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G i v e n TV < G , a subgroup H C G is a c o m p l e m e n t for TV i n G i f / V P = G and N n P = 1, and if /V has a complement i n G , we say that G splits over N . N o t e t h a t i f P complements N i n G , t h e n H = H / { N n P ) = 7 V P / / V = G / / V , a n d thus a l l complements for N i n G are i s o m o r p h i c to G / i V , a n d hence they are i s o m o r p h i c t o each other. A n easy example where a group fails to split over some n o r m a l subgroup is G = G4, the c y c l i c group of
order 4, where N is the unique subgroup of order 2. It is clear t h a t N is not c o m p l e m e n t e d i n G because i f a complement existed, it w o u l d be i s o m o r p h i c to G/N, w h i c h has order 2. B u t N is the o n l y subgroup of order 2 i n G , and of course, N is not a complement for itself.
If P complements N i n G , t h e n each element g € G is of the form g = n h , w i t h n e N a n d h e H , a n d this f a c t o r i z a t i o n is u n i q u e . ( T h e uniqueness follows v i a a n elementary c o m p u t a t i o n , b u t i f G is finite, a n easier argument is t o observe t h a t \ G \ = \ N \ \ H \ a n d so the surjective m a p from pairs (n, h ) to elements n h G G must also be injective.) It is also true, a n d for s i m i l a r reasons, t h a t each element of G is u n i q u e l y of the form h n , w i t h h G H a n d n e J V . (To be specific, suppose t h a t g = n h , a n d we w i s h to w r i t e g = h ' n ' , w i t h ri e N a n d h ' G P . W e have g e N h = h N since J V ^ G , a n d thus we must have /»' = h . T h e equation / i n ' = n h t h e n yields n ' = nh. )
If P complements N i n G , t h e n clearly, every conjugate of H i n G also complements / V . In general, however, a n o r m a l subgroup N of G m a y have nonconjugate complements, a l t h o u g h as we have seen, a l l complements are i s o m o r p h i c . A n example where complements are not conjugate occurs is i n the K l e i n group G = G2 x G2. If TV is any one of the three subgroups of
66 3. S p l i t E x t e n s i o n s
order 2, t h e n each of the other two subgroups of order 2 complements N , b u t these complements are not conjugate i n G since G is abelian.
Our goal i n this section is the following: given groups N a n d H , con- struct a l l (up t o i s o m o r p h i s m ) groups G h a v i n g a n o r m a l subgroup N 0 com-
plemented b y a subgroup H0, where N = N 0 a n d H = H0. (We refer t o such
a group as a split extension of H by N , but unfortunately, the literature is inconsistent o n this p o i n t , a n d t h i s group is sometimes also referred to as a split extension of N b y H . ) O n e such split extension, of course, is the (external) direct p r o d u c t G = N x H , w h i c h is the set of ordered pairs (n, h ) w i t h n € N a n d h € H , a n d where m u l t i p l i c a t i o n is componentwise. Here, we take N 0 = { ( n , 1) | n € N } and H 0 = { ( l , h ) \ h e H } , a n d it is clear t h a t
N0 = N , HQ = H , N o < G a n d H 0 is a complement for N 0 i n G. In this case,
the complement H Q also happens to be n o r m a l , and i n general, it is easy t o
see t h a t i f the n o r m a l subgroup 7V0 has a n o r m a l complement H 0 i n G, t h e n
G is the (internal) direct p r o d u c t of i V0 and H0, a n d so G = N x H . O u r
task, therefore, is t o generalize the direct p r o d u c t c o n s t r u c t i o n sufficiently so t h a t we get complements H 0 for i V0 i n G, where H 0 is not necessarily
n o r m a l i n G. T h e desired c o n s t r u c t i o n is called the "semidirect p r o d u c t " . Before we proceed, however, we digress briefly t o discuss extensions t h a t are not necessarily s p l i t . G i v e n groups N a n d H , a group G is said to be an e x t e n s i o n of H by iV if there exists N 0 < G such t h a t N 0 = N
and G/N0 = H . N o t e t h a t i n our definition, the group preceded b y the
w o r d "of" corresponds t o the factor group, a n d the group preceded b y "by" corresponds to the n o r m a l subgroup. A s we m e n t i o n e d , however, this use of prepositions is sometimes reversed i n the literature, so readers s h o u l d a t t e m p t t o determine the precise m e a n i n g from the context.
