Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
1  
I believe this is still an open question (malnormal implies quasiconvex). One could attempt to prove it by induction for cubulated hyperbolic groups, but even that seems tricky. –  Ian Agol Jun 18 '13 at 20:54
3  
Yes, existence of distorted malnormal subgroups in hyperbolic groups is a known open problem. There are classes of hyperbolic groups which are locally quasiconvex, but this does not use malnormality. –  Misha Jun 18 '13 at 23:59
    
By the way, you should assume that the malnormal subgroup is finitely generated (at least). I think in any hyperbolic group, one can find malnormal infinitely generated (free) subgroups. –  Ian Agol Jun 19 '13 at 4:49
1  
Also, by the technique of combinatorial Dehn filling, the relatively hyperbolic case probably isn't too far from the word-hyperbolic case. –  HJRW Jun 19 '13 at 8:19
1  
Oh, you might want to look at Question 1.8 of Besvina's problem list, which asks whether subgroups of finite width are quasiconvex. Malnormal means width 1. math.utah.edu/~bestvina/eprints/questions-updated.pdf –  HJRW Jun 19 '13 at 8:32

1 Answer 1

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).

share|improve this answer
    
An example of a finitely presented, non-hyperbolic subgroup of a hyperbolic group (hence with no Baumslag--Solitar subgroups) was constructed by Brady. However, if your edge groups were of type F (in the sense of Wall, ie admits a finite Eilenberg--Mac Lane space) then this double would also be of type F, and no such non-hyperbolic groups are known---see Question 1.1 of Bestvina's list. –  HJRW Jul 4 '13 at 9:35
    
Ah, right, I'm aware of that example. I'll say the subgroup is also hyperbolic for simplicity (since this would also follow from quasiconvexity). –  Ian Agol Jul 4 '13 at 14:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.