Chapter 6 Simple induction and restriction pairs

341

Chapter 6 Simple induction and restriction pairs In this chapter we investigate in detail circumstances under which restriction and induction of irreducible modules are completely reducible modules. After presenting some general background pertaining to block theory, we establish a fundamental result due to Green (1959) and Knorr (1976) which characterizes defect groups of blocks in terms of vertices of indecomposable modules. We then turn our attention to the study of simple induction and restriction pairs introduced by Knorr (1977). 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 G is completely reducible for any irreducible FH-module V and M H is completely reducible for any irreducible FGmodule M ; (ii) There exists a normal subgroup N of G such that N H and N has p’-index in G; (iii) J ( F G ) = FG J ( F H ) .

1. Blocks of algebras In this section, A denotes a finite-dimensional algebra over a field F . We begin by recalling the following information. By an F-representation of A (or simply a representation of A ) we understand any homomorphism A EndF(V) of F-algebras, where V is a (finite-dimensional) vector space over F . If n = d i m ~ Vthen ,

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EndF(V) is identifiable with the F-algebra Mn(F). Hence the given representation of A defines a homomorphism A + M n ( F ) ;we shall refer to any such homomorphism as a matrix representation of A. If f : A + EndF(V) is a representation of A, define ICW = f ( z ) v for all x E A,v E V . In this way V becomes an A-module, called the underlying module of f. Conversely, if V is an A-module, define

f : A + EndF(V) by f ( s ) v = xv for all IC E A,v E V . Then f is a representation of A; in case V is the regular A-module, we shall refer to f as the regular representation of A. Two representations f; : A + E n d F ( F ) , i = 1,2, are called equivalent if there exists an F-isomorphism $ : Vl + such that fi(z)= $ ~ I ( I c ) $ - ~ for all x E A It is easily seen that two representations of A are equivalent if and only if the underlying modules are isomorphic. We say that a representation f : A + EndF(V) is irreducible (completely reducible, indecomposable) if the underlying module V is irreducible (completely reducible, indecomposable); the same terminology will be applied to matrix represent at ions.

1.1. Proposition. The following properties hold: (i) There exists a direct decomposition

of A into indecomposable two-sided ideals B;

#

0 with B,Bj = 0 for

i#j (ii) Write 1 = el en with e; E B;. Then the e; are mutually orthogonal centrally primitive idempotents and B; = Ae; = e;A. (iii) Z ( A ) = Z(A)el @ @ Z(A)e, is a direct decomposition of Z(A) into indecomposable ideals and Z ( A ) e ; = B; n Z(A). (iw) J(Z(A)e;) = J ( Z ( A ) ) e ;and Z(A)ei/J(Z(A)e;) is a finite field extension of F . (w) Let F; = Z(A)ei/J(Z(A)e;), let di : Z(A) + 6 be the natural

+ -+

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343

homomorphism and let f; : F; -+ M n i ( F )be the regular representation of F;. Define 7; : Z ( A ) -+ M n i ( F )b y 7; = f;&. Then

Kery; = KerO; = Z( A ) (1 - e ; )

+ J ( Z(A)e;

(1 5 i 5 n )

and (71,. . . ,T,} is a complete set of nonequivalent irreducible matrix representations of Z ( A ) . Each 7; satisfies 7i(e;) = 1 and 7;(ej)= 0 for j # i. Furthermore, i f F is a splitting field for Z ( A ) , then each n; = 1. Proof. (i) and (ii) . Straightforward. (iii) The first statement and the inclusion Z ( A ) e ; E B; n Z ( A ) follows from (ii). Assume that z E B ; n Z ( A ) . Because e; is the identity element of B;, we have z = ze; E Z ( A ) e ; ,proving that B; n Z ( A ) Z(A)e;. (iv) By Lemma 3.18.1, J ( Z ( A ) e ; )= J ( Z ( A ) ) e ; .Because each e; is primitive, Theorem 2.12.4(ii) implies that the Z(A)-module

is irreducible. The latter implies that the commutative algebra

is simple and hence is a field. That this field is a finite extension of F follows from the assumption that A is finite-dimensional over F . (v) It is clear that Kerr; = Kere;. Suppose that z E KerO;. Then ze; = xe; for some z E J ( Z ( A ) ) ,so

