Finding the abelianization of a kernel of group homomorphism Let $G=\langle x,y \; | \; xyx=yxy\rangle$ and $\phi: G\to \mathbb{Z}/2\mathbb{Z}$, $\phi(x)=\phi(y)=1$. Note that I am using the additive convention for $\mathbb{Z}/2\mathbb{Z}$ so 0 is the identity. I am wanting to calculate $\ker(\phi)_{ab}=\ker(\phi)/[\ker(\phi),\ker(\phi)]$ (i.e. the abelianization of $\ker(\phi)$). Here is what I have done: 
I know (I am almost sure) the kernel of $\phi$ is all words in $G$ that have an even number of $x$'s and $y$'s. That is
$$
ker(\phi)=\{x^{\alpha_1}y^{\beta_1} \dots x^{\alpha_n}y^{\beta_n} \in G : \; \sum_{i=1}^n (\alpha_i+\beta_i)\equiv 0 \mod{2}\}.
$$
But then
$$
ker(\phi)_{ab}=\{x^{2n}y^{2m} \in G : n,m\in \mathbb{Z}\}.
$$
Furthermore, when we abelianize the relation $xyx=yxy$, we get $x=y$ and so we should have 
$$
ker(\phi)_{ab}=\{x^{2(n+m)} : n,m\in \mathbb{Z}\}=\{x^{2n} : n\in \mathbb{Z}\}=\langle x^2
\rangle \cong \mathbb{Z}.
$$
Is my process correct here? I know there is really an abuse of notation going on (technically the abelianization is the quotient group and not a subset of the original group) but other than that, is this correct? This comes from my previous post where someone in the comments claimed that $ker(\phi)_{ab}\cong \mathbb{Z}\times \mathbb{Z}_3$ but I don't see where this extra factor $\mathbb{Z}_3$ would have come from. 

Let $K = \ker(\phi)$. Then $|G:K| = 2$, and $1$ and $2=x$ are (right) coset representatives of $K$ in $G$.
So we get the coset table $$K1\cdot x=K1\cdot y = K2,\ \ K2 \cdot x = K2 \cdot y = K1.$$
Now we have $1 \cdot x = 2$ by definition, and then
$$1 \cdot y = a \cdot 2,\ \  2 \cdot x = b \cdot 1,\  {\rm and}\  2 \cdot y = c \cdot 1,$$
where $a = yx^{-1}$, $b = x^2$, and $c=xy$ are the Schreier generators of $K$.
We get defining relations of $K$ on these generators by applying the single group relation $xyx=yxy$ (on the right) to the coset representatives $1$ and $x$. This yields the two relations $c=aba$ and $bab=c^2$, so
$$K \cong \langle a,b,c \mid aba=c,\ bab=c^{2} \rangle \cong \langle a,b \mid (aba)^2=bab\rangle,$$
and it is straightforward to calculate the abelianization of $K$, which is ${\mathbb Z} \times {\mathbb Z}/3{\mathbb Z}$.
