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Given a finitely generated subgroup of a finitely generated hyperbolic group. Is it true that the inclusion of each subgroup is a quasiisometric embedding ?

The first example for a group that does not have this property is a Baumslag-Solitar group $BS(1,m)= \langle a,b| bab^{-1}=a^m\rangle$. We have $a^{m^k}=b^kab^{-k}$ for each $k$. This shows for example that the inclusion of the subgroup generated by $a$ is not a quasiisometric embedding.

Then one can consider the class of groups with the property that the inclusion of any subgroup is a quasiisometric embedding. Has this class been studied?

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3 Answers 3

up vote 8 down vote accepted

The answer to the first question is "no". Look at the Rips' construction http://en.wikipedia.org/wiki/Small_cancellation_theory. The finitely generated normal subgroup there is usually very badly distorted. I do not see many general properties of the class of groups where every f.g. subgroup is undistorted. It contains so diverse groups as free groups and Tarski monsters. Studying hyperbolic groups with this property makes more sense but I doubt anything general is known about this class either.

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For more than you ever wanted to know on the subject, check out http://berstein.wordpress.com/2011/02/23/quasi-convex-subgroups-of-hyperbolic-groups/

and references therein (the short answer is: no. An interesting counterexample is contained in

An example of a non-quasiconvex subgroup of a word hyperbolic group with exotic limit set

by

Ilya Kapovich)

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To add to the examples given, let G be the fundamental group of a closed hyperbolic 3-manifold that fibers over the circle, and let F be the fiber subgroup. Then F is a surface subgroup that is not quasiisometrically embedded.

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(As Richard knows) the famous Virtually Fibered Conjecture is equivalent to: 'Every hyperbolic 3-manifold group is not locally quasiconvex'. –  HJRW Dec 7 '11 at 16:20

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