+ xe; E Z ( A ) ( l - e;) + J ( Z ( A ) ) e ; and thus I c e d ; Z ( A ) (1 - e;)+ J ( Z ( A ) ) e ; .The opposite containment being obvious, we deduce that K e d ; = Z ( A ) ( l - e;) + J ( Z ( A ) ) e ; . z = z(1 - e;)

It follows from (iii) and (iv) that

Z ( A ) / J ( Z ( A ) )= and that each F; is a field. Moreover, if F is a splitting field for Z ( A ) then each F, is isomorphic to F and so n; = 1. Because f; is the only irreducible representation of Fi, the result follows. W

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Simple induction and restriction pairs

We now introduce the important concept of blocks. Let

be the decomposition defined in Proposition 1.1. We shall refer to B; as a block and to e; as a block idempotent of A. We shall also write B = B(e)to indicate that B is a block containing the block idempotent e. The representation : Z ( A ) + M n i ( F )described in Proposition 1.1 is uniquely determined up to equivalence by e;. We call yi the irreducible representation of Z ( A ) associated with e;. By a central character of A, we understand any F-algebra homomorphism Z ( A ) + F . If F is a splitting field for Z ( A ) ,then by Proposition 1.1, ylr. . . ,yn are all central characters of A; we shall refer to y; as the central character of A associated with ei (or Bi).

1.2. Proposition. Let A = B(e1) @ -..@ B ( e n ) be the direct decomposition of A into sum of blocks. (i) For every A-module V , V = $y=leiV is a direct decomposition of V into A-modules. In particular, if V is indecomposabte, then V = e;V f o r a u n i q u e i e {I, ..., n } . (ii) For any given left ideal I of A , we have

nB(e;)) I = @z1(I In particular, each indecomposable left ideal lies in exactly one of the B(e;). (iii) If A1 and A2 are ideals of A and A = A1 @ A2, then A1 and A2 are. direct sums of blocks. In particular, the blocks are the only indecomposable ideals of A. (iv) Let I be an ideal of Z ( A ) and let y; be the irreducible representation of Z ( A ) associated with ei. Then ei E I if and only if y;(I) # 0 . Proof. (i) Since e; is central, eiV is an A-module. Because the ei are orthogonal we also have

1. Blocks of algebras

345

Bearing in mind that

the assertion is proved. (ii) Owing to (i), I = $yZle;A and therefore

e;I 5 I n B(e;)= I f~e ; A Thus a,s

eJ

I = c&(I n B(ei)),

required. (iii) By (ii),

Thus, for all i,

c

a.nd, since B ( e ; )is indecomposable, we have B(e;) Al or B(e;) Hence Ak = @B(eifsAkB(ei) (k = 1,2)

C Az.

as required.

(iv) If ei E I then y;(e;) = 1 # 0 and so n(I)# 0. Conversely, suppose that yi(I) # 0. Since y;(I) is a nonzero ideal of the field yj(Z(A)), we have yi(1) = yi(Z(A))

Thus y;(e;) E ~ i ( 1 ) It . follows, from Proposition l.l(v), that there are elements z E I,y E J ( Z ( A ) )and z E Z(A) such that 2

= e;

+ z(1 - e;) + y

Multiplying both sides by ei gives e; = xe; - ye;. Hence, by raising both sides to the k-th power, we find

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346

with u E I . Because ye; is nilpotent, the result follows. Let el, e2,. . . ,en be the block idempotents of A and let V be an indecomposable A-module. It follows from Proposition 1.2 that there exists i E { 1,.. . ,n} such that e;V= V

and

e j V = 0 for all j

#i

In this case we say that V lies in the block B(e;). Note that if V lies in the block B(e;), then (a) All A-modules isomorphic to V lie in B(e;). (b) v = 1 v = e;v for all v E V. The above gives a classification of all indecomposable (and in particular, all irreducible) A-modules into blocks. Our next aim is to tie together blocks and principal indecomposable modules. Let e be an idempotent of A and let e be the image of e in A = A/J(A). By Lemma 3.5.2, Ae/J(A)e

