# Locally cyclic subgroups of a hyperbolic group

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 at 12:30
@SteveD: Is it immediate that such subgroups are solvable? I don't see why... –  Seirios Jan 22 at 14:32
The group is abelian... –  user641 Jan 22 at 17:01
@SteveD: Of course... Thank you. –  Seirios Jan 22 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 berstein.wordpress.com/2011/02/23/… –  user641 Jan 22 at 17:17
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