Induced Modules over Group Algebras (North-Holland Mathematics Studies)

INDUCED MODULES OVER GROUP ALGEBRAS NORTH-HOLLAND MATHEMATICS STUDIES 161 (Continuation of the Notas de Matematica) ...

Author: Gregory Karpilovsky

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INDUCED MODULES OVER GROUP ALGEBRAS

NORTH-HOLLAND MATHEMATICS STUDIES 161 (Continuation of the Notas de Matematica)

Editor: Leopoldo NACHBIN Centro Brasileiro de Pesquisas Fisicas Rio de Janeiro, Brazil and University of Rochester New York, U.S.A.

NORTH-HOLLAND -AMSTERDAM

NEW YORK

OXFORD TOKYO

INDUCED MODULES OVER GROUP ALGEBRAS

Gregory KARPILOVSKY Department of Mathematics California State University Chico, CA, U.S.A.

1990

NORTH-HOLLAND -AMSTERDAM

NEW YORK

OXFORD TOKYO

ELSEVIER SCIENCE PUBLISHERS B.V. Sara Burgerhartstraat 25 P.O. Box 211, 1000 AE Amsterdam, The Netherlands Distributors for the U.S.A. and Canada: ELSEVIER SCIENCE PUBLISHING COMPANY, INC. 655 Avenue of the Americas New York, N.Y. 10010, U.S.A.

Library of Congress Cataloging-in-Publication D a t a K a r p i l o v s k y . G r e g o r y , 1940Induced modules over group algebras Gregory Karpilovsky. p. cm. -- ( N o r t h - H o l l a n d m a t h e m a t i c s s t u d i e s , 161) Includes bibliographical references. ISBN 0-444-88414-9 ( U . S . ) 1 G r o u p a l g e b r a s . 2. M o d u l e s ( A l g e b r a ) I. T i t l e . 1 1 . S e r i e s . P A 1 7 1 . K 3 7 8 1990 512 .24--dC20 @9-49199 f

CIP

ISBN: 0 444 88414 9 ELSEVIER SCIENCE PUBLISHERS B.V., 1990 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 written permission of the publisher, Elsevier Science Publishers B.V./Physical Sciences and Engineering Division, P.0.Box 103, 1000 AC Amsterdam, The Netherlands. Special regulations for readers in the U.S.A. - This publication has been registered with the Copyright Clearance Center lnc. (CCC), Salem, Massachusetts. Information can be obtained from the CCC about conditions under which photocopies of parts of this publication may be made in the U.S.A. All other copyright questions, including photocopying outside of the U.S.A., should be referred to the publisher. No responsibility is assumed by the publisher for any injury and/or damage to persons or property as a matter of products liability, negligence or otherwise, or from any use or operation of any methods, products, instructions or ideas contained in the material herein. Printed in the Netherlands

To my wife Helen, who typed it all

This Page Intentionally Left Blank

vii

In 1898 Frobenius discovered a construction which, in present terminology, associates with every module of a subgroup the induced module of a group. This construction proved t o be of fundamental importance and is one of the basic tools in the entire theory of group representations. The present monograph is designed for research mathematicia,ns and advanced graduate students and gives a picture of the general theory of induced modules as it exists at present. Much of the material presented has heretofore been available only in research articles. Our approach is not intended to be encyclopedic, but each topic is considered in sufficient depth that the reader may obtain a c1ea.r idea of the major results in the area. A word about notation. As is customary, Theorem 5.4.2 denotes the second result of Section 4 of Chapter 5; however, for simplicity, all references to this result within Chapter 5 itself are designated as Theorem 4.2. The following is a brief description of the content of the book. After establishing algebraic preliminaries (Chapter l),the general facts about induced modules are provided (Chapter 2). Following presentation of some formal properties of induced modules, we examine in detail their annihilators and record a number of applications. Among other results, we prove some classical theorems of Mackey and tie together relative projectivity and induced modules. In Chapter 3 we provide a detailed information on the process of induction from normal subgroups. Among other results, we discover conditions under which induced modules are totally indecomposa.ble or absolutely indecomposable. Special attention is drawn to the study of crossed products over prime rings. The information obtained is then applied to establish a criterion for induced modules to be homogeneous. The chapter ends with the discussion of the Loewy length of induced modules.

viii

Preface

Let H be a subgroup of G, let F be a field and let V be an F H module. It is natural to investigate circumstances under which the induced module V G has projective indecomposable summands. An important contribution to this problem was marked by the appearance of the paper of Robinson (1989) to which Chapter 4 is devoted. Among other results, we prove a useful reciprocity theorem and demonstrate that if V is irreducible, then projective summands of V G (if there are any) have relatively large dimension. The aim of Chapter 5 is twofold: first to present some basic results of Green theory with refinements and extensions achieved in scope of recent developments, and second to provide a number of applications. The chapter ends with a result of Alperin (1986) which relates the Green correspondence with normal subgroups. In Chapter 6 we investigate in detail circumsta.nces under which restriction and induction of irreducible modules are completely reducible modules. As one of the applications, we show that if H is a subgroup of G and F is an arbitrary field of characteristic p > 0, then the following conditions are equivalent: (i) V Gis completely reducible for any irreducible FH-module V and MH is completely reducible for any irreducible FG-module M . (ii) There exists a normal suhgroup N of G such that N H and N has p'-index in G. Chapter 7 provides a detailed account of a distinguished class of induced modules, namely permutation modules. These modules hold much information about the p-modular representations of G, as well as information about the fusion of p-subgroups of G. The main results presented are due to Alperin (1988), Brouk and Robinson (1986), Robinson (1988), Dress (1975), Saksonov (1971) and Brouk (1985). Chapter 8, the final chapter, is based exclusively on an important work of Weiss (1988). A number of applications to the isomorphism problem for group rings is also presented. I would like to express my gratitude to my wonderful wife for the tremendous help and constant encouragement which she has given me in the preparation of this book (as well as in the previous 11 books).

ix

Contents Preface Chapter 1. 1. 2. 3. 4.

5. 6.

Preliminaries Notation and terminology Tensor products Artinian, noetherian and completely reducible modules The radical of modules and rings The Wedderburn- Artin theorem Group algebras and group representations

Chapter 2. 1. 2. 3. 4. 5. 6. 7. 8.

General properties of induced modules . Induced modules, representations and characters Formal properties of induced modules Annihilators of induced modules Clifford’s theorem Dual and contragredient modules Induction, restriction, and outer tensor products Mackey theorems and their applications Counting induced modules and characters 9. Relative trace maps 10. Induction and relative projectivity 11. Unique decompositions 12. Projective covers

Chapter 3. 1. 2. 3.

4. 5. 6. 7. 8. 9.

Induction from normal subgroups Complete noetherian local rings Reduction to G-invariant modules Group-graded algebras and crossed products The endomorphism ring of induced modules Relations between the decomposition of V G and E n d ~ cV( c ) Twisted group algebras over fields Total and absolute indecomposability of induced modules Crossed products over prime rings Homogeneity of induced modules

1 1 4 8 19 29 37

45 45 54 66 74 79 87 89 97 103 107 120 123 131 132 139 141 153 159 167 180 191 208

Contents

X

Frobenius and symmetric algebras Symmetric crossed products EndFc( V') is symmetric Graded modules Induction from irreducible modules and their projective covers Inflated modules over twisted group algebras Induction of absolutely irreducible modules Applications The Loewy length of induced modules

210 220 230 232

Chapter 4. 1. 2. 3.

Projective summands of induced modules The Reynolds ideal Projective summands Applications

273 273 280 295

Chapter 5. 1. 2. 3.

Green theory Vertices and sources The Green correspondence The endomorphism ring of the Green correspondents The Green correspondence and Brauer's induction theorem The Green correspondence and the Brauer lift The Green correspondence and normal subgroups

299 299 309

Chapter 6. 1. 2. 3. 4. 5.

Simple induction and restriction pairs Blocks of algebras Defect groups of blocks Blocks and vertices Simple induction and restriction pairs Complete reducibility of induced modules

34 1 34 1 350 358 365 373

Chapter 7. 1. 2. 3. 4. 5. 6. 7.

Permutation modules Preliminary results Hecke algebras Fusion and permutation modules Complete reducibility of ( 1 ~ ) ~ Induction from Sylow p-subgroups Loewy series for transitive permutation modules Characterizations of p-perrnutaion modules

383 383 392 403 409 416 426 433

10. 11. 12. 13. 14. 15. 16. 17. 18.

4. 5. 6.

238 242 256 26 1 266

318 329 334 337

xi

Contents 8. 9. 10.

Chapter 8. 1. 2. 3. 4. Bibliography Notation Index

The Brauer morphism Scott modules The Brauer morphism and p-permutation modules

438 448 45 1

Permutation lattices Generalized permutation lattices Permutation lattices and normal subgroups Some bimodule isomorphisms Applications

459 459 473 484 489

499 511 516

1

Chapter 1 Preliminaries Our aim here is to review some basic ring-theoretic results and to fix conventions and notations for the rest of the book. Since we presuppose a familiarity with various elementary ring-theoretic terms, only a brief description of them is presented. Many readers may wish to glance briefly at the contents of this chapter, referring back to the relevant sections when they are needed later.

1.

Notation and terminology

All rings in this book are associative with 1 # 0 and subrings of a ring R are assumed to have the same identity element as R. Each ring homomorphism will be assumed to preserve identity elements. Let R be a ring. An element x in R is called nilpotent if xn = 0 for some integer n 2 1. An ideal J in R is nil if every element of J is nilpotent, while J is nilpotent if there is an integer n 2 1 such that J" = 0, where J" is the product of J with itself n times. An element e in R is an idempotent if e2 = e. Two idempotents u,v of R are orthogonal, if uv = vu = 0. A nonzero idempotent is primitive if it cannot be written as a sum of two nonzero orthogonal idempotents. We denote by Z ( R )the centre of R. An idempotent e is called centrally primitive if e is a primitive idempotent of Z ( R ) . The ring of all n x n matrices over R is denoted M,(R). Let R be a ring. From this we can construct a new ring R", called the opposite ring of R. Both the underlying set and the additive structure

Preliminaries

2

of R" are just those of R. But the multiplication, denoted by o is given bY z ~ =y y z for all z,y E R It is easy to verify that R" is a ring with these operations. Clearly, Z ( R ) = Z(R")and R is commutative if and only if R = R". Assume that A is a ring, R a commutative ring, and p : R -+ Z ( A ) a ring homomorphism. The resulting system ( A ,R, p ) is called an Ralgebra. It will be convenient to suppress the p and speak of A as an R-algebra or as an algebra over R. Thus A is an R-algebra (with respect to some p ) if and only if there is an ideal I of R with R / I isomorphic to a subring of Z ( R ) . Setting r a = p ( r ) a , r E R, a E A, it follows that A is an R-algebra such that r(zy) = z(ry) = ( w ) y

for all

r E R,x,y E A

Conversely, if A is an R-module for which the above equalities hold, then the map R + Z ( A ) , r H r 1 is a ring homomorphism and hence A is an R-algebra. By a homomorphism of R-algebras, we understand a ring homomorphism which is also a homomorphism of R-modules. Let (&), i E I , be a family of rings and let R be the direct product set &rRj. We can define addition and multiplication on R by the rules

-

(G) t (Yi) = ( X i

t Y;)

(Z;>(Yi)

= (Wi)

(z;,y; E R;) . An immediate verification shows that R is a ring ; we shall refer to R as the direct product of the f a m i l y (Ri),i E I . We now record some conventions concerning modules. By an Rmodule, we always understand a unital R-module; the symbols RM or M R will be used to underline the fact that M is a left or right Rmodule, respectively. We usually consider left R-modules, and, in this case, speak simply about R-modules. A nonzero R-module V is indecomposable if 0 and V are the only direct summands of V . A nonzero R-module V is irreducible if 0 and V are the only submodules of V . Let V and W be R-modules. We denote by H o r n ~ ( VW , ) the additive abelian group of all R-homomorphisms V + W . The same notation will be used in case V and W are right R-modules. In case V = W ,

1. Notation and terminology

3

we write E n d ~ ( vinstead ) of Horn~(V, V ) . Note that E n d ~ ( vis) a

ring under the multiplication

Let V be an R-module. A submodule W of V is called maximal if W is a proper submodule (i.e.W # V ) and W is not contained in any proper submodule of V . An R-module V is said to have an R-basis (vi), i E I, if there exist elements o; E V such that each o E V can be written as a finite sum = Crp; with uniquely determined coefficients r; E R. Such modules V are called free R-modules. Given a pair of rings R, S, we say that V is an ( R ,S ) - b i m o d u f e if V is a left R-module and a right S-module, with the actions of R and S on V commuting : ( r v ) s = ?-(‘us)

(7.

E R, v E

v,s E S )

Unless explicitly stated otherwise, all groups are assumed to be multiplicative. The multiplicative group of a field F is denoted by F*. If S is a subset of G, < S > will denote the subgroup of G generated by S, while cG(s) and NG(s) is the centralizer and normalizer of S in G, respectively. As usual, Z(G) denotes the centre of G. Let H be a subgroup of a group G. A subset of G containing just one element from each left coset zH is called a left transversal for H in

G, and right transversals are defined correspondingly. Let g be an element of a finite group G and let p be a prime. We say that g is a p e l e m e n t if the order of g is a power of p ; we say that g is a p’-element (or g is p-regular) if its order is prime to p. Each g E G can be written in a unique way g = gpgpt, where g p is a pelement, gpt is a p’-element, and gP and gpt commute. The elements g, and gpt are called the p and p’-parts of g , respectively. A conjugacy class C of G is called p r e g u f a r (respectively, p-singular) if the order of the elements of C is not divisible by p (respectively, divisible by p ) . Let S be an arbitrary set. An action of G on S is a homomorphism

Preliminaries

4

8 of G into the permutation group of S. When the action is understood, we shall suppress the symbol 8 and instead write s H g s for O(g),g E G, s E S. The stabilizer G(s) of s E S is defined by G(s) = { g E G, 's = s} and the orbit of Gs of s E S is defined by G~ = {'slg

E G}

We say that G acts transitively if Gs = S for some (and hence all) s E S. One readily verifies that G(s) is a subgroup of G and that IGsl = (G : G(s))

Throughout the book, S, and A, denote symmetric and alternating groups of degree n, respectively, while Z n is a cyclic group of order n. 2.

Tensor products

Let R be a ring. Given a right R-module V and a left R-module W, the 23-module V @ R W , called the tensor product of V and W , is defined as follows. Let F be a free 23-module with V x W as a basis. Each element of F can be uniquely written in the form

c

Zij(Vi, Wj)

(Zjj

E

z,

vj

E

v,wj E W )

with finitely many zij distinct from 0. Let T be the subgroup of F generated by all elements of the form-

(21,

w1+ w2)- (v,w1) - (21, w2)

(v,r w ) - ( ' U f , 20) with v,v; E V and w,wi E W . Then V @R W is defined as the factor group FIT. The image of (v,w) under the natural homomorphism F + F / T is denoted by v 8 w. Observe that the Z-module V @ R W consists of all finite sums C vi @ w;with vi E V,wi E W . Assume that

2. Tensor products

5

V is an ( S ,R)-bimodule. Then V

@R

W can be regarded as a (left)

S-module by putting s(v@w) = s v @ w

( W E

v , w E w,s E S )

In particular, if V and W are modules over a commutative ring R, then V @ R W is an R-module. Let R be a ring, V,V' two right R-modules, W,W' two left Rmodules and 4 : V + V',$ : W -+ W' two R-linear maps.Then the map

$@$:V@RW+VI@RW' defined by

(4 0$)(W

€3

= +)

@ $(w)

is a 23-linear map. Moreover, if V,V' are ( S ,R)-bimodules and 4 is also S-linear, then $ 8 $ is S-linear. We next record a number of standard properties of tensor products (see Bourbaki (1974), (1972)).

2.1. Proposition. Let V,V',V" be right R-modules, let W,W',W'' be left R-modules and let

be exact sequences. (i) The sequences

w 3' V ~ wR'5 vN@R w o v@Rw'3v @ R w 'Yv 8 R w"+ o

v'@R

4

of 23 -homomorphisms are exact. (ii) The homomorphism g 0t ; V € 3 W ~ -+ V" @ R W" is surjective and its kernel is equal to I m ( f @ 1 ) + I m ( 1 0s ) . In particular, if V' and W' are submodules of V and W , respectively, and f and s are inclusion maps, then the map

(v/v')@R (w/w') ( v@ R w ) / ( I m ( f@ 1 ) + I m ( 1 €3 s))

(v

+ V ' )@ (wt W')

(W

€3 w)t (h(f 6 1)

+ I m p @ s))

Preliminaries

6

is a Z-isomorphism. 2.2. Proposition. Let ( K ) , i E I , be a family of right R-modules and (Wj),j E J a family of left R-modules. Then the map

is a Z-isomorphism. 2.3. Proposition. Let V be an ( S ,R)-bimodule. Then

as S-modules. 2.4. Proposition. If V and W are too free modules over a commutative ring R and (v;),(wj) are R-bases of V and W , respectively, then (v;8 wj) is an R-basis of V @ R W .

2.5. Proposition. Let R be a subring of a ring S and let V be a free left R-module with el,. . . , en as an R-basis. Then S @R V is a free left S-module with 1 8 el,. .. ,1 @ en as an S-basis. 2.6. Proposition. Let V be a right R-module, W a left R-module, V' a submodule of V and W' a submodule of W . If V' is a direct summand of V and W' is a direct summand of W , then the canonical

homomorphism

v' @ R w' v @ R w 4

is injective and the image of V' @R N .

Z-module M

@R

W' is a direct summand of the

Let R be a ring, V a right R-module and W a left R-module. The module V is said to be flat i f for every injective homomorphism f : W' t W of left R-modules, the homomorphism

7

2. Tensor products

is injective. By a projective module, we understand a direct summand of a free module.

2.7. Proposition.Every projective right R-module is pat. Let V be a right R-module and let A be a left ideal of R. The homomorphism V @R A -i V which sends v 8 a to va is called canonical: its image is the subgroup V A consisting of all finite sums C v ; u ; with v; E V,a; E A. The induced homomorphism V 8~A -iV A is called canonical. The canonical map V @ R R + V is an isomorphism. Note that if V is an ( S ,R)-bimodule, then all canonical maps are homomorphisms of S-modules. 2.8. Proposition. Let V be a right R-module. Then the following properties are equivalent: (i) V is flat. (ii) For any finitely generated submodule W' of a left R-module W , the canonical homomorphism l v @ i : V @ R W' + V @ R W (i is the inclusion map) is injective. d (iii) For every exact sequence of left R-modules W' + W + W" the sequence

*

is exact. (iv) For every finitely generated left ideal A of R, the canonical map V @ R A + V A is an isomorphism. Let V be a flat right R-module and let W' be a submodule of a left R-module W . Then the canonical injection V @ R W' + V @ , RW allows us to identify V @ R W' with its image in V @ R W . In the future, we shall always use this identification. 2.9. Proposition. Let V be a flut right R-module. (i) I f f : X + Y is a homomorphism of left R-modules, then

Preliminaries

8

and

Im(1v 8 f ) = v

OR

(Imf)

(ii) If W',W" are two submodules of a left R-module W , then

3.

Artinian, noetherian a n d completely reducible modules

Throughout this section, R denotes an arbitrary ring. Let V be an R-module. We say that V is artinian (respectively, noetherian) if every descending (respectively, ascending) chain of submodules of V terminates. The ring R itself is called artinian (respectively, noetherian) if the left regular R-module R R is artinian (respectively, noetherian). We call V finitely generated if V = Cy=lRvi for some wl, . . . $0, in V . The module V is said to be finitely cogenerated if for every family{ li E I}of submodules of V with niErK= 0, there exists a finite subset J of I such that

3.1. Proposition.For any R-module V , the following conditions are equivalent: (i) V is artinian. (ii) Every nonempty set of submodules of V has a minimal element. (iii) Every factor-module of V is finitely cogenerated. Proof.(i)=+ (ii): Assume that there is a nonempty set S of submodules of V such that S does not have a minimal element. Then, for any given W E S, the set {W' E SlW' c W } is nonempty. Owing to the axiom of choice, there is a function W L-+ W' with W 3 W' for any W in S. By choosing any W in S, we therefore obtain an infinite descending

chain

w 3 W ' 3 W " > ....

of submodules of V. (ii)+ (iii): It suffices to show that if W is a submodule of V and

3. Artinian, noetherian and completely reducible modules

{Eli E I} is

nj&

a family of submodules of V with n;& = W for some finite subset J of I . Let us put

9

= W , then

S = {nkEKvklICI is finite } By assumption, S has a minimal element, say njE& with J C_ I and J finite. It is obvious that W = njEJv,as required. (iii)+ (i): Consider a descending chain & _> & 2 . .. 2 V , 2 . . . of submodules of V . If W = nn>lVn, then by hypothesis V/W is finitely cogenerated. Thus W = V, for some n 2 1. But then V,+j = V, for all i 2 1, as desired. W 3.2. Corollary,Let V irreducible submodule.

#0

be an artinian module. Then V has an

Proof.Apply Proposition 3.1. for the nonempty set of all nonzero submodules of V . W Turning to noetherian modules, we next record

3.3. Proposition. For any R-module V , the following conditions are equivalent: (i) V is noetherian. (ii) Every nonempty set of submodules of V has a maximal element. (iii) Every submodule of V is finitely generated. Proof.The proof of this result is dual to that of Proposition 3.1. and therefore will be omitted. Proposition. If V is a finitely generated R-module, then every proper submodule of V is contained in a maximal submodule ( in particular, if in addition V # 0 , then V has a maximal submodule ). 3.4.

Proof. By hypothesis, V = Ci=lRv; for some v1,. . . ,v, in V . Denote by W a proper submodule of V and by S the set of all proper submodules of V which contain W . Then W E S and if (Wj) is a chain in S , then U;W; is a submodule. If V = U;W;, then vj E W,, for

10

Preliminaries

some nj, 1 5 j 5 r. Let W, be the largest of the modules W,, , . . . ,W,,. Then W contains vl,,. .,u, and hence V = W,, a contradiction. Thus S is inductive and therefore, by Zorn's lemma, it has a maximal element. This gives us a maximal submodule of V containing W , as desired. W

Let 0 --f U --f V -, W --f 0 be an exact sequence ofR-modules. Then V is artinian (noetherian) i f and only i f both U and W are artinian (noetherian). 3.5. Proposition.

Proof. Suppose that V is artinian. Then, because U and W are isomorphic to a submodule and factor module of V , respectively, it is obvious from the definition that both U and W are artinian. Conversely, assume that U and W are artinian. To show that V is artinian, we may assume that U 2 V and W = V/U. Consider a descending chain

v, 2 vz 2 ... 2 v, 2 ...

of submodules of V . Since V/U is artinian, there is an integer rn such that (i = 1,2, . . .) = vm+; u

v, + u

+

Because U is artinian, there is an integer n 2 rn such that

Taking into account modularity and that V, 2 V,+i, we have for each i = 1,2, ...

V, = V , n (v, U ) = V , n (Vn+l

+

=

+U )

K+i + (V, n U )= Vn+i + (K+;n U )

= Vn+i

Hence V is artinian. The proof of the noetherian case is dual. W 3.6. Corollary. Let V = @y=lq.Then V is artinian (noetherian) if and only if each is artinian (noetherian).

P r o o f . The case n = 2 follows from Proposition 3.5. The general case is a straightforward induction on n. '


11

3.7. Corollary. Let R be a ring. Then R is artinian (noetherian) if and only if every finitely genemted R-module is artinian (noetherian). Proof. Assume that R is artinian. If V is a finitely generated R-module, then V is a factor module of a free R-module W of finite rank. By Corollary 3.6, W is artinian, hence so is V , by Proposition 3.5. Conversely, if every finitely generated R-module is artinian, then so is R, since R is generated by 1. The proof of the noetherian case is similar . 3.8. Corollary. Let A be an algebra over a commutative ring R such that A is a finitely generated R-module. If R is artinian (noetherian), then A is artinian (noetherian).

Proof. This is a direct consequence of Corollary 3.7. Let V be a nonzero module. A finite chain of n -t 1 submodules of

V

v = vo 3 v, 3 . . . 3 v, = 0

is called a composition series of length n for V provided that each K-I/V, is irreducible, 1 5 i 5 n , i.e. provided each term in the chain is maximal in its predecessor. The irreducible modules

are called the composition factors of the series. 3.9. Proposition. A nonzero module V has a composition series if and only if V is both artinian and noetherian.

Proof. Suppose that V is both artinian and noetherian. Because any factor module of V is artinian, it follows from Corollary 3.2 that there is an ascending chain

0

c v1 c vz c .

of submodules of V where each term in the chain is maximal in its

successor. But V is noetherian, hence the chain terminates at V after

Preliminaries

12

a finitely many terms. Hence

V has a composition series .

Conversely, assume that V has a composition series. We argue by induction on n, where n is the minimum length of all such series. If n = 1, then V is irreducible and there is nothing to prove. Otherwise, if = vo 3 6 3 . . . 3 =0

v

v,

is a composition series of minimal length for V ,then V/V1is irreducible and V1 has a composition series of length n - 1. Now apply Proposition

3.5. Suppose that and

v = vo 3 6 3 ... 3 v, = 0 v = wo 3 w,3 ... 3 w, = 0

are two composition series for V. We say that these two series are equivalent in case n = rn and there is a permutation r~ of { 1,2,. . . ,n} such that vi/K+l Wo(i)/WC+)+l ( 2 = 1,. . ., n )

V V

3.10. Proposition. (The Jordan-Holder theorem). If a module has a composition series, then every pair of composition series for are equivalent.

Proof. We argue by induction on c ( V ) , where c ( V ) is the minimum length of composition series for V. The case c ( V ) = 1 being obvious, we may assume that c(V)= n > 1 and that any module with a composition series of smaller length has all of its composition series equivalent. Choose a composition series

(1)

v = v o 3 VI 3 . . . 3 v, = 0

of minimal length and fix any other composition series for

(2)

v = wo 3 w13 ... 3 w, = 0

V:


13

If & = W1, then by the induction hypothesis and in view of c ( K ) 5 n - 1, (1) and (2) are equivalent. We may thus assume that V1 # W1, in which case V = V, -t Wl since V, is a maximal submodule of V. It follows that

V, n W1 is maximal in both Wl 3.9, V1 n IV1 has a composition series whence

and V1. Owing to Proposition

and hence

are composition series for V, and W l . Since c(Vl) < n, every two composition series are equivalent. Thus the two series

and

v = VO 2 v1 2 xo 3 . .. 3 XI, = 0

are equivalent. It follows that k < n - 1 and c(W1) < n , proving that every two composition series for Wl are equivalent. Hence the two series = 3 3 ... 3 W, = 0

v wo w,

and

v = w o > w l 2x03 ... > x k = o

are equivalent. Because, by (3), V/Vl Z Wl/XO and V/W1 &/XO, the series (1) and (2) are equivalent, thus completing the proof. Assume that V is both artinian and noetherian. Then, by Propositions 3.9 and 3.10, for such a module V we can define its composition len,gth c(V) unambiguously by c(V) = 0 if V = 0, and c(V) = n if V has a composition series of length n. We next examine direct decompositions of modules.

Preliminaries

14

Let R be any ring and let V # 0 be an Rmodule which is either artinian or noetherian. Then V is a finite direct sum of indecomposable submodules. 3.11. Proposition.

For any nonzero R-module M that is not a finite direct Proof. sum of indecomposable submodules, choose a proper decomposition M = M' @ N' where M' is not a finite direct sum of indecomposable submodules. Assume by way of contradiction that V cannot be written as a finite direct sum of indecomposable submodules. Then

v = V' @ w',V' = v"@ W",.. . is a sequence of proper decompositions. Hence there exist infinite chains

W' c W' @ W" c . . . and V 3 V' 3 V'' 3 . . . proving that V is neither artinian nor noetherian. w

3.12. Proposition. Let R R = & @ . . . @ Vn with K # 0 for i = 1,2,. . . ,n, and write 1 = el - . en with e; E K. Then { e l , . . . e,} is a set of pairwise orthogonal idempotents in R and V , = R e ; , 1 5 i 5 n. Conversely, if { e l , . . . , e n } is a set of pairwise orthogonal idempotents

+ +

in R, then

n

R(Ce;) = @y=lRei i=l

Proof. Given r E R , we have r = r . 1 = 1 . r = re1 Hence, in particular,

+ . - + re,.

which shows that e;ej = &jei for all i , j . Moreover, Rei V , and RR = Re1 @ - . @ Re,, which implies that = R e ; for all i. Conversely, suppose that { e l , . . . ,e n } is a set of pairwise orthogonal idempotents and put e = CZ1e;. Then e2 = e and ee; = e;e = e; for R e ; . If E r i e ; = 0 , then multiplication on the all i. Hence R e = C:=t=, right by ej gives r j e j = 0 for all j. Thus R e = @y=lRe;, as asserted.


15

3.13. Corollary. Assume that R is artinian or noetherian. Then there exists a set { e l , . . . ,en} of pairwise orthogonal primitive idempotents of R with el - - . en = 1 . Moreover

+ +

where each Re; is an indecomposable R-module. Proof.

Apply Propositions 3.11 and 3.12.

We shall refer to the R-modules R e l , . . ., Re, in Corollary 3.13 as the principal indecomposable R-modules. An R-module V is said to be completely reducible if every submodule of V is a direct summand of V .

3.14. Proposition. Let V be a completely reducible R-module. (i) Every submodule ofV is isomorphic to a homomorphic image of V and every homomorphic image ofV is isomorphic to a submodule of

V. (ii) Every submodule and every homomorphic image ofV is comple t ely reducible, (iii) If V # 0 , then V contains an irreducible submodule. Proof. (i) If W is a submodule of V , then V = W @ W’ for some submodule W’ of V . Then W E V/W’ and V/W W‘, as required. (ii) Let W be a submodule of V and let U = V / W . By (i), it suffices to show that U is completely reducible. Assume that U1 = & / W is a submodule of U . We may choose V2 with V = Vl @ V2. Then

U = Ul@

(vz + W ) / W

and therefore U is completely reducible. (iii) Fix a nonzero 2, in V , By Zorn’s lemma, there is a submodule W of V maximal with respect to the property that TI 4 W . Write V = W @ W’ for some submodule PV’ of V . If W’ is not irreducible, then W‘ = W1@W2 for some nonzero submodules Wl, W2of V . Because

(W @ Wl) n (W @I W 2 )= W

16

Preliminaries

it follows that w $ W @ Wi for i = 1 or i = 2. But this contradicts the maximality of W . Thus W’ is irreducible, as desired.

We are now ready to provide two important characterizations of completely reducible modules.

3.15. Proposition. For any nonzero R-module V , the following conditions are eqivalent: (i) V is completely reducible. (ii) V is a direct sum of irreducible submodules. (iii) V is a sum of irreducible submodules.

Proof. (i)Ij(ii) : Consider the collection of sets of irreducible submodules of V whose sum is direct. By Proposition 3.14 (iii), it is nonempty and, by Zorn’s lemma, there is a maximal element, say {K}, in this collection. Let W = and let V = W @ W’. If W’ # 0, then by Proposition 3.14 (ii), (iii), W‘ contains an irreducible submodule V‘. Hence W V’ = V‘ @ (@V,),contrary to the maximality of {K}. Accordingly, W’ = 0 and V = W , as required, (ii)+ (iii) : Obvious. (iii)+ (i) : Let W be a submodule of V . Owing to Zorn’s lemma, there is a submodule W’ of V maximal with the property that WnW’ = 0. Hence we must have W W’ = W @ W’. We are thus left to show that V = W e W’. Assume by way of contradiction that W $ W’ # V and let v E V be such that w $ W @ W‘. By hypothesis, w = v1 . . . w, where v; E V , and V , is an irreducible submodule of V , 1 5 i 5 n. Accordingly, v j 6 W @ W’ for some j and thus

+

+

+ +

V, n (W $ W’) # V, However,V, is irreducible, so

V, n (W @ W‘) = 0 and hence

W 03 W’+ v, =

w @ W‘$ v,

which shows that W n (W’ @ V,) = 0. Since this contradicts the maximality of W’, the result follows. m


17

3.16. Corollary. Let V # 0 be a completely reducible R-module. Then the following conditions are equivalent: (i) V is artinian. (ii) V is noetherian. (iii) V is a finite direct sum of irreducibze submodules. Proof. By Corollary 3.6, if V = @LIK then V is artinian (noetherian), if and only if each V , is artinian (noetherian). Now apply Proposition 3.15 (ii). 3.17. Corollary. If R R is completely reducible, then every Rmodule is completely reducible. Proof. Let V be an R-module and let v E V . Then the map R R --$ Rv, r H r v is a surjective homomorphism of R-modules. Therefore, by Proposition 3.14 (ii), Rv is completely reducible. Because V = CvEVRv,the result follows by applying Proposition 3.15 (iii).

3.18 Lemma. Let V = @K where each Vi is irreducible and let W be an irreducible submodule of V . Then W @I$ where j ranges over those i for which W E V,. Proof. Given a nonzero w E W , we may write w = Cv;with v; E L$ and with finitely many v; # 0. If v; # 0, then the map r w I-+ rvi,r E R, is a nonzero homomorphism from W = Rw to Vl = Rv;.Since W,K are irreducible, W V , as desired.

Let V be a completely reducible R-module. We say that V is homogeneous if it can be written as a sum of isomorphic irreducible submodules. The sum of all irreducible submodules of V which are isomorphic to a given irreducible submodule of V is called a homogeneous component of V .

Let V = @iEzV,, where each V , is irreducible 3.19. Proposition. j E J , are all representatives of the and let J C I be such that the isomorphism classes of V,, i E I . For each j E J , denote b y Wj the sum

K,

18

Preliminaries

of all F with r/: E 6 . Then (i) The Wj, j E J , are all homogeneous components of V . (ii) V = @ j E ~ W j .

Proof. (i) Apply Lemma 3.18. (ii) This follows from the definition of Wj. Let V be any R-module. A submodule W of V is said to be fully invariant in V if it is mapped into itself by all R-endomorphisms of V .

3.20. Proposition. Let V # 0 be a completely reducible Rmodule. A submodule W of V is fully invariant if and only i f W is a sum of certain homogeneous components of V .

Proof. Keeping the notation of Proposition 3.19, fix f E EndR(V). If V , 2 V, then either f ( X ) = 0 or f(V,) Z V , %' vj, since V;: is irreducible. Thus f(Wj) C Wj, so each Wj (and hence the sum of certain Wj) is fully invariant. Conversely, assume that W is fully invariant. It suffices to show that if P is an irreducible submodule of W and Q is an irreducible submodule of V such that Q Z P , then Q C W. If Q = P , this is clear; otherwise P n Q = 0, so P -t Q = P @ Q and thus V = P @ Q @ U for a submodule U of V . Let f : P -+ Q be the given isomorphism and define an endomorphism g of V by

Then W _> g(W) 2 g ( P ) = f ( P )= Q, as asserted. 1 We close by recording the following useful observation. 3.21. Corollary. Let V # 0 be a completely reducible R-module and write V as the direct sum of its homogeneous components: V = @ i ~ W jThen . En&(V) % EndR(Wj)

n

jEJ

4. The radical of modules and rings

19

Proof.

It is clear that any family (fj) with f j E E n d ~ ( W j ) determines a unique endomorphism f of V . The map

is obviously an injective ring homomorphism. Because each Wj is fully invariant (Proposition 3.20), any endomorphism f of V maps each Wj into itself. Setting f, = flWj,j E J , it follows that ( f j ) H f, as desired. 4.

The radical of modules and rings

Throughtout, R denotes an arbitrary ring. Given an R-module V , the radical J ( V ) of V is defined to be the intersection of all maximal submodules of V . If V contains no maximal submodules, then by definition J ( V ) = V . Observe that if V # 0 is finitely generated, then J ( V ) # V by virtue of Proposition 3.4. The Jacobson radical J ( R )of R is defined by

J ( R )= J(RR) Hence, by definition, J ( R ) is the intersection of all maximal left ideals of R. We say that R is semisimple if J ( R ) = 0. 4.1 Proposition. For any nonzero R-module V , the following conditions are equivalent: (i) V is a finite direct sum of irreducible modules. (ii) V is artinian and com.pletely reducible. (iii) V is artinian and J ( V ) = 0.

Proof.

(i)+ (ii): This follows from Corollary 3.6 and Proposition

3.15. (ii)+ (iii): We may clearly assume that V # 0, in which case V is a direct sum of irreducible submodules, by Proposition 5.15. Therefore, by the definition of J ( V ) ,J ( V ) = 0 (iii)+ (i): Owing to Proposition 3.1, V is finitely cogenerated. Since J ( V ) = 0 it follows that f~r=,,K= 0 for some maximal submodules V,,. . . ,V, of V . Thus V is isomorphic to a submodule of (V/K).

ny=,

Preliminaries

20

Hence, by Proposition 3.14 (ii), V is completely reducible. Now apply Corollary 3.16 (iii). 4.2 Corollary.

For any ring R, the following conditions are equiv-

alent:

(i) RR is a finite direct sum of irreducible modules. (ii) RR is completely reducible. (iii) R is semisimple artinian. (iv) Every R-module is completely reducible. Proof. (i)+ (ii): Follows from Proposition 3.15. (ii)+ (iii): Because 1 lies in the sum of finitely many irreducible submodules, it follows that R R is a finite direct sum of irreducible modules. Now apply Proposition 4.1 for V = R R. (iii)+ (iv): By Proposition 4.1 (applied for V = R R),RR is completely reducible. Now apply Corollary 3.17. (iv)+ (i): By assumption, RR is completely reducible. Thus R is semisimple artinian, by the implication (ii)+ (iii). The desired conclusion follows from Proposition 4.1 applied for V = R R. We now proceed to examine J ( V ) in detail.

4.3. Lemma. (i) Let f : W + V be a homomorphism of Rmodules.Then f ( J ( W ) )& J ( V ) with equality i f f is surjective and Ke rf C J ( W ) . (ii) If W is a submodule of an R-module, then J ( W ) C J ( V ) and

J ( V / W )2 ( J ( V )t W ) / W . (iii) If W is a submodule of an R-module V with W 2 J ( V ) , then J ( V / W )= J ( V ) / W . (iv) IfV is an R-module, then J ( V / J ( V ) )= 0 and J ( V ) W for any submodule W ofV for which J ( V / W )= 0 (v) If W is a submodule of an R-module V , then W = J ( V ) if and only ifW J ( V ) and J ( V / W )= 0 . Proof. (i) If V has no maximal submodules, then J ( V )= V and therefore f ( J ( W ) )C J ( V ) .Now assume that M is a maximal submodule of V . Then the map f * : W --+ VIM given by f*(w) = f ( w ) M

+


21

is an R-homomorphism. Since K e r f * is a maximal submodule of W,J ( W ) C K e r f * and so f ( J ( W ) )C M . Hence f ( J ( W ) ) J ( V ) . Suppose that f is surjective and K e r f C J ( W ) . If W has no maximal submodules, then so does V , in which case

f ( J ( W ) )= f ( W )=

v = J(V)

We may thus assume that the set {MiliE I } of all maximal submodules of W is nonempty. By assumption, K e r f Mi for all i E I . If X and Y are the lattices of submodules of W containing K e r f and submodules of V , respectively, then the map U H f ( U ) , U E X , is an isomorphism of X onto Y. Therefore { f ( M i ) l i E I } is the set of all maximal submodules of V and

as asserted.

(ii) and (iii). This follows from (i) applied to the cases where f is the inclusion map and f : V --f V/W is the natural homomorphism. (iv) This follows from (iii) and (ii). (4 Apply (iv>* 4.4.Corollary.

Let R be an arbitrary ring and let I be an ideal of

R. (i) I = J ( R ) if and only if I J ( R ) and J ( R / I ) = 0 (ii) If1 C J ( R ) , then J ( R / I ) = J ( R ) / I . Proof. This is a direct consequence of Lemma 4.3 (v), (iii) and the obvious fact that the radical of the R-module R / I is the same as the Jacobson radical of the ring R / I .

A submodule W of an R-module V is called superfluous if for every submodule X of V , W + X = V implies X = V . 4.5. Proposition. Let V be an R-module. Then J ( V ) is the su'm of all superfluous submodules in V . Futhermore, if V is finitely generated, then J ( V ) is the unique largest superfluous submodule of V .

Preliminaries

22

Proof. Suppose that W is a superfluous submodule of V . To prove that W C J ( V ) ,we may assume that V has a maximal submodule M . If W M , then M + W = V and so M = V , a contradiction. Hence every superfluous submodule of V is contained in J ( V ) . On the other hand, let z E V . If W is a submodule of V with Rz + W = V , then either W = V or there is a maximal submodule M of V such that W & M and z $! M . If z E J ( V ) ,then the latter cannot occur. Thus z E J ( V )implies that Rz is a superfluous submodule of V , proving the first assertion. Suppose that V is finitely generated. By the foregoing, it suffices to show that J ( V )is a superfluous submodule of V . To this end, assume that V = W J ( V ) for a submodule W of V . Owing to lemma 4.3 (ii), J ( V / W )= V / W . Since V/W is finitely generated it follows from Proposition 3.4 that V/W = 0. Hence V = W as required.

+

4.6 Proposition. Let V be an R-module generated b y v l , . . . ,v, and let v E V . Then v E J ( V ) if and only if for all r; E R, 1 5 i 5 n, the elements v; r;v, 1 5 i 5 n, generate V .

+

Proof. Let v E J ( V )and let W be the submodule of V generated by the elements vi + riv, 1 5 i 5 n. Then W + J ( V ) = V so that W = V , by Proposition 4.5. Conversely, suppose that v 6 J ( V ) , Then there is a maximal subW , so for each module W of V such that v $! W . Hence V = Rv i there exists r; E R such that vi E -r;w W , whence v; r;v E W . Thus the elements v; + r;v, 1 5 i 5 n , do not generate V .

+

+

+

Let V be an R-module. The annihilator of V , written ann(V),is defined by

ann(V)= { T

E

RlrV = 0)

It is clear that ann(V) is an ideal and that V may be viewed as an R/ann(V)-module. We say that V is faithful if ann(V)= 0. An ideal I of R is said to be primitive if the ring R / I has a faithful irreducible module. Obviously, 1 is primitive if and only if I is the annihilator of an irreducible R-module.


23

Lemma. (i) An R-module V is irreducible if and only if V E R / X for some maximal left ideal X of R. (ii) Every maximal left ideal of R contains a primitive ideal and every primitive ideal is the intersection of the maximal left ideals containing it. (iii) J ( R ) is the intersection of the annihilators of irreducible Rmodules. In particular, J ( R ) is an ideal of R. 4.7.

Proof. (i) It is obvious that R / X is irreducible. Conversely, suppose that V is irreducible and let 0 # u € V . Then Ru is a nonzero submodule of V and so V = Rv. The map R -+ V,r H ru is therefore a surjective R-homomorphism. Thus R / X E V for some left ideal X of R. Since V is irreducible, X is maximal. (ii) Let X be a maximal left ideal of R. By (i), R / X is an irreducible R-module. Hence the annihilator of R / X is a primitive ideal contained in X . Let I be a primitive ideal in R and let V be an irreducible R-module whose annihilator is I . Given a nonzero IE: in V , put V, = { r E Rlrz = 0). Then V, is a left ideal in R. Since V is irreducible , V = Ra: E R/V, and so V, is a maximal left ideal of R. Because I is the intersection of all V, with 0 # II: E V , (ii) is proved. (iii) Apply (i) and (ii). Let V be an R-module. (i) J(R)V C J ( V ) with equality i f R / J ( R )is artinian. (ii) Assume that R/ J ( R)is artinian. Then V is completely reducible if and only if J(R)V = 0. 4.8. Proposition.

(i) If V has no maximal submodules, then J ( V ) = V and so J(R)V C J ( V ) . Let M be a maximal submodule of V . Then VIM is an irreducible R-module. Because J ( R ) annihilates VIM (Lemma 4.7 (iii)), it follows that J(R)V & M . Thus J(R)V & J ( V ) . Suppose that R / J ( R ) is artinian. Owing to Lemma 4.3 (v), it suffices to verify that J ( V / J ( R ) V )= 0. Because J ( R / J ( R ) )= 0 (Corollary 4.4 (i) and Lemma 4.7 (iii)) and R / J ( R )is artinian, every R / J (R)-module is completely reducible (Corollary 4.2). In particular, the R/J(R)-module V / J ( R ) Vis completely reducible. Hence V / J ( R ) V is a completely reducible R-module and therefore J ( V / J ( R ) V )= 0. Proof.

Preliminaries

24

(ii) If V is completely reducible, then J(R)V = 0 by Lemma 4.7 (iii). Conversely, assume that J(R)V = 0. Then V can be regarded as a module over the semisimple artinian ring R/J(R). Hence, by Corollary 4.2 V is a completely reducible R / J ( R)-module. Thus V is a completely reducible R-module. 4.9. Proposition.

(Nakayama’s lemma.) Let W be a submodule If W J(R)V = V , then W = V .

of a finitely generated R-module V .

+

Proof. Owing to Proposition 4.8 (i), J(R)V C_ J ( V ) and hence W + J ( V ) = V . But, by Proposition 4.5, J ( V ) is a superfluous submodule of V , hence the result. W The rest of this section will be devoted to ring-theoretic properties of J ( R ) . An element x of R is called a left (respectively, right) unit if there exists y E R such that yx = 1 (respectively, xy = 1). By a unit, we understand an element x in R which is both a left and a right unit. Hence x E R is a unit if and only if there exists y E R (denoted by x-l) such that xy = yx = 1 4.10. Proposition. Let x be an element of a ring R. Then x E J ( R ) if and only if for all r E R, 1 - rx is a left unit. In particular

J(R ) contains no nonzero idempotents. Proof. We first note that 1 - T X is a left unit if and only if R( 1 rx) = R. Therefore the first assertion follows from Proposition 4.6 applied to V = R R. Assume that e is an idempotent in J ( R ) . Then, by the above, x(l - e) = 1 for some x E R. Thus e = 1 e = x(1 - e)e = 0, as we wished to show. W 4.11. Proposition. For any ring R, J ( R ) is the unique largest ideal I of R such that 1 - rx is a unit of R for all r E R, x E I .

Proof. Owing to Proposition 4.10, it suffices to show that 1 - x is a unit for all 2 E J ( R ) . Since, by Proposition 4.10, 1 - x is a left unit, we have y ( 1 - x) = 1 for some y E R. Therefore z = 1 - y = -ya:


25

is in J ( R ) and so 1 = y'( 1 - z ) = y'y for some y' E R. Thus y is a unit and 1 - x = y-* is also a unit.

4.12 Corollary. For any ring R, J(R")= J ( R ) and, in particular, J ( R ) is the intersection of all maximal right ideals of R. Proof.

Apply Proposition 4.11.

4.13. Corollary. Let I be a left or right nil ideal of R. Then

I

C J(R).

Proof. Suppose that I is a left nil ideal of R and let x E I . Then, for any r E R,rx E I and so (rx)" = 0 for some n 2 1. Invoking the identity

(1 - y ) ( l

+ y + . . + y"-') *

= (1

+ y + - + yn-')(l * *

- Y) = 1 - y n

we see that 1 - r x is a unit. Thus by Proposition 4.10, I C J ( R ) . If I is a right nil ideal, then the same argument applied to R" yields the

result. W 4.14. Proposition.

For a n y ring R, the following conditions

are equivqlent: (i) R is artinian. (ii) R is noetherian, R / J ( R ) is artinian and J ( R ) is nilpotent. Proof. (i)+ (ii): Put J = J R and consider the chain J 2 J 2 2 . . .. By hypothesis, J k = J"' = . - . for some k 2 1. Setting I = J k we have I 2 = I . Assume by way of contradiction that I # 0. Then there exists a left ideal X in R with I X # 0, for example X = R. Let M be a minimal element in the set of all such X . Then I ( J M ) = ( I J ) M = I M # 0 and, becuase J M C M , we have J M = M . By Proposition 4.9, we are thus left to show that A4 is finitely generated. By assumption, there exists xlinM with I x # 0 and hence I ( R x ) # 0. Since R x C_ M , w e have M = R x , proving that J ( R ) is nilpotent. Since R is artinian, so is R / J ( R )by Proposition 3.5. To prove that R is noetherian, we argue by induction on the nilpotency index n of

Preliminaries

26

J ( R ) . If n = 1, then there is nothing to prove. Consider the exact sequence 0 + J"-1 + R + R/J"-' + 0 Since J ( R / J " - ' ) = J/Jn-' (Corollary 4.4 (ii)), it follows from the induction hypothesis that R/J"-' is noetherian. Hence, by Proposition 3.5, we need only show that Jn-' is noetherian. But J"-' is artinian (as a submodule of an artinian module RR) and J"-' is completely reducible, by Proposition 4.8 (ii). Hence, by Corollary 3.16, J"-l is noetherian, as required. (ii)+(i): We argue by induction on the nilpotency index 7~ of J(R). If n = 1 , there is nothing to prove. By induction hypothesis, R/J(R)"-l is artinian. On the other hand, J(R)"-' is noetherian (as a submodule of a noetherian module RR). Since R / J ( R )is artinian and J(R)" = 0, it follows from Corollary 3.16 and Proposition 4.8 (ii) that J(R)"-' is artinian. Thus, by Proposition 3.5, R is artinian. W 4.15. Corollary. Let R be an artinian ring. Then every finitely generated R-module is both artinian and noetherian. Proof. Apply Proposition 4.14 and Corollary 3.7. W

We next record the following useful observation. 4.16. Proposition,

Let (Ri),i E I , be a family of rings. Then

Proof. An element (r;) E n i E I R is ; a left unit if and only if ri, is a left unit of R; for all i E I . Now apply Proposition 4.10. H

In what follows, we use the standard fact that if el and e2 are idempotents of a ring S , then Sel Sez as left S-modules if and only if elS 2 e2S as right S-modules (see Jacobson (1956)).

s

4.17. Proposition. Let I be a nil ideal of a ring S and, for each E S , let S be the image of s in S = S / I . Then


27

(i) Each idempotent c E S can be lifted to an idempotent e E S , that

e = E. (ii) If e and f are idempotents of S , then e S E fS as right S modules .if and only if & Z $9 as right S-modules, (equivalently, S e E S f as left S-modules aj and only if SZE Sf as left S-modules). (iii) If I = el + €2 + . . . + en is a decomposition of I into orthogonal idempotents in S, then there exist orthogonal idempotents e l , e2,. . . ,en E is

S such that 1 =el+e2+-**+en, Moreover,

~j

e; = E;

is primitive if and only if so is ei.

Proof. (i) Fix an idempotent 6 E S and choose u E S with ii = E . Then u - u2 E I and therefore (u- u2)rn= 0 for some m 2 1. We have

Observe that on the right each term after the first m is divisible by (1 - u ) ~ while , the first rn terms are divisible by urn. Hence if e denotes the sum of the first m terms, then 1 = e + (1 - e)rng,where g is a polynomial in u. Now u ( 1 - u) E I, so

that is = ii = E . Since e ( l - e ) = e ( l - u)"g = 0 , e is a required i dempo tent. (ii) Let e , f be idempotents in S. We first show that e S E f S as right S-modules if and only if there exists a,b E R such that a = e a f , b = fbe,ab=e,

and

ba=

f

(1)

First suppose that 0 : e S ---f fS is an isomorphism that maps e to b and a to f. Then f b = b and O ( e ) e = O ( e ) ,hence be = b, so f b e = b, and similarly eaf = a. Hence e = ( @ - l @ ) ( e )= 0 - ' ( b ) = 0 - ' ( f b ) = O - ' ( f ) b = ab, and similarly f = ba. Conversely, if a and b satisfy (l), then x H bx is a homomorphism from e S to fS with inverse y H ay. Now assume that e S E f S and let a , b E S satisfy (1). Then 2 , b

Preliminaries

28

satisfy the same conditions for the isomorphism ES E fS, so &' E Now assume that d E fS and a, b E S are such that a

G

e a f , b 3 fbe,ab

e , ba

f

fs.

(modl)

and put a1 = e a f , bl = f b e . Then albl = e - z , where z E e l e . Because z is nilpotent, 1 - z has an inverse 1 - z' and so z z' = zz' = z'z. Setting 2" = ez'e, we have z zN = zz" = z"z and it follows that a l b l ( e - z") = e. Now put a2 = a l , b2 = bl(e - z"), then

+

+

a2 3

a , b2

b ( m o d l ) , a2b2 = e

Next write b2a2 = f -y, then y E flf.Because (ha2)' = bzeaz = bza2, we have

f-y=(f

-Y)2=f2-fy-yf+y2=

f -2y+y2

Hence y2 = y and because y is nilpotent, we find that y = 0 and thus bzaz = f. Since a2bz = e and b2a2 = f imply a2 = eu2 f,b2 = f bze, the required assertion follows. (iii) Let e , f be idempotents in S such that ef = f E = 0. We first claim that there exists an idempotent g such that g = f and eg = ge = 0 . To verify the claim, we note that f e is nilpotent and so 1 - f e is a unit. Put

h = (1 - f e ) - ' f ( 1 - f e ) Then h2 = h, he = 0 and

h = f. Now put

g = (1 - e ) h . Then

proving the claim. We now demonstrate that if e l , . . . ,en are idempotents such that Eiej = 0 for all i # j , then there exist idempotents e: such that = ~i and e:es = 0 for all i # j. For n = 1 there is nothing to prove, so we may assume that n > 1 and use induction on n. In other words, we may assume that e;ej = 0 for i # j, i , j > 1. Put e = e2 + e n , then e is an idempotent such that Fli? = EFl = 0. Applying the claim, there exists an idempotent el, such that e: = Cl and eel, = e i e = 0. It follows

4

+

5. The Wedderburn-Artin theorem

29

that e:, e 2 , . . . ,e, are pairwise orthogonal idempotents. By (i) and the above, there exist pairwise orthogonal idempotents e l , e2,. . . ,en such that E; = c i , 1 5 i 5 n. Thus for u = el - - - e, we have U = i, so 1 - u is a nilpotent idempotent, and therefore u = 1. To prove the final assertion, suppose that e E S is a primitive idempotent. Because e # 0, we have e # 0. If 6 = ~t: y for some orthogonal idempotents 2,y E 3, then there exist orthogonal idempotents el, e2 E S with C1 = 1t:,t?2 = y and hence with E = el e2. By (ii), (el e 2 ) S , which is possible only in the case el = 0 or e2 = 0. eS Thus either 5 = 0 or y = 0 and so iii is primitive. Conversely, assume that e is primitive. Then E # 0 and so e # 0. If e = u v, where u , v are orthogonal idempotents, then ?I = U 5 , ii5 = V i i = 0 and so either U = 0 or V = 0. Thus either u = 0 or v = 0, proving that e is primitive. This completes the proof.

+ +

+

+

+

5.

+

+

The Wedderburn-Artin theorem

In this section we provide the ring-theoretic structure of semisimple artinian rings. Throughout, R denotes an arbitrary ring and, for any interger n 2 l , M , ( R ) is the ring of all n x n matrices over R. Let i,j E (1,. . . ,n } and let e;j be the matrix with (i,j)-th entry 1 and 0 elsewhere. The elements eij E M,(R) are called matriz units. As is customary, we shall identify R with its image in M,(R) consisting of all scalar matrices diag(r, T , . . . ,T ) , r E R. As a preliminary to our first result, let us observe that the matrix units satisfy the following properties : ( i ) e;jeks = 0 if j # k and e i j e k , = ei, if j = k. (ii) 1 = ell . . - enn. (iii) The centralizer of { e ; j } in M,(R) is R. (iv) R Z ellM,(R)ell. It turns out that (i)) (ii), and (iii) determine M,(R) up to isomorphism. Namely, we have the following result.

+ +

5.1. Proposition. Let S be any ring containing elements v,j, 1 5 i ,j 5 n , satisfying (i) and (ii) and let R be the centralizer of {vij} in

Preliminaries

30

S . Then the map (aij) H C i , j aijvij

is an isomorphism of rings and R-modules. Furthermore, R Z v l l S q l .

Proof. Note that M,(R) is a free R-module freely generated by the matrix units eij. Hence the map ejj H v;j extends to a homomorphism of R-modules, which clearly coincides with $. Applying (i), (ii), and (iii), we immediately derive that 1c, is also a ring homomorphism. Suppose that Ci,ja;jv;j = 0 and fix k,s E { 1,.. . ,n}. Then, for all t E { I , . ..,n}, 0 = v t k ( x aijvij)vst = aksvtt i,j

and so 0=

c

aksvtt

= a k s ( z vtt) = ak3

t

t

Thus (ajj) = 0 and therefore $ is injective. To show that 1c, is also surjective, fix s E S and for each i,j put

Then, for all

v,t,

we have

and aijvrt

=

VkiSVjkvrt

= VTjSVjt

k

Hence a;j commutes with all vTt and therefore the foregoing, +((Ujj))

=

c i,j

ajjv;j =

c

aij

E R. Moreover, by

v1;svjj

i,j

= (cv;j)s(xvjj) = s

31


proving that 1c, is surjective. Finally, taking into account that

the result follows. For any integer n 2 1 and any (left) R-module V , we write V" for the n-th direct power of V . 5.2. Proposition.

Let V be an R-module and let n

Proof. Fix z,j E { 1,. . . ,n } and define

f&, where

vj

f;j

2 1. Then

E E n d ~ ( v "by )

. . . ,W") = (0,. . . ,v j , o , . . . , O )

is in the i-th place. Then clearly n

fijfks

= ajkfis

and

cfi; = i=l

1

+

If E EndR(V") and $ ( v l , . . . ,on) = ($l ( V l ) , * * * ,+n(vn)), then $ obviously centralizes all f i j if and only if t,bl = $2 = . . . = +". Thus the centralizer of {fij) in EndR(V")is identifiable with EndR(V). The desired conclusion is therefore a consequence of Proposition 5.1. Let V be an R-module and let n 2 1. Then the R-module V" can also be viewed as an M,(R)-module in the following natural way. We visualize the elements of V" as column vectors and, for each A E M,(R) and x E V " , define Az as the matrix multiplication. We now are ready to record the following result. 5.3. Proposition. (i) The map W H W" is an isomorphism from the lattice of submodules of V onto the lattice of M,( R)-submodules of V " . (ii) EndM,,(R)( V " ) EndR(V ).

Preliminaries

32

(iii)The map V H V" induces a bijective correspondence between the isomorphism classes of R ans M,(R)-modules. The inverse of this correspondence is given b y W H ell W . f W" obviously preserves orProof. (i) The correspondence W +

der. Thus we need only show that it has an inverse which is also order-preserving . Consider the projection ?rl : V" + V on the first factor and, for any submodule X of V", put g ( X ) = ?rl(X). Then the correspondence X H g ( X ) is order-preserving and clearly ( g f ) ( W = ) W . From the action of M,(R) on V", we see that X can be written in the form X = V;.. for some submodule 6 of V. Thus ( f g ) ( X )= f(6)= V;2 = X , proving (i). (ii) Put S = M,(R) and choose f E Ends(V"). Then, by the nature of action of S on V", f has the same projections, say X j , on all factors. Conversely, any given 1c, E EndR(V) determines an element of Ends(V") whose projections on all factors are equal to +. Thus the map f H X j provides a desired isomorphism. (iii) It suffices to show that ellVn 2 V and (ellW)" W , where V and W are R and M,(R)-modules, respectively. By the definition of

ell 7

ellVn = V x 0 x . . . x 0 Z V

Observe also that

Becuase eklw = 0 if and only if ellw = 0, 1 5 Ic

5 n, w E W , the map

is well defined and is at least an R-isomorphism of W onto (ellW)". Consequently, we need only show that

Since the latter is a consequence of the action of M,(R) on (ellW)", the result is verified.


33

For any ideal I of R, put

M,(I) = {(a,j)E M,(R)la;j E I

for all i , j E {1,..., n } }

5.4. Proposition. (i) The map I H M,(I) is a bijection between the sets of ideals of R and M,(R). In particular, R is simple if and only i f so is M,(R). (ii) Mn(R)/Ma(I) Mn(R/I). (iii) If R = I1 @ . . . @ I, is a two-sided decomposition of R, then

M,(R) = Mn(Ii)@ . . . @ Mn(Is) is a two-sided decomposition of M,(R). Furthermore, the ideal I; is indecomposable if and only if so is M,(I;).

Proof. (i) It is clear that M,(I) is an ideal of M,(R) and that the given map is injective. Suppose that J is an ideal of R and let I C_ R consist of all entries of elements in J. Then I is an ideal of R such that

J = M,(I). (ii) The natural homomorphism R + R / I induces a surjective homomorphism M,(R) + M,(R/I) whose kernel is M n ( I ) . (iii) This follows from the facts that Z(M,(R)) = Z(R) and that, for any r E R, M,(R)r = M,(Rr). H 5.5. Proposition.

For any ring R and any positive interger n,

J(Mn(R)) = M"(J(R)) Proof. Let V be an R-module and let V" be an M,(R)-module as in Proposition 5.3. Then ann(V") = M,(ann(V)),by the definition of V". Now apply Proposition 5.3 and Lemma 4.7 (iii). H

Let R be a finite direct product of f i l l matrix 5.6. Corollary. rings over division rings. Then R is semisimple artinian. Proof. That R is semisimple follows from Propositions 4.16 and 5.5. To prove that R is artinian, we may assume that R = M,(D)for some n 2 1 and some division ring D. Now apply Proposition 5.3.

Preliminaries

34

Our next aim is to establish the converse of the above result. As a preliminary, we record the following classical observation.

.

(Schur's lemma). Let V be an irreducible module. 5.7. Lemma Then E n d R ( V ) is a division ring. Proof. Let f : V + V be a nonzero R-homomorphism. Since f(V) is a nonzero submodule of V ,f ( V ) = V . Because f # 0, K e r f # V and so K e r f = 0. Thus f is an isomorphism, as required. Let V be an R-module and let e be an idempotent of R. Then HomR(Re, V ) E eV as additive groups. Similarly, i f V is a right R-module, then H o m R ( e R , V ) V e . 5.8. Lemma.

Proof. If f E H o m R ( R e , V ) , then e f ( e ) = f ( e 2 ) = f ( e ) E eV. Therefore the map f H f ( e ) is a homomorphism from HornR(Re, V ) to e V . Conversely, if v E e V , then the map g , : z e H zv is an Rhomomorphism from R e to V . Because the map H gv is an inverse of f H f ( e ) , the first isomorphism follows. The second isomorphism is proved by a similar argument. W 5.9. Lemma. Let e be an idempotent of a ring R. Then

EndR(Re) E (eRe)" and E n d R ( e R ) E eRe

as rings

In particular,

End&R)

E R"

and E n d R ( R R )

R

Proof. Applying Lemma 5.8 for V = R e , it follows that the map f H f ( e ) is an isomorphism of the additive group of E n d R ( R e ) onto the additive group of eRe. Given f , g E E n d R ( R e ) , write f ( e ) = erle and g ( e ) = er2e for some r1, r2 E R. Then ( f g ) ( e )= f(er2e) = erzerle = (erze)(erle)= g ( e ) f ( e ) proving that f H f ( e ) reverses multiplication. Since e is the identity element of the ring eRe, the above map preserves identity elements.


35

This establishes the first isomorphism. The second isomorphism follows from Lemma 5.8 applied for V = eR. W We have now come to the demonstration of the following classical result. 5.10 Theorem.

(Wedderburn-Artin). Let R be a semisimple ar-

tinian ring. (i) There exist only finitely many, say Vi, . . . , V , of nonisomorphic irreducible R-modules and V , 2 Re; for some primitive idempotent e; o f R , 15 i 5 r . (ii) RR =" nl& @ . . .@ n,V, for some positive integers n;, 1 5 i 5 r , where n;T/: is a direct sum ofn; copies of K. (iii) R 2 niz1Mn,(D;),where D;= EndR(x)" S e;Re; is a division ring, 1 5 i 5 r . (iv) The integers n; and r are unique and each D; is determined up to isomorphism.

Proof. (i) and (ii): Owing to Corollary 4.2, RR is a finite direct sum of irreducible modules. Therefore, by Proposition 3.12, there exist idempotents e l , . . . ,en of R such that R R = Rel$. . .$Re, and each Re; is irreducible (in particular, e; is a primitive idempotent of R). We may assume that Rel, .. . , Re,, 1 5 r 5 n , are all nonisomorphic modules of the set { R e l , .. . ,Re,}. If V is any irreducible R-module, then by Lemma 4.7 (i) and Proposition 3.10, V 2 Re; for some i E {1,2,. . . ,r } . This establishes (i) and (ii). (iii) Put Di = EndR(V,)",1 5 i 5 r . Then we have

R"

EndR(RR)

(by Lemma 5.9)

r

!2

nEnd~(n;K)

(by Corollary 3.21)

i=l

n r

Mni(D;o)

(by Proposition 5.2)

i=l

and thus r

Preliminaries

36

(e;Re;)" we have D; Z e;Rei, Because, by Lemma 5.9, EndR(K) proving (iii). (iv) Assume that R z Mk,(Dj), where each Di is a division ring. Then there exist central idempotents u1,. . .,us of R such that

ng=,

Mkj(Di)is a simple ring. Hence Ruj is a direct sum of kj and Ruj copies of an irreducible R-module and Rui, Ruj, i # j , have no common composition factors. Thus r = s and, by renumbering the u j , we have I C3. = n,, 1 5 j 5 r, by Proposition 3.10. Put A = Mkj(Di),D = Di and let V be an irreducible A-module. E n d o ( o D ) . Hence, by Then, by Proposition 5.3(ii), EndA(V) Lemma 5.9, EndA(V) 2 Do. On the other hand, V % V,, hence

This proves that

Di

Dj and the result follows. W

5.11. Corollary. For any ring R, the following conditions are equivalent : (i) R is semisimple artinian. (ii) R is a finite direct product of full matrix rings over division rings. (iii) RR is completely reducible. (iv) Every R-module is completely reducible.

Proof. This is a direct consequence of Corollaries 4.2 and 5.6 together with Theorem 5.10. We close by proving the following classical result. 5.12. Theorem. (Wedderburn.) Let A be a finite-dimensional algebra over a field F and let I be an ideal o f A which has an F-basis consisting of nilpotent elements. Then I is a nilpotent ideal of A .

Proof. We may harmlessly assume that F is algebraically closed. Moreover, because (I J ( A ) ) / J ( A )is an ideal of A / J ( A ) having a

+

6. Group algebras and group representations

37

basis consisting of nilpotent elements, we may assume that J ( A ) = 0. Hence, by Theorem 5.10,

. . ,n,. Let

be the projection of I into M n i ( F ) 1, 5 i 5 r . It obviously suffices to show that 7 r i ( l ) = 0 for all i E ( 1 , . . . , r } . Since 7 r i ( I ) is an ideal of M n i ( F ) ,either 7 r ; ( l ) = 0 or 7 r ; ( l ) = M n i ( F ) . In the latter case, it follows that M n i ( F )has a basis consisting of nilpotent elements. But then the trace of each matrix in M n i ( F )would be zero, a contradiction. Thus 7r;(I) = 0 and the result is established. for some positive integers

6.

121,.

7ri

Group algebras and group representations

Let R be a commutative ring and let G be an arbitrary group. The group algebra RG of G over R is defined to be the free R-module on

the elements of G , with multiplication induced by that in G. More explicitly, RG consists of all formal linear combinations C xg - g, xg E R,g E G, with finitely many xg # 0 subject to for all g E G (i) C xg - g = C yg - g if and only if zg = yg (4 c x g ‘ 9 C Y g Y = C(xg Yg) ‘9 (iii) (C xg . g ) ( C yh - h ) = C zt . t , where t = Cgh=t zg- y h for all r E R (iv) r(C xg - g ) = C ( r x g ). g One easily verifies that these operations define RG as an associative R-algebra with 1 = 1~ l ~where , 1~ and 1~ are identity elements of R and G, respectively. Applying the injective homomorphisms

+

+

we shall in future identify R and G with their images in RG. With

these identifications, the formal sums and products become ordinary sums and products. For this reason, we drop the dot in xg - g. We shall also adopt the convention that RG 2 R H means an isomorphism of R-algebras.