AE

Hence, by Theorem 3.12.4(ii), A E is an irreducible A-module. 1.3. Lemma. Let A = $r==lAe; be a decomposition of A into principal indecomposable A-modules, and let E , be the image ofe; in A = A/J(A). Then, for any A-module V,

dimFe;V = dimFHomA(Ae;,

v) = mdimFEndA(A?!;),

where m is the multiplicity ofthe irreducible A-module A E ; as a composition factor of V .

v)

Proof. By Lemma 1.5.8, the map HomA(Aei, + e;V, f H f (e) is an isomorphism of additive groups. Because this map is obviously F-linear, we have HomA(Ae;, V )

e;V

as F-spaces

The equality dimFHomA(Ae;,V) = mdimFEndA(AE;) is trivial if V is irreducible. We may thus assume that V has a maximal nonzero

1. Blocks of algebras

347

submodule X . Since Ae; is projective, the exact sequence 0 V 4 V / X -+ 0 induces an exact sequence

4

X

+

Therefore we have

If ml and m2 are the multiplicities of Ae; as a composition factor of X and V / X , respectively, then by applying induction on d i m ~ Vwe , have

Adding up these equalities gives

as asserted. 4

1.4. Corollary. Let e be a primitive idempotent of A, let be the image of e in A = A / J ( A ) and let V be an A-module. Then the following conditions are equivalent: (2) AE is a composition factor o f v . (ii) HomA(Ae, # 0. (iii) eV # 0.

v)

Proof. This is a direct consequence of Lemma 1.3. 4 Let U and V be principal indecomposable A-modules. We say that U a.nd V are linked, written U N V , if there exists a sequence

u, = u,u2,..., u,= v of principal indecomposable modules such that any two neighbouring left ideals in this sequence have a common composition factor. It is

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348

clear that is an equivalence relation on the set of all principal indecomposable A-modules. In what follows, we denote by X I , . . . ,X m the equivalence classes of -. N

Let A = $!="=,ei be a decomposition of A into principal indecomposable modules, and for each j E (1,. . . ,m } , put 1.5, Proposition.

Ae; E Xj

(i) I 1 , . . . ,I m are all blocks of A. (ii) Two principal indecomposable A-modules are linked if and only if they belong to the same block. Proof. (i) It is clear that A = $jm=lI,. By Proposition 1.2(iii), it suffices to show that each Ij is a two-sided ideal contained in a block. Il with j # 1. Then Aei and Suppose that Aei C Ij and Aek Aek have no composition factor in common. In particular, Aek has no composition factor isomorphic to A E ; . Hence, by Corollary 1.4, eiAek = 0. This shows that IjIl = 0 for j # 1. Thus

and so Ij is a two-sided ideal of A. If Aei and Aej have the composition factor A e k in common, then by Corollary 1.4,ekAei # 0 and ekAej # 0. By Proposition 1.2(ii), B , Aej C B' and Aek C there exist blocks B , B', B" such that Aei B". It follows that 0 # ekAei & B"B and hence B = B". Similarly, 0 # ekAej C B"B' and so B' = B". Thus Ae; Aej C B and repeated application of this argument shows that Ij is contained in a certain block of A. This establishes the required assertion. (ii) This is a direct consequence of (i).

c

+

We close by recording some properties of separable algebras. An A-module V is said to be separable if VE = E @F V is a completely reducible AE-module for every field extension E / F (here AE = E@=A). Thus a separable module is completely reducible; the converse need not

1. Blocks of algebras

349

be true as we shall see below. For convenience of reference,we now quote the following two standard facts: 1.6. Proposition.

A field F is perfect if and only if every alge-

braic extension of F is separable.

