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H = f . f
+ 1 by d e f i n i t i o n o f (p.
(vi) ^ is the restriction Xi>'-'>X
S
to H of some irreducible
f
character
\j/ of G. L e t
be the distinct i r r e d u c i b l e characters o f G, the t r i v i a l c
being Xi- Since (p* is a class function, we can w r i t e (p* = Y Ui iXi J
character where
G
c =
Xi>G ~ <*A &>G> w h i c h is an integer (by 8.3.3). M o r e
t
over b y the F r o b e n i u s R e c i p r o c i t y T h e o r e m (8.4.4) C
Cl=/<*F,Zl>G-<* ,Xl>G = =
since \j/
/
S
/
Also, 2
Z cf = <
+ i
G
by
c
(v). Therefore Y,i=2 i
i f j > 1. T h u s 9* = fxi
= 1
a
n
d some c = ± 1 , a l l other c-s being 0
± Xi- N o w (l)(p*
f
= (l)(p = 0 b y (iv), whence 0 =
/ ± WXt- T h i s shows t h a t the negative sign is the correct one a n d (l)xt = f Therefore (p* = fxi
— Xi- If x € H, we have (x)x
(x)\j/. Hence (XI)H = *A
t
a
n
= f — (x)(p* = f — (x)(p =
d we can take ij/' t o be Xi-
(vii) L e t *F denote the set o f a l l n o n t r i v i a l i r r e d u c i b l e characters o f H t h a t are constant o n K. Define I = f] K e r if/', the intersection being f o r m e d
250
8. Representations of Groups
over a l l xp i n *F: here, o f course, K e r xp' means the kernel o f the associated representation. C l e a r l y J o G. W e shall show t h a t I = N
thus c o m p l e t i n g
9
the proof. Let xe
x
N a n d xp e ¥. T h e n 0 = (x)
(x)xp' = f since (p* = f% — xp'. L e t p be the representation o f G w i t h charac x
p
ter xp'; then the characteristic r o o t s o f x , w h i c h are roots o f u n i t y , a d d u p to / , the degree o f p. Therefore these characteristic r o o t s a l l equal 1. A p p l y i n g Maschke's T h e o r e m a n d 8.1.6 t o p | thus x e K e r p. Hence x e I a n d N ^ F i n a l l y let x e H\K.
p
, we conclude t h a t x
< x >
= 1 and
I.
T h e n there is a n o n t r i v i a l i r r e d u c i b l e representation
a o f H constant o n K t h a t does n o t m a p x t o 1—otherwise by Maschke's 9
9
T h e o r e m xK
w o u l d belong t o the k e r n e l o f the regular representation,
w h i c h , o f course, is faithful. I f the character o f a is ip then x <£ K e r ip 9
9
9
whence x <£ K e r xp' a n d x$L
T h u s I contains n o element o f (H/K)
for any
g f r o m w h i c h i t follows t h a t I ^ N a n d I = N.
•
9
Frobenius Groups T h e most i m p o r t a n t case o f 8.5.4 is w h e n K = 1. 8.5.5 (Frobenius). If x
HnH
G is a finite
group
with
a subgroup x
= 1 for all x in G\H
9
then N = G\[J (H\l) xeG
group of G such that G = HN and HnN
H such
that
is a normal
sub
= 1.
A g r o u p G w h i c h has a p r o p e r n o n t r i v i a l subgroup H w i t h the above p r o p e r t y is called a Frobenius N the Frobenius
group. H is called a Frobenius
complement
and
kernel. W e shall p r o v e i n Chapter 10 the i m p o r t a n t theo
rem o f T h o m p s o n t h a t N is always n i l p o t e n t . Frobenius groups arise i n a n a t u r a l w a y as transitive p e r m u t a t i o n g r o u p s — f o r example, we observed i n 7.1 t h a t the g r o u p H(q) is a F r o b e n i u s g r o u p . I n fact there is a characterization o f F r o b e n i u s groups i n terms o f p e r m u t a t i o n groups. 8.5.6. (i) If G is a Frobenius
group with complement
cosets of H yields a faithful permutation
H the action of G on the right 9
representation
group in which no nontrivial
of G as a transitive
nonregular
element has more than one
fixed
point. (ii) Let
G be a transitive
nontrivial
but nonregular
permutation
element has more than one fixed
group.
The Frobenius
fixed
points.
kernel consists
group
in which
point. Then G is a
of 1 and all elements
no
Frobenius
of G with no
8.5. Applications to Finite Groups
251
Proof, (i) Suppose t h a t g i n G fixes t w o distinct r i g h t cosets Hx a n d Hy. T h e n Hxg = Hx a n d Hyg = /Jy, equations w h i c h i m p l y t h a t g e H n / J . Since y x /J, this yields g = 1. x
y
- 1
(ii) G\H, a and about
L e t G act o n a set X . Choose a f r o m X a n d w r i t e H = St (a). I f g e t h e n H r\H consists o f the elements o f G t h a t fix the distinct p o i n t s ag. Hence H r\H = 1 a n d G is a F r o b e n i u s g r o u p . T h e statement the F r o b e n i u s kernel follows f r o m the definition. •
EXERCISES
G
9
9
8.5
1. Prove the following generalization of the Burnside p-q Theorem: a finite group with a nilpotent subgroup of prime-power index is soluble. 2. The center of a Frobenius group is always trivial. 3. The dihedral group D
2n
is a Frobenius group if and only i f n is odd and > 1. 3
4. Let P be an extra-special group of exponent 7 and order 7 . Let a and b be generators and put c = [a, b~\. The assignments a\-+a c, b\-+b , ci—>c determine an automorphism of order 3. Show that the semidirect product K P is a Frobenius group with nonabelian kernel. 2
2
4
5. Let G be a Frobenius group with kernel N. Prove that C (x) < N for all 1 ^ x e N. G
6. I f G is a Frobenius group with kernel N, then |G : N\ divides |AT| — 1. 7. Let G be a Frobenius group with kernel N. I f L
CHAPTER 9
Finite Soluble Groups
T h e foundations o f the t h e o r y o f finite soluble groups were l a i d i n a n i n f l u ential series o f papers by P. H a l l between 1928 a n d 1937. After 1950 the subject developed further thanks t o the w o r k o f R . W . Carter, W . Gaschutz, B. H u p p e r t , a n d others. T h i s a c t i v i t y has resulted i n a t h e o r y o f great ele gance. Here we can o n l y present a s m a l l p a r t o f this theory; for a complete account see the recently p u b l i s h e d b o o k [ b l 9 ] .
9.1. Hall 7c-Subgroups L e t G be a g r o u p a n d let n be a n o n e m p t y set o f primes. A Sylow subgroup
n-
o f G is defined t o be a m a x i m a l 7r-subgroup. W h i l e Sylow 7r-sub-
groups always exist, they are usually n o t conjugate i f n contains m o r e t h a n one p r i m e . A m o r e useful concept is t h a t o f a H a l l 7r-subgroup. I f G is a finite g r o u p , a 7r-subgroup H such t h a t | G : H | is a ^/-number is called a Hall
n-subgroup
of G. I t is rather o b v i o u s t h a t every H a l l rc-subgroup is a Sylow 7r-subgroup. I n general, however, G need n o t c o n t a i n any H a l l 7r-subgroups. F o r exam ple, a H a l l { 3 , 5 } - s u b g r o u p o f A
5
w o u l d have index 4, b u t A
5
has n o such
subgroups (why?). W e shall s h o r t l y see t h a t i n a finite soluble g r o u p H a l l 7r-subgroups always exist a n d f o r m a single conjugacy class. N o t i c e t h a t the terms " H a l l p - s u b g r o u p " a n d " S y l o w p - s u b g r o u p " are s y n o n y m o u s for finite groups i n view o f Sylow's T h e o r e m . T h e normal 7r-subgroups o f a g r o u p G p l a y a special role. Suppose t h a t H a n d K are
rc-subgroups
a n d X o G. T h e n clearly HnK
7r-groups, f r o m w h i c h i t follows t h a t HK
252
a n d HK/K
are
is a 7r-group. Consequently the
9.1. Hall 7r-Subgroups
253
s u b g r o u p generated b y a l l the n o r m a l rc-subgroups o f G is a 7r-group. T h i s is the u n i q u e m a x i m u m n o r m a l 7c-subgroup o f G; i t is denoted by 0,(G). T h e f o l l o w i n g properties o f this s u b g r o u p are useful. 9.1.1. Let G be any group and n a set of (i) If H is a subnormal
n-subgroup
(ii) 0 ( G ) is the intersection 7r
primes.
of G, then H <
O (G). n
of all the Sylow nsubgroups
of G.
Proof, (i) B y hypothesis there is a series H = / f o / f o 0
/ < 1, t h e n H < G a n d H < O (G) n
have by i n d u c t i o n o n / t h a t H < O^H^). acteristic i n
• • • o H = G. I f
x
l
by definition. A s s u m i n g t h a t / > 1, we B u t the latter s u b g r o u p is char
a n d hence n o r m a l i n G. T h u s O^H^)
< 0 (G) and H < 7r
0,(G). (ii) L e t R = O (G) n
a n d let S be a S y l o w
rc-subgroup
o f G. T h e n RS is a
7r-group, by the a r g u m e n t o f the p a r a g r a p h preceding this p r o o f ; therefore R < S by m a x i m a l i t y o f S. O n the other h a n d , the intersection o f a l l the Sylow 7r-subgroups is certainly n o r m a l i n G, so i t is contained i n R.
•
I n p a r t i c u l a r i t follows t h a t 0 ( G ) is contained i n every H a l l 7r-subgroup 7r
o f G.
The Schur-Zassenhaus Theorem T h e f o l l o w i n g t h e o r e m m u s t be reckoned as one o f the t r u l y
fundamental
results o f g r o u p theory. 9.1.2 (Schur, Zassenhaus). Let N be a normal subgroup Assume subgroups
that \N\ = n and \G: N\ = m are relatively
of a finite
prime.
of order m and any two of them are conjugate
Then
group G
G.
contains
in G.
W e pause t o i n t r o d u c e a useful piece o f t e r m i n o l o g y . I f i f is a subgroup of a g r o u p G, a s u b g r o u p K is called a complement G = HK
and
of H i n G i f
H n K = l .
T h e reader w i l l recognize t h a t 9.1.2 s i m p l y asserts t h a t complements o f N exist a n d any t w o are conjugate. There is yet another f o r m u l a t i o n o f 9.1.2: i f TZ is the set o f p r i m e divisors o f m, t h e n H a l l rc-subgroups o f G exist a n d any t w o are conjugate. Proof of 9.1.2. (i) Case: N abelian.
L e t Q = G/N.
Since N is abelian, i t can Ng
be made i n t o a Q-module v i a the well-defined a c t i o n a coset x i n Q we choose a representative t , x
9
= a . F r o m each
so t h a t the set {t \x x
e Q} is a
254
9. Finite Soluble Groups
transversal t o N i n G. Since t t x
belongs t o the coset t t N
y
= t N,
x y
there is
xy
an element c(x, y) o f N such t h a t tt x
= t c(x,
y
y).
xy
B y a p p l y i n g this e q u a t i o n a n d the associative l a w (t t )t obtains the r e l a t i o n x y
= t (t t )
z
x
one
y z
z
c(xy, z) • c(x, y) = c(x, yz) • c(y, z\
(1)
w h i c h holds for a l l x, y, a n d z i n Q. N e x t consider the element o f N c
d(y) = n
x
( > y)-
xeQ
F o r m i n g the p r o d u c t o f the equations (1) over a l l x i n Q, we find t h a t z
m
d(z) - d(y) = d(yz) • c(y, z) because N is abelian; thus m
d(yz) = d(yfd(z)c(y,
z)~
(2) m
x
Since (ra, n) = 1, there is a n element e(y) o f N such t h a t e(y) = d(y)~ , (2) becomes e(yz)~ = (e(yfe(z)c(y, z))~ . Hence m
z
e(yz) = e(y) e(z)c(y, W e define s t o be t e(x) x
x
y
z
z).
and compute
z
ss
and
m
= t t e(y) e(z) y z
z
= t c(y,
z)e(y) e(z)
yz
= t e(yz)
= s.
yz
yz
Consequently the m a p p i n g X H S is a h o m o m o r p h i s m 8: Q G. N o w s = 1 implies that t e N a n d x = N = 1 . Hence 6 is injective, I m 6 ^ g , a n d | I m 0| = m. x
x
x
Q
N o w suppose t h a t / f a n d H* are t w o subgroups o f order m. T h e n G = HN = H*N a n d H n AT = 1 = H* n AT. L e t x i n Q m a p t o w a n d w j respectively under the canonical h o m o m o r p h i s m s Q = HN/N H and Q = H*N/N i f * . Then = w a ( x ) where a(x) e N. B u t w j = u*u* = u a(x)u a(y) = u a(x) a(y), whence we deduce the r e l a t i o n x
x
y
y
x
y
xy
y
a(xy) = a(x) a(y).
(3)
a x
Define b = Y\xeQ ( ). F o r m i n g the p r o d u c t o f the equations (3) over a l l x i n Q, we o b t a i n b = b a(y) . Since (m, n) = 1, i t is possible t o w r i t e b = c w i t h c i n N. T h e n the preceding e q u a t i o n becomes c = c a(y) o r a(y) = c~ c. Therefore u* = u a(y) = u c~ c = c~ u c, because c~ = ( c ) . Hence H* = c^Hc. y
m
m
y
y
y
y
y
1
y
_ 1
U y
y
(ii) Existence—The general case. W e use i n d u c t i o n o n | G | . L e t p be a p r i m e d i v i d i n g |iV| a n d P a Sylow p-subgroup o f N. W r i t e L = N (P) a n d C = CP- T h e n L < N (C) = M, say, since C is characteristic i n P. B y the F r a t t i n i a r g u m e n t (5.2.14) we have G = LN a n d a fortiori G = MN. L e t i \ \ denote the n o r m a l subgroup N n M o f M a n d observe t h a t | M : i \ \ | = | G : AT | = m. W e m a y a p p l y the i n d u c t i o n hypothesis t o the g r o u p M/C o n G
G
9.1. Hall 7r-Subgroups
255
n o t i n g t h a t C # 1. L e t X/C
be a s u b g r o u p o f M/C
w i t h order m. Since
\X: C | = m is relatively p r i m e t o | C | , we can a p p l y (i) t o conclude t h a t X has a s u b g r o u p o f order m. (iii) Conjugacy—The
case G/N soluble. D e n o t e by % the set o f p r i m e d i v i
sors o f m a n d w r i t e R = O (G). n
L e t H a n d X be t w o subgroups o f G b o t h o f
w h i c h have order m. T h e n R < HnK
since H a n d X are H a l l
rc-subgroups
of G. E v i d e n t l y we m a y pass t o the g r o u p G/R. Observe t h a t O (G/R)
= 1,
n
so we m a y as well suppose t h a t R = 1. O f course we can also assume t h a t m > 1, so t h a t N ^ G. L e t L / i V be a m i n i m a l n o r m a l s u b g r o u p o f G/N. soluble, L/N H n L and
Since the latter is
is a n elementary abelian p - g r o u p for some p r i m e p i n n. N o w
is a S y l o w p-subgroup o f L because H n L ~ (H n L)N/N | L : / J n L \ = \HL: H\
< L/N
is a p'-number. T h e same is true o f X n L . 9
T h u s Sylow's T h e o r e m m a y be a p p l i e d t o give H n L = (K n L)
9
= K n L 9
for some g i n G. W r i t i n g S for / J n L , we conclude t h a t S o
= J,
say. Suppose t h a t J = G, so t h a t S o G. T h e n , because S is a rc-group, S < K = 1; thus L is a p ' - g r o u p . H o w e v e r this cannot be true since L/N
is a
p-group. I t follows t h a t J ^ G. W e can n o w use i n d u c t i o n o n \G\ t o con 9
clude t h a t H a n d K
are conjugate i n J , whence we derive the conjugacy o f
H and X . (iv) Conjugacy—The order m, then HN'/N'
case N soluble. I f H a n d X are subgroups o f G w i t h 9
a n d KN'/N'
are conjugate by (i). Hence H
< KN'
some g i n G. B y i n d u c t i o n o n the derived l e n g t h o f N we conclude t h a t a n d X are conjugate, being subgroups o f order m i n the g r o u p KN'.
for 9
H
Hence
H a n d X are conjugate. (v) Conjugacy—The
general
case. Since the integers m a n d n are rela
tively p r i m e , at least one o f t h e m is o d d , a n d the F e i t - T h o m p s o n T h e o r e m (see the discussion f o l l o w i n g 5.4.1) implies t h a t either N o r G/N is soluble. T h e result n o w follows f r o m (iii) a n d (iv).
•
N o t i c e t h a t the F e i t - T h o m p s o n T h e o r e m is o n l y required t o p r o v e con jugacy, a n d then o n l y i f we d o n o t k n o w a priori
t h a t either N o r G/N is
soluble. T h e f o l l o w i n g c o r o l l a r y o f 9.1.2 is i m p o r t a n t .
9.1.3. With the notation G with order m
1
Proof. L e t H a n d H
x
T h e n G = HN
of 9.1.2, let m
x
= (H N)
X
9
= ((H N)nH) 1
n (HN) = ((H N)
1
1
x
of m. Then a subgroup
of
in a subgroup of order m.
be s u b g r o u p o f G w i t h orders m a n d m
and H N
t h a t the order o f (H N)nH clude t h a t H
be a divisor
1
is contained
1
n H)N,
respectively. which
shows
equals | H N : N \ = \ H \ = m . B y 9.1.2 we con x
9
< H
x
1
9
for some g i n G; o f course \H \ = m.
•
256
9. Finite Soluble Groups
7r-Separable Groups Let
G be a finite g r o u p a n d let n be a n o n e m p t y set o f primes. T h e n G is
said t o be n-separable
i f i t has a series each factor o f w h i c h is either a n-
g r o u p or a 7r'-group. F o r example, a finite all
soluble group is n-separable
for
n: t o see this one s i m p l y refines the derived series by inserting the n-
component
o f each factor. N o t i c e t h a t ^-separability is i d e n t i c a l w i t h
separability. I t is also very easy t o p r o v e t h a t every s u b g r o u p a n d
n'-
every
image o f a 7r-separable g r o u p are likewise 7r-separable.
The Upper 7r'7r-Series I f G is an a r b i t r a r y g r o u p , the upper applying
n'n-series
is generated by repeatedly
a n d O^. T h i s is, then, the series 1 =
P o ^ N o ^ P ^ N ^ - ' - ^ P ^ N n - ' -
dGfinGd by
N /P = OAG/P,) t
and
t
P /N t+1
=
t
0 (G/N ). %
I t is somtimes convenient t o w r i t e the first few terms N ,
P
0
CMC),
C W G ) ,
t
i V . . . as
l 5
l 5
0,.,AG),....
W h a t we have here is a series of characteristic subgroups whose factors are alternately ^/-groups a n d 7r-groups. 9.1.4. Let X o
G be a finite
•••<] / f
x
w
group and H /K i+1
t
o K
n-separable
group
= G be a n'n-series,
m
is n-group.
terms of the upper n'n-series
and
let
1 = H o
Then H < P and K t
t
of G. In particular,
t
N
m
X o
0
that is, such
H
0
that KJHi
x
< N where P and N t
o
is a n'-
t
are
t
= G.
Proof. Suppose t h a t the i n c l u s i o n H < P has been p r o v e d — i t is o f course t
true i f i = 0. T h e n K PJP i
N
t
h
by 9.1.1(i). T h u s H Ni/Ni i+1
H
< P .
i+1
i+1
a term
is a s u b n o r m a l
rc-subgroup
whence K
t
o f G/N
t
T h e result n o w follows b y i n d u c t i o n .
I t follows t h a t a finite with
t
is a s u b n o r m a l ^/-subgroup o f G/P
i
of
its upper
group
G is n-separable
n'n-series.
• if and only if it
M o r e o v e r the upper
shortest fl/Tr-series; its l e n g t h is termed the n-length
< and
coincides
fl/Tr-series
of G
W e take note o f a simple characterization o f 7r-separable groups.
is a
257
9.1. Hall 7r-Subgroups
9.1.5. The following (i) G is
properties
of the finite
group G are
equivalent:
n-separable;
(ii) every principal factor (iii) every composition Proof, (i)
of G is an or a %'-group;
factor
of G is a n-group
or a
n'-group.
(ii). I f G is ^-separable, then so is every p r i n c i p a l factor; b u t the
latter are characteristically simple, so b y 9.1.4 each one is either a 7r-group or a flZ-group. (ii)
(iii). S i m p l y refine a p r i n c i p a l series t o a c o m p o s i t i o n series.
(iii)
(i). T h i s is obvious.
•
Hall 7r-Subgroups of 7r-Separable Groups The most i m p o r t a n t p r o p e r t y o f ^-separable groups is that H a l l
rc-subgroups
exist a n d are conjugate. 9.1.6 (P. H a l l , C u n i h i n ) . Let n-subgroup
is contained
groups are conjugate
the finite
in a Hall
group
G be n-separable.
n-subgroup
Then
every
of G and any two Hall
n-sub-
in G.
Proof. Since each rc-subgroup is c o n t a i n e d i n a Sylow 7r-subgroup, i t suffices t o prove t h a t a Sylow rc-subgroup P is a H a l l
rc-subgroup
a n d t h a t a l l such
subgroups are conjugate. T h i s w i l l be accomplished by i n d u c t i o n o n | G | , w h i c h we suppose greater t h a n R
1. L e t R = O (G) n
a n d assume
first
that
1. T h e n R < P a n d by i n d u c t i o n P/R is a H a l l 7r-subgroup o f G/R. O f
course i t follows t h a t P is a H a l l 7r-subgroup o f G, then P/R
rc-subgroup
a n d Q/R
o f G. I f Q is another H a l l
are conjugate, whence P a n d Q are
conjugate. Now
assume t h a t R = 1. Since G is ^-separable a n d G ^ 1, we have S =
O ^ G ) ^ 1. O f course PS/S
is a rc-group a n d by i n d u c t i o n i t is contained i n
a H a l l 7r-subgroup Q/S o f G/S. B y 9.1.3 the
rc-subgroup
P is contained i n a
H a l l 7r-subgroup P* o f Q. B u t P is a S y l o w rc-subgroup, so P = P* a n d P is a H a l l 7r-subgroup o f Q a n d hence o f G. I f P is any other H a l l x
of G, then PS/S
a n d P i S / S are conjugate;
rc-subgroup
for these are surely H a l l
n-
subgroups o f G/S. T h u s P f < PS for some g i n G. B u t n o w the S c h u r Zassenhaus T h e o r e m m a y be a p p l i e d t o PS t o show that P a n d P f conjugate.
are •
T h e most i m p o r t a n t case o f 9.1.6 is w h e n G is soluble: this is the o r i g i n a l t h e o r e m o f P. H a l l a n d we shall restate i t as 9.1.7 (P. H a l l ) . If G is a finite tained
in a Hall
conjugate.
n-subgroup
soluble
group,
of G. Moreover
then every n-subgroup all Hall
n-subgroups
is con of
are
258
9. Finite Soluble Groups
I t is a r e m a r k a b l e fact t h a t the converse o f 9.1.7 holds: the existence o f H a l l 7r-subgroups f o r a l l % implies s o l u b i l i t y . 9.1.8 (P. H a l l ) . Let G be a finite there exists a Hall p'-subgroup.
group and suppose
that for every prime p
Then G is soluble.
Proof. Assume t h a t the t h e o r e m is false a n d let a counterexample G o f smallest order be chosen. Suppose t h a t N is a p r o p e r n o n t r i v i a l n o r m a l s u b g r o u p o f G. I f H is a H a l l p'-subgroup HN/N
are H a l l p'-subgroups
o f G, i t is evident t h a t HnN
and
o f N a n d G/N respectively. Therefore N a n d
G/N are soluble groups b y m i n i m a l i t y o f G. B u t this leads t o the c o n t r a d i c t i o n t h a t G is soluble. Consequently G m u s t be a simple g r o u p . 1
h
W r i t e \G\ = pi •"pl
where e >0
and P i , . . . , p
t
are distinct primes.
f c
Burnside's t h e o r e m (8.5.3) shows t h a t k > 2. L e t G be a H a l l p^-subgroup t
of G a n d p u t H = G \G:H\
3
n - n G
i
. T h e n \G:G \=pf .
k
= Yi^ pf\
whence \H\ = p{'p
3
B y 1.3.11 we have
i
e 2
a n d H is soluble, b y Burnside's the
2
o r e m once again. L e t M be a m i n i m a l n o r m a l s u b g r o u p o f H; t h e n M is a n elementary abelian p - g r o u p where p = p
x
e 2
3
\G:HnG \=p '-p \ 2
2
o r p , let us say the former. N o w 2
so \HnG \=p{K
k
T h u s HnG
2
2
is a Sylow p x
2
s u b g r o u p o f H a n d consequently M < H n G < G . Also | i f n G | = p | b y 2
the same reasoning. Hence G = (HnG )G 1
G
2
follows t h a t M
= M°
2
x
b y consideration o f order. I t
2 G
< G < G, a n d M 2
is a p r o p e r n o n t r i v i a l n o r m a l
s u b g r o u p o f G. T h i s is a c o n t r a d i c t i o n .
•
Minimal Nonnilpotent Groups O u r next objective is a t h e o r e m o f W i e l a n d t asserting t h a t i f a finite g r o u p has a nilpotent
Hall
rc-subgroup,
then a l l H a l l
rc-subgroups
are conjugate.
I n order t o prove this we need t o have i n f o r m a t i o n a b o u t m i n i m a l n o n n i l p o t e n t groups. I n d e e d k n o w l e d g e o f the structure o f such groups is useful i n m a n y contexts. 9.1.9 (O.J. Schmidt). Assume that every maximal subgroup of a finite is nilpotent
but G itself
is not nilpotent.
group G
Then:
(i) G is soluble; m
n
(ii) | G | = p q
where p and q are unequal
(iii) there is a unique Sylow p-subgroup Hence
primes;
P and a Sylow q-subgroup
Q is cyclic.
G = QP and P < G .
Proof, (i) L e t G be a counterexample o f least order. I f AT is a p r o p e r n o n t r i v i a l n o r m a l subgroup, b o t h N a n d G/N are soluble, whence G is soluble. I t follows t h a t G is a simple g r o u p .
9.1. Hall 7r-Subgroups
259
Suppose t h a t every p a i r o f distinct m a x i m a l subgroups o f G intersects i n 1. L e t M be any m a x i m a l subgroup: then certainly M = N (M).
I f |G| =
G
n a n d \M\ = m, then M has n/m conjugates every p a i r o f w h i c h intersect t r i v i a l l y . Hence the conjugates o f M
account for exactly (ra — l)n/ra =
n — n/m n o n t r i v i a l elements. Since ra > 2, we have n — n/m > n/2 > (n — l ) / 2 : i n a d d i t i o n i t is clear t h a t n — n/m
— 2
— 1. Since each n o n i d e n -
t i t y element o f G belongs t o exactly one m a x i m a l subgroup, n — 1 is the sum o f integers l y i n g s t r i c t l y between (n — l ) / 2 a n d n — 1. T h i s is p l a i n l y impossible. I t follows t h a t there exist distinct m a x i m a l subgroups M
and M
x
intersection I is n o n t r i v i a l . L e t M
and M
x
m u m order. W r i t e N = N (I). M. x
2
whose
be chosen so t h a t I has m a x i
Since M is n i l p o t e n t , I ^ N (I)
G
t h a t I < Nn
2
b y 5.2.4, so
Mi
N o w I c a n n o t be n o r m a l i n G; thus N is p r o p e r a n d is
contained i n a m a x i m a l s u b g r o u p M . T h e n I < N r\M
< M n M
l
1
?
which
contradicts the m a x i m a l i t y o f | / | . 1
k
(ii) L e t \G\ = pi "-p
where e > 0 a n d the p are distinct primes. As
k
t
t
sume t h a t k > 3. I f M is a m a x i m a l normal subgroup, its index is p r i m e since G is soluble; let us say \G\M\=p .
L e t P be a S y l o w p s u b g r o u p o f G. I f
1
i > 1, then P < t
subgroup P P 1
[P
1 ?
i
t
r
M and, since M is n i l p o t e n t , i t follows that P ; 0 G; also the cannot equal G since k > 3. Hence P P j is n i l p o t e n t a n d thus x
p . ] = l (by 5.2.4). I t follows t h a t A ( P ) = G a n d P ^ G
G. T h i s means
X
t h a t a l l S y l o w subgroups o f G are n o r m a l , so G is n i l p o t e n t . B y this contra 2
d i c t i o n k = 2 a n d | G | = p^pl .
W e shall w r i t e p = p
2
and q =
p. x
(iii) L e t there be a m a x i m a l n o r m a l subgroup M w i t h index g. T h e n the Sylow p-subroup P o f M is n o r m a l i n G a n d is evidently also a Sylow ps u b g r o u p o f G. L e t Q be a Sylow ^-subgroup o f G. T h e n G = g P . Suppose t h a t Q is n o t cyclic. I f g e Q, then w h i c h is cyclic. Hence
P > ^ G since otherwise g ^
G/P,
P > is n i l p o t e n t a n d [g, P ] = 1. B u t this means
t h a t [ P , Q] = 1 a n d G = P x Q, a n i l p o t e n t g r o u p . Hence Q is cyclic.
•
Wielandt's Theorem on Nilpotent Hall 7r-Subgroups I n a n insoluble g r o u p H a l l 7r-subgroups, even i f they exist, m a y n o t be con jugate: for example, the simple g r o u p P S L ( 2 , 11) o f order 660 has subgroups isomorphic w i t h D
1 2
and A \ A
these are n o n i s o m o r p h i c H a l l { 2 , 3}-subgroups
a n d they are certainly n o t conjugate. H o w e v e r the s i t u a t i o n is quite differ ent w h e n a n i l p o t e n t H a l l 7r-subgroup is present. 9.1.10 ( W i e l a n d t ) . Let the finite H. Then every n-subgroup lar all Hall n-subgroups Proof. L e t X be a
group G possess a nilpotent
of G is contained
of G are
rc-subgroup
in a conjugate
Hall
n-subgroup
of H. In
particu
conjugate. o f G. W e shall argue b y i n d u c t i o n o n
\K\,
w h i c h can be assumed greater t h a n 1. B y the i n d u c t i o n hypothesis a m a x i -
260
9. Finite Soluble Groups
m a l s u b g r o u p o f K is c o n t a i n e d i n a conjugate o f H a n d is therefore n i l p o t e n t . I f K itself is n o t n i l p o t e n t , 9.1.9 m a y be a p p l i e d t o produce a p r i m e q i n n d i v i d i n g \K\ a n d a S y l o w ^-subgroup Q w h i c h has a n o r m a l c o m p l e m e n t L i n K. O f course, i f K is n i l p o t e n t , this is still true b y 5.2.4. N o w write H = H
x
x H
2
where
is the u n i q u e Sylow ^-subgroup o f H. 9
Since L ^ K , the i n d u c t i o n hypothesis shows t h a t L
some g e G. T h u s L
9
= H{ x H
because L is a g'-group. Consequently AT =
2
2
for
N (L) G
contains < i f f , X > . Observe t h a t | G : H | is n o t divisible by q\ hence H\ is a x
Sylow g-subgroup o f N a n d by Sylow's T h e o r e m Q < (Hf)* for some x e N. But L = L
x
and, using L < i f f , we o b t a i n X
x
K = QL = QL
9 x
< Hf H
=
2
as required.
•
EXERCISES 9.1
1. Let G be a Frobenius group with kernel K. Prove that every complement of K in G is a Frobenius complement and that all Frobenius complements of K are conjugate in G (see Exercise 8.5.6). *2. Let AT<3 G and suppose that | G : N\ = m is finite and N is an abelian group in which every element is uniquely expressible as an mth power. Prove that N has a complement in G and all such complements are conjugate. [Hint: See the proof of 9.1.2.] 3. Let G be a countable locally finite group (i.e., finitely generated subgroups are finite). Suppose that ATo G and that elements of N and G/N have coprime orders. Prove that N has a complement in G. Show also that not all the complements need be conjugate by considering the direct product of a count able infinity of copies of S . 3
4. If H is a subnormal subgroup of a group G, prove that O (H) = H n 0 (G). n
7C
5. Let H and X be 7r-separable subgroups of a finite group G. I f H is subnormal in G, prove that
r
L
e
t
G
=
p
p
k
a
n
d
t
n
e
8. Let G be a finite soluble group whose order has at least three distinct prime divisors. I f every Hall p'-subgroup of G is nilpotent, show that G is nilpotent. [Hint: Prove that each Sylow subgroup is normal.] 9. (Wielandt). I f a finite group G has three soluble subgroups H ,H ,H with their indices coprime in pairs, then G is soluble. [Hint: Use induction on the order of G. Assume H ^ 1 and choose a minimal normal subgroup N of H : show that N is contained in either H or H . ] l
1
2
3
±
G
2
3
9.2. Sylow Systems and System Normalizers
261
10. A finite group in which every subgroup is either subnormal or nilpotent is solu ble. [Use 9.1.9.] 11. Let G = PQ be a finite minimal nonnilpotent group with the notation of 9.1.9. Derive the following information about G: (a) Frat Q < CG. (b) P = [P, Q] and Frat P < £(G), so P is nilpotent of class at most 2. (c) I f p is odd, P = 1, while P = 1 i f p = 2 [Hint: Prove that [a, x] = 1 or [a, x ] = 1 where a e P and x e Q.] p
4
p
4
12. (Ito). Let G be a group of odd order. I f every minimal subgroup lies in the center, prove that G is nilpotent. [Use Exercise 9.1.11.] 13. (Ito). Let G be a group of odd order. I f every minimal subgroup of G' is normal in G, prove that G' is nilpotent and G is soluble. *14. Let A be a minimal normal subgroup of a finite soluble group G such that N = C (N). Prove that N has a complement in G and all such complements are conjugate. [Hint: Let L/N be minimal normal in G/N. Show that N has a com plement X in L and argue that N (X) is a complement of N in G.] G
G
9.2. Sylow Systems and System Normalizers L e t G be a finite g r o u p a n d let p
l 9
. . . , p denote the distinct p r i m e divisors k
of | G | . Suppose t h a t Q is a H a l l p--subgroup o f G. T h e n the set t
is called a Sylow t h a t a finite
system
{ Q
l
Q
9
k
}
o f G. I t is a direct consequence o f 9.1.7 a n d 9.1.8
group has a Sylow system if and only if it is
soluble.
A S y l o w system determines a set o f p e r m u t a b l e S y l o w subgroups o f G i n the f o l l o w i n g manner. 9.2.1. Let { < 2 . . . , Q } be a Sylow system of the finite 1 ?
(i) If n is any set of primes, particular (ii) The Sylow
soluble group G.
k
P = f] Qj t
j¥:i
subgroups
then ^\ $ Qi Pi
is a Hall
%
is a Sylow p subgroup P
1
?
P
are permutable
k
n-subgroup
of G. In
of G.
r
in pairs,
that is, P Pj t
=
PjP, e k
Proof. L e t | G | = p{'--p H = f] ^ Qi Pi
k
where \G:Q \=
pf*. I t follows f r o m 1.3.11 t h a t
t
has index equal t o YlpitnPi**
n
w h i c h shows t h a t H is a H a l l
7r-subgroup o f G. A p p l y i n g this result t o n = {p
h
K = f) :ijQ k¥
pj}, i ^ j , we conclude t h a t
is a H a l l rc-subgroup w i t h order pppp,
k
c o n t a i n i n g P a n d Pj. t
Since \P P-\ = pf*pp, i t follows t h a t P ^ = K = PjP . t
•
t
A set o f m u t u a l l y p e r m u t a b l e S y l o w subgroups, one for each p r i m e dividing { Q i , Q
the k
}
group
order,
is called
is a S y l o w system
c o r r e s p o n d i n g Sylow basis
a
Sylow
o f finite
= { P
1
?
P
k
}
basis.
B y 9.2.1, i f J
soluble g r o u p
G, there
given by P = f^j^Qj. t
the converse holds: each S y l o w basis determines a S y l o w system.
=
is a
I n fact
262
9. Finite Soluble Groups
9.2.2. / / G is a finite
soluble group, the function
J H-»
is a bijection
between
the set of Sylow systems and the set of Sylow bases of G. Proof. L e t 0* = { P P } be any Ylj&PjSince the Pj are permutable, can tell that Q is a H a l l p--subgroup. system o f G. F i n a l l y one easily verifies l
9
k
t
so that 0*
Sylow basis o f G a n d define Q = Q is a subgroup. F r o m its order we Hence ^ = { Q Q } is a Sylow that t
t
l
9
k
n r u = n and n n e ^ a , i^kj^i k^i j^k a n d J h-» are inverse mappings.
^
•
T w o Sylow systems { Q Q } and { Q Q } o f G are said t o be conjugate i f there is an element g o f G such t h a t Qf = Qj for i = 1, 2 , . . . , k. Conjugacy o f t w o Sylow bases is defined i n the same way. l
9
k
l
9
k
9.2.3 (P. H a l l ) . In a finite soluble group G any two Sylow systems gate, as are any two Sylow bases.
are conju
Proof. D e n o t e b y ^ the set o f a l l H a l l p--subgroups o f G. T h e n G acts o n 9 by c o n j u g a t i o n a n d 9.1.7 shows t h a t this a c t i o n is transitive. Consequently |^| = \G:N (Q )\ where Q -e Sf{. i t follows t h a t | ^ | divides \G:Q \ and equals a p o w e r o f p . N o w G also acts by c o n j u g a t i o n o n the set 9* = ^ x • • • x
G
t
t
t
t
k
l9
t
l
G
9
k
k
= 1
t
9
System Normalizers Let { Q
l
9
Q
k
}
be a Sylow system o f a finite soluble g r o u p G. The subgroup N = 0 i=l
is called a system normalizer m a n y remarkable properties. Sylow basis, an element o f G every P : this i n o n account 9.2.1 a n d 9.2.2). Hence t
N (Qd G
o f G. W e shall see t h a t these subgroups have N o t i c e t h a t i f {P ..., P } is the corresponding normalizes every Q i f a n d o n l y i f i t normalizes o f the r e l a t i o n between the P a n d the Q (see l9
k
t
t
N=f]
N (P ). G
t
i=l a n d the system normalizers can also be o b t a i n e d f r o m Sylow bases.
t
9.2. Sylow Systems and System Normalizers
9.2.4. In a finite
263
soluble group the system normalizers
are nilpotent
and any
two are conjugate. Proof. L e t { P P } be a Sylow basis g i v i n g rise t o a system n o r m a l i z e r N. N o w \N:Nn P \ = \NP : P \ divides \ G:P \ since P ^ o NP . Hence N n P is a Sylow p s u b g r o u p o f N. Also J V n P ^ A . I t follows f r o m 5.2.4 that A is n i l p o t e n t . T h e conjugacy o f the system normalizers is a direct conse quence o f the conjugacy o f the Sylow systems. • 1
?
k
t
t
t
t
t
t
r
Covering and Avoidance Suppose t h a t G is a g r o u p a n d let K o f f < G a n d L < G. T h e n L is said t o cover H/K i f HL = KL, o r equivalently, i f H = K(H n L ) . O n the other h a n d , i f HnL = K n L , t h a t is, i f HnL < K, then L is said t o avoid H/K. 9.2.5. Le£ G be a finite soluble group and let H/K be a principal factor of G which is a p-group. Let M o G and denote by Q a Hall p'-subgroup of M. Then N (Q) covers or avoids H/K according as M centralizes H/K or not. G
Proof. D e n o t e N (Q) b y L . F i r s t o f a l l suppose that M centralizes H/K, so t h a t [ f f , M ] < K. N o w Q = Q [ i f , Q] b y 5.1.6 a n d , because Q < M, i t follows t h a t Q < Q(K n M). I f x e i f , then Q a n d Q are clearly H a l l p'subgroups o f Q(K n M ) a n d 9.1.7 shows t h a t they are conjugate, say Q = Q for some y in K nM. Hence xy' e L a n d x e LK = KL, w h i c h shows t h a t L covers H/K. G
H
H
x
x
y
1
N o w suppose t h a t M does n o t centralize H/K a n d w r i t e C = C (H/K); then D = C n M < M a n d consequently there is a p r i n c i p a l factor E/D o f G such t h a t E < M ; thus E £ C. N o w E / D acts v i a c o n j u g a t i o n o n i f / X . I f £ / D were a p-group, the n a t u r a l semidirect p r o d u c t (E/D) K ( i f /K) w o u l d be a finite p - g r o u p a n d hence n i l p o t e n t , w h i c h w o u l d i m p l y t h a t [ i f , E~]K < i f a n d thus [ i f , E~] < K: however this is false because E £ C. Therefore the p r i n c i p a l factor E/D is a p ' - g r o u p , f r o m w h i c h we deduce t h a t E < DQ: for Q is a H a l l p ' - s u b g r o u p o f M . I t follows t h a t G
[ i f nL,E~]
DQ] < [HnL,
D ] [ i f n L , Q] < K;
for D < C, while [H nL, Q]
H/K
T h i s result is the stepping stone t o a fundamental covering a n d a v o i d ance p r o p e r t y o f system normalizers.
264
9. Finite Soluble Groups
9.2.6 (P. H a l l ) . / / N is a system normalizer of a finite soluble group G, then N covers the central principal factors and avoids the noncentral principal factors ofG. Proof. L e t 1 = G < G < • • • < G = G be a p r i n c i p a l series o f G. L e t the system n o r m a l i z e r N arise f r o m a S y l o w system {Q ..., Q } a n d p u t A^ = N (Qi% so t h a t N = JV n • • • nN . A c c o r d i n g t o 9.2.5 ( w i t h G i n place o f M ) , the s u b g r o u p N covers the central p p r i n c i p a l factors a n d avoids the n o n c e n t r a l p p r i n c i p a l factors (here Q is a H a l l p--subgroup). Hence N certainly avoids the n o n c e n t r a l p r i n c i p a l factors. N o w \G N : GjN \ equals 1 o r \G :Gj\ according t o whether N covers o r avoids G /Gj. I t follows t h a t | G : N \ equals the p r o d uct o f the orders o f the n o n c e n t r a l p p r i n c i p a l factors. Since the | G : N \ are relatively p r i m e , | G : A T | equals the p r o d u c t o f the orders o f a l l the n o n central p r i n c i p a l factors. T h i s implies t h a t \N\ equals the p r o d u c t o f the orders o f the central p r i n c i p a l factors. B u t \N\ equals the p r o d u c t o f a l l the indices \ G nN :GjnN\ where G /Gj is central; for N avoids n o n c e n t r a l factors. Hence | G n N: G n N\ = \G : G \ i f G /Gj is central: this implies t h a t G = Gj(G n N) a n d N covers G /Gj. • 0
x
n
l9
G
X
k
k
t
r
r
t
j+1
t
t
t
j+1
J+1
t
r
j+1
t
j+1
j + 1
j+1
y
j+1
j+1
5
j+1
j+1
A n interesting c o r o l l a r y o f 9.2.6 is the fact t h a t the order of a system normalizer equals the product of the orders of all the central factors in a principal series. As a n a p p l i c a t i o n o f the covering-avoidance p r o p e r t y o f system n o r m a l izers, we shall prove a t h e o r e m o n the existence o f complements. 9.2.7 (Gaschiitz, Schenkman, Carter). Let G be a finite soluble group and denote by L the smallest term of the lower central series of G. If N is any system normalizer in G, then G = NL. If in addition L is abelian, then also N n L = 1 and N is a complement of L . Proof. F o r m a p r i n c i p a l series o f G t h r o u g h L b y refining 1 < L < G. Since G / L is n i l p o t e n t , p r i n c i p a l factors "above" L w i l l be central a n d hence are covered b y N (by 9.2.6). Therefore G = NL. N o w assume that L is a abelian. T h e n i t is sufficient t o p r o v e t h a t n o p r i n c i p a l factor o f G " b e l o w " L is central: for b y 9.2.6 the system n o r m a l i z e r N w i l l a v o i d such factors a n d N n L w i l l be t r i v i a l . W e shall accomplish this by i n d u c t i o n o n | L | > 1. B y the i n d u c t i o n hypothesis i t suffices t o show t h a t L n C G = 1. I f C = C ( L ) , then L < C < G since L = [ L , G ] ^ 1. Hence G/C is n i l potent; we n o w choose a n o n t r i v i a l element gC f r o m the center o f G/C, n o t i n g t h a t [ L , [ g , G ] ] = 1. W e deduce f r o m this r e l a t i o n a n d one o f the fundamental c o m m u t a t o r identities t h a t if ae L a n d xeG, then G
x
La, gY = La ,
x
= [a\ [ x , g ~ ^ = la ,
gl
9.2. Sylow Systems and System Normalizers
265
0
Hence the m a p p i n g 6: L L defined b y a = la, g] is a nonzero G-endom o r p h i s m o f L , a n d K e r # o G. Since L n £ G < K e r 0, we m a y assume K e r 0 ^ 1 , so t h a t ( L / K e r 0) n C(G/Ker 0) is t r i v i a l b y i n d u c t i o n . A l s o L / K e r 6 ~ L , f r o m w h i c h i t follows t h a t L° n CG = 1. N o w 1 ^ L o G a n d ( L / L ) n C ( G / L ) is t r i v i a l b y i n d u c t i o n . Therefore Ln£G = 1, as required. • G
0
0
0
0
9.2.8. Le£ N be a system normalizer of a finite soluble group G. Then N , the core of N in G, equals the hypercenter of G and the normal closure of N equals G itself G
Proof. L e t H be the hypercenter o f G a n d refine the series l < H < G t o a p r i n c i p a l series. B y 9.2.6 every central p r i n c i p a l factor is covered b y A , so H < N a n d hence H < N = K, say. I f H ^ K, there is a p r i n c i p a l factor L/H where L < K. N o w since L/H cannot be central, i t is avoided by A . B u t L < K < N, f r o m w h i c h i t follows t h a t L = H, a c o n t r a d i c t i o n . G
G
I f the n o r m a l closure N were proper, i t w o u l d lie inside a m a x i m a l n o r m a l subgroup o f G, say M . B u t G/M is abelian since G is soluble. Hence A covers G / M a n d G = M A = M , w h i c h cannot be true. • G
Since A is generated by conjugates o f A , we deduce f r o m 9.2.8 the fol l o w i n g fact. 9.2.9. A finite
soluble group is generated
by its system
normalizers.
Abnormal Subgroups 9
A subgroup H o f a g r o u p G is called abnormal i f g e (H, H } for a l l g i n G. F o r example, i t is simple t o show t h a t a n o n n o r m a l m a x i m a l subgroup is always a b n o r m a l . A b n o r m a l i t y is a strong f o r m o f n o n n o r m a l i t y w h i c h leads t o a n interesting characterization o f system normalizers. I m p o r t a n t examples o f a b n o r m a l subgroups are the Sylow normalizers. 9.2.10. Let N be a normal subgroup of the finite group G and let P be a Sylow p-subgroup of N. Then N (P) is abnormal in G. G
9
Proof. L e t H = N (P) a n d p u t K = < f f , H } where g is some element o f G. N o w P a n d P are conjugate i n K n A since they are Sylow p-subgroups of t h a t g r o u p . Therefore P = P for some x in K n N, a n d gx' e H. I t follows t h a t g E K, as requred. • G
9
9
x
1
A n a b n o r m a l subgroup H always coincides w i t h its normalizer: for i f g e N (H), then g e < f f , H } = H. N o w i t is o b v i o u s t h a t any subgroup o f G t h a t contains a n a b n o r m a l subgroup is itself a b n o r m a l . Consequently every 9
G
9. Finite Soluble Groups
266
subgroup t h a t contains f f is also self-normalizing. I n fact the converse o f this statement is true for finite soluble groups. 9.2.11 (Taunt). Let G be a finite H is abnormal
soluble group and let H be a subgroup.
in G if and only if every subgroup
its normalizer
containing
Then
H coincides
with
in G.
F o r the p r o o f we require a simple lemma. 9.2.12. Let G be a finite HN
group and let H < G and N<] G. If H is abnormal in
and HN is abnormal in G, then H is abnormal in G. 9
Proof. I f g e G, then g e {HN, (HN) } 9
x e N a n d y e < f f , H }. 9
1
9
= N{H, H };
where
N o w x e < f f , f f > since H is a b n o r m a l i n HN. 9
Hence x e < f f , H *' }
thus g = xy
x
9
9
< < f f , H } since y e < f f , H }. F i n a l l y g e < f f , H } as
required.
