA.
A A I W for some maximal
PROOF. Clearly A A /W is irreducible. Suppose that V is irreducible. Let E V, v O. Then v E vA. Thus vA is a nonzero submodule of V and so vA = V. Hence V is a homomorphic image of AA. Since V is irreducible the result follows. E LEMMA 6.2. Every maximal right ideal contains a primitive ideal. Every primitive ideal is the intersection of the maximal right ideals containing it.
PROOF. Let M be a maximal right ideal of A. By (6.1) A A /M is irreducible. The annihilator of AA /M is a primitive ideal contained in M Let I be a primitive ideal in A and let V be an irreducible A module whose annihilator is I. For v E V, v 0 let Mu = {a va = 0}. Clearly Mu is a right ideal in A. Since V is irreducible V = vA — AA /Mu for v 0, E V. Thus by (6.1) Mu is a maximal right ideal in A. Furthermore El u.vom, is the annihilator of V.
n
As an immediate corollary to (6.1) and (6.2) we get
COROLLARY 6.3. J(A) is the intersection of all maximal right ideals in A. By (6.3) J(A) consists of all the elements in A which form the radical of the A module A A. An element a in A is right (or left) quasi-regular if 1— a has a right (or left) inverse , a is quasi-regular if a is both right and left quasi-regular. It is easily seen that a is quasi-regular if and only if 1 — a is a unit in A or equivalently 1— a has a unique (two sided) inverse in A. LEMMA 6.4. Let a E A and let N be a right ideal of A. (i) a is right quasi-regularin A if and only if a + b = ab for some b E A. (ii) If every element in N is right quasi-regular then every element in N is quasi-regular.
6]
THE
RADICAL
19
PROOF. (1— a )(1 — b) = 1 if and only if a + b = ab. This proves (i).
If a E N then a +b — ab for some b in A by (i). Thus b is left quasi-regular and b = ab — a E N. Hence b is right quasi-regular in A. Thus 1 — b is a unit in A and (1 — a)=(1—b)-1 . Hence a is quasiregular. El LEMMA 6.5. J(A) consists of quasi-regular elements and contains every right ideal consisting of quasi-regular elements. PROOF. If a E J(A) then 1 = a + (1— a) and so A = J(A)+ (1— a)A. If (1— a )A/ A there exists a maximal right ideal M with (1 — a )A C M. Since J(A)C M by (6.3) this yields that A = M which is not the case. Thus (1 — a)A =A and a is right quasi-regular. Hence a is quasi-regular by
(6.4) (ii). Let N be a right ideal consisting of quasi-regular elements. If N J(A) then by (6.3) there exists a maximal right ideal M such that NZ M. Hence M + N = A. Thus 1 = a + b where a EM, b E N and so a 1—b has an inverse contrary to the fact that a E M and M A. Hence N C J(A). Li The left radical of A is defined analogously to J(A) by using left modules instead of modules. THEOREM 6.6. J(A) contains every right or left ideal consisting of quasiregular elements. Furthermore the following are all equal.
(i) J (A). (ii) The left radical of A. (iii) The intersection of all maximal right ideals in A. (iv) The intersection of all maximal left ideals in A. PROOF. Let J0(A) be the left radical of A. By (6.5) and its left analogue J(A)= Jo(A). The result follows from (6.3) and its left analogue. Li
A is a semi-simple ring if J(A)= (0). LEMMA 6.7. Let I be an ideal of A with I CJ(A). Then J(A11)----- J(A)II. Thus in particular A 1J(A) is semi-simple. PROOF. Immédiate by (6.3). 0
A right (or left) ideal N of A is a nil right (or left) ideal if every element is nilpotent. N is nilpotent if N'' = (0) for some positive integer
CHAPTER I
20
[6
LEMMA 6.8. If N is a nil right or left ideal then N C J(A). PROOF. If a is nilpotent then a = 0 for some positive integer m. Hence (1— a)(1 + a + • • • + 1 and a is quasi-regular. Thus N C J(A) by (6.6). E THEOREM 6.9. Suppose that A satisfies D.C.C. Then J(A) in nilpotent. PROOF. By D.C.C. there exists an integer n 0 such that J(A)" = J(A)"' for all i 0. Let N = J(A)". Assume that N (0). Since N 2 = N/ (0) it is possible by D.C.C. to choose a right ideal M of A minimal with the property that MN/ (0). Thus aN/ (0) for some a E M. aN C M and aNN = aN/ (0). Hence M = aN by the minimality of M. Thus a E aN and so a = ab for some b E N. Thus a (1— b)— 0. Since b is quasiregular this implies that a = 0 contrary to the fact that aN/ (0). Thus J(A)" = N = (0). CI LEMMA
6.10. If A is semi-simple with D.C.C. then every A module is
completely reducible.
PROOF. By (6.3) the radical of AA is (0). Thus by (5.9) A A is completely reducible. The result follows from (5.8). G LEMMA 6.11. Assume that A satisfies D.C.C. Let V be a finitely generated A module. Then V satisfies A.C.C. and D.C.C. PROOF. Let V, = VJ' (A ) for i = 0, 1, . By (6.9) V„ = 0 for some integer n. Since V is finitely generated, V and hence each V, /V,,, satisfies D.C.C. by (3.5). Since J(A) annihilates V, 07,, for each i it follows that V, / V,,, is an AIJ(A) module. Thus by (6.10) each V, / V, ±1 is completely reducible and hence by (5.5) each %NY,' has a composition series. Thus V has a composition series and the result follows from (3.7). COROLLARY 6.12. If A satisfies D.C.C. then A satisfies A.C.C. PROOF. Immediate by (6.11). E
For most of the results in this chapter the assumptions that 1 E A and every module is unital are only a minor convenience. However the reader should be warned that they are essential for the validity of (6.11) and (6.12).
7]
21
IDEMPOTENTS AND BLOCKS
6.13. Suppose that A satisfies A.C.C. and A IJ(A) satisfies D.C.C. Then for every integer m 0 A MAY' satisfies D.C.C.
LEMMA
Induction on m. If m = 1 the result holds by assumption. Since A satisfies A.C.C. J(A)" is finitely generated as a right ideal. Thus J(A)"' MAY"' satisfies D.C.C. since it is a finitely generated AIJ(A) module. Thus if A /J(A) - satisfies D .C.C. so does AIJ(A) - ± I and the induction is established. Ill PROOF.
7. Idempotents and blocks
Let A be a ring. LEMMA 7.1. Let AA = ED • • • el V„ with V,/ (0) for i = 1, . , n. Let 1= e, + • • + e„ with e, E y. Then fe i l is a set of pairwise orthogonal idempotents in A and V, = e,A. Conversely if {e i } is a set of pairwise orthogonal idempotents then (se, )A = e,A.
If a E A then a =l• a = a •1=el a +• • •+ e„a.Thus in particular e, = e,e, +•• + ene,. Hence e,e, = 64e, for all i, j. Furthermore e,A C V, and AA = e i A ED • • e „ A . Hence eiA = V. Conversely let {ei} be a set of orthogonal idempotents. Let e = se,. Thus e 2 = e and ee, = e,e = e, for all i. Then eA = eiA. If Se,a, = 0 then multiplicaton on the left by e, implies that e,a, = 0 for all j. Hence eA .= e,A. LI PROOF.
COROLLARY 7.2. Let e be an idempotent in A. The following are equivalent.
(i) eA is indecomposable. (ii) e is a primitive idempotent. (iii) Ae is an indecomposable left A module.
(ii). This is a direct consequence of (7.1). (i). Let e be a primitive idempotent. Thus A A = (1— e)A e eA. If eA = V 1 6) V2 with V, V2 nonzero then by (7.1) e is not primitive
PROOF. (i)
(ii)
contrary to assumption. An analogous argument shows that (ii) is equivalent to (iii). LEMMA 7.3.
LI
Let A = I, e•••e I„ where each I, is a nonzero ideal of A. Let 1= e,+ • • +en with e, E L. Then {e,} is a set of pairwise orthogonal central idempotents and I = e,A is a ring with unity element e, for each i.
CHAPTER I
22
[7
Conversely if { e,} is a set of pairwise orthogonal central idempotents then (se, )A = e,A where e,A = Ae, is an ideal of A.
PROOF. If a E A then a=1- a= a•l=e,a+•-•+e„a=ae,+•-•+aen.
Hence eia = ae, and eie; = 64e1 for all i, j. Since eiA C /i and A = (13$ e1 4 it follows that eiA = I; and eia = a = aei for all a E L. Conversely by (7.1) (e )A = (13$ ea4 and eiA = Aei since each e, is central. E 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. PROOF. Immediate by (7.3). L LEMMA 7.5. If A satisfies A.C.C. then 1 is a sum of pairwise orthogonal primitive idempotents. 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. E 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) 1= E,n-, e, where e,. , e„) is the set of all centrally primitive idempotents in A. PROOF. (i) By (7.5) 1 = Ef where ffi l is a set of pairwise orthogonal primitive idempotents. Let e be a central idempotent. Then e = Eef and f = ef + (1 — e) f. Thus ef = 0 or ef = f, since f is primitive. Hence e =Ef, where j ranges over those summands with ef = f,. Thus e is the sum of elements in a subset of {f}. This proves (i). (ii) If e i , e2 are centrally primitive idempotents then e l = ele2+ e,(1—e 2). Thus e i = e,e2 or e i e2 = O. Similarly e 2 = e,e2 or e l e 2 = O. Thus either e,e 2 = 0 or e,= e,e2 = e2. (iii) Let { e,, , e„} be the set of all centrally primitive idempotents in A. Then e=E,%,e, is a central idempotent by (ii). If ei 1 then 1 — e is a central idempotent and so by (i) is a sum of centrally primitive idempotents. Since ei (1— e)= 0 for i = 1, n, this is impossible. L
RINGS OF ENDOMORPH1SMS
23
LEMMA 7.7. Assume that A satisfies A.C.C. Then A = Ae, where {e,} is the set of centrally primitive idempotents in A. For each i, Ae, is not the direct sum of two nonzero ideals of A. If A = A where each A, is a nonzero ideal which is not the direct sum of two nonzero ideals (of A or A,) then m = n and A, = Ae, after a suitable rearrangement.
PROOF. This follows directly from (7.3), (7.4) and (7.6). 0 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. THEOREM 7.8. Assume that A satisfies A.C.C. Let V be a finitely generated 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 VEB=B (e) if and only if ye = y for all y E V.
Let { e,} be the set of centrally primitive idempotents in A. By (7.6) 1 = se,. Thus if V is any A module then V = IED Ve,. Hence if V is indecomposable and V# (0) then V = Ve, for a unique value of i. If W is a submodule of V then clearly We, = W, (VI W)e, = Vi W and We, = (0) = (V I W)e, for i j. Clearly ye =v for all v E V= Ve. 0 PROOF.
8. Rings of endomorphisms Let A be a ring. LEMMA 8.1 (Schur). Let V, W be irreducible A modules. Hom, (V, W) = (0) if V W and EA (V) is a division ring.
Then
PROOF. Let f E HomA (V, W). Then either f is an isomorphism or f (V) = (0) since V and W are irreducible. Thus Hom, (V, W)= (0) if V 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. LEMMA 8.2. Let e be an idempotent in A and let V be an A module. Define f :Ve --> Hom,. (eA, V) by f(v)ea = yea. Then f is a group isomorphism. If
24
CHAPTER I
[8
A is an R algebra then Ve and Horn,, (eA, V) are R modules and f is an R-isomorphism. If V = eA then f :eAe —> EA (eA) is a ring isomorphism. A ------ EA (AA). Thus in particular EA (AA)
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. El 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 correspondence 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 C NA n eAe. Since N = Ne and NA n eAe C eAe it follows that NA n eAe = NAe n eAe = NeAe n eAe C 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) imply that eJ(A)e C J(eAe). Let I be a primitive ideal in A and let V be a faithful irreducible Ai/ module. Then Ve is an eAe module. If Ve = (0) then J(eAe)CeAe C i. Suppose that Ve/ (0). Let (0)C W = We C Ve where W is -an eite module. Thus V = WA since V is irreducible. Hence Ve = WeAe C W and so Ve = W. Thus Ve is an irreducible eAe module and so VJ(eAe)= VegeAe)= (0). Hence J(eAe)C I. Since I was an arbitrary primitive ideal in A this yields that J(eAe)C J(A) and so J(eAe)= eJ(eAe)e C eJ(A)e. 0
For any integer n > 0 let A,, be the ring of all n by n matrices with
8] .
RINGS OF ENDOMORPHISMS
25
coefficients in A. In other words A n is an A -free A module with basis Ie„ I i,j = 1, ..., n) where eiy ai,)(E i,j
i,j
ei,bif)=
E eik i,k
con)
This definition is equivalent to the assertion that A n = A Oz Zn where Z denotes the ring of rational integers. It follows easily that for i = 1, , n ec= en and enA„en A. Furthermore A n is also a left A module with ae„ = eqa for a E A. Thus A„ is a two-sided A module. 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 with LEMMA 8.6. Let V be an A module. Assume that V = V i la) » • (1) V„ , n. Then E A (V)— EA (V1)n. V for i, j = 1, ,
PROOF. For i =1, n let ê,, be an isomorphism from V, to V, where 6,, is the identity map. Let "e h be the inverse map sending V, to V. Let 6,1 = e„e„. For i, j = 1, . , n define e,, E EA (V) by e„vs = 6,,(6„vs ) for vs E s = 1, , n. If C EA (V,) define x EE,,( V) by xv, = es ,ie s vs
for s = 1, . . . , n.
(8.7)
The map sending I to x is clearly a monomorphism of EA (V,) into EA (V). Let E denote the image of E A (V) under this map. Let F be the ring generated by E and all e„, i, j = 1, . . . , n. It follows from (8.7) t hat for all j, j and all x E E (e,x)v, = 0 = x(e„v.,) = (xe,i )v.,
for s
(eisx)v., = eiseslie st) = e1xe1 1e,v, = xe,vs . Hence if x 0 xe„ = e yx = 0
for all i, j.
The ring F is an E module generated by e,, i,j = 1, .. then (8.8) yields that for s, t = 1, . . . , n, x i; E E. 0=e
eiix qe„ =
(8.8) n. If e11x11 = 0
26
CHAPTER I
[8
Thus by (8.8) x„ = 0 for all s, t. Hence F is an E-free E module. This implies that F = E. It remains to show that EA (V)C F. Let y E EA (V). Since Eei; = 1 it follows that y = Ei,J eiiye)) . Thus it suffices to show that eiiyei, E F for all i,j. There exists z E EA ( V1 ) such that e„yeb = ei,z. However z = e1 îe 1 for some î E EA (V,). Thus by (8.7) eiiye,i eyce„ E F. E THEOREM 8.9. Assume that V is a completely reducible nonzero A module with a composition series. Let V = ED:1,ER'L, V where each Vi; is irreducible and Vii V„ if and only if i = s. Let D1 = EA O/J O and let = Vi; for all i. Then the following hold. (i) D1 is a division ring for i = 1, . , m. (ii) EA ((J)"' (D1 )„, is a simple ring for i =1, . m. (iii) EA (V) = ED;n=i EA (U,). Every ideal in EA (V) is of the form , m }. EA (U,) where j ranges over a subset of {1, (iv) EA (V) is a semi-simple ring which satisfies D.C.C. and left D.C.C.
ED
PROOF. (i) This follows from (8.1). (ii) EA (VD"' (D)„, by (8.6). It is easily verified that- (Di)„, is a simple ring. (iii) Let ei be the projection of V onto U, for i = 1, .. , m. Thus ei / 0 and ex ) = ke i for all i,j. If x E EA (V) then by (5.11) xe, = eixei for all i. Since Ee1 =1 it follows that x Ei zei . Hence
eix = E epce; = eize i = xe, Thus e, is in the center of EA (V) for all i. Therefore by (7.3) EA (V) = e,EA (V). Clearly e,EA (V) = EA (Us). If / is an ideal of EA (V) then I =1E134:"_, e11 and el is an ideal of e,EA (V). Since e,EA (V) is simple e 11 = (0) or e,EA (V). Hence I = e,EA (V) where j ranges over all i with e,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 IED7=, D, contains only finitely many right or left ideals it satisfies D.C.C. and left D.C.C. By (ii) and (iii) E,, (V) is finitely generated as a module or left module over ED:1, D,. Thus EA (V) satisfies D.C.C. and left D.C.C. CI THEOREM 8.10 (Artin—Wedderburn). Let A be a semi-simple ring with D.C.C. Then A = ED,%, A, where each A, is an ideal of A which is a simple ring with D.C.C. Furthermore the A are uniquely determined.
8]
RINGS OF ENDOMORPHISMS
PROOF. By
(6.10)
AA
is completely reducible. By (8.2) A
27
EA
(A, ). The
result follows from (8.9). E THEOREM 8.11 (Artin—Wedderburn). Let A be a simple ring with D.C.C. Then A D„ for some division ring D and some integer n > 0. 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.
By (8.2) A -- EA (AA), and AA is completely reducible by (6.10). Since A is simple (8.9) implies that A D,, 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 uniquely determined up to isomorphism. By (8.9) AA is a sum of n irreducible submodules. Thus by (1.7) D,„ D„ if and only if m = n. Eli PROOF.
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.
The first statement follows from (8.10). Let V be a right ideal of A. By (6.10) A A = V ED 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. 111 PROOF.
LEMMA 8.13. Let V be a finitely generated free A module. Then E, (V) is a finitely generated free A module. PROOF. EA (V)
V
MAA
A—
for some integer
m. Hence by (8.2) and (8.6)
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 V for some submodule W. Let E, be the subring of ER (F) consisting of all endomorphisms 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
19
ER (F) and E, are R modules. By (8.13) ER (F) satisfies A.C.C. and so E, is
a finitely generated R module and satisfies A.C.C. If f E E, then f induces an endomorphism f of F/ W = V. The map sending f to f is a homomorphism of E, into ER (V). By (4.1) F is projective since it is free. Thus for any g E ER (V) there exists a commutative diagram
where t is the natural projection of F onto V. If w E W then tfw = gtw = 0 and so fw E W. Hence f E E, and f = g. Thus the map sending f to f is an epimorphism of E , 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. CI 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 C J(A). If Visa finitely generated A module then EA (V) is a finitely generated R -algebra.
PROOF. If V is a finitely generated A module then V R 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)E A (V)= EA (V)J(R) = (0). Hence J(R)A annihilates every irreducible A module and so J(R)A C J(A).
9. Completeness Let A be a ring and let V be an A module. Suppose that I is an ideal in A such that VI = (0) where = A. Then for any real number c with 0< c < 1 define the norm II II`, on V as follows
rY,=„
OH; = 0, II v = c'
if y E VI', 't)( VI''.
9]
COMPLETENESS
29
Define d by w)= - w
1.
It is easily verified that for u, y, w E V max{cri (u, w), cr,(y, w)}
and that d is a metric on V. Two such metrics d , cr.,: are equivalent, written as d d if a sequence in V is a Cauchy sequence with respect to cr, if and only if it is a Cauchy sequence with respect to d. It is easily seen that d0< c, c' 0 such that v, - v, E VI' for i, j > n. (ii) lim,— y, = y if and only if for every integer m >0 there exists an integer n > 0 such that v, - y E VI' for i > n.
PROOF. Immediate from the definition.
E
LEMMA 9.2. Let V be an A module. Suppose that d, 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 d, is defined on the A module W and f E Hom, (V, W) then f is continuous.
PROOF. (i) It follows directly from (9.1) (ii) that wi ) = lim
lim
(ii) Let lim yi = y, lim a, = a where y; E V, a, E A for all i. By (9.1) (ii) for every integer m > 0 there exists an integer n > 0 such that y; - y E VI"` and ai - a E / " for i> it Thus
30
CHAPTER I
vo, — va = va — via + va — va = vi (ai — a)+ (v, — v)a E
Hence lim v,a, = va. (iii) Immediate by (i) and (ii). (iv) Suppose that {v,} is a sequence in V with lim v, O. Then (9.1) (ii) implies that for every integer m > 0 there exists an integer n > 0 such that y, = uia, for some 14, E V, a, E I" and all i > n. Hence fy, = (fu,)a i E for i > n. Thus limf/h = O. If lim wi = w this implies that (lim fw, ) — fw = lim(fw, — fw) =limf (w, — w)= O.
Hence f is continuous.
E
9.3. Let V be a finitely generated A module. Assume that A is complete with respect to d, and d, is defined on V. Then V is complete with respect to d,.
THEOREM
PROOF. Let {v,,..., v,} be a set of generators of V. Let {w,} be a Cauchy sequence in V. There exists a sequence {m (n)} of nonnegative integers such that w, — w, E VI -( " ) for all i, j n, m(n) m(n + 1) and
lim„— m (n ) = 00. Let w v rb i , with b,, E A and let w n +1 a„, E r (" ). Define
n —1'1=1 Vtant
with
n -1
6„, = b, +
E
J=1
for all n, t. Then w„ =
v,b„, and
n+14-1
b
„
fk,,
b„, =
E
a,
E i m(") .
Thus for each t { 6„,} is a Cauchy sequence in A. Let 6, =lim„—b„, and let w= I v,b,. Then
lim(v — w„) = E v, {lim(b, — b„, ) } --- O. 1= 1
Thus w = lim w„ and so V is complete.
0
LEMMA 9.4. Let V be an A module and suppose that cl, is defined on V. Let J = J(A). If I C J and A I I satisfies D.C.C. then d, is defined on V and cl, — d,
COMPLETENESS
31
PROOF. By (6.7) J(AII)= JII. Thus by (6.9) J" CI for some integer n > O. Hence RI° VJ c ry:=0 VI' = (0) and d, is defined on V. Since J"" C I'" C J' for every integer m it follows from (9.1) that d, — d. 0 LEMMA 9.5. Suppose that A is complete with respect to d,. Then I C J(A). PROOF. Let a E I. Let b, = 1+ a + • • + a' for i = 0,1, ... . Then b, b, Er for i, j m. Thus 0,1 is a Cauchy sequence in A. Let b = lim b,. Then (1— a)b =
a)b, = lim(1— a 4 ) = 1.
Hence a is right quasi-regular. Since a was arbitrary in I (6.4) and (6.5) imply that I C J. 0
Let J = J(A). A is complete if d, is defined on A and A is complete with respect to d,. The A module V is complete if d, is defined on V and V is complete with respect to d . A is complete on modules if every finitely generated A module is complete. LEMMA 9.6. Suppose that d, 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 d, is defined on V/W, (0) is closed in V/ W. Let f be the natural projection of V onto V/ W. Thus W = r(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 J(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 J(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 WJ(A) = W. Then W = (0).
32
CHAPTER I
(9
PROOF. Suppose that W/ (0). Choose a set of generators w 1 , , with n minimal. Since WJ(A ) = W, w„ w,a, with a, E J(A). Thus w„ (1 — a„)=Z,"=,'w,a,. Since a„ E J(A), 1 — a„ has an inverse in A. Hence w, is in the module generated by w,, w,_, and so W is generated by w,, w,_, contrary to the fact that n was minimal. Thus W = (0). L The proof of the next result is due to I. 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 = fl VI then WI = W.
PROOF. Clearly WI C W. By A.C.C. choose U maximal among all submodules of V such that unw= WI. Let a E I. We will first show that Va - C U for some integer m. For each s let V, = {y vas E U}. Since R is commutative V, is a submodule of V. Clearly V, C V„,, for each s. Hence by A.C.C. there exists an m with V,, = U: 0 V,. Clearly W/ C ( Va + U)n W. Suppose that w E (Va - + U)n W. Thus w = va" + u for some y E V, u E U. Since wa E W/ C U and ua E U this implies that va'' E U and y E V„,. 1 = V„,. Hence va - E U and so w = va + u E U. Thus wEUnW= WI. Therefore (Va - + u)nw= WI. The maximality of U now implies that Va" C U as required. Let a ,, , a„ be a set of generators for the ideal I. Choose m such that Va7 C U for i = 1, . , n. Since R is commutative I — C I,, where Io is the ideal of R generated by a:,". Thus VI — C U and so W C VI" ' C U. Hence
w=unw=w1.
THEOREM 9.10. Suppose that R is commutative and satisfies A.C.C. Let J = J(R). Then d, is defined on every finitely generated R module. In particular = (0).
r-r_,)
PROOF. Let V be a finitely generated R module. Let W = (9.9) WJ = W. Thus W = (0) by (9.8). CI
r-r:=, VJ'. By
THEOREM 9.11. Let A be a finitely generated algebra over the commutative ring R. Assume that R satisfies A.C.C., R I J(R) satisfies D.C.C. and R is complete. Then (i) A satisfies A.C.C., A I J(A) satisfies D.C.C. and A is complete on
modules.
33
LOCAL RINGS
(ii) If V is a finitely generated A module then E A (V) satisfies A.C.C., EA (V)IJ(E A (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 AIJ(R)A is a finitely generated RIJ(R) module it satisfies D.C.C. Thus by (8.15) AMA) 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)) = AJ(R)' for all i. Thus dA, (R) is equal to the metric cb (R) defined on AR as an R module. Hence by (9.6) and (9.10) A is complete with respect to dA , (R) . If V is a finitely generated A module then
n=0
V(AJ(R))'
n
=0
VAJ(R)' =
n
=0
VJ(R)' = (0)
by (9.10). Thus by (9.3) V is complete with respect to dA , (R) . Since A I AJ(R) is a finitely generated R IJ(R) module it satisfies D.C.C. Thus by (9.4) V is complete with respect to c, (4) . Hence A is complete on modules. E 9.12. Let R be commutative. Assume that R satisfies A.C.C., R IJ(R) satisfies D.C.C. and R is complete. Let B Ç A be finitely generated LEMMA
R-algebras with 1E B. Then B
n J(A)C J(B).
PROOF. Let xEBnJ(A). By (9.11) A is complete. Therefore E4' converges in A. Since B is an R submodule of the R module A it is closed by (9.6). As 1 E B this implies that Xo- x E B. Since (1— x)>4' = 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)C J(B). fl 10. Local rings
A 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. 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 IJ(A) is a division ring.
34
CHAPTER I
[10
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. (iii). Clear. (ii) (iii) (iv). Let a denote the image of a in A = AMA). If a 0 then a is a unit in A and so ab = ba =1 for some b E A. Thus ãb = bd 1 and AIJ(A) is a division ring. (iv) (i). Suppose that a0J(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. E 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 I is a commutative ring with no nontrivial ideals. Hence A I I is a field and so A is local by (10.1) (iv). E 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 submodules U, W such that V = U EDW where U is the kernel off'' for all j =0,1, ... and W = f"'`V for all j =0,1, ... . PROOF. Let U; be the kernel of f' and let W, = f' V. Then U, C LI) +, and „ C W, for all j. Let U = W = n:=0 W„. By A.C.C. and D.C.C. there exists an integer n with U. = U and W. = W. It remains to show that V= U (i) W. Suppose that uEUC1W. Then f" u =0 and u = f"v for some vEV. Thus f2"v =0 and 93 v E U. Thus u = f"v =0. Hence u n w = (0) and U + W = U ED W. If v E V then f"v E W = W,„ and so f"v = f 2 "w for some w E V. Hence f"(v — f"w)= 0 and v — f"w EU. Then v = (v — f"w)+ f"w E U+ W. Thus V = Ue W. 0 THEOREM 10.4. Suppose that A is a ring which satisfies A.C.C. Assume that A is complete on modules and A IAA) satisfies D .C.C. Let V be a finitely generated indecomposable A module. Then E A (V) is a local ring. PROOF. Let J = J(A ). Since V/ Ur" is a finitely generated A /J" module
10]
35
LOCAL RINGS
for every integer m >0, (6.11) and (6.13) imply that VIVJ - satisfies A.C.C. and D.C.C. Let f E E, (V). For every integer m 0 fV.T" C VP'. Thus f induces an endomorphism on VI V.1 - . (10.3) applied to VI VT" asserts the existence of an integer n(m) and submodules U„„ W„, of V such that U,,, = {u f" (-) ±iu E VT"). W„, = f"'" )±iV + VT"
for all j
for all j
0,
0
(10.5) (10.6)
and V I VP' U„, I VJ - G w„, /
It may be assumed that n(m +
(10.7)
n(m) for all m. Then
W„,± 1 = f" ("''') V+ VP'±' C f" (-) V + VT" = W„,.
If y E U„, +1 then r(-±1)v E VT"' C V.T" and so f" (»iy E VP' for suitable j 0. Thus y E U,,,. Therefore U„,± 1 C U„„ W„, , I C W,,, and V = U„, + W„, for all m. Let U = fl 11 U„„ W = W„,. We will show that V= Ue Suppose that uEun W. Thus for any m, u E u„, n W„, and so by (10.7) u E VP'. Hence u E fl 11 VT" = (0). Therefore u n w = (0) and U + W = U (i) W.
Let y E V. By (10.7) y = u„, + w„, for every integer m with u„, E U„„ w„, E W„,. Thus for any integers i, j 0= v—v = u— u1 + W — w,. Hence for i, j > m U,
— u, = W 1 — w, E
u,„ n w„, C
Therefore {u,}, { w,} are Cauchy sequences in V. Let u = lim u„ w = lim w,. By (9.6) U,,, and W,,, are closed in V. Since IA E U„, for i > m and w, E W„, for i > m this implies that u E U„, and w E W,„ for any integer m. Hence u E U and w E W. Furthermore y = lim V = lim u, + Ern W 1 = u + W. Therefore V = U (i) W. Suppose that U = (0). Then V = W = W„, for all m and 93 by (10.7) U,,, = VT" for all m. If y E V with fy = 0 then y E U and so y = 0. Thus f is a monomorphism. Let y E V. Then by (10.6) y = fy„, + z„, for all m, with z„, E V'". Hence lim z„, = 0 and so {fy„,} is a convergent sequence. If fu E for some ni then u E U„, = VP' and so {y„, } is a Cauchy
36
[11
CHAPTER I
sequence. Since f is continuous by (9.2) this implies that v = lim fv„, — f (lim v,„). Hence f is an epimorphism and so f is an automorphism. For each f E EA (V) let U = Uf, W = WI be defined as above. Thus either V = Uf or V = Wf. If V = Wf then f is a unit in E, ( V). Furthermore if V = Uf then V = W,- 1 . Hence every element in E,, (V) is either a unit or is quasi-regular. Suppose that EA (V) is not a local ring. Then there exist two nonunits whose sum is a unit. Thus there exist f, g nonunits with f + g = 1. Hence g --- 1—f is a nonunit contrary to the fact that f is quasi-regular. Hence E, (V) is a local ring. CI
11. Unique decompositions
11.1. Let A be a ring and let V be an A module. Suppose that V , V and V W, where each y, vv, is a nonzero indecomposable A module. Assume that E,, (V) and EA (W,) are local rings for all i, j. Then m = n and after a suitable rearrangement V, — W, for i = 1, . . . , m. LEMMA
result is clear since V is indecomposable. It may be assumed that V = (1):1, V = Can= W Let e„ f, be the projection of V onto V, IV, respectively. Thus e, and e l te, are all in EA (V,). Since ei =E; el te, and E,. (V) is local it follows that e l fe, is a unit in EA (V 1) for some j. Hence by changing notation it may be assumed that e l fe is an automorphism of V, = e, V. Thus f, V, C W, and the kernel of f ' on V, is (0). We next show that V = f V ED:12 v. If u E f, v, n ED,72 V, then e l u = 0 and u = el v for some v E V. Thus e l f i e,v = eu = 0 and e i v --- 0 as e, f,e, is an automorphism on V. Hence u = fie, v = 0. Therefore f, V, n ED,12 v = (0). Suppose that v E V. Then e i v E eiV = e, f i e, V. Hence e,v = e, te,w for some w E W. Thus e l (v — f l e,w)— 0 and v — fie l w E V. Hence PROOF. Induction on the minimum of m and n. If m = 1 or n = 1 the
/•
e
=few +(v f,e,w)E
V.
+ i
=2
V. Consequently V = f,V, Since f,V 1 C W, this implies that W, = f, ViG) 04/, n (1):12 VI. As W, is indecomposable and f,V,/ (0) we see that W, = f,V, is isomorphic to V, and V= W, GEB:"_2 V,. Therefore
37
UNIQUE DECOMPOSITIONS
Hence by induction m = n and after a suitable rearrangement V, i = 1, , m. 0
W, 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. W, where each V, w, is indecomposable (ii) If V — I Vi and nonzero then m = n and V1 Wi after a suitable rearrangement.
ex=,
COROLLARY 11.2. 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.
PROOF. Clear by (11.1). 0 THEOREM 11.3. Suppose that A is a ring such that A satisfies A.C.C. and A IJ(A) satisfies D.C.C. Assume that A is complete on modules. Then A has the unique decomposition property.
PROOF. Immediate by (10.4) and (11.2). E THEOREM 11.4 (Krull—Schmidt). Suppose that A satisfies D.C.C. Then A has the unique decomposition property.
PROOF. By (6.11) 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 [1959], Reiner [1961] and Swan [1960] in case R is a complete discrete valuation domain.
THEOREM 11.5. Suppose that R is a complete commutative ring which satisfies A.C.C. and R I J(R) satisfies D.C.C. Let A be a finitely generated R -algebra. Then A has the unique decomposition property. PROOF. Immediate by (9.11) and (11.3). E
38
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12. 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 ï denote the image of x in A = AIL If d is an idempotent then there exists f (t)EZ[t] with no constant term such that f (a) is an idempotent in A and f(a)= a.
Since a (1 - a) E I and so is nilpotent, there exists a positive integer n with a" (1 - a)" = O. By raising la + (1 - a)} = 1 to the (2n - 1)st power we see that PROOF.
a"g(a)+ (1- a)"h(a)= 1,
where g(t), h(t)EZ[t]. Let f(t)= t"g(t). Then f (a) = a"g(a)= a"g(a){a"g(a)+ (1- a)"h(a)} = f (a)2 + a" (1 - a)" g(a)h (a) = f (a)2
Hence f (a) is an idempotent or zero. Thus it suffices to show that f (a) = Since a (a"g(a)+ (1 - a)"h(a)) = a it follows that a"+' g(a) = Since a"' = a" as a is an idempotent this yields that a"g(a) = a. El 12.2. Let I be a nil ideal of A. For x E A let I denote the image of x in A = A IL If e, and e2 are commuting idempotents in A with = e2, then
LEMMA
el
= e2.
Since (1 - e,)e2, (1- e2 )e, E I, they are nilpotent. However {(1 e,)e1} 2 = (1- e, )e, for all i, j. Hence (1- e l )e2 = 0 = (1- e2)e,. Therefore e, = e 2el = e ie2 = e2. D PROOF.
12.3. Assume that A is complete on modules. Let I be an ideal of A with I C J(A). For x E A let denote the image of x mA = AIL If a is an idempotent in A then there exists a sequence of polynomials f, (t) E Z[t] with no constant term such that {f, (a)} converges in A and if e = lim f, (a) then e is an idempotent with e = d.
THEOREM
PROOF. Since J(A)1.1(A)" is nilpotent in AIJ(A)", (12.1) may be applied. Thus there exists f „(t)EZ[t] with f„(a)2 - f„ (a) E J(A)" and f „(a)= a.
12]
CRITERIA FOR LIFTING IDEMPOTENTS
39
Furthermore f, (a) and f, +1 (a) commute and map onto idempotents in A IJ(A)" with f„,i(a)= a = f„ (a). Thus f, (a)— f..,(a)E J(A)" by (12.2). Consequently {f„ (a)} is a Cauchy sequence and so converges as A is complete. This implies the required statement. D THEOREM 12.4 (Brauer, Nakayama). Assume that A is complete on modules. Let I be an ideal of A with I C J(A). For a E A let d denote the image
of a in A = AIL If e 1 , , e„ is a set of pairwise orthogonal idempotents in A then e„. ..,e„ is a set of pairwise orthogonal idempotents in A. If x , x„ is a set of pairwise orthogonal idempotents in A then there exists a set of pairwise orthogonal idempotents e,, , e„ in A such that e, = x, for i =
1,
. , n.
Since / C J(A) it follows that n7=0/ , = (0). Thus if e is an idempotent in A with e = 0 then e = e E I' for all i and so e c ac=„/ , = (0). Hence e 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, e2, , e„ such that e = x + x 2 and e, = x, for i = 3, ... , n. Let x, = 6 and let a = ebe. Thus a = x, and ea = ae = a. By (12.3) there exists an idempotent e o which is a limit of polynomials in a with eo = d. As e commutes with a it follows that eeo = eoe. Hence e, = eeo is an idempotent such that e, = x, and ee l = ei e = e,. Let e 2 = e — e,. Thus e2= x2, e 2e,= 0 = e 1 e 2 and e;= e2 . Therefore if i = 1,2, j = 3, ... , n then e,e, = e,ee, = 0 and e,e, = e iee, = O. Hence e,. , e„ is a set of pairwise orthogonal idempotents with e, = x, for i = 1, . , n. 0 PROOF.
COROLLARY 12.5. Assume that A is complete on modules. Assume further that A IJ(A) satisfies D.C.C. Let N be a right ideal in A. Then either N C J(A) or N contains an idempotent.
Assume that NZ J(A ). Let a denote the image of a in A = Then Si 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 N + J(A) is the inverse image of N in A it may be assumed that x = a 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 {a,} in N such that lim a, = e is an idempotent. By (9.6) N is closed in A and thus e E N. LI PROOF.
A IJ(A).
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CHAPTER I
[12
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. COROLLARY
PROOF. If N does not contain an idempotent then N C J(A) by (12.5). Hence N is nilpotent by (6.9). 1=1 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/ 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 (11.5) as follows: By (11.2) it suffices to show that E A (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 (9.11). 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 d1 . Assume that R satisfies A.C.C., Rh satisfies D.C.C. and R is integrally closed in its quotient field K. For x E R let Î denote the image of x in = R /J(R). Then R is a local ring. If f(t)E R[t] with f(t) monic and f (t)= go(t)h o(t) with (go(t), h o(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 k 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 (9.11) 5 satisfies A.C.C., S/SJ(R) satisfies
12]
41
CRITERIA FOR LIFTING IDEMPOTENTS
D.C.C. and S is complete. Thus by (12.4) S = SISJ(R) contains no idempotent other than 1. Since fz is a field and S = R[t]l(p (t)) the Chinese remainder theorem implies that p(t) is a power of an irreducible polynomial in 1[t]. Now let f(t), go(t), ho(t) be as in the statement of the result. It may be assumed that g o(t), ho(t) are monic. Since R is integrally closed in K, f(t)=11:1, Pi (t) where each Pi (t) is irreducible in K and is monic with p,(t)E R[t]. Hence by the previous paragraph for each i,p,(t)Igo(t) or p,(t)lh o(t). Let g(t) = Hp, Or where j ranges over all values of i with p,(t)Igo(t). Let h(t)= f (t)Ig(t). Then it follows that g(t), h(t) have the desired properties. 111 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 IJ(R) satisfies D.C.C. and R is complete. For x G A let Î 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. 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 a is a central idempotent in A. By (9.11) A is complete on modules. Hence (12.4) implies that a = ë for some idempotent e. We next show that e is central. PROOF.
A = eAee A(1 — e)g) (1— e)Ae (1 — e)A(1— e). This is the Peirce decomposition. Since ë is central in
A
it follows that
JA (1 — e) = (1 — e)A = 0, (1 — e)AJ = (1 — e)J A = 0. Thus eA (1 — e)C J(R)A = AJ(R) and so eA (1 — e)C eA (1 — e)J (R). Since eA(1— e) is an R module, Nakayama's Lemma, (9.8) implies that eA(1— e)= 0. Similarly (1— e)Ae = 0. Consequently A = eAe (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.
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[13
Suppose that J, = j2 for central idempotents e 1 and e 2 . Then e, — e t e 2 E J(A) and (e, — e 1 e 2)2 = e, — e 1 e 2 for i = 1, 2. Thus e, — e,e,= 0 and so e,= et e,= e2.
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 indecomposable, V (0) and V I A A . 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 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 projective module is a principal indecomposable module. Let V be a finitely generated indecomposable projective module. Then mAA for some integer m. Let A A = ED V, where each y is indecomposable. Thus mAA = m V, and so V V, for some i by the unique decomposition property. D
In the rest of this section we will be concerned with rings satisfying the following conditions HYPOTHESIS 13.2. (i) A satisfies A.C.C. (ii) If N is a right ideal of A then either N C 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 IPA) 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
131
PRINCIPAL INDECOMPOSABLE MODULES
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.
LEMMA
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. THEOREM 13.5. Suppose that A satisfies (13.2). Let e be a primitive idempotent in A. Then eJ(A) is the unique maximal submodule of eA and eA leJ(A) is irreducible. If e, is another primitive idempotent in A then eA e,A if and only if eA eJ(A)--- e,A I e,J(A).
PROOF. Let N be a right ideal of A with N C eA, N/ eA. If NZ J(A) then by (13.2)(ii) there exists an idempotent f EN. 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 e is the unique idempotent in eAe. Thus fe = efe = e or fe = efe = O. If fe =e then eA CfA C N which is not the case, hence fe =O. Therefore (e — f) 2 = 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 C 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 e A then since eJ (A), e ,J (A) is the unique maximal submodule of eA, e,A respectively it follows that eAleJ(A)— e,Ale,J(A). Suppose that eAleJ(A)— e,Ale,J(A). Since eA, e i A are projective modules there exists a map f :eA —> e,A with t,f = t where t, t 1 are the natural projections of eA, e,A onto eAleJ(A)— e,Ale,J(A). Since t,e,J(A)= (0) the first statement above implies that f is an epimorphism. Let V be the kernel of f. Then eA/V==eA and so eA — V eeIA since e,A is projective. As eA is indecomposable this implies that V = (0) and eA e,A as required. E 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).
El
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[13
13.7. Assume that A is complete on modules and satisfies A.C.C. Assume further that A IJ(A) satisfies D.C.C. Let I be an ideal of A with I C J(A). For any A module V and y E V let i5 denote the image of v in THEOREM
V = V I VI. Then (i) If W is a finitely generated projective A module then W = .15 for some finitely generated projective A module P. (ii) If P is a finitely generated projective A module then is a finitely generated projective A module. If P is a principal indecomposable A module then .15 is a principal indecomposable A module. (iii) The map sending P to 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.
By (11.3) 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 = ë for some idempotent in A. Thus W = A. (ii) It may be assumed that P eA for some primitive idempotent e in A. Thus 15 — A is projective. 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,, P2 are principal indecomposable modules then P P2 if and only if .15 1 — P2. Since P/Rad P, 15,/Rad P, as AIJ(A)---- AMA) as A modules the result follows from (13.6). D PROOF.
LEMMA 13.8. Suppose that A satisfies (13.2). Let e be a primitive idempotent in A. Then E A (eA leJ(A))— eAe IJ(eAe)= eAe leJ(A)e. If A is a finitely generated R-algebra for some commutative ring R then these isomorphisms are R-isomorphisms.
The second equality follows from (8.4). If f E EA (eA) then since f is an A -endomorphism f (eJ (A)) C eJ(A). Thus f induces an endomorphism f of eA leJ(A). The map sending f to f is clearly a ring homomorphism and an R -homomorphism in case A is an R -algebra. If h E E4 (eA leJ(A)) then h can be viewed as a map h :eA —> eAleJ(A). Since eA is projective this implies the existence of f E E A (eA) with tf = h where t : eA—> eAleJ(A) is the natural projection. Then f= h. Thus the map sending f to f is an epimorphism of EA (eA) onto EA (eA eJ(A)). Since EA (eA I eJ(A)) is a division ring by Schur's Lemma, (8.1) and E (eA)-- eAe is a local ring by (8.2) and (13.4) the result follows. D PROOF.
13]
PRINCIPAL INDECOMPOSABLE MODULES
45
(13.2). Let V be a finitely generated A module with a composition series and let e be a primitive idempotent in A. The following are equivalent. (i) Hom A (e A, V) X (0). (ii) Ve/ (0). (iii) V has a composition factor isomorphic to eA leJ(A).
THEOREM 13.9. Suppose that A satisfies
PROOF. By (8.2), (i) and (ii) are equivalent. (i) (iii). Let f E HomA (eA, V), f/ 0. It may be assumed that f(eA)= V by changing notation. By (13.5) the kernel of f is in eJ(A). Thus V f(eJ(A)) eA I eJ(A).
(iii) (i). There exists a submodule W of V which has eAleJ(A) as a homomorphic image. Since eA is projective there exists a commutative diagram
W -->eAleJ(A)—>0.
Thus f
0, f E HomA (e A, V). 0
Suppose that A satisfies (13.2). Two primitive idempotents eo, e in A are linked if there exists a set of primitive idempotents eo, e t , . . , e„ = e such that for each i = 1, , n there exists a principal indecomposable A e,A) (0). module U, with Hom, e,_, A) (0) and Hom, •
COROLLARY 13.10. Suppose that A satisfies D.C.C. Let e o, e be primitive idempotents in a. Then e and eo are linked if and only if there exists a set of primitive idempotents eo, e,,. , en = e such that e,_,A and e,A have a common irreducible constituent for i = 1,.
PROOF. Clear by (13.9).
El
THEOREM 13.11. Suppose that A satisfies (13.2). Let e o, e be primitive idempotents in A. Then e and eo are linked if and only if eA and e0A are in the same block.
PROOF. Let f be the centrally primitive idempotent with e0 A E B where B is the block corresponding to f. , e„ = e be a set of primitive Suppose eo and e are linked. Let eo, e t ,
46
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idempotents such that for each i = 1, , n HomA (U„ e,_,A )# (0) and Hom 4 (U„ e,A) (0) for some principal indecomposable module U,. To prove by induction that e,A C B it suffices to show that if U, V are principal indecomposable modules with HomA (U, V) # (0) then U, V are in the same block. This follows directly from (7.8). Conversely suppose that eA C B. Thus e o f = e o , ef =e and e o ,e C fAf. If e and eo are not orthogonal then either eAe o # (0) or e o Ae# (0). Thus by (8.2) e and e o are linked. Assume that e and e o are orthogonal. If the result is false then f = eo + • • + e, + e,,, + • • + e„
where { e,} is a set of pairwise orthogonal primitive idempotents in fAf with e = e„ and the notation is chosen so that e, is linked to eo if and only if O i s. Let fl = eo + • + e„ f2= e,+1+ • • • +en . As f2 # 0, f, and J., are orthogonal idempotents in fAf. By (8.2) e,Ae, = e,Ae, = (0) for 0 i s, n. Thus fiAf2 = f2 Afi= (0). Hence fAf i = fl Af i and thus Afi A = fAf I A C fi A. Therefore f l A is an ideal in A. Similarly f2A is an ideal in A. As fA = fl A ef,A and f is centrally primitive (7.7) implies that f l A = (0) or f2A = (0). This contradiction establishes the result.
E
14. Duality in algebras
Throughout this section R is a commutative ring and A is a finitely generated R -free R -algebra. For any A module V define the dual /'1• of V by 17 = HomR (V, R) as in (2.8). Thus is a left A module. If v E V, t E 17 write (v)t = vt for t applied to v. Since R is commutative and V is a two sided R mOdule this does not run counter to the conventions previously introduced. If V, W are A modules and f E Hom A ( W) then J E Horn 4 ( is defined by (14.1)
v(tf)= (fv)t
for u C V, t E If V is a left A module, the A module V is defined analogously. LEMMA 14.2. Let V, W be A modules and let f (i) (V
ED w)
E Hom,, ( V, W). Then
„
w.
(ii) If f is an epimorphism then
f
is a monomorphism.
141 .
PROOF.
47
DUALITY IN Al.GF.13RAS
Clear by definition and (14.1).
D
Let V be an A module. If W is a submodule of V define W` = {t
E 17‘, wt = 0 for all w E
Clearly if W, C W then 14/ 1 C Wit. It follows directly from the definition that W L is a left submodule of the left module V For v E V define i G 1' 13(0 (v)t for all t E V. The map sending v to t is easily seen to be an A-homomorphism. If VR is free with a finite R -basis {v,} define i3, E 17 by (v,)t3, = 8„. Then it follows that {i3,} is a basis of 17 and t, has the same meaning as in the previous paragraph. 03,1 is the dual basis of 0)J. LEMMA 14.3. Let V be an A module such that V, is R -free with a finite R-basis fv,I. If a G A and v,a = Er,,v, with r„ E R then ai = PROOF. For all i, j (v,)af.), =
= (E rikvk) 15, =
Thus ah, = 14.4. Let U, W be finitely generated A modules and let V = U W. Assume that VR is a finitely generated free R module. Then 1^/ = 1471 L L P=1,i'/ and the map sending w to is an isomorphism of W onto ,i/ '1 = W.
LEMMA
a
The first three statements follow from the definitions and (14.2). The last statement follows from (14.3) which implies that the map sending v to i") is an isomorphism from V onto V. 0 PROOF.
In case W is a finitely generated A module such that WR is projective we will generally identify W with W under the isomorphism which sends w to W. LEMMA 14.5. If V, Ware A modules such that VR, WR are finitely generated free R modules and f E Horn (V, W) then f f. If furthermore f is an isomorphism of V onto W then f is an isomorphism of W onto V/.
=
48 PROOF.
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For y E V, t E i;i7 (14.1) implies that
(10t = i3(e) = v(th = (fv)t. If f is an isomorphism and vi is a basis of V then f maps the Thus fv = dual basis {fv,} onto 03,1 and so is an epimorphism. By (14.2) f is an isomorphism. LEMMA 14.6. Let V be a finitely generated A module such that V R is a finitely generated free R module. Then more there exists an exact sequence 0 -) V
VR
AAR
VR OR RAA RAA OR VR.
Further-
1 >
such that 0 -) VR
(VR
is a split exact sequence and
AAR) ±-> WR
WR
is a projective R module.
PROOF. Let {yi } be a basis of V R and let {yd be a basis of RA. Then {y, yi } is a basis of ( VR OR RAA)R and f9; 13 ; } is a basis of (RA A 09kin • R ) R Define •
the R -linear map g: ,A4
(VR OR RAA)
by
(y1 0 i3; )g = y":-F-333;. Clearly g is an R -isomorphism. It remains to show that g is an Ahomomorphism. Let a E A and let yia = 1r,kyk with I-A E R. Then a9i = Erkjk by (14.3). Thus (yi (y, 0 y k ) and a( I) Ø y1 )= Erk,(y, Øy). Hence = {a (Pi
13 ,)}g =
rk,(9k
13) 1 g =
= aer6j0 = a{(9,
E r,(,,) 13,)g}.
Thus g is an A isomorphism and the first statement is proved. Replace V by V in the left analogue of (4.5) with B = R and take duals. Then (14.2) and the previous paragraph imply that V-1
-
vR
is an exact sequence and
.±*
14]
DUALITY IN ALGEBRAS
0 -> VIR
(VR 0
49
->
is a split exact sequence. Hence h is an epimorphism and the result follows. 111 An element t in HomR (A, R) is nonsingular if the kernel of t contains no nonzero left ideal and every element in AA is of the form at for some a E A where b (at) = (ba)t for b E A. An element t in HomR (A, R) is symmetric if t(ab)= t(ba) for all a, b E A. A is a Frobenius algebra if there exists a nonsingular t in HomR (A, R). A is a symmetric algebra if there exists a nonsingular symmetric t in
HomR (A, R). LEMMA 14.7. Let t be a nonsingular element in Hom R (A, R). Then the following hold. (i) The kernel of t contains no nonzero right ideal. (ii) Let f : A A —> A' A be defined by af = at for a EAA. Then f is an isomorphism. (iii) Let g :AA —* A Â be defined by ga = tafor a E. A,. Then g = f, where f is defined in (ii) and g is an isomorphism. (iv) Every element of AA is of the form ta for some a E A.
PROOF. (i) Suppose that (aA)t = 0. Thus a(bt)= 0 for all b E A and so as = 0 for all s E AA. Thus =0 and so a = 0 by (14.4). (ii) Clearly f is an A -homomorphism. By assumption f is an epimorphism. If af = 0 then at = 0 and so (Aa)t = (A)at = 0. Thus Aa is a left ideal in the kernel of t and so Aa = 0. Hence a = 0 and f is a
monomorphism. (iii) If a is identified with for all a E A A then by (14.1) b(ak)= (gb)a = t(ba) for all a, b E A. If t is written on the right then t (ba) = (ba)t and k = f. Thus by (ii) and (14.5) f = g is an isomorphism. (iv) Since g is an isomorphism by (iii), this is clear. D THEOREM
14.8. The following are equivalent.
(i) A is a Frobenius algebra.
(ii) AA AA. (iii) AA = AA.
PROOF. By (14.4), (ii) and (iii) are equivalent.
50
[15
CHAPTER
(ii). This follows from (14.7) (ii). (i). Let f: A A ---> A, be an isomorphism. Let t = (1)f. Then every element in Â,, is of the form at. If (Aa)t = 0 then A (at) = 0 and so af = at = O. Thus a = O. E (i) (ii)
LEMMA 14.9.
Let G be a finite group and let A = R[G]. Define t by
(Ex ec = ri . Then t is symmetric and nonsingular. Thus A is a symmet-
ric algebra.
By definition t E AA. If /rxx/ 0 then ry / 0 for some y E G. Thus = (LEG r„xy - 1 )t O. Hence the kernel of t contains no right or left ideal. It is easily seen that fx -I t x E GI is the dual basis of E GI. Hence every element of ÂA is of the form at for some a E A. Since PROOF.
(y
/„ G rxx)t
{(E Gx)(E s'x)} t =
Gs'
it follows that A is a symmetric algebra.
{( E s'x )( E Gx)} t 0
15. Relatively injective modules for algebras
Throughout this section R is a commutative ring which satisfies A.C.C. and A is a finitely generated R -free R -algebra. Let B be a subring of A with R C B such that B is a finitely generated R -free R -algebra. A finitely generated A module Q is injective relative to B or B-injective if any exact sequence 0—> Q V
W—>0
with V, W finitely generated A modules splits provided o— QB
VB
WB 0
is a split exact sequence. A finitely generated A module Q is injective if any exact sequence 0—>Q—>V—>W—> 0
splits for V, W finitely generated A modules. Clearly injective modules are B-injective for any B. Many rings do not possess any finitely generated injective modules and this concept will not
15.1
RELATIVELY INJECEVE MODULES FOR ALGEBRAS
51
be of great relevance below. We note however that since any exact sequence of vector spaces splits the following holds. LEMMA 15.1. Let R be a field and let Q be a finitely generated A module. Then Q is injective if and only if Q is R-injective.
15.2. Let Q be a finitely generated A module such that OR is a finitely generated free R module. The following are equivalent. (i) Q is R-injective. (ii) 010E oR AAR.
THEOREM
(iii) 6
I 6R OR AAR-
(iv) 6 is R-projective. (ii). This follows from (14.6). (i) (iii). Clear by (14.6). (ii) (iii) (iv). This is a consequence of (4.8). (iv) (i). Let V, W be finitely generated A modules such that
PROOF.
0-->Q±>VW—>0
is exact and 0-->
OR
—> 0
> VR
is a split exact sequence. Then
is a split exact sequence and so
6 ->o
1 V/ 0—> W =>
is also a split exact sequence since 6 is R -projective. Thus I V—> 14/—>0
is a split exact sequence by (14.4). be the map sending x to X' and let k : V —> X>. If tE17 Let g : Q and x E Q then (#)t =.(t'1)= x(th= (fx)t = ^
Thus fg = kf. Since f(Ô) is a component of 12 there exists a homomorph-
[15
CHAPTER I
52
ism cl : such that al is the identity on Ô. Since g is an isomorphism by (14.4) we can define d : V— Q by d = clk. Then df = gldkf = g 1 dfg = g -1 g =1.
