Finite Group Theory (GSM92)

TV. T h e n ip is a crossed h o m o m o r p h i s m i f
If the action of G o n TV is t r i v i a l , w h i c h means t h a t n = n for a l l n G N a n d g G G, then a crossed h o m o m o r p h i s m from G to N is s i m p l y an o r d i n a r y h o m o m o r p h i s m . B u t we w i l l see that there are also interesting examples w h e n the action is n o n t r i v i a l . F o r example, let G be an a r b i t r a r y group a n d suppose TV < G, so that G acts on N by conjugation. N o w fix an element n G TV, and define the map

N by setting i p ( x ) = [ x , n ] . (Recall t h a t [ x , n] is the c o m m u t a t o r of x and n , w h i c h , by definition, is the element x ^ n ^ x n = { n ~ ) n . ) Since n G N < G, it is clear that
x

9

B y analogy w i t h kernels of o r d i n a r y homomorphisms, we define the kernel of a crossed h o m o m o r p h i s m ip : G - > N to be the subset ker^) = { i £ G | (f(x) = 1}. T h e following establishes a few basic facts about crossed h o m o m o r p h i s m s a n d their kernels. 3.6. L e m m a . L e t G a n d N be g r o u p s , a n d suppose t h a t G acts v i a a u t o m o r p h i s m s o n N . L e t ip : G - > N be a c r o s s e d h o m o m o r p h i s m w i t h k e r n e l K . T h e following then hold.

3B

77

(a)
of G.

G, t h e n
(d) \ G : K \

=
if a n d o n l y if K x

=

Ky.

= \
P r o o f . W e have
2

= < p ( i - i ) =
and it follows t h a t 1 = < p ( l ) . nonempty. If k € K , we have

= v(i) ,

T h i s establishes (a) a n d shows that i f is

1

1

i = ^,(1) = ^ ( a r ) = w i k f ' ^ i k - )

1

=

ifik' ), f f

where the last equality holds because

=
- l

y

l = 1,

so xy G K , a n d hence i f is a subgroup, p r o v i n g (b). Now

i f x , y G G and K x = K y , we can write y = kx w i t h k G K . T h e n p(y)

= if(kx)

x

=
=
where the last equality holds because
p(xy- )

= <^(y), we compute that y

= ^(x) ~^(y)

y

= ^(y) ~^(y)

1

= ^ ( y y " ) = p ( l ) = 1,

1

and so xy' G i T . T h u s x G i f y and i f x = i f y , as wanted. F i n a l l y , since

78

3. S p l i t

Extensions

T, a n d we use this bijection to define an element of TV that we call d ( S , T ) . (In some sense, d ( S , T ) measures the difference between S a n d T.) W e set d(S,T)

=

Y[s-H, s=t

where the p r o d u c t runs over a l l s E S a n d i E T such t h a t s = t. O f course, if s = t, t h e n s N = t N , a n d so s ~ H £ TV, a n d hence each factor i n the p r o d u c t lies i n TV. A l s o , because we are assuming t h a t TV is abelian, the order i n w h i c h the factors appear is irrelevant, a n d thus d(S, T ) is a well defined element of TV. Some elementary properties of the function d are given i n the following. 3.7. L e m m a . L e t TV be a b e l i a n a n d n o r m a l i n a finite g r o u p G, a n d l e t S, T a n d U be t r a n s v e r s a l s f o r TV. Then u s i n g the n o t a t i o n defined above, t h e following hold. (a) d ( S , T ) d ( T , U ) (b) d { S g , T g )

= d(S,U).

= d { S , T ) 9 for all g E G .

(c) d { S , Sn) = n l

G : 7 V

l f o r a l l n E TV.

P r o o f . If s = t a n d t = u for s E S, t E T a n d u E U, t h e n TVs — N t = N u , and so s = u. Since ( s " i ) ( r u ) = s~ u a n d the order of factors i n the abelian group TV is irrelevant, (a) follows. 1

1

l

Now let g E G. If s = t, t h e n TVs = N t , a n d so T V s = N t g , a n d sg = t g . factor { s g ) ~ { t g ) of the p r o d u c t defining d { S g , T g ) is equal to ( s ^ i ) , assertion (b) follows. 5

The and

l

9

F i n a l l y , i f n E TV, we have sTV = (sn)TV, a n d so s = sn. E a c h factor - { s n ) of the p r o d u c t defining d { S , Sn) equals n , and so (c) follows because | 5 | = \ G : TV|. • l

S

We are now ready to prove the case of the Schur-Zassenhaus theorem where TV is a n abelian n o r m a l H a l l subgroup of G. P r o o f of T h e o r e m 3.5. F i x an a r b i t r a r y transversal T for TV i n G, a n d define 9 : G -> TV by setting 9 { g ) = d { T , T g ) . W e argue first that 9 is a crossed h o m o m o r p h i s m w i t h respect to the conjugation a c t i o n of G o n TV. If x,y E G, t h e n using L e m m a 3.7(a,b), we compute t h a t d(T,

Txy)

= d(T, T y ) d { T y , T x y ) = d { T , T y ) d ( T , T x )

and so 9 { x y ) = B ( y ) 9 ( x ) v . T h i s , of course, is equal to 9{x)«9{y) abelian, a n d so 9 is a crossed h o m o m o r p h i s m , as claimed.

y

, since TV is

G : 7 V

For n E TV, we have 9 ( n ) = d(T, T n ) = J I b y L e m m a 3.7(c). Since we are assuming t h a t |TV| a n d \ G : TV| are coprime, however, the m a p x i-> \ \ is a p e r m u t a t i o n of the elements of TV. (One way to see this is to observe G : N

X

3B

79

k

that this m a p has an inverse: the m a p x i-> x , where the integer k is chosen so t h a t k \ G : 7V| == 1 m o d \ N \ . ) It follows t h a t 9 maps N onto N , and in particular, 9 ( G ) = N . N o w let H = ker(0), so t h a t i f is a subgroup by L e m m a 3.6(b). A l s o , \ N \ = \ 9 { G ) \ = \ G : H \ by L e m m a 3.6(d), and hence \ N \ \ H \ = \ G \ and \ H \ = \ G : N \ . W e have already observed that since \ N \ and \ G : N \ are coprime, this is sufficient to guarantee that i f is a complement for N , as wanted. Now suppose that K is an a r b i t r a r y complement for N i n G. Since N K = G, we see that every coset of N i n G contains a n element of K , and in fact, each coset of N contains a unique element of K because N C \ K = \. T h u s K is a transversal for N , and we let m = d{K, T ) € N . W e saw that the restriction of 9 to N is the m a p x xl l , w h i c h is surjective, a n d so we can find n € N such that 9 ( n ) = m, a n d thus 9 ( n ~ ) = TO" . G : A f

1

1

To complete the proof, we show t h a t K — H . Since \K\ = \ G : N \ = \ H \ , i t suffices to show that K C i f , or equivalently, t h a t 9 ( k ) = 1 for a l l keK. (Recall that H = ker(0).) F i r s t , observe t h a t i f G AT, then n

n

m and

fc

= d(K,T) k

k

= d(Kk,Tk) 1

n

= d(K,Tk) k

1

so 1 = { m ) - m 9 { k ) = ( m ) - 9 { k ) m .

= d(K, T)d(T, Tk) = W e have

9{k ) = 9{n- kn) = 9(n- k) 9(n) n

1

1

m9(k),

n

= 0(n fc)m _ 1

=

(m ) 0(fc)m, _ 1

f c

where we were able to drop the exponent n i n the t h i r d equality because N is abelian. T h u s 9 { k ) = 1, and we are done. • n

N e x t , we prove the existence part of the Schur-Zassenhaus theorem w i t h out assuming t h a t the n o r m a l H a l l subgroup is abelian. 3.8. T h e o r e m . L e t N < G, w h e r e G i s a finite a r e c o p r i m e . T h e n N i s c o m p l e m e n t e d i n G.

group a n d \N\ and \ G: N \

P r o o f . W e proceed by i n d u c t i o n on \ G \ . Suppose t h a t there exists a subgroup K < G such that N K = G. T h e n \ K : K C \ N \ = \ G : N \ is coprime to | N | , and hence it is also coprime to \ K n N \ . Since K n N < K , the inductive hypothesis guarantees that K n N is complemented i n AT, and we choose a complement H . T h e n | i f | = \ K : K n JVj = |G : 7V| and as we have seen, this guarantees that i f is a complement for N i n G. We m a y thus suppose t h a t there is no proper subgroup K of G such t h a t N K = G, and so i n particular, N must be contained i n every m a x i m a l subgroup of G. I n other words, N C $ ( G ) , the F r a t t i n i subgroup, and so

3. S p l i t E x t e n s i o n s

80

N

is nilpotent. A l s o , we c a n assume that N > 1 or else G is the desired

complement for N i n G. _ N o w let Z = Z ( / V ) , _ a n d ^ b s e r v e t h a t 1 < Z < G. W o r k i n g i n the group G = G / Z , we_see t h a t | G : N \ = \ G : 7V| is coprime to |TV| a n d hence is also coprime t o \ N \ . It follows by the inductive hypothesis t h a t G splits over N , a n d we let K be a complement. T h e n G = N K = N K , and hence G = N K . It follows by the e a r l i e r p a r t of the proof t h a t K cannot be proper, and we have K = G. Because K is a complement for N , however, their intersection must be the t r i v i a l subgroup Z . T h u s Z = N n K

= N n G = N ,

and hence TV = Z is abelian. T h e result now follows v i a T h e o r e m 3.5.

•

A s we said, it is necessary to assume that either N or G / N is solvable in order to to prove the conjugacy part of the Schur-Zassenhaus theorem. Perhaps, therefore, a review of some basic facts about solvable groups w o u l d be appropriate, and so we begin w i t h the definition. A (not necessarily finite) group G is solvable if there exist n o r m a l subgroups N such that t

1 = N

0

C N iC N

2

C ••• C N

r

=

G,

where N / N ^ is abelian for 1 < i < r . (In particular, a l l abelian groups are solvable since i f G is abelian, we can take A o = 1 and TVi = G . ) A finite chain of nested n o r m a l subgroups of G , extending from the t r i v i a l subgroup to the whole group, is called a n o r m a l series for G , and thus G is solvable precisely w h e n it has a n o r m a l series w i t h abelian factors. R e c a l l that the c o m m u t a t o r s u b g r o u p (or derived subgroup) of a n a r b i t r a r y group G is the subgroup G' generated by a l l commutators [ x , y ] = x ~ y ~ x y , w i t h x,y G G . Since [ x , y ] = 1 i f and o n l y i f x and y commute, we see that G is abelian i f and o n l y if G' = 1. A l s o , if tp : G -> H is a surjective h o m o m o r p h i s m , then ip maps the set of commutators i n G onto the set of commutators i n H , and thus
x

The subgroup ( G ' ) ' is usually denoted G", and its derived subgroup, is G'". T h i s n o t a t i o n r a p i d l y gets unwieldy, however, and so we write G ^ for the n t h derived subgroup. I n other words, G ^ = G a n d for n > 0, we define G^ = (G^" ))'. T h e subgroups G ^ are characteristic, and they constitute the d e r i v e d series of G . 1

The

connection w i t h solvability is the following.

3B

81

3.9. L e m m a . A g r o u p G i s s o l v a b l e if a n d o n l y i f G ^ = 1 f o r some m > 0.

integer

P r o o f . If G is solvable, choose n o r m a l subgroups N i such t h a t 1 = N

C ••• C N

0

r

= G,

where N i / N i - i is abelian for 1 < i < r , a n d thus ( N i ) ' C N - i for subscripts i n this range. It is clear, however, t h a t i f i f C G, then i f ' C G ' , and thus since G ' = ( N ) ' C 7 V _ i , we have (if r > 2) t h a t G " C ( 7 V _ i ) ' C i V _ . C o n t i n u i n g like this, we get G ( ) C / V _ for m < r, a n d i n particular, G W C N o = 1. {

r

r

r

r

2

m

r

m

m

Conversely, suppose t h a t G ( ) = 1. Since the subgroups G « are characteristic i n G , they are certainly n o r m a l , a n d we have the n o r m a l series 1 = G

(

m

)

c G

(

m

_

1

)

c ••• c G

( 0 )

= G.

T h e factors here are abelian, a n d so G is solvable.

•

T h e characterization of solvability i n terms of the derived series allows us to prove a number of useful facts. 3.10.

Lemma.

L e t G be a n a r b i t r a r y

(a) I f H Q G ,

then

C G

(b) I f < p : G - > H i s a f o r a l l m > 0.

group.

^ f o ra l l m > 0 .

surjective h o m o m o r p h i s m ,

then < p ( G ^ )

=

(c) If G i s s o l v a b l e a n d H C G, t h e n H i s s o l v a b l e . (d) I f G i s s o l v a b l e a n d N < G, t h e n G / N i s s o l v a b l e . (e) L e t N < G a n d suppose solvable.

that N a n d G / N are solvable.

Then G is

P r o o f . If H C G , then H ' C G' a n d (a) follows b y repeating this observation. Similarly, (b) follows by repeated a p p l i c a t i o n of the fact that H is a surjective h o m o m o r p h i s m . N o w suppose t h a t G is solvable, so t h a t G^ = 1 for some integer n . If H C G , then C G ( ) = 1 by (a), and thus H is solvable, p r o v i n g (c). If

i f is a surjective h o m o m o r p h i s m , then i f ( ) = ( ^ ( G ^ ) = ^(1) = 1, and H is solvable. In particular, if N < G a n d we take

n

For (e), let homomorphism. n , a n d thus G ( ) for some integer is solvable. • n

H = G / N a n d again let y : G -> i f be the canonical Since i f is solvable, i p ( G ^ ) = i f (") = 1 for some integer C ker(y>) = TV. A l s o , since / V is solvable, we have 7V"( ) = 1 m, and thus G ( ) = ( G ^ ) ^ C / V ( ) = 1, a n d thus G m

m + n

m

82

3. S p l i t

Extensions

p

R e c a l l t h a t an abelian p - g m u p P is elementary abelian i f x = 1 for all x £ P . A useful fact about solvable groups is that a m i n i m a l n o r m a l subgroup of a finite solvable group must be an elementary abelian p-group for some p r i m e p . T h e following is a slightly more general version of this. 3.11. L e m m a . L e t M be a s o l v a b l e m i n i m a l n o r m a l s u b g r o u p of a n a r b i t r a r y g r o u p G. T h e n M i s a b e l i a n , a n d if M i s f i n i t e , i t i s a n e l e m e n t a r y a b e l i a n p - g r o u p f o r some p r i m e p . P r o o f . Since M is solvable, its derived series must eventually reach 1, and thus since M > 1, we conclude that M ' < M . B u t M ' is characteristic i n M , a n d hence it is n e r m a l i n G. It follows by the m i n i m a l i t y of M t h a t M ' = 1, a n d thus M is abelian, as wanted. N o w assume t h a t M is finite, a n d let u £ M be an element of p r i m e order, say p . Since M is abelian, the set S = { x £ M | x = 1} is a characteristic subgroup of M a n d hence it is n o r m a l i n G. A l s o , S > 1 since u £ S, and thus S = M by the m i n i m a l i t y of M . T h i s completes the proof. • p

F i n a l l y , we mention that the derived length of a solvable group G is the smallest integer n such t h a t G^ = 1. (Thus, for example, a solvable group has derived length 1 precisely when it is n o n t r i v i a l a n d abelian.) T h e proof of L e m m a 3.10 shows that subgroups and h o m o m o r p h i c images of a solvable group w i t h derived length n have derived lengths at most n . A l s o , if N < G, where N is solvable w i t h derived length m a n d G / N is solvable w i t h derived length n , then the derived length of G is at most m + n . We are now ready to complete our proof of the Schur-Zassenhaus theorem. 3.12. T h e o r e m . L e t N < G, w h e r e G i s finite a n d \ N \ a n d \ G : N \ a r e c o p r i m e , a n d assume either that N is solvable o r that G / N is solvable. Then all complements f o r N i nG are conjugate. P r o o f . W e proceed by i n d u c t i o n on \ G \ . L e t U C G be arbitrary. W e w i l l show t h a t U satisfies the hypotheses of the theorem w i t h respect to its n o r m a l subgroup U n N , a n d also that i f U contains a complement H for N i n G, then H is also a complement for U P i N i n U. It w i l l then follow v i a the inductive hypothesis that i f U is proper i n G a n d contains two complements for N , these complements are conjugate i n U a n d hence i n G, as wanted. N o w \ U n N \ divides \ N \ and \ U : U n N \ = \ U N : N \ divides \ G : N \ , a n d thus \ U C \ N \ a n d \ U : U n N \ are coprime. A l s o , either G / N is solvable, in w h i c h case U / { U n N ) = U N / N C G / N is solvable, or else N is solvable, a n d so U n N is solvable. If U contains a complement H for N i n G, then \ H \ divides |J7| a n d is coprime to \ U n N \ , and hence \ H \ divides \ U : U n N \ . B u t \ U : U n TV| divides \ G : N \ = \ H \ , a n d so we have \ H \ = \U : U n I f |,

3B

83

a n d therefore i f complements U n N i n U. T h i s establishes the assertions of the previous paragraph. Next^we_consider factor groups. Suppose L < G , a n d write G = G / L , so that N < G . Since N is a h o m o m o r p h i c image of N , it is solvable if N is, and its order divides \ N \ . A l s o G / N = G / N L is a h o m o m o r p h i c image of G/N, and_so it is solvable i f G / N is, a n d its order divides \ G : N \ . I n other words, G satisfies the hypotheses w i t h respect to the normal_subgroup N . If_H is a complement for N i n G, t h e n \ H \ _ \ s coprime to \ N \ , and so AT n i f = 1. A l s o , N H = ~NH = G, and thus H is a complement for N i n G. L e t i f and K be complements for N i n G . If L _ > 1, t h e n by the inductive hypothesis a p p l i e d i n G, we deduce t h a t i f a n d K are conjugate i n G , and thus H L and ATL are conjugate i n G. W e can thus w r i t e i f f , = ( K L ) for some element g e G, and hence H L contains b o t h H and K , each of w h i c h is a complement for N i n G. If i f L < G, therefore, it follows by the result of the first paragraph of the proof t h a t i f a n d K are conjugate, and thus i f and K are conjugate, as required. W e c a n thus assume that H L = G for every nonidentity n o r m a l subgroup L of G. 9

9

9

Suppose t h a t N is solvable, and note t h a t we c a n assume t h a t N > 1, or else i f = G = K and there is n o t h i n g to prove. L e t L C N be a m i n i m a l n o r m a l subgroup of G, so that L is abelian b y L e m m a 3.11. B y the result of the previous paragraph, i f L = G, and since L C N a n d \ N \ = \ G : i f |, it follows t h a t N = L , a n d so N is abelian a n d we are done by T h e o r e m 3.5. F i n a l l y , assume that G / N is solvable. W e can assume that N < G, or else i f = 1 = K a n d there is n o t h i n g to prove. L e t M / N be m i n i m a l n o r m a l i n G/N, a n d a p p l y L e m m a 3.11 to conclude t h a t M / N is a p-group for some prime p . A l s o , since p divides \ G : 7V|, we see t h a t p does not divide |7V|. Since N H = G and N C M , D e d e k i n d ' s l e m m a yields M = N ( M n i f ) , a n d it follows that M n i f complements N i n M . I n p a r t i c u l a r | M n i f | = | M : 7V|, a n d so M n i f is a p - g r o u p . A l s o , | M : M n i f | = \ N \ is not divisible by p , a n d we conclude that M n i f is a S y l o w p-subgroup of M . Similarly, M n K is a Sylow p-subgroup of M , and thus by the Sylow C - T h e o r e m , M n i f = ( M n A T ) = M n A T for some element m G M . m

m

m

m

N o w write L — M n i f = M C \ K , and note t h a t L < i f and L o i f since M < G. T h u s i f and A T are complements for N i n G t h a t are contained i n N ( L ) . If N ( L ) < G , we are done by the first part of the proof. W e can assume, therefore, that L < G , a n d since L > 1, we have L H = G. B u t L C i f , and so i f = G a n d thus also K = G and we are done i n this case too. • m

G

G

84

3. S p l i t

Extensions

Perhaps it is of interest to note that the conjugacy part of T h e o r e m 3.5 was used i n the proof of T h e o r e m 3.12 only i n the case where N is solvable. Conjugacy i n the case where G / N is solvable u l t i m a t e l y depends on the conjugacy of Sylow subgroups, but not on T h e o r e m 3.5. Problems 3B 3 B . 1 . Show t h a t a m a x i m a l subgroup of a finite solvable group must have prime power index. H i n t . L o o k at a m i n i m a l n o r m a l subgroup and work by i n d u c t i o n . 3 B . 2 . Suppose 1 = N < N < • • • < N = G, where the factors N i / N ^ are abelian for 1 < i < r . Show that G is solvable. (Note that we are not assuming t h a t the subgroups iVj are n o r m a l i n G.) 0

x

r

3 B . 3 . L e t G be finite a n d let 1 = 7V < N < • • • < N = G, where the factors ZVi/TVi-i are simple for 1 < i < r . Show that G is solvable i f a n d only i f these factors a l l have p r i m e order. 0

x

r

N o t e . R e c a l l t h a t a s u b n o r m a l series such as { N } , where the factors are simple, is a c o m p o s i t i o n series for G, a n d note that if G is finite, such a series necessarily exists. A l t h o u g h G m a y have several different composit i o n series, the J o r d a n - H o l d e r theorem asserts that the factors N i / N ^ are uniquely determined by G (except for the order i n w h i c h they occur). T h i s p r o b l e m says t h a t solvable finite groups are exactly the groups for w h i c h every c o m p o s i t i o n factor has p r i m e order. 3 B . 4 . L e t N < G , where \ N \ a n d \ G : N \ are coprime. L e t U C G, where \ U \ divides \ G : N \ , a n d assume that either N or U is solvable. Show that U is contained i n some complement H for N i n G. 3 B . 5 . Suppose t h a t P e S y l ( G ) and t h a t P C Z ( G ) . Show that the set X of elements of G w i t h order not divisible by p is a subgroup of G, and t h a t G = X xP . p

3 B . 6 . L e t N < G a n d g € G, a n d suppose that w h e n N g is viewed as an element of G/N, it has has order TO. (a) Show t h a t there exists h € N g such t h a t every prime divisor of o { h ) divides m. (b) IfTOand \ N \ are coprime, show that the element h of part (a) must have order TO. (c) N o w assume t h a t N g is conjugate to its inverse i n G / N and that TO a n d \ N \ are coprime. L e t h € N g be as i n (a). Show t h a t h is conjugate to its inverse i n G.

85

Problems 3 B

H i n t . F o r (a), find a cyclic rr-subgroup C such t h a t N C = N ( g ) , where rr is the set of p r i m e divisors of m. F o r (c), let x € G , where ( N g ) = (Ng)-\ Observe t h a t x normalizes N ( h ) . Consider the subgroup ( h ) . N x

x

3 B . 7 . A group G is supersolvable i f there exist n o r m a l subgroups N i w i t h 1 = 7V C TVi C • • • C N 0

r

= G,

a n d where each factor N i / N ^ is cyclic for 1 < i < r. Clearly, supersolvable groups are solvable, and it is routine to check t h a t subgroups a n d factor groups of supersolvable groups are supersolvable. Suppose now t h a t G is finite a n d supersolvable. (a) Show that the order of every m i n i m a l n o r m a l subgroup of G is prime. (b) Show that the index of every m a x i m a l subgroup of G is prime. H i n t . F o r (b), look at a m i n i m a l n o r m a l subgroup and w o r k by i n d u c t i o n . 3 B . 8 . L e t G be finite, and assume t h a t every m a x i m a l subgroup of G has p r i m e index. Show t h a t G is solvable. Show also t h a t G has a n o r m a l Sylow p-subgroup, where p is the largest p r i m e divisor of \ G \ , a n d that G has a n o r m a l g-complement, where q is the smallest p r i m e divisor of \ G \ . N o t e . I n fact, G must be supersolvable, b u t this theorem of B . H u p p e r t is m u c h harder to prove. R e c a l l t h a t a g-complement i n a group G is subgroup w i t h g-power index a n d h a v i n g order not divisible by q. In other words, it is a subgroup whose index is the order of a Sylow g-subgroup. 3 B . 9 . L e t G be finite a n d supersolvable. If n is a divisor of \ G \ , show t h a t G has a subgroup of order n . H i n t . L o o k at a m i n i m a l n o r m a l subgroup a n d work by i n d u c t i o n . 3 B . 1 0 . L e t G be finite and supersolvable, a n d suppose t h a t p is the largest p r i m e divisor of \ G \ . Show that G has a n o r m a l S y l o w p-subgroup. H i n t . L o o k at a m i n i m a l n o r m a l subgroup a n d work by i n d u c t i o n . 3 B . 1 1 . A s usual, let $ ( G ) be the F r a t t i n i subgroup of the finite group G. Show t h a t every p r i m e divisor of | $ ( G ) | also divides | G : $ ( G ) | . 3 B . 1 2 . L e t G be solvable, and assume t h a t $ ( G ) = 1. L e t M be a m a x i m a l subgroup of G , a n d suppose that H C M . Show t h a t G has a subgroup w i t h index equal to | M : H \ . 3 B . 1 3 . Show t h a t every finite group contains a unique largest solvable normal

subgroup.

86

3. S p l i t

Extensions

3 B . 1 4 . L e t F — F ( G ) , where G is finite, and let C = C ( F ) . Z ( F ) is the unique largest solvable n o r m a l subgroup of C.

Show that

G

H i n t . P r o b l e m I D . 19 is relevant. N o t e . If G is solvable, it follows that C = Z ( F ) .

T h i s shows that for

solvable groups, F ( G ) 5 C ( F ( G ) ) . G

3 B . 1 5 . (Berkovich) L e t G be solvable, and let i f < G be a proper subgroup h a v i n g the smallest possible index i n G . Show that H < G . 3C T h e r e exists a n extensive and well developed theory of finite solvable groups to w h i c h we give only superficial mention here. (Readers interested i n pursuing this might w i s h to consult the comprehensive b o o k on solvable groups by D o e r k a n d Hawkes.) U n d e r l y i n g this theory are some results of P h i l i p H a l l analogous to the Sylow E - , C - and D-theorems, but where sets of primes replace i n d i v i d u a l primes. H a l l ' s results are not especially difficult to prove directly, but they also follow easily from the Schur-Zassenhaus theorem, and that is w h y we present t h e m here. R e c a l l that i f vr is a set of primes, then a vr-group is a finite group such t h a t a l l primes d i v i d i n g its order lie i n TT, and a vr-subgroup of a group G is s i m p l y a subgroup t h a t happens to be a vr-group. A H a l l vr-subgroup of a finite group G is a rr-subgroup w i t h index involving no p r i m e of vr. For convenience i n discussing H a l l ' s theorems a n d related topics, we say t h a t an integer m is a vr-number if every prime divisor of m lies i n vr. A l s o , we w r i t e vr' to denote the complement of vr i n the set of a l l p r i m e numbers, so that a vr'-number is an integer w i t h no divisors i n vr. (In the case where TT consists of a single p r i m e p , we generally w r i t e p ' i n place of { p } ' . ) A subgroup i f of G , therefore, is a H a l l rr-subgroup of G precisely w h e n \ H \ is a vr-number and | G : i f | is a vr'-number, or equivalently, \ H \ is the largest vr-number d i v i d i n g | G | . O f course, if vr consists of just one prime, so that vr = { p } , then a H a l l vrsubgroup is just a Sylow p-subgroup, a n d we k n o w by the Sylow E-theorem that such a subgroup a c t u a l l y exists for every finite group. B u t if |vr| > 1, then an a r b i t r a r y finite group G can fail to have a H a l l vr-subgroup. T h e s i t u a t i o n is different, however, if G is solvable. 3.13. T h e o r e m ( H a l l - E ) . Suppose t h a t G i s a finite s o l v a b l e g r o u p , a n d l e t TT be a n a r b i t r a r y set of p r i m e s . T h e n G has a H a l l t r - s u b g r o u p . P r o o f . W e c a n certainly assume that G > 1, and we proceed by i n d u c t i o n on | G | . L e t M be a m i n i m a l n o r m a l subgroup of G a n d let J f be a H a l l vrsubgroup of G = G / M , where H D M . ( O f course, i f exists by the inductive

3C

87

hypothesis.) T h e n \ H : M \ = \ H \ is a Tr-number a n d | G : H \ = \ G : H \ is a vr'-number. B y T h e o r e m 3.11, we know that M is a p-group for some p r i m e p . If p G vr, then since \ H \ = \ M \ \ H : M | , it follows t h a t \ H \ is a vr-number, and thus H is a H a l l vr-subgroup of G, as wanted. If p & TT, on the other h a n d , t h e n | M | and \ H : M \ are coprime, and hence by the Schur-Zassenhaus theorem, M has a complement K i n H . T h e n \ K \ = \ H : M \ is a vr-number and | G : AT| = | G : H \ \ H : K \ = \ G : H \ \ M \ is a rr'-number. I n this case, K is the desired H a l l vr-subgroup of G . • W e used the Schur-Zassenhaus theorem to prove T h e o r e m 3.13, but n o t h i n g t h a t deep is really needed. In fact, the H a l l E - t h e o r e m can be proved using a fairly elementary argument. N e x t , we present ( w i t h a somewhat sketchy proof) the H a l l C-theorem, but we defer the H a l l D - t h e o r e m to the problems at the end of this section. 3.14. T h e o r e m ( H a l l - C ) . Suppose t h a t G i s a finite s o l v a b l e g r o u p , a n d l e t TT be a n a r b i t r a r y set of p r i m e s . Then a l l H a l l T x - s u b g r o u p s of G a r e conjugate. P r o o f . A s i n the previous proof, we proceed b y i n d u c t i o n on | G | . A s s u m i n g t h a t G > 1, choose a m i n i m a l n o r m a l subgroup M of G , w h i c h must be a p-group for some p r i m e p . L e t H a n d K be H a l l vr-subgroups of G , and observe that H a n d K _ a v e Ha]l_ vr-subgroups of G = G / M . B y the i n d u c t i v e hypothesis, therefore, H and K are conjugate i n G , and it follows that H M a n d K M are conjugate i n G . W e can thus w r i t e H M = { K M ) for some element g G G . 9

If p G vr, then \ H M \ and \ K M \ are vr-numbers. B u t \ H \ = \ K \ is the largest vr-number d i v i d i n g | G | , and it follows t h a t H = H M and K = K M , a n d so H = K , as wanted. If p vr, then H a n d K are complements i n H M for the n o r m a l H a l l subgroup M . T h e conjugacy part of the SchurZassenhaus theorem then yields the result. • 9

9

One might wonder how essential solvability is i n the H a l l E - t h e o r e m . C e r t a i n l y , some nonsolvable groups have H a l l vr-subgroups for some interesting sets vr of primes. ( B y "interesting" here, we mean cases where vr contains more t h a n one, but not a l l of the p r i m e divisors of the group order. In other cases, the existence of a H a l l vr-subgroup is either a t r i v i a l i t y , or is guaranteed by Sylow's theorem.) F o r example, the simple group A contains the H a l l {2, 3}-subgroup A , and the simple group of order 168 contains a H a l l {3, 7}-subgroup and a H a l l {2, 3}-subgroup. 5

4

W e say that G satisfies E i f it contains a H a l l vr-subgroup. W e k n o w t h a t solvable groups satisfy E for a l l sets vr of primes, a n d perhaps surprisingly, T

T

3. S p l i t

88

Extensions

the converse is true: if G satisfies E for a l l IT, then G must be solvable. In fact, G is solvable i f it satisfies E for a l l p r i m e sets of the form TT = p ' , where p is an a r b i t r a r y prime. (Recall that p ' is the complement of { p } i n the set of a l l primes.) A H a l l p'-subgroup of G is called a p - c o m p l e m e n t in G, and of course, i f \ G \ is not divisible by p , then G is a p-complement of itself. T h e c o n d i t i o n E > has no content i n this case, and so we can state the converse of the H a l l E - t h e o r e m as follows. v

T

p

3.15. T h e o r e m .

Suppose

that a

finite

g r o u p G has a p - c o m p l e m e n t f o r a l l

p r i m e d i v i s o r s p of \ G \ . T h e n G i s s o l v a b l e . W h a t does this converse of H a l l ' s E - t h e o r e m say i n the case where \ G \ involves just two (or fewer) primes? If \ G \ = p q , where p and q are prime a n d a and b are nonnegative integers, then a Sylow g-subgroup of G is a p-complement a n d a Sylow p-subgroup of G is a g-complement. I n this situation, Sylow's theorem guarantees that the hypotheses of T h e o r e m 3.15 are satisfied, and thus the theorem says that every group of order p q is solvable. T h i s is B u r n s i d e ' s famous pY-theorem, w h i c h we are certainly not prepared to prove at this point. (We present a proof i n C h a p t e r 7, however.) B u t i f we are w i l l i n g to assume B u r n s i d e ' s theorem, it is not h a r d to establish T h e o r e m 3.15 i n its full generality. a

b

a

b

We need a few p r e l i m i n a r y results. T h e first of these is fairly standard, and it also appears i n the gronp-theory review i n the appendix. 3.16. L e m m a . L e t H , K C G, w h e r e G i s a \G : H \ and \G : K \ are coprime.

