Page 51 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Then ab = (hk) (k h ) = h [ (kk ) h ].
-1
-1
-1
-1
-1
1 1 1 1
Now, (kk ) h  KH = HK, Therefore, 3 h k  HK such that (kk )h = h k .
-1
-1
-1
-1
2 2
1
1
1
2 2
1
Then, ab = h(h k ) = (hh )k  HK.
-1
2 2
2
2
Thus, by Theorem 1, HK  G.
The following result is a nice corollary to Theorem 5.
Corollary: If H and K are subgroups of an abelian group G, then HK is a subgroup of G.
3.3 Cyclic Groups
Let us understand the meaning of cyclic group.
Let G be any group and S a subset of G. Consider the family F of all subgroups of G that contain
S, that is,
F = { H | H  G and S  H } .
We claim that F  . Why? Doesn’t G  F? Now, by Theorem 4'(a),  H is a subgroup of G.
H F

Note that
(i) S   H.
H F

(ii)  H is the smallest subgroup of G containing S. (Because if K is a subgroup of G
H F

containing S, then K  F. Therefore,  H  K.)
H F

These observations lead us to the following definition.
Definition: If S is a subset of a group G, then the smallest subgroup of G containing S is called the
subgroup generated by the set S, and is written as <S>.
Thus, <S> =  { H | H  G, S  H }.
If S = , then <S> = {e).
If <S> = G, then we say that G is generated by the set S, and that S is a set of generators of G.

If the set S is finite, we say that G is finitely generated.
We will give an alternative way of describing <S>. This definition is much easier to work with
than the previous one.

Theorem 6: If S is a non-empty subset of a group G, then

n
n
n

a a ..... a |a  S for 1 i  k, n , ..., n  Z .
k
1
2
<S> =  1 2 k i 1 k
a a ..... a |a 

k
Proof: Let A =  1 n 1 n 2 2 n k k i S for 1 i  k, n , ..., n  Z .
1
Since a ,.., a  S  <S> and <S> is a subgroup of G, a  <S>
n
1
1
1
k
 i = 1, ...., k. Therefore, a a n 2 2 ..... a k n k  <S>, i.e., A  <S>.
n
1
1
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