Chapter 8: Operator Groups and Bilinear Forms

Chapter 8

Operator Groups and Bilinear Forms In order not to interrupt the main flow of future exposition, we shall collect in this chapter a number of general properties of operator groups and bilinear forms. The results presented will be applied in our investigation of Schur multipliers of operator groups. In particular, the information obtained concerning alternating bilinear forms will be linked to the triviality of p-primary components of Schur multipliers. Many readers may wish to glance briefly a t the contents of this chapter, refering back to the relevant sections when they are needed later.

1

Operator groups

We shall collect in this section a number of general properties of operator groups and bilinear forms. Let G and H be arbitrary groups. We say that G is an H-group , or that G is an operator group with operator domain H, if H acts on G as a group of automorphisms. This means that we are given a homomorphism

For any h E H and g E G, we denote by hg the image of g under the automorphism corresponding to h. With this convention, the following properties hold : 1

Q=S

for all g E G

(1)

Operator Groups and Bilinear Forms

lyg = "(yg) "(a6) = "ax6

forall x , ~ E H , ~ E G for all x E H , a , b E G

(2)

(3)

Conversely, if for any h E H and g E G, there is a unique element hg E G such that (I), (2) and (3) hold, then G is an H-group and g I+ hg is the automorphism of G corresponding to h. Of course, the reader accustomed to right actions, will use the exponent notation gh to denote the image of g under the automorphism corresponding t o h. In case the H-group G is an additive abelian group (equivalently, G is a ZH-module), we shall write the action of H on G as "left multiplication", i.e. hg will be denoted by hg. Some important examples of operator groups are given below. E x a m p l e 1. Let G be a normal subgroup of H . Then H acts on G by conjugation, i.e. hg = hgh-'

for all h E H,g E G

and in this way G becomes an H-group. E x a m p l e 2. Assume that G is an R-module, where R is a commutative ring, and that H acts on G as a group of R-automorphisms. Then G is an RH-module. Conversely, any given RH-module G determines G as an H-group on which H acts as a group of R-automorphisms. Thus the theory of modules over group algebras can be subsumed into the theory of operator groups. E x a m p l e 3. Assume that G is an H-group and that G is an elementary abelian p-group (for some prime p). Then G is an FPH-module, where F, denotes the field of p elements. Let G be an H-group. An element g E G is said to be H-invariant if h

9=9

for all h E H

The set G~ of all H-invariant elements of G constitutes a subgroup of G. We refer to G~ as the H-stable s u b g r o u p of G. It is clear that G~ is the maximal subgroup of G on which H acts trivially. Let GI,. . . ,G, be H-groups. Then the direct product GI x . . x G, is an H-group via

Thus, by definition

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217

Let G be an H-group. For any subgroup I i of G and any h E H , we put

We say that I i is an H - s u b g r o u p of G if h~ = Ii' for all h E H (equivalently, if hk E Ii' for all h E H , k E I i ) . Thus if I i is an H-subgroup of G , then K is an H-group where H acts on I i by restriction. Let G be an H-group. We say that G is H-simple if 1 and G are the only normal H-subgroups of G. Of course, if G is an additive abelian group, then G is H-simple if and only if G is a simple ZH-module. In particular, if G and H are as in Example 3, then G is H-simple if and only if G is a simple FpH-module. Let G1 and G2 be two H-groups. An H - h o m o m o r p h i s m

is a homomorphism such that

f ( h g >=

h f (57)

forall ~ E H , ~ E G

An H - i s o m o r p h i s m is a bijective H-homomorphism. Let N be a normal H-subgroup of an H-group G. Then GIN is a n H-group where the group H acts on GIN via h

( g N )= ( h g ) ~

forall h ~ H , g e G

L e m m a 1.1. Let G I and G2 be two H-groups and let f : G1 -t G2 be an H-homomorphism. Then (i) Iier f is a normal H-subgroup of G I , f(G1) is an H-subgroup of G2 and the natural map G1/Kerf f ( G l ) , gIier f e f ( g ) is an H asomorphism. f(G1)" and, provided f is an H-isomorphism, f ( ~ r = ) (ii) f (G?)

G? .