Consequently
is a split exact sequence. 0 COROLLARY 15.3. Let Q, Q 1 , 02 be A modules such that Q = Q1e3 0 2 and (Q1)R, (Q2)R are R -free with finite bases. Then Q is R -injective if and only if Q1 and Q2 are R-injective. PROOF. Clear by (15.2)(iv) and (4.9). E COROLLARY 15.4. Let Q be a finitely generated A module such that R-free. The following are equivalent. (i) Q is R -injective. (ii) For every pair of A modules W, V with projective R modules the exact sequence
0—> Q
WR, VR
QR
is
finitely generated
V-- W—>0
splits provided
0—> OR ±-> VR j > WR —* 0 is a split exact sequence.
(i) (ii). Clear. (ii) (i). By (14.6) Q (15.2). 0
PROOF.
QR OR AAR-
Thus
Q
is R-injective
by
THEOREM 15.5. Suppose that A is a Frobenius algebra. Let V be an A module such that VR is R-free with a finite basis. Then V is R -projective if and only if V is R-injective.
PROOF. By (14.8) A Â = AA . Hence (4.8)(ii) and (15.2)(ii).
AAR
-""'" RAA.
The result follows from
COROLLARY 15.6. Suppose that A is a Frobenius algebra and A has the
161
53
ALGEBRAS OVER FIELDS
unique decomposition property. Let V 1 , V 2, W be R -free A modules with no nonzero projective summands such that
—> V, El) P, —> W ({) P —> V2 ED P2 -> 0 is exact for P, P1 , P2 projective. Then there exists a projective module P' such that 0 > 17 ,—> W EDP' —> V2 —>0 —
is exact.
Since P2 is projective, P2 1 W ED P. Thus P P0 e p2 by the unique decomposition property. Hence there is an exact sequence PROOF.
o—.
ED
—> W (1) Po —> V2 -> O.
By (15.5) Pi is R -injective. Hence by the unique decomposition property Po PI EDP and the result follows. 111
16. Algebras over fields Throughout this section F is a field and A is a finitely generated F-algebra. Thus A satisfies D.C.C., A.C.C., left D.C.C. and left A.C.C. If V, W are finitely generated A modules define
/( V, W)= IF (V,
= diMF {HOMA
(VI
w )} .
W) is the intertwining number of V and W. It is clear from the
definition that
/( V, ED V2,
=
VI, W)+ /( V2,
TV),
/( V, WI e W2) = AV, W1)+ I( V, W2).
If V is an irreducible A module then by (8.1) EA (V) is a division ring which is a finitely generated F-algebra. F is a splitting field of V if EA (V) F. F is a splitting field of A if F is a splitting field of every irreducible A module. If V is an irreducible A module then clearly F is a splitting field of V if and only if I(V, V)= 1. If F is algebraically closed then F is a splitting field of A since the only finite dimensional F-algebra which is a division ring is F itself. LEmmA 16.1. Let Z be the center of A and let B be a subalgebra of Z with 1 in B. If F is a splitting field of A then F is a splitting field of B.
CHAPTER I
54
[16
PROOF. By (7.3) it suffices to prove the result in case 1 is the unique central idempotent in A. Thus 1 is the unique idempotent in Z. Hence Z, is indecomposable and by (13.6) Z has a unique irreducible module V. By (13.8) Z I E( V). Thus it suffices to show that Z IJ(Z) F. If W is an irreducible A module there exists a nonzero F-homomorphism of Z onto EA (W) F. Since ZIJ(Z) is an F-algebra which is a field by (8.1) this implies that ZIJ(Z)------ F. Hence B F. E A central character of A is a nonzero algebra homomorphism of the
center of A onto F. 16.2. Suppose that F is a splitting field of A. Let e,,... ,e„ be all the centrally primitive idempotents in A. Then A has exactly n distinct central characters A l , , A.„ and A, (e,)= 6.
LEMMA
Let A be a central character of A. If a is in the center Z of A and a'" = 0 then A (a)'" = 0 and so A (a) = O. Thus J(Z) is in the kenel of A. By (16.1) IEB%, e,F. The result follows easily. E
PROOF.
THEOREM 16.3. Suppose that F is a splitting field of A. Let S be the F -space generated by all elements ab — ba with a, b E A. If char F = p >0 let T = {c E S for some i > 0} . Let k be the number of pairwise nonisomorphic irreducible A modules. The following hold. (i) If A is semi-simple then k = dim, A — dimF S. (ii) If char F = p > 0 then (a + a P" b P" (mod S) f or all a, b E A and all n. Furthermore S C T, T is art F - space and k = dim F A dim F
(i) If A simple then A F„ by the second Artin—Wedderburn Theorem (8.11) for some integer n >0, hence k = 1. It is well known that in this case S consists of all matrices of trace 0. Hence dirn,A — dim, S = 1 proving the result in case A is simple. Now (i) follows from the first Artin—Wedderburn Theorem (8.10). (ii) Suppose that char F =p >0. Let o- = (1, 2, ... , p) be a p -cycle. Let x, y be noncommuting indeterminates. For any monomial t, • • • t„ with t, = x or y for each i, define u- (t, . It is easily seen that t„)= t (I) • • •(p )t, in this way (u) acts as a permutation group on the set of all monomials in x and y of degree p. Furthermore if f(x, y) is such a monomial then o- (f (x, y)) = f (x, y) if and only if f (x, y ) = xi' or f (x, y ) = y". Hence if a, b E A then PROOF.
16.1
55
ALGEBRAS OVER FIELDS
P I
{f (a, b)}
(a + b)P — aP — bP = -0
where f(x, y) ranges over a set of monomials in x and y. Since o- {f (a, b)} = f (a, b) (mod S) this yields (a + b ) a" + bP (mod S). Therefore
-
(ab — ba)P "(ab)
-
(ba)P
a {b (ab)P -I } — fb (ab)" i l a -= 0 (mod S).
Thus if c E S also cP E S. Hence c° more for all i (a + b)°'' —
E
S for all i and so S C T. Further-
(a + b)°' — (aP' + bP' )° {(a
b )" — a"' — b"1"
(mod S).
Hence by induction (a + b )" " = aP" + b"" (mod S) for all n. Thus T is an F-space. Since J(A) is nilpotent it follows that J(A)C T, and if cP' E S + J(A) then c°' E S for some j. Hence by factoring out J(A) it may be assumed that A is semi-simple. Thus by the first Artin—Wedderburn Theorem, (8.10), it suffices to consider the case that A is simple. In this case A = F„ by the second Artin—Wedderburn Theorem (8.11) and S consists of all matrices of trace 0. Since dimF A — dim F S = 1 = k and S C TC A it suffices to show that T/ A. There exists an idempotent e in A whose trace is not O. Hence no power of e is in S and so e T. LI The following notation will be used in the rest of this section. Ul, LA is a complete set of representatives of the isomorphism classes of principal indecomposable A modules. For i = 1, . k set L, = U, /Rad U, where Rad U, is the radical of U. By (13.6) L • ) 1,k is a complete set of representatives of the isomorphism classes of irreducible A modules. V, if V, and If V,, V2 are finitely generated A modules we write V, V2 have the same set of composition factors (with multiplicities). Clearly is an equivalence relation. Define c„ by U, c,,L,. The integers c,, are the Cartan invariants of A and the k X k matrix C = (c,,) is the Cartan matrix of A. If B = B(e) is a block of A where e is a centrally primitive idempotent in A then the Cartan matrix of the algebra eA is called the Cartan matrix of the block B. Of course the Cartan matrix is only defined up to a permutation of rows and columns. ,,
56
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CHAPTER I
16.4. Let V be a finitely generated A module. Suppose that V n,L,. For i =1, ...,k let e, be a primitive idempotent in A with U, e,A. Then for each i
LEMMA
dim F Ve, = I ((J,, V) = nj(L,, Li ). PROOF. The first equation follows from (8.2). By (4.4) and (13.5)
g(Ji, V) = E
niI(U„ L,)=
E niI(L„ L,).
By (8.1) I(I4, L1 )= 0 for i j. This yields the second equation. El COROLLARY 16.5. Let V be a finitely generated A module and let E = EA (V). If e is a primitive idempotent in E then the multiplicity of the irreducible left E module Ee1J(E)e as a composition factor of the left E module V is equal to (1/m)dim F eV where m = I(Ee1J(E)e,EelJ(E)e). PROOF. The left analogue of (16.4) implies the result. E LEMMA 16.6. Suppose that A is a symmetric algebra over the field F. Let e, e,, e, be idempotents in A. Then dimFelAe 2 = dimF e 2Ael. Furthermore eA A. PROOF. By definition there exists t E AA such that no right or left ideal is in the kernel of t and (ab)t = (ba)t for all a, b E A. For a Ee,Ae, define fa E ae, by (fa)(e 2 be 1 )= (ae2 be,)t.
If fa
0 then (be 1 ae2)t = (ae,bel )t = 0
for all b E A and so (Ae l ae2)t = O. Hence a = e1 ae2 F-monomorphism. Hence
= O. Thus f is an
dimF e l Ae 2 .--5• diffiF aè, = dim F e2Ae l . Thus by interchanging e l and e, dimF e,Ae 2 = dimF e 2Ae,. To prove the last statement it suffices to show that f is an Ahomomorphism in case e = e, e2 =1. This follows from {(fa)b}ce = (fa)(bce)=(abce)t = (fab)(ce)
for a, b, c
E A. 0
16]
ALGEBRAS OVER FIELDS
57
16.7. Let 13 1 , , B„, be all the distinct blocks of A and let C(s ) be the Cartan matrix of B5. Let C be the Cartan matrix of A. Then after a suitable rearrangement of rows and columns
THEOREM
C`" • 0
C=
0 \ • C`'" ) )
For no rearrangement of rows and columns is (
C;' )
0 \
0
C 15 )
05) = with matrices CS' ) , C. If furthermore A is a symmetric algebra and F is a splitting field of A then C and each Cs) is a symmetric matrix. PROOF. If U„ U, are in distinct blocks then by (7.8) c,, = O. This proves the first statement. The second statement follows from (13.11) and (16.4). If A is a symmetric algebra and F is a splitting field of A then C and each Cs ) is symmetric by (16.4) and (16.6). El LEMMA 16.8. Suppose that A is a Frobenius algebra. Then for each i, U, contains a unique minimal submodule W. If furthermore A is a symmetric algebra then W, L, and eA/eJ(A) ---.-- (XeP) for any primitive idempotent e in A. PROOF. It may be assume that U, = eA for some primitive idempotent e in A. By (14.7) U, V/ for some principal indecomposable left A module. By the left analogue of (13.5) V contains a unique maximal left submodule W. Then W, = W' is the unique minimal submodule of U. Suppose that A is a symmetric algebra. Then U, = eA Aé by (16.5). Thus W,= (A)e). Since e generates Ae/J(A)e there exists f E
(Xe/. /(Â)é) with f(e)/ 0. Hence fe/ 0 and so W,e/ O. Thus by (13.9) W eA/eJ(A)-- - L,. THEOREM 16.9. Suppose that A is a symmetric algebra over F and F is a splitting field of A. Then (i) Lj= Li ) = I (L„ = 8,,. (ii) A A --(dimpL, )U,.
„e=,
(iii) AA
a= I (diMF Uj)L,.
58
CHAPTER I
[16
(i) follows from (16.8). Let AA ""--- ED a»). By (8.2) I (A A , L,)= dim,L, and so by (i) a, = I(A A , L,) = dim,L,. (iii) is an immediate consequence of (ii) and (16.7). 0 PROOF.
A finitely generated A module is uniserial or simply serial if it has a unique composition series. A is a uniserial algebra or a serial algebra if every finitely generated indecomposable A module is uniserial. LEMMA 16.10. Suppose that A is a serial algebra. Let V 1 , V, be finitely generated indecomposable A modules. The following are equivalent. (i) V, 172. (ii) The composition series of V, and V2 have the same length and Vi /Rad VI V2/Rad V2. (iii) The composition series of V, and V, have the same length and V, and
V2 have isomorphic socles. PROOF. Clearly (i) implies (ii) and (iii). (ii)—> (i). Since A is serial V, is a homomorphic image of a projective indecomposable A module U, for i = 1, 2. Since V,/Rad V, -- V2/Rad V2 it follows that U, U, by (13.6). Since U, is serial every homomorphic image of U, is determined up to isomorphism by the length of its composition series. (iii)—> (i). By (14.4) W W. for every A module W. Thus the dual of a serial module is a serial left module. Hence every left module of A is serial. Thus the left analogue of the previous paragraph implies that 17/ 1 ---' V/2. The result follows from (14.4). E LEMMA 16.11. Suppose that A is a serial algebra. Then A has only finitely many finitely generated indecomposable modules up to isomorphism. PROOF. A serial module is a homomorphic image of AA and so has a composition series of bounded length. As A has only a finite number of irreducible modules up to isomorphism, the result follows from (16.10). Lii THEOREM 16.12. Let A = F[x, y] with x 2 = y 2 = xy = yx = 0 such that 1, x, y are linearly independent over F. Then there exist indecomposable A modules of arbitrarily large dimension.
PROOF. Let n be a positive integer. Let V, W be vector spaces over F with bases {v0,..., v,,I, , wr,} respectively. Define
16]
ALGEBRAS OVER FIELDS
w,x = v, ; w,y =
for 1
i
v,x = 0; v,y = 0
for 0
i n.
59
n
It is easily seen that M= VEDW is an A module. It suffices to show that M is indecomposable. Suppose that M = M, M2 with g nonzero A modules for j = 1, 2. Let t be the projection of M onto W. Let m, = dim,t(M,). Thus m, + m2 n. Suppose that m, = 0 for j = 1 or 2, say m 2 = O. Then t(M,)= W and so V= Wx + Wy CM,. Hence M = M, contrary to assumption. Thus frn, 0 for j = 1,2. Fix j, let s = s, be the largest integer such that t(M3 ) is in the space spanned by {w,..., w„}. Hence Mx is in the space U spanned by {v, , v„} and vs _, E M1y + U. Therefore dimF (M,x + ) 1 + dimF M,x. Since x is a monomorphism of W into V and Mx C M, it follows that dimF M,x = m1. As t(M,x + M,y) C t( V) = (0) we see that dim F M, dim F (Mi x + M,y ) + dinv t(M, ) 2m, + 1.
Therefore 2n + 1 = dim F M = dim F M, + dimFM2= 2(ml + m 2) + 2 2n + 2.
This contradiction establishes the result.
E
The next result yields a partial converse to (16.11) and is essentially due to Michler [1976b]. See also Nakayama [1940], Eisenbud and Griffith [1971].
THEOREM 16.13. Let A be an algebra which has a unique irreducible module up to isomorphism and such that for any extension field K of F the following conditions are satisfied. (I) AK is a direct sum of algebras, each of which has a unique irreducible module up to isomorphism. (II) (A I J(A)) K is semi-simple. (III) There exists an integer depending on K which bounds the dimensions of any indecomposable AK module. Then the following hold. (i) There exists a division algebra D over F and positive integers m, n such that A B,, where B IJ(B) D, B contains an element x with x'" = 0, 0 and J(B)` = Bxs = xs13 for all s. (ii) A is serial. There exist exactly m finitely generated nonzero indecom posable A modules up to isomorphism, v„..., V. For each s = 1, . , m V, D 17,x D • • • D V,x
V,x' = (0)
60
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is the composition series of V. Furthermore V. is projective if and only if s = m.
PROOF. By (13.6) there is a unique principal indecomposable A module U up to isomorphism. Hence AA nU for some integer n. Let B = EA (O. By (8.2) and (8.6) A E4 (A 4 ) B. By (13.8) B/J(B)= D is a division algebra. Thus in particular B has a unique irreducible module up to isomorphism. Let J = J(B). Thus J(A)------J„. Let K be an extension field of F and let M be a B K module. Then nM becomes an AK module in the obvious way. Denote this by /a. Thus nM. Hence if M is indecomposable then every nonzero indecomposable summand of A4 has K-dimension at least dim K M. Thus there is an integer which bounds the K-dimension of every indecomposable B K module. Let É = B/r. For y E B let yv denote its image in Then J is a D module. Let be a D-basis of I Thus ï,".t, = 0 for all i j. We will first show that r 1. Since A B„, B and É satisfy the same hypotheses as A. Thus by hypotheses (I) and (II) there exists a finite extension K of F such that BK = =1 B m , J(1j)K J(EK)-----(131:=,J(B(' )) and B ( ' )/J(B")------- K„, for i = 1, , t. The argument of the first paragraph applied to B (' ) implies that (C')„, for some C(' ) with C ) /J(C) K. By (16.12) and hypothesis III dim K J(C") 1 and so dim K J(B") n:. Hence ,
r dim, D = dimj = dimK J(B) K
E n = dimF D.
Thus r 1 as required. If r = 0 the proof is complete. Suppose that r =1. Let x E J with î = Î. Since J is nilpotent there exists an integer m with f " = 0, # O. (i) By the definition of x, J = xB +.12 . Thus (J/xB)J = J/xB. Hence by Nakayama's Lemma (9.8) J = xB. Since the F-dimension of J/J2 equals the F-dimension of D it follows that Pr is a one dimensional left D module. Thus J = Bx +.12 and so J = Bx as above. Hence F = Bx' = x'B for all s as required. (ii) For s =1,...,m, Jx —I /Jx' ----PR is an irreducible B module since it is of dimension 1 over D. Thus BB is a serial module with composition series BB D Bnx D • • • D Bnx '" = (0). In particular B nx --1 is irreducible and is the socle of B. Similarly B B has an irreducible socle and so BB/Rad(BB) is irreducible. There exists a commutative diagram
16]
61
ALGEBRAS OVER FIELDS
BB
B B /J(B) E,
0
Since BB is projective this implies the existence of a map f: BB which is an epimorphism. As dim, BB = dimFBÊ, f is a monomorphism. Thus BB B B. Since A ---- B„ it follows that A A AA'. Hence A is a Frobenius algebra by (14.8). As AA nU, U is injective by (15.5). Furthermore J(A)S = Ax' -= fA for all s and J(A)f -1 1.1(A)zs -- AMA) is the direct sum of n copies of the unique irreducible A module. Hence U is serial with composition series U D Ux D • • D Ux = (0). We will prove by induction on m that A is serial. If m = 0, A D„ is semi-simple and the result is clear. Let V be a finitely generated indecomposable A module. If Vx —1 = (0) then V is an A /Ax -1 module and so V is serial by induction. Suppose that Vx X (0). Thus there exists v E V with vx # (0). There exists an exact sequence --> WA,,
vA --> O.
# (0), the socle of AA is not in the kernel of f. Hence there exists As vx an indecomposable direct summand u, of AA such that the socle of u, is not in the kernel of f. As U,— U has an irreducible socle this implies that f( U). Hence vA contains a submodule which is isomorphic to U and so V contains a submodule which is isomorphic to U. As U is injective this implies that U V. Since V is indecomposable we see that V ------ U is serial. The remaining statements are immediate consequences of this fact. CI The following result, due to T. Nakayama, was brought to my attention
by D. Burry. THEOREM 16.14. Suppose that every indecomposable projective and injective A module is serial. Then A is a serial algebra.
PROOF. Let V be an A module. We will prove by induction on dim, V that V is the direct sum of serial modules. If dim, V = 1 this is clear. Suppose that dim, V > 1. Let W be a serial submodule of V with dim, W as large as possible. Let X be a submodule of V, maximal with respect to the property that W n x = (0). The maximality of X implies that V/X has an irreducible socle. Thus V/X is isomorphic to a submodule of an indecomposable injective A module and so is serial. Let L ( V/X)/Rad( V/X) and let P
62
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be the indecomposable projective A module with P/Rad P -- L. Then there exists an epimorphism f : P —> V/X. Hence there exists a mapping g : P —> V such that the following diagram is commutative
—> V/X—>0 Thus g(P) is a serial submodule of V. Since W n x = (0) it follows that W is isomorphic to a submodule of V/X. Now the maximality of dim, W implies that g(P) — V/X — W. Therefore g(P)n x = (0) and V = g(P)+ X. Consequently V = g(P)ED X and the result follows by induction. E
If S is a subset of A define /(S)= la aS = 01,
r(S)= {a I Sa = 0}.
Then I(S), r(S) is the left, right annihilator of S respectively. Clearly /(S) is a left ideal and r(S) is right ideal. LEMMA 16A5. Suppose that A is a Frobenius algebra and t is a nonsingular element in Hom, (A, F). Then the following hold. (i) If N is a left ideal of A then r(N)= {a at E NJ l(r(N))= N and ,
dim, N + dim F r (N) = dimF A. (ii) If N is a right ideal of A then l(N)= {a ta E N' }, r(l(N))= N and dim F N + dimF 1(N) = diMF A.
PROOF. Let N be a left ideal of A. By definition r(N)= {a 1 Na = 0}. Since Na is a left ideal for any a E A and t is nonsingular it follows that r(N)= {a l(Na)t = =
(N)at = = {a at E N 1 }.
In particular this implies that dim,N + dim, r(N)= dim, A. Similarly if N is a right ideal then l(N)= {a ta E N il and dim,N + l(N)= dim F A. Hence if N is a left ideal then r(N) is a right ideal and so dim,N = dim F l(r(N)). Since N C l(r(N)), this implies that N = l(r(N)). The analogous statement for right ideals is proved similarly. E COROLLARY 16.16. Suppose that A is a symmetric algebra. Let I be an ideal of A. Then r(I)= 1(I). PROOF. Let t be a symmetric nonsingular element in Hom, (A, F). Then by (16.15)
17]
ALGEBRAS OVER COMPLETE LOCAL DOMAINS
I
63
I
r(I)= {a t(Ia)= 0} = {a t(Ia)= 0} = 1(I). In case A is a symmetric algebra and I is an ideal of A let Ann(/) = r(I) = l(I). In this case Ann(/) is the annihilator of I when I is considered as a submodule AA or a left submodule of A A. The next two results are due to Nakayama [1939], [1941]. See also Tsushima [1971a]. THEOREM 16.17. Let I be an ideal of A. Suppose that A and A
are both
Frobenius algebras. Let s, t be a nonsingular element in Hom F (A F), Hom F (A, F) respectively. Then there exists an element c E A with s = cf -and for any such c, r(I) = cA. If furthermore A and A are symmetric and s and t are symmetric then any such c is in the center of A. PROOF. Since t is nonsingular there exists c E A with s = ct. Thus (k)t = Is = (0) and so k = (0). Hence IcA = (0) and cA C r(I). If x E 1(cA) then xcA = (0) and so xs = (xc)t = O. Thus 1(cA)s = (0). As s is nonsingular in Hom F (A II, F) this implies that 1(cA)C I. Therefore by (16.15) r (I) C r (1(cA))= cA. Consequently r (I) = cA. Suppose now that both s and t are symmetric. Then for all a, b E A
(abc)t = (ab)s = (ba)s = (bac)t = (acb)t. Thus (a (bc — cb))t = 0 for all a, b E A and so A (bc — cb) is in the kernel of t. Therefore bc — cb = 0 for all b E A and so c is in the center of A. 0 LEMMA 16.18. Suppose that A is a symmetric algebra. Let A A =
e,A A ,
where { e,} is a set of pairwise orthogonal primitive idempotents. Then each eA A contains a unique minimal right ideal N, and Ann(J(A )) =
PROOF. By (16.8) each e,A A contains a unique minimal right ideal N. Since N is an irreducible A module it follows that N,J(A)= 0 for each i. Hence N, C Ann(J(A )). By (16.8) dim F e,J(A) + dim,N, = dim F e,A A . As J(A)= e,J(A) this implies that dim F J(A ) + dim F ((BIN) = dim F A. The result now follows from (16.15) and (16.16). E
17. Algebras over complete local domains
The following notation is used throughout this section. R is an integral domain satisfying the following conditions:
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(i) R satisfies A.C.C. and RIJ(R) satisfies D.C.C. (ii) R is complete. (iii) J(R)= (7r) is a principal ideal. By (9.10) and (12.7) R is a principal ideal domain and a local ring. R/(7r) is a field. Thus either R is a complete discrete valuation ring or 71- = O and R is a field. In the latter case most of the results in this section reduce to
trivialities. For our purposes the most important examples of rings R satisfying these conditions are of the following type. Let p be a rational prime. Let R o be the integers in a finite extension field of the p-adic numbers and let R be the integers in the completion of an unramified extension field of the quotient field of R o. K is the quotient field of R. A is a finitely generated R -free R -algebra. Let V be an A module and let y E V. Then f) denotes the image of y in 1-/ = V/(7r) V. Similarly R = RI(7r), A = A/(7r)A. Clearly 17 is an A module and if V is a finitely generated A module then 17 is a finitely generated A module. If S is a commutative (not necessarily finitely generated) R -algebra then As = A OR S and Vs = V OR S for any A module V. Of course As is a finitely generated S-algebra and V s is an As module. If V is a finitely generated A module then Vs is a finitely generated A s module. Of special importance is the case S = K. If V is an R -free A module then we will generally identify V with V OR R C V OR K. An R module V is torsion free if rv = 0 for r E R, v E V implies that r = 0 or y = O. A submodule W of the R module V is a pure submodule if (r)W = W n (r)V for every r E R. The fact that R is a principal ideal domain makes the "fundamental theorem of abelian groups" available. We state some consequences without proof. LEMMA 17.1. Let V be a finitely generated R module. (i) V is free if and only if V is torsion free. (ii) V is projective if and only if V is free. (iii) A submodule W of V is pure if and only if W I V. (iv) An epimorphism of V into V is an automorphism.
LEMMA 17.2. Let V be a finitely generated free R module. Let W be a submodule of V. The following are equivalent. (i) W is a pure submodule of V.
..17]
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ALGEBRAS OVER COMPLETE LOCAL DOMAINS
(ii) (7r)W = (7r)V n w. (iii) (w + ('n-)V) (7r )V W (7r )W (iv) The image of W in V is isomorphic to W (y) V I W is free.
If V is an A module and W is a submodule of V then W is torsion free or pure if that is the case for W considered as an R submodule of the R module V.
THEOREM 17.3. The map sending a to a sets up a one to one correspondence between central idempotents in A and central idempotents in A. If e is a central idempotent in A then e is centrally primitive if and only if e is centrally primitive.
PROOF. Immediate by (12.9). COROLLARY 17.4. There is a one to one correspondence B ---> /3- between blocks of A and blocks of A where B = B(e) for a centrally primitive idempotent e of A implies that f? = B(J). If Vis an A module, V E B, then E B. If W is an A module, W E f?, then W considered as an A module is in B.
PROOF. The first statement follows from (17.3) by defining B(e)= B(e). If V E B(e) then Ve = V and (7r)Ve = (7r)V. Thus Ve Ve V and E B(e). If W E B(e) then We = We = W and so W E B(e). 0 LEMMA 17.5. Let L be a finitely generated A K module. Then there exists a finitely generated A module V which is R-free such that V K = L.
PROOF. Let {v,} be a K-basis of L. Let {ai } be an R -basis of A. Let V be the R module generated by {va,}. Then L = VK and V is a finitely generated A module. V is R -free since V is torsion free. LEMMA 17.6. Let P be a finitely generated projective A module and let V be a finitely generated R-free A module. Then HomA (P, V) is an R-free R module and HornA(P, V) --Hom,(P, V). If rank', {HomA (P, V)} = d then I (15 q= d = IK(PK, VK).
furthermore
,
PROOF. By (13.1) it may be assumed that P = eA for some primitive in A. By (8.2) Horn, (eA, V) --- Ve is R -free and HomA(JA, V)--- Ve = Ve. Thus d = rank ', Ve and
idempotent e
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IK (PK , VK ) = dimKVKe = rank ', Ve = dimA Ve = (P, V).
0
THEOREM
17.7 (Brauer). Let V, W be finitely generated R -free A modules
such that
VK
WK. Then
Let e be a primitive idempotent in A and let m, n be the multiplicity of eAleJ(A) in V, q/ respectively. Then by (16.3) and (17.6) PROOF.
m1(eA I eJ(A), eA I eJ(A)) = I, (ëA, V) = 1K (eAK, VK) -
(eAK, WK) —
(eA, W)
= nI(eA I eJ(A), eA I eJ(A)). Hence in = n. Since e was arbitrary the result follows.
E
Theorem 17.7 is of fundamental importance for the whole subject. We give here another proof which is independent of most of the previous theory. See Serre [1977] p. 125. ALTERNATIVE PROOF OF THEOREM 17.7. Since VK ^r WK it may be assumed that VK = WK and after replacing W by a scalar multiple that W C V. Hence there exist an integer n >0 with 1,-"V C W C V. We will prove by induction on n that if 71-"V C WC V then V. Suppose that n = 1. Then 71-W C rV c W C V. Hence there exists an exact sequence "
0—> 7rV/7rW-- W/71-W--> V/7I-V-- V/ W
Let T = VI W. Then T
O.
71-V/71-W. Thus there is an exact sequence
T—>0. This implies that I? We proceed by induction. Suppose that n > 1. Let M = 7r" -I V + W. Then 'Tr"' VCMC V and T-M C W C M. By induction R. f7 and E R W. Thus V The following notation will be used in the rest of this section. Ui, Uk is a complete set of representatives of the isomorphism classes of principal indecomposable A modules. For i = 1, .. . , k let = U,/Rad U, where Rad U, is the radical of U,. Thus by (13.7) L,. .. , Lk is a complete set of representatives of the isomorphism classes of irreducible A modules.
17)
ALGEBRAS OVER COMPLETE LOCAL DOMAINS
67
X,, , X,, is a set of R -free A modules such that (X 1 ) K , , (X,,) K is a complete set of representatives of the isomorphism classes of irreducible AK modules. Clearly each X, is indecomposable. Such modules exist by (17.5). For s = 1,..., n let X, ds,L,. The nonnegative integers d„ are the decomposition numbers of A. By (17.7) they do not depend on the choice of X,. The n X k matrix D = (d,) is the decomposition matrix of A. If e is a centrally primitive idempotent in A then the decomposition matrix of eAe is called the decomposition matrix of the block B where B = B(e). THEOREM 17.8 (Brauer). Suppose that (L„ L,)
(u,\K
((Xs)K, (X5)K)
AK is semi-simple. Then
d„(X„)K.
Furthermore if fc,,} are the Cartan invariants of A then )
(L
c„ = s
—i IK
( (X K,
d
(Xs)K)
d ' ".
PROOF. Since AK is semi-simple ( U, )K (17.7) we have (L„ L,)ds, = IR(U„
=
((U)K,
d,,(Xs ) K.
POO
=
(UK
By (16.4) and ((X,)K, (Xs)K).
This proves the first statement. The second statement is an immediate consequence of the definition of the Cartan invariants. E
17.9. Suppose that AK is semi-simple. Assume that K is a splitting field of AK and R is a splitting field of A. then (U,) K and C = D'D where C is the Cartan matrix of A. If ,B„, are all the blocks of A and D 1' ) , C" is the decomposition matrix of B„ the Cartan matrix of B, respectively, then after a suitable rearrangement of rows and columns COROLLARY
D1')
0)
D =
and
0
D''D") =
D 1-)
Furthermore for no rearrangement of rows and columns is (
D 1 ') =
D;')
0\
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The first two statements follow from (17.8). By (7.8) ds, = 0 if Xs and L, are in different blocks. The last statement follows from (16.7). 0
PROOF.
LEMMA 17.10. Assume that det C 0 where C is the Cartan matrix of A. Let P, Q be finitely generated pro jective A modules. Then P Q if and only if PK QK. If P Q then clearly PK OK. Suppose conversely that PK Let P = a,U„ Q = =, b,U,. Then by (16.4) and (17.6) PROOF.
oy=, E
(Li , Li ) = (1.1 - ), 15 ) =
((
Uj)K,
QK
PK)
i =1
((Uj)K, QK)
E
1=1
(Li , Li )
for j = 1, , k. Hence E`:=, (a, — b i )c,) = 0 for all j. Thus a, = b, for all i as det O. El LEMMA 17.11. Suppose that A is a Frobenius algebra. Let V be a finitely generated R -free A module such that fi = PI ED W where P, is a projective A module. Then there exists a finitely generated pro jective A module P and a finitely generated R-free A module W with V - P (1)W, 1 5 P, and W,. PROOF. By (13.7) there exists a finitely generated projective A module P with 15 Pl . Since P is projective there exists a commutative diagram
V
V
IP
P,--->0.
Then the image of f (P) in V is f (P). Hence by (17.2) f (P) is a direct summand of V as an R module and f is an isomorphism. Since P is R -injective by (15.5) this implies that P V and so V P W for some A module W. Hence /5 (i) ---- = PI @ W I and so IV WI by the unique decomposition property. E 17.12 (Thompson [1967b]). Let U be a principal indecomposable A module. Let V be an AK module such that V J UK. Then there exists an R -free A module W such that WK"-- V and "/Rad 1,V is irreducible where Rad W is the radical of W Hence in particular W is indecomposable. If A is
THEOREM
".
'18]
EXTENSIONS OF DOMAINS
69
a Frobenius algebra then there also exists an R -free A module W' such that V and the socle of IV' is irreducible.
PROOF. It may be assumed that UK = V El) M. Let L = un M and let W = UIL. It is easily verified that L is a pure submodule of U. Thus L c U and Cif, W. Hence by (13.6) W/Rad W is irreducible. If A is a Frobenius algebra let W' = un V. Then W' is a pure submodule of U and so W' C U. Hence by (14.8) the socle of W' is irreducible. E COROLLARY 17.13. Suppose that
AK is semi-simple. Let V be an irreducible module. Then there exists an R -free A module X such that XK = V and I is indecomposable. AK
PROOF. As VI UK for some principal indecomposable A module, the result follows from (17.12). 0
18. Extensions of domains In this section the same notation is used as in Section 17. An integral domain S containing R is an extension of R if the following hold (i) S is a principal ideal domain and a local ring. (ii) S is free as an R module. (iii) J(S) = J(R)S = (7r)S for some integer e. If S is an extension of R then e is the ramification index of S. S is an unramified extension of R if its ramification index is 1. Observe that if S is an extension of R then SIJ(S) is an extension field of R. If R is a field or equivalently 77. = O then an extension of R is simply an arbitrary field containing R. In this case all extensions are unramified. If S is an extension of R which is a finitely generated R -algebra then S is a finite extension of R.
If L is a finite extension of K then the integral closure of R in L is a finite extension of R. Thus if S, and S, are finite extensions of R, there exists a finite extension S of R with S, C S for i =1,2. LEMMA 18.1. Let B be a subring of A with R C,B. Suppose that AB is a finitely generated free B module and B A is a finitely generated free left B module. Let S be an extension of R and let V be a finitely generated
70 B
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-projective A module. Then Vs is a finitely generated B s -projective A s
module.
PROOF. If V VB OB BAA from (4.8). D
then
Vs
BBs OBs B s
(AS)A s .
The result follows
The next result which is somewhat in contrast to (17.12) and (17.13) shows that extensions intrude themselves in a natural way. See Feit [1967131. LEMMA 18.2. Let M be a finitely generated R -free A module and let C V„ = II be a chain of A modules. Let W, = (0) = Vo C V. V-, and let S be an extension of R with ramification index e n. Let L be the quotient M, and field of S. Then there exists an S -free A s module M' such that AIL
PROOF. For i = 1, , n let d, = dim, W. There exists an R -basis {m11 11 •-•. j d„ 1 i n} of M such that for each k = 1, . , n j 1 i lc} is an /-basis of Vk. Let (H) = J(S). Then (H) = (7r)S and {mg } is an S-basis of Ms . Define mr, in Ms by mf, = 11rn,1 for all i, j and let M' be the A s module generated by {m}. Clearly M' C Ms and ML= ML. If a E A then there exist r,, k, É R such that for each k k
m kt a
d,
E E riAtm,
(mod(7r)M).
i = I 1=l
Thus for a E A s there exist s,,k , E S such that for each k k
mk,a
d,
E E si,k,m,,
(mod(H)Ms ).
i=1 j =1
Therefore k
m'a =
d,
E E s f kH m (modoiy,m , ) i= 1 j
and so dk
mra a
E sk1k,m4
(mod(H)M').
j=1
This implies that M'/J(S)M'
fa= (W1 , . ,•
COROLLARY 18.3. Suppose that 7r
O. Let M be a finitely generated R- free
A module. Then there exists a finite extension 5 of R with quotient field L
18]
EXTENSIONS OF DOMAINS
71
and an S -free A s module M' such that ML M L and 1171' = M'/J(S)M' is completely reducible.
PROOF. For any rational integer n > 0 let L = K (XV/r) and let S be the R -algebra generated by NY7F in L. Then S is a finite extension of R with ramification degree n. The result now follows from (18.2). 0 LEMMA 18.4. Let V, W be finitely generated A modules and let S be an extension of R. Then Hom A (V, W) O R S HomA, (Vs, Ws ) and EA(V)ORS EAs(Vs)• PROOF. Let {,s,} be an R -basis of S. Thus every element in HomA ( V, W)O R S can be written in the form E f, Os, with f, E HomA ( V, W). In such an expression the f, are uniquely determined. Define g : Hom,, ( V, W) OR S —>Hom As (Vs, Ws ) by g(f s,) (y s)= f(v)® s,s. Then clearly g is an S-homomorphism and in case V = W g is a ring homomorphism. If g(Ef s,) = 0 then forai! u E V, E (v) s, = 0. Thus f, = 0 for all i. Hence g is a monomorphism. Let h E HomA, ( Vs, Ws ). For each i define f by h(v 0 1) = f, (v) s, Since V is finitely generated there are only finitely many nonzero f . Clearly f E Hom,, ( V, W) for each i and g(f s,)= h. Thus g is an epimorphism. E If V is an irreducible A module then V = V may be considered as an A module. V is an absolutely irreducible A module if for every unramified extension S of R, Vs is an irreducible A s module.
LEMMA 18.5. Let V be an irreducible A module. The following are equivalent.
(i) V = V is an absolutely irreducible A module. (ii) V is an absolutely irreducible A module. (iii) For every finite unramified extension S of R, Vs is an irreducible A s module. (iv)
EA(V) '••--•
R.
(y) 1 is a splitting field of f/. PROOF. (i) (ii). If S is an unramified extension of R then SIJ(S) is an extension field of 1 and Vs = VS/i(s). (ii) (iii). Clear.
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(iii) (iv). If EA (V)/ R then there exists a monic irreducible polynomial f (t) E R[t] which has a root in E, (V) — R. Let fo(t)E R[t] such that fo(t) is monic and f o(t) = f (t). Let S = R[a] where a is a root of f o(t). Since S/(71-)S is a field S is a finite unramified extension of R and f (t) has a root in S. Thus EA (V) O R S is not a division ring. Hence by (18.4) EA , (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 , Uk be a complete set of representatives of the isomorphism classes of principal indecomposable A modules. By (16.4) there exists j with /A (U„ V) = 1. Let F be an extension field of R. Thus by (18.4) /,((U,),, V,) = 1. Let LI , U„ be a complete set of representatives of the principal indecomposable A, modules. Since (UOF AF this implies that there exists i with /,(U:, V,)= 1. Hence by (16.4) V, is irreducible. 0 An A module V is absolutely indecomposable if for every finite extension S of R, Vs is an indecomposable A s module.
LEMMA 18.6. Let V be a finitely generated indecomposable A module and let E = EA (V). If EIJ(E)----, R then V is absolutely indecomposable. If R is algebraically closed then every finitely generated indecomposable A module is absolutely indecomposable.
PROOF. Since EIJ(E) is a division ring which is a finitely generated 1 -algebra the second statement follows from the first. Suppose that EIJ(E) - ---, ft. Let S be a finite extension of R. By assumption J(R)S C J(S)C J(Es ). Thus J(R)Es C J(Es ). Since J(E), I J(R)Es is a nilpotent ideal in Es I J(R)Es this implies that J(E), C J(Es ). Since E5 1J(E)5 SIJ(R)S it follows that Es 1J(Es ) - ---, SIJ(S) has only one idempotent. Thus by (12.3) Es has only one idempotent and so by (18.4) EA(V s ) has only one idempotent. Hence V, is indecomposable. E 18.7. Let V be a finitely generated indecomposable A module. There exists a finite unramified extension S of R such that Vs = ED,n=, where each V, is an absolutely indecomposable S module and
LEMMA
EA,(V,)1J(EA s (V)) .-----
PROOF. If S is a finite unramified extension of R then V, is the direct sum of n(S) nonzero indecomposable A, modules for some integer n (S) with
181
EXTENSIONS OF DOMAINS
73
n(S) < dimA (V). Choose S so that n(S) is maximum. Clearly every component of V, remains indecomposable when tensored with any finite unramified extension of S. Replacing R by S it may be assumed that V, is indecomposable for any finite unramified extension S of R. For any finite unramified extension S of R let D (S) = EA s (Vs )/J(EA ,(Vs )). Thus D(S) is a finite dimensional g-algebra which is a division ring. If T is a finite unramified extension of S then J(S)T C J(T)C J(EA , (VT)) by assumption and so J(S)EA.,-( VT) C J(EA T ( VT )). Thus since J(EA ,( V, ))71./(S)EA T ( VT) is nilpotent, D (T) is a homomorphic image of D (S)(3), T. Choose a finite unramified extension S of R such that dim§D(S) is minimum. If D (S)/ g there exists a monic irreducible polynomial f (t)E §[t] of degree at least two which has a root in D(S). Choose f o(t)E S[t] such that fo(t) is monic and fo(t)= f (t). Let T = S[a] where a is a root of fo(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)® s T is not a division ring and so dimi-D( T) < dim,-D(S) contrary to the choice of S. Hence D(S) = S' and V, 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 1 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 irreducible if and only if EA (U)/J(EA (U))-- 1. PROOF. By (13.8) EA (L)— EA (U)/J(EA (U)). The result follows from (18.5). 0 COROLLARY 18.9. There exists a finite unramified extension S of R such that S is a splitting field of A. PROOF. By (18.7) there exists a finite unramified extension S of R such that (As , = U, where EA, (U1 )/ J(EA 5 (U)) g for each i. By (18.8) S. is a splitting field of A. LI )A
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[19
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 —> E R (V) is a representation of A define va = vf (a) for y 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 y E V, a E A. Then f is an R -representation of A. Thus there is a natural one to one correspondence 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 adjectives such as irreducible etc. will be applied to the representation f if they apply to the underlying module of f. Two representations f,, f 2 with underlying modules V, V, are equivalent if V, is isomorphic to V,. It is easily seen that f, is equivalent to f, if and only if there exists an R -isomorphism g: V 1 —> V, such that f2= Let V be a finitely generated R -free R module with a basis consisting of n elements. By (8.6) ER (V) R. Thus a representation of A defines an algebra homomorphism of A into R„. 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„. It is clear that tv is independent of the choice of isomorphism mapping ER (V) onto R. 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 = tw . Furthermore if K is the quotient field of R then t,„(a)= t(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 tv,(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 characters. See for instance Curtis and Reiner [1962]. However we prefer to 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 ---- W if and only if tv = tw.
19]
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REPRESENTATIONS AND TRACES
If V------ W then tv = tw . Suppose that W. Let Iv 1 , be the annihilator of V, W respectively. Let I = I n i. It suffices to prove the result for the algebra A I I. Thus by changing notation it may be assumed that I = (0). Hence J (A) = (0). Since A has two nonisomorphic irreducible modules the Artin—Wedderburn theorems (8.10) and (8.11) imply that A= A, where A v and A w are simple rings and V, W is a faithful absolutely irreducible A, A w module respectively. Since V and W are absolutely irreducible it follows from (8.11) that A v F„, and Aw Fn for some m, n and the isomorphisms sending Av to F„, and Aw to F. are equivalent to representations with underlying modules V, W respectively. Choose a E A v such that a corresponds to the matrix
PROOF.
,
e
in F.
Then tv (a) = 1 and tw (a) = O. Hence tv tw. D
Suppose that K is a finite Galois extension of F and cr is an automorphism of K over F. Then o- defines an automorphism of AK by (a .x)" = a Øf' for a E A, x E K. This automorphism will also be denoted by a. If Vis an Ai, module let V' = { v, I v E V} where vn. + w„ = (v + w),
for v, w E
v,,a" = (va),,
for v E V, a E AK
V, .
Clearly V' is an Ai, module and V — ( V') for CT, T automorphisms of K over F. Furthermore { tv - (a )}" = tv (a) for a E AK. Thus tv- = (V, The next result is essentially a corollary of the Artin—Wedderburn theorems (8.10) and (8.11). LEMMA 19.2. Let L be a finite Galois extension of F and let A = L„ for some integer n > 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- ---- e),,EGA„, with A. L„ for all cr E G. (ii) There is an irreducible A, module M such that WL ec,E,;m-,04-} is a complete set of representatives of the isomorphism classes of irreducible A L modules and each M' is absolutely irreducible. Furthermore if t,,,r (a) = tm , (a) for all a E A then o- = T.
76
[19
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PROOF. Let W be an irreducible A module. By (8.11) every irreducible A module is isomorphic to W. It is easily seen that dim, W = n [L : F]. If cr E G then (T defines an automorphism of A = L„ in a natural way. This will also be denoted by u. Let L = F(c) and let 0 b = (0
o
)E
A.
0 Let U be an n -dimensional L space. For ci E G define f, : A L EL (U) by u(f,a)= ua". Thus each f, is an absolutely irreducible representation of A,. Let U„ be the underlying A, module and let t, = tu,r. Then t„ = Hence if a T then t„(b)= c' X c-r = t,(b). Thus Ur,. UT for T by (19.1). Let I be the annihilator of Ui,. By (8.10) and (8.11) A,// „EG A„ where [Jr. is a faithful irreducible A„ module. Hence A, L„ for all ci E G and
ED
ED
dim L A L = n2 (L : F)= dim, ( epi A„) = dimL (AL II). (TEG
Thus / = (0) and (i) is proved. Furthermore every irreducible A L module is isomorphic to some U. and each U, is absolutely irreducible. For some a E G, U. I WL . As IV', = W, the unique decomposition property (11.4) implies that le„E G U. WL. Thus W, ED, E,3 [Jr. as dim, W, = dim, W = n[L : F] = dimL (ED [Ç). ,EG
Define M = U. By (19.1) AT'
= U, for o- E G.
For the next two results we require Wedderburn's theorem which asserts that a finite division ring is a field. See for instance Curtis and Reiner [1962] p. 458 for a proof of this theorem. We will also need other well-known results such as the fact that a finite extension of a finite field is a Galois extension with a cyclic Galois group.
THEOREM 19.3 (Brauer). Let a l , ... ak be an F-basis of A. Let K be an extension field of F and let V be an absolutely irreducible AK module. Assume that F is finite and tv (a,) E F for i =1,... ,k. Then there exists an
19]
77
REPRESENTATIONS AND TRACES
absolutely irreducible A module W such that WK ---- V and tv (a) = tw (a) for all a E A.
PROOF. Let 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 (8.11) and the fact that finite division rings are fields it follows that A ---- L„ for some finite extension field L of F. It may be assumed that A L. By replacing K by KL and V by VKL it may be assumed that L C K. Let M and G be defined as in (19.2). Then V M'Erc for some cr E G. Thus by assumption t,„,- (a) = t,„,= (a) for all a E A and all O T E G. Hence (19.2) implies that G = (1) and so F = L. Therefore W M and ,
WK
V.
D
THEOREM 19.4. Assume that F is finite. Let K be a finite extension of F.
(i) Suppose that V is an absolutely irreducible A K module. Let [F(tv ): F] = m and let (cr) be the Galois group of F(tv ) over F. There exists an irreducible A module W such that WK y with tv, = 6'; . (ii) Let W be an irreducible A module. Then WK ED:=, V, where {V,} is a set of pairwise nonisomorphic irreducible A K modules and s = [K n L : F]. Furthermore there exists an element ci in the Galois group of K over F with V" ----- V and V, V''' for i = 1, , s.
ED:1,
PROOF. (i) Since V is a constituent of AK by (6.1) there exists an irreducible A module W such that V is a constituent of W. It may be assumed that W is faithful. By the Artin—Wedderburn theorems (8.10), (8.11) it follows that A L„, for a finite extension L of F. By (19.2) L = F(t). By (19.3) it may be assumed that K = F (tv ) = L. The result follows from (19.2). (ii) Without loss of generality it may be assumed that W is faithful. Then, as above, A ----- L„ for some finite extension L of F. Let M be defined as in (19.2). Apply (i) to M, and let V be an AK module such that
e
[ICL.K1 VKL
M.
Since dimL M n it follows that dim, VKL = [KL : Kin =[L : K n Lin. Thus if U is an irreducible AKflL module then U, is irreducible. Hence by replacing K by KnL it may be assumed that K=KnLCL.
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Let G = (a) be the Galois group of L over F. If s = [K : F] and H = (as) then G/H is the Galois group of K over F. Since V, W1 it follows that V I WK . Hence V" I WK as WK = W K . It is easily seen that V"' V"' if and only if s — j). Thus WK. Since
e:=,
dimK WK =
it follows that
WK
n[L : F] = ns[L : K] = dinbc V'), ( Et) i=1 fEDI 1 V
CHAPTER II 1. Group algebras All groups are assumed to be finite unless the contrary is explicitly stated. If H is a subgroup of G then a cross section of H in G is a set of elements { x,} in G such that G = U Hx, and Hx,# Hx, for i j. A left cross section of H in G is defined analogously. If x E G and A is a subset of G then A = x'Ax. We will freely use standard terminology and notation from the theory of groups. See for instance Gorenstein [1968], Huppert [1967]. Throughout this chapter R is a commutative ring which satisfies A.C.C. and G is a group. The aim of the chapter is to investigate properties of the group algebra R[G] which depend on the fact that R[G] is a group algebra rather than an arbitrary R -free R -algebra. The structure of the ring R plays a relatively minor role and many of the results in this chapter hold even if R does not satisfy A.C.C. However in some results it is necessary to make some assumptions concerning R. If A is a subset of G define A' in R[G] by A' = E AX.
2. Modules over group algebras
Let H be a subgroup of G and let V be an R [G] module then V, = VR [ H, denotes the restriction of V to R [H] . If W is an R [H] module then W C' = Vit R
RI MR [G1R[GI •
It-fl
80
CHAPTER II
The R[G] module 14/G is said to be Let {x,} be a cross section of H in module with basis {x,}. Thus if V x, where this is a direct sum
ED ,
[2
induced by W. G. Then R[G] is a free left R[H]
is an R[H] module then V`; = of R modules and if v E V, x E G
then (v ®x,)x = y ® x ix = y ®yx; = vy xi where x,x = yx, with y E H. Suppose that u is an automorphism of R[G] such that R = R and G" = G. Thus u defines an automorphism of R and one of G. If H is a subgroup of G and V is an R [H] module define the R [H]" module V' as follows. V" = {v, y E V} where v, + w, = (v + w),
for v, w E V
v,,a" = (va),,
for v E V, a E R[H].
If u is an automorphism of R let V' = 1/"- where cr is the automorphism of R [G] defined by (ExEc; E,,EGr'xrx. Observe that this coincides with the definition in section 19 of Chapter I, where both
definitions apply. If u is an automorphism of G let V" = WI where o- , is the automorphism of R[G] defined by (ExEG Gx) = E G Gx". For x E G define the R [H" I-module V' =V0x={v0xlvE V} where (v 0x )y" = vy x for y E H. Clearly V'---- V" where z" = x -i zx for all z E H. LEMMA 2.1. Let H be a subgroup of G. Let V be an R[G] module and let W be an R [H] module. (i) If V = V, G V 2 then V, = (V,)1, e (V2)„. (ii) If W = w, W2 then W G (iii) If o- is an automorphism of R[G] such that R - = R and G° = G then (W") G (W G )". In particular W G = (W") G for x E G. (iv) If HCA CG for some subgroup A of G then W G (VI / 4 ) G . (y) Let cr be an automorphism of R[G] such that R" =R and G° = G. The map sending U to U" is a one to one map from the set of all submodules U of W onto the set of all submodules II' of W° which preserves inclusion, intersections, sums and direct sums. In particular U is irreducible or
81
MODULES OVER GROUP ALGEBRAS
indecomposable if and only if U' is irreducible or indecomposable respectively. PROOF.
Immediate from the definitions. El
2.2. Let H be a subgroup of G and let G. Let V be an R [G] module. Then V and only if there exists a submodule W, of V= wox, where this is a direct sum of R
LEMMA
{x,} be a cross section of H in for some R [H] module W if V, 1 such that W W , and modules.
If V = W G let Wu = W 01. Then V = e 1/110 x.. Conversely the R -linear map from WV to V which sends w Ø x wx, is easily seen to be
PROOF.
an isomorphism.
E
If U, V are R[G] modules then U OR V is an R module. For u E U, E V, x E G define (u v)x = wc Ø vx. It follows easily that in this way UOR V becomes an R[G] module. This R[G] module is denoted by U0 V. The definition implies immediately that
VOU, (U, (i) U2) 0 V ---- (U, 0 V )(1) U, 0 V). (
2.3. Let H be a subgroup of G. Let V be an R[G] module and let W be an R [H] module. Then
LEMMA
V 0 W G (V, 0 W ) Û . Let {x,} be a cross section of H in G. Define f: VO W G --> (1.7H Ø Wr by
PROOF.
f{v 0(w x,)} = (vx ®w)®x
for v E V, w E W. Clearly f is an R -isomorphism. Let x E G. Suppose that x,x = yx, with y E H. Then f{v 0(w
.x.)}x = (v.xT 1
w)y
x,
=(vxi'y ® wy)0x 1 = ( vxx,
wy)0x1
and
f[fv 0(w ®x)}x]= f fvx 0(wy x,)} = (vxx, '0 wy)0x 1 .
Thus f is an R[G] homomorphism as required.
0
82
CHAPTER II
[2
Let V be an R[G] module. An element v E Visa G -invariant element or simply an invariant element if vx = v for all x E G. Let Inv o (V) denote the set of G-invariant elements in V. Clearly Inv G ( V) is an R module. Let V, W be R[G] modules. For f E Hom, ( V, W) and x E G define fx by (v)(fx)= {(vx 1 )f}x for v E V Clearly ft E Hom, (V, W) and (fx)y = f(xy). In this way Hom, (V, W) becomes an R[G] module. If V, W are finitely generated R -free R [G] modules of rank m and n respectively then Hom, (V, V) R„„ HomR (W, R„ and Hom, (V, W) consists of all m X n matrices with entries in R. For x E G let a„, b„ respectively be the map sending v to vx, w to wx respectively for E V, w E W. Then a„ E Hom, (V, V), b„ E Hom, (W, W) and if f E Hom, ( V, W) then fx = 'fbx. In case W = R with rx = r for all r E R and x E G, Hom R (V, R) = / as an R module. If f E and v E V, (v)(fx) = (vx - 1 ) f = v(x f) for x E G. Hom R (V, R) made into an R[G] module in this way will be denoted by V*(R) or simply V* if R is determined by context. LEMMA
2.4. Let V,
W be R[G] modules. Then Inv G (Hom, ( V, W)) =
H0m, [01 ( V, W). PROOF. By definition f E Inv G (Hom R (V, W)) if and only if f = fx for all x E G. This is the case if and only if (v) f = (v)(fx)= {(vx ')f}x for all v E V, x E G. This last condition is equivalent to the fact that
f E Hom R icd(V, W). D 2.5. Let H be a subgroup of G. Let V be an R [G] module and let W be an R [H] module. Then (i) {Hom R (W, V H )} (3 Hom R (W G, V), (ii) Mom, ( VH , W)l e Hom, ( V, W e ).
