Can torsion be developed in homology of finite index subgroup?

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?

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It might be worth mentioning that $H_1(G) \cong G/G'$. The answer is not necessarily. I can think of an example with $H_1(G) \cong {\mathbb Z}^2$, $|G'|=2$ having an abelian subgroup of index 2 that is not torsion-free. – Derek Holt Feb 1 '13 at 20:29
Thanks Derek - I edited my question to include this. I would be interested in your example. A follow-up question would be- Are there any conditions that would ensure that $H_1(H)$ stays torsion-free? An example would be this trivially holds for free groups. – user60593 Feb 1 '13 at 21:15

$G=\langle x,y,z \mid [x,y]=z, z^2=1, [x,z]=[y,z]=1 \rangle$
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!
I have two probably stupid questions ;-). 1) Is it obvious that H has index 2 in G? 2) I couldn't see how it worked without using the $z^2 = 1$ relation? Thanks again for your time. – user60593 Feb 2 '13 at 16:54
The subgroup $\langle z \rangle$ is equal to $G'$. So, since $z \in H$, $|G:H| = |G/G':H/G'| = 2$, because $G/G'$ is a free abelian group of rank 2. If you leave out the relation $z^2=1$, then you still have $\langle z \rangle = G'$ and $[x^2,y] = z^2$, so $H' = \langle z^2 \rangle$, and $z$ maps onto an element of order 2 in $H/H'$. – Derek Holt Feb 2 '13 at 17:45