Page 115 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Let x  G, x  e and H = < x > . Since x  e, o(H)  1

 o(H) = p.
Therefore,  y  G such that y  H. Then, by the same reasoning, K = < y > is of order p. Both H
and K are normal in G, since G is abelian.
We want to show that G = H × K. For this, consider H  K. Now H  K  H.

 o(H  K) | o(H) = p. o(H  K) = 1 or o (H  K) = p.
If o(H  K) = p, then H  K = H, and by similar reasoning, H  K = K. But then,
H = K.  y  H, a contradiction.
o(H  K) = 1, i.e., H  K = {e}.

So, H  G, K  G, H  K = {e} and o(HK) = p = o(6).
2
 G = H × K  Z × Z .
p
p
So far we have shown the algebraic structure of all groups of order 1 to 10, except groups of order
8. Now we will list the classification of groups of order 8.
If G is an abelian group of order 8, then

(i) G  Z , the cyclic group or order 8, or
8
(ii) G  Z × Z , or
4
2
(iii) G  Z × Z × Z .
2
2
2
If G is a non-abelian group of order 8, then
(i) G  Q , the quaternion group discussed in Unit 4, or
8
(ii) G  D , the dihedral group discussed in Unit 5.
8
So, we have seen what the algebraic structure of any group of order 1, 2, . . . . , 10 must be. We
have said that this classification is up to isomorphism. So, for example, any group of order 10 is
isomorphic to Z or D . It need not be equal to either of them.
10
10
Self Assessment

1. Let a group G be the ................... product of its subgroups H and k. Then hk = kh  h  H,
k  K.
(a) external (b) internal
(c) finite (d) infinite
2. Let H and k be normal subgroups of a group G such that G = H × k. Then G/H  ...................
and G/k  H
(a) k (b) H
(c) H -1 (d) k -1’

3. Let G ................... be and H and k be its subgroup such that G = H × k. Thus O(G) = O(H) o(k).
(a) external (b) internal
(c) finite (d) infinite

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