TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-09-24 (gpt-6-sol)

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Non-elementary hyperbolic groups contain a rank-two free subgroup

Statement

Every non-elementary hyperbolic group contains a free subgroup of rank $2$ .

Facts & Assumptions

Given: A non-elementary hyperbolic group $G$ .

[A1]

Every non-elementary hyperbolic group contains independent infinite-order elements $g, h$ with pairwise disjoint attracting and repelling neighborhoods $U_{g}^{+}, U_{g}^{-}, U_{h}^{+}, U_{h}^{-}$ . Their boundary actions have north--south dynamics: for all sufficiently large $N$ , $g^{\pm N} (\partial G ∖ U_{g}^{\mp}) \subseteq U_{g}^{\pm}, h^{\pm N} (\partial G ∖ U_{h}^{\mp}) \subseteq U_{h}^{\pm} .$ (Kapovich--Benakli, Theorem 2.28, Proposition 4.2, and Theorem 4.3.)

Proof

technique · direct

1.1givenA1choose

By [A1], choose independent infinite-order elements $g, h \in G$ with disjoint attracting and repelling neighborhoods on the boundary.

2.1A1step 1.1choosealgebra∎

Choose $N$ large enough for all four north--south inclusions in [A1]. Write $D_{g^{N}} = U_{g}^{+}$ , $D_{g^{- N}} = U_{g}^{-}$ , $D_{h^{N}} = U_{h}^{+}$ , and $D_{h^{- N}} = U_{h}^{-}$ . If $w = s_{1} \dots s_{k}$ is a nonempty reduced word in $g^{\pm N}, h^{\pm N}$ , choose a letter $t$ distinct from both $s_{1}$ and $s_{k}^{- 1}$ and a point $x \in D_{t}$ . Acting from right to left, [A1] gives successively. [A1, step 1.1, choose] $s_{j} \dots s_{k} x \in D_{s_{j}} (j = k, k - 1, \dots, 1),$ because reducedness says $s_{j + 1} \neq s_{j}^{- 1}$ . Thus $w x \in D_{s_{1}}$ , while $x \in D_{t}$ , and these domains are disjoint. Hence $w x \neq x$ , so $w$ is not trivial. Therefore $g^{N}$ and $h^{N}$ freely generate a free subgroup of rank $2$ .

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Sources

Brian H. Bowditch, A course on geometric group theory, Section 6.11.1 (S4) (standard reference, not scraped)
Ilya Kapovich and Nadia Benakli, Boundaries of hyperbolic groups, Theorem 2.28, Proposition 4.2, and Theorem 4.3 (standard reference, not scraped)