Return to Revisions

The answer for (2) is "no". Take $G$ to be the relatively free group of class $2$ and rank $k$ (isomorphic to $F_k/(F_k)_3$, where $F_k$ is the free group of rank $k$ and $(F_k)_3$ is the third term of the lower central series of $G$). Then $G^{\rm ab}\cong \mathbb{Z}^k$. Let $H=[G,G]$; then $H$ is free abelian of rank $\binom{k}{2}$. Pick $k\gt 3$ to get that the rank of $H/[H,H]$ is greater than the rang of $G/[G,G]$. If you want to exclude the case where $H$ is abelian itself, just add two of the original generators to the commutator subgroup so the abelianization is of rank $2+\binom{k-1}{2}$.