Let $GX$ be a tree group (a right-angled Artin group such that the graph is a tree), such that not all the subgroups of $GX$ are necessarily RAAGs, so the length of the tree is greater or equal than $4$.

Then, $GX= A\ast_{C} B$, where $C=\langle c \rangle$ is a cyclic group. Let $t$ be an hyperbolic element in $GX$. It is known that there is a finite index subgroup $M$ such that $M= B \ast_{C\cap M} \langle t \rangle$ ($M$ is an HNN extension). Hence, $$M= G \ast (M\cap A)\ast (M\cap B) \ast (M\cap A^g) \ast (M \cap B^g) \ast \cdots \ast \langle t \rangle$$ and the relations tell me (apart from the relations in the groups $G,\dots, M\cap B^g$) that $t^{-1}s_{1}t=s_{2}$ where $s_{1}$ is an element in a cyclic edge group. So up to conjugacy, I may assume that $s_{1}$ is $c^n$ for some $n\in \mathbb{N}$.

Now, I would like to find a finite index subgroup such that $s_{1}^m$ is a generator and $s_{1}$ is not in the subgroup. Since $s_{1}$ lies in the edge group, is it enough to find a finite index subgroup in $A$ with that property?