Relative hyperbolicity and splittings A subgroup $H \leq G$ is almost malnormal whenever $|H \cap H^g| <+ \infty$ for every $g \notin H$. For the definition of relatively hyperbolic groups, see for instance there.

Do you know an example of an amalgamated free product $G=A \underset{C}{\ast} B$, where $C$ is almost malnormal in $G$ and $C \neq A,B$, which is not relatively hyperbolic?

My main motivation is the following observation: If $G$ is a finitely generated group with infinitely many ends, then it splits over a finite subgroup according to Stallings' theorem. Now, looking at the action of $G$ on the associated Bass-Serre tree, it follows immediatly from Bowditch's definition of relative hyperbolicity that $G$ is relatively hyperbolic.
Therefore, I wonder if a group splitting over a subgroup which "looks like a finite group" is also relatively hyperbolic.
 A: I maybe found something. Let $G_i$ ($i=1,2$) be the group
$$\langle a_i,b_i,c_i,d_i \mid [a_i,b_i]=[b_i,c_i]=[c_i,d_i]=1 \rangle,$$ (this is the right-angled Artin group associated to a segment of length three) and let $$g_i=a_ib_ic_id_i.$$ Then the subgroup $\langle g_i \rangle$ is malnormal in $G_i$, and I suspect that the same assertion holds in the amalgamated free product $$G= G_1 \underset{g_1=g_2}{\ast} G_2.$$ We can prove that $G$ is not relatively hyperbolic. The key properties of relatively hyperbolic groups I will use are: if $G$ is a group hyperbolic relatively to a finite family of subgroups $\mathcal{H}$, then


*

*a free abelian subgroup of rank at least two is included into a conjugate of one subgroup of $\mathcal{H}$,

*the subgroups of $\mathcal{H}$ are almost malnormal,

*if $H, K$ are conjugates of two different subgroups of $\mathcal{H}$, then $H \cap K^g$ is finite for every $g \in G$.


Suppose $G$ hyperbolic relatively to a finite family of subgroups $\mathcal{H}$. The subgroup $\langle a_1,b_2 \rangle$ is a free abelian group of rank two, so it is included into a conjugate $K_1$ of one subgroup of $\mathcal{H}$. Now
$$\langle b_1 \rangle \subset \langle a_1,b_1 \rangle \cap \langle a_1,b_1 \rangle^{c_1} \subset K_1 \cap K_1^{c_1},$$
so necessarily $c_1 \in K_1$ by almost malnormality of $K_1$. Thus, $\langle a_1,b_1,c_1 \rangle \subset K_1$. Similarly,
$$\langle c_1 \rangle \subset \langle b_1,c_1 \rangle \cap \langle b_1,c_1 \rangle^{d_1} \subset K_1 \cap K_1^{d_1}$$
implies $d_1 \in K_1$. Thus, $G_1 \subset K_1$. With a similar argument, we find a conjugate $K_2$ of a subgroup of $\mathcal{H}$ such that $G_2 \subset K_2$. Because
$$ \langle g_1 \rangle = \langle g_2 \rangle = G_1 \cap G_2 \subset K_1\cap K_2,$$
we conclude that $K_1=K_2=:K$. Thus $$G= \langle G_1,G_2 \rangle \subset K.$$ Therefore, $G$ is not hyperbolic relatively to a finite family of proper subgroups.
