Answer to Question #146439 in Abstract Algebra for J

Question

Assume that (G,*) is a cyclic group and that H is a subgroup of G. Explain why H is a normal subgroup G,and prove that G/H is cyclic.

Answer 1

Each cyclic group $G=\langle a \rangle$ is Abelian. Indeed, $a^ma^n=a^{m+n}=a^{n+m}=a^na^m$. Therefore, $gH=Hg$ for all $g\in G$, and $H$ is a normal subgroup of $G$.

Let $[g]=gH$ be arbitrary element of $G/H$. Since $G=\langle a \rangle$ is cyclic, $g=a^k$ for some $k\in \mathbb Z.$ By defenition of product of elements of $G/H$ , $[g]=[a^k]=[a]^k.$ Therefore, $G/H=\langle[a]\rangle,$ and $G/H$

is cyclic.