LEMMA
PROOF. Let {x i } be a cross section of H in G. (i) If f E {Hom, (W, VH )}e then f = Ef x ; for f E Hom, (W, VH ). Define f EHomR (W G, V) by (Ewi xi )f =Efotofi lx i . Let g be the map sending f to f Clearly g is an R -homomorphism. If f = 0 then = 0 for all w; E W and so fi = 0 for all i. Thus f = O. Therefore g is a monomorphism. If h E Hom R (W G, V) then (/w, ® x)h = E{(wi)hi}x i for some hi E Hom R (W, VH ). Hence h = f with f = Eh; x ; and so g is an
epimorphism. Let x E G. For each i, x ix = yix, where y; E H and x ; —> x, is a permutation of {.0. Thus if f = x; then fx = Efy, Øx. Furthermore
83
MODULES OVER GROUP ALGEBRAS
E w, ox,)(fx)-={(E = {(E wy7 I 0x)f} x Ef(wryTi)f}xix and
(E
w, x,) (f-x- = (
E iv,
x,
E fy,
=
E {(w, y Since y,x,, = xix this implies that g is an R[G] isomorphism. (ii) If f E{Hom R ( VH , W)} G then f = 0x, where f Hom,,(v„, W). Define f E HomR (v, W G ) by (v)! = E(vxT I )f, x,. Let g be the map sending f to f. It is easily seen that g is an R isomorphism. Let x E G. Let xix = yix,, with y, E H for each L Then if f = f .x,
(v)(fx) = {(vx-i)f}x =(vx- ,.)f, Ø x,x =.Ef(vx-ix»)fl y, and
(v)6Z)= (u) (E fyi
= Ef(vicï»yTi)flyi Oxi'.
Since x,»y7 I = x -I xT I this implies that g is an R[G] isomorphism. LI COROLLARY
2.6. Let H be a subgroup of G and let W be an R[H] module.
Then (W*) G (WG)*. PROOF.
Clear by (2.5)(i). 0
LEMMA 2.7. Assume that R satisfies A.C.C. Let H be a subgroup of G. Let Lr, V be finitely generated R[G] modules. The following hold. (i) If V is R -free and U is a projective or free R[G] module then each of U QV, U*, HomR(U, V) and HomR (V, U) is a projective or free R[G] module respectively. V, U*, Hom R (U, V) and (ii) If U is R[H]-projective then U Hom, (V, U) are all R[H]-projective.
84
[2
CHAPTER II
PROOF. (i) Let V( , ) = nR R . By (2.3) V® mR and (2.5) Hom R (mR, V)
mnR
is free. By (1.8.2)
{Hom R (mRR, nRR)} G
mn{Hom R (RR, RR)}' mnR 2 is free and Hom R (V, mR 2)— {HornR (nR R , mR R )} 6 mnR";. is free. The rest of (i) follows from (ii) with H = (1). (ii) By (1.4.8) U I(UH) G . Hence by (2.3) U0 VI(U„) G 0 V
(U„ Ø V„) G ={(UØ V)„} G.
Thus by (1.4.8) U 0 V is R [H]-projective. By (2.5) Hom, (U, V) Hom ((Upir, V)
{Hom R (U,
Hom R (V, U)1Hom R (V, (Uff) 6 ){Homp (V, U)Ii} G .
Thus Hom R(U, V) and Hom, (V, U) are R [H]-projective by (1.4.8). E Osima [1941] first proved (2.7)(i). This answered a question raised in Brauer and Nesbitt [1941]. LEMMA 2.8. Let V be a finitely generated R -free R[G] module and let W be an R[G] module. Then V*0 W Hom, (V, W). PROOF. Let {y,} be an R-basis of V and let WI be the dual basis of V*. Define g: V*0 W—>Hom R (V, W) by (y){g(f0 w)}={(y)f}w for y E V, w E W, f E V*. Clearly g is an R-homomorphism. If g(Ev w,) = 0 then w, = (v,){g(Ey* w,)} = O. Thus g is a monomorphism. If h E Hom R (V, W) then h = g(Ev (Oh). Thus g is an R -isomorphism. If x E G then for y E V, w E W, f E V* (v)[gl(f
w ).0] = (v){g(fx
(v)[{g(f
w)}..z]= [(v.x - '){g(f
wx)} = {v(fx)}wx = {(yx -I )f}wx
and w)}]x = {(v.x -t )f}wx.
Thus g is an R[G] isomorphism. Lii The next two results are due to Mackey [1951].
21
85
MODULES OVER GROUP ALGEBRAS
2.9 (Mackey decomposition). Let H, A be subgroups of G and let {x,} be a cross section of H in G. For any R[H] module W and any (H, A) double coset D define the R[A] module W(D)= ex,E0 W x,. Then W(D)------{W w n A } A for any x E D and THEOREM
=f W(D) -----
{W;r=nAl A
where D and HxA range over all the (H, A) double cosets in G.
PROOF. Let Wo(H, x, A) = { W;pnA} A . If xo E HxA then xo = yxa with y E H and a E A. Thus Wo(H, xo, A) = 1V;:j7.. flAI A {W xf:'"rlA } "
W
Fix (-14 ' } A
WOW x, A ).
Hence Wo(H, x, A) = Wo(HxA) depends only on the double coset HxA. It suffices to show that W(D ) W0(D) for any (H, A) double coset D. Let D = HxA. If x, E D there exists a, E A such that Hx, = Hxa,. Furthermore Hxa, = Hxa, if and only if a,a» E H. n A. Thus { a, } is a cross section of Hx n A in A where i ranges over all values for which E D. Hence
vvo(D) G WL. nA aj, W(D)=
ED
wox,,G
Wxa
x,ED
Define f: W(D)— > W0(D) by f(w xa,)= wx Ø a,. Clearly f is an Risomorphism. Suppose that a E A so- that a,a = x yxa, with x E n A. Thus {f (w
xa, )}a = wx Øa 1a = wyx
a, = f (wy ®xa 1 )
= f(w yxa,)= f(w xa,a)
and f is an R [A ] isomorphism. THEOREM 2.10 (Mackey Tensor Product Theorem). Let H, A be subgroups of G. Let V be an R[H] module and let W be an R[A] module. Then VG W G
® w „..)G
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[2
CHAPTER II
where in G.
x ranges over a complete set of (H, A) double coset representatives
PROOF. By (2.3) and (2.9) VG
® 14/6
)4
Ø
wr e [{ v„}A
0
W.
By (2.3)
{Vw „ } 1 W { VivnA Ø 14/FpnA}'The result follows from (2.1)(iv). 0 Suppose that H < G. Let V be an R[H] module. The inertia group (10 = T( V) of V is defined by
TH
T(V) = {x Vx -----= V, x E G } . Clearly T( V) is a subgroup of G and H C T(V).
2.11. Suppose that H < G. Let W be an R[H] module. Let {x, be a cross section of T(W) in G. Then
COROLLARY
(W G )
te w}.
PROOF. Clear by (2.9). E If V is an R[G] module then the kernel of V is defined as the set of all x E G with vx = v for all v E V. V is faithful if the kernel of V is (1). LEMMA 2.12. Let V be an R[G] module with kernel H. Then H < G. Furthermore V = Inv A (V) for a subgroup A of G if and only if A C H. PROOF.
Clear by definition. Lii
LEMMA 2.13. Let H be a subgroup of G and let W be an R [H] module then the kernel of
W G is contained in the kernel of W.
PROOF. Let A be the kernel of W G . By (2.9) 114/ 4 r is an R[A] submodule of W G and hence is A-invariant. This implies that A = H n A and so A C H. The result follows from (2.11). El
RELATIVE TRACES
87
3. Relative traces Let H be a subgroup of G and let {x,} be a cross section of H in G. For an R[G] module V define Inv H ( V) —> Inv G ( V) by Tr`H(y) = I, vx,. Clearly the definition of Tr".; is independent of the choice of cross section 4,1. Furthermore Tr".; is an R -homomorphism. If HCA CG for some subgroup A then it follows directly from the definition that TrTr','., = If y E InvH ( V) then Tr(v) is the relative G, H trace of v. If y E V then Tr ) (y) is the G-trace of y or simply the trace of v. Let ,S) be a nonempty set of subgroups of G. For an R[G] module V define Fr(G,,, V) = InvG ( V)/
E
Tr`H{Inv H ( V) } .
H E5
= {H } we write H° (G, V) = HAG, H, V). Clearly H° (G, S:), V) is an R module and if V is finitely generated then HAG, V) is a finitely generated R module. If
LEMMA 3.1. Let V,, V2 and V be R[G] modules with V = V, ED V2. Let 5 5 be a nonempty set of subgroups of G. Then the following hold. (i) EHEy5 InVH ( VI) ED EHE.5.5 InVH ( V2) = EHE, Invi4 ( V). EFIEY3 Tr2{Inv,i ( VI)) EDEFieç-, ( V)}. ( V2)) =
Op
G,
Vi )
Tr( G, S:), V2) =
PROOF. Immediate from the definition.
G„ S:), V). E
LEMMA 3.2. Let H be a subgroup of G and let V be an R[G] module. Let x G G. Then y E Inv H ( V) if and only if vx E Inv w ( V). Furthermore Tr2{InvH ( V)) = Tr cH; ‘{InvH , ( V)). PROOF. If lx,) is a cross section of H in G then Ix is a cross section of H" in G. This implies the result. E LEMMA 3.3. Let H be a subgroup of G. Let HE
C {A A is a subgroup of G such that A C H for some x E G).
Then TOInvH(V)} = EA E:1 Tr(A{InvA ( V)} for every R[G] module V and I(G, H, V) = Tr(G, g), V). PROOF. By (3.2)
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CHAPTER II
[3
c E Tr,`,■ {Inv A (V)} C E Tr,`,■ {Inv A (V)}
Tr2{inv,, ( V)}
AE,
= E Tr2Tr{Inv A ( V)} C Tr2{Inv i, ( V)}. Ag hl
The results follow. El
3.4. Let H be a subgroup of G and let W be an R[H] module. Then y E Inv c (W G ) if and only if y = Tr2(w 0 1) for some w E Inv, (W). Furthermore H° (G, H, W G ) = (0). LEMMA
PROOF. If w E (W) then Tr2(w 01)E Inv G (W G ). Let {x,} be a cross section of H in G with x, = 1. Suppose that y = Ew, Øx, EInv G (W G ) for w, E W. Then for each j and each y E H = yx y = w,y Øi+ for suitable w:E W. Thus w,y = w, for all j and all y E H. Hence w, E ( V) and y = W Ø x, = Tr(w, 0 1). E
3.5. Let H be a subgroup of G. Let 21 be a nonempty set of subgroups of H and let 3 ={HC A"1A E2t x E G}. Let W be an R[H] module. Then y E EA ET Tr{Inv A ( W G )} if and only if • = Tr(w 01) for some w c E BETTrLi{Inv B (W)}. Furthermore fr(G,21, Tr(H, T, W). Thus in particular H ( G, (1), W G )-----,LEMMA
,
110 (11, (1), W).
PROOF. Suppose that y = Tr2(w 0 1) with w = Tr(u) where B = H CI A' with A e N and u E InvB (W). Thus y = Tr,( (u 01). Hence by (3.2) E Invw - 'n A (W G ) and u y = Tr;nA , (u
0 1) = Tr";- - ' n A(u
') =
A (u
Ø x )}.
Thus y E EA E91 Tr CA; {inVA (W G )}. y = Tr(u) for some A E Ç21 and Suppose conversely that u EInv A (W G ). Let {x,} be a complete set of representatives of the (H, A) double cosets in G. By the Mackey decomposition (2.9) and (3.4) u = Eu, where u, = x,) for some w, E W. Therefore by (3.2)
= E Trc,:knA(w, oxi) = = Tr?, { E Tr nA „,(w, 001= Trc,-; { E
0 1)
H
,)0 1 1.
89
RELATIVE TRACES
Thus
y
= Tr ii(w 0 1) where w -=
ETrnAv(w)E E BET
TrV,{Inv R ( W)}
as required. Let f: Inv, / ( W)—>Inv G (W G ) be defined by f(w) = Tr(w 01). By (3.4) f is an isomorphism. By the previous paragraph f(E BE 93 TeRifInv R (W)D= EAE2■ Tr`A{InvA (W G )}. Thus f induces an isomorphism from H °(H,Vf, W) onto 1-1`) (G,3, W G ). If A = {(1)} then B = {(1)} and so HAH, (1), H°(G, (1), W G ). El LEMMA 3.6. Let H be a subgroup of G. Let V, W be finitely generated R[G] modules where V is R[11 ] -projective. Then (i) 11°(G, H, V) = (0). (ii) H ( G, H, HomR (V, W))= H( G, H, HomR (W, V)) = (0). PROOF. (i) By (1.4.8) V (V,,) ° . Hence by (3.4) W(G, H, V)1 H ° (G, H,(V H )G )= (0). (ii) Immediate by (i) and (2.7). 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 EHomRIHI(U, V) and g E HomREGI(V, W) then Te(gf)= g Tr".;(f). If f E HomFoGr(U, V) and g E H0mR 1 H I(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 i (ba)=1"1-2(b)a for a E Inv G (A) and b E Inv H (A ). Furthermore Tr(Inv H (A)) is an ideal in the ring InvG (A ). (iii) If Vis an R[G] module then Tr2(EREH 1 V)) is an ideal in EREGi(V). (
PROOF. Immediate by definition and (2.4).
CI
The following fundamental result is essentially due to D. G. Higman [1954]. THEOREM 3.8. Let H be a subgroup of G and let V be a finitely generated R[G] module. The following are equivalent. (i) V is R[H]-projective. (ii) V (VH)G.
90
CHAPTER II
(iii) V W G for some finitely generated R[H] module W. (iv) Hom, R(V, V) is R[H]-projective. (y) HAG, H,Hom R (V, V)) = (0). (vi) There exists f E Homz i Fi l ( V, V) = Inv,{Hom R ( V, V)} such that
Tr2(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 C W such that V, I W,. Thus there exists a projection e of W onto V which is an R [H]-homomorphism. Hence Tr2(ef) E Hom Rici ( W, W). Let fx,} be a cross section of H in G. If w E W then w Tr2(ef) =
E {(wx»)ef}x;
E
E
( Wef)x,
cE
(Vf)x, C V
and if v E V then v Tr(ef)= E {[(vx I )e] flx, =
E {(yx Ti)f}xi
= v E (fx,)
= y Tr(f) = v.
Hence Tr2(ef) is a projection of W onto V and so V W as required. (vii) (ii). Let fx,} be a cross section of H in G with x, = 1. Define g: V (V ii r by gv = E, vx x, . Thus g =Tr2(h) where V 01 with hv = y 01. If gv = 0 then vx Ox, = — O. Hence g is an R[G]-monomorphism. Let W = v, Ox,}. Then W is an R[H] module. Clearly g( V) n w = (0). If Ev, Ø x, c VG then
E y, 0x, = E
yix;i 0x,
= g(v 1 )+
+E (y,
- y 1 x7.)
o
(); - v i x; i)0x, E g(V)-I- W. i
Hence {( V,)9, = g(V) H ED W. Therefore V, If( VF, since V is R[H]-injective. D
)G
and so V (V, )G
COROLLARY 3.9. Let H be a subgroup of G. Let V be a finitely generated R-free R[G] module. Then V is Hu(G, H, V* 0 V) = (0).
R . [H1-projective if and only
if
3]
91
RELATIVE TRACES
PROOF. Clear by
(2.8) and (3.8). 0
COROLLARY 3.10. Suppose that
G : H 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 : H1)1 E Hom Ri pii ( V, V). Then Tr(f) = 1. The result follows from (3.8). CI COROLLARY 3.11. Suppose that 1G1 has an inverse in R. Then every finitely generated R[G] module is R -projective and every finitely generated R -free R[G] module is projective. If furthermore Vis an indecomposable R[G] module and W is a submodule of V with WR1VR then W = (0) or W = V.
PROOF. Clear by (3.10) and (1.4.6). D COROLLARY 3.12 (Maschke). Let F be a field whose characteristic does not divide 1G1. 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 (3.11) 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. E
The following result in case H = (1) is due to Green [19744 LEMMA 3.13. Let H be a subgroup of g. Let V,
W be finitely generated
R-free R[G] modules. Let g E Horn rz[ G 1( V, W). The following are equivalent. (i) g ETI-2(Hom R E H AV, W)). (ii) Suppose that U is a finitely generated R[H]-projective R[G] module such that U L W —>0 is an exact sequence and U,,j-> W,, —>0 is split. Then there exists h EHom R[GI (V, U) with g = fh. (iii) Suppose that U is a finitely generated R[H]-injective R[G] module such that 0— > V L> U is an exact sequence and 0—> V, L> LI, is split. Then there exists h EHom RIGI (U, W) with g = hf.
PROOF. (i)
(ii). There exists t E HomRiii i(W, U) with ft = 1. Let g =
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CHAPTER II
Tr(go ) with go E HomR,H 1 ( Then by (3.7)
W).
[4
Let h =Tr2(tgo)EHom RiGI (V, U).
fh = Tr(ftgo)= Tr(g())= g. (i). By Higman's Theorem (3.8) there exists go E HomR,m(U, U) with Tr(go) = 1. Thus by (3.7) (ii)
TOfgoh)= f Tr2(go)h = fh = g. The dual of the argument above shows that (i) is equivalent to (iii). 0
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 A c (R [G]) is defined as follows. A c (R [G]) 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 ( V, e 172)= (v + ( V2). Multiplication is defined by ( V1)( V2) = V2). It is easily seen that A c (R[G]) is a commutative ring with 1 = (R) where R = Inv R[Gi (R The Grothendieck algebra A`(R[G]) equals A c (R[G])/ I where / is the ideal of A c (R[G]) generated by all (U) (V) + (W) where U, V, W are finitely generated R[G] modules such that there exists an exact sequence )
—
LEMMA 4.1. If R[G] has the unique decomposition property then A c (R [G]) is free as a C module with basis (y) where V- over a complete set of representatives of the isomorphism classes of finitely generated indecomposable R[G] modules.
PROOF.
Clear.
0
LEMMA 4.2. Assume that R is a field. Let C be an integral domain whose quotient field has characteristic 0 and let (V) 0 denote the image of (V) in A`(R[G]). Let LI, ,L, be a complete set of representatives of the isomorphism classes of irreducible R[G] modules. Then (V) ° =1%, c, (L,)° if and only if L, occurs as a composition factor of V with multiplicity c,. In particular (171 )° = ( V 2)° if and only if V, V,.
5] PROOF.
ALGEBRAIC MODULES
Clear by definition.
93
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 is a nonempty set of subgroups of G let A c (R [G.]) be the C submodule of A c (R [GI) generated by all ( V) where Vis R[H1-projective for some H E . LEMMA 4.3. Let
be a nonempty set of subgroups of G. Then A c t,(R[G1) is an ideal in A c (R[G]). If furthermore R[G] has the unique decomposition property then A c t (R[G]) is free as a C module with basis (y) where y 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 PROOF.
Clear by (2.3). E
Let Z be the ring of rational integers. A finitely generated R[G] module V is (R, 5))- projective or simply -projective if V E A 7, t (R[G]). If R[G] has the unique decomposition property then clearly V is S.:)-projective if and only if V = le V, where for each i there exists H, E k, ) such that V, is R[H, Fprojective. For future reference we introduce the following notation. If V and W are finitely generated R[G] modules then V =- W (.Ç5 ) means that (V) — ( W) E A 7, t (R[G]). In particular if V is g-:), -projective then V —= 0 ( ) and conversely.
5. Algebraic modules Throughout this section C is the field of complex numbers. The Green ring A c (R [G1) is generally very large. This section contains the definitions and basic properties of some interesting subrings. These ideas are due to Alperin [1976c], [1976e]. , V. If V is an R[G] module let V" = V • • • An element x E A c (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 a,, + • • • + ak x = O. An R[G] module V is algebraic if V is R -free and ( V) is an algebraic element of Ac(R[G]).
94
CHAPTER
II
[5
5.1. Suppose that R[G] has the unique decomposition property. Let V 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 W,, , W„, such that if W is indecomposable and W IV" for any n, then W = W for some i.
LEMMA
PROOF. (i) (ii). There exist nonnegative integers ao, , ak, bo, . . , b, I,k=c, b, (V)'. with j < k such that ak 0 and a, (V)' = = 0 Thus every indecomposable component of V" is a component of V' for some i 0 the number of irreducible F[G] modules is equal to the number of conjugate classes of G consisting of p'-elements.
Let {C} be the conjugate classes of G. Choose x, E C for each i. Let S be the F-subspace of F[G] generated by all ab — ba with a, b E F[G]. Let PROOF.
so
= E axx E
a„ = 0 for all
EC,
G
If E ax E S o then E ax = E, Ex x — x, E S for x E C. Since
ax (x
(E ax) E bxx) — E ix) (E ax) = (
(
- x,). Thus S o C S since axbx —
axbz„
i)z
=
it follows that S C S o and so S = S. Clearly E, ax, E S o if and only if a, = 0 for all i. Thus dimFF{G} — dim F S is equal to the number of conjugate classes of G. If char F = 0 then F[G] is semi-simple by (11.3.12) and the result follows from (1.16.3) in this case. Assume that char F = p >0. Choose the notation so that CI , ..., Ck are all the conjugate classes of G consisting of p'-elements. Let T = {c I E S for some j}. By (1.16.3) T is an F-space and it suffices to show that k = dimF F[G] — dim F T.
If x E G then x = yz = zy Where y is a p-element and z is a p'-element. Since (z — yz) 0' = z"' — = 0 for sufficiently large j it follows that z — yz E T and so x z (mod T). Since S C T, z x, (mod T) for some i k. Thus x =x, (mod T) for some i and dimF F[G] dimF T k. Suppose that Ei,`=, a,x, E T. Hence by (1.16.3)
E
k
(E ax,)
pi
S
for sufficiently large j. Choose m such that xr = .x, for all i. Thus E S. Hence by the first part of the proof ar = 0 for all i and so a =0 for all i. Hence dimF F[ Gj — dimF T k. D COROLLARY
2.9. Assume that char F = p >0. Let P be a p -group. Then
F(G) MODULES
101
Vo(P) is the only irreducible F[P] module and F[P]F[F] is the unique principal indecomposable F[P] module. Every projective F[P] module is free and F[P] is a local ring.
Let F, be a splitting field of F[P]. By (2.8) V o(P)OF FI is the unique irreducible Fi [P] module. Hence Vo(P) is the unique irreducible F[P] module and F is a splitting field of F[P]. By (1.16.9) this implies that F[P] is indecomposable. Thus Vo(P)------ F[P]FE p l /J(F[P])F[p ] and so dim FF[P]a(F[P])=1. Thus F[P] is a local ring. The remaining statements are clear. 0
PROOF.
COROLLARY 2.10. Assume that char F = p > 0. Let P be a Sr -subgroup of G. If U is a projective F[G] module then UF nF[P]Fip i for some integer n. Thus in particular dimFU 0 (mod I PI). PROOF. Clear by (2.9). 0
2.11. Suppose that H < G. Let W be an irreducible F[H] module with T(W)= H. Then W G is an irreducible F[G] module.
LEMMA
W", PROOF. Let { x,} be a cross section of H in G. By (II.2.11)(W 0 ), = and W", W", if and only if i = j. Let V be a submodule of IV', V (0). By (I.5.11) W", C VH for some i. Hence W", C V, for all i and so (4V G ), = V,. Hence WG = V. Thus IV is irreducible. 0 THEOREM 2.12 (Clifford). Let H < G and let V be an irreducible F[G] module. Then there exists an irreducible F[H] module W and an integer e = e, (V) such that V, e {ED W9 where x, ranges over a cross section of T(W) in G. Thus in particular VH is completely reducible. Let W be a minimal submodule of VH. Then Wx = W' for x E G and so Wx is irreducible. Hence /..G W" is completely reducible. Since E iv" is sent into itself by multiplication by elements of F[G] the irreducibility of V implies that VH = W. Thus if fx,} is a cross section of T(W) in G it follows from (II.2.11) that VH e,Wx.. By (2.5) PROOF.
ED
e,I(W,W)=
Wx.)=
)'3 )= I(V, WG).
I(V,(W
Hence e = e, is independent of i proving the result.
0
The integer e, (V) in (2.12) is the ramification index of V with respect to H.
102
CHAPTER III
2.13. Suppose that char F = p. (i) Let H be a p-group with H V by f (v)= (vy). Clearly f is an R-isomorphism. If z EH then f(v)z = hP(vy)z = hP(vyzY)= hP(vzy)= f(vz). Therefore f E E. By (II.2.11) (V G )H Thus if a E ER (VG ) there exist g,, E p —1, ER (V) for i, j = 0, p — 1 such that for s = 0, a(v
xs)=
h`g,s h - s(v)0 x'. I =0
108
CHAPTER III
Let at = (g,i )E(E R (V))p. The map sending a to (I is clearly an Risomorphism from ER (V G ) to (ER (V))„ If a, a' E ER (V G ) with â p— 1 a'= (g 1 ) then for s = 0, P- 1
ad(v x')= a{ E h ig,h - s(v)Ox'} i=0 1,-1
1=0
{E
( )10 x'
1=0
Thus at' = an' and the map sending a to a is a ring isomorphism. If z EH, a E ER (V) then for s = 0, • • -,p —1 P- I
a(v
x')z =
E thigish-s(v)}zx- oxi P- I
=E
h' ({gish - s (v)}z)
x',
p- I
a(v
' )-= a(vzx — x')= E ,=0
(vz' )0 x'.
Thus a E E R E H )(V G ) if and only if for all i, s, Ig,sh — (v)lz = g,s {h - '(v)z}. Since h is an R-isomorphism this last equation holds if and only if g,s (w)z = g„,(wz) for all w E V, or g,, E E. Therefore E R[H] (V G )= E. by = (h gi,h) where ei = (g,,). Then For a E ER (V G ) define E E REFri (V G ) if and only if a E E RtHi (V G ). Define b E ER (V G ) by b(vOx')=h(v)Ox' ÷' for i = 0,...,p —1. Clearly
and
V = b.
Thus b E E REHI(V G ). Furthermore b(v
x')x = h(v)0 x'' b(v
x)
for i=0,...,p-2;
b(vOxP -1 )x =h(v)OxP +1 =h(vy)Ox = b(vy 01)— b(v Ox"). Thus b E E REG1 (VG ). By (1.9.12) (3.9)
ER,G 1 (vG)n J(Ep )g_ J(E R[G] (V G )).
We next prove two subsidiary results. (3.10)
If g E E
then
gh — g E J(E).
109
GROUP RINGS OVER COMPLETE LOCAL DOMAINS
PROOF. Since E /J(E).--= k there exists r E R such that g — r E J(E). Hence
gh — r = h -l (g — r)h E h -1 .1(E)h = J(E). Thus IC' gh — g E J(E) as required.
(3.11) Let a E E R [ H ]( V G ). Then a E PROOF. Let a = (g,,). For s = 0,
ER[G](
V G ) if and only if b'ab = a'.
p —1
p-I
fba"(v ®x')}x' =
E
(v)Ø x= a{b(o
Ø X ' )x
' }.
r
Hence a h (V 0 X )X -1 = b -l ab{(v Øx' )x'}. This proves (3.11). We return to the proof of (3.8). Let R(b)=1E1;-(1)r,b r, E R). Now A = E R{Gi (V G ). Let B = R(b). Thus P- I IriERIC A.
B i=f1
Observe that B need not be a ring. Since Ep /J(E)E, (E/J(E))p fip it follows that J(E) = J(E)EP . For a E Ep let ô denote the image of a in Ep = EpIJ(Ep). Since E/J(E)-- R there exits s E R, s 0 such that f — s E J(E). Replacing R by a suitable finite extension it may be assumed that f — r° G J(E) for some r G R. Thus
(Y
0
FP
1
.g =
1
•.
Let C be the k -algebra consisting of all p X p matrices with coefficients in R which commute with g. Since (t — F ) is the minimum polynomial of B it follows that C consists of all polynomials in g with coefficients in R. Thus (3.12)
b = c - k[t]/(t - F ) .
By (3.10) and (3.11) A C C = Ê. Since B C A it follows that A = Ê. Thus
=Â = A + J(Ep )/J(Ep )— A /A n J(Ep ). Hence b/J(fI)------ A /J(A) by (3.9). Thus it suffices to show that b/./(B) k. This however is obvious from (3.12). E
CHAPTER III
110
[3
It should be noted that (3.8) is false if H is not normal in G even if R is a field. For instance let G = A, H = A, with p = 5. Let V be an irreducible R[H] module with dimR V = 3. Then V G is projective and dimR V G = 15. However it is not difficult to show that no principal indecomposable R[G] module has dimension 15. Such computations will follow relatively simply from the results in Chapter IV. COROLLARY 3.13. Suppose that H 0 and so V (WR )C1 as ( WOG ( WR contrary to the fact that A is a vertex of V. D 5. The Green correspondence For most of this section we follow Green [1964]. The following notation will be used throughout this section.
116
CHAPTER III
P is a p-subgroup of G and H is a subgroup of G with N G (P)C H. if is a set of subgroups of G write A e G if A E for some x E G. We define the following sets of subgroups.
I(P, H)= { AIACPnP' for some zEG—H},
D = D(P, H)= {A IA CH n Px for some x E G — H}, = W(P, H)= {A IA CP and A
G.1. }•
Observe that since NG (P)C H, consists of proper subgroups of P. Thus P E 91. Clearly 3E C D. The aim of this section is to define a one to one correspondence between indecomposable R[G] modules with vertex in 9f and indecomposable R[H] modules with vertex in 91 which preserves many properties of modules. Such a correspondence can frequently be used to study various properties of an R[G] 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 G X. (ii) A e (iii) A E (iv) A E H D. .
PROOF.
P
n
(ii) (iii) (iv) Thus A
(i) (ii). A'CPnP' for some zEG, xEG—H. Thus AC n P" I . Either z -1 H or xz -1 0 H. Thus A E l.
(iii). Clear since 3E C (iv). Immediate. (i). AY CHnP' for some y E H, xEG—H. C CI HY inP." '. Hence AEI since xy -1 H.
E
LEMMA 5.2. Let W be an R[P ] -projective R[H] module. Then (W G )H = W().
PROOF. There exists an R[P] module U such that W LI". Let UH = W e W. By the Mackey decomposition (11.2.9) ( W G )H = W', W Hence (U G )H = W ■ for some (147),, = W WED W, W' W. Since Hn/;" ED for x H it follows from (11.2.9) that (U G ) F, = W i n. Thus
117
THE GREEN CORRESPONDENCE
W
Therefore W' El)
m .(uG), uH WEJ WO).
,e W
on.
Hence (WG ) H W().
0
5.3. Let V be an indecomposable R[G] module with vertex A C P. Then there exists an indecomposable R[H] module W with vertex A such that V 1W G and V H = W(D). Furthermore V 0(1) if and only if W
LEMMA
00). Let U be an indecomposable R [A] module with V I U G. Let W U" such that V I W G . Hence W has vertex A. Thus VH (W G )H and by (5.2) (W G )H W (J ). Hence V H 0( 9 ) or V H W (J ). If V 0(1) then A E 0 1 and so A E by (5.1). Hence VH 4 / On. If 0(1) then A O G I and so A EH D by (5.1). Thus by (4.6)(ii) VS O (J ) and so VH W 0 (D). E PROOF.
5.4. Let W be an indecomposable R[H] module with vertex A C P. Then there exists an indecomposable R[G] module V with vertex A such that W I V j-j and 1410 V(1). Furthermore W = 0(1) if and only if V 0(1).
LEMMA
PROOF. Let W G = V, where each V, is indecomposable. By the Mackey decomposition (11.2.9) W I (W G ) H and so W (V,)H for some i. Say W I (VI ) H. If W0(1) then WG 0(1) and so V I =- 0 W0 (1). 0(D) by (5.1) and so (VI )H 0(D). Suppose that WX 0(1). Then Since (W G )I-4 (5.2) it follows that ( V,)H by 0(D) for i >1. Thus W(D) by (5.3) VI 0(1) and V, = 0(1) for i >1. Hence WG = V1 0(1). 0
Let V =1ED V, where each V, is an indecomposable R[G] module with vertex in P. Define the R[H] module f(V) as follows: f(V)= @RV), f (V.) = (0)
if V,
0(1),
f(V,)= W
if V1
0(1) where W is defined as in (5.3).
(5.5)
By (5.3) f is well defined on isomorphism classes of indecomposable R[G] modules. By combining (5.2), (5.3), and (5.4) one immediately gets THEOREM 5.6 (Green [1964]). Let f be defined by (5.5). Then f defines a one to one correspondence from the set of all isomorphism classes of
us
CHAPTER III .
[5
indecomposable R[G] modules with vertex in S2X onto the set of all isomorphism classes of indecomposable R[H] modules with vertex in 2T. The mapping f has the following properties. (i) If Vis an R[P] -projective R[G] module then f(V)=. VH(,.9)) and f(V) G =- V(Y). (ii) Let V be an indecomposable R[G] module with vertex in 21. Then V and f (V) have a common vertex and source. If W is an indecomposable R[H] module with vertex in then f(V)= W if and only if VI W G or equivalently W
a
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. LEMMA 5.7. Let V, V,, V2 be indecomposable R[G] modules with vertices in P. Let f be the Green correspondence with respect to (G, P, H). Then the following hold. (i) f (V )0 f (V2) =. f( V, V2 )(1). (ii) f (Hom R (V 1 , V2)) Hom,, (f (V ,), f (V 2))(f). (iii) If o- is an automorphism of R[G] with R" = R and G' = G then f( v = r (V') where f" is the Green correspondence with respect to (G, P", H"). (iv) If V is R-free then V* has vertex in P and f(V*)= f (V)* .
)
V2 or (11.2.7) the vertex of any component of Vi Hom R (V,, V2) is contained in P. Thus f( V, 0 V2) and f(HomR (V', V2)) are defined. By (5.6) and (11.2.7)
PROOF. By
f (V ,
V2) (V ,)H 0 (V2)1, —= f (V) f (V2)(g)),
f (Hom R (V V 2)) -= HomR ((V 1)14, (V2)14) = Hom,, (.f(
f (V2)) ()).
and f(V,)(3 f (V2), f(Hom R (V,, V2)) f(V,O V2), Hom R (f(V,), f (V 2)) are R [P]-projective by (5.3) and (11.2.7), it follows from (5.1) that the above congruences hold modulo X. This proves (i) and
Since
(ii). (iii) Clear by definition.
119
THE GREEN CORRESPONDENCE
(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)* . fl LEMMA 5.8. Let f be the Green correspondence with respect to (G, P, H). Assume that I = {M}. Suppose that Vis an R-free R[G] module which is R[P]-projective and that V is indecomposable. Then f (V).- f( t'). PROOF. If V is projective then f (V) = f(V)= (0). Suppose that V is not projective. By (3.17) V is 1Tt [P]-projective. Thus by (5.6) f(V)=- f(V)(D). Hence f( V) - - f(V)EDIED W, where each W is an indecomposable ft[H] module which is ft [P]-projective such that W = 00)). Let A be a vertex of W with A C P. Since W, = 0(D), there exists z E H with A C H fl P for some x E G - H. Thus A CPn p E I. Consequently A = (1) and W is R -projective. Therefore f (V) f(V)e) U for some projective module U. By (1.17.11), U = (0). 0
be a set of LEMMA 5.9 (Green). Let C be a commutative ring. Let subgroups of G. Then A c (R[G]) is an ideal of A c (R[G]). Furthermore if ,S) is a set of subgroups of P with C then A c i p i (R[G])/ A cs,(R[G])---- A c i p) (R[H])/ A c ,-,(R[H]). PROOF. This follows directly from (11.3.3), (5.6) and (5.7). 0 LEMMA 5.10. Let V, V, V2 be indecomposable R[G] modules with vertices in P. Then the following hold. (i) fr(G, I, V) kr(H,D, fr(H, L, f (V)). (ii) Hom R VI , V2) is R[P]-projective and
HAG, I,Hom R (V 1 ,
HAH,D,HomR (f (V,), f (V 2)) fr(H, L, HomR
(V),
f (V2)).
PROOF. (i) There exists an R[P] module U such that F(V)I U". If B is a subgroup of H then the Mackey decomposition (11.2.9) and (11.3.4) imply that InvB ( U") = EH
Tdr-,R=(InvRnp.(U H )).
Since f( V) LP this implies that InvB (f ( V)) =
z
EEH Trnp= (InvEn-ip=(f( V))-
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120
Consequently if To is a nonempty set of subgroups of H it follows that TI"(H, 93 0, f(V))----- H"(H,T, f (V)) where T = {B n Px B E ) , z E H}. Define the following sets of subgroups of G. 1.0=IAIACHCIPYnP",xEG—H,yEGL 1. 1 --- {AIACHnPY
np.Y CIPz,xEG—H,yEG,zEH}
-- {AIACPY ( -1P"r1Px,xEG—H,yEG,zEH}, g),={AIACHnPx nPz,xEG—H,zEH} = {AIACPx CIPx,xEG—H,z EH}.
then A C PY n P"Y fl PZ for some x E G — H, y E G, z E H. If A E Thus A C PY n Pz and A C P"Y n Px. Either yEG—H or xyEG—H. Thus A E D,. If A E A then A C Px n Pz for some x E G —H, z E H. n Pz and xz' E G — H. Hence A E X,. Therefore Thus A C Px CIP (xx = Di. The first paragraph now implies that 11"(H,D, f (V)) HATI,D„ f(V))- - 11"(H, H"(H,
f(V))
f (V)).
Thus by (11.3.5) H"(G, f(V) G ).
H1H,D, f (V)) ----- fr(H,
Hence by (5.6)(i) and (11.3.6) N"(H, f (V)) —
By (11.3.2) HIH,
(V) G ) H"(G, I, V).
f (V))— fr(H, I, f (V)).
This establishes the second
isomorphism. (ii) By (11.3.8) Hom R (V,, V2) is R [Pi-projective. The result follows from (5.7) and (i). LEMMA 5.11. (i) If V is an absolutely irreducible R[G] module with vertex R. in 91 then I-I"(H,g), Hom R (V), f vertex in 91 such that VK is with (ii) If V is an R-free R[G] module absolutely irreducible then H1H,D,HomR (f (V), f(V))) is a nonzero cyclic R module.
(i) By (11.2.4) Inv G (HomR (V, V)) k. Thus by (4.9) H"(G, Hom, (V, V)) R. The result follows from (5.10). (ii) By (11.2.4) Inv G (Hom R(V, V))R K and so InvG (Hom R(V, V)) is a nonzero cyclic R module. Thus by (4.9) 11() (G,./-,Hom R (V, V)) is a nonzero cyclic R module. The result follows from (5.10). 0
PROOF.
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THE GREEN CORRESPONDENCE
One of the most important problems in the whole theory is to give a characterization of R[H] modules W with vertex in 91 which have the property that W = f (V) for some irreducible R[G] module V or some R-free R[G] module V such that VK is irreducible. Very little is known in this direction. (5.11) gives a rather meager necessary condition. Similarly one may ask for criteria for subgroups and modules to be vertices and sources of irreducible modules. Virtually nothing is known in this connection in general but some results are available if G is solvable. See Chapter X, section 7. The following result illustrates how far the condition of (5.11) is from being sufficient. LEMMA 5.12. Let y, w, U, for i = 1,2 be R-free R[G] modules with (.1; projective such that for i = 1, 2
is an exact sequence. Let (W,e)1 1V,I)e and
be a nonempty set of subgroups of G. Then U, for some projective modules U3, U4
u,-(v,olme
Hom,, ( V,, V2))
, Hom R (
W2)).
PROOF. Since all modules are R -free the following sequences are exact
U,0 W1—> Vi e) 14/ 1;-->0,
0---> W 1 0
0—> V`® V— UØ V,—>W'2`0
e
e
Thus by (1.4.3) ( W, ) U3 (V 1».2' ) U, for some projective modules U3 U,. Thus by (11.2.8) Hom R (W1, HomR (v,, v2) LI,. The result follows from (11.3.6). 0 ,
e
vv,)eu,-
Let H be a subgroup of G and let V be an R[G] module. In general it is difficult to get any information about the structure of Tr2(Inv H ( V)). We conclude this section with some results related to this question. LEMMA 5.13. Suppose that V 1 , V, are R[G] modules which are not projective. Assume that V, is irreducible and V, is indecomposable. Then for {i, j} = {1,2 }
(i) 1-11;,(Hom (V, V, )) = (0), (ii) HAG, (1), Hom, (y, V, )) = Horn,,,,„(v„ y).
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Clearly (i) implies (ii). Suppose that U is a projective R[G] module such that U V, —> 0 is exact. Suppose that g E Tr ) (HomA (V,, V2)) with g O. By (3.13) there exists h E Homfic)( VI , U) with g = fh. As V, is irreducible h(V) is in the socle of U. Thus the socle of U is not in the kernel of f. Hence for some indecomposable direct summand u, of U, the socle of u, is not in the kernel of f. Since u, has an irreducible socle this implies that u, f(U,)C V,. Hence u, V2 contrary to assumption. If g E Tif;) (Hom, (V2 , V,)) then g* E 1-1- ,(HomA (VI, VI)). Thus g: = g = 0 by the previous paragraph. CI PROOF.
LEMMA 5.14. Suppose that P is a p-group and A is a subgroup of P. Let V be an .I[P] module with Tr(Inv A ( V)) # (0). Then there exist submodules W, C W2 C V such that W,/ W , is k[A]-projective and Tr". (InvA W2/ WO) # (0). Furthermore dim, V I P : A.
■
PROOF. Let y E Inv A ( V) with w =Tr(v)#O. It may be assumed that V is generated by y and w generates the socle of V. Thus there exists an exact sequence —> V ±> U with U R[P] tpi. Since _In is a projective R[A] module, every invariant in LJA is a trace. Thus f(v)=Tr ) (u) for some u E U. Since Tr() (u) = f (w) 0 it follows from the structure of U that {ux I x E PI is an 1 -basis of U. Hence if {x,} is a cross section of A in P then {f(v)x,} is a linearly independent set over R and {wf (y)x,} is an 1 -basis of f( V) V. Let Vo be the one dimensional R[A] module with Vo = Inv A ( Vo ). Then it follows that f(V) ---- Vg and so V ---- f( V) is .i[A]-projective. E LEMMA 5.15. Let S'", be a nonempty collection of subgroups of G and let V be an R -free R[G] module. (i) { 711/ + EA EJrcAInvA (V))1/7TV is isomorphic to a submodule of Tr (4'(InvA (17 )). (ii) If HAG, 17) Inv G ( fi) then H"(G,., V) Inv G V. PROOF. (i) Let t : V— \-/- be the natural projection. Since t(Tr,(y)) Tr,(ty) it follows that t maps / A Tr`A; (Inv A ( V)) into the kernel of t restricted to EA EÇ- Trc,, (InvA (17 )). Furthermore Tr`AInvA ( V)). This implies EA Es-, TrcAInvA (V)) is precisely ITV n the result. ■
■
123
DEFECT GROUPS
(ii) This is an immediate consequence of (i).
CI
It should be noted that the converse of (5.15)(ii) need not be true. For instance let V, G, H be defined as in the example after (3.17). Then is 1[H]-projective and so H°(G, {H}, Hom R(V V)) = (0) HornR
G
(
V, V).
However V OR K is absolutely irreducible and V is not R[H]-projective. Thus H°(G, {H},Hom R (V, V)) ----- R Hom R[GI (V, V)
The next result was announced in Feit [1969]. Related results can also be found in Feit and Lindsey [1978]. LEMMA 5.16. Let V be an R-free R[G] module with vertex in Vt such that V, is absolutely irreducible. Assume that {rank R f( V)} 3 < min A El IP : A I and K has characteristic 0. Then f( V) 1 is absolutely irreducible.
PROOF. By (5.14) Tr(Inv A (Hom R (f (V), f(V))))= (0) for A
El.
Thus
H° (H, I, Hom (f (V), f (V))) = Inv,, (HomR (f (V), f (V)).
Hence (5.15)(ii) implies that H° (H, I, Hom (f (V), f( V))) Inv!' (HomR (f (V), f(V))).
By (5.10) H°(H, I, Hom R (f( V), f(V)))------ HAH,AHom R (f(V), f(V))) is a nonzero cyclic R module. As Inv, (Hom R(f (V), R, this implies that Inv, ' (Hom R (f(V), f(V))) --- R. Thus Lc (f( V),,, f(V), ( )= 1. This implies the result as K[G] is semi-simple since K has characteristic 0.L1
6. Defect groups
In this section we follow Green [1962a]. See also Brauer [1956], Osima [1955], Rosenberg [1961]. The following notation and terminology will be used throughout this section. H is a subgroup of G. Two elements x i , x2 of G are H-conjugate if x = )6' for some y in H. Clearly H-conjugation is an equivalence relation. Let L,, L,, . . . be all the
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equivalence classes of H-conjugate elements in G. If G = H then L,, L,, . . . are simply the conjugate classes of G. For x E Lk let a„k be the number of ordered pairs (x,, x, ) with x, E L,, x, c L, and x,x, = x. It is easily seen that a„k is independent of the choice of x in Lk. Furthermore Lu, = Ek Let P be a p-group, P C H. L, is P-defective if X 2 = x for some x E L, and all y E P. Clearly L, is P-defective if and only if there exists x E L, n c, (P). Let x E L, and let P, be a S2 -subgroup of CG (X) n H = CH (x). Then P is a p-defect group of L, or simply a defect group of L,. One immediately sees LEMMA 6.1. Let P, be a defect group of L,. Then (i) L, is R-defective. (ii) If L, is P-defective for some p-group P in H then P
By (6.1) any two defect groups of L, are conjugate in H. Throughout this section P, will denote a defect group of L,. If IP,I= p d, then d, is the p-defect or simply the defect of L,. For x E G define r, E ER (R[G] R[G1 ) by ar, = ax for a E R[G]. The map sending x to r„ defines a ring monomorphism of R[G] into ER (R[GIRE G]). If x, y E G then by definition (see after (11.2.3)) r(,Y ) = (r„ )y. Thus R[G] inherits an R[G] module structure where (x)y = (x 2 ) for x, y E G. Denote this module by E[G]. Hence as rings E[G] = R[G]. Define Z(G:H)= Z(R; G :H)=
(E[G]).
It is easily seen that Z(G : H) is an R -free R -algebra and (L,} is an R -basis of Z(G : H). Clearly Z(G : (1)) = E[G] = R[G] and Z(G : G) is the center of R[G]. LEMMA 6.2 (Brauer). If a,,k0 (mod p) and L k is P-defective for some p-subgroup P of H then L, and L, are P-defective. PROOF.
Choose x
E
Lk nc„. (P). Define
= {(x1 , x,)
Thus I A, =
xi E
L , x; E Li, x 1x 1 = x}.
If y E P define (xi, .02 = (x x ) . Thus P acts as a permutation group on the set Aq. The cardinality of any orbit of P is a power of p. Since 0 (mod p) this implies the existence of an orbit of
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DEFECT GROUPS
length one. Thus there exists (u, CH (V). El
E A,,
with P C CF, (u) and P C
For any p-group P in H define Zp (G : H)=
Er
, r, E R, r,
0 (mod ir)) unless P, C P}
Thus (7r)Z(G : H)C Zp (G : H) and Z0 (G :H)C Zp (G :H) if Q C P. Furthermore if P is a Se -subgroup of H then Z(G : H) = Zp (G :H). LEMMA 6.3 (Osima). Let P, Q be p-groups contained in H. Then Z p (G : H)Z 0 (G : H)
E
ZA
(G : H)
where A ranges over all p-groups in H such that A C HP and A Furthermore Zp (G : H) is an ideal of Z(G : H).
CH
Q.
PROOF. It suffices to show that if P, CH P and P, CH Q then IlL, E /ZA(G:H). If ay k 0 (mod 7r) then a,,k 0 (mod p). Hence by (6.1) and (6.2) Pk CH P. and Pk C, P1 . This proves the first statement. The second statement follows by letting Q be a Sr -subgroup of H. 0
LEMMA 6.4. Let e be a primitive idempotent in Z(G : H) and let 13 be a set of p-subgroups of H such that e E ZpE .4 Zp (G : H). Then e E some Q E PROOF.
Z0
(G : H) for
Since e is an idempotent e E
E
pEsv
eZp (G :H)e C eZ(G : H)e.
As e is primitive it follows from (1.7.2), (1.8.2) and (I.10.4) that eZ(G :H)e is a local ring. By (6.3) each eZp (G :H)e is an ideal eZ(G :H)e. Thus eZo (G : H)e = eZ (G : H)e for some Q E q3. Hence by (6.3) e E eZo (G : H)e C Zo (G : H).
0
LEMMA 6.5. Let e be a primitive idempotent in Z(G :H). There exists a p-subgroup P of H such that (i) e E Zp (G : H). (ii) If e E Zo (G : H) for some p-subgroup Q of H then P C H Q. Furthermore if e = af, then 0 (mod 7r) for some j where L, is P-defective.
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CHAPTER III
PROOF. If Q is a Se -subgroup of H then e E Z(G : H)= Zo (G : H). Choose P minimal such that e EZp(G : H). Then (i) is satisfied. If e E Zo (G : H) then by (6.3)
e = e2 E Z p (G : H)Zo (G : H)
ZA (G : H)
where A ranges over subgroups of H with A C,, P and A c,, Q. By (6.4) e E ZA,,(G : H) for some Ao c,, P and A 0 C,, Q. The minimality of P implies that Ao =HP and so P GHQ proving (ii). If 43 ={P 0 (mod 7T)} then e E EA ZA (G :H). Hence by (6.4) e E Z, o (G : H) for some A o E 43. Thus by (ii) P C,,A 0 proving the last statement. 0 If e is a primitive idempotent in Z(G : H) the group P defined in (6.5) is a p-defect group or simply a defect group of e. By (6.5) any two defect groups of e are conjugate in H. If I Pi= p d then d is the defect of e. LEMMA 6.6. Let P be a p- subgroup
of H. Let Tr',.1 denote the relative trace for the module E[G]. Then the following hold. (i) If a E Zp (G : H) there exists b E Z(G : P) such that a =. Tr(b) (mod(ir)Z(G : H)).
(ii) If e is an idempotent in Zp (G : H) there exists c E Z(G : P) such that ece =c and e = Tep- 1 (c). PROOF. (i) It suffices to prove the result in case a = I:, for some i with P, C P. Choose x E L, nC G (P) and let C = CH (x). Thus E Z(G : P,)
and Tr;.1(Tr(i)) =
(X) =
(T6, CO)
=IC:P,ITT.C(i)=1C:P Since I C: P 'E R the result follows. By (i) there exists b E Z(G : P) such that Tr" (6) = e — d Hence by (11.3.7) TrV(ebe)= e — ede and ede E (7)eZ(G : H)e C J(eZ(G : H)e). Thus ede is a quasi regular element in eZ(G : H)e and so there exists d, E eZ(G : H)e with e = di (e — ede). Let c = di ebe. Therefore ece = c and by (11.3.7) (ii)
with
d E (7)Z(G : H).
TrAc)= e.
0
(Green). Let V be an R[G] module. Let P be a p- subgroup of H and let e be an idempotent in Zp (G : H). Then (VH )e is R[P1-projective.
THEOREM 6.7
DEFECF GROUPS
127
PROOF. Choose c as in (6.6). Define fv = vc for y E V. Then f E E R[ p ] ((V D )e) and Tri(f) = 1 in E R1111 (( V,, )e). Hence by (11.3.8) (V D )e is R[PFprojective. CI
In the rest of this section we will be concerned with the case .1-1 = G. Z(G : G) is the center of R[G]. Thus a primitive idempotent in Z(G : G) is a centrally primitive idempotent in R[G]. If B = B(e) is the block of R[G] corresponding to the centrally primitive idempotent e of R[G] then a defect group of B is a defect group of e and the defect of B is the defect of e. COROLLARY 6.8. Let B be a block of R[G] with defect group D. Let V be an R[G] module in B. Then V is R[D]-projective. PROOF. Let B = B(e) where e is a centrally primitive idempotent in R[G]. Then V = Ve. The result follows from (6.7) with H = G and P = D. Lii COROLLARY 6.9. Suppose that P < G, P a p -group. Then Pis contained in a defect group of every block. PROOF. By (1.17.3) a block B contains an irreducible R[G] module. By (4.13) P is in a vertex of every irreducible R [G] module. The result follows from (6.8). E THEOREM 6.10. Let B = B(e) be the block of R[G] corresponding to the centrally primitive idempotent e. Let À be a central character of R[G] with (e) = 1. Let D be a defect group of B. Then the following hold. (i) If À (i:, ) X 0 then D C G P,. (ii) There exists j such that ) X 0 and D is a defect group of L,. PROOF. If A C G D then eZ A (G: G)e is an ideal of the local ring eZD (G :G)e by (6.3). Since e ZA (G: G) by (6.5) it follows that eZA (G : G)e C J(eZD (G : G)e). Thus A(Z A (G: G)) = A(eZ, (G : G)e) = 0.
If D Z G P, for some i then by (6.3) eZp,(G : G)
CZD
(G : G)Zp, (G : G)
where A ranges over groups with A (eZp, (G : G)) -= 0 proving (i).
Cc
E ZA (G: G) D. Thus A(Z,,(G : G))=--
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Let e =E a,f,,. Since À (e)/ 0 there exists j with a1 X 0 (mod 7r) and (L, )/ O. By (i) D C G P,. Hence D =G P, since aO (mod 77-) and e EZD(G :G).
6.11. Suppose that P is a p-group and P < G. If the conjugate class L, of G is not P-defective then L, is nilpotent in R[G].
COROLLARY
If L, is not nilpotent in R- [G] then it is not nilpotent in S[G] for any finite extension S of R. Thus replacing R by a suitable finite extension it suffices to prove the result in case /:-Z is a splitting field of fi [G]. Let À be a central character of R[G]. Let ë be the centrally primitive idempotent of I[G] with A(e) O. Let D be a defect group of e. By (6.9) P C D. Thus by (6.10)(i) A (f) = O. Thus f, is in the kernel of every central character of R[G]. Hence L, is in the radical of Z (G : G) and so f, is nilpotent. PROOF.
7. Brauer homomorphisms The results in this section are primarily due to Brauer [1956], [1959]. See also Conlon [1964], Green [1962a] and Nagao [1963]. The results in the next 3 sections can also be found in Hamernik [1974b]. The following notation will be used throughout this section. Let P be a p-subgroup of G. Let H be a subgroup of G such that PCG (P) ci H Œ NG (P). Let C,, C2 , ... be all the conjugate classes of G. For each i define C;') = C n CG (P) and C!') = C, — C? ) . Define the R-homomorphism s : Z(G : G)—)Z(H: H) by s(C)= C;' ) . In a natural way s defines an R -homomorphism §:Z(G : G)—> Z(H H). THEOREM
7.1.
is a ring homomorphism.
Let CC, = Ek a,1k0k. Let L,, . . . be all the H-conjugate classes in G and let e;", ) = E, b11, for m = 1, 2. If L, C C;2) for some i then L e is PROOF.
not P-defective. Thus by (6.2)
el2)0
I)
ere12) el2)e12) = E cL
(mod(71- )Z(G : H))
where L e is not P-defective for 0(7r). Hence _ 04; _ ( 02)0 (I) ± ere.12) ±
- E a,,ko(ki) + E
a„kCT) —
E c,f,
(mod(n- )Z(G : H)).
BRAUER HOMOMORPHISMS
7]
129
Every H-conjugate class of G which is contained in C;' ) for some i is a conjugate class of H contained in C c, (P) and so is P-defective. Thus is a linear combination of f, where L, is P-defective. Hence
ever)
s(0,)s(0,)-=- ("=-
E aik0
s() (mod(7)Z(G : H)).