P r o o f . (i) If g E Iier f and h E H , then f ( h g ) = f ( g ) = hl = 1 and so h g E Iier f . Hence K e r f is a normal H-subgroup of G I . For any g E G I , h E H , we have f ( g ) = f ( h g ) which shows that f ( G I )is an H-subgroup of G z . The map 4 is clearly a group isomorphism. Since for all h E H , g E G I ,

$ ( h ( s ~ i ef r) ) = $ ( ( h g ) ~ ef r) = f ( h g ) =

hf(g)=

h $ ( g ~ efr

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218

we see that Il,is an H-isomorphism. (ii) Assume that g E G?. Then, for all h E H , f ( g ) = f ( h g ) = f ( g ) which shows that f ( g ) E f ( ~ , ) ~Suppose . that f is an H-isomorphism and let g E GI be such that f ( g ) E GF. Then, for all h E H , f ( g ) = f ( h g ) = f ( g ) . Hence hg = g for all h E H and g E G r , as required. A sequence of H-groups and H - homomorp hisms

is said to be an H-sequence

.

L e m m a 1.2. Assume that A % B -+P C is an exact H-sequence of finite abelian groups, where H is a finite group of order n coprime to the order of B. ~f = 1 and C H = 1, then B H = 1. Proof. Assume that b E B H . Then, by Lemma 1.1 (ii), P(b) E C H = 1. Hence b = a ( a ) for some a E A. Now the element a' = ha is Hinvariant and so a' = 1. But

nhEH

and therefore a ( a ) = 1, as desired.

.

L e m m a 1.3. Assume that H =< h > is a cyclic group of order two and that A is an abelian H-group. If = 1, then ha = a-l for all a E A.

SO

Proof. For any a E A, h E H , a ha is H-invariant. Hence a ha = 1 and ha = a-l.

L e m m a 1.4. Assume that G is an H-group and let A be an abelian group. (i) Hom(G,A) is an H-group, wherefor any f E Hom(G, A ) and h E H , ( h f )(g)

= f (h-'g)

for all g E G

(zi) If S is an H-subgroup of G , then the restriction map Res : Hom(G, A ) -+ Hom(S,A )

1 Operator groups

is an H -homomorphism. (iii) If S is a normal H-subgroup of G, then the inflation map

Proof. (i) It is clear that If = f for all f E Hom(G,A). Given x, y E H and f E Hom(G, A), we have ( x ~ f ) ( g= ) f(y-lx-'9) = ( ~ f ) ( ~ - l g=) x(yf)(g) for all g E G proving that

"Yf

= "(Yf). Finally, for all a,@E Hom(G,A) and x E H,

proving that "(a@)= "ax@. (ii) Given f E Hom(G, A ) and h E H, we have ~ e sf)(s) ( ~ = ( hf)(4= f(h-ls) = [ h ~ e s ( f ) ] ( ~ ) for all s E S as required. (iii) Given a E Hom(G/S, A) and h E H , we have

as desired. H

Lemma 1.5. Let G1 and G2 be H -groups. Then G1 @G2is an H-group via

h

(giG', @ 92Gb) = ( h 9 1 ) ~ 8 : (hg2)~L

Proof. This is a direct consequence of the fact, that for any h E H, the map g;G: H ( h g i ) ~is: an automorphism of G;/Gi, i = 1 , 2 . Lemma 1.6. Let GI and Gz be H -groups. Then the group P(G1, G2, A) of all pairings of GI and G2 into an abelian group A is an H-group via

220

Operator Groups and Bilinear Forms

for all gl E GI, g2 E G21 f E P ( G I , G ~ , A )h, E I f . Proof. It is clear that f E P(G1, G2,A) and that I f = f for all f E P(G1, G2,A) and h E H. Given x, y E H and f E P ( G l , G2, A), we have y-l x -1 y-lx-l (xYf)(~17g2) =f( 917 92) = x(Yf)(gl)g2) for all gl E GI, 92 E G2. Similarly, x(cr/3) = xcrX/3 for all x E H , P(G1, G2,A ) and the result follows.

cr,P

E

Let G be an H-group. If Ii d L are H-subgroups of G, then L / I i is an H-group, where the group H acts on L / K via

