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

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Hyperbolicity of a finitely generated group is independent of the finite generating set

Statement

Let $G$ be a finitely generated group. If the Cayley graph of $G$ is hyperbolic for one finite generating set, then it is hyperbolic for every finite generating set.

Facts & Assumptions

Given: A finitely generated group $G$ and two finite generating sets $S, T$ .

[L1]

Two finite generating sets of a group give bilipschitz equivalent word metrics (The identity map between the word metrics of two finite generating sets is a bilipschitz equivalence).

[L2]

Hyperbolicity is a quasi-isometry invariant of geodesic spaces (Hyperbolicity is a quasi-isometry invariant of geodesic spaces).

Proof

technique · direct

1.1

By [L1], the identity map on $G$ is a quasi-isometry between the two Cayley graphs $Γ (G, S)$ and $Γ (G, T)$ .

givenL1

2.1

Therefore [L2] transfers hyperbolicity from one Cayley graph to the other. So the definition of a hyperbolic group does not depend on the chosen finite generating set.

L2step 1.1∎

Depends on

Used by

Closed surface groups are hyperbolic Example
FALSE: the same delta works after every finite change of generating set False statement
Free abelian groups of rank at least two are not hyperbolic Proposition

Dependency tree · two levels

16 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 6.3 (standard reference, not scraped)