Then G = H K

finite

g r o u p , a n d suppose

that

a n d \ H : H n K \ = \ G: K \ .

P r o o f . L e t D = H n K , a n d observe that since \ G : D \ = \ G : H \ \ H : D \ , the index \ G : D \ is divisible by \ G : H | . Similarly, \ G : D \ is divisible by \ G : K \ , and since we are assuming t h a t \ G : H \ and \ G : K \ are coprime, we conclude t h a t their product divides \ G : D \ . In particular, \ G : H \ \ G : K \ < \ G : D \ , and hence \ G \ < \ H \ \ K \ / \ D \ = \ H K \ . It follows t h a t H K = G, as wanted. A l s o , since equality holds here, we o b t a i n \ G : K \ = \ G \ / \ K \ = \ H \ / \ D \ = \ H : H n K \ , as wanted. • We m e n t i o n t h a t this proof of L e m m a 3.16 is v a l i d o n l y i f \ G \ is finite, but i n fact, the l e m m a is true even if G is infinite, assuming, of course, t h a t H and K have finite coprime indices. The

following is due to H . W i e l a n d t .

3.17. T h e o r e m . L e t H , K , L C G, w h e r e G i s a finite g r o u p , a n d that the indices \ G : H \ , \ G : K \ a n d \ G: L \ are p a i r w i s e coprime. o f H , K and L is solvable, then G is solvable.

suppose If each

3C

89

P r o o f . Since the result is t r i v i a l i f | G | = 1, we c a n proceed by i n d u c t i o n on \ G \ . Suppose t h a t N < G, a n d consider the subgroups H , K and L of G = G/N. Since these are h o m o m o r p h i c images of the solvable groups H , K and L , respectively, each of t h e m is s o l v a b l e ^ A l s c J G : H | = | G : N ~ H \ = | G : N H \ , w h i c h divides | G : H \ . S i m i l a r l y , | G _ ^ / ^ d i v i d e s | G : K \ and | G : L \ divides | G : L | , a n d ^ o the indices of H , K a n d L are pairwise coprime. If N > 1, therefore, G is solvable b y the i n d u c t i v e hypothesis. If H = 1, then | G | = | G : i f | is relatively p r i m e to | G : K \ , and thus G = K is solvable. W e can assume, therefore, t h a t H > 1, and we choose a m i n i m a l n o r m a l subgroup M of H . B y L e m m a 3.11, we k n o w that M is a p-group for some p r i m e p, and since p cannot d i v i d e b o t h | G : Af| and | G : L | , we can assume that p does not d i v i d e | G : AT|. I n other words, K contains a full Sylow p-subgroup of G , and by Sylow theory, some conjugate of t h a t Sylow subgroup contains the p-subgroup M . T h u s M C K for some element g G G . 9

9

N o w | G : K \ = \ G : K \ is relatively p r i m e to | G : i f | , a n d hence by L e m m a 3.16, we have G = H K . L e t s € G be arbitrary, and w r i t e x = w , where u e H a n d v e K . T h e n M * = M = M C K , where the second equality holds because u £ H and M < H . A l l conjugates of M , therefore, lie i n K , and hence the subgroup M t h a t they generate is contained i n K , w h i c h is solvable. T h e n M is a nonidentity solvable n o r m a l subgroup of G , and also, G / M is solvable by the first paragraph of the proof. It follows b y L e m m a 3.10(e) that G is solvable, as required. • 9

9

U 1 ;

9

v

9

G

9

G

G

P r o o f of T h e o r e m 3.15 (assuming B u r n s i d e ) . W e proceed by induct i o n on the number n of different primes t h a t d i v i d e | G | . If n = 1, then G is a p-group, a n d hence is solvable, a n d i f n = 2, then G is solvable by B u r n s i d e ' s theorem. W e can assume, therefore, t h a t at least three primes p, q and r d i v i d e | G | , and we choose a p - c o m p l e m e n t H , a g-complement K and a n r-complement L i n G . T h e indices of these subgroups are respectively, a power of p, a power of q and a power of r , and so they are pairwise coprime. If we can show t h a t H , K and L are solvable, it w i l l follow by T h e o r e m 3.17 t h a t G is solvable, and we w i l l be done. T h e p r i m e divisors o f \ H \ are just the p r i m e divisors of | G | other t h a n p, and so there are exactly n - 1 of t h e m . L e t s be one of these p r i m e divisors of \ H \ , a n d let Q be an s-complement i n G . T h e n | G : H \ is a power of p and | G : Q\ is a power of s, and so these indices are coprime, a n d it follows by L e m m a 3.16 t h a t \ H : H f ] Q \ = \ G : Q\, w h i c h is the full power of s d i v i d i n g | G | . Since this is also the full power of s d i v i d i n g \ H \ , it follows t h a t H n Q is a n s-complement i n H . T h i s shows t h a t H has a n s-complement for each of the n - 1 primes s d i v i d i n g its order, and so H is solvable by the inductive hypothesis. Similarly, K and L are solvable, a n d the proof is complete. •

3. S p l i t

90

Extensions

W e have seen t h a t for every nonsolvable finite group G, there is a prime set rr for w h i c h G fails to have a H a l l rr-subgroup. H o w about conjugacy? Since H a l l rr-subgroups do sometimes exist i n nonsolvable groups, we can ask whether or not (for a fixed prime set rr) a l l H a l l rr-subgroups of a nonsolvable group G must be conjugate. T h e answer is "no"; they need not even be isomorphic. In the simple group PSL(2,11) of order 660 = 2 - 3 - 5 - l l , for example, there exists a H a l l {2, 3}-subgroup isomorphic to the alternating group A , a n d there is another H a l l {2, 3}-subgroup isomorphic to the d i hedral group D . These groups of order 12 are certainly not isomorphic. W e close this discussion by m e n t i o n i n g (without proof) a lovely theorem of W i e l a n d t w h i c h asserts t h a t if G has a n i l p o t e n t H a l l rr-subgroup, t h e n a l l H a l l rr-subgroups of G are conjugate. 2

A

1

2

Problems 3C 3 C . 1 . L e t U C G be a rr-subgroup, where rr is a set of primes a n d G is solvable a n d finite. Show t h a t U is contained i n some H a l l rr-subgroup of G. H i n t . Use P r o b l e m 3B.4. N o t e . T h i s , of course, is the H a l l D-theorem. 3 C . 2 . L e t 7t be a set of primes, a n d suppose that G has a p-complement H for each p r i m e p e r t . Show that C\ H is a H a l l rr'-subgroup of G. p

pen

v

3 C . 3 . A Sylow system i n a group G is a set S of Sylow subgroups of G, one chosen from S y l ( G ) for each prime divisor p of a n d such t h a t P Q = Q P for a l l P , Q € 5. p

(a) Show t h a t if G has a Sylow system, t h e n it has a H a l l rr-subgroup for every set rr of primes. (b) If G is solvable, prove t h a t it has a Sylow system. H i n t . F o r (b), choose a p-complement H of G for each p r i m e divisor of \ G \ , a n d consider intersections of a l l but one of these p-complements. p

3 C . 4 . L e t M be m i n i m a l n o r m a l i n a finite solvable group G, a n d assume that M = C ( M ) . G

Show that G splits over M a n d that a l l complements

for M i n G are conjugate. H i n t . L e t L / M be m i n i m a l n o r m a l i n G / M , a n d note t h a t L / M is a qgroup for some p r i m e q. Show that q does not divide | M | , a n d consider a Sylow q subgroup of L .

3D

91

3 C . 5 . L e t i f be a m a x i m a l subgroup of G , where G is solvable, a n d assume t h a t c o r e ( i f ) = 1. Show that G has a unique m i n i m a l n o r m a l subgroup M , t h a t H complements M a n d t h a t M = C { M ) . Deduce t h a t i f K is also m a x i m a l i n G w i t h t r i v i a l core, t h e n H a n d K are conjugate i n G. G

G

3 C . 6 . L e t G be finite and solvable, a n d suppose t h a t x , y , z G G have orders t h a t are pairwise relatively prime. If xyz = 1, show t h a t x = 1, y = 1 and z = l. H i n t . W o r k by i n d u c t i o n o n the derived length of G. 3 C . 7 . A C a r t e r s u b g r o u p of a solvable group G is a nilpotent subgroup Csuch thatC = N (C). G

(a) Show t h a t every solvable group has a C a r t e r subgroup. (b) Show t h a t a l l C a r t e r subgroups of the solvable group G are conjugate i n G. (c) If C is a C a r t e r subgroup of G a n d G / N is nilpotent for some n o r m a l subgroup N of G, show t h a t N C = G. N o t e . T h i s p r o b l e m and the next one seem quite a bit harder t h a n most of the problems i n this book. W e present t h e m as challenges to the reader, a n d to give a taste of some of the theory of solvable groups t h a t we have omitted. 3 C . 8 . L e t G be solvable. A nilpotent injector of G is a nilpotent subgroup / c o n t a i n i n g the F i t t i n g subgroup F ( G ) , a n d m a x i m a l w i t h this property. P r o v e t h a t a l l nilpotent injectors of G are conjugate i n G. 3D A finite group G is said to be rr-separable, where rr is some set of primes, if there exists a n o r m a l series 1 = N

0

C TVi C • • • c 7V = G r

such t h a t each factor N . / N ^ x is either a rr-group or a rr'-group. (Observe t h a t a rr-separable group is the same t h i n g as a vr'-separable group.) It is easy to prove, as we shall see, t h a t finite solvable groups are vr-separable for all p r i m e sets rr, and so rr-separability can be viewed as a generalization of solvability. F i r s t , we m e n t i o n t h a t i f G is rr-separable, t h e n so too is every subgroup a n d every h o m o m o r p h i c image. ( T h i s fact is easy to establish, a n d we trust t h a t the reader c a n s u p p l y a proof.) N e x t , we recall t h a t C v ( G ) is the unique largest n o r m a l vr-subgroup i n G , w h i c h may, of course, be the t r i v i a l subgroup. ( T h e uniqueness is an i m m e d i a t e consequence of the fact t h a t i f N

92

3. S p l i t

Extensions

a n d M are n o r m a l rr-subgroups of G, t h e n J V M is also a n o r m a l 7r-subgroup.) Because O ( G ) is u n i q u e l y determined, it is, of course, characteristic i n G. n

In the definition of rr-separability, we can renumber the n o r m a l subgroups N i i f necessary, i n order to eliminate repeats among t h e m . If G is rr-separable a n d n o n t r i v i a l , therefore, we can assume that N i > 1. Since is either a n o r m a l rr-subgroup or a n o r m a l rr'-subgroup of G, we see

Ni

t h a t either C v ( G ) > 1 or C v ( G ) > 1 (or b o t h ) .

I n particular, G has a

n o n t r i v i a l characteristic subgroup N t h a t is either a rr-group or a rr'-group. A l s o , i f N < G, t h e n G / N is rr-separable, a n d so we can find a n o n t r i v i a l characteristic subgroup M / N of G / N t h a t is either a rr-group or a rr'-group. C o n t i n u i n g like this, we c a n construct a series of characteristic subgroups of G whose factors are rr-groups a n d rr'-groups. The

discussion i n the previous paragraph shows that i f we strengthen

the c o n d i t i o n on the n o r m a l subgroups N i n the definition of rr-separability, a n d we require t h e m to be characteristic, a n d not merely n o r m a l , t h e n the same class of groups w o u l d result. Somewhat paradoxically, this observation allows us to prove t h a t a weakened

version of the definition also yields exactly

the same class of groups. It suffices to consider s u b n o r m a l series. 3.18.

L e m m a . Suppose

that G is a 1 = N

i s a series

of subgroups

a it'-group.

Then

such

finite

group a n d that

< N i< • • • < N

Q

t h a t each

r

= G

factor A^/A/j-i is either a it-group o r

G is it-separable.

P r o o f . W e c a n assume t h a t r > 0, a n d we proceed b y i n d u c t i o n on r . Since N - x is rr-separable b y the inductive hypothesis, the preceding remarks T

guarantee t h a t there exists a characteristic series 1 = M

0

C ••• C M

s

= N

r

- i

h a v i n g rr a n d rr'-factors. Since 7 V _ i < G, the characteristic subgroups M i of r

7 V _ i are n o r m a l i n G, a n d hence G has the n o r m a l series r

1 = M

0

C ••• C M

s

= N -x C N r

r

= G

w i t h 7T a n d rr'-factors. T h u s G is rr-separable, as desired. 3.19.

C o r o l l a r y . L e t G be

every

set i t of p r i m e s .

finite

a n d solvable.

Then

•

G is it-separable f o r

P r o o f . C o n s i d e r a c o m p o s i t i o n series for G. (Recall t h a t this is a c h a i n of subgroups 1 = 7V < N 0

x

< • • • < 7V = G r

such t h a t each of the factors N / N - i is simple.)

Since we are assuming

t h a t G is solvable, its c o m p o s i t i o n factors are solvable too, a n d thus each of

3D

93

the simple groups N / N - x has prime order. B u t every p r i m e p is either a vr-number or a vr'-number, depending o n whether p G vr or p 0 vr, a n d thus G is vr-separable b y L e m m a 3.18. • t

t

A somewhat stronger c o n d i t i o n t h a n "vr-separability" is "vr-solvability". A finite group G is vr-solvable if it has a n o r m a l series where each factor is either a vr'-group or is a solvable vr-group. ( O f course, a solvable group is a u t o m a t i c a l l y vr-solvable for every set vr of primes.) W e s h a l l not have m u c h to say about vr-solvable groups except i n the case where vr = { p } consists of a single p r i m e . Since ^ g r o u p s are solvable, it follows t h a t {p}-separability a n d { ^ - s o l v a b i l i t y are the same t h i n g , and groups i n this class are generally described as being > s o l v a b l e " . N o t e that a finite group is solvable i f a n d o n l y i f it is p-solvable for every p r i m e p . A key result about vr-separable groups is the following, w h i c h is a generalization of H a l l ' s E - t h e o r e m , w i t h essentially the same proof. 3.20.

Theorem.

L e t G be a I T - s e p a r a b l e g r o u p .

Then

G has a H a l l vr-

subgroup. P r o o f . W e proceed by i n d u c t i o n o n \ G \ . Suppose first t h a t O ( G ) n

B y the i n d u c t i v e hypothesis, the group G / O ( G ) t

H/O (G), n

> 1.

has a H a l l vr-subgroup

a n d it is immediate that i f is a H a l l vr-subgroup of G . If C v ( G ) =

1, then since we c a n certainly assume t h a t G is n o n t r i v i a l , we have C v ( G ) > 1. W e can thus a p p l y the inductive hypothesis to G / C v ( G ) to produce a H a l l rr-subgroup H / 0 ^ { G ) .

B y the Schur-Zassenhaus theorem, C v ( G ) has

a complement i n H , a n d it is easy to see t h a t such a complement is the desired H a l l vr-subgroup of G .

•

T h e reader might guess t h a t analogs of the C - t h e o r e m a n d D - t h e o r e m also h o l d i n the context of vr-separable groups, a n d indeed they do. T h e i r proofs rely on the conjugacy part of the Schur-Zassenhaus theorem, however, a n d thus they require either some solvability a s s u m p t i o n or else a n appeal to the F e i t - T h o m p s o n theorem. Some of the basic definitions a n d results about vr-separability go back to the work of S. A . C u n i h i n i n 1948, but it was the historic paper of P h i l i p H a l l a n d G r a h a m H i g m a n i n 1956 t h a t established a n u m b e r of deeper properties, at least for p-solvable groups. A basic a n d frequently used result i n the H a l l H i g m a n paper is their L e m m a 1.2.3, w h i c h we state here somewhat more generally: for vr-separable groups. 3.21. assume

Theorem

( H a l l - H i g m a n 1.2.3). L e t G be a i r - s e p a r a b l e g r o u p , a n d

t h a t O / ( G ) = 1. Then t

O (G) D C ( O ( G ) ) . f f

g

t

94

3. S p l i t

P r o o f . W r i t e C = C ( C v ( G ) ) and let B = C n C v ( G ) .

Extensions

O u r goal is to

G

show t h a t 5 = 0 , a n d so we assume t h a t B < C, and we work to derive a c o n t r a d i c t i o n . N o t e t h a t B a n d C are n o r m a l (and i n fact, characteristic) in G and t h a t B is a vr-group. Since C/B

is a n o n t r i v i a l vr-separable group, it has a n o n t r i v i a l charac-

teristic subgroup K / B , where K / B

is either a vr-group or a vr'-group. A l s o ,

since K / B

< G / B , we see that K / B < G/B,

is characteristic i n C/B

hence K < G . Since £? is a vr-group, it follows t h a t i f K / B

and

is also a vr-group,

t h e n AT is a n o r m a l vr-subgroup of G , and hence K C C v ( G ) . T h i s is not the case, however, since £? < K C G and 5 = C n C v ( G ) . W e can thus conclude t h a t K / B is a vr'-group. B y the Schur-Zassenhaus theorem, therefore, there is a complement H for B i n K , a n d we see t h a t H > 1. We

have H C G = C ( C v ( G ) ) C C ( B ) , where the final containment G

G

holds because 5 C C v ( G ) . T h u s £? centralizes

a n d since B H = K , we

see t h a t H < K . T h u s 1 < H C O ^ ( A T ) < G , a n d this is a c o n t r a d i c t i o n since by hypothesis, O / ( G ) = 1, and hence G has no n o n t r i v i a l n o r m a l t

rr'-subgroup.

•

In a group G , a n o r m a l subgroup N t h a t contains its own centralizer can

be thought of as a "large" subgroup, and indeed, i n this s i t u a t i o n , the

i n d e x | G : N \ , is bounded i n terms of N . To see this, recall t h a t for an a r b i t r a r y n o r m a l subgroup N , the factor group G / C ( N )

is i s o m o r p h i c a l l y

embedded i n A u t ( / V ) , and so \ G : C ( / V ) | < |Aut(7V)|.

If C ( N )

G

G

G

C N ,

therefore, we have | G : 7V| < | A u t ( / V ) | . L e m m a 1.2.3 asserts t h a t i n a TTseparable group G for w h i c h C v ( G ) = 1, the n o r m a l subgroup C v ( G ) is large i n this sense. A n o t h e r example of a characteristic subgroup t h a t is guaranteed to be "large" is the F i t t i n g subgroup F ( G ) of a solvable group G.

(See P r o b l e m 3B.14 and the note following it.) I n C h a p t e r 9, we discuss

the "generalized F i t t i n g subgroup" F * ( G ) , w h i c h always contains its own centralizer, w i t h no e x t r a a s s u m p t i o n o n G . We

close this section by showing how the H a l l - H i g m a n L e m m a 1.2.3 can F i r s t , we need a definition. If G is vr-separable, the rr-length

be used.

of G is the m i n i m u m possible number of factors t h a t are rr-groups i n any n o r m a l series for G i n w h i c h each factor is either a vr-group or a vr'-group. For

example, G has vr-length 1 precisely w h e n G is not a vr'-group and there

exist n o r m a l subgroups N C M of G such that N and G / M and M / N 3.22.

are vr'-groups

is a vr-group.

Theorem.

L e t G be a I T - s e p a r a b l e g r o u p ,

I T - s u b g r o u p of G i s a b e l i a n .

Then

a n d suppose

the v r - l e n g t h of G i s a t most

that a Hall 1.

P r o o f . C o n s i d e r the vr-separable group G = G / C v ( G ) , a n d observe t h a t Cv(G)

= 1.

L e t i f be a n abelian H a l l vr-subgroup of G , a n d note t h a t

3E

95

H is a H a l l vr-subgroup of G . _ T h e n H contains every n o r m a l vr-subgroup of G , and i n p a r t i c u l a r , C v ( G ) C H . B u t # is abelian, a n d thus H C C c ( C v ( G ) ) C C v ( G ) , where_the last containment follows by the H a l l H i g m a n L e m m a 1.2.3. T h u s H = 0 * ( G ) , and we have H < G. T h u s 1 C N C T V i i C G is a n o r m a l series for G w i t h just one rr-factor, and the proof is complete. •

Problems

3D

3 D . 1 . L e t G be a p-solvable group, a n d let P G S y l ( G ) . p

(a) If C y ( G ) = 1, show t h a t Z ( P ) C O ( G ) . p

(b) Show t h a t the p-length of G is at most the nilpotence class of P . 3 D . 2 . L e t Z C Z ( G ) and write G = G / Z . Show t h a t C v ( G ) = O ^ G ) for every set rr of primes. 3 D . 3 . L e t G be rr-separable, and assume t h a t C v ( G ) C v ( G ) C Z ( G ) . Show t h a t G is abelian. 3 D . 4 . L e t H act by automorphisms on G , a n d assume t h a t \ H \ and | $ ( G ) | are coprime. Suppose that the induced a c t i o n of # on G / $ ( G ) is t r i v i a l , w h i c h means t h a t each coset of $ ( G ) i n G is setwise invariant under H . Show t h a t the a c t i o n of H on G is t r i v i a l . H i n t . It is no loss to assume t h a t H is a g-group for some prime q. Consider the d e c o m p o s i t i o n of each coset of $ ( G ) into P - o r b i t s . 3D.5.

L e t G be p-solvable, and let P e S y l ( G ) . Suppose t h a t P C N ( K ) , p

G

where K C G is a subgroup of order not divisible by p. Show t h a t i f C Op/(G). H i n t . F i r s t , consider the case where C y ( G ) = I . T o do the general case, consider the group G = G / C y ( G ) .

3E Let A be a group t h a t acts v i a automorphisms o n some group G . B y definit i o n , A permutes the elements of G , b u t because we have a n a c t i o n b y automorphisms, A also permutes m a n y other kinds of group-theoretic objects associated w i t h G . F o r example, A acts o n the set of a l l Sylow p-subgroups of G for any given prime p , a n d A acts on the set of a l l conjugacy classes of G . It is of interest to find, or at least to prove the existence of, ^ - i n v a r i a n t objects of various kinds. Once one has found a n A - i n v a r i a n t object, there are u s u a l l y further actions t h a t c a n be studied. F o r example, i f H is an

96

3. S p l i t

Extensions

A - i n v a r i a n t subgroup of G, then A acts on the set of right cosets or of left cosets of H i n G, a n d on the set of conjugacy classes of H . Of course, the most basic type of A - i n v a r i a n t object i n G is an A¬ invariant element. These A - f i x e d points i n G clearly form a subgroup, w h i c h is denoted C { A ) . T o see w h y this centralizer n o t a t i o n is really quite natu r a l , embed A a n d G i n their semidirect product V = G x A . W e k n o w t h a t the original action of A on G is realized b y conjugation w i t h i n T, a n d thus g = g i f a n d only i f g a n d a commute i n V. T h u s the A-fixed-point subgroup of G is exactly the centralizer of A i n G, a n d this explains the n o t a t i o n C ( A ) , w h i c h is used to denote the fixed-point subgroup even if the semidirect p r o d u c t has not been mentioned. Similarly, w o r k i n g i n the semidirect product, we see t h a t C ( G ) is the set of elements of A t h a t act t r i v i a l l y on G, or i n other words, it is the kernel of the a c t i o n of A on G. T h i s n o t a t i o n also can be used w i t h o u t m e n t i o n i n g the semidirect p r o d u c t . G

a

G

A

It is especially easy to find A - i n v a r i a n t objects i n G i f A a n d G have cop r i m e orders, a n d this is the s i t u a t i o n we study i n this section. F o r example, we w i l l prove the following. 3.23. T h e o r e m . L e t A a c t v i a a u t o m o r p h i s m s o n G, w h e r e A a n d G a r e finite g r o u p s , a n d suppose t h a t ( | A | , \ G \ ) = 1. Assume a l s o t h a t a t l e a s t one of A o r G i s s o l v a b l e . Then f o r each p r i m e p , t h e f o l l o w i n g h o l d . (a) T h e r e exists

a n A - i n v a r i a n t Sylow

(b) / / S a n d T a r e A - i n v a r i a n t Sylow f o r some element c £• C { A ) .

p - s u b g r o u p of G. p - s u b g r o u p s of G, then

S° = T

G

N o t e t h a t T h e o r e m 3.23(a) is a generalization of the Sylow E-theorem, to w h i c h it reduces w h e n A is the t r i v i a l group, a n d similarly, 3.23(b) is a generalization of the Sylow C-theorem. A n o t h e r way to t h i n k about this result is to imagine t h a t we have a pair of magic eyeglasses t h r o u g h w h i c h only A - i n v a r i a n t objects can be seen. T h e o r e m 3.23(a) asserts t h a t even w h e n we look at G t h r o u g h these A-glasses, the Sylow E - t h e o r e m can be seen to hold: there exists a v i s i b l e Sylow p-subgroup for each prime p . The Sylow C - t h e o r e m also remains true when G is viewed t h r o u g h the magic A-glasses. W h a t the usual C - t h e o r e m says, of course, is t h a t given two Sylow p-subgroups of G, say S a n d T , we can find an element g G G such t h a t S = T. N o w suppose t h a t S a n d T are Sylow p-subgroups t h a t we can see after d o n n i n g our glasses. (In other words, they are A¬ invariant.) T h e glasses version of the Sylow C - t h e o r e m says t h a t we can see an element g of G such t h a t S = T. T h e assertion, i n other words, is t h a t S a n d T are conjugate b y some A - i n v a r i a n t element: an element of the fixed-point subgroup C ( A ) . (This, of course, is e x a c t l y the assertion 9

9

G

3E

97

of T h e o r e m 3.23(b).) W e w i l l also prove the glasses version of the Sylow D theorem: i f P is a visible (A-invariant) p-subgroup of G , t h e n P is contained in some visible (A-invariant) Sylow p-subgroup. How c a n the glasses version of the Sylow E - t h e o r e m and similar results be proved? In the case where A is a g-group for some p r i m e q, an easy counting argument suffices, as follows. F i r s t , observe t h a t we c a n assume that A is n o n t r i v i a l , and thus q divides | A | , a n d so q does not divide \ G \ . Since | S y l ( G ) | divides \ G \ , it follows t h a t q does not divide | S y l ( G ) | . N o w A acts on S y l ( G ) , and hence this set decomposes into A - o r b i t s , at least one of w h i c h must have size not divisible by q. B u t we are assuming that A is a g-group, and so a l l A - o r b i t s have g-power size. T h e r e must be a n orbit of size 1, therefore, w h i c h means t h a t the set S y l ( G ) contains a n A - i n v a r i a n t member, as required. p

p

p

p

A similar (but slightly more technical) argument c a n be used to prove 3.23(b) i n the case where A is a g-group, b u t it appears to be u t t e r l y i m possible to make a counting argument like this w o r k if | A | is not a p r i m e power. T h e following result of G . G l a u b e r m a n provides a n effective substitute, however. Unfortunately, the proof of G l a u b e r m a n ' s l e m m a relies on the conjugacy part of the Schur-Zassenhaus theorem, a n d thus we must either appeal to the F e i t - T h o m p s o n theorem, or else assume t h a t something is solvable. (The solvability assumptions t h a t appear i n the statements of T h e o r e m 3.23 and other results about coprime actions are not really necessary, of course, since at least one of | A | or \ G \ is o d d , a n d hence by the F e i t - T h o m p s o n theorem, at least one of A or G must be solvable.) 3.24. L e m m a ( G l a u b e r m a n ) . L e t A a c t v i a a u t o m o r p h i s m s o n G, w h e r e A a n d G a r e finite g r o u p s , a n d suppose t h a t ( | A | , | G | ) = 1. Assume also that a t l e a s t one of A o r G i s s o l v a b l e . Suppose t h a t A a n d G each a c t o n some n o n e m p t y set Vl, w h e r e t h e a c t i o n of G i s t r a n s i t i v e . F i n a l l y , assume the compatibility condition: (*)

(a-g)-a = (a-a)-g

a

f o r a l l a < E V l , a < E A a n d g < E G. T h e f o l l o w i n g t h e n (a) T h e r e exists (b) Ifa,P a-c

a n A - i n v a r i a n t element

hold.

aefi.

