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Suppose $G$ is a finitely generated nilpotent group with abelianization of rank $r$. Does $G$ always have a subgroup $H$ of finite index, such that $H$ abelianized is a free abelian group of rank $r$?


Since this is MathOverflow, I will push the question further - under what conditions can we expect abelianization of a monic map to be monic?



Finite index subgroup with free abelianization
**Warning:** (YCor) the following argument is mistaken as was pointed out by Derek Holt: the assertion that the abelianization of a torsion-free nilpotent group is torsion-free is hopelessly wrong.




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The answer is "yes" because every f.g. nilpotent group has a torsion-free finite index subgroup and because the abelianization of a torsion-free nilpotent group is torsion-free. I assume that by "rank" you meant the torsion-free (${\mathbb Q}$-)rank. 