•
Proof of 9.2.11. O n l y the sufficiency o f the c o n d i t i o n is i n question: assume t h a t H satisfies the c o n d i t i o n . W e shall p r o v e t h a t H is a b n o r m a l i n G b y i n d u c t i o n o n | G | > 1. I f AT is a m i n i m a l n o r m a l subgroup o f G, the h y p o t h e sis o n H is i n h e r i t e d b y HN/N,
w i t h the result t h a t HN/N
is a b n o r m a l
i n G/N b y i n d u c t i o n . O b v i o u s l y this means that i f AT is a b n o r m a l i n G. I f f f AT ^ G, then f f is a b n o r m a l i n HN, b y i n d u c t i o n once again, a n d the de sired conclusion follows f r o m 9.2.12. F i n a l l y , suppose t h a t HN = G. T h e n , since N is abelian, f f n N = 1; we m a y a p p l y 5.4.2 t o show t h a t f f is m a x i m a l i n G. H o w e v e r f f = N (H) b y hypothesis, so f f is a b n o r m a l i n G. G
•
System Normalizers and Abnormality T h e a i m o f the rest o f this section is t o explore the relationship between a b n o r m a l i t y a n d system normalizers, the p r i n c i p a l t h e o r e m (9.2.15) being a characterization o f system normalizers. F o r the latter we shall need t w o pre l i m i n a r y results. 9.2.13. Let M be a nonnormal
maximal
m
subgroup
and let \ G:M\ = p . If Q is a Hall p'-subgroup Proof. I n the first place \G:M\
of a finite
< M/K
G/K
G
is indeed a p o w e r o f a p r i m e p b y 5.4.3.
I n d u c t o n | G | > 1. I f the core o f M i n G — c a l l i t K—is N (QK/K)
soluble group G
of M, then N (Q) < M.
n o n t r i v i a l , then
b y i n d u c t i o n , w h i c h surely implies t h a t N (Q) < M. G
Henceforth we suppose t h a t K = 1. Choose a m i n i m a l n o r m a l s u b g r o u p N of G; then N ^ M , so t h a t G = MN a n d M n J V < G . (Keep i n m i n d that N m u s t be abelian because G is soluble.) I t follows t h a t M n N = 1 a n d | N\ = \G:M\=
m
p.
9.2. Sylow Systems and System Normalizers
267
L e t N be m i n i m a l n o r m a l i n M a n d suppose t h a t N is a p-group. T h e n NN is a p-group a n d hence is n i l p o t e n t ; thus [AT, A ^ ] < N. B u t [AT, A T ^ o MAT = G, so [AT, A ^ ] = 1 a n d 1 ^ A ^ o MN = G. H o w e v e r , since M has t r i v i a l core, this is impossible. I t follows t h a t N has order p r i m e t o p a n d i n consequence A^ < Q. W r i t i n g L for N (Q\ we argue t h a t A/f < N™ = Ni < N N. Therefore JVf <(N N)nQ = N a n d L < N (N ). T h e latter subgroup equals M since J V ^ M a n d M is m a x i m a l : thus L < M . • x
x
1
x
N
G
X
1
x
G
1
9.2.14. 7 / M is a nonnormal maximal subgroup of a finite soluble group G, then every system normalizer of M contains a system normalizer of G. Proof. L e t {Qi, ...,Q } be a Sylow system o f G where Q is a H a l l p-subgroup o f G. T h e index o f M is G is a p r i m e power, say p™. Since a H a l l p i - s u b g r o u p o f M is also a H a l l p i - s u b g r o u p o f G, we m a y assume that Q i < M . I f i > 1, the indices | G : M | a n d | G : <2 | are relatively p r i m e a n d we conclude f r o m Exercise 1.3.8 t h a t G = MQ . Hence \M:MnQ \ = |G:& | , w h i c h is a p o w e r o f p . I t follows t h a t M n is a H a l l p--subgroup o f M and that { g = M n Q MnQ ,.. , Mr\Q } is a Sylow system o f M . Since A ^ Q i ) < M b y 9.2.13, k
f
f
t
t
f
t
l
5
2
t=l
k
i=l
i=l
Hence some system n o r m a l i z e r o f G is contained i n one o f M . T h e required result follows f r o m the conjugacy o f the system normalizers o f M . •
Subabnormal Subgroups A subgroup H o f a g r o u p G is said t o be subabnormal i n G i f there is a finite chain o f subgroups H = H < H < • • < H = G such t h a t i f , is a b n o r m a l i n H . S u b a b n o r m a l i t y is a weaker p r o p e r t y t h a n a b n o r m a l i t y . O u r inter est i n s u b a b n o r m a l i t y stems f r o m the next theorem. 0
x
n
i+1
9.2.15 (P. H a l l ) . The system normalizers cisely the minimal subabnormal
of a finite
soluble group G are pre
subgroups of G.
Proof. F i r s t l y i t w i l l be established t h a t every s u b a b n o r m a l subgroup H contains a system n o r m a l i z e r o f G. B y definition there is a chain H = H < H < " < H = G such that H is a b n o r m a l i n H . Since a d d i t i o n a l terms can be inserted i n the chain w i t h o u t d i s t u r b i n g a b n o r m a l i t y , we m a y assume t h a t H is m a x i m a l i n H . O f course H is n o t n o r m a l i n H . T h u s 9.2.14 implies t h a t each system n o r m a l i z e r o f H contains one o f H , a n d surely H contains a system n o r m a l i z e r o f G. 0
x
s
t
t
i+1
i+1
t
i+1
t
i+1
T o complete the p r o o f i t is sufficient t o prove t h a t every system n o r m a l izer is s u b a b n o r m a l . I f G is n i l p o t e n t , the o n l y system n o r m a l i z e r is G itself:
268
9. Finite Soluble Groups
for this reason we shall suppose t h a t G is n o t n i l p o t e n t . Choose a p r i n c i p a l series o f G a n d let f f be the smallest n o n n i l p o t e n t t e r m i n the series. I f K is the preceding t e r m , then K is n i l p o t e n t and, o f course, H/K
is a p r i n c i p a l
k
factor o f G w i t h order p , say, where p is p r i m e . I f P is a Sylow p-subgroup of f f , then f f = PK and P cannot be n o r m a l i n f f ; otherwise the latter w o u l d be n i l p o t e n t i n view o f 5.2.8. T h u s L = N (P) G
^ G; note t h a t L is a b n o r m a l
i n G by 9 . 2 . 1 0 . A l s o the F r a t t i n i a r g u m e n t ( 5 . 2 . 1 4 ) implies t h a t G = LH = LPK
= We
tem
LK.
show next t h a t any system n o r m a l i z e r A o f L is a u t o m a t i c a l l y a sys
n o r m a l i z e r o f G. B y i n d u c t i o n o n | G | we have t h a t A is s u b a b n o r m a l i n
L a n d hence i n G. A p p l y i n g the result o f the first p a r a g r a p h , we conclude t h a t A contains a system n o r m a l i z e r o f G. Since system normalizers o f G, being conjugate, have the same order, i t suffices t o p r o v e t h a t A is con tained i n a system n o r m a l i z e r o f G. Let
A arise f r o m the Sylow system [Q
u
..., Q } o f L . W r i t i n g K for the k
t
unique H a l l p--subgroup o f the n i l p o t e n t g r o u p K, we observe t h a t K *<\ t
a n d Qf = Q K t
is a p--group. A l s o | G : Qf\ = \LK:
t
since b o t h \L:Q \
a n d \K:K \
t
QiK \
G
is a p o w e r o f p
t
{
are powers o f p . Therefore Qf is a H a l l
t
t
p--subgroup o f G a n d { Q * , Q
k
}
is a Sylow system o f G — w i t h corre
s p o n d i n g system n o r m a l i z e r A * , let us say. Since N (Q ) L
t
< N (Qf), G
we
o b t a i n A < A * as required. F i n a l l y , A is s u b a b n o r m a l i n L , w h i c h is a b n o r m a l i n G; hence A is s u b a b n o r m a l i n G. T h i s completes the p r o o f since a l l system normalizers o f G are conjugate.
•
EXERCISES 9 . 2
1. Locate the system normalizers of the groups S , A , S , SL(2, 3). 3
A
4
2. Let G be a finite soluble group and let n be the set of prime divisors of |G|. Let 7i = K u 7i u • • • u % be any partition of 7i. Prove that there exists a set of pairwise permutable Hall 7i subgroups, i = 1, 2 , . . . , k. 1
2
r
3. (P. Hall). Let G be tinct primes. Prove where m = \GL(d Sylow p subgroup now consider 5^ .] t
h
r
a finite soluble group of order f | * pfs where the p are dis that the order of Out G divides the number YYi=i miPi ~ \ p )\ and d is the minimal number of generators of a of G. [Hint: Let £f be a Sylow basis of G and let y e Aut G: = 1
t
i(ei
t
di
t
7
4. Show that the last part of 9.2.7 need not be true if L is nonabelian. 5. Let G be a finite soluble group in which the last term L of the lower central series is abelian. Prove that every complement of L is a system normalizer. 6. Give an example of a subabnormal subgroup that is not abnormal. 7. Let G be a finite soluble group which is not nilpotent but all of whose proper quotients are nilpotent. Denote by L the last term of the lower central series. Prove the following statements:
9.3. p-Soluble Groups
(a) (b) (c) (d)
269
L is minimal normal in G; L is an elementary abelian p-group; there is a complement X ^ 1 of L which acts faithfully on L ; the order of X is not divisible by p.
8. Suppose that X ^ 1 is a finite nilpotent p'-group that acts faithfully on a simple FX-module L where F = GF(p). Prove that every proper quotient of G = X K L is nilpotent but G is not nilpotent. 9. Let G be a finite soluble group which is not abelian but all of whose proper quotients are abelian. Prove that either G is generalized extra-special (see Exer cise 5.3.8) or G is isomorphic with a subgroup of H(q) where q > 2 (see 7.1).
9.3. p-Soluble Groups I f n is a set o f primes, a finite g r o u p G is called n-soluble
i f i t has a series
whose factors are rc-groups o r 7r'-groups a n d i f the 7r-factors are soluble. E q u i v a l e n t l y we c o u l d have said t h a t each c o m p o s i t i o n factor is either a 7r-group o r a 7r'-group a n d i n the f o r m e r case has p r i m e order. (The reader s h o u l d supply a proof.) 7r-solubility is, therefore, a s t r o n g f o r m
o f %-
separability. E v i d e n t l y a l l finite soluble groups are 7r-soluble. O f p a r t i c u l a r i m p o r t a n c e is p-solubility
( w h i c h is the same as p-separability), a concept
i n t r o d u c e d i n 1956 by P. H a l l a n d G . H i g m a n [ a 8 0 ] i n a paper o f great significance for finite g r o u p theory. T h e f o l l o w i n g result is basic. 9.3.1. If G is a n-separable
group, then C (0 ' (G)/0 (G)) G
7t 7t
< 0 , (G).
JZ
7r
7r
Proof. C l e a r l y we can assume t h a t O ^ ( G ) = 1 a n d p r o v e t h a t C ( O ( G ) ) < G
O ^ G ) . P u t P = O (G)
w
a n d C = C ( P ) . T h e n P n C = C P o G, so t h a t CP <
n
G
O ^ C ) . A l s o O ^ C ) < P n C = CP, a n d i t follows t h a t CP = O ^ C ) . I f C £ P, t h e n O ^ C ) = CP < C. Since C is ^-separable, there exists a characteristic s u b g r o u p L o f C such t h a t O ^ C ) < L a n d L/O (C) n
is a 7r'-group. B y the
Schur-Zassenhaus T h e o r e m (9.1.2) there is a s u b g r o u p K such t h a t L = K(O (C))
a n d K n O ^ C ) = 1. I n fact L = Kx
n
O (C) n
because X < C a n d C
centralizes 0 ( G ) . I t follows t h a t K is n o r m a l i n G, being the unique Sylow 7r
7r'-subgroup o f L . T h u s K < O (G) n
= 1 and L = O ^ C ) , i n contradiction to
the choice o f L .
•
9.3.2 ( H a l l - H i g m a n ) . Let G be a p-soluble denotes O (G), p
then conjugation
group of linear transformations
P
p
p
representation
= 1. If P of G/P as a
of the vector space P / F r a t P.
Proof. Recall t h a t F r a t P = P'P over Z .
group such that O (G)
leads to a faithful
by 5.3.2, so t h a t P / F r a t P is a vector space
T h u s C = C ( P / F r a t P) contains P. I f D = C ( P ) , then D < C a n d G
G
C/D is a p - g r o u p by 5.3.3. A l s o D < O (G) p
C o G, i t follows t h a t C = P.
b y 9.3.1, so C is a p-group. Since •
9. Finite Soluble Groups
270
p-Nilpotent Groups A finite g r o u p G is said t o be p-nilpotent (where p is a p r i m e ) i f i t has a n o r m a l H a l l p'-subgroup, t h a t is, i f O (G) = G. O b v i o u s l y every finite n i l p o t e n t g r o u p is p-nilpotent; conversely a finite g r o u p w h i c h is p - n i l p o t e n t for a l l p is n i l p o t e n t , as the reader s h o u l d check. pp
T h e p r o d u c t o f a l l the n o r m a l p - n i l p o t e n t subgroups o f a finite g r o u p is clearly O (G): this is the m a x i m u m n o r m a l p - n i l p o t e n t subgroup o f G a n d we shall also w r i t e i t pp
Fit (G), the p-Fitting subgroup. T h e f o l l o w i n g characterization o f F i t ( G ) is analogous t o a n already p r o v e n characterization o f F i t G—see 5.2.9. p
p
9.3.3. If G is a finite
group, then F i t ( G ) equals the intersection
izers of the principal
factors
p
of G whose orders are divisible
of the central
by p.
Proof. L e t i f be a n o r m a l p - n i l p o t e n t subgroup o f G a n d let A be a m i n i m a l n o r m a l subgroup o f G whose order is divisible b y p. Assume t h a t [ f f , A ] ^ 1. T h e n H n N ^ l a n d thus A < f f . N o w A ^ O ,(H), so t h a t NnO ,(H) = 1. Hence A is a p-group. A l s o C (N) ^ O ,(H) and f f = H/C (N) is a p-group. T h e n a t u r a l semidirect p r o d u c t f f K A is a p-group, so i t is n i l p o t e n t a n d [ f f , A ] < A . T h i s implies t h a t [ f f , A ] = 1 since A is m i n i m a l n o r m a l i n G. I t follows b y i n d u c t i o n o n | G| t h a t f f centralizes every p r i n c i p a l factor whose order is divisible b y p. p
p
H
p
H
N o w let C be the intersection o f the centralizers o f the p r i n c i p a l factors whose orders are divisible b y p. T h e n F i t ( G ) < C b y the previous para graph. W e are required t o prove t h a t C is p - n i l p o t e n t . L e t A be m i n i m a l n o r m a l i n G. T h e n b y i n d u c t i o n o n the g r o u p order CN/N ~ C/C n A is p-nilpotent. W e m a y therefore assume that C n A ^ 1, so t h a t N < C a n d C/N is p-nilpotent. I f A is a p ' - g r o u p , i t is obvious t h a t C is p-nilpotent. Suppose t h a t p divides | A | ; then [ A , C ] = 1 a n d A < £ C . I t follows t h a t A is a p-group. L e t M/N = O ,(C/N): clearly C/M is a p-group. Because A < C M we can apply 9.1.2 t o show t h a t M has a n o r m a l H a l l p'-subgroup L . B u t C/L is a p-group, so C is p - n i l p o t e n t . • p
p
A n o t h e r analogue o f a result o n n i l p o t e n c y is next (see 5.2.15). 9.3.4. Let G be a finite group and assume that F r a t G < J V < G and A / F r a t G is p-nilpotent. Then N is p-nilpotent. Hence F i t ( G / F r a t G) = F i t ( G ) / F r a t G. p
Proof. L e t F = F r a t G a n d w r i t e O ,(N/F) = factored o u t i f necessary, we m a y assume t h a t F is a subgroup i f such t h a t Q = HF a n d f f n F p ' - g r o u p . I f g e G, then f f a n d H are conjugate p
9
p
Q/F. Since O ,(F) can be is a p-group. B y 9.1.2 there = 1: here o f course f f is a i n Q, b y 9.1.2 again: hence p
271
9.3. p-Soluble Groups
9
x
1
H = H for some x i n Q a n d gx' eL = N (H). I t follows that G = LQ = LF. H o w e v e r F consists o f nongenerators b y 5.2.12. Consequently G = L and G. Since N/H is o b v i o u s l y a p-group, N is p-nilpotent. • G
I n Chapter 10 we shall establish i m p o r t a n t criteria for a g r o u p t o be p - n i l p o t e n t due t o Frobenius a n d T h o m p s o n .
The p-Length of a p-Soluble Group Recall t h a t i f G is a p-soluble g r o u p , the p-length l (G) is the l e n g t h o f the upper p'p-series. W e w i s h t o relate this i n v a r i a n t t o other invariants o f G such as the n i l p o t e n t class o f the S y l o w p-subgroups. T o begin w i t h t w o simple lemmas w i l l be established. p
9.3.5. If G is a finite
group and p is a prime dividing
| G | , then p also
divides
| G : F r a t G|. Proof. Assume t h a t | G : F r a t G | is n o t divisible b y p ; then a Sylow ps u b g r o u p P o f G is contained i n F r a t G. Since the latter is n i l p o t e n t , P < ] G. B y 9.1.2 there is a s u b g r o u p H such t h a t G = HP a n d H n P = 1. B u t , since P < F r a t G, i t follows t h a t G = H a n d P = 1, w h i c h contradicts the fact t h a t p divides | G | . • 9.3.6. If G is a p-soluble
group, l (G) = / ( G / F r a t G). p
p
Proof. I f F denotes F r a t G, then i t is obvious that l (G/F) < l (G). I f l (G/F) = 0, then p does n o t d i v i d e | G : F | a n d 9.3.5 shows t h a t p cannot divide | G | : hence l (G) = 0. Suppose t h a t l (G/F) > 0: n o w F i t ( G / F ) = F i t ( G ) / F b y 9.3.4, t h a t is, O , (G/F) = O {G)/F ^ 1. F r o m this i t follows t h a t the upper p'p-series o f G a n d G / F have same length. • p
p
p
p
p
p
p p
p
pp
T h e fundamental t h e o r e m o n p-length can n o w be established. 9.3.7 ( H a l l - H i g m a n ) . Let G be a p-soluble
group.
(i) l (G) < c (G) where c (G) is the nilpotent class of a Sylow p-subgroup. (ii) l (G) < d (G) where d (G) is the minimum number of generators of a Sylow p-subgroup. (iii) l (G) < s (G) where s (G) is the maximum rank of a p-principal factor of G. p
p
p
p
p
p
p
p
p
Proof, (i) I f c (G) = 0, then o f course G is a p ' - g r o u p a n d l (G) = 0: suppose t h a t c (G) > 0 a n d proceed b y i n d u c t i o n o n c (G). L e t P be a Sylow p-subgroup o f G. T h e n P O ^ G y O ^ G ) is a Sylow p-subgroup o f G / O ^ G ) , p
p
p
p
272
9. Finite Soluble Groups
so i t contains O , (G)/O ,(G). Therefore CP centralizes O . (G)/O .(G) 9.3.1 shows t h a t CP < O , (G). F r o m this i t follows t h a t c (G/O (G)) c (G) - 1; consequently l {G/O {G)) < c (G) - 1. F i n a l l y p p
p
p p
p p
p
p
p
p
p>p
p>p
and <
p
l (G)
= l (G/O . (G))
p
p
+ 1,
p p
so the desired i n e q u a l i t y follows. (ii) T h e p r o o f employs i n d u c t i o n o n d = d (G). N o t i c e t h a t d = 0 is t o be interpreted as P = 1, so t h a t l (G) = 0 i n this case. L e t l (G) > 0. E v i d e n t l y we m a y assume t h a t O ,(G) = 1. L e t P = O (G); then P < P. p
p
p
p
1
p
x
Suppose t h a t P < F r a t P. T h e n P < F r a t G by 5.2.13, w h i c h leads t o l (G/FmtG)
p
p
x
1
p
x
B y the Burnside Basis T h e o r e m (5.3.2) d equals the m i n i m u m n u m b e r of generators o f P / F r a t P, so this g r o u p has order p . N o w F r a t ( P / P ) contains ( F r a t P ) P / P . T h u s the F r a t t i n i q u o t i e n t g r o u p o f P/P is an image o f P / ( F r a t P)P a n d its order is at m o s t p . A p p l i c a t i o n o f the i n d u c t i o n hypothesis t o G/P yields ^ ( G / P J < d — 1 a n d hence l (G) < d. d
x
1
1
1
d - 1
X
1
p
(iii) L e t s = s (G). I f s were 0, t h e n G w o u l d be a p ' - g r o u p a n d / = l (G) w o u l d equal 0; therefore we assume s > 0. L e t H/K be a p - p r i n c i p a l factor of G; then H/K has order p where n < s. N o w G/C (H/K) is i s o m o r p h i c w i t h a subgroup o f Aut(H/K) and Aut(H/K) ^ G L ( n , p) (see Exercise 1.5.11). The set o f (upper) u n i t r i a n g u l a r matrices i n G L ( n , p) is a Sylow p-subgroup a n d its n i l p o t e n t class is equal t o n — 1 (Exercise 5.1.11): i t follows t h a t y P centralizes H/K. Therefore y P centralizes every p - p r i n c i p a l factor a n d y P < F i t ( G ) = O . (G) b y 9.3.3. B y (i) we have p
p
n
G
n
s
s
p
p p
l (G/O . (G)) p
p p
< c (G/O . (G)) p
p p
< s - 1,
f r o m w h i c h i t follows i m m e d i a t e l y t h a t l (G) < s.
•
p
A n a p p l i c a t i o n o f 9.3.7 t o the restricted Burnside P r o b l e m is given i n 14.2.
p-Soluble Groups of p-Length at Most 1 A finite p-soluble g r o u p G has p-length at m o s t 1 i f a n d o n l y i f O . .(G) = G. F o r example, a p - n i l p o t e n t g r o u p has p-length < 1. The next objective is a result o f H u p p e r t t h a t characterizes soluble groups o f p-length < 1 i n terms o f their Sylow bases. T h i s t h e o r e m is preceded b y t w o p r e l i m i n a r y results. p pp
9.3.8. Let G be a p-soluble group and suppose that every proper has p-length < k while l (G) > k. Then: p
image of G
9.3. p-Soluble Groups
273
(i) F r a t G = 1; (ii) F i t ( G ) = O > (G) = N is an elementary abelian p-group; unique minimal normal subgroup of G; (iii) N = C (N) and N has a complement in G. p
p
this is also the
p
G
Proof, (i) follows at once f r o m the e q u a t i o n / ( G / F r a t G) = l (G) (see 9.3.6). (ii) a n d (iii). C l e a r l y O (G) = 1, so N is a p-group. Also F r a t N < F r a t G = 1 b y 5.2.13, w h i c h shows t h a t N is elementary abelian. N o w i f N a n d N are n o n t r i v i a l n o r m a l subgroups o f G such t h a t N nN = 1 then G/N a n d G/N b o t h have p-length < k a n d the m a p p i n g g\-+(gN gN ) is a m o n o m o r p h i s m o f G i n t o G/N x G/N ; however this implies that l (G) < k. I t follows t h a t G has a unique m i n i m a l n o r m a l subgroup L a n d L < N. Since F r a t G = 1, there is a m a x i m a l subgroup M w h i c h fails t o c o n t a i n L . T h u s G = M L a n d M n L < M L since L is abelian. Consequently M n L = 1 a n d M is a complement o f L i n G. A l s o N = N n ( M L ) = (AT n M ) L , a n d AT n M o G since b o t h M a n d L n o r m a l i z e NnM.liNnM^l, then L < AT n M < M , i n c o n t r a d i c t i o n t o the choice o f M . Therefore AT n M = 1 a n d AT = L . p
p
p
x
2
1
x
2
2
9
l9
x
2
2
p
F i n a l l y , suppose t h a t C = C (N) is strictly larger t h a n N. T h e n C = C n ( M N ) = ( C n M)AT a n d l ^ C n M o G . H o w e v e r this implies t h a t AT < C n M , w h i c h is impossible. • G
I t turns o u t t h a t the p r o p e r t y o f h a v i n g p-length at m o s t 1 is closely connected w i t h a curious p e r m u t a b i l i t y p r o p e r t y o f the Sylow p-subgroups of G. T h i s is already i n d i c a t e d b y the next result. 9.3.9. Let G be a finite p-soluble group. Let P be a Sylow p-subgroup Hall p'-subgroup of G. If P'Q = QP\ then l (G) < 1.
and Q a
p
Proof. L e t G be a counterexample t o the assertion w i t h least possible order. Since the hypothesis is i n h e r i t e d b y images, every p r o p e r image o f G has p-length < 1. T h i s is o u r o p p o r t u n i t y t o m a k e use o f 9.3.8. T h u s N = F i t ( G ) is a p - g r o u p a n d N < P. Therefore Nn(P'Q) = NnP'^P. B u t N n ( F 0 < P'Q a n d b y e x a m i n a t i o n o f order G = PQ; hence N n P ' o G. N o w , according t o 9.3.8, the subgroup N is the unique m i n i m a l n o r m a l subgroup o f G, so either N n P' = 1 o r N < P'. I n the first case [AT, P ' ] = 1 because N < P; therefore P' < C (N) = AT, b y 9.3.8(iii). T h i s leads t o the successive conclusions P' = 1, c (G) < 1, a n d l (G) < 1 v i a 9.3.7. I t follows t h a t the other possibility N < P' m u s t prevail. Consequently N < F r a t P. A p p l y i n g 5.2.13(i) we conclude that N < F r a t G. B u t F r a t G = 1 b y 9.3,8, so N = 1, a c o n t r a d i c t i o n . • p
G
p
W e come n o w t o the c r i t e r i o n referred t o above.
p
9. Finite Soluble Groups
274
9.3.10 ( H u p p e r t ) . Let G be a finite Sylow basis of G.
soluble group and let {P ... P } l9
9
be a
k
(i) If P;Pj = PjP; for all i and j , then l (G) < 1 for all p. (ii) Conversely if l (G) < 1 , every characteristic subgroup of P is permutable with every characteristic subgroup of Pf in particular P' Pj = PjP(. p
p
t
t
Proof, (i) L e t Qj be a H a l l pj-subgroup arising f r o m the given Sylow basis (as i n 9 . 2 . 2 ) ; thus Qj = Y\ ^jP . Hence P- surely permutes w i t h Qj b y the k
hypothesis, a n d l (G)
k
< 1 by 9.3.9.
p
(ii) W r i t e H = P P where i ^ j keeping i n m i n d t h a t P a n d Pj permute since they belong t o a Sylow basis. C l e a r l y l (H) < l (G) < 1 , w h i c h shows t h a t the upper pjpy-series o f H has the f o r m l < i V < i NPj^i H where A a n d H/NPj are pj-groups. L e t Kj be characteristic i n Pj. Since the n a t u r a l h o m o m o r p h i s m Pj -> NPj/N is an i s o m o r p h i s m , i t follows t h a t NKj/N is charac teristic i n NPj/N, a n d hence t h a t NKj^i H. N o w N < P so P Kj = Pt(NKj) is a subgroup. B y 1 . 3 . 1 3 we deduce that P Kj = KjP = L , say. N o w l (H) < 1 a n d Pj a n d Kj are respectively a Sylow p a n d a Sylow p - s u b g r o u p o f L . I f K is a characteristic subgroup o f P then K Kj = KjK b y w h a t has already been p r o v e d . • t
j9
9
t
p
p
h
t
t
t
p
r
t
i9
t
t
EXERCISES 9.3
1. A finite group is nilpotent if and only if it is p-nilpotent for all p. 2. Give an example of a finite group that is p-nilpotent and ^-nilpotent for two distinct prime divisors p, q of its order, but is not nilpotent. 3. A finite group is soluble if and only if it is p-soluble for all p. 4. A finite group is p-nilpotent i f and only i f every principal factor of order divisible by p is central. 5. A finite group is p-soluble of p-length < 1 i f and only if the group induces a p'-group of automorphisms in every principal factor whose order is divisible by p. 6. Let G be a finite p-soluble group with O ,(G) = 1. Prove that C (P/Frat P) = P where P = O (G). Deduce that |G| is bounded by a function of |P|. p
G
p
7. Let G be a finite p-soluble group. Prove that G has an abelian Sylow p-subgroup if and only if l (G) < 1 and O ( G ) / O , ( G ) is abelian. p
p>
p
9.4. Supersoluble Groups D u r i n g the e x p o s i t i o n of basic properties o f supersoluble groups i n Chapter 5 i t was s h o w n t h a t a m a x i m a l subgroup o f a supersoluble g r o u p has p r i m e index. O u r object i n this section is t o prove t h a t for finite groups the con-
9.4. Supersoluble Groups
275
verse o f this statement is true, a result due t o H u p p e r t . W e begin w i t h a n a u x i l i a r y result w h i c h is o f independent interest. 9.4.1 (P. H a l l ) . If every maximal subgroup of a finite prime or the square of a prime, then G is soluble.
group G has index a
Proof. L e t AT be a m i n i m a l n o r m a l subgroup o f G a n d denote b y p the largest p r i m e d i v i s o r o f \N\. L e t P be a Sylow p-subgroup o f N a n d p u t L = N (P). I f L = G, then P o G a n d G/P is soluble b y i n d u c t i o n o n | G | ; hence G is soluble. T h u s we m a y assume t h a t L < G a n d choose a m a x i m a l subgroup M w h i c h contains L . T h e n according t o the hypothesis \ G:M\ = q o r q for some p r i m e q. B y the F r a t t i n i argument G = NL = NM a n d \G\M\ = \N: N n M | , f r o m w h i c h we deduce t h a t q divides \N\ a n d conse q u e n t l y q < p. G
2
N e x t , the conjugates o f P i n G account for a l l the Sylow p-subgroups o f N; therefore | G : L | = 1 m o d p b y Sylow's T h e o r e m . F o r the same reason \M: L | = 1 m o d p a n d i t follows t h a t | G : M\ = 1 m o d p. N o w q =fc 1 m o d p because g < p, so we are left w i t h o n l y one possibility, \G:M\ = q = 1 m o d p a n d thus q = — 1 m o d p. H o w e v e r this is possible o n l y i f p = 3 a n d q = 2. T h u s | N : N n M\ = 4 a n d consequently JV has as a n image some n o n t r i v i a l subgroup o f S (see 1.6.6). Hence N > N' a n d AT = 1 b y m i n i m a l i t y . T h u s N is abelian, while G/N is soluble b y i n d u c t i o n . F i n a l l y , we conclude t h a t G is soluble. • 2
4
W e shall also need t w o results o f a m o r e elementary character. 9.4.2. Let N be a minimal normal subgroup of the finite soluble group G and let N < L < G. Assume that L/N is p-nilpotent but L is not. Then N has a complement in G. Proof. Suppose first t h a t N < F r a t G = F . T h e n L F / F is p-nilpotent. H o w ever, b y 9.3.4 this implies t h a t L F , a n d hence L , is p-nilpotent. T h u s there exists a m a x i m a l subgroup M n o t c o n t a i n i n g N. T h e n G = MN a n d M n A T o G since N is abelian. Hence M n N = 1. • 9.4.3. Let G be an irreducible abelian subgroup of G L ( n , p). Then G is a cyclic group of order m where n is the smallest positive integer such that p = 1 m o d m. n
Proof. L e t us identify G w i t h a g r o u p o f linear transformations o f a n nd i m e n s i o n a l vector space V over F = G F ( p ) . T h e n V is a simple m o d u l e over R = F G. Choose a n y nonzero vector v f r o m V. T h e n rt-+vr is a n Rm o d u l e h o m o m o r p h i s m f r o m R o n t o V the kernel I being a m a x i m a l ideal since R/I ~ V. Hence F = R/I is a finite field o f order p . N o w the m a p p i n g gy-±g + I is a m o n o m o r p h i s m c? f r o m G t o F * . Consequently G is cyclic. p
p
9
n
9. Finite Soluble Groups
276
n
n
Also m = \ G\ divides p — 1 a n d p = 1 m o d m. L e t n be a smaller positive integer t h a n n. Since the e q u a t i o n £ = t has o n l y p " solutions i n F a n d since G generates F , we m u s t have g ^ # for some g i n G. Hence p " ^ 1 m o d m. • x
p n i
6
1
pni
1
W e come n o w to the p r i n c i p a l t h e o r e m o f this section. 9.4.4 ( H u p p e r t ) . Let G be a finite group. If every maximal subgroup has prime index, then G is supersoluble. Proof. I n the first place 9.4.1 assures us t h a t G is at least sume t h a t | G | > 1 a n d e m p l o y i n d u c t i o n o n | G | . L e t AT be a subgroup of G; then G/N is supersoluble b y the i n d u c t i o n N is an elementary abelian p - g r o u p , say o f order p . O u r t h a t n = 1. n
soluble. W e as minimal normal hypothesis, a n d task is to prove
L e t L/N = O , (G/N), so L/N is p - n i l p o t e n t . Assume t h a t L is not pn i l p o t e n t . T h e n b y 9.4.2 there is a subgroup X such t h a t G = XN a n d X n N = 1. T h u s \ G:X\ = \N\ = p . B u t i t is clear t h a t X is m a x i m a l i n G, whence i t follows t h a t n = 1. p p
n
N o w assume t h a t L is p - n i l p o t e n t . Consider a p - p r i n c i p a l factor H/K of G such t h a t N < K. Since G/N is supersoluble, \H: K\ = p a n d thus G/C (H/K) is abelian w i t h order d i v i d i n g p — 1. I f g is any element o f G, t h e n g ~ centralizes every p - p r i n c i p a l factor o f G/N. A p p l y i n g 9.3.3 we con clude t h a t g ~ G L . Also G' < L for the same reason, so G / L is abelian w i t h order d i v i d i n g p — 1. G
p
l
v
l
N e x t IN, L ] o G, so either N = [AT, L ] o r [AT, L ] = 1; i n the former event, since L is p - n i l p o t e n t a n d thus L/O (L) is n i l p o t e n t , we s h o u l d have N_ < O ,{L) a n d N = 1. I t follows t h a t [AT, L ] = 1 a n d L < C (N). T h u s G = G/C (N) is abelian w i t h exponent d i v i d i n g p — 1. I t is also i s o m o r p h i c w i t h an irreducible subgroup o f G L ( n , p). W e deduce f r o m 9.4.3 t h a t G is cyclic o f order m where n is the smallest positive integer such t h a t p" = 1 m o d m. H o w e v e r m divides p — 1, so i n fact p = 1 m o d m a n d n = 1. p
p
G
G
• N o t e the f o l l o w i n g consequence o f H u p p e r t ' s theorem. 9.4.5. If G is a finite soluble.
group and G / F r a t G is supersoluble,
then G is super-
EXERCISES 9.4
1. Let G be a finite soluble group which is not supersoluble but all whose proper quotients are supersoluble. Establish the following facts (without using 9.4.4). (a) The Fitting subgroup F is an elementary abelian p-group of order > p. (b) F is the unique minimal normal subgroup of G.
9.5. Formations
277
(c) G = XF and X n F = 1 where the supersoluble subgroup X acts faithfully on F. (d) Frat G = 1. 2. Deduce Huppert's Theorem (9.4.4) from Exercise 9.4.1. 3. Let G be a finite soluble group. Assume that every quotient group G with plength at most 2 is supersoluble for all primes p. Prove that G is supersoluble. 4. Let G be a finite group. Prove that G is supersoluble i f and only i f for every proper subgroup H there is a chain of subgroups H = H < H < • • • < H = G with each index \H : H \ a prime. 0
i+1
1
l
t
9.5. Formations A class o f finite groups g is said t o be a formation^ i f every image o f an g - g r o u p is an g - g r o u p a n d i f G/N n A belongs t o g whenever G/N a n d G / A belong t o g . A c o m p a r i s o n o f this definition w i t h the characterization of varieties i n 2.3.5 m i g h t suggest t h a t a f o r m a t i o n be t h o u g h t o f as a k i n d of finite analogue o f a variety. ( H o w e v e r f o r m a t i o n s are n o t i n general closed w i l l respect t o f o r m i n g subgroups.) 1
2
x
2
Examples o f f o r m a t i o n s are readily found. T h e classes o f finite groups, finite soluble groups, finite n i l p o t e n t groups, a n d finite supersoluble groups are a l l formations. A f o r m a t i o n g is said t o be saturated i f a finite g r o u p G e g whenever G / F r a t G e g . O b v i o u s l y the class o f a l l finite groups is saturated. The n i l p o t e n c y o f F r a t G implies t h a t finite soluble groups f r o m a saturated for m a t i o n . T h e other t w o f o r m a t i o n s m e n t i o n e d above are also saturated i n view o f 5.2.15 a n d 9.4.5.
Locally Defined Formations W e shall describe an i m p o r t a n t m e t h o d o f c o n s t r u c t i n g saturated f o r m a tions. F o r each p r i m e p let g be a f o r m a t i o n o r the e m p t y set. Define g t o be the class o f a l l groups G w i t h f o l l o w i n g p r o p e r t y : i f H/K is a p r i n c i p a l factor o f G whose order is divisible b y p, then G/C (H/K) belongs t o g . p
G
p
I t is clear t h a t the class g is closed w i t h respect t o f o r m i n g images. I f G/N a n d G / A belong t o g d N i n A = 1, a p r i n c i p a l factor o f G is G-isomorphic w i t h a p r i n c i p a l factor o f G/N o r o f G / A . T h i s is a conse quence o f the J o r d a n - H o l d e r T h e o r e m . T h e v a l i d i t y o f the centralizer p r o p erty is n o w apparent. T h u s g is i n fact a f o r m a t i o n , g is said t o be locally defined b y the g . F o r example, we see that the class o f groups o f order 1 is locally defined b y t a k i n g every g t o be empty. a
l
n
2
2
x
p
p
t Some authors allow formations to be empty.
2
9. Finite Soluble Groups
278
9.5.1 (Gaschutz). / / g is locally rated formation.
defined by formations
g , then g is a satu p
Proof. L e t G be a g r o u p such t h a t G / F r a t G e g : we have t o show t h a t G e g . L e t P denote the intersection o f the centralizers o f those p r i n c i p a l factors o f G / F r a t G whose orders are divisible b y the p r i m e p; then G/P e g b y d e f i n i t i o n o f g . N o w P / F r a t G = F i t ( G / F r a t G) = F i t ( G ) / F r a t G b y 9.3.3 a n d 9.3.4. T h u s P = F i t ( G ) a n d 9.3.3 shows t h a t P centralizes every p r i n c i p a l factor o f G whose order involves p. I f H/K is such a p r i n c i p a l factor, then C (H/K) > P a n d therefore G/C (H/K) e g . I t n o w follows that G G g . • p
p
p
p
G
G
p
Remark. P. S c h m i d [ a 185] has p r o v e d t h a t every saturated f o r m a t i o n is locally defined. T h u s a l l saturated f o r m a t i o n s can be constructed i n the above manner. E X A M P L E S . L e t g be the f o r m a t i o n locally defined b y f o r m a t i o n s g where p is p r i m e . (i) I f each g is the class o f u n i t groups, then g is the class o f finite nilpotent groups. (ii) I f q is a fixed p r i m e , define $ t o be the class o f a l l u n i t groups a n d i f p ^ q, let g be the class o f a l l finite groups. T h e n g is the class o f finite groups i n w h i c h every p r i n c i p a l factor w i t h order divisible b y q is central. B y 9.3.3 this is j u s t the class o f finite q-nilpotent groups. p
p
q
p
9.5.2. Let g be a saturated formation and let N be a minimal normal subgroup of the finite soluble group G. Assume that G/N e g G ^ g . T/zen AT /zas a complement in G and all such complements are conjugate. Proof. I f N < F r a t G, then G / F r a t G e g , whence G e g because g is saturated. T h i s is incorrect, so there exists a m a x i m a l subgroup M n o t c o n t a i n i n g N. T h e n G = MAT a n d M n AT o G since AT is abelian. Hence M n AT = 1. F o r the second part, let K a n d K be t w o complements o f N i n G. T h e n K a n d K are m a x i m a l subgroups o f G b y 5.4.2. W r i t e C for the core o f K i n G. T h e n surely CnN =1 a n d , since G g , i t follows t h a t G/C $ g . I f C£K , then G = C K a n d G / C - K / C n X e g because X - G/AT e g . B y this c o n t r a d i c t i o n C < K . Consequently K /C a n d K /C are comple ments o f NC/C i n G/C. M o r e o v e r G/ATC e g a n d G / C £ g , w h i l e AT ATC/C, so t h a t NC/C is m i n i m a l n o r m a l i n G/C. N o w i f C is n o n t r i v i a l , we can a p p l y i n d u c t i o n o n the g r o u p order t o G/C, c o n c l u d i n g t h a t K /C a n d K /C—and hence K a n d K —are conjugate. I f C = 1, then C (N) =1 a n d therefore N = C (N). N o w we m a y a p p l y Exercise 9.1.14 t o o b t a i n the result. • l
l
2
2
x
2
2
2
2
2
2
l
2
G
1
2
x
G
2
Ki
9.5. Formations
279
g-Covering Subgroups I f 5 is a f o r m a t i o n a n d G is a finite g r o u p , let
denote the intersection o f a l l n o r m a l subgroups N such t h a t G/N e g . T h e n G/G e g d this is the largest g - q u o t i e n t o f G. a
n
5
a
n
A subgroup H o f G is called a n ^-covering subgroup i f H e g d if 5 = S ^ / f for a l l subgroups S t h a t c o n t a i n / J : o f course this e q u a t i o n asserts that H covers S/S% a n d hence every g - q u o t i e n t o f S. I t w i l l t u r n o u t that g covering subgroups exist a n d are conjugate whenever G is soluble a n d g is saturated. Before p r o v i n g this we shall r e c o r d t w o simple facts a b o u t g covering subgroups. 9.5.3. Let G be a finite group, H < G and i V < G. Let $ be a
formation.
(i) If H is an ^-covering subgroup of G, then HN/N is an ^-covering group of G/N. (ii) If Hi/N is an ^-covering subgroup of G/N and H is an ^-covering group of H , then H is an ^-covering subgroup of G.
sub sub
1
Proof, (i) L e t HN/N < S/N < G/N a n d p u t RJN then RN/N < RJN. Hence (HN/N)(R /N)
> (HR)N/N
1
= (S/N)%
a n d R = S%;
= S/N,
w h i c h implies t h a t HN/N is a n g - c o v e r i n g subgroup o f G/N. (ii) F i r s t o f a l l observe t h a t H = HN because HJN e g a n d H < H . Assume t h a t H < S < G a n d p u t R = S%. O b v i o u s l y H /N is a n g - c o v e r i n g subgroup o f SN/N a n d SN/RN e g ; therefore SN = H RN = HRN a n d S = S n (HRN) = (HR)(S n N). I n a d d i t i o n SnH =Sn (HN) = H(S n N), so (S nHJR = S. Consequently S n H /R nH ~ S/R e g . Since H is a n g - c o v e r i n g subgroup o f SnH , i t follows t h a t SnH = H(RnHJ and hence t h a t S = (S n HJR = HR, as we w a n t e d t o show. • x
x
l
X
x
x
x
x
x
W e come n o w t o the fundamental theorem o n g - c o v e r i n g subgroups w h i c h yields n u m e r o u s families o f conjugate subgroups i n a finite soluble group. 9.5.4 (Gaschiitz). Let g be a
formation.
(i) If every finite group has an ^-covering subgroup, then g is saturated. (ii) If g is saturated, every finite soluble group G possesses ^-covering sub groups and any two of these are conjugate in G. Proof, (i) L e t G be a finite g r o u p such t h a t G / F r a t G e g . I f H is a n g covering subgroup o f G, then G = H(Frat G), w h i c h implies t h a t G = H b y the nongenerator p r o p e r t y o f F r a t G. T h u s G e g a n d g is saturated.
9. Finite Soluble Groups
280
(ii) T h i s p a r t w i l l be established b y i n d u c t i o n o n | G | . I f G e g , then G is evidently the o n l y g - c o v e r i n g subgroup, so we shall exclude this case. Choose a m i n i m a l n o r m a l subgroup A o f G. T h e n b y i n d u c t i o n G/N has an g - c o v e r i n g subgroup H^/N. Suppose first t h a t G/N g , so t h a t H ^ G. B y i n d u c t i o n i J has an g - c o v e r i n g subgroup H. W e deduce directly f r o m 9.5.3(h) t h a t H is an g covering subgroup o f G. N o w let H a n d / J be t w o g - c o v e r i n g subgroups of G. B y 9.5.3(1) the subgroups H N/N a n d H N/N are g - c o v e r i n g sub groups o f G/N, whence they are conjugate, say H N = H N where g e G. N o w H N ^ G because G / A £ g . Hence i J a n d H , as g - c o v e r i n g sub groups o f H N, are conjugate, w h i c h implies t h a t H a n d / J are conjugate. l
1
x
2
X
2
9
X
2
g
1
x
1
2
1
2
F i n a l l y , assume t h a t G/N e g . T h e n 9.5.2 shows t h a t there is a comple m e n t K o f AT i n G. M o r e o v e r X m u s t be m a x i m a l i n G since N is m i n i m a l n o r m a l . Since G/N e g , we have G% = N a n d G = X G . Therefore, since X is m a x i m a l i n G, i t is an g - c o v e r i n g subgroup o f G. I f H is another such subgroup, then G = HAT, w h i l e H n N = 1 since AT is abelian. A p p l y i n g 9.5.2 we conclude t h a t H a n d K are conjugate. • 5
There is a simple b u t useful a p p l i c a t i o n o f 9.5.4. 9.5.5. Let $ be a saturated formation and G a finite soluble group. If A < ] G then each ^-covering subgroup of G/N has the form HN/N where H is an ^-covering subgroup of G. Proof. L e t K/N T h e n H N/N x
be an g - c o v e r i n g subgroup o f G / A a n d let H
x
be one o f G.
is an g - c o v e r i n g subgroup o f G / A , so / J A / A is conjugate t o x
9
X / A b y 9.5.4. Hence X = (H N) x
= HfN
for some g i n G. Define H t o be
Hf.
•
g-Projectors L e t g be a f o r m a t i o n . A subgroup H o f a finite g r o u p G is called an g projector i f HN/N is m a x i m a l g - s u b g r o u p o f G / A whenever A < ] G. There is a close c o n n e c t i o n between g - c o v e r i n g subgroups a n d g-projectors, as the next t h e o r e m indicates. 9.5.6 (Hawkes). Let $ be a formation (i) Every
^-covering s
(ii) / / g * saturated,
and let G be a finite soluble
subgroup of G in an every ^-projector
group.
^-projector.
of G is an ^-covering
subgroup.
Before c o m m e n c i n g the p r o o f we m u s t establish a n a u x i l i a r y result. 9.5.7 ( C a r t e r - H a w k e s ) . Let $ be a saturated formation and G a finite soluble group. If H is an ^-subgroup such that G = H(Fit G), then H is contained in an ^-covering subgroup of G.
281
9.5. Formations
Proof. L e t us argue b y i n d u c t i o n o n | G | > 1. O b v i o u s l y we m a y assume t h a t G g . L e t AT be a m i n i m a l n o r m a l subgroup o f G. T h e n HN/N inherits the hypotheses o n i f , so b y i n d u c t i o n i t is contained i n some g - c o v e r i n g subgroup o f G/N. T h e latter w i l l b y 9.5.5 be o f the f o r m KN/N where K is an g - c o v e r i n g subgroup o f G. Consequently H < KN. Suppose t h a t KN < G. T h e n b y i n d u c t i o n hypothesis H is contained i n some g - c o v e r i n g subgroup M o f KN. B u t evidently K is an g - c o v e r i n g sub g r o u p o f KN a n d as such is conjugate t o M . T h i s shows M to be an g covering subgroup o f G. W e m a y therefore assume t h a t KN = G. L e t F = F i t G. N o w N < F since G is soluble, a n d indeed N < £ F because l / J V n ( F < G . Therefore K n F < KAT = G. I f K n F ^ l , we can a p p l y the i n d u c t i o n hypothesis t o G/K n F , c o n c l u d i n g t h a t H < T where T / X n F is an g - c o v e r i n g subgroup of G/K n F . T h u s T < X . Also T covers G / ( X n F)N e g , so G = TAT a n d T/TnN ~ K/K n AT e g . T h u s T%< KnN. B u t KnN = 1 since the alter native is AT < X , w h i c h leads t o G = K e g . Therefore T = 1 a n d T e g . I t is n o w clear t h a t T is an g - c o v e r i n g subgroup o f G. 5
5
Consequently we can assume t h a t KnF = 1. Hence F = (KN)nF = N. T h e n G = HN a n d i f is m a x i m a l i n G: for i f ^ G since G £ g . F i n a l l y G = AT since G/N e g . Hence G = / J G ^ a n d / J itself is an g - c o v e r i n g subgroup of G. • 5
Proof of 9.5.6. (i) L e t H be an g - c o v e r i n g subgroup o f G a n d let N o G. I f HN
Carter Subgroups T h e m o s t i m p o r t a n t instance o f the preceding theory is w h e n g is the class of finite n i l p o t e n t groups. T h e n g - c o v e r i n g subgroups a n d g-projectors coincide a n d f o r m a single conjugacy class o f self-normalizing n i l p o t e n t sub groups i n any finite soluble g r o u p . T h e existence o f these subgroups was first established b y R . W . Carter i n 1961.
282
9. Finite Soluble Groups
A Carter
subgroup o f a g r o u p is defined t o be a self-normalizing n i l p o t e n t
subgroup. 9.5.8 (Carter). Let G be a finite
soluble
group.
