Let $G=\langle\ S\ |\ R\ \rangle$ be a finitely presented group. Let $F$ be the free group with generating set $S$. Let $[F,F]$ and $[G,G]$ be the commutator subgroups of $F$ and $G$ respectively. Let $N$ be the normal subgroup of $F$ generated by $R$.
Is it true that $\dfrac{[F,F]}{N\cap[F,F]}=[G,G]$ ?
If $\pi:F\to F/N=G$ is the cannonical map then clearly $\dfrac{[F,F]}{N\cap[F,F]}=\pi([F,F])$ and is a normal subgroup of $G$. To show $[G,G]\subseteq \dfrac{[F,F]}{N\cap[F,F]}$ it is enough to show $G$ mod $\dfrac{[F,F]}{N\cap[F,F]}$ is abelian. But I am unable to do so. Also I have no idea how to show the reverse containment. Is there a counter example for the statement?
Thank you.
A simple explanation:
$$\begin{align}\pi([F,F])&=\pi(\{xyx^{-1}y^{-1}:x,y\in F\})\\ &=\{\pi(x)\pi(y)\pi(x)^{-1}\pi(y)^{-1}:x,y\in F\}\\ &=[\pi(F),\pi(F)]\\ &=[G,G]\end{align}$$
Hope this helps.