T h e r e is a more fundamental a m b i g u i t y here. If G is an extension of H by N , t h e n the n o r m a l subgroup N 0 of G such t h a t N 0 = N a n d G/N0 = H is
not, i n general, u n i q u e l y determined. I n other words, G can be an extension of H b y N i n more t h a n one way. In fact, it can h a p p e n t h a t w i t h one choice of the n o r m a l subgroup N 0 the extension is split, a n d w i t h another it is not.
P r o p e r l y speaking, therefore, we really s h o u l d say s o m e t h i n g like " G is an extension of H b y N w i t h respect to the n o r m a l subgroup 7V0", and then it
w o u l d make sense to ask i f the extension is s p l i t .
Imagine t h a t the "extension p r o b l e m " c o u l d be c o m p l e t e l y solved. Sup- pose, i n other words, t h a t we k n e w how t o construct (up t o i s o m o r p h i s m ) a l l extensions of i f by A for given finite groups H a n d N . (We stress t h a t this seems t o be far from a realistic assumption.) T h e n , given t h a t we k n o w a l l simple groups, we c o u l d recursively construct a l l finite groups. T o do t h i s , assume t h a t we have already constructed a l l groups of order less t h a n n , where n > 1. E a c h group G of order n has some m a x i m a l normal subgroup,
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say N . Since the factor group G / N is s i m p l e , every group of order n c a n be constructed as an extension of some (known) s i m p l e group S b y some (already constructed) group N of order n/\S\. B u t even i f we c o u l d do this, we w o u l d s t i l l be faced w i t h the very difficult p r o b l e m of d e c i d i n g whether or not two groups of order n constructed i n this way are i s o m o r p h i c .
R e t u r n i n g to split extensions now, suppose t h a t G is s p l i t over a n o r m a l subgroup N , a n d let H be a complement. Observe t h a t H acts o n N v i a con- j u g a t i o n i n G, a n d i n fact, conjugation b y h £ H induces an a u t o m o r p h i s m of N . U H < G, t h e n since i V n H = 1, we know that H centralizes N , and so this conjugation a c t i o n is t r i v i a l . If we w i s h to construct split extensions t h a t are not merely direct p r o d u c t s , therefore, we must consider n o n t r i v i a l conjugation actions of H o n N . Before proceeding w i t h our c o n s t r u c t i o n , however, we show t h a t no further i n f o r m a t i o n is needed; the subgroups N a n d H a n d the conjugation a c t i o n of H o n N c o m p l e t e l y determine G u p to i s o m o r p h i s m .
To state the following uniqueness theorem, we i n t r o d u c e some conve- nient, b u t n o n s t a n d a r d n o t a t i o n . C o n s i d e r i s o m o r p h i c groups N and 7V0 )
where some specific i s o m o r p h i s m is given. F o r n £ N , we w i l l w r i t e n0 e iVo
to denote the image of n under the given i s o m o r p h i s m , so t h a t we c a n t h i n k of ( )0 as the name of the i s o m o r p h i s m . W e w i l l use the same name ( )0 to
denote a given i s o m o r p h i s m from H to H 0 .
3.1. L e m m a . L e t G a n d G0 be g r o u p s , a n d suppose t h a t N < G i s c o m p l e - m e n t e d by H , a n d N 0 < G0 i s c o m p l e m e n t e d by H 0 . Assume t h a t N = N 0 a n d H H 0 , w h e r e each of these i s o m o r p h i s m s i s d e n o t e d by ( ) 0 , a n d suppose t h a t ( nh)0 = (no)'1 0 f o r a l l elements n £ N a n d h <E H . T h e n t h e r e i s a u n i q u e i s o m o r p h i s m f r o m G t o G0 t h a t extends t h e g i v e n i s o m o r p h i s m s N -> JV0 a n d H - > H 0 .