(i) The Carter subgroups of G are precisely the covering jectors) for the formation of finite nilpotent groups. (ii) Carter subgroups are abnormal in G.
subgroups
(or pro
Proof, (i) L e t 91 be the f o r m a t i o n o f finite n i l p o t e n t groups a n d let H be a n 9l-covering subgroup o f G. Suppose t h a t H < N (H). T h e n there is a sub g r o u p K such t h a t / f < K a n d K/H has p r i m e order. N o w K = HR where R = Kft. H o w e v e r R < H, so K = H, a c o n t r a d i c t i o n w h i c h shows t h a t H = N (H). Since H e 91, i t is a Carter subgroup. Conversely, let H be a Carter subgroup o f G a n d suppose t h a t H < S < G. W r i t e R = a n d assume t h a t / J & < S. T h e n there is a m a x i m a l sub g r o u p M o f S c o n t a i n i n g HR. Since S/K e 91, we have M o S. I n d u c t i o n o n the g r o u p order shows t h a t / J is a n 9 l - c o v e r i n g subgroup o f M . I f x e S, then i f * is also a n 9l-covering subgroup o f M a n d i t is conjugate t o H i n M by 9.5.4. I t follows t h a t S < N (H)M = HM = M , w h i c h is false, i n d i c a t i n g t h a t H is a n 9 l - c o v e r i n g subgroup o f G. G
G
G
(ii) A g a i n let H be a C a r t e r subgroup. I f H < K < G, then i n fact K = N (K) b y the argument o f the first paragraph. T h a t H is a b n o r m a l i n G is n o w a consequence o f 9.2.11. • G
Since a Carter subgroup o f G is a b n o r m a l , i t has a subgroup w h i c h is m i n i m a l w i t h respect t o being s u b a b n o r m a l i n G. Remembering t h a t the m i n i m a l s u b a b n o r m a l subgroups are precisely the system normalizers (9.2.15), we derive the f o l l o w i n g result. 9.5.9. In a finite normalizer.
soluble
group
every
Carter
subgroup
contains
a
system
I n general the system normalizers are p r o p e r l y contained i n the C a r t e r subgroups (Exercise 9.5.8). H o w e v e r i n the case o f finite soluble groups o f small n i l p o t e n t length quite a different s i t u a t i o n prevails. 9.5.10 (Carter). Let G be a finite soluble group of nilpotent length at most 2. Then the system normalizers coincide with the Carter subgroups of G. Proof. B y hypothesis there exists a n o r m a l n i l p o t e n t subgroup M such t h a t G / M is n i l p o t e n t . I f A is a system n o r m a l i z e r o f G, then A covers every central p r i n c i p a l factor b y 9.2.6 a n d we have G = A M . D e n o t e b y p ..., p the distinct p r i m e divisors o f | G | . L e t N a n d M be the unique Sylow p[subgroups o f the n i l p o t e n t groups A a n d M respectively. T h e n Q = M Ni l9
t
k
t
t
t
283
9.5. Formations
is a H a l l p--subgroup o f G since G =NM. Thus { Q Q } is a Sylow system o f G. N o w M o G; hence N normalizes Q a n d N < N (Qi) for a l l i. Since a l l system normalizers have the same order, being conjugate, i t follows t h a t N = f] i=i N (Qi). I f g normalizes AT, i t also normalizes N a n d hence Q . T h u s g e AT a n d AT is self-normalizing, w h i c h means that AT is a Carter subgroup. Since system normalizers a n d Carter subgroups are conju gate, the t h e o r e m follows. • l
f
t
9
k
G
k
G
t
t
Fitting Classes and g-Injectors A n a t u r a l d u a l i z a t i o n o f the t h e o r y o f f o r m a t i o n s was given b y Fischer, Gaschiitz, a n d H a r t l e y ( [ a 4 7 ] ) i n 1967. A class o f finite groups g is called a Fitting class i f i t is closed w i t h respect t o f o r m i n g n o r m a l subgroups a n d n o r m a l p r o d u c t s o f its members. F o r example, finite soluble groups a n d finite n i l p o t e n t groups f o r m F i t t i n g classes w h i l e finite supersoluble groups d o n o t (see Exercise 5.4.6). A s u b g r o u p H o f a finite g r o u p G is called an ^-injector i f H n S is a m a x i m a l g - s u b g r o u p o f S whenever S is a s u b n o r m a l subgroup o f G. T h u s an g - i n j e c t o r is the n a t u r a l d u a l o f an g - p r o j e c t o r . T h e analogue o f 9.5.4 asserts t h a t i f g is any F i t t i n g class, every finite soluble g r o u p contains a unique conjugacy class o f g-injectors. ( N o extra hypothesis o n g c o r r e s p o n d i n g t o s a t u r a t i o n is required.) W h e n g is the class o f finite n i l p o t e n t groups, i t turns o u t t h a t the g-injectors are precisely the m a x i m a l n i l p o t e n t subgroups w h i c h contain the F i t t i n g subgroup. These usually differ f r o m the g-projectors, t h a t is, f r o m the Carter subgroups.
EXERCISES
9.5
1. Give an example of a formation of finite soluble groups that is not subgroup closed. [Hint: Let g be the class of finite soluble groups G such that no Gprincipal factor in G' is central in G.] 2. I f g is a formation and H is an g-covering subgroup of a finite group G, then H is a maximal g-subgroup. 3. Let g be a formation containing all groups of prime order. I f i f is a subgroup which contains an g-covering subgroup of a finite group G, then G = N (H). Deduce that i f G is soluble, then g-covering subgroups are abnormal in G. G
4. Show that every formation which is locally defined by formations g satisfies the hypothesis of Exercise 9.5.3. p
5. Prove that a finite soluble group has a set of conjugate abnormal supersoluble subgroups. Need all abnormal supersoluble subgroups be conjugate? 6. Identify the Carter subgroups of groups S , A , and S . 3
4
4
7. Prove that S has Carter subgroups all of which are conjugate, but that A no Carter subgroups. 5
5
has
284
9. Finite Soluble Groups
8. Give an example of a finite soluble group i n which the Carter subgroups are not system normalizers. 9. Let G be a finite soluble group with abelian Sylow subgroups. Prove that each Carter subgroup contains exactly one system normalizer. 10. I f G is a finite soluble group of nilpotent length < 2, prove that every subabnormal subgroup is abnormal. 11. A group is called imperfect i f it has no nontrivial perfect quotient groups. Prove that finite imperfect groups form a saturated formation.
CHAPTER 10
The Transfer and Its Applications
T h e subject o f this chapter is one o f the basic techniques o f finite g r o u p theory, the transfer h o m o m o r p h i s m . Since the kernel o f this h o m o m o r p h i s m has abelian q u o t i e n t g r o u p , i t is especially useful i n the study o f insoluble groups. I t w i l l be seen t h a t this technique underlies m a n y deep a n d i m p o r t a n t theorems a b o u t finite groups.
10.1. The Transfer Homomorphism L e t G be a g r o u p , possibly infinite, a n d let i f be a subgroup w i t h finite index n i n G. C h o o s i n g a r i g h t transversal { t Ht g t
= Ht
{i)g
l
9
t „ }
t o H i n G, we have
w i t h g e G, where the m a p p i n g i i—• (i)g is a p e r m u t a t i o n o f the
set { 1 , 2 , . . . , n). T h u s t gt^ t
Suppose that 6: H
g
e H for a l l g i n G.
A is a given h o m o m o r p h i s m f r o m H to some abelian
g r o u p A. T h e n the transfer o f 6 is the m a p p i n g 0*:
G-+A
defined b y the rule
1=1
Since ^ is abelian, the order o f the factors i n the p r o d u c t is irrelevant. 10.1.1. The mapping the choice of
6*: G -> A is a homomorphism
which does not depend
on
transversal.
Proof. L e t us first establish independence o f the transversal. L e t {t[,...,
t'„}
be a n o t h e r r i g h t transversal t o H i n G a n d suppose t h a t Ht
and
t
= Ht[
285
10. The Transfer and Its Applications
286
t[ = h t w i t h h i n H. T h e n , i f x e G, i i
t
a n d therefore, since A is c o m m u t a t i v e ,
8
ft W x t ' o i ) = f l ( ^ < o * ) * • f l W x •
(!)
(
1=1
Now
i=l
1= 1
as i runs over the set { 1 , 2 , n } , so does (i)x, f r o m w h i c h i t fol
l o w s that the second factor i n (1) is t r i v i a l . Hence the uniqueness o f 6* is established. N e x t , i f x a n d y are elements o f G, we calculate t h a t
9
(xy) '
=
ft
(t xyt^J t
1=1 x
— EI
e
(h t(i)x) (hi)xyhi)xyf
i=l
i=l
7=1
a n d 0* is a h o m o m o r p h i s m as claimed.
•
Computing #* W e continue the n o t a t i o n used above. T h e value o f 0* at x can often be effectively c o m p u t e d b y m a k i n g an a p p r o p r i a t e choice o f transversal; this choice w i l l n o t affect 0* b y 10.1.1. Consider the p e r m u t a t i o n o f the set o f r i g h t cosets {Ht
...,Ht }
l9
pro
n
duced b y r i g h t m u l t i p l i c a t i o n b y x e G. A t y p i c a l < x > - o r b i t w i l l have the form li 1
(Hs Hs x,... Hs x - ); i9
li
here x
i
9
(2)
i
li
is the first positive p o w e r o f x such t h a t Hs x
= Hs
t
n
YJ=I k = -
j
The elements s x t
9
i = 1,..., k
i9
a n d o f course
j = 1 , . . . , l — 1 form a right
9
{
e
li
transversal t o H. U s i n g this transversal we calculate x *. Since Hs x t
d
the c o n t r i b u t i o n o f the o r b i t (2) t o x * is i
{{s x)(s r t
i
•••
iX
-1
i
i
i
i
i
>
(s ^ )(s,x '- r (s x 'sr )H , tX
t
0
w h i c h reduces t o (SfX^sj ) . Therefore x<"=fl(s '^f. iX
i=l
W e shall give this i m p o r t a n t f o r m u l a the status o f a lemma.
=
Hs
i9
10.1. The Transfer Homomorphism
10.1.2. Let the x-orbits
287
of the set of right cosets of H in G be 1
1
(HSi^HSiX^.^HSiX *- ),
i = 1 , . . . , k. If 9: H -> A is a homomorphism x°- = n
( s
into an abelian group,
i
X
then
' ^ t
Transfer into a Subgroup T h e most i m p o r t a n t case o f the transfer arises w h e n 6 is the n a t u r a l h o m o m o r p h i s m f r o m H t o / J , t h a t is, x = H'x. I n this case 0*: G - > f f is referred t o as the transfer of G into H. T w o cases of special interest are w h e n H is central i n G a n d w h e n H is a Sylow subgroup o f G. W e consider the central case first. e
a b
a b
10.1.3 (Schur). Let H be a subgroup of the center of a group G and suppose that \G:H\ = n is finite. Then the transfer % of G into H is the mapping x i—• x . Hence this mapping is an endomorphism of G. n
1
Proof. C o n t i n u i n g the n o t a t i o n of 10.1.2, we note t h a t s^s^ li
it is a p r o d u c t of elements of H. I t follows t h a t x SjxV
1
li
r
= x . Finally x
li
=
x
e H
since
e H a n d hence t h a t
n
= x.
•
W e pause t o m e n t i o n a c o r o l l a r y o f 10.1.3 w h i c h w i l l be i m p o r t a n t i n the study of finiteness properties o f a g r o u p t h a t refer t o conjugates (Chapter 14). 10.1.4 (Schur). If G is a group whose center has finite and (GJ = 1.
index n, then G' is finite
Proof. L e t C = CG a n d w r i t e G/C = {Cg ,Cg ). F o r any c i n C we have, o n account o f the fundamental c o m m u t a t o r identities, the equality [c g , Cjgf] = [g , gf\, w h i c h implies t h a t G is generated b y the (£) elements i < j - Since G / G n C ~ G'C/C, w h i c h is finite, we deduce f r o m 1.6.11 t h a t G ' n C is a finitely generated abelian g r o u p . F r o m 10.1.3 we k n o w t h a t the m a p p i n g x i—• x is a h o m o m o r p h i s m f r o m G t o C, a n d since C is abelian, G must be contained i n the kernel; therefore (G') = 1. T h a t G n C, a n d hence G , is finite is n o w a consequence o f 4.2.9. • 1
i
i
n
t
t
n
n
Transfer into a Sylow p-Subgroup I f P is a Sylow p-subgroup o f a finite g r o u p G a n d T: G
P
a b
is the transfer,
then G / K e r T is an abelian p-group. W i t h this i n m i n d we define G'(p)
10. The Transfer and Its Applications
288
to be the intersection o f a l l n o r m a l subgroups N such that G/N is an abelian p-group. T h u s G/G'(p) is the largest abelian p - q u o t i e n t o f G. 10.1.5. Let T : G - > P subgroup
P. Then
restriction
a b
be the transfer
of a finite
G'(p) is the kernel
group
G into a Sylow p -
of x and P n G ' is the kernel
of the
of % to P.
Proof. W r i t e K for K e r T. I n the first place G'(p) < K because G/K is a n abelian p-group. Decompose the set o f r i g h t cosets o f P i n t o x - o r b i t s as i n 10.1.2; then i n the n o t a t i o n o f t h a t result k T
X
= F
n
s ^ v
1
.
N o w G = PG'(p), so we m a y choose the s t o lie i n G'(p). O n the basis o f t
T
n
this e q u a t i o n we m a y w r i t e x = P'x c
where n = | G : P | a n d c e G'(p). T h u s
n
x E K implies t h a t x G P'G'(p) = G'(p). I t follows t h a t K/G\p)
is a p ' - g r o u p ,
a conclusion w h i c h can o n l y mean t h a t K = G'(p). f
F i n a l l y P n K e r T = P n G'(p) = P n G since G'(p)/G' is a p ' - g r o u p .
•
I t follows f r o m 10.1.5 t h a t I m T ~ G/G'(p): o b v i o u s l y the latter is iso m o r p h i c w i t h the Sylow p-subgroup o f G , t h a t is, w i t h P G ' / G ' . Hence a b
I m T - P / P n G'.
(3)
Groups with an Abelian Sylow p-Subgroup These ideas m a y be a p p l i e d w i t h p a r t i c u l a r advantage i n the presence o f a n abelian Sylow p-subgroup. 10.1.6. Let the finite denote N (P).
group G have an abelian Sylow p-subgroup
Then P = C (N) x [ P , i V ] . Moreover,
G
P
if T: G
P and let N P is f/ze frans-
fer, I m T = C (AT) and P n K e r r = [ P , AT]. P
T
Proof. A s i n 10.1.2 we can w r i t e x = ] ^ [ f f
x' . Then y and y s7i
C = C (y ) G
conjugate ^r
1
s r l
1
=1
SfX*^" . L e t x e P a n d w r i t e y for
b o t h b e l o n g t o P. Since P is abelian, the subgroup
contains
Sj71
s r l
s
> , a n d b y Sylow's T h e o r e m P a n d P '
= P
c
1
1
where C G C . T h u s r = s^ c~ eN. i
r l
are
Since
r
= y i we c o m p u t e t h a t ?
(4) where n = | G : P\ a n d n
x
T
= f|f T
= 1
[ x ^ , r j G [ P , AT]. I t follows successively t h a t
G P [ P , AT] a n d P = P [ P , AT] because (n, p) = 1. Suppose next
that
10.1. The Transfer Homomorphism
x
T
G K e r i where XGP; z
289
T
then
T
z
n
T
1 = ( x ) = (x )
because (G') = 1. Conse
T
q u e n t l y x = 1 a n d P n K e r T = 1. Since G = PG'(p), a n d thus I m r = P\ we have P = ( I m T) X [ P , AT]. N e x t we c l a i m t h a t I m T < N . F o r i f x e P a n d y e N, then ( x 7 = f l (t xt^y
= fl
t
i=i
where { t
l
9
T
,
Hence I m x o
y
y
y
t ) is also a r i g h t n
T
T h u s ( x ) = (x ) —see also Exercise 10.1.14.
G
the
1
t „ } is a n y r i g h t transversal t o P. B u t {t{,...,
transversal because y e N (P). We
^ ( ^ r
i=i
N.
deduce t h a t [ I m T, N ] < I m T n [ P , AT] = 1 a n d I m i < C (N).
On
P
other
hand,
i f x e C (N),
then
P
(4) shows
that
T
n
x = x,
which
yields x e I m r a n d C ( i V ) < I m T. Hence C (AT) = I m T. F i n a l l y [ P , AT] < P
P n K e r r , a n d also
P
T
| P : P n K e r r | = | P | = | P : [ P , AT] | since
T
P = P x
[ P , AT]. Therefore P n K e r r = [ P , AT]. 10.1.7 (Taunt). Let G be a finite
• subgroups
are
10.1.5 a n d 10.1.6. Since this is true f o r every p r i m e , i t follows
that
abelian.
group
all of whose Sylow
Then G' n CG = 1 and CG is the hypercenter
of G.
Proof. L e t p be a p r i m e a n d P a S y l o w p-subgroup o f G. T h e n ( G n CG) n P < C (AT (P)) n(PnG') P
by
G'n£G=
G
f
1. F i n a l l y [ C G , G]
= l
a n d therefore CG = C G .
2
2
•
T h e f o l l o w i n g useful result is a n easy consequence o f 10.1.6. 10.1.8 (Burnside). If for
some prime p a Sylow p-subgroup
G lies in the center of its normalizer,
then G is
P of a finite
group
p-nilpotent.
Proof. B y hypothesis P is abelian a n d P = C (N (P)).
W e deduce at once
f r o m 10.1.6 t h a t P n K e r T = 1 where o f course T: G
P is the transfer. T h i s
P
G
means t h a t K e r T is a p ' - g r o u p , w h i c h i n t u r n implies t h a t G is p - n i l p o t e n t since G / K e r T ~ I m T, a p-group.
•
W h i l e m u c h m o r e p o w e r f u l criteria f o r p-nilpotence are available, as the f o l l o w i n g sections w i l l show, 10.1.8 provides significant i n f o r m a t i o n a b o u t the orders o f finite simple groups. 10.1.9. Let p be the smallest Assume cyclic.
prime dividing
that G is not p-nilpotent. Moreover
\ G\ is divisible
the order of the finite
Then the Sylow p-subgroups
group G.
of G are not
3
by p or by 12.
Proof. L e t P be a Sylow p-subgroup o f G a n d w r i t e N a n d C for the n o r malizer a n d centralizer o f P respectively. T h e n C ^ N b y 10.1.8. N o w N/C is i s o m o r p h i c w i t h subgroup o f A u t P , b y 1.6.13. I f P is cyclic, then P < C
10. The Transfer and Its Applications
290
a n d N/C has order d i v i d i n g p — 1 b y 1.5.5. H o w e v e r p is the smallest p r i m e divisor o f | G | , so we are forced t o the c o n t r a d i c t i o n N = C. Hence P is n o t cyclic. 3
N e x t suppose t h a t p does n o t divide | G | . T h e n P m u s t be an elementary abelian p - g r o u p o f order p b y 1.6.15 a n d the n o n c y c l i c i t y o f P. Hence A u t P ~ G L ( 2 , p), w h i c h has order ( p — l ) ( p — p). Since C > P, i t follows t h a t | i V : C | divides (p — l ) ( p + 1), w h i c h , i f p were o d d , w o u l d yield a smaller p r i m e divisor o f | G | t h a n p. Hence p = 2 a n d |iV : C\ = 3, so t h a t | G | is divisible b y 12. • 2
2
2
2
I f G is a finite simple g r o u p o f composite order, t h e n 10.1.9 tells us t h a t | G | is divisible by 12 or the cube o f the smallest p r i m e d i v i d i n g the order o f G. H o w e v e r b y the F e i t - T h o m p s o n T h e o r e m this smallest p r i m e is actually 2. So i n fact the order o f G is divisible b y either 8 o r 12. I n a d d i t i o n the Sylow 2-subgroups o f G cannot be cyclic. F o r a m o r e precise result see Exercise 10.3.1.
Finite Groups with Cyclic Sylow Subgroups W e have developed sufficient m a c h i n e r y t o classify a l l finite groups h a v i n g cyclic Sylow subgroups. The definitive result is 10.1.10 ( H o l d e r , Burnside, Zassenhaus). / / G is a finite Sylow subgroups are cyclic, then G has a presentation m
G = {a, b\a
n
l
= 1 = b , b' ab
=
group all of
whose
r
a}
n
where r = 1 ( m o d m), m is odd, 0
( d _ 1 )
( d _ 1 )
{d
{d
l)
l)
10.1. The Transfer Homomorphism
291
I f Q is a S y l o w ^-subgroup o f G, t h e n 10.1.6 implies t h a t either Q < G o r Q n G = 1: for Q, being cyclic, does n o t a d m i t a n o n t r i v i a l direct decompo sition. Hence q cannot divide b o t h ra = \ G\ a n d n = | G : G | ; consequently these integers are coprime. /
L e t G = a n d G / G = < b G > . T h e order o f fc m u s t be expressible i n 1
the f o r m m n
x
1
where ra divides ra. N o w
l
= fr™ has order n a n d fcG gener
t
ates G/G because ( r a n
integer r satisfies r
1?
n) = 1. Therefore G =
= 1 m o d ra a n d
x
= a
l 9
r
= a where the
1 < r < ra. Suppose t h a t there is a
p r i m e g d i v i d i n g ra a n d r — 1. T h e n r = 1 m o d g a n d i f a have \a \ = q a n d a j =
b
whence a
x
m/q
x
= a ,
we s h o u l d
e G' n CG = 1 b y 10.1.7; b u t this
w o u l d m e a n t h a t \a\ = m/q, a c o n t r a d i c t i o n w h i c h shows t h a t (ra, r — 1) = 1. Since (ra, r) = 1, i t follows t h a t ra is o d d . (ii) Conversely assume t h a t G has the given presentation a n d t h a t P is a Sylow p-subgroup. T h e n o
G a n d G is finite o f order ran, while either
P < <#> o r P n = 1 since (ra, n) = 1. I n either case P is cyclic.
•
P r o m i n e n t a m o n g the groups w i t h cyclic S y l o w subgroups are the groups with square-free
EXERCISES
order: such groups are therefore classified by 10.1.10.
10.1
1. Let H < K < G where G is finite. Denote the transfer of G into K by T that T T = T (with a slight abuse of notation). G K
K
H
Gk
. Prove
G H
lo
2. I f G is a group whose center has finite index n, prove that |G'| divides "[ g2«]-n+2 [Hint: Use Exercise 1.3.4.] n
*3. I f G/CG locally finite 7r-group (that is, finitely generated subgroups are finite ngroups), prove that G' is a locally finite 7r-group. 2
4. I f G is a simple group of order p qr where p, q, r are primes, prove that G ~ [Hint: \G\ is divisible by 12.]
A. 5
5. There are no perfect groups of order 180, (so a nonsimple perfect group of order < 200 has order 1 or 120—see Exercise 5.4.4). 6. Let H be a p'-group of automorphisms of a finite abelian p-group A. Prove that A = [A, # ] x C (H). I f H acts trivially on A[p\ prove that H = 1. A
7. Let N be a system normalizer of a finite soluble group G which has abelian Sylow subgroups. Prove that G = NG' and N nG' = 1. 8. (Taunt). Let G be a finite soluble group with abelian Sylow subgroups. I f L
9. Let G(m, n, r) be a finite group with cyclic Sylow subgroups in the notation of 10.1.10. Prove that G(m, n, t) ~ G(m, n, r) if and only if m = m, n = n and
10. The Transfer and Its Applications
292
10. H o w many nonisomorphic groups are there of order 210? 11. Let G be a finite group with cyclic Sylow subgroups. Prove that every subnor mal subgroup of G is normal. 12. (Burnside). Let n be a positive integer. Prove that every group of order n is cyclic if and only if (n, (p(n)) = 1 where cp is the Eulerian function. 13. What conditions on the integer n will ensure that all groups of order n are abelian? 14. Let H be a subgroup of finite index i n a group G. I f T is the transfer of G into H and y e N (H), prove that (x ) = (x f for all x i n G. x
y
y
G
15. Let G be a group, H a subgroup with finite index i n G, and A any abelian group. Then restriction to H yields a homomorphism Res: Hom(G, A) -> H o m ( # , A). The corestriction map Cor: H o m ( # , A) -> Hom(G, A) is defined by 0 i—• 9* where 0* is the transfer of 0. Prove that Cor is a homomorphism and that Res o Cor is multiplication by |G : H\ i n Hom(G, A).
10.2. Griin's Theorems T w o i m p o r t a n t a n d powerful transfer theorems due t o O . G r i i n w i l l be p r o v e d i n this section. These theorems p r o v i d e us w i t h m o r e useful expres sions for the kernel a n d image o f the transfer i n t o a Sylow subgroup. 10.2.1 (Griin's F i r s t Theorem). Let G be a finite group and let P be a Sylow p-subgroup of G. If N = N (P) and T: G -> P is the transfer, then G
a b
P n K e r r = P n G' =
/ V 0 0 / . I n the first place P n K e r T = P nG b y 10.1.5. Define D t o be the subgroup generated by P n AT' a n d a l l P n (P')* w i t h g e G. T h e n certainly D < P o f f a n d D o P. W h a t we m u s t prove is t h a t P r\G' < D. A s s u m i n g this t o be false, let us choose an element u o f least order i n (P n G ' ) \ D . 7
T
W e shall calculate w b y a refinement o f the m e t h o d o f 10.1.2. F i r s t o f a l l decompose G i n t o (P, P)-double cosets P X J P , j = 1 , 2 , s . N o w a double coset PxP is a u n i o n o f cosets of the f o r m Pxy, y e P, a n d the latter are p e r m u t e d transitively b y r i g h t m u l t i p l i c a t i o n by elements o f P. Hence the n u m b e r o f cosets o f the f o r m Pxy, yeP, w i t h Px fixed, divides | P | a n d equals a p o w e r o f p, say p ' . U n d e r r i g h t m u l t i p l i c a t i o n by u the cosets Pxy fall i n t o o r b i t s o f the f o r m (Pxy
h
Pxy u,..., (
prnil
PxyiU ),
i = 2,...,
r,
(5)
p m i
where y e P a n d w is the smallest positive p o w e r o f u such t h a t PxyiU*™ = P x y . These orbits are t o be labeled so t h a t m <m <"' <m . Replacing x b y xy we can suppose y = 1. The elements xy u , i = 1, t
1
f
1
2
r
j
l9
1
t
10.2.
Gain's Theorems
293
2 , . . . , r, j = 0, 1 , . . . , p ' — 1 f o r m p a r t o f a r i g h t transversal t o P w h i c h m a y be used t o c o m p u t e u\ m x
Let us calculate the c o n t r i b u t i o n o f the w-orbit (5) t o u\ where 1
ptni
v = xy^y^x'
1
= (u [u^\
t
T h i s is
P'v
t
1
yr ])*" ,
(6)
prni xl
an element o f P. T a k i n g i = 1 we deduce that (u ) G P since y = 1. N o w m < m b y o u r o r d e r i n g , so (u ) G P for a l l i a n d i t follows f r o m (6) that c = [w , y f ] * e P. I n a d d i t i o n c G ( P ' ) , w h i c h means t h a t c e D. C o n sequently v = (u ) m o d D . T h e t o t a l c o n t r i b u t i o n t o u o f the double coset PxP is therefore P ' w ( x ) where x
pmi x
x
1
f
p m f
1
- 1
x - 1
pTni xX
x
t
w(x) = t \ v
pt X
i
= [u f
mod D
(7)
i=i mi
since ^ . p = p\ N o w suppose t h a t t > 0. Since w ( x ) G P, we have ( w ) G P n K e r T by (7), a n d b y m i n i m a l i t y o f |w| we o b t a i n ( w ) e D. F o r the same reason u e D. T h u s we certainly have w ( x ) = 1 = u m o d D . =
1
p t
p t
pt
x
1
x _ 1
pt
If, o n the other hand, t = 0, then PxP = Px, w h i c h is equivalent to x G AT. Hence P x P contributes P'xux' = P'u[u x ] t o w : o f course [w, x ] G P n N < D. A g a i n we have w ( x ) = w m o d D . 1
_
1
T
_
1
9
f
p t
T h u s f l 5 = i ( j) — & D where / = Yfj=i P * P is the n u m b e r o f cosets Pxyy i n Px -P. T h e n / = \ G:P\, the t o t a l n u m b e r o f r i g h t cosets o f P i n G. Since ue PnG', we have u = P < D. Hence G D , w h i c h yields u G D since p does n o t divide /. • w
x
u l
m
o
tj
anc
tj
7
x
f
The next result illustrates the usefulness o f G r u n ' s theorem. 10.2.2 (Wong). Suppose that G is a finite group which has a Sylow 2-subgroup P with a presentation
2
b
1 + 2 n
x
Proof. L e t N denote N (P). T h e Schur-Zassenhaus T h e o r e m implies t h a t there is a s u b g r o u p Q such t h a t N = QP a n d QnP = 1; o f course Q has o d d order. H o w e v e r Q/C (P) is i s o m o r p h i c w i t h a subgroup o f A u t P a n d by Exercises 5.3.5 the latter is a 2-group. T h u s we are forced t o conclude t h a t Q = C (P) a n d N = Q x P. A p p l y i n g 10.2.1 we have G
Q
Q
PnG' f
= (Pn
N\ P n (PJ\g
G G>.
2 n _ 1
N o w PnN' = Pn(Q x P') = < a > , a g r o u p o f order 2, w h i c h allows us t o conclude t h a t PnG' is generated b y elements o f order 2. Since a l l the elements o f P w i t h order 2 b e l o n g t o P = < a , fc>, i t follows t h a t PnG < P . 2 n _ 1
0
0
10. The Transfer and Its Applications
294
2
Let L / G ' be the o d d c o m p o n e n t o f G a n d set S = < a , b}L, evidently a n o r m a l s u b g r o u p o f G. N o w P nS = < a , b} (P n L ) a n d P n L = PnG' < P , f r o m w h i c h i t follows t h a t PnS = < a , b} = P say. P is a Sylow 2-subgroup o f S since S o G. F r o m the presentation one sees t h a t P is abelian: indeed i t is the direct p r o d u c t o f a cyclic g r o u p o f order 2 " > 2 a n d a g r o u p o f order 2. Conse q u e n t l y A u t P is a 2-group b y Exercise 1.5.13. T h e a r g u m e n t o f the first p a r a g r a p h n o w shows t h a t P is a direct factor o f A s ( P i ) a n d we deduce f r o m 10.1.8 t h a t S is 2-nilpotent. Since \ G:S\<2, the same holds for G. • a b
2
2
0
1 ?
x
x
_ 1
x
1
There has been m u c h w o r k o n the classification o f finite simple groups of composite order a c c o r d i n g t o the nature o f their Sylow 2-subgroups. I t was r e m a r k e d above t h a t the Sylow 2-subgroups o f such a g r o u p cannot be cyclic; o f course 10.2.2 excludes a further set o f 2-groups. F o r example, we m e n t i o n t h a t Brauer a n d S u z u k i have s h o w n t h a t a (generalized) q u a t e r n i o n g r o u p Q n cannot be the Sylow 2-subgroup o f a finite simple g r o u p . I n another i n v e s t i g a t i o n G o r e n s t e i n a n d W a l t e r have p r o v e d t h a t P S L ( 2 , p ) , p ^ 2, a n d A are the o n l y finite simple groups t o have a d i h e d r a l Sylow 2-subgroup. T h e proofs o f these results are difficult and cannot be discussed here. F o r m o r e i n f o r m a t i o n o n these questions consult [ b 2 6 ] . 2
m
1
Weak Closure and p-Normality I f H a n d K are subgroups o f a g r o u p , t h e n H is said t o be weakly closed i n K i f H < K a n d i f H < K always implies t h a t H = H . Otherwise stated H is w e a k l y closed i n K i f K contains H b u t n o other conjugate o f H. 9
9
I f P is a Sylow p-subgroup o f G a n d i f the center o f P is w e a k l y closed i n P, then G is said t o be p-normal. A m o n g p - n o r m a l groups are groups w i t h abelian Sylow p-subgroups a n d groups i n w h i c h distinct Sylow p-subgroups have t r i v i a l intersection. 10.2.3 ( G r i i n ' s Second Theorem). Let the finite let P be a Sylow p-subgroup of G. If L = N (£P), G/G'(p) ~ L/L'(p). G
group G be p-normal and then P nG' = P nL' and
Proof. I n the first place G/G'(p) - P/PnG' b y 10.1.5 a n d (3). I n a d d i t i o n , since P < L , the s u b g r o u p P is a Sylow p-subgroup o f L a n d L/L'(p) ~ P / P n L ' . T h e t h e o r e m w i l l therefore f o l l o w s h o u l d we succeed i n p r o v i n g t h a t P nG' = P nL'. W h a t is m o r e , b y G r i i n ' s F i r s t T h e o r e m i t suffices to show t h a t P n N' < P n L a n d P n (P') < P n L for a l l g i n G, where N = N (P). Since CP is characteristic i n P, i t is certainly true t h a t N < L and P n AT < P nL'. T h u s we can concentrate o n / = P n ( P ' ) . g
G
9
10.3. Frobenius's Criterion for p-Nilpotence
295
9
9
L e t P = C P a n d M = N (I). T h e n P < M a n d P <M since P = £(P*). L e t P a n d P be Sylow p-subgroups o f M c o n t a i n i n g P a n d P
G
X
0
0
0
2
0
x
1
2
X
x
0
1
1
0
h
x
0
h
1
h
0
P
n
P
L
EXERCISES
s
i
n
c
e
p
L
10.2
1. Show that 10.2.2 does not hold if n = 2. 2. Let P be a finite 2-group such that Aut P is a 2-group and P cannot be generated by elements of order dividing |P'|. Prove that there is no finite simple group whose Sylow 2-subgroups are isomorphic with P. Give some examples. [Hint: Apply Griin's First Theorem.] 3. Prove that a p-nilpotent group is p-normal. 4. Let G be a finite p-normal group. I f P is a Sylow p-subgroup and L = N (£P), prove that LnG'(p) = L'(p). G
10.3. Frobenius's Criterion for p-Nilpotence Here we derive a useful c r i t e r i o n for p-nilpotence, describing i n the sequel some o f its m a n y applications. T h e f o l l o w i n g technical l e m m a is used i n the proof. 10.3.1 (Burnside). Let P and P be Sylow p-subgroups of a finite group G. Suppose that H is a subgroup of P n P which is normal in P but not in P . Then there exists a p-subgroup M , a prime q ^ p and a q-element g such that H<Mandge N (M)\C (M). x
2
x
G
2
1
2
G
Proof. W r i t e K = Np (H) a n d assume that P has been chosen so that K is m a x i m a l subject t o P ^ N (H). Since K < N (H) a n d P is a Sylow psubgroup o f N (H\ there is an x i n N (H) such that K < P b y Sylow's T h e o r e m . N o w H o P implies that i f o P f , w h i c h indicates t h a t we m a y replace P b y P w i t h o u t d i s t u r b i n g the hypotheses. Hence we can assume that K < P so that K < P n P < K a n d K = P n P . Since P P , it follows that K < P a n d K < P . 2
2
2
G
G
x
x
G
G
l
x
1
u
x
x
2
1
2
1
2
2
A p p l y the n o r m a l i z e r c o n d i t i o n t o P i = 1, 2: then K < N < P where N = N .(K). Hence K < L = < i V N }. N o t i c e that L cannot n o r m a l i z e H because, i f i t d i d , N w o u l d be contained i n N (H) = K. h
t
P
1?
t
t
2
2
Pi
N e x t N is contained i n a Sylow p-subgroup o f L , w h i c h i n t u r n is con tained i n some Sylow p-subgroup P o f G. T h e n K < N < Np (H); this, i n x
3
x
3
10. The Transfer and Its Applications
296
view o f the m a x i m a l i t y o f K , implies t h a t P
3
normalizes H. I t follows t h a t
there is a S y l o w p-subgroup o f L w h i c h normalizes H. C o m b i n i n g the results o f the last t w o paragraphs we conclude t h a t there exist a p r i m e q ^ p a n d a S y l o w ^-subgroup Q o f L such t h a t Q does n o t n o r m a l i z e H. L e t g e Q\N (H)
and put M = if
G
< f l f >
. Since H o
<2 < L , we see t h a t M is a p - g r o u p . O b v i o u s l y g e N (M),
L and
but g $
G
C (M) G
since g $ C (H).
•
G
10.3.2 (Frobenius). A finite subgroup Proof.
is centralized
group
G is p-nilpotent
by the p'-elements
in its
if and only if every
p-
normalizer.
Assume t h a t G is p - n i l p o t e n t a n d t h a t P is a p-subgroup. T h e n
all the p'-elements b e l o n g t o O ( G ) , a n d we have [_O (G)nN (P), p
P n O (G)
p
P] <
G
= 1, w h i c h establishes the necessity o f o u r c o n d i t i o n .
p
Conversely assume t h a t the c o n d i t i o n is satisfied i n G a n d let P be a Sylow p-subgroup. T h e p-nilpotence o f G w i l l be established b y i n d u c t i o n o n | G | ; o f course we m a y suppose t h a t | G | ^ 1 a n d | P | ^ 1. W r i t e C = CP a n d L = N (C). G
We
Then 1 / C < L
verify easily t h a t L/C
a n d P / C is a S y l o w p-subgroup o f L / C .
inherits the c o n d i t i o n i m p o s e d o n G; therefore
L / C has a n o r m a l H a l l p ' - s u b g r o u p Q/C. B y the Schur-Zassenhaus T h e o rem there is a c o m p l e m e n t M o f C i n Q. B u t since M normalizes C a n d is a p ' - g r o u p , i t centralizes C; thus Q = M x C. E v i d e n t l y | L : M | is a p o w e r o f p, a n d i n a d d i t i o n M is characteristic i n Q a n d hence n o r m a l i n L . F r o m these properties there follows the e q u a l i t y L = P M . Consequently P n L ' < P n ( P ' M ) = P ' a n d P n L ' = P'. By 10.3.1 a n d the hypothesis a n o r m a l s u b g r o u p o f a Sylow p-subgroup of G is n o r m a l i n every S y l o w p-subgroup t h a t contains i t . I t follows t h a t every Sylow s u b g r o u p c o n t a i n i n g C is itself c o n t a i n e d i n L . B u t L = P M a n d b o t h P a n d M centralize C, so t h a t C < CL. C o m b i n i n g this w i t h the previous sentence we conclude t h a t C lies i n the centre o f every S y l o w psubgroup w h i c h contains i t , a p r o p e r t y t h a t is manifestly equivalent to weak closure o f C i n P, t h a t is, t o p - n o r m a l i t y o f G. W e are n o w i n a p o s i t i o n t o a p p l y G r u n ' s Second T h e o r e m , c o n c l u d i n g t h a t PnG'
= P n L . B u t we have seen t h a t P n L = P', so i n fact P' =
P n G , w h i c h is a S y l o w p-subgroup o f G a n d hence o f G'(p). I f G'(p) were equal t o G, i t w o u l d f o l l o w t h a t P = P' a n d P = 1, c o n t r a r y to assumption. T h u s G'(p) is a p r o p e r s u b g r o u p a n d b y i n d u c t i o n o n | G | i t is p - n i l p o t e n t , w h i c h implies at once the p-nilpotence o f G.
•
Applications of Frobenius's Theorem 10.3.3 (Ito). Let G be a finite mal subgroups
are p-nilpotent.
group which is not p-nilpotent
that \ G : P | is a power of a prime q ^ p. Moreover G is
nilpotent.
but whose
Then G has a normal Sylow p-subgroup every maximal
maxi P such
subgroup
of
10.3. Frobenius's Criterion for p-Nilpotence
297
Proof. Since G is n o t p - n i l p o t e n t , 10.3.2 shows that there exist a p-subgroup P, a p r i m e q ^ p a n d an element # o f order q such that # normalizes b u t does n o t centralize P. N o w P > cannot be p - n i l p o t e n t , b y 10.3.2 again. Consequently G = P> a n d P < G. Hence | G : P | = |#| = q a n d P is a Sylow p-subgroup. L e t H be any m a x i m a l subgroup o f G. Since H is p-nilpotent, H/O ,(H) is nilpotent. I n addition n P is n i l p o t e n t while P n O (H) = 1. I t follows that H is n i l p o t e n t . • m
m
p
p
T h u s i t emerges that a finite m i n i m a l n o n - p - n i l p o t e n t g r o u p is a m i n i m a l n o n n i l p o t e n t g r o u p . F u r t h e r s t r u c t u r a l properties can therefore be deduced f r o m Exercise 9.1.11. 10.3.3 has, i n t u r n , a p p l i c a t i o n to groups whose p r o p e r subgroups are supersoluble. 10.3.4 ( H u p p e r t ) . / / every maximal soluble, then G is soluble.
subgroup
of a finite
group G is super-
Proof. Assume t h a t G is insoluble a n d let p be the smallest p r i m e d i v i d i n g | G | . I f G is p - n i l p o t e n t , then O (G) ^ G, so that O (G) is supersoluble. Since G/O (G) is a p-group, the solubility of G follows. Hence G is n o t p-nilpotent. O n the other h a n d , a m a x i m a l s u b g r o u p o f G is supersoluble a n d hence p - n i l p o t e n t b y 5.4.8. H o w e v e r 10.3.3 shows that G must be soluble. • p
p
p
A g o o d deal o f s t r u c t u r a l i n f o r m a t i o n a b o u t finite m i n i m a l nonsupersoluble groups is available: for details see [ b 6 ] (and also Exercise 10.3.10). O n the basis o f 10.3.4 an interesting characterization o f finite supersoluble groups p e r t a i n i n g t o m a x i m a l chains m a y be established. Here b y a maximal chain i n a g r o u p G we mean a c h a i n o f subgroups 1 = M < M < - - < M = G such that M is m a x i m a l i n M for each i. 0
m
10.3.5 (Iwasawa). A finite chains in G have the same
{
x
i+1
group G is supersoluble length.
if and only if all
maximal
Proof. F i r s t let G be supersoluble. Since a m a x i m a l subgroup o f a supersoluble g r o u p has p r i m e index, the l e n g t h o f any m a x i m a l chain i n G equals the n u m b e r d o f p r i m e divisors o f | G | , including multiplicities. Conversely, assume that G has the chain p r o p e r t y b u t is n o t supersoluble. L e t G be chosen o f least order a m o n g such groups. Since subgroups i n h e r i t the c h a i n p r o p e r t y , G is a m i n i m a l nonsupersoluble g r o u p , so b y 10.3.4 i t is soluble. Therefore a c o m p o s i t i o n series o f G is a m a x i m a l c h a i n a n d the length o f each m a x i m a l chain equals the c o m p o s i t i o n length, w h i c h is j u s t d, the n u m b e r o f p r i m e divisors o f | G|. B y 9.4.4 there exists a m a x i m a l s u b g r o u p M whose index is composite. O n refinement o f the c h a i n 1 < M < G there results a m a x i m a l chain whose length is less t h a n d, a c o n t r a d i c t i o n . •
10. The Transfer and Its Applications
298
EXERCISES
10.3
1. I f G is a finite simple group of even order, then |G| is divisible by 12, 16, or 56. 2. I f G is a finite simple group whose order is divisible by 16, prove that |G| is divisible by 32, 48, 80, or 112. 3. A subgroup H is said to be pronormal i n a group G i f for all g i n G the sub groups H and H are conjugate in (H, H }. I f H is pronormal and subnormal i n G, show that H < G . 9
9
4. Let G be a finite group and let p be the smallest prime dividing |G|. I f every p-subgroup is pronormal i n G, then G is p-nilpotent. [Hint'. Use Frobenius's criterion and Exercise 10.3.3.] 5. (J.S. Rose). Let G be a finite group and p a prime. Prove that every p-subgroup is pronormal i n G i f and only i f each p-subgroup is normal in the normalizer of any Sylow p-subgroup that contains it. [Hint: Use and prove the result: i f P and H <3 P where P is a Sylow p-subgroup of the finite group G, then H and H are conjugate i n iV (P).] 9
9
G
6. Show that Exercise 10.3.4 does not hold for arbitrary primes p. 7. Let P be a Sylow p-subgroup of a finite group G. I f N (H) = P whenever H is a nontrivial abelian subgroup of P, prove that G is p-nilpotent. [Hint: Prove that PnP =1 if # e G \ P ] G
9
8. Let P be a Sylow p-subgroup of a finite group G. Prove that G is p-nilpotent i f and only if N (H) is p-nilpotent whenever 1 ^ H < P. G
9. I f G is a finite group in which every minimal p-subgroup is contained i n the centre of G, then G is p-nilpotent provided p > 2. 10. (K. Doerk). Let G be a finite minimal nonsupersoluble group. Prove that G is p-nilpotent for some p dividing |G|.
10.4. Thompson's Criterion for p-Nilpotence T h e subject o f this section is a very powerful c o n d i t i o n for p-nilpotence due t o T h o m p s o n w h i c h has i m p o r t a n t a p p l i c a t i o n t o F r o b e n i u s groups. A m a j o r role is played b y a rather curious subgroup t h a t can be f o r m e d i n any finite p - g r o u p P. W e define j(p) to be the subgroup generated b y a l l abelian subgroups o f P w i t h m a x i m a l r a n k . O b v i o u s l y J ( P ) is characteristic i n P. 10.4.1 ( T h o m p s o n ) . Let G be a finite p-subgroup of G. Then G is p-nilpotent p-nilpotent.
group, p an odd prime and P a Sylow if and only if N (J(P)) and C ( £ P ) are G
G
10.4. Thompson's Criterion for p-Nilpotence
299
Proof, (i) O f course i t is o n l y the sufficiency o f the c o n d i t i o n s that is i n question. W e assume henceforth that N (J(P)) a n d C (CP) are p - n i l p o t e n t b u t G itself is n o t p-nilpotent. F u r t h e r m o r e let G be a g r o u p o f smallest order w i t h these properties. Observe that any proper subgroup c o n t a i n i n g P inherits the c o n d i t i o n s o n G a n d is therefore p-nilpotent. G
G
T h e p r o o f is b r o k e n u p i n t o a series o f reductions, each one o f itself being fairly s t r a i g h t f o r w a r d . (ii) O .(G) = 1. Suppose that o n the c o n t r a r y T = O (G) ^ 1 a n d write G = G/T a n d P = PT/T O u r immediate object is t o show that the conditions o n G apply t o G. C e r t a i n l y P is a Sylow p-subgroup o f G i s o m o r p h i c w i t h P via the n a t u r a l h o m o m o r p h i s m X K X T . F r o m this we deduce that J(P) = J(P)T/T I f xTeN (J(P)) then J(P) T = J ( P ) T , f r o m w h i c h i t follows by Sylow's T h e o r e m t h a t _ J ( P ) = J(P) for some y i n T ; thus x e N (J(P))T Conse quently N (J(P)) is contained i n N (J(P))T/T w h i c h shows the former to be p-nilpotent. p
p
X
G
9
x
y
G
G
G
9
N o w we must examine C (CP). Clearly £P = (£P)T/T so that i f xT e C (CP), then (CP) T = (CP)T; j u s t as i n the previous paragraph, x e N (£P)T. Clearly we can assume that x e N (CP). T h e n [CP, x ] < T n CP = 1 a n d x e C (CP). Consequently C (CP) = C (CP)T/T is p - n i l p o t e n t F i n a l l y the m i n i m a l i t y o f G allows us t o conclude that G is p-nilpotent, w h i c h implies at once that G is p-nilpotent. (iii) W e i n t r o d u c e the set £f o f a l l n o n t r i v i a l p-subgroups whose n o r m a l izer i n G is n o t p-nilpotent. £f is n o t empty, otherwise the n o r m a l i z e r o f every n o n t r i v i a l p-subgroup w o u l d be p-nilpotent, w h i c h i n view o f the Frobenius c r i t e r i o n (10.3.2) w o u l d i m p l y the p-nilpotence o f G. G
9
X
G
G
G
G
G
G
T h e set is p a r t i a l l y ordered by means of a r e l a t i o n ^ defined i n the f o l l o w i n g manner. I f H H e ¥ then H < H means that either l9
2
(i)
9
1
2
IN^H^KlN^H^
or (ii)
\N (H )\ G
1
= \N (H )\
p
G
2
and
p
\H \ < X
\H \. 2
Here n denotes the largest p o w e r o f p d i v i d i n g n. Choose an element H o f £f w h i c h is m a x i m a l w i t h respect t o this p a r t i a l ordering and write N = N (H). p
G
I f P is a Sylow p-subgroup o f N then H < P since H o N. Replacing P i f necessary by a suitable conjugate, we m a y suppose that P < P a n d thus H < P. Also P < P n N so that P = P nN since P is a Sylow p-subgroup of N. 0
9
0
0
0
9
0
0
(iv) i f = O (G) and G = G/H is p-nilpotent. p
W e shall first establish the fact that N satisfies the conditions o n G. Since H < P, we have [CP, FT] = 1 a n d CP < N , w h i c h implies that CP < P n C ( P ) = CP because P = P n AT. I n consequence C^CPo) < Q ( C P ) , w h i c h demonstrates that (^(CPo) is p-nilpotent. 0
G
0
0
0
10. The Transfer and Its Applications
300
L e t us n o w examine N (J(P )). I f P were equal t o P, this n o r m a l i z e r w o u l d certainly be p - n i l p o t e n t . A s s u m i n g t h a t P < P, we i n v o k e the n o r malizer c o n d i t i o n t o produce a subgroup P n o r m a l i z i n g P such that P < P i < P. Since J(P ) is characteristic i n P , i t is n o r m a l i n P a n d thus P < N (J(P )\ w h i c h shows that \N (J(P ))\ > | P | = \N\ . B y m a x i m a l i t y of H i n £f the g r o u p N (J(P )) must be p - n i l p o t e n t , whence so is N (J(P )). N
0
0
0
l
0
1
G
0
0
0
G
G
0
l
P
0
P
0
N
0
I f N ^ G, then AT is p - n i l p o t e n t b y m i n i m a l i t y o f G. B u t this contradicts the fact that H e £f. Hence N = G a n d H o G, w h i c h leads at once t o H < O (G). Suppose t h a t P /H is a n o n t r i v i a l n o r m a l subgroup o f P/H. T h e n P < AT (P ), f r o m w h i c h i t follows t h a t \N (P )\ = \P\ = \N\ : also | i f | < | P | , o f course. Once again m a x i m a l i t y o f H i n £f enters the argu ment, forcing N (P ) t o be p - n i l p o t e n t . T h i s makes i t clear t h a t P cannot be n o r m a l i n G; since H < O (G) < P, i t follows that H = O (G). p
2
G
2
G
2
p
P
2
G
2
2
p
p
Observe that P = P/H ^ 1 since P cannot be n o r m a l i n G. O n a p p l y i n g the preceding discussion t o P /H = J(P) we o b t a i n the p-nilpotence o f N (J(F)). I n a similar manner, t a k i n g P /H t o be C(P) we find that N (CP) is p - n i l p o t e n t , a n d since C ( £ P ) < N (£P)/H, the p-nilpotence o f C (£P) follows. F i n a l l y | G | < | G | , so G is p - n i l p o t e n t b y m i n i m a l i t y o f | G | . 2
G
2
G
G
G
G
I n the sequel we shall w r i t e K =
O .(G) pp
9
so that G/K is a p-group. Also a "bar" will always denote a quotient group modulo H. (v) P is maximal in G and C (H) < H. Since C ( £ P ) is p - n i l p o t e n t , i t is p r o p e r a n d lies inside a m a x i m a l sub g r o u p M . T h e n P < C ( £ P ) < M , so t h a t M is p - n i l p o t e n t b y (i). W e have to prove that P = M , or equivalently that U = O (M) = 1. N o w [ / < M a n d also H o G a n d H < P < M , so we have [ ( 7 , # ] < U n # = 1 a n d U < C (H). I n a d d i t i o n U < K because G/K is a p-group, a n d therefore U < C (H). G
G
G
p
G
K
N o w consider C (H). Clearly H n C ( / f ) = f r o m w h i c h i t follows that C (H)/£H is a p ' - g r o u p a n d £H has a complement X i n C (H). How ever is o b v i o u s l y contained i n the centre o f C (H\ so t h a t i n fact C (H) = £H x X. Here X = O ,(C (H))^i G because C ( # ) o G. H o w e v e r O , ( G ) = 1 b y (ii); thus X = 1 a n d C ( # ) = C#, a p-group. I t therefore fol lows that U = 1 as required. K
K
K
K
K
K
p
p
K
K
K
F i n a l l y | C ( / f ) : C (H)\ is a p o w e r o f p because | G : K\ is. Since C ( / f ) = £H b y the last paragraph, C (H) is a p-group. Hence C (H) < O (G) = H. (vi) K is a nontrivial elementary abelian q-group q ^ p. I n the first place K cannot equal H otherwise G w o u l d be a p - g r o u p a n d a fortiori p-nilpotent. Choose a p r i m e q d i v i d i n g \K:H\; then certainly q ^ p. I f Q is a Sylow ^-subgroup o f K , the F r a t t i n i argument applies t o give G = N (Q)K f r o m w h i c h i t follows that \G:N (Q)H\ divides | X : i f | a n d is p r i m e to p; thus N (Q)H contains a Sylow p-subgroup o f G, let us say P . T h e n P < N (Q )H and, since Q m a y be replaced b y Q \ we can i n fact assume t h a t P < N (Q)H. G
K
K
G
G
p
9
9
G
9
G
G
9
9
l
9
G
G
10.4. Thompson's Criterion for p-Nilpotence
301
Let <2 denote the s u b g r o u p generated b y a l l the elements o f order q i n 0
the center o f Q. Since Q
0
implies t h a t Q H^ since we agreed G = QP n
=
(QoP)
I t follows t h a t (Q H)P
= QP
0
t h a t P < N (Q)H
0
i n the last p a r a g r a p h .
