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P. 52
Unit 3: Subgroups
Now, let us see why <S> A. We will show that A is a subgroup containing S. Then, by the Notes
definition of <S>, it will follow that <S> A.
Since any a S can be written as a = a , S A.
1
Since S , A .
Now let x, y A. Then x = a a n 2 2 ..... a n k k
n
1
1
y = b b .... b m r a , b S for 1 i k, 1 j r.
m
m
2
1
1
2
r
j
j
Then xy = a a n 2 2 ..... a k n k b b .... b m r 1
m
m
n
-1
1
1
2
1
r
2
1
= a a n 2 2 ..... a n k k b 1 m 1 .... b m 1 1 A.
1 n
1
Thus, by Theorem 1, A is a subgroup of G. Thus, A is a subgroup of G containing S. And hence,
<S> A.
This shows that <S> = A.
Note that, if (G, +) is a group generated by S, then any element of G is of the form n a + n a +
1
2 2
l
..... + n a , where a , a ,....., a, S and n , n , ....., n , Z.
1
2
1
2
r
r
r
For example, Z is generated by the set of odd integers S = { ± 1, f3, ± 5, ......). Let us see why. Let
m Z. Then m = 2 where r 0 and s S. Thus, m <S>. And hence, <S> = Z.
r
s
Definition: A group G is called a cyclic group if G = < {a) > for some a E G. We usually write
< {a) > as < a >.
Note that < a > = { a | n " Z ).
n
A subgroup H of a group G is called a cyclic subgroup if it is a cyclic group. Thus, < (12) > is a
cyclic subgroup of S and 22 = <2> is a cyclic subgroup of Z.
3
We would like to make the following remarks here.
Remark: (i) If K G and a K, then < a > K. This is because < a > is the smallest subgroup of
G containing a.
(ii) All the elements of < a > = { a | n Z) may or may not be distinct. For example, take
n
a = (12) S .
3
Then < (1 2) > = { I, (1 2)), since (1 2) = I, (1 2) = (1 2), and so on.
3
2
We will now prove a nice property of cyclic groups.
Theorem 7: Every cyclic group is abelian.
Proof: Let G = < a > = { a | n Z). Then, for any x, y in G, there exist m, n Z such that x = a , y
n
m
= a . But, then, xy = a . a a m+n = a n+m = a . a = yx. Thus, xy = yx for all x, y in G.
n
m
n
m
n
That is G is abelian.
Note Theorem 7 says that every cyclic group is abelian. But this does not mean that
every abelian group is cyclic.
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