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Let $X=\{x,y\}$ such that $x\ne y$ and denote by $F(X)$ the free group constructed on $X$. Let $\phi_x$ and $\phi_y$ be the canonical injections of $\mathbb{Z}$ into $F(X)$. We have one relator: namely, the commutator $r=[\phi_x(1),\phi_y(1)]$. Let $N_r$ be the normal subgroup of $F(X)$ generated by $r$: i.e. $$N_r=\bigcup_{n\in\mathbb{N}}\left\{y\ |\ (\exists x)(x\in(C\cup C^{-1})^{[1,n]}\land y=x_1\ldots x_n)\right\},$$ where $C=\left\{a\ |\ (\exists g)(g\in F(X)\ \&\ a=grg^{-1})\right\}$.

Let $z_1=(1,0)$ and $z_2=(0,1)$. Then, there exists a unique homomorphism such that $f:F(X)\rightarrow\mathbb{Z}\times\mathbb{Z}$ such that $f(\phi_x(1))=z_1$ and $f(\phi_y(1))=z_2$. Note that $f$ is also surjective.

I have to show that $\text{Ker}(f)=N_r$. The inclusion $N_r\subset\text{Ker}(f)$ is easy to show. I am having difficulties with the other direction.

Attempt:

Suppose $w\in\text{Ker}(f)$. Then $w\in F(X)$ and $f(w)=(0,0)$. Note that $w$ can be uniquely written as $$w=\prod_{\alpha=1}^n\phi_{i_\alpha}(1)^{m_\alpha}$$ for some $n\in\mathbb{N}$, $m\in(\mathbb{Z}-\{0\})^{[1,n]}$, and $i\in\{x,y\}^{[1,n]}$ such that $i_{\alpha+1}\ne i_\alpha$ for $1\leq\alpha\leq n$.

The answer posted here How to prove that $\langle x, y \mid xyx^{-1}y^{-1}\rangle$ is a presentation for $\mathbb{Z} \times \mathbb{Z}$ suggests inducting on $n$. I don't understand the answer properly, so I will try to do that induction here.

Write $x=\phi_x(1)$ and $y=\phi_y(1)$.

Let $n=4$. Either (1) $w=x^{m_1}y^{m_2}x^{m_3}y^{m_4}$ or (2) $w=y^{m_1}x^{m_2}y^{m_3}x^{m_4}$. Wlog let (1) hold. Since $(0,0)=f(w)=(m_1+m_3,m_2+m_4)$, it must be the case that $m_1=-m_3$ and $m_2=-m_4$. Thus $w=x^{-m_3}y^{-m_4}x^{m_3}y^{m_4}$. Somehow, I have to turn $x^{-m_3}y^{-m_4}x^{m_3}y^{m_4}$ into a conjugate of $x^{-1}y^{-1}xy$. Since $m_3,m_4\ne0$, we can write $m_3=k+1$ and $m_4=j+1$ for some $j,k\in\mathbb{Z}$. Then we have $w=x^{-(k+1)}y^{-(j+1)}x^{k+1}y^{j+1}$. I am now stuck. What should I do next?

I don't understand the proof involving the induction step either, so I would appreciate help with that as well.

Thank you

Shaun
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