Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26
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The identity map between the word metrics of two finite generating sets is a bilipschitz equivalence

Statement

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

Facts & Assumptions

Given: The hypotheses of the Statement.

[F1]

The word metric of G with respect to S is dS(g,h)=∣g−1h∣S (The word metric of a group with respect to a generating set).

[L1]

The word length ∣g∣S is the least n such that g is a product of n elements of S∪S−1 (Word length of a group element with respect to a generating set).

[L2]

Word length is defined on every element and satisfies ∣gh∣S≤∣g∣S+∣h∣S, ∣g−1∣S=∣g∣S, and ∣g∣S=0 exactly when g is the identity (Word length is defined on every element and satisfies the subadditivity, inversion and vanishing laws).

[L3]

The word metric is a left-invariant metric and coincides with the path metric of the Cayley graph (The word metric is a left-invariant metric and coincides with the path metric of the Cayley graph).

[L4]

A map is a bilipschitz embedding when c−1d(x,x′)≤d(f(x),f(x′))≤c d(x,x′) for some c>0, and a bilipschitz equivalence when it is a bijective such map with bilipschitz inverse (Bilipschitz embeddings and bilipschitz equivalences of metric spaces).

[L5]
  • d and d′ are topologically equivalent if they have the same metric topology: Td=Td′. - d and d′ are uniformly equivalent if for every real ε>0 there are reals δ>0 and δ′>0 such that, for all x,y∈X, d(x,y)<δ  ⟹  d′(x,y)<εandd′(x,y)<δ′  ⟹  d(x,y)<ε. - d and d′ are Lipschitz equivalent if there are reals α,β>0 with α d(x,y)  ≤  d′(x,y)  ≤  β d(x,y)for all x,y∈X. (Topologically, uniformly and Lipschitz equivalent metrics on a set).
[L6]

A group is finitely generated when some finite subset generates it (Finitely generated groups).

[L7]

A set A is finite when A≈n for some n∈N. (The cardinality ∣A∣ of a finite set).

Proof

technique · direct
1.1F1L1L6L7choose

Let c be the largest word length in the second metric of a member of the first symmetrised set; finiteness of that set is exactly what makes the maximum exist.

2.1F1L1L2L3step 1.1

Expanding an element of length n in the first metric and applying the triangle inequality along the expression bounds its second length by cn.

3.1F1L4L5step 2.1∎

Exchanging the roles of the two sets gives the reverse inequality, so the identity is a bilipschitz equivalence and the two metrics are Lipschitz equivalent.

Depends on

Used by

Dependency tree · two levels

29 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