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

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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 $g \in G$ .

[L1]

For an infinite-order element of a finitely generated hyperbolic group, there is a positive integer $C$ such that $∣ g^{n} ∣_{S} \geq ∣ n ∣ / C$ for all integers $n$ (Infinite order elements have positive stable translation length).

[L2]

Word metrics from two finite generating sets are bilipschitz equivalent (The identity map between the word metrics of two finite generating sets is a bilipschitz equivalence).

Proof

technique · direct

1.1givenL1

By the definition of a hyperbolic group, some finite generating set $T$ has a hyperbolic geometric Cayley graph. Apply [L1] with $T$ : for some $C_{T} > 0$ , $∣ n ∣ \leq C_{T} ∣ g^{n} ∣_{T}$ for every integer $n$ . The proof of [L1] is choice-free.

2.1L2step 1.1∎

By [L2] there is a finite $K > 0$ with $∣ h ∣_{T} \leq K ∣ h ∣_{S}$ for all $h \in G$ . Thus $∣ n ∣ \leq C_{T} K ∣ g^{n} ∣_{S}$ . Take $A = C_{T} K$ and any $B > 0$ . This proves undistortion for the stated arbitrary finite $S$ without importing a choice-dependent hyperbolicity transfer theorem.

Depends on

Used by

Dependency tree · two levels

15 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)