Are the canonical embeddings into $G*_AH$ quasi-isometric? 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 $g\in (G\cup H)\setminus A$.
 A: No. For instance, take two elements $u,v$ of order 2 in $\mathrm{GL}_2(\mathbf{Z})$ whose product has infinite order. Take $G=\langle u\rangle\ltimes\mathbf{Z}^2$, $H=\langle v\rangle\ltimes\mathbf{Z}^2$, and $A=\mathbf{Z}^2$. Then $A$ is undistorted in both $G$ and $H$ (it is a subgroup of index 2 in both), but distorted in $G\ast_A H=\langle u,v\rangle\ltimes \mathbf{Z}^2$.
Example:
$$u=\begin{pmatrix}1 & -1 \\ 0&-1\end{pmatrix},\;v=\begin{pmatrix}1 & 0 \\ -1&-1\end{pmatrix},\;uv=\begin{pmatrix}2 & 1 \\ 1&1\end{pmatrix}$$
(in this case $\mathbf{Z}^2$ is exponentially distorted in the amalgamated product).
A: 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, Int. J. Algebra Comput. 11, No. 2, 185-216 (2001). ZBL1025.20028.
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.
