By Heisenberg group I mean the group with presentation $H$ generated by $x$ and $y$ such that $x$ and $y$ commute with $xyx^{1}y^{1}$. Is there an infinite chain of subgroups $H > H_1 > H_2 > \dots$ such that the index $[H_i: H_{i+1}]< n$ for some $n\ ?$ Thanks

3$\begingroup$ The group is the fundamental group of a 3manifold $M$ which fibers over $S^1$ with fiber a torus. By considering coverings of $M$ which arise by pullpack along finite coverings of degree $2^n$ of $S^1$ by $S^1$, you get such a sequence of subgroups, no? $\endgroup$– Mariano SuárezÁlvarezFeb 5 '10 at 6:08

$\begingroup$ There seems to be a typo in your formula: don't you want $x$ and $y$ to commute with the commutator? Also, is this the same group as the discrete integer Heisenberg group? en.wikipedia.org/wiki/Heisenberg_group $\endgroup$– Yemon ChoiFeb 5 '10 at 6:13

3$\begingroup$ To summarise an underlying theme in the comments and answers: you can do this for any group that surjects the integers. Or even that has a finiteindex subgroup that surjects the integers! $\endgroup$– HJRWMar 2 '10 at 20:03
Assuming this is the discrete Heisenberg group $H=H_3({\mathbb Z})$, as in my comment above, then here is another way of looking at Mariano's answer (I think). Take any sequence of positive integers $n_1 < n_2 < \dots $ where $n_i \vert n_{i+1}$ for all $i$, and put
$$ H_i = H_3(n_i{\mathbb Z}) $$
(Mariano's answer corresponds to taking $n_i = 2^i$.)
If this is in fact the discrete integer Heisenberg group, then can't you just pass to the quotient $H / \langle [x,y] \rangle\cong \mathbb{Z}\times\mathbb{Z}$ and do it there?