Fundamental group of the grid in $\mathbb{R}^2$ isomorphic with commutator subgroup of $\mathbb{Z}*\mathbb{Z}$

Question

I'm having trouble with the following exercise:

Show that the fundamental group of the grid of integer lines $\{(x,y)\in\mathbb{R}^2: x\in\mathbb{Z}\lor y\in\mathbb{Z})$ isomorphic with the commutator subgroup of $\mathbb{Z}*\mathbb{Z}$, the free product of $\mathbb{Z}$ with itself.

I know that the abelianization of $\mathbb{Z}*\mathbb{Z}$ is isomorphic to $\mathbb{Z}^2$, and the abelianization $G_{ab}$ of a group $G$ is $G/[G,G]$.. but I'm not sure what to do with it.

Any advice as to how to approach this problem is appreciated!

Answer 1

Let $X$ be the wedge of two circles, and note that $\pi_1(X) \cong F_2$.

First note that $\mathbb Z^2$ acts transitively on the grid, and that in fact the grid is just $\tilde{X}/[F_2,F_2]$, and in fact this is the Cayley graph for $\mathbb Z^2$.

Then, we know that $[F_2,F_2] \cong \pi_1([\tilde{X}/[F_2,F_2]) \cong F_{\infty}$.

The previous fact can be checked by noting that we have an isomorphism $\pi_1(Y)/p_*(\pi_1(\tilde{X}) \cong\pi_1(Y)\cong G$ where the first isomorphism follows since the universal cover is contractible, and the second follows from basic covering space theory, with reference here.