Preliminaries

38

Let x = C x g g E RG. Then the support of x , Supp(x), is defined by

SUPP(4 = ( 9 E Glx,

# 0)

It is obvious that Supp(x)is a finite subset of G that is empty if and only if x = 0. For convenience of reference, we next record some elementary properties of group algebras. 6.1. Proposition.

Let A be an R-algebra and let

f : G -+ U ( A ) be a homomorphism of G into the unit group U ( A ) o f A . Then the map f* : RG -+ A defined b y

is a homomorphism of R-algebras. In particular, i f f is injective and A is R-free with f(G) as a basis, then RG Z A.

Proof. Because RG is R-free freely generated by G , f* is a homomorphism of R-modules. Let

be two elements of RG. Then

as required.

Let R be a commutative ring and let G be a group. Given a nonzero R-module V , by a representation of G on V we understand a homomorphism p : G + AutR(V)


39

We say that p is faithful if Kerp = 1. By Proposition 6.1, to each representation p of G on V there corresponds an RG-module structure on V given by (v E

V,zgE R,g E G)

Conversely, any nonzero RG-module V determines a representation p of G on V given by p(g)v = gv for all g E G,v E V . Thus there is a bijection between the class of all nonzero RG-modules and the class of all representations of G on R-modules. We say that the representation p of G is irreducible (indecomposable, completely reducible) if the corresponding RG-module is irreducible (indecomposable, completely reducible). Let V be an R-module which is R-free of finite rank n. Then

A u ~ R ( V ) GLn(R) where GLn(R) is the unit group of M,(R). Thus to each representation of G on V , there is a matrix representation p* : G + GL,(R). In particular, if R = F is a field, then all finitely generated F-modules are free of finite rank. In this case the study of FG-modules, representations of G on F-modules, and matrix representations of G over F are essentially equivalent. For the rest of this section, all vector spaces are assumed to be finite-dimentional over a ground field and G is a finite group. Let V be a vector space over a field F having a basis v l , . . . ,v, of n elements, and let f E EndF(V). With respect to this basis, f can be represented as an n x n matrix ( a ; j ) over F . By the trace of f , written tr f, we mean n

t r f = tr(a;j)= Caii i=l

If a different basis for V is chosen, then the two different matrices for f are similar, and hence have the same trace. Thus f H t r f is a well defined F-valued function on EndF(V). The map

Preliminaries

40

g) = t r f . trg. Now suppose that V is an FG-module. Then V determines a homomorphism 1c, : F G + EndF(V) of F-algebras. By the character of V, written x v , we understand the map xv : F G + F given by x v ( x ) = tr+(x) for all x E FG. The map x : G --f F defined by x ( g ) = xv(g) is called the character of G a$orded by V. In view of the equality X V E "99) = %XV(9) = xgx(9) is obviously F-linear, but need not satisfy the equality t r ( f

c

c

the characters x v and x determine each other. Suppose that p : G + GL(V) is a representation of G. By the character of p, (or character of G) we understand the map x : G + F given by x ( g ) = trp(g) for all g E G. It is clear from the foregoing discussion that : (i) We may treat the terms character of G and character of G afforded by an FG-module as interchangeable. (ii) The character of an FG-module V is completely determined by the character of G afforded by V. Let V and W be FG-modules. Define the sum xv xw of the characters of V and W , respectively , by the rule

+

+

Then obviously xV@w = xv xw and therefore the sum of two characters is again a character. Note that if charF = p > 0, then the constant function 0 is the character of the direct sum of p copies of any FG-module V . On the other hand, if charF = 0, then identifying Z with its image in F , we obtain

In this case, we say that x(1) is the degree of x. Let p : G 3 GL(V) be a representation of G and let x be the character of p. Then x is a class function; that is its value is cwstant on any conjugacy class of G. Indeed, for all x , y E G, we have

41


If C is a conjugacy class of G, then we put x(C)= ~ Suppose that the map

pi

( z with ) z E

C.

: G + GL(K),i = 1,2,is a representation of G. Then

P*

@'2

:G

+

GL(V1 @F

VZ)

defined by ( P I @ P 2 ) ( 9 ) = /h(g) @ P 2 ( 9 )

is a.gain a representation of G. We shall refer to pi 8 p2 as the (inner) tensor product of pi and p 2 . Note that if X I , x 2 are the characters of p1, p2, respectively, then the product xlx2defined by

is the character of pi @ p2.

We next record the following classical result. 6.2. Proposition. (Maschke's theorem). Let G be a finite group and let F be a field of characteristic p 2 0 . Then F G is semisimple i f and only i f p does not divide the order of G.

Proof. If p > 0 divides [GI, then z = CSECgsatisfies z E Z(FG) and z2 = IGlz = 0. Hence F G z E J ( F G ) # 0. Conversely, assume that p does not divide [GI, and let W be a submodule of an FG-module V . We may write V = W@W'for some F-subspace W'. Let 8 : V -+ W be the projection map. Define 1c, : V + V by

Because for all v E V and y E G,

1c, is an FG-homomorphism. Now assume that v E V . Then, for any z E G,z-'v E W , so O(x-lv) = z-lv. Accordingly, z8z-l~= v and $(v) = v. Setting

Preliminaries

42

W" = Ker$, it follows that W'' is an FG-submodule of V such that W" n W = 0. Finally, let v E V. Then, by the above, v - $(v) E W", so 21 = $(v) (v - $(v)) E W W'. Thus V = W @ W' and the result follows. rn

+

+

Suppose that A is a finite-dimensional algebra over a field F. Let V be a vector space over F, let E be a field extension of F , and form the tensor products AE=E@FA

and

VE=E@FV

If V is an A-module, then the vector space VE becomes an AE-module under a module action

An A-module V is said to be absolutely irreducible if, for every field extension E/F, VE is an irreducible AE-module. We say that F is a splitting field for A if every irreducible A-module is absolutely irreducible. We say that F is a splitting field for G if F is a splitting field for FG. We close by quoting certain standard facts pertaining to group representations ( see Curtis and Reiner (1962) Dornhoff (1971), Huppert and Blackburn (1982)).

.

6.3. Proposition Let F be an arbitrary field and let ~ 1 , ... ,X , be the characters of G aforded b y all nonisomorphic irreducible FGmodules. Then the characters xl,.. . ,xT are linearly independent (as functions from G to F . )

6.4.Proposition. Let F be a splitting field for G of characteristic 0, and let X I , . . . ,xT be all irreducible characters of G over F. For each i E (1, ...,r } , put

ei = Xi(i)/IGl

c

x;(g-l)g

g€G

Then e l , . . . ,e, is a complete system of centrally primitive idempotents of F G .


43

6.5. Proposition. Let F be a field of characteristic 0, and let V and W be arbitrary FG-modules. Then V 2 W i f and only i f xv = X W .

6.6. Proposition. (Orthogonality relations). Let xv and xw be the characters afforded b y the irreducible FG-modules V and W . Then (i) If F then CgECxv(9)XW(9-1)= 0, (ii) If F is a splitting field for F G and charF does not divide /GI, then

v w,

c xv(s)xv(g-l)

= IGI

g€G

(iii) If F is a splitting field for FG with charF not dividing \GI, CI, . . . ,C, conjugacy classes of G and X I , . . . ,xT irreducible characters of G over F , then

where gj E Cj.

In what follows if p = 0, then by definition all elements of G are p'-element s. Let F be an arbitrary field of characteristic p 2 0, let n be the least common multiple of the orders of the p'-elements in G and let E be a primitive n-th root of unity over F . Denote by I, the multiplicative group consisting of those integers r , taken modulo n, for which E I+ E' defines an automorphism of F ( E )over F . Two $-elements a, b E G are called F-conjugate if z-lbz = a' for some z E G and some r E I,. Thus an F-conjugacy class is a union of ordinary conjugacy classes. 6.7. Proposition. (Witt-Berman theorem). Let F be an arbitrary field of characteristic p >_ 0 and let G be a finite group. Then the number of nonisomorphic irreducible FG-modules is equal to the number of F-conjugacy classes of p'-elements in G.

We close by quoting some standard results on absolutely irreducible modules (see Huppert and Blackburn (1982)).

Preliminaries

44

Let A be a finite-dimensional algebra over a field F , and let V be an irreducible A-module. Then the following conditions are equivalent: (i) V is absolutely irreducible. (ii) If E is the algebraic closure of F , then VE is an irreducible AEmodule. (iii) EndA(V) = F , i.e. each A-endomorphism o f V is a left multiplication b y an element of F . 6.8. Proposition.

6.9. Proposition. Let A be a finite-dimensional algebra over a field F . (i) F is a splitting field for A if and only if

for some r 2 1 and some n; 2 1, 1 5 i 5 r . (ii) If F is a splitting field for A, then for any field extension E I F , E is a splitting field for A E . 6.10. Proposition. (Burnside). Let V be an absolutely irreducible A-module, where A is a finite-dimensional algebra over a field F . Then the homomorphism A EndF(V), a H $a, where $a(v) = av, a E A, v E V , is surjective.

45

Chapter 2 General properties of induced modules In this chapter we record a number of basic results concerning induced modules. After presenting some formal properties of these modules, we examine in detail their annihilators and provide a number of applications. Among other results, we prove some classical theorems of Mackey and tie together relative projectivity and induced modules. The chapter ends with a short discussion of projective covers of induced modules.

1.

Induced modules, representations and characters

Throughout this section, R denotes a commutative ring and G a group. Let H be a subgroup of G. Then the group algebra RH can be identified with the subalgebra of RG consisting of all R-linear combinations of the elements h E H . If V is an RG-module, then we shall denote by VH th RH-module obtained by the restriction of algebra; thus as an R-module, VH equals V , but only action of RH is defined on VH. This process will be called restriction and it permits us to go from any RG-module V to a uniquely determined RH-module V'. There is a dual process of induction. Let V be any RH-module. Since we may consider RH as a subalgebra of RG, we can define an RG-module structure on the tensor product

General properties of induced modules

46

This is the induced module and we denote it by V G . 1.1. Lemma. Let H be a subgroup of G. If T is a left (right) transversal for H in G, then RG is a free right (left) RH-module with T as a basis.

Proof. Suppose that T is a left transversal for H in G. Then, for any t E T , t(RH) is the R-linear span of the coset t H . Therefore, for any tl,. . . ,in E T ,t l ( R H ) -.. t,(RH) is the R-linear span of UY="=,t;H.As is apparent from the definition of RG, if X and Y are disjoint subsets of G, their R-linear spans meet at 0. Thus

+ +

proving that RG is a free right RH-module with basis T . A similar argument proves the case where T is a right transversal. 1.2. Corollary. Let H be a subgroup ofG, let T be a left transversal for H in G and let V be an RH-module. (i) VG = @V (direct sum of R-modules) (ii) V G= (l@V)$(CtET,tgH t@V) (direct sum of RH-modules) (iii) The map V -+ t @ V, v H t @ v is an R-isomorphism, while the map V + 1 8 V , v H 1 @I v is an RH-isomorphism. Proof.

(i) By Lemma 1.1, RG = etETt(RH),hence

as required.

(ii) By (i), it suffices to prove that CtET,tgHt@V is an RH-submodule . h E H and t E T - H. Then ht = t'h' for some h' E H of ( V G ) ~Fix and t' E T - H. Since h(t @ V) = ht @ V = t'h' @ V = t' 8 V, (ii) is established. (iii) The map V -+ R H @wV = 1 @ V, v H 1 @ v is obviously an RH-isomorphism. If t @ v = 0 then 18v = 0, hence v = 0. This shows that V -+ t @ V , v H t @ v is an injective R-homomorphism. Since the above map is obviously surjective, the result follows.

1. Induced modules, representations and characters

47

Let V be an RG-module. By the kernel of V , written I<erV, we understand the kernel of the representation of G afforded by V , i.e. I<erV = { g E Glgv = v for all v E V } 1.3. Corollary. Let H be a subgroup of an arbitrary group G and let V be an RH-module. Then

Proof. Let T be a left transversal for H in G. Then obviously

By Corollary 1.2(i), g E Ii'erVG if and only if g ( t @ v) = t 8 v

for all t E T ,v E V

If g t = t'h with t' E T , h E H , then g ( t @ v) = t ' @ hv. Hence, by Corollary 1.2(i), g ( t 8 v) = t 8 v if and only if g t = t h with hv = v. Thus g E I<er(VG)if and only if g E ntETtKerVt-l,as required. The process of induction is closely related to the concept of imprimitivity of a module. We say that an RG-module V is imprimitive if V can be written as a direct sum

of R-modules r/: with 111 a permutation of the set

> 1 such that for each g E G, V;:

g V , is {r/:li E I } . The latter is equivalent to the assumption that G acts on the set { K J iE I } with each g E G sending V, to g v . We refer to { K l i E I } as a system of imprimitivity for V . The following result ties together imprimitive and induced modules. H

1.4. Proposition. (i) Let H be a proper subgroup of G, let V be an RH-module and let T be a left transversal for H in G . Then { t 8 Vlt E T } is a system of imprimitivity for V G and the action of G on it is transitive. (ii) Let V be an imprimitive RG-module with {r/:liE I } as a system


48

of imprimitivity and let G act transitively on

{xli E I ) . If H is the

stabilizer of any W E { K l i E I } , then W is an RH-module such that VS'G

Proof. (i) This follows from Corollary 1.2(i) and the fact that the left multiplication by g E G induces a transitive action of G on the set

{ t 63 Vlt E T } . ( 2 ) Let T be a left transversal for H in G. Then { K l i E I } = { t W l t E T } and V = etETtW (direct sum of R-modules). Hence, by Corollary l.2(i), the map t 63 W -+ t W , t 63 w H t w induces an Risomorphism W G V . Given g E G, t E T , we may write gt = t'h for some t' E T , h E H . Hence f ( g ( t 8 w)) = f ( t ' h @ W ) = f (t' 63 h w ) = t'hw = gtw = g ( t w ) = g f ( t 8 w ) as required.

Let V be an RH-module and let g E G. We define gV to be the R(gHg-')-module whose underlying R-module is V and on which the elements z E gN9-l act according to the rule 2

* 2, = (g-1zg)v

1.5. Lemma. Let H be a subgroup o f G , let V be an RH-module and let g E G. Then (i) gV E g 8 V and hV V for all h f H . (ii) ( g ~ ) ' z vG. (iii) For all z, y E G, "YV "(YV). (iv) V is irreducible (completely reducible, indecomposible) if and only i f so is gV. (v) If R = F is a field, then for any field extension E / F , the E(gHg-')-modules ( g V ) E and g ( V E ) are isomorphic. . -

Proof. (i) The map f : gV --f g @ V , v H g @ v is an Risomorphism, by Corollary 1.2(iii). Given z = ghg-', h E H , we have


proving that gV

%g @

49

V . If h E H , then

as required. (ii) Choose a left transversal T for H in G with g = t E 2'. Then the stabilizer of t @ V is tHt-'. Hence, by (i) and Proposition 1.4,

as desired. (iii) Applying (i), we have XY

v = (zy) @ v = z(y @ V ) E "(y @ V )z "("),

as required.

(iv) This follows from the fact that W is an RH-submodule of V if and only if W is an R(gHg-')-submodule of gV. (v) Observe that ( ~ V ) and E ~ ( V Ecoincide ) as vector spaces. Let z E gHg-', X E E and w E V . Then the action of z on X 8 v E g(V') is given by 2

* (A @ w)

= g-*zg(X @ v) = X @ g-lzgw

Bearing in mind that the corresponding action in (gV), is given by z(X @ v) = X @ z

*

2)

= X @ g-'zgv,

the result follows. W 1.6. Lemma. Let H be a subgroup of G, let E f F be a field extension and let V be an FH-module. Then

( V G )Z ~(VE)'

as

EG-modules

Proof. Let { v;/iE I }be an F-basis of V and let T be a left transversal for H in G. Then


50

are E-bases of ( V G )and ~ ( V E ) ~respectively. , Hence the map 6' : ( V G )+ ~ ( V E )defined ~ by B ( 1 €3 (t @ v;)) = t @ ( 1 €3 v;) is an Eisomorphism. Fix g E G, t E T and write g t = t'h for some t' E T , h E H . Then, for z = 1 €3 (t €3 vi), we have

as required. W

Let F be a field and let V , W be FG-modules. Then the vector space V @F W is an FG-module where the action of elements g E G is defined by g(v €3 w)= gv €3 gw and then extended to V € 3 W ~ and FG by F-linearity. Note that if pv, pw are representations of G afforded by V , W respectively and d i m ~ V< 00, dirn~W< co,then V @IF W affords the tensor product Pv 63 Pw.

1.7. Lemma. Let EIF be a f i e l d extension and let V , W be FGmodules. Then

(V @ F W ) E% VE@ E WE as EG-modules Proof. Let {v;li E I } and {wjlj E J } be F-bases of V and W , respectively. Then (1 @ (vi @ wj)li E I , j E J }

and

{(I €3 vi) €3 (1 €3 wj)li E I , j E J }

are E-bases of (V @F W ) Eand VE@ E W E ,respectively. Hence the map

extends to an E-isomorphism of (V @ F W ) Eonto VE@ E WE. It is obvious that for all v E V , w E W ,


51

Since for all g E G,

and 9((1@ V i ) 8 (1 63 Wj)) = (g(1 8 v;) 8 g(l €3 Wj)) = (1 @ 9%) €3 (1 8 gwj)

we have +(9(1€3 i.(

€3 Wj))= g+(l C3 (0;€3 Wj)),

as required.

For the rest of this section, G denotes a finite group and all vector spaces over a field F are assumed to be finite-dimensional. 1.8. Proposition. Let H be a subgroup of G, let V be an F H module with F-basis {vl,. . . ,urn} and let p be the matrix representation of H aforded b y V in this basis. If gl,g2,.. . ,gn is a left transversal for H in G , then (91 @ v l , * - - , g l

is an F-basis of V G with respect to which the matrix representation pG aforded b y V G is given b y ~ ~ (=9(Pij(9)) )

(1 L i , j L n>

where p;j(g) = p(grlggj) if gT'ggj E H and p;j(g) = 0 if gr'ggj

9 H.

Proof. That the elements (9;@ vj} constitute an F-basis of V Gis a consequence of Corollary 1.2(i). We may write p(h) = (yij(h)), where hv; = Cj y,;(h)vj. To find the matrix pG(g), fix j E { 1,. . . ,n}and write ggj =gkh(h E H).Then g(gj C3 vt) = ggj 8 vt = gkh C3 vt = gk C3 hvt m

s=l

m s=l


52

Now extend the coefficients -yst to the whole of G by setting yst(z) = 0 for x E G- H. Because for a fixed j there is only one k with gk'ggj E H, we have m n S(9j

63 4 =

y Y s t ( 9 ; 1 g 9 j ) ( 9 i (23 us), s=l i = l

as required.

Let V be an FH-module and let x be the character of H afforded by V . Then the character of G afforded by the induced module V Gis called the character induced from x; we shall use the notation xG for such a character. 1.9. Proposition. Let x be a character of H and let 91,. . . , g n be a left transversal for H in G. Set ~ ( x=) 0 for all x E G - H. (i) X G ( g ) = CZ1 x ( g i l g g i ) and IHIXG(9)= x(z-'gz) (9 E G )

c

XG

(ii)

If charF does not divide \HI, then

XG(d

=

V1-l

c

x(z-19z)

(9 E G )

zEG

(iii) If charF = 0 , then for any given g E G ,

where h l , . . . h , are the representatives for the conjugacy classes of H contained in the conjugacy class Cg of g and, b y convention, X G ( g ) = 0 if H n C, = 0. )

Proof. (i) Let x be afforded by an FH-module V. Then, in the notation of Proposition 1.8, we have trpii (9)= X ( g t l g g i ) and hence i=l

i=l

53


as required.

Since the value of xG does not depend on the basis of V G with respect to which xG is computed, it must also be independent of the choice of the left transversal for H in G. Bearing in mind that G can be written as a disjoint union of IHI left transversals for H in G, it follows from (1) that IHIXG(d =

c

x(s-lg4

(2)

XEG

as asserted. (ii) This is a direct consequence of (2). (iii) If H n Cg= 8, then the assertion follows from (i). Assume that H n Cg # 0. For any h in H n Cg,let X h denote the conjugacy class of H containing h. Then

H n cg = x h l u X h , u . - .u x h , for some h l , . . . ,hm in H . Since x is constant on each

xh,,

we have

Now if h f H n Cg, then as s runs over G, s-'gz = h for exactly C c ( g ) values of 2. Applying (2) and (3), we deduce that

as desired. We now apply Proposition 1.9(iii) to exhibit an orthogonality relation proved by Osima (1952). 1.10. Proposition. Let H be a subgroup of G, let F be an algebraically closed field of characteristic 0 and let X I , . . . ,xr be all irreducible characters of H over F . Then, for all g E G, h E H ,


54

Proof. If H n Cg = 8, then g is not conjugate to h. Hence the result follows from Proposition 1.9(iii). We may thus assume that Cgn H # 8. Denote by h l , hz, . . . ,h , the representatives for the conjugacy classses of H contained in Cgsuch that h E { h l , . . . ,h,} whenever g is conjugate to h. Since the number of conjugacy classes of H is equal to r , we may choose the representatives h,+l,. . . , h, for the conjugacy classes of H not contained in Cgwith h E {h,+l, . . . ,h,} whenever h is not conjugate to g . By Proposition 1.6.6(i), we have

2

Xk(hi)Xk(hi')

= IcH(hi)lhij

k=l

(4)

Now h = h, for some s E { 1,. .. ,r } and h is conjugate to g if and only if 1 5 s 5 m. Applying Proposition 1.9(iii), we derive

and, by (4),the latter is equal to IC,(g)l if 1 5 s 5 rn and 0 if s So the proposition is true. 2.

> rn.

Formal properties of induced modules

Throughout this section, R denotes a commutative ring and G an arbitrary group. 2.1. Lemma. (Transitivity of the induction). If L, H are subgroups of G with H L, then for any RH-module V ,

s

(VL)G

Proof. We have

2VG

2. Formal properties of induced modules

55

as required.

2.2. Lemma.

Let V be an RH-module. Then the map

is an injective homomorphism of RH-modules. Furthermore, the image

1 @ v of V in V G is a direct summand

Of(VG)H.

Proof. Since R H is a direct summand of the RH-module RG, RH @ R H V is a direct summand of the RH-module VG.Since the map

is an isomorphism of RH-modules, the result follows. H

We shall refer to the injective RH-homomorphism V + V G ,v H 1@ 2) as the canonical injection. With the aid of this injection, we now provide a universal characterization of V G . Let H be a subgroup of G, let V be an RHmodule and let f : V + V G be the canonical injection. Then, for a n y RG-module W and for any Q E HomRH(V, W H ) ,there exists one and only one $ E HomRG(VG,W ) such that tp = 1C, o f . 2.3. Proposition.

Proof. Since x@v = z ( l @ v ) , x E R G , v E V ,any RG-homomorphisr V G + W is uniquely determined by its restriction to 1 @ V . Thus there exists at most one such +. Consider the map RG x V + W , (2,v) H zcp(v). Then this map is balanced and therefore determines a 12 -homomorphism V G3 W , x @ v H xcp(v), x E R G , v E V . Because


56

for all z,z1 E RG and v E V, 1c, is an RG-homomorphism. Taking into account that $(l 8 v ) = cp(v) for all v E V the result follows. W

Let S be a subring of a ring L and let M be a (left) S-module. Given 1 E L and 1c, E Homs(L,M), define lcp : L + M by for all x E L

(/y)(x) = cp(z1)

It is straightforward to verify that the additive group Homs(L,M) becomes a (left) L-module. Let H be a subgroup of G and let V be an RH-module. Since R H is a subring of RG, we may apply the foregoing construction so that N o m m ( R G , V ) becomes a (left) RG-module. We shall denote this module by GV and refer to GV as a coinduced module. 2.4. Proposition. Let H be a subgroup of an arbitrary group G, let V be an RH-module, and let W be an RG-module. Then (i) HOmRG(VG, % HOmRH( WH) as R-modules (ii)HomRG(W,G V) S HomRH(WH, V) as R-modules

w)

v,

Proof. (i) Given X E Homm(V, W H )we may, by Proposition 2.3, define A* E HomRG(VG, w) by X*(z €3 v) = .(X(v))

The map X

H

w E V,x E R G

A* is obviously R-linear. Moreover, if A * = 0, then X * ( l @ w) = X(v) = o

for all v E V

and thus X = 0. Finally, given 1c, E HomRG(VG,W), define cp : V + W by cp(v) = $(1 €3 v) for all w E V. Then cp E H o ~ R H ( V WH) , and 1c, = cp* as required. ( V), define y(X) E H O ~ R H WH, ( V ) by (ii) For any X E H O ~ R GW,G


57

Then clearly X H p(X) is an R-linear map. If a E then the map $ ( a ) defined by

[Ilt(a>(w)l(4 = 44

H u ~ R H ( WV H ) ,,

(wE W,sE RG)

belongs to H u ~ R G ( W V ),.~Moreover, ($cp)(X) = X and Hence the map X H $(A) is a required isomorphism. H

((P$J)(CY)

= a.

Let H be a subgroup of an arbitrary group G, let V be an RH-module and let W be an RG-module. (i) If V and W are irreducible, then V is isomorphic to a submodule of WH if and only i f W is isomorphic to a factor module of V G . (ii) If V is irreducible, then for any irreducible factor module X of V", V is isomorphic to a submodule of XH. 2.5. Corollary.

Proof. (i) Note that V is isomorphic to a submodule of W H if and only if H O ~ R H ( W V ,H )# 0. On the other hand, W is isomorphic to a factor module of V Gif and only if H o ~ R ~ (WV) # ~ 0. , Now apply Proposition 2.4 (i). (ii) This is a direct consequence of (i). H We next establish the following result which will allow us to identify, under certain circumstances, induced and coinduced modules.

Let H be a subgroup of G offinite index and let V be an RH-module. Then 2.6. Proposition.

Proof. Let (91, g 2 , . . . ,gn} be a left transversal for H in G. For every $ E" v, put n

d(4)

= csi @ 4(gi') i=l

E VG /'

Then an elementary calculation shows that 8 :G V --+ V G is an Rhomomorphism. To show that 8 is an RG-homomorphism, it therefore suffices to verify that S(g4) = g O ( 4 )

for all g E G, 4 eG V


58

For every g E G, there is an h; E H an a permutation i H r ( i ) of the

index set {1,2, . . . ,n} such that 97'9 = hjg&. It follows that n

n

i=l n

i=l n

=

proving that 8 is an RG-homomorphism. If Cy=lgi 8 $(gi') = 0, then 4(gi1) = 0 for all i E { l , .. . ) n } . Because {gT1,. . . ,g;'} is a right transversal for H in G, given g E G, we have g = hg;' for some h E H and i E (1,. . . ,n}. Hence 4(g) = h4(gT1) = 0 and so q5(Csgg) = C z g 4 ( g ) = 0, proving that 8 is injective. The elements gT',gz',. . . , g i l constitute a basis of a free left RHmodule RG. It follows that for every set {vi E VI1 5 i 5 n } , there is a 4 eG V such that 4(gc') = vi, 1 5 i 5 n. Hence 8 is surjective and the result follows. 2.7. Corollary. Let H be a subgroup of G of finite index, let V be RH-module and let W be an RG-module. Then

Proof. Apply Propositions 2.6 and 2.4(ii). Suppose that U is a submodule of an RH-module V , and let i : V be the inclusion map. Since RG is a free RH-module, RG is flat, so the canonical homomorphism

U

+

is injective. For this reason, from now on we shall identify image in V G .

U G with its


59

2.8. Lemma.

Let &and V2 be submodules of an RH-module V. (i) &G G VF if and only if V1 G V,. (ii) Vp = V : i f and only if V1 = V,.

(iii) (Vl + V2)G= KG + .v : ( i ~ ) (& n v ~= )hG~n ~ 2 ~ . (w) If V = V, @ V,, then VG = V F @ CG.

Proof. We first observe that (ii) is a consequence of (i), that (v) follows from (iii) and (iv) and that (iv) is a consequence of the fact that RG is a free RH-module. To prove (i), note that V1 C V2 obviously implies YGG Vp. Let T be a left transversal for H in G containing 1. Then each element of VG can be uniquely written in the form CtET taut with finitely many vt E V not equal to 0. Therefore if 01 E V1 is such that l @ v l E V p then v1 E V2. Thus qG VF implies & C VZ,proving (i). To prove (iii), observe that the inclusion VF V? C (V1 V2)G follows from (i). The opposite inclusion being a consequence of the

+

+

equality

t 69 (v1

+

v2)

= t €3 211

+ t €3

02

(v1 E

v1,v2

E V2,t E T )

the result follows. The following theorem contains some important formal properties of induced modules.

Let H be a subgroup of an arbitrary group G and let U A V 3 W be a sequence of homomorphisms of RH-modules. Then the following properties hold: (i) The sequence 2.9. Theorem.

is exact if and only if the corresponding sequence of RG-modules

is exact. (ii) Suppose that the sequence (1) is exact. Then (1) splits if and


60

only if (2) splits. (iii) I f U is a submodule of V , then V G / U G (V/U)". Proof. (i) That exactness of (1) implies that of (2) is a consequence of the fact that RG is a flat RH-module (see Proposition 1.2.8(iii)). Conversely, suppose that (2) is exact. By Proposition 1.2.9(i), I ( e r ( l 8 p ) = (I(erp)G

and

Im(1 €9 A) =

Because I(er(1 €9 p ) = Im(1 €9 A), it follows from Lemma 2.8(ii) that K e r p = ImA. A similar argument shows that Ker(1 €9 A) = 0 implies KerX = 0 and that Im(1 €9 p ) = W Gimplies I m p = W . Hence the sequence (1) must also be exact. (ii) Suppose that 0 + U 5 V 1: W -+ 0 is a split exact sequence and let y : W t V be a splitting homomorphism. Then 1 €9 y : WG+ V G is obviously a splitting homomorphism for the induced sequence. Conversely, assume that the exact sequence

of RG-modules splits. Let T be a left transversal for H in G containing 1. Consider the mappings

v

1,

WSWG

VG

W H l ( 8 W

CtfT

t (8 V t

U1

Then c and T are well defined RH-linear maps. Let $I : W G+ V Gbe a splitting homomorphism for (2) and let w E W . Then, on the one hand, for some vt E +(I €9 w)= t B vt

C

t€T

and, on the other hand,

v


Hence

20

61

= p ( q ) and therefore

This shows that ~ $ :0W V is a splitting homomorphism for (1) and the required assertion follows. (iii) Apply (i) to the natural exact sequence 0 -+ U -+ V -+ V / U + 0. H

2.10. Corollary. Let H be a subgroup of an arbitrary group G and let V be an RH-module. I f V G is completely reducible, then so is V.

Proof. Let W be a submodule of V and let

0

-+

w

-+

v

-+

v/w+ 0

be the natural exact sequence. The corresponding sequence 0 + WG+

vG (v/w)G0 --f

-+

splits by the hypothesis. Hence, by Theorem 2.9, the sequence 0+

w v + v/w+ 0 -+

splits, proving that V is completely reducible. Let S be a ring and let V be a left S-module. Given a (left or right) ideal I of S,let IV denote the set of all finite sums of the form

Then IV is an additive subgroup of is in fact a submodule of V .

V and if I is a left ideal, then IV

2.11. Proposition. Let H be a subgroup of an arbitrary group G and let X C Y be left ideals of R H .


62

(i) ( Y / X ) GE A . Y / A- X fii) Let I be a right ideal of RG such that I 2 RG . X . If V is an RH-module, then there is a canonical injective map of additive groups

$J : I ( V G )+ ( X V ) G

Proof. All tensor products below are to be taken over R H . (i) Consider the diagram

0 -+ R G @X

+ R G @Y -+

R G @ ( Y / X ) -+ 0

1s

If 0 --* RG * X

--+

RG. Y

-+

RG * Y / R G* X

+

0

where all maps are canonical. Then the diagram is commutative and since RG is a flat RH-module, both rows are exact, and f , g are isomorphisms (see Proposition 1.2.8). This establishes the required assertion. (ii) Denote by X the natural homomorphism V + V / X V . Then the sequence

o -+ R G B X V --+

R G @v ' 9 R G B (V/XV)

is exact. Let x E I , y E RG, and v E V . Then xy E I can be written in the form

o E RG . X, so xy -+

Thus

and hence I(RG @ V ) C I(er(l@ A), as required. For the rest of this section, G denotes a finite group and F an arbitrary field. All FG-modules are assumed to be finite-dimensional

63


F-spaces.

Let U and V be any FG-modules. Then the intertwinning number for U and V is defined to be

Let U be a completely reducible FG-module, and let V be an irreducible FG-module. We say that a nonnegative interger rn is the multiplicity of V as an irreducible constituent of U if a decomposition of U into irreducible submodules contains exactly rn submodules isomorphic to

V. 2.12. Lemma. Let U and V be completely reducible FG-modules. (i) i(U,V)= i(V,U). (ii) I f V is irreducible and F is a splitting field for FG, then i(U, V ) is equal to the multiplicity o f V as an irreducible constituent of U . (iii) I f F is a splittingfield for F G , then U is irreducible if and only if2(U,U) = 1. Proof. (i) Suppose that U and V are irreducible. If U

V , then

V , then obviously HornFG(U, V ) 2 Horn=~(V, U ) . This proves the case where U and V are irreducible. In the general case, write

If U

where the { U j } and { V k } are irreducible. Because

we have

i ( U 1 e Uz, U3) = i(U1,U3) for any FG-modules U,, Uz, and

U3.

+ i(U2,

Similarly,

u3)

64


Thus

proving (i). (ii) If V is irreducible, then by the above

On the other hand, since F is a splitting field for FG,

proving (ii). (iii) Write U = mlU1 @ - . @ mtUt with U l , . .. ,Ut pairwise nonisomorphic and irreducible. Then, by (2), i(U, Us)= rn, for all s E {1,...,t } . It follows that

i(U,U ) = m:

+ - - + rn:

whence the result.

As a preliminary to the next theorem, we introduce the following definitions. Let A be a finite-dimensional algebra over a field F , and let V be an A-module. By the socle Soc(V) of V , we understand the sum of all irreducible submodules of V. We shall refer to the factor module V / J ( V ) as the head of V . 2.13. Lemma. (i) V / J ( V ) is completely reducible, and for any submodule W such that V/W is completely reducible, J ( V ) W . (ii) Soc(V) is completely reducible, and for any completely reducible submodule W of V, W C Soc(V). (iii) J ( V ) is the intersection of the kernels o j t h e homomorphisms V 4 W , where W ranges ouer all irreducible A-modules. (iv) For any given A-modules V and W , W is isomorphic t o an irreducible constituent of V /J ( V ) if and only if W is isomorphic to an irreducible factor module o j V .


65

Proof. (i) This is a direct consequence of Lemma 1.4.3 and Proposition 1.4.1. (ii) Apply Proposition 1.3.15. (iii) This is a direct consequence of the definition of J ( V ) . (iv) Apply (i). 2.14. Theorem. (Nakayama reciprocity). Let F be an arbitrary field, let H be a subgroup of G and let V and W be irreducible F H and FG-modules, respectively. If m is the multiplicity of W as an irreducible constituent in the head of V G and n is the multiplicity o f V as an irreducible constituent in the socle of W,, then m - i(W,W )= n i(V,V ) In particular, m = n if F is a splitting field for F G and FH

Proof. We first observe that HomFG(U, W )= 0 for all irreducible FG-modules U with U 9 W . Applying Lemma 2.13(i), (iii), we therefore deduce that

HomFc(VG,W ) Z H O ~ F G ( V ~ / J (WV)~ ) , Z horn^^( W,W )@ - . -@ HomFG(W, W ) (m times) Thus i(VG,W ) = m . i(W,W ) . By Lemma 2.13(ii), the image of any FH-homomorphism V -+ WHis contained in the socle S of WH.Accordingly,

and therefore i(V,W H ) = n - i(V,V ) . Bearing in mind that, by Proposition 2.4(i), i(VG,W ) = i(V,W H )the , result follows. 2.15. Corollary. (Frobenius reciprocity). Let H be a subgroup of G , let F be a n algebraically closed field such that charF does not divide /GI, and let V and W be irreducible FH and FG-modules, respectively. Then the multiplicity of V as an irreducible constituent of W H is equal to the multiplicity of W as an irreducible constituent of VG.

66


Proof. Apply Maschke’s theorem and Theorem 2.14. Let F be a splitting field for F G of characteristic 0 and let Cf(G, F ) be the F-vector space of class functions f : G + F . By defining

we obtain a symmetric nonsingular bilinear form Cf(G, F)x C f(G, F ) -+ F, called the inner product. If X I , . . . ,xr are the irreducible characters of G, then X I , . . . ,x r form an orthogonal basis for Cf(G, F ) . This can be easily seen by applying Proposition 1.6.6. Note also that, for any character x of G over F,

and a; =< x,xi >. Applying Corollary 2.15, we immediately derive 2.16. Corollary. (Frobenius reciprocity for characters). Let F be an algebraically closed field of characteristic 0, let H be a subgroup of G and let x and 0 be irreducible characters ofH and G, respectively. Then

< X G , O >=< x,e, >

3.

Annihilators of induced modules

Throughout this section, G denotes an arbitrary group and R a commutative ring. Unless explicitly stated otherwise, RG-modules are assumed to be left RG-modules. Let H be a subgroup of G and let V be an RH-module. We should like to read properties of RG from VGand conversely, properties of V G from RG. One very valuable tool for obtaining some of this exchange of information is propvided by “annihilators”, defined bellow. Let M be an RG-module. Then the annihilator of M , written ann(M),is defined by

a n n ( M ) = {z E RGlzM = 0)

3. Annihilators of induced modules

67

The annihilator of a right RG-module is defined similarly. Note that in both cases the annihilator is an ideal of RG. Our main result in this section describes ann(VG) exclusively in terms of ann( V ) ;most of the material presented is contained in a work of Kniirr(1979 ). Let H be a subgroup of G. Then the natural projection

R:RG+RH is defined by Y

It is clear that T is an R-linear map; in fact property, namely

R

satisfies a stronger

3.1. Lemma. Let H be a subgroup ofG and let R : RG be the natural projection. (i) R is a homomorphism of left and right RH-modules. (ii) For any left ideal I of RG, I E RG R ( I ) . (iii) For any right ideal I of RG, I R ( I ) - RG.

+

RH

-

Proof. (i) Let T be a right transversal for H in G such that 1 E T . Owing to Lemma 1.1, the map t H 0, 1 H 1, 1 # t E T , extends to a homomorphism R' : RG + RH of left RH-modules. Since 7r = R', we conclude that R is a homomorphism of left RH-modules. A similar argument, by taking T to be a left transversal, shows that R is a homomorphism of right RH-modules. (ii) Let T be a left transversal for H in G. By Lemma 1.1, each a E I can be uniquely written in the form

a =xtat

(at E RH)

tCT

with finitely many nonzero at. For each t f T , we have

t-'a E I

and at = r(t-'a) E w(I)

Hence a E RG . R( I ) , proving (ii). (iii) Choose T to be a right transversal for H in G and apply the argument in (ii).


68

3.2. Lemma. Let N be a normal subgroup of G and let T be a transversal for N in G. Then (i) For any ideal A of R N ,

RG - A = $tET t A - t

(direct sum of RN-modules)

where t A = tAt-' (ii) For any family {A,JiE I } of ideals of R N ,

RG - (fI;czAi) = niczRG * Ai (iii) For any ideals A, B of R N , A RG*B.

c

B if and only if RG - A

c

Proof. (i) Since N d G , t A is an ideal of R N . In particular, t A RN and hence, by Lemma 1.1,

C t A * t= $ t c T t A * t t€T

is a left ideal of RG contained in RG-A. Since for t E N , tA.t = t A = A , the required assertion follows. (ii) By (i), we have RG - A; = e t E tA;~* t and therefore

niEzRG - A; = =

t ) = $tET (nicztAi) t (nicZAi) ' t

niEz(etETtAi @tET

= RG * (niEzAi)

as asserted (iii) By (i), we have

-

Hence, if RG A c RG - B , then A c B. Conversely, if A C B , then certainly RG A C RG - B. If RG - A = RG - B , then by the above, A = B , a contradiction. Thus RG A C RG B , as required. H

-

-

For any additive subgroup A of RG, let I d ( A ) denote the sum of all two sided ideals of RG contained in A. Thus Id(A) is the unique largest ideal of RG contained in A.


69

3.3. Lemma. Let H be a subgroup of G, let I be an ideal of R H and let a : RG + RH be the natural projection. (i) Id(RG I ) = ann(RG/RG. I ) = ngEGRG(gI) where gI = gig-'. (ii) Id(RG.I) is the unique largest two sided ideal J of RG satisfying

(iii) Id( I RG) = Id( RG - I ) . (iv) If H is a normal subgroup of G, then *

I d ( RG * I ) = RG( n g E G ’ I ) Proof. (i) Because ann(RG/RG.I ) is an ideal of RG, the equality

Id(RG - I ) = ann(RG/RG - I ) is a consequence of the definition of Id(RG.1). Since I d ( R G - I ) RG-I and I d ( R G . I ) is an ideal of RG, we have

Id(RG.I ) Accordingly,

cg( R G .I ) = RG(’I)

Id(RG - I )

c ngEGRG(gI)

But ngEGRG(gI)is obviously an ideal of RG contained in RG I . Thus

as required.

(ii) Let J be an ideal of RG such that a ( J ) C I . Then, by Lemma 3.l(ii),

J~RG*T(J)GRG.I and so J C I d ( R G . I ) . On the other hand, let J be any ideal of RG contained in RG I . Then T ( J ) 2 a(RG - I ) G I , since a is a homomorphism of right RH-modules (Lemma 3.l(i)) and I is a left ideal of RH.

(iii) Applying Lemma S.l(iii) and the argument of (ii), we conclude that I d ( I RG) is also the unique largest two-sided ideal J of RG

-


70

satisfying ? r ( J ) 2 I . Thus I d ( I *RG) = Id(RG * I ) as required. (iv) This is a direct consequence of (i) and Lemma 3.2(ii). H We are now ready to describe the annihilators of induced modules.

3.4. Theorem. Let H be a subgroup of an arbitrary group G , let V be a left RH-module and let I = ann(V). (i) ann(V G )= Id( RG I ) = ng,GRG(gI). (ii) If H d G, then ann(VG)= RG(flseG”). (iii) If V G is completely reducible, then J ( R G ) G RG(gI) for all g E G. The converse is true provided R G / J ( R G ) is artinian.

-

Proof. It is clear that (iii) is a consequence of (i). Hence, by Lemma 3.3(i),(iv), it suffices to verify that ann(VG)= Id(RG * I ) . To this end, let J C RG.1 be an ideal of RG. Since IV = 0, Proposition 2.11(ii) tells us that J ( V c ) = 0. Hence J & ann(VG)and therefore

Id(RG * I )

ann(VG)

To prove the reverse containment, let T be a left transversal for H in G. Then, by Corollary 1.2, V G = @I V . If 5 E ann(VG),say 5 = &Ttzt, xt E RH, with finitely many xt # 0, then for all o E V ,

Hence each Q E I and therefore 5 E RG.1. It follows that ann(VG)G RG I and thus ann(VG) Id(RG I ) , as desired. H

-

We close by proving a number of applications of the above result. As a preliminary observation, we first record 3.5. Lemma. Let H be a subgroup of an arbitrary group G. Then

J(RG)

RG*J ( R H )

if and only if J ( R G ) 5 J ( R H ) * RG

71


Proof. Apply Lemma 3.3(iii) for I = J ( R H ) . H 3.6. Theorem. Let H be a subgroup of an arbitrary group G and let R be a commutative ring such that R H / J ( R H ) is artinian. Then

the following conditions are equivalent: (i) For a n y irreducible RH-module V , ( V G )is~completely reducible. (ii) J ( R H ) . RG C RG J ( RH ). (iii) J ( R H ) RG = RG J ( RH ) . Moreover, if these conditions are satisfied, then for any given positive integer n, ( R G * J ( R H ) ) " = RG * J ( R H ) " Proof. (i)+(ii): P u t V = R H / J ( R H ) . Since R H / J ( R H ) is semisimple artinian, V is completely reducible, so by assumption ( V G ) ~ is completely reducible; hence J ( R H ) G ann(VG). Now J ( R H ) = ann(V) and thus, by Theorem 3.4(i), ann(V G )= Id( RG - J ( R H ) ) Hence J ( R H ) R G

s I d ( R G - J ( R H ) ) and therefore J ( R H ) - RG

RG-J(RH). (ii)

+ (iii): It follows from (ii) that RG - J ( R H )- R G

C RG

*

J(RH)

whence

RG J ( R H ) *

RG - J ( R H ) * R G G Id(RG - J ( R H ) ) = I d ( J ( R H ) RG), *

by Lemma 3.3(iii). Hence RG. J ( R H ) J ( R H ) - R Gand the required assertion follows. (iii) =$ (i): It is clear from (iii) that R G . J ( R H ) is an ideal of RG. Let V be an irreducible RH-module and let I = ann(V). Then J ( R H ) 5 I and so RG.J(RH)c R G . 1


72

Applying Theorem 3.4(i), we conclude that

J ( R H ) C RG - J ( R H ) C I d ( R G - I ) = a n n ( V G ) Since R H / J ( R H ) is semisimple artinian, the latter implies that (V')H is completely reducible. The final assertion follows by induction on n, using (iii). 3.7. Theorem. Let H be a subgroup o f a group G and let R be a

commutative ring such that both R H / J ( R H ) and R G / J ( R G ) are artinian. Then the following conditions are equivalent: (i) For any irreducible RH-module V , the RG-module V G is completely reducible. (ii) J ( RG) RG J ( R H ) . (ii) J ( R G ) C J ( R H ) - RG. Proof.

That (ii) is equivalent to (iii) is a consequence of Lemma

3.5. (i) =+(ii): Put V = R H / J ( R H ) . Since R H / J ( R H ) is semisimple artinian, V is a completely reducible RH-module, hence by hypothesis V G is completely reducible. Therefore J ( R G ) V G = 0. But, by Proposition 2.ll(i),

VG RG/RG. J ( R H ) and thus J ( R G ) R G - J ( R H ) . (ii) + (i): Let W be any irreducible RH-module and let V be as in (i). Since W is irreducible, W S R H / M for some maximal left ideal A4 of R H . Hence W C V/U where U = M / J ( R H ) . Since W G VG/UG (Theorem 2.9(iii)), we are left to verify that V Gis completely reducible. But J ( R G ) C_ RG J ( R M ) implies that J ( R G ) V G = 0. Since by assumption R G / J ( R G ) is artinian, we conclude that V Gis completely reducible, as required. 3.8. Lemma. Let H be a subgroup of G. If for any irreducible RG-module V , the RH-module VHis completely reducible, then J ( R H )

J ( R G ) . The converse is true under the additional assumption that R H / J ( R H ) is artinian.


73

Proof. Let I be any maximal left ideal of RG and let V = R G / I . By assumption VHis completely reducible. Hence J ( R H ) V = 0 and so J ( R H ) I . Thus J ( R H ) 5 J ( R G ) , as required. Conversely, suppose J ( R G ) and that R H / J ( R H ) is artinian. Let V be that J ( R H ) any irreducible RG-module. Since J ( R G ) V = 0, we must also have J ( R H ) V = 0. But R H / J ( R H ) is artinian, hence VH is completely

s

reducible.

Let H be a subgroup o f a group G and let R be a commutative ring such that both R H / J ( R H ) and R G / J ( R G ) are artinian. Then the following conditions are equivalent: (i) J ( R G ) = RG' - J ( R H ) . (ii) J ( RG) = J ( R H ) - RG. (iii) For any irreducible RH-module V , V G is completely reducible, and for any irreducible RG-module M , MH is completely reducible. Furthermore, under the stronger assumption that G is finite and R is artinian, each o f t h e above conditions is equivalent to (iv) For any irreducible RG-module M , (MH)' is completely reducible. 3.9. Theorem.

Proof. The equivalence of (i), (ii) and (iii) follows from Theorem 3.7 and Lemma 3.8. Furthermore, it is clear that (iii) implies (iv). Assume that G is finite and R is artinian. Then both RG and R H are artinian and hence so are R H / J ( R H ) and R G / J ( R G ) . Let V be an irreducible RH-module. Then V Gis a finitely generated module over an artinian ring RG, hence V Gis artinian. It follows, from Corollary 1.3.2, that V Ghas an irreducible submodule, say M . Then, by Corollary 2.7,

and therefore there is an exact sequence

By Theorem 2.9, the latter gives rise to the exact sequence


74

Hence (iv) implies that V G is completely reducible. Finally, let M be any irreducible RG-module. Then (iv) implies ( M H )is~ completely reducible, hence MH is completely reducible by Corollary 2.10. 4.

Clifford's theorem

Let N be a normal subgroup of an arbitrary group G and let R be a commutative ring. Given an RN-module V and g E G, let gV be the RN-module whose underlying R-module is V and on which N acts accordingly to the rule

n * v = (g-lng)~

(v E V )

where n * v denotes the module operation in gV and nw the operation in V. Such an RN-module gV is called a conjugate of V. 4.1. Lemma. Let V be an RN-module, let U be an RG-module and let g E G. (i) g 8 V is an RN-submodule of(VC)N and g @ V 2 gV. (ii) V is irreducible (indecomposable, completely reducible) if and only if so is gV. (iii) If W is an RN-submodule of UN, then gW is an RN-submodule of UN a n d g W E

gw.

(iv) (")G E V G . (v) 'V E V and "(YV) "YV for all x, y E G, i.e. G acts on the isomorphism classes of RN-modules by conjunction. Proof. It is clear that g @ V is an RN-submodule of ( V G ) ~ . Hence, by Lemma 1.5, we need only verify (iii). It is obvious that gW is an R-submodule of UN. If w E W and n E N, then n(gw) = g(g-'ng)w E gW.Thus gW is an RN-submodule of U N . Finally, the map gW + gW,w H gw is obviously an RN-isomorphism.

Let

V be an RN-module and let H = (9 E GIV 2 W}

4. Clifford's theorem

75

Then, by Lemma 4.l(v), H is a subgroup of G. We shall refer to H as the inertia group of V ;in case H = G we shall say that V is G-invariant. We are now ready to establish the following classical result. 4.2. Theorem.

(Cliflord (1937)). Let N be a normal subgroup of an arbitrary group G, let R be any commutative ring and let U be an irreducible RG-module. If GIN is infinite, assume that UN is artinian. Then (i) UN is completely redzlcible and, in particular, UN contains as irreducible submodule, say V . (ii) If H is the inertia group of V , then (G : H ) < 0;) and there exists a positive integer e such that UN g e

(@tErtv)

where T is a left transversal for H in G and { t V ( t E T } are all nonisomorphic conjugates of V . (iii) If W is the sum of all submodules of UN isomorphic t o V , then W is an irreducible RH-module such that

W N ~ and ~ V U g WG (Here, of course, eV denotes the direct sum of e copies of V ) .

Proof. If G I N is infinite, then by Corollary 1.3.2, U N contains an irreducible submodule V . Hence UN is completely reducible, since U = CsE~gV and each gV is an irreducible RN-module (Lemma 4.1). We may therefore assume that GIN is finite, say GIN = { g l N , . . . , g k N } . Because U is irreducible,

U = RGu = RN(g1u) +

* * *

+ RN(9kt.d)

for any 0 # u E U . Thus U N is a finitely generated RN-module. Applying Proposition 1.3.4, we conclude that UN has a maximal submodule, say M . We now set

MO

= ngEGgM

Owing to Lemma 4.l(iii), each gM is an RN-submodule of U N . Moreover, if g M M' c U for some RN-submodule M' of U , then

A4 5 g - W '

c 9-'u

76


Thus M = g-'M', so M' = gM and therefore gM is a maximal RNsubmodule of U N . Hence each uN/gM is an irreducible RN-module. If g E G, then g = gin for some i E (1,. . . ,k}, n E N and gM = giM. Therefore k Mo = n;=,g;M

But Mo is an RG-submodule of U with Mo C M # U , hence Mo = 0. It follows that there is an injective homomorphism

and consequently UN is completely reducible, as required. (ii) and (iii) . Because CgEGgV is a nonzero RG-submodule of U , we have

u=cgv g€G

where, by Lemma 4.1(ii), (iii), each gV is an irreducible RN-module isomorphic to g V . If GIN is infinite, then by hypothesis there are only finitely many direct summands gV appearing in U N . But every conjugate gV of V is a direct summand of U , hence there are only finitely many nonisomorphic conjugates of V. This shows that (G : H ) = n < 00 and that if { 1 = 91,. . . ,g,} is a left transversal for H in G, then glV,. . . ,gnV are all nonisomorphic conjugates of V . Let W;be the sum of all submodules of U N isomorphic to giV (hence W = W l ) ,1 5 i 5 n. Then, by Proposition 1.3.19(ii),

Hence, to prove (ii), we are left to verify that each W; contains the same number of irreducible direct summands. Now for each g E G, gU = U and the modules gWi, gWj, i # j , have no composition factor in common. Thus W, = giW which proves (ii). It is clear that W is an RH-module and, by the above,

U = W @ g2W

@

* *.

@ g,W

(direct sum of R-modules)

Hence { W,g2W, . . . ,gnW} is a system of imprimitivity for U on which G acts transitively. Because H is the stabilizer of W , it follows from

4. Clifford’s theorem

77

Proposition 1.4(ii) that U Z W G . Finally, assume that X # 0 is an RH-submodule of W . Then

U‘ = x f3 gzx f3 *

*

f3 gnx# 0

is an RG-submodule of U . Therefore U = U’and so X = W , proving that W is irreducible. This completes the proof of the theorem.

4.3. Corollary. Let N be a normal subgroup of G of finite index and let R be any commutative ring. Then J ( R N ) J ( R G ) .

Proof. Apply Theorem 4.2 and Lemma 3.8. Let N be a normal subgroup of a finite group G, let F be a field and let U be an FG-module, dimFU < 00, such that U = CgEGgVfor some irreducible submodule V of U N . Let H and W denote, respectively, the inertia group of V and the sum of all submodules of UN isomorphic to V . (i) UN Z e ( & T t V ) , where e is a positive integer, T is a left transversal for H in G and {“It E T } are all nonisomorphic conjugates of V . (ii) W is an FH-module such that WN 2 eV and U S W G . (iii) U is irreducible if and only if W is irreducible. 4.4. Proposition.

Proof. Except for the irreducibility of W , properties (ii) and (iii) of Theorem 4.2 were deduced from the fact that U = x g E G g V . Hence the properties (i) and (ii) of the present proposition also hold. By the foregoing, we are left to verify (iii). If U is irreducible) then W is irreducible by Theorem 4.2(iii). Conversely, suppose that the FH-module W is irreducible. Let X be an irreducible submodule of U and let Vo denote an irreducible submodule of X N . Because VO is a n irreducible submodule of U N and UN is the sum of irreducible FN-modules gV, g E G, it follows that V, 2 gV

Z

for some g E G

gV

Now X = CgEGgVO,since X is irreducible. On the other hand,

g-lvo2

9-1

v

0 -N 9-1

(”)

v


78

by virtue of Lemma 1.5. Hence, replacing & by g-lVO, we may assume that V E Vo, in which case H is also the inertia group of Vo, Let Wo be the sum of all submodules of X , isomorphic to Vo. By Theorem 4.2(iii), Wois an FH-module such that X E W:. Now Wo is a nonzero submodule of W , so W = Wo, by the irreducibility of W . Thus, by (ii), we have

x 5% w,"= WG5%u Hence U is irreducible and the result follows.

Let N be a normal subgroup of a finite group G, let E I F be a field extension and let U and V be, respectively, an irreducible FG-module, dimFU < 00, and an irreducible FN-submodule of U N . Let H and W denote, respectively, the inertia group of V and the sum of all submodules of UN isomorphic to V . If VE is an irreducible EN-module, then (i) UN E e ( e t 'E V )~where e is a positive integer, T a left transversal fro H in G and {'V(t E T } are all nonisomorphic conjugates of V . (ii) (UE)N e ( e t 'E (VE ~) ) and {'(VE)JtE T } are all nonisomorphic conjugates of VE. (iii) W is an irreducible FH-module and W E is an EH-module such that ( W E ) N 2 eVE , UE (WE)G 4.5. Corollary.

(iv) U E is irreducible if and only if WE is irreducible. In particular, if V and W are absolutely irreducible, then so is U . Proof. Because U = CgEGgVand

it follows that

By Deuring-Noether theorem (see Huppert and Blackburn (1982)), for any 2, y E G , "V E u V if and only if ("V), E (YV),. Because

5. Dual and contragredient modules

79

g ( V ~2 ) ( ~ V ) E(Lemma 1.5(v)), it follows that H is the inertia group of VE. Bearing in mind that W = & H h V , we also have WE = C h E H h Vand ~ the latter is, of course, the sum of all submodules of (UE)N isomorphic to VE. Applying Theorem 4.2 and Proposition 4.4, we conclude therefore that (i)-(iv) hold except possibly for the value of e in (ii). The correctness of value of e being a consequence of UN Z e ( B L E'rV ) ,the equality (UE)N= ( U N ) Eand the isomorphismg(VE) (gV)E, the result follows. 5.

D u a l a n d contragredient modules

Throughout this section, R is a commutative ring, G a finite group and A is an R-algebra. By convention, an A-module means a left A-module. Let V be an A-module. Then the dual $' of V is a right A-module whose underlying additive group is H o r n ~ ( v , R )with the action of A given by (fa)(.)

= f(m) for all

f E H o r n ~ ( V , R ) , Ea A , v E V

Similarly, if V is a right A-module, then the dual $' of V is an A-module whose underlying additive group is H o r n ~ ( VR , ) with the action of A given by (af)(v) = f ( v a )

for all

f E H o r n ~ ( VR, ) ,a E A, v E V

If V, W are A-modules and f E H o r n ~ ( VW , ) ,then is defined by ( j + ) ( v ) = +(f(v>> for all

vE

v,$E r/i/

f* E H o r n ~ ( l / ic) /, (1)

If V , W are right A-modules and f E HomA(V, W ) ,then f^ E H o ~ A (9 ~ ) , is defined analogously. Let U be an A-module. If V is a submodule of U , define

vL= {$ E fil+(V)= 0) Then V L is obviously a right submodule of the right A-module is also clear that if V, C V, then V' 2

K ' .

fi.

It


80

5.1. Lemma. Let Then

U ,V , W be A-modules such that U = V 63 W .

o=W'@V'

and W L g Q ,

vl z2

w

Proof. It is clear that W' n V' = 0. Given f E 0,we may write f = f1 f 2 where f1, f 2 E 0 satisfy f l ( W )= f2(V) = 0. Thus 0 = W' @ V*. Finally, if f E W', then f H f l V is obviously a n isomorphism of W' onto p. Similarly, if f E V I , then f H flW is a n isomorphism of V* onto

+

w.

5.2. Lemma. Let V and W be A-modules. (i) Iff : V -+ W is a surjective A-homomorphism, then f : W -+ 6' is an injective A-homomorphism. (ii) The map V -+ v H v*, where v*($) = $(v) for all 1c, E P, is an A-homomorphism.

(vr,

Proof. (i) This is a direct consequence of the definition o f f given by (1). and (ii) The given map is obviously additive. If ZI E V , It, E a E A, then

as required. In what follows, given a n A-module V , we write VRfor V regarded as an R-module. Assume that VRis free with a finite R-basis vl, . . . ,v,,. Then P is free with a basis 61,.. . ,6, given by 6;(vj) = &j. We shall refer t o C 1 , . . . ,Gn as the dual basis of 01,. . . ,v,. 5.3. Lemma. Let V be an A-module such that VR is free with a finite R-basis vl,. . . ,v,,, and let G1,.. . ,G, be the dual basis of v1,. . . , v,. If a E A and av; = Cj"= rjivj , with rj; E R, then 6 j a = r$j.

cjn=l

81


Proof. For all i, k E { 1 , 2 , . . . ,n}, we have

Thus 6;a =

cy=lr;j6j, as required.

5.4. Lemma. Let V and W be A-modules such that VR and WR are finitely generated free R-modules. (i) The map V --$ v H v*, where v*($) = $(v) for all $ E $' is an A-isomorphism.

($'I,

A

(ii) I f f E HornA(V, W ) ,then j ( v * ) = f(v)* for all v E V . Furthermore, i f f is an isomorphism of V onto W , then f is an isomorphism o j w onto

Q.

Proof. (i) By Lemma 5.2(ii), the given map is an A-homomorphism. Let q , . . . ,v, be an R-basis of V and let Gl,. . . ,f i n be a dual basis of V. Then vT(Gj) = Gj(vi) = 6jj which shows that TI;, . . . ,v;t is an R-basis of

c. Thus the given map is an isomorphism.

(ii) For any given v E V and ?j E W , we have

proving the first assertion. Assume that f is an isomorphism of V onto W . If vl, . .. ,v, is an R-basis of V , then maps the dual basis f ( v l ) ,. . . ,f(vn) onto 61,. . . ,Gn. Thus jis an isomorphism of r/i/ onto h

1

h

v. rn

5.5. Lemma. Let V be a right A-module such that V is R-free. Then ann(V) = ann(C/)

+

Proof. Assume that z E ann(V)and let be an arbitrary element of Then, for all v E V , (z$)(v) = vz) = $(O) = 0, so z E ann(P). Conversely, suppose that z E ann(V). Then, for all $ E and all v E v,

c.

$1

$(4 = (.$)(.I

Q

=0


82

Since V is R-free, we conclude that vz = 0 for all v E V . Hence z E ann(V),as required. Turning our attention to induced modules over group algebras, we now prove 5.6. Proposition. Let H be a subgroup of an arbitrary group G and let V be an RH-module. Then

(vGj 2 (QG Proof. Let T be a left transversal for H in G. By Corollary 1.2(j), a typical element of V G can be uniquely written in the form

Ct8vt

E V)

(Vt

tET

with finitely many ut # 0. Because T-l is a right transversal for H in G, each element of (p)G can be uniquely written in the form

c

$t

8 t-l

($t

E

t€T

with finitely many

gt # 0. Consider the map 8 : (P)" -+(vGj

defined by

Then 8 is obviously an R-homomorphism. If 8 (CtET $t 8 t-') = 0, then $t(vt) = 0 for all t E T , vt E V . Hence each $t = 0 and so 8 is injective. If X E (V"], then

v,

hence t9(C At 8 t - l ) = X where Xt(vt) = X(t 8 vt). But each At is in and so 8 is surjective. We are therefore left to verify that for z = $ @ t - l ,


83

Thus O ( z g ) = O ( z ) g and the result follows. Let V be an RG-module. An element v E V is said to be Ginvariant (or simply an invariant element) if g v = v for all g E G. In what follows Inv(V) denotes the set of invariant elements of V . It is clear that Inv(V) is the largest R-submodule of V on which G acts trivially. 5.7. Lemma. Let V and W be RG-modules. Then the RGmodule H o ~ R ( V W, ) becomes an RG-module under the following action of G:

(sf)(.) = s ( f ( 9 - W

(9 E G, ZI E

v,f E H O ~ R ( Vw) ,

Fu rt h e rmo re, Inv( HomR( V,W ) )= H O ~ R GV,(W ) Proof. The first statement is obvious. To prove the second, note that f E Inv(Horn~(V,W) if and only if f ( v ) = ( g f ) ( v ) = g ( f ( g - ' v ) ) for all v E V , g E G. The last condition is equivalent to f E H o ~ R G ( V W, ) . The special case of the RG-module HornR(V,W ) ,in which W = R and g r = r for all r E R, g E G deserves a separate mention. In this case, HomR(V,R ) = as an R-module and the action of G on is given by

c

( g f ) ( v >= f ( g - l v )

(v E V , g E G,f E

c)


04

We shall denote this module by V*and refer to V' as the contragredient of

v.

5.8. Lemma. Let V be an RG-module which is R-fiee on afinite basis, say vl,. . . , v n and let r : G + G L n ( R ) be the matrix representation of G agorded by V with respect to v1,. . .,vn. Then, with respect to the R-basis $ 1 , . . . ,V;, of V* the matrix representation of G agorded b y V*is given b y g Ht r(g-*) (9 E G) where T ( g - ' ) is the transpose of q g - ' ) .

Proof. Since the left action of g on l;i in V* is the same as the right action of g-' on .L;i in P, the result follows by virtue of Lemma

5.3. Our next aim is to provide an analog of Proposition 5.6 for contragredient modules. This will be achieved as an easy consequence of the following general result.

.

5.9. Proposition Let H be a subgroup of an arbitrary group G, let V be an RG-module and let W be an RH-module. Then (i) om^( W,v)]' HornR( wG,V ) (ii) [ H O ~ ~ ( w)]" V ~ , 2 H O ~ R ( V w") ,

Proof. Let T be a left transversal for H in G. (i) A typical element f E [ H o ~ R ( W VH)]" , can be uniquely written in the form f=Ct@ft

(ft

E H o m ( W 7 VH))

t€T

with finitely many ft # 0. Consider the map

w,V H ) I G + HomR( W G V, )

8 : [HOrnR( defined by


85

(only finitely many wt # 0). Then 8 is clearly an R-homomorphism. If 6(f) = 0, then O ( f ) ( t 8 w)= t f t ( w ) = 0 for all t E T ,w E W . Hence v) each f t = 0 and therefore 0 is injective. If X E H o r n ~ ( W ~ ,then (&T t 8 w t ) = E t E T tXt(wt),where xt(wt) = x(t 8 wt).Since each At is in horn^( VH),we see that 8(f) = for f = &T t 8 At. Hence 6 is surjective. Fix g E G and, for each t E T , write g t = t'ht with ht E H and t I+ t' a permutation of 2'. Hence, if f = CtETt @ f t ,then g f = CtETt ' 8 htft. Moreover,

w,

and

proving (i). (ii) A typical element f E [ H o r n ~ ( WIG v ~ , can be uniquely written in the form

with finitely many

ft

# 0.

Consider the map

8 : [ H O r n R (V H , W)IG--+ H O r n R (

v,W C )


86

defined by O(f>(v) =

ct 8

ft(t-'v)

tCT

Then 8 is obviously an R-homomorphism. If S(f) = 0, then ft(t-'v) = 0 for all t E T , v E V . Hence f t = 0 for all t E T , so f = 0 and thus 8 is injective. If X E H o r n ~ ( VW"), , then X(v) = & - t8 Xt(t-'v) for some At E H o r n ~ ( vW~ ), with finitely many X t # 0. Hence e(z) = X for z = CtET t 8 A t ) proving that 0 is surjective. Fix g E G and, for each t E T , write g t = t'ht with t H t' a permutation of 2'. Then, for all 'u E V , (9W))(.)

= ge(f)(g-''u) =

c

9t 8 ft(t-lg-lv)

t U

and

proving (ii). 5.10. Corollary. Let H be a subgroup of an arbitrary group G and let W be an RH-module. Then

(W*)"c (W")* Proof. This is a special case of Proposition 5.9(i) in which V = R and G acts trivially on V . 5.11. Corollary. The group ring RG is self-contragredient, that is, (RG)*E RG as left RG-modules. Proof. Let H = 1 and let W be the RH-module R on which H acts trivially. Then W" = RG 8~ R E RG and W* E W.Hence, by Corollary 5.10, (RG)*E (W')" Z W GE RG

87

6. Induction, restriction, and outer tensor products

as required. H

Let V be a finitely generated projective RGmodule. Then V* is also projective. 5.12. Corollary.

Proof. By hypothesis, RG @ . . . @ RG = V @ W for some RGmodule W . Hence, by Corollary 5.11, V* is a direct summand of R G @ . . . @ RG. Thus V* is projective. H 6.