P r o o f . E v e r y element g £ G is u n i q u e l y of the f o r m h n , where h £ H and n £ N , a n d so i f we seek an i s o m o r p h i s m 9 : G G0 t h a t extends the
given isomorphisms N - > N 0 and H ->• H 0 , we have no choice b u t to define
9 ( g ) = 9 ( h n ) = 9 { h ) 9 { n ) = h0n0. Since the d e c o m p o s i t i o n g = h n is unique,
9 is well defined, a n d since every element g0 £ G0 has the form h0n0 w i t h
h0 £ H o and n0 £ N0, it is clear t h a t 9 is surjective. It is injective because
the d e c o m p o s i t i o n g0 = h0n0 is unique a n d h a n d n are the unique elements
of H a n d N t h a t m a p to h o and n0 respectively.
It remains to show t h a t 0 is a h o m o m o r p h i s m , a n d so we must c o m p u t e 9 ( ( h n ) ( k m ) ) , where h , k £ H and n , m £ N . W e have
68 3. S p l i t E x t e n s i o n s
a n d so b y definition, a p p l i c a t i o n of 9 to this element yields
{ h k ) o ( nkm )0 = h0k o { nk) o m0 = hQk0{nQ)k°mQ = h0n0k o m o ,
a n d this is e x a c t l y 9 { h n ) 9 { k m ) , as required. •
In order to construct a l l possible split extensions, we need to examine a l i t t l e more a b s t r a c t l y the conjugation a c t i o n of a group H o n a group N . R e c a l l t h a t i f i f acts o n a set ft, t h e n each element h€H induces a m a p o~}i '. ft —> ft, defined b y a ^ a - h . F u r t h e r m o r e , a h is a p e r m u t a t i o n of ft,
a n d the m a p h ^ a h is a h o m o m o r p h i s m from i f i n t o the s y m m e t r i c group
S y m ( f t ) . It s h o u l d be clear t h a t the given a c t i o n of i f on ft is completely d e t e r m i n e d b y this h o m o m o r p h i s m from i f into S y m ( f t ) .
Suppose now t h a t the set ft of the previous p a r a g r a p h happens to be a group, w h i c h we now call N . I n this case, it is n a t u r a l to require t h a t for all elements h e H , the m a p a h : i V -> N is a c t u a l l y an a u t o m o r p h i s m (and
not j u s t a p e r m u t a t i o n ) of N . O f course, since this m a p is a u t o m a t i c a l l y b o t h injective a n d surjective, it suffices to check t h a t it is a h o m o m o r p h i s m , or e q u i v a l e n t s , t h a t { x y ) - h = { x - h ) { y - h ) for a l l x,y € N .
G i v e n groups i f and N , we say t h a t i f acts v i a a u t o m o r p h i s m s o n N if i f acts o n N as a set, and i n a d d i t i o n , ( x y ) - h = { x - h ) ( y - h ) for a l l x,y £ N a n d h e H . J u s t as an a c t i o n of i f o n ft determines and is d e t e r m i n e d b y a h o m o m o r p h i s m i f -> S y m ( f t ) , so too, an a c t i o n v i a a u t o m o r p h i s m s of i f on N determines a n d is d e t e r m i n e d b y a h o m o m o r p h i s m i f -> A u t ( i V ) .
P e r h a p s the most n a t u r a l examples of actions v i a a u t o m o r p h i s m s are where i f is a c t u a l l y a subgroup of A u t ( i V ) a n d n-h is s i m p l y ( n ) h . O t h e r examples of actions v i a a u t o m o r p h i s m s occur w h e n b o t h i f and N are sub- groups of some group G w i t h i f C NG( J V ) , a n d where the a c t i o n is given
by conjugation w i t h i n G. In p a r t i c u l a r , the a c t i o n of an a r b i t r a r y group G on itself b y conjugation is an a c t i o n v i a a u t o m o r p h i s m s .
It is c o m m o n to use e x p o n e n t i a l n o t a t i o n instead of dots for actions v i a a u t o m o r p h i s m s , and so we w r i t e n h instead of n - h , where n G N a n d h € i f .
T h i s is n a t u r a l for conjugation actions, of course, b u t it is p o t e n t i a l l y con- fusing for actions v i a a u t o m o r p h i s m s t h a t are not defined by conjugation inside some given group. T h e following theorem, however, asserts t h a t ev- ery a c t i o n v i a a u t o m o r p h i s m s is essentially a conjugation a c t i o n w i t h i n a n a p p r o p r i a t e group. T o some extent, therefore, this justifies the e x p o n e n t i a l n o t a t i o n for actions v i a a u t o m o r p h i s m s .