Then s(0,)s(0,)— s(0,0)E en - )Z(G : H)n Z(H : H)= (7)Z(H :H).
Therefore §(0, )§(0'; ) =
). E
The mapping s is called the Brauer mapping with respect to (G, P, H) and is called the Brauer homomorphism with respect to (G, P, H). In case G, P, H are determined by the context s is simply the Brauer mapping and is the Brauer homomorphism.
7.2. Let e be a central idempotent in R[G]. If g(e)/ 0 then there exists a unique central idempotent so(e) in R[H] such that §(e) = so(e).
COROLLARY
PROOF. By (7.1) g (e) = 0 or §(e) is a central idempotent in R[H]. The result follows from (1.17.3). 0
In the rest of this section su(e) will be defined as in (7.2) with s(e)= 0 in case §(J) = O.
LEMMA 7.3. Let e be a central idempotent in R[G]. Then e — s o (e)E Z,,(G : H) where Q ranges over p-subgroups of H with PZ Q. 2 )(mod(rr)z(G : H)). By PROOF. Let e =Ea,0,. Then e — s o(e)=- E a definition C;2) is a union of H-conjugate classes of G whose defect group does not contain P. The result follows. E
LEMMA 7.4. Let e be a centrally primitive idempotent in R[G] with defect group D. Let so(e)=E e, where { e,} is a set of pairwise orthogonal centrally primitive idempotents in R[H]. Let D, be a defect group of e, in R[H]. Then D, C G D. PROOF. Let C, be a conjugate class of G with defect group P, C D. If L, is a conjugate class of H contained in C, then the defect group of L, is contained in 1='' n HCBox nH for some x E G. Since e E ZD (G : G) it follows that so(e)E IxE ,, Z L,m H (H: H). Then by (6.3) e, = e‘so(e)E EA EGZ D`r11-1(H : H). Hence by (6.4) and (6.5) D, C H U n H C G D for some x E G. E
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[
7
The next result contains the information from ring theory required for the second main theorem on blocks, (IV.6.1) below. As such it is sometimes called the second main theorem. THEOREM 7.5 (Nagao [1963 1 ). Let e be a central idempotent in R[G]. Let V be an R[G] module with Ve = V. Let 87) be the set of all p-subgroups Q of H with PZ Q. Then V pi VHso(e)(8:3).
PROOF. Let f = e — es o(e). Since e E Z(G : G) either f = 0 or f is an idempotent with so(e)f = fs o(e)= O. By (6.3) and (7.3) f E E oe ,5 Zo (G : H) and VH — VHe = V HeS0(e)
ED V Hf — V HS0(e) vHf.
Let f = f where { f, } is a set of pairwise orthogonal primitive idempotents in Z(G : H). By (6.3) f, = ff E Z0 (G : H). Thus by (6.4) f E Zo (G : H) for some Q E ST). Since VH f, is R[Q]-projectiye by (6.7) this yields that VHf V Hf, 0 ( ). 0 LEMMA 7.6. Let e be a central idempotent in R[G]. Let W be an R[H] module such that W = Ws o(e). Let { x,} be a cross section of H in G with = 1. Define f: W —> ((W" ),.. i )e and g :(W"),.., —> W by f(w)= (w 01)e and g(E w, x,)= w 1 . Let U be the kernel of g. Then W = f(W) and ((IV (' )H)e = f(W)EJ Ue. In particular W ((W") H )e.
PROOF. It suffices to prove the result in case W is indecomposable and so(e) O. By definition ( W G )H = ( W 1) Wo where W0 = ED, W x,. Hence if y E H then ( W 1)y = W 1 and if yEG—H then (W 1)y C Wo . Since e E Z (G : H) and so (e) E Z (H : H) it follows that e = s(e) + a b where a EZ(H:H) and b is an R -linear combination of H-conjugate classes of G which are disjoint from H. Thus ( W 1)(so (e)+ a)C W 01 and ( W 01)6 C W. As so(e)= s(e) the definition of s implies that a is an 1 -linear combination of conjugate classes in H which are disjoint from C o (P). Hence by (6.11) a E J(Z(H: H)) and so a E J(Z(H: H)). Thus there exists c E so(e)Z(H : H) with (so(e)+ a)so(e)c = so(e). Hence W = W(so(e)+ a)so(e)c and so W = W(s o(e)+ a). If w E W then
e
gf(w)=
1)(so (e)+ a)+ (w 01)b} = w(so(e)+ a).
81
R EG>
g as follows. If C is a conjugate class of G then h G (C') = h( ) where 0, ranges over all conjugate classes of Ô with C C. Clearly h ° is an g-linear map. Furthermore if Ô CA CG for some group A then ')G = h G.
(9.1)
Let Li be a block of S[6] and let k be the central character of g[6] corresponding to 13- . The g-linear map kG need not be a central character of g [G]. If however kG is a central character of g [G] we say that Li is defined and let Li be the block of S[G] corresponding to G. In case Li is defined the map sending 1 to PG is called the Brauer correspondence. LEMMA 9.2. Suppose that OCA CG for groups Ô and A. Let Ij be a block of S[6]. If .' is defined and either one of. or O A f is defined then so is the other and (f34)G = jjG PROOF. Clear by (9.1). D LEMMA 9.3. Let a be an automorphism of S[G] such that S° = S and x° = x for all x E G. Let fi be a block of S[6]. Then f i G is defined if and only if (f3°) G is defined. If fi G is defined then (B. °) G = PROOF. Immediate from the definitions.
E
9.4. Let P be a p-subgroup of G and let H be a subgroup of G with PCo (P)C H C N o (P). If B. is a block of S[H] then PG is defined. Let "e, e be the centrally primitive idempotents of S[H], S[G] corresponding to È f3 G and let k be the central character of g[H] corresponding to P. respctivly Then A G = and so(e)è = e where s is the Brauer mapping with respect to (G, P, H). THEOREM
,
THE BRAUER CORRESPONDENCE
PROOF. If C is a conjugate class of G then definition. Thus by (6.11)
137
G (0 7 ) = X(671 -ii) by
À 0 (C)= ,Qc n cp (H)) = X{K, (6}. Thus X' = X • §. Since is a ring homomorphism it follows that X is a ring homomorphism. Since X G (1) = 1, X G X 0 and so X G is a central character of g'[G]. Thus B 0 is defined. Since 1 = (ë) = X(s7 (e)) it follows that §(J) e7 = "é and so so(e)è = J. El LEMMA 9.5. Let 15 C Ô be subgroups of G with CG (15) C Ô. Let 13 be a block of Ô with defect group 15. Then Pi = B #6 for some block B # of N6 (15) and B' =E0 is defined. PROOF. Let H =No (15). Let .§ be the Brauer mapping with respect to (Ô ,15,H). Let ë be the centrally primitive idempotent in S[0] corresponding to E. By (8.8) ,§(J) is a centrally primitive idempotent in S[H]. Let B # = B(1,)( 0). Since
DC o (D)= DC o (D)C H C NG (D) it follows from (9.4) that B" is defined and B' = 13 is defined. Thus by (9.2) EG is defined. 0 LEMMA 9.6. Let Ô be a subgroup of G. Let E be a block of S[6] with defect group 15. Suppose that EG is defined. Let D be a defect group of PG. Then CG D. PROOF. Let X be the central character of g[0] corresponding to E. Thus = is the central character of §f G] corresponding to EG. By (6.10) there exists a conjugate class C of G with defect group D such that A ( 6/0. Thus A (L) X 0 for some conjugate class L of Ô with L C C. Let Do be a defect group of L. Clearly Do CG D. By (6.10) 15 CG Do. Thus c G D. 0 THEOREM 9.7 (First Main Theorem on Blocks ) (Brauer). Let D be a p-subgroup of G. Let G. be a subgroup of G with NG (D) C O. The map sending fi to EG defines a one to one mapping from the set of all blocks of S[6] 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.
138
CHAPTER III
LEMMA 9.8. Let D be a p -subgroup of G and let N = NG (D). If C is a conjugate class of G with defect group D then C n CG (D) is a conjugate 7' class of N with defect group D. Conversely if L is a conjugate class of N with defect group D then L = C n CG (D) for some conjugate class C of G with defect group D.
PROOF. Let C be a conjugate class of G with defect group D. Thus cnC G (D) is a nonempty union of conjugate classes of N. If x e cn CG (D) then D C CG (x) and so D is a Se -subgroup of CG (x) since D is a defect group of C. Let L be a conjugate class of N with L c C nC G (D). Hence L has defect group D. Let x E L and let y ECn CG (D). Thus there exists z E G with y z = x. Hence D, D are S,,subgroups of CG (X ). Therefore there exists u E CG (X ) such that D'u = D Thus zu E N and y = = x. Hence y E L and so L = C n CG (D). Let L be a conjugate class of N with defect group D. Let C be the conjugate class of G with L C C. Choose x E L and let Do be a Sp -subgroup of CG (X ) with D C Do. If D Do there exists D, with D Ls O. As e, = 1 this implies that ts (LOG and so (14 )6 ---- Ls. 0 COROLLARY 4.33. Let G be a subgroup of G. Let B be a block of G. Suppose that (x.) 0 is irreducible for every x. in B. Then (cp,)G is an irreducible Brauer character for every (ID, 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 Sr -subgroup of G and let N = NG (P). Let B be the principal block of G. The following are equivalent. (i) G has p-length 1. (ii) If x. is in B then (x )N is irreducible. (iii) If ço, is in B then (ç', )N is irreducible. PROOF. (i) (ii). Since G = 0,(G) the Frattini argument implies that G = O(G)N. If x„ E B then 0.(G) is in the kernel of x„ by (4.12) and so (x.)N is irreducible. (ii) (iii). This follows from (4.33). (iii) (i). P = 0„ (N) ig in the kernel of irreducible Brauer characters of N and so P is in the kernel of (p, for every cp, in B. Thus P C 0„ , ,„ (G) by (4.12).
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.
166
CHAPTER IV
(I) If IC,Icp,(x,)1 cp, (1) E R is it true that IG I ci,,(x , )1q)1( 1 )= (), 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 1 (x)/ 1 (1) need not be in R. As an example let p = 2 and let G = the smallest Janko group. If x, is an element of order 3 and p chosen suitably then ç, (l) = 56 and cp,(x,) = — 1. Thus cp,(x,)1(p, (1) = I
I C.
-
— 209/2 0 R. See Fong [1974]. (II) Is IG IN, (1) E R for all j?
Willems [1981] has shown that the answer to (II) is yes if and only if I C, (x, )/ I; (1) E R for all i, j. In case G is p-solvable both (I) and (II) have an affirmative answer. See Chapter X, section 6. A result related to these questions can be found in Torres [1971]. Added in proof. J.G. Thackray has shown that McLaughlin's simple group G has an irreducible Brauer character of degree 2° - 7. Since a .52-group of G has order 2 this shows that (II) has a negative answer in general. (III) Does the character table of G uniquely determine (cp, (x 1 )), the table of irreducible Brauer characters? (The uniqueness can of course only be expected after a fixed monomorphism from IC, to the complex numbers has been chosen.) (IV) Does the character table of G uniquely determine the set of integers
Virtually nothing is known about these questions in general. An affirmative answer even to the weaker question (IV) would be very useful in the study of the structure of finite simple groups. Chapter X, section 6 has a related result in case G is p-solvable. (V) Can the number of blocks be characterized in terms of the structure of G?
By using the first main theorem on blocks, (111.9.7), it is possible to describe the number of blocks of positive defect. Thus the question is really about blocks of defect O. A weaker question is the following. (VI) What are some necesary and sufficient conditions for the existence of blocks of defect 0?
167
SOME OPEN PROBLEMS
In connection with problems (V) and (VI) Wada [1977] has proved the following and related results. In case the class C in (5.1) consists of involutions, the result had originally been proved by Brauer and Fowler [1955].
LEMMA 5.1. Let P be a p-group contained in G. Suppose that C is a conjugate class of G such that xy -1 P — {1} for x, y E C. Let 1P1 = p° and let 1C 0 (X ) = p" for x E C. Then there exists an irreducible characterx of G with p°' I x('). If furthermore Pis a S,-group of G then Xis of height 0 in a block B with defect group D, where D is the defect group of C. PROOF.
Let x E C. Define =
x(x)x(x -I
)
X( 1 )
A'
where x ranges over all the irreducible characters of G. By assumption 0(z) = 0 for z E P — {1}. Since OW = CG (X )1 it follows that !cG (x)i - (op, p
1p)p
x(x)x(x-i) 1 ) P. C"P
X
As p°' is the exact power of p dividing the denominator of the left hand side and x(x ), (xp, 1 p )p are algebraic integers, there exists x with x(1).
Suppose that P is a Sr -subgroup of G. Then
E
G : CG (X)1X(X)X(X -I ) (Xp,1p)p X( 1 )
As IG:P XO (mod p ) and I G :CG (x )1 x(x)/x(1) is an algebraic integer there exists x with G:CG (x)lx(x)
x(1)
(x
0
(mod rr)
for
Tr a prime in a suitably large p-adic field. Let B be the block containing x. By (4.4) D is a defect group of B and x is of height 0 in B. 0
In case G is solvable Ito [1951a], [1951b] has also obtained results in this connection. See Chapter X, section 6 for some of these. Related results are proved below. See (VI.5.6), (X.6.3) and (X.6.5). Other conditions for the existence of characters of defect 0 can be found in Ito [1965c], Ito and Wada [1977]. For related results see Gow [1978], Willems [1974
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CHAPTER IV
[5
(VII) Is there a function of p and d which bounds c,, for cp„ cp, in a p-block of defect d? If G has a p -complement, in particular if G is p-solvable, then it follows p°, where p° = Fong [1961] has shown that for from (4.15) that G p -solvable c„ p d See (X.4.6). At one time it had been conjectured that p d for all G. This conjecture was shown to be false by Landrock [1973] for the group Suz(8) and p = 2, where p° = 64 and c,, = 160. The following results proved by Chastkof sky and Feit [1978], [1980a], [1980b] show that Landrock's result is not an isolated example. Let p = 2, let P„, be a S2-group of G„, and let c;;) be the Cartan invariant of G„, corresponding to the principal Brauer character. If G„, = Suz(2"' ) or ) then .
limc;TVI
= 1.
If G„, = SL3(2- ) or SU3(2'" ) then 1P„, =- 8' and I1M
C (IT ) 19 "' %. 1.
For information about the decomposition numbers of Suz(2- ) and 5U3(2") see Burkhardt [1979a], [1979b]. The Cartan matrix of a block with a cyclic defect group can be computed in terms of the Brauer tree. This result will be found in Chapter VII. In general it seems to be difficult to compute Cartan invariants of simple groups even when the character table is known. For instance the Ree groups 2 G2(32 "±') all have a S2-group which is elementary abelian of order 8. Yet the Cartan matrix of the principal 2-block has only recently been computed and shown to be independent of n. This computation involves some very delicate and complicated arguments. See Fong [1974], Landrock and Michler [1980a], [1980b]. For a similar result concerning the smallest Janko group see Landrock and Michler [1978]. A general approach to questions about irreducible and principal indecomposable k[G] modules for Chevalley groups in characteristic p can be found in J.E. Humphreys [1973b], [1976]. However the groups SL2(p") constitute the only class of Chevalley groups for which much is known about Cartan invariants. For instance every Cartan invariant is a sum of at most two powers of 2 and c 1 , = 2" for SL2 (p" ) in characteristic p. See J.E. Humphreys [1973a], Jeyakumar [1974], Srinivasan [1964c], Upadhyaya [1978].
169
SOME OPEN PROBLEMS
The most complete results concerning Cart an invariants for an infinite class of groups of Lie type are due to Alperin [1976c], [1979] for the groups SL2 (2") for p = 2. Before stating these results some notation is needed. Let cp be the Brauer character afforded by the natural 2-dimensional representation of SL2 (2") over the field F of 2" elements. Let o- be the Frobenius automorphism. Let S be the group of integers modulo n. For i E S define ço, = çor''' and for I C S let cp, = R E! ço,. Then every irreducible Brauer character of SL2 (2") is of the form cp, for some subset I of S, and conversely these are all the irreducible Brauer characters of SL2 (2").
THEOREM. CI,/ = 0 unless for each i E S with i EInJ and i +1 I r-1.1 we have i +1 0 I and i +1 0J, in which case c, For decomposition numbers of the groups SL2 (p") see Srinivasan [1964c]; Burkhardt [1976a], [1976b]. Question (VII) is also related to an old conjecture of Frobenius as
follows. LEMMA 5.2. Let IGI=p°go with (p, go) = 1. Suppose that G contains exactly g o p'-elements. If c 1 p° then G has a normal p-complement. PROOF. By the Cauchy—Schwartz inequality and (3.3) 192 = ( 0 1,
711)2
(i3O1)07i,
= cirp'
p 2 ".
Thus î and 0, are proportional. Hence 0,(z)= 0,(1) if and only if z is a p'-element in G. Thus the set of all p'-elements in G is the kernel of o, and this kernel is a normal p-complement. COROLLARY 5.3 (Brauer and Nesbitt [1941]). Let I G I = puer` where p, q, r are distinct primes. If G contains exactly q br` p'-elements then G has a normal p-complement. PROOF. By (4.15)(iv) (VIII) Is
p°. The result follows from (5.2). D
p`l whenever x„, cp, lie in a block of defect d?
By (X.4.6) (VIII) has an affirmative answer for p-solvable groups. (IX) Is it true that every irreducible character in a block B has height 0 if and only if B has an abelian defect group?
170
CHAPTER IV
By (4.18) blocks of defect d 2 have no characters of positive height and so (IX) has an affirmative answer for d 2. In case G is p -solvable Fong [1960] has shown that if the defect group of B is abelian then every character in B has height 0 and the converse holds at least for the principal block. See (X.4.3) and (X.4.5). Knôrr [1979] has recently proved a result which implies that if V is either an R -free R[G] module such that VK is irreducible or V is an irreducible ft [G] module and V lies in a block with an abelian defect group D, then D is a vertex of V. Some consequences of an affirmative answer to (IX) are mentioned in Brauer [1962b]. (X) Let k(B) be the number of irreducible characters in the block B. Let k o(d) be the minimum of k(B) as B ranges over all blocks of defect d in all groups G and let k(d) be the maximum as B ranges over all blocks of defect d in all groups G. (i) Is lim e,— k o(d)= 00? (ii) Is k(d) - -Ç. p a ?
Nothing seems to be known about (X)(i). (4.18) yields an upper bound for k (d) which will be improved below. It can be shown that if the defect group of B is abelian and generated by two elements then k (d) generalizing the known fact that this inequality holds for d 2. See (VII.10.17). P. Fong has observed the following result in which the function k(d) arises naturally.
LEMMA 5.4. Let H be a d-dimensional linear group over the field of p elements. Assume that H is a p'-group and FI X (1). Then H has at most k(d)-1 conjugate classes.
PROOF. Let P be an elementary abelian p -group of order p' 1 . Then P is isomorphic to a d -dimensional vector space over the field of p elements. Let G = HP with P A (x - ')XA (x) for x E G. Then as FIG] modules
V where
E,=-VOV*-=VOV-=E.
Let E, be the set of all scalar matrices in E. Then E2 .-= Vo(G) is a submodule of E. The result follows as El = (ii) Define f: E —> T by f (X) = X + X'. Then f is an of E onto T with kernel S. homrpis (iii)Without loss of generality F may be replaced by its algebraic closure.
190
[11
CHAPTER IV
Thus the map sending a to a' is an automorphism of F. Let x e G and let A (x)= (a,,). Then A (x )' X„A (x ) = (a, for some X E T.
si
=Ea
+X
Cl
Let M = (m11 ) be an upper triangular matrix in E, i.e. mo = 0 for i > j. Define the quadratic form QM on V by QM
Define fM
(v) =
E mixix; where y = (x 1 , ..., x„). by
E Hom, (S, F)
f m ((x; ; ))-
E m,,x,,.
LEMMA 11.4. Let M be an upper triangular matrix in E. Then QM is G-invariant if and only if fm E FlomF[G] (S, F) where F V o(G).
PROOF. Let y = (x,, Qm(vA)=
x„ ) E V and let A = (a,,) E E. Then
E
Thus QM (vA) = Go m (v) if and only if the following two equations are satisfied for all s < t.
E infaisais = mss, E ,.;
+ afs a„) =
(11.5) (11.6)
For all s Vo (G)—> W—> V—>0.
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 0—> To —> T— V0 (G)—>0.
Thus by (11.3)(iii) there exists an exact sequence 0—> Vo (G)—> S IT o —> V (2) —> 0.
As Vo (G) is not a homomorphic image of S it follows that the sequence is not split. Hence there also exists a nonsplit sequence of the required form. 0
COROLLARY 11.9. If Vis not in the principal 2-block then Vis of quadratic type.
PROOF. Clear by (11.8).
LI
COROLLARY 11.10. If G is solvable then V is of quadratic type. PROOF. Induction on GI. If I G I = 1 the result is clear. Suppose that I G I > 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 02,,2 (G) is in the kernel of V by (4.12). As 02,,2(G) (1) the result follows by induction. El
CHAPTER
V
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 REG] corresponding to B. v is the exponential valuation defined on K with v(p) = 1. xi, x2, .. are all the irreducible characters of G. q)2, ... are all the irreducible Brauer characters of G. (0, is the central character of K[G] corresponding to X s. If H is a subgroup of G and h:Z(K;H:H)-->K is linear, define h' : Z(K;G G)—>K by h' (0) = 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 k are splitting fields for every subgroup of G. See for instance Broué [1972], [1973]; Hubbart [1972]; Reynolds [1971]. 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 characterof H. Let e =Es asxs and let co be the central character of K[H] corresponding to Then
E .2„x„ (ow, =
r (ow
G.
SOME ELEMENTARY RESULTS
1]
193
PROOF. Let { L,} be the set of all conjugate classes of H and let z, E L. Let ‘(x) = 0 if x E G — H. If C is a conjugate class of G with z E C the definition of induced characters implies that
Thus
asx, (1)co, (C) =
( E`ICG a IGI (z)I X"z ' s
= IG :1114- (1)
E
L, çC
w(L,)= r(l)co G (C).
LEMMA 1.2. Let H be a subgroup of G and let Bo be a block of H. Assume that B o contains an irreducible character such that r is irreducible. Then Bô is defined and r E Bô .
PROOF. Let 4-G = X s and let co be the central character of K[H] corresponding to By (1.1) cos = co'. Thus 65 0 = Co, is a central character of R[G]. 0 LEMMA 1.3. Let B o be a block of the subgroup H of G for which Bô is defined. Let 4' be an irreducible character in B o and let r =Escio(s. If B is a block of G then
E
avx , ( 1)) > vQ-G (1))
Eel3 avx,(0)= ,(4-- (1))
if B/ if B
Bô,
= K.
xs
PROOF. Let to be the central character of K[H] corresponding to 4- . Let e be the central idempotent in R[G] corresponding to B. Then to, (e)= 1 if xs E B, and cos (e)= 0 if xs 0 B. Also (.5 G (e) = 0 if B B ô and (.5 G (e)= 1 if B = B. The result now follows from (1.1). LI COROLLARY 1.4. Let B o be a block of the subgroup H of G for which Bô is defined. Assume that B o and Bô have the same defect. If 4- is an irreducible character of height 0 in B o then some character x s of height 0 in B ° as a constituent of r with multiplicity a 5 0 (mod p). PROOF. Clear by (1.3).
El
194
CHAPTER
V
[1
THEOREM 1.5. Let B be a block of G with defect group D. Suppose that AB (07 ) 0 for a conjugate class C of G. Then there exists z E C with z C CG (D) such that the p -factor y of z is in Z(D). PROOF. By (111.9.7) there exists a block Bo of NG (D) with defect group D such that Bô = B. Let Ao = À. Since A ô(0)/ 0 there exists a conjugate class Co of NG (D) with CO C C and A 0 (00) # O. Choose z E Co . By z C CG(D). If is an irreducible character in Bo then ‘(z)# O. Hence by (IV.2.4) y is conjugate in NG (D) to an element of D and so y E D. Since D C CG (Z) C CG (y) it follows that y E Z(D). 0 THEOREM 1.6. Let B o be a block of the subgroup H of G for which Bô is defined. Let Do be a defect group of B o and let D be a defect group of Bô . Then (i) (ii) (iii) (iv)
Every element in Z(D) is conjugate to an element in Z(D0). The exponent of Z(D) is at most equal to the exponent of Z(D0). If B q has defect 0 then Bô has defect O. If p IIGI then H# (1).
PROOF. (i) Let x, be a character of height 0 in Bô . By (III.6.10) and (IV.4.8) there exists a p'-element x such that D is a Sr -group of CG (x) and x,(x)(o, (x 0 (mod IT). If y E Z(D) then D is a Sp -group of CG (xy) and x, (xy ) x, (x ) 0 (mod IT). Thus v (a), (xy))= v (0), (x )) = O. Hence U), (xy ) 0 (mod ii- ) for all y E Z(D). Let be an irreducible character in Bo and let to be the central character of K[H] corresponding to Then = Co G. Hence if y E Z(D) there exists a conjugate z of xy in H with (Ti(z)# O. By (1.5) the p -factor of z is conjugate to an element of Z(D0). (ii) and (iii) are immediate by (i). (iv) If H = (I) then Bô has defect 0 by (iii) and so contains exactly one irreducible character xr . Hence if 4' is the unique character in Bo then r = x (1)xs and so by (1.3) .
v(IG1) = v(e (1)) = v(x,(1) 2)= 2v(IG1).
Thus v (IGO= 0 contrary to assumption.
0
LEMMA 1.7. Let B be a block of G and let u be an automorphism of G such that x = x, for all x, E B. If y is an element in a defect group of B then y is conjugate to y in G.
PROOF. By (IV.4.8) there exists a p'-element x and a character xr of height
2]
195
INERTIA GROUPS
0 in B such that x, (x)6), (x -I ) 0 (mod ir) and a S„-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) = x, (x ) 0 (mod 7r). If y and y ° are not conjugate in G then by (IV.6.3)
EE B1,6(yx)1 2
-=
E xxyxnx.(yx)-o
x EB
Since Ix, (yx )1 0 for all s this implies that x, (yx ) = 0 for all x, E B. Hence in particular x,(yx) = 0 contrary to the previous paragraph. E 2. Inertia groups
Throughout this section Ô is a fixed normal subgroup of G. In general a tilde sign will be attached to the quantities associated with Ô. For instance • are all the irreducible characters of Ô. 05s is the central character of K[0] corresponding to is . Let A be a subset of Ô which is a union of conjugate classes of G. Let 6 be a complex valued function defined on A which is constant on the conjugate classes of Ô which lie in A. If z E G define 6 by 6' (x) = 6(x' - ') for all x EA. Then 6' is defined on A and is constant on the conjugate classes of Ô which lie in A. The inertia group T(ô) of 6 in G is defined by T(0) = {z I z E G, 6' = 6 } . Clearly Ô c T (0 ). If iV is an irreducible 17 [ 6] module which affords 0, then it is easily seen that T( W) = T( 1 ). Similarly if is an R -free R[Ô] module which affords L. then it is easily seen that T( 'K) = T However T (V') need not be equal to T( -/-K). Let A be a block of O. The inertia group TO ) of . in G is defined by T(A) = {z I z E =13-}• LEMMA 2.1. Let 1-4 be a block of Ô. If i„ c-,6; T(is )C T(i4 ).
En
then T(0 1 ) C T(ñ) and
PROOF. Clear by (IV.4.9). E THEOREM 2.2. For any group H with Ô CHC G let ÇA, (H) be the set of all irreducible characters ofH whose restriction to Ô have as a constituent. Let 9J(H) be the set of all irreducible Brauer characters ofH whose restriction to have 45 , as a constituent. (i) If T(CP,)C H then the map sending B to 0 G defines a one to one correspondence between 2t(H) and 91 (G).
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[2
(ii) If Vis an irreducible P[G] module which affords a Brauer character in W(G) then a vertex of V is contained in T(Cp,). (iii) If T( S )CH then the map sending 0 to O G defines a one to one correspondence between 2L (H) and ?Is (G). (iv) If x i E 9.i s (G) then there exists an R -free R[G] module which affords xi and whose vertex is contained in T ( s ).
PROOF. (i) It suffices to prove the result in case T(Cp,)= H. Let 0 E 91?(H). By (11.2.9) (e ) ' , = fi + B where Cp, is not an irreducible constituent of Ob. Let i be an irreducible constituent of q1G with yin = i + 0 1 . By (111.2.12) = e(p, for some integer e > 0 and i (1) =G : tpG (1). Hence = i E 9f?(G). Furthermore if 0,, tif2 E W(H) and 0; = 0? then ( 143)ii = (OH and so 01 = 02. Therefore the map sending 0 to O G is one to one. It remains to show that it is onto. Let cp, E 2l?(G). Let V be an R[G] module which affords 4), and let IV, be an 1 [6] module which affords (p,. By (111.2.12) %id is completely reducible and so /(V6, l)/ O. Thus by (111.2.5) /(VH, 0 and so W)#0 for some irreducible constituent W of Ii7 ;1 . Since (1' / 1, 1)6 = : GIW, it follows that Vi7, is a constituent of IV,. Thus if (Is is the Brauer character afforded by W then 0 E 21?(H). Since /( V, W G ) = /(VH, W) #0 the irreducibility of V and WG implies that V = W G and hence cp, = oG as required. (ii) Clear by (i). (iii) If p is replaced by a prime not dividing IG1 this follows from (i). (iv) Clear by (iii). Lii
co=
A block B of G covers a block 14 of O if there exists xs E B and is a constituent of (x)o.
E /E-3
such that
LEMMA 2.3. The blocks of Ô covered by the block B of G form a family of blocks conjugate in G. If I4 is covered by B and xs E B then some constituent of (Xs)6 belongs to P. If t E f3 there exists x s E B such that 1 is a constituent of (xs)o. PROOF. This is a reformulation of (IV.4.10). 111 LEMMA 2.4. xs and xt belong to blocks which cover the same block of Ô if and only if there exists a chain x,, = X s, x12 , =xi such that for any m either x and Xj, are in the same block of G or (x),„)0 and (x,,)G have a common irreducible constituent.
2]
INERTIA GROUPS
197
PROOF. If such a chain exists then by (2.3) consecutive characters belong to blocks which cover a fixed block of Ô. Suppose conversely that x, and xi are in blocks which cover a fixed block of O. Let be an irreducible constituent of (Xs )ô. By (2.3) there exists X, in the same block as x, such that i„, is a constituent of (x,)0. The chain xs,x„x, satisfies the required conditions. 0 THEOREM 2.5 (Fong [1961], Reynolds [1963 1 ). Let fl be a block of Ô. The map sending b. to bG defines a one to one correspondence between the set of all blocks of T (li ) which cover ii and the set of all blocks of G which cover Furthermore if b. is a block of T(B) which covers B the following hold. (i) The map sending 0 to O G defines a one to one correspondence between the sets of all irreducible characters, irreducible Brauer characters respectively, in b and bG. (ii) With respect to the correspondence defined in (i) b. and !I' have the same decomposition matrix and the same Cartan matrix. (iii) b. and b. ° have a defect group in common. (iv) ii is the unique block of 6 covered by Ê.
PROOF. Let b be a block of T(ñ) which covers b and let be an irreducible character in By (1.2), (2.1) and (2.2) (iii) if3 G is defined, E b. ° and ÊG covers ii. Suppose that B is a block of G which covers Let x, E B. By (2.3) (xi )6 has an irreducible constituent is E Û. By (2.1) and (2.2) (iii) xi = r for some irreducible character of T(b) such that is an irreducible constituent of 4"6. Let b be the block of T(A) with 4- E B. is Then Ê covers El and i4 G = B by (1.2). Let 0,,02 be irreducible Brauer characters of T(b) in blocks which cover ii. If Cp E ii and 43 is an irreducible constituent of (e)G then Cp is a constituent of OA )d for i = 1,2. Thus if i= tg then (4/ 1 )0 and (02 )6 have a common irreducible constituent which is in and so 01 = 02 by (2.2). Therefore the map sending 0 to 0G defines a one to one correspondence between the sets of all irreducible characters, irreducible Brauer characters respectively, of T (li ) and G which lie in blocks that cover ii. 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 O of T(b) is in Ê if and only if 0 G E Ê G Thus the map sending Ê to Ê ' is one to one as required. Furthermore (i) and (ii) are proved. (iii) Let 15 be a defect group of Ê. By (111.9.6) 15 C D for some defect group D of Ê. By (i) and !I' have the same defect. Thus D = D.
r
.
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(iv) Since E = E for all z E T(f3) this is clear by (2.3).
[3
D
COROLLARY 2.6. Let B be a block of G and let E be a block of Ô covered by B. Then T(b) contains a defect group of B.
PROOF. By (2.5) B = if3 G for some block E of T(b) which covers E. The result follows from (2.5) (iii). 0
3. Blocks and normal subgroups The notation of the previous section will be used in this section. For the results in this section see Brauer [1967a], [1968]; Fong [1961]; Passman [1969]. LairmA 3.1. Let a E Z (R;G : G)n R[6]. If L is a central character of 1 [6] then (a) = (a).
PROOF. Clear by definition.
D
LEMMA 3.2. Let {B- 1„,} be the set of all blocks of Ô where for some z E G if and only if i = j. Let Jim be the centrally primitive idempotent of R[6] with = B(é,„). Let e, Then fe 1 1 is the set of all primitive idempotents in Z(R; G : G)(-1 R[6]. PROOF. Any idempotent in Z (R ; G: G) n R[6] is a sum of centrally primitive idempotents in R[Ô] and is invariant under conjugation by elements of G. The result follows. D LEMMA 3.3 (Fong [1961]). Let E,„, and
be defined as in (3.2). For each i let be all the blocks of G which cover some and let e,, be the centrally primitive idempotent in R[G] with B„, = B(e,„). Then e, =
I rrt
Jim
= /m
eim
•
By (3.2) {e,} is a set of pairwise orthogonal central idempotents in R [G]. Thus it suffices to show that e,„,e, X 0 for all i, m. Let V be a nonzero R[G] module in B 1,,,. By (2.3) (V 6)e, O. Hence Ve, = Ve,„ e, 0 and so e, X O. D PROOF.
LEMMA 3.4 (Passman [1969]). Let B be a block of G and letE be a block of
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BLOCKS AND NORMAL SUBGROUPS
Ô. Then B covers 1j aEZ(R;G:G)nR[6].
if and
only
if
Ab (a)= AB (a) for
all
PROOF. By (3.2) and (3.3) B covers 1 if and only if Ab (ë) = AB (e) for every idempotent in Z(R;G : G) n 1t[6]. By (I.16.1) ft is a splitting field of Z(R;G : G) n R[6]. The result follows since two central characters of a commutative algebra are equal if they agree on all idempotents. 0 LEMMA 3.5. Suppose that GIG- is a p-group. If E is a block of 6 there is a unique block B of G which covers E. PROOF. Suppose that B, and B 2 are blocks of G which cover E. Let A, = AB, for i = 1,2. If C is a conjugate class of G consisting of p'-elements then E Z(R; G :G)n R[6]. Thus by (3.4) A i (0) = (0= A2(0). Hence by (IV.4.2), (IV.4.3) and (IV.4.8) (iii) B 1 = B 2 . 0
e,) =
A block B of G is regular with respect to 6 if AB ( 0 for all conjugate classes C of G which are not contained in Ô. B is weakly regular with respect to 6 if there exists a conjugate class C of G with C C 6 such that AB (0') / 0 and the defect of B is equal to the defect of C. In case 6 is determined by the context the phrase "with respect to 6" will be omitted. By (III.6.10) a regular block is weakly regular. Furthermore if AB (0') / 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 = EG for every block E of 6 which is covered by B. (iii) B = PG for some block E of Ô. PROOF. (i) (ii). Let B- be a block of 6 which is covered by B. Let C be a conjugate class of G. If CZ 6 then (AA )'(0) = 0 = AB (0). If C C 6 then by (3.4) AiJ (0') = AB (). Hence by (3.1) (AEO G =AB. (iii). Clear. (i). If C is a conjugate class of G with CZG then AB (0) = 00-() = 0 and so B is regular. D (iii)
LEMMA 3.7. Let B be a block of G which is regular and let E be a block of Ô. Then B covers È if and only if B =
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PROOF. If B covers Pi then B = 13 G by (3.6). If B = 13- G then by (3.1) AA (a) = (AO' (a) for all a E Z(R; G G) n R[G]. Thus by (3.4) B covers E LEMMA 3.8. Let 13 be a block of Ô. Then there exists a block of G which is regular and covers 1 if and only if 13 G is defined. In that case 13- G is the unique block of G which is regular and covers
P.
PROOF. If ÉG is defined then 13- G is regular by (3.6) and É G covers 13- by (3.7). The converse follows from (3.7). The last sentence is clear. D 3.9. Let B be a block of G with defect group D. If CG (D) C B is regular.
LEMMA
6 then
PROOF. If AB (C) 0 for some conjugate class of G then by (III.6.10) D CCG(Z) for some z E C. Hence z E CG (D) C 6 and so C C 6. 0 LEMMA 3.10. Suppose that P is a p -group with P< G and CG (P) C Ô. Then every block B of G is regular with respect to Ô. For every block 13 of 6, the block 13 G is defined and is the unique block covering P. PROOF. Let B be a block of G and let D be the defect group of B. By (111.6.9) P C D. Thus C G (D)C CG (P) C 6 and B is regular by (3.9). Let be a block of G. By (3.8) ÉG is defined and is the unique block covering 1. Ll 3.11. Suppose that P is a p-group with CG (P) ci P < G. Then the principal block is the unique block of G.
COROLLARY
PROOF. By (3.10) with 6 = P, 13- G is the unique block of G where 13- is the unique block of O. Hence G has exactly one block which must necessarily be the principal block. 0 THEOREM 3.12 (Fong [1961]). Let 13 be a block of 6 and let B 1 ,13 2 ,... be all the blocks of G which cover B and let B = B, for some j. For each i let e, be the centrally primitive idempotent of R[G] with B, = B(e,) and let e =Ee,. The following are equivalent. (i) B is weakly regular. (ii) A defect group of any block B, is conjugate to a subgroup of a defect group of B. (iii) The defect of any B, is at most equal to the defect of B.
BLOCKS AND NORMAL SUBGROUPS
201
(iv) Let e =Eace where C ranges over all conjugate classes of G and ac E R. There exists a conjugate class C consisting of p'-elements with C C Ô, cic X 0, AB (0) X 0 such that the defect of B is equal to the defect of C. PROOF. (i) (ii). There exists a conjugate class C of G with C C Ô and AB (C_) # 0 such that B and C have a common defect group D. By (3.4) AB, AA AB ) X 0 and so by (III.6.10) a defect group of B, is a subgroup of some defect group of B. (ii) (iii). Trivial. (iii) (iv). Since AB (e) = 1 it follows that d cAB (6) / 0 for some C. By (IV.7.2) C consists of p'-elements. By (3.3) C C O. If e, =Ea,,c e" for each i then for some i a1# O. Hence by (IV.7.3) the defect of C is equal to the defect of B, and so is at most equal to the defect of B. Since A (0) / 0, (III.6.10) now implies that B and C have the same defect. (iv) (i). Trivial. 0
(e) =
(e) =
(e
COROLLARY 3.13. Let Li be a block of O. The following hold. (i) There exists a weakly regular block B of G which covers 1. (ii) Let I be a block of T(13- ) which covers P. Then Li is weakly regular if and only if AG is weakly regular. PROOF. (i) Choose a block B of maximum defect among those which cover A. By (3.12) B is weakly regular. (ii) By (2.5) A and AG have the same defect. Thus Ê has maximum defect among all blocks 1 of T(b) which cover if? if and only if ./f3 G has maximum defect among all blocks Ê. The result follows from (3.12). 0 THEOREM 3.14 (Fong [1961]). Let Li be a block of O and let B be a block of G which covers 1 and is weakly regular. There exists a defect group D of B with D C T (1 ). For any defect group D of 13 with D C T(1) IT(Li): DOI 0 (modp) and D n O is a defect group of t.
T(Li) contains a defect group D of B. Thus by (2.5) and (3.13) (ii) it may be assumed that G = T(A). Let é" be the central idempotent of R EG] with ij = B(é). Since G = T (1 ), =Eace 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 C O, dc X 0, AB (0) X 0 such that D is a defect group of C. Let z E C with D C CG (z). Let L be the conjugate class of O with z G L. Thus C = U Lx where x ranges over a cross section of NG (L) in G. By (3.4) PROOF. By (2.6)
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AR
( 0) = Ao ( 0) =
AB
V
(i')=
AL -1 ( 1 )-
Since T(13- ) = G this implies that
T(fi):No (L)1,ka (Z)= AB (0) O. Thus T(/4 ): NG (L 0 (mod p) and /lb (f..) O. Since NG (L)-- CG (z)Gand D is a Sr -group of CG(Z) it follows that I T(fi):DO I 0 (mod p). Since dcvka (L)= Oit follows from (IV.7.3) that fi and L have a common defect group. Clearly D n G is a Sr -group of Co (z) and so is a defect group of L. E LEMMA 3.15. Let B be a weakly regular block of G and let 1-4 be a block of covered by B. Let Xs be of height 0 in B and let it be a constituent of (X.)6 with it E E. Then the ramification index of xs is not divisible by p height 0 in Pi and IT(b):T(X 7,)1 0 (mod p).
,
has
PROOF. By (2.5) and (3.13) it may be assumed that G = Let D be a defect group of B. Let d = v (I D I) and let d = v (I D n Op). By (3.14) d is the defect of E. By (111.2.12) (x, )d = e 4; where {z} is a cross section of T(it) in G and e is the ramification index of x, Since 6D/0 --- DIG n D it follows from (3.14) that v(I GI)— d=- v(IOD:DO= v(IO: 6-
n D 1) = v(161)- d.
Thus v(xs ( 1 )) v(1 6 1) — d. Since v (it ( 1 )) v — d and x, (1) = eIG:T(X- )lit (1) it follows that e I G : T(x- )I 0 (mod p) and X', is of height 0 in b. O COROLLARY 3.16. Let kJ be a block of Ô. There exists
it E /3- of height 0 such
that T(it ) contains a Se -group of T(13- ).
PROOF. By (3.8) there exists a block B of G which covers E and is weakly regular. Let x, be of height 0 in B. By (2.3) some constituent X", of (xs )ô is in E. The result follows from (3.15). El
4. Blocks and quotient groups The notation of the two previous sections will be used in this section. Furthermore G ° = GIG- and in general a superscript ° will be attached to the quantities associated with G`). Every R [G1 module may be considered
BLOCKS AND QUOTIENT GROUPS
203
to be an REG] module in a natural way. Similarly every character or Brauer character of G' is a character or Brauer character of G. LEMMA 4.1. Let B ° be a block of G'. Then there exists a block of G which
contains B ° and covers the principal block of Ô. Conversely if B is a block of G which covers the principal block of Ô then B contains a block of G'.
Two R[ G °] modules in B ° are linked by a chain of R[G1 modules. Hence they are also linked as R[G] modules and so lie in the same block B of G. Thus B ° C B. If x, E B ° C B then the principal character of Ô is a constituent of (x)6. Thus by (2.3) B covers the principal block of Ô. Conversely suppose that B covers the principal block of Ô. By (2.3) there exists Xs G B such that (x, )d contains the principal character of Ô as a constituent. Thus by (111.2.12) Ô is in the kernel of Let B ° be the block of G' with x, e B ° . Then B ° C B by the previous paragraph. 0
PROOF.
LEMMA 4.2. Let B ° be a block of G' and let B be a block of G with B ° C B.
Let D be a defect group of B. Then D contains a Se -group of 6- and D' = 16contains a defect group of B". In particular ifd is the defect of B and d' is the defect of B ° then d' d — v(1G I). Furthermore if B ° contains an irreducible character or an irreducible Brauer character of height 0 in B then D' = DG - 16 is a defect group of 13'. '
PROOF. Let P be a Se -group of Ô and let P be a p -group with P c P such that P° = 16 is a defect group of B". Let V be an R -free R[G1 module which affords an irreducible character of height 0 in B". Thus P' is a vertex of V. Hence V considered as an R [G] module is R [P]-projective. Thus by (IV.2.2) P is a vertex of V. Hence P C G D. If v(ço, (1)) or v(x, (1)) equals v(I G 1) d then v(1 G 1) d v (I G : Ô 1)- cr and so d') v(1 I). Thus d" = d — v(1 Ô I) by the first part of the statement. E —
—
—
LEMMA 4.3. _Suppose that Ô is a p'-group. Let B, ,
, B„, be all the blocks of G which have Ô in their kernel and let D, be a defect group of B, for j = 1, , m. Let 13 7 , , be all the blocks of G'. Then m = n and after a suitable rearrangement B, = 13; for all j. Furthermore D = D,G - 16 is a defect group of 13; . PROOF. Since Ô is a p'-group the principal character is the only irreducible character in the principal block of Ô. Hence a block of G covers the principal block of Ô if and only if it has Ô in its kernel. Thus by (I V.4.11)
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any two modules in B, are linked by a chain of R [GI modules and so B, is after a suitable a block of G ° . Therefore by (4.1) m = n and B. rearrangement. Since Ô is a p'-group B, and B`i have the same defect and so by (4.2) D(,) is a defect group of B?. 0 LEMMA 4.4. Suppose that Ô is a p-group. Let B be a block of G and let D be a defect group of B. Then Ô C D and there exists a block B ° of G ° such that C B and D ° = D/G- is a defect group of B ° . PROOF Let d = v( D I). Choose cp, E B with v(q), (1)) = v (I G I) — d. By (111.2.1 3) G is in the kernel of cp,. Let B ° be the block of G° with cp, E B ° . By (4.1) B ° C B. Let d ° be the defect of B ° . The result follows from (4.2). CI 4.5. Suppose that Ô is a p-group which centralizes all p'-elements in The inclusion B ° C B defines a one to one correspondence from the set of all blocks of G ° onto the set of all blocks of G. If D is a defect group of the block B of G then D ° = DG- /6 is a defect group of the block B ° of G ° where 13 ° C B. LEMMA
G.
PROOF. In view of (4.1) and (4.4) it suffices to show that if B is a block of G and B?, B? are blocks of G° with 13° C B and B °2 C B then B7 = B ° Let xs e B° and xr e 13'2'. If x is a p'-element in G then Ô c CG (x). Thus CG ()CVO = CG 0 (xl where x ° is the image of x in G° . Since x„ x E B this implies that
I G ° 1 X.±x) _ ICGo(x °) I Xs ( 1 )
_ (x
(x)
_ G° I C G 0(01
(x) xi ( 1) (mod ,n-)
for all p'-elements x in G. Hence by (IV.4.3) and (IV.4.8) B; = .13?. El
4.6. Suppose that D is a p-group and G = DC G (D). Let B be a block of G with defect group D. Then B contains exactly one irreducible character x, which has D in its kernel and B contains exactly one irreducible Brauer character cp,. Furthermore x, (x)= ça, (x) for all p'-elements x in G, has height 0 and (I), (x) = I D I(P, (x) for all p'-elements x in G. COROLLARY
PROOF. By (4.5) there exists exactly one block B ° of G/D with B ° C B. Furthermore B ° is a block of defect O. Let xs, cp, respectively, be the unique irreducible character, irreducible Brauer character respectively, in B ° . Thus x, (x) = (x) for all p'-elements in G. Since every irreducible Brauer
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BLOCKS AND QUOTIENT GROUPS
character in B has D in its kernel, cp, is the unique irreducible Brauer character in B. Then x, has height 0 in B by (4.5). Since (c„) is the Cartan matrix of B, (IV.4.1 6) (i) implies that c„ =ID I proving the last statement. El If G = DCG (D) for some p -group D and B is a block of G with defect group D then the irreducible character Xs constructed in (4.6) is the canonical character of B.
THEOREM 4.7. Suppose that G = DCG (D) for some p-group D. Let B be a block of G with defect group D and let x = x, be the canonical characterof B. Let 4-1 , , be all the irreducible characters of D. Define O. follows : ify0D,
01 (z)= 0 = (Y)X(x)
if Y E D
where y is the p-part of z and x is the p'-part of z. Then {0, j j = 1, .. .,n} is the set of all irreducible characters in B.
PROOF If z E G then 47 = for all j. Thus if xr E B then by (111.2.12) (xt)D = 41 for some j and some integer e > 0. Let z E G, let y be the p-part of z and let x be the p'-part of z. If y D then x, (z ) = 0 by (111.6.8) and (IV.2.4). Suppose that y E D and let H=D x (x) . Then (x( ),, = 6a for some character a of (x). Thus x, (xy ) = (y)a (x) and
"J ( Y )) X, (xY )= ci ( 1 X, (x)=
(Y1 ) d1 (x) =
dt;
w e; (xY
O. where cp; e B. Consequently x, = [cla/6 Let A be the set of all p'-elements in G and let A ° be the image of A in G° = GID. Since x is a character of G° which vanishes on all p -singular elements we see that ( 1 )]
G ,,k-i(y)12,,,,lx(x)1 2 =- I D EG„Ix(z)1 2 =1. - IG1 . 1 ,E Thus
1= 11x, = d:; ) 2 1 0,112 =d( '; ) 2. Hence d, = (1) and so x, = O.
f xr,lx(x)12-
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[5
Therefore there exists a subset S of 11 , , /21 such that {Of jj E SI is the set of all irreducible characters in B. If x, = Of then d,i = 6 (1). Thus I DI=c11= E This implies that S =
E
(1)2.
iEs
,
0
5. Properties of the Brauer correspondence
The results in this section are due to Brauer [1967]. An alternative approach has been given by Passman [1969]. See also Brauer [1971a], [1974a]. The presentation given here is a modification of these two methods. Recently Alperin and Broue [1979], Broil& [1980] have studied the Brauer correspondence in a more general setting. Their approach generalizes and gives an alternative treatment of the first main theorem on blocks and much of the material in this section and the next. P be a p-group, P C G. Let N = NG (P) and let H < N with (P) C H. If Bo is a block of H then Br,' is defined.
LEMMA 5.1. Let CG
By (3.10) Ir is defined. Thus by (111.9.2) and (111.9.4) Br); = (13(;')G is defined. 0 PROOF
p-group, D C G. Let N = NG (D) and let H < N with CG (D) C H. The map sending B o to B 0' defines a one to one correspondence between the blocks of G with defect group D and representatives B o of N-conjugate classes of blocks of H which satisfy (i) D n H is a defect group of B o . (ii) v (1 DII j) v(IT(B 0)1), where T(B 0) is the inertia group of Bo in N. LEMMA 5.2. Let D be a
PROOF. By (111.9.7) it may be assumed that N = G. By (2.3) and (3.10) the map sending B o to BI;I defines a one to one correspondence between blocks B of N and representatives B o of N-conjugate classes of blocks of H. If B = .K has defect group D then by (3.14) D n H is a defect group of Bo and v DH = T(13 0)1)Suppose that Bo satisfies (i) and (ii). Let D, be a defect group of .13 0" . By (3.14) and (i) D n H and D, n H are both defect groups of B0 . Thus By (3.14) and (ii) 7)(1 DH v( T(B0)1)=
5]
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207
I). Thus IDI=1 D, I. Since D1N D CD f . Therefore D = D i . ,
Let D be a p-group, D C G. (D,B) is a block pair in G if B is a block of DCG (D) with defect group D. If (D,B) is a block pair in G then by (111.9.4) B A is defined for every group A with DC G (D) ÇA C G. Let (D,B) and (D*,B*) be block pairs in G. (D*,B*) weakly extends (D,B) if D from the set of all irreducible characters of Go with Go n H in their kernel to the set of all irreducible characters of G0 H with H in their kernel. Thus C--G„ = Let T be the union of all blocks of G which have H in their kernel. (i) Suppose that H is in the kernel of B. Let 4- be an irreducible character in B0 . By (1.3) there exists an irreducible character x in B with xC Thus 4' C xG„ by Frobenius reciprocity (111.2.5). Since Go n H is in the kernel of xG„, it is in the kernel of C. Since C was arbitrary in Bo this implies that G o n H is in the kernel of Bo . Suppose conversely that Go n H is in the kernel of Bo . By Frobenius reciprocity (111.2.5) 4' c Co". Since H < G a cross section of G0H in G is a complete set of representatives of the (Go, H) double cosets in G. Thus the Mackey decomposition (11.2.9) implies that
r.
G G )11 =
= Efc(1)60,-,0H,
=
where x ranges over a cross section of G0 H in G. Therefore (1,,,(C G ) H ), = I G:Go HIC(1)=IG:Go Hl4-- (1)= G (1) .
Thus 4-- is the sum of all the irreducible constituents of C G which have H in their kernel. Hence (C G )T = If B is a block with B CT then v((C G r (1)) v(C G (1)) by (1.3). Since (1) = G0 H: Gole - ( 1) it follows that v(C (1)) = v(4- (1)). Since (C G = 47- this implies the existence of a block B, C T with v((C G )81 (1)) = v(C G (1)). Hence B 1 = .10 by (1.3) and so H is in the kernel of B. (ii) The existence of B- 0 follows from (4.3). Let C be an irreducible character in Bo. Let w, Co be the central character corresponding to respectively. Thus corresponds to B- 0 . Let x i , x2, ... be all the irreducible characters of G and let bs = (xs ,‘ G ). By (i) (C G ) T = c Thus by (1.1) (i)w- = E b. (1)(.0„ G MCC) G = x
E,ET b,x,(1)w s .
Since B O °C T it follows from (IV.4.22) that (C G (1)/e G (1)* G = G . Evaluating at 1 we get that ( z- mu 1 ),= 1. Therefore co = G . Consequently È 1 is defined and fg= B. Furthermore -
IH:
n HI
(
‘G(1)14--
-
(
1) = 1
(mod 71-).
212
Hence
CHAPTER
: Go
n
1 (mod p).
V
[6
111
THEOREM 6.5. Let IT be a set of primes with p E 7r. Let B be a block of G with defect group D. Let N = NG (D) and let B, be the germ of B. Assume that 0,(N) is in the kernel o f B, . Suppose that Go is a subgroup of G, B o is a block of Go with defect group D o, CG C Go and .13 (? = B. Then 0,(G 0 is in the kernel of Bo. )
PROOF. By (111.9.6) it may be assumed that D o C D. By (5.4) Bo = .130 for some block B2 of DOCG (D0). Let d = v(ID I), do = v(jDo I). The proof is by induction on d — do. Suppose that d — do = O. By (111.9.2) .13? = B,?= B and .13? = Thus B t2'j = B, by (111.9.7). By (3.10) B, covers B2. Therefore 0,,(N)= 0,(DC G (D)) is in the kernel of B2. Thus 0.,(G0) n D CG (D ) is in the kernel of B2 and so 0„,(G0) 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 (D o, B2). Let H = D *CG (D o ). By induction 0,(H) is in the kernel of B = B* 11. Since BV covers B2 it follows that 0„ , (H) = 0 , (Do C G (Do)) is in the kernel of B2. As 0„ , (G0) fl DoCG (DOC 0„ ,(D o C G (Do)), (6.4) implies that 0,(G0) is in the kernel of Bo = B. 0 6.6. Let IT be a set of primes with p E 77. Let B be a block of G with defect group D. Assume that OAN G (D)) is in the kernel of the germ of B. Suppose that y is a p-element in G such that CG (y) has a normal S,-subgroup. If x = Xi E B then x(yx)= x(y) for every 7r'-element x in
COROLLARY
CG(Y)•
PROOF. By (IV.6.1) and (6.5) x(yx) = d;cp;'(x) where cp ranges over blocks of CG (y) which have O,. (C G (y )) in their kernel. Thus (p),' (x) = cp; (1) for every 7r'-element x in CG (y). THEOREM 6.7. Let IT be a set of primes with p E
77. Let B be a block of G with defect group D. Let Do be a p-group contained in G and let B o be a block of D o C G (Do) such that .13, = B. (i) Suppose that whenever D o C D' for some z E G, O,' (CG )) is in the kernel of B o. Then 0,(N G (D)) is in the kernel of the germ of B. (ii) Suppose that 0,(CG (Do)) is a S,-group of CG (Do) and 0, (C G (Do)) is in the kernel of B o. Then 0,(N G (D)) is in the kernel of the germ of B.
