# Compactly generated vertex stabilisers in compactly generated t.d.l.c. groups acting on trees

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).

 Castellano, I., Rational discrete first degree cohomology for totally disconnected locally compact groups. arXiv:1506.02310 [math.GR].

• Yes I think it can be proved using plain Bass-Serre theory. I have to think about details... – YCor Sep 10 '19 at 20:14

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.