G f l a r e A - i n v a r i a n t , t h e n t h e r e exists

c G C ( A ) such G

that

= (3.

Before proceeding w i t h the proof of G l a u b e r m a n ' s l e m m a , we show how it can be used to prove T h e o r e m 3.23. P r o o f of T h e o r e m 3.23. T h e set S y l ( G ) is n o n e m p t y b y the Sylow E theorem. E a c h of G and A act on S y l ( G ) , where G acts by conjugation and the a c t i o n of A is induced by the given a c t i o n of A on G . (Since each p

p

98

3. S p l i t

Extensions

element of A induces a n a u t o m o r p h i s m of G, it carries Sylow p-subgroups of G to Sylow p-subgroups.) A l s o , by the Sylow C-theorem, the action of G on S y l ( G ) is transitive. Once we check the c o m p a t i b i l i t y c o n d i t i o n of G l a u b e r m a n ' s l e m m a , it w i l l follow b y 3.24(a) that S y l ( G ) contains an A - i n v a r i a n t member, thereby establishing (a). W e need to check that p

p

9

a

a a

[s ) =

(s y

for S G S y l ( G ) , a G A a n d g G G. (Recall that here, exponentiation by g means conjugation b y g i n G, but exponentiation by a denotes the given action of a . ) Since A acts b y automorphisms, however, we have p

9

(S )

a

a

= (g^Sg)*

1

a

= {g )- S g

a

9

= (S") " ,

as required. A s s e r t i o n (b) is immediate from 3.24(b).

•

In the previous proof, it was t r i v i a l to check the c o m p a t i b i l i t y c o n d i t i o n of G l a u b e r m a n ' s l e m m a . Indeed, it seems to be equally t r i v i a l to check it i n v i r t u a l l y every a p p l i c a t i o n of 3.24. P r o o f of L e m m a 3.24. T o prove (a), let T = G x > A be the semidirect p r o d uct w i t h respect to the given action of A on G, a n d as usual, identify A and G w i t h appropriate subgroups of T. E v e r y element of T is uniquely of the form a g , where a G A and g G G, and so we can unambiguously define a n a c t i o n of r on ft b y setting a - ( a g ) = { a - a ) - g . T o verify that this really is an action, let a , b G A and g , h G G. T h e n for a G ft, we have (a-{ag))ibh)

= (((a-a)- )-6)-/z = 5

b

{{(a-a)-b)-g )-h b

=

a-{(ab)(g h))

=

a-((ag)(bh)),

as wanted, where the second equality holds because of the c o m p a t i b i l i t y c o n d i t i o n , w h i c h we are assuming. Now fix a G ft and let U = T , the stabilizer of a i n T. If x G T, then since by assumption, G is transitive on ft, we can w r i t e a-x = a - g for some element g G G, and thus x g ~ G U. It follows that x G U G , a n d so U G = T. Now G < T , a n d hence U n G < \ U . A l s o , a

l

\ U : U n G \

= \ U G : G \ = \ T : G \ = \A\

is coprime to \ U D G\, and thus the Schur-Zassenhaus theorem applies i n U w i t h respect to the n o r m a l subgroup U n G. L e t H be a complement for U n G i n U, and note t h a t \ H \ = \ U : U n G\ = \ A \ , and thus H is a complement for G i n T. Since A is also a complement for G i n T, the conjugacy part of the Schur-Zassenhaus theorem allows us to write A = H for some element x G T. B u t H C U, and so H stabilizes a . T h e n A = H x

x

3E

99

stabilizes a-x, and thus a-x is a n A - i n v a r i a n t member of ft. T h i s completes the proof of (a). To prove (b), let a a n d P be A - i n v a r i a n t members of ft. Since G is transitive o n ft, the set X = { g G G \ a - g = ¡3} is nonempty, a n d we need to find an. A - i n v a r i a n t element i n X . F o r this purpose, we propose to use part (a) of the lemma, w i t h the set X i n place of ft. If x G X , then a-x = /3, a n d so for a € A , we have a

a-x

= (a-a)-x

a

= ( a - x ) - a = (3-a = ¡3 ,

where the first a n d last equalities h o l d because a a n d ¡3 are A - i n v a r i a n t , and the second is a consequence of the c o m p a t i b i l i t y c o n d i t i o n . T h u s x G X , a n d so the a c t i o n of A on G maps X into (and hence onto) itself, a n d we have a n action of A on X . a

Now let H = G p , the stabilizer of ¡3 i n G. If we a p p l y the above argument w i t h 0 i n place of a and H i n place of X , we conclude t h a t the action of A on G maps H to itself, and this defines a n action v i a automorphisms of A on H . A l s o , since H is a subgroup of G, we see t h a t ( | A | , \ H \ ) = 1 and that at least one of A or H is solvable. Now A acts on H and on X , a n d we show next that H acts t r a n s i t i v e l y on X by right m u l t i p l i c a t i o n . If x G X and h G H , t h e n a-(xh) = (a-x)-h = p - h = P , and thus x h G X a n d H acts on X by right m u l t i p l i c a t i o n , as wanted. I n fact, this a c t i o n is transitive, since if x,y G X , t h e n a r ^ fixes P, a n d so x - ? / G i f , and right m u l t i p l i c a t i o n by x~ y carries x to y . 1

1

l

A t this point we k n o w that A acts v i a a u t o m o r p h i s m s on the group H , t h a t ( | A | , \ H \ ) = 1, and t h a t at least one of A or H is solvable. A l s o , each of A a n d H acts on the nonempty set X , a n d the action of H on X is transitive, and thus we have almost everything we need to a p p l y part (a) of the l e m m a to conclude that X contains an A - i n v a r i a n t element. W h a t remains is to check the c o m p a t i b i l i t y c o n d i t i o n . W h a t we need, i n other words, is ( x h ) = x h for x G X , h G H a n d a G A . T h i s is obvious, however, since A acts on G v i a automorphisms, a n d thus X contains an A - i n v a r i a n t element, as required. • a

a

a

The magic-eyeglasses version of the S y l o w D - t h e o r e m follows easily from T h e o r e m 3.23(a); its proof does not require a direct appeal to G l a u b e r m a n ' s lemma. 3.25. C o r o l l a r y . L e t A a c t v i a a u t o m o r p h i s m s o n G, w h e r e A a n d G finite g r o u p s , a n d suppose t h a t ( \ A \ , \ G \ ) = 1. Assume also t h a t a t least of A o r G i s s o l v a b l e . L e t P C G be a n A - i n v a r i a n t p - s u b g r o u p f o r p r i m e p . T h e n P i s c o n t a i n e d i n some A - i n v a r i a n t S y l o w p - s u b g r o u p o

are one some fG.

3. S p l i t

100

Extensions

P r o o f . R e p l a c i n g P by a subgroup containing it i f necessary, we can assume t h a t P is not contained i n any s t r i c t l y larger A - i n v a r i a n t p-subgroup of G, and we show t h a t i n this case, P £ S y l ( G ) . L e t N = N ( P ) , a n d observe t h a t N is A - i n v a r i a n t since it is uniquely determined b y the A - i n v a r i a n t subgroup P . B y T h e o r e m 3.23(a), we can choose an A - i n v a r i a n t member S € S y l ( / V ) , and we observe t h a t since P is a n o r m a l p-subgroup of N , it must be contained i n the Sylow p-subgroup S. B y our assumption, however, P cannot be p r o p e r l y contained i n S since S is A - i n v a r i a n t , a n d thus P = S. T h u s P is a S y l o w p-subgroup of its normalizer N , a n d it follows t h a t P is a S y l o w p-subgroup of G. (To see this, let P C Q G S y l ( G ) , and suppose t h a t P < Q. T h e n since normalizers grow i n p-groups, Q n N is a p-subgroup of N s t r i c t l y larger t h a n the Sylow p-subgroup P , and this is impossible.) • p

G

p

p

N e x t , we consider conjugacy classes. In general, if i f is a class of G a n d H C G is a subgroup, t h e n i f n f f m a y be empty, and even i f it is nonempty, all we c a n say is t h a t i f n H is a u n i o n of classes of i f . Usually, i f n H is not a single class of H . (Distinct classes of a subgroup H t h a t are contained in the same class of G are said to be fused i n G.) T h e s i t u a t i o n is under m u c h tighter control if the subgroup is the set of fixed points of a coprime action. I n p a r t i c u l a r , there is no fusion i n t h a t case. 3.26. T h e o r e m . L e t A a c t v i a a u t o m o r p h i s m s o n G, w h e r e A a n d G a r e finite groups, and write C = C ( A ) . Suppose that (\A\, \G\) = I , a n d assume t h a t a t l e a s t one of A o r G i s s o l v a b l e . T h e n t h e m a p i f ^ i f n C d e f i n e s a b i j e c t i o n f r o m t h e set of A - i n v a r i a n t c o n j u g a c y classes of G o n t o t h e set of a l l c o n j u g a c y classes of C. G

P r o o f . L e t i f be a n A - i n v a r i a n t class of G. T h e n A acts on i f , and G acts t r a n s i t i v e l y o n i f by conjugation. T h e c o m p a t i b i l i t y c o n d i t i o n of G l a u b e r man's l e m m a is t h a t ( x » ) = ( x ) 3 for elements x G i f , a G A and g € G, and as usual, it is t r i v i a l to see t h a t this holds. B y 3.24(a), therefore, K contains a n A - i n v a r i a n t element, and thus i f n C is nonempty. A l s o , by 3.24(b), a l l elements of i f n C are conjugate i n C, and thus i f n C is a class of C. a

a

a

We now have the m a p i f ^ i f D C from the set of A - i n v a r i a n t classes of G to the set of classes of C; we must show t h a t it is a bijection. If i f and L are A - i n v a r i a n t classes of G, w i t h i f D C = L C ) C , then since i f D C is n o n e m p t y and is contained i n b o t h i f and L , it follows t h a t i f n L is nonempty, a n d thus i f = L . O u r m a p , therefore, is injective. F o r surjectivity, it is enough to show t h a t every element of C lies i n i f n C for some A - i n v a r i a n t class i f of G. O f course, each element c G C lies i n a unique class i f of G, a n d so it suffices to show t h a t i f is A - i n v a r i a n t . If a G A , t h e n i f is a class of G, and since c = c, we see t h a t c G i f n i f " . T h u s the classes i f and i f are a

a

a

3E

101

a

not disjoint, a n d we conclude that K = K , a n d hence K is A - i n v a r i a n t , as wanted. • Now we consider cosets. If A acts on G a n d H C G is A - i n v a r i a n t , t h e n as we observed previously, A permutes the left cosets of H i n G and also the right cosets of H i n G. In the case of a coprime action, we can determine w h i c h left cosets and w h i c h right cosets of H are A - i n v a r i a n t . 3.27. T h e o r e m . L e t A a c t v i a a u t o m o r p h i s m s o n G, w h e r e A a n d G a r e finite g r o u p s , a n d w r i t e C = C ( A ) . L e t H C G be A - i n v a r i a n t s u b g r o u p , a n d suppose t h a t ( | A | , \ H \ ) = 1 a n d t h a t one of A o r H i s s o l v a b l e . Then t h e A - i n v a r i a n t left cosets of H i n G a n d t h e A - i n v a r i a n t r i g h t cosets of H i n G a r e e x a c t l y those cosets of H t h a t c o n t a i n elements of C. G

a

a

a

P r o o f . If a G A and c G C, then ( H c ) = H c = H e , and similarly, ( c H ) = c H , and so cosets of H t h a t contain elements of C are A - i n v a r i a n t . Now suppose t h a t X is an A - i n v a r i a n t left coset of H . T h e n H acts t r a n s i tively by right m u l t i p l i c a t i o n on X and, of course, A acts on X a n d A acts via automorphisms on H . T h e c o m p a t i b i l i t y c o n d i t i o n for G l a u b e r m a n ' s l e m m a i n this s i t u a t i o n is [ x h ) = x h for x G X , h G H a n d a G A , a n d this holds since the given a c t i o n of A on G is an a c t i o n v i a automorphisms. B y 3.24(a), therefore, X contains a n element of C, as wanted. a

a

a

a

To handle the right cosets, it suffices to observe t h a t every right coset of H i n G has the form X " = { x ~ \ x G X } for some left coset X of H i n G. Since X is A - i n v a r i a n t i f a n d only if X " is, the desired result for right cosets is a consequence of t h a t for left cosets. ( A l t e r n a t i v e l y , i f Y is a right coset of H i n G, then y h = h~ y defines a transitive a c t i o n of H on Y, a n d we can a p p l y G l a u b e r m a n ' s l e m m a directly.) • 1

l

1

l

A c t i o n s on cosets are especially interesting for n o r m a l subgroups. L e t A act on G v i a automorphisms, and suppose t h a t N < G is A - i n v a r i a n t . O f course, the cosets of N i n G are the elements of the factor group G/N, a n d since these are p e r m u t e d by A , we see t h a t A acts on G/N. I n fact, this is an a c t i o n v i a automorphisms since {NxNy)

a

a

= {Nxy)

a a

= Nx y

a

= Nx Ny

a

=

a

a

{Nx) (Ny) .

The A - i n v a r i a n t cosets of N i n G, therefore, are e x a c t l y the elements of the fixed-point subgroup C ( A ) . W e can use the bar convention to give a clean restatement of T h e o r e m 3.27 i n this s i t u a t i o n , a statement t h a t can be paraphrased as "fixed points come from fixed points (in coprime actions)". G

/

N

3.28.

Corollary. L e t A a c t v i a a u t o m o r p h i s m s

finite

g r o u p s , a n d l e t N < G be A - i n v a r i a n t . Assume

o n G, w h e r e A a n d G a r e t h a t (|A|, \ N \ ) = 1 a n d

3. S p l i t

102

t h a t a t l e a s t one of A o r N i s s o l v a b l e . C C ( A )

=

W r i t i n g G = G/N,

Extensions

w e have

C G T A ) .

P r o o f . T h e A - f i x e d elements of G are exactly the A - i n v a r i a n t cosets of N , w h i c h by 3.27 are exactly the cosets of N t h a t contain elements of C = C ( A ) . _ B u t the cosets of TV that contain elements of C form the image C of C i n G . • G

3.29. C o r o l l a r y . L e t A a c t v i a a u t o m o r p h i s m s o n G, w h e r e A a n d G a r e f i n i t e g r o u p s , a n d assume t h a t { \ A \ , \ G \ ) = 1. If t h e i n d u c e d a c t i o n of A o n t h e F r a t t i n i f a c t o r g r o u p G / * ( G ) i s t r i v i a l , t h e n t h e a c t i o n of A o n G i s trivial. P r o o f . W e must show t h a t each element a G A lies i n C { G ) , and so it is no loss to assume t h a t A = (a). S i n c ^ A is cyclic, it is certainly solvable, and thus C o r o l l a r y 3.28 applies w i t h G = G / $ ( G ) . W r i t i n g C = C { A ) , A

G

we have G

= Cc(A)

= C =

G$(G),

where the first equality holds since A acts t r i v i a l l y on G . It follows that G = G * ( G ) , and thus G = C by the usual argument. (If C < G , then C is contained i n some m a x i m a l subgroup of G , which necessarily contains $ ( G ) too, a n d this is a contradiction.) • 3.30. C o r o l l a r y . L e t A a c t v i a a u t o m o r p h i s m s o n G, w h e r e A a n d G a r e finite g r o u p s , a n d assume t h a t ( \ A \ , | G | ) = 1 a n d t h a t t h e a c t i o n of A o n G i s f a i t h f u l . T h e n t h e i n d u c e d a c t i o n of A o n G / $ ( G ) i s f a i t h f u l . P r o o f . L e t B = C A ( G / $ ( G ) ) and apply the previous corollary to the action of B on G to deduce that B = l . U G i v e n an a r b i t r a r y group action on a finite set, there is an associated set of positive integers: the orbit sizes. A l t h o u g h there is very little that can be said about the sets of integers t h a t arise i n this way i n general, the s i t u a t i o n is quite different for actions v i a automorphisms on groups, where there is considerably more structure. (See, for example, C o r o l l a r y 8.44.) F o r coprime actions, we can say even more. In t h a t case, it is true that if m a n d n are coprime orbit sizes, then m n must also be an orbit size. T o prove this, we use a remarkable observation of B . H a r t l e y and A . T u r u l l , which allows us to replace the group b e i n g acted on by an abelian group. T h e key step is the next theorem, w h i c h follows fairly easily from results about coprime actions t h a t we have already established. 3.31. T h e o r e m ( H a r t l e y - T u r u l l ) . L e t A a c t v i a a u t o m o r p h i s m s o n G, w h e r e A a n d G a r e finite g r o u p s , a n d suppose t h a t ( \ A \ , | G | ) = 1. Assume also that

3E

103

a t l e a s t one of A o r G i s s o l v a b l e . T h e n A acts v i a a u t o m o r p h i s m s o n some a b e l i a n g r o u p H i n such a w a y t h a t every s u b g r o u p B C A has e q u a l n u m b e r s of f i x e d p o i n t s o n G a n d o n H . A l s o , t h e r e i s a s i z e - p r e s e r v i n g b i s e c t i o n f r o m t h e set of A - o r b i t s o n G t o t h e set of A - o r b i t s o n H . We

extract part of the proof as a separate l e m m a , w h i c h has other

applications. 3.32. L e m m a . L e t A a c t v i a a u t o m o r p h i s m s o n G, w h e r e A a n d G a r e finite g r o u p s , a n d suppose t h a t ( | A | , \ G \ ) = 1. Assume a l s o t h a t a t l e a s t one of A o r G i s s o l v a b l e . L e t C = C { A ) , a n d suppose t h a t P £ Syl (G) i s A - i n v a r i a n t . T h e n P n C G Syl (C). G

p

p

P r o o f . L e t S G S y l ( C ) , a n d note t h a t S is A - i n v a r i a n t . B y C o r o l l a r y 3.25, some A - i n v a r i a n t Sylow p-subgroup Q of G contains S. B y T h e o r e m 3.23(b), we have Q = P for some element c G C, a n d thus p

c

c

Pnc

= Q nc

= (Qr] C)

The result now follows because P n C Sylow p-subgroup S . •

c

c

5

S.

is a p-subgroup of C t h a t contains the

c

The

following is a very general result about groups a c t i n g on finite sets.

3.33. L e m m a . L e t A be a g r o u p t h a t acts o n finite sets ft a n d A , a n d assume t h a t each s u b g r o u p B C A has e q u a l n u m b e r s of fixed p o i n t s i n ft a n d i n A . T h e n t h e r e i s a b i j e c t i o n f : ft A such that f(a-a) = f(a)-a f o r a l i a e V l a n d a G A . I n p a r t i c u l a r , f maps A - o r b i t s o n ft t o A - o r b i t s o n A , a n d so t h e sets of o r b i t sizes i n t h e t w o a c t i o n s a r e i d e n t i c a l . P r o o f . W e prove the existence of / by i n d u c t i o n on |ft|; the assertions about orbits are then immediate. L e t n e U , where U is an A - o r b i t of smallest possible size i n ft. T h e n | A : A ^ | = \ U \ , a n d b y the m i n i m a l i t y of \ U \ , no subgroup properly containing A ^ can fix a p o i n t i n ft. B y hypothesis, therefore, no such subgroup can fix a point i n A . A l s o b y hypothesis, A fixes some point v G A , and it follows t h a t A is the full stabilizer of v i n A , a n d we have A ^ = A . M

M

U

We now define / on U by setting / ( u - x ) = vx for x G A . T h i s is well defined since i f = f i - y , then xy~ G A = A„, and thus vx = vy. A l s o , / is injective on U since i f vx = vy, t h e n xy~ G A = A , and thus H - x = n-y. Furthermore, i i a e U , we can w r i t e a = px for some element x G A , a n d hence if a G A , we have l

M

l

v

f(a-a) = f{fi'{xa))

= v(xa) = (vx)-a =

M

f{a)-a.

Now let V = f ( U ) , so that V is the orbit of v a n d | V | = \ U \ . B y t a k i n g B = 1, we see t h a t |ft| = | A | , and thus if U = ft, we have V = A a n d there is

104

3. S p l i t

Extensions

n o t h i n g further to prove. W e can assume, therefore, that U < Cl and V < A , and we observe t h a t A acts on Q - U and on A - V . We w i l l argue t h a t the hypotheses of the l e m m a are satisfied for the actions of A on Q - U a n d A - V . Once this is established, the inductive hypothesis w i l l guarantee the existence of an appropriate bijection from Cl - U to A - V , and this can be used to extend the definition of / to a l l of Cl. If B C A , we wish to show t h a t B has equal numbers of fixed points i n Q. - U a n d i n A - V , and for this purpose, it suffices to show t h a t B has equal numbers of fixed points i n U and V . T h i s is clear, however, since for each element a £ A and each point a G U, we have f ( a - a ) = f ( a ) - a , and so a fixes a i f and only i f it fixes f { a ) . U

P r o o f of T h e o r e m 3.31. W e proceed i n two steps to prove the first part; the final assertion then follows by L e m m a 3.33. F i r s t , we show t h a t G can be replaced by a nilpotent group, and then we complete the proof by showing t h a t i f G is nilpotent, it can be replaced by an abelian group. For each p r i m e divisor p of \ G \ , choose an A - i n v a r i a n t Sylow p-subgroup of G, and let N be the nilpotent group constructed as the external direct p r o d u c t of the set V of selected Sylow subgroups. Since A acts on P for each member P G V, there is a n a t u r a l action of A on N , and we argue that \ C ( B ) \ = \ C ( B ) \ for each subgroup B C A . Since an element of the direct p r o d u c t N is B - f i x e d if and only i f each of its components is fixed, it is clear t h a t N

G

\C (B)\ N

= Yl

\ C ( B ) \ , P

Pev

and thus for each prime p , the p-part of \ C ( B ) \ is equal to \ C ( B ) \ = \ P C \ C \ , where C = C { B ) . B u t P n C G S y l ( C ) by L e m m a 3.32, and thus \ P n C\ is the p-part of \ C \ . T h i s shows that \ C ( B ) \ and \ C \ have equal p-parts for a l l primes p, and hence \ C { B ) \ = | C | , as wanted. N

G

P

p

N

N

To complete the proof, we can now assume that G is nilpotent, a n d in p a r t i c u l a r it is solvable w i t h some derived length d. W e proceed by i n d u c t i o n on d. If d = 1, t h e n G is abelian, and there is n o t h i n g to prove. We can thus assume t h a t d > 1, and we define the external direct p r o d u c t K = G' x G / G ' , w h i c h has derived length d - 1. Since G' is characteristic i n G, it is A invariant, and thus A acts v i a automorphisms on G' and on G / G ' , and this induces an action v i a automorphisms of A on K . Since |AT| = \ G ' \ \ G / G ' \ = \ G \ is coprime to A , the inductive hypothesis applies, and thus there exists an abelian group H , acted on by A , a n d such that \ C ( B ) \ = | C K ( - B ) | for every subgroup B C A . H

3E

105

It suffices now to show t h a t \ C \ = \ C ( B ) \ , where as before, C = C { B ) . U s i n g C o r o l l a r y 3.28 to o b t a i n the second equality below, we o b t a i n K

G

\C (B)\ = \C K

G / G

,(B)\\C ,{B)\ =

\CG'/G'\\CnG'\

G

=

\C:CnG'\\cnG'\

= \c\, a n d the proof is complete.

•

W e can now prove the result about coprime orbit sizes that we mentioned earlier. 3.34. T h e o r e m . L e t A a c t v i a a u t o m o r p h i s m s o n G, w h e r e A a n d G a r e finite g r o u p s , a n d suppose t h a t ( | A | , \ G \ ) = 1. Assume a l s o t h a t a t l e a s t one of A o r G i s s o l v a b l e . Suppose t h a t t h e r e e x i s t A - o r b i t s of size m a n d n i n G, w h e r e m a n d n a r e c o p r i m e . T h e n t h e r e i s a l s o a n o r b i t of size m n . P r o o f . B y T h e o r e m 3.31, we can replace G by an abelian group of the same order w i t h o u t affecting orbit sizes, and thus we can assume that G is abelian. N o w let X and y be A - o r b i t s i n G of sizes m a n d n , respectively, a n d let x G X a n d y G y. T h e n \ A : A \ = \X\ = m and \ A : A \ = \y\ = n , and since m a n d n are coprime, it follows by L e m m a 3.16 t h a t A = A A and t h a t \ A : A n A \ = m n . N o w write z = xy, a n d observe that A ( ~ ) A C A . We w i l l show t h a t A n A = A , a n d consequently, the size of the A - o r b i t of z is IA : A D A | = m n . W e work i n the next few paragraphs to prove t h a t A Q Ay. S i m i l a r reasoning then shows t h a t A C A , a n d this w i l l complete the proof. x

y

x

x

y

x

x

x

y

y

y

z

z

y

z

z

x

Define the m a p r : G - > G by setting

aeAy

Because G is abelian, the order of the factors i n the product is irrelevant, a n d thus r is well defined. Furthermore, a n d also because G is abelian, r is a group h o m o m o r p h i s m . W e argue t h a t r maps each element of the orbit X into C ( A ) . T o see this, recall that A = A A , a n d so A acts transitively on the set X . If u G X , therefore, then as a runs over the elements of A , each element of X occurs equally often i n the form u . W e conclude t h a t T ( U ) is a power of the product of the elements of the orbit X , and thus r ( u ) is A - i n v a r i a n t , as claimed. G

x

y

y

y

a

Let X = ( X ) and Y = ( y ) , so t h a t X and Y are A - i n v a r i a n t subgroups of G. A l s o , since r is a h o m o m o r p h i s m t h a t maps every element of X into C ( A ) , it follows t h a t T ( X ) consists of A - f i x e d elements. L e t Z = X n Y, and note that Z is also an A - i n v a r i a n t subgroup of G. L e t B C A be the (setwise) stabilizer of the coset Zy, and observe t h a t A C B since G

y

106

3. S p l i t

Extensions

A stabilizes b o t h Z and y. Since B stabilizes the coset Zy, it follows by T h e o r e m 3.27 t h a t Zy contains a B - i n v a r i a n t element c, a n d thus y £ Z c and we can write y = t c for some element i £ Z . y

A

Since c is B - i n v a r i a n t and A C B , we see t h a t r ( c ) = cl «l, and so r ( c ) is /^-invariant. A l s o , since t £ Z C X , we know t h a t r ( i ) is A - i n v a r i a n t , and in particular, it is B - i n v a r i a n t . W e conclude t h a t y \ * \ = r { y ) = r ( i ) r ( c ) is ^ - i n v a r i a n t . Since \ A \ is coprime to \ G \ , however, it follows that y is a power of y l ^ l , and hence y is ^ - i n v a r i a n t and B C A y

A

y

r

We argue next t h a t A C £?, and thus 4 C i T o see this, note that z = xye Xy, a n d thus Xy = Xz, a n d this coset is stabilized by A . Since y £ Y, it is easy to check t h a t n Y = ( X n >")y = Z y , and since A stabilizes b o t h Xy and y , we conclude t h a t A stabilizes Zy, a n d so A C £?, as claimed. W e have now shown t h a t A C A , a n d as we remarked earlier, similar reasoning yields A C A . T h i s completes the proof. • z

r

z

2

z

z

z

z

y

x

We m e n t i o n that i n the proof of T h e o r e m 3.34, the orbit of z = xy is exactly the set xy. T h i s is easy to see, but this observation does not yield a t r i v i a l proof of the theorem because it is not obvious t h a t \Xy\ = \X\\y\. We close this section w i t h one a d d i t i o n a l remark. G i v e n an arbitrary set TV of integers exceeding 1, one can construct a graph w i t h vertex set M by creating an edge j o i n i n g m and n i n TV whenever m a n d n have a c o m m o n prime divisor. A n immediate consequence of T h e o r e m 3.34 is t h a t i f M is the set of n o n t r i v i a l orbit sizes of a coprime finite group action, then the m a x i m u m distance between any two points i n this graph is 2. F o r a notnecessarily coprime group action, it is s t i l l possible to say something about the corresponding graph: it has at most two connected components. (This theorem of the author and C . Praeger is presented i n C h a p t e r 8,)

Problems 3E 3 E . 1 . L e t A act on G v i a automorphisms, and assume t h a t at least one of A or G is solvable, but do not assume that A and G have coprime orders. If G is n o n t r i v i a l , prove t h a t G contains a n o n t r i v i a l A - i n v a r i a n t p-subgroup for some prime p . H i n t . If P < A is a p-subgroup and C { P ) = 1, show t h a t p does not divide \ G \ . Show i n this s i t u a t i o n that for each prime q, there is an A - i n v a r i a n t Sylow g-subgroup i n G. G

3 E . 2 . L e t A act v i a automorphisms on G, and assume t h a t ( \ A \ , \ G \ ) = 1 and t h a t at least one of A or G is solvable. Suppose t h a t G = H K , where

3F

107

H a n d K are A - i n v a r i a n t subgroups. Show t h a t C — ( C ( ~ ) H ) ( C r \ K ) , C = C {A).

where

G

N o t e . W e a r i n g magic glasses, H and K are visible. T h e fact t h a t G — H K means t h a t every element of G is the p r o d u c t of a n element of H w i t h an element of K . T h i s p r o b l e m asserts that this is visible t h r o u g h A-glasses. E v e r y element of G that we can see is the p r o d u c t of a visible element of H w i t h a visible element of K . 3 E . 3 . L e t A act on G v i a automorphisms, a n d let N < G be A - i n v a r i a n t . A s s u m e t h a t [ \ A \ , \ N \ ) = 1 a n d t h a t A acts t r i v i a l l y on b o t h N a n d G/N. Show t h a t A acts t r i v i a l l y on G. H i n t . N o solvability assumption is needed here because it is no loss to assume t h a t A is cyclic. 3 E . 4 . In the usual coprime action s i t u a t i o n w i t h A a c t i n g on G, let C = C { A ) . If H C G is A - i n v a r i a n t , show t h a t \ H : H n G | divides | G : C | a n d t h a t | G : C n B | divides \ G : H \ . G

H i n t . Use L e m m a 3.32. 3 E . 5 . L e t G be p-solvable, a n d assume t h a t C y ( G ) = 1. L e t P = O ( G ) and w r i t e F = $ ( P ) , so that F < G. N o t e t h a t there is a n a t u r a l action of G on P / F , a n d observe that P is contained i n the kernel of this a c t i o n since P / P is abelian. Show that P is the full kernel of this action. p

N o t e . I n other words, G / P acts faithfully on P / F , a n d so G / P is isom o r p h i c to a subgroup of A u t ( P / P ) . A l s o , P / F is an elementary abelian p-group, of order p , say, a n d so it is isomorphic to the additive group of an n - d i m e n s i o n a l vector space over a field of order p . It follows that G / P is isomorphic to a group of invertible n x n matrices over this field. n

3F We close this already long chapter on split extensions w i t h a discussion of certain extensions t h a t are not generally split. G i v e n a group N a n d a positive integer m, we w i s h to construct a l l (up to isomorphism) groups G , where G has a n o r m a l subgroup 7V = N such t h a t G/N is cyclic of order m. A s was the case when we discussed semidirect p r o d u c t s earlier i n this chapter, our "synthesis" w i l l be preceded by some "analysis" that w i l l enable us to understand groups of the type we wish to construct. 0

0

Suppose N < G a n d t h a t G / N is cyclic of order m. L e t g N be a coset t h a t generates the cyclic group G/N, a n d observe t h a t g G N . It should be clear t h a t the element g of N a n d the a u t o m o r p h i s m of N i n d u c e d by g m

m

108

3. S p l i t

Extensions

are relevant to this situation, and i n fact, these two objects, along w i t h the group TV and the integer TO uniquely determine G up to isomorphism. In the following, we w i l l again use the convention t h a t i f N = N , then x £ N 0

and x

0

3.35.

