Proof verification: For isomorphism $\phi : G\to H$, show that if $e_{G}\in G$ and $e_{H}\in H$, then $\phi (e_{G})=e_{H}$.

Question

Could someone please verify my following proof?

For isomorphism $\phi : G\to H$ for groups $G$ and $H$, show that if identities $e_{G}\in G$ and $e_{H}\in H$, then $\phi (e_{G})=e_{H}$.

Proof: Let $e_{G}\in (G,\circ)$ and $e_{H}\in (H,\cdot)$. Then $\phi (e_{G})=\phi (e_{G}\circ e_{G}) = \phi (e_{G})\cdot \phi (e_{G})$. The only idempotent element in the group H is its identity element $e_{H}$. Therefore, $\phi(e_{G})=e_{H}$.

Answer 1

Yep since you know that $\phi (e_{G}) = \phi (e_{G})\cdot \phi (e_{G})$. You can multiply both sides with $\phi (e_{G})^{-1}$ and have your result.