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This list of open problems from http://grouptheory.info/ includes the question:

"Is every biautomatic group which does not contain any $\mathbb{Z} \times \mathbb{Z}$ subgroups, hyperbolic?"

It is credited to Gersten but I don't see any mention of it in the provided reference.

The most similar question there seems to be:

"Let $\phi \in \operatorname{Out}(F)$, where $F$ is a finitely generated free group, and let $G = F \rtimes_\phi \mathbb{Z}$. Is $G$ automatic?

It has also been conjectured that $F \rtimes_\phi \mathbb{Z}$ is word hyperbolic if it contains no subgroup isomorphic to $\mathbb{Z} \oplus \mathbb{Z}$."

Which doesn't even seem that related. I haven't been able to find any other mentions of this question or further work.

Since every hyperbolic group is biautomatic, it seems to me to be a natural question to ask when biautomatic groups are hyperbolic.

Is this question still open? Has there been any related work?

What about characterizing when bicombable groups are hyperbolic?

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    $\begingroup$ I suppose a candidate counterexample is: Recently, Italiano-Martelli-Migliorini found a type $F_\infty$ subgroup of a hyperbolic group that is not hyperbolic: arxiv.org/abs/2105.14795. So in particular this is a non-hyperbolic group with no $\mathbb{Z}\times\mathbb{Z}$ subgroups, and since biautomatic groups must be of type $F_\infty$, this might be one of the only potential counterexamples out there. (I have no idea whether to actually expect this group is biautomatic though; it being of type $F_\infty$ is really my only "evidence".) $\endgroup$ Commented Nov 10, 2021 at 23:33
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    $\begingroup$ This paper is related arxiv.org/abs/1309.5553 $\endgroup$ Commented Nov 17, 2021 at 23:22

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$\DeclareMathOperator\BS{BS}$Since this question goes in several directions, I hope a discursive answer is appropriate.

The question fits into an important family of questions in geometric group theory which look to provide some sort of "algebraic" characterisation of hyperbolic groups. The search for this characterisation starts with two famous properties.

Proposition: Let $\Gamma$ be a hyperbolic group. Then:

  • $\Gamma$ contains no Baumslag–Solitar subgroups; and
  • $\Gamma$ acts properly and cocompactly on an aspherical complex (its Rips complex). In particular, if $\Gamma$ is torsion-free it is of finite type.

(Recall that a Baumslag–Solitar group is given by a presentation $\BS(m,n)=\langle a,b\mid ba^mb^{-1}=a^n\rangle$. In particular, $\BS(1,1)\cong\mathbb{Z}^2$.)

An algebraic characterisation might look like a converse to this proposition. In 1990, the most general form of this question was:

Question 1: If $G$ is a finitely presented group with no subgroups isomorphic to $\BS(1,n)$ for any $n$, must $G$ be hyperbolic?

Question 1 was answered in the negative by Noel Brady, who gave an example of a torsion-free finitely presented subgroup $H$ of a hyperbolic group with $b_3(H)=\infty$; in particular, $H$ couldn't be of finite type. So for the last 20 years, the question has been:

Question 2: If $G$ is a group of finite type with no subgroups isomorphic to $\BS(1,n)$ for any $n$, must $G$ be hyperbolic?

This was the first question on Bestvina's famous problem list, and I think there's a good case to be made that it was the most important question in geometric group theory: when $G$ is a 3-manifold group, the answer is "yes" by geometrisation, so it could be thought of as a geometrisation conjecture for groups.

Very excitingly, as Matt Zaremsky mentions in comments, Question 2 has recently been answered in the negative by Italiano–Martelli–Migliorini, who constructed a normal subgroup $K$ of infinite index in a hyperbolic group, which is both of finite type and $4$-dimensional. It follows that $K$ can't be hyperbolic, by deep theorems of Paulin and Rips.

Nevertheless, Question 2 can, and has been, specialised to any of your favourite classes of finite-type groups: CAT(0) groups, (bi)automatic groups, one-relator groups... As far as I know, it remains open in all these cases. However, as Matt again suggests, the important task remains to figure out whether or not Italiano–Martelli–Migliorini's group $K$ has any of these properties, since it may also provide a counterexample to more refined versions of the question.

Let me quickly finish by noting that there are several classes where positive answers to Question 2 are known. These include:

  1. $\Gamma$ is a 3-manifold group (Perelman).
  2. $\Gamma$ is $F\rtimes\mathbb{Z}$ (Brinkmann).
  3. $\Gamma$ is the fundamental group of a special cube complex (Caprace–Haglund).
  4. Ascending HNN-extensions of free groups (Mutanguha).

So, in summary, the questions you ask about are open (and important) today, but the recent work of Italiano–Martelli–Migliorini raises the hope that more progress might be possible soon.

(By the way, let me make a quick aside about the attribution of the question. These kinds of questions are indeed sometimes attributed to Gersten – the one-relator version of the question is the one I've heard attributed explicitly to him. But it's clear that many of these famous questions must have been asked verbally in the '80s and '90s, but may not have been written down by the authors they're associated with. For instance, the question of surface subgroups for hyperbolic groups is always attributed to Gromov, but I've never found it explicitly anywhere.)

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    $\begingroup$ I'm curious where the fact that $\mathrm{BS}(2,3)$ possesses a $\mathbf{Z}^2$ subgroup can be located. When I was PhD I asked whether it holds to a few people and at least two "famous" people proved me shaking hands that it doesn't. But it indeed does contain such a $\mathbf{Z}^2$ subgroup (as does $\mathrm{BS}(m,n)$ whenever $\min(|m|,|n|)\ge 2$), and this is not hard to check. I'm asking it here because it's implicit in passing the question from $\mathrm{BS}(m,n)$ subgroups (as it appears in Bestvina's list) to $\mathrm{BS}(1,n\ge 1)$ subgroups that the other cases are actually redundant. $\endgroup$
    – YCor
    Commented Nov 11, 2021 at 10:47
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    $\begingroup$ @YCor: it's easy to see: $BS(m,n)$ is an HNN extension of the torus-link group $\mathbb{Z}*_{m=n}\mathbb{Z}$, and one can see this has a $\mathbb{Z}^2$ subgroup in various ways. For instance, the fact that it's the torus-link group proves it immediately geometrically, or one can construct the embedding explicitly. I don't know a reference, though, if that's what you're asking for. $\endgroup$
    – HJRW
    Commented Nov 11, 2021 at 14:46
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    $\begingroup$ Yes, this is the simple observation I had in mind. For some reason it seemed to be unnoticed by several people. $\endgroup$
    – YCor
    Commented Nov 11, 2021 at 15:00

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