GN

are corresponding elements under the given isomorphism.

0

Theorem.

Let G and G

where G / N and GQ/NQ

be g r o u p s ,

0

a r e c y c l i c of

finite

and letN < G and N

o r d e r TO. Suppose

an isomorphism from N to N ,

a n d l e t g N a n d g N r j be g e n e r a t i n g

of G / N a n d G Q / N Q

Assume

0

m

0

n

= (ft))™

is

0

cosets

0

respectively.

(9 )o

< G ,

0

that n ^

that 9

and

{x )

=

0

(x )

9 0

0

f o r a l l elements x G N . Then there is a unique i s o m o r p h i s m 9 : G - > G t h a t extends the g i v e n i s o m o r p h i s m N -> 7V a n d t h a t c a r r i e s g t o g . 0

0

Q

l

P r o o f . T h e distinct cosets of N i n G are g N w i t h 0 < i < TO, and so each element u G G is uniquely of the form u = g x l

for some element x G N and

some integer i i n this range. W e are therefore forced to define 9 by setting i

9(u)

= 9 { g x ) = { g o Y x o , and we must show that 9 is indeed an i s o m o r p h i s m

from G onto G . T h e elements of G 0

l

are uniquely of the form ( g o ) x

0

for

0

0 < i < TO, a n d it follows easily t h a t 9 is b o t h injective a n d surjective. It remains to prove that 9 is a h o m o m o r p h i s m .

e(g x) i

= ( g o Y x o for x e N , and w i t h o u t restriction on the integer i. W r i t e

1 = qm + r, where 0 < r < TO. B y hypothesis, g m

q

a n d thus { { g ) x ) dtfx)

=

= {g )

Q

m g

0

r

m

Since ( x ^ ) i

9{g x)9{giy). Now

m

0

{

0

=

i+j

(x ) 0

m

e N and { g )

0

=

m

{g ) , 0

0

9{g {g Yx) = (g ) ((g yx) r

m

x . W e have

j

as wanted. N o w g x g i y = g x° y N.

W e show first that

{ 9 o ) 3

,

0

= (g )

m q + r

0

x

= (gofxo ,

0

for integers i and j a n d elements x , y G i

it is easy to compute t h a t 9 ( ( g x ) ( g i y ) )

T h i s completes the proof.

=

•

suppose t h a t we are given four ingredients: a group TV, a positive

integer TO, an element a G N a n d an a u t o m o r p h i s m a of N . T h e previous theorem guarantees that up to isomorphism, there is at most one group G h a v i n g TV as a n o r m a l subgroup w i t h cyclic factor group G / N of order TO, and h a v i n g generating coset gN, where g But

m

= a and x

9

= x

a

for a l l

x€N.

given N and TO, not every choice of a G N and a G A u t ( / V ) corre-

s p o n d to an actual group. If the group G exists, then since conjugation by g effects the a u t o m o r p h i s m a on N , and a = g a

a

m

is fixed b y g , we must have

= a . A n o t h e r c o n d i t i o n t h a t clearly must be satisfied is t h a t the m t h

power of the a u t o m o r p h i s m a must be the inner a u t o m o r p h i s m induced by a.

A s the following cyclic extension theorem shows, there are no further

requirements.

109

3F

3.36. T h e o r e m .

L e t N be a g r o u p a n d TO a p o s i t i v e i n t e g e r , a n d l e t a e N

a n d a e A u t ( / V ) . Assume

that a

a

= a

and

= x

a

f o r a l l x E N . T h e n t h e r e exists a g r o u p G, u n i q u e u p t o i s o m o r p h i s m , h a v i n g N as a n o r m a l s u b g r o u p w i t h t h e f o l l o w i n g p r o p e r t i e s . (a) G / N m

(b) g (c) x

a

= (gN)

and

i s c y c l i c of o r d e r TO.

= a. =

x3.

P r o o f . T h e uniqueness follows by T h e o r e m 3.35, and so it suffices to construct G, w h i c h we do by realizing it w i t h i n the s y m m e t r i c group on the set ft of ordered pairs ( i , y ) , where 0 < i < TO a n d y G M . F i r s t , we define an action of N on ft by setting ( i , y ) - x = ( i , y x ) , where 0 < i < TO and x,y G N . It is t r i v i a l to check that this defines a faithful action, a n d so we get a corresponding i s o m o r p h i s m N - > 7V C Sym(ft), where x is the p e r m u t a t i o n effected by x for x G N . F o r convenience, we identify N w i t h N v i a this isomorphism, so that N C S y m ( f t ) . 0

0

0

a

N e x t , we define a function g : ft -> ft b y setting ( i , = (t + l , y ) for 0 < i < m - 1 and y e N , a n d (TO - 1, y ) = (0, a y ) . Since a = a and cr is a n a u t o m o r p h i s m , we see that ( i , a y ) = (i + 1, a y ) for 0 < i < m - 1, and it follows easily that a

a

5

a

?

(i, y ) g

m

m

a

= ( i , ay° )

= { i , a y ) = ( i , y a ) = (i, y ) a

for a l l ( i , y ) £ Vl. (To check the first equality, observe t h a t each a p p l i c a t i o n of g increases the first component b y 1 m o d u l o m. W e pick up one e x t r a factor of a when the first component changes from TO - 1 to 0, and that factor is unaffected by a d d i t i o n a l applications of the function g . ) T h u s g = a , and i n particular, g is a p e r m u t a t i o n of Q. m

a

We show next that x g = gx for a l l x G N . T o verify this, it suffices to check t h a t these functions have the same effect on each element ( i , y ) of ft, a n d so we want ( ( i , y ) x ) g = ((i, y ) g ) x . If 0 < i
((i, y ) ) g = ( i , y x ) g = ( i + l , [ y x f ) x

a

a

= (i+ l , y x ) = {{i, y)g)x

a

,

and also ((TO - l , y ) x ) g = (TO - l , y x ) g = ( 0 , a { y x ) ) = ( 0 , a y x ) = (TO - l , y ) g x a

9

a

a

a

,

a

as wanted. It follows t h a t x = x , a n d i n particular, N < G, where G = ( N , g ) . A l s o , G / / V is cyclic, w i t h generating coset g N . m

A l l t h a t remains to check is that G / N has order TO. Since g = a € N , it suffices to show t h a t g i 0 N for 0 < j < TO. B u t ( 0 , 1 ) ^ ' = (j, 1) and

110

3. S p l i t

Extensions

(0, l ) s = (0, x ) for x E N , a n d thus g* cannot be an element x £ N when 0 < j < m. T h i s completes the proof. • A s an example of the use of the cyclic extension theorem, we give another construction of the generalized quaternion groups. W e saw these previously, i n P r o b l e m 3A.2, where they were constructed as index-2 subgroups of semidihedral groups. G i v e n a positive integer n , divisible by 8, we start w i t h a cyclic group N of order n / 2 . L e t a be the unique element of N of order 2 a n d let a £ Aut(AT) be the m a p x x ~ \ so that a fixes a . T a k i n g m = 2, we see that a is the t r i v i a l a u t o m o r p h i s m of N , a n d so it is equal to the inner a u t o m o r p h i s m induced by a . B y the cyclic extension theorem, therefore, the cyclic group N can be embedded as a n o r m a l subgroup w i t h index m = 2 in a group Q , a n d there exists an element g G Q - N such t h a t g = a and x = x ' for a l l x G N . W e argue that every element of Q - N has order 4, a n d thus a is the only element of order 2 i n Q. T o see this, observe t h a t a l l elements of Q - TV have the form gx w i t h x G N , a n d we have ( g x ) = g x g x = g x x = ax~ x = a , which has order 2. T h u s gx has order 4, as claimed. T h e unique group constructed here is denoted Q ; it is the generalized q u a t e r n i o n group of order n . (See the remark concerning nomenclature following P r o b l e m 3A.2.) m

2

9

1

2

2

9

l

n

Problems 3F 3 F . 1 . L e t G be a group of order n divisible by 8, and suppose that N has a cyclic subgroup C of index 2 such t h a t every element of G - C has order 4. Show t h a t G = Q , the generalized quaternion group. n

3 F . 2 . Show that i n principle, we can construct every solvable group by s t a r t i n g w i t h the t r i v i a l group and repeatedly a p p l y i n g T h e o r e m 3.36. 3 F . 3 . L e t Q = Q . Show that Q has exactly three cyclic subgroups of order 4, a n d t h a t these are t r a n s i t i v e l y permuted by A u t ( Q ) . 8

H i n t . If A and B are cyclic subgroups of Q of order 4, use T h e o r e m 3.35 to construct an i s o m o r p h i s m from Q to Q that carries A to B . 3 F . 4 . L e t S = S L ( 2 , 3), w h i c h is the group of 2 x 2 matrices of determinant 1 over the integers m o d u l o 3. Show t h a t | 5 | = 24 and t h a t it has a n o r m a l Sylow 2-subgroup isomorphic to Q . 8

3 F . 5 . L e t S = 5 L ( 2 , 3 ) as i n the previous problem, a n d note t h a t S has index 2 i n G = G L ( 2 , 3), the full group of invertible 2 x 2 matrices over the integers m o d u l o 3. Show t h a t G-S contains an element g of order 2, but

Problems

3F

111

t h a t it contains no element of order 4. P r o v e t h a t there exists a group H containing S as a subgroup of index 2, where H - S contains an element h of order 4 such t h a t the automorphisms of S i n d u c e d by g i n G and by h i n H are identical. Show t h a t H - S contains no element of order 2 a n d finally, show t h a t b o t h G — S and H - S c o n t a i n elements of order 8.

Chapter 4

Commutators

4A We begin w i t h a b i t of review. T h e c o m m u t a t o r of two elements x and y i n a group G is the element x ^ y ^ x y , w h i c h is denoted [ x , y \ . Since [ x , y ] = { y x ) ~ x y , we can t h i n k of the c o m m u t a t o r of x a n d y as the "difference" between yx and xy, and i n p a r t i c u l a r , [ x , y ] = 1 i f and only i f x a n d y commute. A l t e r n a t i v e l y , we can w r i t e [ x , y ] = a r V ' , so [ x , y ] can also be viewed as the "difference" between x and its conjugate x». l

If H C G and AT C G are subgroups, we w r i t e [ H , K ] to denote the subgroup generated by the set { [ h , k] \ h G H , k G AT} of a l l commutators of elements of J f w i t h elements of K . T h i s subgroup is the c o m m u t a t o r of H a n d AT, and it is t r i v i a l if and only if H and K centralize each other. We stress that, i n general, [ H , K ] is not e q u a l to the set of commutators of elements of H w i t h elements of AT; it is the group g e n e r a t e d by t h e m , or equivalents, it is the unique smallest subgroup of G t h a t contains a l l of these commutators. In p a r t i c u l a r , the subgroup [ G , G ] , w h i c h is generated by a l l commutators i n G, need not consist entirely of commutators. R e c a l l t h a t this subgroup is usually denoted G'; it is the derived subgroup (or c o m m u t a t o r subgroup) of G. T h e derived subgroup G' of G is t r i v i a l if and o n l y i f G is abelian, and as we know, it is the unique n o r m a l subgroup of G m i n i m a l w i t h the property t h a t the corresponding factor group is abelian. T h e r e are a number of useful identities satisfied b y commutators. It is t r i v i a l to check, for example, t h a t [ x , y ] [ y , x ] = 1 for a l l x,y £ G. I n other words, [ y , x ] = [ x , y ] ~ \ and so i f H , K C G, t h e n the generating commutators of [AT, H] are exactly the inverses of the generating commutators of [ H , K ] , and it follows that [ K , H ] = [ H , K ] . A n o t h e r often-used identity,

113

4.

114

Commutators

w h i c h the reader should check, is t h a t [ x y , z] = [ x , z ] y [ y , z] for a l l x , y , z E G. (In the language of Section 3 B , this says that for each element z E G , the map z] from G to G is a crossed h o m o m o r p h i s m w i t h respect to the conj u g a t i o n action of G on itself.) T h i s identity is the key, for example, to p r o v i n g the following fact. 4.1. L e m m a . L e t H a n d K be s u b g r o u p s normalize [ H , K ] , o r equivalently, [ H , K } < by H a n d K .

of a g r o u p G. T h e n H a n d K (H, K ) , t h esubgroup generated

P r o o f . Since [ H , K ] = [AT, H ] , it suffices by s y m m e t r y to prove that H C N { [ H , K } ) . L e t h,x e H and k G K . T h e n [ h x , k ] = [ h , k ] [ x , k ] , and so [h,k] = [hx,k][x,k]e [ H , K ] . In other words, conjugation by x e H maps each of the generators [ h , k ] of [ H , K ] back into [ H , K ] , and thus [ H , K ] C [ H , K ) . It follows t h a t [ H , K ] = [ H , K ) , as wanted. (As usual, we substitute x ' for x to deduce the reverse containment.) • x

G

x

1

x

X

1

Not surprisingly, there is a formula for [ z , x y ] analogous to the identity [ x y , z] = [ x , z ] y [ y , z ] , but it seems pointless to memorize it since it can easily be derived using the formula for [ x y , z ] . T o be specific, we have the following: [z,xy] = [ x y , z ] We

1

l

= ( [ x , z ] * [ y , z\y

1

= [y, z \ - \ [ x , z ] ^ ) " = [ z , y ] [ z , x \ y .

have already said t h a t K centralizes H i f and only i f [ H , K ] = 1.

T h i s can be generalized slightly, as follows. 4.2. L e m m a . L e t N be a n o r m a l s u b g r o u p of a g r o u p G, a n d suppose that H , K C G a r e a r b i t r a r y s u b g r o w p s . W r i t e G = G / N a n d f o l l o w t h e standard " b a r c o n v e n t i o n " , so t h a t H a n d K a r e t h e i m a g e s of H _ a n d K i n G u n d e r the canonical h o m o m o r p h i s m G ^ G . Then [H, K] = [H,K], and i n p a r t i c u l a r , H a n d K c e n t r a l i z e each o t h e r i n G if a n d o n l y if [ H , K ] C N . P r o o f . Since overbar is a h o m o m o r p h i s m , we have [ u , v ] = [ u , v ] for arbit r a r y elements u,v G G, and it follows that [ H , K ] = [ H , K \ .

T h i s subgroup

is t r i v i a l precisely w h e n [ H , K ] C N . • We

can also express the relation "AT normalizes H " i n the language of

commutators. 4.3. L e m m a . L e t H a n d K be s u b g r o u p s of a g r o u p G. T h e n K C N ( J i ) if a n d o n l y if [ H , K ] C H . I n p a r t i c u l a r , H < G if a n d o n l y if [ H , G] C H . G

1

k

P r o o f . L e t h € H and k G AT. Since [ h , k ] = h " ^ , we have h = h [ h , k ] , and thus h G H if and only i f [ h , k] e H . T h e result is now immediate. • k

4A

115

If H and K normalize each other, therefore, t h e n [ H , K ] C H n K . If also H n K = 1 , then [ H , K ] — 1, and so H a n d K centralize each other. (We have seen and used this fact previously.) We can use L e m m a s 4.2 and 4.3 to reexamine the n o t i o n of a central series i n a group. R e c a l l that a series of subgroups 1 = H

0

C Hi C ••• C H

= G

r

is a central series i n G i f a l l H i are n o r m a l i n G, and also

i C

Z ( G / H i - i ) for 0 < i < r . B y L e m m a 4.2, the latter c o n d i t i o n can be restated as [ H i , G ] C for 0 < i < r . Conversely, suppose we have a series of subgroups H as above, and assume t h a t [ H G ] C i i ^ i for 0 < i < r . F i r s t , observe that since H i - i C we have [ i ^ , G ] C H and thus L e m m a 4.3 yields H i < G for a l l i . T h e subgroups thus form a n o r m a l series, and indeed they form a central series by L e m m a 4.2. h

U

i

}

To continue this discussion of central series, it is convenient to define commutators of three or more elements or of three or more subgroups. T h e s t a n d a r d n o t a t i o n a l convention is to use "left association". F o r three elements, for example, the triple c o m m u t a t o r is denned by [ x , y , z ] = [ [ x , y } , z ] , and for three subgroups, the definition is [ X , Y , Z ] = [ [ X , Y ] , Z ] . ( C a u t i o n : in general it is not true t h a t [ X , Y, Z] is generated by the elements of the form [ x , y , z] w i t h x e X , y e Y and z G Z . ) M o r e generally, for n > 2, we define [ x i , X , . . . , X ] 2

and

n

[ [ x

1

, x 2 , . . . , X

suppose t h a t { H i

-

1

] , X

n

] .

| 0 < i < r } is a central series for G , as before.

[G,G]

= [ H , G] C

H

r

r

-

U

thus if r > 2, we have [ G , G , G ] = [[G,G],G] C [ H - G ] r

and

n

s i m i l a r l y for subgroups.

Now Then

and

=

U

C P _ , r

2

i f r > 3, [G, G , G , G] = [[G,

G , G], G] C [ F _ , G] C P _ . r

2

r

3

C o n t i n u i n g like this, we see t h a t i f k < r + 1, then [G, G , . . . , G] C 7 f _ where there are k copies of G i n the c o m m u t a t o r o n the left. r

1

f c + 1

,

Clearly, we need some better n o t a t i o n here, a n d so we write G = G , G = [ G , G ] , G = [G, G , G] and i n general, G = [ G ~ \ G ] for k > 1. ( T h i s n o t a t i o n is not completely standard; some authors w r i t e ( G ) to denote our subgroup G , and one can find other notations i n the literature too.) N o t e t h a t G = [G, G] = G', the derived subgroup, but i n general, G w i t h k > 2 is not one of the groups G^" ) of the derived series, and i n p a r t i c u l a r , 2

3

k

k

f c

7

k

2

k

1

116

4.

2

Commutators

k

G< ) = G" = [ [ G , G ] , [ G , G ] ]

is not usually one of our subgroups G .

(We

shall see, however, t h a t G" C G , and i n general, G ^ C G \ ) 4

2

k

k

N o t e t h a t the subgroups G are characteristic i n G . In particular, G < G , a n d so by L e m m a 4.3, we have G = [ G , G ] C G . T h e subgroups G , therefore, form a descending series. ( A n o t h e r way to see this is by i n d u c t i o n on k. A s s u m i n g t h a t G C G ~\ we get G = [ G , G] C [ G , G] = G . ) k

k

+

l

k

k

k

k

+

l

f c

k

f c _ 1

fc

R e t u r n i n g now to the subgroups E of onr central series for G , we have G C F _ for 1 < k < r + 1, and i n particular, G C H = 1. N o w consider an a r b i t r a r y nilpotent group, w h i c h , we recall, s i m p l y means t h a t G has a central series. It follows t h a t G = 1 for some positive integer m. M o r e precisely, i f G is nilpotent and has a central series of length r , then Qr+i ( f containments.) %

k

r

r

+

1

f c + 1

0

m

=

1

T h e

l

e

n

g

t

h

o f

a

s e r i e g

Conversely, suppose that G

i g

t

m

h

e

n

u

m

b

e

r

0

= 1 for some positive integer m. Consider

the series 1 = G

C G

m

m

_

1

C ••• C G

1

= G.

(Note t h a t the n u m b e r i n g of these subgroups is "backwards", r u n n i n g from high to low.) Since [ G , G ] = G \ it follows t h a t this is a central series, and thus G is nilpotent. Furthermore, we have seen t h a t G C J f _ , where the H i form an a r b i t r a r y central series i n G . Since the G are contained term-by-term i n the subgroups of an a r b i t r a r y central series, the series of G is called the lower central series of G . T h e length of this series is, of course m - 1, where m is the smallest positive integer such t h a t G = 1. k

k

+

k

r

f c + 1

k

k

m

R e c a l l from C h a p t e r 1 t h a t the nilpotence class of a nilpotent group G is the smallest integer r such t h a t G is equal to the t e r m Z of the upper central series of G . W e saw t h a t it is also the smallest integer r such that G has a central series { H i | 0 < i < r } , w i t h H = 1 and H = G. (In other words, the nilpotence class is the length of the shortest possible central series.) W e now k n o w t h a t if G is nilpotent of class r then G = 1. A l s o , it is not possible t h a t G = 1 w i t h m < r + 1 because otherwise, the lower central series of G w o u l d be a central series of length m - 1 < r . T h i s shows t h a t if G is nilpotent of class r , then m = r + 1 is the smallest integer such t h a t G = 1. r

0

r

r

+

1

m

m

O f course, i f G is an a r b i t r a r y (and not necessarily nilpotent) group, one can still consider subgroups of the form G = [G, G , . . . , G ] , where G occurs m times. B y abuse of n o t a t i o n , the series { G \ i > 1} is often called the "lower central series" of G , even though i n the nonnilpotent case, it is not a true central series because the identity is not one of its terms. If G is finite, then since its lower central series cannot involve infinitely m a n y strict containments, there must be some superscript m such t h a t G = G . It is then immediate that G = G for a l l integers n > m, and this final t e r m of the lower central series of G is denoted G ° ° . It is not h a r d to see m

i

m

m

n

m

+

1

4A

111

t h a t if_/V < G , then G / N is n i l p o t e n U f a n d o n l y i f G°° C TV. (To see this, write G = G / N , and observe t h a t G™ = G by repeated a p p l i c a t i o n of L e m m a 4.2.) It follows t h a t if G is finite, t h e n G ° ° is the unique n o r m a l subgroup of G m i n i m a l w i t h the property t h a t the corresponding factor group is nilpotent. m

A s we mentioned i n C h a p t e r 1, the groups of nilpotence class 1 are exactly the n o n t r i v i a l abelian groups. Somewhat more interesting are groups of nilpotence class 2. F o r these nonabelian groups, we have 1 = G = [G, G , G] = [ G , G], so t h a t the derived subgroup G' C Z ( G ) , or equivalently G / Z ( G ) , is abelian. I n this case, a l l c o m m u t a t o r s are central, a n d hence [ x y , z ] = [ x , z \ y [ y , z \ = [ x , z ] [ y , z \ , and so the m a p [ - , z ] defines a homomorp h i s m from G into G ' . T h e following easy consequence of this is sometimes useful. (Before we state our result, we recall t h a t the exponent of a group G is the smallest positive integer n such t h a t x = 1 for a l l x e G . I n other words, the exponent of a group is the least c o m m o n m u l t i p l e of the orders of its elements.) 3

1

n

4.4. L e m m a . L e t P be a p - g r o u p of n i l p o t e n c e class has

e

exponent p .

T h e n t h e e x p o n e n t of P / Z ( P )

2, a n d assume

that P '

e

divides p . I n particular,

if P ' i s e l e m e n t a r y a b e l i a n , t h e n P / Z ( P ) i s e l e m e n t a r y a b e l i a n , a n d $(P)

thus

C Z(P).

N o t e t h a t i f P has class 2, then P ' C Z ( P ) , and so b o t h P ' and P / Z ( P ) are abelian. T h u s P ' is elementary abelian if a n d o n l y if e = 1, and to prove t h a t P / Z ( P ) is elementary abelian, it suffices to show t h a t its exponent divides p , and this is the assertion of the l e m m a w h e n e = 1. T h e last conclusion of the l e m m a , therefore, is an i m m e d i a t e consequence of the characterization of the F r a t t i n i subgroup of a p-group as the unique smallest n o r m a l subgroup h a v i n g an elementary abelian factor group. T h i s fact was mentioned previously, but we never gave a formal proof, a n d so we digress to do so now. 4.5. L e m m a . L e t N < P , w h e r e P i s a p - g r o u p .

Then P / N is elementary

a b e l i a n if a n d o n l y i / $ ( P ) C N . P r o o f . L e t M < P be a m a x i m a l subgroup. Since P is nilpotent, M < P , and we see t h a t P / M has no subgroups other t h a n itself a n d the identity. T h u s P / M has p r i m e order, w h i c h must be p . N o w P / M is abelian, and so P ' C M , and also, given x e P , we have P e M . T h u s P ' and x? lie i n the intersection of a l l m a x i m a l subgroups of G , or i n other words, they lie i n $(P). W e conclude t h a t P / $ ( P ) is elementary abelian, a n d thus also P / N is elementary abelian i f $ ( P ) C TV C P . X

Conversely, suppose P / N is elementary abelian, a n d assume t h a t N < P . B y the fundamental theorem of abelian groups, P / N is a direct p r o d u c t of

4.

118

Commutators

n o n t r i v i a l cyclic subgroups, each of w h i c h must have order p . T h e product of a l l but one of these factors has index p , and so is m a x i m a l i n P / N . T h e intersection of those m a x i m a l subgroups of P / N obtained by deleting one of its direct factors of order p is t r i v i a l , and thus N is the intersection of some collection of m a x i m a l subgroups of P . W e conclude t h a t $ ( P ) C TV, as wanted. • e

P r o o f of L e m m a 4.4. T o prove t h a t P / Z ( P ) has exponent d i v i d i n g p , we must show t h a t x ^ £ Z ( P ) for a l l i e P . W e have [ x P , y ] = [ x , y ] = 1, where the first equality holds since [ - , y \ is a h o m o m o r p h i s m , and of course, the second equality holds because [ x , y ] G P ' , w h i c h has exponent p . T h i s shows t h a t x P commutes w i t h a l l elements y G P , and so x ? G Z ( P ) , as required. B y L e m m a 4.5, the proof is now complete. • e

p B

e

e

e

N e x t , we present a result relating centers and c o m m u t a t o r subgroups. Its proof is based o n a n exception to the general principle t h a t elements of the c o m m u t a t o r subgroup need not be commutators. 4.6.

Lemma.

suppose

L e t A be a n a b e l i a n n o r m a l s u b g r o u p

t h a t G/A

i s c y c l i c . Then

G' = [ A , G ]

of a g r o u p G, a n d

and

G' = A / [ A n Z ( G ) ) . I n p a r t i c u l a r , if A i s

finite,

t h e n \ A \ = \ G ' \ \ A n Z(G)|.

P r o o f . Choose a generating coset A g for G / A ,

and let 9 : A - > A be the

m a p defined by 9 ( a ) = [ a , g ] . (Note that [ a , g] G A since A < G.) If a , b G A , then b

9(ab) = [ab,g] = [a,g] [b,g]

= [a,g][b,g]

= 9(a)9(b),

where the t h i r d equality holds since [a,g] and b lie i n the abelian group A . T h u s 9 is a h o m o m o r p h i s m , and C ( g ) . A l s o , 9 ( A ) is a subgroup, and 9 ( A ) C [ A , G] C G'.

ker(0) =

A

Since A g is a generating coset for G = G / A , we see t h a t J g ) = (g) = G, and thus G = A ( g ) , and it follows t h a t C ( g ) C A n Z ( G ) . T h e reverse inequality is clear, and thus C ( g ) = A n Z ( G ) . T h i s yields 0 ( 4 ) = A / ( A n Z ( G ) ) , a n d so to complete the proof, it suffices to show t h a t 9 ( A ) = G'.

ker(0) =

A

A

We already k n o w t h a t 9 ( A ) is a subgroup of G', and so it suffices to show t h a t 9 ( A ) < G and t h a t G / 9 ( A ) is abelian. F i r s t , since A is abelian and 9 ( A ) C A , it is clear that A C Also, 9 ( a ) = [ a , g } 3 = G [ a 9 , g [ = 9 ( a 9 ) G 9 ( A ) for a l l a G A , and thus G It follows t h a t G G = A( ) C and so 0(A) < G , as required. G

N (0(A)).

N (0(A)),

5

Now

5

9

N (0(A)).