G
is a s u b g r o u p Furthermore
w h i c h we deduce that Q =
2o> thus s h o w i n g Q t o be elementary abelian.
F i n a l l y K = Kn
(QP) = Q(K n P) = QH since K = O (G); pp
K = QH (vii) C (K) G
which
G
G
since P is m a x i m a l i n G, f r o m
0
Q
is characteristic i n Q, i t is n o r m a l i n N (Q\
N (Q)H.
0
and
therefore
K - Q.
(8)
= K.
O f course C (K)
> K b y (vi). N o w K has a n o r m a l c o m p l e m e n t i n
G
a fact that shows C (K)
to be the direct product o f K and PJH
G
I t follows t h a t ? ! < G a n d since P
x
=
C (K) G
O (C (K)). p
G
is a p - g r o u p , P = H, w h i c h yields the x
desired equality. (viii) A t this p o i n t one must observe t h a t J(P) cannot be contained i n H; for i f i t were, since H < P, we s h o u l d have J(P) = J ( / f ) o G, thus forcing G = N (J(P)) G
t o be p - n i l p o t e n t . Consequently there is a n abelian s u b g r o u p
A o f P w i t h m a x i m a l r a n k w h i c h is n o t c o n t a i n e d i n H. T h e n [ K , A~\ ^
H;
for otherwise ^ < K b y (vii), w h i c h w o u l d i m p l y that A < P n K = H. Since K = QH, i t follows t h a t [ Q , A~\ £ H.
(ix) P = AH a n d Q = K is minimal normal in QA. N o t i c e t h a t 2 = K < G a n d t h a t Q is a n abelian Sylow g-subgroup o f QA
i n view o f (vi). A p p l y i n g 10.1.6 t o the g r o u p QA we o b t a i n a direct
decomposition Q = CQ(QA)
x [Q, Q l ]
= C (I) 5
x [Q, I ] .
W e have observed i n (viii) t h a t [ Q , A~] ^ / J ; thus [ Q , A~] choose Q
x
such t h a t Q
x
< Q and
(9)
I a n d we can
is a m i n i m a l n o r m a l subgroup o f QA
c o n t a i n e d i n [ Q , A ] . I t is necessary t o p r o v e that Q = Q. X
T o accomplish this we consider the s u b g r o u p R = < H , A, Q i > = QiAH
=
Q AH X
302
10. The Transfer and Its Applications
w i t h a view t o s h o w i n g t h a t i t satisfies the c o n d i t i o n s o n G. W e recognize AH as a Sylow p-subgroup o f R: for i t is surely a p - g r o u p while AH < R < AK a n d \AK : AH\ is p r i m e t o p. W e shall prove first o f a l l that N (AH) = AH. I f this is false, the g r o u p AH is n o r m a l i z e d b y some element u o f Qx\H; this element m u s t satisfy \_A u] < (AH) n K since u e Q < K. Therefore [A, u] < H a n d uH e CQ(A). T h i s means that uH belongs t o b o t h CQ(A) a n d [ Q , A~] since Q < [ Q , A~] w h i c h contradicts (9). N e x t let x e N (J(AH)). O b v i o u s l y A < J(AH) so we have A < J(AH) < AH a n d x e N (AH) = AH. Consequently N (J(AH)) = AH, a p - g r o u p a n d hence certainly p - n i l p o t e n t . R
9
x
x
R
9
9
x
R
R
N o w consider C ^(AH)\ Since H < P, we have CP < C (H) < H b y (v). Also A < P, w h i c h implies that CP < t{AH) a n d C ^(AH)) < C ^P). Since the latter is p - n i l p o t e n t , so is C (£(AH)). I f R ^ G, the m i n i m a l i t y o f | G | leads to the p-nilpotence o f R; thus R/O (R) is a p-group. B u t O (R) centralizes H b y 10.3.2 a n d C (H) < H b y (v); therefore P is a p - g r o u p a n d Q = 1, w h i c h is c o n t r a r y to the choice R
G
R
R
R
p
p
R
x
o
f
2iHence G = R = Q AH w h i c h implies that Q < Q A. T h u s Q = Q as claimed. F i n a l l y P = : for AH has been seen to be a Sylow p-subgroup of R = G. (x) ^ is cyclic. W e see f r o m (ix) t h a t K is a simple F ^ - m o d u l e where F = GF(q). Also A acts faithfully o n K b y (vii). A p p l y i n g 9.4.3. we conclude that A is cyclic. (xi) W r i t e D = CH a n d consider the g r o u p QD. O b v i o u s l y D is an abelian Sylow p-subgroup o f QD a n d D o QD. T h u s 10.1.6 yields the d e c o m p o s i t i o n X
9
X
l9
q
D = C (QD) D
x [D, g D ] = C ( 0 x [D, Q].
(10)
D
Suppose that [ D , Q ] = 1; then Q < C ( D ) < C (CP) since CP < C ( # ) = D b y (v). N o w C (CP) = P since P is m a x i m a l a n d C (CP) cannot equal G. There results the c o n t r a d i c t i o n Q = 1, w h i c h shows t h a t [ D , Q ] ^ 1. G
G
G
G
G
N o w define F as the subgroup generated b y a l l elements o f order p i n [ D , g ] . N o t i c e that [ D , Q ] = [ D , Q # ] = [ D , K ] o G, so t h a t F < G. O f course V is an elementary abelian p - g r o u p . 2
(xii) | F | < p . L e t A = A n H. Since A/A is cyclic b y (x), we have r(A) — r(A ) < 1. Also ^ F is abelian because F < £H. I n view o f A V < P a n d the m a x i m a l i t y o f A we have r(A V)
0
0
0
0
0
0
0
0
0
9
0
0
x
0
x
G
x
G
X
x
9
D
(xiii) The final
step.
10.4. Thompson's Criterion for p-Nilpotence
303
W e begin b y s h o w i n g that C (V) = H. O b v i o u s l y H < C ( F ) o G, a n d i f H ^ C (V\ then C (V) is n o t a p-group. Hence Q n C (V) 1, w h i c h implies that H < C (V) < K. N o w C ( F ) o G since F < G, a n d yet K is a m i n i m a l n o r m a l s u b g r o u p o f QA b y (ix). Hence K = C (V). H o w e v e r this yields V < C (Q) n [ D , Q] = 1, w h i c h is certainly false. Consequently C (V) = H. G
G
G
G
K
G
K
K
D
G
T h i s result tells us that G is i s o m o r p h i c w i t h a g r o u p o f a u t o m o r p h i s m s of V a n d hence i n view o f (xii) w i t h a subgroup o f G L ( 2 , p). N o w SL(2, p) is n o r m a l a n d has index p — 1 i n G L ( 2 , p), i n d i c a t i n g that a l l p-elements o f G L ( 2 , p) lie i n SL(2, p). H o w e v e r P is m a x i m a l i n G, so we m a y be sure that G, a n d hence G, is generated b y p-elements. Consequently G is i s o m o r p h i c w i t h a subgroup o f SL(2, p). 9
m
Recall f r o m 3.2.7 that |SL(2, p)| = (p - l ) p ( p + 1). T h u s | X | = g divides (p — l ) ( p + 1). Suppose t h a t q > 2, so t h a t g cannot divide b o t h p — 1 a n d p + 1. N o w q = 1 m o d p b y (ix) a n d 9.4.3, w h i c h implies that q > p + 1 a n d q = p + 1. Since p is an odd prime b y hypothesis, p + 1 is even a n d q = 2. B u t an easy m a t r i x calculation—see Exercise 10.4.2—reveals that SL(2, p) has precisely one element o f order 2, namely — 1 , w h i c h is i n the centre. Since K is elementary abelian, K < £(G) a n d (vii) gives K = G a n d hence A < H, c o n t r a r y t o the choice o f A, w h i c h is o u r final c o n t r a d i c t i o n . m
m
m
2
• E X A M P L E . Thompson's theorem is not true if p = 2. F o r let G = S a n d let P be a Sylow 2-subgroup. T h e n P is a d i h e d r a l g r o u p o f order 8 a n d i t is easy t o see t h a t for this g r o u p J(P) = P a n d N (J(P)) = P; i n a d d i t i o n C (CP) = P. O f course P is 2-nilpotent b u t G does n o t have this p r o p e r t y : indeed O ( G ) = 1. 4
G
G
r
Groups with a Nilpotent Maximal Subgroup I n Chapter 9 we p r o v e d the theorem of Schmidt that a finite g r o u p whose m a x i m a l subgroups are a l l n i l p o t e n t is soluble. W e shall use T h o m p s o n ' s c r i t e r i o n t o establish a notable i m p r o v e m e n t o f Schmidt's theorem w h i c h applies to groups i n w h i c h a single m a x i m a l subgroup is n i l p o t e n t . 10.4.2 ( T h o m p s o n ) . If a finite group G has a nilpotent of odd order, then G is soluble.
maximal
subgroup
M
Proof. As usual we suppose the theorem false a n d choose for G a counter example o f smallest order. I f M contains a n o n t r i v i a l n o r m a l subgroup N o f G, then G/N is soluble b y m i n i m a l i t y o f | G | : therefore G is soluble, N being n i l p o t e n t . Hence the core o f M i n G m u s t be 1. Choose a p r i m e p d i v i d i n g | M | a n d let P be the unique Sylow p-sub g r o u p o f M . T h e n P is contained i n a Sylow p-subgroup P of G. Since M is 0
0
304
10. The Transfer and Its Applications
m a x i m a l a n d P is n o t n o r m a l i n G, i t m u s t be true t h a t N (P ) 0
G
implies t h a t N (P ) P
0
= M , which
0
= P n M = P . B y the n o r m a l i z e r c o n d i t i o n P 0
0
= P,
w h i c h means t h a t | G : M | is p r i m e t o p a n d M is a H a l l rc-subgroup where n is the set o f p r i m e divisors o f | M | . W e a i m next t o show t h a t T h o m p s o n ' s t h e o r e m is applicable t o G. Since J ( P ) is characteristic i n P, i t is n o r m a l i n M . N o w J ( P ) c a n n o t be n o r m a l i n G, so i t follows that N (J(P)) G
= M . F o r exactly the same reason C ( £ P ) = M . G
T h u s 10.4.1 assures us t h a t G is p - n i l p o t e n t for every p r i m e p i n n. D e f i n i n g L t o be the intersection o f a l l O (G) for p i n n we see t h a t L is a 7r'-group P
9
a n d G / L a 7r-group. Since M is a H a l l rc-subgroup, i t follows t h a t G =
ML
a n d M n L = 1. Let
Q be any S y l o w s u b g r o u p o f L . T h e n G = N (Q)L
b y the F r a t t i n i
G
argument. T h e Schur-Zassenhaus T h e o r e m provides us w i t h a c o m p l e m e n t of N (Q)
i n N ( e ) , say N (Q)
L
G
G
= XN (Q). L
T h i s gives G = X i V ( g ) L = X L , so L
9
t h a t X is another c o m p l e m e n t o f L i n G a n d X = M
for some g b y the
conjugacy p a r t o f the Schur-Zassenhaus T h e o r e m . I t follows t h a t X is a maximal subgroup and Q o
XQ = G. Hence every Sylow s u b g r o u p o f L is
n o r m a l a n d so L is n i l p o t e n t . Since G / L ~ M , i t follows t h a t G is soluble.
• I n fact 10.4.2 is false i f the m a x i m a l s u b g r o u p possesses elements o f order 2; for the simple g r o u p P S L ( 2 , 17) has a m a x i m a l s u b g r o u p o f d i h e d r a l type D
1 6
.
H o w e v e r Deskins a n d J a n k o have p r o v e d t h a t 10.4.2 remains true i f
the S y l o w 2-subgroup is a l l o w e d t o have class at m o s t 2. F o r details see [b6].
EXERCISES
10.4
1. Let G be a finite group whose order is not divisible by 6. Let P be a Sylow p-subgroup of G. Prove that G is p-nilpotent i f and only i f N (J(P)) and C (£P) are both p-nilpotent. G
G
2. I f p is an odd prime, prove that the only element of order 2 i n SL(2, p) is — 1 . Deduce that a Sylow 2-subgroup of SL(2, p) is a generalized quaternion group and that a Sylow 2-subgroup of PSL(2, p) is dihedral. [Hint: Apply 5.3.6.] 2
3. Let P be a Sylow 2-subgroup of G = PSL(2, 17). Show that P ~ D and that p = N (P). Show also that N (J(P)) = P = C (£P). Thus the p-nilpotence of N (J(P)) and C (£P) does not imply the solubility of G. (Remark: P is actually maximal i n G.) l6
G
G
G
G
G
4. Assume that the finite group G has a nilpotent maximal subgroup M . I f Fit G = 1, prove that M is a Hall 7i-subgroup of G for some n. 5. Let M be a maximal subgroup of a finite group G. Assume that each subgroup of M is normal i n M . Prove that G cannot be simple unless its order is a prime.
10.5. Fixed-Point-Free Automorphisms
305
10.5. Fixed-Point-Free Automorphisms A n a u t o m o r p h i s m a o f a g r o u p G is said to have a fixed point g i n G i f g = g; thus C (oc) is the set o f a l l fixed points of a. I f C ( a ) = 1, a n d 1 is the o n l y fixed p o i n t o f a, then a is called fixed-point-free. A subgroup H of A u t G is said to be fixed-point-free i f every n o n t r i v i a l element o f H is fixedpoint-free. a
G
G
W e shall m a k e use o f T h o m p s o n ' s c r i t e r i o n for p-nilpotency to prove an i m p o r t a n t theorem a b o u t groups w i t h a fixed-point-free a u t o m o r p h i s m of p r i m e order: this can then be applied t o the structure o f Frobenius groups. T h e f o l l o w i n g l e m m a collects together the most elementary properties of fixed-point-free automorphisms. 10.5.1. Let a be be a fixed-point-free let a have order n.
automorphism
of a finite
group G and
l
(i) If (i, n) = 1, then oc is also fixed-point-free. (ii) The mapping — 1 + a which sends x to x~ tion of G.
1+<x
1
<x
= x~ x
is a
permuta
a
(iii) x and x are conjugate if and only if x = 1. (iv) i + « + - + « » i f u Q. 1
x
=
or
a
X
I N
Proof, (i) holds because a is a p o w e r o f oc\ 1 + a
1+
- 1
a
1
(ii) I f x ~ = y~ \ then ( x y ) = x y " a n d x = y. Since G is finite, — 1 + a is a p e r m u t a t i o n . (iii) Suppose t h a t x = x for some g i n G. B y (ii) i t is possible to w r i t e g = y~ for some y i n G. I t follows t h a t x = x = y~ (yxy~ )y a n d hence that (yxy' )* = yxy' . Therefore yxy' = 1 a n d x = 1. (iv) L e t z = i + « + - + « " ~ . e n z = « + « + - + « » - + i * = 1 by a
9
1+<x
a
1
1
9
<x
1
<x
1
1
x
a
t r l
2
x
1
=
z
T
r
m
s
z
(iii).
•
10.5.2. If a is a fixed-point-free automorphism of a finite group G, then each prime p there is a Sylow p-subgroup P such that P = P.
for
a
Proof. L e t P
0
be any Sylow p-subgroup. T h e n P£ = P«? for some g i n G.
A p p l y i n g 10.5.1(h), we m a y w r i t e g = / T
1 + A
for a suitable h: n o w let P =
1
P^" . T h e n a
l
a
p = (p$~r = (p r = 0
_l
= ^o = p-
•
10.5.3. Let H be a group of automorphisms of a finite abelian group A. Sup pose that H is the semidirect product x M where G[S is fixed-point-free of prime order p for every [S in M. Assume also that \A\ and \M\ are coprime. Then M = 1.
10. The Transfer and Its Applications
306
1+
fi+
+
)p
=«
m
i
Proof. L e t ae A a n d e M. T h e n ° '" M ~ = i b y 10.5.1. T a k i n g the p r o d u c t o f these equations for all fi i n M a n d remembering that A is abelian, we o b t a i n a
i = n
n
a i
*
p ) i
n
n
^
•
ai)
i=0 i = l 0eM W e c l a i m t h a t i f i is a fixed integer satisfying 1 < i < p — 1, the elements (<7j8)', J8 e M , are a l l different. F o r suppose t h a t (txjS) ' = (<7j8)'; then
1
J
1
1
Yl
=
^ **fi a
9
so t h a t (11) yields a"
l M l
tfi
= n n ** ' i = l /3eM B u t the r i g h t - h a n d side o f this last e q u a t i o n is clearly fixed b y each element of M; since (\a\, \M\) = 1, we deduce t h a t a is fixed b y each such element. Because a was an a r b i t r a r y element o f A, we conclude t h a t M = 1. • W e come n o w t o the p r i n c i p a l t h e o r e m o f this section. 10.5.4 (Thompson). Let G be a finite group and let p be a prime. If G has a fixed-point-free automorphism oc of order p, then G is nilpotent. Proof. Assume t h a t the t h e o r e m is false a n d let G be a counterexample o f m i n i m a l order. W e see f r o m 10.5.1(h) t h a t — 1 + a is surjective o n G/CG, so the hypotheses o n G are i n h e r i t e d b y G/CG. Hence £G = 1. T h e p l a n of attack is first t o deal w i t h the case where G is soluble, then t o use T h o m p s o n ' s c r i t e r i o n t o reduce the general case, (i) Case G soluble. a
Let 1 G a n d let A be m i n i m a l subject t o A = A. Since (A'f = A\ we see t h a t A is abelian and, since (A ) = A , t h a t A is an elementary g-group for some p r i m e q. N o w G cannot be a g-group because i t is n o t n i l p o t e n t , so there exists a p r i m e r different f r o m q d i v i d i n g | G | . A l s o 10.5.2 tells us t h a t G has an a-admissible S y l o w r - s u b g r o u p R. Observe t h a t r ^ p otherwise a w o u l d have a fixed p o i n t i n R since
G
G
a
q
307
10.5. Fixed-Point-Free Automorphisms
H = <
W e shall show t h a t the c o n d i t i o n s o f
10.5.3 are met i n H. r
L e t g e R a n d let g e M b e the a u t o m o r p h i s m i n d u c e d b y c o n j u g a t i o n z
b y g i n A. W e c l a i m t h a t ag z
ag
is conjugate t o a i n A u t A; this w i l l show t h a t
is a fixed-point-free a u t o m o r p h i s m o f order p. Since a a
y
free o n R, we can f i n d a n r i n R such t h a t g ~ T
_ 1
The automorphism ( r ) 0 r _ 1
a
(rar )
a
a
a
= r a r~ ,
T
= r~ l
1 + a _ 1
- 1
l a
r
f l
r
+ 1
1
a
= g~ a g.
a + 1
.
- 1
sends ae A to b = r~ (rar~ ) r.
so t h a t ft = - i + « ^ - «
is fixed-point-
; hence # = r "
Now (rar )* = T
_ 1
Hence ( r ) o r
T
T
= o-#
as required. W e m a y n o w a p p l y 10.5.3 t o show t h a t M = 1, t h a t is, [A, K] — 1. H o w ever this gives the c o n t r a d i c t i o n 1 ^ A < CG since G = ,4P. (ii) Case G insoluble. L e t g be a n o d d p r i m e d i v i d i n g | G | . B y 10.5.2 there is a Sylow ^-sub a
g r o u p Q such t h a t Q = Q. Hence J(Qf is a-admissible. I f < / ( 0 o G, t h e n G/J(Q)
= J(Q\
w h i c h implies t h a t
N (J(Q)) G
w o u l d be n i l p o t e n t a n d G soluble,
w h i c h b y (i) cannot be the case. Therefore N (J(Q)) G
is a p r o p e r subgroup,
and, being a-admissible, i t is n i l p o t e n t . F o r similar reasons C ( C © is n i l G
potent. T h o m p s o n ' s c r i t e r i o n n o w shows t h a t G is g - n i l p o t e n t a n d G / 0 ^ ( G ) a g-group. Since O ,(G) q
is p r o p e r a n d a-admissible, i t is n i l p o t e n t a n d G is
soluble.
•
I n a d d i t i o n rather precise i n f o r m a t i o n is available concerning the struc ture o f a fixed-point-free a u t o m o r p h i s m g r o u p . 10.5.5. Let H be a fixed-point-free
group of automorphisms
of a finite
G. Then every subgroup of H with order pq where p and q are primes is The Sylow p-subgroups quaternion
of H are cyclic
if p is odd and cyclic
or
group cyclic.
generalized
if p = 2.
Proof. L e t S be a n o n c y c l i c s u b g r o u p o f H w i t h order pq. T h e n b y Exercise 1.6.13 there is a n o r m a l Sylow subgroup, say Q o f order q, a n d clearly S = QP where P is a subgroup o f order p. B y 10.5.4 the g r o u p G is nilpotent: s
let R be a n o n t r i v i a l S y l o w r - s u b g r o u p o f G. C l e a r l y R
= R. Also, i f r = q,
t h e n Q x R w o u l d be n i l p o t e n t a n d have n o n t r i v i a l center, w h i c h w o u l d prevent H f r o m being fixed-point-free. Therefore r ^ q. L e t P = a n d let fi G Q; t h e n |a/?| c a n n o t equal pq since \S\ = pq a n d S is n o t cyclic; thus |a/?| = p. N o w a/? is fixed-point-free, so we m a y a p p l y 10.5.3 t o S as a g r o u p o f a u t o m o r p h i s m s o f CR, o b t a i n i n g at once the c o n t r a d i c t i o n [CP, Q] = 1. T h e structure o f the S y l o w subgroups is n o w a direct consequence o f 5.3.6.
• E X A M P L E . There morphism
is a finite
nonnilpotent
group with a fixed-point-free
auto
of order 4. L e t A = x
of order 49, a n d let X = < x > be a cyclic g r o u p o f order 3. T h e assignments 2
a i—• a
4
a n d b i—• b
determine an a u t o m o r p h i s m o f A w i t h order 3. T h u s
10. The Transfer and Its Applications
308
there is a corresponding semidirect p r o d u c t G = X x A; this is a metabelian g r o u p o f order 147. One easily verifies that the assignments a i—• b, b i—•a" , x i—• x " determine an a u t o m o r p h i s m o f G w i t h order 4. A t y p i c a l element g of G w i l l have the f o r m x a b where 0 < / < 3 a n d 0 < j , k < 7, a n d g is m a p p e d b y a t o x~ b a~ . T h u s # is a fixed p o i n t i f a n d o n l y i f i = j = k = 0, t h a t is, g = 1. Hence a is fixed-point-free, b u t G is n o t n i l p o t e n t . 1
1
l
l
j
j
k
k
Frobenius Groups Recall f r o m Chapter 8 t h a t a g r o u p G is a Frobenius group i f i t has a p r o p e r n o n t r i v i a l subgroup H such t h a t H r\H = 1 for a l l g i n G\H. B y 8.5.5 there is a F r o b e n i u s kernel, t h a t is, a n o r m a l s u b g r o u p X such t h a t G = / J K a n d H nK = 1. 9
T h e f o l l o w i n g fundamental t h e o r e m o n F r o b e n i u s groups is a con sequence o f the m a i n results o f this section. 10.5.6. Let G be a finite Then:
Frobenius
group with kernel K and complement
(i) K is nilpotent (Thompson); (ii) the Sylow p-subgroups of H are cyclic quaternion if p = 2 (Burnside).
if p > 2 and cyclic
or
h
H.
generalized
k
Proof. L e t 1 ^ he H a n d suppose t h a t k = k where k e K. T h e n 1 ^ h = ke H n H , whence ke H nK = 1 b y d e f i n i t i o n o f a F r o b e n i u s g r o u p . Hence c o n j u g a t i o n b y h i n K induces a fixed-point-free a u t o m o r p h i s m i n K. I t follows i m m e d i a t e l y f r o m 10.5.4 t h a t K is n i l p o t e n t . Since C (K) = 1, the g r o u p H is a fixed-point-free g r o u p o f a u t o m o r p h i s m s o f K. T h e second statement is n o w a consequence o f 10.5.5. • k
H
EXERCISES
10.5
1. A finite group has a fixed-point-free automorphism of order 2 i f and only if it is abelian and has odd order. 2. Let G be a finite group with a fixed-point-free automorphism a of order 3. Prove that [ x , y, y] = 1 for all x, y in G. [ F o r the structure of such groups see 12.3.6. Hint: Show first that [ x , x ] = 1 and then prove that [x , x ] = 1 for all x, y.] a
y
a
3. Let a be a fixed-point-free automorphism of a finite group G. I f a has order a power of a prime p, then p does not divide |G|. I f p = 2, infer via the F e i t Thompson Theorem that G is soluble. 4. I f X is a nontrivial fixed-point-free group of automorphisms of a finite group G, then X K G is a Frobenius group. 5. Let G be a finite Frobenius group with Frobenius kernel K. I f | G : K\ is even, prove that K is abelian and has odd order.
10.5. Fixed-Point-Free Automorphisms
309
6. Suppose that G is a finite group with trivial center. I f G has a nonnormal abelian maximal subgroup A, prove that G = AN and A n N = 1 for some elementary abelian p-subgroup N which is minimal normal i n G. Show also that A must be cyclic of order prime to p. [Hint: Prove that AnA = 1 if x e G \ A ] x
7. I f a finite group G has an abelian maximal subgroup, then G is soluble with derived length at most 3. 8. A finite Frobenius group has a unique Frobenius kernel, namely Fit G. Deduce that all Frobenius complements are conjugate (see Exercise 9.1.1). 9. I f a finite Frobenius group G has a Frobenius complement of odd order, then G is soluble. (Do not use the Feit-Thompson Theorem.) 3
10. Let P be a nonabelian group of order 7 and exponent 7. Find a fixed-point-free automorphism of P with order 3. Hence construct a Frobenius group with non abelian Frobenius kernel.
CHAPTER 11
The Theory of Group Extensions
T h e object o f extension t h e o r y is t o show h o w a g r o u p can be constructed f r o m a n o r m a l subgroup a n d its q u o t i e n t g r o u p . I n this subject concepts f r o m h o m o l o g i c a l algebra arise n a t u r a l l y a n d c o n t r i b u t e greatly t o o u r understanding o f i t . T h e necessary h o m o l o g i c a l machinery, i n c l u d i n g the definitions o f the (co)homology groups, is presented i n 11.2. T h e classical theory o f g r o u p extensions was developed b y O. H o l d e r a n d O . Schreier while the h o m o l o g i c a l i m p l i c a t i o n s o f the theory were first recognized b y S. Eilenberg a n d S. M a c L a n e . M u c h o f the version presented here is due to K . W . Gruenberg.
11.1. Group Extensions and Covering Groups I f N a n d G are a r b i t r a r y groups, an extension o f N b y G is, i n familiar parlance, a g r o u p E possessing a n o r m a l s u b g r o u p M such t h a t M ~ N a n d E/M ~ G. F o r o u r purposes i t is best t o be rather m o r e specific. B y a group extension of N by G we shall mean a short exact sequence o f groups a n d homomorphisms 1
• N
—
E
—
e
— +
G
•
1.
T h e m a i n features here are firstly t h a t p is injective a n d s surjective, a n d secondly t h a t I m p = K e r s = M say. T h u s M ~ N a n d E/M ~ G, so E is an extension o f N b y G i n the o r i g i n a l sense. T h e g r o u p N is called the kernel o f the extension. F o r the sake o f b r e v i t y let us agree t o w r i t e >-> t o denote a m o n o m o r p h i s m a n d - » an e p i m o r p h i s m . T h e above extension becomes i n this
310
311
11.1. Group Extensions and Covering Groups
notation N >-^—• E — G . Extensions o f N b y G always exists. F o r example, we m a y f o r m the semidirect p r o d u c t E = Gt<^N corresponding to a h o m o m o r p h i s m £ : G - > A u t N. Define a = ( 1 , a) a n d (g, df = g where a e N, g e G. T h e n M
N >-^—• E
—
G
is an extension o f AT b y G.
Morphisms of Extensions By a morphism f r o m A f A F - ^ G t o A f A F - ^ G i s meant a triple o f h o m o m o r p h i s m s (a, /?, y) such t h a t the f o l l o w i n g d i a g r a m commutes: N
>-^—>
—
P
oc N
E
>^—>
G
y
E —"-^
G.
W e m e n t i o n t h a t the collection o f a l l g r o u p extensions o f N b y G a n d m o r phisms between t h e m is a category, a l t h o u g h this w i l l p l a y n o p a r t i n the sequel. C e r t a i n types o f m o r p h i s m s are o f special interest, foremost a m o n g t h e m equivalences; these are m o r p h i s m s o f the f o r m (1, /?, 1) f r o m N ^ E - » G t o N ^ E - » G. A m o r e general type o f m o r p h i s m is an isomorphism; this is a m o r p h i s m o f the f o r m (a, /?, 1) f r o m N ^ E - ^ G to N ^ E - ^ G where a is an i s o m o r p h i s m o f groups. I t is easy t o see t h a t /? t o o m u s t be an iso m o r p h i s m . I t is clear t h a t equivalence a n d i s o m o r p h i s m o f extensions are equivalence relations.
Couplings Let N A E G be an extension. B y choosing a transversal t o M = I m p = K e r £ i n E one obtains a f u n c t i o n T : G -> E called a transversal function. T h u s i f x e E, the coset representative o f xM is (x ) . U s u a l l y T w i l l n o t be a h o m o m o r p h i s m — a fact t h a t makes extension t h e o r y interesting. B u t T w i l l always have the p r o p e r t y e z
TS =
1.
Conversely any f u n c t i o n T w i t h this p r o p e r t y determines a transversal t o M z
i n G, namely the set {g \g e G } .
312
11. The Theory of Group Extensions
z
W e associate w i t h each g i n G the o p e r a t i o n o f c o n j u g a t i o n b y g i n M . x
Since M ~ N, this leads t o an a u t o m o r p h i s m g (a'> = (0TVGT),
o f N described b y the rule
(aeN,geG).
(1)
I n this w a y we o b t a i n a f u n c t i o n l\ G -> A u t N. T o w h a t extent does I depend u p o n the choice o f transversal f u n c t i o n T ? I f x' is another transversal function, then g a n d g ' differ b y an element of M . Consequently, i f x' leads t o a f u n c t i o n A': G - > A u t AT, then g a n d g differ b y an inner a u t o m o r p h i s m o f N, as one can see f r o m (1). Hence g ( I n n N) = g ( I n n N), w h i c h makes i t reasonable t o define a f u n c t i o n x
T
k
A
A
x
X: G -> O u t AT by the rule A
0* = £ ( I n n N);
(2)
here # does n o t depend o n the transversal function. W h a t is more, x is a homomorphism because (g g Y = GiQi m o d M . T h u s each extension N A E -4» G determines a unique h o m o m o r p h i s m #: G -> O u t AT w h i c h arises f r o m c o n j u g a t i o n i n I m p b y elements o f E. I f G a n d N are a r b i t r a r y groups, we shall refer t o a h o m o m o r p h i s m %: G -> O u t AT as a coupling o f G t o AT, whether or n o t i t arises f r o m an extension. I f C is the center o f AT, a c o u p l i n g # o f G t o N gives rise t o a G - m o d u l e f structure o f C, namely a = a \ this a c t i o n is well-defined because I n n N acts t r i v i a l l y o n C. I n the very i m p o r t a n t case where N is abelian, the cou p l i n g x' G -> A u t AT prescribes a G-module structure for N. 1
2
9
11.1.1. Equivalent
gX
extensions
have the same
coupling.
Proof. L e t ( 1 , 9, 1) be a n equivalence f r o m AT A then there is a c o m m u t a t i v e d i a g r a m N
E
£
G to N
G;
G (3)
N >(Here the left a n d r i g h t vertical maps are identity functions.)
L e t x a n d x be
the respective couplings o f the t w o extensions. Choose a transversal func t i o n T : G -> £ for N ^
£ - » G. T h e n r = T# is a transversal f u n c t i o n for the
second extension: this is because xl = x(9s) = xs = 1 b y c o m m u t a t i v i t y o f (3). f In order to assign a ZG-module structure to an abelian group M it is enough to prescribe the action of the elements of G. We shall therefore treat the terms "G-module" and ZG-module" as synonymous.
11.1. Group Extensions and Covering Groups
313
a
n
n
L e t us use % a n d f t o c o m p u t e the couplings x d X- I the first place = # ( I n n N) a n d g = # ( I n n N) where h G -> A u t AT a n d I : G -> A u t AT arise f r o m x a n d f as i n (1). A p p l y i n g 0 t o (1) a n d keeping i n m i n d t h a t p9 = /Z, b y (3), we o b t a i n (a f = ( # T a ' V ~ ) = (a> *f- Hence g = # a n d g = g , w h i c h shows t h a t x X• x
A
x
x
J
A
g
gk
1
g
x
x
=
T h e p r i n c i p a l aims o f the theory o f g r o u p extensions m a y be summarized as follows: (i) t o decide w h i c h couplings o f G t o N give rise t o a n extension o f N b y G; (ii) t o construct a l l extensions o f N b y G w i t h given c o u p l i n g x\ (iii) t o decide w h e n t w o such extensions are equivalent (or possibly iso morphic). W e shall see t h a t these goals can, i n p r i n c i p l e at least, be attained w i t h the a i d o f c o h o m o l o g y .
Split Extensions A n extension N A E 4> G is said t o split i f there exists a transversal func t i o n T: G -> E w h i c h is a h o m o m o r p h i s m . F o r example, the semidirect p r o d uct extension A T ^ G x A T - ^ G i s split because g i—• (g, 1) is a transversal function w h i c h is a h o m o m o r p h i s m . I n fact this example is entirely t y p i c a l o f split extensions. F o r suppose t h a t N A F 4> G splits v i a a h o m o m o r p h i s m T: G -> F . W r i t e X = F = G . Since rs = 1, we have (x~ xf = x ~ x = 1, so t h a t x ~ x e M = K e r £ a n d F = XM. I n a d d i t i o n X n M = 1 because x e M implies t h a t 1 = ( x ) = x . Hence E = XKM~GKN; this shows t h a t every split extension is a semidirect p r o d u c t extension. £ T
ez
£
£
T
£T
ex
£T
£
£
Complements and Derivations Consider a split extension N >-> F - » G, w h i c h can w i t h o u t loss be taken t o be a semidirect p r o d u c t extension; thus E = G x N. Recall f r o m 9.1 t h a t a s u b g r o u p X w i t h the properties XN = E a n d X nN = 1 is called a comple ment o f AT i n F . O f course G itself—or indeed a n y conjugate o f G—is a complement. I t is a n i m p o r t a n t p r o b l e m t o decide whether every comple m e n t is conjugate t o G. W e shall show t h a t complements o f N i n F correspond t o certain func tions f r o m G t o N k n o w n as derivations. I f X is a n y complement, each g i n G has a u n i q u e expression o f the f o r m g = xa' where xe X a n d ae N: define 3 = 3 : G -> N b y g = a. T h u s gg e X. L e t g e G, i = 1, 2; then X contains the element (g g )(g Q2\ w h i c h equals 0i0 (#iT 02> that the 1
3
3
X
t
3
1
2
1
2
2
s
o
314
11. The Theory of Group Extensions
function 3 has the p r o p e r t y I
(0lff )' = (0l)' 02-
(4)
2
Q u i t e generally, i f N is any G-operator g r o u p , a function 3: G -> N is called a derivation (or 1-cocycle) from G to N i f (4) holds for a l l g i n G. N o t i c e the simple consequences o f (4) t
l' = l
1
and
a
I
1
(ff- )' = ( ( f f r r .
(5)
The set o f a l l derivations f r o m G t o N is w r i t t e n Der(G,iV)
or
Z\G,N).
T h u s far we have associated a d e r i v a t i o n w i t h each complement. C o n versely, suppose t h a t 3: G -> N is a d e r i v a t i o n . T h e n there is a correspond i n g c o m p l e m e n t t o N i n G given b y X = {gg \g e G } . I t follows easily f r o m (4) a n d (5) t h a t X is a subgroup. Clearly E = X N a n d X nN = 1, so X is i n fact a complement. I t s h o u l d be apparent t o the reader t h a t X 3 a n d 3 i—• X are inverse mappings. T h u s we can state the f o l l o w i n g result. s
5
8
d
d
5
X
11.1.2. The mapping X NinE
3
X
d
is a bijection from the set of all complements
of
= G x N to D e r ( G , AO-
Inner Derivations L e t us n o w assume that N is abelian, so t h a t AT is a G-module. W e shall w r i t e A instead o f N. There is a n a t u r a l rule o f a d d i t i o n for derivations, namely a = aa. I t is quite r o u t i n e t o check t h a t 3 + d is a d e r i v a t i o n ; notice however t h a t the c o m m u t a t i v i t y o f A is essential here. F u r t h e r , w i t h this b i n a r y o p e r a t i o n D e r ( G , A) becomes an additive abelian g r o u p . S l + d 2
X
Sl
d2
2
I f a e A, we define a f u n c t i o n 3(a): G -> A b y the rule g
m
= lg, ^
=
a~^\ d2
a
t e
s
u
s
a
t
o
n
c
e
T h e c o m m u t a t o r i d e n t i t y \_g^g ^ = a] [g2> ] ll that 3(a) is a d e r i v a t i o n . Such derivations are called inner (or 1-coboundaries), the subset o f inner derivations being w r i t t e n 2
Inn(G,X) - 1
or
X
B (G,A) 1
Now a f c ] = lg, a ] since ^ is abelian. Therefore 3(ab~ ) = 3(a) — 3(b) a n d consequently I n n ( G , A) is a subgroup o f D e r ( G , A). T h e significance o f the inner derivations is t h a t they determine comple ments w h i c h are conjugate t o G.
11.1. Group Extensions and Covering Groups
315
11.1.3. / / A is a right G-module, there is a bijection between the set of conjugacy classes of complements of A in G x A and the quotient group D e r ( G , ,4)/Inn(G, A) in which the conjugacy class of G corresponds to I n n ( G , A). yia
Proof. Suppose that X a n d Y are conjugate complements. T h e n X = Y = Y for some y e Y, a e A. If g e G, then gg e X where 3 is the d e r i v a t i o n arising f r o m X. T h e n gg = y for some y i n Y. N o w y = v [ v , a], so 99 y[y> w h i c h shows that [ v , a ] = lg, a ] = g because A is abelian. Therefore g(g g~ ) = y e Y. Consequently 3 = 3 — 3(a) a n d 3 = 3 m o d ( I n n ( G , A)). Reversing the argument one can show that i f 3 = 3 — 3(a), then gg * = (gg ) , so t h a t X = Y . T h i s completes the proof. • a
3x
x
X
Sx
dx
a
a
=
m
Sx
S{a)
Y
X
X
Y
s
3r a
Y
X
a
T h u s a l l complements o f A i n G x A are conjugate i f a n d o n l y i f the g r o u p D e r ( G , ,4)/Inn(G, A) is t r i v i a l . W e shall see later t h a t this q u o t i e n t g r o u p can be i n t e r p r e t e d as the first degree c o h o m o l o g y g r o u p H (G, A). X
Factor Sets and Extensions with Abelian Kernel Before proceeding w i t h the general t h e o r y o f extensions we shall consider the special case o f extensions w i t h abelian kernel. Consider an extension
where A is an abelian g r o u p , w r i t t e n additively; let %: G -> A u t A be the c o u p l i n g o f the extension. T h e n % prescribes a G-module structure for A b y conjugation; this is given b y (ax )p = x~ (ap)x where x e E, ae A. As the first step i n the analysis o f the extension, we choose a transversal function r . G -> E; thus TS = 1. N o w T m a y n o t be a h o m o m o r p h i s m , b u t we can w r i t e for x, y i n G e
z
x
xy x
z
1
z
= (xy) ((x, x
where (x, y)(j> e A, since x y a n d (xy) I m p. T h u s we have a function
y)(j))p,
b e l o n g t o the same coset o f K e r s =
G-+A; x
x
x
this is subject t o a r e s t r i c t i o n because o f the associative l a w x (y z ) = (x y )z . S u b s t i t u t i n g for products like x y , we o b t a i n the fundamental equation x
x
x
x
x
(x, yz)(j) + (y, z)(f> = (xy, z)
(6)
w h i c h holds for a l l x, y, z i n G. A f u n c t i o n (f>: G x G -> A satisfying (6) is t r a d i t i o n a l l y called a factor
set; the h o m o l o g i c a l t e r m is a 2-cocycle,
a n d we
11. The Theory of Group Extensions
316
shall w r i t e 2
Z ( G , A) for the set o f a l l 2-cocycles o f G w i t h coefficients i n the G-module A. N o t i c e t h a t Z ( G , A) has the structure o f an abelian g r o u p where the g r o u p opera t i o n is defined b y (x, y)(j> +
1
x
x
x
2
x
2
x
x
x
x
x
x
x
(x, y)(j) = (x, y)(j)' + ( x > # - M*A • y for x, y, z i n G. N o w define
G x G -+ Aby
x
x
(>#
the rule
(x, > # * = ( > # - ( * ) # +
(x)\l/-y.
2
T h e n (j)' = (j) + ij/*, so t h a t i ^ * G Z ( G , A). T h e 2-cocycle ^ * is o f a special k i n d called a 2-coboundary (n-cocycles a n d n-coboundaries are i n t r o d u c e d i n 11.3). I t is easy t o see t h a t the 2-coboundaries this is w r i t t e n B (G, A).
2
f o r m a s u b g r o u p o f Z ( G , A):
2
2
W h a t we have shown is that (j) a n d ' belong t o the same coset o f B (G, A). T h u s the extension determines a u n i q u e element (j) + B (G, A) o f the g r o u p 2
2
2
Z ( G , A)/B (G,
A).
T h i s g r o u p w i l l appear later as the c o h o m o l o g y g r o u p o f degree 2.
Constructing Extension from Factor Sets T h e next step is t o start w i t h a G-module A a n d a factor set (f>: G x G -> A, a n d t o show h o w t o construct an extension o f A b y G w h i c h induces the given G-module structure o f A, a n d w h i c h , for a suitable transversal func t i o n , has (j) as factor set. L e t E((j)) be the set p r o d u c t G x A. A b i n a r y o p e r a t i o n o n E((j)) is defined b y the rule (x, a)(y, b) = (xy, ay + b + (x, y)
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317
verifies t h a t ( 1 , — ( 1 , l)
1
(x , - a x '
- 1
- (1,
- (x, x ) ^ ) ,
as one verifies using the o p e r a t i o n o n E(). Therefore E((f>) is a g r o u p . F i n a l l y we can f o r m a n extension A > - i U Etf) — ^ by
defining
a\i = ( 1 , a — ( 1 , !)) a n d
G
(x, af = x.
For
clearly
Ker s =
{ ( 1 , a)\a G A] = I m fi. A simple c a l c u l a t i o n reveals t h a t (x, ( ^ ( l , a - ( 1 , l ) 0 ( x , 0) = ( 1 , ax - ( 1 , 1 ) 0 (for this one needs (1, x)(j) = ( 1 , l)(j) • x, a n d also the i d e n t i t y (6) w i t h y = x
- 1
a n d z = x). T h i s e q u a t i o n shows t h a t the extension induces the given Gm o d u l e structure i n A. I t is clear t h a t the assignment x i—• (x, 0) is a transversal function T for x
x
the extension; c a l c u l a t i n g w i t h the g r o u p o p e r a t i o n , we find t h a t x y
=
T
(xv) ((x, y)(f>)fi. T h u s (j) is indeed the factor set for the extension w h e n the transversal f u n c t i o n x is used.
Equivalence So far we have seen h o w t o pass f r o m extensions t o factor sets a n d f r o m factor sets back t o extensions. N o w we w i s h t o decide w h e n t w o extensions are equivalent b y l o o k i n g at their factor sets. L e t A be a fixed G-module a n d consider t w o extensions o f A b y G realiz i n g this m o d u l e structure, A >-^U E — ^
G,
t
(i = 1, 2).
Choose transversal functions x a n d let the resulting factor sets be t
F i r s t o f a l l , suppose t h a t the t w o extensions are equivalent, a n d t h a t the diagram A >
> E
» G
x
II
1
e
commutes where 9 is a n i s o m o r p h i s m . N o w T = T 9 is a transversal func 2
tion
for the second
ing
9
to
the
( x v ) ( ( x , y)(f> )fi , 1
extension.
2
equation
T2
2
T l
x v
so t h a t fl
Therefore ^
1
extension because T~ Z = T 0 £
2
T l
2
1
Tl
= ( x v ) ( ( x , y)(/> )fi , 1
1
=
T
2
we
determines the factor set ^
+ B (G,
2
A) = <j> + B (G, 2
2
e
i i — 1- A p p l y obtain for
A) since T
m i n e factor sets b e l o n g i n g t o the same coset o f B (G,
2
A).