In order to state our result, we again use the n o t a t i o n ( )0 to describe
i s o m o r p h i s m s N N 0 a n d i f -> i f0. Technically, this is ambiguous, how-
ever, because we w i l l not assume t h a t N a n d i f are disjoint, a n d so i f x lies i n b o t h of these groups, it m a y not be clear w h i c h i s o m o r p h i s m to a p p l y
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i n order to c o m p u t e x0. Nevertheless, we t r u s t t h a t t h i s a m b i g u i t y w i l l not
cause confusion. 3.2. T h e o r e m . L e t H a n d N be g r o u p s , a n d suppose t h a t H acts o n N v i a a u t o m o r p h i s m s . T h e n t h e r e exists a g r o u p G c o n t a i n i n g a n o r m a l s u b g r o u p N0 = N , c o m p l e m e n t e d by a s u b g r o u p H 0 * H , a n d such t h a t f o r a l l n € N a n d h < E H , w e h a v e ( nh) o = ( n o )ho. H e r e , n h G N i s t h e r e s u l t of l e t t i n g h a c t o n n i n t h e g i v e n a c t i o n , w h i l e ( n o )h o € N 0 i s t h e r e s u l t of c o n j u g a t i n g n 0 by h 0 i n G.
It follows b y L e m m a 3.1 t h a t the group G of T h e o r e m 3.2 is u n i q u e l y d e t e r m i n e d b y N and H a n d the given a c t i o n v i a a u t o m o r p h i s m s of H o n TV; it is called the semidirect product of N by H w i t h respect to the given a c t i o n . A l s o , b y L e m m a 3.1, every group G w i t h a n o r m a l subgroup N a n d complement H is i s o m o r p h i c to a semidirect p r o d u c t of N by H , and so once we prove T h e o r e m 3.2, it w i l l be fair to say t h a t we have constructed all possible split extensions (up to i s o m o r p h i s m ) .
A c o m m o n n o t a t i o n for a semidirect p r o d u c t G of N b y H is G = N x > H . ( T h i s n o t a t i o n is incomplete, of course, because it fails to m e n t i o n the specific a c t i o n v i a a u t o m o r p h i s m s t h a t is essential i n the definition. W h e n we construct a group G = N x < H , we s h o u l d always specify the action.) N o t e t h a t the semidirect p r o d u c t is a direct p r o d u c t i f a n d o n l y i f H 0 < G, a n d
t h i s happens precisely w h e n the conjugation a c t i o n of H 0 o n JV0 is t r i v i a l ,
or e q u i v a l e n t ^ , w h e n the o r i g i n a l a c t i o n of H o n AT is trivial.
We offer a m n e m o n i c for the correct use of the s y m b o l ' V , w h i c h looks like " x " w i t h an e x t r a l i t t l e v e r t i c a l line o n the r i g h t . If G = N x > H , the l i t t l e line reminds us t h a t from the point of v i e w of H , the semidirect p r o d u c t can be different from a direct p r o d u c t since H m a y not be n o r m a l . B u t N is n o r m a l i n G, a n d so from the point of v i e w of N , the semidirect p r o d u c t resembles a direct p r o d u c t .
W e often identify N w i t h N 0 a n d H w i t h H 0 v i a the i s o m o r p h i s m s
x x Q, so t h a t the semidirect p r o d u c t G — N x > H c a n be viewed as a split
extension w i t h N < G, N H = G a n d N n H = 1, and where the original a c t i o n v i a a u t o m o r p h i s m s of H o n N is j u s t the c o n j u g a t i o n a c t i o n w i t h i n G. W e stress, however, t h a t t h a t t h i s i d e n t i f i c a t i o n is not always a good idea, a n d it can lead to confusion. C o n s i d e r , for e x a m p l e , an a r b i t r a r y group G a n d let T = G x G w i t h respect to the n a t u r a l conjugation a c t i o n of G o n itself. (Somewhat remarkably, it t u r n s out t h a t T = G x G for a l l groups G.)
