A possible formulation of the Bestvina-Feighn theorem is as follows (taken from here):

Combination Theorem (Bestvina & Feighn): If $H$ is a malnormal subgroup of hyperbolic groups $G_1, G_2$, then $G_1\ast_H G_2$ is hyperbolic.

I was wondering if the following variation is also true.

Question: Suppose that $G_1,G_2$ are hyperbolic groups with a common subgroup $H$ such that $H \leq G_1$ is malnormal while $H \leq G_2$ is finite index. Then is the amalgamated free product $G_1\ast_H G_2$ hyperbolic? What if we add in the hypothesis that $H$ is also quasi-convex in $G_1$ ?

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    $\begingroup$ You (and your source) are misstating combination theorem: One has to assume in addition that $H$ is quasiconvex in both $G_1, G_2$, while malnormality in, fact, can be weakened to a "flaring condition" (you should read the original paper instead). Once you realize this, you see that malnormality in one of the subgroups is sufficient (but not necessary), since it implies "2-acylindricity" of the action on the tree, which, in turn, implies flaring. $\endgroup$ – Misha Aug 24 '13 at 15:11
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    $\begingroup$ Instead of "One" I should have said "Bestvina and Feighn". Another remark is that, currently, there are no examples of f.g. subgroups of hyperbolic groups which are malnormal and not quasiconvex. See mathoverflow.net/questions/134091 . However, one can prove existence of infinite rank free malnormal subgroups in every nonelementary hyperbolic group. Taking a double along such subgroup, leads to a group which is not finitely presentable and, hence, not hyperbolic. $\endgroup$ – Misha Aug 24 '13 at 15:30
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    $\begingroup$ See also theorem 2 in arxiv.org/pdf/math/9605205v1.pdf $\endgroup$ – Misha Aug 24 '13 at 16:53
  • $\begingroup$ Here's the abstract page for Misha's link: arxiv.org/abs/math/9605205 . (In general, I prefer not to have to download a whole pdf file just to see what the link is.) $\endgroup$ – HJRW Aug 24 '13 at 18:14

As Misha says in comments, as long as H is also quasiconvex in $G_1$ then you will be able to apply the combination theorem. One can deduce this directly from the 'usual' statement that an amalgam of two hyperbolic groups along malnormal, quasiconvex subgroups is itself hyperbolic, as follows.

If $H$ is finite-index in $G_2$ then the amalgam $\Gamma=G_1*_HG_2$ has a finite-index subgroup $\Gamma'$ isomorphic to the amalgamated product of $|G_2:H|$ copies of $G_1$ along $H$. Since this is an amalgam of hyperbolic groups along quasiconvex, malnormal subgroups, $\Gamma'$ is hyperbolic, and hence so is $\Gamma$.


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