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213
PROOF. By (6.4) it may be assumed that D o is a defect group of Bo. (i) Let d = v(ID1) and let do = v ( Do I). The proof is by induction on d — do. If d — do = 0 then Do = I:» and by (3.10) the germ of B with respect to I:» covers B o. The result follows since 0, (C G (Dz ))= 0,(NG (Er )). Suppose that d — do > 0. By (5.4)(ii) there exists a block pair (D*, B*) in G which properly extends (D o, Bo). Let H = D *C G (Do). Suppose that D* C Dz. Let Ho be the kernel of Bo. By (IV.4.11) Ho is a p'-group. Furthermore H o is the kernel of B' = B* 11 and 0(C G (Dz ))C Ho . It follows from (6.4) that 0, , (CG (Dz ))C Ho n D * CG (D*) and so is in the kernel of B*. The result follows by induction. (ii) Since the kernel of Bo contains all 71-'-elements in CG (Do) the assumptions of (i) are satisfied. Thus the result follows from (i). 0 The following example shows that the assumptions in (6.7)(ii) that 0, , (C G (Do)) is a S„.,-group of CG (Do) cannot be dropped even in case
IT = {P }. Let (y ) be a cyclic group of order 4. Let S = S, be the symmetric group on 5 letters and let A = A, be the alternating group. Let G = (y)S where (y ) < G and y z =y for z E A, y z = y ' for z E S — A. Let p = 2. Let 4' be the irreduciblt character of degree 4 of S which is a constituent of the permutation representation of S on 5 letters. Then A is irreducible. Thus 4' A is in a block of defect O. By (3.5) the block of S containing 4' is the only block of S covering that of A. Thus it is weakly regular. Hence by (3.14) there exists an involution x ES—A such that (x) is a defect group of the block of S containing 4'. 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 4' 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 Do CG (Do) = (y ) A and by (3.10) B = B (? for some block Bo of Do CG (D0). 07(130CG (DO) = (1) is in the kernel of B0 . It is easily verified that NG (D)= (x, y) W. Thus W' = 0 2 (NG (D)). Hence by (4.3) the principal block of NG (D) is the only block which has W' in its kernel. Thus by (6.2) 02 , (N G (D)) is not in the kernel of the germ of B. H2 be subgoups of G with LH, C MI-I2 and Hi , H2 p'-groups. Assume that [M,H2] C H2, [L, H,] C HI , L CM and 112 n H, L C H1 . Let B be a block of L with defect group D such that Cm (D) C L. Assume that H, n L is in the kernel of B. Let i be the block of LH, which has H, in its kernel and corresponds to B by the natural
THEOREM 6.8. Let L, M, H,,
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isomorphism from L/L n H, to . Assume furthermore that Ii""2 is defined. Then B " is defined and H2 n M is in the kernel of B". If É " is the block of MI-I 2 corresponding to B " by the natural isomorphism of M/M n H2 onto MI-I2! H2 , then É"", = P. The following diagram illustrates the statement.
= B"H=
MH2
LH,
É
B L PROOF. By (111.9.5) B" is defined. Since M n H2 < M and M n H2 n L C L n Hi it follows from (6.4)(i) that M n H2 is in the kernel of B". Since [L, H2] C H2 and L n H2 ç L n H, is in the kernel of B, there exists a block 1-32 of LH2 with H2 in its kernel such that 112 corresponds to B by the isomorphism L/L n H2 LH 2/H2. By (4.3) 11 2 has D as a defect group. The Frattini argument implies, that C.,/,/,(DH2/1-12)= CM (D)H2/H2 and so C., (D)C (D)H 2 C LH2 . Thus by (111.9.5) 1-3;1 is defined for LH 2 C H C MH2 . We will next show that 21qH
2=
(6.9)
Let 4" E B and let E P2 with 4; L = 4. Let {x s } be the set of all irreducible characters in /3- 12'2 and let as = ( x„ 4' 1' 12).2 . By (1.3) v(4'2(1)) = v (Iasx, (1)). As 4- is the unique irreducible character of LH, with C 4--1 and H, is its kernel and since H, is in the kernel of each x„ it follows from Frobenius reciprocity (111.2.5) that
e).,=
a, = ((xs)L„,, -
((x,)L, ‘), =(( X)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 {(x,),,,,} is the set of all irreducible characters in B'. Hence f3 ' = EM2. However
E as (x,),, (1)) = v (.4
(1)) =,(1mH2: LI-1,1x (1))
) =1)(IM:L14- (1))= 1, (r ( ).
Hence B' = B" by (1.3), and (6.9) is proved. Since H2 n H, L C H, it follows that H, H2 n LH, = H, . Consider the natural isomorphisms
7]
iSOMETRIES
215
L/L n H, LH I /H, LH/LH fl H, H2 LH,/ LH, n H 112 . LH, H 2/ H, H2 Let LI, be the block of LH, H2 with H, H2 in its kernel which corresponds to B. Let A be the block of LH 2 with LH2 n H,H2 in its kernel which corresponds to P3 . Let be an irreducible character in B, let 4 be the unique irreducible character in P2 with 6_ - = and H2 in its kernel. Let 0 be the unique irreducible character of LH, H, with H, H, in its kernel such that OL = ‘. Then 0 E B3 by definition. Furthermore Ou E Since it follows that B = P2. Thus P 2 corresponds to P3. (COL = = Since .r3 is defined, (6.4) implies that f3r , "2 = 13 3 and so by (6.9) " = B- z . By (6.4) fin = /3- ""2 . Therefore 13 M" = B "
r":
In (6.8) it is essential to assume that ii MI!, is defined. This does not follow from the other conditions. For instance let M = L = H, = (1) . Let tp be the central character of (1), let {1} be a conjugacy class of H2 and let C' = {x I x E C}. Then tir", (e')(if" , ((") = 0 but q'(ÊÊ') = C O.
7. Isometries 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 S of blocks of N let V(N, A : S) = 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(/V,N) = V(N). M(N, A : X) = V(/V, A : X) n m(N). Let Let M(N, A : S) --V(N, A : S)n M(N) where. S 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 = Las x., with complex as let a =I,„. 1-a,xs . If Y is the set of all irreducible characters in a union S of blocks of G let a S = Y . It is evident that if a is a generalized character of G then so is a Y. IT is a set of primes and is the complementary set of primes.
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If p E 7 then S„(7r,N) is the union of all p-blocks B of N such that if D is a defect group of B then 0 „. (NN (D)) is in the kernel of the germ of B. If z E G then z, denotes the 7r-part of z. The 7r-section of G containing z is the set of all elements in G whose ii- -part is conjugate to z r . (N , A : X) = {a a E V (N, A : X), a (yx ) = a (y) for any element y E A and any element x E 0, , (C H (y))}. M„ (N, A : X) = V, (N,A :X)n M(N).
Similarly the subscript Tr applied to any of the sets defined above means that it is intersected with V„ (N, A : X). In this section we will be concerned with the following two hypotheses. HYPOTHESIS 7.1. (i) N is a subgroup of G. A is a subset of N which is a union of 7r-sections of N such that any two 7r-elements in A which are conjugate in G are conjugate in N. (ii) If y is a 7r-element in A then CG (y) = C H (y )0, ,(C G (y)).
HYPOTHESIS 7.2. (i) (7.1) is statisfied. (ii) 0 „.(C G (y)) is a S„-group of CG (y) for all 7r-elements y in A.
We first prove some elementary results. LEMMA 7.3. Let H and the result has been proved for m < i n. Thus 13- 2 +,= 1. Since 0 (D B extends (D,„ B,,) it follows from (6.8) that (D„,,,,
extends ( ( )„„ii„,). Therefore B = = K. This proves the statement in the previous paragraph. Since B"tY ) -= B(y) it follows from (6.8) applied to CN (y) and CG (y) that ifi. (Y ) = ij- (y ). Thus b(y)G =
THEOREM 7.9. Let p E Ir. 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 7r-part is conjugate to an element of A. For any p-block B o of N such that V(N, A : B o) (0) and B o C S (7r-,N) let b. ° be defined as in (7.8) (ii). Let B be a p-block of G with B C S, (7r, G) and let /301, B02,... be all the p-blocks of N such that V(N,A (0), B o, CS,(7r,N) and ijg= B. Then the following hold. (i) If a E V, (N, A :U,B0,) then a' E V, (G,{A},: B). (ii) V„ (N, A : U,B 01 c (iii) If a E V„ (N, A :U,130,) then a' -= (a G . (iv) If B C S. (7r, G) then M, (N,A :U,13 0,)" C M„ (G,{A},?:B). )
PROOF. If y is a p-element in G and 0 is a function defined on CG (y) let By be the function on the set of all p'-elements x in CG (y ) defined by PY
(X
) = g (YX )-
+ O. It suffices to show that By = 0 for all p-elements y in G. Let y be a p-element in G. Let S, be the union of all p-blocks Ê of CG (y) with .6 G = B and let S2 be the union of the remaining p-blocks of CG (y). For t = 1,2 let V1(C G (y ): S, ) be the space of all complex linear combinations of irreducible Brauer characters in S. By (IV.6.2) (a")).13 E V' (CG (y): S 1 ) and Oy E V' (C G (y): S2). it suffices to show that a; E V' (CG (y): S1) because then (i) Let a = (a")B
6, = a ; — (a ')1y3 E V'(CG (y): S 1 )
n v'(c G (y): S 2) = (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 -Ea, where a, is a linear combination of irreducible characters in B o,. If y is not conjugate to an element of a defect group of Bo, then (a, ), by (IV.2.4). If x is a p'-element in CG (y) then x = x, x2 where x, C CN (y) and x2 E 0(c. (y)). By (7.6) a ;(x ) = ay (x,). Thus a; = ay is a linear combination of irreducible
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Brauer characters of CG (y )/0„ , (CG (y)) CN (y )/0„ , (CN (y )). Hence (4.3) and (7.8)(iii) imply that a; E V'(C G (y): S i ) 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 and a G vanish on the p-section of G which contains y. Hence by (IV.6.3) (a ) 8 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' and (a G ) 8 agree on the p-section of G which contains y. Let C = CG (y). By the Mackey decomposition (11.2.9) a G (yx) = /, {at,. n }C (yx) for every p' -element x in C, where NzC ranges over all the (N,C) double cosets of G. If yx E cA z then y E c A and so by (7.1)0 there exists z, E N with z, z E C. Thus NzC = NC. Hence {a = {(a Nnc )C l y . Therefore by (IV.6.1) {(a G )i3 } y = ({(a N ,,c) c } s ) y where S is the union of all blocks B, of C with B = B. By (7.4) (a Nnc) y is a linear combination of Brauer characters of N n C all of which have o,,(N n C) in their kernel. Since C/0(C) N n C10(N n C) there exists a linear combination g of Brauer characters of C all of which have 0,(C) in their kernel such that (a N ,-,-) y and p agree as functions on the set of p'-elements in C/0,(C). Since y E Z(C) it follows that {(a Nnc)l y = {(aNn c)y } C . Thus if 0 is the character afforded by an indecomposable projective module of C which has 0,(C) in its kernel then (fa NnO c ly7 1% ((a Nnc),
(1) NnavnC
(0 Nnc, Nnc)vnc — (0, (P)C
where the prime indicates that the summation in the inner product ranges over p'-elements. Thus by (IV.3.3) {(a } y — p is a linear combination of irreducible Brauer characters of C none of which have 0„ , (C) in their kernel. Hence by (6.5) ({(a Nncf} s ) y = ps • By the second main theorem on blocks (IV.6.1) (a N ,-, c)y is a linear combination of Brauer characters in blocks B(y) of N n C with B(y)" C U,1301 . Thus g is the corresponding linear combination of Brauer characters in blocks B(y) of C. Since 13- = B it follows from (7.8)(iii) that 13 s = 6. It remains to show that 0 = (a . Let x be a p'-element in C. Then x = x, x2 with x, a p'-element in N n C and x 2 E 0„, (C). Therefore since B is in S,, (7r, G) it follows from (7.6) that )y
,6(x) = ay (x 1 )= a(yx,)=- a' (yx) = (cr")y (x).
Thus = (a ")y as required. (iv) Clear by (iii).
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223
LEMMA 7.10. Let p E 7T. 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 V(N,A :Bo )/ (0) and B0 CS,(7r,N). Let ijo be defined as in (7.8)(ii). Then f3 (?CS, (7r, 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 B0 . Thus D is a defect group of /Jo . By (IV.2.4) it may be assumed that y E D. Since 0, , (C G (y)) is a S„—group of CG (y) it follows that 0(D CG (D)) is a S„,-group of DC G (D). There exists a block pair (D,B 2) in G with B-^ (D)c ' (E)) = ijo Since b o covers B, it follows that 0, , (D CG (D)) is in the kernel of B2. By (6.7) (ii) /j (? = BC S, (7r, G). 0 •
THEOREM 7.11. Let p E 7T. 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 7r-part is conjugate to an element of A. Let B i , B„ be all the p-blocks of G in Sp (7r, G). For each i let B 1 Bi2, . . . be all the p -blocks of N in Sp (7r, N) such that V(N,A :13 11 )/ (0) and = 131 where is defined as in (7.8)(ii). For each i let Si = U) B ii . Then the following hold. (i) V(N, A :B 11)= V, (N, A : B 11) for all i,j. (ii) V(G, {A } : /31 )= V, (G,{A} :13,) for all i. V, (N, A : (iii) V, (N, A) = V, (N, A : U7- 1 Si) = (iv) V, (G, {A })= V, (G, {A}, : U Bi) = (1)% I V3.3 (G,{A} : (y) V, (N, A : Si ) = V, (G, {A}: BO for i = 1, n. (vi) If {A}, n N = A then M, (N,A :Si ) = M,. (G, {A BO for i = 1 n. (vii) Suppose that for every p-element y in A and every integer h with (h,1 G 1) = 1, y E A. The for i = 1, ... n ,
,
•
I
rank M, (/V, A :S1 ) = rank M, (G,{A} :
= dim V, (G, {A }:B1 ) .
PROOF. (i) and (ii). Since 0, , (C G (z„))C 0, , (C G (4)) for all zE{A} these results follow from (6.5). (iii) The second equality follows from (IV.6.3) and (i). Clearly V,(N,A : Û S i ) C V,(N, A). Suppose that a E V, (N,A). For y a p-element in A, let B be a block of N with (a ')j O. If B i / B2 then by the second main theorem (IV.6.1) no irreducible Brauer character of C ry (y) is a component of both (a B, ), and (a °) ,. By (7.4) (a n ) y is a linear combination of irreducible Brauer
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characters of CN (y ) with 0„,(CN (y)) in their kernels. Hence B C Sp (7r, N) by (6.7) and (7.2). Thus PG cs,(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 VT (N, A). Then m = dim VT, (G, {A}). Since V, (N, A : S ‘ y C V, (G,{A},?: B,) by (7.9)(i), both (iv) and (y ) follow from (iii). (vi) By (7.9)(iv) M,(N, A :S) C M„(G,{A},:B,).
Let a E M, (G, {A}, : B,). Let p = a,. Since {A } fl N = A it follows that /3 E V, (N, A) and a = p-. By (iii) 13 =f3 with 13, E V, (N, A : S,). Thus by (iv) and (v) p, 0 for i/ j. Hence p E V, (N, A : S ‘ ). Since p E M(N) the result follows. (vii) Since A is a union of p -sections it follows that if z E {A} and (h,I G I) = 1 then z h E {A }. Thus by (IV.6.9) rank M, (N, A : S, ) = dim V, (N, A : S,) and rank M, (G, {A},? : B,) = dim V, (G, {A},? : B,). The result follows from (v). COROLLARY 7.12. Suppose that (7.2) holds and then a is a generalized character of G. PROOF. Clear by (7.9)(iii) and (7.11)(iii).
IT
= {p}. If a E M„(N,A)
E
Suppose that (7.2) is satisfied. Let X be a set of characters of N/0, , (N) with M(N, A : X) C M, (N, A). The set X is coherent if T can be extended to a linear isometry from M(N: X) into M(G). If 4- is an irreducible character in X and T is coherent then 11C I12 = 1. Thus -±- Ç is an irreducible character of G. The next result yields some information concerning the values assumed by Ç 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 (7.13) below was omitted in the statement. THEOREM 7.13. Assume that (7.2) is satisfied and 0, , (N) is a S„-group of N. Let X be a set of irreducible characters of N/0„ ,(N). Suppose that the following assumptions are satisfied. (i) If p E IT then the set of p-singular elements in A is a union of p-sections of N. (ii) X is coherent. (iii) V, (N, A :X) = V, (N, N — (N): X). Then the following conclusions hold. (1) If 4- E X then (C,a')G = (ÇN,a), for all a E V, (N,A :X).
7]
225
ISOMÉTRIES
(ii) Let X = { s } and let 6 =16, (1)4's . For each t there exists y, E M(N) such that (y„ 6,)= 0 for all s and an integer a, such that
a, ( 1)
()N
Furthermore there exists y E V(N) and a rational number a such that
C(z)=
(z)+
(1){4(z)+ y(z)}
for all z E A.
PROOF. (i) Let in be the set of all primes p E 7r such that if a E V „(N, A : X) then a vanishes on all p-singular elements in N. Let 7r2 = — in. For each p E 7r2 let S (p) be the union of Sp (1 r, G) and all p-blocks of G of defect O. We will first show that if /- E X and p E 7r2 then EC E S(p) for E = 1 or — 1.
Choose p E 7r2 . Let 4" EX and let E = 1 or —1 such that EC is a character of G. Let k be the set of all p-singular elements in A. By assumption (i) k is a union of p-sections and (7.2) is satisfied if A is replaced by A. Furthermore Vs, (N, Ap) C V„ (N, A) and T defined on (N, AO is the restriction to Vs, (N, A) of T defined on V, (N, A). Let X = { s } with = . For each s let as = (1)6 — . By assumption (iii) {a s } is a basis of V,(N,A : X). Since p E 7r2 there exists t such that a, does not vanish on A. Let a, = 13 + 02 where p, vanishes on N — Ap and 02 vanishes on A. Thus pi 0 and pl -= wy(P) by (7.9)(iii) and (7.11)(iii) applied to (3, E Vs,(N,Ap). Therefore by (IV.3.4) aT -= 0 + il where 0 E V(G:S (p)) and n is a complex linear combination of characters afforded by projective R[G] modules not in S (p). Furthermore 0/ 0 and a; = 4; (1)6T — (1)‘; . Thus either EC E S (p) or (1)C vanishes on all p-singular elements of G. In the latter case C vanishes on all p -singular elements of G and so EC is in a p-block of defect O. Hence in any case EC` E S(p). Let 4- E X. Define tir as follows: ç&(z) = =0
(z)
if z is p -singular for some p E 71-2 and the 7r-section of z in G meets A,
otherwise.
Since EC E U ,,,,S(p) it follows from (6.5) that C is constant on p-singular 7r-sections which meet A for all p E 77-2 . Thus t/f E V,. (G). Hence if a E V,(N,A :X) (7.7) implies that (C, a T)G -= (qr, a )G -= (ON, a )N = (CN , a)N
226
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[8
(ii) Let ()N = 6 + Ea,1 6 + y, where (6, y, ) = 0 for all j and t. Then (i) implies that for all i, j, t 6 (1)au —
(1)aq = ((6`)N, 6 (1)6 — 6(1)6) N — (6, 6 (1)6 — 6(1)6)N
= (41, WC — 6(1)C)a — {6 (1) 8. Therefore 6 (1)a, = =
at;
6(1) 8p} = O.
(1)au for all i, j and t. Hence in particular
6 (1) an for all 6 (1)
j and t.
The first equation in the statement of (ii) follows by setting a, = au. If z E A then 4", (1)6 (z)— 6 (1)C(s) = (1)6 (z)— 6 (1)6 (z) for all s and t. Thus the first equation in the statement of (ii) implies that (1) {
a a ` 6(z)+ y,(z)1 — (1) { 6 (1 ) 6 (1)6(z)+ (z)} s
for all z E A. Thus the second equation in the statement of (ii) follows by setting Y=
1
and
a—
al
6(1)
2.
8. Tr-heights
In this section we will use the notation of Chapter IV, section 8. Let v(1 G i) = a. Let e be a centrally primitive idempotent in R[G]. Let B = B(e) be the block corresponding to e. Let D = D (B) be the defect group of B and let d = d (B) denote the defect. Thus v (I G : D (B )1) = a — d (B). In this section we will primarily be concerned with ChR (G,B) and ZR (G)e. Observe that Chic (G,B) is the dual space of ZK (G)e. See Broué [1976b], [1978b], [1979], [1980] for the results in this section and further related results. LEMMA 8.1. If a E Z R (G)e and O E Ch R (G, B) v(0, (a))
then
P(I G D(B)i).
PROOF. By (111.6.6) e = Tr(c) for some c E R[G]. Since a E Z, (G) it follows that ae = (ac). Let {x,} be a cross section of D in G. Since O is a class function on G it follows that
8]
227
Tr-HEIGHTS
0 (a) = 0 (ae) = 0 (Ix: ' acx;) =IG:D 10 (ac).
This implies the result. 111 For a E ZR (G)e let s (a ) be the largest integer such that 77- ' (° ) I 0(a) in R for all 0 E ChR (G,B). Define the 7r-height of a, h„ (a) by (7r h- (" )) -=- (71-'1G : D
where the parentheses denote a fractional ideal of K. For 0 E Ch R (G,B) let t(0) be the largest integer such that R for all a E Z R (G)e. Define the 7r-height of 0, h,(0) by (7r h- ())= (77- '
0(a) in
G:D1 -1 ).
By (8.1) h„ (a) and h,. (0) are nonnegative integers for a E ZR (G)e and 0 E ChR (G,B). Suppose that x. E B. By (IV.7.1) 1 , (x. ( 1 )) = 1'(X. (e )). For a E ZR (G) X.(a)lx„(1)= co u (a)E R. Therefore (X. ( 1 )) = (X. (e)) (
»IG
(X. ( 1 )).
Thus if ( 71- ') = (P) it follows that h,. ()/m is the height of xu . In particular x„ has height 0 if and only if it has 7r-height O. This argument also shows that h„(e)— O. Since e is a primitive idempotent in ZR (G) it follows that ZR (G)e is a local ring. Thus ZR (G)e J (Z R (G)e) R as K was chosen large enough. Let A be the central character of ZR (G )e. By (IV.4.2) A = rou for any x„ E B. LEMMA 8.2. Let a E ZR (G)e. The following are equivalent. (i) For every 0 E ChR (G,B) with h„ (0)= 0, v(0(a))= v(IG :D I). (ii) There exists 6 E Ch R (G, B) with v(0(a))= v(IG (iii) h„(a)=O. (iv) a J (ZR (G)e).
PROOF. (i) (ii) and (ii) (iii) are immediate. (iii) (iv). Suppose that a E J(ZR (G)e). Then x„(a )Ix u (1) = ro„ (a) = 0 for all E B. Hence 0(a) = 0 (mod 77-p"' ° I)) for all e E Ch R (G, B) and so ii,(a) O. (iv) (i). Let 0 E ChR (G, B) with v (0 (a)) > p', where y = v(I G: D I). Let 0 = Ecuxu. Then
0(a) z cu x,_,(g)_
(x41. ») wr, (a).
228
CHAPTER
[8
V
By assumption (0(a)Ipu ) = O. Since a 0 J (ZR (G)e) it follows that cor (a) = (a) # 0 for all u. Therefore
E c (x“ (1) "
)
=0
(mod 7r)
and so
E cu x“ ( i)
0(1)
Thus h, (0)
O.
0
(mod 7rpu).
0
LEMMA 8.3. Let 0 E ChR (G,B). The following are equivalent. (i) For every a E ZR (G)e with h„ (a)= 0, v(0(a))= v( G: (ii) There exists a E ZR (G)e with v(0(a))= v(1G (iii) h, (0)= O. PROOF. (i) (ii) and (ii) (iii) are immediate. (iii) (i). This follows from the fact that (8.2) (iii) implies (8.2)(i). E LEMMA 8.4. Let 0 E ChR (G,B). Then h,(0)= 0 if and only if v(0(1))= v(IG:D1).
PROOF. If v(0(1)) = v(1 G : D 1) then clearly h, (0) = O. Suppose that h„, (0)= O. Since h,(e)= 0 and v(0(e))= v(0(1)), the result follows from (8.3). 0 LEMMA 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 Z R (G :S)e = (0). (ii) If y is not conjugate to an element of Z(D) then ZR(G:S)e C J(ZR (G)e). (iii) If y is conjugate to ZR (G : S)e J(ZR (G)e).
an
element of
Z(D)
then
PROOF (i) By (IV.2.4) 0(x)= 0 for all x E S and all 0 E Ch R (G,B). Since Chic (G, B) is the dual space of Z R (G)e this implies the result. (ii) This follows from (1.5). (iii) It may be assumed that y E Z(D). By (III.6.10) there exists a p'-element x such that D is a Sr -group of CG (x ) and À 0 if C is the conjugate class containing x and À is the central character of R[G]
(»
8]
229
7r-HEIGHTS
corresponding to B. Let x = x be a character of height 0 in B. Let Co be the conjugate class of G containing xy. Since D is a Sa-group of CG (xy) and x (xy ) x (x) X 0 (mod 1r) it follows that v (x (xy )) = v(x (x))= 0 and SO
Therefore A (0 ) e) = (Ç)= co„ (Co) 00 e E Z R (G : S)e. LI
O. Thus C'o e
J (ZR
(G)e) but
LEMMA 8.6. Define a on G by a(x)= 1 if x is a p-element and a (x)-- 0 otherwise. Then a 8 E Ch R (G,B) and h,(a B )= 0 . PROOF. Let {q,} be the set of all primes distinct from p which divide I GI. Thus q» E R for all i. Hence by (IV.1.3) a E Ch R (G) and so aB E Ch R (G,B). Let 0 E ChR (G,B) with h, (0)= O. Let a = Eo(x - ')x. Let S = G reg and let Ss be defined as in Chapter IV, section 8. Then S s (a)e E ZR (G)e and by (IV.8.5) a B (8,(a)e)-- a(3,(a)e)= a(3,(ae))= a(3,(a))=0(1).
Thus h„(a8 )-= 0 by (8.3).
D
8.7 (Broué [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 =E„ Es a,,x. Then v(a) -- V (I CG (y):D 1). Furthermore v(ay )= v(1C G (y): D I) if and only if a J(ZR (G)e).
THEOREM
PROOF Let a be defined as in (8.6). By (8.1) v(a (a)) v (I G : D I). By (8.2) equality holds if and only if a = ae J (ZR (G)e). Since a B (a)-= a(ae)= a(a)-=a I G :CG equality holds if and only if v(ay ) = v(I CG
): D
I)*
D
COROLLARY 8.8 (Brauer [1976aj). Let B be a block with defect group D. Then
v (E (0 2) = v (E (,o, (1) 4), (1)) = v G :DI IG where x„, cp, ranges over all irreducible, irreducible Brauer characters respectively in B.
230
CHAPTER
V
[9
PROOF. Let e be the centrally primitive idempotent of R[G] corresponding to B. By (IV.7.1) and (IV.7.2) e E ZR (G : Greg) and e = ax with 1Gja, =Ex. (1 = E,p,(1)0,(1). Since eJ(ZR (G)e) the result follows from (8.7). E )
9. Subsections
The results in this section are mostly due to Brauer [1968], [1971a] though the presentation here is based to some extent on Broué [197813]. The notation is the same as in the previous section and in Chapter IV, section 8. Let y be a p-element in G and let B be a block of G. If x. E B and d 1 j 0 then by the second main theorem on blocks (IV.6.1), cp J belongs to a block E of CG (y) with EG = B. For fixed y and j, the column of higher decomposition numbers d, is said to belong to the p -section S which contains y and is said to be associated with the block B. Let E be a block of CG (y) with f3 G = B. The set of columns d ) , as cp; ranges over the irreducible Brauer characters in if? is a subsection as to B or simply a subsection. It clearly depends on y and B. Denote it by S(y,b). A subsection S(y, IJ) is a major subsection if i4 and B = riG have the same defect. LEMMA 9.1. If S(y,IJ) is a major subsection then y E Z(.6) where D is a defect group of E. Furthermore
D
is a defect group of B = fi G
PROOF. Clear by (6.1). 0 Let y be a p-element in G and let if? be a block of CG (y). Define the linear map m'Y " ) :Ch K (G)—> Ch K (G : S) by rn ( “"(0)= bY ({dY (0)}),
for 8 E ChK (G),
where b Y and d are defined in Chapter IV, section 8. Define (y.
B)
Zic (G)—> Zk (G :S)
by tt ( " B) (a)= 0" (é8" (a))
for a
E
ZK (
G ),
where g Y and 5 Y are defined in Chapter IV, section 8 and ë is the centrally primitive idempotent of R [C G (y)] corresponding to
9]
231
SUBSECTIONS
By (IV.8.6) and (IV.8.7) m (Y.B).
9 9 tL (" ) for 0 E
(G).
The definitions of 13 Y and 5 Y imply that if a E ZK (CG (y): CG SY 0 OY (a) = a. Thus re' 6) and i' idempotent. Define x tY,B) —
then
1
oi (uY.
m (Y, f ( xu ) ,
) , eg)
X u (1) ) 6'
•
Thus x („Y'' ) = xu opt, (Y ' B) and co („Y.B) = cou ott (Y.B) . This implies in particular that d
1 (x)
for x E CG (y)„,
wj EB
and if a EZR (G) then f.t4Y. B) (a) E R. Furthermore if x. and x,, are in the same block then co (Y ,3) (a)
co,Y' n) (a) (mod 7r)
for a
E ZR ( G ).
THEOREM 9.2 (Brauer [1968], (4A), (4C)). Let D be a defect group of the block B of G. Let y E Z(D). (i) There exists a major subsection S(y,b) associated to B. (ii) Let S(y,b) be a major subsection associated to If?' =B. If x. E B then v(X (24B) (Y))= vdCG (Y): DO+ h(X.), where h(x.) is the height of Xu.
PROOF. (i) Let e be the centrally primitive idempotent in R[G] corresponding to B. Let MI be the set of all blocks of CG (y) with /f3 = B and let é, be the centrally primitive idempotent in R[C G (y)] corresponding to /i,. Let S be the p -section containing y. By (8.5) there exists a E ZR (G :S)e with a E J(ZR (G :S)e). By (8.7) v(ay ) = v(1C G (y):DI), where a =ax. By (IV.8.7) 8Y (a)= SY (ea) = EASY (a). The coefficient of 1 in 5 Y (a) is ay . Let ii, denote the coefficient of 1 in J,8Y (a). Then ay = Ed,. Thus there exists j with v(ay )= v(IC G (y): D I).
Let b, be a defect group of Thus (8.7) applied to the group
By (6.1) it may be assumed that b, c D. CG (y) with y replaced by 1, implies that
v(ICG(y):DI).--s.v(ICG(y):15,1).--s.v(a,),s.v(ICG(y):DI).
23?
CHAPTER V
[9
Hence equality must hold and so D = A as required. (ii) Let a be the sum of all elements in G which are conjugate to y. Thus a E ZR (G) and V (a) = 1. Hence the coefficient of y in p (''') (a)= 0Y (J) is the coefficient c of 1 in J . Since J J(ZR (C G (y))J), it follows from (8.7) applied to CG (y) with y replaced by 1 that v(c) = V (I CG (y):D 1). Thus ii,") (a) J (ZR (G)e) by (8.7). Hence if x. E B then B)
(a))
o,
(a)
0 (mod 70
which is equivalent to the assertion to be proved.
D
Let y be a p-element in G, let P be a block of CG (y) and let x., x„ be irreducible characters in /3- G . Define (Y . 8 )
m uu
1
-
I I—
(Y. B)
•
(y,b) ,
\*
xEco ( y )„g
where * denotes complex conjugation. In case y = 1 and P is a Sa -group of m (2,; B) = a u,„ where a is defined in Chapter IV, section 4. Thus G then Pi these numbers are a generalization to arbitrary sections of the numbers a 0 . Observe that if a is a field automorphism then (m (.Y,; B) )° = rn (2;» in for suitable y' and B' which do not depend on u or v. LEMMA 9.3. Fix u and v. Then E0,,R ) m („Y,; 13) = „„. PROOF. Clear by definition.
0
THEOREM 9.4. Let y be a p-element in G. Let A be a block of CG (y) and let xu , xu, x,„ E PG. Let d, d be the defect of B, B = .1j G respectively. (i) p d m;') is an algebraic integer and (m (Y ' B) (X.), X.)G =rnI,Y, ; B) =
E
Y“d0 *
where (yriy' is the Cartan matrix of A and * denotes complex conjugation.
(ii)E.m(oy,; B )(m(y;4 B ))*
m
y..v B
(iii) Let 1(.1i) denote the number of irreducible Brauer characters in È.
Then
m (iv) Suppose that x,„ has height O. Let hu , k, denote the height of x„, respectively. If hu then
9]
233
SUBSECTIONS
p d Xu (1)X,, Pd In(u):,B) =
(1)
(y
(mod ir
).
h d and equality can hold only if (y, b) is a Furthermore y (m major subsection and ho = O. —
PROOF. (i) The first equality follows from the definition. Thus (in (") (X“),
= I CG1(Y )1
E ,
p501EB
E60„„ dY (pY (x)
WO* cp,(x - ').
cc
By (IV.6.2) this yields the second equality. Since n has defect cl, (IV.4.16) implies that p ° (y) has integral entries. Thus po d InV3) is an algebraic integer. (ii) By (i) = (in (Y.B) (X.), m (Y.13)(Xw
=E
(Y•B)(xu),x.)G on(Y.B)(x.),xu)t
(iii) By definition m (2',;' ) = (m)*. Thus if M = M" ) = (m (uY,) it follows from (ii) that M2 = M. Hence the trace of M is the rank of M. By definition Eu m(uY,;') is the trace of M. By (i) the rank of M equals the rank of the Cartan matrix of if? which is 1(b). (iv) By definition m (y
= X , (1) 2, IGI
(0“
(C)x. (Yx) *
where C ranges over the conjugate classes in the p -section which contains y and yx E C. As xu and x,, are in the same block this implies that
161 re, X. ( 1 ) —
=
1G1 m o, ij)
(mod 7,-)
(1) "
since 13 ° rn (L in is an algebraic integer by (i). As x, has height 0 this implies that (y. 13) pd m = p d Xu ( 1
) l
elB)
(mod Treu ).
Similarly d (y,
Pm
=P
d Xv ( 1 )
(1)
( B)
Zrn
(mod re-).
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CHAPTER V
[9
The required congruence now follows. Since v (p a m) 0 , (1)X , ( 1) ( y. 6))
v(pd
d
_ J.
Thus the congruence implies that 1)(m,Y,;' ))% h„— d and if equality holds then h„ + h, d — d — d. The result follows as d d. E 9.5 (Brauer [19681(5H)). Let B be a block of G with defect group D. Let y E Z(D). Let S(y,B- ) be a major subsection associated to B. Suppose that x., X G B and x has height O. Let x. have height hu. Then
THEOREM
v(m)
hu - v(I D I).
PROOF. For O E Chic (G) let
=
E O(x)x. GH EG
(0, 02). By (9.4) Thus Ô E Z, (G) and 04 0= (in'Y
' B)
(X.),X.)G = m" ) (X.)(i-)= X. (i-d Y' B
= Xr ) ( X..)
-))
= x ( 1 )(4 B) (X.., )
(WY•13) ( E x`v (x
=
(
1 )x))
_ (E IGI uj"
Xu ( 1 )
Since x,„ has height 0 it follows from (9.2) that v(XV3) (Y -I ))= v(X; B) (Y))= P(ICG(Y):D1)•
Thus by (8.7)
Ex ■■:. B) (X - 1 )X 0
(ZR
(G)). Consequently
=0. Therefore v( 111 ‘,13) )= v(X.( 1 )) — v(I GI) = h. -
1).
9.6. Let B be a block with defect group D. Let B. be a block of DC G (D) with É' = B and let T(f3) be the inertia group of n in N G (D). Let Y CZ(D) be a complete set of representations of conjugacy classes of G
THEOREM
9]
235
SUBSECTIONS
which meet
Z(D).
(T(f3),
n CG (y)) double coset
a complete set of representatives in NG (D). Then each subsection S. S(y,(f3") c. ( Y )) with y E Y and x E My is a major subsection. Conversely every major subsection associated to B is conjugate to exactly one subsection S y.x. NG
(D)
For y E Y let My denote
PROOF. If y E Z(D), then D CG(D)C CG (y). Thus if y E Y and x follows from (111.9.2) that
fo (y))G (fix
G
E
my it
B.
Since D 0 and CG (y) has deficiency class r for every element y of order p in G then G has deficiency class r. Let y be a p -element, let Bo be a nonprincipal block of CG (y) and let d be the defect of B0 . By the third main theorem on blocks (5.4), • = 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 PROOF.
CG (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)(i) there exists a nonprincipal block of CG (y) of defect d and so CG (y) is not of deficiency class r. D 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.4.18) implies the next result.
11]
GROUPS WITH A GIVEN DEFICIENCY CLASS
247
THEOREM 11.6. There exist only finitely many groups G of deficiency class 0 whose S r -group has a given order.
of In case p = 2, Brauer [1974c] has proved the following generalization
(11.6). for the prime 2. THEOREM 11.7. Let G be a group of deficiency class r 0 abelian subgroup of Suppose that a S2- group P of G contains an elementary (a) depending only on a f bound order 2' . Let I P I = 2°. Then there exists a f (a). such that I G
The proof of (11.7) is not very difficult but depends on special properties of involutions. It is not known whether an analogous result is true for odd primes. If y is a p -element in G it is possible to apply (11.4) to CG (y). Thus (11.5) can be used to give an inductive characterization of groups of positive deficiency class. Such a characterization does not seem to be possible for groups of deficiency class 0 since the second statement of (11.5) is false in this case. For instance let q be a prime with q = 1 (mod p) and let G be a Frobenius group of order pq. Then G does not have deficiency class 0 for p but I CG (y)I = p for every element y of order p in G and so CG (y) has deficiency class 0 for every element y of order p in G. This construction shows the existence of infinitely many groups of deficiency class 1 with a Se -group of order p. It follows from (11.6) that an analogous situation
cannot occur for deficiency class O. It has been shown that a simple group of Lie type in characteristic p has deficiency class 1, Dagger [1971], J. E. Humphreys [1971]. This also shows the existence of infinitely many groups with deficiency class 1. Among these there are only finitely many with a given Se -group. A related result of a slightly different sort can be found in Tsushima [1977].
CHAPTER VI
1. Blocks and extensions of R The notation introduced at the beginning of Chapter III will be used throughout this chapter. The following assumptions and notation will also be used. K is a finite extension of Q,,, the field of p-adic numbers. R is the ring of integers in K. By (III.9.10) and (1.19.4) there exists a finite unramified extension k of K such that if T-Z is the ring of integers in k then the following conditions are satisfied. (I) is a splitting field of 14- [H] for every subgroup H of G. k is a splitting field for every p'-subgroup H of G. (II) Let B be a block of R [G] with defect group' D. Let (a) be the Galois group of k over K. Then there exists a block 13' of ./fZ[G] such that if V is an irreducible .g [G] module in B then r -1
v R
=0
V°1 with V E P.
If furthermore fj°) = b.' is the block of 1 [G] which contains %'" is a defect group of kJ ) by (111.9.1 0).
D
LEMMA 1.1. Let B be a block of R[G] with defect group D. Then every R[G] module in B is R[D]-projective and there exists an irreducible k[G] module in B with vertex D. PROOF.
Clear by (111.4.14). 0
LEMMA 1.2. Let B be a block of R[G]. Suppose that Ê contains a unique 248
2]
249
RADICALS AND NORMAL SUBGROUPS
irreducible ïi[G] module up to isomorphism. Then B contains a unique irreducible R[G] module up to isomorphism.
PROOF. Clear by (1.19.4). 0 LEMMA 1.3. Let H < G and let U be a projective indecomposable R[G] module. Then U, where each U, is a projective indecomposable [H] module and for each i there exists x E G with U, = U. PROOF. Immediate by (1.13.6) and Clifford's theorem (111.2.12).
E
Suppose that H < G. The block B of R[G] covers the block b of R[H] if E covers 6(i) for some j. LEMMA 1.4. Suppose that H < 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 DCG (D) C H. Then the following hold. (i) B = b' is the unique block of R[G] which covers b. (ii) D is a defect group of b.
PROOF. (i) Choose the notation so that Ê covers 6, where 6 is defined as in (II). by (V.3.9) fi is regular and so by (V.3.7) 1i-3 = 6G is the unique block which covers 6. Then r3 (' ) is the unique block which covers 6('). 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 6 and so by (II) D is a defect group of b. 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 Kniirr [1976]. Some related results have been proved by Ward [1968], Huppert and Willems [1975], Willems [1975]. LEMMA 2.1. Let V be an k [H]-projective R[G] module and let M= V I VJ(R[G]). Then M is R[H]-projective if and only if VJ(R[G])= VJ(k[H]).
250
[2
CHAPTER VI
PROOF. Let J = J(1[G])and let Jo = J(k[H]). As JoR[G]= R[G]J, vfo is an R[G] module. Suppose that VJ = VJo. Let A = 17z[G], B = 1[I-1] and S = Jo. Then (1.4.11) implies that M is 1[1-1]-projective. Suppose conversely that M is 17z [Il]-projective. Then (V .// VJ o)H —> (V I VJ --->
--->
is a split exact sequence as (V/VJo)H is completely reducible. Thus --> VJ/ VJo --> V/ V.I0 --> M -->
is a split exact sequence as M is 1 [H]-projective. Hence V IV.T o = M e viivio and so VJ I VJo = (Vi VJ0)J = M J + (VJI VJ0)J = (VJI VJo)J.
Thus VJ/VJo = (0) by Nakayama's lemma (1.9.8) and so VJ = VJo .
E
2.2. Let U be a projective indecomposable k[G] module and let M= U IUJ(R[G]). Then H contains a vertex of M if and only if UJ(R[G])= UJ(R[H]). COROLLARY
PROOF. Clear by (2.1).
E
THEOREM 2.3. Let e be a centrally primitive idempotent in k[G] and let B be the block corresponding to e. The following are equivalent. (i) H contains a defect group of B.
(ii) VJ(k[G])" = VJ(R[H])" for all R[G] modules V in B and all integers n O. (iii) J(R[G])e = J(k[H]) R[G]e. (iv) UJ(R[G])= UJ(R[H]) for all projective indecomposable R[G] modules U in B.
PROOF. (i) (ii). By (1.1) every module in B is R [H]-projective. Thus (ii) follows from (2.1) by induction on n. (iii). This is clear as eR[G]= R[G]e is a module in B. (ii) (iii) (iv). Let fz[G]e = U„ where each Lf, is indecomposable. Each projective indecomposable fz[G] module in B is isomorphic to some By the hypothesis UJ(R[G])= k[G]eJ(k[G])= k[G]eJ(R[H])
=
u,J(k [H
])
.
2]
751
RADICALS AND NORMAL SUBGROUPS
As U,J(R[H])C U,J(R[G]) 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 U IUJ(R[G]) for some projective indecomposable module U, the result follows from (2.2). 0
COROLLARY 2.4. Let V be an irreducible R[H] module. Assume that every component of V 0 lies in a block with a defect group which is contained in H. Then V 0 is completely reducible. By assumption there exist centrally primitive idempotents {e,} such that a defect group of each e, lies in H and such that V Ge = 1/0, where e =Ie,. By (2.3)
PROOF.
V GJ(k[G])= VGJ(k[G])e = (V 0 fi[G]) J(k[G])e R[H]
= V 0 J(k[G])e RIF11 = V 0 R[H] =
J(R[H])R[G]e = VJ(R[H]) Ø R[G]e RII-11
(0).
Hence V0 is completely reducible.
0
COROLLARY 2.5. Suppose that p IG : HI. Let V be an irreducible R[H] module. Then V 0 is completely reducible. PROOF.
Since H contains a defect group of every block the result follows
from (2.4). The next result is quite old. See e.g. Green and Stonehewer [1969], Villamayor [1959].
COROLLARY 2.6. J (R[G])= J(R[H])R[G] if and only if p IG :HI. [1-1]If J(k[G])= J(k[H])R[G] then iTZ [G]1.1(R[G]) is projective by (2.1). Thus every irreducible fz [ G ] module is fz [H ]- -projective and so p G : HI by (1.1). If p I G : HI then every R[G] module is R[H] projective. Thus (2.1) implies the result. 0 PROOF.
-
If W is an indecomposable k[G] module let w ( W) = w,, ( W) denote
252
[2
CHAPTER VI
the integer so that WH is a direct sum of w ( W) nonzero indecomposable k [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 projective k [G] module then
w ( U) = w (U1 LJJ(R [G ])) = w (UI U. (R [H ]) )
by Clifford's theorem (111.2.12) and (1.13.7). THEOREM 2.7. Let B be a block of R[G] which covers the block b of R[H]. Assume that H contains a defect group of B. Then the following statements are equivalent. (i) Every indecomposable projective rt[G] module in B is serial. (ii) Every indecomposable projective I7Z[H] module in b is serial. If furthermore (i) or (ii) is satisfied then all irreducible modules in B have the same width.
PROOF. Suppose that (i) is satisfied. Let U be a projective indecomposable k [G] module in B of maximum width. Let UH = P„ where each P, is a projective indecomposable iTt[H] module. Then w = w(U)= w (U/ UJ(k [II])). Let n be a positive integer such that P,J(k [H])" (0). By (1.3) P,J(k [1/])" (0) for i = 1, . , w. Let V = UJ(k[H])" IUJ(172[H])"1 Then
v,, =
P,J(R[H])" I P,J(R[H])"
and so w ( V) w. By (2.3) V is irreducible since U is serial. Let Uo be the projective indecomposable 1:-t[G module which corresponds to V. Then w (U0) = w( V) w (U). Hence the choice of U implies that w (U0 = w. The indecomposability of the Cartan matrix (1.16.7) now implies that every projective indecomposable k[G] module in B has width w and so every irreducible 1:-t[G module in B has width w. Thus (i) implies the last statement. Furthermore P,J(172[H])" IP,J(1-Z[H])" ±1 is irreducible as w( V) = w. Hence each P, is serial. Since w is the width of every projective indecomposable [G] module in B, the previous paragraph implies that if U is any projective indecomposable [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 projective indecomposable R[G] modules in B. Let n be a positive integer such that V = ]
)
]
3]
SERIAL MODULES AND NORMAL SUBGROUPS
253
UJ(.172[G])" IUJ(k[G])"±I X (0). Let UH = 1 p, where each p is a projective indecomposable R[H] module. By (V.2.3) and (1.3) each P, is serial. By (2.3) VH = (UJ(R[H])" IUJ(R[H]) ±1), = e P,J(R[H])" IP,J(R[H])"'I and so w ( V) 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.
It should be mentioned that in general not all irreducible fz[G] modules in V will have the same width. For instance suppose that p = 2, H = SL 2(8), G is the semi direct product of H with a cyclic group of order 3 and rz contains the field of 8 elements. Then the principal block of G contains an irreducible .17Z[G] module of degree 6 and width 3, and of course it also contains the one dimensional module of width 1. COROLLARY 2.8. Let B be a block of R[G] which covers the block b of R[H]. Assume that H contains a defect group of B. Then the following statements are equivalent. (i) Every indecomposable projective k[G] module in B is serial. (ii) Every indecomposable k[G] module in B is serial. (iii) Every indecomposable projective fz[H] module in b is serial. (iv) Every indecomposable k[I-1] module in B is serial.
PROOF. Clear by (1.16.14) and (2.7). 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 0 with e Ii T(6): HI and an irreducible 1 [G] module W such that W, Wa, , Wei' are pairwise nonisomorphic modules in B, W Wa and every irreducible k [G] module in B is isomorphic to some Wa'. Furthermore V s(e;=, Wa'). (ii) If 1 = ft then s = 1 and e = (6): H 1.
Suppose first that k = 1. Clearly T(b) = T(V). Choose x E T(b) so that T (b)= (x, H). Thus the image of x generates the cyclic group T(b)1H. Let s = land let e = IT (b): H . Let A be a representation of R[H] with underlying module V. Since T(b) = T( V), there exists a linear transformation M on V such that A (x -I zx)= M-I A(z)M for all z E H. Thus PROOF.
M—A (z)Me = A (x
= A (x -')A(z)A (x' )
for all z E H. Thus by Schur's Lemma Me = cA(xe ) for some c E k. Let F be a finite extension field of R with co E F such that c. Define A (x) = c 0.1 M. Thus A extends to a representation of T (b) over F. This representation is irreducible as its restriction to H is irreducible. Since R = 1 is a splitting field of T(b) there exists an R[T(b)] module X such that X, ----- V. Consequently Xa' is irreducible for all j and (Xa'),----- V. By (3.2)(i) Xa' is in b T('') for all j. Let [3 denote the trivial k [H] module. By (11.2.3) e
vT( 6,
(x, 0
—I
p T( 6 ) i =0
so
If Y is an irreducible R[T(b)] module in b T(6) then VI Y, by (1.4) and (VT , Y) 0 by (111.2.5). Thus Y Xa' for some j. If Xa' Xa' for some 0---i<j<e then by (111.2.5)
3]
SERIAL MODULES AND NORMAL SUBGROUPS
1 < (V T(b) ,
255
W = X G . Then Wa' =
Xa')= I(V, V) = 1. Let
(Xa' )G by
(V.2.5). The result is proved in case R = P. We will now consider the case of general R. The previous part of the proof implies the existence of an irreducible I [G] module lit in P. so that = (131_, Wa', where ê = T(b):H. Furthermore Wa = Wa' if and only if i j (mod i) and every irreducible module in P. is isomorphic to Wa' for some j. Let (a) be the Galois group of 1 over P. By (1.19.4) VØ1 ---(131:101 17 , where m is the smallest integer with Ver- ------ 1-7 . Let W be an irreducible R[G] module with W I W Ott-1. Let e, t be the smallest positive integer such that Wae W, (V G )'' respectively. Then ell T (6): . Also t f m as (VN ) ()N. Let U be the direct sum of all distinct P[G] modules of the form Wee'a). Then 034
Wal)0/4
Therefore
G
VG R
( 1
m
(v (on wai) t
R)
(V R
1=1
j1
By (2.4) and (3.2) VG is completely reducible. Thus (1.19.4) implies the result. 0
If M is an R [G] module then 1(M) will denote the number of composition factors of M. Let Soc(M) be the socle of M and let Soc" (M) be the inverse image in M of Soc(M/Soc —I (M)) for n 2. 7
COROLLARY 7
3.4. Suppose that V (2,Es (t, (x -1 ))1x = c.
if x is not a p-element, if x is a p -element. The result follows from
(4.3). E
If x is a p'-element of G let c E R[G] be the sum of all the elements in G whose p'-part is conjugate to x. The ideal of k[G] generated by all c is the Reynolds ideal of k[G]. See O'Reilly [1974] for related concepts. THEOREM 4.6. Ann(J(R [G])) is the Reynolds ideal of R[G]. PROOF. Since cx E R[G] and J(14 [G])= J(R[G])OA Ê it suffices to prove the result in case k = 1. By (IV.3.10) there exists for each p'-element x of G an 1-linear combination Ea,t, of t h t2, . . . such that at (x) = 1 and Ea,t,(z)= 0 for all p'-elements z not conjugate to x. If y is a p-element in CG (x) then t, (xy )= t, (x). Thus (4.4)(i) implies that c x E Ann(J(k [GM. For each i, EyeG t, (y = Et, (x -1 )cx . Thus by (4.4)(ii) Ann(J(R [G])) is generated by all c.„ El
LEMMA 4.7. Let H be a subgroup of G and let I be a nilpotent left ideal of i[H]. LetP = Ix and let f = n xEG R[G]Ix. Then the following hold. (i) Ï is a nilpotent ideal of R[G]. (ii) rG(1)= ExEG rH (I)1 [G] and f = la {E.EGrH(IY ft[G ]} , where r, 1 denote right and left annihilators respectively. (iii) If I is an ideal of R[H] then = rLEG ITZ[G].
If y E G then fy C nxEGR[G] --- = Thus f is an ideal of i;[G]. Therefore f - ±1 c (k[G]l)= for any integer m 1. Hence by induction f"'' C PF" for any integer m O. Thus f is nilpotent. (1.16.15) (ii) By I = 1H (r H (I)). This implies that k [G]l = 1G (rH (I)I4G]). Therefore if x E G then PROOF. (i)
17Z [GI = 1G (rH. (I )R[G])= 1G ({rH (I)1"17Z[G]).
4]
THE RADICAL OF
.g [G]
259
Hence
Î=
n
H
R [G])
(ri-i(nrk[G])
Thus by (1.16.15)
(1)= r lQ
E x
EG
r,..,(I)"R[G])1=
E
xEG
rfi(I)R[G].
(ii) By (1.16.16) the left and right annihilators of an ideal in F[G] F[H] coincide. Thus by (ii) f
=
rig (T)R [G]) = rG ( G li,(1)'12[G])
/G
E
=
rG( xEG
n
Or
R[G]h,(.0-)
rG(1-4[Glio-n=
xEG
n
PR [G]. 111
xEG
4.8. For a Se -group P of G let Ep =Ex Ep x. Then Ann(J(k[G]))c EpEA[G], where P ranges over all the Se -groups of G.
THEOREM
Let P be a Se -group of G. Then J(R[P]) = {Eci„xlEa„ =0} and so Ann(J(k[P]))= EpR [P]. Apply (4.7) with I = J(F[P]). By (4.7)(i) PROOF.
J(R[P])C J(R[G]). The result now follows from (4.7)(iii). 0 COROLLARY 4.9 (Lombardo—Radice [1939], [1947]). (i) Let S be the set of all p-elements in G. If >aaxE J(k[G]) then for all x E G
E
yES
axy
E
ay„ =O.
yES
(ii) Suppose that for every Sp-subgroup P of G, all x E G. Then ZEG a„x E J (17Z [G]).
E y EP
=Iy Epay„ = 0 for
Let c be the sum of all the p-elements in G. Then EyEs azy -is the coefficient of z in (ExEG ax x)c = c(E„,Gaxx). Thus (i) follows from (4.6). Let Ep be defined as in (4.8). Apply the previous paragraph to P. Then (ii) follows from (4.8) and (1.16.15). PROOF.
Ey ESay
In general neither of the conditions in (4.9) is both necessary and sufficient for an element of k[G] to be in J(k[G]). This can be seen as follows.