Proof. See Jacobson (1964). 1.7. Proposition. Let V be a completely reducible A-module. Then V is separable i f and only i f for every irreducible submodule W of V the centre of the division algebra EndA(W) is a finite separable field extension of F .

Proof. See Bourbaki (1958). Recall that the algebra A is said to be separable if for every field extension E of F , AE is a semisimple E-algebra. Expressed otherwise, A is separable if the left regular module * A is separable.

The following conditions are equivalent: (i) A / J ( A ) is separable. [ii) For eve y field extension E I F , J(A,y)= J(A),y. (iii) For every field extension E / F , ( A / J ( A ) ) , yE A E / J ( A E ) . In particular, if A/ J ( A ) is separable, then so is AE/J(A,y)for any field extension E / F . 1.8. Proposition.

Proof. Let E / F be a field extension. Then

and the result follows. Let A be an algebra over a field F and let L be a subfield of F . We say that A is definable over L if there exists an F-basis X of A such that for all z 1 , x 2 E X , 51x2 is an L-linear combination of the elements of X. Expressed otherwise, A is definable over L if A 2 F 6 0 A1 ~ for some L-algebra Al.

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1.9. Proposition. Let A be a n algebra over a field F and ass u m e that A is definable over a perfect subfield of F. T h e n A / J ( A ) is separable.

Proof. By hypothesis, A F @ L Al for some L-algebra A1 and some perfect subfield L of F. By Proposition 1.8, we need only show that All J(A1) is a separable L-algebra. The latter being a consequence of Propositions 1.6 and 1.7, the result follows. 1.10. Corollary. Let F be a n arbitrary field. T h e n the F-algebras F G / J ( FG) and Z( FG)/J( Z( FG)) are separable. In particular, f o r every field extension E o f F ,

Proof. Let L be the prime subfield of F. Then L is perfect and

Now apply Propositions 1.9 and 1.8.

2.

Defect groups of blocks

In this section, G denotes a finite group, CI(G)the set of all conjugacy classes of G and F a field of characteristic p > 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

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351

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 = Ci n S 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

+

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

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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

2. Defect groups of blocks

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.

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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

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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

t€T

CFG(S))

where T is a left transversal for H in G. 2.6. Lemma. 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

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)

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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 }.

+ +

2. Defect groups of blocks

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 . by Lemma 2.6(iv), that H c D and that e E T T Z ( C F G ( H ) )Then, 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

Simple induction and restriction pairs

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). 3.1. Lemma. 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.

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), such that e = Trg(w). Consider the map we may choose w E CFG(D) 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 3.3. Lemma.

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

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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<. On the other hand,

and so, by Lemma 3.3, Le is a nilpotent ideal of Z(EG). Hence if ei E L , then ei = eie E L e , a contradiction. W Let S be a subring of an arbitrary ring R. An R-module M is said to be S-projective if any exact sequence

of R-modules and R-homomorphisms which splits as an S-sequence, also splits as an R-sequence.

3.5 Lemma. Let S be a subring of a n arbitrary ring R and let M be an R-module. Then the following conditions are equivalent: (i) M is S-projective.

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(ii) The canonical m a p II, : R @IS M --t M given b y $(r @3m ) = r m is a retraction, i. e. there exists an R-homomorphism cp : M + R 8 s M with II, 0 cp = 1 (iii) There exists an S-module V such that M is a direct summand of

R @sV .

(iv) Consider a n y diagram of R-modules and R-homomorphisms :

with exact row. If there is an. S-homomorphism cp : M + X making the diagram commute, then there is an R-homomorphism $ : M -, X making the diagram commute.

Proof. See Higman (1955, Theorem 4) and Hochschild (1956, Proposition 2). 3.6. Lemma. Assume that the following is a commutative diagram of R-modules and R-homomorphisms.

