# Are all almost virtually free groups word hyperbolic?

Suppose $$G$$ is a finitely generated group with a finite symmetric generating set $$A$$. Lets define Cayley ball $$B_A^n := (A \cup \{e\})^n$$ as the set of all elements with Cayley length (in respect to $$A$$) $$n$$ or less.

Suppose $$R_1, … , R_k$$ are $$k$$ random elements chosen uniformly from $$B_A^n$$. Then we can define a random $$k$$-generated subgroup of $$G$$ as $$H(G, A, k, n) = \langle \{R_1, … , R_k\} \rangle$$.

Now, suppose, $$\mathfrak{X}$$ is some group property closed under finitely-generated subgroups. We say, that a finitely generated group $$G := \langle A \rangle$$ is almost $$\mathfrak{X}$$ iff $$\forall k \in \mathbb{N} \lim_{n \to \infty} P(H(G, A, k, n)) = 1$$.

The following facts are not hard to see:

The definition does not depend on the choice of $$A$$

The property of being almost $$\mathfrak{X}$$ is closed under finitely-generated subgroups

A group is almost almost $$\mathfrak{X}$$ iff it is almost $$\mathfrak{X}$$

Moreover, a following fact was proved by Gilman, Miasnikov and Osin in «Exponentially generic subsets of groups»:

Any word hyperbolic group is either almost free or virtually cyclic

An easy corollary of this statement is:

All word hyperbolic groups are almost virtually free

My question is whether the converse is also true:

Are all almost virtually free groups word hyperbolic?

• I wanted to add the geometric group theory tag, but questions are limited to 5 tags. I think ggt would be more appropriate than "probability", but that's up to you – Alessandro Codenotti Mar 28 '20 at 17:43
• I am certain it is not true although don't have an example that I could prove at the moment. I am guessing relatively hyperbolic groups could have this property. Also some groups which are not relatively hyperbolic―I suspect this is true for mapping class groups of finite type hyperbolic surfaces. Also do you want it to be free on k generators with probability one or just any free group? – user29123 Mar 28 '20 at 20:01
• Also the use of blockquotes looks bad/overused, maybe use bullet points or something if you want that kind of emphasis(that might look overused too). – user29123 Mar 28 '20 at 20:21
• Presumably the infinite cyclic Tarski monsters are counter-examples (finitely generated, non-cyclic groups such that every proper, non-trivial subgroup is infinite cyclic). They are non-hyperbolic. The only only issue is proving that a random set of elements does not generate the whole group. – user1729 Jul 24 '20 at 21:43
• @PaulPlummer Yes, I agree that the Tarski example probably doesn't work. I also don't see why it is closed under f.g. subgroups, because why should the probability aspect fall down to arbitrary f.g. subgroups? And I also agree that this would give you counter-examples. (I also see why the $F_2\times\mathbb{Z}$ example might work. It would work if the derived subgroup of $F_3$ was generic, but unfortunately I don't think it is.) – user1729 Jul 25 '20 at 7:58

The answer is no. The paper Generic free subgroups and statistical hyperbolicity, by Suzhen Han and Wen-yuan Yang, proves almost virtually free for a class of groups which includes relatively hyperbolic groups.

To make sure we are on the same page I will state the result precisely in the case of relatively hyperbolic groups. Define $$U^{(k)}:=\{(u_1,...,u_k) \mid u_i \in U\}$$. Let $$G$$ be a relatively hyperbolic group generated by a finite set $$S$$ and let $$B_n$$ be the ball of radius $$n$$ in the Cayley graph of $$(G,S)$$ centered at the identity. They show

$$\lim_{n \to \infty} \frac{ \left|X \cap B_n^{(k)}\right|}{|B_n^{(k)}|} = 1$$

where $$X \subseteq G^{(k)}$$ is the set of elements $$(g_1,...,g_k)$$ such that $$\langle g_1,...,g_k \rangle$$ is a free group of rank $$k$$ (Corollary of Corollary 1.6). In particular:

• Almost virtually free does not imply hyperbolicity since relatively hyperbolic does not imply hyperbolic (see next bullet point for an example).
• Almost virtually free groups can have subgroups which are not almost virtually free. Note that $$\mathbb{Z}^2$$ is not almost virtually free but can be contained in relatively hyperbolic groups. If $$M$$ is a finite volume hyperbolic three manifold with cusps then $$\pi_1(M)$$ is relatively hyperbolic and contains $$\mathbb{Z}^2$$ subgroups.

I would like to point out that what is shown in Exponentially generic subsets of groups is somewhat different from the result above for hyperbolic groups. Essentially what they prove is that when you look at surjective homomorphism $$F(S) \to G$$, $$G$$ hyperbolic, that tuples of words generically map to tuples of elements which generate a free group. This is somewhat different from the ball model of randomness and I don't believe it follows that you get the almost virtually free property for hyperbolic groups.

If instead you use this model of randomness then your question still has a negative answer. The authors of this paper point out groups which have surjective homomorphisms to non-elementary hyperbolic groups have the "word almost virtually free property". For example you get that $$F_n \times \mathbb Z$$ has this property, witnessed by the projection to $$F_n$$.

• I am actually not entirely convinced that almost virtually free property is preserved by change of generating set, although I have not thought about this much. The basic reasoning is that changing generating sets can introduce multiplicative "errors" and if the ball model converges to 1 slow enough with one generating set it seems possible that a different generating set won't converge to 1. Again I have not thought about this much and there could be a simple argument that answers this question. – user29123 Jul 29 '20 at 17:44