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My curiosity was raised by the following question and the huge variety of comments and suggestions it attracted. I wondered if a converse statement might be equally interesting.

Let $G$ be a finitely presented group which admits a non-trivial graph of groups decomposition. Does it follow that each vertex stabiliser is finitely presented? How about if every edge stabiliser is finite?

Dunwoody comments that if $G$ is finitely generated and each edge stabiliser is finite then each vertex stabiliser is finitely generated.

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    $\begingroup$ A silly example: Take $F= \langle a_1,a_2,\ldots \mid \ \rangle$ a free group of infinite rank, and $f : a_i \mapsto a_{i+1}$ be an isomorphism between the subgroups $\langle a_1,a_2, \ldots \rangle$ and $\langle a_2,a_3, \ldots \rangle$. Then, the associated HNN extension is just $$ \langle t, a_1,a_2, \ldots, \mid ta_1t^{-1}=a_2, ta_2t^{-1} =a_3,\ldots \rangle \simeq \langle t \mid \ \rangle \simeq \mathbb{Z}.$$ $\endgroup$ – Seirios Feb 27 '15 at 17:40
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    $\begingroup$ In my example I assumed the vertex stabilizers to be finitely generated, indeed otherwise there are even easier examples. $\endgroup$ – YCor Feb 27 '15 at 18:55
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Does it follow that each vertex stabiliser is finitely presented: No.

Just notice that there exist infinitely presented groups $H$ with an automorphism such that the corresponding semidirect product $G=H\rtimes\mathbf{Z}$ is finitely presented. This is a non-trivial graph of groups decomposition, namely an HNN-decomposition where the vertex group and edge group coincide; in case you find this too degenerate, just add some further flourish to the graph of groups, so that $H$ still appears as a vertex group and the resulting group is still finitely presented. One example is when $G=BS(1,2)^2=\mathbf{Z}[1/2]^2\rtimes\mathbf{Z}^2$ where the generators of $\mathbf{Z}$ act by the diagonal matrices $(2,1)$ and $(1,2)$, and $H=\mathbf{Z}[1/2]^2\rtimes\mathbf{Z}$ where $\mathbf{Z}$ acts by the diagonal matrix $(2,1/2)$.

On the other hand, for the second question "How about if every edge stabiliser is finite?", the answer is yes. Indeed, write $H$ as a limit of truncated presentations $H_n$ (this limit here means: inductive limit of surjections between f.g. groups). For $n$ large enough (say $n\ge n_0$), all finite subgroups of $G$ occurring in the graph of groups decomposition of $G$ are well-defined, and this allows to write $G$ as (non-attained) limit of corresponding graph of groups $G_n$ for $n\ge 0$, where in the graph of groups, $H$ was replaced with $H_n$). The argument still works if the edge groups are finitely presented instead of finite.

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  • $\begingroup$ Another good example is the finitely presented hnn extension of the grigorchuk group using the L-presentation. $\endgroup$ – Benjamin Steinberg Feb 27 '15 at 17:09
  • $\begingroup$ Thanks @YCor, is there a good reference for this? $\endgroup$ – DavidHume Mar 3 '15 at 8:39
  • $\begingroup$ @DavidHume: no I'm not aware of a reference (especially for the second argument). $\endgroup$ – YCor Mar 3 '15 at 9:37

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