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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-27
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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 gG have infinite order. Then its centralizer

CG(g)={hG:hg=gh}

contains a cyclic subgroup of finite index.

Facts & Assumptions

Given: A hyperbolic group G and an infinite-order element gG.

[A1]

There is L0 such that every geodesic segment from 1 to gn lies in the L-neighborhood of the powers of g.

[A2]

A geodesic quadrilateral in a δ-hyperbolic Cayley graph is 2δ-thin, and every metric ball in a locally finite Cayley graph is finite.

[L1]

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

Proof

technique · direct
1.1

By [L1] and [A1], the powers of g form a quasi-axis: geodesics joining distant powers stay uniformly close to the power orbit.

L1A1
2.1

Let sCG(g) and choose m so large that the two long sides of the quadrilateral with vertices 1,gm,sgm,s have points outside the 2δ-neighborhoods of its short sides. By [A2], some point on [1,gm] is within 2δ of [s,sgm]. Using [A1] on both long sides gives integers i,j with d(gi,sgj)2L+2δ. Since s commutes with g, this says that the coset gs has a representative of word length at most 2L+2δ.

A1A2step 1.1choosealgebra
3.1

The ball of radius 2L+2δ is finite by [A2], so step 2.1 leaves only finitely many cosets of g in CG(g). Hence g has finite index in CG(g), and the centralizer is virtually cyclic.

A2step 2.1

Depends on

Used by

Dependency tree · two levels

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Sources