Chapter 11: Schur's Formula and Applications

Chapter 11

Schur's Formula and Applications This chapter, along with the previous chapter, constitutes the backbone of the theory of the Schur multiplier. It covers a great variety of topics most of which (directly or indirectly) are treated with the aid of Schur's formula. Special attention is drawn t o the problem of effective calculation of Schur multipliers. Further topics arising from the computational workshop will be treated in future chapters. By using a purely group-theoretic approach, we construct dual sequences t o those arising from the treatment of M ( G ) as a cohomology group. One of the most important results presented is a direct proof of the universal coefficient theorem. We then provide a number of important consequences which generalize Schur's theory of covering groups to the infinite case. T h e central roles are played by evaluation maps and differentials. Among other topics, we mention a detailed study of capable and unicentral groups. I t is shown that there is a common approach t o capable and unicentral groups and a close connection with the theory of the Schur multiplier is provided. As one of the applications, we compute the Schur multiplier of extra-special p-groups.

1

Group-theoretic preliminaries

In this section, we record some elementary group-theoretic facts which will be used in subsequent investigations. Let H be a subgroup of a group G. Our first aim is t o exhibit a generating

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376

set for H which will enable us to deduce that if G is finitely generated and H has finite index in G , then N is also finitely generated.

Lemma 1.1. Let H be a subgroup of a group GI let T be a right transversal for N in G with 1 E T and let X be a generating set of G . For any g E G, let A(g) denote a unique element of T with H g = H X ( g ) . Then H is generated by all elements of the form with t E T , x E X

txX(tz)-'

Proof. Given a,b E G, put ( a , b ) = X(a)bX(ab)-l. If t E T, then X(t) = t and so ( t , x ) = t x X ( t x ) - l . Since H a = H X ( a ) , we have HX(a)b = Hab = HX(ab) and so A(a)bX(ab)-l E H

and

X(ab) = X(X(a)b)

for all

a,b E G

It follows that (a, b) E H satisfies

( X ( a ) ,b) = ( a , b) and

( a , b-') = (ab-', 6)-'

for all

a ,b E G

In particular, for t E T and g E G , we have

It therefore suffices to show that each element of H is a finite product of elements of the form ( t ,u ) with t E T and u E X U X - l . An arbitrary element h E H can be written in the form h = ul - . . u , with u; E X U X - ' , 1 5 i 5 n. Put vo = 1, v; = v;-1u; and ti = X(v;), 0 i 5 n. Since vo = 1, v, = ul ...u , = h and 1 E T, we have to = t , = 1, whence h = ( t o ~ l t , ~ ) ( t ~ ~ ~ (tt in'- )l ~ n t , l )

<

NOWX(V;-lu;) = X ( X ( V ; - ~ ) U ; ) ,

SO

.

t ; = X(vi) = X('U,.-~U;) = X ( t ; - l ~ i )and

for all i E (1, . . . ,n). Thus h = (to, zll)(tl,u 2 )

.(t,-1,

u,), as desired.

SO

Corollary 1.2. Let G be a finitely generated group and let H be a subgroup of G of finite index. Then H is finitely generated. Proof. This is a direct consequence of Lemma 1.1. H

1 Group-theoretic preliminaries

377

We now come to the following classical result of Schur (1904). Proposition 1.3. Let Z be a central subgroup of a group G. If (G : Z ) is finite, then G1 is finite. Proof. Setting n = (G : Z), it follows from Lemma 1.3.3(ii) that zn = 1 for all z E GI n Z. Hence GI n Z is a torsion group. Let X I , 5 2 , . . . , x n be a transversal for Z in G, and let gij = [x;,xj]. Fix x, y E G, say x E Zx; and y E Zxj. Then x = ux;, y = vxj with u, v E Z and this easily yields

Hence the elements g;j are all commutators of G. It follows that G' is finitely generated. Taking into account that

and that, by hypothesis GIZ/Z is finite, it suffices to show that G'n Z is finite. But G1nZ is torsion and, by Corollary 1.2, G1nZ is finitely generated. Hence, since GI n Z is abelian, GI n Z is finite. Now assume that G is a finite group. Let us recall the following definition. The Frat tini subgroup @(G)of G is defined to be the intersection of all maximal subgroups of G (if G = 1, then by definition, @(G) = G). Some properties of Frattini subgroups are given by Proposition 8.1.10. We need two additional properties.

Lemma 1.4. Let G be a finite group. Then (i) If X is a subset of G such that G =< X,g

G

=< 3->.

> with g

E @(G), then

(ii) Z(G) n Gf E @(G).

Proof. (i) Let G =< X,g > with g E @(G)and let H =< X >. If H is a proper subgroup of G, then H c M for some maximal subgroup M of G. But then g E M , so G c M , a contradiction. (ii) Put Z = Z(G) n G' and assume by way of contradiction that Z @(GI. Then there exists a maximal subgroup M of G with Z g M. It follows that G = Z M and so a typical element g E G is of the form g = rm with z E Z,m E M . Since Z E Z(G), we have

.

Schur's Formula and Applications

378

whence M d G. But M is maximal, hence it has a prime index in G. Therefore G' C M and so Z C G' E M, a contradiction.

(F'

Lemma 1.5. Let R be a normal subgroup of a free group F. Then n R)/[F, R] is a direct factor of R/[F, R].

Proof. Since F/F1is a free abelian group, so is its subgroup F1R/F'. Since F ' R / F 1 &' R/(F1 n R), it follows that R / ( F 1 n R ) is free abelian. Consider the natural exact sequence of abelian groups 1

-

(F' n R)/[F, R] -+ R/[F, R] + R/(F1 n R )

-t

I

Since R/(F1 n R ) is free abelian, the above sequence splits, a s desired.

.

We close by remarking that in future, when convenient, we shall use a definition of G1 8 G2, for arbitrary groups G1 and G2, which is equivalent t o the one given in Sec.2 of Chapter 1. Namely, we shall regard G1 @ G 2 as the group generated by all

which satisfy the following relations :

for all gl ,gi E GI, g2, gk E G2. Occasionally we shall remind the reader of this convention in our subsequent investigations.

2

Schur's formula and elementary applications

The main result of this section is the cornerstone of the theory of the Schur multiplier. The fundamental formula of Schur which we will prove, will enable us t o compute the Schur multiplier of various groups. The essence of Schur's formula is that, for a finite group G, M ( G ) 2 H2(G, Z)which allows one to bring homology theory into the subject. Note, however, that the isomorphism M ( G ) 2 H2(G, Z) is not natural. In fact, the case m = 2 of Corollary 9.9.4(ii) shows that M ( G ) is dual t o H2(G, Z). We begin by collecting some group-theoretic information pertaining to

2

Schur's formula and elementary applications

379

arbitrary groups. Let G be a group, say G = F I R where F is a free group. Since R/[F, R] is a central subgroup of F/[F,R], the following short sequence 1

-

R/[F,R] -t F/[F,R] -,G -+ 1

(1)

is a central extension of R/[F, R] by G. It depends on a choice of a presentation G = F I R and therefore is not unique. The importance of sequence (I) comes from the following result.

Lemma 2.1. Let 1 -t A + B -t C -, 1 be a central gmup extension and let a : G -t C be a homomorphism. Then there is a homomorphism

which renders commutative the following diagram :

Here y is the restriction o f p to R/[F, R]. Proof. Since F is free, we may certainly choose a homomorphism X : F -t B which renders commutative the following diagram :

Therefore X maps R into K e r ( B + C). We claim that [F,R] C_ KerX. Indeed, given x E F, y E R, we have

380

Schur's Formula and Applications

Bearing in mind that X(y) E A = K e r ( B -t C) and that A is a central subgroup of B , we infer that [X(x),X(y)] = 1. Thus X maps any generator of [F, R] into 1. Hence [F, R] KerX, as claimed. By the foregoing, X induces a homomorphism P : F/[F,R] + B such that the diagram in the statement of the lemma is commutative. As an easy application of Lemma 1.1, we now prove the following property which is the main ingredient in the proof of Schur's formula. Lemma 2.2. Let G = F I R where F is a free group. Then, for a n arbitrary abelian group A on which G acts trivially, the transgression map

associated with the natural exact sequence

is surjective.

Proof. Given cr E z ~ ( G , A ) , let ti E H 2 ( G , A ) be its cohomology class and let l + A + E - +3 G + l be a central extension associated with 6. By Lemma 2.1, there is a homomorphism 8 : F/[F, R] + E such that for y equal to the restriction of d to R/[F, R], the diagram :

commutes ( q is the natural homomorphism). Let p be a section of p. Then, by the commutativity of the diagram, X = 8 o p is a section of $. Thus the

2 Schur's formula and elementary applications

cocycle /3 E Z 2 ( G ,A) defined by

is cohomologous t o a. Taking into account that

for all s , y E G , we deduce that ti is the image of y under the transgression map. We have now come t o the demonstration for which this section has been developed.

Theorem 2.3. (Schur (1907)). Let G be a finite group and let G 2 F I R where F is a free group of an arbitrary rank a. Then (i) M ( G ) 2' (F' r l R ) / [ F ,R ] . (ii) (F' n R ) / [ F ,R] is the torsion subgroup of R / [ F ,R] and the torsionfree factor R/(F' r l R ) is free abelian of rank a. In particular, if a is finite, then R / [ F ,R] is a finitely generated abelian group. (iii) Let S be a normal subgroup of F such that

(by Lemma 1.5, such an S always exists). Then G* = F / S is a covering group of G . (iv) (a) Every covering group of G is a homomorphic image of F / [ F ,R]. (b) For any covering group G* of G and for any exact sequence

with A C Z ( G * )n [G*,G*] and A S M ( G ) , there exists S as in (iii) and an isomorphism F / S 2 G* which carries RIS onto A.

F I R and Proof. ( i ) Put R = R / [ F ,R] and F = F / [ F ,R]. Then G therefore R is a central subgroup of F of finite index. Hence, by Proposition 1.3, [F,F] is finite which implies that R n [F,F ] is also finite. Owing t o Lemma 2.2, the transgression map H O ~ ( R , @ * ) -t M ( G ) associated with the central extension

Schur's

382

Formula and Applications

is surjective. Since R n [ F ,p] is finite, it follows from Proposition 10.1.5(i) (with N = R and G playing the role of F ) that M ( G ) g R n [ F , F ] . But [F,R ] F', so [ F ,F ] = F1/[F,R] and thus

c

R n [ F ,F ] = (F' n R ) / [ F ,R] as desired. (ii) Because F is a free group of rank a , FIF' is a free abelian group of rank a . Now R / ( F 1fi R) 2 ( R F ' ) / F 1is a subgroup of F / F 1 of finite index. Hence R / ( F 1n R ) is free abelian of rank a . Since ( F ' n R ) / [ F R , ] is finite, it , is the torsion subgroup of R / [ F ,R]. Finally, if cr follows that (F' n R ) / [ F R] is finite, then by the foregoing the abelian group R / [ F ,R ] is a n extension of a finite group by a finitely generated group; hence R / [ F ,R] itself is finitely generated. (iii) Setting A = R / S , we have

G*/AE F I R and clearly A

E G,

A

E

(F' n R ) / [ F ,R] E M ( G )

Z(G*). Since

A = R / S = ( R S ) / S c ( F I S ) / S= (FIS)' = [G*,G*] it follows from Theorem 4.2.7 that G* is a covering group of G . (iv) Let F be freely generated by the set X , let n : F -. G be a surjective homomorphism and let R = Kern. Given a covering group G* of G , choose an exact sequence f l-.A+G*-tG -+1

[G*,G*]n Z ( G * ) and A g M ( G ) . Since f is surjective, for any x E X there exists t , in G* such that f(t,) = n ( x ) . Then G* is generated by A and the t,, x E X , and hence only by the t,, x E X , by virtue of Lemma 1.4. Consider the homomorphism $ : F F G* defined by $(x) = t,. Then $ is surjective and n = f o $. Because 1 = T ( R )= f ( $ ( R ) ) ,we have $ ( R ) A, so $([F,RI) = [+CI(F),@)I c [G*,A1 = 1 Thus induces a surjective homomorphism ?I, : F / [ F ,R ] -+G*, proving (a). To prove (b), let a E A and write a = $ ( x ) for some x E F. Then 1 = f (a ) = n ( x ) , so x E R and therefore $ ( R ) > A. B y the foregoing, we must have A = $ ( R ) which implies that with A

c

2 Schur's formula and elementary applications

383

To prove the reverse containment, suppose that z = +(x) = +(y) for some x E F', y E R . Then x-ly E Ker$, so n(x-ly) = 1 and hence x-'y E R . Thus x E R and z E $(F1 n R), proving that A & +(F' n R) and so A = $(F' n R). Consequently, by (i), restricts t o an isomorphism ( F ' n R ) / [ F , R] -, A. Finally, put S = Ker+ n R. Then S/[F,R] is the kernel of the restriction of t o R/[F, R] and the image of this restriction is A because $(R) = A. Thus RI[F, Rl = (F' n R)/[F, Rl x $/IF,Rl

4

4

and therefore, by (iii), F/S is a covering group of G. Let X : F/S -+ G* b e the homomorphism induced by $. Then X is surjective and X(R/S) = A since is surjective and +(R) = A. But I F/SI = IG*1, hence X is an isomorphism and the result follows. H We now provide a number of applications of Theorem 2.3. The following result exhibits an important link between certain central extensions and covering groups. An alternative proof can be found in Yamazaki (1964b), who used a weak version of the universd coefficient theorem.

Corollary 2.4. (Schur (1907)). Let 1 -i Z + E + G 1 be an exact sequence of finite groups with Z C Z ( E ) n El. Then there is a covering group G* ofG such that E is a homomorphic image of G*. Proof. We apply Lemma 2.1 with A = Z , B = E , C = G and a = lc. Z ( E ) n E', Then there is a homomorphism ,O : F/[F,R] + E . Since Z it follows from the commutativity of the diagram and Lemma 1.4 that !?, is surjective. Now let G* = F/S where S is as in Theorem 2.3(iii). Since S/[F,R] is torsion-free, we see that S/[F, R] C Kerp. Hence P induces a surjective homomorphism F/S i E, as desired. N If a finite group is generated by n elements and defined by r relations between them, then by Proposition 9.4.3 we have r 2 n. The groups for which r = n are of particular interest due to the following consequence of Theorem 2.3. In what follows, by convention, the minimal number of generators of the identity group is zero.

Corollary 2.5. (Schur (1907)). Let a finite group G be generated by n elements with r defining relations, and let s be the minimal number of

Schur7sFormula and Applications

384

>

generators of M(G). Then r n + s . In particular, if r = n then M ( G ) = 1 and if r = n 1, then M ( G ) is cyclic.

+

Proof. Assume that G =< xl,x2,. . . ,x,lp; = 1, 1 I i 5 r > and let F be a free group freely generated by the set {xl, . . . ,x,). Then G Ei F I R where R is the normal closure of pl, . . . ,p, in F. Hence the group R is

generated by pl, . . . ,p, and all the conjugates of pl, . . . ,p,. Because R/[F, R] is a central subgroup of F/[F,R], the group R/[F, R] is generated by the images of pl ,. . . ,p, in R/[F, R]. On the other hand, by Theorem 2.3(i), (ii), the minimal number of generators of R/[F, R] is s n. Thus r 2 s n and the result follows.

+

+

Another consequence of Theorem 2.3 provides a practical method for calculation of the Schur multiplier.

Corollary 2.6. (Schur (1907)). Let G = F I R be a presentation of a finite group G as a factor group of a free group of finite rank n . Assume that

where each

Xi

is offinite order. Then

Proof. Put I< =< X I , . . . ,x, > and R = R/[F,R]. Because Ii' is finite, the torsion-free ranks of R and R/I
Proposition 2.7. (Van der Hout (1975)). Let G be afinite group with a presentation

let k; be the order of g;, 1 5 i 5 n , and let the group H have a presentation

2 Schur's formula and elementary applications

Then M(G) 2 H ' n Z where Z is the subgroup of H generated by fj(hl,.

. . , hn), 1 < j I t .

Proof. By Proposition 9.4.1 (with the roles of H and G interchanged), the map H + G, hi H g; extends t o a surjective homomorphism from H to G whose kernel is 2. Hence it suffices t o show that M ( G ) is isomorphic to the kernel of the induced homomorphism cr : HI + GI. By hypothesis, G = F I R where F is a free group on zl, . .. ,x,, R is the normal closure of f i ( x l , . . . ,x,) in F and g; = x;R, 1 5 j 5 t , 1 i 5 n. P u t R = R/[F, R] and F = F/[F,R]. Then the map F -, G, x;[F, R] H g; is a homomorphism which restricts to a homomorphism

<

with li'erp = R n [ F ,F] E M ( G ) by Theorem 2.3(i). By the definition of H , the map X : F + H , z;[F, R] H h; is a homomorphism; its restriction 7 : [ F , F ] -+ H i , [xi, xj][F, R] H [h;, hi] is a surjective homomorphism which renders commutative the following diagram

Hence it suffices to show that KerX n [ F , F ] = 1. Now KerX is the normal closure of xfi[F, R] in F, 1 i n. Furthermore, since = 1, each 2 2 E R , so each x;~'[F,R] is in the centre of F . Since F/F1is free abelian on zl F1,...,o,F' and [F, R] F', each x;~'[F,R] is of infinite order. Thus KerX is an abelian group generated by elements of infinite order and so KerX is torsion-free. Since [ F , F] is finite (see the proof of Theorem 2.3(i)), it follows that KerX n [ F ,F] = 1, as desired.

< <

.

gF

As a further application of Schur's formula we shall prove the following result.

Schur's Formula and Applications

386

Proposition 2.8.

(Jones (1973b)). For any normal subgroup N of a finite group G, there exists a finite group L with a normal subgroup M such that (i) G'n N 2 L/M. (ii) M E M(G). (iii) M ( G / N ) is a homomorphic image of L.

Proof. Let G = F I R be a presentation of as a factor group of a free group F. Then N = S I R for some S d F and so G I N 2 F I S . Hence, setting

M = (F' r l R)/[F, R] and

L = (F'

n S)/[F, R]

we have G'

ilN

= (FIR fl S)/R = (F' E (F'

n S)/(F1 n R)

n S)R/R

(F' n S)/[F, R]/(Ft n R)/[F, R] = LJM 2

Hence (i) and (ii) follow from Theorem 2.3(i). Again, applying Theorem 2.3(i), we obtain M(G/N)

E (F'

n S)/[F, S]

2 ((F'

n S)l[F, RI)/[F, SII[F,Rl

proving (iii) and hence the result. As an easy consequence, we obtain the following properties in which d(G) denotes the minimal number of generators of G and e(G) the exponent of the group G.

Corollary 2.9.

Let N be a normal subgroup of a finite group G. Then (i) IM(G/N)I divides IM(G)I IG'n NI. (ii) e(M(G1N)) divides e(M(G))e(G1n N). (iii) d(M(G/N)) _< d(M(G)) d(G1n N).

+

Proof. Keeping the notation of Proposition 2.8, we see that ILI = n NI, e(L) divides e(M(G))e(Gt n N ) and

IM(G)I IG'

2 Schur's formula and elementary applications

Hence the required properties follow from Proposition 2.8(iii).

.

387

As a preparation for the proof of our next result, we record the following lemma. Lemma 2.10. Let G = F I R where F is a free group, let B = S I R be a central subgroup of G and let A = F / S . Then [ F , S ] / [ FR, ] S 1 is a homomorphic image of A 8 B . Proof. Define the map X : A

xB

+ [F,S ] / [ FR]S1 , by the rule

Because [ F , S ] R , an elementary commutator calculation shows that X is well defined and that for all f , f i , f 2 in F , x,xl,x2 in S ,

and

..

[ f 21x21 -- [ f ,x l ] [ fx2] , (mod[F,R ] S 1 ) 7

Hence X induces a homomorphism from A wished t o show.

@J

B onto [F,S ] / [ FR, ] S 1 ,as we

T h e following result for the case where B is cyclic is due to Green (1956). Proposition 2.11. (Jones (1973b)). Let B be a central subgroup of a finite group G and let A = G I B . Then (i) (M(G)I[GI n BI divides IM(A)I IM(B)I / A@ BI. (ii) d ( M ( G ) )5 d ( M ( A ) ) d ( M ( B ) ) d ( A @ B ) . (iii) e ( M ( G ) )divides e ( M ( A ) ) e ( M ( B ) ) e (@A B ) .

+

+

Proof. (i) Let G = F I R be a presentation of G as a factor group of a finitely generated free group F and let B = S I R . Then [ F , S ] C R. By Proposition 2.8 (with N = B and G I N = A), we have

Since

( [ F ,SII[F,RI)I[F,RISIIIF,RI) 2 IF, SII[F?RIS1

Schur's Formula and Applications

it follows that

On the other hand,

[F,R]S1/[F, R]

2

St/(St n [F, R]) (S'l[S, R1>/(St n [Fl RI)/[Sl RI)

and , since St C IF, S] C R ,

Hence (i) follows by virtue of Lemma 2.10. (ii) For a finite group H , define a positive integer r ( H ) by the following property: every subgroup of H may be generated by r ( H ) elements and there exists a subgroup of H that cannot be generated by fewer than r ( H ) elements. By Proposition 2.8, we have

since M(A) and [F, S]/[F, R] are finite abelian groups by virtue of the fact that [F, Sll Rt [F,Rl

c c

T h e desired conclusion now follows as in (i). (iii) This again follows as in (i). H To provide our final application of Schur's formula, let us recall t h a t the lower central series (Gn) of-a group G is defined by

Proposition 2.12. (Johnson, Warnsley and Wright (1974)). Let G be a finite group and let n be a positive integer such that G, = Gn+l. Then M ( G / G n ) is a homornorphic image of M(G). Proof. Write G = F I R where F is a free group. Then

Gm=RFm/R

forall

m 2 1

3 Computational workshop

Since, by hypothesis, G, = G,+l it follows that RF, = RF,+1 We conclude therefore that M(G/Gn)

= M(F/RFn) 2 (F2 fl RF,)/[F, RFn] (by Schur's formula) = (Fz n RFn)/[F, R][F) F n ] = (F2nRFn+l)/[F)R]Fn+l (by (2) and the definition of Fn+l)

.

= (F2 n R)F,+l/[F, R]Fntl (by the modular law) 2 (F2 n R)/(F2

n R ) n ([F) RIG+,)

which is clearly a factor group of (F2 n R)/[F, R] 2 M(G).

3

Computational workshop

A. Introduction and methods. In almost all our previous discussions we have been concerned in the attempt t o get a clear understanding of the behaviour of Schur multipliers. So far we have established a number of important theorems which throw a good deal of light on this question. However, if one is to make use of the preceding theory, then it is necessary to provide an effective method (or methods) t o identify the isomorphism class of M ( G ) of a group G belonging t o a given class of groups. This is a n exceptionally difficult problem in full generality and any optimism to solve it (even for p-groups of class 2) must be guarded. Nevertheless, due t o relentless efforts of a number of people working in the field, we have accumulated a vast knowledge of Schur multipliers of particular groups. As an example of a spectacular success, we mention that the Schur multipliers of all simple groups have been determined. It should also be pointed out that a complete list of Schur multipliers of groups of order 2", n 5 6, can be found in a work of Sag and Wamsley (1973b). Further note that the Schur multipliers of the groups of order 3", n 5 6, were calculated by computer and it was found that there are eleven nonabelian groups with trivial Schur multiplier, seven of which are

390

Schur's Formula and Applications

metacyclic. The four nonmetacyclic nonabelian groups with trivial Schur multiplier are :

All the information above regarding groups of order 3n, n 5 6, is extracted from Wamsley (1973a) (who relied on a classification by R. James (1968) of groups of order p6, p 2 3). We warn the reader that in the published version of R. James Ph.D. thesis (James (1980)) several errors which appeared in the original list of 3-groups have been corrected. T h e calculational aspects of the theory of the Schur multiplier are as complicated as its classical counterpart. There is a vast literature devoted t o the computation of Schur multipliers of individual groups. In many cases, calculations are more of an art than a science and success often depends on the "right" approach. Even a group of a relatively small size can have the Schur multiplier whose calculation is a formidable task. For example, the group PSL3(4) has Schur multiplier Z4 x Z12 according to a n unpublished result of Burgoyne and Thompson. However, the actual calculation is much more complicated then the calculation of Schur multipliers of some infinite family of simple groups, e.g. PSLn(q) for all n 2 5 and all prime powers q. It is therefore not surprising - in our age of advanced electronics - that attempts have been made to computerize the procedure. An important development in the computational aspects of the theory was marked by the appearance of two papers of Holt (1984, 1985b). The first of these describes an algorithm for calculating the pcomponent of the Schur multiplier of a finite permutation group. This algorithm has been successfully implemented on a machine, and it appears t o give accurate results in a reasonable time for groups of degree up to a few hundred. The second paper provides a computer program for the calculation of a covering group. Historically, P. Hall (1940) was the first t o recognize the significance of Schur multipliers in classification of groups of prime power orders. Following his method and advice, the groups of order 2n, n 6, were classified by M. Hall and J.K.Senior in their classical book published in 1964. T h e notion of an isoclinism introduced P.Hall turned out to be crucial in the classification (note also that it was P. Hall who first observed that two covering groups of

<

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the same group are isoclinic). It will be no exaggeration t o say that the Schur multiplier plays a prominent role in classification of finite simple groups. The contributions in this field are principally the work of three individuals : Schur, Steinberg and Griess (of course, many others lent a hand with specific groups or family of groups). Now given a group G, how does one calculate the Schur multiplier M ( G ) of G ? In order to achieve a better understanding of the problem, we list below some of the methods and provide relevant comments.