It is seldom necessary or appropriate to refer to the p r o o f of T h e o r e m 3.2. In other words, the specific c o n s t r u c t i o n of the semidirect p r o d u c t is largely
70 3. S p l i t E x t e n s i o n s
irrelevant. T h e p o i n t here is t h a t given the three ingredients: a group N , a group H a n d an a c t i o n v i a a u t o m o r p h i s m s of H o n N , there is a group G, w h i c h is unique u p to i s o m o r p h i s m and satisfies the conclusions of T h e o r e m 3.2. It is u s u a l l y unnecessary to k n o w e x a c t l y how G is constructed. (Indeed, there is more t h a n one way to prove T h e o r e m 3.2, w h i c h means t h a t there is more t h a n one possible c o n s t r u c t i o n for the semidirect product.) In order to emphasize this p o i n t , we give a few a p p l i c a t i o n s of semidirect p r o d u c t s before we present the c o n s t r u c t i o n .
R e c a l l t h a t i n C h a p t e r 2, we defined the d i h e d r a l group D = D 2 n as a
group h a v i n g a n o n t r i v i a l c y c l i c subgroup C of order n such t h a t a l l elements of D - C are i n v o l u t i o n s . W e argued using geometric reasoning t h a t such a group a c t u a l l y exists w h e n n > 3. U s i n g semidirect p r o d u c t s , we can give a cleaner c o n s t r u c t i o n of the d i h e d r a l group D 2 n a n d some related groups.
Let A be an a r b i t r a r y a b e l i a n group, and let T = ( t ) be cyclic of order 2. Since t is the o n l y n o n i d e n t i t y element of T , we c a n define a n a c t i o n of T o n A b y setting x t = x ~l for a l l x e A . (Since t2 = 1 a n d ( a ; "1) "1 = x , t h i s is
clearly a genuine action.) Because A is abelian, we have ( x y ) ~l = x ~ly ~l,
a n d so our a c t i o n of T o n A is a n a c t i o n v i a a u t o m o r p h i s m s . N o w let G = A x T w i t h respect to this action, a n d identify (as we may) A and T w i t h the corresponding subgroups of the semidirect p r o d u c t G. T h e n A < G a n d A is c o m p l e m e n t e d by T i n G, and thus \ G : A \ = \T\ = 2, and we have \ G \ = 2 \ A \ . A l s o , the e q u a t i o n x* = x '1, c a n be viewed as a statement about
conjugation i n the semidirect p r o d u c t G. N o w every element g of G - A lies i n the coset A t , a n d so g = a t for some element a G A . In p a r t i c u l a r ,
g 2 = a t a t = aa* = a a '1 = 1, a n d so a l l elements of G - A are i n v o l u t i o n s .
In other words, G is a generalized d i h e d r a l group a n d i n p a r t i c u l a r , i f A is cyclic of order n , t h e n G is d i h e d r a l of order 2 n .
We c a n also use the semidirect p r o d u c t c o n s t r u c t i o n to prove things. A t first glance, the following m a y seem almost obvious, b u t its p r o o f depends on n o n t r i v i a l facts from C h a p t e r 2.
3.3. C o r o l l a r y ( H o r o s e v s k i i ) . L e t a G A u t ( G ) , w h e r e G i s a n o n t r i v i a l
finite g r o u p . T h e n t h e o r d e r o { a ) of a i s less t h a n \ G \ .
P r o o f . L e t A = ( a ) , so t h a t A is a cyclic subgroup of A u t ( G ) and \ A \ =
o ( a ) . Since A C A u t ( G ) , there is a natural action of A on G by automor- phisms, a n d we construct the semidirect p r o d u c t Y = G x > A w i t h respect to this a c t i o n . A s usual, we identify G a n d A w i t h subgroups of T , so t h a t G < T and A is a complement for G. A l s o , the conjugation a c t i o n of A o n G i n T is the o r i g i n a l a c t i o n . F u r t h e r m o r e , since b y definition, n o n i d e n t i t y a u t o m o r p h i s m s of G act n o n t r i v i a l l y , it follows t h a t no n o n i d e n t i t y element of A acts t r i v i a l l y o n G b y conjugation i n T. I n other words, A C \ CV{ G ) = 1.