Induction, restriction, and outer tensor products

The aim of this section is twofold: first, to relate the induction and restriction processes, and second, to prove that the operation of forming outer tensor products commutes with the operation of forming induced modules. Throughout, R denotes a commutative ring and G an arbitrary group. Let V and W be RG-modules. Then the R-module V @ R W is an RG-module, where the action of the elements g,g E G is defined by

(v E v , w E W )

g(v@w) =gv@gw

and then extended to V @ R W and RG by R-linearity. We shall refer to V @ R W as the (inner) tensor product of V and W . It is clear that V @ R W S' W @ R V. Suppose now that G1 and G2 are groups and that & and V, are RGl and RG2-modules, respectively. Then the outer tensor product V, # V2 of V, and V2 is an R(G1 x Gz)-module whose underlying R-module is V, @R V2 with the module operation given by ( z , y ) ( v l8 v2) = X v 1 @ yv2

(z

E Gl7y E

G27v1

E

K,v2

E

b)

and extended to R(Gl x G 2 )and V, @ R V2 by R-linearity. Another interpretation of the outer tensor product V, # V2 comes from the isomorphism RGI @ R RG2 R(G1 x G2). Indeed, V, @ R V, is an RG1 @ RG2-module by means of (U

8 b)(v1 @ ~

2 = ) U V ~8 bv2

(a E

RG1, b E RG2, ~1 E V , 7

~2

E V,)

Hence, if we identify RG1 @ RG2 with R(GI x G2), the module above is nothing else but the outer tensor product of V, and V2.


88

6.1. Theorem. Let H be a subgroup ofG, let U be an RH-module and let V be an RG-module. Then

(U€3 vH)G Proof.

= U G €3 v

as RG-modules.

Let T be a left transversal for H in G. Then the map

f :UG x

v + (U€3RVH)G

given by

is balanced. Hence f determines a unique R-homomorphism

v

€3 G (U €3 VH)G €3 u ) €3 2, H t €3 ( u €3 t-lv)

{ UG (t

Similarly, the map ( U ~ V H -,)UG@V, ~ t@(u@t-lv)H (t@u)@vis an R-homomorphism which is inverse to f*. Thus f* is an R-isomorphism. Fix g E G and write gt = tlh with tl E T , h E H . Then

f*(g[(t€3 U ) €3 v]) = f * ( ( t l€3 hu) €3 gv) = t i €3 (hu 8 t;'gh) and

Hence f* is an RG-homomorphism, as required. 6.2. Theorem. Let Hi be a subgroup of a group G;, i = 1,2, and let V, be an RH;-module, i = 1,2. Then

Proof. We may identify R(H1 x H2) with R H I C ~ R R H and ~ R(G1 x G2) with RG1 € 3 RG2. ~ With these identifications, ( a €3 b ) ( q €3 vz) = av1 @ bv2 for all v1 E V1, v2 E V2, a E RH1, b E RH2 and a similar

7. Mackey theorems and their applications

formuls holds for the action of RGl €3 RG2 on result, we have to establish an isomorphism

89

KG1#bGz. To prove the

V = (RGi € 3 RG2) ~ € ~ ‘ R H ~ @ R(V,#v2) H~ (RGi @ R H ~h)# (RG2 @ R H ~VZ) = W of (RG1 €3 RG2)-modules. Let Ti, i = 1,2, be a left transversal for H;

in G;. Then, by Corollary 1.2,

and

kv = @ ( t l , t z ) ~ ~ l x€3~ Vi) z ( t €3 i ( t 2 8 V2)


f “tl €3 t2) €3 (01 €3 0 2 ) ] = (tl €3 01) 0 (t2 €3 v2) is obviously an R-isomorphism, we see that f induces an R-isomorphism f* : V --f W. We are thus left to verify that for any given 91 E GI, 92

E Gz,

To this end, write g;t; = sihi with

s;

E T,, hi E H , i = 1,2. Then

as required. W

7.

Mackey theorems and their applications

Throughout, unless explicitly stated otherwise) G denotes an arbitrary group and R a commutative ring. Our aim is to prove some classical


90

results of Mackey and to provide some applications of them. In particular, we present a criterion for two induced modules to be disjoint and a criterion for the irreducibility of induced modules. The section ends with the proof that if V and W are irreducible RN-modules, N 4 G, then VG S W Gif and only if V Z gW for some g E G. In what follows, we put t &Ht-'nS = (tV)tHt-lnS

7.1. Theorem (Mackey Decomposition (Mackey (1958)). Let H and S be subgroups of G , let T be a full set of double coset representatives for ( S ,H ) in G and let V be an RH-module. Then

Proof. Let {g;Ii E I} be a left transversal for H in G. Then, by Corollary 1.2 and Lemma 1.5(i),

V G= @iefgi8 V


and gi €3 V

"V

(as R(g;Hg;l-modules)

(1)

Put X = {g; €3 Vli E I } . Then G and, in particular S , acts on X . Moreover, 9; €3 V and gj €3 V lie in the same S-orbit if and only if gi and gj belong to the same double (S,H)-coset. For each t E T , let Wt denote the sum of the gj €3 V for which gj E S t H . Then each Wt is an RS-module and

(vG)s= @tETWt Setting I4 to be the restriction of tV to R ( t H t - l to verify that Wt E (K)". Let J C I be such that

n S ) , we are thus left

Then S acts transitively on the set {gj@VIj E J } and under this action the stabilizer o f t €3 V is {S

E Slst 8

v = t Q V } = {S

E Slt-lst E H } = t H t - l n s

91


Hence, by (1) and Proposition 1.4(ii), Wt Z' (&)' as required. W

7.2. Theorem. (Mackey Tensor Product Theorem (Mackey (1958)). Let H and S be subgroups of G , let V be an RH-module and let W be an RS-module. Then

where T is a full set of double coset representatives for ( S ,H ) in G.

Proof. Invoking Theorems 6.1 and 7.1, we have

On the other hand, by Theorem 6.1,

The desired conclusion is therefore a consequence of the transitivity of induction. In what follows, given an RG-module V , we write I n v ( V ) for the R-module of G-invariant elements of V .

7.3. Lemma. Let H be a subgroup of G of finite index, let T be a left transversal for H in G with 1 E T and let V be an RH-module. Then, the m a p I n v ( V ) + Inv(VG) &Tt

@ 'U

is an R-isomorphism.

Proof. It is clear that if w is H-invariant, then

&Tt

@

v is G-

invariant. Since the given map is obviously an injective R-homomorphism, we are left to verify that any G-invariant element of V Gis of the form


92

CtETt 8 v for some v E I n v ( V ) . So assume that x E V Gis G-invariant and write

x=Ct@vt

(vt

EV)

t€T

Then the equalities t x = z and hx = x ( h E H ) imply, respectively, vt = v1 and hvl = vl, t E T , as required. W In what follows, for any RG-module V , V' denotes the contragredient of V .

7.4. Lemma. Let V be a finitely generated R-free RG-module and let W be an RG-module. Then

V*8 W

HomR(V, W )

as RG-modules

Proof. Let v l , . . . ,v, be an R-basis of V and let v;, . . . ,vz be the dual basis of V*. Define 1c, : V*8 W + H o r n ~ ( VW , ) by

[+(f €3 41(4= [f(v>lw

(v E

v,w E W,f E V*)

Then 1c, is obviously an R-homomorphism. If 1c, (CZ,vr €3 w;) = 0, then wj

C V8 ~

= {IC,

20;

1

}(~j) =

o

proving that 1c, is injective. If X E Horn~(V, W ) ,then X = II, (CZ, v t €3 X(v;) and thus 1c, is an R-isomorphism. Finally, fix g E G , v E V,w E W and f E V*. Then

7.5. Lemma. Let F be a field, let G be a finite group and let V and W be finitely generated FG-modules. Then

dirnFlnv(V* 8 W ) = i ( V , W )


Proof.

93

Applying Lemmas 5.7 and 7.4, we have

as required. W

As an easy application of the above, we now prove

7.6. Theorem. (Mackey Intertwining Number Theorem (Mackey (1958)). Let F be an arbitrary field, let H and S be subgroups of a finite group G and let V and W be finitely generated F H and FS-modules, respectively. Then

where T is a full set of double coset representatives for ( S ,H ) in G.

Proof. We have

where the last equality follows from Lemma 7.5 and the obvious fact that '(V') = ('V)'. In what follows, all modules are assumed to be finitely generated over their ground rings. Let F be a field, let G be a finite group and let V and W be FG-modules. We say that V and W are disjoint if they have no composition factors in common.


94

7.7. Lemma. Let F be a field, let G be a jnite group such that charF does not divide IGl and let V and W be FG-modules. Then V and W are disjoint if and only if i(V,W ) = 0. Proof. By Maschke's theorem, we may write V = CBVj and W = (4) and {Wk} are irreducible FG-modules. Hence

$Wk, where the

i(v,W ) = C i ( 4 ,Wk) j,k

Since y $?! Wk if and only if i ( y ,Wk) = 0 , we see that i(V)W ) = 0 if and only if vj $?! Wk for all j , k. But the latter is equivalent to the disjointness of V and W , hence the result. We are now ready to provide two applications of Theorem 7.6. The first gives a criterion for two induced modules to be disjoint and the second provides a criterion for the irreducibility of induced modules.

7.8. Theorem. Let G be a finite group, let F be a field with charF not dividing IGl and let H and S be subgroups of G. If V and W are F H and FS-modules, respectively, and T is a full set of double coset representatives for ( S ,H ) in G, then V G and W G are disjoint if and only if for all t E T , the F(tHt-' n S)-modules t q H t - l n S and WtHt-lnS are disjoint. Proof. Apply Lemma 7.7 and Theorem 7.6.

7.9. Corollary. Further to the hypotheses and notation of Theorem 7.8, assume that both V Gand W Gare irreducible. Then V G$?! W G i f and only if f o r all t E T , the F(tHt-' fl S)-modules t&Ht-lnS and WtHt-lnS are disjoint. Proof. Since V G and W Gare irreducible, they are disjoint if and only if V G W G .Now apply Theorem 7.8. 7.10. Theorem. Let H be a subgroup of afinite group G, let F be an algebraically closed field with charF not dividing [GI and let V be an FH-module. Denote b y T a full set of double coset representatives for


95

( H ,H ) in G. Then V Gis irreducible if and only i f V is irreducible and for all t E T - H , the F(tHt-' n H)-modules t&Ht-'nH and &Ht-'nH are disjoint. Proof.

Applying Theorem 7.6 for W = V and S = H , we have

Also, by Lemma 2.12(iii), V Gis irreducible if and only if i(VG,V G )= 1. Since for t E H , t K H t - l n H S K H t - I n H (Lemma 1.5), it follows that V G is irreducible if and only if

and

i(V,V ) = 1 Now apply Lemmas 2.12(iii) and 7.7. H

A particular case where H 4 G deserves special attention and is recorded in the following: 7.11. Corollary. Let N be a normal subgroup of a finite group G, let F be an algebraically closed field with charF not dividing [GI and let V be an irreducible FN-module. Then V G is irreducible if and only i f V g V for a l l g E G - N . Proof.

Direct consequence of Theorem 7.10.

Applying Theorem 7.10 to one-dimensional representation of H , we obtain

7.12. Corollary. Let H be a subgroup o f a finite group G , let F be an algebraically closed field with charF not dividing IGJand let p be a one-dimensional representation of pi. Then the induced representation pG is irreducible if and only iJ for each g E G - H , there exists x E gHg-' n H such that p ( z ) # ~ ( g - l z g ) .


96

Proof. Let V be an FH-module which affords p . Then, for any g E G, the representation g p of gHg-' given by "(z) = p(g-'zg) is afforded by g V . Since, for any given g E G, the F(gH9-l n H ) modules '&Hg-InH and &Hg-lnH are irreducible, they are disjoint if and only if they are nonisomorphic. In terms of representations, the latter is equivalent to the existence of z E gHg-' nH such that p ( z ) # g p ( z ) = p(z-'gz). The desired conclusion is therefore a consequence of Theorem 7.10. W We close by presenting two results pertaining to an arbitrary group G and an arbitrary commutative ring R. The following observation will clear our path.

7.13. Lemma. Let S be an arbitmy ring, let V be an S-module such that V = @,EIK whem each V , is irreducible, and let W be an irredu'cible submodule of V . (i) W Z I.j. for some j E I . (ii) If all the are nonisomorphic, then W = I.j. for some j E I . (i) Let J C I be such that the 5 ,j E J are all representatives of the isomorphism classes of V,. For each j E J , denote by Wj the sum of all V; with V; E %. By Proposition 1.3.19(i), the Wj,j E J are all homogeneous cornponenets of V . Hence W Wj for some j E J . Since W is irreducible, W C Xl @ - @ X , for some finitely many S-modules X I , . . . ,X,, isomorphic to 5. Thus W E % by the Jordan-Holder theorem. (ii) If all the V; are nonisomorphic, then Wj = 5 and W C Since the result is established. the latter is possible only in the case W = 6 ,

Proof.

- -

K.

W

7.14. Theorem. Let N be a normal subgroup of an arbitrary group G, let R be a commutative ring and let V be an irreducible R N module. If gV 9 V for all g E G - N , then V G is irreducible.

Proof.

Let W be a nonzero submodule of

VG. Then

W N

is a

8. Counting induced modules and characters

97

submodule of ( V G )and ~ (VC)N

= $ t u t €3

v

(direct sum of RN-modules)

where T is a transversal for N in G. Hence ( V G ) N (and therefore W N ) is completely reducible. Let X be an irreducible submodule of W'. Then X is an irreducible submodule of ( V G )and, ~ by hypothesis and Lemma 1.5(i), the t €3 V,t E T , are nonisomorphic. Thus, by Lemma 7.13(iii), X = t o @ V for some to E T . For any g E G , g X Wwhich implies that t €3 V C W for all t E T . Thus Vc = W as we wished to show. W

7.15. Theorem. Let N be a normal subgroup of an arbitrary group G, let R be a commutative ring and let V and W be irreducible RN-modules. Then V GE W G i f and only i f V 2 gW for some g E G. Proof. If V 2 gW, then by Lemma 1.5(ii), V G (gW)GE WG. Conversely, assume that V G W G .If T is a transversal for N in G, then and (W'), = $t& 8 W (V')N = $t& €3 V Since (VG)' 2 (W C )and ~ 1€3 V is an isomorphic copy of V contained in (VG)', V is isomorphic to a submodule, say X of ( W c ) N . Since, by Lemma 7.13(i) and 1.5(i), X is isomorphic to t €3 W E tW for some t E T , the result follows. w 8.

Counting induced modules and characters

Throughout, N denotes a normal subgroup of a finite group G and F an arbitrary field of characteristic p 2 0. Our first aim is to find the number of nonisomorphic FG-modules induced from the irreducible FN-modules. We begin by recording some preliminary information. All FG-modules bellow are assumed to be finitely generated. Let V be an FG-module. We say that V is a permutation module if there exists a basis {vl,. . . ,vn} of V on which G acts as a permutation group.

8.1. Lemma. Let V be a permutation module and let B = {vl,. . . ,v,} be a basis of V on which G acts as a permutation group.


98

(i) If B1,. . . ,B, are G-orbits of B and H; is the stabilizer of any given element in Bi,l 5 i 5 r then

V

E

$i=l(l~i)G and

r = dirnFInv(V)

(ii) v E V'. (iii) IfB' = {ui,.. . ,vk} is another basis o f V on which G acts as a permutation group, then B' and B have the same number of G-orbits. Moreover, ifcharF = 0 , then for each g E G, the number of elements of B fixed b y g is equal to the number of elements of B' fixed by g . Proof. (i) This is a direct consequence of Proposition 1.4 and Lemma 7.3. (ii) Apply Corollary 5.10 and (i). (iii) The first statement follows from (i). To prove the second, assume that churF = 0 and fix g E G. Denote by M and M' the matrices of the linear transformation g with respect to B and B', respectively. Then M and M' are similar and hence t r M = trM'. Because g permutes the elements of B and B', trM and trM' are, respectively, the numbers of elements of B and B' fixed by g. This completes the proof of the lemma. Given an FN-module V and g E G, denote by x and characters of N afforded by V and gV respectively. Thus

"(4= x(9-lx9)

for all

5

g x

the F -

EN

Let n be the least common multiple of the orders of the p'-elements of G and let e be a primitive n-th root of unity over F . Denote by I, the multiplicative group consisting of those integers p , taken modulo n, for which e H e p defines an automorphism of F ( E )over F . Recall that, by definition, two p'-elements a, b E G are F-conjugate if z-lbx = up for some x E G and some p E I,. Owing to Proposition 1.6.7, the number of nonisomorphic irreducible FG-modules is equal to the number of F-conjugacy classes of p'-elements in G. 8.2. Lemma. Let x be a p'-element of N and let T, and I(, be the F-conjugacy classes of G and N, respectively, with x E T and x E K .


99

Then

Tz = UgEGKg-Izg Proof. Let m be the least common multiple of the orders of the $-elements in N , so that n = mk for some natural number k and S = E~ is a primitive m-th root of unity over F . Assume that s E T,. Then s = g - ' s p g for some g E G and some p E I,. If p = r ( m o d m ) , 0 5 r 5 m - 1, then s = g-ls'g. The automorphism E H cfi of F ( E ) over F induces the automorphism 6 H 6p = 6' of F(6) over F . Hence r E I, and s E Kg-lZg. Since every automorphism of F ( 6 ) over F extends to that of F ( E )over F , it follows that if two elemets of N are F-conjugate in N , then they are F-conjugate in G. This proves the opposite inclusion, as desired. 8.3. Lemma. Let V and W be irreducible FN-modules and let x v and xw be the characters afforded b y V and W , respectively. Then V G W G if and only if x v = gxw for some g E G.

Proof. This is a direct consequence of Theorem 7.15 and Proposition 1.6.3. W Let x be the character of an F-representation p of G. If the $-elements a , b E G are F-conjugate, then x ( u ) = x(b). 8.4. Lemma.

By hypothesis, s-'bs = a' for some x E G and some Proof. r E I,. Let u be the automorphism of F ( E )over F such that a(&)= E'. We may choose a nonsingular matrix M such that p ( u ) = M-'diag(c*',

where k = degp and

~ ( b=) x ( a r ) = as required. W

.. . ,E @ ~ ) M

. . ,cyk are positive integers. Thus + . . . + &*k' - a(&*]+ * . + E a r ) = o(x(a))= x ( a ) ~1,.

Let x be the character of an F-representation of G and let K be an F-conjugacy class. Owing to Lemma 8.4, we may define x ( K ) unambiguously by setting

x(10 = x ( g )

for

g EK


100

Recall that, by Propositions 1.6.7 and 1.6.3, the number of irreducible F-characters of G is equal to the number of F-conjugacy classes of G. We are now ready to prove the following crucial result. 8.5. Theorem. Let xl,. . . ,xm and K1,. . . ,Km be the irreducible F-characters of G and the F-conjugacy classes of p'-elements of G, respectively. Let A be a group acting on {XI, . . . ,xm}and { Kl, . .. ,K m } such that

x;(Kj) = "x;("Kj)

for all

a E A , i , j E { l ) .. . ,rn}

Then the number of A-orbits of { X I , . . . ,Xm} is equal to the number of A-orbits of {KI,. . . ,K,}. Furthermore, if charF = 0 , then for each a E A, the number of elements of {XI,... ,xm}fixed b y a is equal to the number of elements of {Kl, . . . ,K,} fixed by a.

Proof. We denote by V and W the permutation FA-modules corresponding to the action of A on { X I , . . . ,x m ) and { K l , . . . Km} respectively. Let pv and pw be the matrix representations of A afforded by V and W with respect to the bases {xl,.. . ,x,} and {Kl,. . . ,Km}, respectively. Put M = (x;(Kj)), 1 5 i , j 5 rn. By Proposition 1.6.3, the characters X I , . .. ,Xm are F-linearly independent and hence M is nonsingular. Fix a E A and write pv(a) = (a;j),pw(a) = ( b ; j ) . Then, for any # xi. given j E (1,. . .,rn}, a;, = 1 if c x j = xi and a;, = 0 if Similarly, b;j = 1 if "Kj = K; and b;, = 0 if "Kj # K;. Put )

(cij) = pv(a)M

and

(dij) = M p w ( a )

k=l

where the last equality follows from the assumption that x;(Kj) = "xi( " K j ) . On the other hand,


101

which shows that pv(a)M = M p w ( a ) for all a E A. Since M is nonsingular, we conclude that V 2 W . Thus, by Lemma 8.1(i), the number of A-orbits of { X I , . .. ,x,} is equal to the number of A-orbits of {KI,. . . ,K,}. Finally, assume that charF = 0 and let xv and x w be the characters of V and W . Since V W , we have x v ( a ) = Xw(a) for all a E A. But x v ( a ) is the number of elements of {xl,.. . ,x,} fixed by a , while Xw(a) is the number of elements of { K l , . . . ,K,} fixed by a , hence the result. We shall say that two F-characters a,,f?of N are G-conjugate if /? = ga for some g E G.

Let N be a normal subgroup of G and let F be an arbitrary field of characteristic p 2 0. (i) The number of G conjugacy classes of irreducible F-characters o f N is equal to the number of F-conjugacy classes of p'-elements of G contained in N . (ii) If charF = 0 or charF does not divide IN1 and F is algebraically closed, then for any g E G the number of irreducible F-characters x of N with g x = x is equal to the number of F-conjugacy classes I( of p'-elements of N such that gKg-' = I,

= w'@f(V)H

Indeed, if f ( v ) = CtfTt @t-'v E f ( V )n W', then v = 0. On the other hand, if t' E T n H, then

ct

@ vt -

ct

@ t-lt'vt1 E

W',

tET

tfT

thus completing the proof. H We are now ready to prove the following result essentially due to Higman( 1954). 10.3. Theorem. Let G be an arbitrary group, let R be a commutative ring and let H be a subgroup ofG offinite index. Then, for any

RG-module V , the following conditions are equivalent: (i) V is H-projective. (ii) V is isomorphic to a direct summand of(VH)G. (iii) V is isomorphic to a direct summand of W G ,where W is an R H -module. (iv) E n d R ( V ) is H-projective. (v) T r g : EndRH(VH) 4 EndRG(V) is surjective. (vi) There exists $ E EndRH(VH)) such that T r s ( $ ) = 1. (vii) V is H-injective. Proof. (i)+(ii): By Lemma 10.1, there is an exact sequence of RG-modules

0 u (VH)' v 0 such that the associated sequence of RH-modules splits. By hypothesis, V is H-projective, so K e r f is a direct summand of ( V H ) ~ Thus . ( v H ) ~=

K e r f @ V'

111

10. Induction and relative projectivity

where V' 2 V . (ii) + (iii): Obvious. (iii) + (vi): Let us show first that any RG-module V of the form V = W G ,where W is an RH-module, satisfies (vi). To this end, denote by T a left transversal for H in G containing 1 and define 1c, : V + V bv

It is easily verified that 1c, is an RH-homomorphism. Moreover, for all w E W , we have

(t' €3 w)=

c t$(t-'t'

€3 w)= t' €3 w

(t' E T )

t€T

Thus II, satisfies (vi). Turning to the general case, we may harmlessly assume that

W G=

v Cf3 V'

(direct sum of RG-modules)

Let T : W G --+ V be the projection map and let II, satisfy (vi) with respect to W G . Then T 0 II, induces an RH-homomorphism V -+ V , and for all v E V , we have

since 7~ is an RG-homomorphism. The desired implication follows. (vi)+ (vii): Suppose that V is an RG-submodule of the RG-module U such that UH = VH@ W for some RH-submodule W . Let 7r : U,y + VH be the projection map and let 'p E HornRc(U,V) be defined by 'p = TrE(ll,o T ) , where II, E E ~ R H ( V H satisfies ) (vi). If v E V , then ~ ( v= ) v and hence U = V @ Ker'p.

(vii)+ (i): Assume that V is H-injective. Owing to Lemma 10.2, V is isomorphic to a direct summand of ( V H ) G . Applying implication (iii)+ (vi), there exists II, E E ~ R H ( V such H ) that TT$($)= 1. Now assume that f : U + V is a surjective RG-homomorphism for which

U = Kerf

Cf3

W

(direct sum of RH-modules)

112


The restriction f 1 = f IW is an RH-isomorphism of W onto VH. Set cp = and put 8 = Trz(cp$f). Then 8 E EndRG(U) and

fc'

(f 0 e)(4

=

c

ftcp$ft-'u

t€T

= Ctfcp$t-lu tET

= C t $ t - ' f u = f(u) t€T

proving that U = 8 ( U ) 8(u) =

+ K e r f . If 8(u) E 8 ( U ) n K e r f , then

c

c tcp+

ft-lu =

t ET

tcp+t-'fu = 0

tET

Thus U = O(U) @ K e r f , proving (i). (ii)+ (iv): Since V is isomorphic to a direct summand of ( V H ) it~ , follows that EndR(V) is isomorphic to a direct summand of H o m R ( ( v ~ ) ~ , But, by Proposition 5.9(i),

H o m ~( (V H ) ~V, )

H o m ~VH, ( V H )=~(EndR(V ) H ) ~

hence EndR(V) is H-projective, by the equivalence of (i) and (ii). (iv)+ (v): This is a direct consequence of Lemma 9.5 and the implication (i)+ (ii). (v)+ (vi): Obvious. H We now record a number of consequences of the result above. 10.4. Corollary. Let H be a subgroup of G of finite index n such that n is a unit of R. Then any RG-module is H-projective.

Proof. Let V be an RG-module and let $ : V +(v) = n-lv Then

for all

+ E EndRH(VH) and for all v E V,

(E

t$t-l

1

2)

= 2)

2)

--+

E

V

V be defined by


113

where T is a left transversal for H in G. This shows that TT$($)= 1 and the result follows by virtue of Theorem 10.3(vi). W 10.5. Corollary. Let H be a subgroup of G offinite index and let V be an RG-module which is R-free offinite rank. Then V is H projective if and only if

is surjective. Proof.

Owing to Lemma 7.4,

Now apply Theorem 10.3. 10.6. Corollary. Let G be an arbitrarygroup, let H be a subgroup of finite index and let V , W be RG-modules. Then, for any given X E H o m R c ( V , W ) ,the following are equivalent: (i) X E Tr$(HomRH(V, W ) ) . f (ii) Assume that U is an H-projective RG-module and that U -+ f WH + 0 splits. W + 0 is an exact sequence for which UH -+ Then there exists ?,h E HomRG(v with = f o $. (iii) Assume that U is an H-injective RG-module such that 0 -+ f f V U is an exact sequence for which 0 -+ VH -+ UH splits. Then there exists q!~E HomRG(U, with = $ o f .

u)

w)

Proof. We will demostrate that (i) is equvalent to (ii). The dual of our argument will easily establish the equivalence of (i) and (iii). (i)+ (ii): Since UH f, WH-+ 0 splits, there exists p E HomRH(W, such that f p = 1. By hypothesis, there exists Xo E HomRH(V, W )such that X = T r $ ( X o ) . Setting $ = Tr$(pXo) E H o w L R G ( V , U we ) , obtain

u)

f$ = T r g ( f&) = T r g ( X o )= A,

(by Lemma 9.6(i))


114

as required.

(ii)+ (i): Since U is H-projective, it follows from Theorem 10.3(vi) that T r $ ( A,) = 1 for some E EndRH( U ) . Hence

Tr%f A,$>

= f T r Z (A,)$ = f$=A,

(by Lemma 9.6(i), (ii))

as required. We close by recording a number of useful properties of H-projective modules and some applications.

10.7. Theorem. Let S 5 H be subgroups of G offinite index and let V be an RG-module. (i) Let V = @y=lK, where each E is an RG-module. Then V is H-projective if and only if each K is H-projective. (ii) IfV is S-projective, then V is also H-projective. (iii) If V is H-projective and VH is S-projective, then V is Sprojective. (iv) For any RH-module V , the RG-module VG is H-projective. (v) If W is an S-projective RH-module, then any RG-module which is a direct summand of W G is S-projective.

Proof. (i) Suppose that V is H-projective. Then, by Theorem 10.3(vi), there exists $ E E ~ ~ R H ( Vsuch H ) that Tr$($) = 1. Let ~ i 1 ,5 i 5 n, be the projection of V onto T/: defined by the decomposition V = @K. Then K ~ O $E E n d R H ( T / : ) H and obviously Tr$(r;o$) = 1. Thus, by Theorem 10.3(vi), each V , is H-projective. Conversely, if each $; E E ~ ~ R H ( Eis )such H that Tr$($i) = 1, then t,b E EndRH(V'&) defined by $lK = $; obviously satisfies T r $ ( $ ) = 1. Thus V is H projective. (ii) Let 0 + U -+ W + V -+ 0 be an exact sequence of RG-modules such that the associated sequence 0 -+ U, t WH + VH + 0 splits. Then the sequence 0 + US -+ WS -+ VS -+ 0 also splits. Because V is S-projective, the sequence 0 t U ---f W + V + 0 splits and therefore V is H-projective. (iii) Owing to Theorem 10.3(ii), we have (VH)G = V' @ X and


(Vs)H = V" @ Y , where V'

%

115

V and V" 2 VH.It follows that

( vs)G ( v H ) G

@

Y G= v' @ x @ Y G

and thus V is isomorphic to a direct summand of ( V S ) ~Now . apply Theorem 10.3(ii), (iv) This was established in the course of the proof of Theorem 10.3. (v) Owing to Theorem 10.3(ii), ( W S )=~W ' $ X , where W' E W ,so that (Ws)G % W G@ XG.By (iv), (WS)G is S-projective and therefore, by (i) any RG-module which is a direct summand of WGmust be Sprojective. H 10.8. Theorem. Let H be a subgroup of G of finite index n. (i) If V is a projective RG-module, then V H is a projective R H module. The converse is true if n is a unit of R. (ii) If V is an RH-module, then the induced module V G is projective if and only i f V is projective. (iii) If R as a field of characteristic p > 0 and P is a Sylow p subgroup of a finite group G, then an RG-module V is projective if and only i f Vp is projective.

Proof. (i) Let V be a projective RG-module. Then there exists a free RG-module X with X = V @ W for some RG-module W . Since X H = VHfBWHand X H is a free RH-module, we conclude that VHis projective. Conversely, assume that V is an RG-module such that VH is a projective RH-module and that n is a unit of R. Let

be an exact sequence of RG-modules. Then the associated sequence

splits, since VH is projective. By Corollary 10.4, V is H-projective, hence the sequence

O-+U+WtV+O also splits. Thus V is a projective RG-module. (ii) Let T be a left transversal for H in G. Then, by Corollary

116


1.2( ii) ,

(VG),=(l@V)$

c (tET,W

)

t@V

(direct sum of RH-modules)

where, of course, 1 8 V is identifiable with V . Suppose now that VGis projective. Then, by (i), ( V G ) is ~ also projective. Since V S 1 @ V is a direct summand of (VG),, we deduce that V is projective. Conversely, assume that V is projective. If V is a free RH-module, then the RG-isomorphism

implies that VG is a free RG-module. Thus V G is projective. In the general case, we have W 2 V @ U for some free RH-module W and some RH-module U . Then W GS VG$UGand hence V Gis projective. (iii) This is a direct consequence of (i).

A ring S is said to be local if S / J ( S ) is a division ring. In what follows, we write U(S)for the group of units of S. We next record certain properties of local rings, some of which will be required for subsequent investigations. 10.9. Lemma. For any ring S , the following conditions are equivalent: (i) S is local. (ii) S has a unique maximal left ideal.

(iii) J ( S ) = s - U ( S ) .

(iv) The set of nonunits of S is a left ideal.

Proof. (i)+(ii): Let I be a maximal left ideal of S. Then I / J ( S ) is a proper ideal of S / J ( S ) and thus I = J ( S ) . (ii)+(iii): Let J be a unique maximal left ideal of S. Since J ( S ) is the intersection of all maximal left ideals of S , we have J ( S ) = J , and so J is an ideal. The inclusion J ( S ) S - U(S)being obvious, suppose that x E S - U(S).If Sx # S , then Sx lies in a maximal left ideal of S , whence Sx C J ( S ) , so x E J ( S ) . On the other hand,


117

if S x = S then yx = 1 for some y E S . Clearly y 4 J ( S ) , otherwise 1 = y E J ( S ) . Hence S y = S , so zy = 1 for some z E S and hence z = x. Thus z E U ( S ) ,a contradiction. (iii) (iv): 0bvious. (iv)+(i): Let 5 @ J(S)and let I = S - U ( S ) . If M is a maximal left ideal of S , then M C_ I # S , whence M = I = J ( S ) . Hence x is a unit and thus x J ( S ) is a unit of S / J ( S ) . Consequently, S / J ( S )is a division ring as required.

+

+

10.10. Corollary. Let S be a local ring. Then (i) The only idempotents o f S are 0 and 1. (ii) For any ideal I o f S , S / I is a local ring. Proof. (i) Let e # 1 be an idempotent of S. Then s is a nonunit and hence, by Lemma 10.9, e E J ( S ) . Thus, by Proposition 1.4.10, e = 0. (ii) Let J / I be a maximal left ideal of S / I . Then J is a maximal left ideal of S , hence J = J ( S ) by Lemma 10.9. Thus S / I has a unique maximal left ideal and, by Lemma 10.9, S / I is local. 1

We next provide necessary and sufficient conditions under which group algebras of finite groups are local rings. 10.11. Proposition. Let G # 1 be afinite group and let R be a commutative ring. Then RG is local if and only if R / J ( R ) is a field of prime characteristic p and G is a p-group. Proof. Assume that RG is local and put F = R / J ( R ) . Since both R and FG are homomorphic images of RG, they are local by Corollary lO.lO(ii). Now the augmentation ideal I ( G ) of FG is a maximal left ideal, hence I ( G ) = J ( F G ) is nilpotent. By Maschke’s theorem, we conclude that charF = p > 0 and pl /GI. If G has an element g of prime order q # p , then g - 1 E I < g > is not nilpotent by Maschke’s theorem. This is impossible since I < g >C_ I ( G ) , hence G is a p group. Conversely assume that G is a p-group and R/ J ( R ) is a field of chracteristic p . It suffices to show that R G / J ( R G ) R / J ( R ) . If

118


J ( R ) = 0, then R is a field and I ( G ) has an R-basis { g - 111 # g E G } consisting of nilpotent elements. Hence I ( G ) is nilpotent by Theorem 1.5.12. But then J(RG) = I ( G ) and RG/J(RG)Z R, proving the case J ( R )= 0. Turning to the general case, we first note that J(R)G J ( R G ) by Corollary 4.3. Hence, by Corollary 1.4.4(ii), J ( R G / J ( R ) G )= J ( R G ) / J ( R ) G Since RG/J(R)G2 ( R / J ( R ) ) Git, follows from the special case above that RG/J(RG) R / J ( R ) ,as required. The following preliminary observation will allow us to apply the above result.

10.12. Lemma. Let S be a local ring. Then any finitely generated projective S-module is free. Proof. Let V be a finitely generated projective S-module, and let

o1,. . . ,o, be a minimal generating set for V . Assume that we are given

a relation

We claim that all such s; E J ( S ) . Indeed, if si 6 J ( S ) , then it is a unit in S , and we can solve for the corresponding oi as an S-linear sum of the vis. This, however, is impossible since otherwise the generator o; can be deleted. Thus all s; are in J ( S ) , as claimed. Let W be a free S-module freely generated by wl,. . . ,w, and map W onto V by sending w;to 0;. By the foregoing, the kernel of this map, say K ,is contained in J ( S ) W . By hypothesis, V is projective so that W = IC @ V’ with V’ V . Because K 2 W/V’ it must be finitely generated. Moreover, J ( S ) W = J ( S ) K $ J ( S ) V ’ and hence the inclusion I( C_ J ( S ) W forces I( = J ( S ) K . Thus I( = 0 by Nakayama’s lemma (Proposition 1.4.9) and W 2 V , as required. The following result is essentially due to Dickson (1907).


119

10.13. Theorem. Let G be a finite group and let P be a Sylow p-subgroup of G. Assume that R is a local ring such that R/ J ( R ) is of characteristic p and that V is a projective RG-module which is R-free of finite rank n. Then [PI divides n.

Proof. By Theorem 10.8(i), Vp is a projective RP-module. Since Vp is finitely generated and, by Proposition 10.11, R P is local, it follows from Lemma 10.12, that Vp is free. Thus Vp is isomorphic to a direct sum of, say r , copies of R P . But each R P is R-free of rank lPl, hence n = IPIr, as required. We close by recording the following result. 10.14. Theorem. Let R be an arbitrary commutative ring and let N be a normal subgroup of G of finite index n. Then the following conditions are equivalent: (i) J ( R G ) = RG J ( R N ) . (ii) n is a unit of R. (iii) For any irreducible RN-module V , V G is completely reducible. *

Proof. The equivalence of (i) and (ii) follows from Theorems 3.9 and 4.2. (ii)=+(iii): Let T be a transversal for N in G. Then V G= etETt@V where, by Lemma 1.5(i),(iv), each t 8 V is an irreducible RN-module. Let Mr be an RG-submodule of V Gand let 0

+

w + V G+ V G / W+ 0

be the natural exact sequence. Then the sequence

0+

w, + ( V " ) ,

+(VG/W)N +0

splits, since ( V " ) , is a complete reducible RN-module. The desired assertion now follows by virtue of Corollary 10.4. (iii)=+(ii): Let I be any maximal ideal of R. It suffices to show that n = n - 1 I . To this end put V = R / I . Then the irreducible R-module V is also an irreducible RN-module under the trivial action of N . Furthermore, we obviously have

V G (R/I)(G/N)

120


Since, by hypothesis, V Gis completely reducible, it follows that ( R / I ) ( G / N ) is semisimple. Thus, by Proposition 1.6.2, n $ I and the result is established. 11.

Unique decompositions

In this section, we shall collect a number of general results which will be used repeatedly in our subsequent investigations. Throughout, unless explicitly stated otherwise, R denotes an arbitrary ring.

11.1. Lemma. Let V be an indecomposable R-module which is both artinian and noetherian. Then every f E E n d R ( V ) is either a unit or nilpotent.

Proof. Let L, and X n be, respectively, the image and the kernel of f",n 2 1. We assert that

to a sufficient large n; if substantiated, it will follow that f is nilpotent, provided f is a nonunit, hence the result. Since V is both artinian and noetherian, there exists n 2 1 such that Ln = L2n and X n = X2n. Then, for any x E V , we may find y E V for which u(x)= u2(y),where u = f". Thus

Finally, given x E L, n X,, we have u(x)= 0 and x = u(y) for some y E V . Then y E X2,, = X,, whence x = u(y) = 0, proving (1). Let R be an arbitrary ring and let V # 0 be an R-module. We say that V is strongly indecomposable if EndR(V) is a local ring. 11.2. Lemma.

If V is strongly indecomposable, then V is inde-

composable. The converse is true if V is both artinian and noetherian.

Proof. Assume that V is strongly indecomposable and that V = V' $ W is a direct decomposition of V . If .n : V + V' is the projection

121

11. Unique decompositions

map, then x is an idempotent of the local ring E n d R ( V ) , so x = 0 or T = 1, by Corollary lO.lO(i). Hence either V' = 0 or V' = V , proving that V is indecomposable. Conversely, assume that V is indecomposable and both artinian and noetherian. By Lemma 10.9, it is enough to show that every nonunit f E E = EndR(V) is in J ( E ) . Let g be an arbitrary element of E . Then gf is a nonunit and hence is nilpotent, by Lemma 11.1. Thus 1 - gf is a unit of E , so by Proposition 1.4.10, f E J ( E ) as we wished to show. W The following result is Azumaya's generalization of a classical theorem of Krull and Schmidt. 11.3. Theorem.

Let R be a ring, let V be an R-module and let

where each of the R-modules K, Wj is strongly indecomposable. Then m = n and, after possibly reordering the Wj, 2 W; for all i E { l , .. . , m } .

Proof. We argue by induction on min{m,n}. If m = 1 or n = 1, then the result is obvious since V is indecomposable. Let ei : V 3 K and fj : V + Wj be the projection maps, 1 5 i 5 m, 1 5 j 5 n. Note that el = elfjel and that, by hypothesis, E n d ~ ( 6is) local. Thus e l f j e l is a unit of EndR(V1)for some j . By renumbering the Wj, we may therefore assume that e l f l e l is an automorphism of Vl = elV. Thus flV1 C W1 and the kernel of f1 on V, is 0. Our next aim is to show that

@z2K.

To this end, suppose that II: E flVl n Then el2 = 0 and 5 = f i e l y for some y E V . Consequently, e l f l e l y = e1II: = 0 and therefore e1y = 0, since elfie1 is an automorphism on Vl. Hence II: = f l e l y = 0 , proving that flV, n = 0. Now fix x E V . Then e l z E elV = elflelV and hence e l x =

@z2K

122


elflelw for some w E 5

- flelw E

W . It follows that el(z

-

f i e l w) = 0 and

Hence m

5

= flelw

+ (a: - flelw) E f1V + C V , i=2

proving (2). Since flV, C W1, it follows from (2) that

Wl = flvl @ {Wl n @Z1W But W1 is indecomposable and flV,

# 0, so Wl = fl&

S

Vl and

v = Wl @ ( @ 3 3 Consequently,

@Z2K= V/W1 = @jn=2Wj

and therefore, by induction, m = n and after possible reordering the W j , V , S W j f o r a l l i E { 1 , ...,m}. Let V be an R-module. We say that V has the unique decomposition property if the following two properties hold: (i) V is a finite direct sum of indecomposable modules; (ii) If V = = @YzlWj, where each K, Wj is indecomposable, W; for all then m = n and, after possibly reordering the Wj, i E (1,. . . , m } .

@clK

11.4. Corollary. Let R be a noetherian ring such that each indecomposable R-module is strongly indecomposable. Then any finitely generated nonzero R-module has the unique decomposition property.

Proof. Because R is noetherian, it follows from Corollary 1.3.7 that any finitely generated R-module V # 0 is noetherian. Hence, by Proposition 1.3.11, V is a finite direct sum of indecomposable modules. The desired conclusion is therefore a consequence of Theorem 11.3. 11.5. Corollary. (Krull - Schmidt theorem). Let V # 0 be an Rmodule which is both artinian and noetherian. Then V has the unique decomposition property.

12. Projective covers

123

Proof. By Proposition 1.3.11, V is a finite direct sum of indecomposable modules. Since these indecomposable modules are again artinian and noetherian, it follows from Lemma 11.2 that they are strongly indecomposable. Now apply Theorem 11.3.

11.6. Corollary. Let R be an artinian ring. Then any finitely generated R-module V # 0 has the unique decomposition property. Proof. Because R is artinian, the module V is both artinian and noetherian, by virtue of Corollary 1.4.15. The desired conclusion is therefore a consequence of Corollary 11.5. 12.

Projective covers

Our aim in this section is twofold: first to provide some general information pertaining to projective covers of modules and second to record a result concerning projective covers of induced modules. In what follows, R denotes an arbitrary ring and all modules are assumed to be finitely generated left modules. We know that each Rmodule is a homomorphic image of a projective module. For some modules V a stronger assertion is possible. A homomorphism

f:V-+W of R-modules is said to be essential if for every proper submodule V' of V , f(V') # f ( V ) . We say that an R-module P is a projective cover of V in case P is projective and there is an essential epimorphism

P-tV 12.1. Lemma. (i) A homomorphism f : V + W of R-modules is essential if and only i f K e r f is superfluous. (ii) If K , . . . , V, are superfluous submodules of V , then so is Vi *. t (iii) If P; is a projective cover of V;, then P I @ - .$P, is a projective cover of I/, @ , . . @ V, . (iv) A projective cover of an irreducible module is indecomposable.

v,.

+

124


Proof. (i) Assume that f : V + W is an essential homomorphism. If W is a submodule of V such that V = W K e r f , then f(V) = f(W) and so W cannot be a proper submodule of V . This shows that V = W and hence Kerf is a superfluous submodule of V . Conversely, assume that K e r f is superfluous. If V' is a submodule of V with f(V) = f (V'),then V = V' K e r f . Since K e r f is superfluous, we must have V = V' and therefore f is essential. (ii) Suppose that W is a submodule of V such that

+

+

+ +- + + +- + +

Then V1 (V2 - . V, W ) = V and, since & is superfluous, we - * V, W = V . Now apply induction on n. have V2 (iii) If fi : P; -+ is an essential epimorphism, then

is an epimorphism whose kernel @y=llcerfiis superfluous, by virtue of (ii). Hence, by (i), @ fi is essential, which implies that @Pi is a projective cover of @K. (iv) Suppose that $ : P + V is an essential epimorphism, where V is an irreducible R-module. If P = P' @ P", for some proper submodules P' and P", then $(PI) = $(P") = 0 since 0 is the only proper submodule of V . But then V = $(PI) $(PI') = 0, a contradiction. Hence P is indecomposable, as asserted. W

-+

One of the consequences of the following lemma is that if a module does have a projective cover, then it is unique up to isomorphism.

12.2. Lemma. Assume that an R-module V has a projective cover P and let M be an R-module such that V is a homomorphic image of M . (i) If M is projective, then P is isomorphic to a direct summand of

M. (ii) If M has a projective cover L , then P is isomorphic to a direct summand of L. (iii) If M is another projective cover of V , then it4 % P .


125

Proof. (i) Fix an epimorphism cp : M -+ V and an essential epimorphism $ : P + V. Since M is projective, there exists a homomorphism 8 : M + P such that cp = $ 0 8. Because $ is an essential epimorphism and $08 is an epimorphism, 8 is also an epimorphism. But P is projective, hence there is a splitting homomorphism f : P + M and so M = Irn f @ Ker8. Since f is injective, (i) is established. (ii) Assume that L is a projective cover of M and let f : L + M be an epimorphism. Then cpf : L + V is an epimorphism, hence by (i) P is isomorphic to a direct summand of L. (iii) Suppose that M is a projective cover of V and choose 'p in (i) to be essential. It will be demonstrated that 0 is also essential, from which will follow that M = Imf 2 P. Assume that M' is a proper submodule of M and put P' = 6(M'). Then

and so P'

# P.

Hence 8 is essential, as required. 1

From now on, we write P ( V ) for a projective cover of V . Owing to Lemma 12.2(iii), if P ( V ) exists, then it is unique up to isomorphism. 12.3. Theorem. Let R be an artinian ring and let V be an Rmodule. (i) V has a projective cover. (ii) A projective R-module W is a projective cover of V if and only

i f W / J ( R ) Ws V / J ( R ) V . In particular, (a) IfV is completely reducible, then V 2 P ( V ) / J ( R ) P ( V ) . (b) I f U and V are completely reducible R-modules, then U E V if and only i f P ( U )C P ( V ) . (iii) If V is projective and i f V / J ( R ) V= @L1K is a decomposition as a direct sum of irreducible R-modules, then V Z @Y."=,(K)is a decomposition as a direct sum of indecomposable R-modules. (iv) I f V and W are projective R-modules, then V S W i f and only

if V / J ( R ) V2 W / J ( R ) W Proof. (i) Let L be a projective module such that V = L / S for some submodule S of L. For any submodule W of S , let fw : L / W + L / S be

126


the canonical homomorphism. Next choose W to be minimal in S such that fw is essential; such a submodule exists, since fs is essential and S is artinian. We claim that W is a direct summand of L ; if sustained, it will follow that L/W is a projective module and hence that L / W is a projective cover of V . Let W’ be a submodule of L minimai among those whose projection cp : W’ -+ L/W is surjective, and let n : L + L/W be the projection map. Since L is projective, there exists a homomorphism n’: L -+ W’ with ?r = cp o n’,and the minimality of W’ forces ?r’(L)= W’. Denote by L‘ the kernel of n’.The projection f~ factors onto

L/L’+ L/W

-+

L/S

and the two factors are essential. Because L’ is contained in W , the minimality of W implies that W = L’, i.e. that Q is an isomorphism. It follows that L = W @ W’, as required. (ii) Let W be a projective module, let f : W + V be an essential epimorphism and let 71 : V + V / J ( R ) Vbe the projection. Since n is essential, so is nf and thus Kernf E J(R)W (Lemma 12.l(i) and Proposition 1.4.5). On the other hand

and therefore J(R)W

K e r n f . Thus Kernf = J(R)W and so

W / J ( R ) W2 V / J ( R ) V Conversely, assume that W is a projective R-module such that the above isomorphism holds. Let f : V + V / J ( R ) V be the projection map and let g : W + V / J ( R ) Vbe the homomorphism induced by the isomorphism W / J ( R ) WE V / J ( R ) V . Since f is an essential epimorphism and since W is projective, there is an epimorphism g’ : W -+ V with fg‘ = g . Hence

Kerg’ C Kerg = J(R)W is a superfluous submodule of W . Thus W is a projective cover of V , as required.

127


(iii) By Propositions 1.4.5 and 1.4.8, J(R)V is a superfluous submodule of V . Since V is projective, it follows that P ( V / J ( R ) V )= V . Invoking Lemma 12.l(iii) and (iv), we deduce that

where each P( E) is indecomposable. (iv) If V E W , then obviously V / J ( R ) VZ W/J(R)W. Conversely, assume that V and W are projective R-modules such that V / J ( R ) VE W / J ( R ) W .By the Krull-Schmidt theorem, we may write

and

W / J ( R ) W= w163 . . * @ W, where each we have

rZ:, Wi is irreducible and

2 Wi, 1 5

i 5 n. Then, by (iii),

as required. H

Assume that R is an artinian ring and let n 2 1 be such that

where the U; are principle indecomposable R-modules. Owing to the Krull-Schmidt theorem, the Ui are uniquely determined up to isomorphism and the order in which they appear. Furthermore, the above decomposition determines a complete set {el,. . . ,e,} of primitive idempotents in R such that Vi = Re;. Except when J ( R ) = 0, the principal indecomposable R-modules form only a small subclass of all indecomposable R-modules. However, it is the class which plays a very important role due to the following result.

Let R be an artinian ring. (i) The following conditions are equivalent: (a) V is a projective cover of an irreducible R-module.

12.4. Theorem.

128


(b) V is a projective indecomposable R-module. (c) V is a principal indecomposable R-module. (ii) If Rel, , ,. ,Re, are all nonisomorphic principal indecomposable R-modules, then R e l / J ( R ) e l , .. .,Re,/J(R)e, are all nonisomorphic irreducible R-modules. Proof. (i) The implications (a)+(b) and (c)+(a) are consequences of Lemma 12.l(iv) and Theorem 12.3(iii), respectively. Now assume that (b) holds. Since V is projective, there exists n 2 1 such that R" = V @I V', where R" denotes a direct sum of n copies of R. Now the indecomposable components of R" are principal indecomposable Rmodules and V is indecomposable. Hence (c) follows by virtue of the Krull-Schmidt theorem. (ii) Invoking (i), we may find irreducible R-modules Vi, . . . ,V m such that Rei = P ( x ) , in which case K E Rei/J(R)ei by Theorem 12.3(ii). Re, and so i = j . If V is any irreducible RIf then Rei module, then by (i), P ( V ) E P ( K ) for some i E (1,. . . ,n } . Thus, by Theorem 12.3(ii), V Z V , as required.

4,

We close by proving the following result.

12.5. Proposition. Let H be a subgroup of a finite group G, let F be a field and let V and W be FH and FG-modules, respectively. Then (i) P(VG)is isomorphic to a direct summand of P ( V ) G . (ii) P ( W H )is isomorphic to a direct summand of P ( W ) H . (iii) If H is normal, then for all g E G , P(") = " ( V ) . In particular, V is G-invariant if and only if P ( V ) is G-invariant.

Proof. (i) By the definition of P ( V ) ,there is an FH-epimorphism P ( V ) --+ V . Hence, by Theorem 2.9(i), there is an FG-epimorphism P(V)" V G . By Theorem 10.8@), P(V)Gis projective. Thus, by Lemma 12.2(i), P ( V G )is isomorphic to a direct summand of P(V)". (ii) Since there is a homomorphism of P ( W ) onto W and hence of P ( W ) H onto W H ,P(WH)is isomorphic to a direct summand of P ( P ( W ) H ) .Because P ( W ) is projective, so is P(W)Hby Theorem 10,8(i). Hence, by Lemma 12.2(i), P ( W H )is isomorphic to a direct


129

summand of P ( W ) H . (iii) Let cp : P ( V ) + V be an essential epimorphism. Then cp can also be regarded as an epimorphism " ( V ) + gV. Because W is a submodule of P ( V ) if and only if it is a submodule of " ( V ) , the latter epimorphism is essential. It therefore suffices to prove that if V is projective, then so is gV. So assume that V is projective and write FH @I - @I FH = V @ W for some FH-module W . Then

--

and, since obviously

g(

F H ) S F H , the result follows.

This Page Intentionally Left Blank

131

Chapter 3 Induction from normal subgroups In this chapter we provide a detailed information on the process of induction from normal subgroups. Our first major result (Theorem 2.1) reduces the study of V G to the case where V is G-invariant. Under this assumotion we then show that E ~ ~ R G is ( Va crossed ~) product of GIN over E n d ~ ~ ( vThis ) . will allow us to discover circumstances under which V G is totally indecomposable or absolutely indecomposable. As a consequence of our main results, we derive a classical theorem known as the Green Indecomposability Theorem. Special attention is drawn to the study of crossed products over prime rings. The information obtained is then applied to provide a criterion for V G to be homogeneous. Our next major result asserts that if G is a finite group and S a finite-dimensional semisimple algebra over a field F , then any F-algebra which is a crossed product of G over S is symmetric. As an application, it is shown that if V is a completely reducible FN-module ( N is a normal subgroup of G), then the F-algebra E n d r ; . ~ ( Vis~ ) symmetric. This fact will allow us to derive a number of important facts concerning induction from irreducible modules and their projective covers. The chapter ends with the discussion of the Loewy length of induced modules.

132

1.

Induction from normal subgroups

Complete noetherian local rings

Throughout this section, R denotes a commutative ring. We begin by recording the following piece of information. A metric space ( M , p ) is a set M together with a map

p:MxM-+IR such that for all z, y, z E M , (9 P ( Z , Y) = P(Y, 4 2 0. (ii) p ( z , y ) = 0 if and only if x = y. 6;;) P ( 5 , Z ) I P b , Y ) P ( Y , Z ) . A Cuuchy sequence in M is a sequence {z;} of elements of M such that for any e > 0, there exists a positive integer k such that

+

P(2",4

<E

m,n > k

for all

A sequence {xi} is said to converge to provided

lim;,,p(z,z;)

3:

E M , written limxi =

z,

=0

Assume that I is an ideal of a ring S such that

Then, for any distinct z,y E S, there exists a unique m 5 -y

1 0 such that

E I" - Im+l

where by convension Io = S. We define PI : S x S 3 IR by

It is then easily verified that (S,pl) is a metric space. We refer to S as being complete in the I-udic topology if each Cauchy sequence in (S,P I ) converges. Let { s i } be a sequence in S. The following properties are direct consequences of the definitions: (i) {s;} is Cauchy if and only if for any n such that i,j > m implies s; - sj E I".

> 0, there exists m > 0,

1. Complete noetherian local rings

133

(ii) limsi = s if and only if for any n > 0, there exists m > 0 such that i > m implies s; - s E I". (iii) If limsi = s and lims: = s', then lirn(si si) = s s' and

+-

+

lim(sis:) = ss'. 1.1. Lemma. Assume that R is a noetherian ring and that A is an R-algebra which is finitely generated as R-module. If I is an ideal of R such that I J ( A ) , then fl;=,I"A = 0. In particular, if R is a noetherian local ring with maximal ideal P = J ( R), then n =:, P" = 0.

Proof. Let L be an arbitrary ideal of R and let V be a finitely generated R-module. It is well known (see Bourbaki (1961)) that L W = W for W = n;=,L"V. Applying this fact for V = A , L = I and W = nr=,I"A, we conclude that IW = W and hence that J(A)W = W . Because W is a finitely generated A-module, it follows from Nakayama's lemma that W = 0. So the lemma is true. 1.2. Lemma. Let A be an R-algebra which is finitely generated as R-module. Then J(R)A C J ( A ) .

Proof. Let V be any irreducible A-module. Then V = Av for each nonzero v in V , hence V is a finitely generated R-module. Now J ( R)V is an A-submodule of V , hence is 0 or V . But J ( R ) V # V , otherwise by Nakayama's lemma we find that V = 0. Hence J ( R ) V = 0 and so J ( R ) annihilates every such V . Thus the same is true for J ( R ) A and so J ( R ) A J ( A ) .

c

1.3. Lemma. Let R be a local ring with maximal ideal P and residue class field F = R / P . Suppose that A is an R-algebra which is finitely generated as R-module and put A = A / P A . Then (i) J ( A ) = J ( A ) / P A and A / J ( A ) S A / J ( A ) as F-algebras. (ii) For any finite group G , R G / J ( R G ) F G / J ( F G ) .

c

Proof. (i) By Lemma 1.2, P A J ( A ) and so, by Corollary 1.4.4(ii), J ( A ) = J ( A ) / P A . The latter obviously implies the second assertion.

134


(ii) Put A = RG and note that

A = R G / P ( R G ) ( R / P ) G= FG Now apply (i). W

Assume that R is a local noetherian ring with maximal ideal P . Owing to Lemma 1.1, n : y n = 0. We say that R is complete if R is complete in the P-adic topology. 1.4. Lemma. Let R be a complete noetherian local ring with maximal ideal P , and let A be an R-algebra which is finitely generated as R-module. Then, for any ideal J C J ( A ) of A , A is complete in the J-adic topology.

Proof. By Lemma 1.2, P A C J ( A ) ,while by Lemma 1.3(i), J" C_ J(A)" P A for some n 2 1. Thus we may assume that J = PA. Note that, by Lemma 1.1, the intersection of all (PA)" = P"A is zero. Write A = Ro1 . . . Ro,, the o; E A, and let { w ; } be a Cauchy sequence in A. Given an integer n > 0, denote by m(n) the largest integer such that for all i , j 2 n wi - wj E P"(")A

+ +

(if no such m ( n ) exists, set m(n) = n ) . Bearing in mind that {wi} is Cauchy, we have limn-,mm(n)= 00. Let

and write

We then have

135


where

and t=n

so for any t , {bnt} is a Cauchy sequence in R. Since R is complete, limb,, = b, for some b, E R. Setting w = C;=l btvt, it follows that

thus completing the proof.

We have now accumulated all the information necessary to prove the following result. In what follows, we use the standard fact that if el and e2 are idempotents of a ring S , then Sel Z Se2 as left S-modules if and only if e l S 2 e2S as right S-modules (see Jacobson (1956)). 1.5. Theorem. Let R be a complete noetherian local ring, let A be an R-algebra which is finitely generated as R-module and let J C J ( A ) be an ideal of A . For each x E A, let 5 be the image o f x in A = A / J . (i) Each idempotent E E A can be lifted to an idempotent e E A, that is e = E . Moreover, e is primitive if and only if is primitive. (ii) If e l , e2 are idempotents of A, then Ael E Aez as left A-modules if and only if Ael 2 A Z 2 as left A-modules (equivalently, e l A e 2 A as right A-modules if and only i f e l A E e2A as right A-modules). (iii) If I = E~ e2 t.- . en is a decomposition of I into orthogonal idempotents in A, then there exists orthogonal idempotents e l , . . . ,en E A such that 1 = el . . e, and e; = ci

-+

+

+ +

Proof. (i) Given an idempotent e E A, choose x1 E A with 51 = E . Then nl = x: - x1 E J . Once x ; and n; are chosen, put x;+1 = x ;

+ n; - 2x;n;,

n;+l =

2

- xi+l

(1)


136

Then n; E J2iand thus {xi} is a Cauchy sequence. By Lemma 1.4, A is complete in the J-adic topology, so limx; = e for some e E A. Then e2 - e = limn; = o and, moreover, e = 51 = c , so e is a required idempotent of A. Let e be an idempotent in A and let E = €1 6 2 , where €1 and ~2 are orthogonal idempotents in A. It will next be shown that there exist orthogonal idempotents el, e2 in A with

+

To this end, choose a E A such that si = €1 and put z1 = eae. Then 51 = E s i E = €1 so we may consider the Cauchy sequence {z;} given by (1). Since z1 E eAe, each x; E eAe and thus ex; = x; = x;e for all i. Accordingly, setting e; = Zimzi, we deduce that el is an idempotent of A such that E l = €1 and (by taking limits) eel = el = ele. Letting 2 = e2 # 0, 22 = €2: and e1e2 = e2e1 = 0, e2 = e - el, we now have e2 proving (2). If E is not primitive, then so is e, by virtue of (2). The converse being obvious, it follows that e is primitive if and only if so is 2. (ii) Assume that e1,e2 are idempotents in A. Any isomorphism Ael 2 Ae2 of left A-modules carries J e l onto Je2 and induces an isomorphism A E 1 A22 of left A-modules. Conversely, let f : /IE1 --+ Ac!~ be an isomorphism of left A-modules and let 3 = f-'. Because A = Aei @ A ( l - e,), each Ae; is a projective A-module. Thus we can find A-homomorphisms f : Ael -+ Ae2 and g : Ae2 -+ Ael lifting f and g,'respectively. We claim that 6 = gf is an automorphism of Ael,; if sustained, it will follow by symmetry, that fg is an automorphism of Ae2 and hence that f is an isomorphism. therefore (8 - 1)Ael C PAel. Setting It is obvious that 4 lifts p = 1 - 8, we have PkAel C Jkel. Thus 1 p t p2 t . . . is a well defined endomorphism of Ael, and is a two-sided inverse of 1 - p. Because 1 - p = 8, this demonstrates that 8 is an A-automorphism of Ael, as desired. (iii) The case n = 1 being obvious, we argue by induction on n. So assume that n > 1 and that the result holds for n-1. Put 6 = en-l +E,,

sf,

+

137


so that 1 = ~1

+. . . + en-2 + 6 is an orthogonal decomposition. By the

induction hypothesis, we may find an orthogonal decomposition:

- 1 it follows that Applying (2) to the decomposition ?t = 6 = ~ ~ +c,, e= en for some orthogonal idempotents en-l, en in A such that En-1 = En-1 and = c,. Thus 1 = el+- - .+en is the desired orthogonal decomposition.

+

As an easy consequence of the above result, we derive 1.6. Corollary. Let R be a complete noetherian local ring and let A be an R-algebra which is finitely generated as R-module. Then A is local i f and only if 1 is the only nonzero idempotent of A .

Proof. If A is local, then 0 and 1 are the only idempotents of A , by Corollary 2.10.10(i). Conversely, suppose that 1 is the only nonzero idempotent in A . Let P be the maximal ideal of R and put A = A / P A . By Lemma 1.3(i), A / J ( A ) A / J ( A ) and so it suffices to show that A is local. By Theorem 1.5, I is the only nonzero idempotent of A. Since A is a finite-dimensional algebra over a field, the result follows.

The next lemma will enable us to take full advantage of the results so far obtained.

1.7. Lemma. Let R be a noetherian ring and let V be a finitely generated R-module. Then the R-algebra E n d R ( V ) is a finitely generated R-module.

+ +

Proof. We may write V = Rvl Rv, for some vi E V . Let W be a free R-module with basis wl,. . . ,w, let 'p : W 3 V be the R-homomorphism such that 'p(w;) = v,, and put IC = Ker'p. Since E n d R ( W ) Z M,(R), E ~ ~ R (isWa )finitely generated R-module. Put Eo = {f E E n d R ( W ) J f ( K ) I(}. I f f E Eo, then f induces ) an S E E n d ~ ( vby -

f(.

+ IC) = f(.) + K

a

(z E

W)

138


For any given g E E n d ~ ( v write ),

and define g E E n d ~ ( wby )

j=1

If Cj bjwj E I(, then Cj bjvj = 0 and

so

proving that g E Eo. It follows that g H g is a surjective R-homomorphism Eo + EndR(V). Now EndR(W) is a finitely generated R-module and R is noetherian. Hence Eo and EndR(V) are finitely generated Rmodules, as required. W 1.8. Corollary. Let R be a complete noetherian local ring and let A be an R-algebra which is finitely generated as R-module. Then every nonzero finitely generated A-module has the unique decomposition property.

Proof. By Corollary 1.3.8, A is noetherian. Hence, by Corollary 2.11.4, it suffices to verify that any given finitely generated indecomposable A-module V is strongly indecomposable. Because V is inde. to composable, 0 and 1 are the only idempotents of E n d ~ ( v )Owing Lemma 1.7, E n d ~ ( v(and ) hence E n c l ~ ( V )is) a finitely generated Rmodule. The desired conclusion now follows by appealing to Corollary 1.6.

139

2. Reduction to G-invariant modules

2.

Reduction to G-invariant modules

Throughout this section, N denotes a normal subgroup of a finite group G. All modules are assumed to be finitely generated over their ground rings. It will be also assumed that R is a commutative ring of one of the following types: (i) R is artinian. (ii) R is a complete noetherian local ring. These assumptions are made to guarantee that if A is an R-algebra which is finitely generated as R-module, then any nonzero A-module has the unique decomposition property (see Corollaries 2.11.6 and 1.8). Let V be an indecomposable RN-module and let H be the inertia group of V. Our aim is to show that the decomposition of V G is determined uniquely by that of V H .This will allow us to concentrate on the case where V is G-invariant, i.e. gV E V for all g E G. 2.1. Theorem. (Ward (1968) and Willems (1976)). Let V be an indecomposable RN-module, let H be the inertia group o f V and write

where the K are the indecomposable RH-modules. Then (i) V G = KG @ - - .03 Vf, where the yG are indecomposable RGmodules such that qGE yGimplies 2 4. (ii) If, for some i E (1,. . . ,s } , is irreducible, then KG and V are irreducible.

.@KG

Proof. (i) The equality V G= Kc@. . follows from the facts that induction is transitive (Lemma 2.2.1) and preserves direct sums (Lemma 2.2.8). To show that each yGis indecomposable, suppose that

XG=X@Y is is a direct decomposition. We know, from Corollary 2.1.2, that identifiable with a direct summand of ( y G ) ~Hence, . by the KrullSchmidt theorem, we may assume that I$ is a direct summand of X H ,

say

X*=I/:@X'

(2)


140

Observe also that ( V H )is~the direct sum of isomorphic RN-modules of the form h 8 V with h E H . Invoking the Krull-Schmidt theorem, we deduce from (1) that

for some positive integer m;. Accordingly, for any g E G, we obtain RN-isomorphisms (9c3 V t ) N

9 €9

(K)N 2

m;(g €9 V )

which in turn implies

(yG)N

(4)

= ( @ t c T t €9 v i ) N 2 mi(t €9 V )

where T is a left transversal for H in G. Restricting (2) to N, it follows from (3) that X N % mzv @ X & But X is an RG-module, so

proving that for each t E T , t €9 V appears as a direct summand of X N with multiplicity at least mi. Bearing in mind that for distinct t l , t 2 E T , tl 8 V and t 2 8 V are nonisomorphic RN-modules, we de~ isomorphic to a direct summand of XN. duce from (4) that ( q ' ) is But X N is a direct summand of (V;.')N, hence by the Krull-Schmidt theorem, X = KG. This shows that KG is indecomposable. Assume by way of contradiction that KG S '3/1 but V;: V,. Because is identifiable with a direct summand of (K')H, we may write

(KG)* = V , 63 q'

and

( y G =) vj~ @ q'

where K' and are RH-modules. Invoking the Krull-Schmid theorem, we see that K is isomorphic to an RH-direct summand of 5'. It follows that ( K ) N is isomorphic to an RN-direct summand of Applying (3)) we therefore deduce that V is isomorphic to an RN-direct summand of ( 5 ' )and ~ that

(y)~.

(FG)N = ( Y ) N @ ( 6 ' ) N 2 mjV 63 ( 6 ' ) N

3. Group-graded algebras and crossed products

141

However, by (4), the multiplicity of V as an indecomposable direct summand of ( y G is ) ~exactly mj. This contradiction proves the required assertion. (ii) Suppose that there exists an i E (1,. . . ,s} such that r/: is irreducible. Then by Clifford's theorem (Theorem 2.4.2), ( K ) N is completely reducible. Thus, by (3), V is irreducible. Let W be an irre(such W exists, by virtue of Proposition ducible factor module of 1.3.4). Then, by Corollary 1.2.5, V , is isomorphic to an RH-submodule of W,. Consequently, is isomorphic to an RN-submodule of W,. Therefore, by (3), W N contains a submodule X Z m;V. Thus, for any g E G, WN contains the RN-submodule gX Z m;(g@I V ) . Let T be a left transversal for H in G. Then the RN-module t @I V , t E T , are irreducible and nonisomorphic and they appear as direct summands of WN with multiplicity at least mi. On the other hand, by (4), the multiplicity of each t @I V as a direct summand of ( y G ) N is exactly m;. Now W = v G / X for some submodule X of V,' and WN 2 ( K G ) ~ / XThus ~.

