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It is well-known that a subgroup of a hyperbolic group need not be hyperbolic. Let us say that a (finitely generated) group $G$ is locally hyperbolic if all its finitely generated subgroups are (Gromov) hyperbolic. I can think of two families of examples of locally hyperbolic groups:

  1. Free groups; this follows from the Nielsen-Schreier theorem.
  2. Surface groups of genus $g > 1$; this follows from the fact that any subgroup of such a surface group is either a surface group (automatically also of genus $g>1$) or else a free group.

Are there other natural classes? Are there any known properties for locally hyperbolic groups (e.g. decidable subgroup membership problem)?

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    $\begingroup$ All locally quasiconvex hyperbolic groups satisfy this property. Many examples were given by McCammond and Wise. I think the subgroup membership problem for general locally hyperbolic groups is open. It is open even for free subgroups if hyperbolic groups. $\endgroup$
    – Moishe Kohan
    Commented Jul 3, 2022 at 23:05
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    $\begingroup$ @YCor: Isn't standard to call a group locally P if all its finitely generated subgroups satisfy P? For me, a group would be hereditarily P if all its finite-index subgroups satisfy P. $\endgroup$
    – AGenevois
    Commented Jul 4, 2022 at 8:33
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    $\begingroup$ @AGenevois when P is stable under taking subgroups, yes. Otherwise, it can have both meanings (e.g., locally free in a context of other algebras). I view "locally" as a weakening, so well, making it a strengthening makes me confused. $\endgroup$
    – YCor
    Commented Jul 4, 2022 at 8:58
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    $\begingroup$ @YCor: "locally quasiconvex" is another very common example. I'm not saying it's the only standard: I'm saying it's standard enough in GGT that anything else would be confusing. And, with respect, "precisely the point" is that in GGT people are frequently less interested in extending to infinitely generated groups. $\endgroup$
    – HJRW
    Commented Jul 4, 2022 at 9:46
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    $\begingroup$ @YCor There are also locally finite groups, but I suppose that’s less geometric. $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Jul 4, 2022 at 9:48

3 Answers 3

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Many examples can be exhibited using a theorem of Gersten:

Theorem (Gersten): Let $G$ be a hyperbolic group of cohomological dimension 2. Every finitely presented subgroup $H$ of $G$ is hyperbolic.

This is already very interesting, but the question asked for examples where every finitely generated subgroup is word-hyperbolic. Recall that $G$ is called coherent if every finitely generated subgroup is finitely presented.

Corollary: If $G$ is word-hyperbolic, coherent, and of cohomological dimension 2, then every finitely generated subgroup of $G$ is hyperbolic.

We can now search the literature for examples of such groups:

  1. If $\Gamma$ is the mapping torus of a free-group endomorphism then it is 2-dimensional, and coherent by a theorem of Feighn--Handel. It is hyperbolic whenever the endomorphism is atoroidal, by work of Brinkmann (for automorphisms) and Mutanguha (in the general case). (Giles also mentions these in his answer.)

  2. A one-relator group $\Gamma=F/\langle\langle w\rangle\rangle$ has (virtual) cohomological dimension at most 2 by Lyndon and Cockroft, so in the context of your question we need to understand which are hyperbolic and which are coherent. Baumslag conjectured that such $\Gamma$ is always coherent, while Gersten conectured that $\Gamma$ is hyperbolic unless it contains a Baumslag--Solitar subgroup. These conjectures have been confirmed in certain cases:

  • If $\Gamma$ has torsion (i.e. $w$ is a proper power) then $\Gamma$ is hyperbolic by the B.B. Newman Spelling theorem, and is coherent by recent work of Wise and Louder--W.

  • If $\Gamma$ is 2-free, i.e. every 2-generated subgroup of $\Gamma$ is free, then $\Gamma$ is hyperbolic by very recent work of Linton, and coherent by another recent paper of Louder--W.

  1. It is well known that random groups

$\Gamma = \langle a_1,\ldots, a_m\mid w_1,\ldots,w_n\rangle$

are a.a.s. hyperbolic in many regimes. They are also coherent by a recent paper of Kielak--Kropholler--Wilkes whenver $n<m-1$, and coherent with positive probability when $n=m-1$. It is reasonable to conjecture that they are still a.a.s. coherent when $n=m-1$, but a.a.s incoherent when $n\geq m$, and also in the density model.