P

A ~ B - c

0

If Xcp = ly, th.en the following are equivalent: (i) Im(cpXa)G Ima. (ii) There is a 1c, : Z + C such that $a = Pcp and KII, = lz (in particular K is a retraction). Proof. Suppose that Im(cpAa)G Irncr. Since Imcr = li'erp, we

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which shows that Pcpp = 0, by virtue of the fact that p is an epimorphism. Thus P'p(Kera) = &(Imp) = 0 and therefore $ = P'pcr-' is a well defined R-homomorphism from 2 to C with the required properties. Conversely, suppose that (ii) holds. Then

and hence Irn(cpXa) C KerP = Ima. Our final preliminary result below will allow us to take full advantage of the results so far obtained.

3.7. Lemma. Let S be Q subring of an arbitrary ring R, let A C S be a subset such that AR = RA and let U be an S-projective R-module. Then M = U/AU is an S-projective R-module (the fact that AU is a submodule of U is a consequence of the equality R ( A U ) = ARU = AU). Proof. Consider the diagram below where all maps are canonical.

0

Taking into account that

3. Blocks and vertices

363

we have

The desired conclusion now follows by virtue of Lemma 3.6. W

We have now come to the demonstration for which this section has been developed. 3.8. Theorem. (Green (1959), Know (1976)). Let G be a finite group, let F be an arbitraryfield of prime chamcteristicp and let B be a block of F G . If P is a p-subgroup of G , then the following conditions are equivalent: (i) P is a defect group of B. (ii) All indecomposable FG-modules in B are P-projective and there is an irreducible FG-module V in B such that P is u vertex of V .

Proof. For the sake of clarity, we divide the proof into four steps. Step 1. Let D be a defect group of B. We claim that it suffices to verify that there is an irreducible FG-module M in B such that D is a vertex of M . Indeed, assume this to be true. By Lemma 3.1(ii), all indecomposable FG-modules in B are D-projective. Hence (i) implies (ii). Conversely, assume that (ii) holds. Since V is D-projective and P is a vertex of V , P is conjugate to a subgroup of D . On the other hand, A4 is P-projective and has vertex D . Hence D is conjugate to a subgroup of P, proving (i). Step 2. Here we establish the case where F is algebraically closed. Let d be the defect of B and let p" be the order of Sylow psubgroups of G. Owing to Theorem 2.10, there exists an irreducible FG-module M in B such that p"-d is the highest power of p dividing dirnFM. Denote by Q a vertex of M . Because M is indecomposable, M is D-projective and so Q c~ D. On the other hand, by Theorem 5.1.14, (S : Q)divides diinFM where S is a Sylow p-subgroup of G. Because ID1 = p d , this means that Q =G D and thus D is a vertex of M.

364

Simple induction and restriction pairs

S t e p 3. Let E be the algebraic closure of F . Here we demonstrate that every irreducible EG-module V is FG-projective. Indeed, by Corollary 1.10, FG/J(FG) is a separable F-algebra and

Denote by U a projective FG-module such that V EZ U / J ( E G ) U . Then, by (l), J ( E G ) U = J ( F G ) U and so V U/J(FG)U. Applying Lemma 3.7 with A = J(FG), we deduce that V is FG-projective. Step 4. Completion of the proof. Let e be the block idempotent of FG contained in B. Because e is a central idempotent of EG, we may write e = el - .. en

+ +

where the e; are block idempotents of EG. But B' = EGel and observe that, by Lemma 3.4, D is a defect group of B'. Applying Step 2, we deduce that there is an irreducible EG-module V in B' such that D is a vertex of V . Now J(EG)V = 0 and so, by (l), J(FG)V = 0. Thus VFGis completely reducible and therefore

for some irreducible FG-modules M;. Since multiplication by e induces the identity on V , all M; are in B. By Step 3, V is FG-projective and so, by Lemma 3.5, V is a direct summand of

But V is irreducible, so V is a direct summand of EG @FG M for some M E {Ml,. .. , M B ) . Denote by Q a vertex of M . Then M is a direct summand of F G @FQ M and thus V is a direct summand of

Invoking Lemma 3.5, we infer that V is FQ-projective. Hence V is EQ-projective and thus D is conjugate to a subgroup of Q. Now M is an irreducible FG-module in B. Consequently, M is D-projective and so Q is conjugate to a subgroup of D. This shows that D and Q are conjugate, thus completing the proof. W

4. Simple induction and restriction pairs

4.

