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

We now recall that $\varphi \left(g^n\right)=\varphi {\left(g\right)}^n$ so that we obtain $\varphi \left((m,0)\right)=\varphi \left({(1,0)}^m\right)=\varphi {\left((1,0)\right)}^m$ here the power symbol is acting with respect to $\oplus$ and then $+$ respectively. So in multiplicative notation that would mean $\varphi {\left((1,0)\right)}^m=m\varphi \left((1,0)\right)$ (note that multiplication here is just syntactic sugar for it being added to itself $m$ times)

By symmetry it follows that $\varphi \left((0,n)\right)=n\varphi \left((0,1)\right)$ so re-connecting with our original chain of equalities we have $${\displaystyle \varphi \left((m,n)\right)=m\varphi \left((1,0)\right)+n\varphi \left((0,1)\right)}$$

$⟹$ Suppose that $im\left(\varphi \right)$ is abelian, we want to prove that $[G,G]\subseteq ker\left(\varphi \right)$, we already know that $[G,G]$ is the smallest subgroup containing the set $\{xy{x}^{-1}{y}^{-1}:x,y\in G\}$ since we already know that $ker\left(\varphi \right)$ is a subgroup of $G$, then we just have to show it contains the set, and it would automatically become a superset as needed.

Let $xy{x}^{-1}{y}^{-1}$ be in the set, to show that this element is also in $ker\left(\varphi \right)$ we have to show that $\varphi \left(xy{x}^{-1}{y}^{-1}\right)=e_H$, we note that $im\left(\varphi \right)$ was abelian and a homomorphism, thus $$\begin{gathered} {\displaystyle \varphi \left(xy{x}^{-1}{y}^{-1}\right)\&=\varphi \left(x\right)\varphi \left(y\right)\varphi \left(x^{-1}\right)\varphi \left(y^{-1}\right) \\ \&=\varphi \left(x\right)\varphi \left(y\right)\varphi {\left(x\right)}^{-1}\varphi {\left(y\right)}^{-1} \\ \&=\varphi \left(x\right)\varphi {\left(x\right)}^{-1}\varphi \left(y\right)\varphi {\left(y\right)}^{-1} \\ \&=e_H{e}_H \\ \&=e_H} \end{gathered}$$ Note that pulling the power out of a homomorphism is justified on line 2

$⟸$ Now we're assuming that $[G,G]\subseteq ker\left(\varphi \right)$ and we want to show that $im\left(\varphi \right)$ is abelian so let $a,b\in im\left(\varphi \right)$, so that there is some $x,y\in G$ such that $a=\varphi \left(x\right),b=\varphi \left(y\right)$. Let's prove $ab=ba$:

Firstly by definition $[G,G]$ contains the set $\{xy{x}^{-1}{y}^{-1}:x,y\in G\}$ therefore since $ker\left(\varphi \right)$ is a superset of $[G,G]$ it also contains that set, in other words $xy{x}^{-1}{y}^{-1}\in ker\left(\varphi \right)$, that is $\varphi \left(xy{x}^{-1}{y}^{-1}\right)=e_H$, well $\varphi$ is a homomorphism: $\varphi \left(x\right)\varphi \left(y\right)\varphi {\left(x\right)}^{-1}\varphi {\left(y\right)}^{-1}=e_H$, the by multiplying each side on the right we have $\varphi \left(x\right)\varphi \left(y\right)=\varphi \left(y\right)\varphi \left(x\right)$, well that is $ab=ba$ as needed.