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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^-y^-1. Is there an infinite chain of subgroups H > H_1 > H_2 > ... such that the index [H_i, H_i+1] < n for some n? Thanks

edited Mar 2 2010 at 19:53

asked Feb 5 2010 at 6:03

dave
139●4

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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? – Mariano Suárez-Alvarez Feb 5 2010 at 6:08

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 – Yemon Choi Feb 5 2010 at 6:13

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! – HW Mar 2 2010 at 20:03

Yemon Choi · Accepted Answer · 2010-02-05 06:48:41Z UTC

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 $n_i = 2^i$.)

Steve D · Answer 2 · 2010-02-05 06:44:29Z UTC

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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?

answered Feb 5 2010 at 6:44

Steve D
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decreasing chain of subgroups in the Heisenberg group

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