How can we show that locally cyclic subgroups (ie. groups whose finitely generated proper subgroups are cyclic) of a hyperbolic group are cyclic?
|show 1 more comment|
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.