X2
X2
x y
the and T
=
second 2
deter
11. The Theory of Group Extensions
318
2
2
Conversely, assume t h a t <^ + B (G, A) = <j> + B (G, A); then
x
2
2
6
x
(x^iap,))
= x *(a
+
(x)i//)p , 2
(xe G, ae A). T h e n a r o u t i n e c a l c u l a t i o n shows t h a t 9 is a h o m o m o r p h i s m . One verifies easily t h a t p 9 = p —here one has t o note t h a t T = ( 1 , 1)<^ since l l = l ( l , 1)<^. F i n a l l y i t is clear t h a t s = 9s . I t follows t h a t 9 fits i n t o a c o m m u t a t i v e d i a g r a m 1
x
T l
T l
2
T l
x
A >
> E
1
1
»
«
2
G
I!
T h u s 9 is an i s o m o r p h i s m a n d the t w o extensions are equivalent. These conclusions are summarized i n the f o l l o w i n g result. 11.1.4. Let G be a group and A a G-module. Then there is a bijection between the set of equivalence classes of extensions of A by G inducing the given mod ule structure and the group Z (G, A)/B (G, A). Moreover the split extension corresponds to B (G, A). 2
2
2
I n particular, every extension o f A b y G is equivalent t o one o f the con structed extensions A ^ E((f>) - » G. T h i s concludes o u r discussion o f extensions w i t h abelian kernel, w h i c h is essentially Schreier's o r i g i n a l treatment. N e x t we shall show h o w general extensions m a y be constructed s t a r t i n g f r o m a presentation o f the q u o t i e n t group.
Introduction of Covering Groups L e t N a n d G be given groups a n d let G -> O u t N be some c o u p l i n g o f G t o N. W e are g o i n g t o show h o w a l l extensions o f N b y G w i t h c o u p l i n g X—if any e x i s t — m a y be constructed as images o f a "covering g r o u p . " T o start things off we must choose a presentation o f G # >
• F — G .
Here, o f course, F is a free g r o u p a n d # = K e r %. D e n o t e b y v: A u t N -> O u t N the n a t u r a l h o m o m o r p h i s m w i t h kernel I n n N. B y 2.1.6 a free g r o u p has the projective p r o p e r t y ; hence there is a lifting o f nx: F -> O u t AT, t h a t is,
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319
a h o m o m o r p h i s m £: F -> A u t N m a k i n g the d i a g r a m F / / < ^
/
/
/
A u t AT — ^ — > O u t AT n
• 1
x
commute. T h u s £v = 7r#. N o w = (R ) = 1, so < K e r v = I n n N. L e t J: # -> A u t AT be the restriction o f £ t o B y the N i e l s o n - S c h r e i e r T h e o r e m (6.1.1) the g r o u p R is a free group. So it t o o has the projective property. Hence there is a h o m o m o r p h i s m rj: R-+ N such that rjT = J
/
AT — ^ — I n n AT
• 1,
where T : A u t N -> AT is the c o n j u g a t i o n h o m o m o r p h i s m . I t follows that the diagram R >
•
—
F
V
G
£ T
1 v
N — — > Aut N — — » is c o m m u t a t i v e . U s i n g the f u n c t i o n
(7)
Out N
F -+ A u t AT, we f o r m the semidirect p r o d u c t S = F x ^ AT.
T h i s is the covering group f r o m whose q u o t i e n t groups extensions o f N by G w i l l be formed. 11.1.5. Let N and G be given groups, let G -> O u t N be a coupling of G to N and let R ^ F G be a fixed presentation of G. Write S = F x ^ AT w/zere £: F -> A u t AT is a lifting of x as in (7). (i) Every extension of N by G with coupling x is equivalent to an extension N >-> S/M - » G where M is a normal subgroup of S such that MN = M x N = RN. (ii) Conversely N
every
such normal
subgroup
M gives rise
to an
extension
S/M - » G with coupling x-
Proof. Suppose first o f a l l t h a t M < ] S satisfies MN JL\ N -> S/M
and
s: S/M - > G be
the
natural
= M x N = RN.
mappings,
a? = aM
Let and
11. The Theory of Group Extensions
320
(faMf
n
= f,
(ae N,f
e F); then there is an extension N
S/M — ^
G.
(8)
For = MN/M = RN/M = K e r I . D e n o t e the c o u p l i n g o f this extension by x- T o calculate g choose / f r o m F so t h a t f = g a n d conjugate b y / i n S/M. T h i s produces the a u t o m o r p h i s m o f N since S = F N. B y c o m m u t a t i v i t y o f (7) we have x
n
9 a
X —
/f ' ($ I n n N)=f«>=f**
= g
n
Hence x = X d (8) has c o u p l i n g x- T h u s (ii) has been established. I t remains t o p r o v e t h a t an extension N A E 4> G w i t h c o u p l i n g # is equivalent t o an extension o f type (8). T o establish this we begin b y using once again the projective p r o p e r t y o f free groups, this time t o construct a h o m o m o r p h i s m y: F -> E such t h a t ys = n.
E ——+ y
n
N o w (R f picture
= R
y
= 1, so R
G
• 1.
< K e r s = N». T h u s we o b t a i n a c o m m u t a t i v e
R >
• F
—
G (9)
N >K
y
where K is defined b y (r Y = r a n d the r i g h t h a n d m a p p i n g is the i d e n t i t y function. Define x t o be the a u t o m o r p h i s m o f N t h a t arises f r o m conjuga t i o n b y x i n I m \i\ thus (a f = x ^ * , (ae N,x e E). T h e n h E -> A u t N is surely a h o m o m o r p h i s m . Since the c o u p l i n g x arises f r o m c o n j u g a t i o n i n AT, x
xX
- 1
SX =
^v.
Therefore yXv = (ys)x = nx = £v; here we have used the c o m m u t a t i v i t y o f (7) a n d (9). Since K e r v = I n n AT, the elements x a n d x^ differ b y an inner a u t o m o r p h i s m o f AT, say (n£) , n e AT, a n d yX
x
x
yX
x
x
t = x (n£)
y
x
= (x n£)
(10)
for a l l x E E. N o w choose a set o f free generators X o f F. T h e n there is a h o m o m o r p h i s m a: F -> E such t h a t x = x n£ for a l l x i n X. W i t h this a one has x = (x n£) = x = x by (9). Therefore a
ae
y
e
ye
y
n
71 = (78.
11.1. Group Extensions and Covering Groups
321
a
a
lk
N e x t we extend a t o
x
a
a
x a
x
xi
9
y
x a
a
1
a
a
x
aX
x
x
x
7
f f £
1
1
M = { r a | r eR aeN a» 9
= (rT }.
9
Therefore MAT = #AT. M o r e o v e r , i f ra e M n N then r e F n N = 1; hence MAT = M x N. D e f i n i n g a t o be a M a n d (faMf t o be / * we o b t a i n an extension 9
14
N
S / M — i - » G.
T o show t h a t this is equivalent t o the extension we started w i t h , consider the n a t u r a l i s o m o r p h i s m 9: S/M -> E defined b y (sM) = s *. T h e d i a g r a m d
N
S/M
—
G
— -*>
G
a
9 N >-^-->
E d
is c o m m u t a t i v e . F o r a** = (aM) (f a») = f; b u t as = n so (faM) proof. a
e
ae
6e
9
s
a
6e
a
e
= a * = a? a n d (faM) = (fa) * = f = (faMf. T h i s completes n
= the •
W h a t the preceding discussion has achieved is t o reduce the study of extensions t o t h a t o f certain n o r m a l subgroups o f covering groups. There m a y o f course be n o such n o r m a l subgroups, reflecting the fact t h a t exten sions w i t h a prescribed c o u p l i n g d o n o t always exist. H o w e v e r , i f N is abelian, I n n N = 1 a n d R* = 1 b y (7). T h u s [ R , N ] = 1 a n d RN = R x N so t h a t R is a candidate for M . T h e corresponding extension N >-> S/R - » G is equivalent t o the split extension N >-> G x N - » G. 9
Theory of Covering Groups W e propose t o study covering groups i n a somewhat m o r e general context. The outcome w i l l be a classification o f equivalence classes of extensions w i t h given c o u p l i n g .
322
11. The Theory of Group Extensions
L e t there be given groups N a n d U together w i t h a h o m o m o r p h i s m U -> A u t AT, enabling us t o f o r m the semidirect p r o d u c t or covering S = U
group
N.
T h e center o f N shall be denoted b y C. L e t V be a n o r m a l subgroup o f U such that [ C , V~\ = 1. The p r o d u c t FAT is denoted b y L . W e are interested i n the (possibly empty) set M
= Ji{V,
V, AT, f)
of a l l n o r m a l subgroups M o f S w i t h the p r o p e r t y MAT = M x N = L . T h e relative p o s i t i o n o f these subgroups is i n d i c a t e d i n the a c c o m p a n y i n g diagram
S
Each M i n Ji determines a n extension N > - * U S/M —*-»
U/V
(11)
8
where a? = a M a n d (waM) = uV,(ae N,ue U). W h a t is the relevance o f this set Ji t o o u r previous considerations? T h e n o r m a l subgroups M t h a t occur i n 11.1.5 are precisely the elements o f J((F, R, AT, f): notice t h a t [CAT, R] = 1 because < I m N. T h u s we are m o t i v a t e d t o ask w h e n t w o extensions o f type (11) are equivalent.
The Action of Hom^K, C) on Ji I f (p e Hom (V, v
C), so t h a t (p is a U-operator
homomorphism
from
V to C,
define (p'\ L -> L b y the rule (t;a)*' = t;(at;*),
(veV,ae
N).
(12)
11.1. Group Extensions and Covering Groups
323
I t is easy t o check t h a t cp' is a h o m o m o r p h i s m using the fact t h a t belongs t o C a n d [ C , V~\ = 1. T h e inverse o f cp' is clearly ( — (p)\ so cp' is an auto m o r p h i s m o f L . F u r t h e r m o r e q>' is a ^ / - a u t o m o r p h i s m because q> is a 17homomorphism. N o w suppose t h a t MeJi. C l e a r l y L = L * ' = M*' x AT' = A P ' x AT; also M ^ ' o 17. I n other w o r d s M*' e Ji a n d H o m ( F , C) acts o n the set Ji. C o n c e r n i n g this a c t i o n we shall prove the f o l l o w i n g fact. [ /
11.1.6. If Ji = Ji(U, larly on this set.
V, AT, £) is not empty, the group Hom (V,
C) acts
v
regu
9
Proof. Suppose first t h a t M ' = M where M e Ji a n d q> e Hom (V C). I f q>' ^ 1 then q> ^ 0, a n d there is an element v o f V such t h a t ^ 1. O n w r i t i n g v = xa w i t h xe M a n d ae N, we see t h a t M m u s t c o n t a i n x*' = (vaT ) ' = va~ v = xv ; f r o m this i t follows t h a t e M n C = 1, a contra d i c t i o n . I t remains t o prove t r a n s i t i v i t y . v
9
9
1 9
1
9
Choose M a n d M f r o m F o r each u i n 7 there are expressions i; = Xffli, i = 1, 2, where x e M , a e N. Define a f u n c t i o n (p: V-+N by writing = a a ~ . N o w a a n d <s induce the same a u t o m o r p h i s m i n N as v since [ M , AT] = 1; hence e C. I t is o b v i o u s t h a t (p is a h o m o m o r p h i s m : indeed 9 e H o m ^ K C) because (v y = a\a = (v*) i f u e 17. x
2
(
t
t
1
1
2
x
2
f
u
u
u
2
T o complete the p r o o f we verify t h a t M f = M . W e choose x f r o m M w r i t i n g i t i n the f o r m x = va (v e V, a e N). N o w v = m a i = 1, 2, where m e M a e N. Since x = m a a, we have a a = m^x e M n N = 1; thus a = a' . Consequently x ' = (vdf' = vav = vaa a ~ =m eM . Hence Mf < M . Since M f ' e Ji, i t follows t h a t M = M n ( M f ' x N) = Mf. 2
l9
9
i
i9
t
1
1
{
1
i9
i
9
x
9
1
x
1
2
2
2
2
2
2
• Equivalences Classes in ^ ( ( 7 , F , A T , £ ) I n view o f o u r interest i n equivalence o f g r o u p extensions the f o l l o w i n g defi n i t i o n is a n a t u r a l one. T w o elements M a n d M o f Ji are said t o be equiv alent i f the corresponding g r o u p extensions N >-> S / M - » 17/F, i = 1, 2, are equivalent. O b v i o u s l y this is an equivalence r e l a t i o n o n Ji. B u t w h a t are the equivalence classes? I n a t t e m p t i n g t o answer this question one finds t h a t derivations arise i n an essential way. x
2
f
Suppose t h a t 3: U -> C is a d e r i v a t i o n ; denote its r e s t r i c t i o n t o V b y 3. Since [V, C ] = 1, the defining p r o p e r t y o f derivations yields (v v ) = v{v , (v e V); i n short 3: V -> C is a h o m o m o r p h i s m . I f u e U a n d v e V, then b y (4) a n d (5) s
1
t
3
(u-'vu)
3
= ((u-yy^vu)
rlvu
=
1
u
((u*y )~ (v*) u* 3
1
3
3
= (u y (v yu
=
3
(v y.
2
3
2
324
11. The Theory of Group Extensions
u
d u
I t follows t h a t (v f
= {v~ ) , so t h a t 3 e H o m ( F , C). I t s h o u l d be clear t o the [ /
reader t h a t the r e s t r i c t i o n m a p p i n g 3 H-» 3 is a h o m o m o r p h i s m Der(C7, C) -> H o m ^ F , C). T h i s provides us w i t h a n a t u r a l a c t i o n o f Der(C7, C) o n Ji i n w h i c h 11.1.7. / / « ^ is not empty, the equivalence precisely the Der(J7, C)-orbits.
classes
of Ji = Ji(U,
3
=
M '.
V, N, £) are
Proof. Suppose t h a t M a n d M are equivalent elements o f Ji\ then there is a h o m o m o r p h i s m 9 m a k i n g the d i a g r a m x
2
N
>
• S/M
» [7/7
x
e N
>
• S/M
» [7/7
2
c o m m u t a t i v e . W e shall f i n d a 5 i n Der(C7, C) such that M f = M . Let ueU: then by c o m m u t a t i v i t y o f the r i g h t - h a n d square ( w M J maps to wK Hence (uM^) = ucM where c e N is unique. I n fact c e CN = C. F o r i f a G AT, we have ( a W i f = a " M b y c o m m u t a t i v i t y o f the left-hand square i n the d i a g r a m ; thus a M = ((uM ) )~ (aM ) (uM ) , w h i c h becomes aM = aM o n s u b s t i t u t i n g ucM for (uM^) a n d <sM for (aM^) . Hence ( f l " ) f l " e M n N = 1, w h i c h yields (a") = a". Since this is v a l i d for a l l a i n AT, we conclude t h a t c e C. 2
0
6
2
2
u
d
2
u
1
d
1
1
6
2
2
_ 1
d
1
uc
6
2
2
c
c
2
6
s
T h e e q u a t i o n (uM^) = uu M N o w 0 is a h o m o m o r p h i s m , so
therefore determines a f u n c t i o n 3: U - » C.
2
5
( u u f M ) ( u u | M ) = (w w )(w w ) M , 1
2
2
2
U2
w h i c h shows t h a t {u^f T h u s 5 G Der(C7, C).
1
2
1
2
d
= {u{) (u ) —remember
2
here t h a t M
2
2
n C = 1.
T h e next p o i n t t o settle is t h a t 3 maps M t o M under the a c t i o n described above. D e n o t e the r e s t r i c t i o n o f 3 t o V b y 3. I f x e K, then ( x M i f = x x M = x~ 'M b y d e f i n i t i o n o f 5[ see (12)). T h i s also holds i f x e N since i n t h a t case {xM ) = xM a n d x ' = x. T h u s ( x M j f = x ' M is true for a l l x i n L = VN. I f x e M i t follows t h a t x G M a n d M f = M . 1
5
2
s
2
2
e
3
x
< 5
2
2
5
l9
2
2
Conversely, assume t h a t there is a 5 i n Der(C7, C) such t h a t M f = M . Let us show t h a t M a n d M are equivalent. Define 0: S/M -> S / M b y the rules (uM ) = uu M a n d (aM^) = aM where ueU, aeN. I t is simple t o check t h a t ( 1 , 0, 1) is an equivalence f r o m N >-> S / M - » C//F t o N ~ S / M - » U/V. • 2
l
d
1
2
s
1
2
6
2
2
x
2
W e are n o w able t o give a d e s c r i p t i o n o f the equivalence classes o f Ji.
11.1. Group Extensions and Covering Groups
325
11.1.8. There is a bijection between the set of equivalence classes of Ji = Ji(U V N £) and the cokernel of the restriction homomorphism Der(J7, C) -> Hom (V C), provided that Ji is not empty. 9
9
v
9
9
Proof. D e n o t e the image o f the r e s t r i c t i o n m a p b y I . L e t M be any fixed element o f Ji. I f M is any other element o f M there is a unique (p e H o m ( F , C) such t h a t M = M$' ; this is b y 11.1.6. Suppose t h a t M is equiv alent t o M . T h e n M = M~ ' where 3 e D e r ( V , C) b y 11.1.7. Therefore Mfi = M$' *' = M^ ~ since a'jS' = (a + j8)' b y definition (12). I t follows f r o m the r e g u l a r i t y o f the a c t i o n o f H o m ( F , C) t h a t q>^ = (cp + 3)\ where
9
M
M
[ /
d
M
M+
sy
[ /
O V
M
m
F r o m 11.1.5 a n d 11.1.8 i t follows t h a t there is a bijection between the set o f equivalence classes o f extensions o f N by G w i t h c o u p l i n g % a n d the group C o k e r ( D e r ( F , CN) -> H o m ( # , ( N ) )
(13)
F
p r o v i d e d t h a t such extensions exist: here o f course # >-> F - » G is any fixed presentation o f G. W e shall see later t h a t the g r o u p (13) can be identified w i t h the second degree c o h o m o l o g y g r o u p H (G CN). 2
9
EXERCISES
11.1
1. I f (a, /?, y) is a morphism of extensions and a and y are group isomorphisms, prove that p is an isomorphism. 2. Let (a, 0, 1) be an isomorphism from N ^ E - » G to N ^ E - » G. I f these extensions have couplings x d X respectively, prove that x = Z ' where a': Out iV Out N is induced by a: iV iV. a n
a
3. Let G be a group. Prove that G is free i f and only if every extension by G splits. [Hint: Use the Nielson-Schreier Theorem.] 4. Find two isomorphic extensions o f Z b y Z 3
3
x Z which are not equivalent. 3
5. Let A A E G be a group extension with abelian kernel ^4 and G = (g} cyclic of order n. Let g = x with x e £ . A transversal function T: G £ is defined by (g') = x for 0 < i < n. Prove that the values of the corresponding factor set (j) are e
T
l
where
n
a = x e Ker £.
6. Every extension N E G is isomorphic with an extension of the form M E - » G in which M = I m ^ and * is inclusion. 7. Show that there are eight equivalence classes of extensions of Z How many nonisomorphic groups do these give rise to?
2
by Z x Z . 2
2
326
11. The Theory of Group Extensions
8. Let G be an infinite cyclic group and let N be any group. I f yj- G -> Out N is a coupling, prove that some extension of N by G realizes Using the obvious presentation 1 G - » G, show that there is a unique equivalence class of exten sions of N by G with coupling % and that these extensions are split. 9. Let G = <x> be a cyclic group with finite order n and let AT be any group. Sup pose that %: G Out N is a coupling that gives rise to extensions of N by G. Prove that there is a bijection between equivalence classes of extensions of N by G with coupling % and Ker /c/Im v: here v and K are the respective endomorphisms a \-+ ~ and a h-> [a, x ] of 4 = £N. (The action of x on ^4 comes from x) Show also that each such extension is an image of a split extension of N by an infinite cyclic group. 1 + x +
x n l
a
10. Find all equivalence classes of extensions of Q by Z . Identify the groups which arise in this way. 8
2
11. Let N and G be arbitrary groups. For each x i n G let N be a group isomorphic with N via a map a \-+ a . Write £ = Cr N , the cartesian product. I f b e £ and g e G, define b* by the rule ( b ) = b -i. Show that this action of G on £ leads to a semidirect product W = G K £ . (Here is called the standard com plete wreath product N^G and B is the base group of PF). x
x
xeG
x
3
x
xg
12. (Kaluznin-Krasner) Let iV A £ 4> G be any group extension and denote by W the standard complete wreath product N^G. Prove that E is isomorphic with a subgroup of W, so that W contains an isomorphic copy of every extension of N by G. [Hint: Choose a transversal function T: G -> F for the extension. Define y: EPF as follows. I f x e £ , let x = x b(x) where fr(x) is the element of the base group of W given by (b(x)) = ((gx~ ) x(g )~ ) ~\'] y
e
e x
T
1 ,l
g
11.2. Homology Groups and Cohomology Groups T h e purpose o f this section is t o define the h o m o l o g y a n d c o h o m o l o g y groups b y means o f projective resolutions. T h i s m a t e r i a l w i l l be familiar t o those readers w h o have experienced a first course i n h o m o l o g i c a l algebra: they m a y proceed directly t o 11.3.
Complexes L e t # be a r i n g w i t h i d e n t i t y . A right R-complex ^-modules and homomorphisms • C
>C
n+1
•
n
C -i n
C is a sequence o f r i g h t
• • • •,
(n e Z ) ,
infinite i n b o t h directions, such t h a t d d = 0, t h a t is t o say, I m d < K e r d for a l l n. T h e homology H(C) o f the complex C is the sequence o f ^-modules n+1
n
n+1
n
HC n
= KQvd /lmd n
n+u
(neZ),
11.2. Homology Groups and Cohomology Groups
327
w h i c h are usually referred t o as the homology groups o f C . T h u s C is exact i f a n d o n l y i f a l l the h o m o l o g y groups vanish. B y a morphism y f r o m an J^-complex C t o an ^ - c o m p l e x C is meant a sequence o f ^ - h o m o m o r p h i s m s y„: C -> C such that the d i a g r a m n
n
c„
y -i
y +i
n
n
commutes. Define y' \ H C -> H C by the rule (a + I m d )y' = ay + I m d where a e K e r d . N o t i c e that ay e K e r d„ because ay d = ad y -^ = 0 by c o m m u t a t i v i t y o f the d i a g r a m ; also y' is well-defined since ( I m = lm(y d )
n
n
n
n+1
n
n
n
n
n
n
n+1
n
n
n+1
n+1
n+1
n
11.2.1. A morphism y : C - > C of y' : H C -> H C of homology groups. n
n
R-complexes
induces
homomorphisms
n
I t is clear h o w t o define a left R-complex o f left i^-modules. M o s t results w i l l be p r o v e d for r i g h t complexes, b u t they are, o f course, v a l i d for left complexes by corresponding proofs.
Homotopy W e w i s h t o i n t r o d u c e a w a y o f c o m p a r i n g m o r p h i s m s between complexes C a n d C . T w o such m o r p h i s m s y_and £ are said t o be homotopic i f there exist R - h o m o m o r p h i s m s a : C -» C such t h a t n
n
n+1
for a l l z e Z . T h i s m a y be t h o u g h t o f as a sort o f p a r t i a l c o m m u t a t i v i t y o f the d i a g r a m
—check the c o m m u t a t i v i t y statements for the t w o m i d d l e triangles. I t is very easy t o verify that h o m o t o p y is an equivalence relation. T h e f o l l o w i n g fact plays a central role i n the p r o o f o f the uniqueness o f h o m o l o g y a n d c o h o m o l o g y groups.
328
11. The Theory of Group Extensions
11.2.2. / / y: C -» C and £: C -» C are morphisms and £y are homotopic
to identity
morphism for all integers
morphisms,
of R-complexes
such that yt;
then y' \ H C -» H C n
n
n
Proof. B y hypothesis there are ^ - h o m o m o r p h i s m s a : C -> C n
y£ n
-
n
1 = ad n
n+1
+ dG-. N
n
= 1. S i m i l a r l y £ y' n
n
n+1
n
+1
n
n
I f ae K e r d„, then a y £ = a + acr d
N X
I m <9„ . Therefore ay£ + I m <9„ y' ^
is an
iso
n.
+1
n+1
such t h a t = a mod
= a + I m <9„ , w h i c h is j u s t t o say t h a t +1
= 1, so t h a t ^ is the inverse o f y' . n
•
Free and Projective Modules Free i^-modules are defined i n exactly the same w a y as free abelian groups, or free groups for t h a t matter. A n .R-module M is said t o be free on a set X i f there is a m a p p i n g v. X -> M such that, given a f u n c t i o n a: X -> N w i t h N
any .R-module, there is a unique h o m o m o r p h i s m f}\M-+N
such t h a t
a = ip. T h u s the d i a g r a m M
AT commutes. T h e m a p p i n g i is necessarily injective: usually we take i t t o be i n c l u s i o n , so t h a t X c M . T h e f o l l o w i n g statements are p r o v e d i n precisely the same w a y as for abelian groups: a right R-module copies Every
of R , R
is free
if and only if it is a direct sum of
the r i n g R regarded as a r i g h t .R-module by m u l t i p l i c a t i o n .
R-module
is an image of a free-module
A n .R-module M is said t o be projective
(see 2.3.8 a n d 2.3.7). if, given an ^ - h o m o m o r p h i s m
a: M -> N a n d an ^ - e p i m o r p h i s m s: L -> AT, there is an ^ - h o m o m o r p h i s m /?: M -> L such that a = /?£, t h a t is t o say, the d i a g r a m M / /
$ / L commutes. Every
—*—>
a N
free module is projective
• 0 (see the p r o o f o f 4.2.4). A projec
tive m o d u l e is n o t i n general free, b u t merely a direct s u m m a n d o f a free m o d u l e (see Exercise 11.2.6). A c o m p l e x is said t o be free i f a l l o f its modules are free, a n d projective i f all o f its modules are projective.
11.2. Homology Groups and Cohomology Groups
329
Resolutions A complex C is called positive i f C = 0 for n < 0: the complex is then w r i t t e n • • • -» C -» C -» C -» 0. L e t M be a r i g h t P - m o d u l e . B y a ng/z£ Ir resolution o f M is meant a positive r i g h t P - c o m p l e x C and an e p i m o r p h i s m e: C -» M such that n
2
x
0
0
•• •
• C —
d
2
—
C
—
0
M
• 0
is exact. W e m a y abbreviate the r e s o l u t i o n t o C 4 » M . T h e resolution is said t o be free {projective) i f C is free (projective).
11.2.3. Every
R-module
has a free
R-resolution.
Proof L e t M be any P-module. T h e n there exists an e p i m o r p h i s m e: C - » M w i t h C is free. L i k e w i s e there exists a h o m o m o r p h i s m 3 : C -» C where 0
0
X
x
0
I m 3 = K e r e a n d C is free. So far we have an exact sequence C -> C M -» 0. Clearly this procedure can be repeated indefinitely t o produce a free resolution of M . • X
x
x
0
11.2.4. Let P -4» M P -4» M be two projective R-resolutions. If a: M -» M is aw R-homomorphism, there is a morphism n: P -» P swc/z t/zat 7i e = ea. Moreover any two such it's are homotopic. 0
P
— i - »
M
i i
7i
|
a
P — i - » M. Proof. Since P is projective a n d e surjective, there is a h o m o m o r p h i s m 7i : P -» P such t h a t n & = ea. 0
0
0
0
0
M Suppose that we have constructed h o m o m o r p h i s m s %{. P -» P for such that n^i = diit^. T h e n we have 5 7 i 5 = 5 5 7 i _ that Im(<9 7r ) c K e r d = I m <9 . B y p r o j e c t i v i t y o f P there homomorphism 7 i : P -+ P such that 7 r < 9 = d n, as {
n + 1
n+1
n
n
n + 1
n + 1
N
N
n+1
n+1
h
N + 1
n+1
n+1
n+1
n+1
n
n
n
1
i = 1, 2, = 0, so exists a one can
11. The Theory of Group Extensions
330
see f r o m the d i a g r a m
d n + 1
P
>
n+1
0.
lmd
n+1
Hence the f o l l o w i n g d i a g r a m is c o m m u t a t i v e : d n + 1
p
) p
r
n+1
>
r
Pn-l
n
M
••••Pl n
n+1 |
n
1
d
"
P
+
01
v P > r
1
r
Pn-l
r
n +l
0
Pl
0.
M
P
0
Consequently we have defined recursively a m o r p h i s m 7 i : P - » P w i t h the right property. Now
_
suppose t h a t n': P -» P is a n o t h e r such m o r p h i s m . L e t p: P -» P be
the m o r p h i s m given by p = n — n' . T h e n the f o l l o w i n g d i a g r a m commutes n
n
n
— i g n o r e the d i a g o n a l maps for the present:
>
P
n
+
J ^ P
2
n
+
J ^
1
P
P — P
n
t
—^—• • M M
0
,0
/ / /
Pn + 2
Gn+l
/
/
/
/
S
dn+2
p"
v
^
^ +1
Pl Since p e = 0, we have I m p 0
homomorphisms + Gd t
o \ P -+ P {
i+1
t
-» P
0
such t h a t a d
x
0.
By projectivity of P
1
0
• M
^0
< K e r e = Imd .
0
there exists a h o m o m o r p h i s m cr : P pi = did^i
0
Po
/_
/ p
°b
Pl
Pn + 1
/
0
0
= p . Suppose t h a t
1
0
have been constructed i n such a w a y t h a t
i+1
for i = 0, 1 , n .
true w h e n i = 0 i f we i n t e r p r e t a_ G
(Pn + 1 ~~ dn + l n)dn + l
=
(Refer n o w t o the diagram.) T h i s is
as 0. N o w
1
^n + lPn
G
~ ^n + l n^n + l
= dn + lidnVn-l
G
G
+ n^n + l) ~ ^n + l n^n+l
= 0. Hence
Im(p
n + 1
—d a) n+1
n
< Ker d
a
d +i n> n
o
r
Pn+i
=
G
= Im d .
n+1
yields a h o m o m o r p h i s m
a :
n+2
P
n+1
-» P
n+1
G
dn+i n + n+i^n+2
a
n+2
s
The
such
projectivity
that
a
n + 1
d
n + 2
of = p
n+1
P
n+1
—
required. T h u s the o have been c o n n
structed for a l l n, a n d n a n d n' are h o m o t o p i c by d e f i n i t i o n .
•
The Homology Groups H (G, M) n
Let
G be any g r o u p a n d M any r i g h t G-module (that is t o say, r i g h t Z G -
m o d u l e ) . Consider the a d d i t i v e g r o u p o f integers Z regarded as a t r i v i a l left
11.2.
Homology Groups and Cohomology Groups
331
G-module; this means t h a t elements o f G operate o n Z like the i d e n t i t y map. By 11.2.3 there is a left projective r e s o l u t i o n P - • Z . B y tensoring each m o d u l e o f P w i t h M over Z G a n d t a k i n g the n a t u r a l induced maps we o b t a i n a complex o f Z-modules > M(g)
Z G
P
^ ^ M ®
n + 1
Z
G
P
^ - > M ®
n
Z
G
P
n
_ ,
>
w h i c h we call M ( x ) P : here o f course (a (g) b)d' = a (g) (bd ), ae M, b e P . T h i s complex w i l l usually n o t be exact. Define the nth homology group of G with coefficients in M t o be the abelian g r o u p Z G
n
H {G,
M) = H (M®
H
n
n
n
P).
ZG
Since Z has m a n y projective resolutions, the following remark is essential. 11.2.5. Up to isomorphism the homology the projective resolution P -» Z .
groups H (G, M) are independent n
of
Proof. L e t P 4 » Z and P Z be t w o left projective ZG-resolutions o f Z . A p p l y i n g 11.2.4 w i t h 1: Z -» Z for a, we o b t a i n a m o r p h i s m n: P -» P such that 7i e = e. S i m i l a r l y there is a m o r p h i s m 7 i : P - » P such that 7r e = e. T h e n nn: P -» P has the p r o p e r t y (7r 7i )e = e; o f course so does the i d e n t i t y m o r p h i s m 1: P - » P. I t follows f r o m 11.2.4 t h a t nn is h o m o t o p i c t o 1; the same is true o f nn. W e deduce that n'n' and n'n' are h o m o t o p i c t o i d e n t i t y m o r p h i s m s where n': M ( x ) P -» M ( x ) P a n d n'\ M ( x ) P -» M ( x ) P are the n a t u r a l induced m o r p h i s m s i n w h i c h (a (x) b ) ^ = a (x) (bn ) a n d (a (x) b ) ^ = a (x) (b7i ). F i n a l l y 11.2.2 shows that the m a p n' induces an iso morphism 0
0
0
0
Z G
Z G
Z G
Z G
n
n
n
H {M® P)-*H {M® P). n
ZG
n
•
ZG
n
The Cohomology Groups H (G, M) Let G be any g r o u p a n d M any right G-module. L e t P 4 » Z be a ng/tf p r o jective Z G - r e s o l u t i o n o f Z . F o r m the new complex H o m ( P , M), that is, ••• >Hom (P _ A f ) — Hom (P , A f ) — (P„ , M) , •• • G
H o m
G
n
1 ?
G
n
G
+ 1
n
n
where 5 is defined i n the n a t u r a l way, by c o m p o s i t i o n ; thus (<x)5 = d <x where a e H o m ( P _ M ) . Each H o m ( P , M) is a Z - m o d u l e . N o w the c o m plex H o m ( P , M) w i l l usually be inexact. The nth cohomology group of G with coefficients in M is the abelian g r o u p n
G
n
1 ?
G
n
G
n
H (G,
M) = F T ( H o m ( P , M)). n
G
Just as for h o m o l o g y we can prove independence o f the resolution. 11.2.6. Up to isomorphism of the projective resolution
the cohomology P -» M.
n
groups H (G, M) are
independent
332
11. The Theory of Group Extensions
W h e n a projective r e s o l u t i o n o f Z is k n o w n , the h o m o l o g y a n d h o m o l o g y groups m a y be read off f r o m the f o l l o w i n g result. 11.2.7. Let G be a group and M a G-module. tive ZG-resolution
of Z , and denote lm(d :
0
(ii) If P is a right complex sequence n
1 ?
M)
n
>M ®
n
G
n
and M is a right G-module,
• H (G, M)
Hom (P _
that P -» Z is a projec
P -» P „ _ i ) by J .
n
(i) / / P is a left complex quence
Suppose
J
ZG
there is an exact
• M ®
n
and M is a right
G-module,
n
• H (G,
n
n
x
there is an
M)
exact
• 0.
Proof, (i) L e t v: P -» J be the o b v i o u s m a p p i n g a \-+ ad , and let i: J be inclusion. T h e n we have the c o m m u t a t i v e picture w i t h an exact r o w n
n
n
P 1
Pn
N+
—
•
se
P_.
ZG
• H o m ( J , At) G
co-
n
P_ n
1
0
Jn
Tensor each m o d u l e w i t h M, keeping i n m i n d the r i g h t exactness p r o p e r t y o f tensor products. W e o b t a i n an induced c o m m u t a t i v e d i a g r a m w i t h an exact r o w . M® P ZG
- ^ U M® P
n+1
ZG
M® J
n
ZG
> 0
n
M® P . ZG
n
T h u s K e r v' = I m d' . W e c l a i m t h a t ( K e r d' )v' = K e r i'. I f a e K e r d' , then 0 = ad' = av'i by c o m m u t a t i v i t y o f the second d i a g r a m . T h u s av' e K e r i. Conversely let b e K e r i' and w r i t e b = cv' where cs M ® P . Then 0 = bi' = cv'i = cd' , so c E K e r d' a n d b e ( K e r d' )V. Hence V induces an i s o m o r p h i s m f r o m K e r <9 /Im d' t o K e r i'. B u t H (G, M) = K e r d' /Im d' , so we o b t a i n an i s o m o r p h i s m o f H (G, M) w i t h n+1
n
n
n
ZG
n
n
n
n
n
n
n+1
n
n+1
n
Ker(z': M ®
ZG
J
n
• M ®
ZG
P ^\ n
as called for. (ii) T h e p r o o f is similar. Remark:
•
Left modules versus right
modules
I t is also possible t o define H (G, M) as H (P ® M) where P is a right projective r e s o l u t i o n and M is a left m o d u l e . S i m i l a r l y H (G, M) m a y be n
n
n
11.3. The Gruenberg Resolution
333
defined t o be j r 7 ( H o m ( P , M)) where P is a left projective r e s o l u t i o n a n d M is a left m o d u l e . U p t o i s o m o r p h i s m the same groups are obtained. The basis for this is the fact t h a t any left m o d u l e A over a g r o u p G can be regarded as a r i g h t m o d u l e A' over G b y means o f the rule ag = g~ a (ae A,g e G); for details see Exercises 11.2.10 a n d 11.2.11. n
1
EXERCISES
11.2
(In the first five exercises R is a ring with identity.) * 1 . Prove that a right ^-module is free if and only if it is a direct sum of copies of RR.
*2. Any ^-module is an image of a free ^-module. *3. Free modules are projective. 4. Consider an extension of ^-modules, that is, an exact sequence of ^-modules and ^-homomorphisms 0->A-^B->C^>0. (a) Prove that there is an ^-homomorphism y: C B such that ye = 1 if and only if I m ^ = Ker £ is a direct summand of B. (b) Prove that there is an ^-homomorphism /?: B A such that pf} = 1 if and only if I m ^ = Ker £ is a direct summand of B. (The extension is said to split if these equivalent properties hold.) 5. Prove that the following properties of an ^-module M are equivalent: (a) M is projective. (b) Every extension 0 - > ^ 4 - > £ - > M - > 0 splits. (c) M is a direct summand of a free ^-module. 6. Use Exercise 11.2.5 to give an example of a projective module that is not free. 7. I f R is a principal ideal domain, prove that every projective ^-module is free. 8. Let M be a free G-module and let H < G. Prove that M is a free H-module. 9. Prove 11.2.6. 10. Prove 11.2.7(h). 11. Let P be a right G-complex and let M be a left G-module. I f A is a left (right) G-module, let A' be the corresponding right (left) G-module where ag = g~ a (or ga = ag' ). Prove that H (P (x) M ) ^ H„(M' (x) P') where F is the complex whose modules are P^. 1
1
n
ZG
ZG
12. I f P is a left G-complex and M is a left G-module, prove that H „ ( H o m ( P , M ) ) ^ H ( H o m ( F , M')) i n the notation of the previous exercise. ZG
n
Z G
11.3. The Gruenberg Resolution N a t u r a l l y 11.2.7 is o f l i t t l e value u n t i l we have some explicit m e t h o d o f w r i t i n g d o w n a projective Z G - r e s o l u t i o n o f Z . There is a w a y o f d o i n g this whenever a presentation o f the g r o u p G is given.
334
11. The Theory of Group Extensions
Augmentation Ideals L e t G be any g r o u p . T h e n there is an o b v i o u s e p i m o r p h i s m o f abelian groups e: Z G - + Z such t h a t g 1 for a l l g i n G: this is k n o w n as the augmentation o f Z G . I t is easy t o check t h a t e is a r i n g h o m o m o r p h i s m . T h u s the kernel o f e is a n ideal o f Z G ; this ideal, w h i c h is o f great i m p o r t a n c e , is denoted by
h, n
the augmentation ideal o f Z G . O b v i o u s l y I consists o f a l l r = YjgeG Q such t h a t YjgeG( g) = 0- N o w such an r can be r e w r i t t e n i n the f o r m r = Yji^geG g( ~ 1)? conversely any such r has coefficient sum equal t o 0, so i t belongs t o I . T h u s G
g
n
n
a
G
I
G
=
< g - l \ l * g e G \
the additive group generated b y a l l the g — 1 # 0. Indeed i t is easy t o see t h a t I is a free abelian g r o u p w i t h the set {g — 111 # g e G} as basis. T h e f o l l o w i n g p r o p e r t y o f the a u g m e n t a t i o n ideal o f a free g r o u p is fundamental. G
11.3.1. If F is a free group on a set X, then I the set X = {x - l\x e X}.
is free as a right F-module
F
on
Proof. L e t a: X -» M be a m a p p i n g t o some F - m o d u l e M . B y d e f i n i t i o n o f a free m o d u l e i t suffices t o prove t h a t a extends t o an F - h o m o m o r p h i s m fi: Ip -> M. F i r s t o f a l l let oc': F -» F tx M be the g r o u p h o m o m o r p h i s m w h i c h sends x i n X t o (x, (x — l ) a ) . T o each / i n F there correspond f i n F a n d a i n M such that f ' = (f a). N o w i t is clear f r o m the d e f i n i t i o n o f a' that f = f T h u s a function 5: F -» M is determined b y the e q u a t i o n / ' = ( / , f ) . N e x t for any f , / i n F we have l
a
l9
x
a
x
3
2
(/1/2)"' = /i"'/ "' = C/i, /i')(/ , / / ) = (/1/2, (/1V2 + 2
2
3
5
s
o
i n view o f the additive nature o f M . Hence (f f ) = [fi)fi + /2 that d: F M is a d e r i v a t i o n . K e e p i n g i n m i n d that 7 is free as an abelian g r o u p o n the set { / — 111 # / e F } , we construct a h o m o m o r p h i s m fi:I -+M o f abelian groups b y w r i t i n g ( / - I)ft = f . N o w (x - l)P = x = (x - l ) a because x"' = (x, (x — l ) a ) ; thus is an extension o f a t o I . F i n a l l y is an F h o m o m o r p h i s m because 1 2
?
F
F
3
3
F
((/ -
=
-
1) -
(/1 - !))/» =
-
- (/1 -
f
= (jQ i)'-/i' = (/')/i = (/-iWi-
•
11.3. The Gruenberg Resolution
335
As a n a p p l i c a t i o n o f 11.3.1 let us p r o v e t h a t the h o m o l o g y a n d c o h o m o logy groups o f a free g r o u p vanish i n dimensions greater t h a n 1. 11.3.2. If F is a free group and M is any F-module, H (F, n
M)for
n
then H (F, M) = 0 =
alln>l.
Proof. B y 11.3.1 the complex • • • 0 0 I Z F Z is a free Z F r e s o l u t i o n o f Z . U s i n g this r e s o l u t i o n a n d the definitions o f H (F, M) a n d H (F, M) we o b t a i n the result. • F
n
n
Relative Augmentation Ideals L e t G be any fixed g r o u p a n d suppose t h a t N is a n o r m a l subgroup o f G. T h e assignment g \-> gN determines a n e p i m o r p h i s m o f abelian groups f r o m Z G t o Z(G/N) w h i c h is easily seen t o be a r i n g h o m o m o r p h i s m . L e t its kernel be denoted b y
T h i s m a y be t h o u g h t o f as a generalization o f the a u g m e n t a t i o n ideal since
h = L e t / denote the ideal I (ZG) = (ZG)I ; then clearly / < I . W e regard ZG/I as a G-module v i a r i g h t m u l t i p l i c a t i o n . I f x e N, g e G a n d r e ZG, then (r + I)gx = rg + / since x — 1 e L W e m a y therefore t u r n ZG/I i n t o a G/N-module v i a the rule (r + I)gN = rg + I . I t follows t h a t I must act t r i v i a l l y o n ZG/I, o r I < I . There results the equalities N
N
N
N
N
I
= I (ZG)
N
N
=
(ZG)I . N
Hence I is the r i g h t ideal o f Z G generated b y a l l x — 1 where l ^ x e i V . T h e next result generalizes 11.3.1. N
11.3.3. Let R be a normal subgroup of a free group F. If R is free on X, then I is free as a right F-module on {x — 1 | x e X}. R
x
a
Proof. Suppose t h a t Yjxex( ~ ^) x = 0 where a e ZF. Choose a transver sal T t o R i n F; then ZF = Dv (ZR)t, so we m a y w r i t e a = Y.terK,^ where b e ZR. Hence YjteT(Yjxex( — ^)K,t) = ®- O b v i o u s l y this means t h a t Yjxex( — i)b ,t 0 f ° every t. Since I is free o n the set o f a l l x — 1 by 11.3.1, i t follows t h a t b = 0. • x
teT
x
x
t
x t
x
=
r
x
R
XJ
O n e final p r e p a r a t o r y l e m m a is needed. 11.3.4. Let R F G be a presentation of a group G. Suppose that S and T are right ideals of ZF that are free as F-modules on X and Y respectively. Then:
336
11. The Theory of Group Extensions
(i) S/SI is free as a G-module on {x + SI \x e X}; (ii) ST is free as an F-module on {xy\x e X, y e Y} provided R
sided
R
that T is a 2-
ideal n
Proof, (i) I n the first place S/I is a G-module v i a the a c t i o n (5 + SI )f = sf + SI ; this is well-defined because srf = sf + s(r — l)f = sf m o d SI where se S,f e F,r e R. Since S = Dv x(ZF), we have SI = Dr xI , and R
R
R
R
xeX
R
S/SI
R
~ D r x{ZF)/xI
R
xeX
R
~ Dr
xe X
x(Z(F/R)).
xeX
(ii) C l e a r l y
ST = D r x(ZF)T = D r xT = D r xyZF. jce-X"
xe X
•
xe X yeY
11.3.5 (The G r u e n b e r g Resolution). Let R F -» G be a presentation group G. Then there is a free right G-resolution of Z
of a
1
• 7s//r — - hiryhiR — - / r v / s — > • • • • ••—*i\ni—-
irhihii—>i /ii—*
ip/hh—-
R
> Z.
The mappings here are as follows: ZG - » Z is the augmentation, I /I I -» Z G is induced by n: F -+ G and all other mappings are natural homomorphisms. F
F R
Proof. B y 11.3.1 a n d 11.3.3 b o t h I a n d I are free F-modules. A p p l y i n g 11.3.4 we see that the modules I I /I I a n d I /I w h i c h appear i n the complex are free G-modules. N o w check exactness. T h e kernel o f Z G - » Z is I , w h i c h is also the image o f I /I I -+ZG. T h e kernel o f the latter m a p is I /I I since I is the kernel o f ZF - » Z G ; the image o f I /I -»I /I I is also I /I I . A n d so o n . • F
R
+1
F R
G
R
F
F R
F R
+1
R
R
F R
R
R
R
R
F
F
R
F R
There is o f course a c o r r e s p o n d i n g left r e s o l u t i o n i n w h i c h the I appears on the right t h r o u g h o u t . I n the language o f category theory 11.3.5 sets u p a functor f r o m presentations t o resolutions—see Exercise 11.3.8. F
The Standard Resolution Let G be any g r o u p a n d let F be the free g r o u p o n a set {x \l # g e G } . Recall t h a t the assignment x H-> g gives rise t o a presentation R F G called the standard presentation. W e propose t o examine the G r u e n b e r g r e s o l u t i o n t h a t arises f r o m this presentation: i t is k n o w n as the standard resolution. g
g
For convenience define x t o be 1; t h e n the set {x \g e G} is a transversal to R i n F. Since the n o n t r i v i a l x are also free generators o f F, i t is clear t h a t l
g
g
11.3.
The Gruenberg Resolution
337
this is a Schreier transversal i n the sense o f 6.1. I t follows f r o m 6.1.1 (and its p r o o f ) t h a t R is freely generated b y the elements
By
11.3.3 the y
— 1 f o r m a set o f free generators f o r the free F - m o d u l e
g u d 2
I . Define R
(01,
Gl)
=
Xg gi
~
2
X Xg gi
where g belongs t o G. C l e a r l y (g t
=
2
g) e I
l9
2
(1 -
R
y
,g )Xg
g i
2
a n d (g
g ) = 0 i f and only i f g
l9
2
x
or g equals 1. M o r e o v e r (15) shows t h a t the nonzero (g 2
l9
ate the F-module
(15)
i g 2
g )'s freely
gener
2
I . R
N e x t we define for n > 0 the symbols (gi\g \'"\Q2n)
= {QU Gl){Q^
2
Gin) + I R
G^)' " {Qln-U
+
1
and ( G I \ G \ ' ''\Gm-i)
= (1 - x )(g ,
2
gi
These are elements o f P
3
+ 1
=I /I
2N
Observe t h a t P
g )(g ,
2
R
g )"
4
and P
R
(g - ,
5
g -i)
2n 2
2n
1
_ = I IR^ II I
2 n
1
F
hh-
respectively.
F R
a n d P n - i together w i t h P = Z G are the terms o f the
2N
2
0
G r u e n b e r g r e s o l u t i o n (11.3.5). W e deduce f r o m 11.3.4 t h a t P
2N
erated as a G-module b y the {gi\g \"-\g ) 2
2
and P « - i *
2n
as a G-module b y the {g±\g \"'\g -i) (Gi\G2\"'\Gn)
+
s
2
is freely gen
freely generated
where 1 ^ g e G. I t is evident t h a t
2n
t
= 0 i f some & equals 1.
There is a useful f o r m u l a describing the h o m o m o r p h i s m s w h i c h appear i n the standard r e s o l u t i o n . 11.3.6. The homomorphism
d : P -» P _ n
N
N
1
which occurs
in the standard
resolu
tion is given by n-1 (Gl\'"\Gn)8
= (G2\'"\Gn)
n
+
i
Z
(-^) (9l\"'\9i-l\9i9i
+ l\9i +
2\"'\9n)
i=l +
(-mGl\-'\Gn-l)Gn.
