Example 1: Find the proper subgroups of the multiplicative group of the sixth roots of unity.
Solution: From trigonometry we know that six sixth roots of unity are
If and be its proper subgroups, then
Example 2: Find all the subgroup of a cyclic group of order .
Solution: We know that the integral divisors of 12 are 1, 2, 3, 4, 6, 12. Now, there exists one and only one subgroup of each of these orders. Let be the generators of the group and be a divisor of 12. Then there exists one and only one element in whose order is , i.e. .
All the elements of order 1, 2, 3, 4, 6, 12 will give subgroups.
are the required subgroups.
(i) Can an abelian groups have a non-abelian subgroup?
(ii) Can a non-abelian group have an abelian subgroup?
(iii) Can a non-abelian group have a non-abelian subgroup?
(i) Every subgroup of an abelian group is abelian. If is an abelian group and is a subgroup of , then the operation on is commutative because it is already commutative in and is a subset of . Hence an abelian group cannot have a non-abelian subgroup.
(ii) A non-ableian group can have an abelian subgroup. For example, the symmetric group of permutation of degree 3 is non-abelian while its subgroup is ableian.
(iii) A non-abelian group can have a non-abelian subgroup. For example, is a non-abelian group and its subgroup is also non-abelian.