A possible formulation of the Bestvina-Feighn theorem is as follows (taken from [here](http://392c.wordpress.com/2009/04/08/29-doubles-of-free-groups/)):


**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 $G\_1$ ?



Amalgmated free product of hyperbolic groups with one malnormal and one virtual factor is hyperbolic?
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 $\Gamma$.


