Let $G$ be a group, $k$ be a positive integer, $G^k=\{g^k \mid g\in G\}$. Prove that $G$ is cyclic iff every subgroup of $G$ has a set form $G^k$.
Someone gives a proof here: Group $G$ is cyclic $\iff$ every subgroup of $G$ has the form $G^k$, where $G^k=\{g^k\mid g\in G\}$
But I was trying to create a isomorphism between $G$ and $\mathbb{Z}$ or $\mathbb{Z}/n\mathbb{Z}$ depending on whether $G$ is finite as they must be isomorphic if $G$ is cyclic. And the set $H=\{G^k \mid 0\le k\le o(G)\}$ exhausts $G$.
I got lost trying to prove that all the sets $G^k$ are not only groups but abelian ones. Also, if I to compose an isomorphism like $\varphi:G^k\to k \mathbb{Z},g_n\to kn$ when $G$ is an infinite group, how to make $\varphi$ bijective?