TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-09-23 (gpt-6-sol)

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The centralizer of an infinite-order element in a hyperbolic group is virtually cyclic

Statement

Let $G$ be a hyperbolic group and let $g \in G$ have infinite order. Then its centralizer

$C_{G} (g) = {h \in G : h g = g h}$

contains a cyclic subgroup of finite index.

Facts & Assumptions

Given: A hyperbolic group $G$ and an infinite-order element $g \in G$ .

[L1]

Infinite-order elements are undistorted (Infinite-order elements of hyperbolic groups are undistorted).

[L2]

In a finitely generated $δ$ -slim hyperbolic group, if an infinite-order element has a power orbit with quasi-isometry constants $λ, c$ , then its centralizer contains its cyclic subgroup with finite index; each coset has a representative in a ball whose radius depends only on $δ, λ, c$ (Axis fellow travelling controls the centralizer).

Proof

technique · direct

1.1L1given

The standing hyperbolic-group convention supplies a finite generating set whose geodesic Cayley realization is $δ$ -slim. By [L1], the power orbit of $g$ is quasi-isometrically embedded; fix its constants $λ, c$ . The hypotheses of [L2] are therefore met.

2.1L2step 1.1∎

Apply [L2]. It gives finitely many cosets of $⟨ g ⟩$ in $C_{G} (g)$ , with a representative of each in one finite word ball. Thus $⟨ g ⟩$ is a cyclic subgroup of finite index in the centralizer, as claimed.

Depends on

Used by

Abelian subgroups of hyperbolic groups are virtually cyclic Corollary

Dependency tree · two levels

18 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Clara Löh, Geometric Group Theory, Section 6.5.2 (standard reference, not scraped)