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By a mapping tori of $G$, I mean a semidirect product $G\rtimes\mathbb{Z}$, and by a trivial mapping tori I mean one isomorphic to $G\times\mathbb{Z}$.

If $G$ is finitely generated but not finitely presentable, clearly $G\times\mathbb{Z}$ is not finitely presentable. My question is: Does there exist a group $G$, with some property so as to be interesting, and where every other mapping torus is finitely presentable?

More precisely: Does there exist a finitely generated, non-finitely presentable group $G$ such that:

  1. $\operatorname{Out}(G)$ is not virtually cyclic, and
  2. for all $\phi\in\operatorname{Aut}(G)\setminus\operatorname{Inn}(G)$, the mapping torus $G\rtimes_{\phi}\mathbb{Z}$ is finitely presentable?

Condition (1) ensures that any answer isn't just a case of checking finitely many outer automorphisms (and there are examples with $\operatorname{Out}(G)$ finite and with $\operatorname{Out}(G)\cong\mathbb{Z}$, via Bumagin and Wise's version of Rips' construction). Note that $\operatorname{Out}(G)$ is considered rather than $\operatorname{Aut}(G)$, as two automorphisms of $G$ from the same outer automorphism class define isomorphic mapping tori.


My motivation here is that I was idly wondering if a group/class of groups I was looking at has this property. However, I do not know if any groups have this property, so this question is a sanity check (and seems interesting in it's own right). Moreover, I am interested to see what techniques could be used here - so multiple answers using a variety of techniques would be amazing!

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  • $\begingroup$ What happens for the lamplighter group? Does it have any finitely presented mapping torus? $\endgroup$ Commented Sep 13, 2023 at 11:01
  • $\begingroup$ You might look at the comments to mathoverflow.net/questions/104400/… $\endgroup$ Commented Sep 13, 2023 at 13:23
  • $\begingroup$ As a toy model one can ask the same question about nontrivial mapping tori which are all finitely generated. Does it imply finite generation of the original group? $\endgroup$ Commented Sep 13, 2023 at 14:50
  • $\begingroup$ @Carl-FredrikNybergBrodda No, it doesn't have any. $\endgroup$
    – YCor
    Commented Sep 13, 2023 at 15:43
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    $\begingroup$ The non-fp group $\mathbf{Z}\ltimes (\mathbf{Z}[1/2]^2)$, action by diagonal $(2,2^{-1})$ might be more interesting to look at. Of course one can cheat and act by a virtually inner automorphism (e.g., flip of components). But otherwise any "non-inner enough" automorphism is likely to produce a fp group. $\endgroup$
    – YCor
    Commented Sep 13, 2023 at 15:49

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