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CHAPTER VI
[4
Let G be a group with a Se -group of order p" such that p"±1 I (1., (1) for some i. For instance G = SL2(4) A, and p = 2. Then (I),(1)= 8 for some i. Let I be defined as in (4.5). Then J(/72[G]),i / but every element in / satisfies (4.9)(i). Let G = SL2(4) A, and let p = 2. Let P be a S2-group, let Ep = EEpx and let E = V=,(Ep), where Px , , , P", are all the S,-groups of G. Thus E is the sum in k[G] of all elements y in G with y 2 = 1. It is easily seen that Ep is in the annihilator of each V, and so E E J(R[G]). If z is an involution with z P then (zy)2 = 1 for y E P if and only if y = 1. Thus if E = >ax then Ey , aZ = az / O. Hence the condition in (4.9)(ii) is not necessary. However it will follow from (X.3.1) that (4.9)(i) is necessary and sufficient in case G is p -solvable. Section 6 below contains some conditions on G for (4.9)(ii) to be necessary and sufficient. The next result contains some information concerning dimAJ(R [G]).
THEOREM 4.10 (Wallace [1958], [1961]). Let p" be the order of a Sp-group of G. The following hold. (i) P" 1 dim, (J(k[G ]))
G (1— 1/i 1 (1)).
(ii) p" —1= dimA (J(R [G])) if and only if G is a p-group or G is a Frobenius group with a Sp -group as a Frobenius complement. (iii) Suppose that (I), is rational valued. Then dimA(J(R[G]))= IGI(1-1/0,(1)) if and only if G has a normal Se -group.
PROOF. It may be assumed that R = A. Since U,J(R[G])C J(R[G]) and dim, = 1 + dim, U,J(R[G]) it follows that p"
0,(1)-1-dimR(J(k [G])).
(4.11)
If G is a p -group then clearly equality holds in (4.11). Suppose that G is a Frobenius group with Frobenius complement P and Frobenius kernel H, where P is a Se -group of G. Then any irreducible character of G which does not have H in its kernel is of defect 0 and cp, is the only irreducible Brauer character of G which has H in its kernel. Therefore dim, (.1( 1 [G ]) )=IGI — E ,P,(1)2 = p' —1.
Suppose conversely that dim, (J(k[G]))= p"— 1. Then (4.11) implies that dim, (J(k[G])) =00)— 1. Therefore U,J(k[G]) =(0) for i 1 and so cp, is in a block of defect 0 for i 1. Thus in particular cp, is the only Brauer character in the principal block. Hence G =0(G) by (IV.4.12). This proves (ii).
4]
THE RADICAL OF ./2
By (IV.4.15) 0, (1)
IGI= I
[G]
261
,(1)(p, (1) for all i. Thus (1)cp, (1)
,(1)
= cPOA G I — dimfi J (R[G])}.
(4.12)
This shows that the second inequality in (i) holds and completes the proof of (i). Suppose equality holds in (4.12). Then 15, (1) = ,(1)(p. (1) for all i. Therefore 0, = 0,cp, for all i. Since 0, is rational valued (IV.10.1) implies that 0 1 (y1 ) = p", for any p'-element y1 , where p°, is the order of a Se -group of CG (y,). For any p'-element y of G let c (y) be the number of elements whose p'-part is y. Then c(y)--- I 0,(y)1 for every p'-element y in G. Hence
'GI=
E
c(y) --- E1 0 1(Y)1E °I(Y)=1G1( 01,q) , )=IGI.
Therefore in particular c (1) = 0,(1)= p" is the order of a Se -group P of G. Hence P contains all the p -elements in G and so P p". By (3.5) /(1-7,)--.Ç. p' for i = 1,2. Thus there exists a principal indecomposable module P and exact sequences
0—> 17,—)P—)X1 —>0
for i = 1,2. Therefore /(X2)---- /(X,), /(X,)+ /(X2)< p' and T(X1)='—'
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[3
T(X 2). By (111.5.12) and (3.10) HomR 1 0](XI, X2) = ( 0). However (3.9) implies that X, X i /S n (Xi ) for some n contrary to the previous sentence. El
LEMMA 3.12. B contains exactly e irreducible R[G] modules up to isomorphism and B contains exactly elDI nonzero indecomposable R[G] modules up to isomorphism.
The map sending V to 1-7 is one to one from the isomorphism classes of nonprojective indecomposable modules in B to the isomorphism classes of nonprojective indecomposable modules in A. Thus by (3.8) it suffices to show that B contains exactly e irreducible pairwise nonisomorphic modules. By (3.8) and (3.11) B contains at most e pairwise nonisomorphic irreducible modules. For i = 0, , e 1 let V, be an R[G] module such that V, Wa There exists an irreducible k[G] module L, with HomR [Gi ( y, / (0). By (111.5.13) applied to G and 6 and (1.5)(iii) HomAo i (Wa',L,) # (0). Thus Wa ' S(L,). Hence by (3.11) {L O i e 1} is a set of pairwise nonisomorphic irreducible modules in B. D PROOF.
—
—
7
THEOREM 3.13. (i) Let F be a finite extension field of R. Let 0 k a and let V be a nonzero indecomposable ftW k ] module in Bk. Suppose that V, is the direct sum of m algebraically conjugate F[Nk ] modules. Then for any j with 0 • j a and for every indecomposable 11[N1 1 module U in B„ UF is the direct sum of m algebraically conjugate absolutely indecomposable F[N] modules. (ii) Suppose that for some k with 0 k a, Bk contains an absolutely indecomposable nonzero k[Nk I module. Then for all i, every indecomposable R[N] module in B, is absolutely indecomposable.
Clearly (i) implies (ii) so that only (i) needs to be proved. By (3.6) is the direct sum of algebraically conjugate absolutely irreducible modules. Thus it suffices to prove the result for V = W and k = a 1. Suppose first that for some j with 0 j a 1, B, contains a nonzero indecomposable I[N1 ] module U such that U, is the direct sum of in algebraically conjugate indecomposable F[N1 1 modules. By (3.6) S(U) has the same property. By (3.2) every irreducible k[N1 1 module in B, has the required property and so by (3.6) every indecomposable ft[N,] module in B, has the required property. Thus the result holds for j = k = a 1. Suppose that 0 j a — 1. Let y bean indecomposable R[G] module PROOF. WF
—
—
—
3]
287
SOME PRELIMINARY RESULTS
in B, with vertex D, and let U be the 1 [6] module which corresponds to V under the Green correspondence. By (III.5.7)(iii) II, is the direct sum of m algebraically conjugate F[Nk ] modules. Thus the result holds for O j a —1 by the previous paragraph. By (111.5.7) (iii) the result holds for every nonprojective indecomposable R[G] module in B as it holds for every indecomposable 1[6] module in E. It also holds for the indecomposable projective k[G] modules in B as it holds for the irreducible R[G] modules in B. 0 3.14. Suppose that Ip(W,W)= m. Let 0 k a and let V,, V, be indecomposable R[Nk ] modules in Bk. Then b(V,, V 2)=-• 0 (mod m).
LEMMA
7 By (1.19.4) W R = (BC:, Wir; where 'T is an element in the Galois group of 1 over R, W o is absolutely irreducible, WII Vi/r,' if and only if i j (mod m). By (3.6) and (3.13) Vs ØR1 =1(V) for s = 1,2 and some absolutely indecomposable modules (V,)11 . Thus PROOF.
(y,, = i (y ,§) iL y, 1") =
?nip'
P, (v2
3.15. Let M M2 be indecomposable k[0] modules in m = I, (W, W). (i) If 1 (M 1 ).-e then Ifi(Mi,g).-- m for j} = {1, 2 } . (ii) If IA (MI, MI) = m then e.
LEMMA
11)
)„) .
E.
Let
Both statements are immediate consequences of (3.9) and (3.14). 0 PROOF.
3.16. Let V be an irreducible R[G] module in B. Then either l(17)---e or l(17)-?-- p" — e.
THEOREM
PROOF.
Let m = I,( W, W). By (111.5.13), (1.5)(iii) and (3.13) dim, Ir(G, (1), Hom, ( V, V)) = dim, Ir(G, (1), Hom, ( V, V)) = I, (V, V) = m.
Suppose that / ( Thus by (3.10) ./k (1-7 V) = m. Hence / ( e by (3.15). Suppose that l( V ) > There exists a principal indecomposable k[G] module P and an exact sequence ,
0—> -V—>P-1 X—>0.
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[3
Thus /(X)= 1 (P)-1(177)A7f, e,v,EDA->V,IEDA7(1)A->0, where A,, A are projective and A,, A A are sums of indecomposable modules in blocks other than ii. If this sequence is multiplied by the central idempotent corresponding to the block i then (1.15.6) implies the existence of projective 1 [O] modules P, such that
0—> 1.-/ —>ge P, — V1 —› 0 is exact for i = 1, 2. Suppose that /WO + /(1-/)
0 —> 17/
r. Then
o
is exact for i = 1, 2. Thus S(M,) S(M2) and so (3.9). Hence S S ( 72 ). Thus by (3.11) V, ---- V2 contrary to assumption. Suppose that 1(1-/2) + 1(1-i) p°. Since Al, is not projective 1(4)/ p'. Then there exist principal indecomposable modules Pi , P2 in 13 such that
S (Pi ) is exact for i = 1,2. Since Pi is not in the kernel of g, for i = 1, 2. Since f, (1.7) npi/(o) it follows that S (1-/) S (Pi ) for i = 1, 2. Thus S(') ----- T(17,) for i = 1,2 and so TWO— , T(1-/2 ). Hence by (3.11) V, ----- V2 contrary to assumption. CI LEMMA 3.19. Let M be an indecomposable [ G I module with S(M)= Rad(M) such that S(M) is irreducible but T(M) is reducible. Then T(M) is the direct sum of 2 nonisomorp hic irreducible modules. Furthermore l(S(M))/ 1 or pa - 1. PROOF. By (3.17) T(M) is the direct sum of pairwise nonisomorphic irreducible modules. By (3.18) 1(T(M))= 2. For any irreducible constituent V of T(M) 1 p° — 1. Thus if /(S(M))= 1 or p° — 1, (3.18) implies that 1(T(M))= 1 contrary to assumption.
4. Proofs of (2.1)—(2.10)
The proofs of (2.1) and (2.2) will be given simultaneously by induction on GI.
290
CHAPTER
VII
[5
Suppose that G = Co. Then e = 1. By (V.4.6) 60 contains a unique irreducible Brauer character. Thus the hypothesis of section 3 is satisfied and the result follows from (3.12). Suppose that G = Ck for some k with 0 k a — 1. By (1.3) e =- 1. If D = Dk then G = Co and the result follows from the previous paragraph. Suppose that D,,/ D. Let G = GID, By (V.4.5) there is a unique block f3 ° of G ° which is contained in and DID, = D' is the defect group of b`). By (1.3) the index of inertia of b() is 1. Thus by induction b" contains a unique irreducible Brauer character. Since every irreducible Brauer character of G has DK in its kernel it follows that Ê contains a unique irreducible Brauer character. Since the inertial index of /j, is 1 for all i it follows by induction that has a unique irreducible Brauer character for As C, = C , for k i a —1 we see that the hypothesis of section 3 is satisfied. The result follows from (3.12). Suppose finally that G/ C. Hence by induction 1;k contains a unique irreducible Brauer character for 0 k a —1. Hence the hypothesis of section 3 is satisfied and the result follows from (3.12). El —
As a consequence of (2.1) we see that the hypothesis of section 3 is always satisfied. Thus all the statements of section 3 are valid. (2.3) follows from (3.17). (2.4) follows from (3.2) and (3.5). (2.5) follows from (3.11). (2.6) follows from (3.5) and (3.7). (2.7) follows from (3.16). (2.8) follows from (3.9). (2.9) follows from (3.13). By (1.5)(ii)
1 IG:DI
. dim r, V -----
1
IG:DI
dimA V
(mod P")
for any R[G] module in B. Thus (2.10) is a direct consequence of (2.7).
El
5. Proofs of (2.11)—(2.17) in case K = Throughout this section it is assumed that K = k. LEMMA
5.1. In case G = Co, statements (2.11)—(2.17) are true with e =1,
8 1 =1, 80 = —1.
5]
PROOFS OF
(2.11)-(2.17)
IN CASE
K=
PROOF.
This is a direct consequence of (V.4.7). 0
LEMMA
5.2. Statement
291
(2.11) is true.
PROOF. By (5.1), T(b0)1C0 acts as a permutation group with no fixed points on the set of all irreducible K[Co] modules in Bo which do not have D in their kernel. Thus by (5.1) and e = [T(bo): Cd, bJ`b,,) has (p° — 1)/e irreducible characters which do not have D in their kernel. Thus A I = (p° — 1)/e by (V.2.5)(i). Since e (p — 1), I A = 1 if and only if e =p — 1
and
ID =P. LI
LEMMA
5.3. B contains exactly e+IAI irreducible characters.
PROOF. Let y 1. The number of conjugates of yS in D is IN, : CH where y E D, — D, 1. The number of blocks of C, covered by B, is(1/e)IN C, J. Therefore the number of irreducible Brauer characters of CG (y = C, in blocks which are mapped to B by the Brauer correspondence is 1/e times the number of conjugates of y in D. By (2.1) there are exactly e irreducible Brauer characters in B. Hence by (IV.6.6)(ii) B contains exactly e I irreducible characters. 0 LEMMA
5.4. In case G =
the decomposition numbers in B are all 1.
PROOF. By (5.3) there are p° irreducible characters in B = EL, By (2.1) there is a unique irreducible Brauer character in B. The result follows from (V.4.6). El LEMMA 5.5. Let X be an R-free R[C—] module in b such that g is
indecomposable. Then XK has no composition factor with multiplicity greater than 1.
PROOF. By (2.4) ,? is serial. Hence )? /Rad() is irreducible. Thus there exists an indecomposable projective R[C_,1 module P and a commutative diagram
X —> X/Rad(X)—> O.
Since 7rX C Rad(X) it follows that g /Rad(.) X/Rad(X) and Rad(X) is the unique maximal submodule of X. Therefore g is an epimorphism and the result follows from (5.4). D
?92
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VII
[5
The next result was first proved by Thompson [1967b] in a critical special case. LEMMA 5.6. Let 7/ be a character of G such that 0, = +n 1 for some i where = 0 or Tr is a character. Let characters of in Then 1(71c._,, Os) —
0,)1
be all the irreducible
1
for all s,t. IfO and n X 0, then (11c._,, 0s) X(7c, 0,) for some s, t.
PROOF. By (1.17.12) there exists an R -free R[G] module Y such that YK affords n and Y is indecomposable. If Y is projective then n = 0 and the result is trivial since is projective. Suppose that Y is not projective. By (1.5)(i) YG = I.-7E9A,E9A2 where A, is projective and A, is a sum of modules in blocks other than /j. By (111.5.8) and (1.4)(i) -ÇT is indecomposable. If 77, and 77 2 are the characters afforded by (A 1 ) K and (A 2 ) K respectively then ( (772 Os ) = 0 for all s and ((77 1 ) c„_,, 09 ) is independent of s. Hence it may be assumed that G = Ô and Y = 177. It follows from (3.5) that Y is serial and 17-c_, = 631 U', where U is an indecomposable k[C„_ 1 1 module in b„_, and x runs over a cross section of T(bc,_,) in N 1 . Thus Yc = Y0 El) A, where Vo U and A is a sum of modules in blocks other than bd _,. Hence it may be assumed that G = Now (5.5) implies that ( n„ ,, Os ) 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 0, = 2x + n' for some i, where n' = 0 or n' is a character. If 09 is any irreducible character of then (2,vc„ 09) is even contrary to (5.6). El
LEMMA 5.8. Let X be an R-free R[G] module in B such that XK is irreducible and )? is indecomposable. Then one of the following holds. (i) fCic is irreducible. (ii) There exists a principal indecomposable R[G- ] module U and an exact sequence
with YK irreducible.
5]
PROOFS OF
PROOF. /K (XK, XK) = 1.
(2.11)-(2.17)
IN CASE
K=
293
Since X is not projective. this implies that
dim R InG,(1),Hom R (X, X)) = 1.
Thus by (1.5)(iii) dimk H°(G, (1), Hom R (X, X)) = 1. =
Since X = X by (111.5.8) and (1.4)(i), it is serial. Thus there exists a principal indecomposable R[6] module U and exact sequences
As g is R -free, Y is a pure / (U) = p°. Hence /(g)-s_ p° /2 and let Yo = Y if / (V) p°/2. H° (G, (1), Hom R (Y0 , Y0)) is a
submodule of U and so V — V._By (2.6) or /(Y) - s. p° /2. Let_Y0 = g if /(g) p°/2 Thus in any case / (V0 ) p°/2. By (111.5.12) nonzero cyclic R module. By (3.10)
= Tr)(HoniR (Y0, Y0)) = 1. 1".(FlomR (Y0, Y0)).
Thus by (111.5.15) - (1),HomR ( 170, Y 0)) R. H°( G,
H o mREGJ (Y0,
Hence rank R Hom RtG) CY - 0, = 1 and so , - 0, Y Yo is irreducible. 0
(tY OK,
(Y0)1,)
=
1. Therefore
LEMMA 5.9. Let 0 k
a — 1. Suppose that G = Ck . Let G ° = Ck /D.-1. a —1 by 13,C H, and H, 1 Dc, _ 1 = CG o(D, 1 a_ 1 ). Then C, -(1H„ H,/C, is a p- group and TH,(b;)= C, for z e N. The following statements hold. (i) b p. Thus by induction all the results are true for Ck-i and also for G ° = Let ° be the subset of A consisting of all those characters in A which have 13o_ 1 in their kernel. By (V.4.5) there exists a unique block B ° of G ° with B ° C B. By induction there exist irreducible characters xi, x,„ A e A ° in B and these are precisely all the irreducible characters in B which have D, in their kernel. Let 0 1 , OA, A E A be the irreducible characters of Ck_, in bk-I. For any element x in G let xp, x denote the p-part, p'-part of x respectively. Let e,) , , ek be the signs defined in (2.17) for the group G ° . Let Eic—i be the signs defined in (2.17) for the group Ck-i. By induction and (2.15) the values of u = 0, , e are determined in all smaller cases. Let H be defined as in (5.9). By induction there exists y' = ± 1 such that the signs for Lq, multiplied by y' yield the signs for (b T' )° . Thus if a is the nonexceptional character of Hk_i/D-] in (b = (b°k then the generalized decomposition number of x, at for x E D — Dk is y'S times the generalized decomposition number of a, where 8 = 8, for b. If x E D then (5.9)(iv) implies that the sign E which equals the generalized decomposition number of a in (b kH 'f,')° at must equal the generalized decomposition number of a in b:'±V at x. By (5.9)(ii) a = t3 lik ', for f3 E bk _, and so by (2.17) applied to Ck _,, the decomposition number of f3 at x is equal to E. By the observation following (2.17) we may assume that = O. Hence there exists y = ±1 such that E; = ye, for 0 j k — 1. If AEA U {1} then induction applied to Ck-i yields that if is an
5)
PROOFSOF (2.11)-(2.17) IN CASE K =
295
irreducible constitutent of the restriction to D of an irreducible constituent of )t c in bk _, then 0
if x„
O xp)0,-,(x)
if xp E Dk-1,
G,,_, D,
if ; ED ; —
0 i k —2.
As C G (x) C Ck _, for x ED—D k this implies that
ax„)(//,?_1(x) 4- z (xp)tif‘z(xp')
if
( G D,
if x„
E
Dk,
if x„
E
D, — D,,, C
j
k — 1.
If x,, E D, — D,,, for some j k then by induction and (5.9) (iv) E,804- (xp)Ok (xp'),
A'A (x ) =
where 5 0 is defined by induction on G ° . If k < a — 1 then since G ° = Cd)(Dk /D‘,_,), induction implies that 3 0 = 1 and E, = 1 for k i 0. We next compute 11,36,112. Let yS e D, — D„, and let A be the set of all elements in G whose p-part is y '. If i k then EA I )(M. (X )1 2 = Z A I XI( X )1 2 by (5.10). Suppose that i 2 and so Su = 8u . The result follows from (2.15). 0 LEMMA 6.3. Suppose that I A I/ 1. Then Q p (i.), where 1 ,-5. u 6, À E A.
for any automorphism u of
300
[6
CHAPTER VII
PROOF. Suppose that Since o- permutes the irreducible Brauer characters of each CI, it follows from (2.17) that all the higher decomposition numbers of À are -± 1. Then (2.17) implies that p = 2, ê =1 and = Thus if x E D — D i then as 80 + 6, = 0, i(x) = e08,1No : Col (Po(1) = — and so
(x)'.
0
_(x).
Define the Brauer graph of B as follows. There is one vertex for each u = 0, e and one edge for each i = 1, . . . , e. The vertex corresponding to u is on the edge corresponding to i if and only if We will later define the Brauer graphs of a block of F[G] with cyclic defect group, where F is any finite extension of Q,„ and need not satisfy condition (*) of section 2. See section 9 below. LEMMA 6.4. If K = k the Brauer graph of B is a tree, i.e. it contains no closed paths. PROOF. Given i there are exactly two values of u with d, 0 by (2.15). Thus each edge has exactly two vertices. By (1.17.9) the graph is connected. As there are e edges and e +1 vertices there are no closed paths. D LEMMA 6.5. Let T be a connected tree and let o-, p be automorphisms of T which fix no edge. (i) ci fixes at most one vertex in T. (ii) If crp = po- and o- (P)= P, p(Q)= Qfor vertices P and Q then P = Q. PROOF. (i) Suppose that u fixes vertices P and Q of T with P X Q. Then u fixes the unique path from P to Q and so fixes some edge contrary to assumption. (ii) Since crp (P) = per (P) = p(P)it follows from (i) that P = p (P). Thus P=Q by (i) applied to p. THEOREM 6.6. If IA' 1/1 then Q, ("„) =Q. (Or ) for 1 --5 u ê, 1 i ê. If 1=1 then the notation can be chosen so that Q, = Q, (çô, ) for 1 u e, 1 i e. PROOF. By (2.9) M = Q,, ( Y--)P—>k—*0 O --> Y0 -->
—> O.
Hence 1( Y) = 1. By (111.5.12), H 0 (6, (1), Homg ( Yo , Y)) / (0) and so Hom g ( Yo , Y) (0). Thus Y T( Y0). By (2.8) T(Y0) is determined by s(f,). Thus by (2.5), T( Y0) Y is determined by L,. By (2.8) SOO is determined by Y, and hence by L. Thus )? is determined up to isomorphism. Consequently )? is determined up to isomorphism. Thus in particular .k is serial. It remains to show that if )? ,Z„, then X By definition the multiplicity of SOO as a constituent of ,Z is 1 if u 0 and (1 D 1— 1)/e if u = 0. Hence (8.3) implies that
m
if u/ 0, }
IF (X , X) -,
m(IDI— 1) if u = 0,
= au,X “ 112 = rank, Hom REG) (X, X01 ).
This implies that I A (., fC)= rank R Hom R(GI (X, X u1 ). Thus there exists g E Hom REGI (X, X. 1 ) such that g induces an isomorphism g from X to k1. As X„; is serial, Rad X., is the unique maximal submodule of X.» Hence g
9]
SOME PROPERTIES OF THE BRAUER TREE
305
is an epimorphism. Therefore X., X/Z for some module Z. As both X., and X are R -free this implies that rank R Z = rank R X — rank R X„ = O. Consequently Z = (0) and so X X. 111
9.
Some properties of the Brauer tree
Let Ko be an arbitrary finite extension field of Qp. Let K be a field which satisfies condition (*) of section 2 such that Ko C K and K is a totally ramified extension of K 0, i.e. K, and K have the same residue class field. Let Bo be the block of K 0[G] which corresponds to B. Then the Brauer tree of Bo is defined to be the Brauer tree of B. Thus every block of K0[G] with a cyclic defect group has a Brauer tree for any field Ko with [Ko :Qp] finite. It is sometimes convenient to associate the irreducible K 0[G] module to a vertex of the Brauer tree rather than the character afforded by such a module. It is convenient on occasion to label the exceptional vertex if there is one. If x.,x 0 correspond to distinct vertices on the same edge then by (2.19) and (2.25) & + & = O. It follows from the results of section 2 that if G = then the Brauer tree is a star with the exceptional vertex, if any, at the center.
G =N
It will be shown in Chapter X that if G is p-solvable then the Brauer tree of a block B of G with cyclic defect group is a star. The converse of this statement is false. For instance the principal 13-block of Suz(8) has the following tree. The degrees are written by the vertices. 0
1
o
14
64
°14
o
35 ex
CHAPTER VII
306
[9
A more spectacular example is given by the principal 13-block of the automorphism group of Suz(8). 014 0
014
1
014 014 01
If p is any odd prime then the tree for the principal p -block of PSL,(p) looks as follows. There are l(p — 1) edges.
A-
p 1) 1
pl
- -
o
o
" •
4 (p +O l) ex
•
The sign is chosen so that the exceptional character has odd degree. It is not known whether every tree is the Brauer tree for a suitable block of some group. No tree has been shown not to occur though it seems likely that most trees will not occur. (Added in proof By using the classification of the finite simple graphs it can be shown that most trees do not occur as
Brauer trees.) Let e be an integer and let p be a prime with p 1 (mod e). Let G be the Frobenius group of order pe. Then the Brauer tree for the principal p-block of G is a star with e edges. Thus every star occurs as a Brauer tree. The only trees with 1 or 2 edges are stars. Thus they occur as Brauer trees. The principal 7-block of PSL 2(7) has the Brauer tree which is a line segment with 3 edges. Thus every tree with 3 edges is a Brauer tree. The next two examples show that every tree with 4 edges is a Brauer tree. The principal 5-block of S, has the following Brauer tree. 1
4
6
4
1
The principal 5-block of Suz(8) has the following tree. 014
9]
SOME PROPERTIES OF THE BRAUER TREE
307
The next two results were proved by Tuan [1944] in case a =1. See also Yang [1977].
THEOREM 9.1. Suppose that G has a cyclic S e -group. Then every 1i[G] module in the principal p -block is equivalent to an F[G] module where Fe is the field of p elements.
PROOF. As the principal p-block contains the principal character, the result is a direct consequence of (2.9). 0 THEOREM 9.2. The subgraph of the Brauer tree of B consisting of those vertices and edges which correspond to real valued characters and Brauer characters is either empty or is a straight line segment.
PROOF. Let an object be either a vertex or an edge in the tree. Suppose there is a real object in the tree. By (IV.4.9) the set of characters and Brauer characters in B is closed under complex conjugation. Thus complex conjugation defines an incidence preserving map of the tree which sends edges to edges and vertices to vertices. Two real objects in the tree are connected by a path. Thus they are also connected by the complex conjugate path. Since the tree has no closed path it follows that two real objects are connected by a path consisting of real objects. Therefore the set of real objects in the tree form a connected graph. By (2.24) this graph is a straight line segment. E The subgraph of the Brauer tree consisting of real edges and vertices is called the real stem of the tree. The next two results are due to Rothschild [1967] for the case that K = k. He used these methods to prove (2.10). The following notation is needed for these results. T is the Brauer tree corresponding to B. The edge corresponding to L, is denoted by E. The vertex corresponding to x„ is denoted by P. If 1 i e then T {E,} is the disjoint union of two trees. Let these be denoted by To(E,) and T(E) where 70(E, ) contains the vertex Po. n (E,) is the number of vertices in 7,(E,). Thus n (E1 ) is the number of vertices in T which are separated from Po by the removal of the edge E. d(P) is the number of edges on the unique path in T which joins P„
to Po. LEMMA 9.3. (i) For 0
u e, 8„ = (-1)`")80.
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308
(ii) Let 1 j Ç.e and let P = Pu be the vertex on E, which is in T,(E,). Then one of the following holds. (I) 8o(-1)' ) = 8. =1; n(E1 )=1(I:1 ). (II) So( — 1)d ' ) = Su = —1; n(E,)= p`z — l(f ) ). (i) This is an immediate consequence of (2.19) and (2.25). (ii) This is proved by induction on n(E,). If n(E,)= 1 then .g,„ L, where P = Pu. Thus /(f. 1 ) - - S u (mod p') and the result follows from (i). Suppose that n(E,)-1. Let E 1 ,. . . , E. be all the edges in T(E1 ) which have P as a vertex. Hence E 1, , E„ E, are all the edges in T which have P as a vertex. Furthermore n (E,). By (1.5)(i) n (E,) = 1 + PROOF.
S. -=- /(f,)+
2 1(L
1 )
(mod p").
By induction and (i) =
Su
2 n(E,) -
1
Su {fl(E ; )— 1}
(mod p°).
=1
Therefore /(f,,)— Su n(E,)- 0
(mod p°).
The result now follows, as n(E,)--.Ç• e and by (2.7) /(f,) - -s. e or p' —e
e. Choose j with dS 0 and let 9.4. Suppose that ID L1,..., Ls be all the irreducible k[G] modules (up to isomorphism) which are constituents of X 0i. Then one of the following holds. (i) 50= —1. 1(L,)= n(E ; ) , e for 1 - i s and E;=, l(f,,)= e. (ii) 6 = 1. p° — 1(I. 1 )= n(E1 ) - e for 1 i - s and Es,_, fp' — /(f.;)} = e.
LEMMA
E1, . . , E. are precisely the edges of T which have Po as a vertex. Let P1 Po be the other vertex on E for 1 i Thus d(P,)= l and so ( — 1) a(1',) = — 1 for 1 i s. Since E:=, n (E,) = e, the result follows from (9.3)(ii). PROOF.
10. Some consequences 10.1. Suppose that G has a cyclic S„-group (x). Let I(x)I= p". Let V be an indecomposable R[G] module in B and let d be the degree of the minimum polynomial of x acting on V. Then p" - °1(V) - -s. d. If furthermore D .1 G then p (V) = d.
LEMMA
10]
309
SOME CONSEQUENCES
By (1.5)(i) it may be assumed that G = 6- and V = By (3.4) it may be assumed that K = k. By (3.5) it may be assumed that G = Then NG ((x )) = C0 ((x )). Hence Burnside's transfer theorem implies that G = (x)H with H = 0(G). Replacing x by a conjugate it may also be assumed that D C (x). By (111.3.8) every indecomposable R [G] module U in B is of the form U = U,? for some R[DH] module Uo in a block with defect group D. Hence V = V,?, where Vo is an R[DH] module in a block with defect group D. By the Mackey decomposition (II.2.10) (V) > —{(Vø)0 }. Thus the minimum polynomial of y on vo is (Y— 0'0, where d = do. Therefore it may be assumed that V = Vo and x = y. Since H is a p'-group, V,, = M„ where each M, is irreducible. Choose M = M, with MZ (Rad V), Then Ed,=o1 M(y —1) is an 1 [ G ] module which is not in Rad V. B contains a unique irreducible 1 [ G ] module W up to isomorphism by (2.1). V is serial by (2.4). Thus V= M(y — 1)'. Since MI WF., it follows that dim ', M dimA W. Therefore PROOF.
/ ( V)dirru W = dimk V
d dirri M
d dim W.
Hence / ( V) d. Suppose that D O. Let z = r-t . Then (z) = D„ _ 1 and /7e
2 by assumption. This contradicts the previous paragraph.
E
The next result strengthens a Theorem of Lindsey [1974].
COROLLARY 10.4. Suppose that G has a cyclic S p -group (x) with 1(x )1= 19 " =p2 . Assume that (x)fl (x)z = (1) for z E G, z NG (KX». Let V be a faithful irreducible k [G] module. Then dirri i V p" — (p —1). PROOF. If V is projective the result is clear. If V is not projective then by (111.8.14), V is in a block with (x ) as a defect group. Thus the result follows from (10.3). E The next result is a generalization of (5.6) which generalized a result of Thompson [1967b]. In case D j = p this first appeared in Feit [1967b].
THEOREM 10.5. Assume that K = k. Let Y be an R -free R[G] module in B and let 17 be the character afforded by YK Suppose that f7 = MEl) • El) M,,, where each M is a nonzero indecomposable R[G] module. Let 9 ,, , Op . be all the irreducible characters of C- 1 in b-1. Then .
1( 71c.
0, )1
n
for all s, t.
PROOF. Without loss of generality it may be assumed that Y is indecomposable. If f is projective then eric 9,) is independent of s and the result is proved. Thus by (1.17.11) no M, is a projective 1 [G] module. By (1.5)(i)
312
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Y c = (j) A , (j) A 2 where A, is projective and A2 is a sum of modules in blocks other than b. If -rb is the character afforded by (A) R for i = 1,2 then ((-q 2)G„ 0s ) 0 for all s and ((7/ 1 ), 0,) is independent of s. By (111.5.8) and (1.4) 1'-7 — — 2t ED • 11-{„ and each /4, is indecomposable and nonzero. Hence it may be assumed that G = Ô and Y = 177. By (111.4.6) there exists an R[T(6„_,)] module Y. such that Y(? Y, (Y)T( b)= Yo e) A where A is a sum of modules in blocks which do not cover b„_, and Y. is a sum of n indecomposable modules. Thus it may be assumed that G = T(b„_,). By (1.3) every irreducible R[G] module remains irreducible when restricted to C„_,. Thus by (3.5)(ii) each (M,) G._, is indecomposable. Hence it may be assumed that G = It suffices to show that (.71, n for all s. Suppose this is false. Thus there exists 0 = 0, for some t with 0)= n, > n. Let M be a K[G] module which affords 0 and let Y, = y n n,M. Then Y, is a pure submodule of Y and so S(V,)C S(Y). Therefore /(S(Y,)) n since every indecomposable fz[C„_,] module in b„_, is serial. Thus Y, is a sum of at most n nonzero indecomposable modules. Thus it may be assumed that Y = Y. Hence YK affords the character n 1 0. Consequently ( n,
rank', Hom R[G ](Y, Y) =
IK (YK
) = n1.
Since HomR[G] (Y, Y) = Inv G Home ( Y, Y), it is a pure submodule of Hom R ( Y, Y). Hence = dirnk HomR IG] (Y, Y)
IK (YK,
) = n1.
Let W be the unique irreducible R[G] module in B. Then IA (M, Mj) = min{/(M), /(M)} as every R[G] module in B is serial. For 1 s p', O., is irreducible as a Brauer character. Hence /(Y)= n. Thus for 1 i n (M„ Y)=
J=1
(M„M; )%s-
/(M; )= /( Y) = n,. 1=1
Hence Y)=
nn,.
Thus n 1 n. This contradiction establishes the result.
D
The next result is due to Green [1974a] in case K = k. It generalizes an earlier result of Alperin and Janusz [1973]. THEOREM 10.6. Suppose that cl„,/ 0 for some u,i with 0 u e , 1 j There exists an infinite exact sequence
e.
10]
SOME CONSEQUENCES
p
,,
f2
p„_,, • •
313
(10.7)
such that each Ps is a principal indecomposable R[G] module in B and the following conditions are satisfied, where [V] denotes the isomorphism class of R[G] modules which contains V. (i) P — Ps , and fs (Ps )— fs+2e(Ps+2) for all s. dt, 01. (ii) {[fs (Ps )111 _ s -s.2e1 = (iii) {[P 2s_ Ili 1 s el = {[P2] I 1 s el = {[Ui l 11 j e). PROOF. There are exactly 2e modules X, up to isomorphism with du,/ 0 since by (2.19) there exist exactly 2 values of v for each j with du,/ O. Let 01 = a 1 X + a„,,x, and let X = U,I X,,,. Then T(g)----- L, is irreducible. Thus X is indecomposable. Since X i, affords a 1 x it follows from (2.25) that X — X,„, , for some j' with d,,/ O. Thus in particular if L, then 0, = au,x. + a1 X for some v. Therefore the following sequence is exact. U,
—
> Xu,
—
*0 .
(10.8)
Consequently the exact sequence (10.7) exists. Furthermore for all s, X,,, for some v, j with du, X O. As there are exactly 2e modules X, up to isomorphism with d,, X 0, it suffices to prove the result for any one of them. For convenience it may be assumed that /(k) = 1 by (2.25). Furthermore after a change of notation it may be assumed that fC,,, W as defined in (2.4). In view of (1.15.6) and (1.5)(ii) the exactness of (10.8) implies that is exact for some projective module 0 in B. . By (2.14) /(X„,)+ /()Z,,) < 2p". _Hence CI is a principal indecomposable module. Thus there exist principal indecomposable modules i's in E-1 and an exact sequence -->
-L> • • • —> .15
,
-
L.->
—> O.
Furthermore j (i) for all s. By (1.4) and (111.5.8) g,„ X all v, j with d 1 # O. Thus the following sequence is exact
where
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Choose the notation so that W, = Wa 1— for 0 i e 1. Let V(i, 1) denote the i [Ô] module in 13- such that S(V(i, l))= W, and V(i, /)) = 1. By (2.8) the following equations can be read off for all s. —
= f2s(P2..,)'---- V (s, p° — 1),
=
V(S, 1). (10.9) = , Thus {f s (Ps ) 1 s 2e) is a set of 2e pairwise nonisomorphic modules. Hence also {f„ (Ps ) I 1 s 2e) and (Ps )i 1 s 2e} are each sets of 2e pairwise nonisomorphic modules. Thus every module Xu, with du,/ 0 is of the form f, (Ps ) for some s with 1 s 2e. This implies (i) and (ii). Furthermore for each j there exist exactly two values of s with 1 s such that S(fs (Ps ).----- L,. The proof of (iii) will now be completed once it is shown that for 1 j e there exists a unique value of s with 1 s e such that f2s- I (P2,--1)
L.
By (2.5) and (j 0.9) there is a unique value of s with 1 s e such that Hom (L,, f2,-1(P2,-1)) # (0). Thus by (3.10) applied to O and (111.5.13) applied to G and by (1.5)(iii) there is a unique value of s with HomA(1,,,f2s-1(P2s_1))# (0). 0 Suppose that d, # 0 for some u, i. Green [1974a] has defined a function a on 11, , el as follows. Let P, be defined as in (10.6). If P2, Li, then P2s1-1 '''' U). By (10.6)a is a permutation on {1, ..., el. It also follows from (10.6) that if the notation is chosen suitably then there exists an exact sequence ---> U, -->
U,_,—> • • --> U1 ) —> U1—> U (, ) — X, —> 0.
It is clear that l(f,(P,))+
p°
for all s.
The permutation a = o-,,, depends on the choice of u and i. However the definition of a., implies that if du,/ 0 then au, = a„, in case 8. = 8„. If 60 then cr,,,au, = (1, 2 . e). COROLLARY 10.10. Suppose that d.,# 0 for some u, j. Let a = a,„. Assume that d,1 # 0 for some v, j with S. = Su . Then L, o) . PROOF. Clearly Su = 60 if and only if Xu, =f7 ,-(P 2,) for some s. By definition S(f2,,(P2,,1)) S(/52,) and 1 f2, (P2,±)) T(P2, S(P2,1). Thus if P2, U, then S(f2,41(P2,,_,))----- L, and 7tf2 1(P2s Lro). -
10
315
SOME CONSEQUENCES
]
In Green's terminology the ordered sequence /Jo. ( I),
U2,
U0(2) ,
,
U„ U,() ,
describes a circular "walk" around the Brauer tree accomplished in 2e "steps". Every vertex of the Brauer tree is reached at least once and each edge is transversed exactly twice. As Green also points out, a determines the Brauer tree as an abstract tree as follows. Consider the oriented circular graph as shown. The Brauer tree can be derived from this by identifying each pair of edges which carry the same lable in such a way that the orientations cancel out. a- (1) 1
/a(e) ./e . •
This process can be carried out for any permutation but will not necessarily yield a tree. For instance the identity always yields a star. If e =3 and a is a transposition then one gets a straight line segment. However if e = 3 and a = (1, 3,2) then one gets a triangle which is not a tree. LEMMA 10.11. Let B = Ê be an iZ[G] block with a cyclic defect group
D.
ID i= P d
Let C be the Cartan matrix of B and let Q be the quadratic form corresponding to p d C-I . Let q be the minimum value assumed by Q on the set of all nonzero vectors with integral coordinates. Let
.
Then q = e.
PROOF. By (IV.3.11) and (2.12) fx„ 1 u el is a basic set for 13. Let D ° , C° denote the decomposition matrix, Cartan matrix respectively with respect to this basic set. Let Ct be the form corresponding to p d (C1'. Then Q0 is integrally equivalent to Q and q is the minimum value assumed by Q ° on the set of all nonzero vectors with integral coordinates. By (2.12)
D°
[10
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316
where t = (p° —1)/e. Let J denote the matrix all of whose entries are 1. Then = (D')'D° = I + tJ.
Thus p d (Co)-1 = pdI — tJ. Hence if a = (a Q (a, a)=
E ,-1
then (10.12)
+ tE (a, — a) )2 .
Let No be the number of a, = 0 and let N,= e — N o. The last term in (10.12) is at least N oN, and the other term is at least N. Hence if a # 0 then N, 0 and so N,+ NN 1 1‘1,+ No = e.
Thus e q. If a, = 1 for all i then Q (a, a) = e and so q = e. D THEOREM 10.13. Let B be a block of fZ[G] with an abelian defect group D of order p". Let r be the rank of D and let k(B) denote the number of irreducible characters in B. If r = 2 then k(B)< p d. If r = 3 then
PROOF. Let y be an element of maximum order pc in D. By (V.9.2) there exists a major subsection S(y,É) associated to B. By (IV.4.5) there exists a unique block É° of CG (y)/(y) contained in B. Furthermore D/(y) is the defect group of É`) . By (111.2.13) y is in the kernel of every irreducible Brauer character in É. Hence É and fi ° both have /(É) irreducible Brauer characters. If q(ñ) and q(ñ0) are defined as in (10.11) then q (f3)= q (b ()) by (IV.4.27). If r =2 then A° has a cyclic defect group and so q(A)= I(A) by (10.11). Thus k(B)p' 1 by (V.9.17)(i). If r = 3 then 1(170 k (n o) < pd--` by the previous paragraph as A' has an abelian defect group of rank 2. Hence by (V.9.17)(i) D p 2d- c. This implies the result as c The next result was first announced in Brauer and Feit [1959]. THEOREM 10.14. Let B be a block of 1'[G] of defect d. Let k = k(B) denote the number of irreducible characters in B. d = 0, 1, 2 and k p' for d 3.
Then k p d for
PROOF. Induction on d. If d = 0 the result follows from (IV.4.19). If d =1 it follows from (2.1) and if d = 2 it follows from (10.13). Suppose
11]
317
SOME EXAMPLES
that d 3. If B contains an irreducible character of positive height the result follows from (IV.4.18). Thus it may be assumed that every irreducible character in B has height O. Let S(y, fi) be a major subsection associated to B. By (IV.4.5) there exists a unique block of CG (y)/(y) contained in B. Furthermore DI(y) is the defect group of É. By (111.2.13) y is in the kernel of every irreducible Brauer character in A. Hence n and b° both have 1(14.0) irreducible Brauer characters. Thus by (V.9.17)(ii) k 2 p 2d 1 ( n0 ) p 2a k (fo , where k(Jr) is the number of irreducible characters in ife. Since ifi ° has defect at most d — 1, induction implies that k ( bo) p 2(d-1)-2 as d =2d —2 for d = 2. Thus k2
and so k
p 2d+2(d- 0-2 = p 4d-4 p2d-2.
LI
11. Some examples The following result is due to Dade [1966] and shows that every possible sequence of signs Ek can occur in (2.17).
LEMMA 11.1. For O k a —1 let y k
= ±
1. There exists a group G and
a block B of G with defect group D such that if (2.17) then Ek = y k for 0 k a —1.
Ek
is defined as in
PROOF. For 0 k a-1 let qk be a prime with qk = —1 (mod p°) and 2< q,< • • exactly one of W,, say W,„ is the trivial k[N] module. Let W1 = W 1 , for j = 1,2. It remains to show that S(W2) is not the trivial krisil module. Let cp be the Brauer character afforded by W2. No irreducible constituent of (Y"), has D in its kernel hence no irreducible constituent of (Z,), has D in its kernel. Thus 1 + cp(x)= O. As q(x)=- —1 it follows from (2.8) that S(W2) is not the trivial ft[N] module. 0 THEOREM 11.5. Suppose that p/2 and a >1. Let q be a prime with q +1= - p° (modp" 1 ). Let G =PSL2(q). Then a Se -group of G is cyclic
of order p° • G has an irreducible character 0 of degree q (the Steinberg character), which is afforded by a Q[G] module. Furthermore there exists an R-free R[G] module x such that XK affords 0 and .)?= M, M2 where M, is the trivial k[G] module and M2 is an irreducible k[G] module with dimk M2 = q —1.
PROOF. The existence of 0 and the structure of the S„-group of G are well known facts. Let D be a Sp-group of G. Then CG (D) = D x H for some p'-group H and !NQ (D): CG (D)1= 2. In particular NG (D)/H N where N is defined in (11.4). Furthermore D n D (1) for z N G (D). Thus the Green correspondence between G and NG(D) is defined for modules with nontrivial vertex in D. Let k be the R[N G (D)] module with kernel H such that k as an R[N] module is isomorphic to Z. Let X be the R[G] module which corresponds to k. Thus k W1 ED vv2. Hence by (111.5.8) k = M ED M2 where A21, W, for i =- 1,2. Let B be the principal p-block of G. Then the index of inertia of B is 2. Since 0(z)= —1 for z E CG (D) — {1} it follows that 0 is in B, this implies that the Brauer tree for B is 1
q q-1
Let cp,,cp 2 be the irreducible Brauer characters in B where cp, is the principal Brauer character. Thus M, affords cp,. The principal block of
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R[NG (D)] has a unique indecomposable module, namely W2 , which has no invariants and whose dimension over k is congruent to —2 (mod pa). It follows from (III.5.10) and (111.5.14) that a module which affords cp, corresponds to W2 . Thus M2 affords cp 2 . Hence in particular rankR X = q. Since fC has no invariants, neither does X. Thus the character afforded by XK is sum of exceptional characters and possibly B. As rank R X = 0(1) and the degree of every exceptional character in B is q —1 it follows that X, ‹ affords B. 0 The next result which is due to Benard and L. Scott, Benard [1976] will be used in section 13. In some sense it explains why it is necessary to assume that a >1 in (11.3).
LEMMA 11.6. Let R be the ring of integers in an unramified extension of Let X be an R-free R[D] module such that XK iS irreducible. Then
Qp.
.)-C is indecomposable. If furthermore Y is an R-free R[D] module with YK XK then Y----- X.
PROOF. If rank R X = 1 the result is obvious. Suppose that rank R X > 1. Consider the map f :X —> X defined by f(v)= v(1 — y). Then f is an R[D]-monomorphism and IX : f (X)I = det(1 — y). Since X K is irreducible, the characteristic values of y are all the primitive p"th roots of unity for some n. As rank R X> 1, n O. Hence det(1 — y) — = p, where ranges over all primitive p" th roots of 1. Therefore I X : f(X) I = I This implies in particular that .g is indecomposable. Furthermore f(X) is the unique maximal submodule of fC and so f(X) is the unique maximal submodule of X. Thus X is isomorphic to its unique maximal submodule. Suppose that YK X K. It may be assumed that YR = XK. If Y is replaced by a multiple it may be assumed that Y C X. Since PC Y is finite there exists a chain of submodules Y = X,,, C X„,_, C • • • C X 0 = X such that X, is maximal in X,_, for 1 i m. By the previous paragraph X, ----- X,_, for 1 i m and so Y X. LI .
:
12. The indecomposable k[G] modules in B
This section contains a description of all the nonzero indecomposable 1 [G] modules in B in terms of the Brauer tree of B and of its exceptional
12]
THE INDECOMPOSABLE R [G] MODULES IN B
323
vertex. The results in this section were all proved by Janusz [1969b] and Kupisch [1968] [1969] in case fz = fz. Given the results of section 2 for K in general, Janusz's arguments go through without any change. The structure of the principal indecomposable k [G] modules in B is described by (2.20). In this section I D e — 2e pairwise nonisomorphic nonzero indecomposable k [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 I 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 indecomposable 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 a group algebra. (See the remark on p. 306.) denotes the Brauer tree. Vertices of T will be In this section TT denoted by P, P, or Q. The exceptional vertex, if it exists, is denoted by Pe „ . E or E, will denote either an edge of T or O. If L, corresponds to the edge E, write L, E. If E = 0 let (0) E. Set t = (ID I— 1)/e. By (2.13) and (2.19) t is the multiplicity of any L, in ,?0, for do, 0, do, O. If du, 0 let X,,, be defined as before (2.20). If d / 0 then the statement dual to (2.20) implies that T(,)--- L, for some j with d„,/ O. Let P be a vertex of T incident to the edge E. Let L E and let x„ correspond to P. Define V(E, P) = X„, where T(,?„,)----- 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 T and let P, and P, be the vertices incident to E. If E, is incident to P, we will define modules V(E,, E, E,: n) for suitable integers n. We will allow the case that E, = 0 or E, = E for exactly one of i = 1 or 2 if P, = Pex . Several cases will be considered. We begin with some preliminary definitions. Let V, (E)= Rad( V(E, P,)) 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 ( U). Then the dual of (2.20) implies that Rad(Mo (E)) = V 1 (E) ED V,(E). Suppose that E,E,, E, are distinct with E 0 such that
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PI
P2
0
E,
E
E2
is contained in T. Define modules M, (n) for i =1,2 for suitable integers n as follows. If E. 0, M, (1) = V, (E). If E,/ 0 and P Pe. let M (1) be the submodule of V, (E) such that S (V, (E)I M,(1)) 0 this implies that n = m (x)(ps — : Qr]. PROOF.
—
13]
SCHUR INDICES OF IRREDUCIBLE CHARACI- ERS IN B
335
Let g be the character afforded by the irreducible Q, [ID] module of dimension ps — ps' over Qp. Then V, has a pure submodule X which affords g. By (11.6) the minimum polynomial of y on X, and hence on V, has degree at least ps — p". Since D 0. 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(1— 4-1 ) = (0). Let w = v(1 E V(1 - X . By (i) wx = w. By (ii) k = np +s for 0< s
p it follows that p — s < 2(s — 1)+1. Thus n —s — 3)m if s is even and n = (p —s — 2)m if s is odd. Hence in any case n i(p — s — 3)m.
1LN L ",,`, there exists
350
CHAPTER
VIII
[3
Let ./t4 be a direct summand of L, ®LIJ with 1 < /()< d. Since 1 < 414), P is not in the kernel of /a Let M be the k[G] module corresponding to 11-4. The minimality of d implies that MN J%-4. Hence MN has a nonzero projective summand. This implies that LN Ø L I:, has at least n nonzero projective summands. Consequently (p —2s)m — s —3)m. Therefore p 3s —3. As 3 A' p, p 3s — 2. E PROOF OF (3.3). In view of (3.1) cp(1)= 1. Thus m =1 and d = do . Suppose that e ---5,i(p— 1). By (3.1) d > e and so by (VII.2.7) d-?--- p—e>i(p— 1) contrary to assumption. Therefore e i(p —1) and d = p — s with s e. By (2.8) p-i— V(a',2i +1)0) ED V(a',p). LN ®L Since d 1(p + 3). E
For 1 i s —1 let M, be an indecomposable k[G] module with ----- V(a 2i +1). Let dimMi = 2i + 1+ m ip. We need two subsidiary results. LEMMA 3.7. Suppose that m, = 0 for some i with 1 ,5 i s —1. Then i = s —1 and 3s —1% p. PROOF. Since P is not in the kernel of /1;/„ the minimality of d implies that d --52i +1 in case m, = O. If i/ s —1 then d 2(s — 2) + 1 = 2p — 2d —3. Thus 3d 2p —3 h'p. This contradicts the fact that (u +1)1 h'(p — 1). Hence h
O.
LEMMA 4.4. Let n 1 be an integer. Suppose that v is a positive integer such that (vp —1)1(p — 1)(1 + np). Then there exists an integer h 0 such that u = (n — h )/v 0, n = F(p, u, h) and (u + 1)1 h(p —1). If u = 0 then v =
pn — n +1.
PROOF. Let vp —1 = m, m 2 with in , 1 (p —1) and m 2 1(1 + np). Then p = 1, vp —1= 0
(mod m,).
Hence v — 1 0 (mod m i ). Also vp — 1
0,
np +1= -0
(mod m2).
This implies that (n + v)p = 0 (mod m2) and so n + y = 0 (mod m 2). Thus vp — 1 = rn m 2 divides (v — 1)(n + ). 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 ') = 0 where y' = (h — n)/v and so u = — y' is an integer. Since f (1) < 0 and y % 1 it follows that y' < 1 and so y' 0 as y' is an integer. Therefore u O. As f(— u)= 0 it follows that (u + 1)(n — u) = h(up + 1) and so n = F(p, u, h). Furthermore h(p — 1) -= — h(up +1) -= 0
(mod (u + 1)).
If u = 0 then n = h and (4.5) implies that y = pn — n + 1. 111
4]
353
A CHARACTERIZATION OF SOME GROUPS
The proof of (4.1) will now be given. Suppose the result is false. Let G be a counter example of minimum order. 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/ 1, ..., are the exceptional characters in B. Xo, • • •, Xe and 8o, ..., 8, are defined as in (VII.2.12) for the principal p-block of G where x i is the principal character of G. If t = 1 let 4', = xo .
LEMMA 4.6. (i) If 8, = 1 for ,y, an irreducible character then x,(1)= 1 or 1 + np. (ii) If 8, = — 1 for x, an irreducible character then x,(1)= p p(pn — n + 1)-1. (iii) If r1 and 80 = 1 then ga,(1) = 1 + np for 1 i t. (iv) If t 1 and 80 = 1 then t(1)= p — 1 or p(pn — n + 1)-1.
—
1 or
—
PROOF. Let x
1 x(1)
be an irreducible character of G. Then 1 G :CG (Y)1X (Y)
is an algebraic integer. Since 1 G: CG (y)1 = e (1 + np)= (p — 1)(1+ np)/t this implies that nP)(P — 1 )X (Y IX ( 1 )
is an algebraic integer. Suppose that x = x, is a nonexceptional character in the principal p-block. By (VII.2.17) x,(y) = 3, and x, (1) = 8 + mp for some integer m. Thus x(1)1 (p — 1)(1 + np). Suppose that t 1 and x = Then by (VII.2.17) E:_, (y) = 3 o and , 4-,( 1) = so + mp for some integer m. Thus tx(1)1 (p —1)(1 + np). (i), (iii). Let x = x, with 8, = 1 or let x = E:_, where t 1 and 80 = 1. By (4.3) n = F(p, u, h) where x(1) = 1+ up. Since G is a counterexample either h = 0 or u = O. If u = 0 then x(1) = 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 = x, with 3, = — 1 or let x = where t 1 and 80 = — 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
354
[4
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n = F(p, u, 0 ) = u and u = n /y. Thus y = 1 and x(1) = p — 1. If u = 0 then x(1) = p(pn — n +1)— 1 by (4.4). E The next result is a refinement of (4.6). LEMMA 4.7. (i) If 5; =1 for x; irreducible, j 1 then x,(1)= 1+ rip. (ii) If 5, = —1 for x, irreducible then x ; (1)= p —1. (iii) If t 1 and 5,,= 1 then ki (1)=1+ np for 1 i t. (iv) If t 1 and 5,,= —1 then k,(1)= p —1. PROOF. (i) Since G = G', x,(1)>1 for j 1. The result follows from
(4.6)(i). (ii), (iii), (iv). There are at most e values of j with 5; = 1 since the Brauer tree is connected. By (4.6) xi 1 + np if 5, = 1 and x,(1) = 1. Let y o = 1 if 50 = 1 and y o = 0 if 5,, = —1. Let E = 1 — so. Then (VII.2.15)(iii) implies that Ei6(1)+
E >0
f
< e(1+ np)=ÇP t l ) (1+ np) 1 {p(pn —n + 1)— 1 } . Thus x, (1) p(pn — n + 1)— 1 and g", (1) follows from (4.6). n LEMMA 4.8. (i) t
p(pn — n + 1)— 1. The result
1, k,(1)= 1+ np, x,(1)= 1 and x; (1)= p —1 for
(ii) 1 + (1 + np)/t = (p —1)(e —1). (iii) G is simple and C G (y) = y). (
PROOF. Let G, be the subgroup of G generated by all elements of order p in G. Thus G, 3 and G/0(G) is either isomorphic to PSL2 (p) or p = 2 — 1 amd GI/Op (G1)---= SL2(p - 1). Hence G/0(G) has a normal subgroup A such that either A — PSL,(p) or p =2° — 1 and A SL2 (p — 1). Thus G/0(G) is isomorphic to a subgroup of the automorphism group of A. Since G = G' this implies that G/0„, (G) ---, A contrary to assumption. Hence G = G. This in particular implies that G/0(G) is simple.
