In the article cited below, I. Castellano gives a proof for the following result (Proposition 4.1).



> 
> Let $G$ be a compactly generated totally disconnected locally compact group. Suppose that $G$ acts discretely on a tree $\mathcal T$ such that
> 
> 
> 1. the group $G$ is acting without edge inversions;
> 2. the quotient graph $G\backslash\mathcal T$ is finite;
> 3. the edge stabilisers $G\_e$ are compact open subgroups of $G$.
> 
> 
> Then the vertex stabilizers $G\_v$ are compactly generated.
> 
> 
> 


Here acting *discretely* means that the stabilizers are open subgroups of $G$.


The proof as cited is rather involved and uses cohomology arguments, while the proposition looks rather innocent. Is there a more straightforward, elementary proof for this result? Note that the proposition should also hold for trees that are not locally finite (and in this case, the topology becomes a lot more subtle).


[1] Castellano, I., *Rational discrete first degree cohomology for totally disconnected locally compact groups*. [arXiv:1506.02310](https://arxiv.org/abs/1506.02310v3) [math.GR].



Compactly generated vertex stabilisers in compactly generated t.d.l.c. groups acting on trees
As YCor suggests, proceed by Bass–Serre theory. We can write $G$ in the form


$\frac{G\_{v\_1} \ast \dots \ast G\_{v\_m} \ast F(E)}{\langle \langle \overline{e}\alpha\_e(g)e\alpha\_{\overline{e}}(g)^{-1} \; (g \in G\_e), \; e\overline{e}, \; e \; (e \in E') \rangle \rangle}$


where $E$ is a set of representatives for the edges in the quotient graph, $E' \subseteq E$ is a set of representatives for the edges in a spanning tree of the quotient graph, and $\alpha\_e: G\_e \rightarrow G\_{o(e)}$ is the natural embedding of an edge group into the vertex group of the origin of the edge. Note that $E$ is finite and the Bass–Serre relations are gluing together vertex groups along compact open subgroups $G\_e$. (This also ensures that everything is fine in terms of the group topology: you can just start with the topology of one of the edge groups, and then as a group topology, it extends uniquely to the whole of $G$.) Write $F$ for the image of $F(E)$ in $G$.


Let's look at $G\_{v\_1}$. We have a compactly generated open subgroup $H\_0$ of $G\_{v\_1}$ generated by $\alpha\_e(G\_e)$ for all $e \in E$ such that $o(e)=v\_1$. In other words, all the amalgamation of $G\_{v\_1}$ with the other vertex groups happens inside $H\_0$.


Now write $G\_{v\_1}$ as a directed union $\bigcup\_{i \in I}H\_i$, where each $H\_i$ is compactly generated and $0$ is the least element of $I$. Let $K\_i$ be the subgroup of $G$ generated by $H\_i \cup G\_{v\_2} \cup G\_{v\_3}\cup\dots\cup G\_{v\_n} \cup F$. Then $G = \bigcup\_{i \in I}K\_i$; since $G$ is compactly generated, in fact $G = K\_i$ for some $i$. In particular, every $g \in G\_{v\_1}$ can be expressed as a product of elements of $H\_i \cup G\_{v\_2} \cup G\_{v\_3}\cup\dots\cup G\_{v\_n} \cup F$. Using the normal form theorem for graphs of groups, we conclude that $g \in H\_i$. Thus $G\_{v\_1}$ is compactly generated.


