TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-26

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Growth type is independent of the finite generating set

Statement

Let $G$ be a finitely generated group, and let $S$ and $T$ be finite generating sets. Then the growth functions $β_{G, S}$ and $β_{G, T}$ are equivalent under $≃$ . Hence the growth type of a finitely generated group does not depend on the chosen finite generating set.

Facts & Assumptions

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

[L1]

The growth function $β_{G, S} (n)$ counts the elements with $∣ g ∣_{S} \leq n$ , and $β_{G, T} (n)$ is defined similarly (The growth function of a finitely generated group).

[L2]

The relation $≃$ is the mutual comparison relation generated by $f (n) \leq C g (C n + C) + C$ for some natural number $C \geq 1$ (Growth comparison and growth type).

[L3]

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

Proof

technique · direct

1.1

By [L3], choose a natural number $C \geq 1$ such that $d_{T} (g, h) \leq C d_{S} (g, h)$ and $d_{S} (g, h) \leq C d_{T} (g, h)$ for all $g, h \in G$ .

L3choose

2.1

If $∣ g ∣_{S} \leq n$ , then step 1.1 gives $∣ g ∣_{T} \leq C n$ . So every element counted by $β_{G, S} (n)$ is also counted by $β_{G, T} (C n)$ , and therefore $β_{G, S} (n) \leq β_{G, T} (C n)$ . Exchanging $S$ and $T$ yields the reverse inequality.

L1step 1.1

3.1

Because $C \geq 1$ and the growth functions are nondecreasing, step 2.1 implies $β_{G, S} (n) \leq β_{G, T} (C n) \leq C β_{G, T} (C n + C) + C,$ and likewise with $S$ and $T$ interchanged. Thus [L2] gives $β_{G, S} ≼ β_{G, T}$ and $β_{G, T} ≼ β_{G, S}$ . Hence $β_{G, S} ≃ β_{G, T}$ , and the growth type is independent of the finite generating set.

L2step 2.1∎

Depends on

Used by

Polynomial, subexponential, exponential, and intermediate growth Definition
Free abelian groups have polynomial growth of the expected degree Example
FALSE: the growth function is independent of the generating set pointwise False statement

Dependency tree · two levels

11 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

C. Löh, Geometric Group Theory, Sections 5.1-5.3 (standard reference, not scraped)
C. Drutu and M. Kapovich, Lectures on Geometric Group Theory, Chapter 5 (standard reference, not scraped)