Basic Algebra

Solutions to Chapter 3




3.1 (a) Given [s,t] G [S,T] we have [s,*]" 1 = [*,«] G [F,S]. Hence
[5, T1] C [T,JS] and similarly for the reverse inclusion.
(b) We have that

h-1k-1hk = h-1h' eH since H<G

and
h~1k-lhk = k'keK since K<G.
Hence [#, K) < H n K. In the case where H n K = {1} v/e have
[if, if] = {1} and the elements of H commute with those of K.
(c) The first part follows from [xy,z] = y~1x~1z~1xyz and

y~l\x,z]y{y,z\ = y~lx~lz~lxzyy~lz~lyz.

Clearly [H,K] < [HL,K] and [L,K] < \HL,K\, so

\H,K\\LiK\<[HLiK\. '
But [hl,k] = r ^ M H M ] G [F, A"] [L, A"] since [JT, K] < G. Hence the
required equality follows.
(d) This follows immediately on expanding the commutators.
3.2 (
Let Q 8 = (a, 6 | a2 - 62 = (ab)2) and C 2 = (c | c2 = 1). Then
2
(a 2 5 c) and the upper central series is


The derived group of G is

, Solutions to Chapter 3
and the lower central series is


Thus G is nilpotent of class 2 and the upper and lower central series do
not coincide.
Qs has derived group ( a 2 ) and Z(Q%) = (a2). Thus the upper and
lower central series of Qs are both equal to


8.3 Suppose that G is generated by its subnormal abelian subgroups. Then
G = {G\ | G\ subnormal abelian}.
Now since G\ is subnormal in G we have a series
G\ = RQ < R\ < • • • < Rn = G
with Ri normal in i2t-+i for 0 < i < n — 1. But if H is a quotient group
of G then H ~ G/K and so
G^fiT = GxRo GxRi, GxRn = C
K K ~ K ~ '~ K K
a+
is a series of subgroups with —r^- normal in for 0 < i < n — 1.
TVT K K
Now
K ~ GXC\K
which is abelian since G\ is abelian. Hence

—x ianinf).
subnormal abelian
K \ K K
Therefore H ~ G/K is generated by its subnormal abelian subgroups.
Suppose now that G is a nilpotent group and that H < G. Let


be a central series for G. Consider the series
=
H HGr <: HGr—\ fC • • • \ HG\ \ HGQ — G.
Now iTGi < HGi-i, since for /i, /i' G # , ^ G Gt-, &-_i G G8-_i we have
(/i gi—i) hgih gi—\ = g^_ih hg{h gi—\.
But, for any x EG, z#2_i = gi-\xg\ for some g\ G Ga- so it follows that
g~[\h!~xhgib!gi-.\ G HGi as required. Since every group is generated
by abelian subgroups (the cyclic subgroups generated by its elements),
a nilpotent group is then generated by its subnormal abelian subgroups.
63

, Book 5 Groups

S.4 Apply the fundamental isomorphism
H HK
HnK " K
to the case where G = AiH = Af)C and K = B\ then since B < A we
obtain
ADC _ {AC\C)B
Bnc ~ B
Now apply the isomorphism to the case where G — AC,H = A and
K = BC to obtain
A A{BC)
AnBC~ BC '
But AnBC = B(A n C) and A(BC) = AC, and the result follows.
Suppose that G is a soluble group with a series


where each Gi/Gi-i is abelian. Let H < G and consider the series

{1} = Go n H < Gx n H < . •. < Gn n H = H.

We have

i-i n H

and so —~ rr iss abelian
a (being a subgroup of an abelian group).
Gi—i n xz
Also, if K < (7 then

jfiT _ GQK G\K GnK _ G

Now we have
GjK/K GjK
Gi-XK G,-_1(G1- n if)

which is a quotient group of * . Hence —-—/. is abelian (being
Gi GiK/K
a quotient group of an abelian group).
If H < K and both H and K/H are soluble then we have

^ ^ ^
H ~ H ~ ~ H ~H
64