(redefining overbars) write G = G / 9 ( A ) , so t h a t G = A j g ) . Since

[ a , g ] = [a, ] = 9 ( a ) is t r i v i a l for a l l a G A , it follows t h a t the elements o for 5

4A

119

a G A together w i t h g form a pairwise c o m m u t i n g set of generators for G, w h i c h is therefore abelian. T h i s completes the proof. • A s an example of how L e m m a 4.6 c a n be used, we offer the following. 4.7. T h e o r e m . L e t A < P , w h e r e P i s a p - g r o u p , a n d suppose that A is a b e l i a n a n d \ A \ = p . Assume that P / A is cyclic, a n d that \A n Z(P)| = p . T h e n t h e n i l p o t e n c e class of P i s TO. m

P r o o f . L e t Z = A n Z ( P ) , and note t h a t P ' C A a n d \ A : P ' \ = \ Z \ = p by L e m m a 4.6. If TO = 1, then A = Z is central i n P , a n d since P / A is cyclic, it follows t h a t P is abelian, and thus its nilpotence class is 1 = TO, as wanted. We can thus assume t h a t TO > 1, and we proceed b y i n d u c t i o n on TO. Since \ A : P ' \ = p and \ A \ > p , we have P ' > 1, and thus P ' meets Z ( P ) n o n t r i v i a l l y b y T h e o r e m 1.19. T h e n 1 < P ' n Z ( P ) C A n Z ( P ) = Z , and since | Z | = p, we have Z = P ' n Z ( P ) , _ a n d i n p a r t i c u l a r , Z C P ' . N o w w r i t e P = P/Z,_and_observe t h a t ( P ) ' = P ' has index p i n the abelian n o r m a l subgroup A of P . A l s o , ~ P / A = P / A is cyclic, and hence b y L e m m a 4.6, we have \ A n Z ( P ) \ = \ A : ( P ) ' | = p, and thus P satisfies the hypotheses of the theorem. m

Since \ A \ = p " \ we conclude from the i n d u c t i v e h y p o t h e s i s t h a t P has nilpotence class_m - 1, a n d thus i n the lower central series of P , we have (P) = 1 and ( P ) " " ^ 1. Since P = ( P ) = 1, we have P C Z , and thus P = [ P , P ] C [ Z , P ] = 1. m

7

m

+

1

1

m

m

m

m

m

pm

To complete the proof, we must show t h a t P ^ 1. If TO = 2, t h e n _ p i ^ j p t h a t TO > 2, a n d hence P CP = m

>

a n (

g

o

w

e

c

a

n

S

U

p

o

s

_

1

2

e

P ' C A . W e have P ^ = ( P ) ™ " ^ 1, and thus P ™ " ^ = 4 n Z ( P ) . It follows t h a t P ~ % Z ( P ) , and thus P = [ P - \ P ] ^ 1, as required. • 3

m

1

1

1

l

m

m

e

Suppose now t h a t P is a nonabelian p-group of order p . A s s u m e t h a t | Z ( P ) | = p, and t h a t P has an abelian subgroup A of index p (which, of course, is necessarily normal). Since P is n o n a b e l i a n , A is n o n t r i v i a l , and thus A n Z ( P ) > 1, and hence Z ( P ) C A . A l s o , P / A is cyclic, and it follows by T h e o r e m 4.7 t h a t that the nilpotence class of P is e - 1. To put this into context, let P be an a r b i t r a r y p-group of nilpotence class r , and order p , where e > 2. Since P has at least one n o r m a l subgroup of index p and the corresponding factor group is abelian, it follows t h a t \ P : P ' \ > p . N o w consider the lower central series e

2

2

P = P

1

2

> P

2

> P

3

> ••• > P 2

r

+

1

= 1, l

m

and observe t h a t \ P : P | = | P : P ' | > p , and of course, \ P : P | > p for 2 < i < r . Then p = | P : P | is the p r o d u c t of r integers, each of w h i c h is at least p, and one of w h i c h is at least p . It follows t h a t e > r + 1, and e

r

+

1

2

4.

120

Commutators

e

thus the nilpotence class r of a p-group P of order p w i t h e > 2 is at most e - l . If it happens t h a t r = e - 1, we say that P has m a x i m a l c l a s s . e

2

If \ P \ = P > P and | Z ( P ) | = p and P has a n abelian subgroup of index p , we have seen t h a t P has nilpotence class e - 1, and thus P has m a x i m a l class. I n particular, it is easy to see t h a t d i h e d r a l , generalized quaternion and semidihedral 2-groups satisfy these hypotheses, and so a l l of these groups have m a x i m a l class. I n fact, these are the only m a x i m a l class 2-groups, b u t we w i l l not prove t h a t here. A l s o , we m e n t i o n w i t h o u t proof t h a t for p > 2, there exist m a x i m a l class p-groups that fail to have an abelian subgroup of index p . We close this section w i t h a discussion of the set of elements x (E P such t h a t x = 1, where P is a p-group. I n general, these elements do not form a subgroup. (If P = D , for example, there are exactly six such elements i n c l u d i n g the identity, and of course a set of six elements i n a group of order 8 cannot be a subgroup.) T h e group generated by a l l elements x £ P such t h a t x P = 1 is denoted fii(P), and more generally, for integers r > 1, one defines ft (P) = ( x G P \ x " = 1). (There does not seem to be any standard name for this characteristic subgroup, which is usually just called "omega-r" of P . ) p

8

p

r

If p > 2 and P is a p-group of class at most 2, then the elements x G P such t h a t x = 1 a c t u a l l y do form a subgroup. (Equivalently, S l ^ P ) has exponent d i v i d i n g p i n this case.) A s is shown by D , w h i c h has class 2, the c o n d i t i o n p > 2 is essential. B u t the assumption t h a t the nilpotence class is at most 2 is unnecessarily strong; i n fact, it suffices to assume t h a t the nilpotence class is less t h a n the p r i m e p. T h e proof of this result of P h i l i p H a l l uses "commutator collection", which is the basic idea t h a t underlies our proof for class 2 groups. T h e proof of the full result is considerably more complicated, however, and we do not present it here. p

8

p

G i v e n x , y ( E P w i t h P = 1 = y P , we want to show t h a t ( x y ) = 1, and so we investigate how to compute ( x y ) for a r b i t r a r y elements x,y G P and positive integers n . T h e idea here is as follows. W e have X

n

( x y )

n

= x y x y x y

• • • xy

where on the right, there are n copies of x alternating w i t h n copies of y . If P is abelian, it is t r i v i a l to simplify this. W e can move each copy of x as far as necessary to its left, freely passing t h r o u g h the intervening copies of y , thereby collecting the n copies of x at the left, and leaving n copies of y on the right. T h i s , of course, yields the familiar formula ( x y ) = x y for abelian groups. n

yx

n

n

W h a t if P is not abelian? In general, i n any group, i f we have the string and we want to move the x to the left, across the y , we can use the

4A

121

identity yx = x y [ y , x } . I n other words, we c a n s t i l l pass the x t h r o u g h the y , but to do so, we must pay a price. T h e p e n a l t y is t h a t we must insert the c o m m u t a t o r [y, x] to the right of the y . For simplicity, consider the case n = 3. M o v i n g the second x left across the first y (and inserting [ y , x ] a a payment for the privilege of doing this) we have {xyf

= xyxyxy

= xxy[y, x]yxy .

N e x t , we want to move the t h i r d x leftward i n order to j o i n the first two, and we do this i n three steps, first m o v i n g across y , then across [ y , x ] , a n d finally across another y . T o pay for these transits, we must insert [y, x ] , a n d then [ y , x , x ] , and t h e n another [ y , x ] . W e thus have [xyf

= xyxyxy

= xxy[y,

x\yxy

=

xxy[y,x}xy[y,x}y

=

xxyx[y,x}[y,x,x]y[y,x}y

= x x x y [ y , x] [y, x] [y, x , x } y [ y , x ] y , and at this point, we have collected a l l three copies of x at the left, a n d these are followed by a m i x t u r e consisting of three copies of y , three copies of [y, x ] , a n d one copy of [y, x , x } . We can now continue to "simplify" this, b y collecting the three copies of y t o w a r d the left, to a p o s i t i o n just to the right of the three copies of x . I n order to do this, we w i l l need to move the second y across copies of [y, x] a n d [ y , x , x ] , a n d so we must pay penalties of [ y , x , y ] a n d [ y , x , x , y ] . T o collect the t h i r d y , our penalties w i l l be [ y , x , y ] , [ y , x , x , y ] , a n d also [ y , x , y , y ] a n d [y,x,x,y,y\. W h e n this process is finished, we w i l l have ( x y f = x y w , where w is some c o m p l i c a t e d product of one or more copies of each of [ y , x ] , [ y , x , x ] , [ y , x , y ] , [ y , x , x , y ] and [ y , x , x , y , y ] . 3

3

W e could continue. T h e simplest c o m m u t a t o r appearing as a factor of w is [y, x ] , a n d there are three of these. Suppose we collect these leftward, to a p o s i t i o n just to the right of the three copies of y . T o do this, we w i l l need to pay "transit fees" b y inserting commutators like [ [ y , x , x ] , [y,x]] a n d others. T h i s w i l l result i n a formula of the form ( x y ) = x y [ y , x ] w ' , where w' is a p r o d u c t of commutators of "weight" 3 a n d higher. T h i s seems like quite a mess, but it is possible to do a l l of this i n a systematic way, a n d to keep t r a c k of everything, a n d this is the H a l l " c o m m u t a t o r collection process", w h i c h we w i l l not pursue further. 3

3

3

3

If P has class at most 2, everything becomes m u c h easier. I n t h a t case, the c o m m u t a t o r [y, x] is central, a n d so every copy of [y, x] can be moved to the far right, where it w i l l not interfere w i t h further collection. W e s i m p l y need to count how m a n y copies of [ y , x ] we produce as we collect the n

4.

122

Commutators

copies of x leftward i n ( x y f . T h e leftmost x starts i n its desired p o s i t i o n , and it need not be moved. T h e second x moves across one y , i n c u r r i n g one payment of [ y , x ] ; the t h i r d x moves across two copies of y generating two a d d i t i o n a l copies of [y, x } ; the fourth x generates three more copies of [y, x ] , and so o n . I n t o t a l , therefore, we o b t a i n 1 + 2 + • • • + (n - 1) = (n - l ) n / 2 copies of [ y , x ] , a n d this establishes the formula (xy)

n

n

=

n

n

2

x y \ y , x ] ^ - ^ ,

in a nilpotent group w i t h class at most 2. Two

useful consequences of this formula are given i n the following result.

4.8. T h e o r e m . L e t P be a p - g r o u p w i t h n i l p o t e n c e assume t h a t p > 2 . The f o l l o w i n g then hold. (a) The set { x e P \ x P = 1} i s a s u b g r o u p (b) / / [ y , x ] P = 1 f o r a l l x , y E P , p h i s m f r o m P into itself

then

class

a t most

2,

and

of P .

t h e map

x ^

x

p

isa

homomor-

p

P r o o f . F i r s t , assume t h a t [y, x } = 1 for a l l x , y E P . Since the nilpotence class of P is at most 2, we c a n a p p l y our c o m m u t a t o r collection formula to conclude t h a t { x y ) = x y P [ y , x ] , where m = p ( p - l ) / 2 . Since p ^ 2, however, m is a m u l t i p l e of p , a n d thus [ y , x } = 1. It follows that ( x y ) P = x P y P , a n d thus the m a p i ^ ^ i s a group h o m o m o r p h i s m , proving (b). p

p

m

m

Now (without assuming t h a t p t h powers of c o m m u t a t o r s i n P are t r i v i a l ) observe t h a t since a l l c o m m u t a t o r s i n P are central, the m a p [ - , x ] from P to P ' is a group h o m o m o r p h i s m for every element x E P . If y E P a n d y P = 1, therefore, we have [ y , x }

p

p

= [y ,x]

= [l,x]

= 1 for a l l x E P.

In

particular, i f b o t h ^ = 1 a n d / = 1, then c o m m u t a t o r collection yields [ x y ) P = x y P [ y , x ] = 1, where as before, m = p ( p - l ) / 2 is a m u l t i p l e of p because p ^ 2. T h e set { x e P \ x P = 1} is thus closed under m u l t i p l i c a t i o n , and so it is a subgroup. • p

m

Observe t h a t the conclusion of T h e o r e m 4.8(a) is equivalent t o saying t h a t P = 1 for every element x 6 X

fti(P).

Problems 4 A 4 A . 1 . Suppose that G = A B , where A a n d B are abelian subgroups. Show t h a t G' = [ A , B } . 4 A . 2 . L e t H , K a n d L be subgroups of order 2 i n some group G, a n d observe that the set { [ h , k , l ] \ h <E H , k <E K , I <E L } contains at most one n o n i d e n t i t y element, a n d so it generates a cyclic group. N o w let G = A , the 5

Problems 4 A

123

a l t e r n a t i n g group of degree 5. Show t h a t it is possible to choose subgroups H , i f a n d L of order 2 such t h a t [ P , i f , L] = G. H i n t . L e t P C G have order 5. Choose H , i f C N ( P ) w i t h H ^ K , and choose L ^ N ( P ) . Show t h a t [ P , i f ] = P a n d observe t h a t ( P , L ) = G. G

G

N o t e . T h i s p r o b l e m shows that [ P , i f , L] is not necessarily generated by elements of the form [ h , k, 1} w i t h h e H , k e i f a n d / e L . 4 A . 3 . Say that a group Q is q u a s i q u a t e r n i o n i f Q = C U , where C a n d P are cyclic subgroups, w i t h C < Q and C n U = Z ( Q ) . (Note that generalized quaternion a n d semidihedral groups are quasiquaternion.) Suppose t h a t G / Z is quasiquaternion, where Z C G' n Z ( G ) . Show t h a t Z = 1. H i n t . L e t A and B be the subgroups of G = G / Z corresponding to C a n d U of the definition of "quasiquaternion". Show t h a t A n P = Z ( G ) and a p p l y L e m m a 4.6. N o t e . G i v e n any group Q, consider a b e l i a n groups Z t h a t can be isomorp h i c a l l y embedded i n some group G i n such a way t h a t Z C G' D Z ( G ) a n d G / Z = Q. It is a fact t h a t up to i s o m o r p h i s m , there is a unique largest such group Z , depending on Q; it is called the Schur m u l t i p l i e r of Q. T h e h o m o m o r p h i c images of the Schur m u l t i p l i e r are a l l of the groups Z t h a t satisfy our conditions. T h i s p r o b l e m asserts t h a t quasiquaternion groups have t r i v i a l Schur m u l t i p l i e r s . ( A l i t t l e more i n f o r m a t i o n about Schur multipliers appears i n C h a p t e r 5.) 4 A . 4 . A p-group P is said to be e x t r a s p e c i a l i f P ' = Z ( P ) has order p . Show t h a t i f P is extraspecial, then P / Z ( P ) is elementary abelian, or equivalents, Z ( P ) = P ' = $ ( P ) . N o t e . N o n a b e l i a n p-groups of order p

3

are extraspecial.

4 A . 5 . L e t P be a n extraspecial p-group a n d let x , y € P w i t h xy + yx. L e t U = ( x , y )

a n d V = C (U). P

(a) Show t h a t \ P : C ( x ) \ = p . P

3

(b) Show t h a t \ U \ = p . (c) Show t h a t \ P : V \ =

2

p .

(d) If P > U, show that Z ( V ) = Z ( P ) a n d t h a t V is extraspecial. (e) Show t h a t | P : Z ( P ) | = p

2 e

for some integer e.

H i n t . Use i n d u c t i o n on | P | for e. N o t e . F o r every o d d integer n > 1 and every p r i m e p, there are e x a c t l y two i s o m o r p h i s m types of extraspecial p-groups of order p . If p ^ 2, these groups c a n be distinguished by the fact t h a t one of t h e m has exponent p and the other has exponent p . n

2

124

4.

Commutators

4 A . 6 . L e t P be a p-group h a v i n g m a x i m a l class, a n d let N < P , where \ N : P \ > p . Show t h a t N is one of the terms of the lower central series for P , a n d i n p a r t i c u l a r , N is the only n o r m a l subgroup of G h a v i n g order | W | . 2

4 A . 7 . F i x a p r i m e p a n d let A be the external direct p r o d u c t of p copies of a cyclic group C of order p w i t h n > 0. V i e w the elements of A as p-tuples of the form ( c c , . . . , c ) w i t h a E C. L e t U = ( u ) be cyclic of order p , and let U act on A by c y c l i n g the components. I n other words, n

1 ;

2

p

(ci,c ,...,Cp-i,c ) 2

p

u

= (c ,ci,c ,. p

2

. . , C

p

- l )

.

F i n a l l y , let P = A x P be the semidirect p r o d u c t , a n d view A a n d U as subgroups of P . Show that the m a x i m u m of the orders of the elements of P is e x a c t l y p . n

+

1

N o t e . O f course, P is just the regular w r e a t h p r o d u c t P = C \ U . n = 1, we get a n example where

fti(P)

Taking

does not have exponent p .

4 A . 8 . A s s u m e the n o t a t i o n of P r o b l e m 4 A . 7 w i t h n > 0. (a) Show that C ( U ) = Z ( P ) is the set of p-tuples i n A such t h a t a l l A

components are equal. (b) Show t h a t [ A , U] = P ' is the set of p-tuples i n A such t h a t the p r o d u c t of the components (computed i n C ) is the identity. (c) Show | Z ( P ' P ) | = p . (d) If n = 1, show t h a t P has m a x i m a l class, a n d t h a t i n any case, P ' U has m a x i m a l class. 4 A . 9 . L e t G = N A , where N < G a n d A C G, a n d let M be the final t e r m in the series TV D [ N , A ] D [ N , A , A ] D [ N , A , A , A ] D • • •. (a) Show t h a t A «

G if and only if M C A .

(b) If M C A , show that M C A°°. n

4 A . 1 0 . L e t P be a p-group w i t h | P : Z ( P ) | < p . Show t h a t |P'|

<

n

n

1

2

p ( - )/ .

H i n t . Induct o n n . If P is nonabelian, choose Q so t h a t Z ( P ) C Q C P w i t h \ P : Q \ = p . G e t a b o u n d on \ Q ' \ a n d t h e n a p p l y L e m m a 4.6 to the group P / Q ' . N o t e t h a t { P / Q J =

P'/Q'.

N o t e . If the center of a group has s m a l l index, t h e n the group is i n some sense "nearly a b e l i a n " , a n d so it should be no surprise t h a t the derived subgroup is not too b i g .

4B

125

4 A . 1 1 . L e t G = A I H be the regular w r e a t h p r o d u c t of an abelian group A w i t h a n a r b i t r a r y finite group H , a n d view the base subgroup B as the group of functions / : H A w i t h pointwise m u l t i p l i c a t i o n . L e t K C H . Show t h a t [ B , K ] is exactly the set of functions f e B w i t h the property t h a t the product of the values of / over each left coset of K i n H is the identity. Deduce that \ [ B , K } \ = \ A \ \ \ - \ - \ . H

H

K

h

l

H i n t . T h e action of H o n B is given b y f ( x ) = f { x h ~ )

for h € H .

4 A . 1 2 . G i v e n a finite group H , let T ( H ) denote the set of subgroups T of H m a x i m a l w i t h the property that T is abelian a n d c a n be generated by two elements. L e t G = A I H as i n P r o b l e m 4 A . 1 1 , where A is abelian, a n d let B be the base group. (a) Show t h a t if x,y e G a n d [ x , y ] € B , then [ x , y ] € [ B , T ] for some member T E T ( H ) . (b) Show that i f

t h e n G' D B contains elements t h a t are not commutators i n G. (c) If H is not a n abelian group t h a t c a n be generated b y two elements, show that there exists some number m such t h a t the inequality of (b) holds whenever | A | > m. I n p a r t i c u l a r , show that we can take m = 3 if H = D . 8

4 A . 1 3 . L e t G be nilpotent group w i t h class m > 1, a n d let a € G. Show that the nilpotence class of the subgroup H = G ' ( a ) is less t h a n m . 4B

w h i c h holds for every three elements of every group. Its proof, w h i c h proceeds by expansion a n d cancellation, is completely elementary a n d u t t e r l y routine. It is a c o m p u t a t i o n t h a t is fun to c a r r y out, however, a n d we urge the reader to do so. One can t h i n k of the H a l l - W i t t formula as a k i n d of three-variable version of the m u c h more elementary two-variable identity, [ x , y ] [ y , x ] = 1. T h i s observation hints at the p o s s i b i l i t y t h a t a corresponding four-variable formula might exist, but i f there is such a four-variable identity, it has yet to be discovered. We digress briefly to m e n t i o n that the H a l l - W i t t identity is analogous to the somewhat simpler J a c o b i identity for a d d i t i v e commutators i n a ring.

126

4.

Commutators

(We are assuming associativity of m u l t i p l i c a t i o n , of course.) If x,y £ R, where R is a ring, the additive c o m m u t a t o r of x and y is xy - yx, w h i c h is zero precisely w h e n x a n d y commute. Unfortunately, the s t a n d a r d n o t a t i o n for the a d d i t i v e c o m m u t a t o r of x and y i n a ring is e x a c t l y the same as the n o t a t i o n for group commutators: [ x , y ] . B u t we w i l l not refer to additive c o m m u t a t o r s outside of this paragraph, and so we hope t h a t this a m b i g u i t y w i l l not cause any confusion. T h e J a c o b i identity asserts t h a t [ x , y , z\ + [ y , z , x ] + [ z , x , y ] = 0 for a l l x , y , z £ R, where R is a ring, a n d where as w i t h groups, the triple commutators are denned b y left association. W e m e n t i o n t h a t a (nonassociative) r i n g L w i t h m u l t i p l i c a t i o n x-y is a L i e r i n g if x - x = 0 for a l l x £ L a n d the J a c o b i identity ( x - y ) - z + ( y - z ) - x + ( z - x ) - y = 0 holds for a l l x , y , z £ L . In p a r t i c u l a r , every associative r i n g is a L i e r i n g w i t h respect to the m u l t i p l i c a t i o n x-y = [ x , y ] . In fact, L i e rings provide a powerful t o o l for the study of p-groups, but we w i l l say no more about t h a t here. O u r p r i n c i p a l applications of the H a l l - W i t t i d e n t i t y rely on the following l e m m a a n d its corollary, b o t h of w h i c h are referred to as the "threesubgroups l e m m a . " 4.9. L e m m a . L e t X , Y a n d Z be s u b g r o u p s of a n a r b i t r a r y g r o u p G, a n d suppose

t h a t [ X , Y, Z] = 1 a n d [ Y , Z , X ] = 1 . T h e n [ Z , X , Y ] = 1 .

P r o o f . W e want to show t h a t [ [ Z , X ] , Y ] = 1, or equivalently, that every element of the group [ Z , X ] commutes w i t h every element of Y. F o r this purpose, we show t h a t a l l commutators [ z , x] for z £ Z a n d x £ X centralize each element y £ Y. T h i s is sufficient because C ( y ) is a subgroup of G, and so i f it contains a l l of the commutators [ z , x ] , then it contains the subgroup [ Z , X ] generated by these commutators. It is, therefore, enough to show t h a t [ z , x , y ] = 1, for a l l x £ X , y £ Y and z £ Z . ( B u t recall t h a t i n general, these commutators do not generate [ Z , X , Y ] . ) Equivalently, it suffices to show t h a t [ z , x ' \ y ] = 1 for a l l x , y and z i n X , Y a n d Z , respectively. G

1

N o w [ x ^ y - ] £ [ X , Y ] , a n d so [ x , y - \ z ] £ [ X , Y , Z ] = 1, and we have [ x , y ' \ z } y = 1. Similarly, [ y , - \ x ] = 1, and thus by the H a l l - W i t t identity, [ z , x - \ y } = 1. T h u s [ z , x - \ y ] = 1, as wanted. • z

Z

x

4.10. C o r o l l a r y . L e t N be a n o r m a l s u b g r o u p of a g r o u p G, a n d l e t X , Y , Z C G be a r b i t r a r y s u b g r o u p s . I f [ X , Y , Z ] C N a n d [ Y , Z , X ] C N , t h e n [ Z , X , Y] C N. P r o o f . L e t G = G / N a n d follow the s t a n d a r d L e m m a 4.2, we have [ H j C \

= [H,K]

"bar convention".

Then [X,Y,Z]

By

for a l l subgroups H and K of G.

= [[X,Y],Z] = [\X7Y],Z\

= [X,Y,Z]

= 1

4B

127

and and

similarly, [ Y , Z , X ] = 1. B y L e m m a 4.9 then, 1 = [ Z , X , Y] = [ Z , X , Y] thus [ Z , X , Y ] C N . U

A n appeal to the three-subgroups l e m m a i n a proof often makes it possible to avoid tedious and messy elementwise c o m m u t a t o r calculations. U n fortunately, it is not always easy to see how to a p p l y this very useful tool. We give a number of applications here a n d i n what follows, a n d these s h o u l d provide plenty of practice. F i r s t , we o b t a i n some a d d i t i o n a l i n f o r m a t i o n about the lower central series. n

4.11. T h e o r e m . As u s u a l , w r i t e G t o d e n o t e t h e n t h t e r m of t h e l o w e r c e n t r a l series of a g r o u p G. T h e n [ G \ GP] C G i f o r i n t e g e r s i , j > 1. i

+

P r o o f . W e proceed b y i n d u c t i o n on j , w h i c h is the superscript o n the right in the c o m m u t a t o r [ & , & ] . Since G = G, we see that if j = 1, the formula we need is [ G \ G ] C G , a n d this is v a l i d since b y definition, G = [ G \ G ] . W e can assume, therefore, t h a t j > 1, a n d so we can w r i t e 0 = \ G i - \ G \ . Then l

i + 1

i

+

l

j

j

j

[ G \ G ] = [ G , G<] = [ G ~ \ G, G*], and to show t h a t this t r i p l e c o m m u t a t o r is contained i n the n o r m a l subgroup G

i +

i , it suffices by the three-subgroups l e m m a ( C o r o l l a r y 4.10) to show that j

1

[G, G \ G ' } C G We

i + j

j

and

[ G \ G ~\

G] C G

i + j

.

have j

l

[G, G\ G ~ ] = [ G \ G

,

=

i

[ G

+

j

1

\G ~ } C

= G

i + j

,

where the containment is v a l i d by the i n d u c t i v e hypothesis. A l s o , j

[G\ G - \ G ] C [ G

i

+

j

- \ G] = G

i + j

,

where here too, the containment follows from the i n d u c t i v e hypothesis. T h i s completes the proof.

•

Suppose we consider commutators of n copies of G t h a t are not necessarily left associated. F o r example, i f n = 8, we m i g h t have something like [[[G, G], [G, G , G]], [[G, G], G } } . (We refer to such a n object as a c o m m u t a t o r of weight n.) T h e following result shows t h a t the left-associated weight n c o m m u t a t o r of copies of G contains every other such weight n c o m m u t a t o r . (Of course, the left associated weight n c o m m u t a t o r is e x a c t l y G , the n t h t e r m of the lower central series.) n

4.12. C o r o l l a r y . L e t G be a g r o u p . copies o f G is contained i n G. n

T h e n every

w e i g h t n c o m m u t a t o r of

128

4.

Commutators

P r o o f . W e proceed b y i n d u c t i o n on n . T h e only weight 1 c o m m u t a t o r is G itself, a n d so there is n o t h i n g to prove i n this case. A n a r b i t r a r y weight n c o m m u t a t o r w i t h n > 1 has the form [ X , Y ] , where X is a weight i c o m m u t a t o r , Y is a weight j c o m m u t a t o r a n d i + j = n . Since i < n a n d j < n , the inductive hypothesis yields X C G a n d Y C G > , and thus [ X , Y] C [ G \ G ' } C G ^ ' = G by T h e o r e m 4.11. • i

i +

n

Consider the terms G^ of the derived series of G. We see [ G , G] is a weight 2 commutator, G" = [ G ' , G'] = [ [ G , G ] , [ G , G]] 4 c o m m u t a t o r , a n d G'" = [ G " , G " } is a weight 8 c o m m u t a t o r . G » is a weight 2 c o m m u t a t o r for a l l r > 0, a n d this yields the r

that G' = is a weight In general, following.

4.13. C o r o l l a r y . We h a v e G » C G " f o r a l l g r o u p s G a n d i n t e g e r s r > 0. A l s o , if G i s m l p o t e n t of class m, t h e n t h e d e r i v e d l e n g t h of G i s a t most l + log (m). 2

2

P r o o f . Since G^ is a weight 2 c o m m u t a t o r , the first assertion is immediate from C o r o l l a r y 4.12. N o w assume t h a t G is nilpotent of class m , a n d let d be its derived length, so t h a t G^ = 1 a n d d is the smallest integer w i t h this property. W e have r

2

1

G "' d

a n d so 2 ~

l

d

D G^ ~

< m + 1. T h e n 2 ~ d

l

> 1=

l)

G

m

+

1

,

< m, a n d hence d - 1 < l o g ( m ) . 2

•

N e x t , we explore some connections between the conjugacy class sizes and the nilpotence class of a nilpotent group G. Suppose we k n o w the number m of different sizes of conjugacy classes i n G. If m = 1, for example, then since the size of the class of the identity of G is 1, it follows that a l l classes of G have size 1, a n d so G is abelian. ( O f course, we d i d not use the assumption that G is nilpotent here.) T h i s suggests the (still unresolved) question of how the number m of conjugacy class sizes of G affects the structure of G. In p a r t i c u l a r , it w o u l d be interesting to know i f the derived length of G, or perhaps even the nilpotence class of G, is b o u n d e d by some function of m. After m a n y years w i t h o u t progress o n these questions, it was proved by K . Ishikawa t h a t if m = 2, then G has nilpotence class at most 3. (In this case too, it is not really necessary to assume t h a t G is nilpotent; a result of N . Ito shows t h a t a finite group w i t h just two class sizes is necessarily nilpotent.) M o r e recently, A . M a n n proved a stronger result w i t h a n easier proof, and we present t h a t next. G i v e n a finite group G w i t h class sizes 1 = m < n < • • • < n , we write M { G ) to denote the subgroup of G generated by a l l elements l y i n g in conjugacy classes of size at most n . (We w i l l use this n o t a t i o n for the 2

2

m

4B

129

remainder of this section.) N o t e t h a t M ( G ) = G i f m = 2, a n d thus the following result contains Ishikawa's theorem. 4.14. T h e o r e m ( M a n n ) . If G i s a class o / M ( G ) i s a t most 3.

finite

n i l p o t e n t g r o u p , then

the

mlpotence

A c t u a l l y , we can replace the hypothesis t h a t G is nilpotent i n T h e o rem 4.14 b y a m u c h weaker one: t h a t G has a self-centralizing n o r m a l subgroup. ( B y this, we mean a subgroup A < i G such t h a t A = C { A ) . Observe t h a t such a subgroup is a u t o m a t i c a l l y abelian.) G

4.15. T h e o r e m . n o r m a l subgroup. most 3.

Suppose Then

t h a t a finite g r o u p G c o n t a i n s a s e l f - c e n t r a l i z i n g M ( G ) i s n i l p o t e n t , a n d i t s n i l p o t e n c e class is a t

R e c a l l t h a t a nilpotent group G always has a self-centralizing n o r m a l subgroup, a n d so T h e o r e m 4.14 is i m m e d i a t e from T h e o r e m 4.15. ( T h e existence of self-centralizing n o r m a l subgroups i n nilpotent groups is part of P r o b l e m I D . 1 0 , but nevertheless, we give the easy proof here.) 4.16. L e m m a . L e t G be a finite n i l p o t e n t g r o u p , a n d l e t A be m a x i m a l a m o n g n o r m a l a b e l i a n s u b g r o u p s o f G . Then A = C { A ) , a n d hence every nilpotent group contains a self-centralizing n o r m a l subgroup. G

P r o o f . W r i t e G = C ( A ) . T h e n A C C, a n d we assume t h a t this containment is proper. Since A o G , we have C < G, a n d thus we c a n choose a subgroup B C C such t h a t B / A is m i n i m a l n o r m a l i n G / A . Since G/A is nilpotent, it follows that B / A C Z ( G / A ) , a n d thus B / A has p r i m e order. In p a r t i c u l a r , B / A is cyclic, a n d since B C G = C ( A ) , it follows t h a t A C Z ( B ) . W e conclude that B is abelian, a n d since B < G , this contradicts the choice of A . U G

G

The

key ingredient i n the proof of T h e o r e m 4.15 is the following.

