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How can we show that locally cyclic subgroups (ie. groups whose finitely generated proper subgroups are cyclic) of a hyperbolic group are cyclic?

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Such subgroups are solvable, and the result follows from the solvable subgroup theorem. There might be an easier way, though. – user641 Jan 22 '13 at 12:30
@SteveD: Is it immediate that such subgroups are solvable? I don't see why... – Seirios Jan 22 '13 at 14:32
The group is abelian... – user641 Jan 22 '13 at 17:01
@SteveD: Of course... Thank you. – Seirios Jan 22 '13 at 17:04
OK, I now see a much easier proof of this, using the fact that abelian subgroups should be quasi-convex. Basically, take any element in your subgroup; then its centralizer is quasi-convex, and hence hyperbolic. In particular, it is finitely generated, and thus its center - as the intersection of quasi-convex subgroups - is also hyperbolic. This is a finite-generated abelian group, which is either finite or virtually cyclic. Intersecting with your original subgroup finishes. For a reference, see… – user641 Jan 22 '13 at 17:17
up vote 3 down vote accepted

As requested:

To prove this, one can use the fact that abelian subgroups of a hyperbolic group should be quasi-convex; this is because centralizers are quasi-convex. Thus the abelian subgroup is finitely generated, and your subgroup is in fact hyperbolic. I don't want to flesh this argument out too much, because it is already nicely written up here.

I had remarked in my comments above that the Solvable Subgroup Theorem also accomplishes this, but for that to work you need an isometric action on a CAT(0) space.

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