Method 1. For each individual prime p dividing IG(, look a t M ( P ) where P is a Sylow psubgroup of G. If it happens that M ( P ) = 1 (e.g. P is cyclic, generalized quaternion, semidihedral, etc.), then the p-component M ( G ) , of M ( G ) is trivial. Of course, if the same applies t o all primes p dividing [GI, then M ( G ) = 1. Let us illustrate how such a situation can arise. Assume that G = SL2(q) is the group of all 2 x 2 nonsingular matrices of determinant 1 with entries in the field of q elements. If a prime p does not divide q, then by Theorem 2.6 in Chapter 16 the Sylow p-subgroups of SL2(q) are either cyclic or the generalized quaternion groups. Moreover, if q is a prime, then the Sylow q-subgroups of SL2(q) are cyclic. Hence M ( G ) = 1 if q is a prime and M ( G ) is a p-group for q = pn for some prime p and some n 2 1. Method 2. The situation described in Method 1 is of course the most favourable, but unfortunately does not arise too frequently. One of the methods t o examine the pstructure of M ( G ) is to identify the p-stable subgroup M ( P ) ~of P where P is a Sylow p-subgroup of G. This will give us a complete information on M(G), since M(G), 2 M ( P ) G . To achieve our aim, we use the theory of the Schur multiplier of operator groups developed in Chapter 10. For example, if a Sylow 2-subgroup of G is a direct product of two cyclic groups of order 2n and 2" with n 2 m >_ 1, then M(G)2 is cyclic of order 2m (see Proposition 10.5.4). Another typical example arising from the theory of the Schur multiplier of operator groups is as follows. Assume that a Sylow p-subgroup P of G is elementary abelian of order p2 and let p : N G ( P ) -. GL2(p) be the matrix representation of N G ( P ) corresponding to its conjugation action on P.

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Then M(G), E Z, if and only if d e t p ( x ) = 1 for all x E N G ( P ) (see Theorem 10.8.4). A most favourable situation arises when a Sylow p-subgroup P of G is elementary abelian. Then we choose a p'-subgroup H of N G ( P ) such that N G ( P ) = P H . We may, of course, regard P as an FpH-module, where H acts on P by conjugation. It turns out that M(G), is isomorphic t o the additive group of all H-invariant alternating bilicear forms on the IFp-space P (see Theorem 10.8.1). Moreover, if P is a simple FpNG(P)-module, where N G ( P ) acts on P by conjugation, then M(G)p # 1 if and only if P is a symplectic FpNG(P)-module (see Theorem 10.8.6) (in particular, if n is odd, then M (G)p = 1). Let us finally mention another favourable situation. Write N c ( P ) = P H with P n H = 1. If the group H acts on P by conjugation and no H composition factor of the Frattini quotient P / @ ( P ) is contragredient t o an H-composition factor of P, then M(G)p = 1 (see Theorem 10.9.1). Applying this fact, Ward (1968b) showed the triviality of Schur multipliers of most of the simple groups of Lie type, where p is the characteristic of the underlying field.

Method 3. One of the favourite gambits of group theorists in proving that M ( G ) p = 1 is to assume that there is a central extension E of Zp by G with Z, E E' and then to derive a contradiction by using an intimate knowledge of the structure of G. For example, let p be an odd prime and let P be a nonabelian group of order p3 which is a Sylow p-subgroup of G. Assume that g E N G ( P ) is of order two and that g centralizes the centre Z of P and inverts P I Z . Then, by applying the above argument, one can prove that M(G),, = 1 (see Proposition 10.9.3). This property (in the special case where p = 5) was used by McKay and Wales (1971a) to prove that if G is the Higman-Sims simple group, then M(G)5 = 1.

Method 4. This method relies on the existence of a larger group with trivial Schur multiplier. Namely, assume that G is a group with trivial Schur multiplier. If the group H under consideration happens to be isomorphic to G I N for some normal subgroup N of G , then M ( H ) 2 (GI n N)/[G, N] (see Proposition 10.1.5(ii)). This method is especially effective in case G is a p-group. Indeed, by taking N to be any nontrivial central subgroup of G contained in GI, we obtain M ( G / N ) Ei N. In particular, a discovery of a nonabelian group G of order pn with M ( G ) = 1 leads immediately to a

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discovery of a group of order pn-' with M(G) 2 Z, (by taking N to be a subgroup of GI n Z(G) of order p).

Method 5. One of the most popular methods is t o use the Schur's formula M ( G ) 2 ( F ' n R)/[F, R] (see Theorem 2.3) and its various consequences. For example, if it happens that G admits a presentation with the same number of generators and relations, then M ( G ) = 1 (Corollary 2.5). Such groups are said t o have deficiency zero. We shall exhibit a large number of examples arising from this situation. Of course, the easiest examples are cyclic and generalized quaternion groups. A classical method is to use Corollary 2.6 which allows in many cases t o demonstrate that IM(G)I 5 n for some integer n. The next step is t o exhibit a concrete central extension (often arising from matrix groups) t o demonstrate that in fact IM(G)I = n. This method also allows t o identify the isomorphism class of M ( G ) for certain classical groups (e.g. alternating and symmetric groups). Method 6 . This method relies on obtaining a sharp upper bound on the size of M ( G ) by using purely cohomological methods (e.g. the Hochschild-Serre exact sequence and its various extensions). A second step is usually t o construct a covering group of G. For example, we shall apply this method t o compute the Schur multiplier of a group of order p3 and exponent p, where p is an odd prime. Especially important is the term Iil in the sequence of Proposition 10.1.4. By utilizing this term, Holt (1985a) proved that the Schur multiplier of Thompson's group F3 is trivial. It should be pointed out that there is a class of groups, namely unicentral groups (to be discussed in Sec.8) for which the computation of the Schur multiplier can be carried out by using exclusively cohomological methods. It turns out that for any such group G, M ( G ) 2 M(G/Z)/(G1 n 2 ) . Here 2 is a central subgroup of G and G' n Z is identified with its isomorphic copy which is the image of the transgression map

As an application of this method, we shall compute the Schur ~nultiplierof extra-special p-groups. This will be achieved with the aid of a more general fact that if G is a central product of nilpotent groups G I , . . . , G n , n 2 2 with each G: = Z(G;), then M (G) S M(G/G1)/A for some subgroup A

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394

of M ( G I G t ) with A E GI. One of the remarkable properties enjoyed by unicentral groups G is that for any central subgroup Z of G , the inflation map M(G) M(GIZ) is surjective and the coinflation map +

is injective.

Method 7. Sometimes the computation of the Schur multiplier uses the fact that the assignment cr H COG, cr E Z2(G,C*) induces a bijective correspondence between the elements of M ( G ) and the equivalence classes of twisted group algebras of G over C. For example, this method allows t o compute the Schur multiplier of all metacyclic groups in a transparent manner (without using spectral sequences on which some proofs rely).

Method 8. T h e name of this method is "brute force". It was successfully used by Steinberg (1962, 1967), who proved that if G = SL,(q) then, with the exception of certain cases, M ( G ) is always trivial. Namely, we take any central extension 1 + C -+ H + G + 1 and try to prove t h a t the above extension splits. The following modification of this method was used by M. Hall and Senior (1964). Given a group G, one assumes the existence of a group E with a central subgroup Z such that E I Z !2 G. From the defining relations of G, one obtains an upper bound for IZ n E'I and then shows that it is precisely the order of M ( G ) (by effectively constructing a covering group).

Method 9. This is a favourite method for a homologically inclined reader. It is customary to identify M ( G ) with the second integral homology group H2(G,Z) and to use purely homological machinery. I t should be pointed out that, for an arbitrary (i.e. not necessarily finite) group G H ~ ( GC*) ,

Hom(H2(G, Z), C')

i.e. the groups H2(G,C*) and H2(G,Z) are dual t o each other and so the above identification is not natural. We shall have one occasion to apply homological methods. Namely, by using the Ganea map (GIG') 8 Z -+ M ( G )

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we shall demonstrate that if G is a covering group of a nonabelian p-group of order p3 and exponent p (p an odd prime), then M(G) = 1. Finally, we mention that an application of the Schur's formula, amounts t o an identification of the isomorphism classes of M ( G ) and H2(G,Z). Method 10. T h e method mentioned here is based on the study of connections between orthogonal CG-modules, the cohomology group H2(G,Z2) and the Schur multiplier M(G). We shall devote a separate chapter to a detailed investigation of this method. Here we only mention t h a t this approach is very effective in the study of the Zstructure of M(G). For example, it will be shown that if G is a finite group of complex orthogonal matrices with character x and g E G has order two and

then G has a nontrivial double cover. In particular, provided g E G', M ( G ) has even order. As one of the many consequences, we prove the following result. Let G be a group of even order with no subgroup of index 2. Suppose n is a faithful permutation representation of G and that there is an involution g of G such that ~ ( g moves ) m letters, m s 4(mod 8). Then M ( G ) is of even order. T h e results obtained may be used t o show that the Schur multiplier of certain simple groups has even order.

B. Calculations. Here we compute the Schur multipliers of various groups by applying some methods discussed in Sec.A. Example 3.1. n 2 1, i.e.

G

Let G be the generalized quaternion group of order 4n,

=< a , blaZn = 1, b2 = a n , bab-' = a-I >

Our aim is to show that M ( G ) = 1.

.

Since b2 = a n , Proof. The relation bab-' = a-' implies banb-' = it follows that a-n = bb2b-' = b2 = a n . Thus aZn = 1 and therefore G can be generated by two elements with two defining relations. Hence M ( G ) = 1 by virtue of Corollary 2.5.

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Example 3.2.

Let G be the dihedral group of order 2n, n 2 1, i.e.

Our aim is to show that M ( G ) = 1 if n is odd and M ( G )

E Z2

if n is even.

Proof. If n is odd, then all Sylow subgroups of G are cyclic. Hence M ( G ) = 1 by virtue of Proposition lO.l.l(ii). Suppose that n is even. Owing t o Corollary 2.5, M ( G ) is cyclic. On the other hand, by Corollary 10.11.3, M ( G ) has exponent dividing 2. Hence IM(G)I 5 2. Let Q be the generalized quaternion group of order 4 n (see Example 3.1). Then Q / < an >% G and, since n is even, < an > G Z(Q) n Q' Hence, by Corollary 10.1.6, < an > is isomorphic t o a subgroup of M(G). This shows that M ( G ) Z2 (and also that Q is a covering group of G ) . Example 3.3.

Let p be an odd prime and let G be the group with a

presentation

W e wish to show that G is a covering group of the nonabelian group P of order p3 and exponent p and that

Proof. We use a simplified version of a proof due t o Bey1 and Tappe (1982). Another covering group of P was given by P. Hall (1940, p.139). First we show that G has order p5. The relations for G imply ckac-k

= a l - k ~ and

ckbc-k = bl-kp

for all

Observe also that

because

C i = (P - 1

P- 1

i=l

2

) ~and

2

cP = aP = 1

kEZ

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A similar argument shows that b?' is centrd. Setting Z =< up, bP >, we see that Z Z(G) n G' and that [GI 5 p5. Assume by way of contradiction that IGI < p5. Then, upon adding either the relation ap = 1 or bP = 1, we obtain a known group of order p4 (see Huppert (1967, p. 349)). But, if ]GI = p4, then ap = 1 = b p and so G 2 P, a contradiction. Thus IGj = p5 and G/Z S P. Hence, by Corollary 10.1.6, M ( P ) has a subgroup isomorphic to Z,xiZp. On the other hand, p2. This shows that M ( P ) E Zp x Z, by Theorem 10.12.10, IM(P)I and that G is a covering group of P. Finally, for future reference, we note Z(G), then G/Z(G) is abelian. Hence that Z = Z(G). Indeed, if Z c = [a, b] E Z(G) which is impossible since ca # ac.

<

+

.

Example 3.4. (M. Hall and Senior (1964, Theorem 6.2)). Let G be a group with a presentation

a-'ba = bc, a- 1ca = cd, a-Ida = d

>

Then G is a split extension of an elementary abelian group < b, c,d > of order 8 by a cyclic group < a > of order 4 inducing an automorphism of order 4 (in particular, /GI = 32). We wish to show that

Proof. We first define a group G* of order 128 which will turn out to be a covering group of G. Let G* be a group with generators

determined by the relations which express the fact that < P, y, 6, E > is the direct product of the dihedral group y4 = P2 = 1, P-lyP = y-' with the two cyclic groups < 6 > and < E > of order 2 extended by an element a of order 4 which transforms P, y, 6 , into ~ Py, y6 , 6&, E. Then a acts as an automorphism of order 4 on < p, y, 6, E >. Moreover, the centralizer of a in < p , y , S , ~> is < y 2 , &>, which is clearly the centre of G*, since a2 transforms < P , y , 6 , ~> by an outer automorphism. Note also that, by definition of the action of a, we have y2, E E [G*,G*]. It follows that Z(G*) = [G*,G*] and

G*/Z(G*) 2 G

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Since Z(G*) E Z2 x Z2, we deduce (from Corollary 10.1.6) that M ( G ) has a subgroup isomorphic to Z2 x Z2. It will next be shown that IM(G)I 5 4, which will prove that M ( G ) 2 Z2 x Z2 and that G* is a covering group of the group G. Let E be a group with a central subgroup Z such that E / Z E G. We claim that IE' n ZI 5 4. If sustained, it will follow (by taking E to be a covering group of G ) that I M(G)I 5 4. To substantiate our claim, we write E =< Z , a , P > with Z o and Zp corresponding respectively t o a and b. Define y , 6 and E by ~ = [ P , ( r ] , S = [ y , a ] and

E =

[S,a]

Then each of the elements & , P 2 , y 2and S2 is in Z. Since (r-'6a = SE, we have . 1 = [S2,cw] = E ~ Note . also that the commutators of a-lS2(r = S 2 ~ 2Hence p , y , S in pairs lie in Z and so

By a similar argument, we also have 1 = [P,S12 = [y,S12 Also, 1 = [ y 2 , a ] = 6[6,7]6 gives S2 = [y, 61. Similarly, y 2 =

[P,y] so t h a t

However, (r4 E Z and [ P , a 2 ] = y26 and [6,a2] = b2 = 1, commutes with a2. Thus

since y4 = 1. Finally,

SO

that [p, 02]

[P,a2]= y2S and therefore

so that (since y2 E Z and y4 = 1)

Now E' is generated by [P, a] = y and its conjugates in E . It follows from the foregoing relations that E 1 = < y , 6 , & > with

&,y2EZ

.

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and [y,S] = S2 = E~ = 1. Since E'n.2 =< y 2 , >, ~ we see that IE'n 21 5 4, thus completing the proof. Example 3.5.

Let G be the group given by a presentation

We wish to show that M ( G ) E Z2.

Proof. It is well known (e.g. see Huppert (1967)) that G E PSL2(5), where P S L z ( 5 ) denotes the projective special linear group of degree 2 over the Galois field of 5 elements. By the definition of G , G = F I R where F is a free group freely generated by the set {x, y} and R is the normal closure ) F. Let the images of these elements in F / [ F ,R] be u, v of {x5, y3, ( x Y ) ~in and w, respectively. Our first aim is to show that

R] and put xi = x[F, R], To prove this relation, we start from w = (XY)~[F, yl = y[F, R]. Because u, v a,nd w commute with xl and yl, we have

For w3, we inset w between x: and yl. We continue this process t o derive

the last equality being true because yl commutes with w5 and u. For wlO, we insert w between x an y2 t o derive

4 we get A similar argument proves that w15 = uw5 (xlY1xl ~ 1 2 ) ~Finally, . w3" = u12v20, proving (1).

Next we put 3 -5 t = u 6v 10w -15 , T = U2 O W , and

S=UW*W-~

.

Then R/[R, F] =< u,v,w >=< r , s , t > and, by (I), t2 = 1. Applying Corollary 2.6, we infer that M ( G ) Z < t >. Thus IM(G)I 5 2. But the special linear group SL2(5) is perfect, hence IM(G)I 2 by virtue of Corollary 10.1.6. This demonstrates that M ( G ) E Z2, as desired.

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400

Example 3.6. G be defined by

(Neumann ( 1 9 5 5 ~ ) ) .Let p be a prime and let the group

where m > n > 0 if p is an odd prime, and m - 1 > n > 0 if p = 2. It is obvious that G is of order pm+n. Our aim is to show that M ( G ) = 1. Proof. We wish t o demonstrate that G can be defined by the pair of relations : apm = bp" , b- 1 a b - al+~m-n Once this is achieved, the desired conclusion will be a consequence of Corollary 2.5. Because the realtions (2) are satisfied in G, we are left to verify that they imply that upm = 1. To this end, we first note that since

p2m-n

we have a = 1. Thus the order of a is a power of p. Setting s = pn and q = pm-n, we obtain a = bbsabs = a (l+qIs

+

which implies that the order of a divides ( 1 q)' - 1. It will next be shown that if x E pu(modpu+l),then

>

provided u 2 1 for odd p, or u 2 for p = 2. We prove (3) by induction. The induction step itself is obvious upon noting that in the expansion

the first term on the right-hand side is

whereas the further terms are all divisible by p2u+1 a t least, except for the last one, which is divisible by pPu. But 2 u + l u + 2 for u 1 and pu u+2 for p > 2 , u 2 1 or p = 2 , u 2. Hence all terms except the first one are divisible by P " + ~ , which proves ( 3 ) .

>

>

>

>

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Finally, because q = pm-n = pm-n(modpm-n+l), an application of (3), for x = q, inductively n times yields

Therefore no higher power of p than pm divides (1 uprn = 1, as we wished t o show.

+ q)'

- 1, proving that

In what follows, we write

V l ( g l ~ ~ ~ ~ ~ g n ) ~ - ~ ~ >, ~ s ( g l ~ ~ ~ ~ ~ g n )

t o indicate that the group G is generated by gl, . . . ,g, subject to the relations

and where ~ l ( g 1 ,... ,g,),

. . . ,9,(gl,. . . ,g,)

are central.

Example 3.7. (Van der Hout (1975)). Consider the group G with a presentation G =< a , bla2, b3, (ab)12 11 ( ~ b >) ~

Then G is a group of order 48 (see Shephard and Todd (1954, p. 281)). W e wish to show that M ( G ) = 1. Proof. We first claim that G has a presentation

This will follow, provided we show that (4) implies (g1g2)12 = 1. First we note that ( 9 1 9 ~= ) ~g y l (g1g2)3g1 = (g2g1)3 Hence

Similarly, setting g3 = g1g2g1, we have

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402

It follows from ( 5 ) and (6) that

as claimed. The group H in Proposition 2.7 has now a presentation

which is equivalent to

We next claim that

3

2

( h 2 h l ) (h2h1I3= 1

(7)

If sustained, it will follow that (h2h1)3is central, hence H 2 G and, by Proposition 2.7, M ( G ) = 1. To prove (7), note that

and therefore

This implies

= h2hlhihl hihl h2

(hl and Thus

and therefore

3

2

(h2hl) ( h 2 h d 3= ( b h ~ ) ~ ( h l h= ;1) ~ as desired. H In the rest of this section, we record some examples of finite groups for which the number of defining relations is the same as the number of generators. By Corollary 2.5, all such groups have trivial Schur multiplier.

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Example 3.8.

Let a

#

403

1,

P#

1 be integers and let the group G ( a , P )

be defined by G(a,P)

=< a,blc- 1ac = aa,cbc-I = bO >

where c = a-'b-lab

Then G(a,,f?) is a finite group with M(G((r,P)) = 1.

Proof. Owing t o I.D. Macdonald (1962), G(a,,D) is a finite group (in fact, G ( a , p) is a finite nilpotent group of class a t most 8 and its order is a divisor of 27(a - 1)(P - l)d8 where d = ( a - 1,P - I)), see I.D. Macdonald

.

(1962, p. 602)). Since the number of generators is the same as the number of defining relations in the presentation of G ( a , P ) , the desired conclusion follows by virtue of Corollary 2.5. T h e following is an analog of the previous example for 3-generator groups. Example 3.9.

Let n 2 3 be a n integer and let the group G be given by

Then G is a finite group with M ( G ) = 1.

Proof. Owing t o Mennicke (1959), the group G is finite. Hence, by Corollary 2.5, M ( G ) = 1. T h e finiteness of the group G in Example 3.9, follows from the following stronger result which also implies that G is solvable. Example 3.10. (Johnson and Robertson (1979)). integers and let the group G be given by

Let a , b,c 2 3 be

Then G is a finite solvable group with M ( G ) = 1.

Proof. By Corollary 2.5, it suffices t o show that G is finite and solvable. To this end, we first note that the defining relations for G imply that

Schur's Formula and Applications

404

>

for all integers k , t with k 0, together with two cyclic permutants. Next we apply the identity

[x, Y, zxl[z, X , Y ~ I [ Y , ~=, X1 ~ I ([x, y] = ~ - ~ y - ~ x[x, y y,z] , = [[x, y], z], xab-a

bc-b

Y

XY

z

= y-lxy) to deduce that

ca-c

=1

Hence xab-a

Y

bc-b

= -

-

Z-l

(x

ab-a

Y

bc-b

)z

>"

xab-a(x-(ab-a)zxab-a)-lz(ybc-b

(power of z ) . yb(bc-b)

7

applying (8). Thus y(b-l)(bc-b) is a power of z, hence so is y(b-l)(bc-l-l) 2 bc-1'1 = 1 (conjugating by z ) . This (conjugating by z-I), whence y(b-l) ( shows that y is of finite order, and a similar argument shows that x and z have finite order. Because the defining relations may be used t o collect powers of x, y, z in an arbitrary word, it follows that

To prove that G is solvable, we first claim that G' is the normal closure of xa-l, yb-l, zC-l. Indeed, for example, since yb-l commutes with y, is conjugated by z t o a power of itself, and it follows that G' is actually generated by these three elements. A similar argument shows that G" is the normal closure in G' of b-1 c-1] = y(b-l)(bc-l-l) [Y 12 and two similar elements. Hence, since y(b-l)(bc-l-l) is a power of z (as has been shown above), G" is actually generated by these elements and is abelian. Thus G"' = 1 and the result follows. W

Example 3.11. tation

(WamsEey (1973~)). Let G be the group with presen-

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Then G is a finite 3-group with M ( G ) = 1. Proof. By looking a t the normal series

we see that G is a 3-group with [GI 5 35. Let H be the group with presentation

We claim that G r H , if sustained, it will follow (since H has an equal number of generators and defining relations) that M ( G ) = 1. Now let us examine the presentation of H. Since b3 is central and C-'b3c = b-6, we have b9 = 1. Cubing the relation cb-' = a-'b-'a gives ( ~ b - ' ) ~= b-3. On the other hand, since c-'bc = b-2, we have b-'c = cb2. Hence b-3 = = c2bcb-' = ~ ~ ( ~ b - ~ )=b ~- ' ~ b - ~ which shows that c3 = 1. We are left t o verify that d is central. Indeed, if d is central, then cubing cd = a-'ca gives d3 = 1; hence H Gi G as desired. Then we have To prove that d is central, put [x, y] = x-'y-'xy. b-'db

= b-'[c,a]b = [b-'cb,b-lab] = [cb3,ac] = b-3[c,ac]b3[b3,ac] = [c, ac] = c-'[c, a]c = c-'dc

and therefore [cb-',dl = 1 Also we have a3 = b-'a3b = ( ~ c (because ) ~ a3 is central), hence a 2 = cacac = ca2cdc = ca-'ca3dc = c2da2dc a -d-'c ~ or a-'dca = d-'c. But c3 = 1 forces ( ~ d = ) ~ giving t h a t ~ ~ d c = 1, which implies ( d ~ = ) ~1; hence ~ - ' ( d c ) ~ a= ( d - l ~ )or~ a-'c-'d-'a = d-'cd-'c giving

which in turn implies that

[ca-' ,dl = 1

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406

Finally note that

a-'ba = bc-' = c-'b4

whence ba-'b-' = b-3ca-', implying on cubing that a-3 = ca-'ca-' (because b3 = a3 is central) or a-2 = ca-'ca-'c = ~ ~ d a - Thus ~ c . ~ ~ d a - ~=c1a or c2daca-' = 1 and daca-'c-' = 1; hence d = cac-la-' = acdc-la-* and SO

[ac,dl = 1

(11)

Applying (9), (10) and (11))we deduce that d is central, as required. W

Example 3.12. (Neumann (1989)). Let m , n be odd integers, let n

f1 and let the group G be defined by

#

Then G is a finite group of order 2 ( m ( n 2- 1)1 and M ( G ) = 1. Moreover, the group G can be described as an extension of

where the order of a2 is ( 1 / 2 ) ( n 2- I ) , the order of b2 is Iml and the action of Z 2 x Z2 (in terms of cosets of a and b) is given by

Proof. Since the number of generators of G is the same as the number of defining relations, the assertion that M ( G ) = 1 will follow from the finiteness of G (see Corollary 2.5). For convenience of reference, we repeat the defining relations : abab = an+' (12) Step 1. Here we show that

bZm = I

Observe first that an+' commutes with ab and a , hence an+' commutes with b. This demonstrates that an+' is central. Hence, by (12) and ( 1 3 ) ,

an+' = abab = b-mababbm = anbanb

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which implies that

banb = a Because (12) can be simplified t o bab = a n , it follows from (16) that

More generally, we deduce that j

=

{

a an

if if

j j

is even is odd

Since m is odd, it follows from (17) with j = m that

which proves (14). To prove (15), we again transform the central element an+' by bm to obtain = b-man+lbm = an(n+l) by virtue of (13). This obviously implies (15). Step 2. Here we demonstrate that every element of G can be written in the form akbs with 0 k 5 n2 - 1 , O 5 s 5 21ml (18)

<

Indeed, by (17), bja = ab-j if j is even and bja = anb-j if j is odd. Thus, by an easy induction, we see that

with f and g easily computed functions of i, j. Taking into account (14) and (15), it follows that every element of G can be written in the form (18). Step 3. It will now be shown that G is of order 21m(n2 - 1)l. First we note that, by (18), the order of G is at most 21m(n2 - I)(. To prove that this is in fact the exact order, one considers vectors ( f , g ) with f and g integers modulo n2 - 1 and 21m1, respectively, and makes a and b act as permutations on the set of these vectors by the rule :

+ +

" ( f , s ) = ( f 1 7 -9) " ( f d ) = ( f n , -9) b ( f 7 d = ( f , g 1)

+

if if

is even g is odd

g

Schur's Formula and Applications

408

Then one easily verifies that these permutations generate a transitive permutation group on the set of 21m(n2 - 1)1 vectors. It is not difficult t o see that the defining relations (12) and (13) are satisfied by these permutations. Thus G is of order 21m(n2 - 1)l. Step 4. Here we exhibit the stated structure of G. By Step 1, b is of order 21ml and a is of order n2 - 1. H2nce a2 is of order ( 1 / 2 ) ( n 2- I ) , while b2 is of order Iml. Since, by (17), b2a = ab-2 we have

On the other hand, by (17), b-'a = anb and ba = anb-'. Hence

.

