I am interested in the following setting:
$$ 0\rightarrow [G,G]\rightarrow G \rightarrow \mathbb{Z}^r\rightarrow 0, $$ in particular in the case $r=1$. Is $[G,G]$ simple in this case?
I am aware of Thompson's group, but I do not know how to see if its commutator is simple in an easy way and also it is for the case $r=2$ instead of $r=1$ (Commutator Subgroup of Thompson's Group $F$).
Let $F$ be the (absolutely) free group of rank $r$. The lower central series of $F$ is defined recursively by $$\begin{align*} F_1 &= F;\\ F_{n+1} &= [F_n,F]. \end{align*}$$ The group $F/F_{n+1}$ is the relative free nilpotent group of class $n$ and rank $r$. The commutator subgroup is isomorphic to $[F,F]/F_{n+1}$, which is nilpotent and torsionfree, and in particular is not simple. But $F/[F,F]$ is free abelian of rank $r$, hence isomorphic to $\mathbb{Z}^r$.
Similarly, you can define the derived series of $F$, $$\begin{align*} F^{(0)} &= F,\\ F^{(n+1)} &= [F^{(n)},F^{(n)}]. \end{align*}$$ The quotient $F/F^{(n+1)}$ is the relatively free solvable group of length $n$ and rank $r$; it is solvable of length $n$, and the commutator subgroup is solvable of length $n-1$. The abelianization is free abelian of rank $r$.
There is absolutely no reason why the commutator subgroup of your $G$ would be free in such a situation.