An H-series in G is a finite sequence of H-subgroups of G

such that G; is a normal subgroup of Gi+l, 0 5 i 5 n - 1. The G; are called the t e r m s of the series and the quotient groups Gi+1/Gi are called the factors of the series. If all the factors G;+l /Gi are nontrivial H-simple groups, then the series is called an H-composition series and the G;+l/G; are called H-composition factors of G. When H acts as the group of all inner automorphisms of G, an H composition series is called a chief series and its factors are called chief factors . It is clear that in a finite solvable group any chief factor is an elementary abelian p-group for a suitable prime p. Hence such a factor can be regarded as an FpH-module. If S and T are H-series of G, call S a refinement of T if every term of T is also a term of S . If there exists at least one term of S which is not a term of T , then S is called a p r o p e r refinement of T . Two H-series S and T of an H-group G are called H-isomorphic if there is a bijection from the set of factors of S onto the set of factors of T such that the corresponding factors are H-isomorphic. L e m m a 1.7. (Zassenhaus's Lemma). Let A1, A2,B1, B2 be H-subgroups of an H-group G such that A1 a A2 and B1 d B2, and let Cij = A; f l Bj. Then (i) AlC2l d A1C22 and B1C12 a B1C22. (ii) The groups A1C22/A1C21and BIC22/BlC12 are H-isomorphic.

1 Operator groups

221

B2, we have Czl d C22. Since A1 a A2, we deduce that A1C21a A1C22. A similar argument shows that B1C12 a B1C22. (ii) Setting Ii = Ca2and N = A1C21, we have N I i = A1C22. If X , Y , Z are subgroups of G with Y 2 , then the modular law says that

Proof. (i) Because Bl

Hence N n Ii = C12Czlby the modular law. Since N K I N 2 I i / ( N n K) as H-groups, we have A1C22/A1C21 % C22/C12C21as H-groups. Similarly

the result follows. W

Theorem 1.8. (The Schreier Refinement Theorem). Let G be an Hgroup. Then any two H-series of G possess H-isomorphic refinements. Proof. Assume that

and

I = Iba

~ i . .~. a I<, a =G

(5)

are two H-series of G. Define Gij = Gi(G;+l n I<;)

and

I<;i= Iij(Gi n I
Setting A1 = G;, A2 = G;+l, B1 = l C j and B2 = Iij+l,it follows from Lemma 1.7 that j

i

+

i

j 4I

.

, and Gi,i+l/Gij

~ i ~ + ~ , ~ / ~

where the isomorphism is an H-isomorphism. Thus the series {Gij/O _< i 5 n - 1,O 5 j m ) and {IL';ilO 5 i n, 0 5 j 5 m - 1) are H-isomorphic refinements of (4) and ( 5 ) ) respectively.

<

<

Corollary 1.9. (The Jordan-Holder Theorem). Let a finite group G be an H-group. Then G has an H-composition series and any two such series are H -isomorphic.

222

.

Operator Groups and Bilinear Forms

Proof. Since G is finite, it obviously has an H-composition series. Since any such series has no proper refinement, the result follows by virtue of Theorem 1.8. The F'rattini subgroup @(G)of a finite group G is defined to be the intersection of all maximal subgroups of G (by definition, @(G) = 1 if G = 1). The group G/@(G) is called the Frattini quotient of G. In what follows, GP denotes the subgroup of G generated by all gp with g E G.

Proposition 1.10. Let p be a prime and let G be a nontrivial finite p-group. (i) The Frattini quotient of G is elementary abelian and @(G)= GPG'. (ii) If G is an H-group, then each N-composition factor of G is an FpHmodule. Proof. (i) Let G I , . . . ,Gn be all maximal subgroups of G. Then each G; is a normal subgroup of G of index p. Since the natural map

is a homomorphism with kernel @(G) = qL1Gi, G/@(G) is elementary abelian. Since G/@(G) is elementary abelian, we obviously have @(G) G'GP. Setting N = G'GP, we then have @(GIN)= @(G)/N. But G I N is elementary abelian, so @(GIN)= 1 and therefore @(G)= N, as required. (ii) Define G;, i 2 1, by G; = @(G;-l) where by convention Go = G. Then G=Go>G1I.-.2Gn=1

>

>

.

is an H-series of G (for some n I). Since, by (i), each factor GiIG;+l, 0 5 i _< n - 1, is elementary abelian, we see that each Gi/Gi+1 is an FpH-modu1e.B~refining the above series, we conclude that G has an H composition series whose factors are FpH-modules. Hence, by Corollary 1.9, each H-composition factor of G is an FpH-module.

