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?
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$\begingroup$ This question appears to be off-topic in its current form, because an easy counter-example has been provided $\endgroup$– Yemon ChoiAug 20, 2013 at 0:36
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$\begingroup$ thanks, I would like to know with what extra condition, we have $G^{\prime}\cong H^{\prime}$? $\endgroup$– agoalAug 21, 2013 at 5:57
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$\begingroup$ @agoal As the example I gave shows, I think it would be rather difficult to get that kind of condition out of anything that is reasonable. $\endgroup$– Arturo MagidinAug 25, 2013 at 20:05
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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.
answered Aug 19, 2013 at 18:25