KG

(v)~

and, by the foregoing, X = 0. The conclusion is that W = irreducible, as desired. H

YG is

2.2. Corollary. Let V be an irreducible RN-module, let H be the inertia group of V and let ( H : N ) be a unit of R. Then V G is comple t ely reducible .

Proof. Invoking Theorem 2.1, we may harmlessly assume that H = G. Now apply Theorem 2.10.14. H 3. Group-graded algebras and crossed products

This section contains a number of general results which will enable us to scrutinize the structure of the endomorphism ring of induced modules. Our exposition is strongly influenced by a work of Dade (1970a). Throughout, A denotes an algebra over a commutative ring R. If X , Y are R-submodules of A , then X Y denotes the R-submodule


142

of A consisting of all finite sums (2;E

x,y;E Y )

Let G be a multiplicative group. Then A is called a G-graded algebra if there is a family { A h E GI of R-submodules of A indexed by the elements of G such that the following conditions hold: (direct sum of R-modules). (a) A = @gEGAg for all x , y E G. (b) A A , A,, We shall refer to (a) as a G-grading of A and to A, as the g-component of A. When (b) is replaced by the stronger condition, namely (4 A A y = A,, for all x,y E G we say that A is a strongly G-graded algebra. From now on, A denotes a G-graded algebra over R and U ( A )denotes the unit group of A.

c

3.1. Lemma. (i) A1 is a subalgebra of A with 1 E Al. (ii) For each g E G, A, is an ( A l ,Al)-bimodule under left and right multiplication b y the elements of Al. (iii) A is strongly G-graded if and only if 1 E A,A,-l for all g E G.

Proof. (i) By definition, Al is an R-submodule of A and, by (b), Al is multiplicatively closed. It therefore suffices to show that 1 E Al. Invoking (a), we may write

l=Ca, SEG

where a, E A, for all g E G, and all but finitely many of a, are zero. Now fix some h 6 G and aL E A h . Because G is a group, (a) implies that A = @g€GAgh (1) Owing to (b), the product agai lies in Agh for all g E G. Therefore, aI h = l . a h = I

c

,€G

a g 4


143

is precisely the expansion of a; in the decomposition (1). But ah already lies in the summand A1.h = A h in (1). Hence all the agai for g # 1 must be zero and ala: must be u i . This proves that a1 acts as a left identity on A h for all h E G. In view of (a), we deduce that al is a left identity for the algebra A. Thus al E Al is the identity element of A, as desired. (ii) This is a direct sequence of (b). (iii) Suppose that A is strongly G-graded. Because 1 E Al, it follows from (a)that 1 E AgAg-1 for all g E G. Conversely, assume that 1 E AgAg-l for all g E G. Invoking (b), we then have = 1 - A,, = AxA,-~Ax,

A,,

A x A x - ~ z= y A,A, for all x , y E G. Hence A,A, = A,, for all z,y E G, as asserted. W

We refer to a unit u of A as being graded if u E A, for some g E G. In this case, we say that g is the degree of u and write g = deg(u) = degA(u) The set of all graded units of A will be denoted by GrU(A). 3.2. Lemma. (2) I f u E G T U ( A )is of degree g , then u-l is of degree 9 - l . (ii) GrU(A) is a subgroup o f U ( A ) and the map deg : GrU(A)-+ G

is a group homomorphism with kernel U(A1). (iii) The map G r U ( A ) + Aut(A1) u H i,

{

where

i&)

= uzu-l

for all

x E A1


144

is a homomorphism. (iv) Right multiplication b y any u, E A,

n U ( A ) is an isomorphism

of left Al-modules.

Proof. (i) We first write u-l = CxEGu,with a, E A, and with A,A, A,, for all x E G, it finitely many nonzero a,. Since ua, follows that uu-l = C I E G u a ,is a unique expansion for ua-l in the decomposition

c

A = fBxEGAgz = @XEGAX Now, by Lemma 3.1(i), uu-' = 1 lies in A1 = A,A,-1. Hence the az for # 9-l must be zero, and 1 must equal uag-l. Therefore u-' = a,-' lies in A,-], as asserted. (ii) That GrU(A)is a subgroup of U ( A )follows from ( i ) and the fact that A,A, 5 A,, for all x,y E G. Since A,A, A x y ,the given map is a homomorphism with kernel Al n U ( A ) . Owing to (i), A1 n U ( A ) U(Al) and since the opposite inclusion is trivial, the required assertion is established. (iii) By hypothesis, u E A, for some g E G. Thus, by (i) we have

x

uA1u-l 5 A,AlA,-i

A1

Bearing in mind that GrU(A)is a group, (iii) follows. (iv) Since u, is a unit of A , right multiplication by ug is an Alisomorphism of Al onto Alu,. Because u, e A,, we have Alu, C A,. Owing to (i), we also have

A, = A,u,'u,

C AgAg-lUgC Aiug

The conclusion is that A, = Alu, and the result follows. W

We next introduce a very important class of group-graded algebras, namely crossed products. Owing to Lemma 3.2(ii), the sequence of group homomorphisms

1

+ U(A1)+.

GrU(A)%' G + 1

(2)


145

is always exact, except possibly at G. We say that a G-graded algebra A is a crossed product of G over A1, written A = A1 * G, provided the sequence (2) is exact. Thus A is a crossed product of G over A1 if and only if for any g E G, there exists 3 E A, n U(A). In the special case, where (2) is an exact splitting sequence, we shall refer to A as a skew group ring of G over Al. Thus A is a skew group ring of G over A1 if and only if for each g E G, there exists E A,nU(A) such that zy = zy for all z,y E G

A G-graded algebra A is called a twisted group ring of G over A1 if for all g E G, there exists g E A, n U(A) such that g centralizes Al. In a particular case, where Al Z(A) and A, n U ( A ) # 8 for all g E G, we shall refer to A as a twisted group algebra of G over Al. Finally, if Al * G is a twisted group algebra and a skew group ring, A1 * G is notihng else but the group algebra of G over Al. Our next aim is to provide a condition under which a strongly Ggraded algebra A is a crossed product of G over Al. Let A be a strongly G-graded algebra whose subalgebra Al is a local ring. Then A is a crossed product of G over A1 . 3.3. Proposition.

Proof. Since A,A, = A,, for all

J(A1) C A1 = A,A,-1 Thus we may find a, E A, and

2,y

E G, we have

for all g E G

E A,-I for which

Because A1 is a local ring, we have U(Al) = Al - J(Al). Hence

is a right inverse to a, in A. The idempotent a, lies in A,-] A, Because 0 # 1 = l2 = (u,u,.-,)2 = ug(ag-lug)ag-l tI

C Al.


146

we have a;-lag # 0. Therefore a$)-lag= 1 and thus a;-, is a two sided inverse to a, in A. We conclude that

A,

n U ( A )# 8

for all g E G

which shows that A is a crossed product of G over A l . W We now proceed to describe crossed products more explicitly. Let A be a crossed product of G over Al. A map Q

: G -+

Aut(A1)

is called an automorphism system for A if, for all g E G, there exists a unit g of A in A, with I = 1 such that

for all x E Al

a ( g ) ( s ) = gxg-'

(3)

Now suppose that B is an R-algebra. Given maps Q

: GtAut(B)

(Y

:

GxG+U(B)

we say that (G,B , Q, a ) is a crossed system for G in B if, for all 2,y , z E G, and b E B, the following properties hold:

where ,b = a ( g ) ( b )for all g E G, b E B.

Let a G-graded R-algebra A be a crossed product of G over A l . For a n y g E G, fix a unit g of A in Ag with I = 1, let n : G + Aut(A,), be the corresponding automorphism system given b y (3) and let (Y : G x G U(A1) 3.4. Proposition.

--j

be defined b y +,Y)

= ZY"Y

(7)


147

Then the following properties hold: (i) A is a strongly G-graded algebra with A, = A1ij = jA1. (iz) ( G ,A1,o, a ) is a crossed system for G over A1 (to which we refer as corresponding to A) (iii) A is a free (left and right) Al-module freely generated b y the elements g , g E G (iv) For all x,y E G and r1, 7-2 E Al,

Conversely, for any R-algebra B and a n y crossed system ( G ,B , 0,a ) for G over B , the free B-module C freely generated b y the eIements g , g E G with multiplication given b y (8) (with r1,r2 E B ) is a Ggraded R-algebra (with C, = Bg for all g E G) which is a crossed product of G over C1 = B and having (G,B, 0,a ) as a corresponding crossed system.

Proof. Assume that u E A, n U ( A ) for some g E G. Owing to Lemma 3.2(i), u-' E A,-] n U ( A ) and hence 1 = uu-l E A,A,-I. Invoking Lemma 3.1(iii), we conclude that A is a strongly G-graded algebra. By Lemma 3.2(iv), A, = A1g and the argument of that lemma applied to left multiplication shows that A, = gA1, proving (i). Let x,y E G and b E Al. Then

"( Yb) = z(yby-')Z-'

= a ( x ,y)ZybZy-lcu(x, y)-' = ~ ( zy ,) " y b a ( x ,y ) - 1 ,

proving (4). For any x , y , z E G, we also have

(.y). -

.(Y.)

__

= a ( x ,y ) z y z = a ( x ,y ) a ( z y ,2)= q " ( y , z ) y z = "cu(y,z)ZyZ= Z a ( y , Z ) a ( x , y Z ) q z ,

proving ( 5 ) . Since = 1, (6) is also true, proving (ii). Property (iii) is a direct consequence of (i). To prove (iv), fix x,y E G and r l , r2 E Al. Then ( r 1 z ) ( r 2 y= ) rl(%r2ii-')(.y) = r1 " r 2 a ( x ,y)-,


148

proving (8). To prove the converse, it suffices to show that the multiplication given by (8) is associative. To this end, fix x , y , z E G and rl,r2,r3 E B. Then

( r i z )[(r2Y)(r@)]= ri "[r2'r3a(Y,z ) ] a ( zY, z ) m = rl "r2"( Yr3)" a ( y , z ) a ( z ,y z ) ~ (bY(4)) = rl I r z a ( ~ , yzYr3a(x,y)-1 ) =a(y,z ) a ( z ,y z ) z y ~ (by ( 5 ) ) = r1 "r2a(z,y) "Yr3a(z,y)-la(z,y)a(zy, z ) z y ~ =

[ ( T i2) (

f a Y)] (r3z).

as required.

Returning to our study of G-graded algebras, we now introduce the following important notion. Let A be a G-graded R-algebra. An ideal I of A is called a graded ideal if

I = @ g E G ( I n Ag) Assume that B is another G-graded R-algebra. A homomorphism

f:A+B of R-algebras is said to be graded if f ( A g )C B, for all g E G.

3.5. Lemma. (i) If I is a graded ideal of A, then A / I is a G graded R-algebra by setting ( A / I ) , = ( A , I ) / I for all g E G. (ii) I f f : A -+ B is a graded homomorphism of G-graded Ralgebras, then Kerf is a graded ideal. Moreover, if f is surjective, then there is a graded isomorphism A / K e r f --+ B . (iii) An ideal I of A is graded i f and only i f I is the kernel of a graded homomorphism A -+ B of G-graded R-algebras.

+


149

as asserted. (ii) First of all, we know that Kerf is an ideal of A. Let z E Kerf and write x = CgEG x,, xg E A,. Then

0 = f (4=

cf

(z,>

,EG

and

f ( 4E 4

Since B = $,&Ideduce ,, that each z, E Kerf and therefore Kerf is a graded ideal. Now assume that f is surjective and let

f* : A / K e r f + B be the induced homomorphism of R-algebras. Then, for all g E G,

proving that f* is a graded isomorphism. (iii) If I is a graded ideal of A , then the natural map A -+ A / I is a graded homomorphism with kernel I . The converse is a consequence of (ii). H An ideal X of A1 is called G-invariant if

A,XA,-i = X

for all g E G

In the special case where A is a crossed product of G over A1 and ij E U ( A ) n A,, we have A,XA,-I = 3Xg-l. Thus in this case X is G-invariant if and only if gxg-1 =

x

for all g E G

We are now in a position to describe all graded ideals of A . 3.6. Theorem. (Dude (1970~)). Let A be a strongly G-graded R - a lg e b ra . (i) If I is a graded ideal of A and X = I n Al, then X is a Ginvariant ideal of A1 such that

I = AX = X A

and

I n A, = A,X = X A , for all g E G


150

(ii) For any G-invariant ideal X of Al, I = X A = AX is a graded ideal o f A such that

I n A, = AgX = XA,

for all g E G

(iii) J(A1) is a G-invariant ideal of A1 and hence A - J ( A l ) = J(A1)- A is a graded ideal of A. Proof. (i) By hypothesis, I = e g E ~ where I g I, = I n A, for all g E G. In particular, we must have X = 1 1 . Given z,y E G, we have

A,Iy = A,(I n AY)C A,I n A,Ay In&,

c

T

Thus A,-lIacy I y and therefore

Consequently, AxIy = I,, and, in particular, A,I1 = I, for all g E G. A similar argument shows that IlA, = 1, for all g E G. Given g E G, we then also have

AgIIAg-l= IgAg-1= 11, proving that Finally,

11

is G-invariant.

as required. (ii) By assumption, AgXAg-l = X for all g E G, so multiplication on the right by A, gives A,X = XA, for all g E G. Since A = $,EGA,, we have

I = AX = $,,GA,X

= $,EGXA, = X A ,

proving that I is a graded ideal of A such that I n A, = A,X = X A , for all g E G.

151


(iii) Owing to (ii), it suffices to show that J(A1) is a G-invariant ideal of A1. Let V be an irreducible Al-module and let L be the annihilator of V . For any given g E G, it will be shown that A,LA,-I is the annihilator of some irreducible Al-module. Because J ( A l ) is the intersection of all such annihilators, the desired assertion will follow. Choose a maximal left ideal I of A1 such that V E A1/I. Since A,A, = Azy, for all 2,y E G and 1 E Al, the map X + A,X sends bijectively the set of all left ideals X of A1 onto the set of left Alsubmodules of A,, with Y 3 A,-IY as the inverse map. These maps clearly preserve inclusion, hence A,I is an Al-submodule of A,. An elementary calculation shows that the annihilator of the irreducible Al-module A,/A,I is precisely A,LA,-l, as required. H We have now come to the main result of this section. 3.7. Theorem. Let A be a strongly G-graded R-algebra. (i) A/A J(A1) is a strongly G-graded R-algebra by setting +

(A/A - J ( A I ) ) ~= (A,

+ A - J(Al))/A - J(A1)

f o r all g E G

Mo re0 v e r,

(A/A J(A1))l

A,/&%)

and, provided G is finite,

(ii) If A is a crossed product (skew group ring, twisted group ring, twisted group algebra) of G over Al, then A/A J(A1) is a crossed product (skew group ring, twisted group ring, twisted group algebra) of G over Al / J(A,). (iiz) If A is a crossed product of G over A1 and A1 = R. 1~ J(A1), then A/A - J(A1) is a twisted group algebra of G over R . ~ A / (- R 1~n

-

+

J(A& Proof. (i) The fact that A / A . J ( A l ) is a G-graded R-algebra with the given g-component is a consequence of Lemma 3.5(i) and Theorem 3.6(ii), (iii). Because A is strongly G-graded, so is A/A J(Al). By


152

definition,

and therefore

On the other hand,

and therefore

proving that ( A / A - J ( A l ) )2, Al/J(A,). If G is finite, then A.J(A1)C J ( A ) ,by virtue of Corollary 4.2.3 in Karpilovsky (1987a). (ii) Assume that A is a crossed product of G over A,. Then, for any g E G, there exists a unit g of A in A,. Hence ij A . J(A1) is a unit of A / A - J(A1)in ( A / A J ( A l ) ) ,for all g E G. This proves that A / A J ( A l )is a crossed product of G over A l / J ( A l ) ,by applying (i). Suppose that A is a skew group ring of G over A,. Then, for any g E G, there exists a unit g of A in A, such that

+

5g = zy

But then for all

2,y

for all x , y E G

E G,

proving that A / A J(A1) is a skew group ring of G over A,/J(A,). Now assume that A is a twisted group ring of G over Al. Then, for any g E G, we may choose g E U(A)n A, such that 3 centralizes Al. But then, for each g E G, ij+A.J(A,) centralizes (Al+ A . J ( A ) / A - J ( A , ) ) , proving that A / A - J(A1)is a twisted group ring of G over A,/J(Al). Finally, assume that A is a twisted group algebra of G over A,. Then A1 G Z(A) and hence

153

4. The endomorphism ring of induced modules

Thus A / A - J ( A l ) is a twisted group algebra of G over A l / J ( A , ) , as required. (iii) Our assumption guarantees that

and therefore A / A . J ( A l )is a twisted group algebra of G over AlIJ(A1). Since the homomorphism R . 1~ + A l / J ( A l ) ,r - 1 H r 1 J ( A 1 ) is surjective with kernel R 1~ n J ( A l ) ,the result follows.

- +

4.

The endomorphism ring of induced modules

The proof of most of the results of this section can be found in Dade (1980),(1981), although many of the essential ideas are contained in previous works of Conlon (1964) and Ward (1968). Throughout, N denotes a normal subgroup of a finite group G and R an arbitrary commutative ring. Let V be an RN-module. Recall that V is said to be G-invariant if gV Z V for all g E G. If W is another RN-module, we say that V weakly divides W if there exists a positive integer k and an injective homomorphism f : V -+ kW = W

@

W

(k copies)

such that f ( V ) is a direct summand of kW. Weshall say that V and W are weakly isomorphic if each weakly divides the other. Of course, this is an equivalence relation among RN-modules. Finally, we shall refer to V as being weakly G-invariant if it is weakly isomorphic to each g V , g E G. 4.1. Lemma. Let V be an RN-module. (i) The map

is an injective homomorphism of R-algebras whose image consists of all $ E EndRG(V") for which $(1@ V ) C 1 @ V .


154

€3 V is a n RN-isomorphism, t h e n f can be uniquely extended t o a n RG-isomorphism V G+ V G .

(ii) Iff : 1 €3 V

+g

Proof. (i) It is plain that the given map is an injective homomorphism of R-algebras. Moreover, for all f E E n d m ( V ) ,l @f sends 1@V into 18V. Suppose that $ E EndRG(VG)is such that $(1€3V) l @ V . Then there exists f E EndRN(V)such that $ ( l @v) = 1 @ f ( v ) for all v E V . Because $ is uniquely determined by its restriction to 1 €3 V , we have $ = 1 €3 f , as required. (ii) Thanks to a universal characterization of induced modules (Proposition 2.2.3) f can be uniquely extended to an RG-homomorphism $ : VG4 V G .Since for any z E G,

it follows that 1c, is surjective. To show that $ is also injective, denote by T a transversal for N in G. Then, any given w E V G ,can be uniquely written in the form

If $(w)= 0, then

c tf

(1€3 vt) = 0

tCl'

Because t f (1 €3 vt) E tg €3 V and Tg is a new transversal for N in G, we have t f (1 €3 vt) = 0. Thus 1 €3 vt = 0 and therefore w = 0, as we wished to show. Let V be an RN-module. We say that V can be extended to an RG-module if there exists an RG-module V * , whose additive group coincides with that of V while the multiplication

* : RG x V* -, V*

in

V*

satisfies z*v=zv

forall

~ E V , ~ E R(1) N

We are now ready to prove the following fundamental result.


s

155

4.2. Theorem.

Let V be an RN-module, let E = EndRG(VG), = EndRN(V) and, for any g E G , define EgN by

(i) E is a GIN-graded R-algebra with E g as ~ its gN-component, g E G and the identity component of E is identifiable with S . (ii) E is strongly GIN-graded i f and only i f V is weakly G-invariant, while E is a crossed product of G I N over S if and only if V is G -

invariant. (iii) E is a skew group ring of GIN over S if and only if V can be extended to an RG-module. In fact, there is a bijective correspondence between all extensions V* o f V to RG-modules and all splitting homomorphisms y for 1 + U ( S ) + G r U ( E ) 2 G I N + 1, in which V* corresponds to y if and only if 1 @ g * v = y ( g N ) ( g@v)

for all

E V,g E G ( 2 )

-

(iv) If V is weakly G-invariant, then E J ( S ) = J ( S ) .E is a graded ideal of E contained in J ( E ) and such that E I E - J ( S ) is a strongly G/ N-graded R-algebra with ( E / E- J ( S)),, = (E g N +E. J ( S ) ) / E .J ( S ) , g E G . Furthermore, i f V is G-invariant, then E I E . J ( S ) is a crossed product o f G / N over S / J ( S ) . (v) If V is G-invariant and S = R - 1s J ( S ) , then E I E - J ( S ) is a twisted group algebra of G I N over R l s / ( R 1s n J ( S ) ) S / J ( S ) .

+

-

Proof. (i) Observe that if X N = y N , then x @ V = y 8 V which shows that E g N is well-defined. Let T be a transversal for N in G. Then (direct sum of RN-modules) V G= $ t c T t @I V and so if rt : V G+ t @I V is a projection map, then

Now assume that f E E and let gt : 1 @ V + t 8 V be the R N homomorphism that is the restriction of rt 0 f to 1 @ V . Owing to a


156

universal characterization of induced modules (Proposition 2.2.3), there exists ft E E such that

It follows that ft E Et-"

whence f = CtETft. Now assume that then have and

vt

&T

and, applying (3) and (4),we have

CtETft= 0 with

= 0. Thus

ft E

Et-IN. For any v E V, we

vt = f t ( 1 c3 v) E t @ v vt

= 0 and f t = 0 for all

t E T , proving that

I f f E E=N,cp E E,N, then

and thus (fcp)(l@ V)

c f(y-'

63 V ) c y - ' f ( l @ V ) E y-'z-l@

v

It follows that f ( o E E z y and ~ therfore E is a GIN-graded R-algebra with E I l as ~ its gN-component. Finally, by Lemma 4.1(i), the identity component of E is identifiable with S, proving (i). (ii) Thanks to Lemma 3.l(iii), E is strongly GIN-graded if and only if for any given g E G, there exists (o; E E g ~y5i, E E g - l ~1, 5 i 5 n = n(g), such that

It will next be shown that ( 5 ) is equivalent to V and gV to be weakly isomorphic, which will imply the first assertion.

157


If (5) holds, then any map f : l@V + n(g@V),w t-+ ($l(v), . . . ,$ n ( v ) ) is an injective RN-homomorphism. Moreover, the map (wl, . . . ,wn) H ‘pI(w1) * v n ( v n ) is a homomorphism from n(g QD V ) to 1 QD V which is a left inverse to f . This proves that V weakly divides gV. By symmetry, V weakly divides g-’V and thus gV weakly divides V , proving that V and gV are weakly isomorphic. Conversely, suppose that V and gV are weakly isomorphic. Then V weakly divides g 8 V ,so there is a positive integer n and an injective RN-homomorphism $ : 1 @ V + n(g @ V ) such that $(1 @ V ) is a direct summand of n(g 8 V ) . We may write $(z) = ($1(z),. .. ,$ n ( S ) ) , where : 1 @ V + g @ V is an RN-homomorphism, in which case $; can be regarded as an element of E g - l N . Let f : n(g V )-+ 1 8 V be an RN-homomorphism which is a left inverse to $. Then f determines RN-homomorphisms

+-+

+;

‘p;

:g @

v + 1 QD v

(1 5 i 5 n )

(which can be regarded as elements of E g ~such ) that ( 5 ) holds. If V is G-invariant, then E is a crossed product of GIN over S, by virtue of Lemma 4.1(ii). Conversely, suppose that for any g E G there exists fg E U ( E )n E g ~ Then . fg-l : 1 @ V + g @ V is an injective RN-homomorphism such that fg-ifg is an automorphism of 1@ V and hence of g 8 V . Thus fg-1 is an RN-isomorphism, as desired. (iii) Assume that V’ is an extension of V to an RG-module. Then (1) implies that there is a unique RN-isomorphism of g 8 V onto V sending g 8 w into g * w, for all g E G, w E V . This isomorphism obviously depends only on the coset g N . Hence, by Lemma 4.1(ii), there is a unique y(gN) E E g n~U ( E ) such that (2) holds. Now fix z,y E G. Then, for all v E V ,

+N)r(yN)(w @ 4 = 7 ( 4 z 7 ( ” y QD 4 = r(zN)z(lQD Y*4 (by (2))

*

= 7 ( z N ) ( zQD y v) = 1@ z * (y * v) = 1 8 (zy) = 7 ( z y N ) h €3 4

*

2,

(by (2))

which forces r(zN)y(yN) = y(zyN). This demonstrates that y is a splitting homomorphism, the only such homomorphism satisfying (2).


158

Conversely, suppose that 7 : G/N + GrU(E) is a homomorphism . any g E G, v E V , let such that for all g E G, r ( g N ) E E s ~ For g * v E V* be defined by (2). Then for any n E N, 1 8 n * o = r ( N ) ( n8 v) = n 8 v = 1 8 nu and therefore n * v = nv. Furthermore, given s , y E G and v E V , we have

Thus V extends to the RG-module V*,the only such module satisfying (2). (iv) Suppose that V is weakly G-invaraint. Then, by (ii), E is strongly GIN-graded, so by Theorem 3.6(iii), E . J ( S ) = J ( S ) - E is a graded ideal of E . By Theorem 3.7(i), this graded ideal is contained in J ( E ) . The rest of the first statement is a consequence of Theorem 3.7(i). The second statement is a consequence of (ii) and Theorem 3.7(ii) . (v) Apply (ii) and Theorem 3.7(iii). 4.3. Corollary. Let V be a strongly indecomposable RN-module.

Then V is G-invaraint if and only if V is weakly G-invariant.

Proof. Because V is strongly indecomposable, S = EndRN(V) is a local ring. Now suppose that V is weakly G-invariant. Then, by Theorem 4.2(ii), E = EndRG(VG)is strongly GIN-graded. Because S is local, it follows from Proposition 3.3 that E is a crossed product of GIN over S. Thus, by Theorem 4.2(ii), V is G-invariant. Conversely, if V is G-invariant, then V is obviously weakly G-invariant, hence the result. Our next aim is to provide a condition under which J ( E ) n S = J ( S ) , where E = EndRc(VG)and S = EndRN(V). The following simple observation will clear our path. 4.4. Lemma. Let A be an R-algebra which is finitely generated

5. Relations between the decompositions of VG and End&VG)

159

as an R-module and let R / J ( R )be artinian. Then

J(A)" C J ( R ) A

for some n

21

Proof. We know, from Lemma 1.2, that J ( R ) AC J ( A ) . Hence, by Corollary 1.4.4(ii), we have

J ( A / J ( R ) A )= J ( A ) / J ( R ) A But A / J ( R ) Ais an R/J(R)-algebra which is finitely generated as an R/J(R)-module. Hence, by Corollary 1.3.8, A / J ( R ) Ais artinian. Invoking Proposition 1.4.14, we therefore deduce that J ( A ) / J ( R ) Ais nilpotent, hence the result. 4.5. Theorem. Let V be a finitely generated G-invariant R N module, let E = EndRG(VG) and let S = EndRN(V). If R is artinian or a noetherian local ring, then

J ( E )n S = J ( S ) Proof. Owing to Theorem 4.2(iv), J ( S ) C J ( E ) n S . Now V is a finitely generated R-module, hence EndR(V) (and therefore S ) is a finitely generated R-module (Lemma 1.7, Propositions 1.4.14 and 1.3.3, Corollary 1.3.7). Similarly, V Gis a finitely generated R-module and hence E is a finitely generated R-module. Thus, by Lemmas 4.4 and 1.2, J ( R ) S C J ( S ) and J(E)" & J ( R ) E for some n 2 1. Moreover, by Theorem 4.2(i), E = $ z c ~ / ~ E with z El = S. Hence J ( R ) E n S = J ( R ) S and

( J ( E )f l S)" C J ( R ) En S = J(R)S C J ( S ) It follows that ( J ( E )n S ) / J ( S ) is a nilpotent ideal of S / J ( S ) and therefore J ( E )n S J ( S ) ,as required. 5.

Relations between the decompositions of V Gand EndRG(VG)

Throughtout this section, N denotes a normal subgroup of a finite group G and R an arbitrary commutative ring. Given an RN-module

160


V , it will be convenient to identify EndRN(V) with the subring of E n d ~ c ( Vconsisting ~) of all $ E EndRG(VG)for which $(1 @ V ) 1@ V (see Lemma 4.1(i)), In what follows we shall freely use this identification. For convenience of reference, we first record some general observations which provide a link between a module and its endomorphism ring. Let S be a ring, let V be a left S-module and let E = Ends(V). For any subset L of E and any subset W of V , let LW denote the set of all finite sums

It is clear that LV is a submodule of V . Note also that if I is a principal right ideal of E generated by $, then

We may regard V as a left E-module via 'pv = cp(v),cp E The E @ E V becomes a left E-module via

E,v E V .

Moreover, the canonical map

is an isomorphism of E-modules. 5.1. Lemma. Let S be a ring, V a left S-module and I a right ideal of E = Ends(V) which is a direct summand o j E . Then I = Ee for some idempotent e of E c n d (i) IV = e ( V ) (ii) The map I @ E V + IB

is an isomorphism of E-modules.

5. Relations between the decompositions of V G and End&VG)

161

Proof. The fact that I = E e for some idempotent e of E is a consequence of the right analog of Proposition 1.3.12. Hence (i) follows from (1). Because I is a direct summand of E , we may identify I @ E V with its image in E @IE V, in which case (2) maps I @ E V onto IV. This proves (ii) and hence the result.

5.2. Lemma. Let S be an arbitrary ring, let V be a left S-module and let E = E n d s ( V ) . (i) If V = c&V;. and ei : V 3 6 is the projection map, then el,. . . ,e, are mutually orthogonal idempotents of E with sum 1. Conuersely, if el,. . . ,e, are mutually orthogonal idempotents of E with sum 1, then V = c&V, where = e;(V) and ei : V + K is the projection

map.

(ii) K is indecomposable if and only i f e; is primitive. (iii) K Z 4 as left S-modules if and only if e;E Z ejE as right E-modules. be the Proof. (i) Assume that V = @i=lx and let e; : V + projection map. It is plain that e: = ei, 1 5 i 5 T . If i # j , then (e;ej)(V) = e;(y)= 0 , and thus e;ej = 0. Moreover, we obviously have C:==, e; = 1.

Conversely, assume that el,. . . ,e, are mutually orthogonal idempotentsofEwithsum1. Since1 = e l + - - - + e , , w e h a v e V = V,+-..+V,. If wi E V;., 1 5 i 5 r , are such that el(v1) - e,(v,) = 0, then by taking the image of both sides under e;, we obtain e;(v;)= 0, 1 5 i 5 r , Thus V = $r==,K and obviously e; : V + K is the projection map. (ii) If e; = e f , where e, f are orthogonal idempotents, then K = e(V )$ f ( V ) . Conversely, if K = V,' @ Kii and A; : V -+ p; : V + are projection maps, then & , p i are orthogonal idempotents of E with e; = A; pi. Thus V , is indecomposable if and only if ei is primitive. (iii) Assume that eiE 2 e j E as right E-modules and put I; = e;E, 1 5 i 5 r. Then, by Lemma 5.1, IiV = K and V; % V, as E-modules. as S-modules. Hence Q Z Conversely, suppose that 8 : r/: -, vj is an S-isomorphism and fix v E V. Given cp f e;E, we have Imcp C ei(V) = K, so Im(89) C 4.

+- - +

+

+

v',

162


Thus ( ~ Q ) ( v = ) ej(V1) for some ~1 E V, so e j ( b ) ( v ) = ej(V1) = ( b ) ( v ) and therefore

BQ = ej(0cp) E e j E If we now define p : eiE -t ejE by p(cp) = 09, then p is obviously a homomorphism of right E-modules. A similar argument shows that the map X : ejE + eiE defined by A($) = 0-'$ is also an E-homomorphism. Because X is the inverse of p , we deduce that eiE E ejE, as desired. Let V be a free module over a ring S. In case any two bases of V have the same cardinality, we refer to this cardinality as the rank of V. 5.3. Lemma. Let V be a free module over a local ring S. Then any two bases of V have the same cardinality.

Proof. Let {v;li E I} be two bases of V. Denote by V; and tijj the images of vi and wj, respectively, in 3 = V/J(S)V. Then (6ili E I}and {tijjlj E J } are two bases of the S/J(S)-module V . But, by hypothesis, S / J ( S ) is a division ring, hence 111 = lJl, as required. The observation above, will allow us to speak about

"

the rank of

V" whenever V is a free module over a local ring. Let S be a ring and let V be a left S-module. Then V can be regarded as a right Ends(V)'-module by setting VQ

=~ ( v )

for all v E V,cp E Ends(V)'

5.4. Lemma. Let V be a G-invariant RN-module, let S = EndRN(V and let E = EndRG(VG)". For each g E G, choose 'pg E U ( E ) mapping 18 V isomorphically onto g 8 V, put 91 = 1 and define $g : V + V by 'p;l(g 8 V ) = 1 8 $ ~ ~ ( v )Then . V 8 s E is an (RG,E)-bimodule with the actions

5. Relations between the decompositions of V G and End&VG>

163

Proof. Because V is a right S-module and E is a left 5'-module, the tensor product V 8 s E is at least a 2-module. Given s E S, g E G,vE V , we have the following equality in VG:

which implies

We therefore conclude that

Given g E G, we now introduce f : V x E + V 8 s E given by It is clear that f is R-bilinear. Since, for any s E S,

the m a p v 8 e H g(v 8 e) is an R-endomorphism of V 8 s E . We now claim that, given z,y E G and v E V , where s = 'pz'py'p;i E S. Indeed, we have


164

proving (4). It follows that X(Y(V

@ el) = z($J&) @ vve) = $W%/(v)) @ cpxvve

= $z($v(v)) @ ~ ' P q = e 4$Jz($Jv(v))>@ vzve = $z&) @ cpxve (by (4)) = (zY>(v@ 4

Because vl = 1 (hence &(TI) = v for all v E V ) ,we also have l ( v @ e ) = v @ e. Hence V 8 s E is a left RG-module. It is clear that V @sE is a right E-module under the given action of E. Moreover, for all g E G,u E V,e,e'E E ,

[g(v8 e)]e' = $Jg(v) cpgee' = g ( v

eel) = g[(v@ e)er],

thus completing the proof. H The following theorem is the main result of this section.

(Cline (1972), Theorem A). Let V be a G-invariant RN-module, let E = End&VG)' and let S = EndRN(V)' be identified with the subring of E consisting of all cp E E with cp(l@V)C l @ V . Then the map 5.5. Theorem.

PsE + v@e

H

e(l@v) vG

is an (RG,E)-bimodule isomorphism, where the (RG, E)-bimodule structure of V 8 s E is given by Lemma 5.4. Proof. The given map, say f, is obviously R-linear. Since for all seG f (g(v @ 4 ) = f ($Jg(v>@ cpge) = (cpge)(l@$ J g W = (vge)(cp;'(g @ 4) = 4s @ 4 = g e ( l @v) = g f ( v @ e ) ,

it follows that f is an RG-homomorphism. Because for all e,e' E E and v E V f((v

e)e') = f ( v eel) = ( e e ' ) ( l @w) = ( e ( I @v))e' = f ( v e)e',

5. Relations between the decompositions of V G and En&c(VG)

165

we deduce that f is also an E-homomorphism and hence is an (RG,E)bi mo dule homomorphism. Let T denote a transversal for N in G. Then

Because pt(l @ V ) = t @ V , it follows from (5) that f is surjective. Owing to Theorem 4.2(ii) and Proposition 3.4(iii), E is a free left Smodule freely generated by all yt, t E T . Thus any element of V 8 s E can be uniquely written in the form

Because

f ( p w t )= Ct Et TP t ( l @ v t ) and cpt( 1@ vt) E t @ V , it follows from ( 5 ) that if CtETvt €3 pt E K e r f , then pt(l @ vt) = 0 for all t E T . Thus all vt = 0 and therefore C t Evt ~€3 (Pt = 0. Hence f is injective and the result follows. As a second major result, we now prove the following theorem in which the RN-module V is not assumed to be finitely generated.

Let V be a strongly indecomposable G-invariant RN-module and let S = EndRN(V) be identified with the subring of E = EndRG(VG) consisting of all (P E E with p(l @ V ) 1 @ V . (i) I f a right ideal I of E is a direct summand of E, then I is a free S-module offinite rank. (ii) E contains a finite complete set of orthogonal primitive idempotents el,. . . ,en such that for Ij = e;E the following properties hold: I;V where each I;V is an indecomposable RG-module (a) V G= such that (IjV)N is a direct sum o f n ; copies o f V , nj = ranks(I;) (V is identified with 1 8 V ) . (b) I;V IjV if and only i f e j E Z ejE. (c) I f V is R-free offinite rank, then I;V is R-free with 5.6. Theorem.

run kR( 1;V ) = ranks (I;)runkR( V )


166

Proof. (i) By Theorem 4.2(ii) and Proposition 3.4(iii), E is a finitely generated free S-module. Since I is a direct summand of E , I is a finitely generated projective S-module. But, by hypothesis, S is local, hence I is a free S-module, by Lemma 2.10.12. Finally, by Lemma 5.3, I has a uniquely determined rank which is finite since I is finitely generated. (ii) It is a consequence of (i) that E admits a finite decomposition E = $Z,E; where each E; is indecomposable. Hence E; = e;E for some orthogonal primitive idempotents e; of E with el - e, = 1. By Lemma 5.2, we deduce that

+ +

where each ei(VG)is indecomposable and e;(VG)Z ej(VG)if and only if e;E ejE. We now claim that

e;(VG)= I;V

for all i

(6)

Indeed, since V C V G and e;(VG)= I;VG,we obviously have I;V e;(VG).On the other hand, V Gis the sum of all g 8 V,g E G. Hence, to prove (6), we need only verify that ei(g 8 V

) C IiV

for all

g EG

Fix g E G and w E V . Let cp E E map V isomorphically onto g 8 V . Then g @ w = cp(v1) for some 01 E V . Hence ei(g 8 v) = ( e i c p ) ( Y ) E IiV,

proving (6). All the remaining assertions will now follow provided we show that (IjV)N is a direct sum of nj copies of V , n; = ranlcs(1;). Since Ij is a free right S-module of rank n;, E" o e; is a free (left) 9-module of E" 0 e; under the isomorphism rank n;. Moreover, the image of V of Theorem 5.5 is I;V. Hence

which immediately implies the required assertion by virtue of the fact that V @ s o So V .

6. Twisted group algebras over fields

6.

167

Twisted group algebras over fields

In this section, we shall collect a number of general observations concerning twisted group algebras. The results recorded will play a significant role in our subsequent investigations. Throughout, G denotes a finite group and F* the multiplicative group of a field F . Let Z2(G,F*)be the set of all functions

a:GxG-,F* which satisfy the following identities:

We shall refer to the elements of Z2(G,F') as cocycles. Given a , p E Z2(G7F'), define a@by the rule

It is then obvious that a@is also a cocycle and that Z2(G7F * ) becomes an abelian group. Let t : G -+ F* be such that t ( 1 ) = 1 and let 6t : G x G + F* be defined by W(z, 3)=W ( y ) t ( z y ) - ' We shall refer to S t as a coboundary. It is routine to verify that the set B2(G,F') of all coboundaries constitutes a subgroup of Z 2 ( G ,F*). The factor group

H2(G7F') = Z2(G7F * ) / B 2 ( GF, * ) is said to be the second cohomology group of G over F*. The elements of H2(G,F * ) are called cohomology classes ; any two cocycles contained in the same cohomology class are said to be cohomologous. Given (Y E Z2(G7F * ) ,we shall write &. for the cohomology class containing a. Given a E Z 2 ( G 7 F * )we , write F*G for the vector space over F


168

with basis {gig E G} which is in one-to-one correspondence with G. Define multiplication in F"G distributively using

zg = a ( z , y ) w ,

Xa: = a:X

for all z, y E G, X E F. Then F*G is a twisted group algebra of G over F and, conversely, any twisted group algebra arises in this manner. Note that if a ( z , y ) = 1 for all z , y E G, then F a G Z FG. More generally, we have 6.1. Lemma. For any given (Y E Z2(G,F*),the following conditions are equivalent. (i) F"G E F G as F-algebras. (ii) F"G admits an F-algebra homomorphism into F. (iii) a is a coboundary.

Proof. (i)+(ii): This follows from the fact that the augmentation map F G + F is a homomorphism of F-algebras. (ii)+(iii): Let f : F*G + F be a homomorphism of F-algebras. Then, for all x , y E G,

f(a:)f($=

fm)= f ( + , Y ) z y )

= +,Y)f(zy)

and so a = bt, where t : G -+ F*is defined by t ( g ) = f(g). (iii)+(i): Assume that a = St for some t : G --.) F* with t ( 1 ) = 1. Then the map qh : F"G 4 F G , which is the extension of g H t ( g ) g by F-linearity gives the desired isomorphism. H 6.2. Lemma.

(i) Each element of H2(G,F*)has order dividing

IGI-

(ii) If G is a p-group and F a perfect field of characteristic p , then F G for all H2(G,F*) = 1. In particular, b y Lemma 6.1, F*G a E Z2(G,F*). FG (iii) If G is cyclic and F is algebraically closed, then F"G for all a E Z2(G,F*)(in particular, b y Lemma 6.1, H2(G,F') = 1).

Proof. (i) Assume that a E Z2(G,F'). Because 4 3 , Y)+Y,

4 =4 Y ,4 4 x 9Y

4

169


Thus a" = S t , as asserted. (ii) By (i), it suffices to show that if QP is a coboundary, then so is a. To this end, assume that t : G 3 F', t ( 1 ) = 1 is such that

Since F is perfect and charF = p , we have

for some p : G + F' with p(1) = 1. Thus a = Sp, as required. (iii) Let g be a generator of G, say of order n. Then g" = X - 1 for some A E F'. Because F is algebraically closed, X-' = p" for some p E F'. Thus (pij)" = 1. Since the elements I, pij, .. . ,(pij)"-' form an F-basis of F"G, we conclude that FOG Z FG.

6.3. Lemma. Let a E Z2(G,P)and let L be a field extension of F . Then the map

z@g

zg

is an isomorphism of L-algebras. Proof. Let {ijlg E G} be an F-basis of F"G with zij = a ( x , y ) Z y for all x , y E G. For each g E G, put ij = 1 8 9 . Then {jig E G) is an L-basis of L @F F"G and, for all x,y E G,

So the lemma is verified.

170


We are now ready to prove the following useful result. 6.4. Proposition. Let G be a nonidentity group and let CY E Z2(G,F*).If charF = p > 0 and G is a p-group, then F"G is local.

The converse is true if F is algebraically closed.

Proof. Let charF = p > 0 and let G be a pgroup. If E is the algebraic closure of F, then F"G admits an injective homomorphism into E"G (Lemma 6.3). Since E is perfect of characteristic p and G is a pgroup, E"G Z E G by Lemma 6.2(ii). Hence, by Proposition 2.10.11, E"G is local. Thus F"G must also be local. Conversely, suppose that F is algebraically closed and F"G is local. Assume by way of contradiction that G has an element g of order n > 1 where n # 0 in F. Then, for H =< g >, we have F"H Z FH. Since e = (l/n)Ch,=Hh is a nontrivial idempotent of F H , we obtain the desired contradiction. Another useful observation is recorded in the following proposition. 6.5. Proposition. Let F be a field of characteristic p > 0 , let G be a p-group and let Q E Z2(G,F*). Then F"G/J(F"G) is a finite

purely inseparabze field extension of F.

Proof. Let E be the algebraic closure of F . Then, by Lemmas 6.3 and 6.2(iii)

Since G is a p-group and charF = p , we also have E G / J ( E G ) s E . Thus EaG/J(EaG) E E. Now put I = J(EaG) n F"G. Then I is a nilpotent ideal of F"G with F"G/I an F-subalgebra of E. Thus F"G/I is a finite field extension of F and therefore I = J(F*G). Furthermore, the field extension is purely inseparable since it is generated by the images of the element 3,s E G and these satisfy ijpn E F for some n. H Our next aim is to determine explicitly a basis for the centre Z(F"G) Q E Z2(G,F') and a basis ( Z l s E G}

of FOG. To this end, we now fix

6.

171

Twisted group algebras over fields

of F"G such that

iy = a(x,y)zy

for all x,y E G

An element g E G is said to be a-regular if

Expressed otherwise, g is a-regular if and only if

It is obvious that the identity element of G is a-regular. The following elementary properties of a-regular elements are worthwhile mentioning. 6.6. Lemma. (i) An element g E G is a-regular i f and only i f it is @-regular for a n y cocycle @ cohomologous t o a. (ii) If g E G is a-regular, t h e n so is a n y conjugate of 9.

Proof. (i) Let g be an a-regular element and let p be cohomologous to a, say @(x,y) = a ( x , y ) t ( . ) t ( y ) t ( x y ) - l for some t : G + F*. Then, for x E C G ( g ) ,

proving that g is @-regular. The converse is obvious. (ii) Let y E G and let x E C ~ ( y g y - ' ) . Then y-'xy E C G ( ~and ) therefore y - l z y . g = ji . y-1.y Since y - l x y = X ~ - ' C C for ~ some X = X(x,y) E F*, it follows that y-lzys = g y - l x y

Hence i commutes with ygy-'. Again, since ygy-1 = pyijy-l for some p = p ( g , y ) E F*, we see that i commutes with ygy-1. Thus ygy-l is also a-regular.


172

Let Cgbe the conjugacy class of G containing g. We say that Cg is a-regular if g (and hence, by Lemma 6.6(ii), any element of Cg)is aregular. The set of all a-regular elements of G will be denoted by Go. We also say that the cocycle a is standard if it satisfies the following two properties: for all x E G

a ( x , x - ~= ) 1

for all g E G o , x E G

a ( x ,g ) a ( z g ,2 - 1 ) = 1

Expressed otherwise, a is standard if and only if the following two properties hold

___-

xgx

= xgx-1

for all

x E G,g E G"

6.7. Lemma. Let G be a finite group of order n , let F be an algebraically closed field of characteristic p 2 0, and let a E Z 2 ( G ,F*). If p divides n, let P be the largest normal p-subgroup of G. Then, after making a diagonal change of basis, if necessary, the following properties hold: (i) 9-' = ij-' for all g E G (ii) Z g 5 - l = xgx-l for all x E G,g E Go (iii) ~y = ~y for all x , y E P .

Proof. Owing to Lemma 6.2(ii), there exists t : P + F* such that a ( x ,y ) = t(z)t(y)t(xy)-'

for all

2,y

E

P

Let X : G + F* be defined by X(x) = t ( z ) - l if x E P and X(x) = 1 if x $ P . Then, replacing g by X ( g ) g , we may assume that (iii) holds. We now show that (ii) holds for g E P . Indeed, write 5g5-l = pxgx-l for some p E F* and put rn = ]PI. Then, by (iii),

gm whence

= xgx - 1 m = 1


173

Since m is a power of p , p = 1 and therefore (ii) holds. Suppose now that g E G" - P and let Cg = { g = g 1 , g 2 , . - . , g r } . Because g is a-regular, ij has r conjugates in I' = {XglX E F*,g E G}. Choose i j arbitrarily, take g 2 , . . . , g r as its conjugates in I' and leave other basis elements unchanged. Then we have conditions (ii) and (iii) holding. Due to (iii), to prove (i), we may assume that g $ P . Consider the elements of G - P and write G - P as a disjoint union

where S1 consists of the a-regular elements. For one of each pair 9,g-l E Sz with g # 9-', leave g unchanged and replace by g-' = a(g,g-')-'g-l. For each g E S2 with g = g-' replace i j by a ( g , g-')-'/Zg. Now consider g E S1 and write Cg = ( 9 = 91,. . . , g p ) . We still have above the choice of g at our disposal. If Cg # Cg-l, we choose ij,g-' asin the case g# 9-'. If Z g 2 - l = gi, then 2ij-lz-l = gi-' = g i ' , by choice of gi,gy'. Finally, let Cg = Cg-1. Thus 9-' = ygy-' for some y E G and 9-' - Y9Y (1)

----'

Replacing g by a(g,g-')-'12ij, and so all g; by c ~ ( g , g - ' ) - ' / ~ g ; , we still have (1) and also 9;' = g i - ' . This completes the proof.

6.8. Lemma. Let a E Z 2 ( G , F * ) ,let (91,. ..,st} be a full set of representatives for the a-regular classes of G and let Ti be a left transversal for Cc(g;)in G . Then the elements CzETi ZgiZ-', 15 i 5 t f o r m an F-basis for Z(F*G). In particular, if a is standard and C; is the conjugacy class of g i , then the elements

form an F-basis for Z(F*G). 7

Proof. Setting sgiz-1 = Zg;Z> z E Ti, 1 5 i 5 t , and = i j if g ----I is not regular, we have zgz - zgz-l for all z E G and all a-regular


174

g E G. Thus we may assume that 3 g Z - l = xgz-' for all x E G and all a-regular g E G and we must show that the elements of (2) form an

F-basis for Z(F"G). To this end, put ki = EgECig, 1 5 i _< t . Let y = Eygg E F"G be the elements of F* such that and let

Then y is central if and only if for all x E G

9

9

Now put Suppy = (9 E GI?, # 0) and fix t E Suppy,a E Cc(t). If y is central, then comparing the coefficients of f, we get yt = ytX,,t. Hence = 1 and therefore t is a-regular. Since Zij5-l = zgz-l for all z E G and all a-regular g E G, it follows from (3) that y is central if and only if (i) Suppy consists of a-regular elements (4 r g s = 7gzg"-' = 7z-lgz9- for all z E G or, equivalently, if and only if y is an F-linear combination of k;,1 I: i 5 t. Since the elements Ici are obviously linearly independent, the result follows.

c

c

c

Let A be an F-algebra and define [A,A ] ,the commutator subspace of A , to be the F-linear span of all Lie products [x,y] = zy -yz, z, y E A. The following observation illustrates how the commutator subspace can be brought into argument. 6.9. Lemma. Let A be an algebra over a field of characteristic

P>O cz1 ay"(mod[A,A ] ) (i) ( c E 1 a*)'" (ii) I f s E [ A ,A ] , then s p E [A,A]

(a; E A , n E N)

Proof. (i) By induction argument, it suffices to show that (a

+ b)P" E a'" + bP"(mod[A,A ] )

175


for all a, b E A. Suppose first that n = 1. Expanding by the distributive law, we have (a+

b ) P = up

+ bp +

&la2

(4)

*"UP

where the sum is over all products a1 - + - a of p p terms, a; E { a , b } , not all equal to a or b. With each word a1a2 - - up associate its cyclic permutations

-

U1U2

*

- - U p , U2U3

* *

a

UpUl,

* *

,U p U l - - U p - l *

All these products are congruent modulo [A,A ] . This is so since for x = a;,a;, - . . u ; ~y , = a; a;,+,

-

1

a;,ql

- . . a;3-1

we have x - y = [y, 61 E [A,A ] , where y = a;, a,,

-

*

a;J-l and

S = aiJa;j+l

- a;,

Therefore the sum of these cyclic permutations is p x 1 - x p modulo [ A , A ] ,and hence it belongs to [ A , A ] .Applying (4),it follows that

and hence that

(ab - b ~ E) (~~ 6 ) ' - (ba)P(mod[A,A ] ) G [a,(ba)P-'b](rnod[A, A]) Accordingly

The required assertion now follows from (5) and (6) by induction on n. (ii) Apply (6). We next describe the commutator subspace of FaG. 6.10. Lemma. Let a be a standard cocycle and let y = E y Z z E FOG. Then y E [FOG,F"G] if and only if the sum of the coeficients yz over each a-regular class of G is zero.

176


Proof. Denote by C1, ,C, the a-regular classes of G and let : F*G -+ F x F x - * x F (s times) be the F-linear mapping given

t,b

by

c

-

7 3 H (Ply ' ' ' P a )

where

Pi

=

c

7x

XECi

xEG

We must show that [F"G,F"G] = Kert+b. Observe that [F"G,F"G] is spanned by all Sy - yii with z,y E G. First assume that zy is not a-regular. Then so is yz since zy and yz are conjugate. But then

zy - yz = a ( x ,y) zy - a ( y , z)yz which shows that [ z , y ] E K e r $ . Suppose next that z y is a-regular. Then, bearing in mind that a is standard, we have

zy - yz = 5 y - z-l(zy)z = a ( z ,y )z y - a@,y)5-1"Yz = Q(Z, y ) ( a - z-'(zy)z) Hence [Z,y] E Kerzl, and therefore [F"G,F"G] KerG. To prove the opposite containment, we first demonstrate L a t i g E G is not an a-regular element, then g E [F*G,F*G].Indeed, let g be a nonregular element so that Zg # 33 for some x E cG(g). Because z commutes with g, we have zg2-l = X i j for some X E F* with X # 1. In view of the identity

g - 235-1 = (33-1 )z- - "35-1)

(7)

we deduce that (1 - X)g, and hence g, belongs to [F*G,F"G]. We are thus left to verify that if xi = CIECi y&i and if CIECi~i = 0, then x; E [ P G ,FaGI. The latter being a consequence of (7) and a vector space identity

if

xy=lX i = 0, the result follows. H As a generalization of Maschke's thpwern) we next prove

6.11. Proposition. Suppose that the characteristic of the field F does not divide the order of G. Then FOG is semisimple.

177


Proof. Let W be a submodule of an PG-module V . Since V is a vector space over F , its subspace W has a complement in V, say V = W@W'.Let 8 : V + W be the projection map, and let $ : V + V be defined by

$(v) = IGI-'

C z8s-lv

(v E V )

xEG

Because for all v E V,y E G,

$(yv)

C ZOZ-lijv = 1GI-l CjE6jjZ-1yv 1GI-l C i j Z ~ r ( yz, ) - ~ ~ c Y~ )( Z~ -, ~ i j - ~ i j v 1GI-l C i j Z 8 Z - l ~

= 1GI-l

xEG

=

ZEG

zEG

=

zEG

= Y+(v>,

it follows that $ is an FQG-homomorphism. Now assume that v E W . Then, for any x E G,Z-'v E W , so O(Z-lv) = Z-lv and thus ZOii-'v = v. Consequently, $(v) = v for all v E W . This implies that W is a direct summand of V and the result follows. H The next lemma is a final step in the preparation for the proof of the main result. 6.12. Lemma. Let A be a finite-dimensional algebra over an algebraically closed field of characteristic p > 0 and let S = [A,A].

Then

T = (5 E AlzPmE S

for some integer m 2 1)

is an F-subspace of A and the number of nonisomorphic irreducible A-modules equals dimF(A/T). Proof. Owing to Lemma 6.9(ii), a E [A,A]implies ap E [A,A]. Thus S C 2'. By Lemma 6.9(i), if a , b E T , then for sufficiently large m,

(a

+ b)pm = apm+ bp'"

3

0

(mod S)

178


+

which shows that a b E T . Clearly, Xu E T for all X E F and a E 2'. Hence T is an F-subspace of A. Since the Jacobson radical J ( A ) of A is nilpotent, it follows from the definition of T that J ( A ) & T . Now put A = A / J ( A ) and T = T / J ( A ) .If r is the number of nonisomorphic irreducible A-modules, then there exists positive integers n l , . . . ,n, such that P

A 2 IIM,;(F) i=l

+

Now put = (S J ( A ) ) / J ( A ) Then . clearly S is the F-subspace of spanned by all ab - bu ( a ,b E A ) and

T = {z E A l s p "

ES

for some integer

A

rn 2 I}.

Let Si, 2';be the analogous F-subspaces defined for the F-algebra A; = M n i ( F ) .Then P

d i r n ~ ( A / T=) d i m ~ ( A / T=)

C dirn~(A;/T;) i=l

and therefore it suffices to verify that

Let est be the ni x ni-matrix with (s, t)th entry equal to 1 and all other entries 0. Then, for any s # t , est = esjejt - ejtesj E Si

and ess - ett = estets - etsest E S;

It follows that S; contains the n' - 1 linearly independent elements est(l 5 s , t 5 ni,s # t ) and ess - e l l ( s = 2 , . . . ,ni). Because S; C T.j, this shows that dirn~(A;/Ti)= d i m ~ ( A i-) d i m ~ ( T ;5) n: - (ns - 1) = 1 On the other hand, the elements of 5'; are all n; x n;-matrices with trace 0. In particular, ell # S;. But eyy = ell for all m 2 1, so

179


4

T,. Thus Ti # A;, so from dirnF(A/T;) 5 1, we deduce that dirn~(A;/T;) = 1, as desired. ell

In what follows, if p = 0 then by definition all elements of G are p'-elements. 6.13. Theorem. (Asano-Osima-Takahasi (1937)). Let G be a finite group, let F be an algebraically closed field of characteristic p 2 0 and let a E Z2(G,F * ) . Then the number of nonisomorphic irreducible F"G-modules is equal to the number ofa-regular classes of p'-elements of G.

Proof. Let r be the number of nonisomorphic irreducible F*Gmodules. If charF = 0 , then FaG is semisimple (Proposition S.ll), hence r = ( Z ( F " G ): F ) This proves the case charF = 0, by applying Lemma 6.8. Now assume that charF = p > 0 and put S = [F*G,F"G] and

T = { x E F"G1xPm E S

for some integer

rn 2 1)

Choose an F-basis a l , . . . , a t of T . We may find the integer m so large that afm E S for each i and pm is at least as large as the order of a Sylow p-subgroup of G. Owing to Lemma 6.9, we have

for all A; E F . Hence x = equivalently

xET

Cxgg E T if and only if xPm E S or,

if and only if

c

x;mXSpmgpmE

s

where, by our choise of rn, each gPm is a $-element and for each g E G, A, E F* satisfies g P m = A:mgpm. Let C1, Cz, . . . ,C, denote the a-regular classes of p'-elements of G, and Put Cz*= (9 E GlgPmE C;} (1 5 i 5 s )


180

By Lemma 6.7, we may assume that CY is a standard cocycle. Applying (8) and Lemma 6.10, we therefore deduce that x = C xga E T if and only if /

\

(c

g€C?

pm

= XgXg)

c

(XgXg)Prn

=0

S€CY

for each i E (1,. . . ,s}. Thus xgg E T

if and only if

c

xgXg = 0

(9)

gECf

for all i E (1,. . . ,s}. Now define 1c, : F"G + F x the F-linear map given by

x F (s times ) as

with ,8; = C g E C f zgXg. Then 1c, is clearly surjective and, by (9), T is the kernel of +. Thus, by Lemma 6.12, r = dirnF(F"G/T) = s

and the result follows.

7.

Total and absolute indecomposability of induced modules

The notion of indecomposability of a module can be strengthened in two different ways. The first is absolute indecomposability in the sense of Huppert and the second is absolute indecomposability in the sense of Green. Since we shall use both notions, it will be convenient to rename the stronger one (namely absolute indecomposability in the sense of Huppert) as total indecomposability. Throughout this section, N denotes a normal subgroup of a finite group G, R a commutative local ring and A an R-algebra. Let V be an A-module and let S = EndA(V). We say that V is totally indecomposable if = R * 1s J ( S )

s

+

7.1. Lemma. The following conditions are equivalent: (i) V is totally indecomposable.

7. Total and absolute indecomposability of induced modules

181

(ii) J ( R ) S E J ( S ) and S / J ( S ) is ofR/J(R)-dimension I . (iii) J(R)S C J ( S ) and S / J ( S ) E R / J ( R ) as R/J(R)-algebras. (Note that if R is noetherian and V finitely generated, then J ( R ) S J ( S ) b y Lemmas f.7and 1.2).

+

Proof. (i)+(ii): By hypothesis, S = R 1s J ( S ) ,hence

S / J ( S ) 2 R . i s / ( R . is n J ( S ) )

-

Since R is local, so is its homomorphic image R 1s and hence Re 1s n J ( S ) J ( R . l s ) . Thus, by Corollary 1.4.4(i), R . l s n J ( S ) = J ( R - I s ) . SInce J ( R 1s) = J ( R ) ls, it follows that J ( R ) S C J ( S ) . Thus S / J ( S ) is an algebra over the field R / J ( R ) having dimension 1 since = R 1s -t J ( S ) . (ii)=+(iii): Obvious. (iii)+(i): Obvious.

-

-

s

It follows from Lemma 6.1 that if V is totally indecomposable, then

V is strongly indecomposable and hence indecomposable. Moreover, as a partial converse, we have

7.2. Lemma. Let V be finitely generated, let R be a complete noetherian local ring and let the jield R/ J ( R ) be algebraically closed. If V is indecomposable, then V is totally indecomposable.

Proof. Let V be indecomposable. By the proof of CaroDary V is strongly indecomposable, hence S / J ( S )is a division ring. By Lemma 1.7 and 1.2, J ( R ) S J ( S ) and so S / J ( S )is an R/J(R)-algebra. Furthermore, by Lemma 1.7, S is a finitely generated R-module. Hen= S / J ( S ) is a finite-dimensional algebra over the field R/J(R). Since R / J ( R ) is aIgebraically closed, it follows that S / J ( S ) is of R / J ( R ) dimension 1. Hence, by Lemma 1.7, V is totally indecomposable. W The next observation will justify our concentration on totally indecomposable modules.

7.3. Lemma. Let V be a totally indecomposable G-invariant R N module, let S = EndRN(V) and let E = EndRG(VG). Then

182


(i) E - J ( S ) J ( E ) and A = E / E - J ( S ) is a twisted group algebra of G / N over the field R / J ( R ) . (ii) V G is strongly indecomposable if and only if A is local. (iii) V G is totally indecomposable if and only i f A/ J ( A ) is of R/ J ( R ) dimension 1. Proof. (i) This is a direct consequence of Theorem 4.2(iv), (v) and Lemma 7.l(iii). (ii)and (iii) Since E - J ( S ) 2 J ( E ) , we have E / J ( E ) E A / J ( A ) as R/ J ( R)-algebras. Hence E is local (respectively, E / J ( E ) is of R/ J ( R)dimension 1)if and only if A is local (respectively, A / J ( A ) is of R / J ( R ) dimension 1). Invoking some general facts on twisted group algebras, we now record the following consequences of the results so far obtained.

7.4. Theorem. Let V be a totally indecomposable G-invariant RN-module, let the field R / J ( R ) be of prime characteristic p and let GIN be a p-group. Then V G is strongly indecomposable. Proof. Let A be as in Lemma 7.3(i). Then, by Proposition 6.4, A is local. Hence, by Lemma 7.3(ii), V Gis strongly indecomposable.

7.5. Theorem. Let V be a totally indecomposable G-invariant RN-module. If V G is totally indecomposable, then either G = N or c h a r R / J ( R ) = p > 0 and G / N is a p-group. The converse is true if the field R/ J ( R ) is perfect. Proof. We may harmlessly assume that G # N . Let A be as in Lemma 7.3(i), let F = R / J ( R ) and let H = G / N . Then A F"H for some a E Z 2 ( H , F * )and, by Lemma 7.3(iii), V G is totally indecomposable if and only if

dimFF"H/J(F"H) = 1

(1)

Assume that (1) holds. Then, by Lemma 6.1(ii), F"H Z F H and hence F H is local. Since H # 1, we conclude from Proposition 2.10.11 that charF = p > 0 and H is a pgroup.


183

Conversely, assume that F is perfect, charF = p > 0 and H is a F H . Hence (1) holds and pgroup. Then, by Lemma 6.2(ii), F a H the result follows.

7.6. Theorem. Let R be a complete noetherian local ring, let V be a finitely generated totally indecomposable RN-module and let H be the inertia group of V . If charR/ J ( R ) = p > 0 and H / N is a p-group, then V G is indecomposable.

Proof. Owing to Theorem 2.1, we may assume that G = H , in which case the result follows by virtue of Theorem 7.4. H The result above can be strengthened under the additional assumption on R. This is the content of the following theorem.

7.7. Theorem. Let R be a complete noetherian local ring such that the field R/ J ( R ) is algebraically closed. If V is a finitely generated indecomposable RN-module and H is the inertia group of V , then the following conditions are equivalent: (i) V G is indecomposable. (ii) V G is totally indecomposable. (iii) H = N or c h a r R J J ( R )= p > 0 and HIN is a p-group. Proof. b y Lemma 7.2,(i) and (ii) are equivalent. Moreover, by Theorem 2.1, V G is indecomposable if and only if V H is indecomposable. Hence we may assume that G = H , i.e. that V is G-invariant. By Lemma 7.2, V is totally indecomposable. Moreover, since R is a complete noetherian local ring, V G is indecomposable if and only if V G is strongly indecomposable (see proof of Corollary 1.8). Hence, by Lemma 7.3(ii), V G is indecomposable if and only if the twisted group algebra A of that lemma is local. The desired conclusion is therefore a consequence of Proposition 6.4. Assume that V is a finitely generated totally indecomposable Ginvariant RN-module, where R is a complete noetherian local ring. Put F = R / J ( R ) ,E = EndRG(VG)and S = EndRN(V). Then, by


184

Lemma 7.3(i),

E / E . J ( S )E F " ( G / N ) and

for some

CY

E Z 2 ( G / N ,F*)

c

E * J(S) J(E) By Theorem 1.5, any decomposition F"(G/N) = L1@

@ L,

into indecomposable right ideals of F " ( G / N ) can be lifted to a decomposition E = II@*..@I, into indecomposable right ideals of E. We are now ready to prove the following result essentially due to Conlon (1964). 7.8. Theorem. Let V be a finitely generated totally indecomposable G-invariant RN-module, where R is a complete noetherian local ring, Then, in the notation above, the following properties hold: (i) V G = @r=lI;V is a decomposition into indecomposable RGmodules such that

( I i v ) 2~ e;V

with e; = ranks(Ii) = dimFL;

(ii) I;V E IjV as RG-modules if and only i f L; % Lj as right F"(G/N)-modules. In particular, the number of nonisomorphic indecomposable direct summands ofVG is equal to the number of nonisomorphic irreducible Fa(GIN)-modules. (iii) If V is R-free, then each I;V is R-free and rankR(I;V) = (dimFL;)(rankR(V))

(1 5 i 5 n )

Proof. All of the stated properties will follow from Theorems 5.6, 1.5 and 2.12.4 provided we show that ranks(I;) = dimFL;

(1 5 i 2 n )

-

Because F a ( G / N )is identifiable with E / E J ( S ) ,the latter will follow provided we show that for e; = ranks(1;)

+

dimF(1; E J ( S ) ) / E- J ( S ) )= e;

(2)


185

To prove (2), fix i E { 1,. . . ,n } and put ei = e. Then choose an S-basis { d l , . . . ,d e } of Ij and extend it to an S-basis

(this is always possible because all direct summands of E are free Smodules by Theorem 5.6(i)). Then we have

which implies that

Ij f l E J ( S ) = $;=,d;J(S)

+

-

Thus each element in (I; E - J ( S ) ) / E J ( S ) is a unique S / J ( S ) linear combination of the images of d l , . . . ,d , in ( I ; E J ( S ) / E J ( S ) . Invoking Lemma 7.1, we conclude that (2) holds and the result is established.

+ -

*

7.9. Corollary. Let R be a complete noetherian local ring such that the field F = R / J ( R ) is algebraically closed of characteristicp > 0 . If V is a finitely generated indecomposable RN-module and H is the inertia group of V , then f o r

E = EndRH(VH)

and S = EndRN(V)

the following properties hold: (i) E / E . J ( S ) 2 F " ( H / N ) for some LY E Z 2 ( H / N , F * ) . (ii) The number of nonisomorphic indecomposable direct summands o f V G is equal to the number of a-regular classes ofp'-elements of H / N . Proof. (i) By Lemma 7.2, V is totally indecomposable. Now apply Lemma 7.3(i). (ii) By Theorem 2.1(i), we may assume that H = G, i.e. that V is G-invariant . The desired conclusion now follows by applying Theorems 7.8(ii) and 6.13.

