# 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

\[\cos \left( {\frac{{n\pi }}{3}} \right) + i\sin \left( {\frac{{n\pi }}{3}} \right)\] i.e. \[{e^{\frac{{n\pi i}}{3}}}\] where $$n = 0,1,2,3,4,5$$

\[\therefore \,\,\,G = \left\{ {{e^{\frac{{n\pi i}}{3}}},\,n = 0,1,2,3,4,5} \right\}\]

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

\[{H_1} = \left\{ {{e^{\frac{{n\pi i}}{3}}},\,n = 3,6} \right\} = \left\{ {{e^{\pi i}},\,{e^{2\pi i}} = 1} \right\} = \left\{ {{e^{\pi i}},\,1} \right\}\]

and

\[{H_2} = \left\{ {{e^{\frac{{n\pi i}}{3}}},\,n = 2,4,6} \right\} = \left\{ {{e^{\frac{{2\pi i}}{3}}},\,{e^{\frac{{4\pi i}}{3}}},1} \right\}\]

__Example 2__**:** Find all the subgroups 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 group 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-abelian 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 abelian.

**(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.