Finally, an' = a by (15). Hence, by (19),

as desired.

Example 3.13. (Neumann (1989)). Let m , n be arbitrary odd integers and let the group G be defined by

Then G is a finite group of order 8 ) m n Jand M ( G ) = 1. Moreover, the group G can be described as an extension of < a4 > x < b4 > by the quaternion group of order 8 generated by am and bn (here a4 is of order Iml and b4 is of order In\). Proof. Again the assertion that M ( G ) = 1 will follow from the fact that G is finite. For convenience of reference, we repeat the defining relations

abab = a2m

Step 1.

Here we show that

a4* = 1

(20)

3 Computational workshop

409

Note first that, by (20), a2m commutes with ab. Since a2m also commutes with a , we see that a2m is central. Hence, by (21),

which proves (22). Now ( ~ b =) a'~"

is central. Hence transformation by bn yields abab = b-nababbn = a-lba-'b

where the last equality follows from (21). Consequently

Since, by hypothesis, m is odd, m - 1 is even and repeated application of (24) shows that am-lbam-l -b using (20). Hence a-"barn = b-I and therefore abab = a-mababam = ab-lab-' which implies that

b2a b2 = a

Because, by hypothesis, n is odd, repeated application of (25) yields

= b-nabn = b-2n-klab which implies that b2n-1a-1 = ab. Hence, by (20),

proving (23). Step 2. Completion of the proof. We wish t o show that each element of G can be written in the form

To this end, we first observe that

Schur's Formula and Applications

Thus inductively

bia =

,2m-lb-j

if

j

is odd

ab-j

if

j

is even

Similarly, and more generally, it follows that

for some functions f and g of i, j. This proves (26), by applying (22) and

(23).

By (26), the order of G is at most 81mnl. An easy verification shows that G has, in fact, precisely this order. The last assertion regarding the structure of G is a straightforward application of (21), (22) and (23). H

4

Schur's formula and operator groups

In this section we are going t o take a closer look at Schur's formula and especially a t the precise nature of the isomorphism between the groups M ( G ) and ( F ' n R)/[F, R] where G = F I R and F is a free group. All terminology pertaining to operator groups is contained in Sec.1 of Chapter 8. Some general background concerning the Schur multiplier of operator groups was presented in Sec.2 of Chapter 10. Let G be a finite group, say G = F I R where F is a free group and let

T T : Hom(R/[F, ~ R], C*)

-+

M(G)

(1)

be the transgression map associated with the central extension

We know, by Theorem 2.3(ii), that ( F ' n R)/[F, R] is the torsion subgroup of R/[F, R]. Moreover, by Lemma 1.5, ( F ' n R)/[F, R] is a direct factor of R I [ F , Rl, say R I P , Rl = (F' n R)I[F, Rl x SI[F, Rl (2)

4 Schur's formula and operator groups

41 1

By ( 2 ) , we may identify H o m ( ( F Jn R ) / [ F ,R ] , C * ) with the subgroup of H o m ( R / [ F ,R ] ,C*) consisting of all homomorphisms which vanish on S / [ F ,R ] . It follows that the transgression map ( 1 ) restricts t o a homomorphism

T r a : H o m ( ( F 1n R ) / [ F ,R ] , @ * )+ M ( G )

(3)

T h e following lemma will enable us t o motivate the discussion of this section.

Lemma 4.1.

With the notation above, the map (3) is an isomorphism.

Proof. By Proposition 10.1.4, there is an exact sequence H o m ( F / [ F ,R ] ,C * ) % H o m ( R / [ F ,R ] ,C * ) 2 M ( G ) Hence K e r ( T r a ) = I m ( R e s ) . On the other hand, I m ( R e s ) consists precisely of all homomorphisms R / [ F ,R ] + C* which vanish on ( F ' n R ) / [ F ,R]. Hence map (3) is an injective homomorphism. Since, by Theorem 2.3(i), both sides of ( 3 ) are of the same order, the result follows. H It should be pointed out that the isomorphism ( 3 ) is independent of the decomposition (2). Indeed, by the proof of Lemma 4.1, (3) can be obtained by an arbitrary extension of any given x E H o m ( ( F 1n R ) / [ F ,R ] , C * ) t o an element of H o m ( R / [ F ,R ] , C * ) and by applying (1). Now assume that G is an H-group. We know, by Lemma 10.2.l(i), that M ( G ) is also an H-group. It is therefore appropriate to endow ( F 1 n R ) / [ FR, ] with a structure of an H-group such that ( 3 )is an isomorphism of H-groups. This is precisely the aim of the present section. We begin by recording the following useful observation.

Lemma 4.2. Let R be a normal subgroup of a free group F and let A : F I R -, F I R be a homomorphism. Then (i) There is a homomorphism A* : F -, F such that A( f R ) = A*( f ) R for all f E F . (ii) The homomorphism : (F' n R ) / [ F ,R] -+ (F' n R ) / [ F ,R ] given by

is independent of the choice of A* satisfying A( f R ) = A*( f ) R for all f E F .

Schur's Formula and Applications

412

Proof. (i) Let F be freely generated by a set X. Then, for each x E X, there exists A, E F such that A ( x R ) = A,R. Define a homomorphism A* : F -t F by A*(x) = A,. Then A(xR) = A*(x)R for all x E X and hence A ( f R ) = A * ( f ) R for all f E F , as required. (ii) If f E R then clearly A * ( f ) E R . Hence the map is indeed a homomorphism. To prove that is independent of the choice of A*, assume that p : F -t F is any homomorphism such that X ( f R ) = p ( f ) R for all f E F . Then p( f ) = A * ( f ) r ( f )for some r( f ) E R . Hence for any f l , f2 E F , ~ ( [ ff 2l ),l

=

.

[ ~ ( f i ) , ~ ( f 2 )=1 [ A * ( f l ) ~ ( f i )A*(f2)r(f2)1 , [X*(fl),

=

A * ( f i ) l = A * [ f l 7 f 2 l mod[F,Rl

which shows that X*(t) p(t)mod[F,R] for all t E F'. In particular, if r E F' n R , then A*(r)[F,R ] = p(r)[F,R ] , as required. Let G = F I R where F is a free group and let G be an H-group. Since for each h E H , the map is a homomorphism, we may define (by Lemma 4.2(i)) for any h E H a homomorphism

h*:F-+F such that

h * ( f ) R = h ( f ~ ) forall ~

E

F

(4)

Lemma 4.3. Let G = F I R where F is a free group and let G be an H-gmup. Then (F' n R ) / [ F ,R ] is an H-group via

Moreover, the action of H on (F' n R ) / [ F ,R ] is independent of the choice of the homomorphism h* : F + F satisfying (4).

Proof. By Lemma 4.2(ii), the right-hand side of (5) is independent of the choice of h*. If h = 1, then we may choose h* to be the identity map; hence ' v = v for all v E ( F ' n R ) / [ F , R ] . Given h , s E H, we may choose (hs)* to be h*s*. Indeed,

4

Schur's formula and operator groups

Hence, for any

T

E F1fl R, we have

Finally, given TI,r2 E R, we have

as required. Assume that a finite group G is an H-group. Then M(G) is also an Hgroup (see Lemma 10.2.l(i)). Let us recall an explicit action of H on M(G). For any a E Z2(G,C*) and h E H , define

by

h

Q(X,Y)

= a(

h-15,

h-I

y)

forall x , y E G

Then M(G) is an H-group via

Finally, recall that Hom(G, C*) is an H-group via

for all x E Hom(G,@*), h E H, g E G (see Lemma 8.1.4(i)). Thus, by (5) and (71, Hom((F1n R)/[F, R], C*) is an H-group via

for all x E Hom((F1 n R)/[F, R],C*), h E H, T E F' n R. We have now come to the demonstration for which this section has been developed.

Schur's Formula and Applications

414

T h e o r e m 4.4. group and let

Let G be a finite group, say G = F I R where F is a free T r a : Hom(R/[F, R], C*) -, M ( G )

be the transgression map associated with the central extension

Identify Hom((F1 n R)/[F, R],@*) with the subgroup of Hom(R/[F, R],@*) consisting of all homomorphisms which vanish on S/[F, R] (see (2)). Then the tmnsgression map above restricts to a n H-isomorphism T r a : Hom(F'

n R)/[F, R], C*) -, M ( G )

Here the actions of H on the left and right-hand sides are given by (8) and (6), respectively. Proof.

By Lemma 4.1, the given map is a group isomorphism. Fix E L where L = ( F ' n R)/[F, R], and put y = (h-I)*. For each g E G, let p(g) E F be such that g = p(g)R and let p(1) = 1. For any x , y E G, define r ( x , y ) E R by

h E H,

x

Let R I [ F , R] -, L, r[F, R] H r'[F, R] be the projection map corresponding t o the direct decomposition (2). Then, by our convention, x(r[F, R]) = x(rt[F, R])

for all r E R

Hence, by the definition of the transgression map, T r a ( x ) = fi where 6 is the cohomology class of a E Z2(G, C*) given by

Therefore, by the definition of ha,

On the other hand, T r a ( h X ) = P where ,O E Z2(G,C*) is defined by

5 Some exact sequences

415

where the last equality follows from (8). We now show that 0 is cohomologous to h a , which will complete the proof. By (4) and definition of p, ~ ( p ( g ) ) = R h-lg = P ( ~ - ' S ) R Hence, . for each g g G , there exists rg E R such that p("lg) = y ( p ( g ) ) r g . I t follows from p("' x)p("' y) = r(h-' x , h-' y ) / ~ ( ~( -X 'Y ) ) that

Hence TXTY

- r(h-'x, =

h-I

Y)rxyY(r(x,y11-l (mod[F,RI)

and therefore

r:rb

E

r("'x,

h-'

y)1r:y7(r(x,y)')-' ( m o d [ F R , ])

Setting t ( x ) = x(r;[F, R ] ) and applying x together with ( 9 ) and ( l o ) , we have t ( x ) t ( y )= h a ( x ,Y ) ~ ( x Y ) P ( y)-l x, Hence

h a(.,

Y ) = t ( x ) t ( y ) t ( x ~ ) - l pY( )~ ,

as required.

5

Some exact sequences

Schur's formula allows us to treat M ( G ) in a homological way. Thus we can construct dual sequences to those arising from the treatment of M ( G ) as a cohomology group. It is the purpose of this section to exhibit a number of such sequences by using purely group-theoretic approach. An advantage of this approach is that it is direct, does not use any homological machinery and provides an explicit description of the maps involved in the sequences. It will be convenient to extend the definition of M ( G ) to an arbitrary (i.e. not necessarily finite) group G . Given a presentation G E F I R of a group G as a factor group of a free group F , we define M ( G ) by

M ( G ) = (F' n R ) / [ F ,R ]

Schur's Formula and Applications

416

Owing to Hopf's formula (see Theorem 9.7.1) the isomorphism class of M ( G ) is well defined, i.e. does not depend on a choice of a presentation of G. Hence this extended definition amounts to defining M ( G ) as H2(G,Z) (see Theorem 9.7.1). With this extended definition, it follows from Corollary 5.3 below that H2(G,C*) 2 Hom(M(G), C*).

A. Universal coefficient theorem. Our aim here is to provide a direct proof of the universal coefficient theorem in dimension two (see Theorem 9.9.2). Let G = F I R where F is a free group. Then, by Lemma 2.2, for an arbitrary abelian group A on which G acts trivially, the transgression map T r a : Hom(R/[F, R], A ) + H ~ ( G A) , associated with the natural exact sequence

is surjective. Consequently, by Corollary 1.1.13 (with N = R/[F, R], G = F / [ F ,R]), there is an exact sequence Hom(F/[F, R], A) % Hom(R/[F, R], A)

2 H ~ ( GA) , +1

(1)

Let a group G act trivially on an abelian group A. With the notation above, for each c E.H2(G,A) choose X, E Hom(R/[F, R],A) such that Tra(x,) = c and let e, E Hom(M(G), A) be defined by e, = xcIM(G). Then e, is independent of the choice of X, and the map Lemma 5.1.

is a homomorphism. Moreover, if

is a central extension corresponding to c, then

5 Some exact sequences

417

Proof. Assume that A, p E Hom(R/[F, R], A) are such that Tra(A) = Tra(p). Then, applying exactness of sequence ( I ) , we have

Taking into account that M ( G ) = ( F ' fl R)/[F, R] = (F/[F,R])' fl R/[F, R] it follows that (A-lp)(x) = 1 for all x E M(G). Hence A(x) = p(x) for all x E M(G) and therefore e, is independent of the choice of x,. Assume that c,d E H 2 ( G , A ) .Since

it follows that

Thus the map c I-+ e, is a homomorphism. Assume that (2) is a central extension corresponding to c. Then, as we have shown in the proof of Lemma 2.2, there exists a homomorphism 8 : F / [ F , R] -+ E such that X, = y , where y is the restriction of 9 to R/[F, R], and such that the diagram

commutes ( 9 is the natural homomorphism). Now E = A . Im(9) and so, since A Z ( E ) ,it follows that E' = (Im(9))' = 9(Ff/[F,R]). Hence Ime, = 9((F1n R)/[F, R]) G E'

nA

On the other hand, if a E E' n A, then a = 9(x) for some x E F 1 / [ FR]. , Hence 1 = $(a) = +9(x) = q(x)

.

418

Schur's Formula and Applications

and so x E R / [ F ,R ] n F1/[F,R] = (F' n R ) / [ F ,R]. Thus a E Ime, and the result follows. The homomorphism

6 = e,: M ( G )

+

A

in Lemma 5.1 is called the differential of c (or of the central extension corresponding t o c) while the homomorphism

sending c E H 2 ( G , A )to 6 = e, is called the evaluation map make use of this map to prove the following important result.

.

We shall

Theorem 5.2. (Universal Coefficient Theorem). Let an arbitrary group G act trivially on an abelian group A and let f : Ext(G/G1,A ) -t H 2 ( G ,A ) be the restriction to Ext(G/G1,A ) of the inflation map

Then 1

-

Ext(G/G1,A )

H?(G,A ) 5 H o m ( M ( G ) ,A )

-

1

is a splitting exact sequence.

-

Proof. By Theorem 1.2.10, f is an injective homomorphism whose image is the subgroup H ~ ( GA,) of H 2 ( G , A ) consisting of all cohomology G -, 1 with classes c associated to central group extensions 1 -t A + E E' n A = 1. Since, by Lemma 5.1, e, = 1 if and only if E' f l A = 1, we see that I m f = Iiere. We are therefore left t o construct a homomorphism y : H o m ( M ( G ) ,A ) + H ~ ( GA,) such that ey = 1. By Lemma 1.5, M ( G ) = ( F ' n R ) / [ F ,R] is a direct factor of R / [ F ,R]. Hence there is a homomorphism

which is the identity map on M(G). For each y, E H o m ( M ( G ) ,A) define

5 Some exact sequences

419

by xl(x) = ~ ( " ( x ) )for all x E R/[F, R]. Since i~ is the identity map on M , M ( G ) = X. Finally, define y : Hom(M(G), A) + H ~ ( GA) , by

x'I

.

Y(X)= Tra(x1)

Then y is a homomorphism such that

as desired.

e ( r ( x ) > = e(Tra(xt)) = x1IM(G) = X ,

The map e, in Lemma 5.1 is the same as the cotransgression map introduced in the proof of Theorem 5.10. Using this observation and the calculus of induced extensions, an alternative (but less transparent) proof of Theorem 5.2 can be found in Bey1 and Tappe (1982, p.34). Note also that a proof based on the Gruenberg resolution is contained in Eckmann, Hilton and Stammbach (1972a, p.106). Corollary 5.3. Let a n arbitrary group G act trivially on a n abelian group A. If G is perfect or A is divisible, then the evaluation map

.

is a n isomorphism.

Proof. If G is perfect then obviously Ext(G/G1, A) = 1. On the other hand, if A is divisible, then Ext(G/Gt,A) = 1 by Lemma 1.1.8. Hence the desired conclusion follows by virtue of Theorem 5.2. Corollary 5.4. Let G be a finite group acting trivially on a finite abelian group A. Then

H ~ ( G , A )E [(GIG') @ A] x [M(G) 8 A] Proof. Since G is finite, so is M(G). Hence, by Lemma 1.3.l(iv),

EX~(G/G',A)

(GIG') 8 A

and

Hom(M(G),A ) G M ( G ) @ A Now apply Theorem 5.2.

Schur's Fortnula and Applications

420

B. Stem extensions and stem covers. In this section, by applying the universal coefficient theorem, we extend Schur's theory of covering groups to the general case of not necessarily finite groups. All the proofs are self-contained and rely extensively on the properties of differentials and evaluation maps. A central extension

is called a stem extension of G if

and it is called a stem cover of G if its differential

is an isomorphism. By Theorem 5.2, every homomorphism 6 : M ( G ) is the differential of a suitable central extension

-t

A

Hence, by taking A = M ( G ) and 6 the identity map, we see that a stem cover always exists. Corollary 5.5.

A central extension

is a stem extension of G if and only if its differential 6 : M(G)

-t

A

is a surjective homomorphism. In particular, every stem cover of G is a stem extension of G. Proof. Let c E H 2 ( G , A ) correspond t o the above extension and let 6 = e,. Then, by Lemma 5.1, Im6 = E ' n A and the result follows. H

T h e above shows that if G is finite, then the group E appearing in the definition of a stem cover of G is nothing else but a covering group of G. Two stem covers of G :

5 Some exact sequences

421

are called isomorphic if they are isomorphic as central extensions, i.e. if : A1 -+ A2 such that f l = there exist isomorphisms f : El -, E2 and f2 f and f ( a ) = +(a) for all a E A. Thus the notion of isomorphism of central extensions is a generalization of the notion of equivalence of central extensions (in which Al = A2 and is the identity map).

+

+

Proposition 5.6. cover

Every stem cover of G is isomorphic with a stem l+M(G)+E+G-.l

whose diBerential6 : M(G) -t M ( G ) is the identity map. 6

Proof. Let 1 + A -, El -5 G -+ 1 be a stem cover of G with differential S1 : M ( G ) -. A. Write G = F I R where F is a free group. Then, by definition, there exists x E Hom(R/[F, R],A) such that T r a ( x ) = c and S1 = xIM(G), where c E H ~ ( G , A )corresponds to the given central extension. Put X = 6;lX and let 6 be the restriction of X to M(G). Then X E Hom(R/[F, R], M(G)) and S is the identity map. Consider the central extension l - , M ( G ) - + E - , Gf - , l

corresponding to d = Tra(X). Then, by definition, 6 is the differential of this extension. Consider the natural exact sequence

and choose a section p : G

-, F/[F, R]

of FI[F,R] -t G. Define

and p:GxG-,M(G)

422

Schur's Formula and Applications

Then El consists of all pairs (a,g), a E A, g E G with multiplication

while E consists of all pairs (b,g), b E M ( G ) , g E G with multiplication

where {(a, 1)la E A ) and {(b, 1)lb E M ( G ) ) are identified with A and M ( G ) , respectively, and fl((a, x)) = x, f(b,x) = x for all a E A, b E M ( G ) , x E G. Finally consider the map 8 : E + El defined by

Then 0 is a n isomorphism which restricts t o an isomorphism Since f = fie, the result follows. W

+ : M(G)

+

A.

Corollary 5.7. There is a bijection between the set of isomorphism classes of stem covers of G and Ext(G/Gt, M(G)). In particular, G has a unique isomorphism class of stem covers if and only if

Proof. By Proposition 5.6, it suffices to consider all stem covers of G of the form 1 + M(G) + E -, G -, 1. But any two such stem covers are isomorphic if and only if they are equivalent. The desired conclusion is therefore a consequence of Theorem 5.2. W Corollary 5.8. If G is finite, then G has a unique isomorphism class of stem covers if and only if (IG/GII, (M(G)I) = 1. Proof. This follows from Corollary 5.7 and the fact that

by Lemma 1.3.l(iv). W

Proposition 5.9. For any stem extension

5 Some exact sequences

of G , there exists a stem cover

of G and surjective homomorphisms 6 : M ( G ) render commutative the following diagram :

+

A, II, : El

+

E which

Proof. (Rinehart, see Stammbach (1973, p.116)). Let 6 : M ( G ) + A be the differential of the given stem extension of G. Consider the following commutative diagram : 1, Ext(G/G1,M(G))

2 H 2 ( G , M ( G ) ) 2 H o m ( M ( G ) , M ( G ) ) -.1

Here f and f' are induced by the corresponding inflation maps, e and e' are the evaluation maps and the maps 6, are induced by the surjective homomorphism 6. By the universal coefficient theorem (Theorem 5.2) both rows are exact. Moreover, all vertical maps are surjective since so is 6. Let c E H 2 ( G , A ) correspond t o the given stem extension of G. Then, by definition, e1(c) = 6 = 6,(1M(G)). NOWchoose p E H2(G, M ( G ) ) such that e(p) = lM(q.Then

Hence 6,(p) = cfl(x) for some x E Ext(G/G1,A), say x = 6,(y) for some y E Ext(G/G1, M(G)). Since f'(x) = f1(6,(y)) = 6,( f (y)), we see that

Schur's Formula and Applications

424

6,(p) = c6,(f(y)). Setting X = f(y)-lP, it follows that

&(A) = c and

e(X) = lM(q

Consider the central extension

the above extension is a stem cover corresponding to A. Since e(X) = lM(G), of G. Since &(A) = c, one easily verifies that there is a homomorphism $I : El -+ E which makes commutative the diagram in the statement of the result. Since 6 is surjective, so is $I and the result follows.

C. The 5-term homology sequence. Let N be a normal subgroup of G. Then the proof of Theorem 5.10 below introduces two maps :

We refer t o map (3) as the coinflation map and t o map (4) as the cotransgression map . This terminology, borrowed from Evens (1972, 1968), reflects the dual situation in which inflation and transgression maps appear by reversing the arrows Inf :M(G/N)

-+

M(G)

T r a : Hom(N/[G, N], C*) + M ( G / N ) (see Proposition 10.1.5). Theorem 5.10. tension

(The 5-term Homology Sequence). For any group ex-

l - - + N + G - f+ Q - + l

(5)

there is a natural exact sequence M(G)

-+

M(Q)

-t

N/[G, N ]

-+

GIG'

-+

QIQ'

-+

1

(6)

The naturality means that for any morphism (a,@, y) from (5) to a n extension

~+N+SJ'-+Q-+~

5 Some exact sequences

425

there are induced homomorphisms a,,d,, y, which make the diagram

commutative. Proof. Of course, N -, G is assumed to be the inclusion map. The last two maps in (6) are given by N/[G,N] n[G,N]

+

H

GIG' nG'

xG1

f(x)Q1

Exactness at GIG' and Q/Q' is therefore obvious. We now proceed to construct the other two maps and verify exactness at M(Q) and N/[G, N]. Let T : F -+ G be a surjective homomorphism, where F is free, and let . f n : F -+ Q is a surjective homomorphism whose kernel R = l i e r ~ Then S contains R and S I R E N. By definition, M(G) = (F' n R)/[F, R] and M(Q) = (F' n S)/[F, S]. Consider the following homomorphism :

Then the image of this homomorphism is obviously (GI n N)/[G, N] which is the kernel of N/[G, N] + GIG'. This proves exactness at N/[G, N]. Finally, we define M(G) -. M(Q) by means of the natural homomorphism

Because the kernel of M ( Q ) + N/[G, N] is (F1n R)[F,S]/[F,S], we deduce that the sequence is also exact at M(Q). The proof of naturality of the sequence is left as an exercise. W

Schur's Formula and Applications

426

Corollary 5.11.

