Are there two infinite and non abelian finitely generated groups $G$ and $H$ such that $\frac{G}{ G^{\prime}} \cong \frac{H}{H^{\prime}}$ and $G^{\prime}$ is finite but $H^{\prime}$ is infinite?
closed as offtopic by Benjamin Steinberg, Kevin Ventullo, Yemon Choi, Theo JohnsonFreyd, John Pardon Aug 20 '13 at 2:23This question appears to be offtopic. The users who voted to close gave this specific reason:



You may want to put more conditions. Otherwise, take $G=\mathbb{Z}^2\times S$ where $S$ is a nonabelian finite simple group, and $H=F(x,y)$, the free group of rank $2$. Then $G/G' \cong H/H' \cong \mathbb{Z}^2$, $G'\cong S$, and $H'$ is free of countable rank. 