Proof. Consider the statement
w h e n n = 1. B y d e f i n i t i o n (g) = 1 — x +
I I ,
agrees w i t h
F
g
so (g)d = 1 — g, w h i c h
R
1
a
(9r\'"\9 )
s
1
s
whenr >
the f o r m u l a
5.
N e x t consider the case n = 2. N o w d : P -+ P 2
x x g
g i
+ II t o x X
g i S 2
- xx gi
2
X
maps ( # i l # ) = x 2
g i 9 2
—
+ I I . The identity
g2
F
R
!
X
M2 ~ **1**2 = ( - J - (! -
shows t h a t ( ^ l ^ ) ^ = ( # ) - {g^g ) 2
Now
i f we i n t e r p r e t
+ U " **X
2
+ ( # i ) # , as predicted.
2
2
let n = 3. B y d e f i n i t i o n <9 : P -» P maps 3
3
2
(01I02I03) = (1 - ^ H ^ , G3) + V * to
(1 — x )(g , gi
2
g) + I 3
R
= (1 — x )(g \g ). gi
2
3
O n the other h a n d , the value
11. The Theory of Group Extensions
338
predicted by the f o r m u l a is (02I03) -
(QiQilds)
+
(dildids)
~
(di\g )93, 2
w h i c h is j u s t X
( 9203
~~
X
X
X
92 d3^
~~ ( 9 l 9 9 3
X
~
2
X
X
9l92 03^
( 919293
~~
After cancellation this becomes (1 — x )(x gi
(1 -
X
~
X
( 9i92
X
~
9 l
—xx)
g2g3
g2
X
9i 9293^
X
9
2
)
+ I
93
X
9 3
+
or
R
x )(g \g ). gi
2
3
I f n > 3, i n d u c t i o n m a y be used t o reduce t o one o f the cases already dealt w i t h — f o r details see [ b 2 9 ] .
•
Cocycles and Coboundaries Let P - » Z be the standard Z G - r e s o l u t i o n o f Z . L e t us use this r e s o l u t i o n t o n
calculate H {G, M) where M is any r i g h t G-module. O n e has t o take the h o m o l o g y o f the complex H o m ( P , M ) ; i n this the h o m o m o r p h i s m s are the G
n+1
S : Now
H o m ( P „ , M)
>Hom (P
G
G
,M).
B + 1
P is the free G-module o n the set o f a l l (g \g \'" n
1
\g )> Qi ^ 1- T h u s a
2
n
if/ i n H o m ( P , M) is determined b y its value at {gi\g \"'\9n)> G
n
conversely
2
these values m a y be chosen a r b i t r a r i l y i n M t o produce a if/. Hence elements \j/ o f H o m ( P , M) correspond t o functions cp: G x • • • x G -» M such t h a t G
n
n
(g ...,
g )cp = 0 i f some g equals 1; the correspondence is given b y
l9
n
t
(dl\"-\9n)^
=(9l,~.,9n)
Such functions
corresponds t o (il/)S 0> 2
n + 1
n+1
= d \l/. =
l9
g )cp = 0 i f some n
o n H-cochains is easy t o discover:
n+1
(
U s i n g 11.3.6 we conclude t h a t
n+1
• • • > 9n+i)(p$
N o t e t h a t (g ...,
(9 >
• ••,
2
Qn+i)(p
n
+
Z
(~
+ (-l)
• • • > Qi-n 9i9i+u B + 1
• • • , #n+i)
(0i ...^»)^^» i. J
(16)
+
I f we w r i t e n
n+1
Z (G, M ) = Ker d
n
and
n
£ (G, M ) = Im d,
then n
n
/ T ( G , M ) ~ Z ( G , M)/B (G,
M).
(17)
11.3. The Gruenberg Resolution
339
n
n
Elements o f Z ( G , M) are called n-cocycles
w h i l e those o f B (G, M) are
n-coboundaries. l
F o r example, suppose t h a t n = 1. I f cp e Z (G, M ) , then (16) shows t h a t 0 = (g )
o
r
+ (0i)
2
=
(0i02)
an element a o f M , then {g)(pS
1
= a ( l — g). T h u s the 1-coboundary {^S
is
j u s t a n inner d e r i v a t i o n . W e have therefore made the f o l l o w i n g identifica tion: X
H (G,
M) = D e r ( G , M ) / I n n ( G , M).
(18)
I f cp is a 2-cochain, the c o n d i t i o n f o r cp t o be a 2-cocycle is (0i,020s)
= (9i9 ,93)
2
for a l l
+ (0i,0 )
2
2
i n G. W e recognize this as the factor set c o n d i t i o n (6) encountered
i n 11.1. 11.3.7. Let G be a finite
group of order m. Suppose
that M is any
G-module.
n
Then m • H (G, M) = 0 for all n > 0. Proof. L e t cp. G x • • • x G -» M be a n y H-cochain. There is a corresponding n
(n — l ) - c o c h a i n \j/ given b y (02,
= Z
(x,g ...,g )(p. 29
n
xeG
Sum the f o r m u l a (16) over a l l 34 = x i n G t o get Z
(x,0 , ...,# 2
n + 1
)(?(S
n + 1
=m((0 ,...,0 2
B + 1
)<jo)
xeG n
+ Z
(-^J(92,"^9i-u9i9i u9i 2,"^9n M +
+
+
i=2 n+1
0n + l W + ( - l) (02,
~ (03, • • •, n
n
•••,
0„W * 0„+l.
n
I f (/> e Z ( G , M ) , this becomes mcp = t/f(S , so t h a t mcp e B (G, M) a n d n
m-H (G M)
= 0.
9
•
T h i s has the f o l l o w i n g useful c o r o l l a r y . 11.3.8. Let G be a finite which is uniquely
group
of order m. Suppose
that M is a
G-module
n
divisible by m. Then H (G, M) = 0 for all n > 0. n
n
Proof. L e t cp e Z (G, M ) : then mcp = \j/d for some (n — l ) - c o c h a i n \j/ b y 11.3.7. Since M is u n i q u e l y divisible b y m, i t is meaningful t o define \j/ t o be (l/m)ij/.
n
T h e n \j/ is a n (n — l ) - c o c h a i n a n d cp = tyd" e B (G, M). Thus
# « ( G , M ) = 0.
•
11. The Theory of Group Extensions
340
The corresponding results for h o m o l o g y are also true b u t require a dif ferent p r o o f (see Exercise 11.3.10).
EXERCISES
11.3
* 1 . For any group G show that I {g-\\\±geG}.
G
is a free abelian group on the set G — 1 =
2. Let G = <#> be an infinite cyclic group. Using the presentation 1 >-> G - » G write down the left and right Gruenberg resolutions for G. I f M is a G-module, show that H°(G, M) ~ M ~ H ( G , M) and H\G, M)~M ~ H (G, M) where M = {ae M\ag = a} and M = M/M(gr - 1). G
t
G
0
G
G
3. I f F is a free group and M is an F-module, use the Gruenberg resolution to calculate H ° ( F , M ) and H\F M). 9
4. The same question for H (F, M) and H ( F , M ) . 0
t
5. Let G = <#> be a cyclic group of finite order m. I f F is an infinite cyclic group, show that the presentation F >-> F -» G determines a resolution m
•••
• ZG-?->
TG^->
ZG-^->
ZG^->
ZG—U Z
where a and /? are multiplication by g — 1 and 1 + # + • • • + 0
m _ 1
• 0, , respectively.
6. Use the previous exercise to calculate the (co)homology of a cyclic group G of order m. I f M is any G-module, then for n > 0 2n
l
H ~ (G,
M) ~ Ker jff/Im a
2n
and
H (G,
M) ~ Ker a/Im fi.
Also H (G, M ) ^ H
W + 1
n
( G , M).
7. Let G be a finite cyclic group and let M be a finite G-module. Assume that H\G, M) = 0 for some fixed i > 0. Prove that H"(G, M) = 0 for all n > 0. [Use Exercise 11.3.6.] 8. (a) Let R ^ F -» G, i = 1, 2, be two presentations of a group G. Prove that there is a morphism (a, /?, 1) from R ^ F G to R ^ F G. (b) Prove that any such morphism of presentations of G determines a morphism of the correspondence Gruenberg resolutions. (c) The association of a Gruenberg resolution with a presentation determines a functor from the category of presentations of G to the category of free Gresolutions of Z. t
t
x
1
2
2
9. Let G be the union of a countable chain of groups G < G < • • •. Let M be a G-module such that H ( G , M ) = 0 = H (G M) for all f and for some fixed n. Prove that H (G, M) = 0. t
n
2
n+1
£
i9
n+1
10. Let G be a finite group of order m and let M be any right G-module. By adopting the following procedure (due to R. Strebel) prove that m • H (G, M) = 0 i f n > 0. (a) Let F be a free left ZG-resolution of Z. Define morphisms of complexes a: M ® F -+ M ® F and 0: M ® F -+ M ® F by {a ® tya, = X ^ G ^ ® g r ^ and (a (g> b)p = a<S)b where a e M and b e F . n
Z G
z
z
Z G
1
t
t
11.4.
341
Group-Theoretic Interpretations of the (Co)homology Groups
(b) Show that a/3 induces homomorphisms H (M (x) F) H (M (x) F) which are simply multiplication by m on each module. (c) Prove that M (x) F is exact, noting that if A B - » C is an exact sequence of free abelian groups, then M®A^M(g)B-»M(g)C is an exact se quence (with the obvious maps). (d) Deduce from (b) and (c) that m • H (G, M) = 0. n
ZG
n
ZG
z
n
11.4. Group-Theoretic Interpretations of the (Co)homology Groups T h e group-theoretic significance o f the h o m o l o g y a n d c o h o m o l o g y groups i n l o w dimensions w i l l n o w be discussed.
The Groups H ( G , M) and H°(G, M) 0
T o c o m p u t e H (G, obtaining 0
•••
M)
we tensor the left G r u e n b e r g r e s o l u t i o n by M ,
• M ®
(I /I I )
ZG
F
• M ®
R F
ZG
ZG
• 0.
Now M ® Z G S M v i a the m a p p i n g a g ) g H-> ag. Therefore b y definition of the h o m o l o g y groups H (G, M) is i s o m o r p h i c w i t h M / M / , the largest G-trivial quotient of M; this is usually w r i t t e n M . I n a similar m a n n e r i t m a y be s h o w n t h a t Z
G
0
G
G
FT°(G, M)~{ae
M\ag = a, V# e G } , G
the set of G-fixed points of M , w h i c h is often w r i t t e n M n o r m a l closures being unlikely). T h u s we have 11.4.1. / / G is a group and M a right G-module, FT (G, M ) - M 0
and
G
(confusion w i t h
then
FT°(G, M ) -
G
M.
The Group if^G, M) 11.4.2. / / G is a group and M a right G-module, then H (G, M) is isomorphic M ® I -*M in which X
with the kernel of the homomorphism a®{g
-
ZG
1)
M>
G
a(g - 1).
This follows at once w h e n 11.2.7 is a p p l i e d t o the G r u e n b e r g resolu t i o n . There is a neat f o r m u l a for # i ( G , M ) w h e n G operates t r i v i a l l y o n M. I n order t o derive this we take note o f a result w h i c h has some interest i n itself.
11. The Theory of Group Extensions
342
11.4.3. If G is any group, then I /I G
^ G
G
a b
.
Proof. Since I is free o n {g — 111 # g e G} as a n abelian g r o u p , the m a p p i n g x — 1 i—• xG' determines a h o m o m o r p h i s m I - » G . N o w the i d e n t i t y (x — l ) ( j / — 1) = (xy — 1) — (x — 1) — (y — 1) implies that I is m a p p e d to 0. Hence there is a n induced h o m o m o r p h i s m q>: I / I G ~ + G ((x — 1) + I )(p = xG'. O n the other hand, x H-> (X — 1) + I is a h o m o m o r p h i s m f r o m G t o I /I b y the same i d e n t i t y . Therefore i t induces a h o m o m o r p h i s m G -» 7 / w h i c h is clearly inverse t o cp. • G
G
a b
G
S U C N
G
T N A T
AB
G
G
G
a b
G
G
G
11.4.4. / / M is a trivial right G-module, then H (G, M)
G
X
a b
.
Proof. B y 11.4.2 we have / ^ ( G , M ) ~ M ( x ) I . I t is seen t h a t the assign ment a (x) (x — 1) i—• a (x) ((x — 1) + I ) yields a n i s o m o r h i s m Z G
G
G
M ®
Z
G
2
{I II \
I ~>M® G
G
Z
G
The result n o w follows f r o m 11.4.3.
•
The Group H \ G , M) X
A n i n t e r p r e t a t i o n o f H (G, M) as the q u o t i e n t g r o u p D e r ( G , M ) / I n n ( G , M) has already been m e n t i o n e d d u r i n g o u r discussion o f the standard resolu t i o n . A n o t h e r approach t o this i m p o r t a n t result w i l l n o w be given based o n a l e m m a for w h i c h we shall find other uses. 11.4.5. / / G is a group and M a G-module,
then D e r ( G , M ) ^ H o m ( / , M ) . G
G
Proof. I f S e D e r ( G , M ) , we m a y define a h o m o m o r p h i s m 8*: I -+ M b y G
(g — 1)(S* = gb, keeping i n m i n d t h a t I set {g - 1|1 =£geG}.
For g
l9
G
is free as a n abelian g r o u p o n the
g i n G we have (g - l)g 2
x
= {g g
2
1
2
- 1 ) -
(Qi ~~ 1)> w h i c h implies t h a t l
( ( ^ i - )QiW
= (9i9 )S 2
- #2<S = (9iS)g
2
= ((9i -
1)<**)02-
Hence 8* e H o m ( / , M ) . Conversely, i f 9 e H o m ( / , M ) , define g9* = (g — 1)6; a similar c a l c u l a t i o n shows t h a t 6* e D e r ( G , M ) . O f course 3 H-> S* a n d 9 \-+ 9* are inverse mappings. I t is equally clear that 8 \-+ 8* is a, h o m o m o r p h i s m o f abelian groups. Thus 8 i—• 8* is a n i s o m o r p h i s m . • G
G
G
11.4.6. / / G is a group and M a G-module, H\G,
G
then
M) ~ D e r ( G , M ) / I n n ( G , M ) .
Proof. A p p l y i n g 11.2.7 t o the G r u e n b e r g r e s o l u t i o n we o b t a i n the exact sequence H o m ( Z G , At) G
• H o m ( / , At) G
G
1
• H (G, At)
• 0.
(19)
11.4.
Group-Theoretic Interpretations of the (Co)homology Groups
343
A c c o r d i n g t o 11.4.5 the m i d d l e g r o u p is i s o m o r p h i c w i t h D e r ( G , M). N o t e t h a t cp e H o m ( Z G , M ) is completely determined b y (l)cp = a i n M ; for then G
(g)cp
= ag. T h e image o f cp i n H o m ( / , M ) is the restriction o f cp t o 7 , G
G
G
w h i c h corresponds t o the d e r i v a t i o n g H-> (g — l)cp = a(g — 1), t h a t is, t o a n inner d e r i v a t i o n . T h u s I n n ( G , M ) corresponds t o the image o f the left h a n d m a p p i n g i n (19). I t follows t h a t the sequence 0
• Inn(G, M )
1
> Der(G, M )
> H (G, M )
>0
(20)
is exact. T h e required i s o m o r p h i s m is a consequence o f (20).
•
C o m b i n i n g 11.4.6 w i t h 11.1.3 we o b t a i n the next result. 11.4.7. Let G be a group
and M a G-module.
the
product
associated
X
H (G
semidirect
All complements
G K M are conjugate
of M in
if and only
if
M) = 0.
9
MacLane's Theorem We
come n o w t o the c o n n e c t i o n between the second c o h o m o l o g y g r o u p
a n d extension theory. First, however, a simple observation must be made. 11.4.8. Let R
F -» G be a presentation
i—• (r — 1) H~ I F I R is a G-isomorphism
rR'
ah
(f~ f)R'
R
R
F
rir
2
R
h
to
I /I I . R
F R
is a G-module v i a c o n j u g a t i o n i n F : thus (rR')
where r e R a n d / e F . L e t T = I / I I
r K (r - 1) + I I
a
fn
Proof. O f course R lr
of a group G. Then the mapping from
F
=
a n d consider the m a p p i n g
R
f r o m R t o T. N o w i f r , r are elements o f R, x
2
- 1 = ^ - 1 ) + ( r - 1) + (r - l ) ( r - 1) 2
=
(r
x
-
1) + ( r
This shows t h a t r K (r - 1) + I I F
R
2
x
-
1)
2
mod
I I . F
R
is a h o m o m o r p h i s m o f groups. Since T
is abelian, there is a n i n d u c e d h o m o m o r p h i s m 9: R
ah
-» T.
Let us check t h a t 9 is a G - h o m o m o r p h i s m ; f
((rR') y
= (f-hfRf = f~\r
= (f-'rf
- 1) +
- \)f + l l F
= (r-l)f+I I
F R
R
=
F R
II
((rR'f)r.
F i n a l l y , i n order t o show t h a t 9 is a n i s o m o r p h i s m we shall produce an inverse. L e t R be free o n X. B y 11.3.3 f o r left ^ - m o d u l e s we have I
R
I (x F
Z,
= Dr ZF(xxeX
1), so
that
T =I /I I R
F
equals
R
— 1) a n d hence is i s o m o r p h i c w i t h Dv {ZF/I )(x xeX
we conclude
that
T is the free
abelian
Dr
x e X
( Z F ) ( x - 1)/
— 1). Since ZF/I
F
F
group
^
o n the set o f a l l
344
11.
(x — 1) + I I , xe F
R
The Theory of Group Extensions
X. Hence there is a g r o u p h o m o m o r p h i s m
such t h a t ((x — 1) + I I R ) ^ F
11.4.9 ( M a c L a n e ) . Let R F ^ G be a presentation any right G-module. Then there is an exact sequence 1
H (F,
M)
T -»
R
ah
= xR'. O b v i o u s l y \j/ is the inverse o f 9.
• Hom (£ G
a b
of a group G. Let M be
2
, M)
•
• # (G, M)
• 0.
%
Proof. Here M is an F - m o d u l e v i a 7i, so t h a t qf = af where a e M , / e F. A p p l y 11.2.7 using the G r u e n b e r g r e s o l u t i o n associated w i t h the given pre sentation. There results an exact sequence Hom (I /I I , G
F
F R
M)
2
• Hom (/,// /„ M) G
> H (G,
f
M)
• 0.
(21)
Recall t h a t the left-hand m a p is i n d u c e d b y the i n c l u s i o n I / I I -»I /I I . By 11.4.8 we have I / I I ^ K , so the m i d d l e g r o u p i n the exact sequence is i s o m o r p h i c w i t h H o m ( K , M ) . W e claim that R
R
F
R
G
R
F
F
R
a b
Hom {I /I I , G
F
a b
F
M) - H o m ( / , M ) .
F R
F
F
To see this let a: I M be an F - h o m o m o r p h i s m . T h e n a maps I I t o 0; for, o n the basis o f the t r i v i a l a c t i o n o f R o n M , we have ( ( / — l ) ( r — l ) ) a = ( / — l ) a • (r — 1) = 0. Hence a induces a h o m o m o r p h i s m a: I / I I -» M . I t is clear t h a t a i—• a is an i s o m o r p h i s m , so the assertion is true. N e x t D e r ( F , M ) ^ H o m ( / , M ) b y 11.4.5. Hence (21) becomes F
F
F
F
Der(F, M )
F
R
R
F
2
• Hom (K , M) G
• H (G, M)
a b
• 0.
(22)
W e have, o f course, t o keep t r a c k o f the left-hand m a p p i n g ; i t is the obvious one d i—• d' where d' is i n d u c e d i n R b y S. I f S is inner, there is an a i n M such t h a t (rR'f = a( — r + 1) = 0 for a l l r i n this is because R operates t r i v i a l l y o n M . Consequently I n n ( F , M ) maps t o 0. T h e result n o w follows f r o m 11.4.6. • ah
The Second Cohomology Group and Extensions Let G a n d N be groups a n d let %: G -» O u t AT be a c o u p l i n g o f G t o N. L e t us assume t h a t there is at least one extension t h a t realizes Recall f r o m 11.1.8 t h a t there is a bijection between the set o f equivalence classes o f extensions o f N by G w i t h c o u p l i n g x d the cokernel o f D e r ( F , C) -» H o m ( # , C); here C is the center o f N. N o w H o m ( # , C) ~ H o m ( # , C) a n d R acts t r i v i a l l y o n K ; thus H o m ( K , C) ^ H o m ( K , C). W e have t o deal w i t h the cokernel o f the obvious m a p p i n g D e r ( F , C) -» H o m ( K , C); b y (22) this is i s o m o r p h i c w i t h H (G, C). a
n
F
a b
a b
F
G
a b
G
2
W e have p r o v e d a fundamental theorem.
a b
11.4. Group-Theoretic Interpretations of the (Co)homology Groups
345
11.4.10. Let G and N be groups and let % be a coupling of G to N. Assume that x is realized by at least one extension of N by G. Then there is a bijection between equivalence classes of extensions of N by G with coupling x d elements of the group H (G, C) where C is the center of N regarded as a G-module via xan
2
W h e n N is abelian, this reduces 11.1.4, w h i c h was p r o v e d b y using factor sets.
Extensions with Abelian Kernel I t has been p o i n t e d o u t t h a t w h e n N is abelian, every c o u p l i n g o f G t o N is realized b y some extension, for example the split extension, w h i c h cor responds t o 0 i n H (G, N). I t is w o r t h w h i l e being m o r e explicit i n this i m p o r t a n t case. 2
11.4.11. Let G be a group, A a G-module and R F G any presentation of G. Let A e H (G, A) and suppose that cp is a preimage of A under the mapping Hom (R , A) -» H (G, A) of MacLane's Theorem. Then 2
2
G
ah
A > is an extension whose equivalence
which induces
> (F K A)/R?' the prescribed
class corresponds
» G G-module
structure
in A and
to A.
Here A is regarded as an F - m o d u l e v i a n.F-*G. T o comprehend 11.4.11 i t is necessary t o l o o k back at the p r o o f o f 11.1.8. W e take R t o be the fixed element o f M i n t h a t proof, w h i c h is possible since A is abelian. T h e equiva lence class o f R? corresponds t o cp + / where / is the image o f D e r ( F , C); this cp + I corresponds t o A. T h e next result is essentially the abelian case o f the Schur-Zassenhaus T h e o r e m (9.1.2). 11.4.12. Let G be a finite group of order m. Suppose that A is a such that each element of A is uniquely divisible by m. Then every of A by G splits and all complements of A are conjugate.
G-module extension
T h i s follows directly f r o m 11.3.8, 11.4.7, a n d 11.4.10.
Central Extensions A n extension C A E - » G is called central i f I m p is contained i n the centre of E. I n this case G operates t r i v i a l l y o n C, so t h a t C is a t r i v i a l G-module.
346
11. The Theory of Group Extensions
Such extensions are frequently encountered; for example every n i l p o t e n t g r o u p can be constructed f r o m abelian groups by means o f a sequence o f central extensions. Suppose t h a t C is a t r i v i a l G-module a n d let R F - » G be a presenta t i o n o f G. L e t us interpret M a c L a n e ' s T h e o r e m i n this case. O f course F , like G, operates t r i v i a l l y o n C, a n d D e r ( F , C) = H o m ( F , C) ~ H o m ( F , C). a b
Also Hom (£ G
a b
, C) - H o m ( K / [ F , K], C)
because [ F , R]/R' must m a p t o 1 under any G - h o m o m o r p h i s m f r o m R t o C. As a consequence o f these remarks M a c L a n e ' s T h e o r e m takes the following form. ah
11.4.13. If C is a trivial G-module and R group G, t/zere is aw exact sequence H o m ( F , C) Thus
2
• H o m ( K / [ K , F ] , C)
a b
each central
V'.R/tR,
F - » G is a presentation
extension
• # ( G , C)
of C by G arises
from
of the
• 0.
a
homomorphism
F]-+C.
Abelian Extensions A central extension o f one abelian g r o u p by another is usually n o t abelian —consider for example the q u a t e r n i o n g r o u p Q . Suppose t h a t G is an abelian g r o u p a n d C a t r i v i a l G-module. C a l l the extension C E -» G abelian i f E is an abelian g r o u p . There are, o f course, always abelian exten sions, for example the direct p r o d u c t extension, C C x G ^» G. 8
T h e equivalence classes o f abelian extensions correspond t o a certain subset E x t ( G , A) o f t f ( G , A). L e t us use 11.4.11 t o determine w h i c h central extensions are abelian; i n the sequel G is an abelian g r o u p . As usual choose a presentation R F -» G here o f course F' < R. B y 11.4.11 a central extension o f C by G is equiva lent t o one o f the f o r m 2
C >
> ( F x Q/R*
» G 9
where cp e H o m ( K , C). T h i s extension is abelian i f a n d o n l y i f F' < R . Let / e F' a n d r e R; i f / = r ^ ' = r r ^ , we observe t h a t r* = r~ f e F n C = l ; thus f = r a n d f = r = 1. Hence (/> maps F ' t o 1. Conversely, i f
a b
l
G
a b
2
11.4. Group-Theoretic Interpretations of the (Co)homology Groups
11.4.14. Let G and C be abelian groups. sentation
Suppose
that R
F - » G is a pre
of G. Then there is an exact sequence of abelian H o m ( F , C)
• Hom(R/F\
a b
C)
347
groups
• E x t ( G , C)
• 0.
This follows o n c o m b i n i n g the preceding remarks w i t h 11.4.13.
The Schur Multiplicator Let G be any g r o u p a n d let Z be regarded as a t r i v i a l G-module. F o r brevity it is customary t o w r i t e HG
= H,{G, Z ) ,
n
the integral homology group o f degree n. F o r example H G by 11.4.4. The g r o u p
~ Z (x) G
1
M(G) =
a b
~ G
a b
HG 2
is k n o w n as the Schur multiplicator o f G. I t plays a p r o m i n e n t role i n Schur's theory o f projective representations. W e shall see that i t is also relevant t o the t h e o r y o f central extensions. There is an interesting f o r m u l a for M ( G ) . 11.4.15 ( H o p f ' s F o r m u l a ) . If R then
F -» G is a presentation
M ( G ) ~ F'nR/[F, In particular
this factor
of a group
G,
R].
does not depend on the
presentation.
Proof. A p p l y 11.2.7 t o the left G r u e n b e r g resolution. There results M(G) - Ker(Z ®
Z G
(I /I I ) R
>Z ®
R F
(I /I I )).
Z G
F
N o w I /I I is i s o m o r p h i c as a left G-module w i t h R by of 11.4.8—here R is t o be regarded as a left G-module f (rR') = {frf-^R'. Also Z ( x ) R ~ R/[F, R] via n
R F
ab
ab
R
Z G
ZG
F
F
R F
ab
F
the left version via the a c t i o n ^ r [ F , R]. I n 11.4.3 we have n
a b
M ( G ) - K e r ( / ? / [ F , /?] the m a p F'nR/\_F,R\.
R F
being the
obvious
one,
r [ F , R]
• F ), a b
rF'.
The
kernel is clearly •
I n the case o f abelian groups there is a simpler f o r m u l a for the Schur m u l t i p l i c a t o r . I f G is an abelian group, define G
A
G = (G
348
11. The Theory of Group Extensions
where D = (g (x) g\g e G>. T h i s is called the exterior denotes g ® g l
Qi
A
+ D, t h e n (g
2
l
a
Qi = ~(QI
+ g)
A (g
2
square o f G. I f g
A g
l
+ g ) = 0, w h i c h
l
2
shows
2
that
T h u s G A G m a y be regarded as a skew-symmetric
tensor p r o d u c t . 11.4.16. / / G is aw abelian group, then M ( G ) ~ G A G. Proof. and [/
Let K
F - » G be a presentation o f G. Since G is abelian, F' < R
M ( G ) ^ F y [ F , .R] b y H o p f ' s
1 ?
/ ] [ F , K ] , (feF\
f o r m u l a . The m a p p i n g
(fiR,f R)^ 2
is well-defined a n d bilinear. Consequently there is
2
an i n d u c e d h o m o m o r p h i s m {F/R) (x) {F/R) -» F' [ F ,
by the fundamental
m a p p i n g p r o p e r t y o f the tensor p r o d u c t . W h a t is m o r e , fR (x) fR maps t o 1, so there is i n d u c e d a h o m o m o r p h i s m 9: (F/R) A (F/R) -» F ' / [ F , f , R
fR
A
is sent t o [fu
2
fi\[F,
i n which
R].
T o construct an inverse o f 9 choose a set o f free generators { x for F . B y Exercise 6.1.14 the g r o u p F'/y F tors
the
(j) : F'/y F 0
/IJ f
E
/I-R
where
3
i < j.
Hence i ?
2
there
is
X / ] y F to xR 3
(
a
homomorphism
A XJR. N O W for any
F we see from b i l i n e a r i t y o f the c o m m u t a t o r t h a t ( [ / i , / 2 ] y 3 ^ ) ^ ° =
2 A
[X-, Xj]y F,
- » ( F / K ) A (F/R) sending [ x
3
x ,...}
1 ?
is free abelian w i t h free genera
3
fiR-
Hence <^ maps [ F , R]/y F 0
t o 0, a n d so there is an i n d u c e d
3
h o m o m o r p h i s m (j>: F ' / [ F , R] -» ( F / K ) A ( F / K ) . Clearly 0 a n d <j> are inverse functions.
•
F o r example, i f G is an elementary M ( G ) ~ G A G is an elementary
abelian p - g r o u p o f r a n k r, then
abelian p - g r o u p o f r a n k ( ) 2
(Exercise
11.4.9). H o p f ' s f o r m u l a can be used t o associate an exact sequence o f h o m o l o g y groups w i t h any extension. T h i s w i l l be applied i n Chapter 14 t o the study o f one-relator groups. 11.4.17 (The F i v e - T e r m H o m o l o g y Sequence). Corresponding extension
N
M(E) This
sequence
E -4>> G there is an exact M(G) is natural
N/IE,
making the
AT| •
in the following
from N>-^E^»GtoN>-^E^»G
group
1.
E» sense. Given
a morphism
there are induced homomorphisms
(a, /?, y) a^,
diagram
M(E)
M(G)
N/IE,
AT|)
M(E)
M(G)
N/{E,
AT|)
commutative.
to any
sequence
J
ab
^ab
1
11.4. Group-Theoretic Interpretations of the (Co)homology Groups
349
Proof. Here o f course we can take AT t o be a subgroup o f £ as a matter o f convenience. T h e mappings A / [ F , AT] - » F a n d F -» G are the obvious ones, x [ F , AT] H-> xE' a n d xE' i—• x G ' respectively. Exactness at F a n d G is easily checked. I t remains t o construct the other t w o mappings a n d check exactness at M ( G ) a n d A / [ F , AT]. a b
a b
a b
£
a b
a b
Let K F £ be some presentation o f F . T h e n ns: F -» G is a presen t a t i o n o f G; let K e r TIE = 5. B y H o p f ' s f o r m u l a M ( F ) ~ F' n K / [ F , # ] a n d M ( G ) ~ F ' n S / [ F , S ] . N o w S is the preimage o f A under 7i, so i n fact R < S a n d S/K ~ A . The relevant subgroups are situated as follows:
Define M ( G ) -> A / [ F , A ] b y means o f x [ F , 5 ] ^ A ] and M ( G ) ~ F ' n S / [ F , S ] . The image o f this m a p p i n g is clearly E' n A / [ F , A ] , w h i c h is the kernel o f A / [ F , A ] -» F . T h i s establishes exactness at A / [ F , A ] . a b
F i n a l l y , define M ( F ) -» M ( G ) b y means the n a t u r a l h o m o m o r p h i s m f r o m F'nR/[F, K] t o ( F n # ) [ F , S ] / [ F , S ] , together w i t h H o p f ' s f o r m u l a . The kernel o f M ( G ) -+ A / [ F , A ] corresponds t o ( F n # ) [ F , S ] / [ F , S ] , so the se quence is also exact at M ( G ) . W e shall n o t take the space t o establish n a t u r a l i t y (see however Exercise 11.4.16). • r
T o conclude this discussion o f the Schur m u l t i p l i c a t o r we m e n t i o n an i m p o r t a n t result t h a t illustrates the relationship between the m u l t i p l i c a t o r a n d the theory o f central extensions. A p r o o f is sketched i n Exercise 11.4.5. 11.4.18 ( U n i v e r s a l Coefficients Theorem). / / G is a group and M a G-module, there is an exact sequence Ext(G , M ) > a b
2
> H (G,
M)
trivial
» Horn(M(G), M ) .
T h e m a p p i n g o n the r i g h t shows t h a t every central extension o f M by G gives rise t o a h o m o m o r p h i s m f r o m M ( G ) t o M . A n i m p o r t a n t special case occurs w h e n G is perfect; for then E x t ( G , M ) = Oand a b
2
H (G,
M) ~ H o m ( M ( G ) , M ) ,
11. The Theory of Group Extensions
350
so equivalence classes o f central extensions o f M a n d G stand i n one-to-one correspondence w i t h h o m o m o r p h i s m s f r o m M ( G ) t o M.
The Third Cohomology Group and Obstructions F i n a l l y we shall address the t h i r d m a j o r p r o b l e m o f extension theory. G i v e n groups A a n d G together w i t h a c o u p l i n g %: G - • O u t A , w h e n does there exist a n extension o f A by G w h i c h realizes the c o u p l i n g W e shall w r i t e C for the centre o f A , keeping i n m i n d t h a t x prescribes a G-module structure for C. Choose any presentation R
F
G o f G. Just as i n 11.1—see e q u a t i o n
(7)—we can find h o m o m o r p h i s m s £ a n d n w h i c h m a k e the d i a g r a m R
>
•
F
*
—
G
*
*
A — — » Aut A — O u t
(23)
A
c o m m u t a t i v e : here T is the c o n j u g a t i o n . h o m o m o r p h i s m a n d v the n a t u r a l homomorphism. I f r e R
?
a n d / e F , the element
1
(r* )~ ((r
/ _ 1
?
/
y )*
surely
belongs t o A . L e t us a p p l y the f u n c t i o n T t o this element, observing t h a t a
T
T
a
rjr = <J by c o m m u t a t i v i t y o f the d i a g r a m , a n d t h a t ( x ) = ( x ) i f a e A u t A . We obtain 1
/
1
1
( r « r ( ( r " ) ^ = ( r « r ^ = 1, so t h a t the element 1
/*r
/
1
= (rT ((r " r)
/ {
(24)
belongs t o K e r T, t h a t is t o C, the center o f A . U s i n g the d e f i n i t i o n (24) i t is completely s t r a i g h t f o r w a r d t o verify the formulae f*(r r ) 1
= (f*r )(f*r )
2
1
(25)
2
and ,
(fif2)*r Let
= (f *r)(f *r'*- yi. 2
(26)
1
us n o w consider the G r u e n b e r g r e s o l u t i o n t h a t is associated
with
the chosen presentation o f G. Suppose t h a t F is free o n a set X a n d R is free o n a set Y. Since Iplz/Iplz (1 — x)(l
— y) + I I R , (xe F
is free as a G-module o n the set o f a l l
X,ye
Y\ b y 11.3.3 a n d 11.3.4, the assignment
(\-x)(\-y)
+ I Il\
>x*y
P
determines a G - h o m o m o r p h i s m 2
ils:I I /I I F R
F
>C.
R
N e x t one observes t h a t ((1 — x ) ( l — r) + I I R ) I I / = x*r, F
(X e X,r
e R). T h i s
11.4. Group-Theoretic Interpretations of the (Co)homology Groups
351
follows b y (25) a n d i n d u c t i o n o n the l e n g t h o f r, together w i t h the e q u a t i o n
(1 _
x
)
i _ r)
(
ri
= (1 -
2
X
) ( ( l - rj + (1 - r ) - (1 - r )(l - r )). 2
t
2
I n a similar w a y one can prove t h a t ( ( l - / ) ( l - r ) + /
F
7 i W = / T
(27)
for a l l / i n F a n d r i n R; this m a y be accomplished by using i n d u c t i o n o n the l e n g t h o f / , e q u a t i o n (26) a n d also the e q u a t i o n (1 " / i / ) ( l " r) = (1 - / ) ( 1 - r) + (1 - / , ) ( 1 2
2
l
r* )f . 2
1
1
1
I f ^ i ? , the definition (24) tells us t h a t r * r = ( r j ) " " ^ ) ' ) ^ = (rlY ^ = 1. E q u a t i o n (27) n o w shows t h a t ^ maps I /I I t o 1. Conse q u e n t l y \j/ induces a h o m o m o r p h i s m
2
1
R
F R
q> e Uom (I I /ll G
F R
R
C)
F R
the obstruction determined b y the c o u p l i n g %. T h e o b s t r u c t i o n q> w i l l i n general depend o n the choice o f £ a n d n i n (23); for these are n o t unique. H o w e v e r , a c c o r d i n g t o 11.2.7, there is an i s o m o r p h i s m o f fl" (G, C) w i t h the cokernel o f the h o m o m o r p h i s m 3
Hom (I /ll G
R
C)
• Hom (I I /ll G
F R
C).
(28)
3
T h u s x determines an element A o f H (G, C). T h e i m p o r t a n t facts a b o u t A are as follows.
11.4.19. Let x be a coupling of G to N. Let n be homomorphisms as in (23) leading to an obstruction
F o r a p r o o f o f 11.4.19 we refer the reader t o [ b 2 9 ] , §5.5. I t is n o w possible t o give a f o r m a l c r i t e r i o n for a c o u p l i n g t o be realiz able i n an extension.
11.4.20. Let G and N be groups and suppose that x is a coupling of G to N. Then there is an extension of N by G with x as its coupling if and only if X corresponds to the zero element of H (G, C) where C is the centre of N regarded as a G-module by means of x3
Proof. Suppose t h a t % is realized by an extension N E - » G. As usual let R F G be a fixed presentation. There is a h o m o m o r p h i s m o: F -• E
11. The Theory of Group Extensions
352
t h a t lifts 7 i ; thus R
>
• F
N
>
>E
—
G
» G
is a c o m m u t a t i v e d i a g r a m . Define F - • A u t A t o be the composite o f a w i t h the c o n j u g a t i n g h o m o m o r p h i s m E - • A u t A , a n d let n be the restric t i o n o f £ t o R. T h e n the d i a g r a m R >
•
F
n A
G
{
z
— — » Aut A — O u t
is commutative. This £ and I f / e F a n d r e i ^ , then f*r
—
(29) A
w i l l do very well to construct an o b s t r u c t i o n cp.
= {r°y\{rf-yy
=
(rT^f'THr'")'/* 1
= (r*)- ^
= 1. 3
Hence cp = 0 a n d # corresponds t o the zero element o f FT (G, C). (Here one s h o u l d remember t h a t the element o f FT (G, C) determined by x does n o t depend o n £ a n d n b y 11.4.19). Conversely suppose t h a t cp corresponds t o 0 i n FT (G, C). T h e n 11.4.19(h) shows that £ a n d n m a y be chosen so t h a t (29) is c o m m u t a t i v e a n d f*r=l for a l l feF a n d r e R. T h i s means t h a t r = ((r ~ ) ) \ w h i c h , w h e n r is replaced b y r , becomes 3
3
n
f
1 T1 f
f
( /)n r
=
( iy\
(30)
r
1
N o w define M = { r f r " " ) " ^ e R}, a subset o f F ^ A = 5. B y using is easy t o show that the m a p p i n g r \-+ r ( r ~ ) ' is a h o m o m o r p h i s m operator groups. Hence M < i FM. A l s o for x i n A we have x ~ " = x * (rr-^ | ^ j i V ] = 1. Therefore M < i F A = S M A = M x N = RN, so M e Jt{F, R, A , £). Finally A S / M - » G is an extension w h i c h has c o u p l i n g x1
?
r(r
x
=
x
s
n c e
=
t
l u s
1}
r
(30) i t of F~ = and
(r
1){
•
Extensions of Centerless Groups 3
Consider the case where A has t r i v i a l centre. T h e n FT (G, C) = 0 a n d every c o u p l i n g o f G t o A arises f r o m some extension o f A b y G. M o r e o v e r u p t o equivalence there is o n l y one such extension since FT (G, C) = 0. O f course 2
11.4. Group-Theoretic Interpretations of the (Co)homology Groups
353
each equivalence class o f extensions gives rise t o a unique c o u p l i n g , as we saw i n 11.1.1. W e sum u p o u r conclusions i n the f o l l o w i n g f o r m : 11.4.21. Let N be a group with trivial center and let G be any group. Then there is a bijection between the set of all equivalence classes of extensions of N by G and the set of all couplings of G to N. A t the other extreme is the case where N = C is abelian a n d x is essen t i a l l y a h o m o m o r p h i s m f r o m G t o A u t N. W e can define £ = nx and n = 1, o b t a i n i n g i n (29) a c o m m u t a t i v e d i a g r a m . U s i n g this £ a n d n i n the defini t i o n we get / * r = 1 for a l l / a n d r. T h i s means t h a t every c o u p l i n g x corre sponds t o the zero element o f # ( G , C), a conclusion t h a t is h a r d l y surpris i n g since x is realized b y the semidirect p r o d u c t extension. 3
W e m e n t i o n w i t h o u t g o i n g i n t o details the f o l l o w i n g a d d i t i o n a l fact. Sup pose t h a t G is a g r o u p a n d A a G-module. I f A e H (G, A), there exists a g r o u p N whose center is i s o m o r p h i c w i t h A, a n d a c o u p l i n g x'. G - • O u t N w h i c h is consistent w i t h the G-module structure o f A a n d corresponds t o A as i n 11.4.19. O f course x is realizable b y an extension o f N by G i f a n d o n l y i f A = 0. A fuller account o f the t h e o r y o f obstructions m a y be f o u n d i n [b29]. 3
W h i l e group-theoretic interpretations o f the c o h o m o l o g y groups i n dimensions greater t h a n 3 are k n o w n , n o really c o n v i n c i n g applications t o g r o u p t h e o r y have been made.
EXERCISES
11.4
1. Let G be a countable locally finite group and let M be a G-module which is uniquely divisible by every prime that divides the order of an element of G. Prove that H (G, M) = 0 if n > 1 [use 11.3.8 and Exercise 11.3.9]. Show that H ^ G , M) need not be 0 by using 11.4.7. n
2. Let N < ] E and assume that C (N) = ( N = C say. Put G = E/N and write e for the extension N >-> E - » G. Denote by A(N) the set of all automorphisms of E that operate trivially on N and G. (a) Prove that A{N) is an abelian group which is isomorphic with Der(G, C). [Hint: Show that if y e A{N), then [ £ , y] < C ] (b) Prove that A(N) n I n n E maps to Inn(G, C) under the above isomorphism, and deduce that A(N)/A(N) n I n n E ~ H ^ G , C). E
3. I n the notation of the preceding problem let Aut e denote the group of auto morphisms of E that leave N invariant. Put Out e = Aut e/Inn E. Show that there is an exact sequence 0
• H ^ G , C)
• Out e
• Out N.
[Hint: Let y e Aut e and consider the restriction of y to N . ]
354
11. The Theory of Group Extensions
4. (R. Ree). Let £ be a noncyclic finitely generated torsion-free nilpotent group. Prove that Out £ contains elements of infinite order. [Hint: Let N be a maxi mal normal abelian subgroup of £ : then N = C (N). Assuming the result false, argue that H (G, N) is a torsion group where G = E/N. N o w find an M such that N < M and E/M infinite cyclic. Show that H (E/M, N ) is torsion and apply Exercise 11.3.2.] E
1
l
M
5. I f M is a trivial G-module, establish the existence of the Universal Sequence 2
E x t ( G , M) >—> H ( G , M) ab
Coefficients
» Hom(M(G), M).
Show also that the sequence splits [Hint: Choose a presentation of G and apply 11.4.13 and 11.4.14.] 2
6. I f G is a finite group, show that M(G) is finite and that M(G) ~ H ( G , C*) where C* is the multiplicative group of nonzero complex numbers operated upon trivially by G. [Hint: Use Hopf's formula to show that M(G) is finitely generated. Apply Exercise 11.3.10.] 7. Let G be an abelian group. Prove that every central extension by G is abelian if and only if G A G = 0. Prove also that a finitely generated group has this prop erty precisely when it is cyclic. n
*8. I f G is an elementary abelian group of order p , then M(G) is elementary abelian of order 9. I f G is free abelian of rank n, then M(G) is free abelian of rank ( 2 ) . 10. According to 11.4.18 a central extension C >-> E - » G determines a homomor phism 5: M(G) -> C called the differential. Prove that I m 5 = E' n C. Deduce that 5 is surjective i f and only if C < E'. (Such an extension is called a stem extension.) [Hint: The given extension is equivalent to one of the form C >-> (C x F)/^' - » G where R >-> £ - » G is a presentation of G and ^ e H o m ( # , C).] F
11. Prove that Schur's theorem (10.1.4) is equivalent to the assertion that M(G) is finite whenever G is finite. (Use Exercise 11.4.10.) 12. A central extension C >-* E - » G is called a stem ccwer of G i f its differential is an isomorphism. Prove that every stem cover of G is isomorphic to a stem cover M(G) >-> £ - » G with differential equal to 1. [ H w t : Let ^ e Hom(R/[R, £], C) determine the given stem cover, R >-> £ - » G being a presentation of G. I f 5 is the differential of a stem cover C >-> £ - » G, consider e Hom(#/[\R, £ ] , M(G)).] 13. Prove that every stem extension is an image of a stem cover (i.e., there is a morphism (a, /?, 1) from a stem cover with a and /? surjective). 14. Prove that there is a bijection between the set of isomorphism classes of stem covers of G and E x t ( G , M(G)). Construct four non-isomorphic stem covers of Z x Z . [Use Exercise 11.4.13.] ab
2
2
15. A finite group G has a unique isomorphism class of stem covers i f and only if (|G,„|,\M(G)\) = 1.
11.4. Group-Theoretic Interpretations of the (Co)homology Groups
355
*16. (a) Establish the naturality of the 5-term homology sequence (see 11.4.17). [Hint: Start with presentations R >-> F -» E and R >-> F - » £ and lift y: G G to £ : F - > F . ] (b) Prove also that (77)* = 7*7* where 7: G G and 7: G G are homo morphisms. (In the language of category theory this says that M( —) is a functor.) [Hint: Show that 7* does not depend on the particular lifting £.] a n
17. Let G be the group with generators x x , x d defining relations ^ 2 ] = [ * 2 > ^ 3 ] = •*• = [ x „ - i , x„] = 1. Using Hopf's formula, show that M(G) is free abelian of rank n — 1. l9
2
n
CHAPTER 12
Generalizations of Nilpotent and Soluble Groups
I n Chapter 5 we f o u n d numerous properties o f finite groups w h i c h are equi valent nilpotence—see especially 5.2.4. F o r example, n o r m a l i t y o f a l l the Sylow subgroups is such a p r o p e r t y . W h e n applied t o infinite groups, these properties are usually m u c h weaker, g i v i n g rise t o a series o f wide general izations o f nilpotence. F o r soluble groups the s i t u a t i o n is similar. T h e a i m of this chapter is t o discuss the m a i n types o f generalized n i l p o t e n t a n d soluble groups a n d their interrelations.
12.1. Locally Nilpotent Groups I f & is a p r o p e r t y o f groups, a g r o u p G is called a locally 0>-group i f each finite subset o f G is contained i n a ^ - s u b g r o u p o f G. I f the p r o p e r t y & is inherited b y subgroups, this is equivalent t o the requirement t h a t each finitely generated subgroup have 0>. O u r first class o f generalized n i l p o t e n t groups is the class of locally nilpotent groups. I t is easy t o see t h a t images a n d subgroups o f a locally n i l p o t e n t g r o u p are locally n i l p o t e n t . There are certain properties o f n i l potent groups w h i c h are o f a local character i n the sense that they are state ments a b o u t finite sets o f elements; such properties are i n h e r i t e d b y locally n i l p o t e n t groups. F o r example, there is the f o l l o w i n g result. 12.1.1. Let G be a locally nilpotent group. Then the elements of finite order in G form a fully-invariant subgroup T (the torsion-subgroup of G) such that G/T is torsion-free and T is a direct product of p-groups.
356
12.1. Locally Nilpotent Groups
357
T h i s follows i m m e d i a t e l y f r o m 5.2.7. N o t e that infinite p-groups are n o t i n general n i l p o t e n t (Exercise 5.2.11), or i n fact even locally n i l p o t e n t b y a famous example o f G o l o d [ a 5 9 ] . One rather obvious w a y o f c o n s t r u c t i n g locally n i l p o t e n t groups is t o take the direct p r o d u c t o f a family o f n i l p o t e n t groups: i f the n i l p o t e n t classes o f the direct factors are u n b o u n d e d , the resulting g r o u p w i l l be a n o n - n i l p o t e n t l o c a l l y n i l p o t e n t group.
