Solutions to Chapter 4
4..1 The first part is achieved by a standard matrix reduction using the
elementary operations of the forms
(a) add an integer multiple of one row/column to another;
(b) interchange two rows/columns;
(c) multiply a row/column by —1.
In this case the reduction begins and ends as follows :
37 27 47" 1 0 0
52 37 67 ~ ... ~ 0 35 0
59 44 74 0 0 0
This then shows that
G~{x,y,z
If we now add the relation a3b2c4 = 1 to those of G then a corre-
sponding matrix reduction gives
37 27 47 1 0 0
52 37 67 0 7 0
59 44 74 ~ '"~ 0 0 0
3 2 4 0 0 0
Thus adding the relation a3b2c4 = 1 to the relations of G changes G to
the group CV x C^. Hence the relation a3b2c4 — 1 cannot hold in G.
However, it follows immediately from the first two relations of G that
, Book 5 Groups
a i5 6 io c 2o = x a n d s o t h e o r d e r o f a 3 6 2 c 4 divides 5. As the order is not
1, it must then be 5.
From the second and third relations it is clear that (abc)7 = 1 and so
abc has order 1 or 7. Adding the relation abc = 1 to those of G gives a
matrix that reduces to
"1 0 0"
0 5 0
0 0 0
0 0 0_
which corresponds to the group C 5 x Coo- Thus we see that abc ^ 1 in
G and that consequently abc has order 7.
Now in an abelian group, if x has order m and y has order n then xy
has order l.c.m.(m, n). Thus we deduce that
a4b3c5 = a3b2c4 • abc
has order 5 • 7 = 35.
4-2 (a) The relation matrix for G reduces as follows :
2 3 6 1 0 0
4 9 4 0-2 0
Consequently, G ~ (x, t/, z \ x = y 2 = z° = 1) c^ C^ x
(b) The relation matrix for G reduces as follows :
2 3 6 1 0 0
4 9 4 ~ 0 1 0
3 3 2 0 0 -66
Consequently G ~ CQQ,
2
4*$ Froml abab = 1 we have bob = b~1a~1. Hence babab — 1 and so
x
bab~ a~ = 1, from which it follows that G is abelian.
G is the infinite cyclic group. In fact, a relation matrix for the abelian
group G is
[2 3]~[2 l ] ~ [ 0 l ] .
Alternatively, it is easy to see that G = {ab) since b = [ab)~2 and
a = {ab)3.
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4..1 The first part is achieved by a standard matrix reduction using the
elementary operations of the forms
(a) add an integer multiple of one row/column to another;
(b) interchange two rows/columns;
(c) multiply a row/column by —1.
In this case the reduction begins and ends as follows :
37 27 47" 1 0 0
52 37 67 ~ ... ~ 0 35 0
59 44 74 0 0 0
This then shows that
G~{x,y,z
If we now add the relation a3b2c4 = 1 to those of G then a corre-
sponding matrix reduction gives
37 27 47 1 0 0
52 37 67 0 7 0
59 44 74 ~ '"~ 0 0 0
3 2 4 0 0 0
Thus adding the relation a3b2c4 = 1 to the relations of G changes G to
the group CV x C^. Hence the relation a3b2c4 — 1 cannot hold in G.
However, it follows immediately from the first two relations of G that
, Book 5 Groups
a i5 6 io c 2o = x a n d s o t h e o r d e r o f a 3 6 2 c 4 divides 5. As the order is not
1, it must then be 5.
From the second and third relations it is clear that (abc)7 = 1 and so
abc has order 1 or 7. Adding the relation abc = 1 to those of G gives a
matrix that reduces to
"1 0 0"
0 5 0
0 0 0
0 0 0_
which corresponds to the group C 5 x Coo- Thus we see that abc ^ 1 in
G and that consequently abc has order 7.
Now in an abelian group, if x has order m and y has order n then xy
has order l.c.m.(m, n). Thus we deduce that
a4b3c5 = a3b2c4 • abc
has order 5 • 7 = 35.
4-2 (a) The relation matrix for G reduces as follows :
2 3 6 1 0 0
4 9 4 0-2 0
Consequently, G ~ (x, t/, z \ x = y 2 = z° = 1) c^ C^ x
(b) The relation matrix for G reduces as follows :
2 3 6 1 0 0
4 9 4 ~ 0 1 0
3 3 2 0 0 -66
Consequently G ~ CQQ,
2
4*$ Froml abab = 1 we have bob = b~1a~1. Hence babab — 1 and so
x
bab~ a~ = 1, from which it follows that G is abelian.
G is the infinite cyclic group. In fact, a relation matrix for the abelian
group G is
[2 3]~[2 l ] ~ [ 0 l ] .
Alternatively, it is easy to see that G = {ab) since b = [ab)~2 and
a = {ab)3.
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