Is there an example of a group $G$ that has the properties

  1. the cohomological dimension of $G$ is infinite: $\operatorname{cd}(G) = \infty$,
  2. $G$ is torsion-free,
  3. $G$ is of type $F_\infty$,
  4. $G$ does not contain (an isomorphic copy of) Thompson's group $F$?

Those who don't know about Thompson's group can replace 4 by the stronger requirement

4'. $G$ does not contain an infinite-rank free abelian group.

Motivation: there are various known groups that satisfy 1-3 but all of them contain $F$. If one is bold one might wonder whether this because of what we know or because of $F$.

On a less speculative level, this is a variation of this question: two good reasons for 1 are for $G$ to have torsion or to contain infinite-rank free abelian groups. So how to assure it if the two are ruled out? The cited question did not require 3 although in comments IgorBelegradek and IanAgol speculate in that direction.

  • $\begingroup$ There are some recent papers on variants on Thompson's group using the group $\mathbf{Z}+\alpha\mathbf{Z}$, $\alpha$ irrational, in lieu of $\mathbf{Z}[1/2]$ (arxiv.org/abs/1806.00108, arxiv.org/abs/2006.02401). Might they be candidates? (Btw for a group of homeomorphisms, "contain" $F$ has its literal meaning distinct from the one you have in mind, so I'd rephrase 4. as "...not contain any isomorphic copy of Thompson's group $F$".) $\endgroup$
    – YCor
    Jun 20, 2020 at 9:55
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    $\begingroup$ @YCor: I doubt it, it is hard to keep $F$ out of a Thompson-like group, see for example Theorem 1.1 of arxiv.org/abs/1610.04099. But more to the point, 4 is an attempt at formalizing "is not related to Thompson groups", so if there was a Thompson-like construction that avoided $F$ I would alter my question accordingly ;-). I don't see a big risk of misunderstanding "contain $F$" since my question is about abstracts groups, but I'll modify it nonetheless. $\endgroup$ Jun 20, 2020 at 10:21
  • $\begingroup$ Very interesting reference! I ignored that $F$ was so unavoidable. $\endgroup$
    – YCor
    Jun 20, 2020 at 11:06
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    $\begingroup$ Did you try central extensions of $F$? $\endgroup$
    – Tox
    Jun 21, 2020 at 15:50
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    $\begingroup$ @Tox: The answer to you suggestion is: the central extensions of $F$ contain $F$, so no candidates there. The infinite version of the above presentation has the relations $X_i^{-1} X_{i+1} X_i = X_{i+2}Z^r$ but $X_i^{-1}X_jX_i = X_{j+1}$ for $j > i+1$. As a consequence the subgroup generated by the $X_{2i}^2$ is isomorphic to $F$. And then it contains a copy of $F$ that lies in the commutator subgroup. [Both of these facts were pointed out by Matt Zaremsky.] $\endgroup$ Sep 22, 2020 at 18:38


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