For the rest of this section, we assume that R satisfies the following two conditions:

186


(i) R is a complete local ring. (ii) R is a principal ideal domain. (for example, R can be a field, or the ring of P-adic integers, P a prime ideal of an algebraic number field). If R is not a field and R satisfies (i) and (ii), then R is said to be a complete discrete valuation ring. An integral domain S containing R is an extension of R, written SIR, if the following hold: (i) S is a principal ideal domain and a local ring. (ii) S is R-free. (iii) J ( S ) e= J ( R ) S for some integer e 2 1. We refer to e as the ramification index of SIR. In case e = 1, we say that SIR is unramified. Note that if SIR is an extension of R, then S / J ( S ) is a field extension of R / J ( R ) .Of course, if R is a field, then an extension SIR is simply an arbitrary field extension. Also, if R is a field, then all extensions S/R are unramified. We say that the extension SIR is finite if S is a finitely generated R-module. In this case, if R is a complete discrete valuation ring, then so is S. This is a consequence of a classical theorem of Cohen (1946). Let A be an R-algebra and let SIR be an extension of R. Then we Put As = S @ R A It is clear that A s is an S-algebra. Now assume that A is R-free of finite rank and let { q ,. . . ,a , } be an R-basis of A. Then the map

is an injective homomorphism of R-algebras. Identifying A with its image in As, it follows that each element of A s can be uniquely written in the form EKl X;a;, X i E S. If V is an A-module, then V . = S @ R V becomes an As-module under a module action

We say that V is absolutely indecomposable if for every finite extension SIR, Vs is an indecomposable As-module.


187

7.10. Proposition. Let A be an R-algebra which is R-free of jinite rank, let V , W be finitely generated A-modules and let SIR be an extension of R. Then

If V = W , then the two sides of this are isomorphic S-algebras. Proof. Let { s;} be an R-basis of S . Each element in HornA (V,W ) , can be written uniquely in the form Csj 8 fi with fj E H o ~ A ( V , W ) and with finitely many f; # 0. Consider the map given by

+

HomA(V,w)S

HOmAs(VS,WS)

Then clearly 1c, is an S-homomorphism and, in case V = W , 1c, is also a ring homomorphism. If $(Cs; @I f j ) = 0, then Cs; 8 f;(v) = 0 for all v E V . Thus all f; = 0 and therefore 1c, is injective. Now fix X E H o ~ A ~ ( Ws). V S , For each i, define f; by

Because V is finitely generated, there are only finitely many nonzero fi. Clearly, f; E HornA( V,W ) for each i and $(C s; 8 fi) = A. Thus 1c, is surjective and the result follows. H 7.11. Lemma. Let A be an R-algebm which is R-free of finite rank and let SIR be a finite extension of R. Then, for E = EndA(V), (i) J ( E ) s J ( E s ) . (ii) E s / J ( E ) s ( E / J ( E ) ) sas S-algebras.

c

Proof. (i) By Lemmas 1.7 and 1.2, E is a finitely generated Rmodule with J ( R ) E 5 J ( E ) ,so E / J ( R ) Eis a finite-dimensional algebra over the field R / J ( R ) with J ( E / J ( R ) E )= J ( E ) / J ( R ) E .Hence J(E)" 5 J ( R ) E for some rn >_ 1. If sl,. . . ,,s E S and x1).. . ,x, E J ( E ) ,then we have


188

and thus J ( E ) s / J ( R ) E sis a nilpotent ideal of Es/J(R)Es. Hence

and therefore J ( E ) s E J(Es). (ii) We may write S = Rsl @ Then

... @ Rs, for some sl,. . . ,s,

E S.

which implies that

as required.

7.12. Lemma. Let D be a division ring and let F be a subfield of Z ( D ) such that dimFD < 00 and D # F . If Xo E D - F , let f ( X ) be the rnonic irreducible polynomial over F with root Xo) and let K be a field extension of F by a root of f(X). Then K @ F D is not a division ring. Proof. By hypothesis, deg f ( X ) = n > 1 and

Since f ( X ) is reducible over K 8 1 %! K , we may write

and g and h are of degree at least 1. Owing to (3), we have

Since


189

it follows that K 693 D has zero divisors. 7.13. Proposition. Let A be an R-algebra which is R-free of finite rank, let V be a finitely generated indecomposable A-module and

let E = EndA(V). (i) If V is totally indecomposable, then for a n y finite extension S I R , Vs is totally indecomposable. (ii) I f V is totally indecomposable, then V is absolutely indecomposable. (iii) There exists a finite unmmified extension S I R such that

where each Proof. 7.11,

I$ is a totally indecomposable As-module. (i) By hypothesis E / J ( E ) Z R / J ( R )and so, by Lemma

E s / J ( E ) sz S 6 9 R~/ J ( R )2 S / J ( R ) S Hence

Es/J(&)

=S/J(S),

proving (i) by applying Proposition 7.10. (ii) Direct consequence of (i). (iii) Given a finite unramified extension S I R , it follows that Vs is a direct sum of, say n ( S ) ,nonzero indecomposable As-modules, where 1 5 n ( S ) 5 dirnR,JcRl(V/J(R)V).We may therefore choose S so that n ( S ) is maximum. It is obvious that every component of Vs remains indecomposable when tensored with any finite unramified extension of S. Replacing R by S, we may thus assume that Vs is indecomposable for any finite unramified extension SIR. For any finite unramified extension S I R , put

Then D ( S ) is a division ring and a finite-dimensional S/J(S)-algebra. If T / S is a finite unramified extension, then T / R is a finite unramified extension. We now show that D ( T ) is a homomorphic image of T @s


190

D W

Owing to Lemma 7.11(i), we have

VT

= ( t l @V S ) €B

..-€B ( t , €3 VS)

Consider the map X : T x EndA,( Vs) + EndA, (V T )given by

Then X obviously induces a well defined S-bilinear map

T x D ( S )+ D ( T ) and a homomorphism

Since by Proposition 7.10,

it follows that

is surjective. Thus D ( T ) is a homomorphic image of

T 63s D ( S ) . Choose a finite unramified extension K / R such that d i m ~ , ~ ( q D ( I < ) is minimum. If D ( K ) # l " / J ( K ) , then there is a monic irreducible polynomial f ( X ) E K / J ( K ) [ X ]of degree at least two having a root in D ( K ) . Choose fO(X) E K [ X ] such that the image of f o ( X ) in K / J ( K ) [ X ]is f ( X ) . Let S = K[b]where b is a root of f o ( X ) . It is easily verified that S / K is a finite unramified extension. By Lemma 7.12, S @K D ( K ) is not a division ring, hence D ( S ) is a proper homomorphic image of S € 3 D~ ( K ) and dirnsp(s)D(S)< d i r n ~ l ~ ( ~ ) D ( K ) , a contradiction. Thus D ( K ) = K / J ( K ) and the result follows. N

Lemma. For any two finite extensions S I / R and SZ/R, there is a finite extension T / R such that T is an extension of both S1 and S2. 7.14.

8. Crossed products over prime rings

191

Proof. This is an easy consequence of basic results in valuation theory. For a self-contained proof refer to Dornhoff (1972, p.328). W We are now ready to prove the following fundamental result.

7.15. Theorem. (Green (1959)) Feit (1962)). Let R be a complete local ring and a principal ideal domain. Let V be a finitely generated absolutely indecomposable R N - m o d u l e , let the field R / J ( R ) be of prime characteristic p and let GIN be a p-group. T h e n V Gis absolutely indecomposable.

Proof. Owing to Proposition 7.13(iii), we may choose a finite extension S I R such that every indecomposable component of (VG)s% ( Vs)" is totally indecomposable. Again, by Proposition 7.13(iii), there is a finite extension T / S such that VT is totally indecomposable. By Proposition 7.13(i), every indecomposable component of ( V G )=~( ( V G ) s ) , is totally indecomposable. But, by Theorem 7.6, (V"), is indecomposable. Hence ( V"), is totally indecomposable. If the theorem is false, then there is a finite extension K / R such that ( V G )is~decomposable. Invoking Lemma 7.14, we may choose a common finite extension L of T and K . By Proposition 7.13(i), (VG), is totally indecomposable. But K L , so (VG)Lis decomposable, a contradiction. 8.

Crossed products over prime rings

Throughout this section, G denotes a finite group, S a ring and S * G a crossed product of G over S. &call, from Proposition 3.4, that if (G, S,0,a ) is a crossed system of S * G, then S * G is a free (left and right) S-module freely generated by the elements ij,g E G, with multiplication given by

~ " s ~ ~ ~ , Y )( z~ Y1 ~ E3 2S, Z , Y E G) where a ( z , y ) E S and Is2 = a(z)(sz) E S. We now introduce some termonology. First of all, recall that an ideal f of S is said to be G-invariant if ( S I ~ : ) ( S ~= Y )s

gig-1 = I

for all g E G


192

A G-invariant ideal I of S is said to be G-prime if III2 C I for Ginvariant ideals I; of S imply 11 E I or I2 E I. The ring S is said to be prime if for all nonzero ideals A, B of S we have AB # 0. An ideal P of S is called a prime ideal if SIP is a prime ring. A minimal prime ideal of S is a prime ideal P of S such that P is not properly contained in any other prime ideal. Our aim is to prove some purely ring-theoretic results concerning the structure of S * G, where S is a prime ring. These results will be applied in the next section to investigate the homogeneity of induced modules. 8.1. Lemma. If P is a prime ideal of S * G, then P n S is a G-prime ideal of S . Conversely) if A is a G-prime ideal of S ) then there exists a prime ideal P o f S * G such that P n S = A.

Proof. Observe that if I is an ideal of S * G, then since I is Ginvariant so is I n S. Conversely, if A is a G-invariant ideal of S , then by Theorem 3.6(ii), A . (S * G ) = ( S * G ) A = A * G is a two sided ideal of S * G with (A * G) n S = A. Assume that P is a prime ideal of S * G. Then, by the above, P n S is at least G-invariant. Now suppose that A and B are G-invariant ideals of S with AB P n S. Then (A * G ) ( B* G) = ( S * G)AB(S* G)

c c

( S * G)(P n S)(S * G) p,

c

so the primeness of P ensures that A * G P or B * G P. Thus A P n S or B C_ P n S, proving that P n S is G-prime. Conversely, let A be a G-prime ideal of S. Then A is a G-invariant ideal of S and so A*G is an ideal of S*G with (A*G)nS = A. Applying Zorn’s lemma, we may thus choose an ideal P of S * G maximal with respect to P nS = A. Let I , J be ideals of S* G properly containing P. Then I n S and J n S are G-invariant ideals of S properly containing A. Because A is G-prime, this impiies that

( I nS)(Jn S ) g A

193


and hence I J

9 P.

Thus

P is prime as required.

In what follows, given a subset T of G, we write S * T for the set of elements of S * G with support in T . 8.2. Lemma. Let I be a nonzero left and right S-submodule of S* G and let T = {gl,g2,. . . , g n } be a subset of G with g1 = 1. Assume

that and I n ( S * T ' ) = O forall T ' c T

Iil(S*T)#O

For each i E { 1,. . . ,n } , define B; b y n

B; = { s E SI them exists p = C s j i j j E I

with s = sj}

j=1

Then (i) Each B; is a nonzem ideal of S . (ii) There exists a natural bijection f; : B1 4 B; such that for all s,t E S,a E B1 (a) f;(sat) = sfj(u)git (b) fl is the identity map. (iii) The elements of I n ( S * T ) are precisely the elements of the form:

c n

Q

=

f;(.)ji

with u E B1

i=l

Proof. Since I is a left and right S-submodule of S*G, it is clear that each B; is an ideal of S. Furthermore, the minimality of T implies that each B; is nonzero. Observe also that for each b; E B;, the minimality of T ensures that there exists a unique ,O = C s j i j j E I with si = b;. This is so since if p and p' are two such elements, then p - p' E I is an element of smaller support and so /? - p' = 0. By the foregoing, for each a E B1, there exists a unique CY = C;=,b& E I with bl = a . We may thus define fi : B1 4 Bi by setting f , ( a ) = b;. By the definition of f;,fl is the identity map and


194

Conversely, it is clear that each element of I n (S * 2’) is of this form. Furthermore, each fi is clearly an additive bijection. Finally, let a E B1 and cy be as above and let s , t E S. Then

and since S; = 1, this implies that fj(sat) = sb;”t = s f ; ( a ) ” t

thus completing the proof. H For the rest of this section, S denotes a prime ring. In order to make further progress, we briefly discuss a certain ring of quotients M = Qo(S) which is defined in Martindale (1969). Consider the set X of all left S-module homomorphisms

f:A+S where A ranges over all nonzero ideals of S. We refer to two such homomorphisms f : A + S,g : B 3 S as equivalent if there exists a nonzero ideal C of S with C C A n B and with f(c) = g(c)

for all c E C

This relation is obviously reflexive and symmetric. It is also transitive, because for all nonzero ideals A, B of S, A n B is a nonzero ideal of S. We have thus defined an equivalence relation on the set X . For each f E X , let [f,A] (or f if no confusion can arise) denote the class of the function f above. We can define an arithmetic on the set &o(S) of these classes by

if,A1

[9, BI = [fs,BAI where fg is the composition first f then g . In this way we obtain a ring, called the Martindale ring of quotients of S. *

195


The following five lemmas provide some basic properties of the ring Qo(S)* 8.3. Lemma. For any f : A + S in X , f = 0 if and only if f = 0. In particular, i f g : B + S is another element o f X , then f = ij

ifand o n l y i f f = g o n A n B . Proof. It is obvious that f = 0 implies f = 0. Conversely, suppose that f = 0. Then there exists a nonzero ideal B of S contained in A such that f ( B )= 0. Note that B A G B and so 0 = f ( B A )= B f ( A ) Because S is prime, it follows that f ( A ) = 0 and therefore f = 0. The second assertion follows from the first and the fact that f = g if and only if f - g = 0.

For every a E S, let

a, E

Homs(S,S) be defined by

a,(z) = sa.

8.4. Lemma. (i) The map S + Qo(S) d efined b y a H a, is an injective ring homomorphism. (ii) Iff : A + S is an element of X , then iirf = f ( a ) , .

+

Proof. (i) If a , b E S, then T, = 1, arb+- = (ab), and a, br = ( a b),. Hence the given map is a ring homomorphism. If a, = 0, then a, = 0, by Lemma 8.3. Thus a = 0 and (i) follows. (ii) For every a E A, Sa G A, hence f is defined on Sa. Therefore, for every x E S ,

+

proving that

- = f ( a ) r *W

~ , =f a,f

Thanks to Lemma 8.4(i), we may identify S with its image in Qo(S). With this identification, Lemma 8.4(ii) tells us that for f : A + S in

x,

af = f (4

for all a E A, f E QO(S)

(1)

196


8.5. Lemma. Let M = Qo(S) be as above. (i) If m E M and Am = 0 for some nonzero ideal A of S , then

m=0. (ii) If ml,m2,...,mk E M , then there exists a nonzem ideal A of S with each Am; in S . (iii) M is a prime ring. (iv) If Q is an automorphism of S , then 0 extends uniquely t o an automorphism of M . (v) Z ( M ) = CM(S)and Z ( M ) is a field. Proof. (i) Let m E M with Am = 0. If m = f,then (1) shows that ! ( a ) = 0 for all a E A. Thus f vanishes on an ideal in its domain and therefore m = f = 0. (ii) Let ml,m2,. . . ,mk E M with m; = &. We may harmlessly assume that all the fi are defined on the common domain A. Then, by (11,

-

am; = afi = f;(a) E S

for all a E A. (iii) Let I be a nonzero ideal of M . If 0 # m E I , then by (ii) and (i), 0 # Am E I n S for some nonzero ideal A of S. Hence every nonzero ideal of A4 meets S nontrivially. Because S is prime, so is M . (iv) Let Q be an automorphism of S and let f : A + S. Then the map f":A"+S given by

mu) = f (4"

is clearly a left S-module homomorphism. From this we easily deduce that the map f H f" gives rise to an automorphism of M extending 0. To establish uniqueness of extension, it suffices to verify that if T is an automorphism of M fixing S elementwise, then T = 1. To this end, let m E M and let A be a nonzero ideal of S with Am C S. Then, for all a E A, we have

am = (am). = armT= am7 Thus A(m - m') = 0 and (i) implies that m7 = m. (v) Suppose that m E CM(S) and that m # 0. Because m cen-

197


tralizes S, it is clear that T = {t E Sltm = 0) is a two-sided ideal of S and so we must have T = 0 by (i). Owing to (ii), we may choose a nonzero ideal A of S with Am C S. Then the map f : A -+ S defined by f(a) = am is an injective homomorphism of left S-modules. If B = f ( A ) ,then B = Am = mA since m E CM(S).Now there exists an inverse map g : B + S so that f g = 1 A . Since B is an ideal of S , ij E M and g is an inverse of f = rn. Moreover, because rn centralizes S, we must have 7n-l E C M ( S )and therefore C M ( S )is at least a division ring. Finally, conjugation by m induces an automorphism of M which is trivial on S. Hence, by (iv), the automorphism must also be trivial on M . Thus m E Z ( M ) and the result follows. H The preceding results indicate that the structure of Qo(S) is very close to that of S. Our next lemma will indicate that Qo(S) is large enough to contain certain needed additional units. 8.6. Lemma. Let CT be an automorphism o f S and let a , b E S be fixed nonzero elements. If for all s E S

as6 = bs'au

then there exists a unit m E M = Qo(S) such that b = am such that m-'sm = so

for all s E S

Proof. Put A = SaS,B = SbS and define the maps f : A + B and g:B+Aby

To prove that f is well-defined, it suffices to verify that C z i a y i = 0 implies Cx;by,O= 0. To this end, suppose that C x i a y i = 0. Then, for all s E S, the formula asb = bs'au yields

198


and so C xiby: = 0 since a" # 0 and S is prime. Similarly, if C xiby; = 0, then for all s E S, we have

and we conclude that Cx;ayp-' = 0. Thus both f and g are welldefined and since they are clearly homomorphisms of left S-modules, we have f = m E M and ij E M . Furthermore, fg = 1~and g f = lg, so g = rn-l and m is a unit of M . Note that a,f is defined on S and for all x E S we have

Thus ii,f = b,, or equivalently, am = b. Finally, let c E S. Then gcrf is defined on B and for all xby E B we have

Thus gC,f =

and so rn-lcrn = c", as required.

8.7. Lemma. Let M = &O(S), let c be an automorphism of S and let A,B be nonzero ideals of S . Suppose that f : A + B is an additive bijection which satisfies f ( s a t )= s f ( a ) t " for all s,t E S and a E A . Then m = f is a unit an M , m-lsm = s" for all s E S and f ( a ) = am for all a E A .

Proof. Because f is an additive bijection, its inverse g : B + A is also an additive bijection. Furthermore, for all s , t E S and b E B , g(sbt) = sg(b)t"-' Observe also that f and g are left S-module homomorphisms so that m = f and g are elements of M . Moreover, because g is an inverse cif


199

f,we have g = m-l. Let q E S. Then g q r f is defined on B and for all b E B , we have

Thus g q r f = q: and this yields rn-lqrn = q' for all q E S. Finally, because m = f we have f(a) = a m for all a E A. Let CT be an automorphism of S. Following Kharchenko (1975), we say that is X-inner if it is induced by conjugation by a unit of M = Qo(S). In other words, these automorphisms arise from those units m E M with m-lSm = S.If ml and m2 are two such units, then clearly so is mlm2. Applying Lemma 8.5(iv), we deduce that the set of all X-inner automorphisms of S is a normal subgroup of A u t ( S ) . Now let S * G be given and let G* be the group of graded units of S * G. Then G* acts on S by conjugation and the elements of U ( S ) surely act as X-inner automorphisms. Taking into account that G ' / U ( S ) 2 G, we conclude that

Ginn = { g E GIs I+

is an X-inner automorphism of

gs

S}

is a normal subgroup of G.

By Lemma 8.5(iv), the automorphism s H g s of S extends to a unique automorphism of M which we denote by the same symbol. We now define M*G to be the free M-module freely generated by (31s E G } and with multiplication given by (az)(@) = a"ba(x, y)zy

for all a , b E M and x , y E G. Here of course a : G x G + U ( S ) is given by

a ( x , y ) = zyzy- 1

The first part of the lemma below demonstrates that M G over M extending S * G.

* G is in fact

a unique crossed product of

Let M = Qo(S) and let F = Z ( M ) . (i) There exists a unique crossed product M * G extending S (ii) If E = C M * G ( M ) , then E C M * G;,,, M * Ginn = M

8.8. Lemma.

* G. @F

E


200

and E = FtG;,,, some twisted group algebra of Gin,, over the field F . (iii) If H is a subgroup of Gin,, then

M

*H

=

( E n M * H ) ) and E n ( M * H ) = F t H

(iv) If L is a G-invariant ideal o f E , then L ( M * G ) = ( M * G ) L is an ideal ofM*G. Furthermore, (M*G)L considered as a left M-submodule of M * G is a direct summand of M * G . Also, (M*G)Ln(M*Gi,,) = M L and if L # E , then ( M * G ) L n M = 0. Proof. (i) Owing to Lemma 8.5(iv), the automorphisms of S extend uniquely to those of M . Thus, the above definition of M * G is the only possible extension of S * G. We are therefore left to verify the associativity of the multiplication. Let G* be the group of graded units of S * G. observe that G* acts on S and therefore the uniqueness of extension ensures that we obtain a group action of G* on M . This fact will be used implicitly in the computations below. Given a , b, c E M and x, y, z E G, we have

-

[ ( a i ) ( b i j ) ] ( c .= ~ )mlxyz

and

( a i )[ (bij)(cz)] = m a w for some ml,m2E M . An easy computation shows that ml = a"ba(x, y)"Yca(x, y)-lsl

and m2

= a"b"( YC)SZ

where s1 and s2 are elements of S independent of a , b, and c. Because

the a , b,and c terms in the two expressions are equal. But in the special case when a = b = c = 1 the above products belong to S * G and so s1 = s2. Hence ml = m2 and M * G is associative. (ii) Let y E E and let x E Suppy, say y = rnI .. By Lemma

+-

201


8.5(i),(ii), there exists a E S with ay E S Because y commutes with sa, we have asay = aysa

* G and with x E S u p p a y .

for all s E S

We deduce therefore that

asamii = amiisa = am"s"aii Since b = am is a nonzero element of S, we see that the identity

asb = bxsxa holds for all s E S. Applying Lemma 8.6, we deduce that s H "s is an X-inner automorphism of S and thus x E Ginn. This proves that E C M * Gin,. For each g E G;,,, choose a unit m, E M inducing the automorphism s H g s on S and put ij = mi's. Then the elements 6 for all g E Gin, form an M-basis for M * G;,,. The elements # , g E Ginn are obviously F-linearly independent and we claim that they form an F-basis for E. To this end, observe that each ij is a unit in M * G which acts by conjugation on M centralizing all of S. Invoking Lemma 8.5(iv), we conclude that j must centralize all of M , i.e. j E E . Suppose that y E E C M * Gin,. Then we can write y = C-y,ij with y, E M and it is clear that each ygij centralizes M . Because ij is a unit of E , we have y, E M r ) E = F and we infer that the elements form an F-basis for E . Furthermore, because F C Z ( E ) we must have M * Gin, = M @= E . To establish the Iast assertion, note that E is an associative Falgebra with basis (519 E Gin,}. Furthermore, for s,y E G;,,, itg E E and itij = m ( z , y ) @ for some m ( z , y ) E M . It follows that E is isomorphic to FtG;,,, some twisted group algebra of Gin, over the field F . (iii) This is a direct consequence of the way the algebra FtG;,, is constructed. (iv) Let L be a G-invariant ideal of E. Then, by definition of E , we have LM = M L and the G-invariance yields L j = gL for all g E G.


202

We deduce therefore that L(M * G) = ( M * G)L is an ideal of M

* G.

Let Y be a transversal for Gin, in G and let L' be the F-complement for L in E. Because M * Ginn = M @gF E = M L @ ML', we have

M

*G=CM YEY

L @~

CML'~ YEY

Moreover, since L is G-invariant,

( M * G)L = C

( M * G;,,)Ly

=

Y€Y

C MLij Y €Y

which implies that ( M * G)L is an M-direct summand of M thermore, C y e yMLjj is clearly a direct sum and so

* G.

Fur-

( M * G)L n ( M * Gin,) = M L If L # E , then we can choose L' above to contain the identity element 1. But then CyEY ML'Y contains M and hence ( M * G)L n M = 0, as required. From now on, we shall employ the following notation:

M = &o(S),E= CM*G(M) and F = Z ( M ) We wish to apply the foregoing results in order to establish a bijective correspondence between the prime ideals P of S * G satisfying P n S = 0 and the G-prime ideals of E. Recall that the group G acts on E by the rule: g x = gxg-1 for all x E E , g E G and that, by definition, a G-invariant ideal A of E is G-prime if A1 A2 A for G-invariant ideals A; of E implies that A1 C A or A2 E A. If L is a G-invariant ideal of E , then we put L" = L ( M

* G) n S * G

so that L" is an ideal of S * G by Lemma 8.8(iv). For any ideal I of S * G we put

I d = { 7 E El Ay 5 I

for some nofizero ideal A of S}

203


We claim that I d is a G-invariant ideal of E . Indeed, assume that y1,y2 E I d with Alyl I , A272 I and let t E E. Owing to Lemma 8.5(ii), there exists a nonzero ideal B of S with B t C S*G. Since y 1 , ~ 2 and t centralize S, we have

AIB(7lt) = (AlTl)(Bt)c I and

BAl(tT1) = (Bt)(Al71)c I

+

Thus 71 y 2 , y l t , t ~ 1E I d so I d is an ideal of E which is clearly Ginvariant, as claimed. Let I be an ideal of S * G. Following Lorenz and Passman (1979), we say that I is S-cancelable if for every x E S * G and any nonzero G-invariant ideal A of S , Ax I implies that x E I . It is obvious that any prime ideal P of S * G satisfying P n S = 0 is S-cancelable.

c

Let L, Ll,and L2 be G-invariant ideals of E . Then (i) L" is S-cancelable. (ii) L;"Ly (L1L,)".

8.9. Lemma.

Proof. (i) Suppose that A z C L" for some x E S * G and some nonzero G-invariant ideal A of S. By definition of L", we have Ax C ( M * G)L. By Lemma 8.8.(iv), ( M * G)L is a left M-module direct summand of M * G and hence

M

* G = ( M * G ) L@ K

for some M-submodule K of M ( M * G ) L ,x 2 E I = v(wj 8 f i v ) = ~

( f i8~n ui ) = ~ ( i i i ( w8 j v)) m

= ncp(w; @I v) =

fi(Cw; 8 Qj;v) j=1

m

=

c w ;8 f i Q j j ( V ) j= 1

and thus 'pjj(fiv)= ncpj;(v),as required. (ii) Let Q , $ E E ~ ~ F - ~ N ( (@IWVIN))N and let O(Q) = ( $ j i ) . Then

(vji),O($) =


252

which shows that O(cp$) = O(cp)O($). (iii) Fixing g E G, cp E H o ~ F ~ G@(FW V,~W, @ F V ) ,we have

m

m

j = 1 k=l

m

/ m

and

k=l

whence

m

m

aji(9)vkjgv j=1

=

bkj(g)&'jiv j=1

Substituting v for ij-lv, we therefore derive m bkj(g)gyji(g-'v)

253

15. Inflated modules over twisted group algebras

thus completing the proof. We have now accumulated all the information necessary to prove the following result essentailly due to Huppert and Willems (1975, Satz. 2.)

15.8. Theorem. Let N be a normal subgroup of a finite group G, let F be an arbitrary field and let V be an FaG-module. Then for Q = GIN and for FPQ-modules U, U1 and U2 the following properties hold: (i) If VN is totally indecomposable and U is indecomposable, then

i n f ( U )@ F V is an indecomposable FrG-module, where y = (inf pja. (ii) If V N is absolutel~irreducible and U is irreducible, then inf (Uj @ F V is an irreducible FYG-module. (iii) If V , is totally indecomposable, then inf ( & ) @ F V = inf ( U ~ ) @ F V implies U, U2.

Proof. We put W = inf(U) and W; = inf(V;),i = 1,2. (i) Suppose that cp is an idempotent of EndF7G(W @ V ) . It suffices to verify that cp = 0 or cp = 1. Choose an F-basis ( ~ 1 , .. . ,wn}of W and write m

Swi = C a j i ( g ) w j

(9 E G,aji(g) E F )

(6)

j=1

In the notation of Lemma 15.7(iii) with 1 5 i 5 rn, we have O(p) = ( c p j ; ) , where m

U1

=

U2 = U and w;

= wi,

m

C aji(g)Ykj = jC akj(9)v:; =1

(7)

j=l

Applying Lemma 15.7(iii)) ( c p j ; ) is an idempotent of (EndFaN(VN)),. Bearing in mind that VN is totally indecomposable, we have

EndpN(VN) = F - l v

+ J(EndpN(VN))


254

Thus we may write

+

= X j i IV

vji

with

Xji

E

(1 5 i , j

xji

5 m)

E J ( E ~ ~ F ~ N ( VByN Lemma )). 15.6, it follows that

F , ~ j i

~ 3' 9

=. X j i * l v

+

for all 9 E G

~ ; i

where x;i f J ( E ~ ~ F ~ N ( VComparing N)). the coefficients of Iv in (7), we derive m m j=1

j=1

Applying (6), it now follows easily that the map cp' : W -+W given 1 5 i 5 n, is an F'"fflG-homomorphism. by cp'(w;) = & X j i w j , Note that, by Lemma 15.4, W is indecomposable. Observe also that the matrix ( X j i ) of cp' is obtained from ( c p j i ) by reduction modulo J(Endp,( V N ) ) .Since ( c p j i ) is an idempotent, so is 'p' and hence cp' = 0 or cp' = 1, by the indecomposability of W . Replacing cp by 1 - cpl if necessary, we may therefore assume that cp' = 0, in which case Qji

EJ(E~~F~N(VN for) )all

;,j

Since J(EndpN(V,)) is a nilpotent ideal, there is a positive integer s such that ( c p j i ) s = 0. Thus cp = pa = 0 and i n f ( V )8~V is indecomposable. (ii) Let M # 0 be an FYG-submodule of W @ F V and let u be a nonzero element of M . Then we may write k

=CS~@V (Vi ~ E V , SE~ W ) i= 1 s1 # 0 and 211,. . . ,vk linearly independent over F . u

with Since V , is absolutely irreducible, E n d p , ( V ~ ) = F . Hence, by the Jacobson's density lemma, given v E V ,we may find a = CxEN uzZ in F*N, ax E F such that avl = v and a V i = 0 , 2 5 i 5 k. Since the elements 5 , s E N , operate.trivial1y on W , we obtain a(si

@ Vi) =

c c

axZ(si

@ Vi)

ZEN

=

a,(& @ Z V J

SEN

= s; @ a v i

15. Inflated modules over twisted group algebras

255

and therefore k

We conclude that

s1

k

@

V 2 M and hence, for all g E G,

However, FaGsl is a nonzero submodule of W and W is irreducible, by Lemma 15.4(i). Thus M = W @F V and therefore W @ F V is irreducible. (iii) Assume that cp : Wl @ F V -+ W2 @ F V is an FYG-isomorphism. Then we have dirnFWl = dirn~W2and so we may apply Lemma 15.7(iii). If 6(cp) = ( c p j ; ) then in the notation of Lemma 15.7(iii), we have m m aji(g)Pkj

=

j=1

C

bkj(g)vTi

j=1

By repeating the argument of (i), we derive m

m

j=1

j=1

for some Xj; E F . The latter implies that the map cp' : W1 defined by

c

+

Wz

cp-l,

we

m

cp'(W1)

=

xjiw;

j=1

is an F'"fPG-homomorphism. Applying the same procedure to deduce that cp' has an inverse and therefore

Thus, by Lemma 15.4(ii), U 1 2 U2 as FPQ-modules. This completes the proof of the theorem.

In the notation of Theorem 15.8, assume that is absolutely irreducible. Then inf (U)@ F V is absolutely irreducible if and only if so is U. 15.9. Corollary.

VN


256

Proof. Let E / F be a field extension. Applying Lemmas 15.4(iv) and 15.3, we have

( i n f ( U )@F

V)Eg

inf(U)E @ E V , g inf(UE)8, VE

(8)

Assume that i n f ( U ) @F V is absolutely irreducible. Then, by (8)) inf(UE) is irreducible. Hence UE is irreducible, by Lemma 15.4, and therefore U is absolutely irreducible. Conversely, assume that U is absolutely irreducible. Then UE is irreducible and since ( V E ) N = ( V N ) E , (V')N is absolutely irreducible. Applying (8) and Theorem 15.8(ii), we conclude that ( i n f ( U )@F V ) , is irreducible. Thus inf(U )@F V is absolutely irreducible, as we wished to show.

16. Induction of absolutely irreducible modules Throughout this section, N denotes a normal subgroup of a finite group G and F an arbitrary field. Given an absolutely irreducible FN-module V , our aim is to provide a detailed analysis of the decomposition of VG.

16.1. Lemma. Let A be a finite-dimensional algebra over F , let V be an absolutely irreducible A-module, and let W = U @ F V , where U is some finite-dimensional vector space over F . For each a E A, let cp, E End=(V) be defined b y cp,(v) = uv,v E V . (i) I f 6 E E n d ~ ( wis) such that

6(l@ 9,) = (1 @ ya)6

for all a E A

then 6 = II, @ 1 for some II, E EndF(U). (ii) If6 E GL(W) and T E GL(V) are such that 8-l(1 @ cpa)6 = 1 @ T - ' ~ , T

for all a E A

then 0 = $ 8 T for some II, E GL(U).

Proof. (i) Let {ul,. . . ,u m } be an F-basis for U . Then, for all v E V , we have m

o(ui 8 v) =

C uj 8 Oji(v) j=1

(1)

16. Induction of absolutely irreducible modules

257

for some 8ji E E n d ~ ( v )Given . a E A, we have

m

j=1

Because the {uj} are linearly independent, we infer that

or, equivalently, that 0,; E E n d ~ ( v ) . Since V is absolutely irreducible, it follows that Bj; = X j i - l v for some Xi; E F . Now define T,!J E EndF(U) by

Then II, E GL(U)and, by (l),8 = $ 8 1, proving (i). (ii) Setting 7 = 1 87,we have

and therefore, by (i), 87-' = II, 8 1 for some II, E G L ( U ) . Hence

as required.

We are now ready to prove the following result partially due to Clifford (1937). 16.2. Theorem. Let F be an arbitrayfield, let V be an absolutely irreducible G-invariant FN-module, and let w = wo(V) E Z2(&,F*),


258

Q = GIN,be an obstruction cocycle of V . Let ezt(V) denote any extension of V to an FOG-module, where a = infw ( b y Theorem 15.2,such an extension always exists). '1 {Ul,Uz,.. . ,Un)is a f i l l set of nonisomorphic irreducible (rspectively, absolutely irreducible) F"-' &-modules, then {inf(Ui)@F ezt(V)I 1 5 i 5 n} is a full set of nonisomorphic irreducible (respectively, absolutely irreducible) FG-modules whose restriction to N is a direct sum of copies of v.

Proof. By Corollary 15.9, it suffices to treat the case where all

U;are irreducible. Let M denote an irreducible FG-module such that MN is a direct sum of copies of V and put S = ezt(V), Si = inf (Ui),1 5 i 5 n. We may clearly assume that M = W @ V , where W is an F-space such that for all n E N,w E W and v E V ,

the

If 7 : N

G + G L ( M ) and T : G + G L ( S ) are the representations afforded by V,M and S, respectively, then by (2) and by the equality SN = V, we obtain + GL(V),p :

p(n) = 1 @ 7 ( n ) and ~ ( n=r)(n) ) for all n E N

(3)

Also if ng = gnl for some n,nl E N and g E G, then T(n)T(g) = T(g)T(nl) because a(n,g)= a(g,n)= 1 for all n E N,g E G. Thus

and hence, by (3), we have

16. Induction of absolutely irreducible modules

259

Invoking Lemma 16.1(ii), for each g E G there exists $ ( g ) E GL(W) such that

P ( d = $(s>@ 4 7 ) Setting gw = $(g)w,w E W , it follows that W is an Fa-lG-module such that M g W ezt(V) Moreover, since M is irreducible, so is W . Finally, since $(n) = 1 for all n E N, we conclude that W i n f ( U ) for some irreducible F"-'Qmodule U . Thus

M

inf(Vi)@ ezt(V)

.

for some i E {I,. . ,n }

Conversely, assume that U is an irreducible F"-'Q-module and let {ul,. . . ,uk} be an F-basis of U . By assumption, V is absolutely irreducible. Because V = e z t ( V ) N , it follows from Theorem 15.8(ii) that

inf(U ) 8 ezt ( V ) is irreducible. Finally, write

i n f ( U )8 ezt(V)= u 1 @V

$ - a .

$uk @ V

where each ui @ V is an FN-module, by the definition of i n f ( U ) . Since the map v H u, @ o is an FN-isomorphism of V onto u; 8 V , we deduce that ( i n f ( U )@ ezt(V))N is a direct sum of copies of V . Since, by Theorem 15.8(iii),

the result is established.

16.3. Corollary. Let N be a normal subgroup ofG and let W be an irreducible FG-module such that some irreducible direct summand V of WN is absolutely irreducible. Let H be the inertia group ofV and let w = W H ( VE) Z 2 ( H / N ,F*)be an obstruction cocycle of V . Then (i) There exists an irreducible F"-'(H/N)-module U such that

w 2 ( i n f ( U )8 e z t ( V ) ) G


260

(ii) I f U is absolutely irreducible, then so is W .

Proof. (i) This follows from Theorems 2.4.2(iii) and 16.2 and transitivity of the induction. (ii) By Theorem 16.2, i n f ( U ) @ e z t ( V ) is absolutely irreducible. Now apply Corollary 2.4.5(iv). We now come to the main result of this section. 16.4. Theorem. Let F be an algebraically closed field, let V be an irreducible FN-module, let H be the inertia group of V and let charF not divide ( H : N ) . Denote by w = O H ( V E ) Z 2 ( H / N ,F') an obstruction cocycle o f V and by e x t ( V ) an extension ofV to an F"Hmodule with Q = infw. Let U1,. . . ,U,, be all nonisomorphic irreducible F"-'(HIN)-modules and, for each U;,let inf (U;) be the Fa-' H-module

injlatedfim U;. Then the

are nonisomorphic irreducible FG-modules such that for m; = dimFU;,

Proof. By Theorem 2.1 and transitivity of the induction, it suffices to prove that the inf(U;) @ e z t ( V ) are nonisomorphic irreducible FHmodules such that

To this end, we first invoke Theorem 2.10.14 to deduce that V H is completely reducible. Thus there exist integers t l , t Z , . . . ,t , and nonisomorphic irreducble FH-modules W1,W2,.. . ,W, such that V H2 @&W; By Theorem 2.2.14, ti equals the multiplicity of V as an irreducible constituent of (W;)N.Invoking Theorem 16.2, we may thus assume that W; % i n f ( U j ) e z t ( V ) (1 5 i 5 s )

17. Applications

261

Note that the multiplicity of V as an irreducible constituent of ( i n f ( U i ) @ e s t ( V ) ) N is dimFU; = m;. Accordingly,

By Proposition 6.11, F"-'(H/N) is semisimple, because charF does not divide ( H : N ) . Taking into account that F is algebraically closed, we have n

Thus

dimFVH = ( g m : ) dimFV, proving that n = s and hence the result.

17.

Applications

In this section, we provide a number of applications of the preceding results. Throughout, F denotes a field and G a finite group.

17.1. Lemma. Let F be an algebraically closed field such that charF does not divide [GI, and let V be an irreducible module over a twisted group algebra F"G. Then dimFV divides the order of G.

Proof. By making a diagonal change of basis {glg E G} of FOG, we may replace cu by any cohomologous cocycle. Hence, we may assume that cy is of finite order, say m. Let E be a primitive m-th root of 1 in F and let G' =< e'glg E G, 1 5 i 5 m >. Then the map f : G' + G given by f ( E i g ) = g is a surjective homomorphism whose kernel is a central subgroup < E >. Moreover, if p : G --t G L ( V ) is an arepresentation of G afforded by V , then p* : G + G L ( V ) defined by p*(ciij) = Eip(g) is easily seen to be an irreducible representation of G'. Because charF does not divide ]GI and G*/ < E >%G, the result


262

follows by applying the following standard fact (see Curtis and Reiner (1981)): If F is an algebraically closed field, charF does not divide IGl and 2 is a central subgroup of G , then the dimensions of irreducible FGmodules divide ( G : 2). Let p be a prime. A group G is said to be p-solvable if the composition factors of G are either pgroups or p’-groups. It is an immediate consequence of the definition that (i) Any extension of a psolvable group by a psolvable group is psolvable. (ii) Subgroups and homomorphic images of psolvable groups are psolvable. We are now ready to establish the following important result. 17.2. Theorem. Let N be a normal subgroup of G , let F be an algebraically closed field and let V be an irreducible FG-module. If charF = p > 0 divides ( G : N ) , assume that GIN is p-solvable. Then dimFV divides ( G : N ) d , where d is the dimension of an irreducible constituent of VN.

Proof. We first demonstrate that the result holds under either of the following hypotheses: (i) charF does not divide ( G : N ) . (ii) charF = p > 0 and G / N is a p-group. Denote by W an irreducible constituent of VN. Let H be the inertia group of W and let w E Z 2 ( H / N ,F*)be an obstruction cocycle of W . By Corollary 16.3(i), there exists an irreducible F”-’(H/N)-module S such that

v

( i n f ( S )C ~ ezt(W))G I

Because dimFS = dirnFinf ( S ) and dirnFW = dirnFezt(W), we have

dimFV = (dirnFS)(dirnFW)(G: H ) In case (i), dimFS divides ( H : N ) , by Lemma 17.1, so dirnFV divides

( G : N)dimFW. In case (ii), F”-l(H/N) is a local ring by Proposition 6.4. Hence

17. Applications

263

dimFS = 1 and so dimFV again divides ( G : N)dimFW. Turning to the general case, we use induction on /GI. Suppose that the result is true for groups of lower order than G. By the above, we may assume that charF = p > 0 divides ( G : N ) , in which case GIN is psolvable, by hypothesis. Because GIN is p-solvable, there exists a proper normal subgroup M of G containing N and such that G / M is either a p or p'-group. Denote by W an irreducible constituent of VM and let S be an irreducible constituent of WN. Then S is clearly an irreducible constituent of V N . By the foregoing, dimFV

divides (G : M)dimFW

and, by induction hypothesis,

dimFW

divides

( M : N)dimFS

Thus dimFV divides ( G : M ) ( M : N)dirnFS = (G : N ) d i m ~ S as , required. 17.3. Corollary. (Dude (1968), Swan (1965')). Let F be an algebraically closed field of characteristic p > 0 and let A be a normal abelian subgroup of a p-solvable group G. Then the dimensions of irreducible FG-modules divide ( G : A).

Proof. Note that all irreducible FA-modules are of dimension 1. The desired conclusion is therefore a consequence of Theorem 17.2.

As a prepararion for the proof of our next result, we now record the following properties. 17.4. Proposition. Let N be a normal subgroup ofG such that (G : N ) = pn, p prime, and let F be an algebraically closed field of characteristic p. If V is an irreducible FN-module, then

where U is an irreducible FG-module such that V is a constituent of UN.


264

Proof. We know, from Theorem 2.12.4, that P ( V ) is an indecomposable FN-module. Hence, by Theorem 7.5, P(V)" is indecomposable. It follows from Proposition 2.12.5(i) that P(VG)E P(V)". By Corollary 2.2.5(i), there is an FG-epimorphism V G + U. Thus, by Lemma 2.12.2(ii), P ( U ) is isomorphic to a direct summand of P(V"). But, by the above, P(V") is indecomposable, hence the result. H Let N be a normal subgroup ofG and let V be an irreducible FG-module. If(G : N ) # 0 in F , then 17.5. Proposition.

Proof. By Theorem 2.10.14, J(FG) = F G J ( F N ) and so

where

with nonisomorphic irreducible FG-modules 2.12.3(iii)) F G = $idiP(x)

x. Then, by Theorem

and so (FG)N = $ i d ; P ( V i ) ~ (3) Bearing in mind that F G is a free (hence, projective) FN-module, we also have (FG)N = P((FG)N) = P ( (FG)N/ J ( FN)(FG)N) = ~ ( ( F G / J ( F G ) ) N ) (by (1)) = $;d;(P((K)N)) (by (2) and Lemma 2.12.l(iii)) Thus the desired conclusion follows by virtue of (3) and Proposition 2.12.5(ii). H Given a natural number n and a prime p , we denote by np the highest power of p dividing n and put np:= n/np.

17. Applications

265

Let G be a p-solvable group, let F be an algebraically closed field of characteristic p , and let V be an irreducible FG-module. Then (i) (Fong (1961)) dirnFP( V ) = /GIp(dirnFV),I. (ii) (Willems (1980)). If V = W Gfor some FH-module W , H is a subgrozlp of G, then P( V ) s P( W ) G 17.6. Theorem.

Proof. (i) We argue by induction on IGl. Let N be a maximal normal subgroup of G. Because G is psolvable, either [ G I N [= p" for some n 2 1 or G / N is a p'-group. Owing to Clifford's theorem, there is a decomposition

with nonisomorphic irreducible FN-modules K of equal dimension. Furthermore, by Theorem 17.2, es divides IG/NI. Suppose first that IG/NI = p". Because dirnFV = esdirnFV1, we have

(dirnFV),t = (dirnF&),t Note also that, by Proposition 17.4,

(5)

Thus

as required.

Next assume that GIN is a p'-group. Owing to Proposition 17.5, P ( V , ) S P(V)N and so dirnFP(V) = dirnFP(VN). On the other hand, by (4>1 ~(VN = )e ( P ( l 4 )CB @ P(K))


266

and thus

where the last equality follows from the fact that [NIP = ]GIp and esl(G : N ) . Hence (i) is established. (ii) Owing to Proposition 2.12.5(i), P ( V ) is isomorphic to a direct summand of P ( W ) G .On the other hand, by (i), we have

Hence P ( V ) "= P(W)Gand the result follows. H 18.

The Loewy length of induced modules

Throughout this section, A denotes a finite-dimensional algebra over a field F . Unless explicitly stated otherwise, all A-modules are assumed to be left and finitely generated. The same assumption is made on FG-modules, where G is a finite group. Let V # 0 be an A-module. Then the descending chain

v 2 J(A)V 2 J(A)2V2 . . . of submodules of V is called the (lower) Loewy series of V . Since J ( A ) is nilpotent, there is an integer k, called the Loewy length of V such that J(A)"-'V # 0 but J(A)"V = 0 Observe that if J(A)'V = J(A)'+lV for i < k, then

J(A)"'V = J(A)k-i-'J(A)'V= J(A)"-'-'.J(A)'+'V = J(A)kV= 0,

18. The Loewy length of induced modules

267

which is impossible. Thus, if k is the Loewy length of V , then

V

J(A)V 3

18.1. Lemma. Let e

#0

-

a

-

3 J(A)k-lV 3 0

be an idempotent of a ring R. Then

J(eRe) = eJ(R)e Proof. Let I be a primitive ideal in R and let V be a faithful irreducible R/I-module. Then eV is an eRe-module. If eV = 0, then J ( e R e ) eRe C 1. Assume that eV # 0. Let 0 c W = e W eV where W is an eRe-module. Then V = RW since V is irreducible. Thus eV = e R e W W and so eV = W . It follows that eV is an irreducible eRe-module, hence J ( e R e ) V = J ( e R e ) e V = 0. Consequently, J ( e R e ) 5 I for any primitive ideal I in R. Thus J ( e R e ) J ( R ) and J(eRe) = eJ(eRe)e eJ(R)e. Assume that a E J ( R ) . Then eae E J ( R ) and so, by Proposition 1.4.10, ( 1 - b ) ( l - eae) = 1 for some b E R. Thus beae = eae b and multiplying by e on both sides yields eae + ebe = ebeae. Therefore ( e - ebe)(e- eae) = e so that e - eae is a left unit. Because e J ( R ) e is an ideal of eRe, it follows from Proposition 1.4.10 that e J ( R ) e C J ( e R e ) , as required.

c

c

+

If W # 0 is an A-module and E = EndA(W),then W will also be regarded as an E-module via 'pw = 'p(w) for all 'p E E , w E W . 18.2. Lemma. Let e # 0 be an idempotent of A, let W = A e and let E = EndA( W ) . Then

for all i 2 1

J(E)'W = Ae(eJ(A)e)'

In particular, the Loewy lentgh of W as an E-module is equal to the nilpotency index of J ( E ) . Proof. For each x E eAe, let By Lemma 1.5.9, the map

E E be defined by f3:(w) = wx.

f3:

eAe + E


268

is an anti-isomorphism of rings. Moreover, by Lemma 18.1, J(eAe) =

eJ(A)e and so fz E J(E)' if and only if x E (eJ(A)e)i.Thus

J(E)'W = Ae(eJ(A)e)' Finally, J(E)'W = 0 if and only if (eJ(A)e)' = 0. Thus the Loewy length of the E-module W is equal to the nilpotency index of J(eAe) = eJ(A)e. Because eAe Z E", the result follows. We have now accumulated all the information necessary to prove the following result in which L ( V G )denotes the Loewy length of the induced FG-module V G .

(Clarke (1972)). Let F be a field of charucteristic p > 0, let N be a normal p'-subgroup of G , let H be the inertia group of an irreducible FN-module V and let E = EndFG(VG). Then (i) J(FG)"VG = FG J(FH)"VH = J(E)"VG for all n 2 1. (ii) L ( V c ) is equal to the nilpotency index o f J ( E n d F H ( V H ) ) . 18.3. Theorem.

Proof. We first show that (ii) is a consequence of (i). Indeed, we may take V = F N e for some primitive idempotent e of F N , in which case V G = FGe. Hence, by Lemma 18.2, the Loewy length of V Gas an E-module is equal to the nilpotency index of J(E). Thus (ii) is a consequence of (i) and Lemma 12.1. To establish (i), we write

as a sum of primitive idempotents of F N with e = el. Then we have

J ( F G ) V G = J(FG)e = FG(J(FG)e) = FGe(J(FG)e)t - - - FGe,(J(FG)e)

+

(1)

as left FG-modules, where the sum is not necessarily direct. For each u E e;FGe, let f a E HOmFG(FGe;,FGe) be defined by fa(.) = xu for all x E FGei. It follows, from Lemma 1.5.8, that the map

1

e;FGe a

HomFG(FGei,FGe)

+ H

f

a


269

is an F-isomorphism. In particular, if FGe; E FGe, then there is an a E e;FGe such that fa : FGe; + FGe is an FG-isomorphism. For the sake of clarity, we divide the rest of the proof into three steps. Step I. Here we demonstrate that J(FG)e = FGeJ(FG)e. Because FGeJ(FG)e J ( F G ) e ,it follows from (1) that it suffices to show that

e;J(FG)e ejFGeJ(FG)e

(1 5 i 5 m )

(2)

Let f; be the centrally primitive idempotent of F N with e;fi = e;, 1 5 i 5 m, and let f: be the sum of G-conjugates of f;. Then f: is clearly a central idempotent of FG. Next observe that if and thus

f1

and f; are not G-conjugate, then

frf1

=0

e;FGe = e;f;f;*FGf,*flel= 0 We may therefore assume that f1 and f; are G-conjugate, say g - l f j g . Then ( g - l e ; g ) f l = g - l ( e i f i ) g = g-leig

f1

=

and therefore the irreducible FN-modules FN(g-'e;g) and F N e are in the same block F N f l . However N is a p'-group, so F N e Z FN(g-'e;g) and thus

FGe 2 FG(g-'e;g) E FGe; By the foregoing, we may find a E e;FGe such that

fa

: FGe; + FGe

is an FG-isomorphism. Consequently, there exists b E eFGe; with f;'(y) = yb for all y E FGe. Accordingly, zab = z for all z in FGe;. Thus e; = eiub = (e;a)b = ab

and therefore for any c E eiJ(FG)e, c = e;c = (ub)c= a(bc) E e;FGeJ(FG)e

This proves (2) and hence the required assertion. Step 2. Our aim is to prove that J(FG)"VG = J(E)"VG for all n 2 1. Owing to Lemma 18.2, we need only show that

J(FG)ne= (FGe)(eJ(FG)e)"

for all n 3 1


270

The case n = 1 being a consequence of Step 1, we argue by induction on n. Thus assume that

J(FG)ke= (FGe)(eJ(FG)e)k for all k 5 n Multiplying (3) on the left by J ( F G ) gives

(3)

J(FG)"+'e = (J(FG)e)"+' for all k 5 n whereas multiplying (3) on the right by J(FG)e gives

(4)

( J ( F G ) k e ) ( J ( F G ) e=) (FGe)(eJ(FG)e)"'

for all

k 5n

(5)

We obtain therefore

J(FG)"+'e = (J(FG)e)"+l (using (4)with k = n ) = (J(FG)e)"(J ( F G ) e ) = (J(FG)"e)(J(FG)e) (using (4) with k = n - 1 ) = (FGe)(eJ(FG)e)"+' (using ( 5 ) with k = n )

+

which establishes (3) for k = n 1. Step 3. We now complete the proof by demonstrating that

J(FG)"VG= FG - J ( F H ) " V H

for all n

21

We keep the notation of Lemma 12.1 and put El = EndFH(VH).Then k

i=s+l

k

= J(FH)"VH

+C i=s+l

= FG * J ( F H ) " V H ,

gjJ(El)"(1 @ V )

(by Step 2)


271

as desired. Our next aim is to provide a sufficient condition under which L ( V G ) is equal to the nilpotency index of F P , where P is a Sylow psubgroup of the group H in Theorem 18.3. 18.4. Theorem. Let F be an algebraically closed field of characteristic p > 0 , let N be a normal p'-subgroup of G and let H be the inertia group of an irreducible FN-module V . If P is a Sylow p subgroup of H , assume that P N d H . Then (2) The Loewy length L(VG)o f V G is equal to the nilpotency index of J ( F P ) . (ii) n(p - 1) + 1 5 L(VG)5 p", where pn = IPI. (iii)L(VG)= n(p - 1) 1 i f and only if P is elementary abelian. (iu) L ( V G )= p" if and only if P is cyclic. (v) I f P is abelian, say P = l-If=l < gi > with I < g, > I = p";, then

+

L ( V G )= 1 - k t

k

Cpmi i=l

Proof. By Theorems 4.2 and 18.3(ii), L ( V G )is equal to the nilpotency index of J ( F * (H / N ) ) for some cy E Z 2 ( H / N ,P). Put = H / N and P = P N / N . By hypothesis, is is a normal Sylow p-subgroup of H and, by Lemma 6.2, F a p F P . Furthermore, since N is a p'-group, p E P and hence F a p 2 F P . We now use the following standard property of crossed products: if N is a normal subgroup of G of index n, S an arbitrary ring such that n is a unit of S and S * G is a crossed product of G over S, then

J ( S * G) = ( S * G ) J ( S * N )

(6)

(see Theorem 3.16, p.87 in Karpilovsky (1987a)). Since ( S * G ) J ( S* N ) = J ( S * N ) S * G by Theorem 3.6(iii), it follows from (6) that the nilpotency index of J ( S * G) is equal to that of J ( S * N ) .

Applying the general observation above to the twisted group algebra PI?, we conclude that the nilpotency index of J ( F " H ) is equal to that of J ( F P ) , proving (i). The rest follows from some general results on the Jacobson radical of group algebras, namely from Proposition

272


1.10, p.313, Theorem 3.2, p.323 and Corollary 2.5, p.122 in Karpilovsky (1987b). H

273

Chapter 4 Projective summands of induced modules Let H be a subgroup of G, let F be a field and let V be an FH-module. It is natural to investigate circumstances under which V Ghas projective indecomposable summands. An important contribution to this problem was marked by an appearance of paper of Robinson (1989) to which this chapter is devoted. Among other results, we prove a useful reciprocity theorem and demonstrate that if V is irreducible, then projective summands of V G(if there are any) have relatively large dimension. Some of the results presented place restrictions on the size of Cartan invariants in terms of group-theoretic structure. Finally, we demonstrate that if there is only one (H,H)-double coset representative, say g, for which 1gHg-' n HI is not divisible by p , then ( l H ) G contains at most one indecomposable projective summand. Furthermore, if there exists such a summand, then it must be irreducible. 1.

The Reynolds ideal

Throughout this section, A denotes a finite-dimensional algebra over a field F . Recall that if $ E H o m ( A , F ) is such that Ker$ contains no nonzero right ideals (equivalently, Ker$ contains no nonzero left ideals), then the pair ( A , $ ) is said to be a Frobenius algebra. If, in addition, $(zy) = $(yz) for all z,y E A , then we refer to ( A , $ ) as a symmetric algebra. Some basic properties of Frobenius and symmetric

Projective summands of induced modules

274

algebras were established in Sec.10 of Chapter 3. We begin by recording certain additional properties of these algebras. For each subset X of A, we denote by I ( X ) and r ( X ) the left and right annihilators of X defined by

Z(X) = { a E AlaX = 0 } , r ( X ) = { a E AlXa

=0)

It is clear that 1(X) and r ( X ) are left and right ideals of A, respectively. Recall that if ( A ,+) is a Frobenius algebra and X is a subset of A , then the subsets ' X and X I of A are defined by

'X X'

= { a E A I $ ( u X )= 0 } = { a € A I $ ( X a >= 0 }

1.1. Lemma. (i) Suppose that ( A , $ ) is a Frobenius algebra. Then, for any subset X of A, l ( X ) is the largest left ideal of A contained in

LX. (ii) Let [ A , A ]be the commutator subspace of A. If ( A , $ ) is a symmetric algebra, then Z ( A ) = [A,A]*. Proof. (i) By the definition of ' X , we have Z(X) C ' X . Let L denote any left ideal of A contained in ' X . Then

+ ( A L X )= $ ( L X ) = 0 so L X = 0 and hence L Z(X), as asserted. (ii) Given z E A, z E [A,A]* if and only if for all 2,y E A +((EY - y+)

=

$(w) - $(yzz) = $(w> - $(w) =0

or if and only if

$(A(yz - zy)) = 0

for all y E A

Because the latter is equivalent to yz - zy = 0 for all y E A , the result follows. rn From now on, we assume that charF = p

T ( A )= { a E AlaPn E [A,A]

> 0 and put

for some n 2 1)

1. The Reynolds ideal

275

Then T ( A )is an F-subspace of A containing [A,A].

1.2. Theorem. (fr'iilshammer(1981)). Let A be a Jinite-dimensional algebra over a j e l d F of characteristic p > 0 . (i) T ( A ) = [A,A] J ( A ) and J ( A ) is the largest left ideal of A contained in J ( A ) . (ii) If A is a Frobenius algebra, then SocA = T(A)'A. (iii) If A is a symmetric dgebra, then T(A)' = Z ( A ) n SocA and, in particular T(A)' is an ideal of Z ( A ) .

+

Proof. (i) We first assume that A is a simple algebra. Because for any z E Z ( A ) and 2,y E A ,

both [A,A] and T ( A ) can be regarded as vector spaces over the field Z ( A ) . We may therefore assume that Z ( A ) = F , i.e. that A is a central simple F-algebra. Let E be the algebraic closure of F , so that

, and for some n 2 1. Then obviously T ( A E )= [ A EAE]

Taking into account that

we deduce that dzmF[A,A] = dzmFA - 1. If it were true that A = T ( A ) ,then since T ( A E )is an E-subspace of AE we would have AE = T ( A E ) ,a contradiction. Thus A # T ( A ) and hence, by the above, T ( A ) = [ A , A ] . Since A is simple, ( A , + ) is a symmetric algebra for some t,h E HomF(A,F) (Proposition 3.10.9). But then [ A , A )C Ker+ and therefore T ( A ) contains no nonzero left ideals of A. This proves the case where A is a simple algebra. Turning to the general case, we put A = A / J ( A ) and write

A = A1 $ * * * $ A ,


276

where the Ai are pairwise orthogonal simple F-algebras. Then T ( A ; )= [A;,Ai],1 5 i _< rn, by the special case proved above. It follows that

+

Clearly, [ A , A ] J ( A ) T ( A ) and T ( A ) / J ( A )C_ T ( A ) . Thus T ( A )= T ( A ) / J ( A )and therefore T ( A )= [A,A] J ( A ) . Let L be a left ideal of A contained in T ( A ) . Then ( L + J ( A ) ) / J ( A ) is a left ideal of A contained in T ( A )and therefore in [A,A ] . Since A is semisimple, A is a symmetric algebra (Proposition 3.10.9). Thus, by the previous argument, [A,A]contains no nonzero left ideals of A. Hence L C J ( A ) and the required property follows. (ii) Assume that A is a Frobenius algebra. By Theorem 3.10.11, we have

+

' ( T ( A ) * )= T ( A )

(1)

r[Z(T(A)*A)]= T(A)'A (2) By (i) and (l),J ( A )is the largest left ideal of A contained in ' ( T ( A ) * ) . Applying Lemma 1.1(i), we deduce that J ( A ) = Z(T(A)') = Z(T(A)'A)

(3)

Thus, by (2) and (3), it follows that

SocA = r ( J ( A ) )= T(A)'A, as required.

(iii) Assume that A is a symmetric algebra. By (ii) and Lemma l.l(ii), it suffices to show that

T(A)' = [ A ,A]' n T(A)'A Because Z(T(A)'A) is a left ideal of A , we have

[Z (T(A )* A)]' = r [I( T (A)' A]

(4)

277


by Theorem 3.10.11. Consequently,

as we wished to show.

Let A be an F-algebra and let Soc(A) be the left socle of A . We shall refer to the ideal Z ( A ) n SocA of Z ( A ) as the Reynolds ideal of A . In what follows, we put

Rey(A) = Z ( A ) f l SocA The name is partially justified by the fact that Reynolds (1972) introduced this ideal for the special case where A is a group algebra. Our next aim is to determine Rey(FG) exclusively in terms of F and G which will give us the original ideal constructed by Reynolds. In what follows, G denotes a finite group, F a field, and Cl(G)the set of all conjugacy classes of G. 1.3. Lemma.

We have

[FG,FG] = { c z g g E FGI

c

xg = o for all C E CZ(G)}

g€C

Proof. This is a special case of Lemma 3.6.10. An easy proof can also be obtained from the following elementary argument. Fix C = { g l , . . .,gn} in CZ(G) and assume that z = Cy=lA;gi where A; E F and CZl A; = 0. Then z = Cyr: X;(g; - g n ) , and so in view of the identity g - tgt-' =

(gt-1)t - t(gt-1)

(t,g E G )

we have z E [FG,FG]. Since [FG,FG] is spanned by all z y - y z with z,y E G, the result follows by virtue of zy - yz = zy - z-l(xy)z. W

1.4. Lemma. Let tr : FG + F be defined b y tr

Then ( F G , t r ) is a symmetric algebra.

(C,,,

zgg)= 2 1 .


278

Proof. It is an immediate consequence of the definition that tr is an F-linear map. Fix z = CgEGzggand y = CgEGyggin FG. Then we have t r ( z y )= zgyg-l = YgZg-1 = t r ( y 4

c

c

g€G

g€G

and tr(g-'z) = zg

for all g E G

In particular, if tr(FGz) = 0 then z = 0 as required. For the rest of this section, we assume that F is a field of characteristic p > 0. Given g E G, we write gp and gp1 for the p and p'-parts of 9 , respectively. We shall refer to C E C1(G) as being p-regular if all elements of C have order not divisible by p . For any subset X of G, we Put

x-'

= {z-llz E X }

and denote by F X the F-linear span of X . Let C1 = { l}, C2,.. . ,C, be all p-regular classes of G. Then the sets

are called p-regular sections of G. It is an immediate consequence of the definitions that (i) G = U;==,S; with S; n Sj = 0 for i # j and each Si is a union of conjugacy classes of G, S1 is a union of conjugacy classes of p-elements of G. (ii) Let X be a subset of G and let z = CSEG zgg E F G . Then

t r ( z X + )=

c

zg

g€X-1

and, in particular,

tr(zS?) =

c gES,-'

zg


279

1.5. Proposition. Let S1, Sz, . . . ,S, be all p-regular sections of G. Then (i) T ( F G ) = {x = C x g g E FGIC,,, xg = 0 for all i E { l , .. . , r } }

Proof. (i) We first claim that there exists n gpm = ypt

2 1 such that

for all rn = O(rnodn) and all g E G

(5)

Indeed, write (GI = p"k with ( p , k) = 1. Since (p,k) = 1, there exists a positive integer n such that pn 5 l(rnodk). Replacing n by its multiple, if necessary, we may assume that n 2 a . Let rn be divisible by n and m let g E G. Then gEm = 1 and, since p" l(rnodl),gg, = gpf. Hence P"

9

- P" -Yp

Pm

gp, = gp'7

proving ( 5 ) . Let x = C g E G x g g E FG. Then, by (5) there exists rn 2 1 such that gPm = gpt for all g E G and such that x E T ( F G ) if and only if xp" E [FG7FG].Owing to Lemma 3.6.9(i), xpm E

C x;"gpm

(mod[FG,F G ] )

9EG

T ( F G ) if and only if EgEGxgPmgpt E [FG,FG]. Applying Lemma 1.3, we deduce that x E T ( F G ) if and only if CgESi xi"' = 0 for all i E (1, . . . , r } . Because so 2 E

the required assertion follows. (ii) By Theorem 1.2(iij), it suffices to show that T(FG)' = Ci==, FS? By (i), we have

2 FS? C T(FG)' i=l


280

Conversely, suppose that z = CgEGzgg E T(FG)* and let a , b E S;. Then and b-' belong to the p-regular section Sr' and so, by (i), a-' - b-' E T ( F G ) . But then tr(za-' - zb-l) = 0 which implies that

as required. 2.

Projective summands

Throughout, H denotes a subgroup of a finite group G and, unless explicitly stated otherwise, F is an algebraically closed field of characteristic p > 0. All modules are assumed to be finitely generated over their ground rings. Our main theme here is the study of projective summands of modules of the form V G ,where V is an FH-module. Among other results, we prove a useful reciprocity theorem and demonstrate that if V is irreducible, then projective summands of V G (if there are any) have relatively large dimension. Recall that a conjugacy class C of G is said to be p-regular if the order of the elements of C are not divisible by p. In what follows, we fix the following notation: C1, (72,. . . ,CT are all p-regular classes of G. Si = { g E Gig,, E Ci}, 1 5 i 5 r , are all p-regular sections of G.

si+ = CsES, 3.

Rey(FG) = Z ( F G )n Soc(FG) is the Reynolds ideal of FG. Note that, by Proposition 1.5(ii), T

Rey(FG) = C i=l

while by Theorem 1.2(i),(iii), Bey( F G ) = ( [ F G FG] ,

2.1. Lemma. Let

+ J ( FG))'

x be an F-character of G .

Then, for all g E G,

281

2. Projective summands

Proof. We may choose an integer m 2 1 such that g p m is p-regular. In a representation of G with character x, the linear transformation . say, in F and x(g) = corresponding to g has eigenvalues ~ 1 , ~ 2. ,. ,en, m E~ + . . + en. Similarly, x(gPm) = E; - - - &Ern. Since F has characteristic p , the binomial theorem shows that

+ +

x(g)Prn= &;m

+ + &Krn = x(gPrn) ' * *

Bearing in mind that gPrn= (gpt)pm,we have x(g)pm = x(gPrn)= x((gPdprn)= X(S,l)""

and therefore

Thus x ( g ) = x(gpt),as asserted. 2.2. Lemma. Let K , . . . , V , be all nonisomorphic irreducible F G modules and let x; be the character of G aforded b y V,, 1 5 i 5 r . Then 1< - i 5 r, form an F-basis (i) The elements z; = ,&Gx;(g-l)g, of Rey(FG). (ii) For any u E FG, i E (1,. . . ,r } , uz; = C g E G xi(ug-')g. (iii) An element u E F G annihilates V; if and only if uzi = 0,

l 0. Let A be an F-algebra. Recall that the commutator subspace [A,A] of A is defined to be the F-linear span of all Lie products. [ z , y J = z y - ya: with z , y E A.

