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

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    $\begingroup$ 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 $e_H$." $\endgroup$
    – user357980
    Mar 3, 2018 at 3:14
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    $\begingroup$ This is true for any homomorphism, your proof is correct. See math.stackexchange.com/questions/647697/… $\endgroup$ Mar 3, 2018 at 3:14
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    $\begingroup$ @user357980 This is often proved as a separate exercise in introductory group theory courses. $\endgroup$ Mar 3, 2018 at 3:16
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    $\begingroup$ 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. $\endgroup$
    – user357980
    Mar 3, 2018 at 3:17
  • $\begingroup$ Would it be better if I follow the answer and multiply both sides by the inverse? $\endgroup$
    – user482939
    Mar 3, 2018 at 18:53

1 Answer 1

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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.

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  • $\begingroup$ And since $\phi (e_{G})\cdot \phi ^{-1}(e_{G})$ is equal to the identity in $H$, it must be that $\phi (e_{G})=e_{H}$? $\endgroup$
    – user482939
    Mar 3, 2018 at 18:54
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    $\begingroup$ exactly, sorry for the late reply. $\endgroup$ Mar 9, 2018 at 21:05

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