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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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closed as off-topic by Benjamin Steinberg, Kevin Ventullo, Yemon Choi, Theo Johnson-Freyd, John Pardon Aug 20 '13 at 2:23

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Benjamin Steinberg, Theo Johnson-Freyd, John Pardon
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This question appears to be off-topic in its current form, because an easy counter-example has been provided –  Yemon Choi Aug 20 '13 at 0:36
    
thanks, I would like to know with what extra condition, we have $G^{\prime}\cong H^{\prime}$? –  agoal Aug 21 '13 at 5:57
    
@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. –  Arturo Magidin Aug 25 '13 at 20:05

1 Answer 1

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.

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