🏗️ $\Theta \rho ϵ\eta \Pi \alpha \tau \pi$🚧 (under construction)

We know $e_G=e_G{e}_G$, therefore $\varphi \left(e_H\right)=\varphi \left(e_H{e}_H\right)=\varphi \left(e_H\right)\varphi \left(e_H\right)$, but also $\varphi \left(e_H\right)=e_G\varphi \left(e_H\right)$ since $e_G$ is the identity, by right cancellation this yields $e_G=\varphi \left(e_G\right)$ as needed.

We first show it for all $n\in {\mathbb{N}}_1$ via induction. For $n=1$ we can see that $\varphi \left(g^1\right)=\varphi {\left(g\right)}^1$ trivially holds.

Ho we let $k\in {\mathbb{N}}_1$ and assume that the statement holds for $k$ we'll now show it must hold for $k+1$, to do this we note the following : $$\begin{gathered} {\displaystyle \varphi \left(g^{k+1}\right)\&=\varphi \left(g^k{\star }_{G}g\right) \\ \&=\varphi \left(g^k\right){\star }_H\varphi \left(g\right) \\ \&=\varphi {\left(g\right)}^k{\star }_H\varphi \left(g\right) \\ \&=\varphi {\left(g\right)}^{k+1}} \end{gathered}$$ Where we've used our induction hypothesis going from the 2nd to the 3rd line.

The above holds us over for ${\mathbb{N}}_1$, but it turns out we can now extend this to ${\mathbb{Z}}^{-}$ quite easily, given some $z\in {\mathbb{Z}}^{-}$ we first note that $e_H=\varphi \left(e_G\right)=\varphi \left(g^z{\star }_G{g}^{-z}\right)=\varphi \left(g^z\right){\star }_H\varphi {\left(g\right)}^{-z}$, note that we justify the last equality because $-z\in {\mathbb{N}}_1$ and then use our previous paragraph's proof. Finally by multiplying on the right by $\varphi {\left(g\right)}^z$ then we obtain $\varphi {\left(g\right)}^z=\varphi \left(g^z\right)$ as needed.

We could stop there and we've left nothing out. Well indeed we have left nothing out, that is we've left $0$ out, well if $\varphi \left(g^0\right)=\varphi \left(e_G\right)=e_H=\varphi {\left(g\right)}^0$ as needed.