2

Bilinear forms

Let G1 and G2 be H-groups and let A be an abelian group. Then, by the definition of the action of H on P(Gl,G2,A) given in Lemma 1.6, we see

2 Bilinear forms

that f E P(G1, G2,A) is H-invariant if and only if

We now examine the special case of P(G1, G2,A) where G1 = V and G2 = W are vector spaces over a field F and A is the additive group of F . Then, switching to additive notation, P(G1,G2,A) becomes the F-space B(V, W) of all bilinear forms f:VxW+F If V and W are FG-modules, then by Lemma 1.6, B(V, W) is an FG-module via

( s f )(v, w) = f (g-lv, g-lw) for all g E G, f E B(V,W), v E V, w E W. Then, by definition, f is G-invariant if and only if f ( v , w ) = f(gv,gw)

for a11 v E V,wE W,gE G

In what follows, given an FG-moudle V, we write V* for the contragredient of V. Recall that V* = HomF(V, F ) and the action of g E G on f E V* is given by (gf)(v) = f(g-lv) for all V E V Lemma 2.1. Let G be a finite group, let F be a field and let V, W be finitely generated FG-modules. Then the map

given by +(fl @ f2)(v, W)= fl(V)f2(~)

(v E V, w E W, fi E V*, f2 E W*)

Proof. It is straightforward to verify that 4 is an F-isomorphism. Given v E V, w E W, fl E V*, f2 E W* and g E G, we have

Operator Groups and Bilinear Forms

224

as desired. W

Lemma 2.2. Let G be a finite group, let F be a field and let V, W be finitely generated FG-modules. Given f E B(V, W), define the F-linear map

by f*(v)(w) = f ( v ,w). Then (i) f * is an FG-homomorphism if and only if f is G-invariant. (ii) f * is an FG-isomorphism if and only i f f is G-invariant and nonsingular. In particular, V S W* if and only if there exists a nonsingular G-invariant bilinear form on V x W . (iii) There exists a nonzero FG-homomorphism V -, W* i f and only if there is a nonzero G-invariant bilinear form on V x W . Pro,of. (i) Taking into account the following equality,

the desired assertion follows. (ii) By definition, f is nonsingular if and only if f * is an F-isomorphism. This proves the first assertion, by applying (i). Assume that y5 : V t W * is an FG-isomorphism. Then TJ = f * where f E B(V, W ) is defined by f (v, w) = $(v)(w). Hence f is nonsingular. The converse being a consequence of the first assertion, (ii) is established. (iii) This is a direct consequence of (i).

.

For future use, we shall record some further properties of bilinear forms. Let V be a vector space over a field F . A bilinear form

is said t o be alternating if

f ( v ,v) = 0 Taking into account that

for all v E V

2 Bilinear forms

225

it follows t h a t every alternating bilinear form is also skew symmetric , i.e.

Conversely, i f f is a skew symmetric bilinear form, then putting vl = v;! = v in (2), we have 2f(v, v) = 0 for all v E V. Hence, if c h a r F # 2, then f must be alternating. It follows that in case c h a r F # 2, there is no need t o distinguish between alternating and skew symmetric forms. However, if c h a r F = 2, this reasoning does not apply. Because we do not wish t o restrict the characteristic of the field F, we shall make the stronger assumption (1). Lemma 2.3. Let V be a finite-dimensional vector space over a field F a n d let f : V x V + F be a nonsingular alternating bilinear form. Then dimV = 2n for some n 2 1 and there exists a basis u l , . . . ,u,, vl, . . . ,v, of V such that f(u;,v;) = - f(v;,u;) = 1 (1 i 5 n)

<

and f ( x , y) = 0 for all other basic vectors x,y. Proof. Take x , y E V such that f ( x , y ) # 0. Then on dividing x or y by f (x, y), we obtain vectors u l , vl such that f (ul,vl) = 1. Note t h a t ul and vl are linearly independent. Indeed, if vl = Xul say, then f (ul, vl) = X f ( u l , u l ) = 0, a contradiction. Let Vl be the subspace spanned by u1 and vl and let

Then U is a subspace of V and any v E V can be written as

.