4]
A
CHARACTERIZATION OF SOME GROUPS
355
Let G" = GIO„,(G). Every irreducible character in B has O„ (G) in its kernel and so may be identified with a character of G. Suppose that there exists an irreducible nonprincipal Brauer character cp in B with cp (1) (p — 1). As G" is simple, (3.1) implies that 6 ° — (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". By (4.7) there exists j with x, (1) = p — 1. Thus (x,), is irreducible and so (xj, ) is the character afforded by the regular representation of (x). Hence the linear transformation corresponding to x in the representation which affords x, has determinant — 1, contrary to the fact that G = G'. Thus t 1. If 5,, = — 1 then by (4.7) 4", (1) (p — 1) which is impossible. Thus 5,, = 1. By (4.7), (4"1 (1) = 1 + np. Suppose that 8„ = 1 for some u with 2 u e. For any j let P, denote the vertex on the Brauer tree corresponding to ) 6. The path from Po to P„ contains a vertex P, with 5, = — 1. Since P, is an end point of the tree, P, does not occur on this path. Thus x, has at least two nonprincipal Brauer constituents. By (4.7) ,vs (1) = p — 1. Hence there exists an irreducible nonprincipal Brauer character cp with ça (1) (p — 1) which is not the case. Consequently 5, = — 1 for 2 j e. The result follows from (4.7). (ii) This is a direct consequence of (i) and (VII.2.15)(iii). (iii) Since G" is simple it follows from (3.2) that a Se -group of G" is self centralizing. Hence I G"I= pP---1-) (1 + nop) where G" has 1+ nop Se -groups. Thus 1 + nop 1 + np and (1 + np)/t = 6(1) is an integer. By (ii) p — 1 and (1 + np)/t are relatively prime. It follows that (1 + np)/t (1+ nop)/t. Thus 1 + np 1 + flop and so n = no. The minimality of G now implies that G = G. E Let 0 1 , 02, ... be all the irreducible characters of G which are not in B. Then each is in a block of defect O. Let pa,. Let q be a prime with q I e. Let x be an element of order q in N
a
am=
LEMMA 4.9. (i) t = p — 2 — n. (ii) PROOF. (i) Since te
(ii) By (4.8)(i)
E
= n.
= p — 1 this follows from (4.8)(ii).
356
CHAPTER VIII
p
[4
(p — 1) (1 + np)= I GI = ‘; (02+i xi (1)2+E t 1=1 1=1 (np +1)2 =1+ (p 1) 2(e —1)+ E a
Direct computation using (i) yields the result. CI 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) (z = 1 for 1 i t. )
—
PROOF. (i) By (4.8)(ii) (p — 1) is relatively prime to (1 + np)/t. As GI = p(p — 1)(1 + np)/t, this implies that x, is in a block of defect 0 for q, where 2 i e. The result follows. (ii) This follows from (i) and (VII.2.15)(iii). fl LEMMA 4.11. 0.
ai for all i.
PROOF. Let F = + x, —E;,x,. By (VII.2.15)(iii) F(z) = 0 for z E N, z/y with 1 By (VII.2.17) F(y) = e — A (y') for all i, where A is a faithful irreducible character of N. Thus
E
F(z ) = ep —
ZEN
E
i=1
(y ) = ep.
Hence (FN , 1 N ) = 1. Since (4.10)(i) holds for a n arbitrary prime divisor q of e it follows that if 2 j e then p -1
E xi (z) = i x,(y') = 0. zEN Thus ((x) N , 1 N ) = O. Hence also ((
)N, 1N ) = 0 for 1 i t by the previous paragraph. Let p be the character afforded by the regular representation of G. Then (pN,1N)= 1+ np. Hence ( E (i)(6,) N, iN) = np and so
E a, ((0,)N, 1 N )=
e, E a, (0,)N, 1 N ) = n.
(4.12)
If 0 is a principal indecomposable character of N then 0 = q + E A where cp, is an irreducible character of NIP and A ranges over all the faithful irreducible characters of N. For each j )N is a sum of a, principal 1 indecomposable characters of N. Thus (z)— co(z) for z E N — P. In ,
(a
4]
A CHARACTERIZATION OF SOME GROUPS
357
particular this implies that (0, 1 N ) 1 and (0, 1N ) = 1 if and only if 45(z) = 1 for all z E N — P. This last condition holds if and only if = 00 , the principal indecomposable character which corresponds to the principal Brauer character. Hence ((a )N , 1 N a, for all i. Thus (4.9)(ii) and (4.12) imply that ((a )N, 1N ) = a, for all i. Therefore (0,) N = a,00 and so
61,(x)= a,. CI LEMMA 4.13. e = 2. p = 2 + 1 for some a >1. Every element of even order in G is conjugate to x. A S 2-group of G has order p —1. PROOF. Let {Tb} be the set of all irreducible characters of G. If z is a q -singular element then (4.10) yields that 0
=E
,(1)77,(z)= 1 —(1 + np)+ pE ai01 (z).
Thus E a,O, (z)= n. Suppose that z' is not conjugate to x. Then (4.10) and (4.11) imply that
0=E
(z)
= 1 + t +E
01(x)0,(z)
= 1 + t +E afi, (z)= 1+ t + n>0. 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 CG (X ) is a Sq -group of G. By (4.10) and (4.12)
IcGool E I 77, (x)1 2 = 1+t+Ea=1+t+n. Hence by (4.9)(i) I CG (X)I = p —l is even. Thus q = 2 and p = 2 + 1. If a = 1 then a S2-group of G has order 2 contrary to the simplicity of G. Since N I P is cyclic of order e it follows that e = 2. 0 LEMMA 4.14. Let S be a S 2-group of G. Then order (p — 1)(p — 2).
NG
(S) is a Frobenius group of
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, NG (S) is a Frobenius group with Frobenius kernel equal to S. By (4.13) S I = p — 1 and any two involutions in G are conjugate. Hence by a theorem of Burnside any two involutions in NG (S) are conjugate. 0
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358
t5
PROOF OF (4.1). By (4.13) t =(p — 1). Thus by (4.9)(i) n = (p — 3). Hence = p(p — 1)(p — 2). By (4.14) G has a permutation representation on p letters in which NG (S) is the subgroup leaving a letter fixed. Since NG (S) is a Frobenius group, any faithful permutation representation of NG (S) on G has a triply transitive permutation representation on p letters. As G = p (p — 1)(p — 2) the assumptions of (4.2) are satisfied and so by (4.2) G ---- SL2 (p —1). E 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 CG (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 Se -group of G is assumed to be cyclic but not necessarily of prime order.
THEOREM 5.1. Suppose that G = G' and a Se -group P of G has order p. Let 1+ np be the number of Se -groups in G. Let H = 0(G). Then one of the following holds. (i) p > 3, G/H ----.PSL2(P)(ii) p = 2' + 1 > 3, G/H ----.5L - 2 (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 F(p, u, h) =
hup + u2 + u + h u +1
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). 111 COROLLARY 5.2. Suppose that p is a prime factor of 1G1 and p> G. Assume further that the following conditions are satisfied.
(i) G = G'. (ii) A Se -group of G is not normal in G.
51
SOME CONSEQUENCES OF
(4.1)
359
(iii) G has no normal subgroup of order 2. Then p >3 and either G ----- PSL,(p) or p —1=2" and G ----- SL 2 (p —1).
PROOF. Let P be a Sp -group of G. If then IG:Pl
3 and G ----- PSL2 (p) simple. It is easily seen and well known that every p'-element in G is conjugate to its inverse. Thus every irreducible Brauer character of G is real valued. Thus by (VII.10.6) the Brauer tree of G is a straight line segment. Hence x has at most two irreducible Brauer constituents. Since (po is a Brauer constituent of x the result follows. Li
THEOREM 6.2. Let G be a transitive permutation group on p letters where p is a prime. Let 00 be defined as in (6.1). Then (Po is the character afforded by any permutation representation of G on p letters.
PROOF. Let H be a subgroup of G of index p. Then p HI. The character afforded by the permutation representation of G on the cosets of H is (1,)'. The result follows from the Nakayama relations (111.2.6). D A group G may have two inequivalent transitive permutation representations on p letters. This is the case precisely when G has two nonconjugate subgroups of index p. However (6.1) implies that any two such permutation representations afford the same character and hence are equivalent as complex linear representations. The study of this phenomenon leads to the study of certain symmetric block designs. See Ito [1967b].
THEOREM 6.3 (Burnside). Let G be a transitive permutation group on p letters for some prime p. Assume that a Sr -group of G is not normal in G. Then G is doubly transitive.
PROOF. By (6.1) and (6.2) the permutation representation of G on p letters affords the character 10 + x for some irreducible character x. Thus G is doubly transitive. E THEOREM 6.4 (Neumann [1972a]). Let G be a transitive permutation group on p letters for some prime p. Assume that a Sp-group P of G is not normal in G and IN G (13 )1 is even. Then G is triply transitive.
Before proving (6.4) we will prove a general result about doubly transitive permutation groups which is due to Frobenius. Let G be a doubly transitive permutation group on the set {1, ..., nl. Let F be a field of characteristic 0 and let V be an F[G] module such that V
6]
PERMUTATION GROUPS OF PRIME DEGREE
363
has an F-basis ..., v„1 where for z E G, v,z = v„. Let 10 + x be the character afforded by V. Let VO V=A ,F,E1)A where A is the space of symmetric tensors and A is the space of skew tensors. Let Ho be the subgroup of G consisting of all z in G which fix v, and v2. Let H be the subgroup of G consisting of all z in G which fix the set 4 1 , v21. Thus I H :Ho l = 2 as G is doubly transitive. Let 0 0 = 1,, 131 be the irreducible characters of H which have Ho in their kernel. 6.5. (i) A', affords the character PW +10 ± (ii) A affords the character 13 (iii) (x, ) O.
LEMMA
PROOF. (i) {V, v, v, ®VJ 1 j p} is an F-basis of A on which G acts as a permutation group. Since G is doubly transitive on 4,1 there are two orbits: {2v Ø v, I1 i pl and 4, 0 y + v, v, Ii i < j pl. The first orbit yields an F[G] module isomorphic to V and the second orbit has isotropy group H. Hence A affords the desired character. (ii), (iii) 0 v, — v, v, I 1 i < j pl is an F-basis of A on which G acts as a monomial group. Let V,, be the one dimensional F-space spanned by v, v, — v, y. Since G is doubly transitive on 4,1 it follows that G is transitive on the set of spaces V1 and H is the subgroup of G consisting of all elements which leave V12 fixed. Since Ho is the subgroup of G consisting of all elements which leave v, 0 v2 — v2 0 v, fixed it follows that A affords the character [3 This proves (ii). For 1 i p let w, = E,,, (y, - v, v, ). Let W be the F-space spanned by {w}. If z E G and v,z = vk then w,z = wk. Thus W is an F[G] module and the linear map f: V—* W defined by f(v,)= w, is an F[G]homomorphism. Since E;', w, = 0 it follows that E';_, v, spans the kernel of f. Thus W affords the character x. Hence by (ii) (x, ?) O. 0 ,
From now on in this section assume that G satisfies the hypotheses of (6.4).
N = NG (P) is a Frobenius group. As N/ G it follows that p/ 2. By assumption there exists an element x of order 2 in N. By (6.3) G is doubly transitive. Let H, Ho, PO, PI be defined as above. We will use the notation introduced in section 1. Let X be the R -free R[G] module obtained by tensoring the permutation module of degree p with R. By (6.1) and (6.2) X affords the character 1 0 + x, where x is irreducible. By (6.1) x = ço o + SO as a Brauer character where so is an irreducible Brauer character. Let 00, 0 be the character
364
[6
CHAPTER VIII
afforded by the principal indecomposable R[G] module which corresponds to (po, cp respectively. Let X0X=A + 9A - where A + is the space of symmetric tensors and A is the space of skew tensors. The group P( ,,, has exactly 2 irreducible Brauer characters. let no, n , be the characters afforded by the corresponding principal indecomposable R[P(x)] modules, where no corresponds to the principal Brauer character. LEMMA
6.6. (i) A +
is projective and
1(13 + 1 ) 71°(i i) A is projective and
(A +)no affords the character
affords the character -.(p —1)n i .
PROOF. Since X is projective and InvG X/ (0) i t follows from section 1 that XN V( 1 , p). Hence by (2.9) (p -lo ( p-3)/2
A
V(a 2 ',p),
A
ED
V(cr 21 ,p).
Since a (x) = — 1 it follows that (A +) p(„) , ( A - ) p(„) affords the Brauer character (p + 1)n o , (p — on, respectively. Since A, A are direct summands of X Ø X they are both projective. This implies the result. D LEMMA
6.7. (x, (3 (?) = 1.
By (6.5) A affords the character I3. By (6.6) A is projective. Thus = a 0 00 + (1,0 + 01 where (x, 00= O. The Frobenius reciprocity theorem implies that (0 1 G ) = 0 and so a o = O. By (6.5)(iii) al =([3, x) X O. By (6.6) (0 ) p(„) = cob for some positive integer c, and so POO = cri, for some positive integer c. By (6.5) A + affords the character 0,Y + 00 . By (6.6) A + is projective. Thus (+ 00 = b000 + + 0, where (x, Bo) = (1G, Bo) = O. By (6.6) (f3+ col() for some positive integer c o . If 6 1 X 0 this implies that P(x) = C ' no for some positive integer c' contrary to the previous paragraph. Therefore /30G+ o = bo00 + O. Consequently by the Frobenius reciprocity theorem ( 13 x) = (j3 c?,1 G )= 1. D PROOF.
-
PROOF OF (6.4). Choose the notation so that G acts on {1, ...,p} as a transitive permutation group and x leaves 1 fixed. Let G, be the subgroup of G consisting of all elements which leave 1 fixed. Thus x E GI . The permutation representation of G on the set of all unordered pairs { i, j} is transitive and the subgroup consisting of all elements which leave 0,21
7]
CHARACrERS OF DEGREE LESS THAN p — 1
365
fixed is H. Thus this permutation representation affords the character p c?. By (6.7) and Frobenius reciprocity ((t3 ô)G, , 1 G,) = (Pc? ( 1 G) G ) = ()V') ± X) = 2 -
Hence G, has exactly 2 orbits on the set of unordered pairs {0}. These must necessarily be A i = {{1, i }2 i and A = {{i, i/1, j/ 1 } . Thus in particular G, is transitive on A. Suppose that the result is false. Then G, is not transitive on the set of ordered pairs (i, j) with 2 i/ j p. Since G, is transitive on A this implies that there are 2 orbits F 12 and (i,j) E 11 if and only if (j, E F2. As x E G, and x interchanges some pairs the last possibility cannot occur. E 7. Characters of degree less than p —1
Suppose that a Sr -group of G has order p and G is not of type L 2 (p). Let x be a faithful irreducible character of G. By (3.3) x(i) i(p — 1) for p 13. This already strengthens earlier results of Brauer [1942b] and Tuan [1944] in case p 13. Sections 7-10 are devoted to the proof of the following three results which strenghen the above mentioned inequality. The first of these includes the results of Brauer [1942b] and Tuan [1944]. The second includes earlier results of Brauer [1966c] and Hayden [1963]. For the third result we will follow the proof given by Blau [1975b]. Actually (7.1) is a consequence of (8.1) below which is a more general result.
THEOREM 7.1 (Feit [1967b]). Let G be a finite group whose center Z has odd order. Let p >7 be a prime and let P be a S r -group of G. Assume that I P1= p and P sel G. Let be a faithful irreducible character of G. Then one of the following occurs. (i) G / Z PSL2(p), (ii) p — 2.
an= (p -± 1).
THEOREM 7.2 (Blau [1971b]). Let G be a finite group and let Z be the center of G. Let p >7 be a prime and let P be a Sp-group of G. Assume that P=p and P sei G. Let be a faithful irreducible character of G and let t denote the number of conjugate classes of elements of order p. Then one of the following occurs. (i) G/Z PSL2 (p), "(1) = (p -1- 1), t = 2. (ii) “1)% p — 1. (iii) al) p + — p + p t 2 3t +1.
366
CHAPTER VIII
[7
THEOREM 7.3 (Feit [1967b], [1974]). Let G be a finite group and let Z be the center of G. Let p > 5 be a prime and let P be a S„-group of G. Assume that IP I = p, P sti G and G has a faithful irreducible character 4- with 4' (1)= p - 2. Then p = 2° + 1 and G -1)x Z. By using the results of Feit [1964] it can be shown that once (7.1), (7.2) and (7.3) are proved then the same conclusions hold even when the assumption that IPI=p is dropped. In case p = 2 or 3, (7.1), (7.2) and (7.3) are true and trivial. If p = 5 (7.2) and (7.3) are false as a covering group Â6 of A, has a faithful irreducible character of degree 3. This group has 2 classes of elements of order 5 but 5 > 22 - 3.2 + 1. In case p = 7, (7.1) and (7.2) are false as a covering group A, of A, has a faithful irreducible character of degree 4. This group has 2 classes of elements of order 7 but 7 > 22 - 3.2 + 1. As a consequence of (7.3) we can get the following result which strengthens (6.4) in a special case. THEOREM 7.4 (Ito [1960a]). Let p > 2 be a prime. Let G be a permutation group on p letters and let P be a Se -group of G. Assume that I NG (P)I = 2p. Then either P < G or p - 1= 2° and G - 1). PROOF. If p = 3 or 5 this can easily be verified. Suppose that p > 5. The principal p-block B of G contains irreducible characters of degree 1 and p - 1. Since I NG (P): P =2 these are all the nonexceptional characters in B. Thus by (VII.2.15)(iii) an exceptional character in B has degree p - 2. The result follows from (7.3). E Throughout the remainder of this chapter the notation introduced in section 1 will be used. Also the following notation will be used. = N = NG (P). C = CG (P).
e t is the number of conjugate classes of elements of order p in G. Thus te = p - 1. We will be concerned with groups G which satisfy the following conditions. (i) G = G' and G IZ is simple where Z is the center of G. (ii) p >5. 1 <e
7 or to (2° ) for 2° = 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 PSI_,(p), (2° ) for 2° = p —1> 4. None of these groups admit 5L2 (p) for p 7 or 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 GIZ 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) C = P X Z. Hence by (VII.2.1 6) “1)= p — e and (iv) is satisfied. As G has a faithful irreducible character, Z is cyclic. Furthermore I Z (1) = p — e. Hence (1Z e)= 1 and (iii) is satisfied. Since G = G' 1< ( 1) V(11" tta', 2i +1)—> V(71a',2e) —> V(77u.a - ',2(e —1— i)+1)—› O.
As a' = 1 it follows from (1.18.2) 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,
V(-71"a, 2i +1)e
Let M(i, u) be defined by M(i,
v (1-r,aa
2(e —1— i)+ 1).
Y(i, u). The result follows from
(111.5.8). The next result is implicit in Brauer [19664
LEMMA 7.14. Suppose that u,v, w ---51Z1-1 with w u + v (modIZI). Assume that the following conditions are satisfied. (i) 8 (u)= 8(v) = 8(w) = 1. (ii) 4-Sm ) (1)< p — 1 for m = u, v, w. Let a') 41' ) t. hok
hiika" ) + F, where F is orthogonal to each 41" ) . Then
Let (p i , cp, be all the irreducible characters of E where cp, = Then 0, = cp;'" for j =1, .. e are all the principal indecomposable characters of PE. By abuse of notation let (A, )pE = For fixed m, U1'1 is a set of irreducible characters, any two of which agree on p'-elements. Thus for m = u, I) or w (), = Of( „, ) — A,, where f(m) does not depend on i. Hence PROOF.
8]
PROOF OF (7.1)
371
(41" ) 41")PE = OR.)0f(u) — Ai`Pfm AJO.ro.o+
(7.15)
Let PE, PEP be the character afforded by the regular representation of E, EP
respectively. Then çf
= {(A1)EVs
} PE
= (pE) PE =
PPE.
Similarly cl'f(0
4)f(u)
{( cl'f(0)EfPf(o} PE — {(tPE
(Pro.))(PRolPE
— (tp E ) PE (cp f( ,f () ) PE = tPPE ±
For each i, (A, )p is a sum of e distinct nonprincipal irreducible characters of P. Thus ((AA,),
1 p) =
((A,
)p)
e.
Hence A,A, = 0, + 02 where 0, is a sum of at most e principal indecomposable characters of PE and 0, is a linear combination of the (A, )PE. Since ik,A, vanishes on E — {1} it follows that 0, = 0 or 0, = ppE. Therefore (7.15) becomes ()PE
=
(t
—2 + 8)p„ +
f(u), f ()+ 02
(7.16)
where 5 = 0 or 1. The set {0,} U {A., } 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 D. is nonnegative for .all s. By (7.16) the coefficient of any 0, in W“ ) C;') )pE is at most t. Since (4)1, E = 0R.„ ) — Ak the result follows. D
8. Proof of (7.1)
This section contains a proof of the following result which implies (7.1). 8.1. Suppose that condition (*) of section 7 is satisfied with p >7 and 4- a faithful irreducible character. Assume that there exists u, y, w with O u, y, w J Z — 1 such that w = u + y (mod IZ1) and the following conditions are satisfied.
THEOREM
(i) 8(u) = 6(y)= 6(w) = 1.
(ii) = 41" , Then e = 2.
0 ) ( 1)< p —1
and ‘I" ) ( 1)< p —1.
LEMMA
[8
CHAFFER VIII
372
8.2. Statement (7.1) follows from (8.1).
PROOF. Suppose that (8.1) has been proved. In proving (7.1) it may be assumed that (*) is satisfied by (7.5). Let u = v. Then 17 2 " has the same order as Ti " since I Z1 is odd. Hence there is a field automorphism which sends 71" to -ri 2". Thus the exceptional character in B 2 is algebraically conjugate to and so 8(2u) = 8(u)= 1. Now (7.1) follows from (8.1). El 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 ,C,'" ) be an R-free REG] module which affords 0,'" ) for m = u, u, t such that X; -) is indecomposable. Let A, = À . Thus ';' ) (y) = a' ) (y Furthermore there exist irreducible characters 1,, of N IC such that (X) N V(71"v, p - e),
p - e).
(X) N
LEMMA 8.3. C (1 "' ) ( 1)= p - e. PROOF. By (2.7) e— I
(X ")
V(Ti" vvia k ,2k +1)8) A, k
=o
where A is a projective 1 [1•1] module. Let = defined as in (7.13). Thus
and let W(k,w) be
e—I W(k, W )IEBIA k
for some projective R[G] module A'. Suppose that C;"' ) (1) p - e. Thus C;(1) 2p - e. By (7.12) and (7.13) dimk W(k, w)+ dimk W(e - 1 - k, w)
(t -2)a") (1)+ p -1 (t -2)(2p - e)+ p -1
for 0
k
-1). If e is even then dimA W(e /2, w)
(t -
(t - 1)(2p - e).
If e is odd then dimk WO(e -1),
(t - 1)C;"' (1) + p -1 (t - 1)(2p - e)+p -1.
8]
PROOF OF
(7.1)
373
Furthermore e-I
p 2 — e (2p — e) = (p — e) 2 = dimk ,C" ) ,Cu )
E
dim, W(k, w).
k
Suppose that e is odd. Then
dim A {W (k, w)03 , W(e — k
p 2 — e(2p — e)%
w)}
k =0
+ dimg
(e —1), w)
—1)(t — 2)(2p — e)+(e —1)(p —1) +(t—1)2pe .
Hence (2p — e){e + iet
— e +1+ t —1} +(e +1)(p — 1)
= (2p — e)d(e +1)+ (e +1)(p —1) =
+ 1){2pt — te +p — 1 } = pt(e +1).
Thus p%t(e+1)=te+t=p—l+t
and so t
1 contrary to the fact that e < p — 1. Suppose that e is even. Let f = dim k W(e —1, w). Then p 2 — e(2p — e)
dimg { W(k, w )
W(e — 1 — k, w)}
k
+ dimg W(1 e, w)+ f —1)(t — 2)(2p — e)+ ( ir e —1)(p —1) +(t—1)2pef = (2p — e)fiet — e — t + 2 + t — + (ie — 1)(p — 1) + f.
Hence pet + p =
(2p — e)(4et + 1) +(e— 1)et + f
= pet — e 2 t + le2 t — et + f + 2p — e.
Therefore 0?---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
[8
CHAPTER VIII
374
LEMMA 8.4. There is an irreducible Brauer character cp, such that 41" ) = cp, as a Brauer character.
PROOF. Since 5(w) = 1, the principal Brauer character does not occur as a constituent of 6" ) . If the result is false then some nonprincipal irreducible Brauer character has degree at most i(p — e)< (p — 1). Thus by (3.1) G is of type L 2 (p) contrary to assumption. D LEMMA 8.5. Let cp be defined as in (8.4). Let 0 be the unique nonexceptional irreducible characterin B. which has cp as a constituent. Then 0 (1)
2p — 1.
PROOF. If the result is false then 0 = p — 1 as 0(1) = — 1 (mod p). Thus = + cp, as a Brauer character and cp 1 (1) = e — 1. The principal Brauer character can occur as a constituent of 0 with multiplicity at most 1. Since e > 2, q(1)> 1 and so there exists a nonprincipal irreducible Brauer character in B, of degree at most e — 1 < ;(p — 1). Thus by (3.1) G is of type L 2 (p) contrary to assumption. 0 Let
41") 0)= a0 + E
hi,kck")+
F;
h=
E
hijk,
where F is orthogonal to 0 and to all a" ) . LEMMA 8.6. (i) h
t.
(ii) If e is odd, a +h (iii) If e is even, a +h
(e — 1)(t — 2) + t — 1. (t — 2) + 1.
PROOF. (i) This follows from (7.14) and (8.3). Let cp be defined as in (8.4). For any R -free R [G] module Y let n( Y) be the multiplicity with which cp occurs as a constituent of the Brauer character afforded by Y. Thus a + h = n(X X). Let W(k, w) be defined as in (7.13). Thus by (2.7) e —I
W(k,
B1 A
k
for some projective k[G] module A. (ii) By (7.7) and (7.13) {n(W(k,w))+ n(W(e — k —1, w))1
a+h k —0
+ n(W ((e — 1), w)) (e — 1)(t — 2)+ (t
—
1).
8]
PROOF OF
(7.1)
375
(iii) By (7.7) and (7.13) en-2
a+h
{n(W(k, w))+ n(W(e —
k1, w))1+ n(W(e, w))
k =0
—1)(t —2)+ (t —1)=- (1e)(t —2)+1.
D
LEMMA 8.7. If 1 k p —I then (p —1)F(y k ) a(p —1)— he + ep — e 2 .
PROOF. By
(7.6) ‘;' ) (y ) ) (y) = As (y)A, (y -I ) for suitable s. Thus )= e + g(y), where g(y) is a sum of e 2 — e primitive pth roots
of 1. Let Tr denote the trace from the field of pth roots of 1 to the rationals. Since Fz =F(1)î and F is orthogonal to all a" ) it follows that F(y) is rational. As 0(ys)= —1 it follows from (7.6) that e(p —1)— e 2 + e = Tr{0 ) (y )ny = — a (p — 1) + he + (p —1)F(y). This implies the result.
D
LEMMA 8.8. h + e = 1 (mod t). PROOF. Divide the equation in (8.7) by e and read modulo t.
D
LEMMA 8.9. F(1) > (p +1){a + e —1— (e + h —1)11} %(p +1)a. PROOF. Since F., = F(1)-ri' and F is orthogonal to B and all 417) it follows that = E' ckxk + I" ckxk + r(),
where the first sum is over characters xk in B, with xk (1)=. 1 (mod p), the second sum is over characters x k in B,„ with xk (1)- — 1 (mod p) and F, is a sum of characters in blocks of defect O. Therefore if 1= s p —1 1- (ys)=
ck — E" Ck
c
Thus by (8.7) I'ck F(y)=a + e +— ti (1— e — h).
CHAPTER VIII
376
[8
As the principal character of G occurs in 41"41" ) with multiplicity at most 1
this implies that 1(1)=
ckx k (1) {(E' c k )— 11(p + 1) + 1 > (p + 1){a + e — 1+1( (1— e — h)}.
This proves the first inequality. If the second inequality is false then a > a + e —1+ (110(17 e — h).
Hence by (8.6)(i) 1 (e — 1) < 1— (e + h — 1) = — (e — 1) + t t t
t
(e — 1) + 1.
Thus ( t — 1)(e — 1)< t. Since e 3 this implies that t < 2 and so t = 1 contrary to the fact that e
t — 2. Thus e = t — 1 and a k = v 2. Let )? )?' = A ±e A - where A + is the space of symmetric tensors and A is the space of skew tensors. By (2.9), W(i, 2) and W(e — 1 — i, 2) are both summands of A + in case i is odd and are both summands of A - in case i is even. Since e = t — 1, V(772 , p) occurs exactly twice as a summand of each W(i, 2)e W(e —1— i, 2) and exactly once as a summand of WO (e — 1), 2). Therefore V(71 2a k, p) occurs as a summand of A strictly more times than it occurs as a summand of A. This contradicts (2.9). Thus e t —2. Since e is odd and t is even, e t —3. Therefore p = et + 1 (t — 3)t + 1 = t2 — 3t + 1.
(9.2)
Furthermore e (p — 1)/e — 3 and so e 2 + 3e — (p — 1) O. Hence e - 3 + V p + Since 4"(1)= p — e the result follows from this inequality and (9.2). D
10. Proof of (7.3)
Suppose that the assumptions of (7.3) are satisfied. Assume furthermore that condition (*) of section 7 is satisfied. As (1)= p — 2, it follows that e = 2. Thus Z is odd as (e, I Z 1) = 1. Hence ri is algebraically conjugate to 77 2. Since 4" is in 13 1 this implies that 412) (1) = "(1) = p — 2 for i = 1, . . t. In particular 8(2) = 8(1) = 1. Thus 412) does not have the principal Brauer character as an irreducible constituent. Since G is not of type L 2(p) it follows from (3.1) that "(, 2) is irreducible as a
Brauer character. The Brauer tree for
B2
looks as follows
Then x = ço is irreducible as a Brauer character. Hence cp (1) 1 (mod p). Observe that a 2 = 1 as e = 2. Let X be an R -free R [G] module which affords ‘. Let X (2) be an R -free R[G] module which affords 4.(12). Let L be an R -free R [G] module which affords x. Thus f affords cp. By (1.17.12) there exists an R -free R [G] module Y which affords - ;2) such that Y is indecomposable. Let X0X =A + 0A -, where A " is the space of symmetric tensors and A is the space of skew tensors.
382
[10
CHAPTER VIII
LEMMA 10.1. (1 ) iC(2) V (71 2 a, p — 2) (ii) L V(71 2 ,1). PROOF. (i) Suppose the result is false. Thus (A - )N V(î1 2 3)6)4(p — 5) V (71 2 a, p). By (2.7)
V(71 2 , p —
2). By (2.9)
,
.11°(N, (1), Hornfi (5-C (2) , A ) ) .11 °(N, (1), fC (2) * 0 A
-) / (0).
Thus by (III.5.10) H°(G, (1), Hom k (5-0 2) , )) # ( 0) and SO HOIrik[G] (g (2) , A -) (0). Since 5-0 2) is irreducible this implies that fr) is isomorphic to a submodule of A -. Consequently V(1 2, p —2) = .5-C (, ) is isomorphic to a submodule of A N. This is however impossible as InvEp (A Ep)= (0). This proves (i). Suppose that (ii) is false. Then t V(1 2 a, 1) and SO by (III.5.10) and (111.5.13) HOMA [N ] (E, .k (2) (0). Thus Homp EG O, .k(2)) j (0) which is not the case. El LEMMA 10.2. A - ---. Y PROOF. As t = 4(p — 1) it follows that
i= I
= — 2(t — 1) =- 3
(mod p).
V(.7) 2 cy k, 3) for k = 0 or 1. By definition HomE [G] (i, y) --- (0). Hence ( N, (1), HOr (f i.7 )) = ( 0). (N, (1), Hence by (III.5.10) if.f,1 ) A. Since Thus by (10.1)(ii) k 0 and so i" V(i 2a, 3). Hence by (2.9) Y IA
t-
dimk Y =4 (p — 2)(p — 3) = dimA A it follows that A - ---- . Y.
E
LEMMA 10.3. x (1) = 1 + np with n(p - 3). 4. ( 1 ) = — 1 + (n + 1 )P. PROOF. Clearly x(1) = 1 + np for some n. Thus 4. (1)= p — 2 + 1 + np = — 1+ (n + 1)p. By (2.9) and (10.1) L IA'. Consequently 1 + np 4(p — 1)(p — 2). This implies that n 4(p — 3). CI LEMMA 10.4. Let x be a {2, p } ' element in G, x 0 Z. Then ‘"(x ) = O. PROOF. Let 0 + , 0 - be the character afforded by A ±, A respectively. Then it is well known that 0 + (x)= 0 - (x)+ (x 2). Hence by (10.2) 0 + (x ) = 1-(p — 3 ) ;2) (x ) + (x 2 ) and so
101
PROOF OF
4' 2(x)= (P
(7.3)
383
3 )4 (2) (x)±4- (x 2).
Let u be the automorphism of the field of 1 G 1 th roots of 1 over the rationals which fixes all 2 th roots of 1 and squares all m th roots of 1 for m odd. Thus 4- (x 2) = (x) and ";2) (X ) = ( x). Hence 4-2 (x ) (p — 2)4'" (x). Choose k so that er " = 1. Thus 0. (x) = (p — 2)2k(X ) = p — 2)2" /"(x ). k
(
Hence if 4' x)# 0 then 4' 2k-1 (X ) = (p — 2)2" and so 10x )1= p — 2 = Hence x E Z contrary to assumption. D
10.5. (i) 1G :Z112bp(p — 2) 2 for some positive integer b. (ii) If I Z I = 1 then I G I = 2bp (p — 2) for some positive integer b.
LEMMA
PROOF.
Let g be a prime with g # 2, p. Let Q be a Sq -group of G. By (10.4).
11‘0112 = FIGTI
n z1(1)2 = !
JP ,
Thus IQ : Q n Z11(p — 2)2 . This proves (i). If 1 Z = 1 then ( o , 1 Q ) = ( 1)/1 Q I by (10.4). Thus 1 Of (p — 2). This proves (ii). D PROOF OF
(7.3). By (7.5) it may be assumed that condition (*) of section 7 is
satisfied. Suppose first that 1 Z I = 1. By (10.5)1 G I = p2" (p — 2) for some positive integer b. Since G is not of type L 2 (p) the result follows from (5.3). Thus it may be assumed that 1Z1> 1. Hence x(1) # 1 as G = G' and so
n >0. Suppose that n is even. Then by (10.5) 1 + np 1(p — 2)2. Let d = (1 + np, (p — 2)). Thus 2n + 1 0 (mod d). Hence 1 + np 1(2n + 1)2 . If 1 + np = (2n + 1)2 then p = 4(n + 1) which is not the case. Thus 1 + np 1(2n + 1)2. This implies that p 2n + 2 contrary to (10.3). Suppose that n is odd. Hence by (10.3) and (10.5) {(n + 1)p — — 4"(1)1 (p — 2)2 . If d = (p — 2, (n + 1)p — 1) then 2n + 1 0 (mod d) and so {(n + 1)p — 1}1(2n + 1) 2. If (n + 1)p — 1 = (2n + 1) 2 then p=
4n 2 + 4n +2
= 4n+
2 n + 1 n+1
Hence n = 1 as (n + 1)1 2 and so p = 5 contrary to assumption. Therefore {(n + 1)p — 1} (2n + 1)2 and so p 2n + 1 contrary to (10.3). 0
384
CHAPTER VIII
[11
11. Some properties of permutation groups
The material in this section, except for (11.8), is independent of the previous results of this chapter and is included here 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 12, let ax denote the image of a under the action of x for a E fl, x E G. If A is a subset of G let G, denote the subgroup of all elements in G which leave every element of A fixed. In other words G, = fx x E G, ax = a for all a E A 1. If A = {a} let G„ = Gi„) . For a E f2 let a' = fax x e GI. A proof of the following result can be found in Wielandt [1964] (13.1). THEOREM 11.1 (Jordan). Let G be a primitive permutation group on D. Let A be a subset of 12 with 1- f21— 1. If G, is transitive on fl — A then G is doubly transitive on D.
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 12 x D. Clearly G C Suppose that G is a permutation group on D. Let A be a subset of D and let H be a subgroup of G such that ax E A for a E i , x E H. For x E H define the permutation x ° as follows If a
E
ax ° = ax,
if a E f2 —
ax = a.
Let H ={x ' I x EH}. The following result is due to Wielandt [1969] (6.5). THEOREM 11.2 (Dissection Theorem). Let G be a permutation group on fl. Let A be a subset of 12 and let H be a subgroup of G such that ax E A for aEd, xEH. Assume that for all aEA, bE12— A, H = H„Hb. Then C H° x H PROOF. We will first show that if x E H then x ° E H(2) . It suffices to prove that if (a, b)E 12 x12 then there exists z E H such that (a, b)z = (a, b )x ° . There are three cases. If a, b E d let z = x. If a,bED—d let z =1. Suppose that a E i, b E D — A. By assumption x = x„xb with x„ E Xb E Hb. Let z = Xb• Then
11]
SOME PROPERTIES OF PERMUTATION GROUPS
385
(a, b).x" = (ax, b)= (axaxb, b)= (axb, b)= (a, Oxb.
This proves that x ° e Let x E H then x = -`). Since x e H(2) it follows that x" -° E Thus H° 1/ 2-° = H x H" -° C H. E 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 fl. Assume that I Q l is even. Let 1 G + x + 0 be the character afforded by the permutation representation of G on fl where x, 0 are irreducible characters of G. Since ID l is even, x(1)/ 0(1) and so x, 0 are rational valued. For a e fl let la 1, d (a), 1(a) be the orbits of G„ where i(ax) = d (a )x and F (ax) = F (a)x for x e G. Let k = 1 d (a)1, 1 =1F(a)1. Since I GI is even, kj 1 and so A and F are self paired in the sense that 3 (a) = fax 1 ax' E d (a)},
F(a) = lax I ax' E F(a)}.
Define the integers ix, A by {A
if b
Ed
if b
E
(a), (11.3)
F(a).
Then A, ti, are independent of the particular choice of a, b. Furthermore { / — k + IL
-
1
if b
E
F(a),
1F(a)n F(b) 1=
(11.4)
1 —k+A+1 if b EA(a). The following conditions are satisfied. Let d = (A — + 4(k — g). Then d is a perfect square and 0(1)
4._ 2k +(À — tt)(k + 1)
k+1
X (1)
2V'
2
(11.5)
If furthermore s, t are the characteristic values of the incidence matrix associated to the orbit A then
sT _ (À— p4 ± V d (11.6) 2 tJ 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, and S, acting on the set of 2-element subsets of {1, ..., 5 } . All other known examples of
386
[11
CHAPTER VIII
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. THEOREM 11.7 (Wielandt [1956], [1964]). Let p be a prime. Let G be a primitive permutation group on 12 where If2 I = 2p. Assume that G is not doubly transitive. Then 2p = m 2 + 1 for some integer m. Furthermore G has rank 3 on fl and the following conditions are satisfied, where the notation is that introduced above.
(i) X ( 1 ) = 0 ( 1 ) = P 1 (ii) k = m (m — 1), 1 = m (m + 1). (iii) s + t = — 1.
THEOREM 11.8 (Ito [1962c1). Let p, G, D be as in (11.7). Let P be a Sr -group of G. Assume that p> 3 and NG (P)! = 2p. Then p = 5 and G — A,. PROOF. The Brauer tree of the principal p-block of G is as follows •
1p—1
•
Thus there exists an irreducible character of G with (1)= p — 2. Since Op• (G)= (1) it follows that is faithful. By (7.3) p = 2 + 1 for some a and G----- SI-2 (p — 1). Since G has a subgroup of index 2p the known properties of PSL2 (p — 1) imply that p = 5 and G ---- SI-2 (4) A,. 111 LEMMA 11.9 (Ito [1967a]). Let p, G, D be as in (11.7). Then = (m + 1)(m — 3),
1.1. = À + 1 = i(m — 1) 2.
PROOF. By (11.6) and (11.7)(iii) A—ii = —1. By (11.5) and (11.7) = -± Vd(x(1) — 0(1)) = 2k — (k + 1) Hence
4(k — )= d — (À — p.) 2 = m 2 — 1
and so by (11.7)(ii) Thus
=4k + 1 — m 2 = m 2 — 2m +1 = (m _1)2 . 4A = — 4 = m 2 — 2m — 3 = (m + 1)(m — 3). E
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 = (1) for p > 5.
12]
PERMUTATION GROUPS OF DEGREE
2p
387
LEMMA 11.10 (Ito [ 1967a]). Let p > 3, G,,f2 be as in (11.7). Then the following hold. (i) G„ is faithful in its action on F(a). (ii) Let H be the kernel of the action of G„ on ,(a). Then H is an elementary abelian 2-group and every orbit of H on F(a) has cardinality 1 or 2. PROOF. (i) Let H be the kernel of the action of G„ on F(a). Let {c,, be an orbit of H in 3(a). Since H C („) it follows that
c„}
F* = 1(a) n 1(c i ) = • • • = 1(a) n F (c„).
By (11.4), (11.7) and (11.9), 1E1* =m +A + 1 = m + Let i# j. Then I' CF(c) fl 1(c1 ). By (11.4), (11.7) and (11.9) 1F(c,)n m + g. Hence n F(c,)= F* for i j. Thus {F(c,)— F*} is a collection of pairwise disjoint subsets of 3(a) and so
no
n{im(m +1)—m —i(m — 1)21= n{im(m +1)—m —III =im(m — 1).
-
This implies that n 2m1(m + 1) 5, m >3. Thus n 2m/(m — 1) 3 then m is not a
prime.
We will give Scott's proof as presented in [1970], [1972] which is a simplification of an earlier proof given in [1969]. This proof depends on the results in section 11 as well as on the results in Chapter VI. In particular (12.1) implies that if p < 313 then p = 5,41 or 113. The cases p = 41 or 113 have been shown to be impossible in Scott [1976] by special arguments. Thus it is known that if G exists with p >5, then 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 11. Since G is not doubly transitive on fl it follows that G') is not doubly transitive on (2. Thus G may be replaced by G (2) . Hence it may be assumed that G = G") . Let Q be a -group of G. LEMMA 12.2. I Q I
q.
PROOF. Suppose that the result is false and I Q > q. Choose a e 12 with Q C G. Since the orbits of G„ on D — {a} have cardinalities q(q — 1) and q(q + 1) by (11.7), it follows that a is the only point in D fixed by Q. Furthermore every orbit of Q on fl — {a } has cardinality q since lq(q+1) 3. Thus either a,(1) 0 (modp) or 0(1) 0 (mod p). By (13.6) and Frobenius reciprocity (111.2.5) g (1) # p. Hence a (1) = p and p (1) = p + 1. Let 0 = a, Xk = p. By (13.6) and Frobenius reciprocity ON = e. Direct computation yields ((A, — ) G )N = Al+A (14+14j ).
■j—
Hence (xk )N = 1 + ON - ((Al — i.LIE))N is as required. LEMMA
D
13.9. t = O.
Suppose not. Let 0, 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
PROOF.
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 4) be an irreducible Brauer character with cp (1) xk (1) = (p + 1) which is a constituent of x k . Since x k (1)=• 1 (mod p) 4) is not the principal Brauer character. Hence cp is faithful by (13.4). If p >7 then (p +1)1, p > 5. Suppose that p = 7. Then e = 3 and T looks as follows 0-,D-C,
8 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 —PSL 2 (7) contrary to assumption. Hence e is even. Therefore s is even and so s 2. By Frobenius reciprocity (p, xk )= 1 and (A xk ) = 0 for all i. Hence (A2 — = Xk + a -± 13 for irreducible characters a, 13 of G. Suppose that (À 2 — ) G = — xk + a + 13. Then a (1) + (1) = p + 1 and so it may be assumed that a (1) (p + 1). Since (1 G, ( A2 - 1.4) G ) = 0 it 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 1
a
) G
4
s +2t = e
a(1)(p + 1)
it follows that p
7. Suppose that p = 7. Since e l(p — 1) this implies that 6 = e a (1) 4, which is not the case. Suppose finally that (À 2 — /Lk ) G = — xk + a — f3. If a(l), [3 (1) 0 (mod p ) then a (1), OW ± 1 (mod p) as (A 2 — ) is p-rational. Hence —1 -± 1 -±1 0 (mod p) and so p 3 contrary to the fact that e s + 2t 4. By (13.6) l3(1)/ p. Hence a (1) = p and so 13(1) = 1 contrary to (13.4). E
LEMMA 13.10. 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/ 2 then (x) is a characteristic subgroup of H. Hence NG(H) C NG ((x)) =H. Thus H is a S2-group of G and Burnside's transfer theorem implies that G has a normal 2-complement contrary to (13.4). E LEMMA 13.11. There exist pairwise distinct nonprinci pal irreducible characters a, 0, y of G such that
13]
CHARACTERS OF
(A
— A 2) 0 = 1 G + a —
DEGREE p
395
y.
PROOF. Since E is a T.I. set it follows that II (A — A2) G 112 = 11(A , — )1 2 )"11 2 = 4. The result follows as (A, — A 2)G (1) = O. CI PROOF OF (13.1). Suppose first that one of a, g, or y in (13.11) is an exceptional character in the principal block. Since (A — A 2 )G is p -rational this implies that -;•(p — 1) = (p —1)/e 2 and so p = 5. The Brauer tree of the principal block is
1 a+1 a
Thus if a, g, 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, 0, 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, g, y is exceptional in the principal block. Since e = 2 by (13.10) there are only 2 nonexceptional characters in the principal block unless p = 3 in which case there are 3 nonexceptional characters. If p = 3 then one of a, 13, 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 at least 2 of the degrees a(1), 0(1), y(1) are p. Thus exactly two of them are p and the other degree is 1, 2p — 1 or 2p + 1. The case of degree 1 is impossibly by (13.4). Hence the possible Brauer trees are as follows 1 2p + 2 2p + 1
1 2p — 1 2p — 2
If 0 is an irreducible character with 0(1)= p then (0, (A — A 2) G ) 0 by (13.6). Hence G has exactly 2 irreducible characters of degree p. Since (2p + 2E ) I ! GI for E = 1 or —1 this implies that 0
I GI = 2p2 +1+ (2p + e)2 + (p —1)(2p +25)2 4 (mod p + e).
Hence p = 5 and e = — 1 as p > 3. Therefore GI= 50+1 + 81 + 128 = 260. However 2p + 2E = 8 and so (2p + 2s) lishes the result. E
! GI. This contradiction estab-
CHAPTER IX
Throughout this chapter R is either a field of characteristic p or the ring of integers in a finite extension K of Q,. Let 14 denote the residue class field. Thus 14 = 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 A, (R [G]) was defined in Chapter II, section 4. We will frequently write A c (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 be a nonempty set of subgroups of G. Let A (1) (G) = A,„t,(R [G]) be defined as in (11.4.3). If H is a subgroup of G let A, (G) = A oi) (G). By (11.4.3) 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 S) then {( V)} is a C-basis of i(G). It is natural to ask for conditions under which A, (R [G]) is semi-simple. In section 2 it is shown that this is the case if a Sa -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 A' 1G then A, (R[G]) is also semi-simple. The reader is referred to Conlon [1965], [1966], [1969]; Donovan and Freislich [1976], and earlier papers of Basev [1961] and Heller and Reiner [1961] which contain a classification of k [P] modules in case p = 2 and P is a noncyclic group of order 4. For related results see Wallis [1969], Hernaut [1969], Muller [1974a], [1974b], Erdmann [1979a], Donovan and
I
Freislich [1978]. In this connection we mention some other known results which will not be treated here. If R = Z, is the ring of p -adic integers and G has a cyclic Sr -group then A, (Z, [G]) is semi-simple. See Reiner [1966b]. If G has noncyclic Sa-groups then A, (Z [G])contains nonzero nilpotent elements. See 396
1]
THE STRUCTURE OF A
(G)
397
Reiner [1966a], Gudivok, Gon'earova 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 [1973]. If R = 14 is a field and p/ 2 it has been shown by Zemanek [1971] that if a Sr -group of G is not cyclic then A c (R[G]) contains nilpotent elements. Zemanek [1973] has also investigated the case that G is of order 8. See also Yamauchi [1972], Bondarenko [1975], Ringel [1975]. Various types of representations and modules have been studied. See for instance Carlson [1974], [1976a], [1976b] or Janusz [1970], [1971], [1972] or Johnson [1969a], [1969b]. 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 dim, A (G) is finite if and only if G has a cyclic S r -group. This section contains some results concerning the structure of A (G). Let P be a p -group in G. Let (P) be the set of all proper subgroups of (NG (P)). P. Define Wp (G) Ap (G)/AE, ( p ) (G). By (111.5.9) W( G)=
THEOREM 1.1 (Conlon [1967], [1968]). A
wp (NG (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 [1964]). 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 [1968] in case R is a field. Related results may be found in Conlon [1967], [1968]; Wallis [1968].
THEOREM 1.3. (i) Let P be a p- group in G. Then Ap (N G (P)) is isomorphic to a subalgebra of ED Ap (H), where H ranges over all subgroups of NG (P) such that P C H and H / P 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 NG (P) with H/P a cyclic p'-group such that A (H) contains a nonzero nilpotent element.
398
CHAPTER IX
[1
For future reference the following elementary result is included here.
LEMMA 1.4. Assume that R = 1 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)0 A (H). PROOF.
Clear by (111.3.7). E
The rest of this section is devoted to the proofs of (1.1) and (1.3). Some preliminary results are proved first. Let A' ) (R[G])= f4(R[G]) be the Grothendieck algebra. Then A '(/72 [G]) 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 0,, be the Brauer character afforded by V. Define the algebra homomorphism 0 : A (R[G]) 4 /0(R[G]) by 0 (( V)) = 0, for an R[G] module V. -
A ( r ) (R[G]).
LEMMA 1.5. A°(1 [G])
PROOF. The second isomorphism follows from (1.13.7). Let {cp,} be the set of Brauer characters afforded by the irreducible R[G] modules. Let 0, be the Brauer character afforded by the principal indecomposable R[G] module which corresponds to cp,. Given i, then by (1.19.3) there exist irreducible Brauer characters co,, such that cp, = E, . Thus 0, = where 0,, corresponds to cp,,. By (IV.3.3) this implies that (cp,, 0 ' = 5,km where m 0 depends on i. This implies that {0,} is a basis of A' )(R[G]). Hence the restriction of /3 to A () (1 [G]) is an isomorphism from A ( , ) (R[G]) to A°(1 [G]). 0 (A ,
)
The next result which generalizes (111.2.7) first appeared in Conlon [1967] in case R is a field.
LEMMA 1.6. Let T < G. Let Ube an R- free R[T]-projective R[G] module. Let V be an R -free R[G IT] module and let V I be a pure submodule of V. Then U V(U®V,)(1)(U V IVI).
Since T is in the kernel of V and V, is a pure submodule, the sequence PROOF.
-> VOT -> VT -> V/ VI»
is a split exact sequence. Tensoring with UT shows that 0 —(U® V,),- (
V) 7 (U 0 V/ V)T --> 0
1]
THE STRUCTURE OF A
is split. Since UOV,IU0v.
U® V,
(G)
399
is R [ 71-projective it follows
that
1.7. Let T 1 and let IP : PI I= p. If Mi and M2 are indecomposable F[P] modules then M1 Ø M2 = v, where M o is F[Pi ] -projective, each V, is indecomposable and MO
eED:._,
dim, V, / dim, V, for i/ j. If IPI=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
2]
A
(G)
IN CASE A Sp -GROUP OF
G
IS CYCLIC AND
R
IS A FIELD
403
information. By using these methods Rudko [1968] was able to prove the following result. See also Renaud [1978]. THEOREM 2.4. Let P be a cyclic p-group. Assume that p / 2. Let C = Q and let L be the real subfield of the field of pth roots of 1 over Q. Then A (P)- QED A where A is the direct sum of 2(I P/ — 1)/(p —1) copies o f L. We will not pove (2.4) here but in case IPI =p it follows from (VII.2.7). This can be seen as follows. Let e be a primitive (2p)th root of 1. By (VII.2.7) the linear map sending V(1, t) to (et — - ')/(s — e -1 ) is a ring homomorphism. The result is a simple consequence of this. Related results can be found in Gudivok and Rudko [1973], Butler [1974], Carlson [1975] and Jakovlev [1972]. See also Lam and Reiner [1969], Santa [1971]. Let P be a cyclic p-group. For 1 i let V, be the indecomposable F[P] module with dim F V, = i. Then (VI), • • • , ( YF0 is a basis of A (G). Each V, is serial and V, Rad(V,,,) for 1 i 1P I — 1. The proof of (2.2) depends on two elementary lemmas. LEMMA 2.5. For 1 s, t 1P I, VOV, is the direct sum of min(s, t) nonzero indecomposable F[P] modules.
y be the direct sum of n nonzero indecomposable PROOF. Let V. modules. Since 1/--- V, it follows that n = dim, Inv, (V,
0
The structure of V. min(s, t). E LEMMA
= dimF Hom,m(Vs,
V,
shows that dim, Hom, t ,j ( Vs , V,)=
2.6. Let m be an integer and let f(X)=1,7 =, min(i, j)X,X, be a real
quadratic form. Then f(X) is a positive definite form.
