Why do non-elementary limit groups over hyperbolic groups contain a non-abelian free subgroup?

Question

I found the above statement in the following article on page 8:

https://arxiv.org/pdf/2002.10278.pdf

I looked at some sources and did not find an answer and I also was not able to prove it by myself.

Does anybody have an idea?

If you have an answer it would be great if you could give me a reference for the proof or a similar statement because I need this for my master thesis.

Edit: My tutor said that it would be a possibility to construct an epimorphism from the limit group to a non-abelian free group. To "get back" from the free group to the limit group should be easy then, she said. But I dont know how to construct such an epimorphism.

Answer 1

Firstly: Your tutor's idea does not work. Every subgroup of a hyperbolic group $\Gamma$ is a $\Gamma$-limit group. As these can be two-generated but non-free, they do not in general surject onto $F_2$.

Instead, we should use the characterisation Proposition 1.18 in Sela's paper Diophantine geometry over groups. VII. The elementary theory of a hyperbolic group, Proc. Lond. Math. Soc. (3) 99 (2009), no. 1, 217–273.

Let $\Gamma$ be a non-elementary torsion-free hyperbolic group. A finitely generated group $G$ is called $\omega$-residually $\Gamma$ if for any finite set of elements $\{g_1,\ldots,g_n\}\in G$, there exists a homomorphism $h:G\to \Gamma$ that maps these elements into distinct elements in $\Gamma$.

Proposition 1.18. Let $\Gamma$ be a non-elementary torsion-free hyperbolic group. A finitely generated group $G$ is $\omega$-residually $\Gamma$ if and only if it is a $\Gamma$-limit group.

This mirrors the situation for the classical limit groups, as fully residually free groups.

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We will use this result combined with the following two standard results:

Theorem (Gromov). Let $\Gamma$ be a torsion-free hyperbolic group. For any two non-commuting elements $x, y\in\Gamma$ there exists an integer $n>0$ such that $\langle x^n, y^n\rangle$ is a non-abelian free group.

Lemma. Let $\phi:A\to B$ be a homomorphism. If the image $\phi(A)$ contains a non-abelian free subgroup, then $A$ contains a non-abelian free subgroup.

I'll now prove the result assuming everything is torsion-free. I'll leave you to complete the "with torsion" case.

Theorem. Let $\Gamma$ be a torsion-free hyperbolic group, and let $G$ be a non-cyclic torsion-free $\Gamma$-limit group. Then $G$ contains a non-abelian free subgroup.

Proof. As $G$ is non-cyclic and torsion-free, there exists a pair of elements $a, b\in G$ such that $[a, b]=1$. By taking the set $\{a, b, [a, b], 1\}$ in Sela's Proposition 1.18, there exists a homomorphism $h: G\to\Gamma$ such that $h(a)\neq1$, $h(b)\neq1$ and $h([a, b])\neq1$. In particular, the elements $x=h(a)$ and $y=h(b)$ do not commute in $\Gamma$. By the above theorem, it follows that there exists some $n>0$ such that $\langle x^n, y^n\rangle$ is a non-abelian free group. In particular, $h(G)$ contains a non-abelian free subgroup. By the lemma, $G$ itself contains a non-abelian free subgroup as required. QED