365

Simple induction and restriction pairs

In this section, H denotes a subgroup of a finite group G, F an arbitrary field of characteristic p > 0 and e and f central idempotents of F G and FH, respectively. Recall that, by definition, the zero module will be regarded as completely reducible.

Let V C W be FG-modules. Then eV FG-modules such that eW/eV E e ( W / V ) . 4.1. Lemma.

eW are

Proof. Since e is a central idempotent of FG, eV 5 eW are certainly FG-modules. Let f : W + W/V be the natural homomorphism. Then f restricts to a surjective homomorphism

f’ : eW with K e r f * = eW

-, e ( W / V )

n V . But eW n V = eV, hence the result.

Following Knorr (1977), we define simple induction and restriction pairs as follows. We say that ( e , f ) is a simple induction pair if eVG is a completely reducible FG-module for all irreducible FH-modules 1’ = f V . The pair ( e , f ) is called a simple restriction pair, if fMH is a completely reducible FH-module for all irreducible FG-modules A d = e M . Let us first show that the above definitions may be rephrased in terms of J ( F G ) and J(FH). 4.2. Proposition. (i) The following conditions are equivalent: (a) ( e , f ) is a simple induction pair. (b) e J ( F G ) f C F G . J(FH).

-

(c) f J ( F G ) e J ( F H ) F G . (ii) ( e , f) is a simple restriction pair if and only if

e J ( F H )f

C J(FG)

(iii) Let E F be the smallest field containing the coeficients of e and f . Then ( e , f) is a simple induction (respectively, restriction) pair for F G and F H if and only if it is for E G and E H .

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Proof. (i) Put V = F H f / J ( F H ) f . Since V = fV is completely reducible and contains a copy of every irreducible FH-module S = fS, it follows that (e,f) is a simple induction pair if and only if eVG is completely reducible. Invoking Proposition 2.2.1 l(i), we have

V GE F G f / F G - J ( F H ) f and hence, by Lemma 4.1,

e V G Z e F G f / e F G -J(FH)f Consequently, eVG is completely reducible if and only if e J ( F G ) f C_ eFG - J ( F H ) f . Because the latter is equivalent to e J ( F G ) f C F G J ( F H ) , the equivalence of (a) and (b) is established. Assume that (b) holds. Then a

e J ( F G ) = e J ( F G ) f @ eJ(FG)(1- f ) E FG.J(FH)tFG(l-f) = F G [ J ( F H ) F H ( 1 - f)]

+

By applying Lemma 2.3.3(iii), we have

Hence f J ( F G ) e C J ( F H ) - F G , proving (c). The converse follows by symmetry. (ii) Put M = FGe/J(FG)e. Because A4 = eM is completely reducible and contains a copy of every irreducible FG-module W = eW, we deduce that ( e , f ) is a simple restriction pair if and only if fMH is completely reducible. Bearing in mind that, by Lemma 4.1,

fM f F G e / f J ( F G ) e

as FH-modules

it follows that ~ M isHcompletely reducible if and only if

fJ(FH)FGe

fJ(FG)e

4. Simple induction and restriction pairs

367

Because the latter containment is equivalent to e J ( F H ) f J(FG), (ii) follows. (iii) This is a direct consequence of (i), (ii) and Corollary 1.10. Note that if e f = 0 , then obviously (e, f) is both a simple induction arnd restriction pair. From now on, we investigate the case where ef # 0. The following preliminary observation will clear our path.