2.1. Lemma.

W e have

[FG, FG] = { C z g g E FGI

C zg= 0

for all C E CI(G)}

g€C

Proof. This is a special case of Lemma 3.6.10. W Let H be a subgroup of G. Then the natural projection x:FG-+FH

351

2. Defect groups of blocks

defined by

is obviously an F-linear map. In what follows, for each subset X of G, we define Xt E FG to be the sum of all elements in X . The next result due to Brauer, is of fundamental importance. 2.2. Theorem. Let P be a p-subgroup of G, let S = CG(P)and let H be a subgroup o f G for which S C_ H 5 N G ( P ) . Then the natural

projection T : F G --t FS induces a ring homomorphism of Z(FG) into Z(FH) whose kernel is the F-linear span of all C+ with C E C l ( G ) and C n S = 0.

Proof. Let GI,. . . ,C, be all conjugacy classes of G, let X i = and assume that Xr, # 0 for k E (1,. . . , t } and X k = 0 for k E ( t 1,. . .,r } . We know that the elements C t , . . . ,C: form an Ci

nS

+

F-basis for Z ( F G ) . Note also that

and that X 1 , .. . ,X t are mutually disjoint. Moreover, each X i , 1 5 i 5 t , is a union of conjugacy classes of H . Hence T induces an F-linear map from Z ( F G ) to Z ( F H ) whose kernel is of the required form. We are thus left to show that for all i, j E (1,. . . , r } ,

n(C;+Cjf)= T(C;+)T(Cjf) Given s E S , we put T, = { ( x , y ) l z E Ci,y E Cj and xy = s} and denote by T the union of all 2' with s E 5'. If T = 0,then either Xj = 0 or X j = 0, in which case T(C:C:) = 0 = T(C;')T(C~'). Now assume that T # 0. Then 7r(C;'C,')=

c

x y = c

(",Y)ET

c

zy

8ES ( ~ , Y ) E T ~

Because for all h E P and s E S , hsh-' = s, it follows that P acts as a permutation group on T, via h ( x , y ) = (hxh-',hyh-l). Since

Simple induction and restriction pairs

352

churF = p, the sum of all elements xy where (z,y) ranges all orbits of T, of size # 1, is equal to 0. On the other hand, an orbit T, has size 1 if and only if it is of the form ((5, y)} with 5,y E S. Thus

as required. The homomorphism Z ( F G ) + Z ( F H ) defined in Theorem 2.2 is called the Bruuer homomorphism.

2.3. Theorem. (Osima (1955)). Let e be a nonzero centruk idempotent of FG. Then Supp e consists of p'-elements. Proof. (Passman (1969)). Assume by way of contradiction that z E Suppe with z = zy = yz where x # 1 is a p-element and y is a $-element. Let P be the subgroup of G generated by x. Then, by Theorem 2.2, r(e) is a central idempotent of F S , where S = CG(P). Moreover, since z E S, we also have z E Supp?r(e). Hence we may assume that x E Z(G). Applying Lemma 2.1, we deduce that 5

E Suppv

for all v E [FG,FG]

(1)

Choose an integer n with pn 2 [GI and with p" f l(rnodq), where q is the order of y. Put t = y-*e and write t = Ct,g,t, E F . Then, by Lemma 3.6.9, tP" ZE ti"gP" (rnod[FG,FGI) Because gp" is a p'-element, it follows from (1) that x @ SupptP". On the other hand, since e is a central idempotent and since p" = l(rnodq), we have tP" = y+'e = y-*e = t But, by definition of t, x E Suppt, a contradiction. So the theorem is true. Let C be a conjugacy class of G and let g E C. A Sylow psubgroup of CG(g) is called a defect group of C (with respect to p ) . Thus all defect

353


groups of C are conjugate and so have a common order, say p d . The integer d is called the defect of C. We denote by S(C)any defect group of C. If HI and H2 are subgroups of G, we write HI LG H2 (respectively, HI CG H 2 ) to indicate that Hl is G-conjugate to a subgroup of H2 (respectively, to a proper subgroup of H2), while HI =G Hz will mean that HI is G-conjugate to H2. Let e be a block idempotent of F G and let B = B ( e ) . Because e E Z ( F G ) we have

Supp e = C1U C2U . . . U Ct for some C; E CI(G). The largest of the defect groups of C;, 1 5 i 5 T , denoted by 6 ( e ) (or S(B))is called a defect group of e (or of B ) . It will be demonstrated (Proposition 2.5) that all defect groups of e are conjugate and so have a common order, say p d . The integer d is called the defect of e (or of B ) . 2.4. Lemma. Let D be a p-subgroup of G and let C1, C2,. . . , Ct be all elements of Cl(G) with S(C;) CC D . Then the F-linear span ID(G) of C:, C t , . . . ,C t is an ideal o f Z ( F G ) .

Proof. Fix i E { 1,. , . ,t } and denote by Cj any conjugacy class of G. It clearly suffices to show that C;'CT E ID(G). We may assume that C;'C,' # 0, in which case we may choose g E SuppC:C:. Bearing in mind that

we have g = uv for some u E C; and v E Cj. Let of the conjugacy class containing g and let

P

be a defect group

Then P acts on X by conjugation, so by the argument employed in the proof of Theorem 2.2, we have Ci n &(P) # 8. Thus P GG 6(C;) and so P GG D. Hence C;'C[ E ID(G),as required.


354

If D is a defect group of C E CI(G), we shall write I [ C ]instead of ID(G). The fact that I[C]is an ideal of Z ( F G ) will be needed in the proof of the following result. 2.5. Proposition. Let e be a block idempotent of F G and let y be the irreducible representation of Z ( F G ) associated with e . (i) Suppe = C1 u Cz u . . . u Ct u Ct+l u . . . u Cr. where Cl, . . . ,C, are p-regular classes of G such that (a) For all i E { 1,. . . ,t } , e E I[C;]and S(C;) =G S(Cj) for all i , j E (1,..., t}. (b) S ( C k ) cc S(Ci) for all k E {t 1,. . . ,r } and i E (1,. . . , t } . (ii) If C E CZ(G) is such that 6(C) C G S(C;)for some i E { 1, ... ,t } , then y(C+) = 0. (iii) There ezists i E { 1,. . . ,t } such that 7(C:) # 0. (iv) i f C E CZ(G) is such that -y(C+)# 0, then 6 ( e ) CC S(C).

+

Proof. (i) By Theorem 2.3, e = XC ; : for some nonzero X i in F and some pregular classes C1, . . .,C, of G. Becuase y( 1) = y( e) # 0, there is some k E (1,. . .,r } with y(C$) # 0. But C$ E I[Ck], so we must have y(I[Ck])# 0. Invoking Proposition 1.2(iv), we deduce that e E I[Ck]. By renumbering C1,. . . C, in such a way that {Cl,. . . ,C t } is the subset of {Cl,.. . ,C,} consisting of all Ci with S(Ci) = S ( C k ) , we deduce that (a) and (b) hold. (ii) By (i), 6(C) cc S ( C k ) . Assume that y ( P ) # 0. Because C+ E I [ C ] ,we have y(I[C]) # 0 and so, by Proposition 1.2(iv), e E I[C]. Thus 6(Ck) CG 6(C),a contradiction. (iii) This was established in (i). (iv) By the argument of (ii), e E I[C]. The desired assertion is therefore a consequence of the definition of I[C]. We next provide a useful characterization of defect groups in terms of relative trace maps. Let R be an arbitrary commutative ring and let H be a subgroup of G. Then the centralizer C m ( H )of H in RG is defined by

CRG(H)= {z E RG(& = hz for all h E H } In particular, CRG(G)= Z(RG). Consider the RG-module V = RG


355

where G acts on V by conjugation . Then

I ~ ~ ( V=HCRG(H) ) Hence, if S C H are subgroups of G, then the corresponding relative trace map

T ~ :FCRG(s)

-+

cRG(H)

is given by

T r f ( z ) = Ctzt-'

(z E

CFG(S))

t€T

where T is a left transversal for H in G.

Let S C H be subgroups of G and let T be a left transversal for S in H . (i) Tr? : CRG(S)-+ CRG(H) is a homomorphism of CRG(H)bimodules. In particular, Trf(CRG(S)) is an ideal Of CRG(H). (ii) If D is a p-subgroup of G, then 2.6. Lemma.

h ( G ) = Trg(CFG(D))

where ID(G)is defined in Lemma 2.4.

Proof. (i) We know that T r f is an R-linear map. Fix y E CRG(H) and z f CRG(S). Then, for all z E T , we have yz = zy and so TrF(yz) =

C zyzz-l = y C zzz-l xET

= y~r:(z)

%EX

A similar argument shows that T r F ( z y ) = T r F ( z ) y ,as asserted. (ii) Let z = C+ E I D ( G ) ,where C is a conjugacy class of G. Fix g E C, put L = & ( g ) and choose a Sylow psubgroup Q of L. Since z E I D ( G ) ,we may assume Q C D. Furthermore, by the definition of L, we must have z = T r f ( g ) . Applying Lemma 3.9.2(i) to compute T r z in two different ways, we find

TrLG (Tr&+) = TrE (Tr&))

(2)

Because Tr$(g)= ( L : Q)g and ( L : Q ) is prime to p , we may define

w = ( L : Q)-'Tr;(g)


356

Then w E CFG(D)and, by (2)) T r g ( w )= z , proving that

ID(G) E Trg (CFG(D)) Let I( be the D-conjugacy class containing g E G and let C be the Gconjugacy class containing 9. Since the elements K+ form an F-basis for CFG(D),it suffices to show that T r g ( K + )E ID(G).But

Trg(Ic+)= ( c G ( g ) : c D ( g ) ) c + and therefore either Trg(I(+)= 0 or CD(9) is a defect of C. Hence in both cases TrE(K+)E ID(G),as desired. H

Proof. (i) This is a special case of Lemma 3.9.6. (ii) It follows from (i) and Lemmas 2.6(i) and 3.9.2(i) that TrE(a)TrE( b ) = T r g (TrE(a )b)

as asserted. (iii) This is a direct consequence of (i). (iv) Apply (ii). 2.8. Lemma. Let A be a jinite-dimensional algebra over F and let e be a primitive idempotent of A . If I l , . . . ,In are ideals of A such that e E Il - - - I,,, then e E Ii for some i E { 1,. . . ,n }.

+ +


357

Proof. Without loss of generality we may assume that n = 2. Since e is primitive, e is the identity element of the local ring eAe and e E 11 + 1 2 implies e E eI1e + e12e. Thus we may assume that A is local and that e = 1. If 1 = a + b with a E I l , b E 12,then not both a and b lie in J ( A ) , say a # J ( A ) . Since A is local, a has inverse a - l , which gives 1 = aa-' E 11,as required. W It is now an easy matter to provide the following useful characterization of defect groups of blocks. 2.9. Proposition. Let B = B ( e ) be a block of FG. Then the following conditions are equivalent: (i) D is a defect group of B. (ii) D is a minimal element in the set of all subgroups H of G such that e

E T r E (CFG(H))

Proof. Assume that D is a defect group of B. Since e E ID(G), it follows from Lemma 2.6(ii) that e E T T E ( C F G ( D ) ) .Now assume that H c D and that e E T T Z ( C F G ( H ) )Then, . by Lemma 2.6(iv), e E IH(G)contrary to the assumption that D is a defect group of e. Thus (i) implies (ii). Assume that H is a subgroup of G such that e E T T ~ ( C F G ( Hand )) let D be a minimal element in the set of all subgroups H of G. By Lemma 2.7(iv), (with K = G), we have

Therefore, by Lemma 2.8, we have

Then the minimality of D implies that gHg-' 2 D. Now suppose that H (and hence gHg-') is a defect group of B. Then gHg-' 2 D and, by the implication (i)+(ii), we obtain gHg-l = D ,as required. H


358

We close by recording the following deep result due to Brauer (see Curtis and Reiner (1962)). The proof will be omitted since it relies on modular character theory, a topic we do not touch upon in this book. 2.10. Theorem. Let F be an algebraically closed field of characteristic p > 0, let G be a group of order pnm with (p,m) = 1 and let B be a block of F G with defect d. Then d is the smallest integer such that pn-d divides the dimensions of all irreducible FG-modules in B.

3.

Blocks and vertices

In this section, G denotes a finite group and F an arbitrary field of characteristic p > 0. For any psubgroup D of G, ID(G)denotes the ideal of Z ( F G ) defined in Lemma 2.4. All conventions and notations adopted in the preceding section remain in force. Our aim is to examine the vertices of the irreducible FG-modules which lie in a given block B of FG. It will be proved that the defect group of B can be characterized as the maximum among these vertices. A special case of this result, where F is algebraically closed, was established by Green (1959a), while its extension to an arbitrary F is due to Knorr (1976).

Let D be a p-subgroup of G and let e be an idempotent in ID(G) (e.9. e is a block idempotent of F G with D as a defect PUP). (i) For a n y FG-module V , eV is D-projective. (ii) IfV is an indecomposable FG-module in a block B = B ( e ) with defect group D , then V is D-projective. 3.1. Lemma.

Proof. If V satisfies (ii), then eV = V and hence (ii) follows from (i). To prove (i), we employ the relative trace map. By Lemma 2.6(ii), we may choose w E CFG(D) such that e = Trg(w). Consider the map 8 : eV t eV defined by 8(z) = w z for all 5 E eV. Since w commutes with all elements of g with g E D, we have 8 E EndFD(VD). Let T

3. Blocks and vertices

359

denote a left transversal for D in G and let x E eV. Then

and the result follows by virtue of Theorem 2.10.3. H 3.2. Lemma. Let B = B ( e ) be a block of F G and let I be an ideal of Z ( F G ) . I f e 6 I , then Ie is a nilpotent ideal of Z ( F G ) .

Proof. Let y be the irreducible representation of Z ( F G ) associated with e. Owing to Proposition l . l ( v ) Kery = Z ( F G ) ( l- e) + J(Z(FG))e We claim that I s Kery; if sustained it will follow that

l e C (Kery)e= J(Z(FG))e as required. Assume by way of contradiction that y(I) # 0. Because y ( Z ( F G ) ) is a field and 0 # y(I) is an ideal of y ( Z ( F G ) ) ,it follows that

y(Z(FG)) = Y ( I ) Hence y ( e ) = y(x) for some x E I and therefore e - x E Z(FG)(l - e)

+ J(Z(FG))e

Thus e-x = re+z(l-e) for some r E J ( Z ( F G ) )and some z E Z ( F G ) . It follows that e = xe+re and e--ze = re E J ( Z ( F G ) ) . Now J ( Z ( F G ) ) is a nilpotent ideal of Z ( F G ) and Z ( F G ) is a commutative algebra over the field F of characteristic p > 0. Therefore there exists a positive integer n such that

o = (re)P" = ( e - xe)P" = ep" - (xe)P" = e - (xe)P" Thus e = (xe)P" E I , a contradiction. H

Let e be a block idempotent of F G with deject group D and let j D ( G ) be the F-linear span of all C+,C E Cl(G), with 3.3. Lemma.


360

6(C) CG D. Then ID(G)is an ideal o f Z ( F G ) and jD(G)e is a nilpotent ideal of Z ( F G ) . Proof. By Lemma 2.4, I D ( G )is an ideal of Z ( F G ) . Since j o ( G ) is the sum of all IH(G) with H c D,we see that ~ D ( Gis)an ideal of Z ( F G ) . By the definition of the defect group of e, e $ i D ( G ) . Thus f D ( G ) e is a nilpotent ideal of Z ( F G ) ,by virtue of Lemma 3.2. Let e be a block idempotent of FG with defect group D , let E be any field extension of F and write 3.4. Lemma.

as a sum of block idempotents of EG. Then, for all i E (1,. . . ,n } , ei has D as a defect group.

Proof. Let K and L be the counterparts of I D ( G ) and io(G), respectively, in EG. Fix i E (1,. . , ,n } and observe that, by the definition of the defect group of e;, it suffices to verify that e; E K - L. Since e E ID(G) C I< and since I< is an ideal of Z ( E G ) , we have e; = e;e E I']H. It follows from the Mackey decomposition that W is a direct summand of [(gv)HngQ,-l]H

for some g E G

and consequently

as asserted. R

Two remarks are now in order. 4.5. Remark.

The following assertions are direct consequences

of the above proof. (i) Assume the conditions of Theorem 4.4(i). Then

SoC(fVH)2 f VH/J ( F H )f VH for all irreducible FG-modules V = eV. (ii) Assume the conditions of Theorem 4.4(ii). Then Soc(eWG)%! eWG/J(FG)eWG for all irreducible FH-modules W = fW.

370


4.6. Remark. Write e = Ce; and f = Cfj as a sum of block idempotents of F G and FH, respectively. Then, by the definition, ( e ,f ) is a simple restriction (induction) pair if and only if ( e i ,f j ) are simple restriction (induction) pairs for all i and j .

The rest of the section will be devoted to recording a number of consequences of the results so far obtained. As a preliminary result, we first establish the following useful property.

4.7. Lemma. Let H S be subgroups of G and let W be an irreducible FS-module. (i) If ( e ,f) is a simple induction pair and f w H # 0, then eWG is completely reducible. (ii) If ( e ,f) is a simple restriction pair and eWG # 0 , then fWH is completely reducible.

Proof. (i) By assumption, f WH # 0 and hence we may choose an irreducible FH-module V in Soc( f W H ) .Then f V = V and

Since W is irreducible, there is an exact sequence

vs-+w+o Thus there exists an exact sequence

Because ( e ,f ) is a simple induction pair, eVG is completely reducible. Hence eWG is also completely reducible. (ii) The proof is analogous and therefore will be omitted. W

Let H 2 S be subgroups o f G and let e and f be central idempotents of F G and FH, respectively. Let { e i } denote all block idempotents of FS such that E ; f # 0. Put E = x c ; and assume that ( E , f) is a simple induction pair. Then the following conditions are equivalent: 4.8. Corollary.

4. Simple induction and restriction pairs

371

(i) ( e , f) is a simple induction pair. (ii) ( e ,E ) is a simple induction pair. Assume that W = EW is an irreducible F S module. By Remark 4.6 and Theorem 4.4(i), fW, # 0 since (e,f) is a simple induction pair. Thus, by Lemma 4.7(i), eWG is completely reducible which means that ( e ,E ) is a simple induction pair. (ii) +(i): Let V = f V be an irreducible FH-module. Then e V S and e(eVS)Gare completely reducible, since (E, f) and ( e ,e ) are simple induction pairs. Applying Lemma 4.3, it follows from the construction of E that V s = eVS. Thus eVG is completely reducible and so ( e , f ) is a simple induction pair. f l

Proof. (i) +(ii):

Let H S be subgroups o f G and let e and f be central idempotents of FG and FH, respectively. Let { ~ i } be all block idempotents of FS with t;;e # 0. Put c = C ei and assume that ( e ,E ) is a simple restriction pair. Then the following conditions are equivalent: (i) ( e ,f ) is a simple restriction pair. (ii) ( e , f) is a simple restriction pair. 4.9. Corollary.

Proof. The proof is analogous and therefore will be omitted. H Let N be a normal subgroup of G and let f be a block idempotent of F N . It is clear that the set is a subgroup of G containing N . We refer to G ( f )as the inertia group of f (or b = b ( f ) ) . Note that if T is a left transversal for G ( f ) in G, then the tft-’,t E T , are all distinct conjugates o f f and

f* = Ctft-l tET

is an idempotent of Z( F G ) n Z( F N ) . We say that a block idempotent e E F G covers f (or B = B(e) covers b = b ( f ) ) if e occurs in the decomposition of j” into the sum of block idempotents of FG.

4.10. Lemma. A block B = B ( e ) o f F G covers the block b = b ( f ) of F N if and only i f e f # 0.

372


Proof. It is clear that e f # 0 if and only if ef* # 0. Since the latter holds if and only if e covers f, the result follows. W 4.11. Theorem. (Know (1977)). Let N be a normal subgroup of G , let b = b( f) be a block of F N and let B; = B;(e;),1 5 i 5 m, be the block of F G covering b. Let Vl,. , .,& be all nonisomorphic irreducible FN-modules in b and let Hi be the inertia group of V;., 1 5 i 2 t . (2) rn 5 min((H1 : N ) , ( H 2 : N ) , . . . , ( H t : N ) } . (ii) If m = ( G ( f ) ): N ) , where G(f) is the inertia group of f, then

FG J ( F N )

J ( FG)f

*

and

b(f) Proof. (i) Let V be a n irreducible module in b and let H be the inertia group of b. Because N d G, it follows from Clifford’s theorem that (1, 1) is a simple restriction pair. Thus, by Remark 4.6, ( e i ,f) is a simple restriction pair, 1 5 i 5 rn. Note also that, by Lemma 4.10, e;f # 0, 1 5 i 5 m. Applying Theorem 4.4(ii), we deduce that Wi = e;VG # 0 , 1 5 i 5 m. Let ci > 0 be the composition length of W; and Mjj its composition factors. Since +i)

(VC)N

=G

= @gETg €3

v

where T is a transversal for N in G, it follows from Clifford’s theorem that

dimFM;j = r;j(G : H)dimFV for some integers

and

rij.

Accordingly,

373

5. Complete reducibility of induced modules

as asserted. (ii) Because H C G(f), it follows that rn = (G(f) : N ) 2 ( H : N ) 2 rnin((H1 : N ) , . . . ,(Ht : N ) } 2 rn and hence ( H : N ) = rn. It follows from (1) that c; = 1 for all i, so V G= $e;VG is completely reducible. Thus (1,f) is a simple induction pair and hence, by Proposition 4.2, J ( F G )f FG J ( F N ) . The assertion regarding the defect groups follows from Theorem 4.4 and Remark 4.6. W

-

5.

C o m p l e t e reducibility of induced modules

Throughout this section, F denotes an arbitrary field of characteristic p > 0 and G a finite group. Our aim is to prove a number of results pertaining to complete reducibility of induced and restricted modules. By the principal block of FG, we understand the block which contains the trivial FG-module, i.e. the irreducible FG-module on which G acts as the identity operator.

5.1. Lemma. The Sylow p-subgroups of G are defect groups of the principal block of F G .

Proof. Let B = B(e) be the principal block. By Theorem 2.3, we may write e in the form e = AICf

+ - + Arc,? *

(0 # A; E F )

for some p-regular classes C1,. . . ,C, of G. Denote by V the trivial FG-module. Since V belongs to B , we have ev = v for all v E V . Hence = (X1IC1I+...+XrIC,l)v for all v E V

which implies that p does not divide ICI for some pregular class C in Suppe. Hence a Sylow p-subgroup of G is a defect group of C, as required. W An application of the above lemma together with Theorem 3.8 now easily yield the following two results.


374

(Knorr (1977)). Let H be a subgroup of G. Assume that V G is completely reducible for any irreducible FH-module V . Then H is ofp'-index in G. 5.2. Proposition.

Proof. Let B = B(e) be the principal block of FG. By Lemma 5.1, a Sylow p-subgroup P of G is a defect group of B. By Theorem 3.8, we may choose an irreducible FG-module U in B such that P is a vertex of U. Let V be an irreducible submodule of UH.Since eU = U , we have

0 # HOmFH(V, UH) HOmFc(VG, u)

HOmFG(eVG,u )

But eVG is a submodule of V G and VG is completely reducible by assumption. Thus eVG is completely reducible and therefore U is a direct summand of VG. Hence U is H-projective and so H contains a conjugate of P . This shows that H is of p'-index, as desired. W

Let H be a subgroup ofG. Then the following conditions are equivalent: (i) All irreducible FG-modules are H-projective. (ii) H is ofp'-index in G. 5.3. Proposition.

Proof. (i) +(ii): Let B = B(e) be the principal block of FG. By Theorem 3.8 and Lemma 5.1, we may choose an irreducible FGmodule V in B such that a Sylow p-subgroup P of G is a vertex of V. By assumption, V is H-projective and so H contains a conjugate of P. Thus H is of $-index in G. (ii) +(i): This is a consequence of Corollary 2.10.4. Our next aim is to strengthen Theorem 2.3.9 under the additional assumption that R is a field. We need a number of preliminary results contained in a work of Knorr (1977). 5.4. Proposition. Let H be a subgroup ofG and let e and f be central idempotents of FG and F H , respectively. Then the following

conditions are equivalent: (i) (eVG)H is completely reducible for all irreducible FH-modules V.

375


(ii) e J ( F H ) F G C F G . J ( F H ) . (iii) e J ( F H ) F G = F G . J ( F H ) e . Moreover, if these conditions are satisfied, then

e J ( F H ) " F G = ( e J ( F H )- FG)"

for all n 2 1

Proof. (i) +(ii): Put W = F H / J ( F H ) . Then W is completely reducible, so by assumption (eWG)H is completely reducible. Thus e J ( F H ) ann(WG).However, by Theorem 2.3.4(i),

c

ann( W G )= I d ( F G - J ( F H ) )

-

and so e J ( F H ) F G C F G J ( F H ) . (ii) +(iii): It is obvious from (ii) that

eFGJ( F H ) F G C F G - J ( F H ) and therefore, by Lemma 2.3.3(iii),

eFG J ( F H )

c

eFGJ(FH)FG Id(F G * J (F H ) ) = I d ( J ( F H )* F G )

It follows that eFG - J ( F H ) e J ( F H ) F G . The opposite containment is obvious from (ii). (iii) +(i): It is clear from (iii) that F G J ( F H ) e is an ideal of FG. Let V be an irreducible FH-module and let A = a n n ( V ) . Then J ( F H ) C A and FG J ( F H ) e C F G A. Applying Theorem 2.3.4(i), we conlude that

-

J ( F H ) eC F G . J ( F H ) e2 Id(FG * A ) = ann(VG) Hence J ( F H ) ( e V c ) = 0 and so ( e V G ) H is completely reducible. The final assertion follows by induction on n , using (iii). W 5.5. Proposition. Let H be a subgroup o f G and let e and f be central idempotents of F G and F H , respectively. Then the following conditions are equivalent: (i) (fMH)G is completely reducible for all irreducible FG-modules


376

M. (ii) V G is completely reducible for all irreducible FH-modules V = f V , and fMH is completely reducible for all irreducible FG-modules M. (iii) J ( FG) f = FG - J ( F H )f . (iv) f J ( F G ) = f J ( F H ) F G . Proof. Condition (ii) may be reformulated as: (1,f) is a simple induction pair and a simple restriction pair. The equivalence of (ii), (iii), and (iv) is therefore a consequence of Proposition 4.2. Since (ii) obviously implies (i), we are left to show that (i) implies (ii). Let M be an irreducible FG-module. Then, by assumption, (fM H ) ~ is completely reducible and hence, by Corollary 2.2.10, fMH is completely reducible. Let V = f V be an irreducible FH-module and let M be an irreducible FG-module in Soc(VG). Then

0#

M,VG)

H O ~ F G (

HomFH(MH,

v)

f M H ,v )

H O ~ F H (

Hence there is an exact sequence

which gives rise to an exact sequence

Since (fM H ) is~completely reducible, so is V Gas required. 5.6. Lemma. Let H be a subgroup o f G and let e be a central idempotent of FG. Then the following conditions are equivalent: (i) For any FG-module M = e M , if MH is completely reducible, then so is M . (ii) F G . J ( F H ) F G 2 J(FG)e.

Proof. (i) *(ii): Put M = F G e / F G . J ( F H ) F G e . Then M = eM and M H is completely reducible. Thus, by hypothesis, M is completely reducible. Hence

J ( F G ) e C F G . J ( F H ) FGe 5 F G - J ( F H ) . F G ,

377


as required.

(ii) +(i): Let M = eM be an FG-module such that MH is completely reducible. Then J(FH)M = 0 and if F G J ( F H ) . FG 2 J ( F G ) e , then b

J(FG)M = J ( F G ) e M F G - J ( F H ) - FGM = FG-J(FH)M = 0,

proving that M is completely reducible. 5.7. Lemma. Let H be a subgroup of G and M be an FG-module such that M / J ( F G ) M is H-projective. Then

FG * J(FH)M 2 J(FG)M

+

Proof. Put X = F G - J(FH)M. Then Y = M / ( X J ( F G ) M )is a direct summand of the completely reducible and H-projective module M / J ( F G ) M , so Y is completely reducible and H-projective. Consider the exact sequence

0

+

(X

+ J(FG)M)/X+M / X +Y

4

0

(1)

Because J ( F H ) M 2 X , it follows that ( M / X ) H is completely reducible. Hence (1) splits as an FH-sequence. But Y is H-projective, so (1) splits as an FG-sequence. Thus

MIX 2 [(X + J ( F G ) M ) / X ] CB Y a.nd multiplying both sides by J ( F G ) gives

( J ( F G ) M t X)/X 2 J ( F G ) [ ( Xt J(FG)M)/X] since

Y is completely reducible. Hence (X t J ( F G ) M ) / X = 0

and so J ( F G ) M

X, as required. H


378

5.0. Lemma. Let H be a subgroup o f G and let e be a central idempotent of F G . (i) If J ( F G ) e F G . J ( F H ) , then all irreducible FG-modules M =

eM are H-projective. (ii) If all irreducible FG-modules M = eM are H-projective, then V H completely reducible implies V is completely reducible for a n y FGmodule V = eV. Proof. (i) Owing to Proposition 4.2(i), (e, 1) is a simple induction pair. Write e = C e; as a sum of block idempotents of FG. Then, by Remark 4.6,each ( e i , l ) is a simple induction pair. Therefore, by Theorem 4.4(i), b(e;) S G H. Let M = eM be an irreducible FGmodule. Then M = e;M for some i and so, by Theorem 3.8, M is b(ei)-projective. Hence M is H-projective, by virtue of Theorem 2.10.7( ii). (ii) Suppose that all irreducible FG-modules M = eM are H projective. By Lemma 5.6, it suffices to show that F G J ( F H ) - F G 2 J ( F G ) e . By assumption and by Theorem 2.10.7(i), F G e / J ( F G ) e is H-projective. Hence, by Lemma 5.7,

-

F G - J ( F H ) FG 2 F G - J ( F H ) F G e 2 J ( F G ) e as asserted. 1 5.9. Lemma. Let H be a subgroup o f G and let e be a central idempotent of F G . If (eVG)H is Completely reducible for all irreducible

FH-modules V , then J ( F H ) e C J ( F G ) . Proof. Suppose that for each irreducible FH-module V , ( e V G ) H is completely reducible. Then, by Proposition 5.4, e J ( F H ) F G is a nilpotent ideal of FG. Accordingly,

e J ( F H )C e J ( F H ) F G C J ( F G ) as required. W

We are now ready to prove our first major result.


379

5.10. Theorem. (Knorr (1977)). Let H be a subgroup o f G and let e be a central idempotent of FG. Then the following conditions are

equivalent: (i) J ( F G ) e = F G - J ( F H ) e . (ii) J ( FG)e = J ( FH)FGe. (iii) eVG is completely reducible for every irreducible FH-module V and M H is completely reducible for any irreducible FG-module M = eM. (iv) eVG and ( e V G ) H are completely reducible for all irreducible FH-modules V . (v) (e V G ) H is completely reducible for every irreducible FH-module V and all irreducible FG-modules M = eM are H-projective. (vi) (eVG)H is completely reducible for any irreducible FH-module V and MH completely reducible implies M completely reducible for any FG-module M = e M . Proof. Condition (iii) may be reformulated as follows: (e, 1) is a simple induction and a simple restriction pair. Thus, by Proposition 4.2, (iii) is equivalent to either of the following conditions:

J ( F G ) e C F G . J ( F H ) and J ( F H ) e C_ J ( F G )

(2)

J ( F H ) - F G and J ( F H ) e C_ J ( F G )

(3)

J ( FG)e

But (2) is equivalent to (i) and (3) is equivalent to (ii), hence (i), (ii) and (iii) are equivalent. The implications (iv)+(v)=+(vi) follow by Lemma 5.8 and Proposition 4.2, while the implication (iv)+(iii) is a consequence of Lemma 5.9 and Proposition 4.2. We are thus left to show that (vi) implies (iv) a.nd (ii) implies (iv). (vi) +(iv): Owing to Proposition 5.4, F G . J ( F H ) e = J ( F H ) F G e and, by Lemma 5.6

F G . J ( F H )- FG 2 J(FG)e Consequently,

J ( F G ) e C F G . J ( F H ) .FGe = F G - J ( F H ) e & F G - J ( F H )

380


and therefore, by Proposition 4.2, eVG is completely reducible for any irreducible FH-module V . (ii) +(iv): This is a direct consequence of the implications (ii) +(iii), (ii) +(i) and Proposition 5.4. H As an application of the above theorem, we now establish our second major result. 5.11. Theorem. (Knorr (1977)). Let H be a subgroup of G and let F be a field of characteristic p > 0. Then the following conditions

are equivalent: (i) There exists a normal subgroup N of G such that N C_ H and N has p'-index in G. (ii) V G is completely reducible for any irreducible FH-module V and MH is completely reducible for a n y irreduicble FG-module M . (iii) V G and ( V G )are ~ completely reducible for all irreducible F H modules V . (iv) ( V G )is~completely reducible for all irreducible FH-modules V and all irreducible FG-modules are H-projective. (v) ( V G ) H is completely reducible for any irreducible FH-module V and MH completely irreducible implies M completely reducible for any FG-module M . (vi) ( V G )is~ completely reducible for all irreducible FH-modules V and H is of p'-index. (vii) ( M H ) Gis completely reducible for any irreducible FG-module M. (viii) J ( F G ) = F G . J ( F H ) . (ix) J ( F G ) = J ( F H ) F G .

Proof. Applying Proposition 5.5 for f = 1 and Theorem 5.10 for e = 1, it follows that the conditions (viii), (ix), (ii), (iv), (v), and

(vii) are equivalent. Moreover, by Proposition 5.3, (iv) and (vi) are equivalent. It therefore suffices to show that (i) is equivalent to one of the conditions (viii), (ii), (vi). For the sake of clarity, we divide the rest of the proof into two steps.

Step 1. Suppose that for any irreducible FH-module V , (V"), is


381

completely reducible. Our aim is to show that for any given g E G,

J ( F H ) = F H . J ( F L ) where L = H n gHg-' Put V = F H / J ( F H ) . Then V is a completely reducible FH-module, hence so is ( V G ) by ~ ,hypothesis. Setting W to be the restriction of gV to F L , it follows from Mackey's decomposition that W His completely reducible. Put A = ann(W) and S = F L / A . Owing to Theorem 2.3.4(i), J ( F H ) C F H A and hence, by Proposition 2.2.11(i), SH is completely reducible. Hence, by Corollary 2.2.10, S is completely reducible and therefore J ( F L ) C_ A. But

A = F L n g(annV)g-l = F H n F(9Hg-l) n gJ(FH)g-' = F H n gJ(FH)g-l C J(FL) since F H n gJ(FH)g-' is a nilpotent ideal of FL. Thus

A = F H n gJ(FH)g-l = J ( F L ) and hence

J(FH)C FH * J(FL)

Replacing g by g - l , it follows by the same argument that

F H n g-'J(FH)g = J ( F ( H n g-'Hg)) Because

9FHg-l = F(9Hg-l) and gJ(FH)g-l = J(F(gHg-')) the latter implies that

Hence J ( F H ) 2 J ( F L ) and thus J ( F H ) = F H - J ( F L ) , as required. Step 2. Completion of the proof. Suppose that (vi) holds. To prove

382


(i), we argue by induction on [HI. If H d G then we are done with N = H . Assume that L = H n g H 9 - l is a proper subgroup of H for some g E G. Because (viii) and (vi) are equivalent, it follows from Step 1 that

J ( F G ) = FG * J ( F H ) = FG * F H - J ( F L ) = FG - J ( F L ) Hence, by induction, there is a normal subgroup N of G such that N E L and N has $-index in G. Since L E H , (i) is established. Finally, suppose that (i) holds. Let V be an irreducible FH-module. Because ( V G )is~a direct sum of FN-modules of the form 9V,g E G, ( V G )is~completely reducible. Since N has p'-index in both H and G, it follows from the equivalence of (viii) and (vi) that

J ( F G ) = F G . J(FN) and J ( F H ) = F H - J ( F N ) Hence

FG. J ( F H )= F G . J(FN) = J(FG), proving (viii) and the result follows. 1

383

Chapter 7 Permutation modules This chapter provides a detailed account of a distinguished class of induced modules, namely permutation modules. These modules hold much information about the p-modular representations of G, as well as information about the fusion of p-subgroups of G. The main results presented are due to Alperin (1988), Broue' and Robinson (1986), Robinson (1988), Dress (1975), Saksonov (1971) and Broud (1985). We give a detailed account of the Brauer morphism which enables us to provide a comprehensive coverage of an important type of ppermutation modules, the so called Scott modules. These modules were first discovered by L.L. Scott, and independently by J.L. Alperin. Our method is based on a work of Broud (1985) who employed a systematic use of the Brauer morphism. The chapter ends with the study of ppermutation modules via the Brauer morphisrn. 1.

Preliminary results

In this section, R denotes an arbitrary commutative ring and G a finite group. All RG-modules are assumed to be finitely generated. Given a subgroup H of G, we write 1~ for the trivial RH-module, i.e. 1~ = R as R-module and

hr = r

for all r E R , h € H

Let V be an RG-module. We say that V is a permutation module (respectively, transitive permutation module) if V is R-free with an R-

Permutation modules

384

basis on which G acts as a permutation group (respectively, G acts as a transitive permutation group). Given an RG-module V , we write as usual Inv(V) for the R-module of G-invariant elements of R. 1.1. Lemma. (i) An RG-module V is a transitive permutation module if and only if V Z ( l H ) G for some subgroup H of G. (ii) An RG-module V is a permutation module if and only if it is a direct sum of transitive permutation modules. (iii) Let V be a permutation module and let B be a basis of V on which G acts as a permutation group. If B1,. . . ,B, are G-orbits of B , then Inv(V) is a free R-module with basis Bf,. . . ,B:, where B: = CIEBi 2, 1 I iI r.

Proof. (i) Assume that an RG-module V is a transitive permutation module. Then there exists an R-basis, say v1,. . . ,vn, of V such that G acts as a transitive permutation group on {vl,. . . ,vn}. Set= Rvi it follows that V is an imprimitive RG-module with ting {Kll I: i 5 n } as a system of imprimitivity and with G acting transitively on {Kll 5 i 5 n}. If H is the stabilizer of V,, then V s V y by Proposition 2.1.4(ii). Since & E l ~we, conclude that V E ( 1 ~ ) ~ . Conversely, if V E ( l ~ )then ~ ,by Proposition 2.1.4(i) and Corollary 2.1.2(iii), V is a transitive permutation module. (ii) This is a direct consequence of (i) and the definition of a permutation module. (iii) Straightforward. H For any subset S of G, we define S+ E RG by

s+=cs sES

1.2. Corollary. Let H be a subgroup of G and let Cl, . . . , C, be all H-conjugacy classes of G. Then CRG(H)is a free R-module with basis

c:,

*

. . ,c;.

Proof. We put V = RG and note that V is a permutation RHmodule, where H acts by conjugation on the R-basis G of V . Since the H-orbits of V are precisely the H-conjugacy classes of G and since

1. Preliminary results

385

I n v ( V ) = C R G ( H ) ,the result follows by virtue of Lemma l.l(iii).

Let H be a subgroup of G. (i) 1~ ES R H + . (ii) ( 1 ~E)RG ~ - H+. (iii) If n = [HI is a unit ofR, then e = n-lH+ is an idempotent of RH such that ( 1 ~E)RGe ~ 1.3. Lemma.

(iv) RG+ C RG H+ (hence

1G

is isomorphic to a submodule of

(lH)G).

(v) Let (G : H ) be a unit of R. Then cp E EndRG(RG H + ) given b y v ( H + ) = (G : H)-lG+ is a nonzero idempotent. In particular, (a) RG - H+ = RG+ @ RG(H+- (G : H)-'G+). (b) lc is isomorphic to a direct summand of (1~)'. (vi) If H # G and ( 1 ~ is) indecomposable, ~ then (G : H ) is a nonunit of R . Proof. (i) This is obvious. (ii) Apply (i) and Proposition 2.2.11(i). (iii) This is a direct consequence of (ii). (iv) Let T be a left transversal for H in G. Then

as required.

(v) Consider the RG-homomorphism cp : RG. H+ + RG- H+ given by

cp(gH+) = (G : H)-'G+

By (iv), we have

and hence cp is a nonzero idempotent of EndRG(fiG- H + ) , as required. (vi) Direct consequence of (v). W

386

Permutation modules

We now introduce the following important concept. A G-algebra over R is an R-algebra A on which G acts as a group of R-algebra automorphisms, i.e. g E G acts on a E A to give ga E A , and this G-action transforms A into a left RG-module, and

for all a,b E A and g E G. Let A be a G-algebra. Then A is an RG-module and, for any subgroup H of G, we write AH for the R-subalgebra of H-invariant elements, i.e.

AH = { a E A1 ha = a for all h E H } For example, if A = RG where G acts by conjugation, then A is a G-algebra and

AH = CRG(H) for any subgroup H of G Another typical example of G-algebras is the R-algebra

A = EndR(V) for any RG-module V. Here for any g E G, f E En&(V), gf = gfg-' where (gfg-l)(v) = g(f(g-'v)) for all EV Observe also that for any subgroup H of G, we have

AH = EndRH(V)

If H , I< are subgroups of G such that H C I 0, let H be a p-subgroup of G and let &, . . . ,V , be all nonisomorphic irreducible RG-modules. Denote b y n; the multiplicity of F/: as a composition factor of (WG. (i) Each E is a homomorphic image of ( 1 ~ ) ~ . (ii) Each ni 2 1 and & n;dimV, = (G : H). (iii) If R is a splittingfield for RG, then n; = dimRP(K)/lHI where P ( K ) is the projective cover of q, 1 5 i 5 r .

Proof. It is clear that (ii) is a consequence of (i). To prove (i), let V be an irreducible RG-module. If W is an irreducible submodule of V H , then W Inv(V'), since charR = p and H is a pgroup. Hence I ~ v ( V H#) 0 and so, by Proposition 1.7,

Since V is irreducible, this shows that V is a homomorphic image of ( l ~ ) ' ,as desired. To prove (iii), note that by Lemma 6.1.3,

On the other hand, by Corollary 2.2.7,

and so n; = dimRHomRH(P(F/:),1 ~ )Now . P ( K ) is RH-free of rank

Thus n; = tidimRHomRH(RH, 1 ~=)ti as required.

(1 5 i

5 r)

1. Preliminary results

391

Let R be a field of characteristic p > 0 or a complete discrete valuation ring with c h a r R / J ( R ) = p > 0 . Let H be a subgroup of G, let P be a p-subgroup of G and let T be a full set of double coset representatives for ( P ,H ) in G. Then 1.9. Proposition.

a n d each ( ltHt-lnp)P is an absolutely indecomposable FP-module.

Proof. Setting V = l ~we, have t K H t - l n p = l t ~ t - ~ n p ,Et 2'. Now apply Theorem 2.7.1 and Corollary 5.1.12. Given a finite set X on which G acts as a group of permutations, we denote RX the corresponding permutation RG-module. If G acts transitively on X and H is the stabilizer of x E X , then obviously R X E ( 1 ~as) RG-modules. ~ For any subgroup S of G, we put

Cx(S) = {z E X ( s z = x for all

s E

S}

(i) R C x ( S ) is a permutation RNG(S)-module. (ii) If G acts transitively on X and H is the stabilizer o f x E X , then 1.10. Lemma.

where T is a complete set of those double coset representatives t for ( N G ( S )H, ) in G for which S C t H t - l .

Proof.

(i) Given g E NG(S) and x E C x ( S ) ,it suffices to show

that

s(gz) = gx

for all s E S

Since s ( g z ) = g(g-'sg)z = gz, (i) is established. (ii) Let g l , . . . ,gn be a left transversal for H in G containing 2'. We may assume that X = ( g I H , . . . ,g n H } and G acts on X by left multiplication. Then

Permutation modules

392

and, in particular, {tHlt E T } & C x ( S ) . If g E N G ( S ) and t E T , then gtH = t H if and only if g E tHt-' n N G ( S )= N t ~ + ( S ) . Hence N t ~ t - ~is(the S )stabilizer of t H in N G ( S ) . Thus we are left to verify that {tHlt E T } is a complete set of representatives of NG(S)-orbits of

CX(S).

H NG(S)tH for some Assume that giH E Cx(S).Then N G ( S ) ~ ; = t E T . Hence gi = 9th for some g E N G ( S ) ,h f H , and therefore 9iH = 9 ( t H )

Finally, assume that t l , t 2 E T are such that gtlH = t2H for some g E N G ( ~ ) Then . N G ( S ) t l H = N ~ ( s ) t 2 Hand hence tl = t 2 , as required. 1.11. Corollary. (Scott (1973)). Let R be a complete noetherian local ring with c h a r R / J ( R ) = p > 0 , let P be a p-subgroup of G and let X be a finite set on which G acts as a group of permutations. Then the Green correspondence with respect to ( G ,P, N G ( P ) ) gives a multiplicity-preserving bijection between the nonisomorphic indecomposable components of R X with vertex P and the nonisomorphic indecomposable components of RCx(P ) with vertex P .

Proof. It suffices to prove the result when G is transitive on X , since then the result can be applied to each transitive component. Since R X S' ( l H ) G , where H is the stabilizer of x E X in G, the desired conclusion follows by applying Lemma l.lO(ii) (with S = P)and Theorem 5.2.9(ii) (with V = 1 ~ ) . 2.

Hecke algebras

Throughout this section, R denotes an arbitrary commutative ring and H a subgroup of a finite group G. Invoking Lemma 1.3, we shall identify the transitive permutation module ( 1 ~with ) ~the left ideal RG H+ of RG where H+ = &H h E RG. If T is a left transversal for H in G, then

-

{tH+It E T } is an R-basis of RG H + which is permuted transitively by G via left multiplication.

393

2. Hecke algebras

The R-algebra E n d ~ c (1 (~ ) ' ) is called the Hecke algebra associated with the triple (G, R, H ) . Our subsequent investigations will show that, in the special case where R is a field of prime characteristic p, this algebra holds much information about the fusion of p-subgroups of G, as well as information about the p-modular representations. Our aim here is to examine the structure of E n d R G ( ( l H ) G ) . The results presented will play a fundamental role in our future study of permutation modules. Given g 1 , g 2 E G and f E EndR( ( 1H ) ' ) , we define

by ( g 1 f s 2 > ( 4=

sl(f(s24)

(v E

(WG)

Recall that E n d R ( ( 1 ~ is ) ~a)G-algebra with the G-action given by

More generally, end^( ( l H ) G ) can be regarded as an R(H x H)-module via

( h l ,h 2 ) f = hlfh,'

(hl,h2 E H,fE E n d R ( ( l H ) G )

Let T be a left transversal for H in G. For any given x E RG, define Q H ( ~ E ) E ~ ~ R ( ( Iby H)~)

a H ( x ) ( t H + ) = xH+ if t E H if t 6H aH(x)(tH+) = 0 Since {tH+lt E T } is an R-basis of ( l H ) ' , a ~ ( x )is indeed a uniquely determined element of E T Z ~ R ( ( ~ HIt) ~is) clear . that c u is~ an R-linear map.

2.1. Lemma. The R-linear map a H : RG + EndR((1H)') satisfies the following properties: (i) aH(x)(gH+)= xH+ i f g E H and aH(Z)(gH+)= 0 i f 9 $ H for all x E RG. (ii) a&zdi) = a H ( Z ) , ha&)h1 = c r H ( h Z ) f o r all h , hl E H , x E RG.

Permutation modules

394

(iii) CYH is a homomorphism of R ( H x H)-modules. (iv) I f g E G and h E gHg-l, then a ~ ( g=) aH(hg) and so a H ( 9 ) E EndRpmgHg-l,((lH)G) for all 9 E G

Proof. (i) This is a direct consequence of the definition of CYH. (ii) The first equality is a direct consequence of the definition and the fact that z h H + = z H + for all h E H , z E RG. To prove the second, it suffices to show that for all g E G, 2 E RG, h , hl E H ,

If g $ H , then both sides of (1) are zero. If g E H , then H+ = gH+ = h l ( g H + ) and so

a H ( h x ) ( g H + )= h x H t = h a H ( z ) h l ( g H + ) , as required. (iii) Given

2

E R G , h l , h2 E H , it follows from (ii) that

W ( ( h 1 ,h2)z) = QH(hl&l) = QH(h1Z:) = hlaH(z)h,' = &I, h Z ) W ( Z ) , as required. (iv) Fix g E G and h = ghlg-' for some hl E H . Then, by (ii),

Hence, if h E H n g H g - ' , then by (2) and (ii),

as asserted.

2.2. Lemma.

erties:

The elements a ~ ( g g) ,E G, satisfy the following prop-

2. Hecke algebras

395

(i) .H(g)(H+) = ( H g H ) + . (ii) For all x,y E G, a ~ ( x=) a ~ ( y if) and only if H x H = H y H . (iiz) The set of distinct a H ( g ) 's is an R-basis of E ~ R G ( 1( ~ ) ~ ) .

Proof. (i) Let T be a left transversal for H

H g H = UteJtgH

n gHg-'

in H . Then

(disjoint union)

and therefore

as required.

(ii) Since a ~ ( x E) E ~ R G ( ( I H ) ~ it ) , is uniquely determined by u ~ ( x ) ( H + The ) . desired conclusion now follows by virtue of (i). (iii) Let T be a full set of double coset representatives for ( H ,H ) in G. Then, by (ii), { a ~ ( t ) lEt T } is the set of all distinct a H ( g ) ' s . If

then by (i),

0=

r,(HtH)+

rtaH(t)(Ht)= tCZ'

t€T

which implies that all rt = 0. Thus the set { a ~ ( t ) lEt T } is R-linearly independent. ) ~ claim ) . that Let f be an arbitrary element of E n d ~ c ( ( 1 ~We

which, by (i), will imply f = C t E T r t a H ( t )hence , the result. Let g l , . . . ,gn be a left transversal for H in G. Then f(Ht) can be

396

Permutation modules

uniquely written in the form n

f ( H t ) = C X ; g ; H + for some A; E R i=l

(4)

Hence, to prove (3), it suffices to show that if HgkH = H g j H , then Ak = Aj. so assume that gj = hgkhl for some h , hl E H . Then gjH = hgkH and

By comparing (4)and ( 5 ) , we see that

Ak

= X j , as required. H

Let H be a subgroup of G and let T be a left transversal for H in G. Since ( 1 ~is )a ~ free R-module with basis {tH+It E H } the R-module E n d ~ ( ( 1 has ~ ) ~an) R-basis

given by

f t s ( x H + )= for all

2

if x H = t H if x H # t H

E G.

2.3. Lemma. For any t , s E T and g E G, we have gfts

= fx,.

where X,p E T are defined b y p H = gsH,XH = g t H . G(ft,) = tHt-' n s H s - l . (ii) Proof. (i) This is a direct consequence of the definition of fts and the action of g E G on f t s . (ii) By (i), g f t s = fts if and only if t H = gtH and sH = g s H , i.e. if and only if g E tHt-l n s H s - l , as required. 2.4. Proposition. Let T be a left transversal for H in G , let S T be a complete set of double coset representatives for ( H ,H ) in G andlethETnH.

397

2. Hecke algebras

(i) EndR( ( 1 ~ ) is~a )permutation G-algebra with permutation basis {f&S E T). are all distinct representatives of G-orbits of { f t s l t , s E (ii) { fhsls E T }. (iii) For all s E S , a H ( s ) is the sum of all elements in the G-orbit of f h s (iv) G(fh,) = H n sHs-'.

s}

Proof. Properties (i) and (iv) follow from Lemma 2.3. By Lemma 2.2, (iii) implies (ii). To prove (iii), note that fhs = a ~ ( s )Hence, . by

(4,

G

= T r H n , H ~ - l ( f h s ) = Trg(fhs)(fhs)7 i.e. a H ( s ) is the sum of all elements in the G-orbit of fh,. H aH(S)

2.5. Corollary. Let S be a complete set of double coset representatives for ( H ,H ) in G, let I( be a subgroup of G and, for each s E S , let m, = g.c.d.{(H

n sHs-') : (9ICg-l n H n s H s - ' ) J g E G }

Then the nonzero elements of the set {msaH(s)lS

E

s}

f o i m an R-basis of T r g ( E n d R K ( l H ) G ) . Proof. Apply Propositions 2.4 and 1.5. H 2.6. Corollary. In the notation of Corollary 2.5, the RG-module (iff)G is I(-projective if and only if each m,, s E S , is a unit of R.

Proof.

Apply Theorem 2.10.3 and Corollary 2.5. H

2.7 Corollary. Let a prime p be a nonunit of R and let H be a p-subgroup of G . If I( is any subgroup of G , then ( 1 ~ is )K-projective ~ if and only if H C 9 K g - l for some g E G . Proof. We keep the notation of Corollary 2.5. Since H is a p group, for any s E S , m, is a nonnegative power of p . Hence, by

Permutation modules

398

Corollary 2.6 and the assumption on p , ( 1 ~is )K-projective ~ if and only if each rn, = 1. Since the latter is equivalent to the requirement that H 9Kg-l for some g E G, the result follows. H The ordinary trace map

tr : ~ n d R ( ( 1 H )-, ~ )R restricts to the R-linear map

tr : E

~ ~ R G ( ( I H -+ ) ~R )

which is given by

Indeed, if g E H , then by Lemma 2.2(i), a ~ ( gis) the identity map and therefore t r a H ( g ) = (G : H ) ( = R-rank of ( l H ) G ) . On the other hand, if g 6 H then for all t E G,

.H(9>(tH+) = t ( H 9 H ) + and the coefficient of t H + in aH(g)(tH+)is zero. Thus in this case we indeed have traH(g) = 0. Keeping the above notation, we now record the following result. 2.8. Lemma.

For all z,y E G ,

Proof. Let T be a left transversal for H

n z H z - l in H . Then

H z H = U t E ~ t z H (disjoint union) and therefore

399

2. Hecke algebras

Hence, if X is the coefficient of H + in a ~ ( y ) a ~ ( x ) ( Hthen +),

( H :( H nxHx-')) 0

A={

if if

H x H = Hy-lH HxN#Hy-'H

Since X is also the coefficient of u ~ ( 1in) a s ( y ) a ~ ( xand )

the result follows by virtue of (6). H We denote by I ( H ) the augmentation ideal of R H , i.e. the kernel of the augmentation map aug : RH -+ R given by a u g ( C ~ h h= )

2.9. Lemma.

C X ~(

~ Eh R, h

E H)

The left annihilator of H+ in RG is RG - I ( H ) .

Proof. Let T be a left transversal for H in G. Since

it follows that

RG - I ( H ) = $ t e ~ t- I ( H )

(8) By ( 7 ) , a typical element x of R G can be written uniquely in the form z = C t E T t x t with x t E R H . Since xtH+ = aug(xt)H+,it follows that zl3' = 0 if and only if xt E I ( H ) for all t E T . The desired conclusion is therefore a consequence of (8). From now on, we write

for the natural homomorphism of R-algebras given by ~ H ( Z ) Z= I

m

for all

x E RG,v E ( 1 ~ ) ~

Given x = C xgg E RG, xg E R,g E G, we also put

x* = C x g g - l

(9)

Permutation modules

400

Then x

H

z* is an antiautomorphism of

RG. Finally, we set

NRG(H+)= (x E RGIH+x* E R G - H') It is clear that N R G ( H + ) is a subalgebra of RG. Furthermore, if H then obviously NRG(H+)= RG. Consider the R-linear map U H : NRG(H+)

Q

G,

EndRG((lH)G)

determined by

It is clear that

The basic properties of the map

are recorded in the proposition below, which is extracted from a work of Broue and Robinson (1986). WH

2.10. Proposition. The map W H : NRG(H+)-+ EndRG((1H)') satisfies the following properties: (i) W H is a surjective homomorphism of R-algebras and Ii'erwH =

RG . I(H ) . (ii) R N G ( H )C NRG(H+)and d H ( g ) = aH(g-') f o r allg E N G ( H ) . (iii) CRG(H) NRG(H+)and for g E G, we have WH

( T r g H ( g ) ( g )=) ( ( H n g H g - ' ) : C H ( g ) ) a H ( g - ' )

(I1)

In particular, WH(X)=

T r $ ( a H ( x * ) ) f o r all x E CRG(H)

(iv) W H ( Z ) = ( T H ( z * ) for all z E Z(RG), where (v) If C is a conjugacy class of G, then

UH

(12)

is given b y (9).

Ic n gHlaH(g)

wH((c-')+) = 9

where g runs over a complete set of representatives of ( H ,H)-double cosets of G.

401

2. Hecke algebras

Proof. (i) Given z, y E NRG(H+) and g E G, we have WH(xy)(gH+) = gH+(zy)*= gH+y*z* = wf(+JH(Y)(YH+)

(by (10)) which shows that W H is a homomorphism of R-algebras. For any x E RG', we have H+x* = ( z H + ) *and thus H+x* = 0 if and only if z H + = 0. Hence, by Lemma 2.9, we have KerwH = RG * I ( H ) . Owing to Lemma 2.2(iii), to prove surjectivity it suffices to exhibit for any given g E G an element of NRG(H+)whose image is u ~ ( g ) To . this end, choose a left transversal tl, . . . ,t , for H n gHg-' in H , and a right transversal ul, . . . ,u, for H n gHg-' in H . Setting

X one readily verifies that

= {tig-luill

5i

n}

Permutation modules

402

proving (12) for x = Tr,!&(ST')(g-l). Since, by Corollary 1.2, the set of all such x forms an R-basis of C R G ( H ) ,(12) must hold for all 2 E

CRG(H ) .

(iv) If z E Z ( R G ) and v E ( l ~ ) ' ,it follows from (10) that WH(Z)V

= vz* = z*v

and thus w H ( z ) = o~(z*). (v) To find the coefficient of aH(g) in w ~ ( ( C - l ) + ) note , that

where X(C,g) is the coefficient of g in the element H+C+ (expressed as a linear combination of the natural basis elements of RG). Hence

and the result follows. H

(Broue' and Robinson (1986)). Assume that R is an arbitrary commutative ring of prime characteristic p and H is a p-subgroup of G . If C is an H-conjugacy class of G and WH(C+)# 0, then for all g E C , C H ( g ) = H n gH9-l 2.11. Corollary.

and WH(C+)

Proof. If g

= aH(s-')

c C , then C+ = Tr&(,)(g)

and therefore, by (ll),

Since H is a p-group and charR = p , it follows that HngHg-l = C H ( g ) , as required. H

403

3. Fusion and permutation modules

3.

Fusion and permutation modules

In this section, F is an arbitrary field of prime characteristic p and G is a finite group. Our aim is to demonstrate that permutation modules ( l p ) ' , P a p-subgroup of G, hold much information about the fusion of p-subgroups of G, as well as information on FG-modules. For convenience of reference, we first recall some notation of the previous section. Let H be a subgroup of G. Then

denotes the natural homomorphism of F-algebras given by a H ( x ) = zu

for all

z E RG,u E ( 1 ~ ) '

Recall, from Proposition 2.10, that the map

defined by

is a surjective homomorphism of F-algebras and KerwH = F G - I ( H )

(by Lemma 2.9, Ii'erwH is the left annihilator of H + in FG). 3.1. Lemma. For anyp-subgroup P ofG, (i) K e r a p C_ J ( F G ) . (iz) Ii'erwp n C F G ( P )C J(CFG(P)).

Proof. (i) If z E Ii'erap, then x annihilates ( l p ) ' . Hence, by Corollary 1.8(i), x annihilates every irreducible FG-module. Thus x E J ( F G ) , proving (i). (ii) It suffices to verify that ICerwp n C F G ( P contains ) no nonzero idempotents. Our assumptions on F and P ensure that FP is local, hence any projective FP-module is free. Assume that e is a nonzero

Permutation modules

404

idempotent of CFG(P).Then eFG # 0 is a left projective (hence free) FP-module. Hence P+eFG is not zero : its dimension is the FP-rank of eFG. Thus eP+ = P+e # 0 and so e # KerwH. W 3.2. Corollary. (Broue' and Robinson (1986)). Let P be a normal p-subgroup of G. Then

C F G ( P ) / ( ~ (n~C~F~GP( P ) ) F ( C c ( P ) / Z ( P ) ) as F-algebras. In particular, C F G ( P )is local if and only if C G ( P )is a p -9ro up. Proof. If C is a P-conjugacy class of G and g E C, then by Proposition 2.10( ii), (iii),

Hence WP(CFG(P)) = W P ( F C G ( P ) )Since . the map 49-7

-+

9Z(P),

9 E CG(P),

induces an F-algebra isomorphism

the first assertion is established. The second assertion follows from the first and Lemma 3.1(ii). Following Brou6 and Robinson (1986), we say that the subgroup

H of G controls the G-fusion of its p-subgroups if whenever P is a p-subgroup of H and g is an element of G such that 9-%7

c H,

then 9 E CG(P)H

(Broue' and Robinson (1986). Let H be an arbitrary subgroup of G. Then

3.3. Theorem.

WH(CFG(H))= EndFG((1H)')

(1)

3. Fusion and permutation modules

405

if and only if H controls the G-fusion of its p-subgroups.

Proof. It follows from Proposition 2.lO(iii) and Lemma 2.2(ii), (iii) that the equality (1) holds if and only i f (a) For any g E G, there exists g1 E HgH such that ((Hng1Hg;') : C H ( ~ is~ not ) ) divisible by p. But if gl = h'gh for h, h' E H , then

and thus (a) is equivalent to: (b) For any g E G, there exists h E H such that ( ( H n 9Hg-l) : C ~ ( 9 h - l )is) not divisible by p. We now claim that (b) is equivalent to: ( c ) For any g E G and any Sylow psubgroup P of H ngHg-l, there exists h E H such that P CH(gh-l). It is clear that (c) implies (b). Conversely, assume that (b) holds. Then, given g E G and a Sylow p-subgroup P of H n gHg-', there exists h E H such that CH(gh-') contains hlPh,' where hl E H n 9Hg-l. If hl = gh29-l with h2 E H , then

which proves (c). By the foregoing, we are left to verify that ( c ) holds if and only if H controls the G-fusion of its p-subgroups. First assume that (c) holds. Let P be a psubgroup of H and let g-'Pg C H for some g E G. Then P C H n gHg-'. Denote by Q a Sylow p-subgroup of H n gHg-' containing P. By (c), we may find h E H such that Q 2 C~(9h-l) Setting z = gh-', we then have z E Cc(Q). Hence z E CG(P) and g = z h proving that H controls the G-fusion of its p-subgroups. Conversely, assume that H controls the G-fusion of its p-subgroups. Let g E G and let P be a Sylow p-subgroup of H n gHg-'. Since g-'Pg G H , we then have g = z h for some z E C c ( P ) , hE H . Since z = gh-', we conclude that P C CH(gh-'), proving (c) and hence the result. H

Permutation modules

406

3.4. Corollary. Let P be a p-subgroup of G which controls the

G-fusion of its subgroups. Then ( 1 ~is)indecomposable ~ if and only if CFG(P)is local.

Proof. This is a direct consequence of Theorem 3.3 and Lemma 3.l(ii). We say that a group G is p-nilpotent if G has a normal p'-subgroup N , called the normal p-complement, such that GIN is a p-group. Thus G is p-nilpotent if and only if G = N S where N is a normal p'-subgroup of G and S is a Sylow p-subgroup of G. We close by providing a proof, due to Robinson (1988), of an extension of a classical result, known as Frobenius normal p-complement theorem. This extension provides a characterization of p-nilpotent groups by the surjectivity of the map US

: CFG(S)

EndFG((lS)G).

3.5. Theorem. Let S be a Sylow p-subgroup of G. Then the following conditions are equivalent: (i) Whenever two elements o f S are conjugate in G, they are conjugate in s. fii) Whenever Q is a p-subgroup of G, NG(Q)/CG(Q)is a p-group. (iii) Whenever Q is a subgroup o f G with Q, g-lQg S, then g = cs where c E C c ( Q ) and s E S. (iv) US(CFG(S))= E n d F G ( ( l s ) G ) . (v) G is p-nilpotent.

Proof. (i) + (ii): Assume by way of contradiction that (i) holds, but that (ii) does not. Then we may find a subgroup Q of S of minimal order subject to Nc(Q)/CG(Q)is not a p-group. Let z E N G ( & ) - CG(Q) be a p'-element. Then, by (i), we must have

Because S is nilpotent, [Q,S]2 Q, so

407

3. F'usion and permutation modules

Due to our choice of Q , NG([Q,z ] ) / c ~ ( [ 21) & , is a pgroup and thus [ Q , s , x ]= 1. Hence, by Theorem 5.3.6 in Gorenstein (1968), we must have [ Q , x ]= 1, a contradiction. (ii) +(iii): We argue by induction on ( S : Q). If Q = S , then N c ( Q ) = QCG(Q)and (iii) holds. Now assume that Q c S and that (iii) holds for subgroups R of S such that ( S : R ) < (S : Q). Choose g E G with gQg-' c S and put U = N s ( Q ) . Then U 3 Q. Let T be a Sylow p-subgroup of NG(Q) with U C T and let V be a Sylow p-subgroup of G with T V . Since z - l V z = S for some x E G, we have U,z-'Uz S,and by the induction hypothesis, x = cs for some c E CG(U),SE S. Then z-'Qz = s-'Qs and Ns(z-lQz) 2 z-lTz, so Ns(s-lQs) is a Sylow p-subgroup of NG(s-'&s) and N s ( Q ) is a Sylow p-subgroup of NG(Q).Similarly, Ns(gQg-l) is a Sylow psubgroup of NG(gQ9-l). Then gNS(g-'Qg)g-l is a Sylow p-subgroup of NG(Q),so

c

S , so by the for some h E NG(Q). Thus Ns(Q),g-lh-'Ns(Q)hg induction hypothesis, hg = clsl for some c1 E C G ( N S ( Q ) ) , SE~ S. Then

gs-' E NG(Q)= CG(Q)NS(Q) since N s ( Q ) is a Sylow p-subgroup of N G ( Q ) and NG(Q)/CG(Q)is a p-group. Thus g E C G ( Q ) S ,proving (iii). (iii) +(iv): Let T be a complete set of double coset representatives for (S,S) in G. By Lemma 2.2, { a s ( t ) l t E 2') is an F-basis of E n d F G ( ( l s ) G ) . Let us first show that for any z E G, there exists c E SzS such that n CSC-* = c,(c) (2)

s

Let Q = S n x5'z-l. Then Q, s-lQx some c E C c ( Q ) , u E S. Then

Sncsc-l

S , so we may write x = cu for

=

sn~ss-l= Q

c

cs(c)

S~CSC-~,

proving (1). Let C be the S-conjugacy class of c, and let X be a left transversal

Permutation modules

408

for S n cSc-l in S. Then

c+s+= c scs+ = (SCS)+ SEX

and therefore a&')

= ws(C+),

proving (iv), by applying (1). (iv) j (v): We first note, that by Lemma 1.3(v), 1~ is a direct summand of ( 1 ~ which ) ~ corresponds to the idempotent, say e, of E n d ~ G ( ( l s )with ~ ) e(S+) = ( G : S)-'G+. By hypothesis, there is an idempotent f of C F G ( S )with w s ( f ) = e. Then

and therefore

S+f = fS+ = ( G : S)-'G+

Now f E CFG(S)and f F G is a direct summand of FG as an FSmodule. Hence f F G is a projective left FS-module, so is free (since S is ap-group). Because S + f F G = FG+, it follows that d i r n ~ ( f F G= ) IS[. Similarly dirnF(FGf) = ISI. On the other hand,

as a left FS-module. Thus f F G f is a projective left FS-module, and d i r n ~ ( f F G f2) ISl. Hence f F G f = f F G = FGf and

proving that f E Z(FG). Since dirnF(FGf) = IS1 and any direct summand of FG f is projective, FGf is an indecomposable left FG-module. Hence f is a primitive idempotent of FG. Let x be a p'-element of G of order n, let X =< 2 > and let e, = n-'X+. Then e: = ex and e,fG+ = exG+ = G+, so e,f # 0. Because f is primitive in FG with e,f = fe,, we have e,f = f . But f E Z ( F G ) and xe, = e, so that xu = u for all u E FGf. Because FGf is a free left FS-module, no nontrivial p-element of G acts as the identity on FGf and thus G has a normal pcomplement,

4. Complete reducibility of ( 1 ~ ) ~

409

as required.

