The first homology of a group G, denoted $H_1(G)$ is just the abelianization of G, i.e. G/[G,G].
Suppose that G is a group with $H_1(G)$ torsion-free. If H is a finite index subgroup, is $H_1(H)$ torsion-free?
Thanks for your time!
My example was the class 2 nilpotent group
$G=\langle x,y,z \mid [x,y]=z, z^2=1, [x,z]=[y,z]=1 \rangle$
with $H = \langle x^2,y,z \rangle$, but in fact it works just as well without the relation $z^2=1$ of $G$, which gives a torsion-free example. So it's hard to think of conditions on $G$ that make it work!