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$(G, * )$ and $(H, \star )$ : Group
$\rm \bf Group\ homomorphism$ is a function $\phi :G \to H$ s.t. \begin{equation} \label{def_group_hom} \phi(g_{1} * g_{2})=\phi(g_{1}) \star \phi(g_{2}),\ \ \forall g_{1},\forall g_{2}\ \in G \end{equation} and satisfies following conditions:
the identity element:$e_{G} \in G, e_{H} \in H$ \begin{equation} \label{group_hom_identity} \phi (e_{G})=e_{H} \end{equation} $\forall g \in G$ \begin{equation} \label{group_hom_inverse} \phi(g)^{-1}=\phi(g^{-1}) \end{equation}
(Proof \ref{group_hom_identity})
$\phi(e_{G})=\phi(e_{G}*e_{G})=\phi(e_{G})\star \phi(e_{G})$
then,
${(\phi(e_{G}))}^{-1} \star \phi(e_{G})={(\phi(e_{G}))}^{-1}\star(\phi(e_{G})\star \phi(e_{G}))$
$e_{H}=\phi(e_{G})$ is identity element of $H$.

(Proof \ref{group_hom_inverse})
$\forall g,g' \in G:\phi(g*g'^{-1})=\phi(g)\star\phi(g'^{-1})$
Putting $g=g`$,
$\phi(g * g^{-1})=\phi(e_{G})=e_{H}$
$(\phi (g))^{-1} \star (\phi(g) \star \phi(g^{-1}))=(\phi (g))^{-1} \star (\phi(g * g^{-1}))=(\phi (g))^{-1} \star e_{H}$
$\phi ({g^{ - 1}}) = (\phi (g))^{ - 1}$