For any normal subgroup N of G , there is a natural

exact sequence

Proof.

The kernel of the natural map

is (G' n N ) / [ G ,N ] . Hence the desired conclusion follows from the exact sequence (6). H Blackburn and Evens (1979) have added a further t e r n on the left of the above sequence. Since the corresponding proof is so easy, we shall not appeal t o Corollary 5.11.

(Blackburn and Evens (1979)). Suppose that N is a normal subgroup of G. If F is a free group, R is a normal subgroup of F for which G E F I R and S is Q normal subgroup of F for which S R I R corresponds to N , then there is an exact sequence T h e o r e m 5.12.

Proof.

M ( G / N ) + (G' n N ) / [ G ,N ]

-

1

Because G I N % E'ISR, we have

M ( G / N ) = (F' n S R ) / [ F ,S R ] Taking into account that

it suffices to give an exact sequence

-

(F' n S R ) / [ F ,S R ]

-

( F I R n S R ) / [ F ,S I R

-

1

But this sequence with the natural mappings is exact, hence the result. H

5 Some exact sequences

For any group G, we denote the lower central series of G by

Recall t h a t rz(G) = [G, GI

and

y;(G) = [G, yi-I (G)]

( i > 2)

+

If G is nilpotent, then y,(G) = 1 for some m 2 1. If n 1 is the least value of m satisfying this condition, than n is called the class of G. As is t o be expected, nilpotent groups enjoy rather special properties that are by no means true in general. For example, we have the following interesting fact. Corollary 5.13. Let G be a nilpotent group of class c 2 2 and let G = F I R be a presentation of G as a factor group of a free group F. Then

there is an exact sequence

Proof. Since

-

M(GIyc(G))

--+

yc(G)

-, 1

We apply the exact sequence of Theorem 5.12 with N = y,(G).

rc(G) ?c(F)RIR we have S = yc(F), in which case [F,S ] = yC+l(F). Since G is nilpotent of class c, y c + l ( F ) G R. Thus R fl [F,S]= Y , + ~ ( F ) . Finally, because c 2 2, we have yc(G) n G' = yc(G), which shows that (G' fl N)I[G, Nl = rc(G) since [G,yc(G)] = yC+i(G)= 1. This concludes the proof. W

D. The Ganea map. In the special case where N is a central subgroup of G, Ganea (1968) has added a further term on the left of the 5-term homology sequence. We shall present an elementary approach to the Ganea extension due to Eckmann, Hilton and Stammbach (1972a). Theorem 5.14. (Ganea (1968)). Let N be a central subgroup of an arbitrary group G. Then there is a natural exact sequence

(GIG') @ N

-+

M(G)

+ M(G/N)

3

G'n N

-+

1

Schur's Formula and Applications

428

Proof. Write G = F I R for some free group F and choose S d F such that N = S I R . Since N is central, [F,S]C R. Hence, if

is the coinflation map, then by definition (see proof of Theorem 5.10) we have Kera = [F, S]/[F, R] Our next objective is to find a natural surjection

Once this is accomplished, the required sequence will be exact by Corollary 5.11. Since the functoriality of the sequence is a consequence of the universal property of free groups, the result will follow. Define a map f : F x S -, [F,S]/[F, R] by

Since [ab, c] = a[b, c]a-'[a, c] and [a, be] = [a, b]b[a, c]b-l, we have

f (xx', S) = x f (XI,s)x-l f ( x , S) f ( x , SS') = f ( x , s ) s f ( x , sl)s-l

(x,xl E F , s E S ) (x E F, s, s1E S )

However [F,S] C R and [F,S]/[F,R] is commutative. Hence

f (xxl, s ) = f (5, s ) f (XI,8) f (x, ss') = f (x, s ) f (x, s') for all x , x l E F , s,sl E S. Since [F,S]/[F,R] is commutative, it follows that f (F1 x S) = 1. On the other hand, it is clear that f( R x S) = 1, f ( F x R ) = 1. Thus f induces a pairing of abelian groups

and hence a homomorphism (GIG') 8 N

-

[F, SI/[F, R]

which is clearly surjective. This concludes the proof.

5 Some exact sequences

429

The homomorphism (GIG') 8N the Ganea map. Corollary 5.15.

that IM(G/Z)I = IG'

-t

M(G) in Theorem 5.14 will be called

Let Z be a central subgroup of a finite group G such Then the Ganea map

n ZI.

(GIG') 8 Z

-t

M(G)

is surjective. Proof. Consider the exact sequence in Theorem 5.14 :

n Z -+ 1 = IG' n ZI, the map M(G/Z) + G' n Z is an isomorphism.

(GIG') @ Z

-t

M(G) -t M(G/Z)

Since IM(G/Z)I Hence the sequence

(GIG') 8 Z is exact and the result follows.

-t

M(G)

-t

-

G'

1

The following example due t o Wamsley (1973b) (see also Bey1 and Tappe

(1982)) illustrates the use of Corollary 5.15 and the Ganea map. Example 5.16.

Let p be an odd prime and let G be the group with a

presentation Then [GI = p5 and M(G) = 1. Proof. By Example 3.3, IGI = p5. Setting Z =< UP,bP >, it follows from Example 3.3 that Z Z(G) n G' (so that IG' f~ZJ = p2) and that IM(G/Z)I = p2. Hence by Corollary 5.5 the Ganea map (G/Gf)@Z-t M(G) is surjective. In particular, M(G) has exponent at most p. In the rest of the proof, we show that the Ganea map is trivial, which will imply the required assertion. Let G = F I R for some free group F. By Theorem 2.3,

c

where S/[F, R] is free abelian. Because M(G) has exponent at most p, we deduce that Rp[F, R] n ( F' n R) = [F, R]

Schur's Formula and Applications

430

Next we define elements u , v , w of R by = a ~ Z , 2=, a - l - ~c-1 ac, w = cP

+

and put p = 1 p. Then an easy induction shows that

for dl k

> 0.

To examine the case k = p, we put

and note that

Hence it follows that

-

a r w-law = c-pacP = altpava(mod[F, R ] ) with cr E O(modp) and 1+ p a that u

1 +p2(modp3). Inserting up2 = u , we deduce

= O(modRp[F,R ] )

Next we apply (1) and (2) to find that

b-'aPb

= (b-'ab)P = (ac)p . . .c - f ac - aCPC-(P-l)aCP-l -

=

w a a v X ( m o d [R~ ], ) waP(modRPIF,R ] )

which implies

~ - ~ b - ' ar ~ bw = cP(modRPIF,R ] ) By the same argument, applying the equality

(2)

5 Some exact sequences

we deduce that

431

= w = cP(modRPIF,R]) the Ganea map x : (GIG') 8 Z

a-lb-paP

By the direct description of Theorem 5.14, its image is generated by the cosets of [a, bP] [b,ap]-l. Because x is a pairing, we have [ap,bp] = [a, bP]P

(4) -t M(G) in and [aP,b] =

= [a, bp2] = O(mod[F,R])

(5)

Now the elements cac-lap-', cbc-'bp-l and [up,b-l] lie in R and hence are central in F / [ F ,R]. Consequently,

and therefore Again, because

[b-', up] = [am', bp] (mod[F, R ] )

x is a pairing, the combination of (3)-(6) w2 = O(modRPIF,R])

(6) gives

and hence w r O(modRP[F,R]). It follows that w E [F,R] and therefore is trivial, as required.

x

Corollary 5.17. Let p be an odd prime and let P be the group of order P4 with a presentation 2

P =< a , b, clap = P = CP = 1, b-lab = ac, c-lac = al+p, c-lbc = b > Then M ( P ) E 25,.

Proof. Consider the group G in Example 5.16 and let N =< bp >. Then N C Z(G) n G' and so G' n N = N is of order p. It is clear that G I N E P. Since M ( G ) = 1, it follows from Proposition 10.1.5(ii) that

Schur's Formula and Applications

432

as asserted. H We close by remarking that a number of applications of the Ganea map can be found in a work of Eckmann, Hilton and Stammbach (1973). In particular, it is shown (see Corollary 2.5 of their paper) that if H1 and H 2 are finite groups, 2 a central subgroup of H = H1 x Hz, G = H I Z , 2; the image of Z under the projection of H onto Hi, i = 1,2, then is a homomorphic image of M(G). A special case of this result is due to Wiegold (1971b), who employed a different method.

E. Applications t o nilpotent groups. Our principd objective here is to provide an application of Corollary 5.13 and t o record a number of related results. Given x , y E G , l e t [x,y] = x-'y-lxy be their commutator. If X , Y are subsets of G , then we write so that [X,Y] is always a subgroup of G. We now extend the definition of the commutators [x, y] and [X, Y] to an arbitrary number of elements or subsets of a group G. If x; E G , 1 i n, n 2 2, we define [xl, x2,. . . ,xn] recursively to be

< <

Similarly, if Xi, 1 5 i t o be

5 n, n 2 2, are subsets of G, we define [ X I , X 2 , . . . ,Xn] [[XI, . - ,Xn-11, Xn]

Finally, for x, y E G, define xY t o be y-'xy. Lemma 5.18.

(Witt identity). Let x , y , z be elements of a group G.

Then we have [x, y-l, z]Y[y, z-l, xIZ[z,2-I, yIX = 1 Proof.

Set a = xzx-'yx, b = yxy-'zy, and c = zyz-'xz. [x,y-l,z]Y

= y-l[x-lyxy-l,z]y = Y-l(Yx-lY-lx)z-'(x-lyxy-l)*y x-l y -1 X ~ - ' X - ' ~ X ~ = - ~a-'b ~ ~

-

Then

5 Some exact sequences

433

Because b and c are obtained by a cyclic permutation of the elements x , y, z , we also have

[ y ,z -1 ,xIZ = b-lc Hence applying the identity

the desired conclusion follows.

.

and

[ z ,x-', ylX = c-la

Lemma 5.19. Let A , B and C be subgroups of a group G. If N such that [ B ,C, A ] C N and [ C ,A , B ] C N , then [ A ,B , C ] C N .

a G is

Proof. Assume that a E A , b E B and c E C . Then, by Lemma 5.18 and the assumption that N is normal,

.

[a,b-l, elb = ( [ c ,a-l, bIa)-'([b, c-l, ale)-l E N It follows that [a,b-*, c] E N for all a E A , b E B, c E C and therefore [A,B , C ] N , as asserted.

Lemma 5.20.

Let H and S be subgroups of a n arbitrary group G and

let

H = Ho

> HI >

. a .

be a chain of normal subgroups of H such that [Hi,S ] C H;+l for all i = 0,1,2,. . .. Let us put Sj =

{S

E SI[H;,S ]

C H;+j

for all i)

Then each S j is a subgroup of G and, for all i ,j, m [ S j ,Sm]

Sj+m

and

[Hi,7 j ( S ) ]G Hi+j

Proof. Because [ H i ,S ] C Hi+l, we have S assume that x, y E S j and let z E Hi. Then

and so x y E S j . Note also that

N G ( H j ) for all i. NOW

Schur's Formula and Applications

434

whence [z,

2-7

: = ( [ z ,21-1 )x - I E:;H

= Hi+j

proving that x-l E Sj. Hence each Sj is a subgroup of G. We may harmlessly assume that G is generated by H and S. Since S E NG(Hi), each Hi is normal in G. Applying the definition of Sj, we deduce that [[Hi, Sj],Sm] G [Hi+j,Sm] C Hi+j+m

Consequently, by Lemma 5.19,

and so, by the definition of Sj+,

,

In particular, we must have [Sj, S1]= [Sj, S] C_ Sj+1. Tehrefore the S; form a central series of S , which in turn implies that yj(S) C Sj. The conclusion is that [Hi,yj(S)] E [Hi, Sj] C Hi+j as required.

.

For any group G, the upper central series of G

is defined by Z1(G) = Z(G) and for i > 1, Z;(G) is the inverse image in G of Z(G/Z;-1(G)). Let GI and G2 be arbitrary groups. Recall that the following definition of G1 8 Gz is equivalent to the one we employed earlier i,e.

Namely, GI 8 G2 can also be defined as the group generated by all gl 69 g2 (91 E G I , 92 E G2) which satisfy the following relations :

5 Some exact sequences

435

for all g1,g: E G1, g27g; E G2. We now apply the group theoretic results proved above to establish the following useful property.

Proposition 5.21. (Jones (1974)). Let G = F I R be a presentation of a nilpotent group G of a class c 2. Then [ F , y , ( F ) R ] / [ FR] , is a homomorphic image of (G/ZC-l(G))63

>

Proof. Put Z k ( G )= Y k / R for 0 Consider the chain

5k

< c so that Yo = R and Yc= F .

of normal subgroups of F. Because [ Y k F , ] C Yk-l, we may apply Lemma 5.20 for H = S = F . Hence we must have

Next we introduce the following map

To see t h a t 0 is well defined, suppose that fl = fy and xl = xr for y in YCv1and T in R. Because x E y c ( F ) and y E Yc-l, it follows from (7) that [ y ,x ] E [F,R ] and [ry,[ f ,XI-'1 E [F,R]. Accordingly

which shows that 0 is well defined. The conclusion of the proposition now follows from the fact that for all f , f i , f 2 E F and x,x1,52 E T Y ~ ( F ) ,

and as required.

.

If,~ 1 x 2 51 [ f ,x l ] [ f 2, 2 1 (mod[F,R ] )

We are now ready to provide the following application of Corollary 5.13.

Schur's Formula and Applications

436

Theorem 5.22. (Blackburn and Evens (1979)). Let G be a nilpotent group of class c 2 and let G = F I R be a presentation of G as a factor group of a free group F . Then, upon identifying G 8 yc(G) with ( G I G ' ) 8 yc(G), the following properties hold : (i) The map

>

is a surjective homomorphism (f E F , x E y c ( F ) ) . (ii) Let X = IierX and let cp : (GIG') 8 yc(G) -+ M ( G ) be the composition of X with the homomorphism y c + l ( F ) / ( [ FR, ] n y,+l(F)) -, M ( G ) in Corollary 5.13. Then the sequence

is exact. (iii) In the notation of (ii), M ( G ) is an extension of ( ( G I G 1 ) @ y c ( G ) ) / X by I [ e r ( M ( G l y c ( G ) )-,7c(G)). Proof. We first note that (iii) follows from (ii), while (ii) is a direct consequence of (i) and Corollary 5.13. Owing to Proposition 5.21, to prove (i) it suffices to show that

To this end, fix f E F , f l E yc(F), r E R and put fo = [ f ,fl]. Then [ f y f i ~ ]

= [ f , r ] [ f , f i ] [ f o , r ]f0 (mod[F,R ] )

which shows that

The opposite inclusion being obvious, the result is established. We close by providing two related results. In what follows, e ( G ) and d ( G ) denote respectively the exponent and the minimal number of generators of a finite grodp G .

Theorem 5.23. (Jones (1974)). Let G be a finite nilpotent group of class c _> 2 and let G = F I R be a presentation of G as a factor group of the

5 Some exact sequences

free group F . Then (i) Irc(G)l IM(G)I = IM(GIT~(G))II[F,rc(F)RIIIF,RII. (ii) d(M(G)) I d(M(Glyc(G)) d([F,yc(F)RIl[F, RI). (iii) e(M(G)) I e(M(Glyc(G))e([F,7c(F)RII[F, RI).

+

Proof. (i) Setting Q = G/yc(G), we have

and

Q

F/yc(F)R

It follows that

and therefore

Taking into account that

property (i) follows. (ii) and (iii). For any finite group H, let a positive integer r ( H ) be defined as follows : every subgroup of H may be generated by r ( H ) elements and there is at least one subgroup that cannot be generated by fewer than r ( H ) elements. Applying (8), we have

and e(M(G)) I e(M(Q))e([F,yc(F)RII[F, RI), as required. W

Schur's Formula and Applications

438

Corollary 5.24. (Jones (1974)). Let G be a finite nilptent group of class c 2. Then (i) l~c(G)lIM(G)l 5 IM(GIIIc(G))I IGIZc-l(G) @ yc(G)I. (ii) d(M(G)) 5 d(M(Glyc(G)) d(GIZc-l(G) @ yc(G)). (iii) e(M(G)) 5 e(M(G/yc(G)))e(G/Zc-l(G) @ IIc(G)).

>

+

Proof.

6

Apply Theorem 5.23 and Proposition 5.21.

.

Isoclinisms and covering groups

We have already examined a number of properties of covering groups of finite groups. For example, we know that two covering groups of the same group need not be isomorphic (e.g: the quaternion and dihedral groups of order 8 are both covering groups of Z2 x Z2). On the other hand, if the orders of GIG' and M(G) are coprime, then the group G has a unique covering group up to isomorphism (see Corollary 4.4.5). We have also examined a relation between M(G) and M(G*), where G* is a covering group of G. In particular, we have demonstrated that if the orders of GIG' and M(G) are coprime, then M (G*) = 1 (see Proposition 10.1.12). In this section we investigate some necessary conditions which two groups G1 and G2 must satisfy in order t o be covering groups of the same finite group G. This is a classical problem going back to Schur (1904) who established the isomorphism of Gi and G;. Subsequently, Gaschiitz, Neubiiser and Ti Yen (1967) showed that GI/Z(Gl) 2 G2/Z(G2). Unfortunately, they overlooked an even stronger result due to P. Hall (1940, p.138) who demonstrated that GI and G2 are isoclinic. This fact unites and generalizes the above results of Schur (1904) and Gaschiitz, Neubiiser and Ti Yen (1967). Our principal goal is to prove Hall's theorem by using an approach due to Jones and Wiegold (1974). We also establish a related result due to Read (1977). In order to construct p-groups of a given order, it is necessary as a preliminary to split the problem up by introducing some system of classification. A systematic classification theory was introduced and developed by P. Hall (1940). He introduced the notion of isoclinism on which his system of classification was based. Let G and H be two arbitrary groups. Following P. Hall (1940, p.133), we say that G and H are isoclinic if there exist two isomorphisms :

6 Isoclinisms and covering groups

such that if

cr(xZ(G))= x l Z ( H ) , a ( y Z ( G ) )= y l Z ( H ) then

P ( [ x ,Y I ) = [ X I , y11

where [ x ,y] = x-'y-lxy. It is quite possible for two groups to have isomorphic central quotient groups and isomorphic commutator subgroups without being isoclinic. Indeed, it may not be possible to choose the isomorphisms cr and ,B in (1) to satisfy the consistency condition (2). It is clear that G is isoclinic t o the identity group if and only if G is abelian. It is also clear that the relation of isoclinism is, in fact, a n equivalence relation. How to recognize that two groups are isoclinic? An important device for doing so is via the notion of isoclinic epimorphism due to Neumann (195513, p.80). Let G and H be groups and let f : G + H be an epimorphism. Then f is called an isoclinic epimorphism if

Kerf Lemma 6.1.

n GI = 1

(Neumann (19553, p.81)).

If there exists an isoclinic

epimorphism f:G-+H then the groups G and H are isoclinic. Proof. We first claim that f - ' ( Z ( H ) ) = Z ( G ) . Indeed, assume that g E G is such that f ( g ) E Z ( H ) . To show that g E Z ( G ) , we argue by contradiction. If g &I Z ( G ) , then there is an element x E G such that [g,x ] # 1. But then [g,x ] &I Iier f since K e r f n GI = 1. Thus

and therefore f ( g ) &I Z ( H ) , a contradiction. By the foregoing, f induces an isomorphism a : G / Z ( G ) + H / Z ( H ) given by cr(xZ(G))= f ( x ) Z ( H ) for all x E G Consider the induced homomorphism ,B : GI

p(x) = f(x)

-+

HI given by

for all x E GI

440

Schur's Formula and Applications

Since f is an epimorphisrn with Iierf n G' = 1, we see that isomorphism. Finally, since for all x , y E G ,

P

is in fact an

and

a ( x Z ( G ) )= f ( x ) Z ( H ) , a ( y Z ( G ) )= f ( y ) Z ( H ) , we see that the pair a,P determine an isoclinism of G and H . W To provide an application of the main result (Theorem 6.4 below), it will be convenient to record the following basic property of isoclinism.

Lemma 6.2 Assume that G and H are isoclinic groups via isomorphisms cr : G / Z ( G ) + H / Z ( H ) , P : G' -t H' and let the group G / Z ( G ) act on G' and H' by conjugation where we identify G / Z ( G ) and H / Z ( H ) via a. Then P is a G/Z(G)-isomorphism. Proof. Given g E G , write cr(gZ(G)) = g l Z ( H ) . It suffices t o show that for all x , y E G

Since g-'[x,y]g = [g-lxg,g-lyg] and P ( x , y ) = [ x l , y l ] ,we have

as desired. W

Corollary 6.3. Let i > 1 and let G ; be the i-th term of the lower central series of a group G (i.e. G I = G , G ; = [ G i - l , G ] , i > 1). If H is a group which is isoclinic to G, then G; 2 H;. Proof. Let ,i3 : G' = G 2 i H' = H2 be the isomorphism in Lemma 6.2. Since [ G f , G ]is the minimal normal G/Z(G)-subgroup N of G' such that G / Z ( G ) acts trivially on G 1 / N ,we see that P maps G3 = [ G 2 , G ]onto

6 Isoclinisms and covering groups

HB= [Hz,HI. It follows by induction that P maps G; onto Hi for all i as required. H

441

> 1,

With the aid of Lemma 6.1, we now easily derive our main result.

Theorem 6.4. (P. Hall (1940, p.138)). Let G be a finite group. Then any two covering groups of G are isoclinic. Proof. (Jones and Wiegold (1974)). Let G = F I R be a presentation of G as a factor group of a free group F. It will be shown that every covering group G* of G is isoclinic to F/[F,R], from which will follow that all covering groups of G are isoclinic. By Proposition 10.1.11, there is an exact sequence

with A Z(G*) n [G*,G*] and A 2 M(G). Since cp is surjective, it induces a surjective homomorphism [G*,G*] -i [G, GI whose kernel is [G*, G*] n l i e r v = [G*,G*]n A = A Thus [G*,G*] has order equal to IG'J IM(G)J. Consider the exact sequence

where R = R / [ F ,R] and F = F/[F,R]. Then,

where the last equality is a consequence of Theorem 2.3(i). Hence

On the other hand, according t o Theorem 2.3(iv), there is a surjective homomorphism f : F + G*. By (3), Kerf n [F, F ] = 1 and hence f is a n isoclinic epimorphism. Thus, by Lemma 6.1, G* is isoclinic t o F and the result follows.

Schur's Formula and Applications

442

Corollary 6.5. Then

Let G and H be covering groups of a given finite group. G; E H;

for all i > 1

.

where G; and Hi are the i-th terms of the lower central series of G and H , respectively. Proof.

Apply Theorem 6.4 and Corollary 6.3.

We note that the converse of Theorem 6.4 is decidedly false. Namely, given a finite group G and a covering group G* of G, it is not always true that every group isoclinic to G* is also a covering group of G (even if it is of the same order as G*). The following simple example is due to Jones and Wiegold (1974). Suppose that n 2 3 is an integer. Denote by G* the generalized quaternion group of order 2n and let D be the dihedral group of order 2n. Then M(G*) = 1 by Example 3.1, so G' is a covering group of itself. But D is isoclinic to G* and is clearly not a covering group of G*. From now on, we concentrate on the centers of covering groups. T h e o r e m 6.6. (Read (1977')). Let Gi, i = 1,2 be two covering groups of a finite group G and let

be exact sequences with A;

Z(G;)

n G: and A; E M(G), i = 1,2. Then

Proof. Let G = F I R be a presentation of G as a factor group of a free group F. Fix an arbitrary covering group G* of G and consider an exact sequence

I - - + A + G * ~ G + ~ with A Z(G*) fl [G*,G*] and A S M(G). Our aim is to demonstrate that the isomorphism class of Z(G*)/A is determined by the persentation G = F I R . Once this is accomplished, the result will follow. For the rest of the proof, we fix S satisfying property (b) of Theorem 2.3(iv). Owing to Theorem 2.3(iv), we have

6 Isoclinisms and covering groups

443

Let L/[F, R] be the center of F/[F,R]. We claim that Z(F/S) = LJS; if sustained, it will follow from (4) that Z(G8)JA 2 LJR. Because the latter group is determined by the presentation G = F I R , the result will follow. To substantiate our claim, we first note that

and therefore L / S C Z(F/S). To prove the opposite containment, suppose that XSE Z(F/S). Then we have

.

whence x[F, R] E Z(F/[F, Rj). It follows that x E L, Z(F/S) C L / S and hence that L / S = Z ( F / S ) . This concludes the proof of the theorem. In view of Theorem 6.6, it is natural to ask : Is it true that Z(G1) E Z(G2) whenever G1 and G2 are covering groups of the same finite group G ? The following example provides a negative answer.