4.17. L e m m a . L e t K be a n a b e l i a n n o r m a l s u b g r o u p of a finite g r o u p G, a n d l e t x E G be n o n c e n t r a l . Then \ C ( x ) \ < \ C ( y ) \ , where y = [k,x] and k E K is arbitrary. G

G

P r o o f . Observe that y <E [AT, G] C AT. If y = 1, t h e n C { y ) = G > G

C (x), G

a n d there is n o t h i n g further to prove, so we assume that y ^ 1. B o t h x a n d y lie i n the subgroup H = K C ( x ) , a n d of course, C { x ) G

C ( v ) Q H

Now have

H

= C (x) G

and

C { y ) . It suffices, therefore, to prove t h a t | C ( a ; ) | < \ C ( y ) \ . G

H

let 6 : K

H

AT be the m a p defined by 6 { t ) = [ t , x ] . F o r s,t€K,

6(st) = [st,x] = [ s , x ] % x ]

= [s,x][t,x]

=

e(s)6(t),

we

130

4.

Commutators

where the t h i r d equality holds because b o t h t a n d [ s , x] lie i n the abelian group AT. T h u s 6 is a h o m o m o r p h i s m , a n d 6 ( K ) is a subgroup, a n d we have \ 9 { K ) \ = \ K : ker(0)| = \ K : K n C { x ) \ = \ H : C { x ) \ . H

H

We argue next that each of the subgroups AT and C ( x ) normalizes B ( K ) , a n d thus B ( K ) < K C ( x ) = H . C e r t a i n l y , AT C N ( 0 ( A T ) ) since 9 { K ) C K a n d AT is abelian. A l s o , since 0(AT) is uniquely determined by the n o r m a l subgroup AT a n d the element x , we see that C ( x ) C N ( 9 ( K ) ) . Now 0(AT) < H , a n d y is a n o n i d e n t i t y element of 6 { K ) , and thus the entire i f - c o n j u g a c y class Y of y consists of nonidentity elements of 6 { K ) . T h e n |JJ : C ( y ) \ = | ^ | < |#(AT)| = \ H : C { x ) \ , as required. • G

G

G

G

H

G

H

4.18. C o r o l l a r y . L e t K be a n a b e l i a n n o r m a l s u b g r o u p of a n a r b i t r a r y g r o u p G, a n d w r i t e M = M ( G ) . Then [AT, M ] C Z ( G ) .

finite

P r o o f . W r i t e G = G / Z ( G ) . It suffices to show t h a t [AT~, M ] = 1, a n d for this purpose, it is enough to check that x centralizes K for a l l elements x i n some generating set for M . It suffices, therefore, to observe that [ k , x ] e Z ( G ) for a l l k <E K , where x lies i n a conjugacy class of size 1 or n i n G. O f course, i f x e Z ( G ) , there is n o t h i n g to prove, and if x lies i n a class of size n , then b y L e m m a 4.17, the element [ k , x ] lies i n a smaller class, and hence [ k , x ] e Z ( G ) , as required. • 2

2

P r o o f of T h e o r e m 4.15. L e t K < G be self-centralizing, and write M = M(G). B y C o r o l l a r y 4.18, we have [AT, M ] C Z ( G ) , a n d thus [AT, M , M ] = 1 a n d [ M , A T , M ] = 1. B y the three-subgroups l e m m a , [ M , M , K ] = 1, a n d thus [ M , M ] C C ( A T ) = AT. T h e n M = [Af, M , A f , Af] C [ K , M , M ] = 1, a n d it follows t h a t A f is nilpotent w i t h class at most 3. • 4

G

We clearly cannot drop the assumption i n T h e o r e m 4.15 t h a t G contains a self-centralizing n o r m a l subgroup since, for example, i f G is nonabelian and simple, t h e n M ( G ) = G is certainly not nilpotent. W e do, however, o b t a i n the following very general result. 4.19. M

Theorem.

= M ( G ) . Then

L e t G be a

finite

the nilpotence

group,

class

and

w r i t e F = F(Af),

of F i s a t most n

P r o o f . L e t n be the nilpotence class of F , so t h a t F > 1 a n d F A s s u m i n g , as we may, t h a t n > 3, consider the characteristic F ~ of G . If this subgroup is abelian, then by C o r o l l a r y 4.18, [ F ~ , M ] C Z ( G ) , a n d thus F = [ F " , F , F ] C [ F " , M , F ] = is a c o n t r a d i c t i o n . W e conclude t h a t F ~ is nonabelian, a n d 1 < [F ' ,F ' } C F by T h e o r e m 4.11. B u t F = 1, 2n - 4 < n + 1. T h i s yields n < 4, as wanted. • n

2

n

2

n

n

2

n

n

n

2

n

2

2

n

4

where

4.

2

2

n

+

l

n

+

1

= 1. subgroup we have 1, w h i c h therefore and thus

4C

131

Problems 4B 4 B . 1 . L e t G be nilpotent of class exceeding 2. Show t h a t G has a characteristic abelian subgroup that is not central. 4 B . 2 . L e t G be nilpotent. Show t h a t G has a characteristic subgroup K such t h a t K D C ( K ) a n d the nilpotence class of K is at most 2. G

H i n t . Use P r o b l e m 4 B . 1 . 4 B . 3 . L e t { Z j \ j > 1} be the upper central series for an a r b i t r a r y group G. Show t h a t [ G \ Zj] C Z j - i for a l H > 0 a n d a l l j . I n p a r t i c u l a r , [ G \ Z \ = 1 for i > 0. {

4 B . 4 . L e t X , Y C G be a r b i t r a r y subgroups, a n d suppose t h a t Y centralizes [X,Y]. (a) P r o v e t h a t Y' centralizes X . (b) N o w assume that X < G. Show t h a t [ X , Y) is abelian. 4 B . 5 . L e t G be supersolvable. Show t h a t M ( G ) is nilpotent w i t h class at most 3. 4C Let A act on G v i a automorphisms. Since A and G c a n be viewed as subgroups of the semidirect product r = G x A , the c o m m u t a t o r [ G , A ] can be c o m p u t e d as a subgroup of T. B u t G < T, a n d so [G, A] C G , a n d i n fact it is possible (and often convenient) to t h i n k of [G, A ) as being c o m p u t e d entirely w i t h i n G , thereby rendering the semidirect product irrelevant. T o do this, observe t h a t i f g e G and a G A , t h e n [ ,a] = ^ a ^ a = $"V, where we c a n view # either as the result of conjugating g by a i n T, or alternatively, as the result of a p p l y i n g a to g i n the o r i g i n a l action. T h e c o m m u t a t o r [G, A ] , therefore, is the subgroup { g ~ g \ g <E G , o G /4), and this w o u l d make sense even i f we d i d not k n o w about semidirect products. Nevertheless, to prove facts about [G, A ] and related objects, we w i l l appeal to semidirect products when it is convenient to do so, and i n particular, we w i l l a p p l y the three-subgroups l e m m a to subgroups of the semidirect product. 5

a

l

a

W e can a p p l y L e m m a 4.1 (in the semidirect p r o d u c t ) to deduce t h a t [G, A ] < G. A l s o A normalizes [G, A ) i n G x A , w h i c h means t h a t [G, A ] is invariant under conjugation by elements of A , or equivalently, it is invariant under the original action of A on G . I n general, i f H C G is an a r b i t r a r y A - i n v a r i a n t subgroup, we have a n a t u r a l action of A on H , and sometimes i n this context, we say that H a d m i t s the a c t i o n of A . I n particular, since

132

4.

Commutators

[ G , A ] admits the a c t i o n of A , we can compute [ G , A , A ] , w h i c h is n o r m a l in [ G , A ] , but not necessarily i n G. (It is, of course, s u b n o r m a l i n G.) A l s o , [ G , A , A ] admits A , a n d thus we can compute [ G , A , A , A ] , and we can continue like this. A l l of these repeated commutators, of course, a d m i t A and are s u b n o r m a l i n G. R e c a l l t h a t i f A acts v i a automorphisms on G and H C G is A - i n v a r i a n t , then A permutes the set of right cosets of H i n G a n d also the set of left cosets of H i n G. I n p a r t i c u l a r , i f H < G, we have an action of A on G / H , and it is easy to see that this is an action v i a automorphisms. ( T h i s is sometimes called the i n d u c e d a c t i o n of A_on G / H . ) F o r g G G a n d a G A , we have ( g H ) = g H , a n d so if we write G = G / H we have ( g ) = g , and hence a

a

a

1

[9, a] = g ' g

a

1

= ( g - ) ^ ) = (g)~H9)

a

5

= foa] •

It follows t h a t [ G , A ] = [ G , A \ . The

following characterization of [ G , A ] is sometimes useful.

4.20. L e m m a . L e t A a c t v i a a u t o m o r p h i s m s o n G, w h e r e A a n d G a r e g r o u p s . T h e n [ G , A ] i s t h e u n i q u e s m a l l e s t A - i n v a r i a n t n o r m a l s u b g r o u p of G such t h a t t h e i n d u c e d a c t i o n of A o n t h e f a c t o r g r o u p i s t r i v i a l . P r o o f . Suppose t h a t N < G, where N is A - i n v a r i a n t , and w r i t e G =

G/N.

T h e n A acts t r i v i a l l y on G / N if and only i f 1 = [ G , A ] , w h i c h by the previous c o m p u t a t i o n is equivalent to 1 = [ G , A ] .

T h i s , of course, is the same as

[ G , A ] C N , and the result is now immediate.

•

4.21. C o r o l l a r y . L e t A a c t v i a a u t o m o r p h i s m s o n G, w h e r e A a n d G a r e groups, and l e t H Q G . (a) E v e r y r i g h t coset (b) E v e r y left

coset

Then the f o l l o w i n g are equivalent. of H i n G i s a n A - i n v a r i a n t of H i n G i s a n A - i n v a r i a n t

subset. subset.

(c) [ G , A ] C H . I n p a r t i c u l a r , [ G , A] i s t h e u n i q u e s m a l l e s t s u b g r o u p of G w i t h t h e p r o p e r t y t h a t a l l of i t s r i g h t cosets a r e A - i n v a r i a n t , a n d s i m i l a r l y if " l e f t " r e p l a c e s "right". l

1

P r o o f . F i r s t , observe that i f X is a subset of G and we write X ~ = { x ' \ x G X } , then the m a p X ^ X ~ takes the right cosets of H to the left cosets of H and v i c e v e r s a . It follows that an a u t o m o r p h i s m of G fixes a l l right cosets of H i f a n d o n l y i f it fixes a l l left cosets of H . W e see, therefore, t h a t (a) and (b) are equivalent. l

Now assume (b), so that every left coset of H is A - i n v a r i a n t . T h e n for g G G and a G A , we have g G ( g H ) = g H , and so [g, a] = g ' g G H and (c) follows. a

a

1

a

4C

133

F i n a l l y , assume (c), so t h a t [ G , A ] C H . E a c h right coset of H i n G , therefore, is a u n i o n of right cosets of [G, A ] i n G . B u t a l l cosets of [G, A ] in G are A - i n v a r i a n t b y L e m m a 4.20, a n d so unions of such cosets are also A - i n v a r i a n t , and (a) follows. • R e c a l l that an action of a group A on a set ft is faithful i f the i d e n t i t y of A is the only element that fixes every p o i n t . I n general, the kernel of an a r b i t r a r y action of a group A is the (necessarily normal) subgroup B of A consisting of the elements of A that act t r i v i a l l y . I n this situation, a l l elements of each coset B a of B i n A mduce the same p e r m u t a t i o n on ft, a n d so we have a well denned action o f A = A / B on ft, where a - a = a - a . T h i s action of A / B is clearly faithful. Now suppose that A acts on G v i a automorphisms. W o r k i n g i n the semidirect p r o d u c t G x > A , we see that the kernel of the action of G on A is C { G ) . ( A s we remarked previously, it is c o m m o n to w r i t e C ( G ) to denote the kernel of the action of A on G even if we do not e x p l i c i t l y construct the semidirect product.) T h i s kernel is the largest subgroup B C A such t h a t [ G , B ] = 1. If B = C { G ) , then B < A , a n d w r i t i n g A = A / B as before, we see t h a t the n a t u r a l action of A on G is given by g = g for g <E G a n d a E A , a n d it follows t h a t a subgroup H C G is A - i n v a r i a n t if a n d only if it is A - i n v a r i a n t . (In other words H admits A if a n d only i f it admits A . ) A l s o , [ g , a \ = [g, a] for g € G and a G A , and so if H admits A and A , we have [ H , A ] = [ H , A \ . A

A

A

u

a

W h a t can we say about the structure of a group A i f we k n o w t h a t it acts via automorphisms on some group G i n such a way t h a t [ G , A , . . . , A ] = 1, where there are (say) m copies of A i n this c o m m u t a t o r ? T h e answer is t h a t we can say precisely nothing. T h i s is because an a r b i t r a r y group A can act t r i v i a l l y on G , i n w h i c h case [G, A ] = 1. M o r e generally, i f B is the kernel of the action of A on G , then B and G cannot "see" each other, a n d so the fact t h a t [ G , A , . . . , A ] = 1 tells us n o t h i n g about B . T o draw a conclusion about the structure of A from some assumption about how G sees the action of A , we must therefore assume that the action is faithful. To facilitate the discussion, we introduce some (nonstandard) n o t a t i o n . We w r i t e [ G , A , . . . , A ] to denote the c o m m u t a t o r w i t h exactly m copies of A . T h e n [ G , A , . . . , A ] i is s i m p l y [G, A ] and i n general [[G, A ] , A , . . . , A ] = [G,A,...,A] . m

m

m

+

1

4.22. T h e o r e m . L e t A a n d G be g r o u p s . Suppose t h a t A acts faithfully o n G v i a a u t o m o r p h i s m s , a n d assume t h a t [ G , A , . . . , A ] = 1. T h e n A i s s o l v a b l e , a n d i t s d e r i v e d l e n g t h i s a t most m - 1 . m

P r o o f . W e begin by reformulating the result i n such a way that we need not assume t h a t the action is faithful. I n the general case, we cannot hope

134

4.

Commutators

m

t o prove that A is solvable, b u t we show b y i n d u c t i o n o n m t h a t A ^ ~ ^ C C ( G ) , where we recall t h a t C ( G ) is the kernel of the action. I n the original situation, where A acts faithfully, we have C { G ) = 1, a n d it w i l l follow t h a t A ^ - ) = 1, a n d so A is solvable w i t h derived length at most m - 1, as required. A

A

A

1

If m = 1, then [ G , A ] = 1, a n d thus A = C ( A ) a n d there is n o t h i n g more to prove. A s s u m e then, that m > 2, a n d let H = [ G , A ] , so that H admits A a n d [ H , A , . . . , A ] _ i = [ G , A , . . . , A ] = 1. B y the inductive hypothesis applied to the action of A o n H , therefore, we have A ^ " ) C C ( H ) , a n d thus 1 = [ H , A ^ - ^ ] = [ G , A , A ^ ~ \ Since A D A ( " ) , it follows t h a t [ G , A ^ ~ \ A ^ ~ ^ ] = 1, a n d thus also [ A ^ ~ \ G, A ( " ) ] = 1 since interchanging the first two entries i n a m u l t i p l e c o m m u t a t o r of subgroups has no effect. T h e three-subgroups l e m m a now yields G

m

m

2

m

m

m

2

A

m

2

m

2

m

m

2

m

2

[A^ - \A^ - \G}

2

m

2

=l ,

a n d since b y definition, j ^ m - l ) m

l

we have [ A ^ ~ \ G \

_ ^ ( m - 2 ) ^(m-2)j m

1

= 1. T h u s A ^ ' ^ C C ( G ) , as wanted. A

•

W h e n m > 2, the step i n the proof of T h e o r e m 4.22 where A is replaced by A ( " ) throws away information, a n d this suggests t h a t perhaps more is true t h a n we proved. I n fact more is true: P . H a l l showed t h a t under the hypotheses of our theorem, A is a c t u a l l y nilpotent, a n d not merely solvable. A l s o , the nilpotence class of A is at most m { m - l ) / 2 . R e c a l l that b y C o r o l l a r y 4.13, the derived length of a nilpotent group is l o g a r i t h m i c a l l y b o u n d e d i n terms of its nilpotence class, a n d therefore it follows b y H a l l ' s theorem t h a t the linear b o u n d o n the derived length of A that we established in T h e o r e m 4.22 is too weak; it is of the "wrong" order of magnitude. m

2

m

2

If m = 2, then A ^ ~ ^ = A ^ = A , a n d we lose n o t h i n g i n the replacement step of the previous proof. I n this case, H a l l ' s result coincides w i t h T h e o r e m 4.22 since a n o n t r i v i a l group is solvable w i t h derived length 1 i f a n d o n l y i f it is abelian, w h i c h is exactly when it is nilpotent w i t h class 1. We list this case as a separate corollary. 4.23. C o r o l l a r y . L e t A a n d G be g r o u p s . Suppose t h a t A acts f a i t h f u l l y o n G v i a a u t o m o r p h i s m s a n d t h a t [ G , A , A] = 1. T h e n A i s a b e l i a n . P r o o f . T h i s is just the case m = 2 of T h e o r e m 4.22.

•

W i t h o u t using a d d i t i o n a l machinery, it seems to be difficult to establish H a l l ' s upper b o u n d o n the nilpotence class of A . B u t it is not too h a r d to show t h a t A is nilpotent i n the s i t u a t i o n we have been discussing, a n d the proof offers more opportunities for using the three-subgroups l e m m a .

4C

135

4.24. T h e o r e m . L e t A a n d G be finite g r o u p s . Suppose t h a t A acts faithf u l l y v i a a u t o m o r p h i s m s o n G, a n d assume t h a t [ G , A , . . . , A ] — 1. T h e n A is n i l p o t e n t .

P r o o f . A s before, we drop the assumption t h a t A acts faithfully. W e prove by i n d u c t i o n on \ G \ that i n general, the last t e r m A°° of the lower central series of A acts t r i v i a l l y on G. Once this is established, we see t h a t if the action of A is faithful, then A°° = 1, and hence A is nilpotent. Since we can assume that G > 1 and we k n o w that [ G , A , . . . , A ] = 1, it follows t h a t [ G , A ] < G. A l s o , [ G , A ] admits the action of A and [ [ G , A ] , A , . . . , A ] = 1. B y the inductive hypothesis, therefore, A°° acts t r i v i a l l y on [ G , A ] , and hence [ G , A , A°°] = 1, or equivalently, [ A , G, A } = 1. O u r goal w i l l be to show that [ G , A°°,A] = 1 since, i f we can accomplish that, the three-subgroups l e m m a w i l l y i e l d [A°°,A,G] = 1. B u t [A°°,A] = A°° because A°° is the last t e r m of the lower central series of A , a n d it w i l l follow that [A°°, G] = 1. T h e proof w i l l then be complete. 0 0

O u r strategy for showing that [G,A°°,A] = 1 is to find a n o n t r i v i a l n o r m a l subgroup C of G on w h i c h A acts t r i v i a l l y . Once we have found C , we can a p p l y the inductive hypothesis to the i n d u c e d action of A on G = G / C . (This is v a l i d since we have [ G , A , . . . , A ] = [ G , A , . . . , A ] = 1.) B y the i n d u c t i v e hypothesis, 1 = [G,A°°] = [G,A°°], a n d so [G,A°°] C C. Since we are assuming that A acts t r i v i a l l y on C, this yields [ G , A°°,A} = 1, w h i c h is what we want. We can assume t h a t [ G , > 1, or else there is n o t h i n g to prove. Take C = C ( A ) , so t h a t A certainly acts t r i v i a l l y o n C, as required. T o see t h a t C > 1, observe that [ [ G , A°°],A, . . . , A ] = 1 , a n d thus if we start w i t h the n o n t r i v i a l group [ G , A } and repeatedly compute commutators w i t h A , we get a sequence of subgroups of [ G , A°°] t e r m i n a t i n g at the identity. T h e last nonidentity t e r m i n this sequence, therefore, is centralized b y A , and this shows t h a t C > 1, as required. l G i A o o ]

0 0

To complete the proof now, it suffices to show t h a t C < G, and our argument for this involves two further applications of the three-subgroups l e m m a . F i r s t , we argue that [G,A°°] centralizes [ G , A } . W e have already mentioned that A°° acts t r i v i a l l y on [ G , A ] , so t h a t [[G,A\,A°°} = 1, and thus we have [[G,A],A°°,G] = 1. A l s o , [ G , A ] < G, a n d thus [ G , [ G , A } } C [G, A ] , and we have [G, [G,A],A°°] C [[G, A ] , A°°] = 1. N o w we apply the three-subgroups l e m m a to deduce that [A°°, G , [G, A}] = 1, and thus [G, A] centralizes [A°°, G] = [G, A ) , as claimed. I n p a r t i c u l a r , since C C [G, A } , we have [ G , A , C ] = 1. A l s o , of course, [ A , C , G ] = 1 since A centralizes C. B y the three-subgroups lemma, once again, we have [G, G , A ] = 1, and so A centralizes [G, G ] . 0 0

0 0

4.

136

Commutators

R e c a l l now t h a t C C [G,A°°] < G, and thus [ C , G ] C [G,A°°]. W e just showed, however, t h a t A centralizes [ C , G ] , and thus [ C , G ] C C =o](A) = C , where the equality holds by the definition of C. W e have now shown t h a t [ C , G] C C , or equivalently, that C < G, as wanted. T h e proof is now complete. • [ G i X

Let us continue to assume that [ G , A , . . . , A ] = 1 for some unspecified number of copies of A . W e have seen that i f the action is faithful, then A must be nilpotent. It is impossible, however, to draw any conclusions about the structure of G since, for example, it could be that A is t r i v i a l . B u t although we can say n o t h i n g about G itself, we shall see that the subgroup [ G , A ] is under control. W e begin w i t h a special case. 4.25.

Lemma.

L e t A a n d G be g r o u p s .

a u t o m o r p h i s m s , a n d assume

Suppose

t h a t A acts

o n Gv i a

t h a t [ G , A , A ] = 1. T h e n [ G , A ] i s a b e l i a n .

P r o o f . Since [ [ G , A } , A ] = 1, we certainly have [ [ G , A ] , A , G ] = 1. A l s o , [ G , A] < G, and thus [ G , [ G , A ] , A ] C [[G, A ] , A] = 1. B y the three-subgroups l e m m a , we have 1 = [A,G, [G,A\] = [[G,A],[G,A]] = [G, A}', and so [ G , A ] is abelian. 4.26. T h e o r e m .

•

L e t A be a p - g r o u p t h a t acts

g r o u p G, a n d suppose

via automorphisms o na

finite

t h a t [ G , A , . . . , A ] = 1. T h e n [ G , A ) i s a p - g r o u p .

P r o o f . W e proceed by i n d u c t i o n on \ G \ . W e can certainly assume that G > 1, and thus since [ G , A , . . . , A ] = 1, it follows that [ G , A ] < G. For convenience, write N = [ G , A ] , and observe that N < G and that N admits the action of A . Since N < G , the inductive hypothesis applies i n N , and we deduce t h a t [ N , A ] is a p-group. A l s o , since [TV, A ] < N , we have [AT, A ] C U, where U = O ( N ) is characteristic i n N < G, a n d thus U < G . p

Now

consider the induced action of A on G = G / U .

Since [TV, A ] C U,

we see t h a t A acts t r i v i a l l y on N . A l s o , [G, A ] = [ G , A \ = N , and thus A

acts t r i v i a l l y on G/N.

W e Jiave J G , A , A] = [ N , A ] = 1, and thus by

L e m m a 4.25, we conclude t h a t N = [ G , A ] is abelian. B u t O ( N ) p

is t r i v i a l ,

a n d since N is abelian, it follows that N is a p'-group. Since A is a p-group and N is a n o r m a l p'-subgroup of G, we can apply C o r o l l a r y 3.28 to the action of A on G. T h a t corollary tells us that "fixed points come from fixed points" ,_and_ since N is contained i n the fixed-point subgroup and every element of G / N is A - f i x e d , we conclude t h a t the whole group G consists of fixed points. T h e n 1 = [ G , A ] = [ G , A ] , and hence [ G , A ] C U. Since U is a p-group, we are done. •

137

4 D

In the s i t u a t i o n of T h e o r e m 4.26, we can conclude t h a t [G, A] is nilpotent even w i t h o u t assuming that A is a p-group. 4.27. T h e o r e m . L e t A a n d G be finite g r o u p s . Suppose t h a t A acts v i a a u t o m o r p h i s m s o n G, a n d assume t h a t [ G , A , . . . , A ] = 1. T h e n [ G , A ] i s nilpotent. P r o o f . W e proceed by i n d u c t i o n on \ A \ . ( O f course, i f A is t r i v i a l t h e n [ G , A ] = 1 and there is n o t h i n g to prove.) Suppose that B is an a r b i t r a r y proper subgroup of A . T h e n [ G , B , . . . , B ] C [ G , A , . . . , A ] = 1, and so by the i n d u c t i v e hypothesis, [ G , B ] is nilpotent. Since [G, B ] < G , we have [G, B ) C F ( G ) , the F i t t i n g subgroup, and thus B acts t r i v i a l l y on G / F ( G ) . We now k n o w that every proper subgroup of A is contained i n the kernel of the i n d u c e d action of A on G / F ( G ) , and so i f A is generated by its proper subgroups, then this kernel must be a l l of A . I n this case, [ G , A ] C F ( G ) , and hence [G, A ] is nilpotent, as wanted. W e can assume, therefore, that A is not generated by its proper subgroups. E v e r y finite group, however, is generated by its Sylow subgroups, a n d thus some Sylow subgroup of A is not proper. I n other words, A is a p-group for some p r i m e p , and hence [G, A ] is a p-group by T h e o r e m 4.26. In particular, [G, A ] is nilpotent. •

Problems 4C 4C.1.

L e t A act o n G v i a automorphisms, a n d suppose t h a t 1 = H

0

C H i C ••• C H

m

= G

is a chain of A - i n v a r i a n t subgroups such t h a t each right coset of H ^ i n H i is A - i n v a r i a n t for a l l i w i t h 0 < i < m. I n the older literature, this s i t u a t i o n was described by saying that A stabilizes the chain { H i } . If A stabilizes a chain of subgroups i n G , show that it stabilizes a c h a i n of s u b n o r m a l subgroups. 4 C . 2 . A s s u m e that A acts faithfully on G v i a automorphisms, and assume t h a t A stabilizes a chain { H i | 0 < i < m } of n o r m a l subgroups of G . Show t h a t A is nilpotent w i t h class at most m - 1. 4 C . 3 . L e t A act v i a automorphisms on G , and suppose t h a t N < G and t h a t A acts t r i v i a l l y on N . Show that N centralizes [G, A ] , 4D We continue our study of commutators that arise w h e n A acts v i a automorphisms on G , but now we focus on coprime actions, where ( | G | , \ A \ ) = 1.

138

4.

Commutators

The following u l t i m a t e l y relies on G l a u b e r m a n ' s l e m m a , and t h a t is the reason for the solvability assumption, which, as usual, is not really necessary i f one is w i l l i n g to appeal to the F e i t - T h o m p s o n theorem. 4.28. L e m m a . L e t A a n d G be finite g r o u p s . L e t A a c t v i a a u t o m o r p h i s m s o n G, a n d suppose t h a t ( \ G \ , \ A \ ) = 1 a n d t h a t one of A o r G i s s o l v a b l e . ThenG = C (A)[G,A\. G

P r o o f . W r i t e G = G / [ G , A ] . Since fixed points come from fixed points i n coprime actions ( C o r o l l a r y 3.28) we have C ( A ) U

= C (A). G

B u t A acts

t r i v i a l l y on G / [ G , A ] , and so the left side of this equation is the whole group G.

Thus C (A)[G,A]

= C ( A ) = G,

G

and

G

it follows t h a t [ G , A ] C ( A ) = G by the correspondence theorem. G

•

To demystify this proof somewhat, we observe that what is going on here is the following. E a c h coset of [ G , A] i n G is A - i n v a r i a n t , and so by a fairly standard G l a u b e r m a n ' s l e m m a argument, each such coset contains an A - f i x e d element. The

following result is used frequently.

4.29. L e m m a . L e t A a c t v i a a u t o m o r p h i s m s o n G, w h e r e A a n d G a r e finite g r o u p s , a n d suppose t h a t ( \ G \ , \ A \ ) = 1. T h e n [ G , A , A ] = [ G , A \ . P r o o f . O f course [ G , A , A] C [ G , A ] , and so it suffices to prove the reverse containment. F o r this purpose, we show that [g, a] G [ G , A , A ] for a l l g G G and a G A . T h e result w i l l then follow since [ G , A ] is generated by these commutators [ g , a ] . Suppose first t h a t A is solvable. L e m m a 4.28 yields G = C ( A ) [ G , A ] , and so i f g G G, we can w r i t e g = cx, where c G C ( A ) and x G [ G , A } . N o w let a G A , and observe t h a t [c, a] = 1 . T h e n G

G

x

[g, a] = [ c x , a] = [c, a ] [ x , a] = [ x , a] G [ G , A , A] , as wanted. T h i s proves t h a t the result is true i f A is solvable. Now

i n the general case, again let g G G and a G A . T h e n [g, a] G [ G , (a)} = [ G , ( a ) , (a)} C [ G , A , A]

since the cyclic group ( a ) is certainly solvable. T h i s completes the proof.

