[Math] Distortion of malnormal subgroup of hyperbolic groups

geometric-group-theorygr.group-theoryhyperbolic-geometry

Let $G$ be a countable, Gromov-hyperbolic group.

We say that $H$ is hyperbolically embedded in $G$ if $G$ is relatively hyperbolic to {$H$} (in the strong sense). This definition is due to Osin.

A theorem of Bowditch says that infinite, finitely generated, almost-malnormal and quasi-convex subgroups of $G$ are hyperbolically embedded in $G$. Later Osin has proved that the conditions are necessary (even in the wider context of relatively hyperbolic groups).

Quasi-convex subgroups are not necessary malnormal but they have always finite height by a result of Gitik, Mitra, Rips and Sageev. The height of $H\subset G$ is defined to be the maximal $n$ such that there exist $g_1,\ldots,g_n\in G$ with $g_1Hg_1^{-1}\cap\ldots\cap g_nHg_n^{-1}$ infinite (but all the $g_iHg_i^{-1}$ different).

I would like to know how distorted a malnormal subgroup can be in $G$.

Is there some class of groups for which malnormal implies quasi-convex? Examples?

What about the relatively hyperbolic case?

Best Answer

There's a weak sort of quasiconvexity proved by Kapovich: if one has an acylindrical graph of hyperbolic groups, with finitely generated edge groups which embed q.i. into the vertex groups, then the edge groups embed q.i. into the graph of groups. Acylindrical here is weaker than malnormality: it says that the stabilizers of the action on the tree have bounded length, whereas malnormality of the edge groups would correspond to stabilizers having length 1.

Incidentally, if one had a finitely-generated malnormal hyperbolic subgroup of a hyperbolic group which was not quasiconvex, then the double of the group along the subgroup is finitely presented and does not contain any Baumslag-Solitar subgroups. However, this group is not hyperbolic. This would be quite interesting, as I believe there are no known examples of such groups which are of type F (acting properly cocompactly on a contractible locally compact complex). As Henry points out in the comments, there are finitely presented examples of Brady which are not hyperbolic, but which are also not type F (they occur as normal subgroups of a hyperbolic group, so are far from being malnormal subgroups).

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