for a unique vector v'. It is clear that v' E U, whence V = Vl $ U. If U = 0, then there is nothing to prove. If U # 0, then the restriction of f t o U x U is nonsingular, since f is nonsingular and any vector in U is orthogonal t o any vector in Vl. Since dimU = dimV - 2, the result follows by induction on dimV. By a symplectic space , we understand a pair (V, f ) where V is a finite-dimensional vector space over a field F and f is a nonsingular bilinear alternating form defined on V. A basis of V described in Lemma 2.3 is called

Operator Groups and Bilinear Forms

226

symplectic . By an isometry of V, we understand any nonsingular linear transformation X of V such that

The isometries of V constitute a group, called the symplectic group and denoted by Sp2,(F), where dimV = 2n. Hence Sp2,(F) may b e interpreted t o consist of all linear transformations of V transforming a given symplectic basis into another such basis. Such linear transformations correspond to 2n x 2n-matrices A over F such that

AJA' = J

(3)

where A' is the transpose of A and

where I is the identity n x n-matrix. Any matrix A satisfying (3) is called symplectic . Let G be a finite group and let V be a finitely generated FG-module. We say that V is symplectic if there exists a G-invariant nonsingular alternating bilinear form f on V. Since (V, f ) is a symplectic space, any such V determines a homomorphism from G to Sp2,(F), where d i m V = 2n. Conversely, any homomorphism of G into Sp2,(F) determines a symplectic FG-module V of dimension 2n. Let f : V x V -t F be an alternating form defined on V and let vl,. . . ,v, be a basis of V. Then f is specified by its matrix A = (aij), where

Owing t o (1) and (2), we have a;; = 0,

a3'2. - - a '23 .

We refer to any such matrix A as alternating . Lemma 2.4. Let A be a nonsingular alternating m x m matrix over a field F. Then m is even and det(.4) = X2 for some X E F.

Proof. By Lemma 2.3, m = 2n for some n 2 1 and A = BJB' for some matrix B (where J is given by (4)). Since d e t ( J ) = 1, we obtain det(A) = det(B)2, as required.

2 Bilinear forms

227

Our next aim is to investigate determinants of m x m alternating matrices over an arbitrary field F. If m is odd, then by Lemma 2.4, any such matrix has determinant zero. For this reason, from now on we assume that m is even and fix m once and for all. Lemma 2.5. Let x;j, 1 5 i < j 5 m, be m(m - 1 ) / 2 independent , .. ,x,-~,,) be the field obindeterminants over Q , let li = Q ( x 1 2 x13,. tained by adjoining to Q all the indeterminants x;j, i < j , and let R = 2 [ x I 25, 1 3 , . . . ,~ ~ - l be, the ~ polynomial ] ring i n the xij, i < j , over Z.Define the m x m matrix ( x ; j ) over Ir' by x;; = 0 and x;j = -xj; for i > j . Then there exists a uniquely determined polynomial P f = P f ( x i j )E R, called the Pfafian of degree m, such that (i) d e t ( x i j ) = [ Pf ( x ; ~ ) ] ~ . (ii) The value of P f ( x i j ) is equal to 1 if each xij is replaced by a;j, where ( a i j ) is the matrix J given in (4).

Proof. We first note that K is the quotient field of R and R is a unique factorization domain. Since ( x ; j )is a nonsingular alternating matrix over Ir', it follows from Lemma 2.4 that d e t ( x i j ) = f 2 / g 2 for some coprime f , g E R. Hence f 2 = g 2 d e t ( x i j ) and so g divides f . Since f,g are coprime, g is invertible and so g = f1. Hence g2 = 1 and therefore d e t ( x i j ) = f 2 . It is clear that f = f ( x i j ) is uniquely determined up to sign. Since d e t ( J ) = 1, there is a uniquely determined f ( x i j ) which in addition takes value 1 when the x;j are replaced by the coefficients of J. Setting P f ( x i j ) = f ( x j j ) , the result follows. W Let F be an arbitrary field and let Z [ x ; j ]be the integral polynomial ring in the m ( m - 1 ) / 2 commuting indeterminants x i j , 1 i < j 5 m. Given a n alternating m x m matrix A = ( a j j )over F , consider the homomorphism

<

which sends x;j t o a;j and define P f ( A ) by

P f ( A )= 4 P f (

~ i))j

Denote by y' the induced homomorphism of m x m matrix rings.