(v) + (i): Let N be a normal p-complement of G. Assume that x,g-lxg E S for some g E G. Since G = S N , we have g = s u for some s E S,u E N . Then z - l g -1 xg = 5 - ~ ( s - ~ 5 s ) ( s - ~ 5 s ) - ~ u - ~ ( s - 1 5 s ) u and ( ~ - ~ x s ) - ~ u - ~ ( s - ~E xSs n ) uN = 1 Thus g - l s g = S-'ZS

and the result follows.

Corollary. Let S be a Sylow p-subgroup of a p-nilpotent group G. Then 3.6.

CFG(S)/(CFG(S) n F G - I ( S ) )2 E ~ ~ F G ( ( ~ s ) ~ ) Proof. Direct consequence of Theorem 3.5 and Proposition 2.10. Complete reducibility of ( 1 ~ ) ~

4.

Let F be a field of characteristic p > 0 and let P be a psubgroup of a finite group G. By Corollary 1.8, each irreducible FG-module is a homomorphic image of (lp)'. It is therefore appropriate to investigate is completely reducible. circumstances under which ( 4.1. Proposition. The following conditions are equivalent: (i) ( 1 ~ is) completely ~ reducible. (ii) J ( F G ) FG - I ( P ) , where I ( P ) is the augmentation ideal of

FP.

Proof. If V = F P / I ( P ) , then V '2.2.11 (i),

!Z l p

and, by Proposition

VG FGfFG * I ( P ) Hence

V G is completely reducible if and only if J ( F G ) 5 F G . I ( P ) ,as

required. Our next observation shows that complete reducibility of (lp)' depends only on the characteristic p of the field F .

Permutation modules

410

Let E be a field extension of F. Then ( 1 ~ p ) ~ is completely reducible i f and only if ( 1 . g ~ is ) ~completely reducible. 4.2. Proposition.

Proof. By Proposition 4.1, it suffices to show that J(FG) E F G I ( P ) if and only if J(EG) EG I ( P ) . We know, from Corollary 6.1.10, that J(EG) = E @F J(FG) Hence, if J(FG) C F G - I ( P ) , then J(EG) C EG . I ( P ) . Conversely, assume that J(EG) E EG - I ( P ) . Then, by Lemma 2.9,

J ( F G ) 5 EG * I ( P )n F G = FG * I ( P ) as required.

Let U and V be finitely generated FG-modules. Recall that the intertwinning number for U and V , written i ( U , V ) is defined to be

i(U,V ) = dirnFHomFc(U,V ) It is clear that

Let { K, . . . ,V,} be a full set of nonisomorphic irreducible FG-modules. Given an FG-module V , we write

k=l

if

vk

appears ak times as a composition factor of V .

4.3. Lemma. Let V x c;=,akVk and W M C;=,b k v k be two FG-modules, where F is assumed to be a splittingfield for FG. Then (i) i(V,W ) 5 EL=,akbk and, in particular, z(V,V ) 5 a:. (ii) V is completely reducible i f and only i f i(V,V ) = a:. Proof. (i) We argue by induction on the value of C;=,ak. If C;=,ak = 1, then V is irreducible and so V % V, for some j E


411

{ 1,. . . ,r } . Therefore

and so, by (l),

Now assume that Vo is a submodule of V . By looking at the exact

sequence

we deduce that

If Vo is a nontrivial submodule of V with

then by the induction hypothesis

and

c T

i(V0,W)I

(4)

k=l

as required. (ii) If V is completely reducible, then by (1) i(V,V ) =

C;=,

Permutation modules

412

since z(vk,vk) = 1 for each k E { 1,. . . ,r } . Conversely, assume that a: and write i ( V ,V ) = f

T

VIsoCv M k=l

ikv,,s o c v M

alvk

k=l

Then, by (i), we have

k=l

and

i ( s o c v , v)5

c T

a$ak

k=1

Hence, applying (2), we derive

and i(S0C v,v )=

c T

aiak

k=l

Bearing in mind that r

i ( S o c V , V )= i ( s o c v , s o c v ) = c ( a ; ) z k= 1 it follows from (6) that a” - ak if a! # 0. kll Finally, assume that ak = 0 and ak # 0. Then ak = a; # 0 and so, by ( 5 ) , i ( V / S o c V , V ) # 0. Hence there exists a submodule L with SOCV G L G V such that V / L is isomorphic to a submodule of V . The latter implies the existence of j E (1,. . . , r } with us # 0 and a; # 0.

413


But if a; # 0, then aj = a; and so a; = 0, a contradiction. Hence a k = a; for all k E { 1,. . . , r } and thus V = Soc V as required. W

(Sabonov (1971)). Let F be a splitting field for FG, let { &, . . . ,V,.} be a full set of nonisomorphic irreducible FGmodules and, for each k E (1,. . . ,r } ) let U k = dimFP(Vk). If d p is the number of double ( P ,P)-cosets of G, then 4.4. Corollary.

T

c ( u k / I p 1 ) 22

dP

k=l

with equality if and only if (lp)' is completely reducible. Proof. By Corollary l.g(iii), we have (IP)'

k(uk/lPl)vk

k=

1

On the other hand, by Lemma 2.2, i ( ( l p ) G ,( 1 ~ )=~d p) . The desired conclusion is therefore a consequence of Lemma 4.3. W 4.5. Lemma. Let A be a finite-dimensional algebra over a field

and let V be a finitely generated A-module. Then

c

I = (9 E EndA(V)lY(V) J ( A ) V } is a nilpotent ideal of EndA(V).

Proof. It is clear that I is a subspace of the finite-dimensional

7

Permutation modules

414

and so cp" = 0. Thus I is nilpotent, by virtue of Theorem 1.5.12. H 4.6. Proposition.

For any p-subgroup P of G, the following con-

ditions are equivalent: (i) ( is completely reducible. (ii) EndFG( ( 1 ~ )is~semisimple. ) (iii) N&P+)/FG I ( P ) is semisimple. Proof. The equivalence of (ii) and (iii) is a consequence of Proposition 2.10. Since (i) obviously implies (ii), we are left to verify that (ii) implies (i). Put V = ( 1 ~and ) ~assume that J ( F G ) V # 0. Then J ( F G ) V has an irreducible submodule W . By Corollary 1.8(i), there is a surjective FG-homomorphism cp : V + W

J(FG)V

Thus cp is a nonzero element of EndFG(V) such that cp(V) C J ( F G ) V . Hence, by Lemma 4.5, Endf;.G(V)is not semisimple, as required. H 4.7. Theorem. (Saksonov (1971)). Let P be a p-subgroup of G, let r be the number of p-regular classes of G and let d p be the number of double ( P ,P)-cosets in G. Then d p 2 r and, under the assumption that F is a splitting field of FG, the following conditions are equivalent:

(i) d p = r. (ii) ( 1 ~ ) V, ~@ V, @ - - @ V,, where {h, . . . ,V , } is a full set of nonisomorphic irreducible FG-modules. (iii) P is a Sylow p-subgroup of G, ( 1 ~ is) completely ~ reducible and dimFP(K) = IPl for all i E {l,.. . , r ] . Proof. Let F be an algebraically closed field of characteristic p . Then, by Theorem 3.6.13, there exist exactly r , say V,,. . . , V , of nonisomorphic irreducible FG-modules. Put V = ( 1 ~and ) ~write V / J ( F G ) V E $;=,akVk, Soc V Then we have

E $;=lbkVk

(7)

7-

i ( V / J ( F G ) VSOC , V )=

C akbk

k=l

(8)


415

where each ak 2 1, by Corollary 1.8(i). Note also that, by Corollary 2.2.7, bk. = q k . , SOCV) = i(Vk,V ) = i ( ( & ) p , lp) # 0 and hence

Ckzla k b k 2 r .

Since

and, by Lemma 2.2, d i m F E n d F G ( V ) = d p , it follows from (8) that dp

2 T.

In what follows, we assume that F is a splitting field for FG and prove the equivalence of (i), (ii) and (iii). (i) j (ii): Assume that d p = r . Then, by the above,

and ak

for all k E {1,...,r }

= bk = 1

(9)

Hence, by (7), V / J (FG)V 2 Soc V and so E n d ~ cV( ) 2 E n d ~ cSoc ( V). But Soc V is completely reducible, hence E n d ~ c ( Vis>semisimple and therefore, by Proposition 4.6, V is completely reducible. Thus V = SOCV and the required assertion follows by virtue of (7) and (9). (ii) +(iii): Let Q be a Sylow p-subgroup of G containing P. Then, by Corollary 1.8, Ci==ldimV, 5 (G : Q ) . On the other hand, by hypothesis, P

C d i m V , = ( G : P ) 2 ( G :Q ) i=l

Hence Q = P and so P is a Sylow psubgroup of G. The remaining assertion follows from Corollary 1.8(iii). (iii) +(i): If u k = d i m ~ P ( V k )then , (uk/lPI) = 1 for all Ic E { 1, . . . ,r } . Since ( l p ) G is completely reducible, it follows from Corollary 4.4 that T

dP

= c(%/l= p f ,I )2 k=l

as required.

Permutation modules

416

5.

Induction f r o m Sylow p-subgroups

Throughout F denotes a field of characteristic p > 0 and Syl,(G) the set of all Sylow psubgroups of G. Given P E Syl,(G), our aim is to examine circumstances under which ( l p ) G is completely reducible. Following Motose and Ninomiya (1975), we say that G is p-radical if (lp)' is completely reducible. Note that, by Proposition 4.2, the notion of a p-radical group depends only upon the characteristic p of a field F (and not on the field F itself). For convenience, we divide this section into two subsections.

A. General results In what follows, given a subgroup H of G , we write I ( H ) for the augmentation ideal of F H . If X is a subset of G, then X + E F G is defined to be the sum of all elements in X . Our aim is to establish some general results on p-radical groups and provide their various characterizations.

5.1. Lemma. Let H be a subgroup of G, let I be a nilpotent lefl ideal of FH and, for each g E G, let 1 9 = g-'Ig. Then J = ngEGFG-Ig is a nilpotent ideal of FG. Proof.

Given x E G,we have

proving that J is an ideal of FG. It follows that for any integer rn 2 1, Jmt'C - Jm(FG.I)= Jm-I

and hence, by induction,

417

5 . Induction from Sylow psubgroups

for any integer rn 2 0. Because I is nilpotent, we conclude that J is a nilpotent ideal of FG. H

5.2. Corollary.

Let P be a Sylow p-subgroup o f G . Then

is a nilpotent ideal o f F G .

Proof. Note that, for any given g E G , I(P)g = I(P9). Since I ( P ) is a nilpotent ideal of F P , the result follows by virtue of Lemma 5.1. H

We are now ready to provide the following characterizations of pradical groups contained in the work of Motose and Ninomiya (1975). 5.3. Theorem.

The following conditions are equivalent:

(i) G is p-radical. (ii) For any subgroup H ofp'-index in G , J ( F G ) FG - J ( F H ) . (iii) J ( F G ) C FG . I ( P ) for some (and hence all) P E SyZ,(G). (iv) J ( F G ) = nSESyl,(G)FG ' I ( s ) * Proof. The equivalence of (i) and (iii) is a consequence of Proposition 4.1. Since (ii) obviously implies (iii), we are left to verify that (iii) =+(iv) and (iv) =+(ii). (iii) +(iv): If J ( F G ) E FG . I ( P ) for some P E Syl,(G), then

J ( F G ) = J ( F G ) g FG - I ( P ) g= FG - I ( P g ) for all g E G and therefore

This proves (iv), by applying Proposition 5.2. (iv) + (ii): Let H be a subgroup of G of p'-index. Then there exists S E Syl,(G) such that S H . By hypothesis, J ( F G ) F G I ( S ) , so V G is completely reducible for any irreducible FS-module V , by Theorem 2.3.7. Since V c E (VH)",it follows from Corollary 2.2.10

-

Permutation modules

418

that V H is completely reducible, for every irreducible FS-module V . Again, by Theorem 2.3.7, J ( F H ) 2 F H . I ( S ) and therefore, by the implication (iii)j (iv),

Thus

as required. 5.4. Corollary. Let H be a subgroup of G of p'-index.

If G is

p-radical, then so is H . Proof. Let S be a Sylow p-subgroup of G with S H . Then, by the proof of the implication (iv) =+ (ii) of Theorem 5.3,

J ( F G ) C FG * I ( S ) j J ( F H )

c FH

*

I(S)

Hence, by Theorem 5.3, if G is pradical, then so is H . The next result, with the exception of the first property of (i), is due to Khatri (1973). 5.5. Theorem. Let N be a normal subgroup of G. (i) If G is p-radical, then so are N and G I N . (ii) If N is a p-group and GIN is p-radical, then G is p-radical. (iii) If GIN is a p'-group, then G is p-radical if and only if N is

p-radical. Proof. (i) Suppose that G is p-radical and let P E Syl,(G). We may choose a Sylow p-subgroup Q of N with Q C P. By Corollary 2.4.3, J ( F N ) 5 J ( F G ) and so, by Theorem 5.3,

J(FN)

c F N n FG - I ( P )

5 . Induction from Sylow psubgroups

I f x E FN

419

n F G . I ( P ) ,then by Lemma 2.9,

But x Q + and x ( P - Q)+ have disjoint supports, so z&+ = 0 and thus, by Lemma 2.9, z E F N I ( & ) . Hence J ( F N ) F N . I ( & ) and so, by Theorem 5.3, N is pradical. I f H = P N , then H is of $-index in G, so by Theorem 5.3,

-

c

J ( F G ) C FG * J ( F H ) Hence, by Theorem 2.3.7, V Gis completely reducible for all irreducible FH-modules V . In particular, ( 1 ~is )completely ~ reducible. Hence ( ~ H I N is) completely ~ ~ ~ reducible. But HIN is a Sylow p-subgroup of G I N , so GIN is pradical. (ii) Assume that N is a normal psubgroup of G such that GIN is pradical. Since N C P E Syl,(G), we have HIN = P I N . By hypothesis, ( l p / ~ ) is~ completely / ~ reducible. Thus ( 1 ~is )completely ~ reducible and G is p-radical. (iii) Assume that GIN is a #-group. If G is p-radical, then so is N , by virtue of Corollary 5.4. Conversely, suppose that N is p-radical. Since GIN is a p'-group there is a P E Syl,(G) with P 5 N . By Theorem 5.3, J ( F N ) F N - I ( P ) . On the other hand, by Theorem 2.10.14, J ( F G ) = FG J ( F N ) . Hence

J(FG)

FG * F N - I ( P ) = FG - I ( P )

and so G is pradical, by applying Theorem 5.3.

(Khatri (1973)). If G = G1 x Gz, then G is p-radical if and only if so are G1 and G2. 5.6. Theorem.

Proof. Suppose that GI and G2 are pradical. Then, by Theorem

5.3,

J(FGi)

FGi . I(Pj)

where P; E SyZ,(Gi),i = 1,2. As is well known (see Motose (1974a)),

Permutation modules

420

and so

+

J ( F G ) FG - I ( Pi) FG - I ( P2) If we put P = Plx P2,then I(Pi) C P and so by the above J ( F G ) C FG * I ( P ) Consequently, G is p-radical, by Theorem 5.3. The converse being a consequence of Theorem 5.5(i), the result follows.

If X is a subset of G, we write r ( X ) = ~ G ( Xand ) l ( X ) = IG(X) for the right and left annihilators of X in FG, respectively. 5.7. Lemma. Let H be a subgroup of G , let I be a left ideal of FH and let J = flgEGFG- 19. Then (i) J = IG{&GrH(I)’FG}. (ii) ~ G ( J=) CgEG r~(1)gFG. (iii) If I is an ideal of F H , then J = rlgEGIgFG.

Proof. (i) By Theorem 3.10.11, I = ZH(TH(I))which implies that

FG * I = ZG(TH(I))FG since FG is a free FH-module. Hence, if g E G , then

It follows that J

= ngEG1G({rH(I)}’FG)

as required. (ii) Applying (i) and Theorem 3.10.11, we have rG(J)

=

fG{lG(

rH(1)’FG))

’a

= xr~(l)’FG, ’€G

5 . ,Induction from Sylow p-subgroups

421

as asserted.

(iii) By Theorem 3.10.11, the left and right annihilators of an ideal in FG (or F H ) coincide. Applying (i), we therefore derive

n,EGIgFG, as desired. W 5.8. Lemma. Let H be a subgroup of G. Then for all g E (2) F G * I ( H )= lc(H+)= {C zgg E FGI C h E H z g h = 0

GI. (ii) I ( H ) - F G= r c ( H + )= { C g E G zgg E FGI

C h E Hz h g

=0

for all g E

GI. Proof. The first equality of (i) is a special case of Lemma 2.9. The second equality is a consequence of the fact that ( C z g g ) H += 0 if and only if ChEH X g h = 0 for all g E G. The proof of (ii) is similar. 5.9. Lemma. Let P be a Sylow p-subgroup of G. Then

ngE~FG * I(Pg) =

x x { axzl

xEG

=

{C

axy = 0 for all

z

E G , S E Syl,(G)

YES

aXzl

xEG

C ayx= 0

for all z E G , S E Syl,(G)

YES

Proof. Put X = ng,GFG- I ( P g )and Y = ngEGI(Pg)FG.Then, by Lemma 5.7(iii),

x =y =

nS€SyZ,(G)FG

'

I(s)= n s ~ S y t , ( G I) ( S ) * FG

and so the desired conclusion follows by virtue of Lemma 5.8.

Permutation modules

422

We are now ready to provide some further characterizations of pradical groups.

Proof. The equivalence of (i),(ii),(v) and (vi) follows from Lemmas 5.9 and 5.7(iii), and Theorem 5.3. That (ii) is equivalent to (iii) follows by taking annihilators. Similarly, since (ii) is equivalent to J ( F G ) = nSESylp(G)FGI ( S ) ,it follows that (ii) is equivalent to (iv) by taking annihilators. W

B. S o m e classes of p-radical groups.

A transitive permutation group G in which only the identity element fixes more than one letter, but the subgroup fixing a letter is nontrivial, is called a Frobenius group. In what follows we shall use the following standard facts for the proof of which we refer to Gorenstein (1968). (a) (Frobenius). Let G be a Frobenius group and let H be the subgroup fixing a letter. Then the subset of G consisting of the identity element together with those elements which fix no letters, forms a normal subgroup N of G of order (G : H ) . (b) (Thompson). Let a group G admit an automorphism of prime order which fixes only the identity element of G. Then G is nilpotent. We shall refer to the normal subgroup N and the subgroup H of G in (a) as the Frobenius kernel and Frobenius Complement, respectively. 5.11. L e m m a . Let G be a Frobenius group with complement H and kernel N . Then

5 . Induction from Sylow p-subgroups

423

(i) G = N H with N n H = 1. (ii) Every h # 1 in H induces b y conjugation an automorphism of N which fixes only the identity element of N . (iii) I HI divides IN I - 1. (iv) N is nilpotent. (9) If g E G - N , then g lies in a conjugate of H . (vi) For every x # 1 in N , C G ( ~ ) N . Proof. (i) It follows from the definition of N that N n H = 1. Since, by (a), IN1 = (G : H ) , we must have IGI = (NHI. Hence G = N H , as asserted. (ii) Because G is transitive and H is the subgroup fixing a letter, our representation is equivalent to that on the cosets of H . Owing to (i), we can take the elements of N themselves as a transversa1 for H in G. Assume that h-lnh = n for some 1 # h E H and 1 # n E N . Then it is immediate that h fixes the coset H n as well as the coset H . But, by definition of a Frobenius group, only the identity fixes more than one letter. Thus h-lnh # n for any 1 # h f H and 1 # n f N , proving (ii). (iii) For a fixed 1 # n E N , the set rn = {h-lnhlh E H } must consist of m = (HI distinct elements of N . But clearly for x,y in N = or n = 0. Thus with x # l , y # 1, we have either IN - {1}1 is a multiple of m and (iii) follows. (iv) Because H # 1, we may find an element h E H of prime order. By (ii), h induces an automorphism of N which fixes only the identity element of N . Thus, by (b), N must be nilpotent. (v) If g E G - N , then g fixes a letter. But the conjugates of H are simply the subgroups fixing a letter. Thus g lies in a conjugate of H . (vi) Owing to (v), N consists precisely of the elements of G which lie in no conjugate of H - (1). Thus CG(x) N for 1 # x E N , as asserted.

ry rX rY

5.12. Lemma. Let F be a field of characteristic p > 0 , let N be a normal p'-subgroup of G and let e = 1NI-l EXEN x. Then e is a central idempotent of F G such that FGe F ( G / N ) .

Permutation modules

424

Proof. It is clear that e is a central idempotent of FG. Let f : FG

+ F(G/N)

-

be the natural homomorphism. Since K e r f = F G I ( N ) and F G = FGe @ F G ( l - e ) , it suffices to verify that FG I ( N ) = F G ( l - e). Because f ( e ) = 1 , we have 1 - e E K e r f and so F G ( l - e ) F G . I ( N ) . Given n E N , we also have ne = e so that ( n - l ) e = 0 and F G . I ( N ) e = 0. Thus F G . I ( N ) F G ( l - e ) and the result is established.

-

s

5.13. Lemma. Let F be a field of characteristic p > 0, let N be a normal p'-subgroup of G and let e = IN[-' CnEN n. If H is a subgroup of G such that G = N H and N n H = 1 , then (i) J ( FG)e = J ( F H ) e . (ii) dimf;.J(FH)e = dim^ J ( F H ) .

Proof. By Lemma 3.18.1 J ( F G e ) = J ( F G ) e , while by Lemma 5.12, FGe 2 F ( G / N ) 2 F H . The latter implies that

{

F H + FGe x H x e

is an isomorphism of F-algebras. Hence J ( F G e ) = J ( F H ) e and

J(FG)e = J(FGe) = J(FH)e, J ( F H )% J(FH)e proving (i) and (ii).

(Motose (1974b)). Let G be a Frobenius group with complement H and kernel N and let F be a f i e l d of characteristic p > 0 . (i) I f p divides IN1 and P is a Sylow p-subgroup of N (and hence of G , by Lemma 5.11(iii,)), then J ( F G ) = F G - I ( P ) (ii) If p does not divide INJ, then 5.14.

Proposition.

J ( F G ) = J ( F H ) e and d i m F J ( F G ) = dirnFJ(FH) where

5 . Induction from Sylow p-subgroups

425

Proof. (i) Suppose that p divides IN1 and let P be a Sylow psubgroup of N . We know, from Lemma 5 . 1 1 ( i ~ that ) ~ N is nilpotent. Thus P d N, so P d G and therefore J ( F G ) = F G I ( P ) . (ii) Assume that p does not divide IN[. Owing to Corollary 6.1.10, we may assume that F is algebraically closed. Note that F N = F N e @ F N ( 1 - e) and FNe 2 1 ~ Hence . F N ( l - e) is a direct sum of nontrivial irreducible FN-modules. Therefore by Theorem 2.8.8(i) and Lemma 5 . 1 1 ( ~ i )F~G ( l - e) E ( F N ( 1 - e))G is completely reducible. Thus J ( F G ) ( l - e) = 0 and so J ( F G ) = J(FGe). Now apply Lemma 5.13. H Turning to p-radical groups, we now prove our first major result. 5.15. Theorem. (Khatri 19?'3), Motose and Ninomiya (1975)). Let G be a Frobenius group with kernel N and complement H and let F be a field of characteristic p > 0. (i) If p divides the order of N , then G is p-radical. (ii) If p divides the order of H, then G is p-radical if and only if so is H.

Proof. (i) By Lemma 5.11(iv), N is nilpotent and so, by Theorem 5.3(iii), N is p-radical. But, by Lemma 5.ll(iii), G/N S H is a p'-group. Hence, by Theorem 5.5(iii) G is p-radical. (ii) Assume that H is pradical. By Theorem 5.5(i), it suffices to show that G is p-radical. If P is a Sylow p-subgroup of H, then J ( F H ) F H . I ( P ) by Theorem 5.3. By Proposition 5.14(ii), J ( F G ) = J ( FH)Nt and therefore J(FG)

FH * I(P)(N+)C F G . I ( P )

Since P is a Sylow p-subgroup of G, it follows from Theorem 5.3 that G is pradical, as required. H The next result provides another class of p-radical groups. 5.16. Theorem. (Tsushima (1986)). Let G be a p-nilpotent group with abelian p-complement N. Then G is p-radical.

Permutation modules

426

Proof. By Proposition 4.2,we may harmlessly assume that F is algebraically closed. Let P be a Sylow psubgroup of G and let

be a decomposition of 1 as a sum of orthogonal primitive idempotents of F N . Because F is a splitting field for F N and N is abelian, each e;FN is one-dimensional. Hence eiFG (e;FN)G is of a dimension lPl. Now G = N P , so FG = ( F N ) ( F P )and thus

e,FG = ( e ; F N ) ( F P )= e;FP It follows that the map

FP

+

z

H

e;FG e;z

is a surjective homomorphism of right FP-modules of the same Fdimension. Consequently,

FP

E e;FG

as FP-modules

(1 5 i

5 n)

and therefore e ; F G . I ( P )is a unique maximal FP-submodule of eiFG. But e;J(FG)is a proper FP-submodule of e;FG, hence

e;FG I ( P ) 2 eiJ(FG)

(I 5 i 5 n )

and thus FG - I ( P ) 2 J ( F G ) . Hence, G is p-radical, by applying Theorem 5.3. H 5.17. Remark. (i) Any p-radical group is p-solvable (see Okuyama (1986)). (ii) It can be shown (see Tsushima (1986)) that a p-nilpotent group is p-radical if and only if [O,! ( G ) ,D]nCo,,(G)(D)= 1 for any p-subgroup D of G (here O,t(G) is the maximal normal p’-subgroup of G ) . 6.

Loewy series for transitive permutation modules

Throughout this section, F denotes a field of characteristic p > 0, H a subgroup of a finite p-group P and P / H the set of all left cosets of

6. Loewy series for transitive permutation modules

427

H in P . We denote by F ( P / H ) the F-linear span of all left cosets of H in P . Since P acts on the set P / H by left multiplication, it is clear that F ( P / H ) is a permutation FP-module which is isomorphic to ( 1 ~ ) To ~ .state our main objective, we first record the following piece of information. Let A be a finite-dimensional algebra over a field and let V # 0 be an A-module. Recall from Sec.18 of Chapter 3 that the descending

chain

V 2 J ( A ) V 2 J ( A ) 2 V2

* * -

of submodules of V is called the lower Loewy series of V . The upper

Loewy series

0 = So(V) G S'(V)

c S2(V) c ...

= Soc(V/S,-l(V)) for n > 0. The Loewy length of V is defined to be the integer k 2 1 such that J(A)"-'V # 0 and J ( A ) k V = 0. The reader may readily verify that

of V is defined by S,(V)/S,-l(V)

V 3 J ( A ) V 2 . * . 3 J(A)k-'V 3 J(A)kV = 0 and

0 = & ( V ) c S,(V)

c

* *

a

c &(V)

=

v

and

Sn(V)= {v E VIJ(A)"v = 0}

( n > 0) Our aim in this section is twofold: first to provide a basis for each I ( P ) " F ( P / H ) / I ( P ) " + ' F ( P / H ) and second to prove that the upper and lower Loewy series for F ( P / H ) coincide. Here and in what follows I ( P ) denotes the augmentation ideal of F P (which coincides with J ( F P ) since P is a p-group and churF = P ) . All the information recorded below may be found in Huppert and Blackburn (1982) or Karpilovsky (198713). The Bruuer - Jennings - Zussenhuus series for P is defined inductively by M'(P) = P and for n 2 2,

M 7 p ) = [Mn-I(P),P]M(P)p where i is the smallest integer satisfying ip 2 n and M,(P)p is the subgroup generated by all X P with x E M ; ( P ) . The subgroups M,(P)

428

Permutation modules

satisfy the following properties: (i) [Mi(P),Mj(P)IG M+j(P) (4.i 2 1). (ii) Mi(P)/Mj+l(P) is elementary abelian ( i 2 1). For each i 2 1, choose elements x;, of P (with s running over an initial sequence of positive integers, depending on i ) lying in M i ( P ) such that the cosets of M;+1(P)which contain them, form a basis for M;(P)/Mi+l(P).If P has order p n , then there are p" products, lexicographically ordered, of the form

where the product with all exponents zero is taken to be the identity element of P . The weight of such a product is defined to be C ia;s. A classical theorem of Jennings asserts that the products of weight rn lie in I(P)" and form a basis of I(P)" modulo I(P>"+l. We begin by proving the following modification of Jennings theorem due to Alperin (1988). 6.1. Theorem. Suppose that we choose an arbitrary linear order for the xis (instead of the lexicographing ordering). Then the p" products

have the property that the ones of weight m lie in I(P)" and form a basis of I(P)" modulo I(P)"+' (xi8 - 1 still has weight i, ( x i s - l)ais has weight iajs, and the weight of a product, no matter what the order of the factors, is the sum of the weights of the factors).

Proof. Since x;,- 1 E I ( P ) ,it is clear that the products of weight rn lie in I(P)". We now claim that every element of P is expressible (and hence uniquely so) as a product nx;;, with the factors in the given order. Choose j with M j ( P ) # 1 and Mj+l(P>= 1. Since the subgroups M i ( P ) are invariant under epimorphisms, it follows by induction on j and passage to P / M j ( P ) ,that if g E P then g = x;js .z with z E M j ( P ) , 1 5 i 5 j - 1. Bearing in mind that z can be expressed as a product of factors xjs and these are central, the claim is established. By the foregoing, the p" products nx;ia (in the given order) span F P . We next claim that this implies that the pn products n ( x i , - l ) a i a

n


429

span F P . Indeed, if u1,. . . ,u, are elements of any ring, then it is easily proved by induction that u1 . . . u, - 1 = C ( U i l - l)(ua2- 1) *

-

*

(Uik -

1)

where the sum is over all nonempty subsets I = {il, . . . ,ik} of { 1,.. . ,T } and il < iz < ... < ik. Thus, since the product n ( x i S - 1)' has been taken as 1, it follows that the product nx;:, can be expressed unitri- l ) a i s , as claimed. angularly in terms of the products

n(q,

We shall conclude the proof by demonstrating that the products n(xc;, - l ) a i s of weight rn are a basis of I(P)" modulo I(P)"+'. We argue by downward induction on rn. If I ( P ) k + l = 0 and I ( P ) k # 0, then I ( P ) k c_ SocFP = {ACT E PIX E F } X

and hence dirnFI(P)k= 1. We also have n(z:,-' - 1) E I ( P ) k since k is the product of p - 1 and the sum of the products of i and the number of subscripts s for that i. This product is not zero, since otherwise the remaining pn - 1 products would span F P , (which is impossible) so the assertion holds for rn = k. Assume that our assertion holds for integers greater than rn. Note that the number of our products of weight rn is the same as the number of products of weight rn in lexicographic ordering. Hence if the products in the new ordering are linearly independent modulo I ( P)"+', then they form a basis for I(P)"/I(P)"+'. If these products of weight rn are linearly dependent, then one of them is a linear combination of the others plus (by the induction hypothesis) the products of weight > rn. But this implies that the pn - 1 products (obtained by deleting the expressed product) span F P , which is impossible. The proof is therefore complete. For each i 2 1, let us choose elements x;, in a particular manner: select sets of elements y;, and z;, whose union will be the x;, but with the y;, chosen so that their images in M ; ( P ) / ( Hn M;(P))M;+l(P) are a basis of that group and so that the z;, lie in ( H n M ; ( P ) ) M ; + l ( P ) n H and their images in ( H n M ; ( P ) ) M c ; + l ( P ) / M ; + lform ( P ) a basis of that group.

Permutation modules

430

6.2. Theorem.

(Alperin (1988)). Let f : F P F ( P / H ) be the FP-homomorphism sending each element of P to the left coset containing it. Then the images f(n(y;,-l)"is) ofthe products, lexicographically ordered with 0 5 a;, < p , are a basis of F ( P / H ) such that the images of the products of weight m lie in I ( P ) m F ( P / H )and form a basis of I ( P)"F( P / H ) modulo I ( P)"+' F ( P / H ) . .--)

Proof. It is a consequence of Theorem 6.1 that the products

where the factors yi, - 1 and the z;, - 1 are in lexicographic order, which are of weight m lie in I(P)" and form a basis modulo I(P)m+l. The homomorphism f sends each ziS- 1 to 0 since z;, E H . Therefore, since any product in (1) which has some b;, # 0, can be expressed in the form I C ( Z ; , - 1) with x E F P we have that

f(z(z;, - 1)) = 2 f(z;, - 1) = 0 Thus F ( P / H ) is spanned by the images of all products in the yi, - 1. We claim that these images form a basis. To substantiate the claim, it suffices to verify that there are pn-k of them, where p k = ]HI. The latter will follow provided we show that the total number of y;, - 1 is k. But

and the subgroups H n M ; ( P )obviously filter H , hence the claim. The image under f of a product in the y;, - 1 of weight m lies in I ( P ) " F ( P / H ) , since the product lies in I(P)". We are therefore left to verify that the images of the products of weight m are linearly independent modulo I(P)"+'F( P / H ) . Assume by way of contradiction that this is not the case. Then the image of one of the products of weight m is a linear combination of images of other products in the y;, - 1 of weight m and images of such products of weights > m , since

V ) " F ( P / H ) = f (W") is spanned by the images of all the products in the y;, - 1 of weight of at least m. In particular, deleting this image, that we have just expressed

431


as a linear combination, from the set of such images, we would still have a spanning set for F ( P / H ) . But this contradicts the fact that the number of products in the yis - 1 is pk which is the dimension of F ( P / H ) . So the theorem is true. H

As a preliminary to our final result, we first record the following property. 6.3. Lemma.

Let G be a finite group and let V be an FG-module.

Then

where V* is the contragredient of V . Proof. We claim that for all k 2 0,

sk(v*) = (J(FG)"v)'

(2)

where for any submodule W of V , W' = {f E V*If ( W ) = 0). Since

the result will follow by applying (2) for k = i and k = i - 1. We argue by induction on k 2 0. If k = 0, then both sides of (2) are zero. Assume that (2) is true for k. Then SOC(V*/(J(FG)'V)')= Sk+l(V*)/(J(FG)kV)'

(4)

and Sk+l(V*)= W' for some submodule W of V . Since

+

is completely reducible, it follows from (3) (applied to i = k 1) that (J(FG)"'V)'/( J(FG)kV)' is also completely reducible. Hence, by

(4),(J(FG)"'V)'

C Sk+l(V*).On the other hand, by (4),

WL/ ( J (FG)'V)

'

( J (FG)kV /W ) *

Permutation modules

432

is completely reducible. Hence J ( FG)’V/W is completely reducible and therefore J(FG)k+’V C W . Thus

(J(FG)k+’V)’ 2 W’ = Sk+l(V*) and the result follows.

We are now ready to prove the following result. 6.4. Theorem. (Alperin (1988)). Let F be a field of characteristic p > 0 , let P be a finite p-group and let V be a transitive permutation FP-module. Then the upper and lower Loewy series of V coincide.

Proof. We may of course identify V with F ( P / H ) for some subgroup H of P . For each j 0, put % = I ( P ) j F ( P / H )and keeping the notation of Theorem 6.2, denote by t the weight of n ( y i , - 1)P-I. Then, by Theorem 6.2, V, is of dimension 1 and = 0. Note that the weights of the “complementary” elements n ( y ; , - l)a:s and n ( y i s - l)P-l-a:s add up to t. Hence, for all j E {0,1,. . . ,t }

>

dimFT/,/T/,+, = dimFvt-j/V,-j+l (5) This means that the dimensions of the quotients of the lower Loewy series read in reverse order and are the same as read in the usual order. Owing to Lemma 6.3, the quotients of the upper Loewy series of V* are the contragredients of the quotients of the lower Loewy series of V . But V E ( l H ) G , so V* E V and therefore

dim^ (Si+l( V )/ Si ( V )) = dim^ (V ,/ V,+,) for all i E {0,1,. . . ,t } . Hence, by (5), we also have

(6)

dimF(Si+l(V)/Si(V))= dimF(&-i/&-i+l) (7) for all i E {0,1,. . . ,t } . Moreover, we certainly have S o ( V ) = K+’, S l ( V ) 2 V,,S 2 ( V )2 &-’ and so on, which implies dirnF(Si(V))2 dimF(vt-i+l) Applying (6) and (7), we therefore conclude that dimF(S;(V))= dimF(V,-;+I) which in turn implies that S ; ( V ) = V,-i+l, as required. H

7. Characterizations of p-permutation modules

433

7. Characterizations of p-permutation modules In this section, G denotes a finite group and R either a field of characteristic p > 0 or a complete discrete valuation ring such that charR/J(R) = p > 0. All RG-modules are assumed to be finitely generated. We say that an RG-module V is a p-permutation module if the restriction Vp of V to a Sylow psubgroup P of G is a permutation RP-module. Thus V is a p-permutation module if and only if V is R-free of finite rank with an R-basis on which a Sylow p-subgroup of G acts as a permutation group. The following result is essentially due to Dress (1975).

7.1. T h e o r e m . Let V = V1 @ - - .@ V, be a decomposition of an RG-module V into indecomposable submodules. Then the following conditions are equivalent: (i) V is a p-permutation module. (ii) For any p-subgroup Q of G, VQ is a permutation RQ-module. (iii) V is a direct summand of a permutation module. (iv) There exist subgroups H I , . . . ,H , of G such that, for any i E (1,. . . ,n ) V , is isomorphic to a direct summand of ( 1 ~ ~ ) ~ . (v) Each V; is a p-permutation module. (vi) Each V; has trivial source.

Proof. (i) + (ii): Let P be a Sylow p-subgroup of G such that Vp is a permutation RP-module. Then Q C gPg-' for some g E G. If u l , . . . ,vh is a permutation basis for Vp, then gvl,. . . ,gvk is a permutation basis for 4 p S - 1 and hence for VQ,as required. (ii) + (iii): Let P be a Sylow p-subgroup of G. By hypothesis, V p is a permutation RP-module. Hence (Vp)" is a permutation RGmodule. But V is P-projective, so V is isomorphic to a direct summand of (Vp)", proving (iii). (iii) j (iv): By hypothesis, there is a permutation RG-module M such that each V; is a direct summand of M . Since V; is indecomposable, it follows from the Krull-Schmidt theorem that V; is isomorphic to a direct summand of ( 1 ~for~ some ) ~ subgroup H; of G, 1 5 i 5 n, proving (iv). (iv) + (v): Since ( l H i ) G is a permutation module, it is also a p-

Permutation modules

434

permutation module. Hence we need only verify that a direct summand N of a p-permutation module M is a p-permutation module. If P is a Sylow psubgroup of G such that M p is a permutation RP-module, then N p is a direct summand of Mp. But M p is a direct sum of modules of the form (1~)' where Q is a subgroup of P. By Corollary 5.1.12, (19)' is indecomposable. Hence, by the Krull-Schmidt theorem, Np is a direct sum of modules of the form (1~)'. Thus N is a p-permutation module. (v) (vi): Put M = V;; and let P be a Sylow p-subgroup of G such that M p is a permutation RP-module. If Q & P is a vertex of M , then M is Q-projective and so M is isomorphic to a direct summand of ( M Q ) ~By . hypothesis, MQ is a direct sum of modules of the form (1s)Qwhere S is a subgroup of Q. Thus M is a direct summand of (1s)' for some subgroup S of Q. Since M is S-projective and S C Q, S is a vertex of M and hence M has trivial source. (vi) + (i): It suffices to show that each V;; is a p-permutation module. By the implication (i) + (iii), V;; is a direct summand of a permutation (hence p-permutation) module. Hence, by the proof of the implication (iv) j (v), V;; is a p-permutation module.

+

7.2. Corollary. Let H be a subgroup of G and let V and W be RG-modules. (i) If V = V, @ . . - @ V , for some RG-modules K:, then V is a ppermutation module if and only if each T/: is a p-permutation module. (ii) If V and W are p-permutation modules, then so is V @R W . (iii) If V is a p-permutation module, then VH is a p-permutation R H -module. (iv) If U is a p-permutation RH-module, then U G is a p-permutation RG-module.

Proof. (i) Direct consequence of Theorem 7.1. (ii) If {vi} and {wj}are R-bases of V and W , respectively, which are permuted by a Sylow p-subgroup of G, then the same is true for the R-basis {TJ; 8 wj} of V 8~ W . (iii) If Q is a Sylow p-subgroup of H ,then VQ = (VH)Qis a permutation RQ-module, by virtue of Theorem 7.1(ii). (iv) This is obvious.

7. Characterizations of p-permutation modules

435

To provide some further characterizations of p-permutation modules, we need to introduce some terminology. Let F be a field of characteristic p > 0 and let V be an FG-module. Then V is said to be monomial if there exists a basis q ,. . . ,TI, of V such that G permutes the subspaces F v l , . . . ,Fv, of V . Expressed otherwise, V is monomial if and only if V is a direct sum of modules of the form W Gwhere W is a one-dimensional FH-module for some subgroup H of G. Following Dress (1975), we say that V is virtually monomial if there exist monomial FG-modules V, and V2 such that V @ V’ &. It is clear that the direct sums of monomial (virtually monomial) modules are again monomial (virtually monomial) modules. Our second major result will be proved with the aid of the following theorem.

=

7.3. Theorem.

(Dress (1975)). Let F be an algebraically closed field of characteristic p > 0 and let V be an indecomposable F G module with vertex Q and a trivial source. Then there exist subgroups H I , . . . ,H, of G and one-dimensional FHi-module K, 1 5 i 5 n, such that (a) V @ qG@ - - .@ Vf 2 V g l @ . @ V,G (for some li < n). (b) Each H;,1 2 i 5 n, has a normal SyEow p-subgroup P; such that H;/P; is elementary (i.e. a direct product of a cyclic group and a q-group for some prime q) and such that g;P;g,yl C Q for some 9;E G ( l 5 i 5 n ) . m .

Proof. We argue by induction on IQI. First assume that Q = 1 and denote by {E;liE I } the set of elementary subgroups of G. Since V is projective, it follows from Swan’s extension of Brauer’s induction theorem to modular representations (see Swan (1960)) that for each i E I , there exist FE;-modules M; and N; such that

have the same composition factors (counting multiplicities). Since, for any FG-module W , the FG-module V @ F W is projective, we may tensor both modules in (1) with V to obtain the following isomorphism

Permutation modules

436

of FG-modules:

(see Theorem 2.6.1). Hence, since VEi @ M i and V E 8 ~ N; are projective FE;-modules, we may harmlessly assume that G is an elementary group; If p does not divide the order of G, then there is nothing to prove, since in this case the representation theory of G over F is the same as over the complex numbers (see Curtis and Reiner (1981)) and any elementary group is known to be an M-group (see Curtis and Reiner (1962)). Otherwise, we have G = P x H , where P is a p-group and H is an elementary p’-group. Let &, . . . , be all nonisomorphic irreducible FH-modules. Then V y ,. . .,KG are easily seen to be all nonisomorfor some phic indecomposable projective FG-modules. Hence V i E { 1,. .. ,t } and we are again reduced to the case where p does not divide the order of G. This completes the proof of the case where Q = 1. Now assume that IQI > 1 and let H = NG(Q). By the Green correspondence, there exists an indecomposable FH-module W with the same vertex Q and trivial source such that

KG

wG ~ U , @ U , @ . . * @ U ,

(U; is indecomposable, 0 5 i 5 m )

with Uo E V and \&I < I vertex (U;)l for all i E (1,. . . , m } . Because W is a direct summand of a permutation FH-module, W Gis a direct summand of a permutation FG-module. Hence U,, . . . , Urn are direct summands of permutation FG-modules. Thus, by Theorem 7.1 and the induction hypothesis, the theorem holds for each FG-module U;, 1 5 i 5 m. Therefore it suffices to prove the result for W , i.e. we may assume that Q d G. Since Q d G, Q acts trivially on (lQ)G and thus on its direct summand V . Hence V is inflated from an F(G/Q)-module where is also projective. Thus we may apply the case Q = 1 to deduce that V satisfies the conclusion of the theorem (with G replaced by GI&). This means that there exist subgroups Hl/Q,. .. ,Hn/Q of G/Q and one-dimensional F(Hi/Q)-modules E, 1 5 i 5 n, such that (k < n ) . (c) @ @ . . . @ vp @ . .. @ (d) Each Hi/&, 1 5 i 5 n, is elementary of order not divisible by

v,

v :v

= vcl

v;

v

7. Characterizations of ppermutation modules

437

P. Thus if V , is the FG-module inflated from G, then HI, ...,Hn and 6,. . . ,V, satisfy the conclusion of the theorem. This completes the proof. H 7.4. Corollary. (Dress (1975)). Let F be an algebraically closed field of characteristic p > 0 and let V be a projective FG-module. Then there exist subgroups H I , . . . ,H, of G and one-dimensional FH,-module V,, 1 5 i I. n, such that

and such that each H; is an elementary p'-group. Proof. Apply the special case of Theorem 7.2 where Q = 1 to each indecomposable direct summand of V.

7.5. Corollary. (Dress (1975)). Let F be an algebraically closed held of characteristic p > 0 . Then there exist subgroups H I , . . . ,H,, of G and one-dimensional FH;-module V;, 1 5 i 5 n, such that (a) 1~@ V y @ - @ V j 2 V j l @ - @ V,G (for some k < n). (b) Each Hi, 1 5 i 5 n, has a normal Sylow p-subgroup P;,such that H;/P; is elementary. -

a

Proof. Let Q be a Sylow psubgroup of G. Then Q is a vertex of 1~ and 1, is a source of 1 ~Now . apply Theorem 7.3. H

7.6. Corollary. (Dress (1975)). Let F be an algebraically closed field of characteristic p > 0 . Then, for any FG-module V , there exist subgroups H I , .. . ,H , of G and FHi-modules Wi,1 5 i 5 n, such that (a) V @ WF @ , - WkG-N Wt+l @ - - - @ W: (for some k < n). (b) Each H , has a normal Sylow p-subgroup with elementary factor group. Proof. Tensoring the isomorphism of Corollary 7.5 with V, we obtain v @ @i"=l(v 63 !3 @;&+l(V €4

YG)

YG)

Permutation modules

438

Since, by Theorem 2.6.1,

v @ Fc

E)‘

5 n) the result follows by setting Wi = VHi8 V,, 1 5 i 5 n. (vHi @

(1 5 i

7.7. Theorem. (Dress (1975)). Let F be a field of characteristic p

> 0 and let V be an FG-module. Then the following conditions are

equivalent: (i) V is a p-permutation module. (ii) V is a direct summand of a monomial FG-module. Furthermore, under the assumption that F is algebraically closed, each of the above conditions is equivalent to: (iii) V is virtually monomial. Proof. Since each permutation FG-module is a monomial FGmodule, the implication (i) j (ii) follows from Theorem 7.1. The implication (iii) j (ii) follows from the definition of “virtually monomial”. Note also that if F is algebraically closed, then (i) implies (iii), by Theorems 7.1 and 7.3. By the foregoing, we are left to verify that (ii) implies (i) for an arbitrary F . To do this we may, by Theorem 7.1, assume that V 2 W G where W is a one-dimensional FH-module for some subgroup H of G. But W is a p-permutation module, since a Sylow p-subgroup of H acts trivially on W .Hence V is a p-permutation module, by virtue of Corollary 7.2 (iv). 8.

The Brauer morphism

In this section, p denotes a prime, I< a subgroup of a finite group G and R a commutative local ring with c h a r R / J ( R ) = p . In what follows we put F = R / J ( R ) . Let V be an RG-module. Given a subgroup H of I(, we put

and

V ( K )= Inv(VK)/

+ J(R)lnv(V-)

8. The Brauer morphism

439

where H runs over the set of all proper subgroups of K with the convention that for K = 1, &V," = 0 (and hence V ( K )= V / J ( R ) V ) . We remark that in most papers on the subject Inv(V'&) is denoted by V H ;we shall use this notation only in the case V = A is a G-graded R-algebra so as not to conflict with our notation for induced modules. Following Brou6 (1985), we refer to the natural surjection

as the Brauer morphism. Our aim is to record some basic properties of the Brauer morphism with an eye to future applications in the next section. Most of the results recorded are contained in BrouC (1985) and

Brou6 and Puig (1980). 8.1. Lemma. With the notation above, the following properties hold: (i,~~ n v ( v K )is an R(NG(I ,

proving that f s u = g f u . Hence u H f u is an RG-homomorphism. A similar argument shows that fg. = gfv for all g E G,vE V and the result follows.

Permutation modules

442

8.4. Lemma. Let U,V and

W be RG-modules and let f

:U x

V -t W be a G-stable R-bilinear map. Then, for any subgroup K of G, the map f K : U ( K )x V ( K )+ W ( K ) given b y

frc (Br;(u),Br;(v)) = W ( f ( . , V ) )

(u

E Inv(W,v E I~v(vK))

is a well-dejned (Nc(IC)/lC)-stableR-bilinear map. Proof. If u E Inv(U~),vE 1nv(V~), then for all g E IC,

proving that f ( u , v ) E I n v ( w ~ ) To . prove that f K is well-defined, it suffices to show that B r F ( f ( u , v ) ) = 0 if u E Ker(Br;) or w E K e r ( B r L ) . Assume that u E Ker(BrK). then

for some X E J ( R ) ,z E I ~ w ( U K ) , U HE U E , where H runs over the set of proper subgroups of K(if (I = 1, then put all U H = 0). Hence

and therefore it suffices to show that Brg(f ( u ~w)), = 0 for all H c IC. Now U H = & t y for some y E Inv(UH),where T is a left transversal for H in K. Since w E 1nv(V~), y E Inv(UH) and f is G-stable, we have f ( y , v) E Inv(lV~).Hence

A similar argument shows that if v E K e r ( B r g ) ,then BrE(f(u,v)) = 0, and thus f is well-defined.


443

It is clear that f is an R-bilinear map. Furthermore, for any g E

NG ( K )7 f K ( g B r $ ( u ) , g B r m ) = fK(Br;(P), BrL(gv)) = B.KW(f(P,gV))

= sBrKW(f(., 4) = SfK(Br;w, B d m )

by applying G-stability of f together with Lemma 8.1(ii). This completes the proof of the lemma.

Let V be an RG-module, let P be a p-subgroup of G such that Vp is a permutation RP-module with an R-basis X permuted b y P , and let V = V /J ( R)V. Denote b y x1,x2,. . . ,x, all distinct elements o j X fixed b y P. Then the following properties hold: (i) The set {BrK(x;)Il 5 i 5 n } is an F-basis for V ( P ) . (ii) The F(NG(P)/P)-modulesV ( P ) and v ( P ) are canonically isomorphic. (iii) I f V * is the contragredient ofV and the basisX* = { x * ( xE X } o f V * is dual to X , then there is an F(NG(P)/P)-isomorphism 8.5. Lemma.

V * ( P )-+ V ( P ) * which sends the basis {Br~'(x*)ll5 i 5 n } o f V * ( P ) onto the dual basis o f { B r ~ ( x ; ) 5 ( li 5 n } . Proof. (i) Let x l , x 2 , . . . ,x,, x,+~,. . . ,xt be all representatives for the P-orbits of X, and let Qi be the stabilizer of xi in P , 1 5 i 5 t. By our choice of the xi,we have Q1 = Q2 = = Qn= P and &%+I,. . . ,Qt are all proper subgroups of P . By Lemma l.l(iii), {Trgi(x;)Il5 i 5 t } is an R-basis for Inv(Vp). It therefore suffices to show that for any proper subgroup Q of P , VJ is contained in M , where M is the sum of J ( R ) l n v ( V p )and the R-linear span of T r c k ( x k ) , n 1 5 k 5 t . To this end, fix x E X and denote by H and L the stabilizers of x in Q and P, respectively. Since the elements w E V of the form w = T r QH ( z ) form an R-basis for 1nw(V~), it suffices to verify that for any such w, TT;(V)E M . Since a .

+

~ r g ( ~ r Q H (= x )T) r g ( x ) = Tr,P(Trf;(x))

Permutation modules

444

and Trfi(x) E J(R)lnv(VL)whenever L # H , the required assertion follows. (ii) We first note that V p is a permutation FP-module with F-basis = {?(a: E X},?= x J ( R ) V . Hence, by (i),

x

+

{ ~ r ; ( ~ i )5l il 5 n }

is an F-basis for V ( P ) . Thus Brg(x;) I+ BY:(?;) is an F-isomorphism of V ( P ) onto V ( P ) . Since this isomorphism obviously preserves the action of N G ( P ) / P , the assertion follows. (iii) Let a , b E X and let g a = b for g E P . Then ga*(b) = a * ( g - l b ) = a*(a) = 1 and, for x # b,x E X,ga*(x) = a*(g-'x) = 0. Thus ga* = b*, and similarly if ga* = b', then g a = b. In particular, this implies that x;,.. , ,xi are all distinct fixed points of X * in P . Hence, by (i), {BrF*(xTIl5 i 5 n } is an F-basis for V * ( P ) . It follows that the map

{ B&$;

:

is at least an F-isomorphism. Let F : V x V* + R be defined by f(v, cp) = cp(v) for all v E V,cp E V*. Then, regarding R as the trivial RG-module, f becomes a G-stable R-bilinear map. Since R ( P ) = R / J ( R ) = F, it follows from Lemma

8.4 that the map f-

: V ( P )x

V * ( P )+ F

defined by f P ( ~ 4 4B.pv'(cp)) ,

= CpW

+J(R)

(v E Inw(Vp),cp E Inv(V,*)) is an (NG(P)/P)-stable F-bilinear map. Hence, by Lemma 8.3, the map

V * ( P )+ V ( P ) *

+

defined by B ~ ; * ( ( P H) f,, where f,(Br;(w)) = ~ ( v ) J ( R ) , is an F(NG(P)/P)-homomorphism. Setting cp = x:, we see that this homomorphism coincides with $ and the result follows.


445

8.6. Lemma. Let H be a subgroup of G and let V be an H projective RG-module. Then, for any subgroup K of G, V ( K ) = 0 unless K is G-conjugate to a subgroup ofH. In particular, if W is any RH-module, then W G ( K )= 0 unless I( is G-conjugate to a subgroup ofH.

Proof. that

Since V is H-projective, it follows from Theorem 2.10.3

T r g (EndRH( V ) )= EndRG( V ) Hence, by Lemma 2.9.6(ii) l v E TrE(EndRH(V))

TrgngHg-I (EndR(KngHg-l)(v)) g€G

For each g E G, we may therefore choose and thus K that l v = C s E G TrKngHg-l($g)

Inv(Vh-)

c

$g

E EndqKngHg-i)(V)such

Tr$ngHg-l($g)(lnv(vK)) g€G

But for any

z1

E I n z 1 ( V ~ we ) , have $ ~ ~ ( vE)Inv(VKngHg-l)and K

TrKngHg-l ($g)(')

' h

= T r K gHg-' ~

($g

(' )

which implies that

Inv(VK)

v/ngHg-l

i

9EG

thus completing the proof. 8.7. Lemma. Let P be a p-subgroup of G , let H be a subgroup of G such that P is G-conjugate to a subgroup o f H and let T be a set of all representatives t for the double cosets NG(P)tH for which P tHt-'. Then

1

(lRH)G(P) @tET [ l F ( N t H t - l ( P ) N G ( p ) as F(NG(P)/P)-modules Proof. Let X 2 T be a full set of representatives for ( N G ( P )H), double cosets in G. Then, by Mackey decomposition,

Permutation modules

446

where v, = [1R(N2Hz-1(P))

On the other hand, by Lemma 8.6,

V,(P) = o if P

xHx-'

(4)

Furthermore, if P C_ x H x - l , then P acts trivially on V, and thus, by Lemma 8.l(v),

The desired conclusion is therefore a consequence of (3), (4), ( 5 ) and Lemma 8.l(iii), (iv). Given a G-algebra A over R and subgroups H C K of G, we put

A; = T r $ ( A H ) where AH is the subalgebra of H-invariant elements of A and

"

TrH ; AH + A" is the trace map introduced in Sec.1. Note that

Trz(ab)= Trg(a)b for all a E A H ,b E AK

(6)

and

TrE(bu)= bTr$(a) for all a E AH,bE A" (7) It follows from (6) and (7) that A; is an ideal of A". Note also that = A ~ ~ s - for ' all g E G

(8)

Since A is an RG-module, we may define A ( K ) as we did for any RGmodule V , i.e. by putting

A ( K )= AK/ ( C A Z H

+ J(R)AK

8.

The Brauer morphism

447

where H runs over the set of all proper subgroups of K with the convention that for K = 1,& AZ = 0. In particular, the natural surjection

B r i : AK t A ( K ) gives us the Brauer morphism with respect to the RG-module V = A. 8.8. Lemma. Let K be a subgroup ofG and let A be a G-algebra over R. Then (i) Ker(Br;) is an ideal of A K . (ii) AK is an (NG(K)/K)-algebraouer R, and A ( K ) is an ( N G ( K ) / K ) algebra ouer R annihilated b y J ( R ) . (iii) Br; is a homomorphism of (Nc(K)/K)-algebras, i.e. a homomorphismof R-algebras perserving the actions of NG(K ) / K on AK and A(IC).

Proof. (i) This is a consequence of the fact that each A$ and J ( R ) A K are ideals of AK. (ii) The assertion regarding A" follows from (8) applied to H = K . The remaining assertion is a consequence of (9) and the definition of A( K ) . (iii) Direct consequence of (i) and Lemma 8.1(ii). W We close by recording the following observation.

Let A = R X be the group algebra o f a group X over R and let G act on X as a group of automorphisms of X . Then A becomes a G-algebra in the natural way. Let P be a p-subgroup ofG and let H be the subgroup of all elements o f X fixed b y P . Then (i) The map 8.9. Lemma.

is an isomorphism of F-algebras. (ii) If R = F, X = G and G acts on itself b y conjugation, then H = C G ( P )and, upon the identification of FG(P) with FCG(P),the restriction of BrFG to Z(FG) is the classical Brauer homomorphism (see Theorem 6.2.2).

Permutation modules

448

Proof. (i) This is a direct consequence of Lemmas 8.8(iii) and 8.5(i). (ii) Let C1,. . . ,C, be all P-conjugacy classes of G. If IC;l > 1, then BrcG(C:) = 0, while Br;"(h) = h for all h E CG(P).This obviously implies the required assertion. 9.

Scott modules

Let G be a finite group, p a prime and R either a field F of characteristic p or a complete discrets valuation ring with F = R / J ( R ) . All RG-modules are assumed to be finitely generated. Our aim is to examine an important type of indecomposable p permutation RG-modules, the so called Scott modules. Such modules were first discovered by L.L.Scott, and independently by J.L.Alperin; the first published account is in Burry (1982). Our method is based on an,important work of Broud (1985) who utilized a systematic use of the Brauer morphism. We begin by recording the following definition due to Puig (see Brou6 (1985)). Let V be a ppermutation RG-module and let P be a psubgroup of G. Then the Scott coeficient of V associated with P , written s p ( V ) is defined by

Note that, by Lemma 8.2(ii),

s p ( V ) = dim= ( V ( P ) p ( p ) ' p ) 9.1. Lemma. Let P be a p-subgroup of G.

(i) If V and V' are two p-permutation RG-modules, then

(ii) If V is u p-permutation RG-module and V* is the contragredient ofV then s p ( V ) = s p ( V * ) . (iii) For any subgroup H of G, sp [ ( ~ R H ) = ~ ] 1 i f P is G-conjugate to a Sylow p-subgroup of H and sp [(lRH)G] = 0 otherwise.

9. Scott modules

449

Proof. (i) Apply Lemma 8.l(iii) and ( 1 ) . (ii) Owing to Lemma 8.5(iii), s p ( V * ) is equal to the rank of the endomorphism T r NG(p)'p , of V ( P ) * which , is equal to the rank of its transpose, i.e. the endomorphism T r NG(p)'p , of V ( P ) . Thus s p ( V ) = s p ( V * ) , as asserted. (iii) By Lemma 8.6, ( l m ) G ( P )= 0 unless P is G-conjugate to a subgroup of H . By Sylow theorems in H , if g-lPg is a Sylow psubgroup of H for some g in G , then gTIPgl is a Sylow psubgroup of H only for those gl E G for which g1 E NG(P)gH.By Lemma 8.7, it ) ? dimension 1 or 0 according therefore suffices to show that (( ~ F H ) ~ has to whether H is, or is not a p'-group. By Lemma 1.3, we may identify ( ~ F H ) ' with the submodule F G - H + of FG, in which case

as required.

We are now ready to prove the following result. (Scott, Alperin). Let P be a p-subgroup of G. (i) There exists a unique (up to RG-isomorphism) indecomposable p-permutation RG-module Sp(G, R ) such that sp(Sp(G,R ) )# 0 . Furthermore, s p ( S p ( G ,R ) ) = l, and 9.2. Theorem.

Sp(G,R)

[SP(G,R)]*

(ii) If H is a subgroup of G, then Sp(G,R ) is isomorphic to a direct summand of ( ~ R H if) and ~ only if P is G-conjugate to a Sylow p-subgroup of H . In particular, (by taking H = P ) , S p ( G , R ) has vertex P. (iii) If H is a subgroup of G such that P is G-conjugate to a Sylow p-subgroup of H , then (a) H O ~ R G ( ~ SRPG( G , , R ) )% H o ~ R G ( S P ( G , R~)R, G % ) R. (b) S p ( G , R ) is the unique indecomposable direct summand V of ( l R H ) G such that H o ~ R G ( ~ R G#, V 0.)

Permutation modules

450

(c) Sp(G,R) is the unique indecomposable direct summand V of ( 1 ~ ) such ' that HomRc(V,1 R G ) # 0. (d) Sp(G, R)/J(R)Sp(G, R) g Sp(G, F ) . (iv) For any given p-subgroup Q of G, SQ(G,R ) E Sp(G, R ) if and only if Q is G-conjugate to P .

Proof. (i) By Theorem 7.1 (vi), any indecomposable ppermutation RG-module V is a direct summand of ( ~ R Q ) ' for some p-subgroup Q of G. If sp(V) # 0, then by Lemma 9.l(i), (iii), it follows that P is G-conjugate to Q. Furthermore, since ~ p ( ( l R p ) = ~ ) 1, we see that V is the unique indecomposable direct summand of ( 1 ~ p ) ' such that sp(V) # 0. Hence sp(V) = 1 and, by the unicity of V , it follows from Lemma 9.l(ii) that V E V'. (ii) It follows from (i) and Lemma 9.l(i) that Sp(G, R ) is isomorphic to a direct summand of ( ~ R H if) ~and only if s p ( ( 1 ~ ~ ) '#) 0. Hence, by Lemma 9.l(iii), P is G-conjugate to a Sylow psubgroup of H if and only if Sp(G, R ) is isomorphic to a direct summand of ( ~ R H ) ' . (iii) Assume that P is G-conjugate to a Sylow psubgroup of H . By Proposition 2.2.4(i) and Corollary 2.2.7,

HomRG (IRG,

(lRP)G)

R

HomRG

((lRF')G, 1 R G )

Since Sp(G,R) 2 Sp(G,R)*, to prove (a), (b) and (c), it suffices to show that HomRG (IRG, SP(G, R ) ) # 0 (2) To this end, note that by Lemma 8.2(ii), (Sp(G,R))$ is mapped onto ( W G , R))(P)lNQ(p)'p; this last module is not zero by definition of Sp(G, R) since its dimension is precisely sp(Sp(G,R)). Hence

proving (2). Finally, let V be any p-permutation RG-module and let = V / J ( R ) V . It follows from Lemma 8.5(ii) that sp(V)= s p ( v ) . Hence, by the characterization of Sp(G, R) given above, we deduce that

v

10. The Brauer morphism and ppermutation modules

451

proving (d). (iv) This is a direct consequence of (ii). W We shall refer to the RG-module Sp(G,R) as the Scott module of G associated to P. Two extreme cases for R = F are worth noticing: SI(G, F ) is a projective cover of ~ F G while , if P is a Sylow psubgroup of G, then Sp(G,F) 2 ~ F G . The latter follows from Lemma 1.3(v) and Theorem 9.2(iii)(b), while the former is a consequence of Theorem 9.2(iii ) (c). 10. T h e Brauer morphism and p-permutation modules All the notations and conventions adopted in the previous section remain in force. Our aim is to provide a number of important properties of the Brauer morphism applied to ppermutation modules. We begin by recording two auxiliary results, both extracted from a work of Brou6 (1985). 10.1. Lemma. Let V be a p-permutation RG-module and let P be a p-subgroup of G. Then V ( P ) is a p-permutation F ( N c ( P ) / P ) module.

Proof. Let Q be a Sylow psubgroup of NG(P).Then, by Theorem 7.1, VQis a permutation RQ-module. Choose an R-basis X of V which is permuted by Q. Then the F-basis for V ( P )given by Lemma 8.5(i) is permuted by Q / P , as required. W 10.2. Lemma. Let P be a p-subgroup ofG, let V be a p-permutation RG-module, and let A = EndR(V). Then the map

where A$(a) (Br,V(s))= BrpV(a(z))

( a E AP,x E Inv(Vp),

is an isomorphism of F(NG(P)/P)-algebras.

452

Permutation modules

Proof. The natural bilinear map f : A x V G-stable. Hence, by Lemma 8.4,the map

+V

f p :A ( P )x

V(P)

, ( a , v ) H a ( v ) is

V(P)

given by

f P ( B r m , B r 3 x ) ) = B.Pv(4.N is a well defined (Nc(P)/P)-stable F-bilinear map. Thus, by Lemma 8.3, the given map is a homomorphism of F(NG(P)/P)-algebras. Let X be an R-basis of V which is permuted by P.Then the set

where uz,y E A is defined by az,y(.z)= SY,+xfor x,y , z in X is an R-basis of A which is permuted by P . Let C x ( P )be the set of fixed points of X under P. Then we obviously have

S ~ ~ , ~=( ZS g) , z ~ for all

x,y , z E C x ( P )

which, by Lemma 8.5(i), shows that the given homomorphism is an isomorphism. H We are now ready to prove the following result. 10.3. Theorem. (Broue' (1985)). Let V be an indecomposable p-permutation RG-module. (a) The vertices of V are the maximal p-subgroups P of G such that V ( P )# 0. (ii) A p-subgroup P of G is a vertex ofV if and only if V ( P )# 0 and v ( P ) is a projective F(NG(P)/P)-mod.ttle. (iii) The map V H V ( P )induces a bijective correspondence between the isomorphism classes of indecomposable p-permutation RG-modules with vertex P and the isomorphism classes of indecomposable projective

10. The Brauer morphism and p-permutation modules

453

F ( NG (P)/P)-modules. In particular, (Sp(G , R ) ) (P ) is the projective cover of ~ F ( N G ( P ) / P ) . (iv) Let M be a p-permutation RG-module, let E be an indecomposable projective F(NG(P)/P)-module, and let M ( P ,E ) be the corresponding p-permatation RG-module with vertex P . Then M ( P , E ) is a direct summand of M if and only if E is a direct summand of M ( P ) .