Example 6.7. (Read (1977)). Let G Z 2 4 x 2 2 . Then there exist covering groups GI and G2 of G such that Z(G1) Z Zq and Z(G2) 2 Z2 x 2 2 (in particular Z(Gl) Z(G2)). Proof. The group G has a presentation

Owing to Corollary 10.7.2, M ( G ) S iZz. Let G1 be the group defined by

Then G1 is obviously a covering group of G and Z(G1) = {l,~~,b:,cb:) =< b: >%

24

Now define the group G2 by

.

Then G2 is also a covering group of G. However, in this case we do have

as desired.

Z(Ga) = { l , c z , b ~ , c g b ~Z) 2 2 x

2 2

Schur's Formula and Applications

444

7

Quasisimple and perfect groups

In what follows, unless explicitly stated otherwise, no assumption on finiteness of a group is made. A group G is said t o be perfect if G = GI. We say that G is quasisimple if G is perfect and G/Z(G) is simple. A typical example of a quasisimple group is G = SL2(q), q odd, q > 3. Here Z ( G ) is of order two and is generated by diag(-1, -1). As we shall see below, a nonabelian finite group G with trivial Schur multiplier is quasisimple if and only if G is a covering group of a simple group. It is therefore not surprising that quasisimple groups play an important role in the analysis of simple groups. Our first aim is to examine Schur multipliers of finite perfect groups. The assumption that a group G is perfect is so strong that it allows us to obtain most of the properties in a self-contained fashion without using heavy guns of previously developed machinery. Let G be an arbitrary group and let be a central extension. Given another central extension by G, say

we say that E covers (respectively, uniquely covers ) El if there exists a homomorphism (respectively, unique homomorphism) which makes commutative the following diagram :

Here and in what follows all unmarked vertical maps are restriction homomorphisms. We shall say that the central extension E above is universal if it uniquely covers any central extension by G. Hence the central extension 1

l + l + G z G + l

7 Quasisimple and perfect groups

445

is universal if and only if every central extension by G splits (or, equivalently, if and only if H 2 ( G , A ) = 1 for any abelian group A on which G acts trivially). The following two lemmas collect together some elementary properties of universal central extensions. Lemma 7.1. Let E and El be central extensions given by (1) and (2).

Then (i) If E and El are universal central extensions, then there is an isomorphism H + H1 which carries A onto A1. (ii) If E is a universal central extension, then both H and G are perfect. (iii) Assume that H is perfect. Then E covers El if and only if E uniquely covers E l . G -+ 1 is a universal central extension, then so is (iv) If 1 + 1 + H i - + i + ~ ~ ~ + i .

Proof. (i) Because E and El are universal central extensions, there exist homomorphisms 8:H

-. H1

and

el : H I

-t

H

satisfying 1C, o 8 = cp and p 0 81 = $. Then the mapping 81 o 6 : H -. H is such that 9 o (el o 8) = p. Hence el o 8 = l ~by, the uniqueness of 8 in the definition of unique covering. A similar argument shows that 8 o 81 = lH,. We deduce therefore that 6 is an isomorphism of H onto H1 which carries A onto A1, as required. (ii) Consider the central extension

where p!((a, b)) = p ( a ) for all a E A , b E H/H1. Define the homomorphisms

by & ( h ) = ( h , 1) and B2(h) = ( h , hH1). Then we have Il,o 8; = 9 , i = 1,2, whence el = 82 since E is a universal extension. Hence H / H 1 = 1 and therefore H is perfect. But then G is perfect since G is a homomorphic image of H . (iii) Of course, we need only verify sufficiency. So assume that E covers El. To prove that E uniquely covers El, suppose that 0; : H -t H I , i = 1,2, = Il,o 81 = Il,o 82. Then, for all x E H , are homomorphisms such that

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01(x)02(x)-I E Z(H1). It follows that the map x I+ 01(x)02(x)-1 is a homomorphism of H into an abelian group. But H is perfect by hypothesis, hence 01(x)02(x)-1 = 1 for all 2 E H and therefore O1 = 02, a s desired. (iv) Assume that 1 -t 1 H 3 G -+ 1 is a universal central extension. As has been observed earlier, it suffices to show that any central extension

.

splits. By hypothesis, there exists a homomorphism a : H + G* such that y = Il,o a. Setting P = a o y-l, it follows that P is a homomorphism from G t o G* such that Il,o O, = lG.Thus the extension 1 + A +- G* + G -t 1 splits.

*

Lemma 7.2. Assume that we are given two central extensions

and let A = I i e r x , where a = Il,o y . Then (i) If G is perfect, then

is a centrul extension. (ii) If El is universal, then so is Eg. Proof. (i) For each a E A, let A, : G --, G be defined by X,(x) = [a, XI, x E G. Then y([a, XI) = [y(a), y(x)] = 1 since y(a) belongs t o the centre of L. Hence, for all x E G, X,(x) E Z(G) and therefore [a,

"YI

= [a, yly-'[a,

XIY = [a,

y] for all x, Y E G

Thus A, is a homomorphism from G to Z(G). Because G is perfect by hypothesis, we infer that X,(x) = 1 for all x E G. Hence a E Z(G) and therefore A Z(G), as required. (ii) Suppose that the extension El is universal. Then, by Lemma 7.l(ii), G is perfect. Hence, by (i), E3 is a central extension. Because G is perfect, t o prove that E3 is universal it suffices (by Lemma 7.l(iii)) t o show that E3 covers any given central extension of the form

7 Quasisimple and perfect groups

To this end, we put

Then T is a group and the natural projection X : T -t L is a homomorphism whose kernel I x D is a central subgroup, Because El is universal, there exists a homomorphism a : G 4 T with p = X o a. Let y : T + S be the natural projection and let ,O = 7 0 a. Given g E G , put a ( g ) = (lg,sg). Then we have P(g) = s, and

whence (P oP)(g) = cl(sg) = W g ) = (+ O v)(g) = r ( g ) This shows that El is universal, as desired.

Lemma 7.3. A covering group of a finite perfect group is also perfect.

Proof. Let G* be a covering group of a finite perfect group G. Then there exists an exact sequence

Z(G*) n [G*,G*]. Since G is perfect, we have G* = [G*,G*]A. with A Hence G* = [G*,G*], by virtue of Lemma 1.4. As a preliminary t o our main result below, let us record the following piece of information. Let G = F I R be a presentation of a perfect group G as a factor group of a free group F and let

be the natural exact sequence. Since Fr/[F,R] is the commutator subgroup of F / [ F , R] and G is perfect, we see that G is the image of F'/[F, R]. Hence the restriction of F/[F, R] t o F1/[F, R] induces a central extension 1 -+ ( F '

n R)/[F, R]

+

F1/[F,R]

+

G

+

1

With this information a t our disposal, we now prove the following result.

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Theorem 7.4. Let G be a finite perfect group and let G = F I R where F is a free group. Then (i) F1/[F,R] is a covering group of G and the central extension 1 + (F' n R)/[F, R]

-

-

F'/[F, R]

-

G+1

is universal, G* + G + 1 is a universal central extension, then (ii) If 1 + A A E M ( G ) and G* is a covering group of G.

Proof. (i) By Schur's formula, we have M(G) S (F'

n R)/[F, R]

Hence, t o prove that F1/[F, R] is a covering group of G, it suffices t o show R] is perfect, i.e. that F N [ F ,R] = F'. that F1/[F, T h e inclusion F1'[F, R] F' is obvious. To prove the opposite inclusion, we first note that a typical element of F is xr with x E F' and r E R , since G' = F I R / R = G. Hence, given y, = x;r;, i = 1,2, s; E F', ri E R , we need only show that [yl, y2] E F"[F, R]. To do this, we apply the identities

[a,bcl = [a,c l b , bl[[a, bl, cl t o deduce that

--

Thus [yl, yz] E FN[F, R], proving that F"[F, R] = F'. Suppose now that 1 + A H G + 1 is a central extension. Then, by Lemma 2.1, it is covered by the natural exact sequence

-

Since G is perfect, it must also be covered by the induced natural exact sequence 1 (F' n R)/[F, R] -t F1/[F, R] -+ G + 1 T h a t the latter extension is universal is now a consequence of Lemma 7.l(iii) and the fact that F1/[F, R] is perfect.

7 Quasisimple and perfect groups

.

449

(ii) By (i) and Lemma 7.l(i), there is an isomorphism F1/[F, R] + G" which carries ( F ' nR)/[F, R] onto A. Hence, by Schur's formula, A E M (G). Since, by Lemma 7.l(ii), G* is perfect, it follows that G* is a covering group of G.

Corollary 7.5. Let G be a finite perfect group and let M ( G ) = 1. Then (i) For any abelian group A on which G acts trivially, H2(G, A) = 1. (ii) If Z is a central subgroup of G, then Z S M(G/Z) and G is a covering group of G / Z . In particular, (i) and (ii) hold i n case G is a covering group of a finite perfect group. Proof. (i) Since M ( G ) = 1, i t follows from Theorem 7.3(i) that the central extension 1+ 1- F1/[F,R]+G+ 1 is universal. Hence, by Lemma 7.l(iv), 1 + 1 i G 2 G + 1 is a universal central extension. Because the latter implies that any central extension of A by G splits, the required assertion follows. (ii) Owing to (i), 1 i 1 -+ G 9 G + 1 is a universal central extension. Invoking Lemma 7.2(ii), we deduce that the central extension

.

is also universal. Hence, by Theorem 7.3(ii), Z 2 M ( G / Z ) and G is a covering group of G / Z . Finally, assume that G is a covering group of a finite perfect group. Then, by Lemma 7.3, G is perfect. Moreover, by Proposition 10.1.12, M ( G ) = 1. Now apply (i) and (ii). It should be pointed out that Corollary 7.5(ii) holds under slightly more general circumstances. Namely, by Proposition 10.1.5(ii), if G is a finite group with M ( G ) = 1 and Z is a subgroup of G with Z C Z(G) n GI, then Z 2 M ( G / Z ) and G is a covering group of G/Z. An alternative proof of this assertion can be found in Harris (1977, Corollary 1.2). Observe also that Corollary 7.5(i) is a special case of Corollary 5.4 whose proof relies on the universal coefficient theorem. An advantage of the proof of Corollary 7.5(i) is that, in the special case of perfect groups, it uses minimal machinery and can be easily understood by any reader with no cohomological background.

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Corollary 7.6. Let G be a nonabelian finite group with trivial Schur multiplier. Then G is quasisimple i f and only i f G is a covering group of a simple group. Proof. Assume that G is quasisimple. Then, by Corollary 7.5(ii), G is a covering group of the simple group G/Z(G). Conversely, assume that G is a covering group of a simple group H. Then there is an exact sequence with Z C Z(G) n GI. Since G / Z 2 H is simple and G is nonabelian, we have Z = Z(G). Again, since G is nonabelian and H is simple, H must be nonabelian and hence perfect. Thus, by Lemma 7.3, G is perfect and the result follows. For convenience of reference, we next record the following lemma.

Lemma 7.7. Let G be a finite group such that (IG/GII, IM(G)I) = 1 and let G I , G2 be two covering groups of G with corresponding central extensions l - + Z i + G ; +f G - + l (i=1,2) i

Then then: exists an isomorphism f : G1 -+ G2 which carries Z1 onto Z2 and makes commutative the following diagram

Proof. Apply the "if" part of Corollary 5.8.

.

The following result shows that the converse of Theorem 7.4(ii) is also true.

Corollary 7.8. Let G* be a covering group of a finite perfect group G and let E:l+Z-+G'-,G+l

7 Quasisimple and perfect groups

451

.

be a corresponding central extension. Then E is a universal central extension. Proof. Apply Theorem 7.4(i) and Lemma 7.7.

As a further simple application of Lemma 7.7, we prove the following result.

Corollary 7.9. (Harris (1977)). Let G be a finite group such that (IG/G1(, IM(G)I) = 1, let G* be a covering group of G and let

be a corresponding central extension. Then, for any automorphism y of G, there exists an automorphism q* of G* such that n o q* = y o n.

f2

Proof. Apply Lemma 7.7 with Z1 = Z2 = Z, G1 = G2 = G*, fi = y a , = n and f = q*. H

A special case of the above result in which G is a finite perfect group is due t o Alperin (see Griess (1973a)). We close the section with two results re1,ating the automorphism group of G t o the automorphism group of a covering group of G under fairly general hypotheses on G (which are automatically satisfied if G is perfect).

Theorem 7.10. (Harris (1977)). Let G be a finite group such that (IG/GII, IM(G)I) = 1, let G* be a covering group of G and let

be a corresponding central extension. Put A = { f E Aut(G*)lf ( Z ) = Z) and for each f E A, let f be the induced automorphism of G = G*/Z. Then (i) The map A + Aut(G), f I+ f is an isomorphism. (ii) z(G) = Z(G*)/Z (hence Z(G) = n(Z(G*))). Proof. (i) It is clear that the map in (i) is a homomorphism. Assume t h a t f = 1. Then x-'f(x) E Z Z(G*) n [G*,G*]for all x E G*. Hence, for any XI,x2 E G*,

Scl~ur'sFormula and Applications

452

Therefore the map

iX2 G*

Z x-lf(x]

is a homomorphism. Because Z G' M ( G ) and GIG' G' G*/[G*,G*], we have

Thus 1C, is trivial, i.e. f = 1 and so the map f I+ f in (i) is injective. To prove surjectivity, assume that cp E A U ~ ( G ) .Consider the natural exact sequence 1-+z+G*1:G+1 Then, by Corollary 7.9, there exists an automorphism cp* of G* such that p o p* = p o p. Hence q* E A maps to p, as required. E Z for all y E G*. Setting (ii) Let X Z E z(G). Then x-'y-'xy fy : G* G* to be fy(t) = y-'ty, we see that f y E A and x-' fy(x) E Z. Since the map 1C, in (i) is trivial, we have x = fy(x) = y-'xy. Because y is arbitrary, we deduce that x E Z(G*) and the result follows.

.

-+

Corollary 7.11. Let G be a finite perfect group with Z ( G ) = 1, let G* be a covering group of G and let

be a corresponding central extension. Then Z(G*) = Z

Proof. Since G*/Z

and Aut(G*) 2 Aut(G)

G and Z(G) = 1, we have Z(G*/Z) = 1. Hence, by Theorem 7.10(ii), Z(G') = Z. It follows that A = Aut(G*), where A is as in the statement of Theorem 7.10. Thus, by Theorem 7.10(i), Aut(G*) 2 Aut(G). H

8

S

Capable and unicentral groups

A. Introduction. Up to this point we have been discussing some very general ideas and it would be possible t o continue in this way, laying broad foundations. We shall, instead, consider some questions of a more specialized nature which

8 Capable and unicentral groups

are concerned with the study of capable and unicentral groups. Throughout, G denotes a finite group. With the exception of Proposition 8.11, all other groups are also assumed to be finite. Following M. Hall and Senior (1964), we say that G is capable if

for some group H. The significance of capable groups was emphasized by P. Hall (1940) who showed that their characterization is an important step in classifying groups of prime-power order. Another important notion, namely that of a unicentral group, was introduced by Evens (1968). We shall demonstrate that there is a common approach t o capable and unicentral groups and provide a close connection with the theory of the Schur multiplier. As an application, we compute the Schur multiplier of extra-special pgroups. The main significance of unicentral groups lies in the fact that, for any unicentral group G and any central subgroup Z of G, the inflation map

is surjective, which is also equivalent to the requirement that the coinflation map M(G) + M(G/Z) is injective (see Theorem 8.13). In practice, this allows us to identify the isomorphism class of M ( G ) in terms of M(G/Z). Indeed, by applying the Hochchild-Serre exact sequence, the above conditions are equivalent to the requirement that M (G) S M (G/Z)/(G1 n Z) where we have identified G' n Z with the image of the transgression map

Now what does it mean for G to be unicentral ? We shall demonstrate that the formal definition which will be introduced can be restated in terms of a certain characteristic subgroup Z*(G) of G which lies in Z(G). It turns out that G is unicentral if and only if Z*(G) = Z(G) and that G is capable if and only if Z*(G) = 1. This gives us a common approach t o the study of capable and unicentral groups. As in the case of the Schur multiplier M(G) of G, one can provide a

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454

precise formula for Z*(G) in terms of any given presentation G = F I R of G as a factor group of a free group F. It turns out that if

is a natural exact sequence, then

It will be clear that if one is t o make use of the theory of capable and unicentral groups, then it is necessary t o be able t o recognize these groups. One of our main results asserts that if G is a central product of groups G;, i E I, 111 2 2 such that G: = Z(Gi) for all i E I, then G is unicentral. This result implies that for any such G,

which will allow us to compute the Schur multiplier of extra-special p-groups. It should be pointed out that certain results presented below hold for arbitrary (i.e. not necessarily finite) groups. The corresponding proofs require minor modifications and use the notion of a stem cover instead of covering group (an interested reader should consult Beyl, Felgner and Schmid (1979)). Finally, we might remark that except for abelian and metacyclic groups, very little seems t o be known about the isomorphism class of Z*(G)in terms of G.

B. Capable groups. We remind the reader our convention that, unless explicitly stated otherwise, any group is assumed t o be finite. As a point of departure, let us record some results which show certain relationships of capable groups with projective representations and Schur mulitipliers. To make the formulation more concise, let us say that G is strongly capable if G 2 H / Z ( H ) for some group H such t h a t the socle of H is generated by a single conjugacy class of H. Of course, a strongly capable group is always capable. Note also that if G is nilpotent, then G is strongly capable if and only if G E H / Z ( H ) for some group H with Z ( H ) cyclic. Proposition 8.1. A group G is strongly capable if and only ifG has a faithful irreducible projective representation over C.

8 Capable and unicentral groups

. .

455

Proof. This is a special case of Theorem 4.3.1 in which F = C.

Assume that a group G is strongly capable. Then, for any covering group G' of G, Proposition 8.2.

M ( G ) E Z(G*) Proof.

This is a special case of Theorem 4.3.1 in which F = C.

.

Proposition 8.3. Assume that a group G is strongly capable. Then the group H in the definition of strong capability can be chosen such that Z ( H ) E H' and Z ( H ) is isomorphic to a subgroup of M(G). Proof.

Apply Theorem 4.3.1 with F = C.

To show that the class of capable groups is rather restrictive, we next record the following facts (properties (i) and (ii) below are contained in Beyl, Felgner and Schmid (1979)). Proposition 8.4.

Let G be a capable group. Then (i) The exponent of Z(G) divides the exponent of GIG'. (ii) If G has a central eleinent of order n , then G has an abelian factor group of order n . (iii) If G # 1 and G is nilpotent, then M ( G ) # 1. Proof.

(i) By hypothesis, there is an exact sequence ~--+Z(H)+HZG-+I

Let { t x l x E G ) be a transversal for Z ( H ) in H with ~ ( t , )= x for all x E G. Note that the commutators

[t,, t,] = t,'t,'tXt, are independent of the chosen transversal. Furthermore, the map

is a pairing, i.e. the maps

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456

are homomorphisms for all choices of x E Z(G), y E G. Put

R = nxEz(GIIierpx

L = nyEGKerX, and

Since [t,,ty] = 1 for all y E G implies t , E Z ( H ) , we see that z = ~ ( t , )= 1 and hence L = 1. On the other hand, since each px is a homomorphism into an abelian group, we have R G'. Observe also that

>

for all x E Z(G), y E G and n E Z. Finally, let e be the exponent of GIG'. Then f(xe,y) = f(x,ye) = 1

for all y E G

Hence, since L = 1, xe = 1 as desired. (ii) This is a direct consequence of (i). (iii) By hypothesis, G E H / Z ( H ) for some group H . Since G # 1, H is nonabelian. Because G is nilpotent, so is H . Hence, by Corollary 10.1.8, M ( G ) # 1..

Corollary 8.5. if Z(G) = 1.

Let G be a perfect group. Then G is capable if and only

Proof. If Z(G) = 1, then G

G/Z(G) is capable. The converse follows

from Proposition 8.4(i). W Of course, we could have defined the notion of a capable group for an arbitrary (i.e. not necessarily finite) group. Certain properties of capable groups hold under this more general setting. A typical example is the following proposition contained in Beyl, Felgner and Schmid (1979).

Proposition 8.6. Let {N;li E I) be a family of normal subgroups of a group G. If each GIN; is capable, then so is G/ niE1N;. Proof. For each i E I , let the exact sequence

exhibit the assumption that GIN; is capable. Put

1 =

( E ) iEZ

N = niaN;

8 Capable and unicentral groups

457

niEI

niEI

and let H be the subgroup of Ei consisting of all (e;) E Ei for which there exists g E G with +;(e;) = gN; for all i E I. It is clear that I< C H. Given g E G, choose a n element e,,; E E; such that $;(e,,;) = gN;. Then e, = (eg,;) is an element of H which belongs to I< if and only if g E N. It follows that the map

GIN gN

Hlli e,li

-t

H

is an isomorphism. Since K = Z(H), the result is established. H

C. Unicentral groups. In what follows, G denotes a finite group. With the exception of Proposition 8.11, for any central extension 1 -t A -t E -. G -t 1, the group E is assumed t o be finite. Following Evens (1968), we say that G is unicentral if for any central extension

*

l-tA+E+G+l we have

+(Z(E)) = Z(G) To provide a useful characterization of unicentral groups, we need to introduce a certain characteristic subgroup Z*(G) of G. Let us fix a covering group G* of G and a surjective homomorphism

with A = l i e r f and A C Z(G*) fl [G*,G*]. Define a central subgroup Z*(G) of G by Z*(G) = f(Z(G*)) It will be shown beIow (see Theorem 8.7 and Corollary 8.8) that Z*(G) is a characteristic subgroup of G which depends neither on the choice of G* nor on the choice of f . Given a E z ~ ( G C*) , and g E G, recall that g is said to be a-regular if

a ( ~ , =~a(g, ) x) for all x E C G ( ~ ) T h e following result ties together Z*(G) and a-regularity.

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Theorem 8.7. (Read (1977)). The group Z*(G) consists precisely of those central elements of G which are a-regular for all a E Z2(G, C*). Proof. Let a E Z2(G,C*) and let p be an a-representation of G. We know, from Lemma 2.6.l(i), that g E G is a-regular if and only if g is ,f3regular where ,f3 is any cocycle cohomologous to a . Hence, replacing a by a cohomologous cocycle, if necessary, we may assume that there is a section p o f f and an ordinary representation I? of G* such that

If g E Z8(G), say g = f ( z ) for some z E Z(G*), then z = ap(g) for some a E A. Hence p(g) E Z(G*) and therefore

It follows that for all x E G

proving that g is a-regular. Conversely, suppose that g E Z(G) is a-regular for all a E Z2(G,@*). Then, p(g)p(x) = p(x)p(g) for all x E G and hence

.

It follows that r(p(g)) permutes with all r(g*), g* E G*. Since the latter is true for any choice of irreducible representation r of G*, we conclude that p(g) E Z(G*). Hence g = f (p(g)) E f (Z(G*)) = Z*(G), as required.

Corollary 8.8. The group Zt(G) is the intersection of all subgroups of the form $(Z(E)), where $ : E -. G is a surjective homomorphism (of finite groups) with I i e r $ C Z ( E ) . Proof. By the definition of ZW(G),it suffices t o verify t h a t , given a central extension 1-C-E4G-1

*

we have Z*(G) & $(Z(E)). To this end, let p be a section of $ and let g be an arbitrary element of Z*(G). Then, by Theorem 8.7, g is a-regular

8 Capable and unicentral groups

459

for all a E Z2(G,C*). Let I? be any irreducible @-representation of G and let p(g) = r(p(g)) for all g E G. Then p is an a-representation of G for some a E Z2(G,C*). Since g is a-regular and I' is an arbitrary irreducible C-representation of G, it follows from the last paragraph in the proof of Theorem 8.7 that p(g) E Z(E). Hence

= +(~(9)) E +(Z(E)) and therefore Z*(G) C $(Z(E)), as required. 1 Corollary 8.9.

A group G is unicentral if and only if

.

Proof. Assume that G is unicentral. Then, by definition and Corollary 8.8, Z*(G) = Z(G). Conversely, if Zt(G) = Z(G), then by Corollary 8.8, +(Z(E)) = Z(G) for any surjective homomorphism : E -+ G with Ker$ C Z ( E ) . Thus G is unicentral.

+

Applying Corollary 8.9 and a theorem of Harris (Theorem 7.10(ii)), we now provide a class of unicentral groups. Corollary 8.10. Let G be a group such that (IG/GfI, IM(G)I) = 1. Then G is unicentral (in particular, every perfect group is unicentral). Proof. lary 8.9.

