TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-27

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Infinite-order elements of hyperbolic groups are undistorted

Statement

Let $G$ be a hyperbolic group and let $g \in G$ have infinite order. Then the cyclic subgroup $⟨ g ⟩$ is undistorted in $G$ : for some constants $A, B > 0$ ,

∣ n ∣ \leq A ∣ g^{n} ∣_{S} + B

for all $n \in Z$ , where $∣ \cdot ∣_{S}$ is word length with respect to a finite generating set $S$ of $G$ .

Facts & Assumptions

Given: A hyperbolic group $G$ , a finite generating set $S$ , and an infinite-order element $g \in G$ .

[A1]

In a hyperbolic group, the orbit map $n \mapsto g^{n}$ is a quasi-isometric embedding of $Z$ into the Cayley graph whenever $g$ has infinite order.

[L1]

Morse stability controls quasi-geodesics in hyperbolic spaces (Morse stability of quasi-geodesics).

Proof

technique · direct

1.1

The source fact [A1] says that the orbit map $n \mapsto g^{n}$ is a quasi-isometric embedding into the Cayley graph of $G$ .

givenA1

2.1

A quasi-isometric embedding gives the displayed linear lower bound on $∣ g^{n} ∣_{S}$ in terms of $∣ n ∣$ , while [L1] explains geometrically that the powers of $g$ stay near a quasi-axis. Therefore $⟨ g ⟩$ is undistorted.

L1step 1.1∎

Depends on

Used by

Dependency tree · two levels

13 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 6.5.1 (standard reference, not scraped)