Proof. (i) By Lemma 8.6, it suffices to show that if P is a vertex of V , then V ( P )# 0. Suppose V is a component of ( l ~ p ) ' . Then, by Proposition 1.9, Vp is isomorphic to a direct sum of modules of type

where g runs over a certain subset S of G. But V is a component of (Vp)', and since P is a vertex of V it follows that S n NG(P)# 8. Hence 1 ~ isp a component of Vp and so V ( P )# 0. (ii) Assume that V ( P )# 0 and that V ( P )is a projective F ( N G ( P ) / P ) module. Setting A = E n d ~ ( v )it, follows from Lemma 10.2 and Theorem 2.10.3 that

But, by Lemma 8.2(ii), ( A ( P )N) G , ( p ) ' p is the image of A: under the Brauer morphism B r s . Since l v is the unique nonzero idempotent of AG, we deduce that l v E A:, proving that V is P-projective. Hence, by (i), P is a vertex of V . Conversely, assume that P is a vertex of V . Then 1" E A: and so

Thus, by Lemma 10.2, V ( P )is a projective F(NG(P)/P)-module. (iii) It follows from Lemma 8.7 that

The indecomposable p-permutation RG-modules with vertex P corresponds to the indecomposable components of (lRp)Gwith vertex P. Note also that the indecomposable projective F(NG(P)/P)-modules

Permutation modules

454

correspond to indecomposable components of ( l ~ ) ~ G ( ' ) l ~Put . A = f ? ~ ~ d R ( ( l R p ) ~Then, ). by Lemma 10.2,

A ( P ) 3 EndF ( ( 1 ~ ) ~ ~ ( as ~ F(Nc(P)/P)-algebras ) / ~ ) ) ~ vertex P corNow the indecomposable components of ( l ~ p with respond to the primitive idempotents of AF whose image in A ( P ) is nonzero. Since, by Lemma 8.2(ii), Br; maps A: onto A ( P )N,G ( P ) I P

and since AG = AS, we have

(A(P))NG(p)'p = ( A ( P ) N) ,G ( P ) / P Thus the first assertion follows from the standard facts about lifting idempotents: an indecomposable component V of ( 1 ~ p with ) ~ vertex P corresponds to a primitive idempotent e of A; such that Br$(e) # 0, which corresponds to the primitive idempotent Br$(e) of (A(P))NG(P)IP, which in turn corresponds to the direct summand

BrpA(e)

*

( l F ) N ~ ( p ) ' p=

B ~ ~ V ( V ) = V(P)

of ( ~ F ) ~ G ( ' ) / The ~ . assertion regarding Sp(G,R) follows from the fact that by definition, (Sp(G, B)(P))NG(p)Ip # 0 (alternatively, it follows from the equality S p ( G , R ) ( P )= S I ( N G ( P ) / PF, ) . (iv) Let E be any indecomposable projective F ( NG(P)/P)-module. Then, by definition, the RG-module M ( P ,E ) is the indecomposable ppermutation RG-module determined by the condition ( M ( P ,E ) ) ( P )= E . It is obvious that if M ( P , E ) is a direct summand of M , then E is a direct summand of M ( P ) . Conversely, let M be a ppermutation RG-module such that E is a direct summand of M ( P ) , and let A = E n d ~ ( A 4 )Owing . to Lemma 10.2,

A(P)

E~F(M(P))

and therefore, by hypothesis, ( A ( P ) N) ,G ( p ) / pcontains a primitive idempotent e such that e M ( P ) E E. Using the standard facts about lifting idempotents together with Lemma 8.2, it follows that A; contains a primitive idempotent f such that Br$( f) = e. Hence M is an indecomposable component of M with vertex P , and (f M ) ( P )=

-

f

a

10. The Brauer morphism and p-permutation modules

e - M ( P ) Z E , which implies proof of the theorem. W

f-M

455

M ( P ,E ) . This completes the

10.4. Corollary. (Broue' (1985)). The reduction module J ( R ) defines a bijection between the isomorphism classes of p-permutation RG-modules and the isomorphism classes of p-permutation FG-modules.

Proof. By Lemma 8.5(ii), V ( P ) E V ( P ) where V and P are as in Lemma 8.5. The desired assertion is therefore a consequence of Theorem 10.3(iii). 10.5. Corollary. (Broue' (1985)). There is a bijection between the isomorphism classes of indecomposable p-permutation RG-modules and the G-conjugacy classes of pairs ( P ,E ) , whem P is a p-subgroup of G and E is an indecomposable projective F(NG(P)/P)-module.

Proof. Apply Theorem 10.3(iii). W The next observation illustrates that V ( P ) is a familiar classical object for the case where V is an indecomposable p-permutation FGmodule with vertex P . 10.6. Proposition. (Broue' (1985)). Let V be an indecomposable p-permutation FG-module with vertex P . Then the Green correspondent ofV is the FNc(P)-module V ( P ) .

Proof. Let W be the Green correspondent of V with respect to ( G , P , N G ( P ) ) .Since, by Theorem 7.1, V has trivial source, W must also have trivial source (see Theorem 5.2.4(iii)). Hence W is isomor, P acts trivially on W and phic to a direct summand of ( l p p ) N ~ ( p )so thus W ( P ) = W . But, by Theorem 5.2.4, V N G ( p ) E W @ W ' , where W' is a direct sum of indecomposable F(NG(P))-modules with vertex strictly contained in P. It follows from Lemma 8.6 that W ' ( P ) = 0 and therefore, by Lemma S.l(iv)?

'(') as required.

= 'NG(P)(')

W?

Permutation modules

456

We close by presenting some applications of Theorem 10.3. For the rest of this section F denotes an arbitrary field of characteristic p > 0.

10.7. Proposition. (Cabanes (1988)). Let V be a p-permutation FG-moduEe and let P be a normal ~ - s u b g r o uof~ G. Let U be the direct sum of those indecomposable components of V whose vertex contains P and let W be the sum of the others. Then ( i ) V = U @ W , V ( P ) Z UandW(P)=O. (ii) Inv(Up) = U and Inv(Wp) = Ii'er(BrF). (iii) For A = EndF(V),Ii'er(Bri)= { f E E n d ~ p ( V ) l f ( U ) W } . Proof. (i) and (ii). It is clear that V = U @ W . Furthermore, by Theorem 10.3(i), W ( P )= 0 and, by the definition of U , Inv(Up) = U . Since Inv(Up) = U , we have Ker(BrE) = 0 (hence U ( P )= U), while since W ( P )= 0 we have Ii'er(Brr) = Inv(Wp). Thus

V ( P )c U ( P )@ W ( P )c U ( P )= u and

I<er(BrL) = K e r ( B r g )

Ii'er(BrpW) = ~ n v ( ~ p ) ,

proving (i) and (ii). (iii) Let Q be a proper subgroup of P and let f E E ~ F Q ( V To ). prove the "L"containment, it suffices to show that

Since u E Inv(Vp), we have T r g ( f ) ( u ) = T r % ( f ( u ) )with f ( u ) E ITZ~(VQ But ) . then

as required.

To prove the reverse containment, let e : V + U be the projection map, and let 1 - e = C7.1ei be the decomposition of 1 - e associated with a decomposition of W as a direct sum of indecomposable FGmodules whose vertices do not contain P. Let P; be a vertex of e;(W), 1 5 i 5 n. Then Pi does not contain a G-conjugate of P and e; E AFi,

10. The Brauer morphism and ppermutation modules

1 5 i 5 n. But if i E (1,. . . , n } , then gPig-' n P ei E AE

c c A;Eg-I"P

c P for all g

457

E G, so

c - Ker(Br$)

g€G

and therefore 1- e E Ker(Br$). Hence, iff E E n d ~ p ( Vwith ) f(U) C W , then efe = 0 and

f = (1 - elf

+ ef(1-

e) E

Icer(BrAp),

as asserted. 1

As an application of the above, we now prove the following result due to Puig (see Cabanes (1988)). 10.8. Theorem. Let V be a p-permutation FG-module and let P be a p-subgroup of G . Then

( E n d F ( V ) )( P )2 End=(V(P))

as

Nc(P)-aZgebms

Proof. We first note that Inv(Vp) and Ker(Br;) are stable under . the natural action of E n d ~ p ( Von ) every element of E n d ~ p ( V ) Thus V (P ) induces a homomorphism $h : End=p(V) -+ E n & ( V ( P ) ) of NG(P)-algebras. It will be shown that this homomorphism is surjec-

tive and has the desired kernel. Write VNG(p)= U @ W , where U is the direct sum of those indecomposable components of V N c ( p ) whose vertex contains P and where W is the sum of the others. We know, from Lemma 8.1(iv), that V ( P ) = v ~ ~ ( p ) ( P )Hence, . by Proposition 10.7(i), V ( P ) 2 U . It is then clear that the homomorphism 4 corresponds to the map

{ En&p(V)f

--f

I-+

E&(U) efe

where e : VNG(p)+ U is the projection map, by identifying E n d ~ ( u ) with {f E Endp(V)lf(W) = 0 and f(V) U }

c

Permutation modules

458

Hence

Icer+ = {f E End~p(U)lf(U) GW} proving, by Proposition 10.7(iii), that Ker$ = Ker(Br$) for A = hdF(V). We are left to verify that is surjective. By Proposition 10.7(ii), we have U = Inv(Up) and so E n d ~ ( u= ) E n d ~ p ( U ) .Now fix X E E n d ~ ( uand ) denote by A’ an element of E n d ~ p ( Vby ) extending X to V with 0 on W . Then it is clear that 1c, maps A’ to X and the result follows.

+

459

Chapter 8 Permutation lattices This chapter is based exclusively on an important work of Weiss(1988). We are concerned with the detailed study of permutation Z,P-lattices where P is a finite p-group and Z pis the ring of p-adic integers. One of the main results asserts that V is a permutation lattice for P provided there is a normal subgroup N of P such that VNis free and l n v ( V ~ ) is a permutation lattice for PIN. The last section is devoted t o a number of applications of the results obtained. Among them, we prove that if G is a finite nilpotent group, then Z G determines the isomorphism class of G. We also show that if A is an abelian normal subgroup of a finite group G such that G / A is a p-group, then Z G determines the isomorphism class of G. 1.

Generalized permutation lattices

Throughout this section, we fix a prime p and a finite p-group P. We denote by Z p the ring of p-adic integers and put R = Z,[E] where E is a primitive p-th root of 1. We also put

so that F is the field of p elements. Each RP-module below will be assumed to be R-free of finite rank. If V is an RP-module, then we

Permutation lattices

460

Put

v = V/(1 - &)V

Thus V is an FP-module. As usual, we write l p for the trivial R P , Z p P or FP-module (the coefficient ring will always be understood from the context). The following terminology, extracted from Weiss (1988) is nonstandard, but will be very useful in avoiding excessive verbosity. By a permutation lattice for P, we understand any permutation %$-module. Thus, by Lemma 7.1.1, V is a permutation lattice for P if and only if V is a finite direct sum of Z,P-modules of the form (1s)' for some subgroup Q of P . A generalized permutation lattice for P is defined to be a finite direct sum of RG-modules of the form xG for some homomorphism x : Q +< c > of a subgroup Q of P . Here we write xG for R P @RQ R with Q acting on R via x. By a permutation module for P (or a permutation P-module), we understand any permutation FP-module. Our aim is to prove that if V is an RP-module such that V is a permutation module, then V is a generalized permutation lattice. Let M be a generalized permutation lattice for P . Following Weiss (1988), by an eigenfactorof M , we understand a split RP-monomorphism

f:E+M where E is a generalized permutation lattice for P such that the indecomposable summands of E have vertex P and indecomposable summands of Cokerf have vertex not equal to P. Thus, if

M for some homomorphisms

$rx,

P

xr : P, +< c >, then

An eigenfactor of a permutation module for P is defined in a similar fashion. Note that if f : E --+ M is an eigenfactor of the generalized permutation lattice M for P , then the induced map f : E -+ A? is an = (1~)' eigenfactor of the permutation module A?. This is so since

1. Generalized permutation lattices

461

x

for any homomorphism : Q +< c > of a subgroup Q of P. Let V be an FP-module. Given a subgroup Q of P , we let

be the relative trace map and put

v;

= Tr;(Inv(vQ))

and

where Q runs over the set of all proper subgroups of

P

1.1. Lemma. Let Q , S be subgroups o f P and let V be an F P module isomorphic to (1s)'. Then

v;

=

{

I n v ( V ) if gSg-' Q for some g E P 0 otherwise

Proof. We may regard V as an F-space with basis gS,g E P and with P acting by g(g'S) = gg'S, g,g' E P . Let T be a full set of double coset representatives for (Q, S ) in P . Then, given t E T , there exists a set X ( t ) 2 Q such that

QtS = U,,x(t)xtS

(disjoint union)

and hence Q can be written as a disjoint union

t , constitute an F Observe that the elements b, = C z E X ( t ) ~ t SE, T basis for Inv( VQ).Furthermore, TrQp(b,)

=

TrgPnts,-l (ts) tst-1

= TrsL-IT'rQ",St-I ( t S ) = TrlpSt-l ((tSt-l : Q n tSt-*)tS) =

(t5't-l

:Q

ntSt-')Trc(S)


462

with T r c ( S )# 0 in I n v ( V ) . Since (tSt-l : Q fl tSt-') = 0 in F unless tSt-l & Q and dirnpInv(V) = 1, the result follows.

1.2. Lemma. Let V be a permutation module for P and let Q be a subgroup of P. (i) VJ # 0 if and only if V has a direct summand isomorphic to (1s)' for some subgroup S of Q . (ii) Suppose that Q is minimal with respect to V[ # 0 , and choose w E Jnv(VQ)be such that Tr;(w) # 0 . Then the map

{

(lQ>' g @ l H gv

is a split monomorphism of FG-modules. Proof. (i) By assumption, there exist subgroups PI, P2,. . . ,P, of P such that v 2 @=l(lP,)' Owing to Lemma 1.1, V$' has a one-dimensional contribution precisely from those Pi which are P-conjugate to a subgroup of Q. Because vertices are well-defined up to G-conjugacy, the required assertion follows. (ii) Write w = o1 - .. o, with wj E ( l ~ ~ 1 5) i ~5 ,r. Then

+ +

T

i=l

implies that TrG(wj) # 0 for some j E (1,. . . ,r } . Applying (i) and minimality of Q, we conclude that Pj is P-conjugate to Q. We may therefore assume that Pj = Q, in which case there is a projection $ : V (19)' with $(v) = oj. Now f is obviously an FP-homomorphism and therefore $ O f : (lQ)' (1Q)' is also an FP-homomorphism. We claim that $ 0 f is injective; if sustained, it will follow that $ o f is an automorphism, say A, of ( 1 ~ ) ' and A - l 0 $ is a desired splitting of f . Assume by way of contradiction that K e r ( $ o f ) # 0. Since P is a p-group, it follows that K e r ( $ o f) contains a nonzero fixed point of P . --f

463


But Inv( (1~)') is one-dimensional with basis CteTt 8 1, where T is a left transversal for Q in P . Hence

a contradiction. 1.3. Lemma.

Let 8 : E + V be an eigenfactor of the permutation P-module and let Br; : I n v ( V ) + V ( P ) be the Brauer morphism. Then the map BrgoO:E+V(P) is an FP-isomorphism. Proof. Since O(E) I n v ( V ) ,the composite map Brg 0 0 is welldefined. By Lemma 1.1, if Q is a submodule of P and W = ( l ~ ) ' ,then W ( P )= 0 unless Q = P , in which case B r r : I n v ( W ) + W ( P )is an FG-isomorphism. Now there is a splitting exact sequence

O+E&V+U+O

(1)

of FG-modules. By the first paragraph and the definition of eigenfactors, we have U ( P )= 0. Hence the sequence (1) yields an isomorphism

e :E(P)

--f

V(P)

By the same argument,

BrpE : E = ~ n v (+~E)( P ) is also an isomorphism. Taking into account that Br;

the result follows. I

o

8=

e

o

Br;,

1.4. Corollary. Let V be a permutation P-module and let Q be a subgroup of P which is minimal with respect to V l # 0 . If 8 : E + VQ is an eigenfactor for the permutation Q-module VQ,then


464

Proof. By Lemma 1.3, we have

On the other hand, the minimality of Q implies that Vg = 0 for all proper subgroups S of &. Hence

T r G ( K e r ( B r F ) )=

C v,P = o

(3)

SCQ

The desired assertion is therefore a consequence of (2) and (3). W

Let G be an arbitrary group and let A be a left G-module, that is, A is an additive group on which the elements of G act from the left as additive homomorphisms such that 1 . a = a and

z(ya) = ( 2 y ) a

for all a E A , z , y E G

A map f : G + A is called a derivation (or crossed homomorphism) if

If f and g are derivations, then their sum f ig defined by

is again a derivation. It follows easily that the set Z'(G,A) of all derivations of G into A is an abelian group. The derivations fa,a E A, given by fa(x) = 2a - a (x E G) are called principal derivations and form a subgroup B'( G, A ) of 2' (G, A ) . The factor group

H'(G, A ) = Z1(G, A)/B'(G,A ) is called the first cohomology group of G with coefficients in A. In what follows, we use some standard properties of cohomology groups (see Atiyah and Wall (1967)).


465

1.5. Lemma. (i) If V and W are generalized permutation lattices for P , then so is HomR(V,W ) .

(ii) If V and W are permutation lattices for P , then so is H o m z (V,W ) . P

fiii) I f V is a generalized pernutation lattice for P , then (1 - € ) H ' ( P ,V ) = 0

(iv) If V is a permutation lattice for P , then

H'(P,V) = 0 Proof. (i) Applying additivity of both variables, we may harmlessly assume that V = p', W = t+bp where p : Q +< E > and 1c, : S +< E > are homomorphisms for some subgroups Q and S of P. Let V* = H o m ~V, ( R ) be the contragredient of V . Then

by Frobenius reciprocity and Mackey decomposition. Thus HomR(V,W ) is a generalized permutation lattice for P . (ii) Repeat the proof of (i) with p, +, R replaced by l ~I s , Z p re, spect ively. (iii) and (iv). Applying additivity, we may assume that V = pp where cp : Q +< E > is a homomorphism and Q a subgroup of P . In voking Shapiro's lemma, we obtain

This allows us to assume that V has R-rank 1 and that P acts by p. Put I( = K e r p and note that H'(Ii', V ) = H o m ( K , V ) = 0 since V has no torsion. Thus, by the inflation-restriction sequence, we have

H' ( P ,V ) z H'( P / I - ,V )


466

with P / K acting faithfully by 'p. In particular, if 'p = 1 then P / K = 1 and H1(P,V) = 0, proving (iv). If p # 1, then P / K is cyclic of order p , say generated by 9. Thus

{w E V l ( l + g

H l ( P / K ,V)

+ +9 * * *

Choosing g such that 'p(g) = E , we have p(1- 9) = 1 - E . But then

P ( P , V)

'p( 1

q v = 0}/(1 - 9)V

+ g + . . . + gp-')

= 0 and

= H l ( P / K ,V) z V/(1 - €)V,

proving (iii). 4 1.6. Lemma. Let G be a finite group, let S be a commutative

ring and let

o-tw+xAv-to

(4)

be an exact sequence of SG-modules. Assume that (4) is S-split and let h : V -, X be an S-homomorphism such that f o h = 111. (i) The map G + Homs(V, W),g H xg where

is a derivation. (ii) The sequence derivation.

(4) is SG-split if and

only if g H

xg is a principal

Proof. (i) Given g E G and w E V, we have f ( h ( v ) g h ( g - ' v ) )= - 9 g - l ~= 0

which shows that xg(v) E W . Given z , y E G, we also have xZy(v) = h(u) - ryh(y-lz-lw) = [h(v)- zh(z-'v)]

=

+ [zh(z-'w) - zyh(y-'z-' 41

x&) + ( z c X y ) ( 4 ,

proving that g H xg is indeed a derivation. (ii) Suppose that the map g H xg is a principal derivation. This

467


means that there exists cp E Horns(V, W ) such that all g E G. Hence

h ( v ) = gh(g-'w) = gcp(g-'v) - cp(w) and therefore, for all

for all

II

xg = gcp - cp for E V,g E

G

E V,g E G,

+

Setting X(v) = h(v) cp(w), for all w E V , it follows that X is an SGhomomorphism such that f 0 X = l v . Conversely, if X : V + X is an SG-homomorphism such that f o x = l v , then for all g E G , x g = gcp - 'p where cp E Horns(V, W ) is given by cp(v) = X(w) - h ( v ) . The proof is therefore complete. W 1.7. Lemma. RP-modules

Suppose that we are given an exact sequence of

o+w+xtv+o

(5)

such that: (i) V and W are generalized permutation lattices. (ii) The reduction rnod(1 - E )

splits. Then (5) also splits.

Proof. Put A = H o r n ~ ( VW , ) .Since all RP-modules are assumed to be R-free, the sequence (5) is R-split. Hence, by Lemma 1.6, the sequence (5) is RP-split if and only if a suitable 1-cohomology class Ax E H 1 ( P , A ) vanishes. Because A = H o r n ~ ( V , the reduction mod( 1 - e ) of (5) is similarly determined by xx = Ax E H'(P, A ) , where = 0 by virtue of (ii). Now the exact sequence 0 + A % A -+ A + 0, where a ( a ) = (1 - & ) a ,a E A , induces an exact sequence

w),

x,

H 1 ( P ,A ) t H 1 ( P ,A ) t H 1 ( P ,A )


468

It follows from (i) and Lemma 1.5 that the homomorphism H1(P,A ) --+ H'(P, A ) is injective. But AX E H1(P,A ) is mapped to xx = 0, so Ax = 0 as asserted. As a final preparatory result, we now prove the following lemma. 1.8. Lemma. Let V be a n R P - m o d u l e such that P acts faithfully o n V and trivially o n = V/(1 - E)V. T h e n P is elementary abelian.

v

Proof. We argue by contradiction. Choose P to be a minimal counterexample. Since the hypotheses of the lemma carry over to the subgroups of P,it follows that every proper subgroup of P is elementary abelian but P itself is not. For the sake of clarity, we divide the rest of the proof into four steps. Step l.Our aim here is to demonstrate that P is one of the following groups. (a) P is a cyclic of order p 2 . (b) P =< X , T J , Z ~ X = ~ y P = ZP = ~ , z z z - ' = z = YZY-',TJXY-' = 2 5 > , p > 2. Suppose that P is not of type (a). Then P contains no elements of order p 2 , so P is nonabelian of exponent p . Hence the centre 2 of P is a proper subgroup, so the pgroup P/Z has a central element zZ of order p . Since z 6 2, yxy-' # x for some y E P . But ZZ is central in P/Z,so yzy-' = ZIC for some 1 # z E 2. Because < z,y,z > is nonabelian of exponent p and order p 3 , the minimality of P forces that P =< z, y, z >. Finally, the case (b) occurs only for p > 2. Indeed, if p = 2 then z = ( y ~ =) 1, ~ which is impossible. Step 2.Here we assume that P is of type (a) or (b) satisfying the hypothesis of the lemma, where V is chosen to have minimal R-rank. Put M = I( @ R V where I( is the quotient field of R. Our aim is to show that the KP-module A4 is irreducible. Assume by way of contradiction that N is a nontrivial KP-submodule of M . Then V' = N n V , V'' = V/V' are RP-modules of smaller R-rank. The RP-exact sequence


469

is R-split, hence its reduction mod(1 - e) remains exact. Because P

acts trivially on V , it must also act trivially on V ' a n d V . Applying minimality of V , we deduce that P does not act faithfully on V',V''. However, in both cases (a) and (b), P has a unique minimal normal subgroup Po # 1 and so Po acts trivially on both V' and V". Taking the Po-fixed points in (6)) we obtain an exact sequence 0+

v'+ I1ZD(Vpo)+ v"+ H'(P0, V')

Because H1(Po,V') = Horn(P0, V ' ) = 0, it follows from (6) that

1n0(Vpo)=

v

Thus Po acts trivially on V , a contradiction.

Step 3.0ur aim here is to eliminate the case where P =< g > is cyclic of order p 2 . The field 1((6), where 6 P = e, is totally ramified of degree p over K , and R[6]is integrally closed, hence a valuation ring. Observe also that (6- l ) P / ( 1 -E) is a unit of R[6].Now every irreducible KP-module on which P acts faithfully is of the form M j = [ K ( S ) ] ,as a I, every RP-module of R-rank 1 is given by a homomorphism cp : P -+< E >, i.e. P acts on R via cp. We divide the rest of the proof into two steps.

Step 1.Here we treat the more difficult case where Pi # P for some i E { 1,. . . ,r } . Let Q be a subgroup of P minimal with respect to V$ # 0. Owing to Lemma 1.2(i) and the assumption that Pi # P , it follows that & is a proper subgroup of P . Now (VQ) = ( V ) , is a permutation module, hence by induction VQ is a generalized permutation lattice for &. Let 0 : E 4 VQ be an eigenfactor of VQ. Then, as has been VQ observed at the beginning of the section, the induced map 8 : E is an eigenfactor for the permutation module VQ. Applying Corollary 1.4, we therefore infer that

v;

= Tr;(

e(E ) )

(7)


472

We may choose an R-basis of E consisting of eigenvectors e for Q in the sense that for all q E Q,qe = cp(q)e for a suitable homomorphism ‘p :-Q +< E >. Reducing this basis modulo (1 - e ) gives an F-basis of E . Hence, applying Tr; 0 8 to this basis, we obtain a spanning set of # 0, by virtue of (7). Therefore E contains an eigenvector e for Q such that Trc(8(E))# 0 in I n v ( v ) . In what follows, we write cp : Q +< E > for the homomorphism corresponding to e . Consider the RP-homomorphism f : cpp --+ V given by

v[

If

f : ( 1 ~ ) ‘-+ 3 is the reduction of f

modulo 1 - E , then

Because Trg(g(E))# 0, it follows from Lemma 1.2(ii) that f is a split monomorphism of FG-modules. In particular, setting R = 1 - e, we have a commutative diagram

v0-

7t

-

-0

with exact rows and with K e r f = 0. Here T and - denote, respectively, multiplication by R and reduction modulo R. It follows from the snake lemma t.hat R : K e r f + K e r f is surjective and R : Cokerf 3 Coker f is injective. Because R is a valuation ring, we deduce that f is injective and Cokerf is an RP-module, say L , which is R-free of finite rank. Consequently, we obtain an exact sequence

2. Permutation lattices and normal subgroups

473

of RP-modules whose reduction modulo 7r splits. It follows that 1 is a direct summand of the permutation module and so, by the KrullSchmidt theorem, is a permutation module. But L has smaller Rrank than V , so L is a generalized permutation lattice, by the induction hypothesis. Applying Lemma 1.7, we deduce that (8) splits. Thus V L @ (pp is a generalized permutation lattice for P , as required.

v

Step 2.Here we treat the case where all Pi = P (hence P acts trivially on V ) . By induction, we may assume that P acts faithfully on V . Hence, by Lemma 1.8, P is an elementary abelian pgroup. Let K be the quotient field of R and put M = I( @ R V . Then all irreducible KG-modules are of the form K @ R (pp for a suitable homomorphism cp : P -+< E >. Hence M has such a constituent, say N . It follows that V n N is an RP-module isomorphic to 'pp and that V / ( Vn N ) is an RP-module which is R-free of finite rank. This, as in (8), yields an exact sequence 0 + 'pp -+ v -+ L + 0

whose reduction modulo 7r splits, since V is trivial. The desired conclusion now follows by the argument following (8). 2.

Permutation lattices and normal subgroups

All conventions, notation and terminology introduced in the previous section remain in force. In particular, p is a prime, P a finite p-group and Z,the ring of p-adic integers. Let N be a normal subgroup of P and let V be a finitely generated Z,P-module. Our aim is to exhibit circumstances under which V is a permutation lattice for P. Namely, we prove that V is a permutation lattice for P , provided the following two conditions hold: (i) The Z,N-module VN is free. (ii) I ~ w ( V is N a) permutation lattice for P I N . 2.1. Lemma. Let A , B and C be algebras over the same commutative ring S and let cy : A + C and p : B -+ C be algebra homomorphisms. Then there exists an 5'-algebra D , unique up to isomorphism, and there are algebra homomorphisms y : D + A , 6 : D + B such that


474

(i) the diagram D

r

-A

a

6

is commutative; and (ii) if

(2)

is a commutative diagram, then them is a unique homomorphism cp : D'+ D s u c h t h a t y o c p = y ' a n d 6 o c p = 6 ' . Proof. Given a,@,define D by

D = { ( a , b ) l aE A , ~ B E , a ( a ) = @(b)} and let y : ( a , b) H a

(3)

and S : ( a ,b) w b

be the corresponding projections. Then (1) obviously becomes commutative.


475

Suppose that (2) is a commutative diagram, and define cp : D' + D by

cp(d') = (~'(d'),S'(d')) for all

d' E

D'

(here (y'(d'), S'(d')) E D since cu 0 7' = ,B 0 6'). Clearly, rcp(d') = ~ ' ( d ' ) and Scp(d') = S'(d') for every d' E D',and therefore 'p is of the stated kind. It follows from (3) that Kery = (0, K e r p ) and

K e r b = (Kercu,0)

(4)

Therefore, if cp' : D' + D also satisfies y 0 cp' = 7' and S o cp' = S', then

Kery n K e r b = 0. This shows that cp - cp' = 0, and thus Irn(cp - cp') and cp is unique. The uniqueness of D can be established by considering a D with the same properties; we denote the corresponding maps in (1)by 7 : D + A and 8 : D + B. Then we have unique homomorphisms cp : D + D,Cp : D + 0 , such that

Hence ycpq = y and Sp(P = 6,that is 'pp : D + D preserves both the A and B-coordinates and so cpp = 1 on D. Hence Cp(D)= Cpcp(q(D))is a direct summand of the additive group of D , D = p ( D )@ X . Since in (ii), a unique map D' + D was required for every D',X must vanish. Thus (/3 is an isomorphism and the result is established. W We next present some applications of the above lemma. It will be convenient to introduce the following standard terminology. A commutative diagram (1) satisfying (ii) is called a pullback square (or pullback diagram) of algebras. The algebra D defined in (3) is called the fibre product of the pair ( a ,p}. Observe that Lemma 2.1 is valid in case A , B and C are modules and a : A + C, ,G' : B + C are module homomorphisms. For this reason, we use the same terminology for modules, namely pullback square of modules and the fibre product of the pair ( a ,p } . The following is a typical example of a pullback square of

4 76


algebras. Let I and J be ideals of an algebra A. We construct a pullback square of algebras, where all the arrows in the diagram below represent natural homomorphisms. We leave it to the reader to verify that A / ( I n J ) is isomorphic to the fibre product of the pair { a ,p}.

In what follows, p is a prime, F is a field of p elements and R = Z,[E], where E is a primitive p t h root of 1. 2.2. Lemma. Let C =< g > be a cyclic group of orderp, let @ : Z,C t R be the homomorphism of Z,-algebras induces b y g H E , and let a : R 4 F , ,8 : !Up --+ F be reductions modulo T = 1 - c and p , respectively. Then

is a pullback square of !U,-algebms. Proof. Owing to Lemma 2.1, we need only verify that Z p C is identifiable with the fibre product of { a ,p}. Consider the homomorphism


477

A;gi(A; E Z,), then x H (Eyzt A ; E ' , , ~ ~A;).~ Because 1 E - cP-' = 0 and l , ~. .,. , F 2is a 23,-basis of R, it follows that the given map is an injective homomorphism of Z,-algebras. Moreover, we obviously have

If

J:

=

+ +

+

: A;&; E R(A; f Z,) and let A E Z,be such Conversely, let r =::C that a ( r ) = p(X). Then A; = A(modpZ,) and so::C : A; = A+pp for some p E Z p . Setting J:

= (A,

-p)

*

1

+ (A,

-p)g

+ + * * *

(Ap-2

- p)gp-2 - pgp-l,

it follows that $(x) = r and aug(z) = A, as required. 2.3. Lemma. Let C be a cyclic normal subgroup of order p in a finite p-group P , and let V be a Z,P-module such that the Z,C-module Vc is free. Put V, = V/I(C)V,Vz = V/C+V and V3 = V/(C+V I ( C ) V ) , where C+ = CzECx and I ( C ) is the augmentation ideal of Zp C . Then

+

V

is a pullback square of Z,P-modules, where all the maps are natural horno mo rp hisms.


478

Proof. Consider the homomorphism

{

v

vz x v1

+

(r(v),W)

If ~ ( v = ) S(v) = 0, then v E C+V f~I(C)V = 0 since C+I(C) = 0. Hence the given homomorphism is injective. Moreover, we obviously have ar(z))= PS(w) for all w E V Conversely, let (v2,vl) E Vz x V, be such that a(v2) = P(v1). Denote by a l , . . . , a , a Z,C-basis of V. Because a ( q )= P(vl), we may write v2

= A1a1

and

+ - + A,a, + c+v *

*

+

+

p l a l + * * . prar I ( C ) V for some Ai,pi E Z p C such that x;=:=,(Xi-pi)ai E C + V + I ( C ) V (hence A; - pi E I ( C ) Z p C + ) .Because Z,C+ and I ( C ) are ideals of Z p C with intersection 0, there is a pullback square ~1 =

+

where all the arrows represent natural homomorphisms. Therefore there exist t l , . . . , t r in Z p C such that t i = Ai(rnodZ,C+), t; pi(rnodI(C)). Setting v = &tia;, it follows that y(v) = v2 and 6(v) = vl, as asserted. W

=

Let S be a local commutative ring and let A be an S-algebra which is finitely generated as S-module. Then, for any ideal I of A , the natural map U ( A ) + U ( A / I ) is surjective. 2.4. Lemma.


479

Proof. Owing to Lemma 3.1.3(i), A / J ( A )is artinian. The desired conclusion is therefore a consequence of some standard facts on lifting units (see Corollary 7.1.5 in Karpilovsky (1988)). 2.5. Lemma. Let N be a normal subgroup of a finite group G and let S be a commutative ring. If V is an SG-module such that the SN-module VN is free, then

{

V/I(N)V v+I(N)V

+

I~V(VN) N+v

where N + = C I E N x, is an isomorphism of SG-modules.

Proof. If n E N and v E V , then n ( N + v ) = N+v and so N+v E 1 n v ( V ~ )Since . N + I ( N )= 0, the given map is well-defined. Moreover, it is an SG-homomorphism since N + is central in SG. Let B be an SN-basis of V . If v E I ~ ~ ( V N then ) , v = zlbl . - . x,b, for some bl,. . . ,b, in B and some zl,. . .,x, in I n v ( ( S N > N ) Since .

+ +

I ~ ~ ( ( S N= ) NS )- N+ the given map is surjective. Finally, let v E V be such that N+v = 0. We may write v = xlbl * . - xmbm for some xi E S N , b; E B , in which case N+x; = 0 for all i E { l , . .. , m } . Because N+x; = aug(z;)N+, it follows that xi E I ( N ) and hence v E I ( N ) V , as required.

+ +

We have now come to the demonstration for which this section has been developed. 2.6. Theorem. (Weiss (1988)). Let p be a prime, let Z pbe the ring of p-adic integers and let N be a normal subgroup of a finite p group P . Assume that a finitely generated ZpP-module V satisfies the following two conditions: (if The Z,N-module VN is free. (ii) Inv(VN) is a permutation lattice for P I N . Then V is a permutation lattice for P .

Proof. It will be convenient to divide the proof into two steps.


480

Step 1.Reduction to the case where N = C is cyclic of order p and central in P. Suppose that the result is established under the above hypothesis. The case IN1 = 1 being trivial, we argue by induction on IN[. Assume that 1 # N CI P satisfies (i) and (ii) and choose a central subgroup C of P of order p such that C C N . Put W = Inv(Vc),P = P / C and N = N / C . Then W is a Z,P-module and N is a normal subgroup of p . Because V is a free Z,N-module, Inv(Vc) is a free Inv((Z,N)c)-module. But, by Lemma 2.5,

Z,N g Z , N / I ( C ) Z , N g I n u ( ( Z , N ) c ) and therefore Wjq is a free Z,N-module. Because

Inv(WN) = Inv,,C(Inv(VC)) = Inv(V,>, we see that 1 n v ( W ~is>a permutation lattice for P I N = P / N . Bearing in mind that IN1 < I N ] , it follows from the induction hypothesis that W is a permutation lattice for P = P / C . Since restricting a free Z , N module V , to C trivially yields a free Z,C-module? it follows from our assumption that V is a permutation lattice for P , as required.

Step 2.Here we complete the proof by treating the crucial case where N = C is cyclic of order p and central in P . Then the hypotheses of Lemma 2.3 are fulfilled and we use the notation of that lemma. Owing to Lemma 2.5, V, Z Inv(Vc) as Z,P-modules. Thus hypothesis (ii) says that V, is a permutation lattice for P . The images of P ( V / ~ ( C ) V=)(PV W V ) / W ) V

+

and

(C+V

+ I(C)V)/I(C)V

under the isomorphism of Lemma 2.5 coincide. Hence

and therefore & / p x E V3. Thus V3 is permutation module for P (over the field F of p elemets). But now C+ annihilating Vz implies that we may consider V2 as an RP-module, via the isomorphism 1c, : Z,C/Z,Cs -+ R, since C C Z ( P ) ensures that the R and P-actions


481

commute. Then V./rV2 E V,,where r = 1 - E , so Theorem 1.9 implies that V2 is a generalized lattice for P . By the following, we may write

V, E @

k ( l ~ ~ (as ) ~

Z,P-modules)

(5)

= @TVT

(as RP-modules) (6) for some subgroups Q k of P and some homomorphisms 9, from a subgroup P, of P into < .E >. Bearing in mind that v 2

v, E V1/pK 2 V,/rv, we deduce that @ k ( l ~ ~ ) ' 3 @,(lp,)' as FP-modules. Since our permutation modules are indecomposable, it follows from the KrullSchmidt theorem that we can re-index the k's as T ' S in such a way that (1Qr)' 2 ( I P ~ ) for all T Moreover, uniqueness of vertices up to P-conjugacy ensures that we can replace Q, by P,,i.e. we can choose our initial isomorphisms ( 5 ) and (6) as V, Z @ , ( I P , ) ~ (as Z,P-modules) (7)

v, ="

@T(PT

(as RP-modules)

(8)

Now C acts on V, via II, : Z,C + R, by the definition of the R-action, so every (P, has the same restriction II, to C. Hence Q , = Kercp, has index p in P, 2 C, Q, n C = 1. Thus P, = C x Q, for all T and therefore by Lemma 2.2, we have the Z,-algebra pullback square

On applying induction from P, to P , this becomes a P-module


482

pullback square

Letting W = mT(lQr)', as Z,P-module, the direct sum of the above squares over T can be viewed as a pullback square for Wi

with

WI = W/I(C)W,w 2 = W/C+W

and w 3

+

= W/(C+W I ( C ) W )

and all the arrows being the natural homomorphisms. By construction of W , the isomorphisms (7) and (8) may be viewed as P-module isomorphisms f : W2 4 V, and h : Wl + V, over R and Z,, respectively. We therefore obtain the following diagram with f,h isomorphisms and W,V faces pullback squares, all of P-modules (the maps fi and f are induced by h and f in the obvious way and we seek to construct A). Suppose that fi = f,i.e. there is only one arrow W3 + V,. Then the above diagram commutes and so a diagram chase shows that X exists


483

and is uniquely determined by the diagram. Thus X : W -+ V is an isomorphism of Z,P-modules, proving that V is a permutation lattice for P.

I\

'\

I

A

h

is a P-module Because i,fare isomorphisms, the map ,f3 = f=' o automorphism of W3.We claim that there exists a P-module automorphisn p of Wl such that jl = ,B; if sustained, we simply replace h by h o p-' in the above diagram, in which case h

( h 0 $1

= i0 ($1

*

=h0

A

(h)-lo

f=f

But then the result follows by the preceding paragraph, so we are left to prove the claim. Wl/pWl. Put By construction, ,f3 is a P-automorphism of W3 E = E n d Z (Wl). Then I n v ( E ) = Endzp,(W1) and

EndFG(Wl/PWl) = fnv(End~(Wi/pW~)) = Inv(E/pE) Thus we are left to show that the ring homomorphism

is surjective on units. Because Wl is a finitely generated Z,P-module, it follows from Lemma 3.1.7 that Endzp,(Wl) = I n v ( E ) is a 22,algebra which is finitely generated as a Z,-module. Therefore, by


484

Lemma 2.4, it suffices to show that the ring homomorphism (9) is surj ect i ve . Now Wl is a permutation lattice for P, hence by Lemma 1.5(ii), (iv), H ' ( P , E ) = 0. But then the exact sequence

0 + E -+ E

E/pZ

i

+0

induces an exact sequence

0

+ I n v ( E ) + In?@) + I n v ( E / p E )+ 0

as desired. 3. Some bimodule isomorphisms

Throughout this section, unless explicitly stated otherwise, G and H denote arbitrary groups and R an arbitrary commutative ring. If V is a finitely generated (left or right) free RH-module, then any two bases of V have the same number of elements (called the rank of V ) . This is so since RG admits a homomorphism into a field. Given a positive integer n, we write M,(RH) for the ring of all n x n-matrices over R H . The unit group of M,(RH) will be denoted by GL,(RH). We say that V is an (RG, RH)-birnodule if V is a left RG-module and right RH-module such that (xv)y = x(vy)

for all x E RG, y E RH, w E V

and rv = vr

for all

T

E R, w E V

Two (RG, RH)-bimodules V and W are called isomorphic if there exists an isomorphism f : V + W of additive groups such that f ( m y ) = xf(v)y

for all

5

f RG, y E

RN,w E V

It is straightforward to verify that V E W if and only if there exists an R-isomorphism f : V + W such that

f(gvh)=gf(v)h

for all g E G, h E

H,vE V

3. Some bimodule isomorphisms

485

Observe also that if V is a left RG-module, then V can be viewed as an (RG, R)-bimodule by setting v r = rv for all r E R,v E V . Furthermore, if W is another left RG-module, then V 2 W as RGmodules if and only if V % W as (RG, R)-bimodules. Let S be a subgroup of G and let V be an (RS,RH)-bimodule. Because RG is an (RG, RS)-bimodule, RG @RS V is an (RG, R H ) bimodule, where the bimodule structure is given by t(x 8 v ) s = tx 8 vs

(t, x E RG, v E V,s E R H )

We refer to R G @ , R ~as V the induced bimodule and denote it by V G .The special case where H = 1 corresponds to the usual notion of induced module. Let V be an (RG, RH)-bimodule and let < V >= { < 2) > 1v E V } be an R-module copy of V . Then < V > can be viewed as an R ( G x H ) module by setting ( g , h )< v

>=< gvh-l >

(g E G , h E

H,v E V )

Let V and W be two (RG, RH)-bimodules. Then V S W as (RG, RH)-bimodules ifand only if< V >2< W > as R(Gx H ) -

3.1. Lemma.

modules.

Proof.

Let $ : V

+W

be an R-isomorphism such that

f ( g v h ) = gf(v)h

for all g E G, h E H , v E V

Then the map $* :< V >+< W an R-isomorphism such that $*((g, h) < v >) =< f(gvh-l)

> given by $*(< v >) =< f ( v ) > is

>=< g f(v)h-l >= (9,h)+*(< ?>) I

for all g E G , h E H,v E V . Because the argument is obviously reversible, the result follows.

A matrix representation of G over RH is a group homomorphism p :G

--f

GL,(RH)

(1)

486


for some n 2 1. We refer to n as the degree of p. Two such representations p1 and p2 of degree n are said to be equivalent if there exists A E GL,(RH) such that

Given p as in (1))define p* : RG + M,(RH) by

Then p* is a ring homomorphism. Hence, if V, is the free right RHmodule of all n x 1-matrices over RH, then the formula

defines Vp as an (RG,RH)-bimodule. We refer to V, as the (RG, RH)bimodule corresponding to p. It is a standard fact that one may treat the terms "matrix representation of G over R' and "R-free left RG-modules of finite rank" as interchangeable. However, a modification is required when R is replaced by RH. Namely, the following property holds.

3.2. Lemma. The map p H Vp induces a bijection between the equivalence classes of matrix representations of G over RH and the isomorphism classes of (RG, RH)-bimodules which are free of finite rank as right RH-modules.

Proof. Let Q and p be matrix representations of G over RH of degree n. A map f : V, --t Vp is an isomorphism of right RH-modules if and only if there exists A E GL,(RH) such that f ( v ) = Av for all v E V,. On the other hand, for any such f , f(gv) = gf(v) for all v E V,g E G if and only if P ( g ) = Aa(g)A-' for all g E G. This demonstrates that (Y is equivalent to p if and only if V, E Vp. Let V be an (RG,RH)-bimodule which is free of finite rank as a right RH-module. Because (gv)x = g(vs)for all g E G, v E V, x E RH, the map pg : V + V given by pg(v) = gv is an automorphism of V . Thus, by identifying V with the right RH-module of n x 1-matrices E GL,(RH). It follows over RH, we have pg(v) = rgvfor a unique rg

487

3. Some bimodule isomorphisms

that p : G + G L n ( R H ) ,g that V, = V , as required.

I+

rS is a representation of G over R H such

We are now ready to prove our main result, in which p denotes a prime and Z pthe ring p-adic integers. 3.3. Theorem. (Weiss (1988)). Let G and H be finite p-groups GLn(ZpH) be a matrix representation of G over Z p H . let p : G Define p* : G + G L n ( Z p )b y composing p with the homomorphism GL,(Z,H) + GLn(Zp) induced b y the augmentation map. Let M be the (Z,G,Z,H)-bimodule corresponding to p and let N be the Z,Gmodule corresponding to p*. If

for some subgroups GI,. . . , G, of G, then ME@T r=l

M piG

as (Z,G, Z,H)-bimodules

for suitable homomorphisms pi : G; + H , where Mpi is the ( Z p G i ,Z , H ) bimodule corresponding to p;.

Proof. By definition, M is viewed as n x 1 column vectors with entries in Z p H on which G acts by (left) matrix multiplication via p and Z , H acts by (right) scalar multiplication. Let K =< M > be the Z p ( Gx H)-module defined by (2,y

)

< M >=< p(x)rny-l >

(z E

G, y E H , rn E M )

Then, by Lemma 3.1, the isomorphism classes of the ( Z p G , Z p H ) bimodule M and the Z p ( Gx H)-module < M > determine each other. Put P = G x H and L = 1x H . Because M is a free right Z,H-module, A'L is a free Z,L-module. Therefore, by Lemma 2.5, K / I ( L ) K Inv(l{~)as Z,P-modules. Now I% N as Z,(P/L)-module, via the identification P / L = G induced by the

projection G x H

--+

G. It follows, from our hypothesis on N , that

K/I(L)I< E $i==l(lGi)G

via

P/L =G

(2)


488

Invoking Theorem 2.6, we deduce that

K 2 $$=l(lpj)p

(3)

(Pjis a subgroup of P) is a permutation lattice for P.

Because L a P and KL is a free Z,L-module, it follows from (3) and Mackey decomposition that

Pjn L

for all j E { I , .

=1

. . ,k}

(4)

Observe also that

and

Z P ( P / L )@ Z p P ( l P J P = %(P/L) @ Z p P ( Z P P @ Z p P 3%> Z P ( P / L )@ Z P P j ZP

= 2

PIL Iq LIL

(by (4))

Applying (2) and (3) and using identification P / L = G, we conclude that G @:=l(lG,)G

(lpr~(P>))

Invoking the Krull-Schmidt theorem, we have r = k and, after suitable renumbering, (lGi)'

= ( l V G ( p i ) )G

for all i E ( 1 , . . . , r }

Moreover, replacing P;by a P-conjugate, we may assume that prG(Pi)

= G;

for all i E { 1 , ..., r }

(5)

Therefore, by (4)and ( 5 ) , composing the inclusion Pi 4 P with the projection p r G induces an isomorphism P;--+ G;. Thus, for each z E G;, there exists a unique p;(z) E H such that ( z , p ; ( z ) ) E Pi. It follows that p; : G; --+ H is a group homomorphism such that

4. Applications

489

Let Mp, be the (Z,G;, Z,H)-bimodule corresponding to pi. Then the induced (Z,G, Z,H)-bimodule ME corresponds to the Z p ( G x H ) module < M f > satisfying

with P, determined by (6), as is easily verified. Consequently, by (3) and the fact that k = T ,

(e3L1M.)

=

e3;=l(lPi)p

E

< M >

as Z,(G x H)-modules. Invoking Lemma 3.1, we finally deduce that

as ( Z p G ,Z,H)-bimodules, thus completing the proof. W

4.

Applications

In this section, we apply Theorem 3.3 to obtain some results pertaining to the isomorphism problem for group rings described below. Let RG be a group ring of a finite group G over a commutative ring R. There has been a considerable amount of work over the years dedicated to the following problem: To what extent does RG determine the group G? A favourite gambit of group ring theorists has been to impose some conditions on the ring R in the expectation that RG determines G up to isomorphism. There is a striking example of Dade (1971) of two nonisomorphic metabelian groups G and H such that such that for all choices of the field R , RG and RH are R-isomorphic. Therefore, generally speaking, a field is not a suitable candidate for R. The rings R for which the group ring RG yields the most information on the structure of G are integral domains of characteristic 0 in which no rational prime divisor of the order of G is invertible. Typical examples of such rings are as follows: (a) The ring R of algebraic integers in some finite extension of the rationals (in particular, the ring Z of rational integers). (b) The ring Z(n)= { a / b J a ,b E Z , b # 0,n = /GI, (b,n) = l}.


490

(c) The rings 22, and Z(,in )case G is a p-group. Of special interest is, of course, the case where R = Z . The isomorphism problem in this case may be stated as follows: The isomorphism problem. Is it true that Z G 2 Z H implies

G Z H? To present some results pertaining to the above problem, we first record a number of preliminary observations. A homomorphism from the group ring RG1 to the group ring RG2 is, by definition, a ring homomorphism which is also an R-homomorphism. The augmentation ideal I ( R G )is the kernel of the homomorphism from RG to R induced by collapsing G to 1. Explicitly, I ( R G ) consists of all 2

=L 9 ,

x9 E R,g E G

for which

aug(x) = E x g = 0 We shall write I ( G ) instead of I ( R G ) when there is no danger of confusion. A unit u in RG is called trivial (respectively, normalized) if u = u g g for some ug E U ( R ) ,g E G (respectively, if aug(u) = 1). A normalized group basis of RG is, by definition, a group basis consisting of normalized units. We shall write RG = R H to indicate that H is a normalized group basis of RG. Note also that if H1 is another group basis of RG, then

RG = RH

where H = {aug(t-l)tlt E H I }

and H E HI. Therefore the isomorphism problem may be stated as follows: Does R G = RH imply G Z H ? Let J be an ideal of RG. Then the multiplicative kernel of the map G RGIJ is G n (1 t J ) = { g E GJg- 1 E J } --f

and G + J will denote the image of G under this map. In other words,

G + J = ( g t J J gE G }

4. Applications

491

With this notation, we obviously have

G / ( G n (1

+J))2 G +J

(1)

Finally, note that if R G = R H then aug = aug' where aug' is a homomorphism RG + R induced by collapsing H t o 1. Hence I ( G ) = I ( H ) and every unit normalized with respect to G is also normalized with respect to H . 4.1. Lemma. Let a be an algebraic number and let n be a natural number such that na is an algebraic integer. If (a1 = a,a2,. . . ,at} is the set of all Q -conjugates o f a , then either a is an algebraic integer or in the ring 5% [al,a2, . . . ,at]at least one rational prime divisor of n

is invertible. Proof. Assume that a is not an algebraic integer. Then there exists an elementary symmetric function f of t variables such that f (a1,. . . at) @ i2 . Since na is an algebraic integer )

f ( a l , a 2 ., . , a t )= a j b

for some a , b E Zt

such that ( a ,b) = 1,b > 1 and all prime divisors of b are divisors of n. If p is one of these divisors, then because of (a,p) = 1, there exist c, d E Z such that ac dp = 1. It is clear that

+

and hence

a l p E 5%

[a19a2,

- ,at] *

But then

as required. H

4.2. Theorem. (Saksonov (1972)). Let R be an integral domain of characteristic 0 in which no rational prime divisor of IGl is a unit. If u = Cu,g(u, E R,g E G ) is a unit of a finite order in R G with


492

u1 # 0, then u = u1 . 1. In particular, all central units offinite order in RG are trivial.

Proof. Let z = C z g g ( z gE R,g E G) be a central unit of finite order. Then zt # 0 for some t E G and therefore u = zt-' is a unit of finite order with u1 = zt # 0. Hence it suffices to show that if urn= 1 and u1 # 0 then u = u1 1. To prove the required assertion, let t r ( z ) be the trace of z E RG in the regular representation of RG. Then the matrix of u is conjugate to diag(e1,. . . ,E,), n = IGl, and

-

-

t r ( u ) = u1 n = E~

+ + . - .+ ~2

E~

where E ; , 1 5 i 5 n , is an rn-th root of unity (belonging to a sufficiently large field containing R). By looking at the tr(u') where ( r , m ) = 1 , we deduce that the set {PI = u l ,p2,.. . ,pa}of all (Q -conjugates of u1 is a subset of R and thus

Because lGlu1 is an algebraic integer, Lemma 4.1 may be employed to infer that u1 = (el . e n ) / n is an algebraic integer. Since u1 # 0, the latter implies ~1 = e2 = ... = cn, hence the result. 1

+ +

4.3. Proposition. Let R be an integral domain of characteristic 0 in which no rational prime divisor of [GI is a unit. (i) I f H is a torsion subgroup of normalized units of RG, then H is a linearly independent set (in particular, H is finite) and the order of H divides the order ofG. (ii) Let RG = RH, let N be a normal subgroup of G and let N* = H n ( l + R G - I ( N ) ) .Then R G . I ( N ) = R G . I ( N * ) , I N I = IN*[ and

R(G/N) R(H/N*) (iii) If RG = RH and G is p-nilpotent, p a prime, then H is p nilpotent and RP RQ where P and Q are Sylow p-subgroups of G and H , respectively.

4. Applications

493

Proof. (i) Assume by way of contradiction that C;"=lcvihi= 0 where hi E H , a ; E R,i E (1,...,n } and cuj # 0 for some j E { l , ..., n } . Then aj - 1 = -C'Y;(h;h;') i#j

and if we express the elements hihj', i # j , in terms of the elements of G then at least one of them, say hkhj', k # j , has a nonzero coefficient of 1. Hence, by Theorem 4.2, h&J' = a * 1 for some cu E R. Since a u g ( h k ) = aug(hj) = 1 we have h k = hj, a contradiction. Put e = IHJ-' ChEH h. Then e is obviously an idempotent of FG, where F is the quotient field of R. For any 2 = Cz,g E RG, put t r ( z ) = xl. If h # 1, then since aug(h) = 1, we have t r ( h ) = 0 by Theorem 4.2. Thus ire = /HI-'. On the other hand, it is easy to see that t r ( e ) = IGI-'rn for some rn 2 1. Hence /GI = lHlm as asserted. (ii) Let RG = R H and let T : RG + RG, G = G / N , be the natural homomorphism. Then, by (i), the image H of H in RG is a linearly independent set in RG which obviously spans RG, hence RG = RH. Moreover, since T can be regarded as the extension of the epimorphisn H + H (whose kernel is N * ) by R-linearly, we have

K e r T = RG . l ( N ) = RH I ( N * ) Furthermore, since IG/NI = [GI = = IH/N*I, we have IN1 = IN* which proves the required assertion. (iii) Let N be a normal p'-subgroup of G such that GIN 2 P . Then, by (ii), N* Q H is such that [NI = IN*] and H / N * 2 Q, where Q is a Sylow p-subgroup of H . Since, by (ii), R ( G / N )E R ( H / N * ) ,the result follows. 4

Let R be an integral domain of characteristic 0 in which no rational prime divisor of /GI is a unit. If S is a subgroup of G , then for I ( G ) = I ( R G ) and I ( S ) = I ( R S ) , 4.4. Lemma.

G n (1 + I ( G ) I ( S ) )= S' *

Proof. We first consider the special case where S = G. By taking the case n = 2 in Theorem 2.1 of Sandling (1972) we see that

s n (1 + I ( S ) ~=)S'


494

whenever Tp(SmodS') = S for all primes p for which peR = p"+'R for some nonnegative integer e(Tp(SmodS') denotes the subgroup of S generated by all elements of S some p t h power of which is in 9).It is clear that Tp(SmodS') = S'

if p does not divide 15'1. If p is a prime such that p"R = pe+lR for some e, then p"( 1- p z ) = 0 for some z E R and since R has no zero divisors, p is a unit of R. This shows that p does not divide IS1 and completes the proof of the special case. To prove the general case, let T be a left transversal of S in G containing 1 and let g = t s be a typical element of G ( t E T ,s E S). Consider the R-linear map

(p:RG+RS which is the R-linear extension of g z E I ( S ) and the equality

H

s. Then p(z) = z for any

shows that cp(I(G) I ( S ) )= I(S)'. Consequently, a

I ( G ) I ( S ) n I ( S ) = I ( S)' which obviously implies

+

G n (1 I ( G ) - I ( S ) )= S n (1 + I ( S ) ' ) , thus completing the proof by applying the special case established above. H

Let m be the exponent ofGIGI and let R be any commutative ring such that R/mR Z / m Z . Then, for any t E I(RG), there ezists g E G szlch that t = g - l(rnodI(RG)2). 4.5. Lemma.

Proof. The identities

+

+

~b - 1 = ( a - l ) ( b - 1) ( a - 1) ( b - 1) ( a , b E RG) [ a ,b] - 1 = a-lb-' [ ( a- l ) ( b - 1) - ( b - l)(a - l)] ( a , b E U(RG))

4. Applications

495

show that the map p : G -+ I ( R G ) / I ( R G ) 2given by

+

‘ p ( g ) = ( g - 1) I ( R G ) 2

for all g E G

is a homomorphism with G’ Kerp. Since gm E G‘ for any g E G, we have I ( R G )= ~ ‘ p ( g m ) = mcp(g)= m(g - 1) I ( R G ) ~

+

and therefore mr(g - 1) E I(RG)2for all r E R,g E G. Since a typical element of R is r = t - 1 mrl where rl E R and t E {0,1,. . . ,m - l}, we have p ( g t ) = r(g - 1) I ( R G ) ~

+

+

This shows that

‘p

is surjective and completes the proof.

4.6. Lemma. Let A be an abelian normal subgroup of G and let R be an integral domain of characteristic 0 in which no rational prime divisor of [GI is a unit. Assume further that R/mR E Z / m Z where m is the exponent of A . If H is a normalized group basis of RG and the image of each h E H in R ( G / A ) is a trivial unit, then G E H .

Proof.

By Lemma 4.4,we have

+

G n (1 I ( G ) I ( A ) )= 1 and therefore, by (I), G + I ( G ) . I ( A )E G. If A* is the normal subgroup of H corresponding to A (see Proposition 4.3(ii)), then RG . I ( A ) = RH I ( A * ) and hence I ( G ) * I ( A ) = I ( H ) . I ( A * ) . The latter (as the reader may easily verify) implies

H n ( l + I ( G ) . I ( A )= 1 It follows that H

+ I ( G )- I ( A )Z H . H

It will next be shown that

+ I(G) I(A)C G + I(G) *

which will complete the proof. Fix h E H . By hypothesis, h Since

*

I(A),

= g(modRG. I ( A ) )for some g

+

RG I ( A ) = I ( A ) I ( G ) * I ( A ) *

E G.


496

= +

-

we have h g t ( m o d l ( G ) I ( A ) )for some t E I ( A ) . Furthermore, by Lemma 4.5, t G a - l(modl(A)’) for some a E A. Hence

h

= g + ( a - 1 ) = (1 - g ) ( a - 1) + ga G g a ( m o d l ( G )- I ( A ) ) ,

thus completing the proof.

4.7. Corollary. (Whitcomb (1968)). Let G be a finite metabelian group. Then ZIJG determines the isomorphism class o f G .

Proof. Apply Lemma 4.6 and Theorem 4.2. From now on, we concentrate on some applications of Theorem 3.3, which will provide us with another class of groups determined by their integral group rings.

Lemma. Let R be a commutative ring and let G and H be finite groups such that G is a subgroup of U ( R H ) . If V is the 4.8.

(RG,RH)-bimodule corresponding to the inclusion G + U ( R H ) , then G is conjugate to a subgroup of H if and only i f there exists a homomorphism u : G + H such that V E V, (equivalently, b y Lemma 3.1, < V >E< V, > as R(G x H)-modules). CY : G t H is a homomorphism such that V,. The inclusion G --t U ( R H ) and CY : G t U ( R H ) are matrix representations of G over R H of degree 1. Hence, by Lemma 3.2, these representations are equivalent. Thus there exists a unit u E R H such that a ( g ) = ugu-* for all g E G. This shows that G is conjugate to the subgroup u(G) of H . Conversely, assume that there exists a subgroup S of H and u E U ( R H ) such that S = uGu-l. Define u : G + H by u ( g ) = ugu-l for all g E G. Then u : G + U ( R H ) and the inclusion G + U ( R H ) are equivalent matrix representations of G over R H . Hence, by Lemma 3.2, V E V, as required. 4

Proof. Suppose that

V

Z

Let G be a finite group and let R be a commutative ring. The set of all normalized units of RG constitutes a subgroup of U ( R G ) which

4. Applications

497

we denote by V ( R G ) . Since U(RG) = U ( R ) x V ( R G ) ,the study of U(RG)is equivalent to that of V ( R G ) .

(Weiss (1988)). Let p be a prime, let Z,be the ring ofp-adic integers and let P be a finite p-group. If Q is a finite subgroup o f V ( Z , P ) , then 4.9. Theorem.

uQu-'

P

for some u E V ( Z p P )

Proof. Owing to Proposition 4.3(i), Q is also a pgroup. We now apply Theorem 3.3 for n = l , Q = G , P = H and p : Q + V ( Z , P ) being the inclusion map. Then N = l ~so, T = 1,G; = G and hence, by Theorem 3.3, M E Adpl. Thus, by Lemma 4.8, zQ2-l P for some 5 E U ( Z , P ) . Setting u = aug(z-l)x, the result follows. 4.10. Corollary. (Roggenkamp and Scott (1987')). Let p be a prime, let G be a finite p-group and let H be a subgroup of V ( Z p G ) with \HI = /GI. Then H = u-lGu for some u E V ( Z , G ) .

Proof.

This is a direct consequence of Theorem 4.9.

4.11. Remark.The above result was established by Roggenkamp and Scott under more general circumstances. Namely, instead of Z p one can take a local or semilocal Dedekind domain of characteristic 0 with a unique maximal ideal containing p . a

4.12. Corollary.

Let G be a finite nilpotent group. Then

ZGZ ZH

implies G % H

Proof. If G is a finite p-group, then the result follows from Corollary 4.10. The general case follows from this special case by applying Proposition 4.3(iii). 4.13. Theorem. Let A be an abelian normal subgroup of a finite group G such that GIA is a p-group for some prime p. If Z G % ZH , then G E H .

498


Proof. Let n = (GI and R = { a / b l a , b E Z , b # O,(b,n) = 1). Then RG E RH and we show that G E H . We may harmlessly assume that H is a normalized group basis of RG. It is clear that R is an integral domain of characteristic 0 in which no rational prime divisor of n is a unit. Moreover, for any natural number rn dividing n, R / m R E Z/ m Z . Consider the natural map f : RG + R(G/A). Then, by the proof of Proposition 4.3(iii) RG = RH, where G = G / A and H = f ( H ) . Hence, by Remark 4.11, H = u-lGu for some u E U(RG). Since R is semilocal, there exists a unit v E U(RG)such that f(v) = u. Hence, replacing H by uHu-', if necessary, we may assume that H = G. The desired conclusion is now a consequence of Lemma 4.6. We close by remarking that Roggenkamp and Scott (1987) announced that any finite abelian-by-nilpotent group is determined by its integral group ring. A more general version of Theorem 4.13 can be found in Roggenkamp and Scott (1987, Corollary 5 ) .

499

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511

Not ation N u m b e r Systems

the natural numbers the rational integers the padic integers the rational numbers the real numbers the complex numbers the integers mod rn Set theory C

E

1x1

x-Y

proper inclusion inclusion the cardinality of the set X the complement of Y in X

Number theory a divides b the greatest common divisor of a and b the p-part of n the p’-part of n

512

Not a tion

Group Theory the multiplicative group of a field F the subgroup generated by X the cyclic group of order n direct product of GIand G2 N is a normal subgroup of G the centralizer of X in G the normalizer of X in G - x-l -1 Y XY the commutator subgroup of G the centre of G = {9"l9 E GI = {g E Glg" = 1) the group of all nonsingular linear transformations of V '

the the the the

image of J: under the action of g E G stabilizer of J: E X automorphism group of X p-component of A

Rings and modules the F-dimension of V projective cover of V the opposite ring the centre of R tensor product the characteristic of R direct sum direct product of rings polynomial ring on X over R the Jacobson radical of a ring R the radical of an R-module V the submodule of invariant elements

Not at ion

VH VG

513 the restriction to H induced module relative trace map dual module contragredient module outer tensor product the algebra of R-endomorphisms of V the principal ideal generated by r n x n-matrices with entries in R

=L@Fv =L@FA the Loewy length of V the annihilator of M G-conjugate of V the intertwinning number inflated module extension of V the matrix unit n-th direct power of V support of z the g-component of A the crossed product of G over R the unit group of A the degree of u graded units of A the inner automorphism induced by u crossed system the class sum the sum of all elements in X the trivial RH-module = a for all h E H} = { u E A1 = Trg(AH) right annihilator of X left annihilator of X

514

Not ation the group algebra of G over R the augmentation ideal of RG the group of normalized units of R G the augmentation map the socle of V = TrE(Inv(VH)) = WVK)/(CHCK VHK J ( R ) W V K ) ) the Brauer morphism the Martindale ring of quotients of S the twisted group algebra of G over F the block containing e the commutator subspace HI is G-conjugate to a subgroup of H2 HI is G-conjugate to Hz the defect group of e

+

Cohomology theory the group of all A-valued 2-cocycles of G a coboundary the subgroup of ZZ(G,A ) consisting of all = Z2(G,A)/B2(G, A ) coboundaries the cohomology class of f = Z'(G, A)/B'(G,A) the corestriction map the restriction map obstruction cocycle inflation map Field theory field extension degree of F over I< Galois group of F over Ii the smallest subfield containing S and Ii the smallest subfield containing Ii and Ql,.

. . ,Q,

Notation

515 the algebraic closure of F the norm of p over F the trace map

516

Index algebra 2 definable over a subfield 349 graded 142 homomorphism of 2 strongly graded 142 Alperin 299,330,333,337,338,383, 428, 430, 432,448, 449 annihilator 22 A sano-0sima-Takahasi theorem 179 Atiyah 464 basis 3 bilinear form 210 associative 21 1 nonsingular 21 1 symmetric 211 bimodule 3 Blackburn 42, 43,210, 427 block 344 block idempotent 344 Bourbaki 5 Brauer 332, 337, 358 Brauer’s induction theorem 332 Brauer-Jennings-Zassenhaus series 42 7 Brauer lift 337 Brauer morphism 439 BrouC 383,400, 402,404,439,448, 451, 452, 455 Burnside 44

Burry 314, 315, 317, 448 Cabanes 456,457 canonical injection 55 Carlson 314 Cauchy sequence 132 centralizer 3 character 40 degree of 40 module of 40 characteristic polynomial 222 Clarke 268 class function 40 Clifford 75, 245, 257 Clifford’s theorem 75 Cline 164 coboundary 167 cocycle 167 Cohen 186 cohomology class 167 cohomologous cocycles 167 crossed homomorphism 464 commutator subspace 174 complete discrete valuation ring 186 composition length 13 conjugate characters 101 conjugate modules 74 Conlon 153, 184 control 404 crossed product 145

INDEX crossed system 146 Curtis 42, 102, 222, 223, 358, 436 Dade 141, 149, 153, 229, 235, 263, 489 derivation 464 Dickson 118 direct product 2 disjoint modules 93 Dornhoff 42, 191 Dress 383, 433, 435, 437, 438 eigenfactor 460 extension of domains 186 extension of modules 154 Feit 191 fibre product 475 finite extension 186 first cohomology group 464 Fong 265 Frobenius 422 Frbenius algebra 212 Frobenius complement 422 Frobenius kernel 422 Frobenius reciprocity 65 Frobenius reciprocity for characters 66 fully invariant submodule 18 G-algebra 386 generalized permutation lattice 460 generalized I

Induced Modules over Group Algebras (North-Holland Mathematics Studies)

Induced modules over group algebras

Induced modules over group algebras