Owing to Theorem 7.10(ii), Z(G) = Z*(G). Now apply Corol-

By applying a more elaborate approach we shall strengthen the above result by showing that G is unicentral provided the exponents of M(G), Z(G) and GIG' are coprime (see Corollary 8.15). We need another useful description of Z*(G) contained in Beyl, Felgner and Schmid (1979). Proposition 8.11. Let G = F I R be a presentation of G a free group F, let R = R/[F, R], F = F/[F,R] and let

be the natural exact sequence. Then

QS a

factor of

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460

Proof. We begin by choosing an arbitrary central extension

(i.e. E is not necessarily finite). Owing t o Lemma 2.1, there exists a homomorphism $ : F + E which renders commutative the following diagram

Here the unmarked vertical map is the restriction of II,t o R. Note that E = A$(F). Because $(Z(F)) centralizes A and $(F), we obtain $ ( z ( F ) ) Z ( E ) . It therefore follows that

Hence T ( z ( F ) ) can be characterized as the intersection of all subgroups of G of the form y ( Z ( E ) ) , where : E -t G is a surjective homomorphism with K e r y 2 Z ( E ) . Thus n ( z ( F ) ) is independent of the chosen free presentation of G. For any subgroup X of the free group F with [F,R] X, let x = X / [ F ,R]. Choose S a F with S _> [F, R] such that

c

R = ( F ' n R) x S (see Theorem 2.3). Then, by Theorem 2.3(ii), F/S is a covering group of G with corresponding central extension induced by ?r :

Let Z ( F / S ) = T / S . It is clear that z ( F ) C T. On the other hand,

and so

T

= z(F). Hence

8 Capable and unicentral groups

461

as required. W Our next result provides a link between Z*(G)and the notion of capability.

Theorem 8.12. (Beyl, Felgner and Schmid (1979)). For any finite group G, the following properties hold : (i) Z*(G) is the smallest central subgroup of G whose factor group is capable. I n particular, G is capable if and only if Z*(G)= 1. (ii) Z*(G) is the intersection of all subgroups N of G such that GIN is capable.

Proof. It suffices to show that G/Z*(G)is capable and that Z*(G)E N for any normal subgroup N of G such that GIN is capable. 4 Let 1 -t C + E -+ G -t 1 be a central extension. Since

the group G / $ ( Z ( E ) )is capable. Hence, by Corollary 8.8 and Proposition 8.6, G / Z * ( G )is capable. Assume that N is a normal subgroup of G such that GIN is capable. Choose a central extension

1 + Z ( H ) -t H

2 GIN + 1

and put

E = ( ( 9 , h )E G x HlgN = $ ( h ) ) If n denotes the projection ( 9 ,h ) H 9, then

is a central extension. Assume that ( g ,x) E Z ( E ) and let h E H. If gl E G is such that $ (h) = gl N , then ( g l ,h ) E E and therefore x E Z ( H ) . It therefore follows that g E N and hence n ( Z ( E ) )2 N. Applying Corollary 8.8, we deduce that Z*(G)E N , as required.

D. The main result. Our main result will use the properties of certain maps introduced earlier. For convenience, let us recall them.

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462

-

Let Z be a central subgroup of a finite group G. Then the coinflation map M(GIZ) M(G) was introduced in the proof of Theorem 5.10. Of course, we also have a map in the opposite direction, namely the inflation map

which was used repeatedly in our previous discussions. Recall also that upon identification of P(G, Z, C*) (in Theorem 4.5.1) with G 8 Z = (GIG') 8 2, there is an exact sequence M(G/Z)

5 M(G) 5 (GIG') 8 Z

The Ganea map (GIG') 8 2

-

(1)

M (G)

was defined in the proof of Theorem 5.14. Finally, by Corollary 10.1.6, G'n Z is isomorphic to the image of the transgression map Hom(Z, C*) + M(G/Z) In what follows, for convenience, we identify G' fl Z with its image in M(GI-0 Theorem 8.13.

(Read (1977), Beyl, Felgner and Schmid (1979)). For any central subgroup Z of a finite group G, the following conditions are equivalent : (i) M (G) 2 M (G/Z)/(G' n 2 ) . (ii) Z Z*(G). (iii) The coinflation map M(G) + M(G/Z) is injective. (iv) The inflation map M(G1.Z) + M(G) is surjective. (v) The map 0 in (1) is trivial. (vi) The Ganea map (GIG') @ Z M(G) is trivial. In particular, by Corollary 8.9, if G is unicentml then each of the above properties holds for any central subgroup Z of G.

-

Proof. The equivalence of (iv) and (v) follows from exactness of sequence (I), while the equivalence of (iv) and (i) follows from the exactness of the sequence Hom(Z, C*) 5 M(G/Z) M(G)

8 Capable and unicentral groups

463

which constitutes a part of the Hochschild-Serre exact sequence (see Corollary 1.1.13). By Theorem 5.14, there is an exact sequence (GIG') 8 2

-

M(G)

+

M(G/Z)

where the first map is the Ganea map and the second is the coinflation map. Hence (iii) is equivalent t o (vi). By Theorem 4.5.1, 8 is trivial if and only if for any z E Z and any a E Z2(G,@*), z is a-regular. Hence, by Theorem 8.7, (v) is equivalent t o (ii). By the foregoing, we are left to verify that (ii) is equivalent t o (iii). Write G = F I R for some free group F and choose S 4 F such that Z = S I R . If f : M ( G ) + M ( G / Z ) is the coinflation map, then

K e r f = [F,S]/[F,R] (see proof of Theorem 5.14). Therefore we need only verify that [F, S] = [F, R] if and only if

Z & Z*(G)

P u t R = R/[F, R], F = F/[F,R] and S = S/[F, R]. Then [F,S]= [F,R] if and only if C_ Z ( F ) . Let

be the natural exact sequence. Then, by Proposition 8.11,

Hence we deduce that r ( S ) 5 Z*(G) if and only if

S

Z(F)

Bearing in mind that n ( S ) = Z , the result follows. H Corollary 8.14. The gmup Z*(G) is the maximal subgroup of Z(G) for which any (and hence all) of the conditions (i), (iii) - (vi) i n Theorem 8.13 hold. Proof.

This is a direct consequence of Theorem 8.13.

Corollary 8.15. (Read (1977)). Suppose that the exponents of the groups M ( G ) , Z(G), and GIG' are coprime. Then

464

Schur's Formula and Applications

(i) G is unicentral. (ii) M ( G ) r M ( G / Z ) / ( G tn Z ) for any central subgroup Z of G.

Proof. By Theorem 8.13, it suffices t o show that for Z = Z ( G ) , B : M ( G ) -, (GIG') 8 Z is the trivial map. Let e ( G ) denote the exponent of G . Observe that e ( ( G / G t )@ Z ) divides ( e ( Z ) e, ( G / G t ) ) . Thus if c E M ( G ) , the order of B(c) must divide

( e ( Z ) ,e(G/Gt),e ( M ( G ) ) ) Because the latter is possible only in the case B(c) = 1, the result follows.

.

Corollary 8.16. The group Z * ( G ) consists precisely of all z E Z ( G ) for which the coinflation map

is injective. In particular, G is capable if and only if for any 1 # z E Z ( G ) the coinflation map above is not injective.

.

Proof. T h e first statement follows from the equivalence of (ii) and (iii) in Theorem 8.13. The second assertion follows from the first and the fact that, by Theorem 8.12(i), G is capable if and only if Z 8 ( G )= 1. Corollary 8.17. (Beyl, Felgner and Schmid (1979)). Let

be the adjoint homomorphism of the Ganea map

Then Z 8 ( G )= K e r x . In particular, G is capable if and only if x is injective.

Proof. By the equivalence of (ii) and (vi) in Theorem 8.13, we see that z E Z * ( G ) if and only if the Ganea map y<,> : ( G / G t ) @< z >+ M ( G ) is

8 Capable and unicentral groups

465

trivial. Fix z E Z ( G ) and let i :< z >--t Z ( G ) be the inclusion map. Then, by the definition of the Ganea map, the diagram below commutes

This implies the required assertion. H

Corollary 8.18. (Bey1 and Tappe (1982)). Let G be a capable group. (i) If(IG/G'I, IM(G)I) = 1, then Z ( G ) = 1. (ii) The exponent of Z ( G ) divides the exponent of M ( G ) . Proof. T h e desired assertions are immediate consequences of Corollary 8.17. H We next compute Z * ( G ) for an abelian group G.

Theorem 8.19.

(Read (1977)). Let G be a finite abelian group, say

and let m = n l / n 2 . Then Proof.

Z * ( G ) 2 Z,

Owing to Proposition 10.7.1, we have

< <

i t. where Z;,-*) denotes the direct product of i - 1 copies of Zni, 2 Because G' = 1, Z * ( G )is the maximal subgroup Z of G such that M ( G / Z ) 2 M ( G ) (see Corollary 8.14). If Z is a subgroup of G , then M ( G / Z ) S M ( G ) if and only if G / Z S Z,, x Z,, x x Z,, where ml is a positive integer satisfying n21mllnl. For Z to be maximal with respect to this property it is necessary and sufficient that ml = n2. Thus Z * ( G ) 2 Z, as required.

Schur's Formula and Applications

466

The capability criterion below can be found in the book of M. Hall and Senior (1964, p.13)

Corollary 8.20. Let G be a finite abelian group, say

Then (i) G is capable if and only if nl = nz. (ii) G is unicentral if and only if G is cyclic.

.

Proof. This is a direct consequence of Theorems 8.19 and 8.12(i) together with Corollary 8.9. As a preliminary to the next result, let us recall the following definition. We say that G is a central product of groups G;, i E I , if

and, for all i # j , [Gi,Gj] = 1 and

G i n G j = Z(G)

It is clear from the above definition that Z(G;) = Z(G) for all G'

iEI

(2)

=< G:li E I >

(3)

We are now ready to provide the following class of unicentral groups.

Theorem 8.21. (Beyl, Felgner and Schmid (1979)). Suppose that a finite group G is a central product of nilpotent groups G;, i E I, 1 1 1 2, such that G: = Z(G;) for all i E I

>

Then G is unicentral. Proof. It follows from (2), (3) and the hypothesis G: = Z(Gi) that G' = G:

for all i E I

(4)

8 Capable and unicentral groups

Let E be a covering group of G and let

be a corresponding central extension. Let E; be the inverse image of G; in Z ( E ) n El and E = A < Eili E I >, it follows from E, i E I. Since A Lemma 1.4 that E =< Eili E I >. For all i j , n([Ei,E j ] )= [G;,G j ]= I and so

+

This implies that

[Ei,E j , E;] = [ E j 7Ei, Ei] = 1 and hence, by Lemma 5.9 (with N = 1, A = B = Ei, C = E j ) , we have

[E;',Ej] = 1

for all

i#j

(5)

Moreover, by ( 4 ) and the fact that A C El, we have

E1=AEi

forall i E I

Now fix i E I. Since 111 2 2, we may choose j E I with j El = AE; and so [E;,El] = [E;,AE;] = 1 by virtue of (5). Since [Ei,El] = 1 for all i E I and E deduce that El Z ( E ) . Hence

(6)

# i. Then, by =< E;li

E I

(6),

>, we

where the last equality follows from ( 4 ) ,( 2 ) and the assumption that GI = Z(G;). Since n ( Z ( E ) ) Z ( G ) , we deduce that

Hence, by Corollary 8.9, G is unicentral.

Corollary 8.22. Suppose that a finite group G is a central product of nilpotent groups G;, i E I , 111 2, such that G: = Z(Gi)for all i E I . Then there is a subgroup A of M(G/G1)with A "? GI such that

>

Schur's Formula

468

and

Applications

In particular,

Proof. As we have seen in the proof of Theorem 8.21,

Consequently, the desired conclusion now follows by applying Theorem 8.13 for Z = G'.

E. Extra-special p-groups. Our aim here is twofold. First, to compute the Schur multipliers of extraspecial pgroups and, second;to examine capability of extra-special pgroups. All groups considered below are assumed t o be finite. Let p be a prime. A p g r o u p P is called extra-special if

PI = Z ( P ) , IP'I = p and PIP1 is elementary abelian Recall that, by Theorem 21.2.17 in Vol.1, if P is an extra-special p g r o u p , then there exists r 1 such that [PI = p2'+' and P is the central product of r nonabelian groups of order p3.

>

Theorem 8.23. (Bey1 and Tappe (1982), Blackburn and Evens (1979)). Let P be an extra-special p-group of order p2'+', r 2 1. (i) If T > 1, then P is unicentral and M ( P ) is an elementary abelian p-group of order p2'2-f-1. (ii) Suppose that [PI = p3 and p is odd. The

M(P)l

Z p x Zp

if if

P P

is of exponent is of exponent

p P2

(iii) The quaternion group of order 8 has trivial Schur multiplier, whereas the Schur multiplier of the dihedral group of order 8 is of order 2. Proof. (i) We know that P is a central product of r > 1 of nonabelian groups G; oT order p3. Each such G; evidently satisfies GI = Z(G;). Hence, by Theorem 8.21, P is unicentral. Since [PI = p2'+' and IP'I = p, we see that PIP' is elementary abelian of order p2'. Hence, by Proposition 10.7.3, M ( P / P 1 )is elementary abelian

9 Schur multiplier and commutator relations

of order Of order

469

Therefore, by Corollary 8.22, M ( P ) is elementary abelian - 2r2-T-1 -P (ii) Suppose first that P is of exponent p2. Then P is metacyclic and has the presentation pT(2T-1).

pT(2~-l)-l

Hence, by Corollary 10.1.27, M ( P ) = 1. Suppose now that P is of exponent p. Then, by Example 3.3, M ( P ) 2 Z p x Z p . (iii) Apply Corollary 10.1.27. From now on, we turn our attention t o capability. Assume that P is a p-group. Then Z ( P ) # 1 and so, if P is unicentral (i.e. if Z 8 ( P ) = Z ( P ) by Corollary 8.9), then P cannot be capable by Theorem 8.12(i). Using this observation, we now prove the following result. Theorem 8.24. (Beyl, Felgner and Schmid (1979)). An extra-special p-group P is capable if and only if it is either dihedral of order 8 o r of order p3 and exponent p > 2. Proof. If IPI > p3, then by Theorem 8.23(i), P is unicentral. Hence P cannot be capable. Hence we may assume that [PI = p3. If P is of order p3 and exponent p > 2, then the covering group G of P exhibited in Example 3.3 satisfies G/Z(G) % P. Hence P is capable. If P is of exponent p2, p odd, then M ( P ) = 1 by Theorem 8.23(ii). Hence, by Proposition 8.4(iii), P cannot be capable. Since the quaternion group of order 8 has trivial Schur multiplier, it cannot be capable by Proposition 8.4(iii). Finally, the dihedral group of order 8 is capable since it is isomorphic to G/Z(G) where G is a generalized quaternion group of order 16.

.

9

Schur multiplier and commutator relations

Let G be an arbitrary group and let < G , G > be the free group freely generated by all pairs < x, y > with x, y E G. Put [x, y] = xyx-'y-' and observe t h a t there is a natural epimorphism

Schur's Formula and Applications

470

In what follows, we denote by C(G) the kernel of this homomorphism and refer t o the elements of C(G) as the commutator relations of G. To explain our aim, we shall introduce the following universal commutator relations in the sense that they belong to C(G) for any choice of a group G :

(x, y, z E G, < y, z >"=< yx, zx >=< xyx-l, xzx-I >. The normal closure in < G, G > of the above universal commutator relations will be denoted by B(G). It is clear from the definition that B(G) C C(G). We define the group H (G) by putting

Our aim is twofold : first, to demonstrate that B(G) consists of all universal commutator relations of G and, second, to show that H(G) E (F' n R)/[F, R] where G = F I R is a presentation of G as a factor group of a free group F. The latter implies that, in case G is finite, we have

M (G)

H (G)

by virtue of Schur's formula. Thus, for any finite group G, M ( G ) g'lves a measure of the extent to which relations among commutators in G fail t o be consequences of universal relations. In order not to interrupt future discussion at an awkward stage, we first introduce the notion of a free product. Let {Gala E I ) be a family of groups. A free product of the G, is a group G having the following properties : (i) For each a E I, there is an injective homomorphism n, : G, -t G. (ii) For every group H and every family { f,la E I ) of homomorphisms f, : G, -t H , there is a unique homomorphism f : G -, H which for all a E I renders commutative the following diagram :

9 Schur multiplier and commutator relations

The definition of a free product implies that (if it exists) it is unique, up to isomorphism. If we identify G, with its image in G, condition (ii) says that for any family {f,le E I}of homomorphisms f , : G, + H (where H is any given group), there is a unique homomorphism f : G + H such that

for all cr E I. The following result shows that a free product does exist and G,. We remark that for the free product of finitely we shall denote it by many groups G I , G 2 , . . ,Gn we use the notation GI * G2 * .-.* G,.

n: .

Proposition 9.1. For any family {Gala E I } of groups, there exists their free product G = G,. The group G contains an isomorphic copy of each G, and each element g E G can be uniquely written as a finite product of the form = glg2'''gn

n:

>

for some n 0 and some g; E G,,, g; interpreted as g = 1).

#

1 with cr;

#

(the case n = 0 is

Proof. We begin by observing that if G satisfies the foregoing property, then it obviously satisfies the properties in the definition of a free product. Indeed, if f, : G, + H , cu E I, are homomorphisms then we define the homomorphism f : G -t H by

It is clear that f is a required homomorphism. Now we demonstrate that G exists. Let W be the set of all words, including the empty word, of the form

Schur's Formula and Applications

472

for some n and some g; E G,,, g; # 1 with a; # a;+l. Denote by S the group of all bijections of the set W. Given g E G,, define ij E S by I = 1 and, for g # 1, g(w) = g(gig2 ' ' 'gn-1gn) =

Q1S2 "'Qn-1 9192 ' .gn-lg' 9192 ' ' ' gn-lgng

if if if

a, = a and g,g = 1 a, = cxandgng = 9' an # a

#

It is obvious that ij : W + W is a bijection. Moreover, if x, y E G,, then it is plain that = 3 y. Hence the map G, -+ S , g H g is a homomorphism, G o , g H g is whose image we shall denote by G,. In fact, the map G, an isomorphism. This is so because if g E G,, g # 1 and if wo is the empty word, then g(wo) = g # wo and g # 1. Denote by G the subgroup of S generated by all G,, a E I. If g E G, then clearly g can be written as a suitable finite product

with g; E G,, , g; # 1 and a; # a;+l. Hence we are left t o verify uniqueness of expression. To this end, let wo be the empty word. Then we easily see that g(wo) is the word g(w0) = 9192 ' ' ' gn and therefore g l , ga, . ..,g, aae uniquely determined. This concludes the proof. W We now end this digression and return to the study of commutator relations. Let G be an arbitrary group and let B(G), C(G) be the groups introduced at the beginning of the section. In what follows, we write [w] for the image of w E< G, G > in [G, GI under the homomorphism

For convenience, let us use the symbol to denote congruence in < G , G modulo B(G). With this convention, we have the following relations :

< x, y >N< y,x >-I < xy, Z >N x < 2, Z >

>

1

9 Schur multiplier and commutator relations

< y,z

> x ~
473

>< y , z >

(4)

where x , y and z range over G . It will be convenient to record a number of consequences of the above relations. Lemma 9.2.

With the notation above, the following relations hold :

< x , y > < a * b > ~x<, y

(5)

(x,y,a,b E G )

[< X , Y >,< a,b >I N < [x,yl,[a,bI >

< b,bl >< ao, bo >

(6)

< [b,b1],ao>< ao, [b,bl]bo>< b,bl > (b,bl,aO,bo E G )

(7)

< b, b' >< bo,ao >N< [b,b1]b0,ao>< ao, [b,bl] >< b, b' > < b,bl >< a,al >N< [b,b1],[a,a']>< a,al >< b,bl > < x n , x S >N 1 ( x E G , n , s E Z)

(8) (9) (10)

Proof. To prove ( 5 ) , we first invert both sides of ( 3 ) and apply ( 2 ) t o obtain < x , y ~ > ~ < x , y > < x , z > ~( x , y , z ~ G ) (11) Next we expand < a x , by



N

< a x , by >

> in two ways, using ( 1 1 ) and ( 3 ) : <~x,~>~ < x , b >a< a , b >< x , y >ba< a , y >b < x , by >a< a , by > <~,b>~~~~

Comparing, we find that

or

< a,b >< x , y

Finally, replacing x and y by

>ba<

a,b and

~ ( ~ ~ 1 - l

>-IN<

y(ba)-l

x,y yields

>ab

Schur's Formula and Applications

proving (5). Note that (6) is a consequence of (4) and (5), because

< [x, y], [a, b] >< a , b >< a , b >-I

(by (4))

= < [x, yl, [a, bl > Relation (7) is verified by expanding

< ao, [b, b1]bo >

N N

< ao, [b, bt]bo > by

< ao, [b,b'] > < a07 bo < ao, [b, b'] > < ao, bo

(11)) which yields

>ib'b'l >
by (5))

Substitution in the right member of (7) gives the equality (7). Relation (8) is proved similarly, while relation (9) is a restatement of (6). Finally, relation (10) is proved, for nonnegative n and s , by induction on n s, by applying (3) and (11). When n t s = 1, say n = 0 and s = 1, setting x = z and y = 1 in (3) gives the result. The general case follows from the nonnegative case by applying (3).

+

Lemma 9.3. Let G be generated by its subgroups A and B. Then, for a n arbitrary w E < G, G >, 'W

N

app

f o r s o m e a E < A,A >, P E < B , B > a n d p is in the subgroup of generated by all < a , b > with 1 # a E A, 1 # b E B .

Proof. Let < x, y

> be a generator

x = albl

-

4

.

a,b,

and

of

< G,G>

< G , G >, with

y = til bl ...ti, b,

where a;, 7ij E A, bi,&j E B . By a repeated application of the product rules (3) and (11)' it follows that < x, y > is congruent modulo B ( G ) t o a product of elements of the form

< a , a' >', < b, b' >', < a , b >'

nad

< b, a

>2

9 Schur multiplier and commutator relations

475

where a , a' E A , b, b' E B, z E G. Each element of this form can in turn be broken down into a product of terms of the same type, without the exponent z appearing. This can be achieved by repeated use of the rules

< a , b > a O ~ < b, a0 > < a , b > b O ~ < bo,a >< a,bob >

(13)

(14) and three similar rules, obtained from these by interchanging a with b, a0 with bo and a' with b'. Observe that (12) is a restatement of (4), and (13) and (14) are restatements of (3) and ( l l ) , using < c,d > - I N < d , c > by rule (2). Consequently, we deduce that < x, y > (and hence any element w E< G,G >) is congruent t o a product 7 ~ .of terms < a , a l >, < b,b' >, < a , b > and < b,a >. To complete the proof, we perform the following. First, take each term < b, b' > in n and "commute" it to the right (beginning with the farthermost right one and proceeding one a t a time) via (7), (8) and (9). It follows, for a n arbitrary element w E < G , G >, that with p a product of terms < b,bf >, and n' a product of terms < a,a' >, < a , b >, and < b, a >. Next, take each term of the form < a, a' > in n' and commute it t o the left via the rules d u d t o (7) and (8) (obtained from them by inversion and interchanging a and b). This gives us with n u involving only terms < a , b > and < b,a >, and a a product of terms < a , a' >. Finally, by replacing each < b, a > in T" by < a , b > - I , we replace ?rN by p belonging to the subgroup of < G , G > generated by dl This gives w

N

and

l#aEA,l#bEB

apP, as we wished t o show. W

As a final preparatory step, we establish the following lemma.

Lemma 9.4. Let G = A * B be the free product of some groups A and B and let M be the subgroup of < G, G > generated by all

with

l#aEA,l#bE B

Schur's Formula and Applications

476

If p E M is such that [p] = 1, then p = 1.

Proof. We may write p as a reduced word in the free group M :

Then, by induction on k, it follows that [p] can be written as a reduced word in the free product G = A * B in which the last two entries are bklakl if ~k = -1, or a k l b i l if ck = 1. Hence, if p is not the empty word, then [ p ] # 1 as required. Before stating the next result, it will be convenient t o introduce the following notation. If cp : G1 t G2 is a group homomorphism, we define

cp* :< GI, GI >+< G2, G2 > by the rule

Then y" carries C(Gl) into C(G2) and B(G1) into B(G2), inducing a homomorphism

cp* : H(G1) which satisfies

+

H(G2)

( d l * = v*$*, o* = 0, 1* = 1,

where 0 is a zero homomorphism, O(x) = 1, and 1 is the identity homomorphism l ( x ) = x. We have now accumulated all the information necessary t o prove the following result.

Theorem 9.5.

(Miller (1952)). If F is a free group, then H ( F ) = 1.

Proof. First assume that F is not finitely generated. If u E H ( F ) , then u E i,(H(Fl)), where Fl is a subgroup of F on finitely many generators and i : Fl -t F is the inclusion map. Hence we may assume that F is finitely generated. If F has one generator, then H ( F ) = 1 by virtue of rule (10). We are therefore left to verify that if G = A * B with H ( A ) = H ( B ) = 1, then

9 Schur multiplier and commutator relations

H ( G ) = 1. To this end, fix w E C(G). Owing to Lemma 9.3, we have w ~ a p pforsome ~ E < A , A > , P E < B , B > , ~ E M where M is the subgroup of < G , G > generated by all 1 # a E A, 1 # b E B . Since [w] = 1, we.have

<

a,b

>

with

and projecting into A and B , we have [a] = [PI = 1 so that [p] = 1. It follows that a N 1, ,f3 1 and, by Lemma 9.4, p 1. Thus w 1 and the result is established. N

N

N

Corollary 9.6. For an arbitrary group G, B(G) consists of all universal commutator relations of G. Proof. It suffices t o show that if w E < G , G > is a universal commutator relation in G , then w E B(G). To this end, write w in the reduced form w =< x i , y1

>"l

. . -< x,, y, >""

( E= ~ *l, xi, yi E G)

Let F be the free group freely generated by X i , . . . ,X, , Yl, . . . ,Y,, and let cp : F -t G be the homomorphism given by

Then the induced homomorphism cp* carries < X i , Yl of < F , F > into w. Hence it suffices t o verify that

>"I

. . < Xn7Yn >""

Because w is a universal commutator relation, < X I , Yl >"' . . . < X,, Y, >"" E C ( F ) . Bearing in mind that, by Theorem 9.5, B ( F ) = C ( F ) , the result follows. T h e following theorem is the main result of this section.