Products of Normal Locally Nilpotent Subgroups Recall that the p r o d u c t o f t w o n o r m a l n i l p o t e n t subgroups is n i l p o t e n t — this is F i t t i n g ' s T h e o r e m (5.2.8). The corresponding statement holds for locally n i l p o t e n t groups and is o f great importance. 12.1.2 (The H i r s c h - P l o t k i n Theorem). Let H and K be normal
locally
potent subgroups
nilpotent.
of a group. Then the product
J = HK is locally
nil-
Proof. Choose a finitely generated s u b g r o u p o f J , say {h k , ...,/z /c > where h e H a n d k e K. W e must prove that J is n i l p o t e n t . T o this end we i n t r o d u c e the subgroups X = < / z . . . , h } and Y = < / q , . . . , fe >, and also Z = < X , 7>. Since J < Z , i t is enough t o show that Z is n i l p o t e n t . 1
t
1
m
m
t
1 ?
m
m
L e t C denote the set o f a l l c o m m u t a t o r s [h kj], i, j = 1 , . . . , m; then C ^ H nK since H a n d K are n o r m a l . Hence {X, C> is a finitely generated subgroup o f FT, so i n fact {X, C> is n i l p o t e n t . Since finitely generated n i l potent groups satisfy the m a x i m a l c o n d i t i o n (5.2.17), the n o r m a l closure C is finitely generated, as w e l l as being n i l p o t e n t . M o r e o v e r C < H n K, so that < y , C } < K. Therefore < y , C } is n i l p o t e n t and finitely generated, whence i t satisfies the m a x i m a l c o n d i t i o n . N o w [ X , y ] = C b y 5.1.7. Thus, using 5.1.6, we have h
x
x
x
x
XY
x
XY
=
X
9
y]> = Y. Y
I t follows t h a t Y is n i l p o t e n t , and by symmetry X is n i l p o t e n t . F i n a l l y Z = <X, y> = XY is n i l p o t e n t b y F i t t i n g ' s T h e o r e m . • Y
X
12.1.3. In any group G there is a unique maximal normal locally nilpotent subgroup (called the Hirsch-Plotkin radical) containing all normal locally nilpotent subgroups of G. Proof. I t is easy t o see that the u n i o n o f a chain o f l o c a l l y n i l p o t e n t sub groups is l o c a l l y n i l p o t e n t . T h u s Z o r n ' s L e m m a can be applied t o show t h a t each n o r m a l l o c a l l y n i l p o t e n t subgroup is contained i n a m a x i m a l n o r m a l locally n i l p o t e n t subgroup. I f H a n d K are t w o m a x i m a l n o r m a l locally n i l p o t e n t subgroups o f G, then HK is l o c a l l y n i l p o t e n t b y 12.1.2. Hence H = K. •
12. Generalizations of Nilpotent and Soluble Groups
358
T h e reader w i l l observe t h a t the F i t t i n g subgroup is always contained i n the H i r s c h - P l o t k i n radical. O f course these subgroups w i l l coincide i f the g r o u p is finite.
Ascendant Subgroups Before proceeding further w i t h the t h e o r y o f locally n i l p o t e n t groups we shall i n t r o d u c e a very useful generalization o f a s u b n o r m a l subgroup called an ascendant subgroup. By an ascending
series
i n a g r o u p G we shall mean a set o f subgroups
{ i f j a < /?} indexed b y ordinals less t h a n an o r d i n a l /? such that: (a) H
ai
(b) H
0
(c)
< H^if
CL < oc ; 2
x
= 1 and G =
\j H ; a
a
H ^H ; a
(d) H
k
a+1
= (J
a < A
H i f X is a l i m i t o r d i n a l . a
C o n d i t i o n (d) is inserted t o ensure completeness o f the series under unions. I t is often convenient t o w r i t e the ascending series i n the f o r m 1 = FT
•• H
0
O f course the H
a
factors:
are the terms
= G.
p
o f the series, while the F T / F T are the a+1
the o r d i n a l f$ is the length o r ordinal
a
type. S h o u l d ft be finite, the
ascending series becomes a familiar object, a series o f finite length. Some times i t is convenient t o speak o f an ascending series beginning group K: i n this case H
0
at a sub
= K b u t (a)-(d) are otherwise unchanged.
A subgroup w h i c h occurs i n some ascending series o f a g r o u p G is called an ascendant
subgroup; this is an evident generalization o f a s u b n o r m a l sub
g r o u p . I t m a y be as well t o give an example at this p o i n t . Let G = X K A be a so-called locally dihedral 2-group; 00
x
is o f type 2 , while X = < x > has order 2 a n d a
this means t h a t A
- 1
= a , (a e A). L e t A be the t
unique (cyclic) subgroup o f A w i t h order 2 \ T h e n \_A ,X~\
= Af
i+1
2
since [ a , x] = a~ . series X < a XA
Consequently XA <\ t
XA ^\ 2
• • -
XA
a n d there is an
i+1
= G; notice t h a t {J XA i<(0
G
Hence X is ascendant i n G. O n the other h a n d , X XA
+1
= XA
{ A
= X
= A
t
ascending here.
= X\_A, X~\ =
= G, so X is n o t s u b n o r m a l i n G. R e t u r n i n g n o w t o locally n i l p o t e n t groups, we shall show t h a t the H i r s c h
- P l o t k i n radical contains m a n y m o r e t h a n j u s t the normal locally n i l p o t e n t subgroups. 12.1.4. / / G is any group, the Hirsch-Plotkin dant locally nilpotent Proof.
radical
contains
all the
ascen
subgroups.
L e t H be an ascendant l o c a l l y n i l p o t e n t subgroup o f G. T h e n there
is an ascending series H = FT < i F ^ < i • • • H 0
p
= G. Define H
a
t o be
Ha
H ;
12.1. Locally Nilpotent Groups
359
then one q u i c k l y verifies t h a t
is an ascending series. W e argue b y transfinite i n d u c t i o n t h a t H is locally n i l p o t e n t . I f this is false, there is a first o r d i n a l a such that H is n o t locally n i l p o t e n t . I f a were a l i m i t o r d i n a l , H w o u l d equal \J H and, H being locally n i l p o t e n t for y < a, i t w o u l d f o l l o w t h a t H is locally n i l p o t e n t . Hence a cannot be a l i m i t o r d i n a l , a — 1 exists a n d H _ is locally n i l p o t e n t . N o w (H _ ) = (H * ) * = H = H : moreover for any x i n H we have < i i f * = H . Consequently H is a p r o d u c t o f n o r m a l locally n i l p o t e n t subgroups a n d is therefore locally n i l p o t e n t b y 12.1.3. B y this c o n t r a d i c t i o n Hp = H is locally n i l p o t e n t , w h i c h shows t h a t FT , a n d hence FT, is con tained i n the H i r s c h - P l o t k i n radical o f G. • a
a
a
y
y
y
a
a
Ha
0(
H
l H
x
Ha
1
a
a
a
a
G
G
Maximal Subgroups and Principal Factors in Locally Nilpotent Groups W e recall t w o k n o w n properties o f n i l p o t e n t groups: m a x i m a l subgroups are n o r m a l a n d p r i n c i p a l factors are central (5.2.4 a n d 5.2.2). L e t us show t h a t these h o l d for locally n i l p o t e n t groups. 12.1.5 (Baer, M c L a i n ) . / / M is a maximal group G, then M is normal in G. Equivalently
subgroup of a locally G' < F r a t G.
nilpotent
Proof. I f M is n o t n o r m a l , then G' £ M a n d there is an element c i n G'\M. T h e n G =
u
n
n
F o r finite groups the c o n d i t i o n G' < F r a t G is equivalent t o nilpotence. H o w e v e r for infinite groups this is a very weak p r o p e r t y because an infinite g r o u p m a y n o t have any m a x i m a l subgroups a n d G = F r a t G is a real pos sibility. F o r example, let G be the standard w r e a t h p r o d u c t o f groups o f type p a n d q™: i t is an easy exercise (see Exercise 12.1.8) t o show that G = F r a t G, so t h a t certainly G' < F r a t G. H o w e v e r G is n o t locally n i l p o t e n t i f p # q. Thus G' < F r a t G does n o t i m p l y local nilpotence. 0 0
12.1.6 (Mal'cev, M c L a i n ) . A principal central.
factor
of a locally nilpotent group G is
12. Generalizations of Nilpotent and Soluble Groups
360
Proof. L e t AT be a m i n i m a l n o r m a l s u b g r o u p o f G. I t suffices t o prove that N is central i n G. I f N ^ CG, there exist a in N a n d g i n G such t h a t b = [ a , g ] # 1. Since b e AT, we have N = b by m i n i m a l i t y o f AT. Hence a e (b \ b } for certain ^ i n G. L e t H = (a, g, g , g } , a n i l p o t e n t g r o u p , a n d set >1 = a . T h e n b e [A, FT], so t h a t b ^ e [ A , i f ] a n d consequently a e [A, FT]. Hence >1 = [ A , FT] a n d A = [A, H] for a l l r. Since FT is n i l p o t e n t , i t follows that A = 1 a n d a = 1. B u t this means that b = [a, g\ = 1. • G
9
dn
x
n
H
r
Locally Nilpotent Groups Subject to Finiteness Conditions Sometimes w h e n a finiteness restriction is placed u p o n a l o c a l l y n i l p o t e n t g r o u p , the g r o u p is forced t o become n i l p o t e n t . A n obvious example is: every finitely generated l o c a l l y n i l p o t e n t g r o u p is n i l p o t e n t . Here is a less t r i v i a l result o f the same k i n d . 12.1.7 ( M c L a i n ) . / / the locally nilpotent group G satisfies the maximal condi tion on normal subgroups, then G is a finitely generated nilpotent group. Proof. Clearly G / G satisfies m a x - n a n d hence m a x since i t is an abelian g r o u p . Therefore G = XG' for some finitely generated s u b g r o u p X. T h e n X is n i l p o t e n t o f class c, say. L e t "bars" denote q u o t i e n t groups m o d u l o y G . T h u s _ G = XG' and G is _nilpotent. Therefore G' < F r a t G by 5.2.16 a n d G = X ( F r a t G). B u t F r a t G is finitely generated, by 5.2.17, a n d its genera tors are nongenerators o f G by 5.2.12. Hence G = X, w h i c h means t h a t G has n i l p o t e n t class at most c a n d y G = y G. W r i t i n g L for y G, we have L = [ L , G ] . I f L # 1, then by m a x - n there is a n o r m a l s u b g r o u p N of G w h i c h is m a x i m a l subject t o N < L . B u t L/N is m i n i m a l n o r m a l i n G/N, so i t is central by 12.1.6. Thus [ L , G ] < N < L , a c o n t r a d i c t i o n . Hence L = 1 and G is n i l p o t e n t . • c + 2
c+1
c+2
c+1
N o t i c e the c o r o l l a r y : m a x a n d m a x - n are the same p r o p e r t y for l o c a l l y n i l p o t e n t groups. I f one imposes m i n - n , the m i n i m a l c o n d i t i o n o n n o r m a l subgroups, o n a locally n i l p o t e n t g r o u p , h o p i n g for nilpotence, one is disappointed. F o r example, let G = X tx A be a l o c a l l y d i h e d r a l 2-group. This g r o u p is l o c a l l y n i l p o t e n t a n d even satisfies m i n . H o w e v e r G' = A = [A, G ] , so the l o w e r central series terminates at G' a n d G c a n n o t be n i l p o t e n t . Nevertheless a fairly g o o d description o f the structure o f l o c a l l y n i l p o t e n t groups w i t h m i n - n can be given. 12.1.8 ( M c L a i n ) . A locally nilpotent group G satisfies the minimal condition on normal subgroups if and only if it is the direct product of finitely many Cernikov p-groups for various primes p.
12.1. Locally Nilpotent Groups
361
Proof. L e t G satisfy m i n - n . A c c o r d i n g t o 5.4.22 there exists a unique m i n i m a l subgroup F w i t h finite index i n G. F u r t h e r m o r e F satisfies m i n - n b y 3.1.8. N o w i f F is abelian, i t satisfies m i n a n d b y the structure theorem for abelian groups w i t h m i n (4.2.11), F is d i v i s i b l e — b e a r i n m i n d t h a t F cannot have a p r o p e r subgroup w i t h finite index. I n this case G is a C e r n i k o v group; also G is the direct p r o d u c t o f its p-components by 12.1.1. Hence f o r t h assume t h a t F is n o t abelian. B y 12.1.6 m i n i m a l n o r m a l subgroups o f F are contained i n £ F ; thus m i n n guarantees t h a t ( F # 1. M o r e o v e r ( F ^ F, so t h a t ( F < £ F b y the same argument. L e t x e ^ F a n d y e F ; then x = xz for some z i n £ F . N o w ( F has m i n because F has m i n - n , a n d £ F / £ F has m i n for the same reason. Therefore x a n d z have finite orders: w h a t is more, |z| divides Regard x as fixed a n d y as variable. Since ( F contains o n l y finitely m a n y elements o f each given order, there are o n l y finitely m a n y conjugates o f x i n F . Hence | F : C (x)\ is finite, F = C (x) a n d x e ( F . H o w e v e r this contradicts ( F < £ F. 2
y
2
2
F
F
2
T h e converse is left t o the reader as an exercise.
•
I t is an i m m e d i a t e c o r o l l a r y t h a t m i n a n d m i n - n are the same p r o p e r t y for locally n i l p o t e n t groups.
McLain's Characteristically Simple Groups W e shall describe next some famous examples due t o D . H . M c L a i n o f locally n i l p o t e n t groups t h a t are characteristically simple. These groups are perfect a n d have t r i v i a l center, w h i c h s h o u l d convince the reader t h a t locally n i l p o t e n t groups are far r e m o v e d f r o m the familiar realm o f n i l p o t e n t groups. I n essence M c L a i n ' s groups are infinite analogues o f u n i t r i a n g u l a r m a t r i x groups (see 5.1). W e begin w i t h Q, the ordered set o f r a t i o n a l numbers, a n d any field F , a n d we f o r m a vector space V w i t h c o u n t a b l y infinite d i m e n s i o n over F : let {v \X e Q } be a basis for V. I f X < p, denote by e the usual elementary linear t r a n s f o r m a t i o n o f V: thus x
Xll
= *V
and
ve v
Xfl
= 0
(v # X).
The standard m u l t i p l i c a t i o n rules h o l d for these e : Xfl
^
f
i
y
= e
and
Xy
e e^ Xfl
= 0
(/* # v).
2
(1)
1
I n p a r t i c u l a r e = 0, f r o m w h i c h i t follows t h a t (1 + ae^Y = 1 — ae^ for a i n F . U s i n g this rule for inverses a n d also (1) we calculate t h a t kll
[ 1 + ae , Xfl
1 + be^~\ = 1 + abe
ky
and [ 1 + ae , 1 + be ^] = 1 Xfl
v
if
and
] V ;U#v.J
(2)
362
12. Generalizations of Nilpotent and Soluble Groups
M c L a i n ' s g r o u p is the g r o u p o f linear transformations o f V generated b y all 1 + ae where a e F a n d X < p e Q. L e t this be w r i t t e n Xfl
M = M ( Q , F). One sees f r o m the m u l t i p l i c a t i o n rules (1) t h a t every element o f M can be w r i t t e n u n i q u e l y i n the f o r m 1 + X A < ^ A ^ A ^ where a e F a n d almost a l l of the a are zero. Conversely any element x o f this f o r m belongs t o M . T o prove this we assume t h a t x # 1 a n d choose p e Q as large as possible subject t o a # 0 for some X . W r i t e u = a e a n d v = x - u - 1. T h e n o f course x = 1 + u + v = (1 + u)(l + v); for uv = 0 because a^ = 0 i f p < jx . B y i n d u c t i o n o n the n u m b e r o f nonzero terms i n x we have 1 + v e M , so t h a t x e M a s claimed. Xfi
kll
0
XollQ
0
XoflQ
XoflQ
ofii
0
1
The f o l l o w i n g theorem lists the essential properties o f M c L a i n ' s g r o u p . 12.1.9 ( M c L a i n ) . Let M = M ( Q , F ) . (i) M is the product of its normal abelian subgroups, so M is locally nilpotent. (ii) / / F has characteristic p > 0, then M is a p-group: if F has characteristic 0, then F is torsion-free. (iii) M is characteristically simple. (iv) CM = \ andM = M. Proof, (i) B y the c o m m u t a t o r relations (2) each conjugate o f 1 + ae belongs t o the subgroup generated by a l l 1 + be^ where b e F a n d v < X < p < (; therefore a l l conjugates o f 1 + ae c o m m u t e a n d thus (1 + ae ) is an abelian g r o u p . Clearly M is the p r o d u c t o f a l l the (1 + ae ) . Thus M equals its H i r s c h - P l o t k i n radical a n d is locally n i l p o t e n t . Xfl
M
Xfl
Xfl
M
kll
(ii) If xe M, there exists a finite set o f elements X ..., X i n Q such t h a t X < - - • < X a n d x belongs t o the s u b g r o u p FL generated b y a l l 1 + cte . ^ i = 1 , . . . , n — 1, a e F. Clearly the m a p p i n g 1 + ^x x • 1+ E establishes an i s o m o r p h i s m between FL a n d the u n i t r i a n g u l a r g r o u p U(n, F); b o t h assertions n o w f o l l o w f r o m the discussion o f the u n i t r i a n g u l a r g r o u p i n 5.1. u
l
n
n
k ki+
a
t
i+l
i i + 1
(iii) Suppose t h a t N is characteristic i n M a n d N # 1. W e need t o prove t h a t N = M. The first step is t o show t h a t N contains a generator o f M . I f 1 # x e N, then x e FL where FT is a u n i t r i a n g u l a r g r o u p as i n (ii). Also 1 # FT n AT
T
N
N o w let X < p be a pair o f r a t i o n a l numbers. I t is a w e l l - k n o w n fact t h a t there exists an order-preserving p e r m u t a t i o n a o f Q such t h a t Xoc = X a n d pa = X . M o r e o v e r a determines an a u t o m o r p h i s m o f M given by 1 + be £ i—• 1 + be £ . Since N is characteristic i n M , i t follows t h a t Af con tains 1 + ae for a l l X < p i n Q. F i n a l l y , i f b e F , then N also contains
x
n
y
yoC
a
kvi
[ 1 + ae
X/l9
1 + a^be^
= 1+
be
ky
for a l l X < v i n Q. Therefore N = M a n d M is characteristically simple.
363
12.2. Some Special Types of Locally Nilpotent Groups
(iv) B y (iii) C M = 1 or M : o b v i o u s l y M is n o t abelian, so C M = 1. F o r a similar reason M' = M. •
EXERCISES
12.1
1. A soluble p-group is locally nilpotent. 2. A group is called radical i f it has an ascending series with locally nilpotent factors. Define the upper Hirsch-Plotkin series of a group G to be the ascending series 1 = R < R < • • • i n which R /R is the Hirsch-Plotkin radical of G/R and R = \J xR f° ordinals L Prove that the radical groups are pre cisely those groups which coincide with a term of their upper Hirsch-Plotkin series. 0
1
a+1
r n
x
a<
n
n
a
a
t
a
3. Let G be a radical group and H is Hirsch-Plotkin radical. (a) I f 1 # N o G, show that H n N # 1. (b) Prove that C (H) = C#. G
4. Show that a radical group with finite Hirsch-Plotkin radical is finite and soluble. *5. Write H asc K to mean that H is an ascendant subgroup of a group K. Establish the following properties of ascendant subgroups. (a) H asc K and K asc G imply that H asc G. (b) H asc K < G and L asc M < G imply that H n L asc K n M. (c) I f H asc K < G and a is a homomorphism from G, then H asc X . Deduce that HN asc KNifN^ G. a
a
6. A n ascending series that contains all the terms of another ascending series is called a refinement of it. A n ascending composition series is an ascending series with no refinements other than itself. Characterize ascending composition series in terms of their factors and prove a version of the Jordan-Holder Theorem for such series. 7. (Baer). Show that a nilpotent group with min-n is a Cernikov group whose center has finite index. Show also that the derived subgroup is finite. (See also 12.2.9). 8. Let G be the standard wreath product of two quasicyclic groups. Prove that G = Frat G. 9. Establish the commutator relations (2). 10. Prove that M ( Q , F) has no proper subgroups of finite index and no nontrivial finitely generated normal subgroups. Identify the Frattini subgroup. 11. Need a cartesian product of finite p-groups be locally nilpotent?
12.2. Some Special Types of Locally Nilpotent Groups I n this section we shall consider some n a t u r a l generalizations o f nilpotence t h a t are stronger t h a n local nilpotence.
364
12. Generalizations of Nilpotent and Soluble Groups
The Normalizer Condition Recall that a g r o u p satisfies the normalizer condition i f each proper sub g r o u p is smaller t h a n its normalizer. I t was s h o w n (in 5.2.4) t h a t for finite groups the n o r m a l i z e r c o n d i t i o n is equivalent t o nilpotence. I t is n o t diffi cult t o see t h a t the n o r m a l i z e r c o n d i t i o n can be reformulated i n terms o f ascendance. 12.2.1. A group G satisfies group is ascendant.
the normalizer
condition
if and only if every
sub
Proof. L e t G satisfy the n o r m a l i z e r c o n d i t i o n . F o r any subgroup H one m a y define a chain o f subgroups {FT } by the rules a
H
0
= H,
= N (H ),
H
a+1
G
and
a
H=
[j
x
H
p
where a is an o r d i n a l a n d A a l i m i t o r d i n a l . I f H < H for a l l a, i t w o u l d f o l l o w that | G | is n o t less t h a n the c a r d i n a l i t y o f a for a l l a, o b v i o u s l y absurd. Hence H = F T for some a, so t h a t H = G by the n o r m a l i z e r con d i t i o n . Since the FT 's f o r m an ascending series f r o m FT, i t follows that H is ascendant i n G. Conversely assume that every subgroup o f G is ascendant a n d let H < G. T h e n there is an ascending series H = FT
a
a+1
a+1
a
a
0
2
1
p
x
G
G
N o w that we have a clearer idea o f w h a t the n o r m a l i z e r c o n d i t i o n entails, let us relate this p r o p e r t y t o l o c a l nilpotence. 12.2.2 ( P l o t k i n ) . If a group locally nilpotent.
G satisfies
the normalizer
condition,
then it is
Proof. L e t g e G; then <#> is ascendant i n G by 12.2.1, while by 12.1.4 the subgroup <#> is contained i n FT, the H i r s c h - P l o t k i n radical o f G. Therefore H = G a n d G is locally n i l p o t e n t . • M c L a i n ' s g r o u p M ( Q , F ) is an example o f a locally n i l p o t e n t g r o u p that does not satisfy the n o r m a l i z e r c o n d i t i o n (Exercise 12.2.8). A simpler exam ple is W = H where \H\ = p a n d K is an infinite elementary abelian p-group; here K is self-normalizing.
Hypercentral Groups A n ascending series 1 = G < i G < i • • • G = G i n a g r o u p G is said t o be central i f G < i G and G / G lies i n the center o f G / G for every a < /?. A g r o u p w h i c h possesses a central ascending series is called hypercentral. 0
a
a + 1
x
a
p
a
365
12.2. Some Special Types of Locally Nilpotent Groups
This n a t u r a l class o f generalized n i l p o t e n t groups can terized i n terms o f the transfinitely extended upper central defined i n the f o l l o w i n g manner. I f G is any g r o u p a n d a terms £ G o f the upper central series o f G are denned by the a
C G=1
and
0
Ca iG/CaG
=
+
also be charac series, w h i c h is an o r d i n a l , the usual rules
C(G/CaG)
together w i t h the completeness c o n d i t i o n
CxG = U £*G a< A
where X is a l i m i t o r d i n a l . Since the c a r d i n a l i t y o f G cannot be exceeded, there is an o r d i n a l ft such t h a t CpG = Cp+iG = , etc., a t e r m i n a l subgroup called the hypercenter o f G. I t is sometimes convenient t o call £ G the acenter o f G. a
12.2.3. A group is hypercentral
if and only if it coincides
with its
hypercenter.
The easy p r o o f is left t o the reader. As examples o f hypercentral groups we cite the l o c a l l y d i h e d r a l 2-group a n d any direct p r o d u c t o f n i l p o t e n t groups. H o w is hypercentrality related t o l o c a l nilpotence? The answer is given by 12.2.4. A hypercentral locally nilpotent.
group G satisfies
the normalizer
condition
and hence is
Proof. L e t { G J a < /?} be a central ascending series o f G a n d let H < G. Since G / G is central i n G, we have FTG
a
a
0
A+1
P
F o r m a n y years i t was n o t k n o w n i f a g r o u p w i t h the n o r m a l i z e r c o n d i t i o n was a u t o m a t i c a l l y hypercentral. This was finally s h o w n t o be false by H e i n e k e n a n d M o h a m e d i n 1968—see [ a 8 8 ] a n d [ a 8 3 ] . 12.2.5. A locally nilpotent groups is hypercentral.
group with the minimal
condition
on normal
sub
This follows f r o m 12.1.6.
Baer Groups and Gruenberg Groups The next t w o types o f generalized n i l p o t e n t groups are characterized by the s u b n o r m a l i t y or ascendance o f finitely generated subgroups. The f o l l o w i n g result is basic.
12. Generalizations of Nilpotent and Soluble Groups
366
12.2.6. Let H and K be finitely
generated
nilpotent
subgroups
of a group G
and write J =
in G, then J is ascendant
nilpotent. (ii) (Baer). If H and K are subnormal in G, then J is subnormal and
and
nilpotent.
Proof, (i) B y 12.1.4 b o t h H a n d K lie inside R, the H i r s c h - P l o t k i n r a d i c a l o f G: hence J < R. Since J is p l a i n l y finitely generated, we conclude t h a t i t is nilpotent. T o prove t h a t J is ascendant i n G is harder. L e t H = FT < i Ff\ < i • • • H = G be an ascending series w i t h a > 0. K e e p i n g i n m i n d t h a t J is a finitely generated n i l p o t e n t g r o u p , we note t h a t H is finitely generated. I f a is a l i m i t o r d i n a l , i t m u s t f o l l o w t h a t H < H for some /? < a. Transfinite i n d u c t i o n o n a yields the ascendance o f H i n H a n d hence i n G. If, o n the other h a n d , a is n o t a l i m i t o r d i n a l , t h e n H < H _ ^ H = G and H < H^: again transfinite i n d u c t i o n leads t o H being ascendant i n G. W h a t this argument demonstrates is t h a t H can be replaced b y H . I n short we can suppose t h a t FT < i J a n d J = HK. 0
a
J
J
fi
J
fi
a
J
x
a
J
J
N e x t we pass t o H = H = H
x
2
Ha
a m o d i f i e d ascending series, defining H t o be H ; then < i • * * H = H is an ascending series. B u t w h a t is really ascending series, w i t h X - a d m i s s i b l e terms. Such a series writing a
G
a
keK G
I t is fairly clear t h a t H = H$ = Hf a n d H = H*; also Hf^ Hf . How ever t o conclude t h a t the i/jf's f o r m an ascending series we m u s t p r o v e c o m pleteness, +1
m
=
u
H
f
for l i m i t ordinals L O n e inclusion, Hf > [jp Hf, is o f course obvious. T o establish the other i n c l u s i o n choose x f r o m Hf. C l e a r l y <x, K} < (H , X > , w h i c h is contained i n the H i r s c h - P l o t k i n radical o f G (since H a n d K are). N o w <x, K} is finitely generated, so i t is n i l p o t e n t a n d x is finitely generated. B u t x < (Hf) = Hf < H = {J H a n d the H / s f o r m an ascending series. Consequently x < H for some f$ < X\ therefore x < Hf a n d i n p a r t i c u l a r x e Hf. T h i s settles the p o i n t at issue. <x
G
x
K
K
K
A
p<x
p
K
p
K
The remainder o f the p r o o f is easy. F o r each j ? < a w e have Hfi < i Hfi K a n d K is ascendant i n Hf K; hence HfiK is ascendant i n Hfi K b y Exer cise 12.1.5. P u t t i n g these relations together for all /?, we deduce that J = H$K is ascendant i n H*K = H K a n d hence i n G since H K is ascendant i n G. +1
+1
G
+1
G
(ii) The same m e t h o d applies i n this case, b u t i t is easier since, o f course, we need n o t consider l i m i t ordinals. • T h e f o l l o w i n g statements are i m m e d i a t e corollaries o f 12.2.6.
12.2. Some Special Types of Locally Nilpotent Groups
12.2.7. If every cyclic generated
subgroup
subgroup is ascendant
367
of a group is ascendant, and
then every
finitely
then every
finitely
nilpotent.
12.2.8. / / every cyclic subgroup of a group is subnormal, generated subgroup is subnormal and nilpotent.
Definitions. A g r o u p is called a Gruenberg group i f every cyclic subgroup is ascendant a n d a Baer group i f every cyclic subgroup is s u b n o r m a l . O b v i ously every Baer g r o u p is a G r u e n b e r g g r o u p , a n d by 12.2.7 every G r u e n berg g r o u p is locally n i l p o t e n t . N o t i c e also t h a t a g r o u p w i t h the n o r m a l i z e r c o n d i t i o n is a G r u e n b e r g g r o u p i n view o f 12.2.1. The structure o f Baer g r o u p s — a n d hence o f n i l p o t e n t g r o u p s — w i t h m i n - n is described by the f o l l o w i n g result: 12.2.9. / / G is a Baer group satisfying the minimal condition groups, then G is nilpotent and its center has finite index.
on normal
sub
Proof. W e k n o w f r o m 12.1.8 t h a t G is a C e r n i k o v g r o u p . L e t N denote the smallest n o r m a l subgroup o f finite index i n G. O f course, AT is a divisible abelian g r o u p . I t suffices t o p r o v e t h a t N is contained i n the center o f G. I f this is false, we can find an element g such t h a t [AT, g~] # 1. N o w <#> is s u b n o r m a l i n G, so there is a series <#> = F T < i H < i • • < i H = G. T h e n [ A , g~] < H _ [ A , g~\ < H _ , etc., a n d finally [ A , # ] < <#>. Hence [ A , i # ] = 1. I f r is the smallest such integer, then r > 1. Let M = [ A , _ i # ] a n d p u t m = \g\. Since [M, g, g~] = 1, we can use one o f the fundamental c o m m u t a t o r identities t o show t h a t [ M , g ] = [ M , g ] = 1. B u t the m a p p i n g a i—• [ a , g ] is an e n d o m o r p h i s m o f N; therefore [ M , g~] = [ A , g ] is divisible. Consequently, [ M , g~] = 1 i n c o n t r a d i c t i o n t o the choice o f r. • 0
r
u
2
r
1
2
r
r
r +
r
m
r
m
r
T w o further classes o f locally n i l p o t e n t groups w i l l be briefly m e n t i o n e d . A g r o u p G is a Fitting group i f G = F i t G, t h a t is, i f G is a p r o d u c t o f n o r m a l n i l p o t e n t subgroups. F o r example, M c L a i n ' s g r o u p is a F i t t i n g g r o u p . I f x e G = F i t G, then x lies i n a p r o d u c t o f finitely m a n y n o r m a l n i l p o t e n t subgroups, a n d hence i n a n o r m a l n i l p o t e n t subgroup by F i t t i n g ' s theorem. T h u s we have an alternative description o f F i t t i n g groups. 12.2.10. A group G is a Fitting group if and only if every element is contained in a normal nilpotent subgroup. Every Fitting group is a Baer group. Lastly, a class o f groups a b o u t w h i c h very l i t t l e is k n o w n , groups i n w h i c h every subgroup is s u b n o r m a l . O b v i o u s l y such groups are Baer groups a n d by 12.2.2 they satisfy the n o r m a l i z e r c o n d i t i o n . W e m e n t i o n i n this connection a notable theorem o f Roseblade [ a l 7 4 ] ; if every subgroup of
12. Generalizations of Nilpotent and Soluble Groups
368
a group is subnormal in a bounded number of steps, the group is nilpotent. T h i s contrasts w i t h the example o f a n o n - n i l p o t e n t g r o u p w i t h every sub g r o u p s u b n o r m a l constructed by H e i n e k e n a n d M o h a m m e d [ a 8 8 ] . F i n a l l y M o h r e s has recently s h o w n t h a t i f every subgroup o f a g r o u p is s u b n o r m a l , then the g r o u p is soluble.
Diagram of Group Classes o locally n i l p o t e n t
normalizer condition
hypercentral
nilpotent
I t is k n o w n t h a t a l l these eight classes are distinct—see [ b 5 4 ] for details.
EXERCISES
12.2
1. I f G is a hypercentral group and 1 ^ J V < G , then N n (G # 1. 2. (Baer). A group is hypercentral if and only if every nontrivial quotient group has nontrivial center. 3. A nontrivial hypercentral group cannot be perfect. 4. I f A is a maximal normal abelian subgroup of a hypercentral group G, then A = C (A). G
5. (P. Hall). The product of two normal hypercentral subgroups is hypercentral. 6. I f G is hypercentral and G is a torsion group, show that G is a torsion group. Does this hold for locally nilpotent groups? ab
7. Give an example of a hypercentral group that is not a Baer group.
12.3.
369
Engel Elements and Engel Groups
8. Prove that M ( Q , F) does not satisfy the normalizer condition. Deduce that Fitting groups need not satisfy this condition. [Hint: Consider the subgroup generated by all 1 + ae where ae F and either X < \i < 0 or 0 < X < X]X
9. A countable locally nilpotent group is a Gruenberg group. (This is false for uncountable groups—see [b54].) 10. Prove 12.2.6(h). 11. What is the effect on the diagram of classes of locally nilpotent groups i f the condition max-n or min-n is imposed on the groups? *12. Every group has a unique maximal normal Gruenberg (Baer) subgroup which contains all ascendant Gruenberg (subnormal Baer) subgroups. 13. Prove that a group G is hypercentral i f and only if to each countable sequence of elements # > • • • there corresponds an integer r such that [_g g , g l — 12
u
2
r
14. (Cernikov). A group-theoretical property 0* is said to be of countable character if a group has & whenever all its countable subgroups have Prove that nil potence and hypercentrality are properties of countable character. [Hint: Use Exercise 12.2.13.] 15. (Baer). Prove that the normalizer condition is a property of countable charac ter. [Hint: Assume that every countable subgroup of G satisfies the normalizer condition but G has a proper subgroup H such that H = N (H). Choose a countable subgroup X satisfying 1 < X n H < X. I f x e X\(X n H), there is an x* i n H such that (x*) and (x*) do not both belong to H. Define X* = <X, x*\x e Xy. N o w X = X and X = X*. Consider the union U of the chain X < X < • • •.] G
x
x
t
t
1
i+1
2
*16. I f N o G and N is hypercentral, prove that N' < Frat G. [Hint: Let M be a maximal subgroup of G not containing N. Prove that J V n M < G and N/N n M is a principal factor of G.]
12.3. Engel Elements and Engel Groups I n this a n d the f o l l o w i n g sections generalized n i l p o t e n t groups w h i c h are not locally n i l p o t e n t w i l l be considered. A m o n g the best k n o w n groups o f this sort are the so-called Engel groups. This is a subject whose origins lie outside g r o u p theory, i n the t h e o r y o f L i e rings.
Engel Elements A n element g o f a g r o u p G is called a right Engel element i f for each x i n G there is a positive integer n = n(g, x) such t h a t [g, x] = 1. N o t i c e t h a t the variable element x appears o n the right here. I f n can be chosen indepen dently o f x , then g is a right n-Engel element o f G, or less precisely a bounded right Engel element. The sets o f r i g h t a n d b o u n d e d r i g h t Engel elements o f G n
12. Generalizations of Nilpotent and Soluble Groups
370
are w r i t t e n R(G)
and
R(G).
Left Engel elements are defined i n a similar fashion. I f g e G a n d for each x i n G there exists an integer n = n(g, x) such t h a t [ x , g ] = 1, then g is a Ze/t Engel element o f G. Here the variable x is o n the left. I f n can be chosen independently o f x, then g is a /e/t n-Engel element o r bounded left Engel element. W r i t e n
L(G)
and
L(G)
for the sets o f left a n d b o u n d e d left Engel elements o f G. W h i l e i t is clear t h a t these four subsets are i n v a r i a n t under a u t o m o r phisms o f G, i t is u n k n o w n i f they are always subgroups. W h a t inclusions h o l d between the four subsets? 12.3.1 (Heineken). In any group G the inverse of a right Engel element is a left Engel element and the inverse of a right n-Engel element is a left (n + 1)Engel element. Thus 1
RiG)'
c L(G)
and
^(G)"
1
c 1(G).
Proof. L e t x a n d g be elements o f G. U s i n g the fundamental c o m m u t a t o r identities we o b t a i n [x,
g]
n+1
1
g
= [ [ x , g\ g~\ = [ [ g f , x~] , g] n
=
xl
x
Hence [_g~ , g~] = 1 implies t h a t [ x , g] follow. n
n+1
n
gJ
n
= 1. B o t h parts o f the result n o w •
I t is still an open question whether every r i g h t Engel element is a left Engel element. T h e t w o sets o f left Engel elements are closely related t o the H i r s c h P l o t k i n r a d i c a l a n d the Baer r a d i c a l respectively, the latter being the unique m a x i m a l n o r m a l Baer subgroup ( w h i c h exists i n any group—see Exercise 12.2.11). M o r e o v e r i t turns o u t t h a t the t w o sets o f r i g h t Engel elements have m u c h t o do w i t h the hypercenter a n d the co-center. 12.3.2. Let G be any group.
Then:
(i) L ( G ) contains the Hirsch-Plotkin radical and L ( G ) the Baer radical; (ii) R(G) contains the hypercenter and R(G) contains the co-center. Proof, (i) L e t g b e l o n g t o the H i r s c h - P l o t k i n r a d i c a l H a n d let xeG. Then lg, x ] e H a n d thus K = (g, [ x , # ] > < H. I t follows t h a t K is n i l p o t e n t a n d [x, „ # ] = 1 for some n > 0, so t h a t g e L(G). N e x t suppose t h a t g belongs t o
371
12.3. Engel Elements and Engel Groups
the Baer radical; then <#> is s u b n o r m a l i n G, a n d there is a series o f finite length (g) for
any
= G ^ G < i • • - < i G = G. Clearly [ x , # ] e G „ _ 0
x
x i n G, a n d
n
so
on; finally
[x, #] e G n
0
1 ?
= <#>.
[x, # ] e G„_ 2
2
Consequently
[*, „+i0] = 1 and # e L ( G ) . (ii) L e t g belong t o the hypercenter o f G a n d let xeG. Suppose t h a t [#> „ * ] ^ 1 f ° H - N o w g e ( G for some first o r d i n a l a, w h i c h cannot be a l i m i t o r d i n a l ; for i f i t were, ( G w o u l d equal {Jp CpG. Hence ge ( C G ) \ ( C _ i G ) . I t follows t h a t [ g , x ] e C«-iG, so t h a t by the same argument [ ( / ^ ] e ( ( , ' G ) \ ( U G ) where a' < a. S i m i l a r l y x] e ( f ^ G M ^ G ) where a" < a' < a. Since this process cannot terminate, i t leads t o an infinite descending chain o f ordinals • • • < a" < a' < a; this cannot exist. Hence g e R(G). F i n a l l y , i f g e ( G a n d x e G , then [ g , x ] = 1 a n d g e R(G). r
a
n
a
a
a
<0C
a
2
n
n
• The m a j o r goal o f Engel theory is t o find c o n d i t i o n s w h i c h w i l l guarantee t h a t L ( G ) , L ( G ) , R(G\ a n d R(G) are subgroups w h i c h coincide w i t h the H i r s c h - P l o t k i n radical, the Baer radical, the hypercenter a n d the co-center respectively. T h a t equality does n o t always h o l d is s h o w n b y a famous example o f G o l o d [ a 5 9 ] (see also [ b 3 3 ] ) o f a finitely generated infinite p - g r o u p G such t h a t G = L ( G ) = R(G). T h i s g r o u p does n o t equal its H i r s c h - P l o t k i n radical, otherwise i t w o u l d be finite.
Engel Groups F o r any g r o u p G the statements G = L ( G ) a n d G = R(G) are clearly equi valent, a n d a g r o u p w i t h this p r o p e r t y is called an Engel group. B y 12.3.2 (or directly) we see t h a t every locally n i l p o t e n t g r o u p is an Engel group. G o l o d ' s example m e n t i o n e d above shows t h a t Engel groups need n o t be locally n i l p o t e n t . T h u s Engel groups represent a rather wide generalization of n i l p o t e n t groups. B y an n-Engel group is meant a g r o u p G such t h a t [ x , y] = 1 for a l l x, y e G; t h a t is, every element is b o t h left a n d r i g h t n-Engel. Thus the class of n-Engel groups is the variety determined b y the l a w [ x , y] = 1. F o r example, a n i l p o t e n t g r o u p o f class n is an n-Engel group. O n the other h a n d , n-Engel groups need n o t be nilpotent—see Exercise 12.3.1. A g r o u p is a bounded Engel group i f i t is n-Engel for some n. n
n
Engel Structure in Soluble Groups The sets L ( G ) a n d L ( G ) are well-behaved i f G is a soluble group. 12.3.3. (Gruenberg). Let G be a soluble
group.
(i) L ( G ) coincides with the Hirsch-Plotkin radical and is a Gruenberg Thus a soluble Engel group is a Gruenberg group. (ii) L ( G ) coincides with the Baer radical. is a Baer group.
Thus a soluble bounded Engel
group. group
372
12. Generalizations of Nilpotent and Soluble Groups
Proof, (i) L e t g e L ( G ) : b y 12.2.7 i t suffices t o prove t h a t
Since A is abelian, the m a p p i n g a i—• [ a , g ] is an e n d o m o r p h i s m £ o f A . I f F 7^ 1 is a finite subset o f A , there is an integer n such that [ x , g ] = 1 for a l l x i n F; hence F^ = 0, w h i c h clearly implies t h a t C (g) # 1. N o w define sub groups 1 = A , A . . . b y the rules A /A = C (g) and A = [jp< A where a is an o r d i n a l a n d X a l i m i t o r d i n a l . Since we can always find a n o n t r i v i a l element t h a t is centralized b y g i n a n o n t r i v i a l q u o t i e n t o f A, there is an o r d i n a l y such t h a t A = A. N o w >l > l + i > because g centralizes A /A . I t follows t h a t the > l > f o r m an ascending series f r o m <#> t o A > a n d <#> is ascendant i n G. n
n
A
0
u
a+1
a
A/Aa
Y
a+1
x
a
a
x
p
a
a
(ii) N o w suppose t h a t g is a left n-Engel element. K e e p i n g the same n o t a t i o n , we have A} s u b n o r m a l i n G b y i n d u c t i o n o n d. H e r e [ a , g ] = 1 for all a i n A , so A > is n i l p o t e n t a n d <#> is s u b n o r m a l i n A > a n d hence i n G. Therefore g is i n the Baer radical. • n
O n the other hand, quite simple examples show t h a t R(G) m a y be larger t h a n the hypercenter even w h e n G is soluble (Exercise 12.3.1). The f o l l o w i n g was the first t h e o r e m t o be p r o v e d a b o u t Engel groups. 12.3.4 (Zorn). A finite
Engel group is
nilpotent.
Proof. Suppose t h a t this is false a n d let G be a finite Engel g r o u p w h i c h has smallest order subject t o being n o n n i l p o t e n t . T h e n every p r o p e r subgroup of G is n i l p o t e n t a n d G is soluble b y Schmidt's theorem (9.1.9). B y 12.3.3, G equals its H i r s c h - P l o t k i n radical, w h i c h means t h a t G is n i l p o t e n t since i t is finite. •
2-Engel Groups O b v i o u s l y a 0-Engel g r o u p has order 1 a n d the 1-Engel groups are exactly the abelian groups. Greater interest attaches t o the class o f 2-Engel groups; this, i t turns out, includes a l l groups o f exponent 3. 12.3.5. A group of exponent Proof. {xy' )' 1
1
3 is a 2-Engel
group. 1 2
L e t G be a g r o u p o f exponent 3 a n d let x, y e G. T h e n (xy' ) = yx' . P o s t - m u l t i p l i c a t i o n b y y yields 1
x
xy~ xy
2
x
2
2
= yx~ y
1
1
= y~ x~ y~
1
1
1
1
= y~ (y~ x~ y~ )
=
Therefore x
x(y~ xy)
1
1
= y^xix^y' )'
=
1
(y~ xy)x.
1
1
1 2
y~ x(x~ y~ ) .
=
12.3. Engel Elements and Engel Groups
373
y
y
I t follows t h a t x commutes w i t h x
a n d hence w i t h x x = [ y , x ] . Thus
[ y , x, x ] = 1.
•
W e shall n o w establish the basic result o n 2-Engel groups. 12.3.6 (Levi). Let G be a 2-Engel Then: (i) (ii) (iii) (iv)
G
x is [ x , y, [ x , y, [ x , y,
group and let x, y, z, t be elements
of G.
abelian; z ] = [ z , x, y ] ; z ] = 1; z, t ] = 1, 5 0 that G is nilpotent of class < 3. 3
y
Proof, (i) W e have [ x , x ] = [ x , x [ x , y ] ] = [ x , [ x , y ] ] . N o w x commutes w i t h [ y , x ] a n d hence w i t h [ x , y ] . Therefore x a n d x c o m m u t e , a n d i t follows b y c o n j u g a t i o n t h a t any t w o conjugates o f x c o m m u t e . Hence x is abelian. y
G
G
(ii) L e t A = x , an abelian g r o u p . T h e m a p p i n g a i—• [ a , y ] is an endo m o r p h i s m o f A w h i c h w i l l be w r i t t e n y*. Since (y*) sends a t o [ a , y, y ] = 1, we have ( y * ) = 0. 2
2
F r o m the elementary c o m m u t a t o r formulae for [ x , y z ] a n d [ x , y o b t a i n the results (yz)* = y* + z* + y*z*
-
1
] we (3)
and
(j,-!)* Now (yz)
- 1
=
( 4 )
commutes w i t h [ a , y z ] , so b y (3) a n d (4) we have 1
1
0 = (yz)*(z~ y~ )*
= ( y * + z* + y*z*)( — z* — y * + z*y*) = — y*z* — z*y*.
Therefore y*Z* =
(5) - 1
- 1
w h i c h tells us t h a t [ x , y, z ] = [ x , z, y ] . Since A is abelian, [ x , z, y ] = [ [ x , z ] , y ] = [ z , x , y ] ; hence [ x , y, z ] = [ z , x, y ] as required. (iii) U s i n g (ii) we o b t a i n [ x , y , z ] = [ x , y , z ] a n d thus [ x , y , z ] = [[X j]" , = [x, y, z ] " . N o w a p p l y the H a l l - W i t t i d e n t i t y (5.1.5) a n d (ii) t o get the result. (iv) B y (5) we have y*(zt)* + (zt)*y* = 0. E x p a n d i n g this w i t h the aid o f (3) a n d (5) one obtains - 1
- 1
1
y
- 1
- 1
y
1
0 = y*z* + y*£* + y*z*£* + z*y* + £*y* + z*£*y* = 2y*z*£*. 2
Hence [ x , y, z, t] [ x , y, z, r ] = 1.
= 1. B u t
also
(iii) implies t h a t
3
[ x , y, z, t ] = 1,
so •
374
12. Generalizations of Nilpotent and Soluble Groups
G
N o t i c e that G is a 2-Engel g r o u p i f a n d o n l y i f x is abelian for a l l x i n G. W h i l e 3-Engel groups are less well-behaved, i t is true t h a t G is 3-Engel i f a n d o n l y i f x is n i l p o t e n t o f class < 2 for a l l x i n G ( K a p p e a n d K a p p e [ a l l l ] ) . I n general 3-Engel groups are n o t n i l p o t e n t , b u t they are always locally n i l p o t e n t (Heineken [ a 8 7 ] ) . F o r a clear a n d concise account o f the t h e o r y o f 3-Engel groups see G u p t a [ b 3 0 ] . G
Engel Structure in Groups with the Maximal Condition A c c o r d i n g t o 12.3.4 a finite Engel g r o u p is n i l p o t e n t . C a n we weaken the hypothesis o f finiteness here? C e r t a i n l y finitely generated w i l l n o t d o — G o l o d ' s example tells us t h a t — b u t is there any hope for the m a x i m a l con dition? The answer turns o u t t o be affirmative; i n fact groups w i t h m a x have excellent Engel structure. 12.3.7 (Baer). Let G be a group which satisfies the maximal condition. Then L ( G ) and L ( G ) coincide with the Hirsch-Plotkin radical, which is nilpotent, and R(G) and R(G) coincide with the hypercenter, which equals ( G far some finite m. In particular, if G is an Engel group, it is nilpotent. m
M o s t o f the l a b o r o f the p r o o f resides i n establishing the f o l l o w i n g spe cial case. 12.3.8. / / G satisfies nilpotent.
G
m a x and aeL(G),
then a
is finitely
generated
and
Proof. Assume t h a t the statement is false. (i) L e t {a } denote the set o f a l l conjugates o f a i n G. A subgroup X o f G w i l l be called a-generated i f i t is generated b y those conjugates o f a t h a t i t contains, i n symbols X = {X n { a } > . G
G
(ii) / / X and Y are nilpotent a-generated subgroups such that X < Y, then N (X) contains at least one conjugate of a which does not belong to X. Y
Because Y is n i l p o t e n t , X is s u b n o r m a l i n Y a n d there is a series X = I < l ! < —=a X = Y. Since X ^ Y a n d Y is a-generated, Y\X must c o n t a i n a conjugate o f a. Hence there is an integer i such t h a t 0
s
G
Xn
{a }
= Xn t
G
{a }
^ X
i+1
n
G
{a }.
