Could someone please verify my following proof?
For isomorphism $\phi: G\to H$ of the groups $G$ and $H$, show that $|\phi (a)|=|a|$ for all $a\in G$.
Proof: Let $a\in G$ with $|a|=n$. Then $a^{n}=e_{G}$, the identity of $G$. Let $|\phi(a)|=m$. Then $(\phi (a))^{m}=e_{H}$, the identity of $H$, or $\phi (a^{m})=e_{H}$. Since $\phi(e_{G})=e_{H}$, it must be that $a^{m}=e_{G}$. Then it must be that $m=n$. Therefore, $|\phi (a)|=|a|$ for all $a\in G$.