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Since G is n o n t r i v i a l a n d A is c y c l i c , L u c c h i n i ' s t h e o r e m (2.20) applies i n T , a n d we deduce t h a t \ A : K \ < \T : A \ , where K = c o r er( A ) . B u t
K n G C A n G = 1, and b o t h AT and G are normal i n T, and thus K C
A n Cr( G ) = 1. T h u s o(<r) = \ A \ = \ A : K \ < \ T : A \ = | G | , as required. •
Somewhat similarly, we have the following.
3.4. Corollary. L e t P be a n a b e l i a n p - s u b g r o u p of A u t ( G ) , w h e r e G i s a
finite g r o u p of o r d e r n o t d i v i s i b l e by t h e p r i m e p . T h e n P has a r e g u l a r o r b i t o n G. I n p a r t i c u l a r , if G i s n o n t r i v i a l , t h e n \ P \ < | G | .
R e c a l l t h a t i f a group P acts o n a set f l , t h e n a P - o r b i t A i n f l is regular if | A | = | P | , or e q u i v a l e n t s (by the f u n d a m e n t a l c o u n t i n g principle) the stabilizer i n P of a n element of A is t r i v i a l . A l s o , i n the s i t u a t i o n of the corollary, i f P > 1 a n d A is a regular P - o r b i t , t h e n A does not c o n t a i n the i d e n t i t y of G , a n d so | G | > 1 + | A | > | P | . T h i s shows t h a t the last sentence of C o r o l l a r y 3.4 w i l l follow once the existence of a regular P - o r b i t is established.
P r o o f o f C o r o l l a r y 3.4. Since P C A u t ( G ) , there is a natural action of P o n G v i a a u t o m o r p h i s m s . L e t T = G x P w i t h respect to t h i s a c t i o n , a n d as u s u a l , identify P a n d G w i t h subgroups of T. T h e n |T : P | = | G | is not d i v i s i b l e b y p , a n d so P € S y lp( r ) . A l s o , the c o n j u g a t i o n a c t i o n of P o n G
i n r is the o r i g i n a l a c t i o n , a n d since P consists of a u t o m o r p h i s m s , o n l y the i d e n t i t y i n P acts t r i v i a l l y . It follows t h a t P n Cr( G ) = 1.
N o w Op( r ) C P , and so Op( r ) D G = 1. A l s o , b o t h G and Op( r ) are
n o r m a l i n T, a n d hence they centralize each other, a n d we have Op( r ) C
P n Cr( G ) = 1. Since the Sylow subgroup P is abelian, it follows by
B r o d k e y ' s t h e o r e m (1.37) t h a t Op( T ) is a n i n t e r s e c t i o n of two S y l o w p -
subgroups. R e p l a c i n g t h e m by conjugates i f necessary, we c a n assume t h a t one of these S y l o w subgroups is P , a n d the other is P x for some element
x € T. W e thus have P n P x = Op( T ) = 1.
W e c a n w r i t e x = u g , w i t h u G P and g € G , and thus P x = P u 3 = P » ,
a n d we have W e c l a i m t h a t the P - o r b i t i n G t h a t contains g is regular, a n d this w i l l complete the proof. It suffices to show t h a t the stabilizer i n P of g is t r i v i a l , b u t since the a c t i o n of P o n G is c o n j u g a t i o n i n r , w h a t we want to establish is t h a t CP( g ) = 1. W e have CP( g ) = CP( g ) 9 ,
a n d hence CP( g ) C P n P 9 = 1, as w a n t e d . •
A c o m p a r i s o n of the previous two corollaries suggests t h a t perhaps w h e n - ever A i s & cyclic subgroup of A u t ( G ) , there is always a regular A - o r b i t i n G . If t h a t were true, it w o u l d be a strong f o r m of C o r o l l a r y 3.3, b u t alas, it is false.
