Let $G$ and $H$ be groups and $\varphi:G \to H$ a group isomorphism. Show that if $H$ is cyclic, then $G$ is cyclic also.
As $\varphi$ is a group isomorphism, then so is $\varphi^{-1} :H \to G$. Let $a \in H$, then $a=h^m$ for some $m \in \Bbb Z$ (as $H$ is cyclic). Since $\varphi$ is bijective there exists $g \in G$ such that $a = h^{m}=\varphi(g)$. Now applying $\varphi^{-1}$ we have that $$\varphi^{-1}(h^m) =(\varphi^{-1}(h))^m=g$$ since $g$ is arbitary we have that $\langle \varphi^{-1}(h)\rangle$ generates $G$.
Is this correct reasoning? I feel like this could be done without the usage of the inverse map.