•

We can a p p l y this to the situation that we studied i n the previous section. 4.30. C o r o l l a r y . L e t A a c t f a i t h f u l l y v i a a u t o m o r p h i s m s o n G, w h e r e A a n d G a r e finite g r o u p s , a n d assume t h a t [ G , A , . . . , A ] = 1. T h e n every p r i m e d i v i s o r of \ A \ a l s o d i v i d e s \ G \ .

4D

139

P r o o f . L e t P € S y l ( A ) where p is a p r i m e t h a t does n o t d i v i d e | G | . T h e n [ G , P] = [ G , P , . . . , P ] = 1, where the first equality holds by repeated a p p l i c a t i o n of L e m m a 4.29. Since the action of A is faithful and P acts trivially, we deduce that P = 1, and thus p does not d i v i d e \ A \ . • p

N e x t , we prove the so called P x Q l e m m a of T h o m p s o n , and we show how it is relevant to the study of local subgroups. 4.31. T h e o r e m ( T h o m p s o n ) . L e t A be a finite g r o u p t h a t acts v i a a u t o m o r p h i s m s o n a p - g r o u p G, a n d suppose that A = P x Q is a n internal d i r e c t p r o d u c t of a p - g r o u p P a n d a p - g r o u p Q. Suppose that Q fixes every element of G t h a t P fixes. Then Q acts t r i v i a l l y o n G. 1

Before we begin the proof of T h o m p s o n ' s theorem, we establish some easy facts about p-groups acting on p-groups. 4.32. L e m m a . L e t P be a p - g r o u p t h a t acts v i a a u t o m o r p h i s m s t r i v i a l p - g r o u p G. Then [ G , P ] < G a n d C ( P ) > 1.

o na non-

G

P r o o f . T h e semidirect product r = G x P is a p-group, and hence it is nilpo¬ tent. It follows that some repeated c o m m u t a t o r of the form [T, T , . . . , T] is t r i v i a l , a n d thus we have [ G , P , . . . , P ] = 1. Successive c o m m u t a t i o n w i t h P , therefore, yields a descending sequence of subgroups s t a r t i n g w i t h G > 1 and ending w i t h 1. T h u s [ G , P ] < G, as wanted. A l s o , i f H is the last nonidentity group i n this sequence, then [ H , P ] = 1, and so 1< H C C { P ) . • G

A s s u m i n g t h a t the p-group G is n o n t r i v i a l i n the s i t u a t i o n of the P x Q l e m m a , it follows by L e m m a 4.32 that C ( P ) > 1. B y hypothesis, however, C ( P ) C C ( Q ) , and therefore Q has at least some n o n t r i v i a l fixed points in G. W h a t we need, however, is that Q fixes every element of G, and the proof of this, while not hard, is a little more subtle; it gives us another o p p o r t u n i t y to use the three-subgroups l e m m a . G

G

G

P r o o f of T h e o r e m 4.31. W e can assume t h a t G > 1, and we proceed by i n d u c t i o n on | G | . W e have [G, P] < G by L e m m a 4.32. A l s o , [ G , P ] is A¬ invariant since b o t h G and P are n o r m a l i z e d by A i n the semidirect p r o d u c t G x > A . T h e hypothesis that Q centralizes every element of G centralized by P is inherited by the a c t i o n of A on [G, P ] , and so the i n d u c t i v e hypothesis applies to this action. W e conclude that Q acts t r i v i a l l y on [G, P ] , and hence [ G , P , Q ] = 1. A l s o , [ P , Q , G ] = 1 since P and Q centralize each other i n A . It follows by the three-subgroups l e m m a that [ Q , G , P ] = 1, and thus P acts t r i v i a l l y on [ Q , G ] = [ G , Q ] . B y hypothesis, Q acts t r i v i a l l y on this subgroup too, and so [G, Q , Q ] = 1. F i n a l l y , since Q is a p'-group and G is

4.

140

Commutators

a p-group, we have [G, Q, Q] = [ G , Q] by L e m m a 4.29, and so [G, Q] = 1, as required. • A s an a p p l i c a t i o n of the P x Q l e m m a , we prove the following "local to global" result for p-solvable groups. (Recall from Section 2 C t h a t i f p is a prime, a subgroup H C G is said to be p-local i n G i f H = N ( P ) , for some n o n t r i v i a l p-subgroup P of G.) G

4.33. T h e o r e m . L e t G be a finite p - s o l v a b l e g r o u p f o r some O , ( H ) C C y (G) / o r every p - l o c a l s u b g r o u p H C G .

prime p.

Then

p

P r o o f . L e t Q = C y ( P ) , where i f is p-local i n G. F i r s t , we consider the case where C y ( G ) = 1, and where our goal, therefore, is to show that Q = l . Since H is p-local, we can write H = N ( P ) for some p-subgroup P C G , a n d we observe t h a t b o t h P and Q are n o r m a l i n P . Since P is a p-group and Q is a p'-group, we have P n Q = 1, and thus P Q is the direct product of P a n d Q . G

Now let U = O p ( G ) , so that U is a p-group, and observe that P Q acts on U b y conjugation since U < G. W e argue that Q fixes every element of U that is fixed by P , or equivalently, that Q centralizes Q / ( P ) . T o see this, observe that C { P ) C P n N ( P ) = P n P . B u t P n P is a n o r m a l p-subgroup of H , a n d Q is a n o r m a l p'-subgroup of P , and so Q centralizes U n P , a n d hence it centralizes C [ / ( P ) , as wanted. W e conclude by the P x Q l e m m a t h a t Q C C ( P ) . B u t G is p-solvable and we are assuming that C y ( G ) = 1, a n d thus U contains its own centralizer i n G . (This is L e m m a 1.2.3 of H a l l and H i g m a n , w h i c h is our L e m m a 3.21.) W e now have Q Q C ( P ) C U, and since Q is a p'-group and U is a p-group, we deduce that Q = 1, as required i n this case. V

G

G

G

In the general situation, let TV = C y ( G ) and w r i t e G = G/N, Cy(G)

=

1.

so that

If H C G is p-local, then since \ N \ is not divisible by p,

L e m m a 2.17 guarantees that H is p-local i n G , and thus by the first part of the proof, Op/(IF) = 1. N o w Q is a n o r m a l p'-subgroup of P , and so Q Q O / ( P ) = 1. F i n a l l y , since Q = 1, we have Q C N , and the proof is p

complete.

•

We return now to the s i t u a t i o n considered at the beginning of this sect i o n . W e h a d A acting c o p r i m e l y on G v i a automorphisms, and either G or A is solvable, and we showed i n L e m m a 4.28 that G = C ( A ) [ G , A ] . A s the following result of H . F i t t i n g shows, more is true i f G is abelian. G

4.34. T h e o r e m ( F i t t i n g ) . L e t A a c t v i a a u t o m o r p h i s m s o n a n a b e l i a n g r o u p G, a n d assume t h a t A a n d G a r e finite a n d t h a t ( | G | , \ A \ ) = 1. T h e n G = C { A ) x [G, A ] . G

4D

141

Proof. that G and so and so Let

A s G is abelian, b o t h factors are n o r m a l , a n d so it suffices to show = C ( A ) [ G , A ] and t h a t C ( A ) n [ G , A ] = 1. O f course, G is solvable, we can appeal to L e m m a 4.28 to deduce the first of these two facts, it suffices to prove the second. G

G

9 : G -> G be the m a p denned by 0(g) =

H g

b

,

beA

and note t h a t since G is abelian, the order of the factors i n this p r o d u c t is irrelevant, and hence 9 is well defined. A l s o , i f x , y e G , then ( x y ) = x y for a l l b e A , and so again using the fact t h a t G is abelian, it is easy to see t h a t 9 ( x y ) = 6 ( x ) 9 { y ) , a n d thus 9 is a h o m o m o r p h i s m . b

b

b

b

A

If g € C ( A ) , then g = g for a l l b e A , a n d thus 9 ( g ) = g \ \ . A l s o , since \ A \ is coprime to | G | , we see t h a t g \ \ ^ 1 i f g + 1, a n d thus C ( A ) n k e r ( 9 ) = 1. T o complete the proof, therefore, it is enough to show t h a t [G, A] C ker(0), and since [ G , A] is generated by commutators of the form [g, a ] , w i t h g e G and a e A , it suffices to prove t h a t each of these commutators lies i n ker(0). G

A

G

If g e G a n d a e A , then

o(g ) a

= X[g

a b

= e{g),

beA

where the second equality holds because a b runs over A as 6 runs over A . It follows t h a t e([g,

1

a

l

a

1

a}) = 9 ( g - g ) = 9 { g - ) 9 { g ) = 9 ( g ) - 9 ( g )

T h i s completes the proof.

= 1.

•

A n easy a p p l i c a t i o n of F i t t i n g ' s theorem is the following. 4.35. C o r o l l a r y . L e t A a c t o n G v i a a u t o m o r p h i s m s , w h e r e G i s a n a b e l i a n p - g r o u p , a n d assume t h a t A i s a finite p ' - g r o u p . If A fixes every element of o r d e r p i n G, t h e n A acts t r i v i a l l y o n G. P r o o f . B y F i t t i n g ' s theorem, G = C { A ) x [ G , A ] , a n d by hypothesis, C ( A ) contains every element of order p i n G . Since C ( A ) and [G, A] have o n l y the identity i n c o m m o n , it follows t h a t [ G , A ] contains no elements of order p . B u t [ G , A] is a p-group a n d every n o n t r i v i a l p-group contains elements of order p. W e conclude t h a t [ G , A ] = 1, and the proof is complete. • G

G

G

Surprisingly, the hypothesis t h a t G is abelian i n C o r o l l a r y 4.35 is unnecessary if p > 2.

142

4.

Commutators

4.36. T h e o r e m . L e t A a c t o n G v i a a u t o m o r p h i s m s , w h e r e G i s a p - g r o u p w i t h p > 2 , a n d assume t h a t A i s a finite p ' - g r o u p . If A fixes every element of o r d e r p i n G, t h e n A acts t r i v i a l l y o n G. The idea of the proof of T h e o r e m 4.36 is quite simple. W e show t h a t if G is a m i n i m a l counterexample, then there is an abelian counterexample of the same order, and this contradicts C o r o l l a r y 4.35. It is straightforward to show t h a t a m i n i m a l counterexample w o u l d have nilpotence class at most 2 (and t h a t does not require that p > 2). B u t getting from class 2 to abelian seems to require a new idea: a remarkable trick due to R . Baer. 4.37. L e m m a (Baer t r i c k ) . L e t G be a finite n i l p o t e n t g r o u p of o d d o r d e r a n d n i l p o t e n c e class a t most 2. T h e n t h e r e exists a naddition operation x + y d e f i n e d f o r elements x,y G G such t h a t G i s a n a b e l i a n g r o u p w i t h r e s p e c t to addition. Also, the f o l l o w i n g hold. (a) If xy = yx f o r x,y

<EG, then x + y = - x y .

(b) T h e a d d i t i v e o r d e r of each order. (c) E v e r y a u t o m o r p h i s m

element

o f G i s equal to its m u l t i p l i c a t i v e

of G i s a l s o a n a u t o m o r p h i s m

of t h e

additive

g r o u p (G,+). Before we begin the proof of Baer's l e m m a , we introduce a bit of suggestive n o t a t i o n . If m is an integer and G is a group of order relatively prime to m, then the m a p x i-> x has an inverse, namely the m a p x ^ x , where n is chosen so t h a t r a n = 1 m o d \ G \ . G i v e n x G G, therefore, there is a unique element y G G such that y = x , a n d i n fact, y is a power of x . In the case where | G | is o d d a n d m = 2, we write y = ^/x~ to denote the unique element w i t h square x , and we observe that if H C G is any subgroup, then V h e H for every element he H . A l s o , it is easy to see that V o F = ( v ^ ) " and that if x a n d y commute, then y / x y = y / x ^ y . m

n

m

1

P r o o f of L e m m a 4.37. for

a l l x,y

Let x + y = x y ^ ^ x ] .

e G, we must check that y x y ^ ~ y ]

alent to y - ' x ^ y x

1

=

/ravrar -

But

1

T o show t h a t x + y = y + x = x y ^ / \ y ~ r ] . T h i s is equiv-

(^iSl)"

1

= v l ^ / F = 2

y/[y,x],

and so the identity we must establish is [ y , x ] = ( \ / { y , x \ ) , w h i c h

is clearly true. Observe that if x a n d y commute, then [ y , x ] = 1, a n d since = 1, it follows t h a t x + y = xy, a n d this establishes (a). In p a r t i c u l a r , since 1 and x commute, we have 1 + x = l x = x for a l l x G G, and thus 1 is an additive identity. A l s o , x a n d x ~ commute, and hence x + x ~ = xx~ = 1, and thus x ~ is the additive inverse of x . l

l

l

l

4D

143

We

have x

+ ( + ) = y

z

x

+ yzy/^y]

=

x y z ^ y \ s ) [ y z ^ y \ , x \ .

Since we are assuming that G has class at most 2, a l l commutators i n G are central, a n d thus the m a p [ - , x ] is a h o m o m o r p h i s m . It follows that [ y z ^ [ z ~ y ] , x] = [y, x] [ z , x] [ y / & y ] , x] , a n d we observe that the t h i r d factor is t r i v i a l since [ z , y] € Z ( G ) , and hence y / [ z ~ y ] € Z ( G ) . W e now have x + (y + z) = x y z y f [ z ~ y ~ W { y , x } [ z , x ]

=

xyzy/[z,y][y,x][z,x],

where the second equality holds because [ z , y ] , commutes w i t h O n the other hand, (x + y ) + z = x y ^ x ]

+ z = xy^x\zyJ[z,xy^/[yTx\}

[y,x][z,x].

,

a n d this can be simplified i n a similar manner. (Here, we use the fact t h a t [ z , •} is a homomorphism.) W e o b t a i n (x + y ) + z =

xyzy/[y,x][z,x][z,y],

w h i c h is equal to the result of our previous c o m p u t a t i o n since the c o m m u tators commute. T h i s shows t h a t a d d i t i o n on G is associative, a n d hence (G, +) is a group. We have already proved (a). For (b), we use the customary notation for a d d i t i v e l y w r i t t e n groups, and we w r i t e nx to denote the sum of n copies of x , where n is a positive integer a n d x <E G. W e argue by i n d u c t i o n on n t h a t nx = x , w h i c h is certainly v a l i d for n = 1. F o r n > 1, we have n

nx

n

= x + ( n- l)x = x + x ~

l

n

l

= xx ~

= x

n

,

where the second equality holds by the inductive hypothesis a n d the t h i r d holds by (a), since x and x ~ commute i n G . It follows t h a t nx = 1 if a n d o n l y if x = 1, and since 1 is the additive identity (as well as the m u l t i p l i c a t i v e identity), it follows that the additive a n d m u l t i p l i c a t i v e orders of x are equal. n

l

n

F i n a l l y , the a d d i t i o n i n G is uniquely determined by the m u l t i p l i c a t i o n , a n d thus any p e r m u t a t i o n of the elements of G t h a t preserves the m u l t i p l i cation w i l l also preserve the a d d i t i o n . A s s e r t i o n (c) follows a n d the proof is complete. • P r o o f o f T h e o r e m 4 . 3 6 . W e can assume t h a t G > 1, a n d we proceed by i n d u c t i o n on G . If H < G and H admits the a c t i o n of A , then since every element of order p i n H is fixed by A , the i n d u c t i v e hypothesis guarantees that A acts t r i v i a l l y on H , and hence [ H , A] = 1. I n particular, if [G, A ] < G, then since [G, A ] admits A , we have 1 = [G, A , A ] = [G, A ] , a n d there is

144

4.

Commutators

n o t h i n g further to prove i n this case. (The second equality holds, of course, by L e m m a 4.29, w h i c h applies since = 1.) The derived subgroup G' is characteristic i n G, and hence it admits A . A l s o , the p-group G is certainly solvable, a n d thus G' < G. W e deduce that [ G ' , A] = 1, a n d so [
CZ(G).

In other words, the nilpotence class of G is at most 2, a n d we can a p p l y the B a e r trick to construct the abelian group structure ( G , +) on the u n d e r l y i n g set of G. Now an element a G A induces an a u t o m o r p h i s m of G, and we conclude by L e m m a 4.37(c) that this p e r m u t a t i o n of the elements of G is also an a u t o m o r p h i s m of ( G , +). In other words, A acts v i a automorphisms on the abelian group ( G , + ) . B y L e m m a 4.37(b), the elements of order p i n ( G , + ) are e x a c t l y the elements of order p i n G, and by hypothesis, they are a l l fixed by A . B y C o r o l l a r y 4.35, therefore, A acts t r i v i a l l y on ( G , + ) , and hence its action on G is t r i v i a l . • A similar argument allows us to prove a stronger version of the T h o m p son P x Q l e m m a i n the case where p is o d d . R e c a l l that i n T h e o r e m 4.31 we h a d a group A = P Q acting on a p-group G, where P is a p-subgroup and Q is a p'-subgroup of A , and b o t h are n o r m a l i n A . ( A n d we assumed t h a t C ( P ) C C ( Q ) . ) W e now drop the assumption that P < A . G

G

4.38. Theorem. Let A p - g r o u p G, w h e r e p > 2 , is a p - g r o u p a n d t h a t Q every element of G fixed

be a finite g r o u p t h a t acts v i a a u t o m o r p h i s m s o n a a n d l e t P a n d Q be s u b g r o u p s of A . Suppose that P < A a n d Q i s a p ' - g r o u p . Assume also that Q fixes by P . T h e n Q acts t r i v i a l l y o n G.

P r o o f . F i r s t , observe t h a t b o t h G and Q are n o r m a l i z e d by A i n the semidirect product G x > A . Therefore [ G , Q] is A - i n v a r i a n t , and it admits the action of A . Now assume t h a t G is abelian, so that by F i t t i n g ' s theorem, G = C { Q ) x [ G , Q ] . If [ G , Q ] > 1, then by L e m m a 4.32 a p p l i e d to the act i o n of P on [ G , Q ] , we have C ( P ) > I . T h u s P has n o n t r i v i a l fixed points i n [ G , Q ] , and by hypothesis, these P - f i x e d points are also Q-fixed points. T h i s is a c o n t r a d i c t i o n , however, since [ G , Q ] n C ( Q ) = 1, and it follows t h a t [ G , Q] = 1 i n this case, as wanted. G

[ G i 0 ]

G

Now for the general case, we can assume t h a t G > 1. W e proceed by i n d u c t i o n on G, observing t h a t if P < G and H admits the action of A , then the inductive hypothesis yields that [ H , Q] = 1. In particular, since [ G , Q] admits the action of A , we see that if [ G , Q] < G, then 1 =

Problems

4D

145

[G, Q, Q] = [ G , Q ] , a n d we are done. W e c a n assume, therefore, t h a t [ G , Q] = G. N o w G' < G admits the a c t i o n of A , a n d hence [ G ' , Q] = 1, a n d we have [ G ' , Q , G ] = 1 a n d [ G , G ' , Q ] C [ G ' , Q ] = 1. T h u s [ Q , G , G ' \ = 1, a n d since [ Q , G ] = [ G , Q ] = G, we conclude t h a t G' C Z ( G ) . T h u s G has nilpotence class at most 2, a n d we can a p p l y the B a e r t r i c k . Now

A acts o n the abelian group ( G , +) a n d every element fixed b y P

is fixed by Q. B u t we have already proved the t h e o r e m for a b e l i a n groups, and

so we deduce that Q acts t r i v i a l l y o n ( G , +), a n d hence also o n G .

Problems

•

4D

4 D . 1 . L e t A act v i a automorphisms o n G , a n d assume t h a t N < G admits A a n d t h a t N D C ( 7 V ) . A s s u m e that (|AT|, G

= 1.

(a) If A acts t r i v i a l l y on N , show t h a t its a c t i o n o n G is t r i v i a l . (b) If A acts faithfully o n G , show t h a t its a c t i o n o n N is faithful. H i n t . F o r (a) show t h a t [G, A ] C N . C o n s i d e r C ( N ) , r

where T = G x A

N o t e . If G is solvable a n d AT = F ( G ) , t h e n N D C ( N ) .

A l s o , if G is is

G

p-solvable w i t h C y ( G ) = 1 a n d N = O ( G ) , t h e n TV D C ( 7 V ) . T h u s i n p

G

either of these cases, if A acts faithfully a n d c o p r i m e l y o n G , t h e n A acts faithfully o n N . 4 D . 2 . L e t G be nilpotent w i t h class at most 2, a n d assume | G | is o d d . F o l l o w i n g the proof of L e m m a 4.36, construct a n a d d i t i v e structure o n G , and

(as u s u a l for a d d i t i v e l y w r i t t e n groups) w r i t e x - y to denote the s u m

of x a n d the additive inverse of y . Show t h a t xy - yx = [ x , y } . 4 D . 3 . L e t A act v i a automorphisms o n G , where ( | G | , \ A \ ) = 1, a n d assume t h a t one of A or G is solvable. Suppose t h a t A acts t r i v i a l l y o n every A¬ invariant proper subgroup of G , but that the a c t i o n of A o n G is n o n t r i v i a l . P r o v e a l l of the following. (a) G is a p-group. (b) G' C Z ( G ) . (c)

N o A - i n v a r i a n t subgroup H exists w i t h G' < H < G.

(d) G I G '

is elementary abelian.

(e) G' is elementary abelian (and possibly t r i v i a l ) . (f)

If p > 2, t h e n

(g) If p = 2, t h e n x

= 1 for a l l x G G . 4

= 1 for a l l x e G .

H i n t . F o r (c), a p p l y F i t t i n g ' s theorem to the a c t i o n of A o n

G/G'.

4.

146

Commutators

4 D . 4 . L e t A act v i a automorphisms on G , where G is a 2-group and A has odd order. Show t h a t i f A fixes every element x G G such that x = 1, then the a c t i o n of A on G is t r i v i a l . 4

4 D . 5 . L e t A be solvable w i t h derived length n , a n d suppose that A acts faithfully v i a automorphisms on an abelian group B , where ( \ B \ , \ A ^ ~ ^ \ ) = 1. Show that G = B x > A has derived length n + 1. Deduce that there exist solvable groups w i t h a r b i t r a r i l y large derived lengths. n

H i n t . F o r the second part, suppose A has derived length n a n d let C be cyclic of p r i m e order p not d i v i d i n g \ A < - - % Show t h a t if G = G U is the regular w r e a t h p r o d u c t , then G has derived length n + 1. n

4 D . 6 . L e t # : G -> G be the m a p that appeared i n our proof of F i t ting's theorem (Theorem 4.34.) Show that (in the n o t a t i o n of t h a t theorem) 0 ( G ) = C ( A ) a n d ker(0) = [G, A ] . G

4 D . 7 . L e t G be p-solvable, and suppose t h a t a Sylow p-subgroup of G is cyclic. L e t K C G be a p'-subgroup, and suppose that p divides | N ( A " ) | . Show that K C C y ( G ) . G

H i n t . Induct on | G | . If C y ( G ) > 1, a p p l y the i n d u c t i v e hypothesis to G = G/Cy

(G).

Otherwise, show that G has a n o r m a l Sylow p-subgroup.

Chapter 5

Transfer

5A We seek results t h a t can be used to prove that a group is not simple. In other words, we want to find techniques t h a t can produce n o r m a l subgroups of a group G , or equivalently, h o m o m o r p h i s m s 9 from G into some group H , a n d we want to find hypotheses sufficient to guarantee that k e r ( 9 ) is neither t r i v i a l nor is the whole group G. A s we shall see, the "transfer" h o m o m o r p h i s m , w h i c h we discuss i n this chapter, is a powerful t o o l for p r o v i n g such "good" theorems. W h a t groups H are available to serve as targets for homomorphisms from G ? One possibility, of course, is to take H to be a s y m m e t r i c group S. (Recall that this was how we proved T h e o r e m 1.1 a n d its corollaries.) A n o t h e r useful and well studied s i t u a t i o n is where H is the group of i n vertible n x n matrices over a field F . (Recall t h a t this group is the general linear group, denoted G L ( n , F ) . ) A h o m o m o r p h i s m from G into G L ( n , F ) is called an F - r e p r e s e n t a t i o n of G , a n d representation theory has proven to be a fertile source of good theorems, especially w h e n the field F is taken to be the complex numbers C . W e digress to m e n t i o n t h a t a n F-representation 9 of G determines a function x • G - > F , called the character associated w i t h 9, w h i c h is defined by setting x { s ) to be the trace of the m a t r i x 9 ( g ) . Perhaps surprisingly, it turns out t h a t i n the case where F = C , it suffices to study these characters i n order to determine most of the interesting properties of the corresponding representations. I n p a r t i c u l a r , it is not t e r r i b l y difficult to prove that ker(0) = { g e G | x { g ) = x ( l ) } , a n d so characters can be used as proxies for C-representations i n order to prove good theorems. Indeed, it is character theory that p r o v i d e d W . B u r n s i d e w i t h the key to his n

147

148

5.

Transfer

p Y - t h e o r e m , w h i c h asserts t h a t a group whose order has exactly two prime divisors cannot be simple. We can also choose subgroups of G t o serve as targets for homomorphisms denned o n G. T h a t is approximately what the transfer is: a certain h o m o m o r p h i s m from G into an arbitrary subgroup H . A s we shall see, however, technical considerations necessitate that the target group for the transfer h o m o m o r p h i s m should be abelian, a n d so instead of a subgroup H , we use its c o m m u t a t o r factor group H / H ' as the target. Let G be a n a r b i t r a r y (not necessarily finite) group, a n d let H C G be a subgroup of finite index. I n this section, we define the transfer m a p v : G -»• H / H ' , a n d we prove t h a t it is a h o m o m o r p h i s m . W e start b y fixing a set of representatives for the right cosets of H i n G. Such a set, called a r i g h t t r a n s v e r s a l for H i n G, is s i m p l y a set of elements of G chosen so t h a t each right coset of H contains exactly one member of the set. If T is a right transversal for H i n G, therefore, the right cosets H t for t G T are distinct, a n d they are a l l of the right cosets of H i n G. W e c a n thus use the elements of T as labels for the right cosets of H , a n d i n particular, \T\ = \ G : H \ . R e c a l l that G acts b y right m u l t i p l i c a t i o n o n the set of right cosets of H , a n d so we c a n write ( H x ) - g = H x g for each right coset H x of H i n G a n d each element g G G. Since the elements of the right transversal T can be viewed as labels for the right cosets of H , the action of G o n the set { H t | t G T } of right cosets uniquely determines a n action of G on T, and we define t - g accordingly. I n particular, ( H t ) - g = H ( t - g ) , or i n other words, if t G T a n d g G G, then t - g is the unique element of T that lies i n the coset ( H t ) - g = H t g . Since our "dot" action of G o n T is really just the action of G o n the right cosets of H , we see, for example, t h a t the stabilizer i n G of an element t G T is exactly the stabilizer of the coset H t , w h i c h we know is H . M o s t importantly, since we know t h a t right m u l t i p l i c a t i o n on right cosets really is an action, it follows t h a t the dot a c t i o n of G o n T is a true action, a n d thus, for example, ( t - x ) - y = t - ( x y ) for t G T a n d x,y G G. l

Now suppose that t G T a n d g G G. T h e n t g G H t g = H ( t - g ) , a n d hence there is a unique element he H such that t g = h ( t - g ) . Since t g { t - g ) ~ = h , it follows t h a t for a l l elements t G T a n d g G G, we have t g { t - g ) ~ G H . G i v e n g G G, we w o u l d like t o m u l t i p l y a l l of the elements t g i t - g ) - for i e T t o o b t a i n a n element of H depending o n g . There is an obvious difficulty here, however, since i f H is nonabelian, the order i n w h i c h this m u l t i p l i c a t i o n is carried out m a y matter. In order t o resolve fhis issue, we suppose that some specific (but arbitrary) ordering of the elements of T has l

l

1

5A

149

been declared, and we let V : G - > H be the function defined by the formula

teT

where i n c o m p u t i n g the product, the elements t E T are taken i n the specified order. A l t h o u g h this nomenclature is not standard, we refer to any m a p V : G —> H constructed i n this way as a pretransfer m a p . T h e r e are usually m a n y different pretransfer maps from G to each subgroup H because, in general, the m a p depends b o t h on the choice of the right transversal T and on the ordering of T. G i v e n a pretransfer m a p V : G - > H , where H C G is an a r b i t r a r y subgroup, we can define a new m a p v : G -> H / H " by composing V w i t h the canonical h o m o m o r p h i s m h -> 7i from H t o H = H / H ' . I n other words ufo)

7

=

V ^, 1

and we observe t h a t since H / H ' is abelian and each factor t g i t - g ) - i n the product defining V { g ) lies i n I f , it follows t h a t the m a p v is independent of the specific ordering o n T t h a t we used to define V. A s we shall see presently, v is also independent of the choice of the right transversal T, and so for each subgroup H C G, the m a p v is u n a m b i g u o u s l y defined. It is the transfer m a p from G to H / H ' , w h i c h is often inaccurately referred to as the transfer from G to H . (Perhaps we should m e n t i o n t h a t it is customary to use the letter "v" for the transfer because i n G e r m a n , "transfer" is "Verlagerung"). Before we present any theorems or proofs, we introduce some convenient n o t a t i o n . G i v e n two elements x a n d y i n our subgroup H , we can assert t h a t x and y have the same image i n H / H ' b y w r i t i n g H ' x = H ' y or x = y . Unfortunately, if either x or y is some c o m p l i c a t e d expression such as the p r o d u c t defining V { g ) above, the coset n o t a t i o n is somewhat unwieldy, and overbars are even worse. F o r that reason, we introduce the alternative n o t a t i o n x = y to mean the same t h i n g : t h a t x = y . (It w o u l d be more correct to w r i t e x = y m o d H ' , but we w i l l usually suppress explicit m e n t i o n of the m o d u l u s w h e n no confusion is likely to arise.) 5.1. T h e o r e m . L e t G be a g r o u p , a n d suppose t h a t H i s a s u b g r o u p of finite i n d e x . T h e n t h e t r a n s f e r m a p v . G - > H / H ' i s i n d e p e n d e n t of t h e c h o i c e of t h e r i g h t t r a n s v e r s a l used t o d e f i n e i t . P r o o f . L e t S and T be two right transversals for H i n G, a n d observe t h a t for each element t E T, there is a unique element s E S that lies i n the same right coset of H . W e can thus write s = h t for some uniquely determined element h E H , and we have t

t

t

t

S = {

St

It E T }= {h t \t E T} . t

150

5.