PROOF. It is easily seen that f(x)= (X t'.. + x„,)2 + (x2+ • • • +X„, )2
+ • • +
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 = Eb, (Vs ). Let (V,)( V, ) = a,„ (V,). By (2.5) E, as, = min(s, t). Thus 0= u2=E bsbtas„(V,) S, I
404
CHAPTER IX
[2
and so b,bra„; = 0 for all j. By summing over j this implies that min(s, t)b,b, = O. Hence by (2.6) b, = 0 for all s and so u = O. 0 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 < G where P is a cyclic p-group with 'PI> 1 and H is a cyclic p'-group. Let : Pl i= p. (i) For 1 ,51 P and cp an irreducible character of H there exists an indecomposable F[N] module V(cp, 1), unique up to isomorphism, such that cp is afforded by the socle of V and dim, V(cp, 1) = 1. Every nonzero indecomposable F[N] module is isomorphic to some V(cp,1). V(cp,1) is F[Pd-projective if and only if 1 0 (mod p) and V(cp, 1) ------ V (cp,1)* if and only if cp* = cp where a is defined as in Chapter VII, section 1. Furthermore V(cp, Op is indecomposable. (ii) If M, and M2 are indecomposable F[N] modules with M* M for i = 1,2 then MI v, Mo where Mo is F[P,]-projective and M2 = each V is indecomposable with V*V. Go Suppose that 2 I CH (P)!J. For each 1 with 1 1 P there exist exactly 2 irreducible characters cp of H with cp* al = cp. Each such cp has Ho in its kernel where CH (P): Ho l= 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) m10m2= v, Mo where Mo is F[P i ]-projective and dim, V,/ dim, V, for i/ j. As (MI V. Thus M2 this implies that V, — M2) * — V — VI by the unique decomposition property and the fact that dim, y dim, 1P, for i/ j. (iii) As H is cyclic, cp* = cp - ' and so cp *a' = cp if and only if q' 2 = cr 1-1 . Since =1 there are exactly two choices for cp. Clearly H o is in the kernel of cp. El PROOF OF (2.1). It may be assumed that C is the field of complex numbers. Let F, be the algebraic closure of F. Then A (F[ G]) is a subalgebra of A (F ,[G]). Thus it may be assumed that F = F , is algebraically closed as A (F,[G]) has finite C-dimension. Furthermore it suffices to show that A (F[ G]) has no nonzero nilpotent elements. By (1.3) it may be assumed that G = PH with P < 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) =
3]
PERMUTATION MODULES
405
A (P) 0 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'. Define
= (x, y
= x" = 1, x -1 Yx = Y')-
Then G is a homomorphic image of G , and so A (G) ÇA (G). Hence it suffices to show that A (GO is semi-simple. Therefore by changing notation it may be assumed that 211 CH (P)I The proof is by induction on I P . If 1P = 1 then A (H) - = C[H] is semi-simple. Suppose that I P>1. Let I P : P =p. Let I = Ap,(G). It suffices to show that I and A (G)/I are semi-simple. The map sending ( V) to (1/p)( VG ) for any F[PH] module V is easily seen to induce an isomorphism from A (P G) onto I. Thus by induction I is semi-simple. It remains to show that A (G)/I is semi-simple. Let S be the subspace of A (G) spanned by all ( V) with V indecomposable and V* — V. By (2.7)(ii) S + III is a subalgebra of A (G)11. If 1 I I P then (2.7)(ii) and (2.7)(iii) imply the existence of an indecomposable F[G] module V with dim, V = 1 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 + I)I I) A (H) onto A (G)I I. Since A (H) is semi-simple it suffices to show that (S + I)I I is semi-simple. Let Z = CH (P). By (2.7)(ii) and (iii) the map sending V to V c G (P) defines a one to one linear map of S onto A (P X Z). Since V is F[P] -projective if and only if VcG (P) is F[Pd- projective, it follows that (S + I)II----- A (P x Z)1Ap i (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). D 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 V 0(H) denote the R -free R[G] module of R -rank 1 with V 0(H) = Inv,, ( 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 V 0(H) K affords the trivial character. An R[G] module is a transitive permutation module if it is isomorphic to V 0(H) Gfor some subgroup H of G. A direct sum of transitive permutation modules is called a permutation module.
406
CHAPTER IX
[3
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. E
LEMMA 3.2. Let V be a permutation module. Then the following hold. (i)
(ii) V is an algebraic module. (iii) If M is a subgroup of G then V,,, is a permutation module. (iv) If G is a subgroup of M then V" is a permutation module. PROOF. (i) Clear by (11.2.6). (ii) This follows from (11.5.3). (iii) Clear by the Mackey decomposition (11.2.9). (iv) Immediate from the definition. See (II.2.1)(iv). Li LEMMA 3.3. Let V and W be permutation modules. Then the following hold. (i) V ED w is a permutation module. (ii) V 0 W is a permutation module. PROOF. (i) Immediate by definition. (ii) This follows from the Mackey tensor product theorem (II.2.10). E 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 (111.3.8). (ii) This is a consequence of (i). D 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 (V 1 . By the Mackey decomposition (II.2.10) V I (V") 0. Hence V is a permutation module by (3.4). Li ') 0
4]
LEMMA
ENDO-PERMUTATION MODULES FOR
p- GROUPS
407
3.6. Let V be a transitive permutation module. Then the following
hold.
(i) rankR (Inv G ( V)) = 1. (ii) V = W I W2 with W indecomposable such that Inv, (14/2) = (0) (0). and InvG (
e
PROOF. (i) This follows from (11.3.4). (ii) Immediate by (i). LI
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. VAN) is defined as in the previous section. An endo -permutation R[P] module is an R -free R[P] module V such that V*0 V Horn R ( V, V) is a permutation module. An endo -trivial R[P] module is an R -free R[P] module V such that V* 0 V — Hom R ( V, V) -- vo(P)ED U, where U is a projective R [P]
module. Since any projective R[P] module is a permutation module it follows that an endo-trivial R[P] module is an endo-permutation R [P] module. Clearly a permutation module for R [P] is an endo-permutation R[P]
module. These concepts were introduced by Dade [1978a], [1978b], who also classified the endo-permutation R [P] modules in case P is abelian. See also Carlson [1980a]. We will here only present basic elementary properties of endo-permutation R [P] modules and compare some of these with the corresponding 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 = 0(G) and Pis a S„-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 is an endo -permutation F[P] module.
408
CHAPTER IX
[4
The existence and uniqueness of W follows from (111.3.16). Under the action x —> yxy F[H] becomes a permutation module M for F[G]. Let e be the centrally primitive idempotent of F[H] corresponding to e. Then V*0 V eF[H]l MH . Since T(V)= G, eF[H].-----W W* as F[G] modules and W*0 W M. Thus 1,1/'; Wp Mp and so VI/Ø Wp is a permutation module for F[P] by (3.4). Ill PROOF.
,
LEMMA 4.2. Let 0—> V—> U—> W—>0 be an exact sequence of R -free R[P] modules with U projective. Then V is an endo-permutation (endotrivial) module if and only if W is an endo -permutation (endo -trivial) module. PROOF.
Immediate by (111.5.12). 111
It follows from (4.2) that for instance every factor in a projective resolution of Vo(P) is an endo-permutation module. In this way it can be shown that if P is neither cyclic nor a generalized 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 4 V*. This is in contrast to (3.1) and (3.2)(i), (ii). Do there exist only finitely many isomorphism classes of endo -permutation V*?
R[P] modules with V
This is an open question. It follows from Dade's results that the answer is affirmative if P is abelian. LEMMA 4.3. Let V be an endo -permutation R[P] module such that V" (V")* for some positive integer n. Then V is algebraic. PROOF. As V2" ----- V" Ø ( V" )* is a permutation module the result follows from (3.2)(ii). CI
LEMMA
4.4. If V, W are endo-permutation R[P] modules then so are V*,
V® W and Hom R (V, W) V'® W. PROOF.
Clear by definition.
LEMMA
4.5. Let V be an endo -permutation R[P] module. If W V then W
CI
is an endo -permutation module. PROOF.
Since W*0 W V* 0 V the result follows from (3.4). (11
4]
ENDO-PERMUTATION MODULES FOR p -GROUPS
409
LEMMA 4.6. Let V be an endo-permutation R[P] module and let Q be a subgroup of P. Then V0 is an endo-permutation R[Q] module. PROOF.
Clear by definition. Li
In general the direct sum of endo-permutation modules need not be an endo-permutation module in contrast to (3.3)(i). Two endo-permutation R[P] modules V, W are compatible if V 83, w is an endo-permutation R[P] module.
LEMMA 4.7. Let V, W be endo-permutation R[P] modules. The following are equivalent. (i) V and W are compatible.
(ii) V* 0 W Hom, (V, W) is a permutation module. (iii) W*0 V----- Hom, ( W, V) is a permutation module.
PROOF. Since (V*0 W)* W*0 V, (ii) and (iii) are equivalent. Since (VED W)0(V*(1)W*) (VØ V*)(VØW*)E(WØ V*)(!)(W0 W*), the equivalence of (i) with (ii) and (iii) now follows from (3.3)(ii). D
LEMMA 4.8. Let V be an indecomposable endo -permutation R[P] module with vertex P. Then V0(P)1 V* 0 V.
PROOF. Let
be the set of all subgroups Q of P with Q# P. By (111.4.9) 1- °(P, , V* 0 V)# (0). Thus there exists an indecomposable component W of V*0 V with 1-1n (P„S), W) (0). Since V is an endo-permutation module, W Vo(Q) for some subgroup Q of P. By (11.3.4) 1-1(P, Q, W)= (0). Hence W)= (0) if Q Thus Q0 and so Q = P. D The next result indicates that compatibility of two endo-permutation modules is a rare phenomenon.
LEMMA 4.9. Let V, W be indecomposable endo -permutation R[P] modules 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 be the set of all subgroups Q of P with QX P. There exist nonnegative integers ao, 1)0 for Q C P such that
and W are coMpatible. Let
410
[4
CHAPTER IX
V*
W= apV0(P) oEv
v.— bpv0(p)
e b.( 0E„,
vom).
Thus apv
e OE ,
ao (Vor V
V*0 W bpW
e b0(W0)G. OE
Hence any indecomposable component of V® V*0 W with vertex P is isomorphic to V. By (4.8) ap #0 and so V --- W. E In general endo-permutation modules do not behave well with respect to induction. The next result is concerned with this situation. LEMMA 4.10. Let Q be a subgroup of P and let V be an endo -permutation R[Q] module. The following are equivalent. (i) V P is an endo-permutation R[P] module. (ii) The endo-permutation R[Qx fl Q] modules Vo.no and Vb. compatible for all x e P.
(70
are
PROOF. The Mackey tensor product theorem (11.2.10) implies that V P 0 (V P )* is a permutation module if and only if ( Vô-no Vt,n o) P is a permutation module for all x G P. Thus (i) is equivalent to (ii) by (3.5) and (4.7). 0
CHAPTER
X
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 [1960], [1961], [1962]. 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 N„ 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 [1978c]. LEMMA 1.1. Let H G by f ((x, z))= x. Thus there exists an exact sequence (1)—}N,—> d(F1 )i> G —> (1). Since N, consists of scalars, it follows that N, is a cyclic group in the center of G and IN, I 4- (1)1 H 1. Let Z = N, and let If = {(y, A (y )) I y E HI. Thus H i 6(F1 and f(1:-/)= H. Furthermore Z n = 0,1)). Thus ZI:-/ = Z x For v E V', (x, z)E (F i) define v (x, z)= vz. Let 't-/' denote the [6(F1 )] module constructed this way. For F, =F' let V, = V'. For F = F' let V = V'. Then V — V1 0F, F. If y E H then v(y, A (y)) = vy. Hence f(VH ).--= V. Let F", F(P) be algebraically closed fields of characteristic 0,p respectively where p I H . To complete the proof of (i) it suffices to show that O(F")— Let K be the extension of Q generated by a primitive I H12 root of 1. Thus (K)--, (Fr). Let R be the ring of integers in K and let fz be the residue class field. Then (R)= 6(F?'). Let X be an R -free R[6(K)] module such that XK \-/ - 1 . Since p !HI, )?E-1 is irreducible. From this it follows easily that (K)— 6(.1). This completes the proof of (i). (ii) Let W be an irreducible F[G] module such that V is a constituent of WH. Thus W is an F[ O] module where w (x, z)= wx for w E W. Define the vector space W = HomF[H, ( V, W). )
own.
[1
CHAPTER X
414
Since V may be identified with V- it follows that if v E V then (x, z)= vz. By Clifford's Theorem (111.2.12), Wt., nV for some integer n. Thus the linear map g: 1-7 —> W defined by g(v 0h)= vh is an isomorphism. (x, z) EÔ define h(x, z) by v[h (x, z)} = [(vz For h e }x for vEV.Thenfor vEV,hEW {g(v
(x, z)= (vh )(x, z)= (vh)x
and so g{(v h)(x,
= g{v(x, z)
h(x, z» = evz Oh (x, z
= (vz){h(x, z)} = (vh)x = {g(v
h)}(x, z).
Since 1-7 and W are F[G] modules, this implies that W is an F[6] module and g is an F[6]-isomorphism. (y, A (y))) -= {(vy - ')h}y = If (y,A(y))EI then for v E 1Ç h e viz. Thus W has H in its kernel and so W is an F[6/1I] module. W is irreducible as W 1-7 W is irreducible. (iii) For any module X let p, denote the Brauer character afforded by X. Then 13w = Ovf3“, by (ii). Suppose that W and Vi» both have H in their kernel and 043* = If 0“,/ Ow there exists x E G such that f3v (xy) = 0 for all y EH. Therefore A (x )A (y ) has trace 0 for all y E H. Since {A (y )13/ E H} spans the full matrix ring this implies that A (x) = 0 which is impossible. Thus is uniquely determined by ow. Hence Ow -> Piiv defines a one to one mapping from 3 (F) to L (F). It only remains to show that if U is an irreducible F[6/1--11 module such that Z is in the kernel of 17 U then 17 U is irreducible. Suppose that W1 ..... are the composition factors of 17 0 U. Then V is a constituent of ( W,), for all i. By (ii) W, = where each W, is irreducible. This implies that 18 you = PV ' I OW, and so Pi, = u i f3 y Hence m = 1 by (IV.3.4) and so 1-7 U is irreducible. 0 The group 6/1---/ defined in (1.1) is called the representation group of the character
1.2 (Fong [1961 1 ). Let p be a prime. Let H< G with p HI. Let B (G) be a block of G and let B (H) be a block of H which is covered by B(G). Let 4- be an irreducible character of H in B(H) and let Ô,II Z be defined as in (1.1) corresponding to the group T(4'). Let G(4') be the THEOREM
1]
GROUPS WITH A NORMAL p' -SUBGROUP
415
representation group of 4'. Then there exists a block B of G(0 such that the following conditions are satisfied. (i) There exists a block B (T()) of T(4) such that if 0 is an irreducible character or an irreducible Brauer character in B () then 0 = Oô for some irreducible character or irreducible Brauer character respectively in B (T( ) ). (ii) Let n -i) denote the one to one mapping defined in (1.1) for char F = O. Titus 0 —> Ô defines a one to one mapping from the set of all irreducible characters in B (G) onto the set of irreducible characters in B. This mapping preserves heights and sends the set of all p -rational irreducible characters in B (G) onto the set of all p -rational irreducible charactersin B. (iii) Let n —> 7) denote the one to one mapping defined in (1.1) for char F = p. Thus 0 —> 0, defines a one to one mapping from the set of all irreducible Brauer characters in B(G) onto the set of all irreducible Brauer characters in B which' preserves heights. (iv) With respect to the one to one mappings defined in (ii) and (iii), B(G) and B have the same decomposition matrix and the same Cartan matrix. (y) B and B(G) have isomorphic defect groups.
PROOF. Since p I H 4- is the unique irreducible character in B (H) and = is irreducible as a Brauer character. Furthermore T( ) = T (B (H)). A field automorphism which preserves all p'th roots of unity preserves blocks. Thus by (V.2.5) it suffices to prove the result in case G = T ( ) since p -conjugate characters of T(4') induce p -conjugate characters in G. Let 3, = 1 (F), A o = zi(F) be defined as in (1.1) where char F = O. Let 3„ = 3 (F), 4, = (F) be defined as in (1.1) where char F = p. Let V afford ep in case of char F = 0, p respectively. Thus 0 (x)= (x) for p'-elements x in O. Furthermore eo is p -rational by (1.1)(ii). Hence 0= 6,6
if 0 E Ao,
0= ei,
if 0 E dp.
-6
(1.3)
If 0 E 2io then 0 is p -rational if and only if Ô is p-rational as 4,- 0 is p-rational and the map 0 —> O is one to one. If 0 is an irreducible character or an irreducible Brauer character in B(G) then by (V.2.5)(iv) 0 E do , 3, respectively. Let S=
I 0 is in B(G), 0 E 210 U
If ,y e 0, (p e Lip then by (1.3), d(X, (P) = 45) where d denotes the appropriate decomposition number. This and (IV.4.2) imply that S consists of all the irreducible characters and irreducible Brauer characters in a set
CHAPTER
416
X
[1
of blocks 13 1 , . . B„, of G(). Since the decomposition matrix of B (G) is indecomposable by (1.17.9) it follows that m = 1. Let B = B 1 . Let p° be the order of a Sr-group of G. Then p° is the order of a Sr -group of G(0. Let d be the defect of B(G). Since al) 0 (mod p), (1.3) and (IV.4.5) imply that d is the defect of B. Thus (1.3) implies that if 0 e Ao u A, and 0 is in B(G) then 0 and 6- have the same height. Let Y be an irreducible 1 [G] module in B (G) which affords a Brauer character of height O. Thus a defect group D of B (G) is a vertex of Y. As Yaffords a Brauer character of height 0, a defect group /5 of B is a vertex of Y. As Y = Y it follows that Y is R [15]-projective. Hence D is conjugate to a subgroup of 15 in G. As IDI=I 15I = p ° this implies that D=13. 0 Let G, H = 0(G), be as in (1.2). Thus a Sr-group of G ( ) has order at most that of a Sr -group of G. Furthermore if G is p-solvable and H = 0(T()) then either p I G()1 or (G()) = Z x P for some p-group P/ (1) since Z = 0(G(0) is in the center of G ( ). These two facts will make it possible to prove some results about p-solvable groups by induction on the order of a Sr -group.
LEMMA 1.4 (Fong [19621). Let p be a prime and let G be a p-solvable group. be an irreducible character of H such that Let H = 0(G) and let H = 0(T()). Let p° be the order of a Se -group of the representation group Then every block of G ( ) has defect a. GG) of
PROOF. If a = 0 the result is trivial. Suppose that a / O. Let P = Or (G(‘)). Thus P/ (1). As Z =0(G()) is in the center of G ( ') it follows that C G(c) (P)= P X Z. By (IV.4.17) G() has exactly I Z I blocks of defect a. By (V.3.6) and (V.3.10) G(4') has exactly I Z I blocks. 0 The following result is somewhat simpler and appears to have been known for some time though it first appeared explicitly in Fong and Gaschiitz [1961] as did (1.6), both for solvable groups.
THEOREM 1.5. Let p be a prime and let G be a p- solvable group. (i) The principal block is the only block of G if and only if 0(G)= (1).
(ii) An R [G] module V is in the principal block of G if and only if 0(G) is in the kernel of V.
PROOF. Let B be the principal block of G. By (IV.4.12) 0,,(G) is in the
1]
GROUPS WITH A NORMAL
p ' -SU BGROUP
417
kernel of B. Thus (ii) follows from (i). It only remains to show that if Op (G)= (1) then B is the unique block of G. Since 0„ , (G) = (1) it follows that CG (Op (G)) C Op (G). The result now follows from (V.3.11). Lii
As an immediate Corollary of (1.5) one gets the following result which like (1.5), is false for groups in general. COROLLARY 1.6. Let p be a prime and let G be a p-solvable group. Let V, and V, be R[G] modules in the principal block of G. Then every direct summand of V, 0 V2 is in the principal block of G.
THEOREM 1.7 (Ricfien [1972]). Let p be a prime. The following statements are equivalent. (i) Every irreducible character of G is irreducible as a Brauer character in characteristic p. (ii) G = O(G)P, where P is an abelian p-group.
PROOF. Let H = Op, (G). (i) (ii). By (1.17.9) each block contains a unique irreducible Brauer character. Thus G has a normal p -complement by (IV.4.12) and so G = HP for P a Sr -group of G. Since the hypotheses are satisfied by P G I H it follows that every irreducible character of P has degree 1 and so P is abelian. (ii) (i). Let O be an irreducible character of G and let be an irreducible constituent of OH. Let Ô be defined as in (1.1). By (1.2) it suffices to prove that O is irreducible as a Brauer character. Since G()1 Z P for a central p'-group Z in G ( ") it follows that G(4) -- Z X P is abelian and so every irreducible character of G(‘) is irreducible as a Brauer character. D See Osima [1942] for results related to (1.7). THEOREM 1.8 (Hamernik and Michler [1972]). Let p be a prime and let G be a p -solvable group. Let F be an algebraically closed field of characteristic p. Let W be an irreducible F[G] module. Let cp be the Brauer character afforded by W and let P be a vertex of W. III G Ip = p", I P = p" and cp (1) p = pc then a — b = c.
PROOF. Induction oni G :0p, (G)I . If G = O(G) then a = b = c =O. Thus it may be assumed that G/ Op , (G). By definition a — b c, thus it suffices to show that c a — b.
418
CHAPTER
X
[1
Suppose that Op (G)= (1). By (111.4.13) W is an F[GID] module with vertex PID. The result follows by induction. Suppose that Op (G) = (1). Thus H = 0(G)/ (1). Let V be an irreducible F[H] module with V I WH. By induction and (V.2.5) it may be assumed that T(V)= G. By (1.1)(ii) W 17 03) W for an irreducible F[6/171] module 14-/ and dim, = dim, V 0 (mod p). Since (6/1-7/)/ (1), induction and the previous paragraph applied to G- 47/ implies that pc = (dimFW), = (dimF
=13'0,
where J0I = p and Po is a vertex of W. Let X be a source of W. By Frobenius reciprocity (111.2.5) W 17 Ø X' ----- (171o Ø X ) . Hence W is F[Pd-projective and so b b 0 . Thus c = a — b0 a — b. E For a related result see Cliff [1979]. It follows from (1.8) that question (II) of Chapter IV, section 5 has an affirmative answer for p-solvable groups. However the next section contains stronger results in this direction. For groups in general (1.8) is false. For instance if G = J1, the smallest Janko group and p = 2 then I G1 2 = 8 but G has a nonprojective irreducible Brauer character of degree 56. See Fong [1974]. As another example let G = 5 L2(4) and p = 2. Let W be the natural 2-dimensional representation of G in characteristic 2 and let P be a S2-group of G. Since W, is faithful and P is abelian, WF is not induced from a representation of a proper subgroup. Thus P is a vertex of W by (111.3.8). The statement of (1.8) asserts that if G is p-solvable then the vertex of an absolutely irreducible F[G] module is as small as possible. If R is the ring of integers in a p-adic number field K then the analogous statement for R[G] modules W with WK absolutely irreducible is not true. In answer to a question raised in an earlier version of this material Cline [1971], [1973] suggested the following counterexamples. Let p be a prime. Let P be an extra special p -group of order p3 which admits a cyclic group E of automorphisms with IE J = p + 1 such that [E,Z(P)1= (1) and E acts transitively on the set of all subgroups of index p in P. Thus P is of exponent p if p/ 2 and P is the quaternion group if p = 2. Let G = PE. Then G has a faithful irreducible character x with x(1) = p. Let V be an R -free R[G] module which affords x. We will show that P is the vertex of V. Without loss of generality it may be assumed that R contains a primitive I th root of 1. V is absolutely indecomposable as x is irreducible. If the
2]
BRAUER CHARACTERS OF
p - SOLVABLE GROUPS
419
result is false then a vertex D of N has index p in P. Since VI( VD) 0 it follows from (111.3.8) that V, — 14/ P for an R[D] module W of rank 1. Then D is in the kernel of V, VV,. This is however impossible since P is the smallest normal subgroup of G which contains D.
2. Brauer characters of p-solvable groups
Let cp be an irreducible Brauer character of a p-solvable group G. Swan [1960] observed that it is a very simple consequence of Fong's work that there exists an irreducible character x of G such that x =cp as a Brauer character. This result is now known as the Fong—Swan Theorem. The following refinement is due to Isaacs [1974], though the proof is different from his and is essentially the same as Swan's proof of the Fong—Swan Theorem.
THEOREM 2.1. Let G be a p-solvable group. Let cp be an irreducible Brauer character of G. There exists a p-rational irreducible character x of G such that x = cp as a Brauer character.
PROOF. Induction on 1G :0(G)1. If G = (G) the result is clear. Let H = 0p , (G). Since p HI, ( — y2, v 1 ). It is easily seen that J2 = — I and f = 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
e
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 C 0(f). Assume that P acts irreducibly on V. If x E Z(P)— {1} then vx # for all v E V, V # O. PROOF.
Clear.
El
LEMMA 7.7. Let q be a prime and let Q be an extra-special q-group with = q2"+1 . Let Z = Z(Q) and let V = Q I Z. Let p be a prime distinct from q and let F be an algebraically closed field of characteristic p. (i) Let f (x, y)= [x, y]. Then f defines a nonde generate 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 (Q) denote the group of all outer automorphisms of Q and let A 0(Q) denote the subgroup consisting of all automorphisms which fix all the elements of Z. Then A o(Q)< A (Q) and A (Q)1A1AQ) --' Aut(Z) is cyclic of order q —1. If q 2 then A o(Q) Sp(f) Sp2,, (9). If q = 2 then Ao(Q)= 0(f)= 02„ (2) is an orthogonal group. In any case a subgroup H of Q is abelian if and only if the image of H in V is isotropic. (ii) Let P be a p-group with P C A 0(Q) and let G = QP be the semidirect product. If A is a linear characterof Z with A X 1, then (up to isomorphism) there exists a unique irreducible F[G] module X, such that A is a constituent of the character afforded by (X,),. Furthermore (X,) 0 is irreducible and every irreducible F[Q] module which does not have Z in its kernel is isomorphic to some (X, ) o. If H is a maximal abelian subgroup of Q then H = Ho x Z, I H o l = q" and '1.° =(X,) 0 where (hz)= (z) for h E Ho. (ii) For i =1,2 let Q, be extra special with 4 = Z(Q,). Then Z, Z 2 = Z. Let c E IC and let A I , A 2 be linear characters of Z with A, = A 2 '. Let Q=Q1> 0 and let G, = SL2(p - ). Each G,. has a faithful F-representation of degree 2 but a normal abelian subgroup of G,„ has order at most 2 while I G,,, 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) = f, (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'" be the order of a Se -group of G. Then G has a normal abelian subgroup A with IG 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* 0 V® V* 0 V with L in B(G) and L L,(G). (iii) Let P be a .50 -group of G. Let N = N G (P). There exists an irreducible constituent L of V* 0 V with dim F 0 (mod p) and L L,(G) such that if Ho is the kernel of 11H then 1H :H o t< J (n), where J (n) is defined by (1.1).
PROOF. Assume that G satisfies neither (i) nor (ii). Let L, = L,(G), , L, denote the distinct irreducible constituents of V*0 V. Let W be an F[G] module, all of whose irreducible constituents are constituents of V* 0 V 0 V* 0 V. We will show that the multiplicity of L,(G) in W is equal to dim, InvG W. It clearly may be assumed that W is indecomposable. If W is not in B(G) the result is trivial. Suppose that W
446
[1
CHAPTER XI
is in B,(G). Since (ii) is excluded, every composition factor of W is isomorphic to L,(G). If x is a p '-element in G then Woo 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 dim, W = 1 as W is indecomposable and the result is proved. This fact will be applied to several modules. Let W L, L, with 1i, j s. Thus Inv, W Hom olGi (L„ L) ). For i = j, Schur's Lemma implies that dim, Inv G W = 1. Thus by (111.2.2) and the previous paragraph dim, L, 0 (mod p) for 1 i s. For i # j, Schur's Lemma implies that Inv, W = (0). Since (ii) is excluded it follows that no irreducible constituent of W is in B I (G). Hence by (IV.4.14), L,, . . . , L, lie in s distinct p-blocks. Let Y be an indecomposable direct summand of V* 0 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 L, for some i. Let b be the length of a composition series of Y. Thus L I (G) occurs with multiplicity b in L* 0 Y. Since every irreducible constituent of L 0 Y is a constituent of L L, and so of V* 0 V 0 V* 0 V, it follows that b = dimF Hom F[Gi(L„ Y). Let Yo be the socle of Y and let b,, be the length of a composition series of Yo . The previous argument applied to Yo shows that bo = dimF Floin,[G](L„ Yo). However Hom olGI (L„ Y) Hom,[G] (L,, Yo). Thus b = and so Y = Y o. As Y is indecomposable this implies that Y L,. Consequently V*0 V is completely reducible. Let
V*0 V --- ----
(IL,.
(1.4)
By Schur's Lemma a, = 1. Hence $y (111.2.2) dim, VO (mod p). Thus P is a vertex of V. Hence (111.5.7), (V.6.2) and (1.4) imply that
e S,
17* 0 17
(1.5)
where each indecomposable direct summand of S has a vertex properly contained in P. By (111.7.7) ii L(N) and I., is not in B(N) for i 2. By (111.3.7) 17,, I (U, X,), where X, , X, are distinct irreducible F[H] modules and U,, . . . , U, are F[P] modules which are conjugate under the action of N. Thus
07 * 1-0. e =( 1' i,j
uJ)
I
—4631 (U® i=1
PO
x; )
AN ANALOGUE OF JORDAN'S THEOREM
447
where S' is an F[P X H] module which does not contain L,(P x H) as a constituent. By (1.5) L,(N)I 17* 0 1-/. Thus L,(P x H)1 ( 17* 0 f/)p)
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 1 G0 : A o h(T(G o, XG0)). Let A = nxec; x'A ox. Then G : A < g(m, n)h(T(G o, XG0))g ("'" ). This implies that 1G:A 1 0 has exactly m conjugates under the action of N = NG (P). The notation will be chosen so that {0,11 j r} is a complete set of . . .
representatives of the orbits of N. By definition a ( p;)= a(0,) for all j and x E N. 4.6. Suppose that (4.4) is satisfied. There exists an integer s m such that the principal p-block B o of G contains exactly r + s irreducible characters x, =1 G,...,xs . If y is the p-part of z in G and y E P — {1 } then THEOREM
6(z)- d + 5E q,;(y ) 1j 1
u
where x ranges over a cross section of and au O. Moreover
(d — 5) 2 + (r — 1)d 2 +
CG
r, s, (P) in N; 5 = -1- 1; d, a 1 , . . . , a E Z
2
a „ = m + 1.
(4.7)
If furthermore t is a p'-element then ‘",(t)— Mt) for all j and (rd — 5)‘, (t) +
ax (t) = 0.
(4.8)
PROOF. Let b, = a(0( — ) = a(ti/o)— a(e/r, ). By (4.3) and (4.5) (b„ b,) — Let c, = b, — b,. Then (c„ c, ) = 1 + m+ The result now follows from a standard argument in character theory. See e.g. Feit [1967c]. El
CHAPTER XII
462
[5
Observe that in case P is cyclic (4.6) is a very special case of the results in Chapter VII.
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, a, = 5 and 6 (1) = x2(1) + 8. (ii) If m = 3 then the notation may be chosen so that in (4.6) d = 0, s = 3 and a,, = ±1 for all u. If furthermore G contains an involution which is in no proper normal subgroup then exactly two of 8, a,, a, are equal to —1 and (1)x 2(1)x,(1) is the square of a rational integer. COROLLARY
PROOF. d1 by (4.7) as m 3. Hence either d = 5 or d = 0 by (4.7). Suppose that d = 5. If s = 1 then (4.8) implies that (r —1)56(1) = — 1 and so 6 (1) — 1 which is not the case. Thus s 2. As m it follows that r 2. Hence (4.7) implies that 2 r m + 2 — s m. Thus r = 2 if m — 2. If in = 3 then 2 r 3 and so I P I = 5 or 7. Thus P = 7 as in I — 1. Therefore r = 2 in any case. Hence if j' j then (4.6) implies that
6.(z)= 8 ( 1
E (Kcy))= —SE
(y)= —6(z).
Thus the notation may be chosen so that d = 0 in all cases. (i) By (4.7) s = 2 and a2 = -± 1. Thus (4.8) implies that — (1) + 1 + a2x2(1) = O. Hence (1) = 8a2x2(1) + 6 and so 6 a 2 >0. (ii) By (4.7) s = 3 and a = -1- 1 for all u. Suppose that G contains an involution which is in no proper normal subgroup. Let 0 = {(00 — (/J1) N }p. By (3.7) exactly two of 5, a2, a, are equal to — 1. Let V be a 4-dimensional vector space over Q with diagonal quadratic form a2
ai
X2( 1 )
X1( 1 )
x-2,, ±
8 6(1)
x2 4
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 6,(1)x2(1)x3(1) is a square in Q. El The situation described in (4.10) occurs infinitely often. For instance, let G,(q)= PSL 3(q) and G(q)= PS11 7(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 k = (q 2 + eq + 1, 3). The degrees of the given characters are q3, q 2 + eq and (q 2 — 1)(q — E).
5]
GROUPS WITI I AN ABELIAN S,-GROI. PP OF TYPE (2,2)
463
Another example comes from A, for p = 7. The corresponding degrees are 6, 10, and 15.
5. Groups with an abelian S2 -group of type (2'", )
This section contains a proof of the following result. See Brauer [1964b] Section VI.
THEOREM 5.1. Let G be a group whose S 2-group Pis abelian of type (2- , ) with r n 2. If 0 2, (G)= (1) then P < G, CG (P)= Pand IG : P I is 1 or 3. PROOF. The proof is by induction on 'GI. Let G be a counterexample of minimum order. Then 02, (G)= (1). Let C = C G (P), N = N G (P). Let t,, t 2, I- , be the involutions in P. If N = C then Burnside's transfer theorem implies the result. Since N/C acts faithfully on P by conjugatioh it follows that IN:Cl= 3 and N/C permutes It, t 2, t3) cyclically. If y E P - {1} then CG (y) has a normal 2-complement by induction. Thus (IV.4.12) implies that (C G (y)) = (P)=
for y
I = 22"
E P - {1}.
(5.2)
Let 00 = 1, 0,, ... be all the irreducible characters of P. Let r = i(22"' - 1) 5. Then the notation may be chosen so that tifo,. .., 0, is a complete set of representatives of the orbits of N on NO. Let U = P - {1} in (3.6) then w = 3/IN : PI . Hence (3.6) implies that
(a(tif, -
a(0,)) = 8„
(a(
a(0 ( )) = —3 for 1
—
for 1
i, j i
Define b, = a(1,G, - 00) = a0,1A a( 0() for 1 (b„ b, ) = 3 + 5,, 1i, j r. Let c•
b, - b, for 1 (c„ c,)= 1 + 5,,
r,
r. • r. Then (5.3)
i r - 1. Then
for 1
i, j
r - 1.
A standard argument (see e.g. Feit [1967c], §23) implies that the nonzero coefficients of the r -1 columns c, appear in r rows and if the rows are suitably arranged, the matrix in the first (r -1) of the rows is E/ with E = ±1 while all coefficients in the rth row are - E. By (5.3) (b„ c,) = - 1. Thus b, has the same coefficient b in each of the first r - 1 rows and the coefficient b + E in the rth row. There occur further
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464
[5
rows in which all the coefficients of c1 ,. c,_, vanish. If the coefficients of b, in these rows are 80, 6 . .. then (5.3) implies that
E 5; + (r — 1)b 2 + (b +
4=
(5.4)
.
For the row corresponding to xo= 1 we have ao(00)— 1, a 0(0,)= 0 for i > 1. Hence the coefficient of b, in this row is — 1, while the coefficient of each c, vanishes. This shows that one 8, has the value — 1. Thus b = 0 in (5.4) as r — 1 4. Thus if the rows are arranged suitably the nonzero coefficients of the columns bi , b2, , b, appear as follows (),
• • • ,
30
5 8 2,
• • • 7
52
3 0, 6
27
(5.5)
where 80 = — 1, xo = 1 and x, denotes the irreducible character corresponding to the (i + 1)st row. In particular a, (0,) — a, NO = 5, for i = 0,1,2 and j = 1, . . . , r. Hence a, (to— a, (0 )) = a, for every nonprincipal irreducible character of P and i = 0,1, 2. It follows from (3.5), the definition of a, (0, that if y E P then
for O j 2 a, (0)0(Y).
X, (Y) = Thus if y
EP - 111, 00=
O j
2 then
E
la, (0) — ai (00)10(y) = 8 ; xipo tlf (y)= — 8, .
(5.6)
Since ((x0p, 1) is an integer this implies that X, ( 1)
—8, (mod 22"),
i = 1,2.
(5.7)
Choose 01 as a nonprincipal character of P with (14= 00 . Then all elements of order less than 2' belong to the kernel of 0, and so
ai (01 00) = T137 yEP
(Y)(01(Y) 00(Y))
.E, (Y)(01(Y) — 040), = pi ,E
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BLOCKS WITH SPECIAL DEFECT GROUPS
465
where S is the set of all elements in P of order 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 t,. Hence (3.7) may be applied with U = S and B = — tifo . By (5.5) and (5.6) with y = t = —
1+
a, + 32 + EX302 ( — 0, x1(1) x2(1) x 3 (1)
—1 + 8,x,(1)+ 82x 2(1)+ Ex3(1)= 0, —3+ ex,(t)
O.
(5.8) (5.9) (5.10)
Suppose that x,(1) 1 and x 2 (1) 1. By (5.7) x,(1) 15 and x2(1) 15. Hence (5.8) and (5.10) imply that 9e /x3 (1) 13/15. Thus E = 1 and X3(l) 10. By (5.9) x 3 (1) =- 3 (mod 16) and so x3(1) = 3 contrary to (5.8). Hence x, (1) = 1 for i = 1 or 2. By (5.7) 8, = — 1 and so P is in the kernel H of x, by (5.6). Thus H is a proper normal subgroup of G and 02, (H) C 02, (G) = (1) and so by induction P < H. Hence P < G. The remaining statements are immediate as 02, (G) = (1).
6. Blocks with special 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 [1964b] 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 [19644 [1966a]; Landrock [1976]; Erdmann [1977b]. For the general case of D dihedral see Brauer [1971a], [19744 Erdmann and Michler [1977]; Donovan [1979]. For quasi-dihedral,
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[6
quaternion and similar 2-groups see Brauer [1966a]; Olsson [1975], [1977]; Erdmann [1979b]; Kiilshammer [1980]. Olsson [1977] also has some results for p 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 [1980a], [1980b]. For the case of ID = 8 and B arbitrary see Landrock [1981].
In this section we will only prove some very elementary results which will be needed for the next section. See Brauer [1964b].
LEMMA 6.1. Let G have a S 2- 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 B o of G are the four characters of degree 1 of G 102,(G) .--- P. In the latter case the principal 2-block B o contains exactly four irreducible characters xo =1, Xi, X27 XI. For u = 1, 2, 3 there exists Eu = 1 such that x,, (y) = e,, for every 2-singular element y in G. Furthermore X.
(1)
-= E. (mod 4) for u = 1, 2, 3.
1 + e lx,(x )+ E 2x2 (x)+ E 3x 3(x) = 0
(6.2) (6.3)
for any 2'-element x in G.
PROOF. If NG (P)= CG (P) then G has a normal 2-complement by Burnside's transfer theorem and the result follows. Suppose that NG (P» CG (P). Then I NG (P): CG (P)I= 3 and G has only one conjugate class of involutions. Let t be an involution in G. Then CG (t) has a normal 2-complement and so the principal Brauer character is the unique irreducible Brauer character in the principal block of CG (t). By (V.6.2) xu (Y) = for all u. By (IV.6.2) Z4-0 I 401 2 = c '00 = 4. Since x. does not vanish on all 2-singular elements 4 0 # 0 for all u. As 40 is rational for all u and doe = 1, it follows that B o contains exactly 4 irreducible characters. Furthermore E. = C v.o = ±1 for all u. As E yE px. (y) 0 (mod 4), (6.2) holds. Equation (6.3) follows from (IV.6.3)(ii). COROLLARY 6.4. Suppose that G has exactly one class of involutions in (6.1). Then the notation can be chosen so that e, = 1, e, = — 1. In that case 1, XI, E2X2 is a basic set for B o and the decomposition matrix and Cartan matrix with respect to this basis are as follows.
GROUPS WITH A QUATERNION S,-GROUP
7]
467
100 01
D
0 0 s2 1 1 1
21 1 1 2 1 = (1 + 3, ). 11 2
C =-
PROOF. Let x = 1 in (6.3) then the notation can be chosen so that El = 1 and E3 = 1. By (6.3) 1, x i , E2x2 is a basic set and D has the required form. Thus C = D'D as required. CI 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 =
( 1 0 0 1 1 1
0 1 0 0 0 1
1 0 0 1 1 1 D— ( 1 1 0 ' 1 0 1
C=
21 1 1 2 i). 11 2
(6.5)
C =
4 2 2 2 2 1). 2 1 2
(6.6)
Both of these cases occur infinitely often. A direct computation shows that if q is a prime power and G = PSL2(q) then (6.5) occurs in case q =. 3 (mod 8) and (6.6) occurs in case q 5 (mod 8). If q 3E (mod 8) with E = ± 1 then
xo(1)= 1, xi(1)q, x2(1)= xi(1)=i(q — c). If cpo,
17 ÇO2
are the irreducible Brauer characters in Bo then cp 0(1)=1 and
cp,(1) = 402(1) =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 S2-group THEOREM 7.1. Suppose that 02, (G)= (1) and a S 2-group T of G is a (generalized) quaternion group. Then the center of G has order 2.
468
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Brauer and Suzuki [1959] 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 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 [1974] has given a proof in case I T I = 8 which uses only ordinary characters. The proof given here is that in Brauer [1964b]. It uses results proved earlier in this chapter. See also Dade [1971b] for the results in this section and the next. PROOF OF (7.1). The proof is by induction on T = (y, z y 2 " = 1,
I GI. Let I T I = 2 —I with
= z 2, z'yz
Let t = y 2' 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 S2-group T 0 of H contains only one involution and so is either cyclic or a quaternion group. Furthermore Oz (H) C 02, (G)= (1). If To is cyclic then H has a normal 2-complement and so To < H. Hence To < G and so (t)< G. Thus t E Z, where Z is the center of G. If To is a quaternion group then (t)< H by induction and so t is the unique involution in H. Hence (t) < G and so t E Z. Hence t E Z in any case. Since 02,(G) = (1) and (t) is the center of T it follows that (t)= Z. Thus it suffices to show that t is contained in a proper normal subgroup of G. Assume first that I Tr > 8 and so n >2. Let N = T and P = (y). Let U be the set of all elements of order 2" or 2' >2 in P. We will show that the assumptions of (3.6) are satisfied with w = 1. If s E U then C G (S cannot contain a S2-group of G since the center of T has order 2. Thus P is a S2-group of CG (s) and since P is cyclic CG (s) has a normal 2-complement. Hence À (C G (s))= 2" =ICN (s)I and w = 1. If s and s, are conjugate in G with s, s, E U then s, = sk for some odd integer k. Hence if x I sx = s, then the 2'-factor of x commutes with s and so it may be assumed that x is a 2-element. Thus (x, s) is a 2-group and so is a subgroup of a quaternion group of order 2". Thus s, = s' and (3.6)(i) is )
satisfied. Let tlf denote a linear character of P with /'(y) = V — i. Let 0 = — 1. Then 0 vanishes on P — U. Hence (3.6) implies that
(a(0), a(0)) =
—1, ql — 1 +
lif z —
1) = 3.
Since a(0) is a column with integral coefficients this shows that there are exactly 3 nonzero coefficients. No 2-element of G of order at least 4 is inverted by any involution since an involution is in the center of any
GROUPS WITH A QUATERNION S 2 -GROUP
7]
469
2-group which contains it. Hence (3.7) implies that (t) is contained in a proper normal subgroup of G. Thus it may be assumed that n = 2 and T is the ordinary quaternion group. The conjugate classes of T are represented by 1, t, y, z, yz. If no two distinct elements in the above list are conjugate then G has a normal 2-complement by a theorem of Frobenius and the result is trivial. Suppose that two of these elements are conjugate. Say x'yx --= z. Thus x 'N G (())))x = NQ ((z)). Since T is a S2-group of NQ ((y)) and of NG ((z)) it may be assumed that x E NG (T). Hence x induces an automorphism of odd order of T and so x'zx = (yz). Thus G has only one conjugate class of elements of order 4 as each of y, z, yz is conjugate to its inverse in T. Furthermore any two elements of order 4 are conjugate in CG (t). The group CG (y) has a cyclic S 2-group of order 4 and so has a normal 2-complement. Thus the principal Brauer character ‘N is the only irreducible Brauer character in the principal block of CG (y) by (IV.4.12). Let x (y ), bô be the columns (x. (y)), (dY“ () respectively. By (IV.6.1) and (V.6.2)
(7.2)
X(Y)—
By the second main theorem on blocks (IV.6.1) (bÔ, bô) = 4,
(bô, x (1)) = 0,
(7.3)
where x(1) is the column (x.(1)). 2) is a S2-group of H with )72 2 2 = 1. Let II = CG (t)/(t). Then 1" = Furthermore fl contains only one class of involutions. Choose a basic set 02 for the principal block of it as in (6.4). Thus the Cartan tifo = 1, invariants are 1 + 6,. Then thr di 2 is also a basic set for the principal block of CG (t) but the Cartan invariants are twice those for FL Let {d } be the set of higher decomposition numbers for this basic set and let b; be the column (4). Thus in particular 2 (7.4) (t) = (1).
E
Furthermore (b;,
= c;, = 2(1 + 80,
(x(1), b ,i )
(b6, to) = 0.
For i = 0, 1,2 0,(1) is odd. If
)--=.
at) (mod 2). Thus x. (y )=-
(7.5) (7.6)
is an irreducible character of T then (t) (mod 2) for all u. By (7.2) and (7.4)
CHAPTER XII
470
b,; + b`o + b; +
0
(mod 2).
Define 2D, = b(Y) + 2D2 =
b; — b; +
2D3 = b(); + + b; —
Then D,, D2 , D, are columns with integral entries. By (7.3) (7.5) and (7.6) (x(1), D, ) = 0, (b(Y), D, ) = 2, (D,, D, ) = 1 + 25, .
Thus in particular each D, has 3 nonzero coefficients and these are -±- 1. A coefficient 1 occurs in the first row corresponding to xo = 1. Choose the notation so that the other two nonzero coefficients 5,, 52 of D i occur in the second and third row. It may also be assumed that the coefficients of tg in the first three rows are 1, 5,, 0 as (N, D,) = 2. If D, for j = 2,3 has a nonzero coefficient in the second or third row it follows from (D, i) = 1 that D I - t, only one nonzero coefficient contrary to (x(1), D, — Di ) = O. Hence the coefficients of D, and Di in the second and third row vanish. Thus (7.2) and (7.4) imply that .
— — b;1 3 =
1 5, 252
,
[bl,— b; + b'2 ],, = [bl,+ b; —
la = (
1) 51 0
—
where [ ] , means that we only look at the first three rows. Therefore Xi(t) = — 5 1 (1 +
0,(1)+ 02(l)), x 2(0= - 52(tp 1 (1)+
Hence 1+ 806(0— 8 2x 2 (t)= 0.
(7.7)
Since (x(1), D,) = 0 it follows that 1 + 5, x,(1) + 52x2(1) = 0.
(7.8)
No 2-singular element in G is the product of two involutions as t is the unique involution in T. Hence if s is a 2-singular element X.(1)
8]
THE Z * -THEORFM
471
as x„ ranges over all the irreducible characters of G. Thus by (IV.6.3) x„ (s)x„ (t)2 0
X. ( 1 ) for all 2-singular elements. Let n, = (vu ). Then n, is a linear combination of the columns (x„ (s)) with s E T—{1}. Thus (02 x»f3
X.
_ 0.
( 1 )
Therefore
1 + 3ix.(02+ 82x,(02= 0. X1( 1 ) X2(1) If (7.7) and (7.8) are substituted into (7.9) we see that 81{x1(1) — xl(t)} 2
0
X1(1){ 1 + 8 1x1( 1)} and so x1(1) = xi(t). Thus t is in the kernel H of x i . Since xi / 1 it follows that H/ G. D
8. The Z*-theorem If G is a group let Z(G) denote the center of G and let Z*(G) denote the inverse image of Z(G10 2,(G)) in G. If 0,(G)= (1) then of course Z*(G) = Z(G). THEOREM 8.1. Let t be an involution in G and let T be a S2- group of G which contains t. The following are equivalent. (i) x'txt has odd order for all x E G. (ii) No element in T — {t} is conjugate to t in G.
(iii) t E Z* (G). This result, which is known as the Z*-theorem, is due to Glauberman [1966a]. It is of vital importance for many results connected with the classification of finite simple groups. Glauberman [1966b] has also used it to derive some results about automorphism groups of simple groups. A generalization is due to Goldschmidt [1971]. If T is a quaternion group then (8.1)(ii) is obviously satisfied. Thus (7.1) is a direct consequence of (8.1). However (7.1) is needed for the proof of (8.1). We will follow Glauberman's proof quite closely. We begin with a well known result about involutions.
472
CHAPTER
XII
[8
By definition a dihedral group H of order 2 is defined for n >1 by H = (u, x 1 u 2 =x = 1, uxu =
Thus the noncyclic group of order 4 is the dihedral group of order 4. 8.2. Let u following hold. LEMMA
y be involutions in G. Let H = (u, y). Then the
(i) 1H :(uv)1= 2 and H is a dihedral group. (ii) A S 2-group of H either has order 2 or is a dihedral group. IH then u is conjugate to y in H. (iii) If 4 (iv) If 411H1 then 1H : H'1= 4 and u is not conjugate to v in H. (y) If 411H1 then either 1H1= 4 or 1Z(H)1= 2 and u,v are both not in Z(H). (i) This follows as u -I (uv)u = vu = (uv) -I . (ii) Let x = uv and let x have order mn where m is odd and n is a power of 2. Then (u, x'") is a S2-group of H. As ux 'flu = x - '" it follows that either x'" = 1 or (u, x '") is a dihedral group. (iii) As 4 A' I HI, (u) and (v) are S2-groups of H and so are conjugate in H. Since u, y is the unique element of (u), (y) respectively of order 2 u is PROOF.
conjugate to v. (iv) (uv)2 = uvu H'. As 4111/I it follows that : ((uv ) 2 )1 = 4. Since ((uv)2) O. Since u -I x k U = x -k the only power of x in the center of H is the involution x'u. 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.
(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 - ' = t. (ii) (i). Let x E G and let s = x -I tz. Suppose that st = x -I txt has even order. Thus 4 Ii (s, t) 1. By (8.2)(v) there exists an involution z s, t with z in the center of (s, t). Let D be a S2-group 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 t. Choose x E G with x -1 Dx C T. Then x -I tx E T and x -1 (y -I sy)x E T. Hence by assumption x -I tx = t = x -1 (y -1 sy )x and so t = y sy contrary to what has been shown above. 0
PROOF.
8]
THE T-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 tZ 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 I H where t H. LEMMA 8.5. If H < G then 02 ,(H)= (1). PROOF. The minimality of G implies that 02 (G)= (1). Hence 02 (H)C 02, ( G)= (1). E
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 - ' txt E D and x I txt has odd order. Hence x I txt = 1. E 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). D 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(ts o) and so E CG (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 S2-group 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 T. Thus Toi = 4 by (8.2)(v). Let so be a conjugate of s in (s, t) which lies in To . Then so z, t, so E CG (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 CG (x) that X(ts)= X(zY)=X(z)= x(rso).
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' = tx then t's' — x -1 (txsix -I )x. Thus it may be assumed that t' = t as x is a class function on G. By (8.8) there exist conjugates so, 4 of s in CG (t) such that x(ts)= x(ts 0), x(ts') = x(ts).
(8.10)
Let x -I sx = so. Then t,x -I tx are both in CG (S0). By assumption x -I txt has odd order and so by (8.2)(iii) there exists y E (t,x -I tx)C CG (So) with y ty = x 1 tx. Thus yx E CG (t). Therefore
(yx -I )-1 (ts0)(yx -1 ) = t(yx -1 )-1 s 0(yx -1 ) -= txs ox
= ts().
Hence x(ts o) = x(ts) and the result follows from (8.10). 0 LEMMA 8.11. Let s be an involution in T with s# t. Let x be an irreducible character in the principal 2-block of G with x # 1. If x(s)#O then x(t)= — x(1). PROOF. By (8.3) s is not conjugate to t in G. Let Co, C,,... be all the conjugate classes of G where Co = 111, t e C , s G C2. Let 0, = E„E c, x in the complex group algebra of G. Let 0,02 = Ea. Thus
IctlIcd=Eadcil.
(8.12)
Let x, E C, for all j. If co is the central character corresponding to x then w(C,) IC, I x(x,)/x(1). Thus
CJ x (t) 1c21x(s) x(i) x(1)
E n Icilx(x,) ,
x(i)
For x E G define a (x)= x (x)I x (1). By (8.9) x(x,) = x (ts) whenever a, # O.
Hence iCila(t)1C21a(s)=E ailCila(ts)=a(ts)E
CC I.
8]
THE V-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/ 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)/ O and x / 1. Then a(s) O and so a(t)2 = 1. Hence a (t)= ±1. Thus X(t) — ± x ( 1 ). Suppose that x(t)= x(1). Then t is in the kernel H of x. As x / 1, H G. Thus the minimality of 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 -I txt) 2 = 1. Thus x I txt = 1 as x -I txt has odd order. Consequently t E Z(G). Lii (8.1). By (8.7) T contains an involution s/ t. By (8.3) s is not conjugate to t in G. By (IV.6.3) PROOF OF
O=
E x(s)x(0= E
(s)x. (1)
where x„ ranges over all the irreducible characters in the principal 2-block of G. Thus (8.11) implies that if xo = 1 then 2 = xo(s)(Xo(t)+ 1) =
= This contradiction completes the proof.
0
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SUBJECT INDEX
completely primary, 33 component, 2 composition series, 3 equivalent composition series, 3 constituent, 3 irreducible constituent, 3 cover, 169, 249 cross section, 79
Algebra, 3 finitely generated, 3 free, 3 Frobenius, 49
R-algebra, 3 serial, 58 symmetric, 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, 124, 126, 127 defect group, 124, 126, 127 deficiency class, 246 descending chain conditicin, 5 dual basis, 47
BG is defined, 136 basic set, 148 basis, 1 block, 23 block pair, 207 extend, 207 properly extend, 207 weakly extend, 207 Brauer character, 142 ' Brauer correspondence, 136 Brauer graph, 300 Brauer homomorphism, 129 Brauer mapping, 129 Brauer tree, 301, 305
elementary subgroup, 141 exceptional character, 277, 279 extension, 69 finite extension, 69 unramified extension, 69 First Main Theorem on blocks, 137 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, 113 Grothendieck algebra, 92 group algebra, 4
500
SUBJECT INDEX
Hall-Higman Theorem B, 309 H-conjugate, 123 height, 151 Hensel's Lemma, 40 higher decomposition number, 172 with respect to a basic set, 451 Higman's Theorem, 89 idempotent, 1 centrally primitive, 1 primitive, 1 inertia group of a block, 195 of a character, 195 of a module, 86 inertial index, 235, 272 intertwining number, 53 invariant, 82 inverse endomorphism ring, 4 Jacobson radical, 18 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, 91 McKay conjecture, see Alperin-McKay metric complete, 31 complete on modules, 31 defined on a module, 29 equivalent metrics, 29 module absolutely indecomposable, 72 absolutely irreducible, 71 A-faithful, 17 algebraic, 93
Artinian, 6 B-projective, 11 compatible 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 5-3-pr o jectiv e, 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, 188 periodic, 96 permutation, 405 principal indecomposable, 42 principal series, 427 projective, 8 reducible, 2 regular, 2 relatively injective, 50 relatively projective, 11 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, (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, 177 P-defective, 124 p-radical group, 266
501
502
SUBJECT INDEX
p-rational character, 178 p-section, 172 7r-height, 227 7T-section, 216 primitive ideal, 17 principal block, 154 Brauer character, 154 character, 154 projective resolution, 95 pure submodule, 64, 65
Schur's Lemma, 23 Second Main Theorem on blocks, 172 section, 172 semi-simple, 19 socle, 16 source, 113 splitting field of an algebra, 53 of a module, 53 subsection, 230 with respect to a basic set, 451 symmetric element in Hom,(A,R), 49
quasi-regular, 18 radical of a module, 16 of a ring, 18 ramification index, 69 of a module, 101 real stem, 307 regular block, 199 representation, 74 equivalent representations, 74 representation algebra, 92 representation group of a character, 414 Reynold's ideal, 258
Schanuel's Lemma, 9 Schreier's Theorem, 3 Schur index, 185
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 L 2(p), 347 underlying module, 74 unique decomposition property, 37 vertex, 112 weakly regular block, 199 width, 252