72 3. S p l i t E x t e n s i o n s
F i n a l l y , we are ready to prove T h e o r e m 3.2 by a c t u a l l y c o n s t r u c t i n g the semidirect p r o d u c t T = N x H . A c o m m o n way to do this, b y analogy w i t h the c o n s t r u c t i o n of the e x t e r n a l direct p r o d u c t , is to define an appropriate m u l t i p l i c a t i o n o n the set of ordered pairs ( h , n ) w i t h h £ H a n d n £ N . O n e tedious, b u t necessary, step using t h i s c o n s t r u c t i o n is to check t h a t the denned m u l t i p l i c a t i o n is associative. W e choose to use a different approach, w h i c h avoids checking associativity. T h i s is a c c o m p l i s h e d b y w o r k i n g i n an a p p r o p r i a t e s y m m e t r i c group, where, of course, a s s o c i a t i v i t y is a u t o m a t i c . P r o o f o f T h e o r e m 3.2. L e t ft be the set of ordered pairs ( k , m ) , where
k € H a n d m € N . It is t r i v i a l to check t h a t actions of N a n d of H o n ft are denned b y the following:
( k , m ) - n = ( k , m n ) and { k , m ) - h = { k h , m h) ,
where ( k , ra) £ ft a n d n e N a n d h £ H . Here, of course, mh is the result
of l e t t i n g h act o n m i n the given a c t i o n v i a a u t o m o r p h i s m s of H o n N . Since N acts o n ft, we have a n a t u r a l h o m o m o r p h i s m n i-> n0 from N
into the s y m m e t r i c group S y m ( f t ) , where n0 is the p e r m u t a t i o n a ^ a-n for
a £ ft. W r i t e N 0 = { n0 \ n £ A/'}, so t h a t N 0 C Sym(ft) is a subgroup a n d
( )o is a surjective h o m o m o r p h i s m from N to A ^ . If n0 = 1, t h e n ( k , m ) - n =
( k , m ) for a l l k £ i f a n d m £ TV, a n d i n p a r t i c u l a r , (1,1) = (1, l ) - n = (1, n ) . It follows t h a t n = l , a n d so the h o m o m o r p h i s m ( )0 has t r i v i a l kernel. It
is thus a n i s o m o r p h i s m from N onto N 0 .
S i m i l a r l y , i f h £ H , w r i t e h 0 £ Sym(ft) to denote the m a p a H * a - h ,
a n d let H 0 = { h0 \ h £ H } . T h e n i T0 C Sym(ft) is a subgroup a n d ( )0 is a
h o m o m o r p h i s m from H onto 7 J0- ( A s before, we are using the n o t a t i o n ( )0
for two different maps.) If h o = 1, t h e n (1,1) = (1, l ) - h = ( h , lh) , a n d thus
h = 1 a n d ( )0 is an i s o m o r p h i s m from H onto H0.
N e x t , we argue t h a t ( nh)0 = (n0)h° for a l l n £ N a n d h £ H . Since
t h i s is a statement about elements of S y m ( f t ) , it suffices to check t h a t b o t h { nh) o a n d {n0)h° have the same effect o n each element of ft. W e have
( k , m ) ( nh) o = { k , m ) - nh = ( k , m nh) . A l s o , ( k , m ) ( n0)h o = { k , m ) { h o ) -ln o h o = { { { k , m ) - h -l) - n ) - h
= ( O ^ - V ^ V H
= ( k h - \ m h~ \ ) - h = ( k , m nh) ,where the last equality holds because ( mh~1n )h = m nh. ( T h i s is v a l i d
P r o b l e m s 3 A 73
t h a t ( nh)0 = ( n0)h o as c l a i m e d , a n d i n p a r t i c u l a r , H 0 C NS y m ( Q )( 7 V0) . W e
can now define G = N0H0, so t h a t G is a subgroup of S y m ( f t ) , a n d N 0 < G.
A l l t h a t remains is to show t h a t H 0 is a c o m p l e m e n t for N 0 i n G. Since by
definition, G = N0H0, it suffices to check t h a t N 0 n H 0 = 1. T o this end,
suppose t h a t n0 = h 0 w i t h n G TV a n d / i G H . T h e n
(/i, 1) = (1, l ) - h = (1, l ) / i0 = (1, l ) n0 = (1, l ) - n = ( l , n ) ,
and it follows t h a t / i = 1. T h u s n 0 = h 0 = l , a n d the p r o o f is complete. •
We close this section w i t h a d e s c r i p t i o n of the w r e a t h p r o d u c t , w h i c h is useful for c o n s t r u c t i n g examples a n d counterexamples. T h e ingredients here are two gronps G a n d H , a n d a set ft, acted o n b y G. L e t B be the set of a l l functions from ft into H , a n d make B i n t o a group b y defining m u l t i p l i c a t i o n pointwise. T h u s i f / , g G B , t h e n f g is defined by the f o r m u l a ( f g ) ( a ) = f ( a ) g ( a ) for a G ft. It is t r i v i a l to check t h a t B is a group, a n d in fact, B is really j u s t the e x t e r n a l direct p r o d u c t of |ft| copies of H .
