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

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, $\phi(e_{G})=e_{H}$.

• Seems good, but since this is a simple problem I would recommend explaining why "The only idempotent in the group $H$ is its identity element $e_H$." Mar 3, 2018 at 3:14
• This is true for any homomorphism, your proof is correct. See math.stackexchange.com/questions/647697/… Mar 3, 2018 at 3:14
• @user357980 This is often proved as a separate exercise in introductory group theory courses. Mar 3, 2018 at 3:16
• Was this a theorem or a statement that you know that you can use? I just thought that one was something that was on the same level as what you were trying to prove. Mar 3, 2018 at 3:17
• Would it be better if I follow the answer and multiply both sides by the inverse?
– user482939
Mar 3, 2018 at 18:53

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.
• And since $\phi (e_{G})\cdot \phi ^{-1}(e_{G})$ is equal to the identity in $H$, it must be that $\phi (e_{G})=e_{H}$?