G
Suppose t h a t y e X n {a } a n d y $ X . T h e n y normalizes X so t h a t (Xn {a }) = {X n {a }) = X n {a } = I n {a }. Since Xn {a } gener ates X, the element y normalizes X. i+1
G
y
t
G
y
t
and
(iii) There V.
exist
G
i9
G
G
{
two distinct
maximal
a-generated
nilpotent
subgroups
U
Consider the set Sf o f a l l a-generated n i l p o t e n t subgroups. B y the m a x i m a l c o n d i t i o n each element o f Sf is contained i n a m a x i m a l element. I f there
12.3.
375
Engel Elements and Engel Groups
were j u s t one m a x i m a l element o f say M , then i t w o u l d c o n t a i n ; for e £f. B u t M < G since conjugates o f M belong t o £f\ thus a < M a n d a is n i l p o t e n t , a c o n t r a d i c t i o n . G
G
(iv) Consider the subgroup G
/ = (U
nVn{a }).
Here we can suppose t h a t U a n d V have been chosen so t h a t / is m a x i m a l . N o w define W=(N (I)n{a }y. G
v
F r o m its d e f i n i t i o n we see t h a t / is a-generated a n d also that / # U since I < V. B y (ii) we conclude that I < W. F o r the same reason N (I) n {a } has an element v t h a t does n o t belong t o / a n d hence n o t t o U. G
V
Suppose t h a t v e N (W). N o w m a x shows t h a t W can be finitely gener ated, say b y a ,..., a . K e e p i n g i n m i n d t h a t v is conjugate t o a a n d there fore belongs t o L(G), we find an n > 0 such that [a \ v\ = 1 for i = 1 , . . . , m. Let H = , W}; then clearly W'<\ H a n d H/W is n i l p o t e n t o f class < n. B u t W is n i l p o t e n t since U is, so we can a p p l y H a l l ' s c r i t e r i o n (5.2.10), con c l u d i n g t h a t H is n i l p o t e n t . Since W is a-generated a n d v is conjugate t o a, we see t h a t H is a-generated. Therefore H e 9* a n d H is contained i n a m a x i m a l element T o f ^ . N o w v e T\U, w h i c h shows that T # U. Also W < FT < F, so A ^ n ^ } c t / n Tn{a } and / < < <JU n F n { a } > . B u t this contradicts the m a x i m a l i t y o f F G
01
9rn
g
n
G
G
(v) I t follows f r o m the preceding a r g u m e n t t h a t v 4 N (W). Since I
G
V
n
k
G
z
_1
z
fc
G
z
G
G
z
By c o n s t r u c t i o n u a n d v belong N {I\ so that z = [ f , _iM] e N (I). Since w is conjugate t o a, we o b t a i n u e N (I) n { a } c PF by definition o f VT. A l s o w e VF, so we have u e W . I t follows t h a t u e U n W n { a } . N o w / < ^ a n d / = l < W , so t h a t I <{UnW n {a }). T h u s W is con tained i n a m a x i m a l element o f Sf w h i c h m u s t equal 1/; otherwise the m a x i m a l i t y o f / w o u l d again be contradicted. Consequently W < U a n d (N (I) n {a }) is contained i n N (I) nUn {a } = N (I) n {a } because z e A ( 7 ) . Hence W < W. Since # ^ i t follows t h a t W < W; con j u g a t i n g b y negative powers o f z, we o b t a i n W < W ' < W ~ < • •, w h i c h contradicts max. • G
z
fc
z
z
z
G
G
V
?
z
z
z
z
z
G
G
z
z
G
z
Z
V
G
G
G
V
z
z
G
2 1
z
2
G
Proof of 12.3.7. Here G is a g r o u p w i t h max. I f a e L ( G ) , then a is n i l p o t e n t by 12.3.8, so t h a t a is contained i n the H i r s c h - P l o t k i n radical H. M o r e G
376
12. Generalizations of Nilpotent and Soluble Groups
over H is finitely generated a n d n i l p o t e n t b y max. Hence L ( G ) = H c o i n cides w i t h the Baer r a d i c a l a n d hence w i t h L ( G ) b y 12.3.2. The statement a b o u t r i g h t Engel elements requires a little m o r e atten t i o n . L e t aeR(G); then a " e L ( G ) b y 12.3.1 a n d ( a ) = a = A is n i l potent b y 12.3.8. L e t xeG. Since A is generated b y finitely m a n y r i g h t Engel elements, [A, x] < A' for some positive integer k. Hence <x, A}/A' is n i l p o t e n t , w h i c h implies t h a t <x, A} is n i l p o t e n t . Refine the upper central series o f A t o a G-admissible series whose factors are elementary abelian o r free abelian o f finite r a n k . L e t B be such a factor; then x acts u n i p o t e n t l y o n B. B y 8.1.10 the a c t i o n o f G o n B is u n i t r i a n g u l a r a n d i n consequence A < C G for some 5 . 1
_ 1
G
G
ah
fc
S
I t follows t h a t R(G) = ( G for some r i n view o f max. F i n a l l y R(G) = ( G by 12.3.2. • r
EXERCISES
r
12.3
1. Let G be the standard wreath product of a group of order p and an infinite ele mentary abelian p-group. Prove that G is a (p + 1)-Engel group, yet £G = 1. Deduce that a 3-Engel group need not be nilpotent (see 12.3.6). 2. I f G is a locally finite group, show that L(G) equals the Hirsch-Plotkin radical and that R(G) is a subgroup of L(G). {Hint: Use 12.3.4.] 3. Let G be a soluble group with a normal series of finite length whose factors are abelian groups of finite rank with finite torsion-subgroups. Prove that L(G) = L(G) and R(G) = R(G). Show that these conclusions are not valid for ar bitrary soluble groups. 4. A soluble p-group of finite exponent is a bounded Engel group. 5. A group G is a 2-Engel group if and only i f the identity [ x , y, z] = [ y, z, x ] holds in G. 6. Let x,y be group elements satisfying [ x , „y] = 1. Prove that < x > generated.
< y >
is finitely
7. (Plotkin). I f G is a radical group (see Exercise 12.1.2), prove that L(G) coincides with the Hirsch-Plotkin radical and that R(G) is a subgroup of L(G). [Hint: Let {G } be the upper H i r s c h - P l o t k i n series. I f X is a finite subset of L(G)nG , prove that <X> is nilpotent, using Exercise 12.3.6.] a
2
12.4. Classes of Groups Defined by General Series There are numerous interesting classes o f generalized soluble a n d n i l p o t e n t groups w h i c h are defined b y means o f a series of general order type, a con cept w h i c h w i l l n o w be explaind.
12.4.
377
Classes of Groups Defined by General Series
Definition. L e t G be an Q-operator g r o u p . B y a (general) Q-series o f G we shall m e a n a set S o f subgroups, called the terms o f S, w h i c h is linearly ordered b y i n c l u s i o n a n d w h i c h satisfies the f o l l o w i n g conditions: (i) I f 1 # x e G, there are terms o f S w h i c h d o n o t c o n t a i n x a n d the u n i o n of a l l such terms is a t e r m V of S. (ii) I f 1 # x e G, there are terms o f S w h i c h c o n t a i n x a n d the intersection of a l l such terms is a t e r m A o f S. (iii) V ^iA . (iv) Each t e r m o f S is o f the f o r m V or A for some x # 1 i n G. x
x
x
x
x
x
Thus x belongs to the set A \V ; the corresponding quotient g r o u p AJV is a factor o f 9C. N o t i c e t h a t n o t e r m o f S can lie strictly between V a n d A . Hence i f x , y # 1, then either A < V or A < V . X
X
X
x
x
y
y
x
x
T h i s enables us t o linearly order the factors o f S b y the rule t h a t A /V precedes A /V i f A < V . The order-type o f S is the order-type o f the set o f all factors o f S. I t follows easily f r o m the definition t h a t x
y
y
x
x
y
V =[jA x
and
y
A =f]V x
where the u n i o n is f o r m e d over a l l factors A /V the intersection over a l l factors t h a t succeed AJV . properties o f the series. y
y
X
y
t h a t precede A /V and These are completeness x
x
I f S has finite order-type, i t is clearly j u s t a series o f finite length: the smallest V w i l l equal 1 a n d the largest A w i l l equal G. I f S has the ordertype o f an o r d i n a l n u m b e r /?, the series w i l l be an ascending series. O n the other h a n d , suppose that the order-type o f S is the reverse o f an o r d i n a l n u m b e r /?, t h a t is, the set o f ordinals a < ft in descending order. T h e n S is a descending series; this can be w r i t t e n i n the f o r m x
x
-
H
2
where H
a+1
Note that
< i H and H a
A
= f] H y<x
y
0
= G
w i t h a an o r d i n a l a n d X a l i m i t o r d i n a l .
(] H =l. a
a
Composition Series I f S a n d S* are Q-series i n G a n d i f every t e r m o f S is a t e r m o f S*, then S* is said t o be a refinement o f S, i n symbols S <^ S*. I f 1 # x e G, then, w i t h the obvious n o t a t i o n , V < V* < A J < A . A n Q-series w h i c h has n o refine m e n t other t h a n itself is called an Q-composition series. W h e n Q is e m p t y , we speak o f a series a n d a composition series: w h e n Q is the g r o u p o f inner a u t o m o r p h i s m s , we speak o f a normal series a n d a principal series. General c o m p o s i t i o n series have the definite advantage that they exist i n any g r o u p . x
x
12. Generalizations of Nilpotent and Soluble Groups
378
12.4.1. Every
Q-series can be refined to an Q-composition
series.
Proof. L e t S be a given Q-series i n an Q - g r o u p G. Consider the set 3C o f all refinements o f S . The r e l a t i o n <^ is clearly a p a r t i a l o r d e r i n g o f 9C. L e t = {S \y e T } be a c h a i n i n 9C\ we shall construct an upper b o u n d for # 0
0
iy)
(y)
I f 1 # x e G, define 7 = ( J F and A = a n d A ^ are terms o f S . C e r t a i n l y V <\ A . A J for each y i n T. Hence the set S = { A , by inclusion. I t is clear that S is a series w h i c h X
7)
y
x
x
(y)
x
(
x
}
x
y)
f] A™ where, o f course, V± I n addition < V < A < 1 # x e G } is l i n e a r l y ordered is an upper b o u n d for y
x
x
W e m a y n o w a p p l y Z o r n ' s L e m m a t o produce a m a x i m a l element o f 3C. B u t this is s i m p l y a c o m p o s i t i o n series o f G. • T h e reader s h o u l d observe t h a t n o analogue o f the J o r d a n - H o l d e r T h e o rem exists for general series. F o r example, Z has the t w o c o m p o s i t i o n series • • • 8 Z < 4 Z < 2 Z < Z a n d • • • 27Z < 9 Z < 3 Z < Z , b u t these have noniso m o r p h i c factors.
Groups with a Central Series Suppose that the g r o u p G has a central series S, t h a t is, each t e r m is n o r m a l a n d each factor AJV is central i n G. T h i s represents a generalization o f nilpotence since G w o u l d be n i l p o t e n t i f S were finite. G r o u p s w i t h this p r o p e r t y are sometimes called Z-groups. X
12.4.2. If G is a locally nilpotent group, then G has a central
series.
Proof. B y 12.4.1 there is a p r i n c i p a l series i n G. The factors o f this series are p r i n c i p a l factors o f G a n d b y 12.1.6 they are central i n G. • G r o u p s w i t h a descending central series are examples o f Z - g r o u p s — these are sometimes called hypocentral groups a n d are characterized b y the fact t h a t their l o w e r central series reaches the i d e n t i t y subgroup w h e n con t i n u e d transfinitely (Exercise 12.4.1). A m o n g the m o s t c o m m o n l y encoun tered h y p o c e n t r a l groups are the residually nilpotent groups. E v e n this class is very extensive, c o n t a i n i n g groups w h i c h m i g h t be regarded as h i g h l y n o n n i l p o t e n t , for example, free groups b y 6.1.10.
Serial Subgroups A subgroup w h i c h occurs i n some series o f a g r o u p G is called serial. T h i s m u s t be regarded as a very b r o a d generalization o f s u b n o r m a l i t y a n d ascendance. I t is quite possible for a serial subgroup t o be self-normalizing or t o have the whole g r o u p as its n o r m a l closure (Exercise 12.4.4).
12.4. Classes of Groups Defined by General Series
379
Here we are interested i n groups h a v i n g a l l their subgroups serial; n a t u r a l l y these include a l l n i l p o t e n t groups. G r o u p s o f this type can be charac terized i n terms o f the F r a t t i n i properties o f their subgroups. 12.4.3. Every all H
subgroup
of a group G is serial if and only if H' < F r a t FT for
Proof. O f course, the c o n d i t i o n o n H is equivalent t o the n o r m a l i t y o f a l l its m a x i m a l subgroups. Suppose that every subgroup o f G is serial a n d let M be m a x i m a l i n H. There is a series i n G w h i c h includes M . Intersect the terms o f this series w i t h H t o get a series i n H w h i c h also includes M . B u t M is m a x i m a l i n i f , so M a n d H m u s t be consecutive terms i n the second series a n d M < i H. Conversely, assume that the subgroups o f G have the p r o p e r t y stated. L e t L < G a n d define t o be the set o f a l l chains o f subgroups t h a t are refinements o f 1 < L < G a n d t h a t satisfy a l l the c o n d i t i o n s i n the definition of a series except perhaps the n o r m a l i t y c o n d i t i o n (iii). As i n 12.4.1 we use Z o r n ' s L e m m a t o construct a m a x i m a l element S o f J f . I f X a n d Y are consecutive terms o f S, then X m u s t be m a x i m a l i n Y b y m a x i m a l i t y o f S. Hence I < 7 a n d S is a series. Consequently L is serial i n G. • I n view o f this result a n d 12.1.5 we have the f o l l o w i n g interesting p r o p erty o f locally n i l p o t e n t groups. 12.4.4. Every
subgroup of a locally nilpotent group is
serial
O n the other h a n d , i t has been s h o w n b y W i l s o n [ a 2 2 3 ] t h a t local n i l potence is n o t a consequence o f the seriality o f a l l subgroups o f a group.
Generalized Soluble Groups The concept o f a series permits the creation o f m a n y classes o f generalized soluble groups. W e m e n t i o n briefly some o f the most i m p o r t a n t . A g r o u p w h i c h possesses a series w i t h abelian factors is called an SNgroup; this is an immensely w i d e generalization o f solubility. F o r example, i t is k n o w n t h a t there are simple S A - g r o u p s w h i c h are n o t o f p r i m e order ([b54]). Somewhat n a r r o w e r is the class o f Si-groups, groups w h i c h possess a n o r m a l series w i t h abelian factors. W e shall s h o r t l y see that locally soluble groups are S F g r o u p s (12.5.2). A n i m p o r t a n t subclass o f SI is the class o f groups w h i c h have an ascend i n g n o r m a l series w i t h abelian factors; these are called hyperabelian groups. Being m u c h closer t o soluble groups this is a relatively tractable class. W e conclude w i t h a result w h i c h illustrates the power o f the m i n i m a l condition.
12. Generalizations of Nilpotent and Soluble Groups
380
12.4.5 ( C e r n i k o v ) . An SN-group Cernikov
G satisfies
min if and only if it is a soluble
group.
Proof. W e need o n l y prove t h a t i f G satisfies m i n , i t is a soluble C e r n i k o v group. L e t F denote the unique m i n i m a l subgroup w i t h finite index i n G (see 5.4.22). T h e n F m a y be assumed t o be n o n t r i v i a l . N o w G is an SNgroup, so i t has a series w i t h abelian factors. B y m i n the members o f this series are well-ordered b y set i n c l u s i o n : thus G has an ascending series w i t h abelian factors. Because thus conclusion applies equally t o G / F , this finite g r o u p is soluble. Since F # 1 , there is a smallest t e r m o f the ascending series h a v i n g n o n t r i v i a l intersection w i t h F , say G ; here the o r d i n a l a is n o t a l i m i t o r d i n a l , so G _ ! n F = 1 and G nF ~ (G n F ) G _ / G _ . T h u s G n F is abelian; i t is also ascendant i n F and contained i n the H i r s c h - P l o t k i n r a d i c a l H o f F b y 12.1.4. N o w H is hypercentral b y 1 2 . 2 . 5 , so 1 # ( f f < a G a n d F con tains a n o n t r i v i a l n o r m a l abelian s u b g r o u p A o f G. T h e rest o f the p r o o f is j u s t like t h a t o f 5.4.23 and is left t o the reader as an exercise. • a
a
a
a
a
1
a
1
a
EXERCISES 12.4
1. Define the transfinitely extended lower central series of a group G by the rules y G = G, y G = [ y G , G] and y G = f)p ypG where a is an ordinal and X a limit ordinal. Prove that G is residually nilpotent i f and only if y^G = 1, and G is hypocentral if and only if some y G = 1. 1
a+1
a
x
<x
a
2. A subgroup is called descendant i f it is a member of some descending series. Find all the descendant subgroups of D^. Show that descendance is not pre served with respect to taking quotient groups. 3. Prove that
is residually nilpotent.
4. Find a serial subgroup which coincides with its normalizer and whose normal closure is the whole group. [Hint: Consider McLain's group M ( Q , F).] 5. I f every subgroup of a group G is serial and G satisfies min, prove that G is locally nilpotent. 6. A group G is said to be residually central i f for each nontrivial element x there is a normal subgroup N such that N # xN e £(G/A). Prove that G is residually central i f and only i f x $ [G, x ] whenever \ ^ xeG. Show also that Z-groups are residually central. (The converse is false—see [al55].) 7. (Ayoub, Durbin). Let G be a residually central group. (a) Show that each minimal normal subgroup is contained i n the centre of G. (b) I f H is the hypercenter, prove that G/H is residually central. (c) Prove that a residually central group with min-n is a hypercentral Cernikov group. 8. A group is hyperabelian if and only i f each nontrivial quotient group has nontrivial Fitting subgroup.
12.5. Locally Soluble Groups
381
9. By an SN*-group is meant a group which has an ascending series with abelian factors. Show that hyperabelian implies SN* implies radical. Give an example of a finitely generated SN*-group that is not hyperabelian. [Hint: Let M = M(Q , F ); this is McLain's group with the ordered set Q °f ^ rationals of the form ml", m,neZ, and F = GF(p). The assignments 1 + e i-> 1 + e and 1 + e i—• 1 + e determine automorphisms a and P of M respectively. Consider the group G =
2
p
2
p
Xfl
X/t
x+l
M+1
2X2ll
10. Prove that a group G is an SN*-group if and only if it has an ascending series whose factors are Gruenberg groups. 11. Assume that G is a group with a normal series (of general order type) whose factors are cyclic. Prove that G' is a Z-group. *12. Complete the proof of 12.4.5. 13. A nontrivial group which has no proper nontrivial serial subgroups is called absolutely simple. Prove that a series is a composition series precisely when all its factors are absolutely simple. 14. A finitely generated simple group is absolutely simple. (Note: Nonabsolutely simple groups exist [b54].) 15. A n SN-group is a group such that the factors i n every composition series are abelian. Prove that a group G is an SN-group i f and only if every image of a serial subgroup of G is an SN-group. 16. A n Si-group is a group such that all factors in every principal series are abelian. Prove that a group G is an Sl-group if and only i f every quotient group of G is an Sl-group.
12.5. Locally Soluble Groups A g r o u p is locally soluble i f every finitely generated subgroup is soluble. Generally speaking, this type o f g r o u p is m u c h harder t o deal w i t h t h a n locally n i l p o t e n t groups, essentially because the finitely generated subgroups need n o t satisfy max. F o r example, there is n o analogue o f the H i r s c h P l o t k i n Theorem—see [ b 5 4 ] , §8.1. Here is one o f the few positive results that have been p r o v e d about locally soluble groups. 12.5.1 ( M a l ' c e v , M c L a i n ) . / / G is a locally factor of G is abelian.
soluble
group,
every
principal
Proof. O b v i o u s l y i t is enough t o p r o v e t h a t a m i n i m a l n o r m a l subgroup N o f G is abelian. Suppose t h a t this is false a n d let a, b be elements o f N such t h a t c = [a, b ] ^ 1. Since c e N, we must have N = c , so t h a t there are elements g . . . , g o f G such t h a t a, b e ( c , c } . Set FT = G
9
l 9
m
1
9
r
n
382
12. Generalizations of Nilpotent and Soluble Groups
H
m
H
12.5.2. Every locally soluble group G is an SI-group. soluble group has prime order.
Thus a simple
locally
Proof. B y 12.4.1 the g r o u p has a p r i n c i p a l series a n d 12.5.1 shows t h a t the factors are abelian. • O n the other hand, infinite simple S A - g r o u p s were s h o w n t o exist by P. H a l l (see [ b 5 4 ] , §8.4). W e note another simple a p p l i c a t i o n o f 12.5.1. 12.5.3. A locally soluble groups is hyperabelian.
group
with
the minimal
condition
on normal
sub
I n general a locally soluble g r o u p w i t h m i n - n is n o t a C e r n i k o v g r o u p . M o r e o v e r the classes o f locally soluble groups a n d hyperabelian groups are incomparable. F o r m o r e o n these matters consult [ b 5 4 ] .
Locally Soluble Groups with the Maximal Condition on Normal Subgroups 12.5.4 ( M c L a i n ) . Let G be a locally soluble group with the maximal condition on normal subgroups. Then to each integer p > 0 there corresponds an integer m = m(p, G) such that G < y (• • • y ( y ( G ) • • •) for every sequence of p posi tive integers r r , . . . , r . ( m )
r2
u
2
ri
p
Proof. I f we define m(0, G) = 0, the assertion is vacuously true for p = 0. Assume that m = m(p, G) has been p r o p e r l y defined. N o w G / G is finitely generated because i t is a soluble g r o u p w i t h m a x - n (see 5.4.21). Hence G = XG for a suitable finitely generated subgroup X. T h e n X is soluble, w i t h derived l e n g t h d, let us say. N o w choose any sequence o f p + 1 positive integers r , . . . , r . F r o m G = XG i t follows via Exercise 5.1.7 t h a t G - Xy (G ). But X = 1, so ( m + 1 )
(m+1)
(m+1)
1
p + 1
(m)
(d)
rp+i
()
G d
< ^y JG^) X
rp
F i n a l l y define m(p + 1, G) t o be d.
< y Jy rp
rp
• • • y ( G ) • • •)• ri
•
U s i n g this l e m m a an interesting c r i t e r i o n for a locally soluble g r o u p w i t h m a x - n t o be soluble m a y be established.
12.5. Locally Soluble Groups
383
12.5.5 ( M c L a i n ) . Let G be a locally soluble group with the maximal condition on normal subgroups. Then G is soluble if and only if finitely generated sub groups of G have bounded nilpotent lengths. Proof.
O n l y the sufficiency requires any discussion. Assume t h a t every
finitely generated s u b g r o u p o f G has n i l p o t e n t length at m o s t /. Just as i n (m+1)
the p r o o f o f 12.5.4 we have G = XG
for some finitely generated
g r o u p X. Since X has n i l p o t e n t l e n g t h < /, there exist integers r t h a t y (--y (X)-') ri
sub
..., r such
u
t
= 1. N o w b y 12.5.4 there is an integer m = m(/, G) such
ri
that +1
G<*> < y (• • • y ( G ) • • •) < y ,(• • • y (X)• ri
ri
r
+1
• • )G*" > = G<* >.
ri
( m )
Hence L = G satisfies L = L . I f L = 1, then G is soluble. Otherwise b y m a x - n we can choose a n o r m a l s u b g r o u p M o f G w h i c h is m a x i m a l subject t o M < L . B u t then L/M is a p r i n c i p a l factor o f G a n d 12.5.1 shows t h a t L/M is abelian, w h i c h conflicts w i t h L = U. • 12.5.6 ( M c L a i n ) . A locally supersoluble mal subgroups is supersoluble.
G with the maximal
condition
on nor
Proof. I f X is a finitely generated s u b g r o u p o f G, then X is supersoluble and X' is n i l p o t e n t by 5.4.10. T h u s X has n i l p o t e n t length 2 or less. N o w a p p l y 12.5.5 t o conclude t h a t G is soluble. B y 5.4.21 the g r o u p G is finitely gener ated a n d hence supersoluble. •
E X A M P L E . There is a locally soluble group with m a x - n which is not soluble and not finitely generated. F o r each positive integer i we construct a finite soluble g r o u p G w i t h a unique m i n i m a l n o r m a l subgroup N . T o start the c o n s t r u c t i o n let G be the symmetric g r o u p o f degree 3 a n d N the alternat i n g subgroup. Suppose t h a t G has been constructed. I f p is a p r i m e n o t d i v i d i n g \ G \, there exists a faithful irreducible module N for G over GF(p) (Exercise 8.1.4). Define G t o be the semidirect p r o d u c t o f N and G. Then N is the u n i q u e m i n i m a l n o r m a l subgroup o f G since i t is selfcentralizing i n G . t
t
x
x
{
t
i+1
t
i+1
i+1
i+1
t
i+1
i+1
Define G t o be the u n i o n o f the c h a i n o f groups G < G < • • • . T h e n G is a locally soluble g r o u p a n d i t is also a t o r s i o n g r o u p . C e r t a i n l y G is n o t finitely generated—otherwise G = G for some i. F i n a l l y we show that G has max-n. L e t 1 ^ A < a G; then N n G ^ l for some j . N o w J V n G < G ; thus N < N i f Nn G ^ 1 since A is the unique m i n i m a l n o r m a l subgroup o f G . Therefore N contains
2
t
;
t
t
t
f
£
j + 1
384
12. Generalizations of Nilpotent and Soluble Groups
EXERCISES 1 2 . 5
1. The class of locally soluble groups is not closed with respect to forming exten sions (see Exercise 12.4.9). 2. A locally soluble group need not contain a nontrivial normal abelian subgroup. (Hence locally soluble does not imply hyperabelian.) 3. Let H and K be normal locally polycyclic subgroups of a group. Then the prod uct J = HK is locally polycyclic. Deduce that every group has a unique maximal normal locally polycyclic subgroup and this contains all ascendant locally poly cyclic subgroups. [Hint: Imitate the proofs of 12.1.2-12.1.4.] 4. I f G is a locally soluble group with max-n, some term of the (transfinitely extended) derived series equals 1. (Such groups are called hypoabelian.) 5. (McLain). A principal factor of a locally polycyclic group is elementary abelian. Deduce that a locally polycyclic group with min-n is a torsion group. 6. (McLain). Let G be a locally polycyclic group with min-n. Prove that G is a Cernikov group i f and only i f there is an upper bound for the rank of a principal factor of a finite subgroup. [Hint: Let r be this upper bound. Suppose that N is a minimal normal subgroup which is an infinite elementary abelian p-group. Choose a linearly independent subset {a a , a + i } °f A and put A = < a . . . , a }. Find a finite subgroup H containing A such that L == A = a for all 1 # a e A. Choose M maximal i n L subject to M < H and a $ M , and show that MnA = 1.] l9
2
r
H
l 9
H
r+1
1
7. (McLain). A locally supersoluble group with min-n is a Cernikov group. (Note: There exist locally soluble groups with min-n which are torsion groups but which are not Cernikov groups—see M c L a i n [a 140].)
CHAPTER 13
Subnormal Subgroups
A l t h o u g h s u b n o r m a l i t y is a very n a t u r a l generalization o f n o r m a l i t y , i t re ceived n o a t t e n t i o n f r o m g r o u p theorists u n t i l 1939 w h e n Wielandt's funda m e n t a l paper [ a 2 1 5 ] appeared. H o w e v e r there has been m u c h activity i n this field i n recent years. F o r a full account o f the subject see [ b 4 2 ] . W e shall often w r i t e FT sn G to denote the fact t h a t FT is a s u b n o r m a l subgroup o f a g r o u p G. T h e most elementary properties o f this r e l a t i o n are o u t l i n e d i n Exercise 3.1.8.
13.1. Joins and Intersections of Subnormal Subgroups A useful t o o l i n the study o f s u b n o r m a l i t y is the series of successive normal closures. I f X is a n o n e m p t y subset o f a g r o u p G, a sequence o f subgroups X\ i = 0, 1, 2 , . . . , is defined b y the rules G,
G
X >°
= G
and G,i
Thus X is contained i n every X G 2
G 1
G
i+1
XG
X>
=X '\
and G 0
••'X ' *aX -
= G. G
O f course, X is j u s t the n o r m a l closure X . I t should be clear t o the reader h o w t o extend the series transfinitely—see Exercise 13.1.12. The significance o f this series for s u b n o r m a l i t y is made apparent b y the f o l l o w i n g result.
385
13. Subnormal Subgroups
386
13.1.1. Let H sn G and suppose
that H = H ^i
H_
n
Gl
nite series from H to G. Then H
n
1
H .
t
Gi
Proof The assertion is true for i = 0. I f H < H then H H? \ = H and the result follows by i n d u c t i o n o n i.
G
h
+
= G is a fi
0
G,n
< H and hence H =
i
+
1
RGi
= H
< •
i+1
G,n
Consequently, a subgroup H is subnormal in G if and only if H = H for some n < 0. M o r e o v e r , i f H is s u b n o r m a l i n G, i t also follows f r o m 13.1.1 that o f a l l series between H a n d G the series o f successive n o r m a l closures is shortest. The length o f this shortest series is called the subnormal H i n G. This w i l l be w r i t t e n 5 ( G : FT).
index or defect o f
O b v i o u s l y s ( G : FT) equals 0 precisely w h e n H = G, w h i l e s ( G : H) = 1 i f and o n l y i f G and H ^ G. A n o t h e r evident fact is this: i f H sn K sn G, then H sn G and 5 ( G : H) < s(G : X ) + s(K : H ) . a
I n addition, i f H sn X < G and a is a h o m o m o r p h i s m f r o m G, then FP sn K and 5 ( X : H ) < s(K : H ) . a
a
As regards the d i s t r i b u t i o n o f defects o f a g r o u p there are basically t w o situations w h i c h can arise. 13.1.2. / / a group G has a subnormal subgroup with positive defect i , it has a subnormal subgroup with defect i — 1. Hence either there is an integer s > 0 such that G has subnormal subgroups with defects 0, 1 , 5 but none of de fect greater than s, or else all nonnegative integers occur as defects of sub normal subgroups of G, 01 1
Proof. I f H sn G and s(G : H) = i > 0, then s(G : H ' ) n o w follows.
= i -
1. The l e m m a •
Some interest attaches t o groups which have bounded defects; these i n clude, o f course, a l l finite groups a n d also a l l n i l p o t e n t groups (see the p r o o f of 5.2.4). F u r t h e r examples w i l l be encountered i n 13.3. G,i
There is a useful f o r m u l a for Gi
13.1.3. If H < G, then H
H .
= FT[G, FT] for all i > 0. f
Proof. This is t r i v i a l i f i = 0. A s s u m i n g the result for i a n d using 5.1.6, we argue that H
G , i
+
l
=
HO»
H
=
H
[ G ,
l
H
]
=
H
[
G
?
D
This f o r m u l a gives another p r o o f o f the result: i f G is a n i l p o t e n t g r o u p o f class c and FT < G, then FT sn G and s(G : FT) < c.
13.1. Joins and Intersections of Subnormal Subgroups
387
Intersections of Subnormal Subgroups I t is easy t o see t h a t the intersection o f any finite set o f s u b n o r m a l sub groups is itself s u b n o r m a l . M o r e generally there is the f o l l o w i n g fact. 13.1.4. Let {H \XeA} be a set of subnormal subgroups of a group G such that s(G : H ) < s for all X. Then the intersection I of the H s is subnormal in G and s(G : I) < s. k
k
Proof.
k
If x e I
G , S
1 3 . 1 . 1 , so t h a t I
G
,s
, then clearly x e Hf s
for a l l X i n A ; therefore x e H
k
= L
by •
Nevertheless the intersection o f an a r b i t r a r y collection o f s u b n o r m a l sub groups m a y well fail t o be s u b n o r m a l . x
- 1
2
E X A M P L E . Consider the infinite d i h e d r a l g r o u p G = <x, a\a = a , x = 1 > a n d set H = <x, a }. T h e n H ^ H since [a \ x~\ = a~ \ Consequently H sn G . H o w e v e r H r\H r^"= <x>, a subgroup t h a t coincides w i t h its n o r m a l i z e r i n G . Hence the intersection is n o t s u b n o r m a l i n G . 21
2
t
i+1
t
1
2l+
t
2
Joins of Subnormal Subgroups A n altogether m o r e subtle p r o b l e m is t o determine whether the j o i n o f a p a i r — o r m o r e generally o f any s e t — o f s u b n o r m a l subgroups is s u b n o r m a l . I f turns o u t t h a t t w o s u b n o r m a l subgroups m a y well generate a subgroup t h a t is n o t s u b n o r m a l . The f o l l o w i n g fact is basic. 13.1.5. Let H sn G and K sn G, and assume that K normalizes
Proof. I n the first place, i f H = H \
then H = H [ G , H\ b y 1 3 . 1 . 1 3 . There
t
fore K normalizes H . N e x t H ^H K and elementary properties o f s(H K : H K) < s(G : K). s(G : H\ i t follows t h a t J sn
H. Then J =
t
t
t
i+1
t
i+1
t
K sn H K. Therefore H KsnH K by the s u b n o r m a l i t y already m e n t i o n e d , a n d indeed Since H K = G a n d H K = HK = J i f r = G a n d s ( G : J) < s(G : H) s(G : K). • t
i+1
0
t
r
T h i s result can be used t o reformulate the j o i n p r o b l e m for a pair o f s u b n o r m a l subgroups. 13.1.6. Let H sn G, K sn G and J =
Then the following
statements
13. Subnormal Subgroups
388
(i) J sn G; K
(ii) H sn G; a n d (iii) [ # , X ] sn G. Proof. The i m p l i c a t i o n s (i) - • (ii) - • (iii) are t r i v i a l consequences o f the rela tions [ t f , X ] < a H ^ J. N o w suppose t h a t (iii) holds. Since H = X] and FT normalizes [FT, X ] , we deduce f r o m 13.1.5 t h a t H snG. Similarly J = H K and 13.1.5 implies t h a t J sn G. • K
K
K
K
A g r o u p is said t o have the subnormal joint property (SJP) i f the j o i n o f every p a i r — a n d hence o f every finite s e t — o f s u b n o r m a l subgroups is sub n o r m a l . N o w the set o f a l l s u b n o r m a l subgroups o f a g r o u p is a p a r t i a l l y ordered subset o f the lattice o f a l l subgroups, a n d i t is closed under finite intersections. Hence a group has the SJP exactly when the set of all its sub normal subgroups is a sublattice of the lattice of subgroups. Some examples of groups w i t h the SJP are given b y 13.1.7. Every group with nilpotent derived subgroup has the subnormal property. In particular this conclusion applies to metabelian groups.
join
T o p r o v e this one observes that, i n the n o t a t i o n o f 13.1.6, the subgroup [FT, X ] is s u b n o r m a l i n G' because the latter is n i l p o t e n t . Hence [FT, X ] sn G and 13.1.6 gives the result. The next result is o f quite a different character, asserting that i f a g r o u p is sufficiently finite i t has the SJP, whereas i n 13.1.7 the hypothesis is a f o r m o f c o m m u t a t i v i t y . Indeed i t is the i n t e r p l a y between finiteness a n d c o m m u t a t i v i t y w h i c h makes the SJP such an elusive p r o p e r t y . 13.1.8 (Robinson). Let G be a group whose derived subgroup satisfies maximal condition on subnormal subgroups. Then G has the subnormal property.
the join
Proof. L e t H sn G, X sn G, a n d J = < H , X > . P u t s = s(G: H). I f 5 = 0, then H = J = G a n d a l l is clear; assume therefore t h a t 5 > 0. L e t { x x , . . . , x } be a given finite subset o f X a n d p u t 1 ?
2
n
X
L =
X
H \...,H »}
Xi
Since s(H : H } = 5 — 1 , repeated use o f an i n d u c t i o n hypothesis o n 5 gives L sn H a n d hence L sn G. N o w the e q u a t i o n h = h[h, x] implies t h a t < H , H ) = < H , [ H , x ] > . Consequently L = < H , M > where M is the sub g r o u p generated b y [H, x ],..., [H, x ~\. Since H normalizes [FT, x j , i t normalizes M , f r o m w h i c h i t follows t h a t M < L a n d thus M sn G. T h a t M sn G' is an i m m e d i a t e consequence. O n the basis o f max-s we can find a subgroup M w h i c h is m a x i m a l o f the above type. B u t M m u s t equal [FT, X ] since one can always a d d another [FT, x J t o M. Hence [FT, X ] sn G, w h i c h by 13.1.6 implies t h a t J sn G. • G
x
x
l
n
13.1. Joins and Intersections of Subnormal Subgroups
389
T h e f o l l o w i n g special cases are n o t e w o r t h y . 13.1.9 ( W i e l a n d t ) . A group subgroups
satisfying
has the subnormal join
the maximal
13.1.10 (Wielandt). In a group G with a (finite) all subnormal
subgroups
= (H \A
subnormal
series the set of
sn G. I f A
K
0
subgroups.
is a finite subset
e A > sn G b y 13.1.9. Since G satisfies max-s, there is
X
a m a x i m a l subgroup J
on
of the lattice of
e A > where H
X
A Q
composition
is a complete sublattice
Proof of 13.1.10. L e t J = (H \A of A , t h e n J
condition
property.
0
A Q
. B u t clearly J
equals J a n d J sn G. T h e argument
A o
for intersections is similar.
•
An Example of a Group Without the Subnormal Join Property L e t Sf denote the set o f a l l subsets X o f the integers Z such t h a t X contains all integers less t h a n some integer l(X) a n d none greater t h a n some
L(X)
where l(X) < L(X). T h u s , r o u g h l y speaking, Sf consists o f subsets t h a t con t a i n a l l large negative integers b u t n o large positive ones. W i t h each X i n Sf we associate symbols a
x
a n d b . L e t A a n d B be elementary abelian 2-groups x
h a v i n g as bases the sets {a \X
e Sf)
x
a n d {b \X x
e Sf)
respectively. N o w
f o r m the direct p r o d u c t M
x B
= A
T h e next step is t o define suitable a u t o m o r p h i s m s o f M. Define a * x
a
xu{n}
if
n
n
t o be
4 X a n d 1 i f n e X: a similar c o n v e n t i o n applies t o the b's. F o r
each integer n, a u t o m o r p h i s m s u a n d v o f M act a c c o r d i n g t o the f o l l o w n
n
i n g rules. F i r s t l y u acts t r i v i a l l y o n B a n d v acts t r i v i a l l y o n A ; secondly n
n
u: a
t-+a b * ,
v :b
t-+a * b .
n
x
n
x
x
x
x
n
n
x
I t is easy t o see t h a t these are i n fact a u t o m o r p h i s m s o f M. Consider the subgroups o f A u t M H = (u \neZ}
and
n
K =
and put J
=
F i n a l l y f o r m the semidirect p r o d u c t G = J K
M.
C o n c e r n i n g the g r o u p G we shall p r o v e the f o l l o w i n g .
(v \neZ}, n
390
13. Subnormal Subgroups
13.1.11. The subgroups
H and K are subnormal
3, but J is not subnormal
in G. Thus
in G with defect
equal
to
G does not have the subnormal
join
property. Proof. T h e f o l l o w i n g equations are direct consequences o f the definitions: [a ,
w j = b*,
x
x
[b , w j = 1,
n
(1)
x
and b
l"x> *>»] = 1,
t x,
*>„] = a * . x
(2)
n
I t is also easy t o check f r o m the definitions t h a t u u m
2
= uu
n
n
and u
m
= 1,
relations w h i c h show H t o be a n elementary abelian 2-group; o f course K too is o f this type. Now
write Z
mn
=
W
C m? ^ti\-
U s i n g the H a l l - W i t t i d e n t i t y (5.1.5) n
ax
[ u , v~\ a Y \v , m
x
1
a \ u ~\ [a ,
n
x
m
u " , v Y™ = 1,
x
n
together w i t h (1) a n d (2), we o b t a i n [_a , z x
] =
m n
a* *l x
m
n
similarly Z
b
[ ^ J mn\ ~ X*m*nBy the H a l l - W i t t i d e n t i t y once again Zrnn
1
l"x, Zml, Ui~] lz ,
l
ax
u f , a J [u
mn
x
a \ z ]
h
x
= 1.
mn
E v a l u a t i n g this w i t h the a i d o f (1) a n d (2) we deduce t h a t [a , x
w i t h a c o r r e s p o n d i n g result for b . Hence [ z x
= 1. W h a t these equations show is t h a t z
mn
m n
[z
m n
, w j ] = 1,
, u{\ = 1 a n d likewise [ z
m n
, v{\
belongs t o the center o f J. T h u s
[ J ' , J ] = 1 a n d J is n i l p o t e n t o f class 2. E q u a t i o n s (1) a n d (2) i m p l y t h a t [ H , A] < B a n d [ K , B] < A. Ii X e a n d n is the largest integer i n X, w ] = b* n
We
Y
= b , w h i c h implies t h a t [FT, ^4] = B. S i m i l a r l y [ K , 5 ] = A.
n
x
G
G
< H , 5 > . Therefore H M
9
we have
are n o w i n a p o s i t i o n t o calculate the successive n o r m a l closures o f
H a n d K. I n the first place H = H
then, o n w r i t i n g Y = X\{n},
K
= (H ,
M
K
A K
= (H ) K
B}
since H sn J a n d s ( J : H) <2.
A
= (H ) ; K
= H M.
also H
=
Hence we have H
Thus H
G
G
1
RK M
=
2
(H ) G
' = < H , B}. F i n a l l y F T '
3
=
FT because [FT, 5 ] = 1. N a t u r a l l y a similar a r g u m e n t applies t o K. G
However J
G
= G because J
A
contains b o t h H
=
B
and K
< X , ^4>. Consequently J is n o t s u b n o r m a l i n G.
= •
Joins of Infinitely Many Subnormal Subgroups I f the j o i n o f an a r b i t r a r y set o f s u b n o r m a l subgroups is always s u b n o r m a l , the g r o u p i n question is said t o have the generalized erty. T h i s is a m u c h stronger p r o p e r t y t h a n the SJP.
subnormal
join
prop
13.1. Joins and Intersections of Subnormal Subgroups
l
1
391
1
E X A M P L E . L e t G = (x a \af = 1 — xf, a? = a^ } be a d i h e d r a l g r o u p o f order 2 ; then H = (x } is s u b n o r m a l i n G w i t h defect t. N o w consider the direct products t
h
t
m
t
G = G
(
1
x G
{
2
x -
and
H = H
x
x H
2
x • •.
T h e n o b v i o u s l y H sn G a n d the H generate FT. I f H were s u b n o r m a l i n G a n d s ( G : H) = r, i t w o u l d f o l l o w t h a t s(G : H ) < r + 1 for a l l i, a contradic t i o n . T h u s G does n o t have the generalized SJP. H o w e v e r G does have the SJP because i t is a metabelian g r o u p (see 13.1.7). t
t
t
t
There is a c r i t e r i o n for a g r o u p t o have the generalized SJP w h i c h p r o vides some insight i n t o the nature o f t h a t p r o p e r t y . 13.1.12 (Robinson). A group G has the generalized subnormal join property and only if the union of every chain of subnormal subgroups is subnormal.
if
Proof. T h e c o n d i t i o n stated is surely necessary. L e t us assume therefore t h a t i t is satisfied i n G. T h e first step is t o p r o v e that G has the SJP. T o this end let H sn G, K sn G a n d J = < i f , K}: we shall proceed b y i n d u c t i o n o n 5 = s ( G : H), w h i c h can be assumed positive. T h e elements o f K m a y be w e l l ordered as {x \oc < y} where y is some o r d i n a l . F o r ft
Xa
K
y
K
p
p
p
1
p
G
1 ?
Xpi
G
p
x
N o w let Sf be a possibly infinite set o f s u b n o r m a l subgroups o f G. W e need t o prove t h a t the j o i n o f a l l the members o f Sf is s u b n o r m a l . Since G has the SJP, i t is permissible t o assume that Sf is closed under the f o r m a t i o n o f finite j o i n s . T h e given c h a i n c o n d i t i o n a n d Z o r n ' s L e m m a can be used t o produce a m a x i m a l element J o f Sf. I f H e Sf, then < i f , J > e Sf, whence FT < J. Consequently J is the j o i n o f all the members o f Sf. B u t J sn G, so G has the generalized SJP. • Some classes o f groups t h a t have the generalized SJP can be read off f r o m the next result. 13.1.13 (Robinson). L e t i V < i G and assume that N has the generalized sub normal join property while G/N satisfies the maximal condition on subnormal subgroups. Then G has the generalized subnormal join property. Proof. L e t { f f j a e ^4} be any c h a i n o f s u b n o r m a l subgroups o f G a n d let U denote the u n i o n o f the chain. B y 13.1.12 i t is enough t o prove t h a t U sn G. N o w max-sn implies t h a t I W = H N for some a i n A. Hence U = a
392
13. Subnormal Subgroups
U n (H N)
= HJJJ
a
{H nN\(xe
A}
a
n N).
O b v i o u s l y U nN
a n d H r\A
certainly UnNsnG.
is the
u n i o n o f the
sn A . T h u s by hypothesis
a
F i n a l l y UnN^iU,
UnNsnN
so i t follows f r o m
chain and
13.1.5 t h a t
U sn G.
•
F o r example, finitely generated metabelian groups have the generalized SJP, whereas this is n o t true o f a r b i t r a r y metabelian groups, as we saw i n the example above. F o r m o r e classes o f groups w i t h the SJP, see [ b 4 2 ] .
EXERCISES
13.1
1. Find all subnormal subgroups of the groups S Z)„, D^. n9
2. Let H sn G, K sn G and J =
i
f
t
i+1
{
4. There exists a finitely generated soluble group which does not have the SJP. [Hint: I n the notation of 13.1.11 let t e Aut M be defined by a \->a b \-+b where X + 1 = {x + 1 |x e X}. Let L = <J, t} K M and prove that L is finitely generated.] x
x+l9
x
x+1
5. Let D = D r G where G ~ G. I f D has the generalized SJP, prove that there is an upper bound for subnormal defects i n G. A
x
A
< x >
6. Let H < G and X c G. Show that for any positive integer i the equation H = H [H, < X > ] holds where S - l u l u ( X X ) u • • • u ( X • • • X ) . [ i f m t : Use 5.1.6.] Si
f
7. I f H < G and X is a finitely generated subnormal nilpotent subgroup of G, then H is generated by finitely many conjugates of H i n K. K
8. Let H sn G, X sn G and assume that H. I f H and X belong to a class of groups X which is closed with respect to forming normal subgroups and finite normal products, prove that J belongs to X. Give some applications. 9. Generalize the previous exercise to the case where J = HK = KH. 10. (J.E. Roseblade and S.E. Stonehewer). Let X be a class of groups which is closed with respect to forming normal subgroups and finite normal products. Assume that H sn G, K sn G and J =
13.2. Permutability and Subnormality
393
12. Show how to extend the series of successive normal closures of a subgroup to transfinite ordinals. Use this to give a criterion for a subgroup to be descendant, that is, a term of a descending series.
13.2. Permutability and Subnormality Recall t h a t subgroup H is said t o be permutable whenever K < G. W e shall sometimes w r i t e
i n a g r o u p G i f HK = KH
i f per G to denote this r e l a t i o n . O f course every n o r m a l subgroup is permutable, w h i c h m i g h t lead one t o hope t h a t s u b n o r m a l subgroups also have this p r o p e r t y . H o w e v e r any such hope is soon dispelled. I f G is a d i h e d r a l g r o u p o f order 8, i t can be gener ated by t w o subgroups H a n d K each o f order 2. N o w H sn G a n d K sn G since G is n i l p o t e n t . H o w e v e r \HK\ = \H\- \K\ = 4, so t h a t G HK, a n d HK^KH by 1.3.13. O n e m a y a d o p t the opposite p o i n t o f view, asking whether permutable subgroups are s u b n o r m a l . H e r e there is an encouraging answer for finite groups at least, as we shall soon see. 13.2.1 (Ore). If H is a maximal H^ G.
permutable
subgroup
of a group
G, then
Proof Suppose t h a t this is false; then there is a conjugate K o f H such that K^H. N o w HK < G a n d clearly HK per G. Therefore G = HK a n d K = H for some he H,ke K. H o w e v e r this implies t h a t H = K. • hk
This has immediate application to permutable subgroups o f finite groups. 13.2.2 (Ore). / / H is a permutable subnormal in G.
subgroup
of a finite
group G, then H is
Proof Refine 1 < H < G t o a c h a i n 1 = G < G < • • • < G = G such t h a t Gi is a m a x i m a l p r o p e r p e r m u t a b l e subgroup o f G . B y 13.2.1 we have G;
x
n
i+1
i+1
W h i l e i n general a p e r m u t a b l e s u b g r o u p o f an infinite g r o u p need n o t be s u b n o r m a l (Exercise 13.2.3), such subgroups are i n v a r i a b l y ascendant. T o p r o v e this i t is necessary t o establish a technical lemma. 13.2.3. Let G = HK where H per G and K =
