Page 50 - DMTH403_ABSTRACT_ALGEBRA

Unit 3: Subgroups

The whole argument of Theorem 4 remains valid if we take a family of subgroups instead of just Notes
two subgroups. Hence, we have the following result.

H
Theorem 4 (a): If   is a family of subgroups of a group G, then  H H is also a subgroup
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
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
of G.
Now question arises that does the union of two or more subgroup is again a subgroup. Lets see
its true or not. Consider the, two subgroups 2Z and 3Z of Z. Let S = 2Z  32. Now, 3  32  S,
2  22  S, but 1 = 3 – 2 is neither in 2Z nor in 3Z. Hence, S is not a subgroup of (Z, +). Thus, if A
and B are subgroups of G, A  B need not be a subgroup of G. But, if A  B, then A  B = B is a
subgroup of G. The next exercise says that this is the only situation in which A  B is a subgroup
of G.
Let us now see what we mean by the product of two subsets of a group G.
Definition: Let G be a group and A, B be non-empty subsets of G.

The product of A and B is the set AB = { ab | a  A, b  B).
For example, (2Z) (3Z) = { (2m) (3m) | m, n  Z)
= { 6mn | m, n  Z }
= 62.
In this example we find that the product of two subgroups is a subgroup. But is that always so?
Consider the group
S = {I, (1 2), (1 3), (2 3), (1 2 3), (1 3 2)}, and its subgroups H = { I, (1 2) } and K = { I, (1 3)).
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 1 2 3   1 2 3 
(Remember, (1 2) is the permutation  2 1 3  and (1 2 3) is the permutation  2 3 1  .
   
Now HK = {I × I , I × ( 1 3 ), (1 2) × I,(1 2) × (1 3)}
= { I, (1 3), (1 2), (1 3 2) }
HK is not a subgroup of G, since it is not even closed under composition. (Note that (1 3)  (1 2)
= (1 2 3)  HK.)
So, when will the product of two subgroups be a subgroup? The following result answers this
question.

Theorem 5: Let H and K be subgroups of a group G. Then HK is a subgroup of G if and only if HK
= KH.
Proof: Firstly, assume that HK  G. We will show that HK = KH; Let hk  HK. Then

(hk) = k h  HK, since HK  G.
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Therefore, k h = h k for some h  H, k  K. But then hk = (k h ) = k h  KH. Thus,
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HK  KH.
Now, we will show that KH  HK. Let kh  KH. Then (kh) = h k  HK. But HK  G. Therefore,
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((kh) )  HK, that is, kh  HK. Thus, KH  HK.
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Hence, we have shown that HK = KH.
Conversely, assume that HK = KH. We have to prove that HK  G. Since e = e  HK, HK  .
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Now, let a, b  HK. Then a = hk and b = h k for some h, h  H and k, k  K.
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