I'm currently working on understanding the accessibility results of Dunwoody [1] and Bestvina-Feighn [2]. Both of these papers seem to state different Dunwoody's accessibility result differently. In particular, we have the following definition.
Definition. We say that a group $G$ is accessible if there is a $G$-tree $T$ such that the edge stabilisers of $T$ are finite and the vertex stabilisers have at most one end.
In [1], Dunwoody shows that all finitely-presented groups are accessible. In [2], the authors restate Dunwoody's result as the following.
Theorem. Let $G$ be a finitely-presented group. Then there exists a positive integer $\zeta(G)$ such that if $T$ is a reduced $G$-tree with finite edge stabilisers, then $T/G$ has at most $\zeta(G)$ vertices.
By a reduced $G$-tree, we mean a minimal $G$-tree such that the vertex group of every vertex of valence two in $T/G$ properly contains the edge groups of the abutting edges.
I can see from the argument in [1] that one can deduce the above statement by carefully examining the argument used, however I cannot tell if this statement follows from accessibility in general. Precisely, my question is the following.
Question. Let $G$ be a finitely-generated group, and suppose that $G$ is accessible. Is it true that there exists a positive integer $\zeta(G)$ such that if $T$ is a reduced $G$-tree with finite edge stabilisers, then $T/G$ has at most $\zeta(G)$ vertices?
The reason that I ask this is that in [3], Dunwoody seems to use the affirmative of the above question to prove that a certain finitely-generated group is inaccessible (unless I am misunderstanding this paper).
I'm feeling a little bit lost with these results right now, so any references/explanations would be appreciated.
References
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Dunwoody, Martin J. "The accessibility of finitely presented groups." Inventiones mathematicae 81.3 (1985): 449-457.
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Bestvina, Mladen, and Mark Feighn. "Bounding the complexity of simplicial group actions on trees." Inventiones mathematicae 103.1 (1991): 449-469.
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Dunwoody, Martin J. "An inaccessible group." Geometric group theory 1 (1993): 75-78.
NB: I posted a similar question earlier today, but deleted it after realising my question wasn't quite formulated correctly.
Best Answer
Let $G$ be an accessible group. Fix a reduced decomposition $G= \pi_1(\mathbb{A})$ as a graph of groups whose edge-groups are finite.
Step 1. We begin by "blowing up" the vertex-groups of $\mathbb{A}$ that have more than one end, replacing them with reduced decompositions as graphs of groups with finite edge-groups and with vertex-groups that have at most one end. The construction is described in Dick and Dunwoody's book Groups acting on graphs (Section III.3). Thus, one obtains a new reduced decomposition $G=\pi_1(\mathbb{A}')$ where the number of vertices (resp. edges) of $\mathbb{A}'$ is at least the number of vertices (resp. edges) of $\mathbb{A}$.
Now, I want to simplify $\mathbb{A}'$. Namely, I want the graph of groups to satisfy: $(\ast)$ for every edge $[u,v]$, if $G_u = G_{[u,v]}$ then $u=v$. This condition is stronger than being reduced. Observe that, if $[u,v]$ is an edge with $G_u = G_{[u,v]}$ and $u \neq v$, then one can collapse this edge. So, after collapsing some edges, one obtains a new graph of groups $\mathbb{A}''$. Reversely, $\mathbb{A}'$ is obtained from $\mathbb{A}''$ by expanding edges. But, because $\mathbb{A}'$ is reduced, we have to avoid creating vertices of degree two. Typically, the picture is
As a consequence, the quantity $$\sum\limits_{v \text{ vertices}} \left( \mathrm{deg}(v)-1 \right)$$ decreases after each expansion. So $$v(\mathbb{A}) \leq v(\mathbb{A}') \leq v(\mathbb{A}'')+ \sum\limits_{u \text{ vertex of } \mathbb{A}''} \left( \mathrm{deg}(u)-1 \right)= 2 e(\mathbb{A}'')$$ where $v(\cdot)$ and $e(\cdot)$ denote the number of vertices and edges under consideration.
Since $G$ surjects onto the fundamental group of the graph underlying $\mathbb{A}''$, which has rank $e(\mathbb{A}'')-v(\mathbb{A}'')$, we conclude that $$v(\mathbb{A}) \leq 2 v(\mathbb{A}'') + 2\mathrm{rank}(G).$$ In the next step, we show that $v(\mathbb{A}'')$ depends only on $G$.
Step 2. In order to conclude the proof, we observe that:
So let $\mathbb{B}, \mathbb{C}$ be two such graphs of groups and let $\mathcal{B}$ (resp. $\mathcal{C}$) denote the collection of vertex-groups in $\mathbb{B}$ (resp. $\mathbb{C}$). We are going to prove that $\mathcal{B}$ and $\mathcal{C}$ have the same size.
Write $\mathcal{B}= \{B_1,\ldots, B_r\}$ and $\mathcal{C}= \{C_1,\ldots, C_s\}$. For every $1 \leq i \leq r$, there exist $g_i \in G$ and $1 \leq n(i) \leq s$ such that $B_i \leq g_i C_{n(i)}g_i^{-1}$; and similarly, for every $1 \leq i \leq s$, there exist $h_i \in G$ and $1 \leq m(i) \leq r$ such that $C_i \leq h_i B_{m(i)} h_i^{-1}$. Of course, we have $$B_i \leq g_ih_{n(i)} B_{m(n(i))} h_{n(i)}^{-1}g_i^{-1} \text{ and } C_j \leq h_j g_{m(j)} C_{n(m(j))} g_{m(j)}^{-1}h_j^{-1}$$ for all $1 \leq i \leq r$ and $1 \leq j \leq s$. Now, apply the following lemma in order to conclude that $m(n(i))=i$ and $n(m(j))=j$, i.e. $m$ and $n$ are bijections.
Lemma: Let $G$ be group that decomposes as a reduced graph of groups whose edge-groups are finite and whose vertex-groups have at most one end. Let $\mathcal{H}$ denote the set of vertex-groups. Then, for all $H,H' \in \mathcal{H}$ and $h \in H$, if $H \leq hH'h^{-1}$ then $H=H'$.
Considering the action of $G$ on the Bass-Serre tree, $G_u$ and $hG_vh^{-1}$ stabilise two vertices $a$ and $b$. If $G_u \leq hG_vh^{-1}$, then $G_u$ stabilises the geodesic $[a,b]$, which implies that $G_u$ coincides with the stabiliser of each edge of $[a,b]$. By applying property $(\ast)$ step by step along the vertices in $[a,b]$, we conclude that $u=v$.