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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

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    $\begingroup$ The group is the fundamental group of a 3-manifold $M$ 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? $\endgroup$ Feb 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 Choi
    Feb 5 '10 at 6:13
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    $\begingroup$ 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 finite-index subgroup that surjects the integers! $\endgroup$
    – HJRW
    Mar 2 '10 at 20:03
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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 $n_i = 2^i$.)

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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?

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