Theorem 2.6. Let F be an arbitmryfield, let m 2 1 be an even integer and let A = ( a ; j ) be an alternating m x m matrix over F . Then

Operator Groups and Bilinear Forms

228

(i) det(A) = (Pf (A))2. (ii) Pf ( J ) = 1 where J is as in (4). (iii) Pf(BAB1) = det(B)Pf(A) for any m x m matrix B over F. (iv) det(B) = 1 for any symplectic m x m matrix B over F. Proof. (i) Setting X = (xij), we have A = p*(X). Hence

det(A) = detv*(X) = p(det(X))

= ~(pf(x))~

(by Lemma 2.5 (i))

as required. (ii) This is a direct consequence of Lemma 2.5 (ii). (iii) First suppose that B is a matrix with indeterminate coefficients bij. Then det(BAB1) = [ d e t ( ~ ) ] ~ d e t ( ~ ) Hence applying (i) and taking square roots on both sides, we find

Pf (BAB') = ~ d e t ( B )fP(A) where E = &I. This is a polynomial identity in the bij; taking B = I, we see that E = 1. Hence (iii) holds for any m x m matrix B over F. (iv) Let B be a symplectic matrix. Then, by definition, B J B ' = J. Hence, taking A = J in (iii) and applying (ii), we obtain det(B) = 1. H We next show that the converse of Theorem 2.6 (iv) holds provided m = 2. Proposition 2.7. Let B be a 2 symplectic if and only if det(B) = 1.

x 2 matrix over a field F. Then B is

Proof. Let V be a two-dimensional vector space over F with a nonsingular alternating form f defined on V. Fix a basis {vl, v2) of V and let p be the linear transformation of V with matrix B in this basis. We have to show that p is isometry if and only if det(B) = 1. Write p(vl) = alvl a2v2, p(v2) = blvl b2v2 with ai,bi E F. Then

+

+

2

Bilinear forms

On the other hand, for x = xlvl have

+ x2v2, y = ylvl + y2v2 with xi, y; E F, we

.

which shows that p is isometry if and only if f(p(vl),p(vz)) = f ( v ~ , v z ) . Since, by (6), the latter is equivalent t o det(B) = 1, the result follows. We close by proving the following result. Proposition 2.8. (Arad and Glauberman (1975)). Let p be a prime, let V be a n IFp-space of dimension m 2 1 and let G be a n abelian group of automorphisms of V (hence V is a n FpG-module). If V is a simple symplectic FpG-module and E is the ring of endomorphisms of V generated by the elements of G, then m is even, E is a finite field of pm elements and IGI divides 1 pm/2.

+

Proof. By Lemma 2.3, m is even. Put K = EndFPG(A). Since V is a simple FpG-module, li is a finite-dimensional division algebra over lFp. Hence Ii is a finite division algebra and therefore li is a field. Since G is abelian, it is clear that E is a subfield of Ii. Hence E* is cyclic and, since G C E*, G is cyclic. We may regard V as a vector space over E. Then 'I/ is a direct sum of one-dimensional subspaces over E. But G C E* and V is a simple FpG-module, hence V is one-dimensional over E. Therefore, IEI = IVl = pm. Since V is a symplectic FpG-module, there is a nonsingular G-invariant alternating bilinear form f : V x V -t F,. Choose a generator a of G. Define g(x) to be the minimal polynomial of g over Fp. Then g(x) can be expressed

By the elementary field theory, the roots of g(x) over E are distinct and are precisely a , a P , . . . ,dm-'. Now choose a nonzero v E V and put v' = g(a-')(v). Then, for all

Operator Groups and Bilinear Forms

Since f is nonsingular, we deduce that v' = 0 and therefore g ( c r - l ) = 0. It follows that a-I = a p i for some i E {0,1,. . .,m - 1). If i = 0, then cr2 = 1, contrary t o the fact that m 2 2 and a # c r p . Thus i E {I,. .. ,m - 1). Finally, note that

Because a generates E and [El = pm, 2i is a multiple of m. Consequently, i = m/2. Setting n = m/2, we then have a-l = a p " , and a l + p n = 1. Thus

+

/GI divides 1 pn and the result follows. W