Transfer

Now let s G S be arbitrary, and let g G G. T h e n s = s for some element i 6 T a n d s - g is the unique element of S i n the coset H s g = H t g = H { t - g ) . It follows t h a t s - g = s . = h . ( t - g ) . W e thus have t

t g

sg(s-g)Writing V have

s

and V

T

Vs(g)

= J]

1

t

g

1

l

= h tg{h . {t-g))- = ht{tg{t-g)- )K;} . t

t

g

g

to denote pretransfers constructed from S and T , we

^(s-g)-

1

= [J

1

htitgit-g)- )^.]

Here, the first congruence is not necessarily an equality because we made no assumption about how the a r b i t r a r y orderings on S and T are related. The second congruence is v a l i d because we are w o r k i n g m o d u l o H , and so we are free to rearrange factors t h a t lie i n H . F i n a l l y , t - g runs over a l l of T as t does, a n d thus the first and t h i r d factors on the right cancel (modulo H ' ) , and we conclude that V ( g ) = V { g ) , as required. • 1

s

T

A l t h o u g h it m a y be reassuring to k n o w that the transfer m a p from G to H is u n a m b i g u o u s l y defined, as we have just shown, the really v i t a l fact about the transfer is that it is a h o m o m o r p h i s m , and the proof of that fact does not rely o n T h e o r e m 5.1. 5.2. T h e o r e m . L e t G be a g r o u p , a n d suppose t h a t H i s a s u b g r o u p of index. Then the transfer m a p v :G - > H / H ' is a h o m o m o r p h i s m .

finite

P r o o f . L e t T be a right transversal for H i n G, and let V be a pretransfer associated w i t h T. F o r t G T , we have t - { x y ) = ( t - x ) - y , a n d thus tixy^Hxy))-

1

l

l

= {tx{t-x)- ){{t-x)y{{t-x)-yy ).

Since b o t h factors o n the right lie i n H , we have V(xy)

= l[t(xy)(t-(xy))-

1

= JJ t ^ i - x ) " =

1

n^xM(i-x)-y)-

1

V{x)V{y),

where the last congruence holds because t-x runs over a l l of T as i does. It follows t h a t v ( x y ) = v ( x ) v ( y ) , and the proof is complete. • A l t h o u g h T h e o r e m 5.2 is not very deep, it enables us to prove the following, w h i c h is our first a p p l i c a t i o n of transfer theory. 5.3. T h e o r e m . Suppose t h a t G i s a finite g r o u p , a n d l e t p be a p r i m e d i v i s o r of \ G ' n Z ( G ) | . T h e n a S y l o w p - s u b g r o u p of G i s n o n a b e l i a n .

151

5A

P r o o f . L e t P G S y l ( G ) , and assume that P is abelian. L e t T be a right transversal for P i n G , and consider the transfer h o m o m o r p h i s m v : G -> P . Since P is abelian, we have u = V, where V is the pretransfer c o m p u t e d using T. p

Now G' n Z ( G ) is a n o r m a l subgroup of G w i t h order divisible by p, and so its intersection w i t h the Sylow subgroup P is n o n t r i v i a l . W e can thus choose a nonidentity element z G P D G' n Z ( G ) , a n d we compute v { z ) . If t G T , t h e n since z G Z ( G ) , we have P t z = P z t = P t , where the last equality holds because z G P . T h u s i is the element of T i n the coset P i s , and we conclude t h a t t-z = t. T h e n t z { t - z ) ~ = tzt~ = z, where the last equality follows because z is central. W e conclude t h a t v { z ) = V { z ) = z\ \ = z\ \. l

l

T

G:P

O n the other hand, v is a h o m o m o r p h i s m from G into the abelian group P , a n d hence the derived subgroup G' is contained i n ker(w). Since z G G ' , we have 1 = v ( z ) = z\ \, a n d thus z has order d i v i d i n g the p'-number | G : P | . B u t z G P , and so z has p-power order, and we conclude that z = l . T h i s c o n t r a d i c t i o n shows t h a t P must be nonabelian. • G:P

5.4. C o r o l l a r y . L e t Z C Z ( r ) n T ' , w/iere r i s a finite g r o u p . T h e n a S y l o w p - s u b g r o u p of T / Z i s n o n c y c l i c f o r every p r i m e d i v i s o r p of \ Z \ . P r o o f . L e t P G S y l ( r ) . W e are assuming that p divides \ Z \ and that Z C Z ( r ) n r ' , so it follows by T h e o r e m 5.3 t h a t P is not abelian. Since P n Z C Z ( P ) a n d P is nonabelian, we conclude t h a t P / ( P n Z ) cannot be cyclic. B u t P / ( P n Z ) = P Z / Z G S y l ( r / Z ) , a n d thus T / Z has a noncyclic Sylow p-subgroup. • p

p

We digress to provide some context for C o r o l l a r y 5.4. G i v e n a finite group G , consider pairs of groups ( T , Z ) , where Z C Z ( T ) and T / Z = G . (We say t h a t T is a central extension of G i n this situation.) O f course, it is t r i v i a l to find central extensions of G since we could s i m p l y take T = G x Z , where Z is any abelian group. T o prevent this, a n d to make the s i t u a t i o n more interesting, we insist that Z should be contained i n the derived subgroup T' as well as i n the center Z ( r ) . W i t h this a d d i t i o n a l requirement, the p r o b l e m is m u c h more restrictive. Indeed, one can prove that a m o n g all pairs ( T , Z ) , where Z C Z ( T ) n T' a n d T / Z = G , there is a unique (up to isomorphism) largest group Z t h a t can occur as a second component, and t h a t every other possible second component is a h o m o m o r p h i c image of this largest one. T h e largest possible second component of a p a i r (T, Z ) associated w i t h a given group G is called the Schur multiplier of G , and it is denoted M ( G ) . If (T, Z ) is one of our pairs i n w h i c h Z = M ( G ) , then, of course, | T | = \ G \ \ M ( G ) \ , and i n this case, T is said to be a Schur representation g r o u p for G . If the group G is perfect, w h i c h , we recall, means t h a t G' = G ,

5.

152

Transfer

t h e n it has a unique (up to isomorphism) Schur representation group, but in general, a given group G can have more t h a n one i s o m o r p h i s m type of Schur representation group. F o r example, if G is noncyclic of order 4, it is not very h a r d to show t h a t \ M ( G ) \ = 2, and thus b o t h the dihedral group D g and the quaternion group Q are Schur representation groups for G. 8

C o r o l l a r y 5.4 can be restated this way: If p is a prime divisor of \ M ( G ) \ , t h e n a Sylow p-subgroup of G must be noncyclic, and i n particular, it is n o n t r i v i a l . F o r example, the only p r i m e p for w h i c h a Sylow p-subgroup of the alternating group A is noncyclic is p = 2, and it follows that M ( A ) is a 2-group. I n fact, the Schur m u l t i p l i e r of A has order 2. (We cannot resist mentioning that \ M ( A ) \ = 2 for a l l integers n > 4 except n = 6 and n = 7; i n those two cases, the Schur m u l t i p l i e r has order 6. T h i s is another example of what seems to be c o m m o n phenomenon i n finite group theory: general patterns w i t h finite lists of exceptions.) h

5

b

n

Problems 5A 5 A . 1 . Suppose that G is abelian and H C G has index n . Show that the transfer h o m o m o r p h i s m from G to H is the m a p g g . n

5 A . 2 . Show that the transfer h o m o m o r p h i s m v : G ->• G / G ' is just the canonical h o m o m o r p h i s m from G to G / G ' . 5 A . 3 . L e t H C K C G, where \ G : H \ < oo, and suppose that S is a right transversal for H i n K and that T is a right transversal for K i n G. W r i t e ST

— { s t \ s e S, t e T } . (a) Show t h a t each element of ST is uniquely of the form st w i t h s G S and i G T. Show that S T is a right transversal for H i n G. |G:H| = \ G : K \ \ K : H \ . Let

Deduce that

s G -S and t G T, and suppose that g E G. Show t h a t

(si).gr =

_1

S'(i3(i«5) )(i-g).

(d) L e t V r : G —> i f be a pretransfer c o m p u t e d w i t h respect to T , and let v : A" -»• Tf/Tf' be the transfer h o m o m o r p h i s m . Show t h a t the composite m a p w : G TJ/TJ' denned by w ^ ) = v { V ( g ) ) is the transfer h o m o m o r p h i s m . T

5 A . 4 . L e t H C Z ( G ) , and assume t h a t | G : H \ = n < oo. L e t v : G -> H be the transfer h o m o m o r p h i s m . (a) I f h e H ,

show that

n

= h .

5B

153

(b) N o w assume that H is finite a n d t h a t \ H \ is coprime to n . Show that G = H x ker(v). N o t e . T h e fact t h a t H is a direct factor of G i n (b) can also be proved by an appeal to the Schur-Zassenhaus theorem. 5A.5.

L e t G be finite, and suppose C is a cyclic n o r m a l subgroup such that

G / C is cyclic. Show that \ M ( G ) \ divides \ C : G'\. H i n t . A p p l y L e m m a 4.6 to a group Y c o n t a i n i n g a subgroup Z such t h a t Y/Z

= G and Z C T' n Z ( T ) .

5 A . 6 . L e t D be dihedral of order 2 m u l t i p l i e r M ( D ) has order 2.

n

where n > 2. Show that the Schur

5 A . 7 . L e t B and C be cyclic subgroups of a finite group G, and suppose t h a t B C = G and B n C > 1. A s i n P r o b l e m 5A.5, assume that C « G and t h a t | C : G'\ = n . Show that the order of M ( G ) is s t r i c t l y less t h a n n . N o t e . I n p a r t i c u l a r , semidihedral and generalized quaternion groups have t r i v i a l Schur multipliers. 5 A . 8 . L e t A and B be a r b i t r a r y finite groups. (a) Show t h a t \ M ( A x £ ) | >

\M(A)\\M(B)\.

(b) A s s u m i n g that | A | and | B | are coprime, show t h a t M ( A x B ) = M ( A ) x M(£). 5B It is often difficult to o b t a i n useful i n f o r m a t i o n about the transfer m a p i f we work d i r e c t l y from the definition. B u t if we appeal to the transfer-evaluation l e m m a , w h i c h we present i n this section, c o m p u t a t i o n s become easier, and transfer becomes more useful as a t o o l for p r o v i n g theorems. We continue to allow our groups to be infinite, but of course, i n order for the transfer from a group G to a subgroup H to be defined, we must assume t h a t the index \ G : H \ is finite. 5.5. L e m m a (Transfer E v a l u a t i o n ) . L e t G be a g r o u p , a n d l e t H be a s u b g r o u p w i t h finite i n d e x n . L e t V : G - > H be a p r e t r a n s f e r m a p c o n s t r u c t e d u s i n g a r i g h t t r a n s v e r s a l T f o r H i n G, a n d fix a n element g G G. Then t h e r e e x i s t a subset T C T and positive integers n for t e T (depending o n g ) such t h a t t h e f o l l o w i n g h o l d . 0

(a) t g ^ t -

1

t

6 H for all t G T. 0

(b) V { g ) =. n ten

n

l

tg n~ .

0

5.

154

(c)

Transfer

£ TH = n . ier 0

(d) If g has f i n i t e o r d e r m, t h e n n

t

i s a d i v i s o r of m f o r a l l i G T . 0

O f course, the p r o d u c t i n assertion (b) is not well defined unless some order i n w h i c h to take the factors is established. T h i s is really irrelevant, n

x

however, since we are w o r k i n g m o d u l o H ' and by (a), each factor t g n ~

lies i n H . If H happens to be abelian, so that H ' = 1, then of course, assertion (b) says that v ( g ) = n*0 * > nt

_1

w

h

e

r

e

v

i s

t

h

e

transfer map.

P r o o f of L e m m a 5.5. T h e cyclic subgroup (g) C G acts on T v i a our "dot" action, and thus T decomposes into (#)-orbits. L e t T C T be a set of representatives of these orbits (so that each orbit contains exactly one member of T ) a n d let n be the size of the orbit containing t € T . O f course, the s u m of the orbit sizes is | T | = n , and this establishes (c). A l s o , (d) is immediate since if o ( g ) = m < oo, then \ ( g ) \ = m, and so the size of each (#)-orbit on T divides m. 0

0

Now

t

0

let (!) be a (#)-orbit on T, and let t G T lie i n O, so that \ 0 \ = n . 0

t

Since O = { t - x \ x £ (g)} = {t-g* \ i E Z } , l

we observe t h a t the elements t - g must be distinct for 0 < i < n , and that these are exactly the elements of O. A l s o , it should be clear that t - g = t. t

n t

1

R e c a l l that for every element s € T , we have s g ( s - g ) € H , and the p r o d u c t of these elements (in some order) is equal to V ( g ) . T a k i n g s = t - g \ we have 1

{t-g^git-g^ )for

a l l integers i. Since t - g

n t

1

1

- sg(s-g)-

€ H

= t, we see that the p r o d u c t of the elements

of this form w i t h 0 < i < n - 1 (taken i n order of increasing i ) is precisely t

tg

n

t

t - \ w h i c h therefore lies i n H . T h i s proves (a), and it shows that except

possibly for the order of the factors, the element t g

n t

t~

l

is the c o n t r i b u t i o n

to the p r o d u c t denning V ( g ) c o m i n g from the elements of T that lie i n O. Since T is the u n i o n of the various orbits and T

0

of these orbits, assertion (b) follows.

is a set of representatives

•

A s our first a p p l i c a t i o n of the transfer-evaluation l e m m a , we compute the transfer into a central subgroup. 5.6. T h e o r e m . L e t G be a g r o u p , a n d suppose that Z is a central subgroup of f i n i t e i n d e x n . T h e n t h e t r a n s f e r f r o m G t o Z i s t h e m a p g ^ g . T h i s m a p i s therefore a h o m o m o r p h i s m f r o m G into Z . n

N o t e t h a t since Z is abelian, the transfer to Z really does m a p to Z ; there is no need to work m o d u l o anything. A l s o , observe t h a t T h e o r e m 5.6

5B

155

generalizes b o t h P r o b l e m 5 A . 1 and P r o b l e m 5 A . 4 ( a ) . F i n a l l y , we m e n t i o n t h a t since, i n the s i t u a t i o n of the theorem, G / Z is a group of order n , it is obvious that g G Z . W h a t is not at a l l obvious, however, is t h a t the m a p g i-> g is a h o m o m o r p h i s m , so that { x y ) = x y for a l l x,y£G. n

n

n

n

n

P r o o f of T h e o r e m 5.6. Choose a right transversal T for Z i n G and let g G G . Choose a subset T C T and integers n for i G To, as i n the transferevaluation l e m m a (5.5). F o r t G T , therefore, we have t g ^ t ' G Z , and since Z C Z ( G ) , we have 0

4

1

0

t^'f-

1

1

= (i^T )' = t ^ i t g ^ t - ^ t = g

n t

.

B y L e m m a 5.5(b), the product over i G To of these elements is v ( g ) , where v :G Z is the transfer. Since £ n = n , however, this p r o d u c t is g . • n

t

We digress from our study of the transfer to discuss a n a p p l i c a t i o n of T h e o r e m 5.6. 5.7. T h e o r e m (Schur). Suppose T h e n G' i s a

finite

t h a t Z ( G ) has

finite

i n d e x i n a g r o u p G.

subgroup.

We need a few p r e l i m i n a r y results. F i r s t , we show t h a t i n the s i t u a t i o n of Schur's theorem, there are only finitely m a n y commutators i n G . O f course, this falls far short of the assertion of the theorem since i n general, not every element of G ' is a c o m m u t a t o r . 5.8. L e m m a . L e t T be a r i g h t t r a n s v e r s a l f o r Z ( G ) i n a g r o u p G. Then every c o m m u t a t o r i n G has t h e f o r m [ s , t ] , w h e r e s,t G T . I n p a r t i c u l a r , if | G : Z ( G ) | < oo, t h e n t h e r e a r e o n l y finitely many different commutators. P r o o f . T h e second statement follows from the first since \T\ = \ G : Z ( G ) | . To prove the first assertion, observe t h a t every element of G can be w r i t t e n i n the form xs, where x G Z ( G ) a n d s G T. To complete the proof, therefore, it suffices to show that [ x s , yt] = [ s , t ] , where x , y G Z ( G ) and s,teT. W e have [ x s , y t ] = [ x , y t } [ s , y t ] = [ s , y t ] , where the second equality holds because [ x , y t ] = 1 since x is central. A l s o , [ s , y t ] = [ i / M ] = ([y, s ] % s])" = [i, s } ' = [ s , t ] , where the t h i r d equality holds since y is central. • s

- 1

1

1

We

also need the following easy consequence of T h e o r e m 5.6.

5.9. C o r o l l a r y . Suppose that G i s a g r o u p a n d t h a t \ G: Z(G)| = n . t h e n t h p o w e r of every c o m m u t a t o r i n G is the identity.

Then

n

P r o o f . T h e m a p g ^ g is a h o m o m o r p h i s m from G into the abelian group Z ( G ) , and thus the c o m m u t a t o r subgroup G' is contained i n its kernel. • The final ingredient i n the proof of T h e o r e m 5.7 is the following somewhat more general fact due to A . P . D i e t z m a n n .

5.

156

Transfer

5 . 1 0 . T h e o r e m ( D i e t z m a n n ) . L e t G be a g r o u p , a n d l e t X C G be a finite subset t h a t i s c l o s e d u n d e r c o n j u g a t i o n . Assume that there is a positive i n t e g e r n such t h a t x = 1 f o r a l l x e X . T h e n ( X ) i s a finite s u b g r o u p of G. n

P r o o f . L e t S be the subset of G consisting of a l l products of finitely m a n y members of X , allowing repeats, and allowing the "empty" p r o d u c t , w h i c h is equal to 1. T h e n S is closed under m u l t i p l i c a t i o n , a n d also, since x = 1 for each element x G X , we see that x ~ = x ' lies i n S. It follows t h a t S is closed under inverses, and hence S = ( X ) . E v e r y element of ( X ) , therefore, is a p r o d u c t of members of X , and we w i l l show t h a t it is never necessary to use more t h a n (n - 1 ) \ X \ factors. Since X is finite, it w i l l follow t h a t ( X ) is finite, as wanted. n

l

7 1

1

Consider an a r b i t r a r y p r o d u c t g = x i x - - - x of TO not necessarily distinct elements of X , and suppose t h a t some p a r t i c u l a r element x € X occurs at least k times a m o n g the factors x where k > 1. W e argue that g can be rewritten as a p r o d u c t of m elements of X i n such a way that the first factor is x , and there are still a t o t a l of at least k factors equal to x . T o see w h y this is true, suppose that t is the smallest subscript such that x = x . W e c a n assume t h a t i > 1, and we observe t h a t 2

m

u

t

X \ X

2

• • •X

t

- \ X

t

=

X \ X

2

• ••X

t

- \ X = =

X ( X \ X X ( X

1

)

•••

2

X

{ X 2 )

X X

- l )

t

X

•••{x - ) t

X

l

.

T h e expression on the right is a product of t elements of X because X is closed under conjugation by x , and of course, its first factor is x . A l s o , there are at least k - 1 copies of x among the elements x w i t h t < i < TO, a n d so i f we m u l t i p l y the above expression on the right by x i • • • x , we get a new representation of g as a product of TO elements of X , and these include a t o t a l of at least k copies of x . T h e first factor i n this p r o d u c t is x , as desired. {

t +

m

A s s u m i n g now (as we may) that x = x , suppose t h a t k > 2. W e can repeat the above argument for the product x x • • • x , a n d so we can assume t h a t x = x . C o n t i n u i n g like this, we see that we c a n assume t h a t g is a p r o d u c t of m elements of X such that the first k factors are a l l equal to x . x

2

3

m

2

N o w fix an a r b i t r a r y element g € ( X ) and w r i t e g = x x x • • • x , where the factors 6 X are not necessarily distinct, a n d where m is as s m a l l as possible. S u p p o s i n g t h a t m > ( n - l ) \ X \ , we derive a c o n t r a d i c t i o n . B y the "pigeon-hole" principle, at least one element x € X must occur at least n times among the factors x for 1 < i < TO, and by the above reasoning, it is no loss to assume that the first n factors are a l l copies of x . B u t x = 1, x

2

%

m

X i

%

n

5C

157

and

thus g = x

n

+

l

x

n

+

• ••x

2

m

is a p r o d u c t o f m - n elements of X.

the desired c o n t r a d i c t i o n by the assumed m i n i m a l i t y of m .

T h i s is

•

F i n a l l y , we assemble the pieces to o b t a i n a proof of Schur's theorem. P r o o f of T h e o r e m 5.7. L e t X be the set of a l l c o m m u t a t o r s i n G, a n d observe t h a t \X\ is finite by L e m m a 5.8. x,

A l s o , since [x,y]*>

9

= [x°,y }

for

y a n d g i n G, we see t h a t X is closed under conjugation. N o w w r i t e

n = \ G : Z ( G ) | , so t h a t by C o r o l l a r y 5.9, we have x result now follows by T h e o r e m 5.10.

n

= 1 for a l l xeX.

The

•

Problems 5B 5 B . 1 . L e t G be finite, a n d suppose t h a t P G S y l ( G ) a n d that g G P has p

order p . If g G G', but g £ P ' , show that ^

G P ' for some element i G G

with i 0 P . H i n t . L e t u be the transfer h o m o m o r p h i s m from G to P / P ' a n d observe t h a t g G ker(u), a n d thus V ( g ) G P ' , where V is a corresponding pretransfer map. A p p l y the transfer-evaluation l e m m a , a n d note t h a t each integer n is either 1 or p . t

5 B . 2 . I n the s i t u a t i o n of T h e o r e m 5.10, suppose | A | = m.

Show t h a t

m

\ ( X ) \ < n . H i n t . Write X

= { x e

has the form ( x i ) * ( x ) 2

x ,..., x

u

2

e 2

m

}

a n d show t h a t every element of ( X )

e

• • • ( x ) ™ , where the e m

%

are integers satisfying

0 < e* < n . 5 B . 3 . L e t G be an a r b i t r a r y group. Show t h a t a n element x G G lies i n some finite n o r m a l subgroup of G i f a n d o n l y if x has finite order a n d its conjugacy class i n G has finite size. 5C We

r e t u r n now to finite groups. If we want to use transfer theory to prove

t h a t some group G is not simple, we need to choose a subgroup P and H/H'

C G,

we must show t h a t the kernel of the transfer h o m o m o r p h i s m v : G is neither t r i v i a l nor the whole group G. If P is proper i n G, t h e n

automatically, k e r ( v ) > 1 since i n t h a t case, v cannot be injective. t h a t we are assuming that G is finite.)

(Recall

T o prove "good" n o n s i m p l i c i t y

theorems, therefore, it is sufficient to find conditions t h a t w i l l guarantee t h a t ker(?j) < G, or equivalently that there exists at least one element x G G such t h a t v ( x ) ^ 1.

158

5.

Transfer

G i v e n a n element x G G , it is sometimes easier to compute v ( x ) if x happens to lie i n the transfer target H . O f course, i f we l i m i t our attention to elements of H and we can show that not every element of H lies i n ker(u), this is good enough. (Obviously, i f H % ker(u), then ker(?j) < G, and i n that case, G is not simple.) I n fact, i f H is a Sylow subgroup of G, or more generally, if H is a H a l l subgroup, then the converse of this assertion is true too: if k e r ( v ) < G, then H % ker(v). I n the s i t u a t i o n where H is a H a l l subgroup, therefore, we lose n o t h i n g by c o m p u t i n g the transfer only on elements of H . T h e following easy l e m m a proves somewhat more t h a n this. 5.11. L e m m a . L e t H be a H a l l s u b g r o u p of a f i n i t e g r o u p G, a n d l e t v : G - > H / H ' be t h e t r a n s f e r h o m o m o r p h i s m . T h e n v { H ) = v { G ) , a n d so \ H : H D k e r { v ) \ = |G:ker(u)|. P r o o f . W e k n o w that \ G : ker(7j)| = \ ( G ) \ , and since v { G ) C H / H ' , we deduce that \ G : ker(v)| divides \ H \ . T h e n \ G : ker(v)| is coprime to \ G : H \ , and so k e r ( v ) H = G by L e m m a 3.16. T h e result now follows. • V

A g a i n let H C G, and consider the transfer h o m o m o r p h i s m v : G - > H / H ' . W h e n we a p p l y the transfer-evaluation l e m m a to compute v ( x ) for some element x G G, we are led to consider elements of H of the form t x t ~ for certain elements i e G and positive exponents n . U x e H , w h i c h as we now k n o w is often the case of interest, then also x G H . I n this situation, the elements x and t x t ~ b o t h lie i n H , and they are clearly conjugate in G ( v i a the element i ) , but of course, they need not be conjugate i n H . n t

1

t

n

n

t

n t

t

l

M o t i v a t e d by this, we consider the intersection of a conjugacy class X of G w i t h a subgroup H , where we assume that X n H is nonempty. Clearly, if x G A n J f and y lies i n the conjugacy class of H c o n t a i n i n g x , then y £ X D H , a n d this shows t h a t I n i i i s a u n i o n of conjugacy classes of H . W e say t h a t classes Y\ and F of H are fused i n G i f Y and F are contained i n the same class of G . A l s o , if H C AT C G , we say that AT controls G - f u s i o n i n H if every two classes of H t h a t are fused i n G are already fused i n K . Equivalently, AT controls G-fusion i n H if and only i f every pair of G-conjugate elements of H are ^ - c o n j u g a t e . I n particular, G always controls G-fusion i n H , and H controls G-fusion i n itself precisely when no two distinct classes of H are fused i n G . A s we shall see, the n o t i o n of "control of fusion" is v i t a l for understanding transfer, especially transfer to Sylow subgroups, and more generally, to H a l l subgroups. 2

We

x

2

begin our s t u d y of fusion w i t h the following general result.

5.12. L e m m a . L e t P G S y l ( G ) , w h e r e G i s a f i n i t e g r o u p . controls G-fusion i n C (P). p

G

Then N (P) G

5C

159

P r o o f . A s s u m i n g t h a t x , y G C { P ) are conjugate i n G , we must show t h a t in fact, these elements are conjugate i n N ( P ) . W r i t e y = x w i t h g G G, a n d observe t h a t since x G C ( P ) , we have y = x G C { P ) = C {P ). T h e n P C C ( y ) , a n d hence P is a S y l o w p-subgroup of C ( y ) . Also y G C ( P ) , and so P C C ( y ) , and hence P is also a Sylow p-subgroup of C ( y ) . W e can now a p p l y the Sylow C - t h e o r e m i n the group C { y ) to the two Sylow p-subgroups P and P , and we deduce t h a t ( P ) = P , for some element c € C ( y ) . T h e n #c G N ( P ) and x ^ = y = y , and so x a n d y are conjugate i n N ( P ) , as required. • G

9

G

9

9

G

9

G

9

G

9

G

G

G

G

G

G

9

9

c

G

c

c

G

G

We can now prove B u r n s i d e ' s n o r m a l p-complement theorem. (Recall t h a t i f p is a prime, then a n o r m a l p-complement i n a finite group G is a n o r m a l subgroup w i t h index equal to the order of a Sylow p-subgroup. E q u i v a l e n t s , it is a subgroup N < G such t h a t \ N \ is not divisible by p and \ G : N \ is a power of p. In s t i l l other words, a n o r m a l p-complement is a n o r m a l H a l l p'-subgroup. A group t h a t has a n o r m a l p-complement is said to be p - n i l p o t e n t . ) 5 . 1 3 . T h e o r e m ( B u r n s i d e ) . L e t P G S y l ( G ) , w h e r e G i s a finite a n d suppose P C Z ( N ( P ) ) . Then G has a n o r m a l p - c o m p l e m e n t . p

group,

G

P r o o f . I n this s i t u a t i o n , P is abelian, so P C C ( P ) . B y L e m m a 5.12, therefore, every two elements of P t h a t are conjugate i n G are conjugate i n N ( P ) . B u t P is central i n N ( P ) , so no two distinct elements of P can be conjugate i n N ( P ) . It follows t h a t no two distinct elements of P can be conjugate i n G . G

G

G

G

Now consider the transfer h o m o m o r p h i s m v : G -> P , and recall t h a t in this case, where P is abelian, there is no d i s t i n c t i o n between the transfer and pretransfer. L e t x G P , and a p p l y the transfer-evaluation l e m m a to compute t h a t n

v ( x ) = J]

l

tx H~ ,

teT

0

where T is some subset of a right transversal T for P i n G . A l s o , £ n = \ G : P | , a n d we k n o w that each of the factors t x ^ t ' lies i n P . Since tx n~ and x are G-conjugate elements of P , the first paragraph of the proof guarantees t h a t they are equal, a n d thus x = t x n ~ for a l l t G T . Then 0

t

1

n

l

n

t

n

t

n

l

0

v(x)

=Y[x

n

t

=

x

| G : F |

,

ten

and so if v ( x ) = 1, t h e n x has order d i v i d i n g \ G : P | . B u t x G P , and so the order of x also divides | P | . It follows t h a t x = 1, a n d hence P n k e v ( v ) = 1. B y L e m m a 5.11, we have | G : ker(u)| = | P | , a n d so ker(?j) is the desired n o r m a l p-complement i n G . •

5.

160

Transfer

5.14. C o r o l l a r y . L e t P e S y l ( G ) , w h e r e G i s a finite g r o u p a n d p i s t h e s m a l l e s t p r i m e d i v i s o r of \ G \ , a n d assume t h a t P i s c y c l i c . T h e n G has a normal p-complement. p

P r o o f . W e k n o w t h a t N ( P ) / C ( P ) can be i s o m o r p h i c a l l y embedded i n A u t ( P ) , w h i c h , because P is cyclic, has order < p ( \ P \ ) , where