Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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)=g1hS (The word metric of a group with respect to a generating set).

[L1]

The word length gS is the least n such that g is a product of n elements of SS1 (Word length of a group element with respect to a generating set).

[L2]

Word length is defined on every element and satisfies ghSgS+hS, g1S=gS, and gS=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 c1d(x,x)d(f(x),f(x))cd(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,yX, 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 αd(x,y)    d(x,y)    βd(x,y)for all x,yX. (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 An for some nN. (The cardinality A of a finite set).

Proof

technique · direct
1.1

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.

F1L1L6L7choose
2.1

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

F1L1L2L3step 1.1
3.1

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.

F1L4L5step 2.1

Depends on

Used by

Dependency tree · two levels

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Sources