4.3. Lemma. The following conditions are equivalent: (i) e f # 0. (ii) There exists a projective indecomposable FH-module V = fV such that eVG # 0 . (iii) There is a n FH-module V = f V such that eVG # 0. (iv) There exists a projective indecomposable FG-module P = e P such that f P # 0 . (v) There exists a n FG-module M = eM such that f M # 0 . Proof. (i) +(ii): By Proposition Z.Z.ll(i), we have e ( F H f ) GS eFGf = F G e f

#0

Therefore there exists at least one indecomposable component V of F H f such that eVG # 0. (ii) +(iii): Trivial. (iii) +(iv): Let M be an irreducible FG-module in Soc(eVG)and let P = P ( M ) be the projective cover of M . Because V = f V , we deduce that from Corollary 2.2.7 that

# HOmFG(P,VG) and hence f P # 0. 0

HOmFH(PH,V)

(iv) =+(v): Trivial. (v) +(i): By assumption, 0 a.s required. W

#fM

HomFH(fPH,V)

= f e M = e f M and so ef

#

0,

If e and f are block idempotents, then S(e) and S(f) will denote the defect groups of the corresponding blocks; the vertex of a module M will be denoted by v z ( M ) . We are now in a position to establish our first main result.

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(Knorr (1977)). Let e and f be block idempotents of F G and F H respectively, and let ef # 0. (i) If ( e ,f) is a simple induction pair, then f M H # 0 for all FGmodules 0 # M = e M ; moreover 6(f) >G 6(e). (ii) I f ( e , f ) is a simple restriction pair, then eVG # 0 for all F H modules O # V = f V ; moreover, S(f) CG b ( e ) . 4.4. Theorem.

Proof. (i) Let V = eV be an irreducible FG-module such that f VH = 0. We first show that f PH = 0 for the projective cover P of V. Assume the contrary. Because e P = P , it follows from Corollary 2.2.7 that there exists an irreducible FH-module W = fWsuch that

0 # H o ~ F H ( PW H ,) E HomFc(P, W G )Z H o ~ F G ( Pe W , G) However, eWG is completely reducible by hypothesis, so 0

# HonzFG(V, wc)

Hu~FH(VH w) , = HUrnFH( f V H , W ) = 0

a contradiction. Now suppose that

eM = M # 0 and fMH = 0 for some FGmodule M.Then f VH = 0 for any composition factor V of M . Because e is a block idempotent, all indecomposable projective modules in the corresponding block are linked (Proposition 1.5). Thus, by repeated use of the above argument, f P H = 0 for all projective modules P = e P , contradicting Lemma 4.3(v). This proves the first assertion. Owing to Theorem 3.8, there exists an irreducible FG-module V = eV such that vz(V) =G S ( e ) Then, by the foregoing, f v H # 0 and therefore there exists an irreducible FH-module W in Soc( f VH).Because eV = V , it follows from Proposition 2.2.4(i) that 0

#H

v)

o m ~ ~ (VH) w , 2! H o m ~ c ( W ~ Zi , HomFc(eWG,

v)

However, eWG is completely reducible by hypothesis, and e W G is a direct summand of W G ,hence V is a direct summand of W G . Thus, by Theorem 3.8

4. Simple induction and restriction pairs

369

as required. (ii) The first statement is an easy adaptation of the ideas in the proof of (i). To establish the assertion regarding the defect groups, we choose an irreducible FH-module W = fWsuch that

v x ( W ) = H b(f)

Let V = e V be an irreducible FG-module in Soc(eWG). Then

Because f VH is completely reducible and ~ V isHa direct summand of VH, we deduce that W is a direct summand of VH. Denote by Q a vertex of V . Then V is a direct summand of (VQ)', so VH is a direct ] Hthus W is a direct summand of [(VQ>']H. It summand of [ ( V Q ) ~and 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.

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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.

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(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.

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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,

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373

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.

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(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.

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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

Simple induction and restriction pairs

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 ,

5. Complete reducibility of induced modules

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

Simple induction and restriction pairs

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.

5. Complete reducibility of induced modules

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

Simple induction and restriction pairs

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

5. Complete reducibility of induced modules

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, ~ ,hypothesis. Setting W to be the restriction of gV hence so is ( V G ) by 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

Simple induction and restriction pairs

(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