Suppose $A,G,H$ are finitely generated groups and $A$ is quasi-isometrically embedded into $G$ and $H$. Does it follow that the natural embeddings of $G$ and $H$ into $G\*\_AH$ are quasi-isometric?


I can more or less see that this is true if $A$ is finite (in particular, for the free product), or, more generally, if $G$ and $H$ have finite generating sets which are invariant under conjugation by $A$, but I can't quite solve the general case.


This seems like something that should be well-known (one way or the other), but I could not find any information on this in the literature (not even for the case of the free product). If this is true, I would appreciate a reference. (If necessary, I don't mind assuming that $A,G,H$ are hyperbolic, but this seems like something that should be true in general.)


Edit: I am also happy to assume that the embedding of $H$ into $G$ and $H$ is conjugate-separated i.e. that $A\cap A^g$ is finite for $g\in (G\cup H)\setminus A$.



Are the canonical embeddings into $G*_AH$ quasi-isometric?
In the setting when $G, H$ are hyperbolic groups and $A$ is almost malnormal in $G, H$ (the one you are actually interested in, per your comments), the positive answer is given in


*Kapovich, Ilya*, [**The combination theorem and quasiconvexity**](http://dx.doi.org/10.1142/S0218196701000553), Int. J. Algebra Comput. 11, No. 2, 185-216 (2001). [ZBL1025.20028](https://zbmath.org/?q=an:1025.20028).


On the other hand, if you only assume hyperbolicity of $G, H$ and $G\*\_AH$, the subgroups $G, H$ need not be quasiconvex in $G\*\_AH$.


Examples that I know come from topology. Suppose that $S$ is a closed nonorientable hyperbolic surface, $F\to S$ is its orientation double-cover, $C(F\to S)$ the mapping cylinder. The latter is a 3-dimensional manifold with boundary (homeomorphic to $F$). Take two copies $M\_1, M\_2$ of $C(F\to S)$ and a pseudo-Anosov homeomorphism $h: \partial M\_1\to \partial M\_2$ (this makes sense due to canonical hoomeomorpisms of these boundaries to $F$). Glue $M\_1, M\_2$ via $h$. The result is a closed hyperbolic 3-manifold $M$. Of course,
$$
\pi\_1(M)=\pi\_1(M\_1)\*\_{\pi\_1(F)} \pi\_1(M\_2)
$$
and amalgam of two hyperbolic groups over a hyperbolic subgroup of index $2$. At the same time, $M$ admits a 2-fold covering $p: N\to M$ such that $N$ is fibered over the circle, such that the fiber corresponds to a connected component of $p^{-1}(F)$. The fundamental group of the fiber is not quasiconvex in $\pi\_1(N)$, hence, $\pi\_1(F)$ is not quasiconvex in $\pi\_1(M)$.


I am not quite sure what happens if you drop the hyperbolicity assumption but keep (almost) malnormality. I think the answer is still positive.