Theorem 9.7.

(Miller (1952)). Let G = F I R be a presentation of an arbitrary group G as a factor group of a free group F. Then

H(G) 2 (F'

n R)/[F, R]

Schur's Formula and Applications

478

and, in particular, H ( G ) 2 M ( G ) if G is finite.

Proof. Consider a central group extension

We define a homomorphism < G , G >-, E by mapping a generator < x , y > of < G, G > into [3, ij], where f (3) = x and f (ij) = y. This is independent of the choices 3 and y because Z Z(E). Note that the above homomorphism carries C ( G ) onto Z n E' and carries B(G) onto 1, and hence induces a surjective homomorphism

Moreover, it is readily seen that the sequence

is exact. Now put E = F/[F,R], Z = R/[F, R] and let f : E -, G be the natural homomorphism. Then we have a central extension 1 -, Z + E G -, 1 such that Z n E' = ( F 1n R)/[F, R] Hence, by (15), it suffices to verify that f, = 0. Let X : F -, E be the natural homomorphism and let

Then, by definition, [w] = [xl, yl] .. [xk,yk] = 1 E E. Choosing 3i, yi E F such that A(?;) = xi and X(y;) = y;, we see that the element

satisfies X*(w) = w, X([w]) = [w] = 1, and hence [w] E [F, R]. Accordingly,

Lw1

= [fi TI]' ' ' [ft ,rt] for some f; E F and

ri E R

But F is free, so H ( F ) = 1 and B ( F ) = C ( F ) by Theorem 9.5. Thus

9 Schur multiplier and commutator relations

and therefore

Hence f, = 0 and the result follows. 1 Let A be an arbitrary abelian group. Recall that the exterior square A A A of A is defined by

The following result may be regarded as a generalization of Corollary 1.3.7 which treats the case where A is finite. Theorem 9.8.

(Miller (1952)). For any abelian group A,

Proof. Since A is abelian, C(A) =< A,A >. On the other hand, C(A)/B(A) = H ( A ) is abelian by Theorem 9.7. Hence we see that for the abelian case only, we may as well have taken < A, A > to be the free abelian group on the pairs < x, y >. Writing both A and < A, A > additively, the defining relations for B(A), together with the consequent relation (11) become

Observe that (20) is a consequence of (19) by setting y = 0 in (19). Note also that < x y , x f y > N 0 by (16), and expanding this by (18) and (19) shows that (17) is a consequence of (16), (18) and (19). The conclusion is that B(A) can be defined by (16), (18) and (19), which demonstrates that H(A) 2 AA A. H

+

Turning to the free products, we next establish the following result.

Schur's Formula and Applications

480

Theorem 9.9. (Miller (1952)). For any groups G1 and G 2 ,

Proof. Let i : G1 + G1 * G2 and j : G2 -+ G1 * G2 be the natural injections and let p : G1 * G2 -t G1 and q : G1 * G2 -+ G2 be the natural projections. Consider the induced maps :

Then it is clear that i, and j, are injective homomorphisms and that i,(H(G1)) and j,(H(G2)) have trivial intersection. Because H(G1 * G2) is abelian, it suffices to show that

To this end, put G = G1 * G2 and denote by M the subgroup of generated by all

< G, G >

To prove (21), it suffices to verify that, for any given w E C(G), w N some a E < G1,G1 >, p E < G2,G2 > with [a]= 1 and [p] = 1. Invoking Lemma 9.3, we may write

.

w ~ a p pforsome a €
> , p ~ Gg,G2 < >,pE

ap for

M

Since [w] = 1, we have [a][p][p]= [w] = 1, and projecting into G1 and G2, it follows that [a] = 1 and [PI = 1. Thus [p] = 1 and so, by Lemma 9.4, p = 1. This concludes the proof. Let G1 and G2 be arbitrary groups. Recall that we have previously defined G1@G2by G1 @G2= (G1/G{)@z(G2/G$). AS we mentioned before, it is sometimes convenient to use the following equivalent definition. Namely, we define G1 @ G2 as the group generated by all g1 @ g2 (gl E G l , ga E G2) which satisfy the following relations :

9 Schur multiplier and commutator relations

481

for all gl,g; E G1, g279; E G2. The following result can be regarded as a homological analog of Theorem 1.2.3. Theorem 9.10.

(Miller (1952)). Let GI and G2 be arbitrary groups.

Then H(G1 x G 2 ) r H(G1) x H(G2) x (GI @ G z ) Proof. Let i : G1 + GI x G:! and j : G2 + G1 x G2 be the natural injections and let p : GI x G2 --, G1 and q : GI x G2 -+ G:! be the natural projections. Denote by T a homomorphism from GI 8 G2 into H(Gl x G2) defined by requiring that r(gl @ g2) is the image in H (G) of < gl ,g2 > The fact that r is a well-defined homomorphism follows from the fact that H (G) is abelian, the congruences

.

and the dual congruence

Let a : H(Gl x G2) -+ G1 @ G2 be induced by

where

Then, by applying properties of induced homomorphisms i,, j,, p, and q, together with Lemma 9.3, it follows that i,, j, and T are injective and that

This completes the proof of the theorem. H

Schur's Formula and Applications

10

Cyclic extensions

Let N be a normal subgroup of G such that GIN is cyclic. In this section, we provide a link between the Schur multipliers of M ( N ) and M ( G ) by exhibiting an appropriate exact sequence.

A. Computation of H2(G,Z) via bar resolution. We begin by recalling the notion of the standard resolution (which is also called the bar resolution). All the relevant details can be found in Chapter 9 of Vol.1. Let RG be the group algebra of an arbitrary group G over a commutative ring R. For any given n > 0, define Qn as the free RG-module with basis consisting of all n-tuples [ g l , .. . ,gn] of elements of G. Define Qo to be the free RG-module on the single generator [ 1. T h e augmentation

is defined by E([

and the homomorphism

is given by

In particular,

Then, with the notation above,

I) = 1

10 Cyclic extensions

483

is a free resolution of the trivial RG-module R (called the bar resolution of

R).

There is a useful modification of the bar resolution called the normal. This resolution is defined as follows. Consider the RG-submodule P, of Q, generated by all [gl, 92,. . . ,g,] with at least one g; = I , and put Q, = Q,/P,. An easy calculation shows that ized bar resolution

and therefore 8, induces a homomorphism

We have Qo = Qo so we use the same augmentation E : Qo reasoning applies to the contracting homotopy, so that

+

R . Similar

is a free resolution of R. We now specialize t o the case R = Z. Recall that, in order t o evaluate H 2 ( G , Z )a t the bar resolution, we form the induced sequence Z @ ~ G Q ~ ~ * Z @ Z G Q ~ ~ ~ Z @ ~ : Q ~ where Z is the trivial right ZG-module. Then, by definition,

(by Proposition 6.8.2 in Vol.1, the right-hand side is independent of the choice of the projective resolution of Z, up to a functorial isomorphism). Lemma 10.1. Let Ai(i = 1,2,3) be the free abelian group o n i-tuples [gl,. . .,g;] with all gl, ... ,g; E G. Then the maps

defined by

Schur's Formula and Applications

484

are homomorphisms with I m d s C lierd2 and

.

Proof. This follows by identifying Z (11, ( 2 ) and (3).

Q iwith A; and by applying

Remark 10.2. The conclusion of Lemma 10.1 holds if A; is the free abelian group on i-tuples [gl, . . . ,g;] where the gj, 1 j 5 i, run through the nonidentity elements of G and [gl,. . . ,g;] = 1 whenever some g j = 1. Indeed, this follows by applying the normalized bar resolution.

<

B. Cyclic extensions. In what follows, N denotes a normal subgroup of a finite group G such that G I N is cyclic of order n generated by g. Thus gn = t

for some t E N

Given x E G, we put xg = g - l x g (thus tg = t ) . Denote by A the abelian group with generators c ( x , y), b ( x ) ( x E N, y E N ) and defining relations

Theorem 10.3. (Blackburn and Evens (1979)). With the notation and assumption above, let V denote the free abelian group with basis { v ( x ) l l # x E N ) , and put v ( 1 ) = 0. Then the map 0 : A -, V given by

is a homomorphism such that

10 Cyclic extensions

485

+

Proof. Given x , y E G, put c'(x, y) = v ( x ) v ( y ) - v ( x y ) and bt(x) = v ( x )- v(x9). Denote by H the subgroup of A generated by the c l ( x ,y), bl(x) for all x , y E N . Then the generators ct(x,y) and bl(x) satisfy relations ( 1 ) - (5) and hence 0 is a homomorphism, For the sake of clarity, we divide the rest of the proof into two steps.

Step 1. Here we construct a suitable free presentation of G. We regard A as a subgroup of A* = A@ < d >, where < d > is an infinite cyclic group. Owing t o relations (1) and ( 2 ) , there is a group

E = {(x,a)lxE N , a E A*) with

+

( x , a ) ( y , b )= ( x Y , ~ b+ ~ ( X , Y ) ) Observe that the mapping 1C, : E + E given by $ [ ( x ,a)] = ( x g ,a

+ b(x))

is an automorphism. Hence there exists a cyclic extension D =< s , E > in which E d D, the automorphism of E induced by s is Il, and s" = ( t , d ) . Setting B = ((1,a)la E A*) we obtain the central extension with B E A* and G E D/B. Let F be a free group with basis { e ) U { e ( x ) l l # x E N ) , and put e ( 1 ) = 1. Then there is a homomorphism X : F -+ D given by

Given x , y E N , put

C(x,Y) = e ( x y ) - l e ( x ) e ( y ) b ( x ) = e(xg)-'e-'e(x)e 2 = e(t)-'en Then

Schur's Formula and Applications

and therefore B C ImX. It follows that

and so X is a surjective homomorphism. P u t T = I i e r X and denote by R the inverse image of B. Then R I T E B Z A * and

FIRSG

Hence, by Schur 's formula,

M(G)

(F' n R)/[F7R]

and therefore it suffices to verify that Icer 8 2' (F' n R)/ [F, R]

(6)

Step 2. To prove (G), we first construct an isomorphism from B onto R/[F, R]. Since B C Z(D), we have [F,R] E T. Hence X induces a surjective homomorphism p : F / [ F , R] + D. An easy calculation shows that the elements F(x7 Rl7 &(X)[F,Rl satisfy the defining relations (1) - (5) of A. Hence there exists a homomorphism $ : B + R/[F, R] given by

From the definition of $, it follows that pII, = 1. Now Il,can be extended to D by putting $(si(x, 4 )= ( e i e ( x ) [ ~~, 1 ) ( $ ( 1a, ) ) T h e conclusion is that

10 Cyclic extensions

487

proving that q!y = 1. Hence p and $J are isomorphisms. Thus 1C, gives a n isomorphism of B onto R / [ F ,R ] . P u t S = (F' n R ) / [ F R , ] . Then S can be viewed as the kernel of the restriction to R / [ F ,R ] of the natural epimorphism

or, of the epimorphism

, H v ( x ) and e[F,R ] I+ v. where < u > is an infinite cyclic group, e ( z ) [ F R] Consequently, c ( z , Y ) IF, Rl 6 ( x )[F,R] d [ F ,R]

+

H H

v ( z ) U ( Y ) - v(x9) v ( x ) - v(xg) nv - v ( t )

and thus S is the kernel of the corresponding homomorphism of A' into V @< v >. Because d H nu - v ( t ) , this kernel is contained in A and is in fact Ir'er 6. This proves (6) and hence the result. H We have now come t o the demonstration of the main result.

Theorem 10.4. (Blackburn and Evens (1979)). Let N be a normal subgroup of a finite group G and let G I N be cyclic of order n, say, generated by g N . Denote by C N I N l ( g )the subgroup of elements of N / N 1 fixed under conjugation by g, and let L be the subgroup of C N I N t ( g )generated by gnN' and all elements of the form

Then there is an exact sequence

In particular, (i) If M ( N ) = 1 , then M ( G ) 2 CNIArt(y)/L. (ii) If N is perfect, then M ( G ) is a homomorphic image of M ( N ) .

Schur's Formula and Applications

488

Proof. We keep the notation employed in the proof of Theorem 10.3. As a point of departure, we note that there is a homomorphism

given by

where N = N / N 1 and 2 = xN'. Denote by C the subgroup of A generated by all c ( x ,y) with x , y E N . For the sake of clarity, we shall divide the rest of the proof into two steps.

Step 1. Our aim here is to demonstrate that the restriction of /? to Iier 0 induces a surjective homomorphism

Iier

o -Z CNIN1(g)/L

Let a be the homomorphism of V into N given by a ( v ( x ) )= Z. If a=

jZYc(z, y)

+

M ( x ) E Ber 0

then we have

L Applying a , it follows that n z k z E CNIN1(g),SO P(a) E C N I N ~ ( g ) /and therefore P ( I i e r 0 ) CG/G1(9)/L

c

To prove the opposite containment, suppose that Z E CGIG'(g). Then g-lxg = x z with t E N'. Hence v ( z ) E U , where U denotes the subgroup of V generated by all v ( x l ) v ( x 2 )- v ( x l x 2 ) . Consequently,

+

Since U = O(C), there exists c E C such that O(b(x)) = O(c). But then b ( x ) - c E Iier 0 and, since we also have P(b(x) - c ) = LZ, we deduce that

10 Cyclic extensions

489

as claimed. Step 2. Completion of the proof. We wish to show that C = I i e r p which will imply by Step 1 that the sequence

C n I i e r 0 + l i e r 0 2 CNIN,(g)/L

+

1

is exact. Since, by Remark 10.2 and Proposition lO.l.l(iii), C n K e r 0 is a homomorphic image of M ( N ) the result will follow by virtue of Theorem 10.3. To prove that C = I i e r p , we first observe that, because of relation (3), any element of A can be written in the form a = b(x) c for some c E C. Hence p ( a ) = p(b(x)) = 1 if and only if

+

(where t = gn E N ) . If so, there exists u E U such that

It follows that b(x) = b(t)j t x ( b ( y )

+ b(yg) t . . . + b(ygn-l))ky+ u',

where U'

E< b(xlx2) - b(x1) - b(x2)Ix1, 2 2 E N

>

However, applying relations (5), (4) and (3), the right hand side of the above equation is in C, as required. In what folIows, d(G) denotes the minimal number of generators of G. Corollary 10.5. (Jones (1974)). Let G be a finite group and N a normal subgroup such that GIN is cyclic. Then (i) I M(G)I divides IM(N)I I NIN'I. (ii) d(M(G)) I d ( M ( N ) ) d(N/N1).

+

Proof. This is a direct consequence of Theorem 10.4.

Schur's Formula and Applications

490

Corollary 10.6. (Jones (1974)). Let G be a finite p-group (p prime) with a maximal subgmup N such that M ( N ) = 1. Then M ( G ) i s elementary abelian of order at most pd(NIN').

Proof. First we note that N is a normal subgroup of G of index p (in particular, G I N is cyclic). By Theorem 10.11.2, M ( G ) is elementary abelian. The desired assertion is therefore a consequence of Corollary 10.5(ii).

11

Deficiency, efficiency and Schur multipliers

A. The Reidemeister-Schreier theorem. In this section, we present a method of writing down a presentation of a subgroup when a presentation of the group is known. Recall that, by writing,

we mean that G = F ( X ) / N where F ( X ) is a free group freely generated by X and N is the normal closure of R in F ( X ) . The elements of R are called defining relators for G.

Theorem 11.1. (Reidemeister-Schreier). Let H be a subgroup of a n arbitrary group G =< X I R >. Write H = K I N for some subgmup K of F ( X ) , let T be a Schreier transversal for Ii' i n F and, for each f E F ( X ) , l e t p ( f ) E T be defined by Ii'f = K p ( f ) . Then

where Y = {txp(tx)-'It E T ,x E X , t x 6 T ) is a set of free generators for Ii' and S = {trt-'It E T , T E R ) . Moreover, if both (G : H ) and 1x1 are finite, then lYl = (1x1- l ) ( G :H ) 1

+

Proof. Owing t o Nielsen-Schreier theorem (Theorem 9.5.5), Ii' is a free group freely generated by Y. Since S C N C f i , the first assertion will follow provided we show that N is the normal closure of S in K. Ii', we have Let L denote the normal closure of S in Ii. Since S C N L C N . To prove the opposite inclusion, note that any f E F ( X ) can be written as f = k t with I; E K and t = p ( f ) E T . Hence a typical generator

11 Deficiency,

efficiency and Schur multipliers

491

f r f - ' of N is equal t o k(trt-')k-l, which belongs to the normal closure in K of S. Thus N & L and so L = N, proving the first assertion. Assume that both (G : H) and are finite. Put

1x1

n=1 x1 and m = ( G : H) Then m = ( F ( X ) : I i ) = (TI since (G : H) = ( F ( X ) : l i ) . Now put

Then, by the first paragraph, I{ is a free group of rank mn - s, where s is the number of At,, equal to 1. Choose any 1 # t E T and write t = uxE, where x E X, E = f1, and ux' is a reduced word in X. To each right coset #t(l # t E T) with t written as above, we now associate a pair (t', x) = y ( I i t ) where t' = t if E = -1 and t' = u E T if E = 1. Observe that if E = -1, then At, = 1, and if E = 1, then ,A, = 1. I t will next be shown that all these associated pairs are distinct, i.e. that y is injective. Assume that ~ ( I i a = ) y(1ib) for some nonidentity a, b E T. Then, by the definition of y , both a and b have the same last letter, say x. Hence a = u x E and b = v x 6 (~,S=fl) where ux' and vx6 are reduced words in X. If E = -1 = 6, then y(Ir'a) = ( a , x ) and y(Kb) = (b,x) so that a = b and K a = Kb. If E = 1 = 6, then y(1ia) = ( u , x ) and y(Iib) = (v,x) so that u = v, a = b and Ir'a = l i b . If E # 6, say E = -1 and 6 = 1, then y ( I i a ) = (a,x), y(Iib) = (v,x) and a = v. Therefore b = vx = ax = (ux-')x = u so that the last letter of u is x. This contradicts the fact that a = ux-' is a reduced word in X. We have therefore exhibited m - 1 A t , equal to 1. Finally, every (t, x) with At,, = 1 must arise from a coset of K via 9, for A t , = 1 merely says that p(tx) = tx. Hence

as desired. H

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492

B. The deficiency and efficiency o f a finite g r o u p . Although with certain modifications the discussion below can be carried out for finitely presented groups, we consider exclusively finite groups. In what follows, G denotes a finite group and M ( G ) the Schur multiplier of G . Assume that G has a presentation with n generators and r defining relators. Then, by Proposition 9.4.3, r - n 2 0. We refer t o r - n as the deficiency of such a presentation. If r > 0, then it is possible t o add further relators that are consequences of the orginal ones, thus giving a presentation of a larger deficiency. Hence it is natural to seek presentations with as small deficiency as possible. With this in mind, we define the deficiency of the group G written de f ( G ) , t o be the minimum of r - n taken over all finite presentations of G . I t is clear that de f ( G ) is a nonnegative integer. Let d ( M ( G ) ) denote the minimal number of generators of M ( G ) . To illustrate the relevance of deficiency to the theory of the Schur multiplier, we first reformulate Corollary 2.5. Lemma 11.2.

With the notation above, we have

.

In particular, if de f ( G ) = 0 then M ( G ) = 1 and if de f ( G ) = 1, then M ( G ) is cyclic. Proof.

This is a direct consequence of Corollary 2.5.

T h e lemma above motivates the following definition due t o Epstein (1961). We say that the group G is efficient if

Lemma 11.3. Let G be a finite abelian group, say

Then de f ( G ) = d ( M ( G ) )= ( 1 / 2 ) k ( k - 1 ) In particular, G is efficient.

11 Deficiency, efficiency and Schur multipliers

493

Proof. If k = 1, then G is cyclic and hence de f ( G ) = d(M(G)) = 0. We may therefore assume that k 2 2, in which case

) the direct product o f t copies by virtue of Theorem 1.5.12 (here H ( ~denotes of a given group H ) . Hence

On the other hand, G has a presentation with k generators, b power relators and (1/2)k(k - 1) commutator relators. Hence

and the desired conclusion follows by comparing (1) and (2) and applying Lemma 11.2. We close by remarking that all finite metacyclic groups are efficient. This result was established by Wamsley ( 1 9 7 0 ~ )(and subsequently by Bey1 (1973)). C. Swan's example. Neumann (1955a) raised the following problem. Is it true that any finite group G with trivial Schur multiplier has zero deficiency ? Our aim is to present a n example, due t o Swan, which provides a negative answer. We begin by recording the following useful observation contained in Swan (1965). Lemma 11.4. Let H be a subgroup of a finite group G . Then

de f ( H )

+ 1 5 ( G : H)(def (G) + 1)

Proof. Let G =< XIR > be a finite presentation of G with def ( G ) = IRI - 1x1. Then, by Theorem 11.1, H has a presentation H =< YIS > such that IS1 = (G : H)IRI and IYI = (1x1 - l ) ( G : H)

+1

Schur's Formula and Applications

Hence

.

def(H)

as required.

IS1 - IYI = ( G : H)IRI - (1x1- 1 ) ( G : H ) - 1 = ( G : H ) ( d e f( G ) 1 ) - 1 ,

I

+

>

Example 11.5. (Swan (1 965)). For each integer k 1, there exists a group G k of order 7 k 3 such that (i) M ( G k )= 1 for all k 1. (ii) de f (Gk) > 0 for k > 3 (actually, de f ( G k ) -t oc for k + m).

>

Proof. Let G k be a semidirect product of a normal elementary abelian group P of order 7 k and a cyclic group < h > of order 3 with the corresponding action of < h > given by hxh-' = x2 for all x E P. Then, by Theorem 10.8.7, M ( G k ) = 1 for all k 2 1. By Lemma 11.3, we have de f ( P ) = ( 1 / 2 ) k ( k - 1 )

(1)

On the other hand, by Lemma 11.4, de f ( G k )

>

(Gk : P ) - ' ( d e f ( P ) 4- 1 ) - 1 = ( 1 / 3 ) ( d ef ( P ) 1 ) - 1

Applying ( 1 ) and ( 2 ) , it follows that for all k

as required.

+

(2)

> 3,

.

D. Another example of zero deficiency. The importance of finite groups with zero deficiency stems from the fact that they have trivial Schur multiplier. Furthermore, if a group is a factor group of a finite group with zero deficiency, then one can easily compute its

11 Deficiency, efficiency and Schur multipliers

495

Schur mu1tiplier. Indeed, if N is a normal subgroup of a finite group G with def (G) = 0, then M ( G / N ) 2 (G' n N)/[G, N ] by virtue of Proposition 10.1.5(ii). To the best of our knowledge, the problem of determining all groups which are factor groups of finite groups with zero deficiency has not been approached so far, Nevertheless, there is a large literature devoted t o the study of finite groups with zero deficiency. A number of such groups were exhibited in Sec.3. Our aim here is to present another family of finite groups with zero deficiency.

