# Examples Subgroup of Cyclic Groups

Example 1: Find the proper subgroups of the multiplicative group $G$ of the sixth roots of unity.

Solution: From trigonometry we know that six sixth roots of unity are

i.e.

where $n = 0,1,2,3,4,5$

If ${H_1}$ and ${H_2}$ be its proper subgroups, then

and

Example 2: Find all the subgroup of a cyclic group of order $12$.

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 $a$ be the generators of the group and $m$ be a divisor of 12. Then there exists one and only one element in $G$ whose order is $m$, i.e. ${a^{\frac{{12}}{m}}}$.

All the elements of order 1, 2, 3, 4, 6, 12 will give subgroups.
$\therefore \,\left( {{a^{12}}} \right) = \left\{ e \right\},\left( {{a^6}} \right),\left( {{a^4}} \right),\left( {{a^3}} \right),\left( {{a^2}} \right),\left( a \right)$ are the required subgroups.

Example 3:
(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?

Solution:
(i) Every subgroup of an abelian group is abelian. If $G$ is an abelian group and $H$ is a subgroup of $G$, then the operation on $H$ is commutative because it is already commutative in $G$ and $H$ is a subset of $G$. 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 ${P_3}$ of permutation of degree 3 is non-abelian while its subgroup ${A_3}$ is ableian.

(iii) A non-abelian group can have a non-abelian subgroup. For example, ${P_4}$ is a non-abelian group and its subgroup ${A_4}$ is also non-abelian.