Finite index subgroup with free abelianization This question has been asked on MathExchange to no avail.
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?
 A: This seems surprisingly difficult! Let's try and do it by induction on the nilpotency class of $G$. The result is clear for abelian groups.
Let $Z$ be the last nontrivial subgroup in the lower central series of $G$. So $Z \le G' \cap Z(G)$. By induction, $G$ has a finite index normal subgroup $H$ containing $Z$ such that the abelianization $H/ZH'$ of $H/Z$ is free abelian.
Since $G$ is nilpotent, $|G:H|$ finite implies $|G':H'|$ finite. So $|Z:Z \cap H'| = |H'Z:H'|$ is finite, and $H'Z/H'$ has a free abelian normal complement $K/H'$ in $H/H'$ with $|H:K|$ finite.
But $H = KZ$ and  $Z$ is central imply $K' = H'$, and hence $K/K' = K/H'$ is free abelian.
A: Here's an alternative proof (at least, an alternative wording), making $H$ very explicit.
For $r$ the rank, let $G\to\mathbf{Z}^r$ be a surjective homomorphism (e.g., take the abelianization homomorphism and mod out by the torsion subgroup). Lift the $r$ generators to $r$ elements of $G$: they generate a subgroup $H$. So $H$ is generated by $r$ elements, and surjects onto $\mathbf{Z}^r$, and hence the abelianization of $H$ is isomorphic to $\mathbf{Z}^r$.
To conclude, we need to check that $H$ has finite index in $G$. For this I claim more generally that 

if $G$ is a nilpotent group and $H$ is a subgroup such that $H[G,G]$ has finite index in $G$, then $H$ has finite index in $G$. 

If $G$ is abelian the result is trivial. Otherwise, let $G^k$ be the last nontrivial term in the lower central series, so $G^k$ is central and contained in $[G,G]$. By induction on the nilpotency class, we obtain that $G^kH$ has finite index. The $k$-fold commutator map induces a surjective homomorphism $\Lambda^kG_{\mathrm{ab}}\to G^k$. By an easy argument, the finite index inclusion $H\to G$ induces a homomorphism $\Lambda^kH_{\mathrm{ab}}\to \Lambda^kG_{\mathrm{ab}}$ whose image has finite index. Hence the image of $\Lambda^kH_{\mathrm{ab}}$ in $G^k$ has finite index in $G^k$. This image is precisely $H^k$. Hence the inclusion $H\subset G^kH$ has finite index. Since we obtained that $G^kH$ has finite index in $G$, we conclude that $H$ has finite index.
[Actually this argument through $\Lambda^k$, maybe suitably rephrased, is needed to prove the fact asserted by Derek that a finite index inclusion of nilpotent groups induces a finite index inclusion of the derived subgroups. It's true for all terms of the lower central series, and in a sense the most basis step is that of the last term (done above), where things can be linearized.] 
PS. A better version of all this would be to control the index of $H$ in terms of the size of the torsion subgroup of $[G,G]$, and the Hirsch length of $G$ (or, more optimistically, the size of the torsion subgroup of $G$ and the nilpotency class of $G$ modulo its torsion subgroup).
A: 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.

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. 