>

Theorem 11.6. (Campbell, Robertson and Thomas (1989)). Let m, n 1 be coprime integers a n d let k = 3" 1 if m is odd and k = 3" - 1 if m is even. Then the group G given by the presentation

+

is of order 4nmk. The group G has a normal subgroup N =< a , bn > of index n and N has a normal subgroup M of index 2, where M is a metacyclic group of order 2mk with a presentation

Proof. Put c = bn and d = bm. Our first aim is to show that G has a presentation

G =< a , c, dla2 = 1, dad-' = ac-'ac"a,

[c, dl = 1, cm = dn

>

(1)

By introducing new generators c and d, we first obtain

G =< a,b,c,dla2 = 1, dad-' = ac-lac-'a,

[c,d] = 1, c = bn, d = bm

Since ( m , n ) = 1, we may choose integers u, v such that um

>

+ vn = 1. Then

and we have G =< a , b, c, dla2 = 1, dad-' = ac-lac-'a,

[c,d] = 1, c = bn, d = bm, b = d"cv >

Hence we may delete the generator b using the last relation and so

G =< a,c,dla2 = 1, dad" = ac-lac-'a,

[c,d] = 1, c = ( d " ~ " ) ~d ,= ( d " ~ ' ) ~>

Schur's Formula and Applications

496

Now the relation

is equivalent to

(dnc-m)u = 1 since [c,d] = 1. On the other hand, the relation

d = (ducv)m = dumcvm = dl-vn

C urn

is equivalent to

(d-ncrn)v = 1 and hence, after inversion and conjugation, t o

Since (u, v) = 1, the relations

(dnc-m)u = 1 and

(dnc-m)u = 1

are equivalent to dnc-m = 1, and hence to cm = dn. Thus G has a presentation given by ( 1 ) . Setting N =< a,c >, it follows from ( 1 ) that N is a normal subgroup of G of index n. Our next task is to prove that N has the following presentation :

If we choose the coset representatives 1, d , d 2 , . . . ,dn-I for N , we obtain the following generators for N :

a , a1 = dad-', a2 = d2ad-2,. . . ,an-l = dn-lad-(n-l) 2 = p - l c d - ( n - l ) , t = dn C , c1 = dcd-l, c2 = d2cd- ,. . . ,c,-1 and the following relations for N : a2

= al2 = a 22 =

... = an-*

= 1,

a1 = ac-lac-'a, a2 = a l c ; l a l c ~ l a l , . . . , -1 -1 an-l = an-2Cn-2an-2Cn-2an-2, tat-'

= an-lCn-lan-lCn-lan-l -1 -1

11 Deficiency, efficiency and Schur multipliers

497

By eliminating the generators c l , c 2 , .. . ,cn-1 and t , we obtain the following presentation for N :

Now the relation a; = 1 is equivalent t o (ac-lac-' a ) - 1 , i.e.

which shows that a: = 1 is equivalent to

ac2a = c-2

We may rewrite the relations a1 = (ac-1 )2 a , a2 = ( a l e- 1 )2 a l , . . . ,a,-1 = ( a , - 2 ~ - ~ ) ~ ~ , - 2

wher

T

= 3n-1. The relations

are then redundant via ac2a = c - ~ .Thus we may successively delete the generators a l , a 2 , . . . ,an-1 to obtain

If m is even, then the second relation is equivalent t o ~~~a = ( a ~ ) ~ ' + ~ a which in turn is equivalent to c~~ = ( a ~ ) ~ ' + ~

If m is odd, then the second relation is equivalent t o CmaC-m+l -( a ~ ) ~ " ~

Schur's Formula and Applications

+

+

Since k = 3r 2 if m is even, and k = 3r 4 if m is odd, the desired presentation (2) for N follows from (3), (4) and (5). P u t e = aca. Then a-lea = c, a-'ca = e and so M =< e,c > is a normal subgroup of index 2 in N =< a , c > with presentation :

Given the first and third relations, we have

which shows that the second relation is redundant. Thus

Finally, let us introduce f = ec and delete e = fc-'.

Then

M =< c , flc2m = f k l 2 , c2 = c f - l c f - '

Because c-' f c = f

>

-' , we have

so that f k i 2 = ff-lc/' and hence f k = 1. Thus we have

This is a presentation for the metacyclic group of order 2 m k . Hence JGJ= 4 m n k and the result follows. 1

Corollary 11.7. Let m 2 1 be a n odd integer and let the group G of order 16m have presentation

11 Deficiency,

efficiency and Schur multjpliers

Then G has zero deficiency (and hence M ( G ) = 1).

Proof. Let G be the group given by Theorem 11.6, in which n = 1 and 2 1 is an odd integer (hence k = 4). Then, in the notation of the proof Theorem 11.6, G = N =< a, c > and M =< c, f > is a normal subgroup G of index 2. More precisely, a2 = 1 and

Since M has presentation

it follows (by setting f = x, c = y, a = 2) that G has presentation given in the statement of the corollary. W

We close by quoting the following three examples of finite groups with zero deficiency :

< x,y~xyx-lyxy-l = 1, xy 2 x4 (y2 x 2 y2 )2 x4 y2 -- 1 > G2 = < x, y~xyx-'yxy-l = 1, x4 (y3 x3 y2 x3 y2 )2 y = 1 > = 1, (xy)4x 3 y4 x 2 = 1 > G3 = < x,Y~xYx-lYxY-l G1 =

T h e group GI is of order 43008, G2 is of order 115248 and G3 is of order 15000 (for details, refer t o Campbell and Robertson (1986)).

E. Efficiency of direct powers of groups. In what follows, given a finite G and an integer n 2 1, we write

Gn = G x

.. . x G

( n times)

The problem of finding def ( G n ) , even in the case def(G) = 0 and n = 2, is extremely complicated (e.g see Wiegold (1982,p.149)). Here we are concerned with a related problem. Namely, our aim is to provide a class of groups G such that Gn is efficient for all n 2 1. As a candidate for G we choose the dihedral group Dam of order 2m, m 2 1. It turns out, as we shall see below, that D L is efficient for all n 2 1. Some further information on the efficiency of direct powers of groups can be found in the paper of Campbell, Robertson and Williams (1990).

Schur's Formula and Applications

500

Lemma 11.8. Let G and H be arbitrary groups with presentations :

G = < XIR >

and H

=< Y J S >

where X, Y are disjoint sets. Then the direct product G x H has the presentation < X u Y ( R u S u [ X , Y ]> (1) where [X, Y] denotes the set of all commutators [x, y] with x E X , y E 1'.

Proof. Let K be the group with presentation (1) and let F ( X ) , F(Y) and F ( X U Y )be free groups freely generated by X , Y and XUY, respectively. The map x I+ x, x E X, extends to a homomorphism F ( X ) -t F ( X U Y). We denote by CY : G -t K the induced homomorphism. Similarly, the map y H y, y E Y , extends to a homomorphism F ( Y ) -t F ( X U Y ) and we denote by ,f3 : H -t K the induced homomorphism. The relators [ X , Y] guarantee that [a(g),P(h)] = 1 for all g E G , h E H Hence we have an induced homomorphism 7 : G x H

-t

li given by

Since X and Y are disjoint, there is a homomorphism F ( X U Y) -t F ( X ) induced by x I+ x, y H 1, x E X , y E Y. Then this homomorphism induces a homomorphism S : K -t G and, similarly, we obtain a homomorphism cp : K -t H. Since the homomorphism Ii i . G x H , k H (6(k), cp(k)) is the inverse of 7 , the result follows. I

Theorem 11.9. (Campbell, Robertson and Williams (1990)). Let D2, be the dihedral group of order 2m, m 1. Then, for any integer n 2 1,

>

Proof. For the sake of clarity, we divide the proof into a number of steps. In what follows, the presentation of the direct product G x H given by Lemma 11.8 will be calIed standard. Step 1. Here we show that the result will follow provided we demonstrate that, for m odd, the following properties hold : (a) 0 2 2 , has a presentation with 2 generators and 3 relators.

11 Deflciency, efficiency and Schur multipliers

(b) D:, has a presentation with 3 generators and 6 relators. Let d(G) denote the minimal number of generators of a finite group G. By Proposition 10.6.1,

and, by Example 3.2, 1 if mis odd z2 if mis even

It follows, by an inductive argument, that (1/2)n(n - 1) if m i s odd n(2n - 1) if m i s even Assume that m is even. To prove that D L is efficient, it suffices t o exhibit a presentation with 2n generators and 2n n(2n - 1) = n(2n 1) relators. We argue by induction on n. The case n = 1 being obvious, assume that we have a presentation for D ; ~with 2k generators and k(2k 1) relators. x D2, has 2k 2 generators and Then, the standard presentation for k(2k 1) relators from the first factor, 3 relators from the second factor and has a presentation with 2(k 1) generators 4k commutators. Thus D:$' and (k 1)(2(k 1) 1) relators, as required. From now on, we assume that m is odd. To prove that D L is efficient, it suffices t o exhibit a presentation with n generators and (1/2)n(n+ 1) relators for n 2. Suppose that such a presentation exists for D i m and D;,, i.e. assume that (a) and (b) hold. Then we consider D L for n 4 and argue by induction. If n = 2k, then D,", = D2k, x D;, and the standard presentation for x D$, has 2k generators together with (1/2)k(k 1) relators from the first factor, (1/2)k(k+ 1) relators from the second factor and k2 commutators. Hence the total number of relators is

+ +

+

+

DL

+

+

+

+ +

>

>

+

DL

as required. If n = 2k 1, then D;, = D;, x and the standard presentation has 2k 1 generators together with (1/2)k(k 1) relators for D$, x D*:; from the first factor, (1/2)(k 1)(k 2) relators from the second factor and k(k 1) commutators. Hence the total number of relators is

+

+

+

~5:'

+

+

+

Schur's Formula and Applications

502

as required.

Step 2.

We know that D2, has the presentation

Here we show that Dim has the presentation

Indeed, the standard presentation for Dim is

D:,

=< a, b,c,dla2

= bm = (ab)2 = c2 = dm = ( ~ d ) ~

= [ a , ~= ] [a,d] = [b,c] = [b,d] = 1 > Put x = ad, y = bc and note that since, by hypothesis m is odd, a = x m , d = ~ ~ + ~ , mc =and y

b=yrn+l

Hence we have

I

(xm+' ym)2= [xm,ym] = [xm+l, y m + l = 1 > yvhich immediately simplifies to the presentation (2).

Step 3. We next provide an alternative presentation for 0 22, by showing that

All we have t o do is to show that (c) The relations x2m = y2m = = 1 in (2) imply [xm,ym] = 1. (d) The relation (xy)' = 1 holds in (2). ~ 1 implies xmyxm = y-l. Hence, using To prove (c), note that ( ~ " y ) = x2m = 1, it follows that for any integer t ,

11

Deficiency, efficiency and Schur multipliers

503

Taking t = m and using x2" = y2" = 1, we see that (c) holds. ) ~1 and y2" = 1 imply To prove (d), note that the relations ( X Y ~ =

and so raising t o the power m

+ 1, we obtain

Hence, since [xm+17ym+l] = 1, we have

A

x-m-l

x-m-l

-1

-m-1

Y Y - ~ x - ~ Y - ~ x ~

Y

X

(since ( ~ " y = ) ~1 and

xm = x-")

which proves (d). Thus Dim has the presentation (3). Step 4. Our next aim is to show that the relations x2m = 1, y2" = 1, = 1 and ( ~ y=) 1~imply that

= 1, we see that ( 4 ) holds. To prove (6), note Since x2m = y2" = = 1 implies ~ - m - l y - l x - l y - l x m = 1. Hence, applying (4) with that t = 1, we have x - m - l y - l x - l x m y = 1. Therefore

By squaring (8) and using x2" = 1, it follows that y - ' ~ - ~ y= x2 which implies ( 6 ) . To prove ( 7 ) , we apply ( 4 ) with t = m - 1 to obtain

Since m is odd, m

- 1 is even

and so, using (6) the equality ( 9 ) reduces to

Schur's Formula and Applications

504

It follows that (xym)2 = xy m-lX-1 Y m-1

(since ( x Y ) = ~ 1)

proving (7). Applying (3), (6) and (7), we have now proved that 2 D2m =< x, y1x2m= Y2m = ( x Y ) = ~ ( X ~ Y= )1~>

since rn

+ 1 is even and [xm",

ym+l]

(11)

= 1 is a consequence of [x2,y2]= 1.

Step 5 . Here we demonstrate that

which will show that 022, has a presentation with 2 generators and 3 relators (thereby proving property (a) in Step 1). Denote temporarily by G the group with presentation (12). Then, by ( l l ) , it suffices to prove that Y2m = 1 in G. To this end, note that G =< y,xy > so y2m E Z(G). Also y2 = 1 in G/G1 and therefore Y2m E Z(G) n GI. But, by (ll),

so y2" E ~ ( ~ 2 2 , and ) thus y4m = 1. On the other hand, y2m = ( x Y ) implies ~ (using ( X ~ Y= )1 ~) that

Hence, since

y2m

E Z(G), raising (13) t o the power of m gives

But rn - 1 is even and x2m = 1, SO yields y2m = 1, as desired.

2

Y2m

= 1 which, taken with

Step 6. As a preliminary step, we now show that D;, following presentation :

y4m

= 1,

( m o d d ) has the

Dim =< x, Z,alx 2rn = 1, (xm-l zm )2 - 1, z~~ = ( X - ~ Z ~ ) ~ ~ , a2 = 1, p ( m - 1 ) = 1( a 2m-1 12 = I , [x, a] = [x, i m - l ] = [a,zm] = 1 > (14)

11

Deficiency, efficiency and Schur multipliers

505

Indeed, the standard presentation for 022, x DZm using presentation ( 1 2 ) for 022, is

Hence, setting z = xyb-l and eliminating b = z m - l , y = x - l z m , we obtain the desired presentation (14) for 023,.

Step 7.

Here we show that in ( 1 4 ) , we have

zZm

(15)

is centrd

Because [ a , z m ] = 1, we have [ a , ~ =~ 1. ~ ] Observe also t h a t z2" = ( X - ~ Z ~ SO ) ~ [x-'zm, ~ , ( X - ~ Z ~ )=~ 1~ implies ] [ X - ~ Z ~ , X= ~ ~1.] Thus [ x , z Z m ]= 1, proving (15). To prove ( 1 6 ) , we apply ( X ~ - ' Z ~=) 1~ to deduce (using [ x ,zm-'1 = 1 ) that z x m - l -1 - -2m -m+l Z -z x (17)

By raising ( 1 7 ) t o the power m and using the facts that m - 1 is even and x Z m = 1, we obtain z2m2 = 1. Hence, applying zm(m-l) = 1, we obtain z2m = 1, proving (16). From the above, we have the following presentation for D;, ( modd) :

Step 8 . Our aim here is t o demonstrate that D;, (m odd) has the following presentation :

It suffices t o show that in (18), we have

(xm-1zm)2 = 1 is equivalent t o z-'xm-'

2

= Xm+l

(20)

Schur's Formula and Applications

= 1 is equivalent to

( a ~=) 1~

[ x ,zm-'1 = 1 is equivalent to x-l *

= zm+l

(21) (22)

Property ( 2 0 ) follows from [x,zm-'1 = 1 and z2m = 1, while ( 2 1 ) is a consequence of [a,zm] = 1 and a2 = 1. Finally, ( 2 2 ) follows by using z2m = 1. Step 9. In this final step, we demonstrate that for m odd, Dim

= 1, ( X Z m )2m -1 m+l =x , x -1 t 1-m x = zm+l

=< a , x , zla2 = I , [ x ,a] = 1,

> (23)

which will show that 02, has a presentation with 3 generators and 6 relators (thereby proving (b) in Step 1 and thus completing the proof). To prove ( 2 3 ) ,it suffices to show that the final three relations in presentation ( 1 9 ) are redundant. Raising aza = z-I to the power of m, we see that [a,zm] = 1 is a consequence of ( ~ z=)1,~a2 = 1 and z~~ = 1. It will next be shown that in ( 2 3 ) we have

which will complete the proof. To prove ( 2 4 ) , we apply z-'xm-'z

= x m f l to obtain

Now use ( a ~=) 1~t o write (28) as

which gives

x m+lz

xm-l

Taking the product of ( 2 9 ) with Z - ' X - ~ + ' = x-m-l gives ( 2 4 ) . - xm-l . Hence Because m+ 1 is even, applying ( 2 4 ) t o ( 2 9 ) gives x - ~ - ' -

11

Deficiency, efficiency and Schur multipliers

x2" = 1, proving (25). Applying X - ' Z ' - ~ X = zm+' , we have

Now apply (24) t o x-2z'-mx2 (using the fact that 1 - m is even) t o obtain

Hence, taking the product of (30) with z ' + ~= X - ' Z - ~ + ' X , gives (26). Finally, apply (26) to (30) (noting that m 1 is even) to obtain (27). This completes the proof of the theorem.

+

As an application of Theorem 11.9, we finally prove Corollary 11 .lo. (Campbell, Robertson and Williams (1990)). Let m be an odd positive integer. Then Dim has a covering group G which is of zero deficiency (and hence, by Lemma 11.2, G is efficient).

Proof. Consider the group G with presentation

We show that G is a finite group and that G is a covering group of Dgm, which will complete the proof. First, we note that, by Step 5 in the proof of Theorem 11.9, D i m has the presentation :

Now let us take a closer look at the presentation of G. Since x2m = ( x ~ ~ ) ~ , we see that x2" E Z(G). Also x2 = 1 in GIG', so x2" E G'. Thus x2m E Z(G) n G'

and

GI

< x2m > Z D2,2

Setting N =< x2m >, it follows from Lemma 1.3.3(ii) that N is finite. Since G I N 2 D i m , we see that G is finite. Hence M ( G ) = 1, since G is of zero deficiency (Lemma 11.2). But then, by Proposition 10.1.5(ii),

which implies that G is a covering group of D;,

(Theorem 4.2.7).

.

Schur's Formula and Applications

508

Covering groups of direct products

12

Given any two finite groups A and B , our aim is t o construct a covering group of their direct product A x B. It will be shown that if A* and B* are covering groups of A and B , respectively, then the second nilpotent product A* o B* of A* and B* (defined below) is a covering group of A x B. Assume that A and B are (not necessarily finite) groups. As usual, we write A * B for the free product of A and B introduced in Sec.9. If A, B are subgroups of a group G, then [A,B] denotes the subgroup of G generated by all commutators [a,b]= aba-lb-' with a E A, b E B . The second nilpotent product A B of arbitrary groups A and B is defined by Ao B = ( A * B ) / [ [ A , B ] , A B] * In what follows, we put

D = [ [ AB, ] , A* B ]

T h e fact that D is a normal subgroup of A

* B is a consequence of

and Lemma 12.1 below. It is clear that D is the smallest normal subgroup of A * B such that [A,B ] / D is central in ( A * B ) / D . Lemma 12.1. Assume that G is a group such that G

=< A, B > for

some subgroups A, B of G. Then [A,B I d G Proof.

Given al, a2 E A and b l , b2 E B , we have

a2[a1,blla,'

= [a2a1,b1][a2, b ~ ' 1E [A,Bl

and

b2[al,bl]bz1= [al,bz]-'[al,b2bl]E [ A , B ] Hence

[ A ,B ] a < A, B

>= G

as asserted. Lemma 12.2. Let A and B be arbitrary groups. Then the subgroup [ A ,B ] of A * B is a free group freely generated by the set

12 Covering groups of direct products

Moreover, An[A,B]= Bn[A,B]= 1

Proof. For brevity, let us refer to the elements of the form [a,b] or [b,a] = [a,b]-l with 1 # a E A, 1 # b E B, as elementary commutators. Then every element x of [ A ,B] is a product of some n 2 0 elementary commutators. We wish t o demonstrate, by induction on n, that the last two letters remain unchanged in the unique reduced form of x given by Proposition 9.1. This will clearly yield the result. The case n = 1 being trivial, suppose that w, is a product of n elementary commutators, where each elementary commutator is not followed by its inverse. Then we may write w, in one of the forms :

Assume that w, = -..[a,, b,]. Then, by induction, w, reduces t o w, = ...b*a;lb,l

If

=

for some b* E A U B, b* # a,

bn+l], then no cancellation is possible. ~ , [ b , + ~ , a , + ~with ] b,+l # b,, then w,+l reduces t o Wn+l

W,[U,+~,

If w,+l

=

proving that the last two letters remain untouched. Finally, if wn+l = wn[bn,a,+l], then an+l # a, by our specification. Therefore w,+l reduces to . .b*(a,lan+l)b,la~~l and again the last two letters remain unchanged. This establishes the case where w, = . . [a,, b,]. The case where w, = . . [b,, a,] is proved by symmetry. W Lemma 12.3. Let f : A * B -+ A x B be the homomorphism induced by the natural projections A * B -. A and A +. B + B . Then I i e r f = [A, B] and, in particular, (A * B)/[A, B ] r A x B

Proof. By the definition of f , we have f (ab) = (a, b) for d l a E A, b E B . Therefore,

f ([a, b]) = f (ab)f (a-lb-') = ( 1 , l )

Schur's Formula and Applications

510

which shows that [A, B] K e r f . By Lemma 12.1, [A, B] a A a B . Let Q

:A

--, (A

* B)/[A, B]

and

P :B

+

(A * B)/[A, B]

be the compositions of the natural injections A + A * B and B i A* B, respectively, with the natural homomorphism A * B + (A * B)/[A, B]. Because for all a E A, b E B , a(a)P(b) = P(b)a(a), the map A x B (a,b)

5

(A*B)/[A,B] (ab) [A,Bl

is a homomorphism. Since y is the inverse to the homomorphism

induced by f , the result follows. The following result is contained in the works of MacHenry (1960) and Wiegold (1959).

Theorem 12.4. Let A, B be arbitrary groups, let G = A o B and let be the images of A, B respectively, under the natural homomorphism A * B 4 A o B. Then the following properties hold : (i) t? S A, B Z B and [A,3 1 is central in G. (ii) Every element of G can be uniquely written in the form zyz with x E 2, y E B and z E [A, B]. (z'z'z') [A,B] S A @ B.

2, B

Proof. (i) The second statement in Lemma 12.2 ensures that and B Z B . Because [A, B] = [A, B ] / D

A

Z A

we see that [A, B] is central in G, by the definition of D. (ii) We keep the notation of Lemma 12.3. For each u E A * B, let .li be the image of u under the natural homomorphism A * B + A o B . If f (u) = ( a , b) for some a E A, b E B , then f (u) = f ( ab) and so u = abc for some c E [A, B] = I i e r f . Hence ii = xyz with x = Si E A, y = b E B and z = E E [A,] = [A,B]. To show uniqueness, suppose that xlylzl = 229222 with xi E A, yi E B

12 Covering groups of direct products

511

and z; E [A,B ] ,i = 1,2. We may write z; = iii, y; = b; and z; = ci for some a; E A, b; E B and c; E [A,B ] ,i = 1,2. Then we have

and therefore f ( a l b l c l ) = f (a2b2c2). Thus ( a l , b l ) = (a2, b2) and so X I = 2 2 , y1 = y2 and zl = 2 2 , as desired. (iii) Let us consider the map

{ A ~ B 5 [A,B] ( a , b ) ++ [a,&] Given a , a l , a2 E A, b, bl, b2 E B , we have

$(Z, & I & ) = [a,blb2D] = 1% bil [a,b21 [ [ a ,b2]-I ,611 D = [a,b1][a,ba]D = $(a, &)$(a, 6 2 ) and

+(al a2,6)

=

[ala2,bl D

=

[a2,b][[az,b]-',al][al,bID

=

[ a l , b ] [ a 2 , b ] D = [ ~ l , b [Z:,,b] ] (since [A,B] is central in G )

=

+(al, b)$(a2,b)

The map q!~factors over A x B -t ( A / A ' )x (BIB') and the universal property of tensor products gives a surjective homomorphism

We are thus left t o exhibit a homomorphism

which is a left inverse to A. To this end, we first apply Lemma 12.2 to define a homomorphism

Schur's Formula and Applications

512

by

p([a, b]) = a[A, A] 8 b

[ ~B],

(this formula extends to the cases a = 1 and b = 1). Hence it remains to prove that D l i e r p or, equivalently, that p([a, b],c]) = 1 for all a E A, bE B a n d c ~ A * B . Fix a , x E A and b, y E B and note that

and y[a,bIy-l = [a, yl-'[a, ybl Hence we must have p(x [a, b] x-')

= ( 5 a [A, A] 8 b [B,B])(? = "[A,A] @ b [ B , B ]

[A, A] 8 b [B, 31)-I

= cp([a,bl)

and similarly v(Y[., bly-') = p([a, bl) We conclude therefore that p(auz-') = p(u) for all u E [A, B] and all E A * B. Hence, for all a E A, b E B and c E A * B ,

z

as desired. W We have now come to the demonstration for which this section has been developed.

Theorem 12.5. (Wiegold (1971a)). Let A, B be any finite groups and let A*, B* be covering groups of A, B, respectively. Then the second nilpotent product A* o B* is a covering group of A x B. Proof. We shall refer to the form exhibited in Theorem 12.4(ii) as normal. Since A* is a covering group of A there is a subgroup U of [A*, A*]n Z(A8) such that U E M(A) and A*/U 2 A. Similarly, there exists a subgroup V of [B*,B*] n Z(B*) such that V E M ( B ) and B*/V 2 B.

12 Covering groups of direct products

513

By Theorem 12.4(i), we may identify A* and B* with their images in A* o B*. With this identification, we put L =< U, V, [A*,B*] > Owing t o Theorem 12.4(i), [A*,B*] is central in A* o B*. Moreover, U centralizes A* by assumption, and it centralizes B* because U C [A*, A*] and [A*, B*] is central. Hence U is central, and similarly V is central, so that L is central. We deduce therefore that L

c Z(A* o B*) n [A*

o

B*, A* o B*]

which is part of what we need. It will next be shown that

(A* o B*)/L

%

AxB

(1)

and that L 2 M(A x B), which will complete the proof. To prove (I), note that every element of A* a B* has normal form xyz for x E A*, y E B* and z E [A*, B*]. Hence A*nL=U

and

B*nL=V

It therefore follows that

and similarly B8L/L % B. It is clear that (A* o B*)/L is generated by A*L/L and B8L/L and that these groups centralize each other. Another consequence of the normal form mentioned above is that

and this implies (1). By the uniqueness of normal form, any element of L is represented uniquely in the form uvz with u E U, v E V and z E [A*,B*]. Observe also that A*/[A*, A*] E A/[A, A]

Schur's Formula and Applications

514

and

B*/[B*,B*]Z B / [ B ,B] Applying Theorem 12.4(iii)and Proposition 10.6.1, it follows that

L 2 UxVx[A*,B*] % M ( A ) x M ( B ) x (A* @ B*) = M ( A ) x M ( B ) x ( A * / [ A * , A * ]B@* / [ B * , B * ] )

M ( A ) x M ( B ) x ( A @B ) r M(A x B), S

proving (2). This concludes the proof of the theorem. H