The a c t i o n of G o n ft induces an a c t i o n v i a a u t o m o r p h i s m s of G o n B . If we v i e w B as a direct p r o d u c t , the a c t i o n of G c a n be described s i m p l y as a p e r m u t a t i o n of coordinates, b u t it is useful t o be more precise. G i v e n / G B a n d x G G, the function f x o n ft is defined so t h a t fx( a - x ) = / ( a ) ,
or e q u i v a l e n t s , fx( a ) = f { a - x ~l) . (It is r o u t i n e to show t h a t this defines
an a c t i o n v i a a u t o m o r p h i s m s , a n d we o m i t the proof.)
Now let W = B x < G w i t h respect to t h i s a c t i o n . T h e n W is the w r e a t h p r o d u c t of H w i t h G, a n d B , viewed as a subgroup of W, is called the base g r o u p of the w r e a t h p r o d u c t . It is c o m m o n to w r i t e W = H I G, b u t of course, this n o t a t i o n for the w r e a t h p r o d u c t is defective i n t h a t it does not m e n t i o n the set ft or the a c t i o n of G o n ft. If G is given as a n abstract group, a n d no set ft is m e n t i o n e d , t h e n it is u n d e r s t o o d t h a t H I G is c o n s t r u c t e d u s i n g the "regular" a c t i o n of G. T h i s means t h a t we take ft = G, w i t h G a c t i n g b y right m u l t i p l i c a t i o n . Sometimes, this is referred to as the regular w r e a t h p r o d u c t of H b y G.
P r o b l e m s 3 A
3 A . 1 . L e t C be a c y c l i c group of order n d i v i s i b l e b y 8, a n d let z be the unique i n v o l u t i o n i n C.
(a) Show t h a t C has a unique a u t o m o r p h i s m a such t h a t ca = c~lz
for every generator c of C, a n d show t h a t a has order 2.
(b) L e t S = C x ( o - ) , so t h a t |5| = 2 \ C \ . Show that half of the elements o i S - C have order 2 a n d t h a t the other half have order 4.
(c) S h o w t h a t the elements of order 2 i n S - C form a single conjugacy class of S, a n d s i m i l a r l y for the elements of order 4.
74 3. S p l i t E x t e n s i o n s
N o t e . T h e group S is the s e m i d i h e d r a l group S D2n, a l t h o u g h i n the
literature, the w o r d "semidihedral" is u s u a l l y reserved for the case where n is a power of 2, a n d there seems to be no s t a n d a r d name for other members of t h i s f a m i l y of groups.
3 A . 2 . L e t S a n d C be as i n the previous p r o b l e m , a n d let B be the subgroup of i n d e x 2 i n G . Show t h a t the elements of order 4 i n S - C form a coset of B , a n d let Q be the u n i o n of t h i s coset a n d B . Show t h a t Q is a subgroup of order n .
N o t e . Since a l l elements of Q - B have order 4, the i n v o l u t i o n i n B is the unique i n v o l u t i o n i n Q. T h e group Q is the generalized q u a t e r n i o n group Qn. ( T h e phrase "generalized q u a t e r n i o n " is often restricted to the
case where the order n is a power of 2. T h e undecorated w o r d "quaternion" is u s u a l l y reserved for Q8. )
3 A . 3 . L e t p be a p r i m e , a n d suppose t h a t m is a d i v i s o r of p-1 w i t h m > 1. Show t h a t there exists a group G of order pm w i t h a n o r m a l subgroup P of order p , a n d such t h a t G/P is cyclic a n d Z ( G ) = 1.
3 A . 4 . L e t q be a power of a p r i m e p . Show t h a t there exists a group G of order q ( q - 1) w i t h a n o r m a l elementary a b e l i a n subgroup of order q a n d such t h a t a l l elements of order p i n G are conjugate.
H i n t . L e t F be a field of order q a n d observe t h a t the m u l t i p l i c a t i v e group