As far as I know, the membership problem remains open in all these cases. It certainly isn't known for general locally hyperbolic groups.

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    $\begingroup$ So every finitely generated subgroup of a one-relator group with torsion is hyperbolic! In particular, every such subgroup has decidable conjugacy problem. This is a really nice (recent) extension of Bill Newman’s old result. $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Jul 4, 2022 at 9:26
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    $\begingroup$ @Carl-FredrikNybergBrodda: Sure. The 2-free examples (we call them "negative immersions") can also be seen as a much deeper, and much more generic, extension of Newman's theorem. $\endgroup$
    – HJRW
    Commented Jul 4, 2022 at 9:30
  • $\begingroup$ How "effective" is this decidability? For example, if $G = \langle a, t \mid (t^{-1} a^{-1} t a^2)^2 = 1 \rangle$, and $H$ is the subgroup of $G$ generated by $\{ t, a^2, ata^{-1} \}$, how hard is it to decide (in practice) whether $ata^{-2}t^{-1}a^{-1}t^{-1}$ is conjugate to $a^{-4}t^{-1}$ in $H$? (Example chosen at "random", but spoiler: the answer should be "yes, they are conjugate". However, $H$ is a torsion-free one-relator group, so we cannot rely on Newman's theorem). $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Jul 4, 2022 at 12:25
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    $\begingroup$ @Carl-FredrikNybergBrodda: It's unclear if our coherence result is effective. This is also something I would be very interested to know! It certainly doesn't follow immediately from our proof. I think the same is true for Scott for 3-manifolds and for Feighn--Handel for ascending HNN extensions of free groups, but would be interested to hear otherwise. In the 3-manifold context, the presentation problem is known to be solvable (see Remark 4.19 of arxiv.org/abs/1405.6274) but this uses a lot more machinery. $\endgroup$
    – HJRW
    Commented Jul 4, 2022 at 12:51
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    $\begingroup$ For the benefit of future readers: Baumslag's conjecture that one-relator groups are coherent is now a theorem of Jaikin-Zapirain and Linton arxiv.org/abs/2303.05976 $\endgroup$
    – Giles Gardam
    Commented Nov 13, 2023 at 16:53
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  1. Finite groups.

  2. ...

  3. ...

  4. Fundamental groups of closed connected hyperbolic three-manifolds.

Here is a proof of the latter using far too many tools from kleinian groups. Fundamental groups of closed connected hyperbolic three-manifolds are quasi-isometric to hyperbolic three-space and thus are Gromov hyperbolic. This deals with all finite index subgroups of $\pi_1(M)$.

Suppose instead that $N$ is an infinite cover of $M$, with $\pi_1(N)$ finitely generated. Applying the tameness theorem (Agol, and also Calegari-Gabai) and the covering theorem (Canary) we have that $N$ is either completely degenerate (and homotopic to a closed surface) or has flaring ends. In the latter case $\pi_1(N)$ is quasi-convex in $\pi_1(M)$ and thus is Gromov hyperbolic.

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A locally hyperbolic group is in particular coherent (i.e. locally finitely presented), which is already a special property. To add to the examples already given by Sam Nead: ascending HNN extensions of free groups (a.k.a. mapping tori of free group endomorphisms), such as free-by-cyclic groups, were shown to be coherent by Feighn and Handel. Recently, Mutanguha has shown that indeed if such a group is hyperbolic, then it is locally hyperbolic (see Corollary 7.6 in his paper).

Feighn, Mark; Handel, Michael, Mapping tori of free group automorphisms are coherent, Ann. Math. (2) 149, No. 3, 1061-1077 (1999). ZBL0938.20022.

Mutanguha, Jean Pierre, The dynamics and geometry of free group endomorphisms, Adv. Math. 384 (2021), ZBL07415144 arXiv:2005.11896.

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  • $\begingroup$ Crossed answers! In fact, Gersten's theorem means that local hyperbolicity already followerd from Feighn--Handel. $\endgroup$
    – HJRW
    Commented Jul 4, 2022 at 9:24
  • $\begingroup$ Excellent point! So the relevance of Mutanguha's result to this question is really about establishing hyperbolicity in the first place, as you wrote in your answer. $\endgroup$
    – Giles Gardam
    Commented Jul 4, 2022 at 9:49
  • $\begingroup$ That's right.${}$ $\endgroup$
    – HJRW
    Commented Jul 4, 2022 at 9:53

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