Finite generation of vertex groups of a cyclic splitting of a hyperbolic group and generalisations of Grushko Theorem Let $G$ be a finitely generated word hyperbolic group. Suppose $G$ acts non-trivially (without a global fixed point) on a tree without inversions and with cyclic edge stabilizers. Is it true that the vertex stabilizers are finitely generated? Can we say the same without hyperbolicity?
In the case of a general f.g. group acting on a tree with trivial edge stabilizers, Grushko's theorem tells us that the vertex groups must be finitely generated. Although I suspect there are other ways to approach my question (especially with the extra assumption of hyperbolicity), are there generalisations of Grushko's theorem to amalgamated free products?

Theorem: If $G$ is a finitely generated group acting non-trivially and without inversions on a tree $X$, and if the edge stabilizers are finitely generated, then the vertex stabilizers are also finitely generated.
The theorem follows by combining  three fairly elementary lemmas which I state below. The proof of these can be found in the following book:
D. E. Cohen, Combinatorial Group Theory: A Topological Approach. Cambridge etc.: Cambridge University Press (1989); ZBL0697.20001.

We say a graph of groups is finite if the underlying graph is a finite graph.
Lemma 1: If $G$ is a finitely generated group acting non-trivially and without inversions on a tree $X$, then $G$ is the fundamental group of a finite graph of groups $\mathcal{Y}$. If $G_v$ is a vertex stabilizer under the action of $G$ on $X$, then either it is isomorphic to an incident edge stabilizer or it is conjugate to a vertex group of $\mathcal{Y}$.
Lemma 2: Let $I$ be a set and $H$ an HNN extension of the form $$ H = \langle A, t_i \mid t_iB_it_i^{-1} = B_{-i} \, \forall i\in I\rangle. $$ If $H$ is finitely generated then I is finite and $A$ is finitely generated.
Lemma 3: Let $H = A \ast_C B$. If $H$ and $C$ are finitely generated then so are $A$ and $B$.
