Basic Algebra

Solutions to Chapter 1




1.1 Let G be partitioned by the set {Hxa \ a G A} of cosets of H, and let
H be partitioned by the set {Kyp \ f$ G B} of cosets of K. Suppose
that g G G. Then we have g G jETaja for some a £ A and so # = hxa
for some (unique) h G if. But & / G if ^3 for some /? G JB and so we
have that # = kypxa for some A; G if. Thus we see that every element
of G belongs to a coset Kypxa for some ft £ B and some a G A. The
result now follows from the fact that if Ky@xa = Ky^xa> then, since
the left hand side is contained in the coset Hxa and the right hand side
is contained in the coset Hxa/, we have necessarily xa = xa', which
gives Kyp = Ky^ and hence yp = y$>.
Now observe that (HnK)x = HxnKx for all subgroups H and K of
G. Then, if H and K have finite index, the fact that there are finitely
many cosets Hx and Kx implies that there are only finitely many cosets
of Hf] K, so Hn K is also of finite index. The result for the intersection
of a finite number of subgroups now follows by induction.
The result is not true for an infinite number of subgroups each of finite
index. To see this, consider the additive group Z. The subgroup nZ has
index n; but Hn>i n ^- = i®} w n i ° n 1S n °t °f finite index in Z.
1.2 If xH is a subgroup of G then we have 1 G xH which gives x~x G
x~lxH = H and hence x G H, so that xH = if.
That (p is a mapping (or, as some say, is well-defined) follows from
the observation that

,Book 5 Groups

It is clear that <p is a bijection.
ip on the other hand is not a mapping. To see this, take for example
G = GL(2, Q) and
[a b
ac
0 c
We have that
1 0

which is immediate from the observation that if
1 0 D 1 1
A=
1 1 1 2
then
-1 D _ 1 1
A~lB eH.
0 1
However,
i oi, ri
" i lj7 ii
since equality here would give the contradiction
0 1
-1 2
eH.
1.8 Consider the subgroups H and K of the group G = GL(2, Q) given by

0
It is readily seen that
ab b
HK =

But ^TJFC is not a subgroup of G since, for example, the matrices

1 lj 1 [o 1
are each in HK but the product
[2 l ] [ l l ] = [ 2 3
[l l j [ o l j [ l 2
is not.
30

, Solutions to Chapter 1
1.4 The right cosets are


#(13) = {(13), (132)}
#(23) = {(23), (123)}.

The left cosets are

(13)1/= {(13), (123)}


It is clear from this that H is not normal in G.
1.5 Since m and n are coprime there exist integers a and b with am + bn = 1.
Then, using the fact that gn = 1, we have

g = g1 = 0a™+fc" = 0 o m / w = (0 m )°(0*) 6 = ($ m ) a .

Since it is given that #m G # we have that (# m ) a € if and hence g £ H.
1.6 (i) If H is a subgroup of G then clearly i T ^ C H\ and, since every
subgroup contains the identity element, we have H — 1H C HH.
(ii) Given x E l w e have zX C XX = X. Since y H-> a;y is injective
(by the cancellation law) we deduce that \xX\ = |X| and hence, since
X is finite, that xX = X. Consequently, x = see for some e G X. The
cancellation law gives e = 1, and so we have that 1 G X. We now observe
from 1 G xX that 1 = xy for some y G X, which gives x" 1 = y G X. It
now follows from the fact that XX C X that X is a subgroup of G.
That (ii) no longer holds when X is infinite may be seen, for example,
by taking G to be the additive group of integers and X to be the set of
non-negative integers.
1.1 Suppose that t G HxK n yK / 0. Then (with a notation that is self-
explanatory) we have t = /izA; and £ = yk\. If now 5 G yif then

5 = yk2 = tk~[lk2 = hxkk^xk2 G iiTziif,

and hence yif C HxK.
Suppose now that t G # z i f D HyK ^ 0. Then we have t = /ixx^i
and i = h2yk2. Ii now s G # £ # then

a = hxk = h{h~[ltk^l)k = hh
Thus Fxiir C jyyiT, and similarly ^yiif C HxK.
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