I'm trying to solve the following extra (non-homework) problem from my lecture notes:
Let $G$ be the group $\langle x, y \mid x^3y^3\rangle$ Let $\varphi: G \to \mathbb{Z}/3\mathbb{Z}$ be the homomorphism that sends $x$ and $y$ to the generator of $\mathbb{Z}/3\mathbb{Z}$. Find a finite presentation for the kernel of this homomorphism.
I know that in theory a finite index subgroup of a finitely presented group is again finitely presented, but I have no idea how to do this for a particular example and the notes don't give a similar example.
Does anyone have any suggestions?
I come up with a solution using hint in the comment of this question:
$G$ is the fundamental group of a cell complex $X$ obtained from the wedge of two circles by attaching the 2-cell $x^3y^3$.
Below is a 3-fold cover $\tilde X$ of $X$, the fundamental group of $\tilde X$ is $\ker\varphi$.
(3 vertices correspond to 3 cosets of $\ker\varphi$.)
Two $x$ edges form a spanning tree of $\tilde X$.
The graph remains one $x$ edge and three $y$ edges. So the loops\begin{array}l x_1:=x^3,\\ x_2:=yx^{-1},\\ x_3:=xyx^{-2},\\ x_4:=x^2y\end{array}form a generating set of $\ker\varphi$, so $\ker\varphi$ has a presentation $$\langle x_1,x_2,x_3,x_4\mid x_2x_3x_4\rangle$$ We can verify the number of generators using Schreier index formula:
the number of generators of $G$ is $2$,
the index $[G:\ker\varphi]$ is $3$,
then the number of generators of $\ker\varphi$ is $1+3(2-1)=4$.