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PropositionStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26
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The quotient map by a finite normal subgroup is a quasi-isometry of word metric spaces

Statement

The quotient map by a finite normal subgroup is a quasi-isometry of word metric spaces.

Facts & Assumptions

Given: The hypotheses of the Statement.

[F1]

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

[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 of G with respect to S is dS(g,h)=g1hS (The word metric of a group with respect to a generating set).

[L4]

A map is (L,C)-coarse Lipschitz when d(f(x),f(x))Ld(x,x)+C, and an (L,C)-quasi-isometric embedding when in addition L1d(x,x)Cd(f(x),f(x)) (Coarse Lipschitz maps and quasi-isometric embeddings).

[L5]

A subset is coarsely dense when every point of the space is within a fixed distance of it, and a quasi-isometry is a coarse Lipschitz map admitting a coarse Lipschitz quasi-inverse (Coarsely dense subsets, quasi-inverses and quasi-isometries).

[L6]

The quotient group, or factor group, G/N has the left cosets (The quotient group G/N and coset product (gN)(hN)=ghN).

[L7]

The subgroup N is normal in G when (Normal subgroup: invariance under conjugation).

[L8]

A set A is finite when An for some nN. (The cardinality A of a finite set).

Proof

technique · direct
1.1

The image of a finite generating set generates the quotient, so the quotient map does not increase word length and is one-Lipschitz.

F1L1L2L3L4L6L7
2.1

Let M=max{nS:nN}, which exists because N is finite. For each coset gˉG/N, choose a representative s(gˉ)G of minimal word length in that coset, breaking ties lexicographically in the fixed finite alphabet SS1; then the quotient map q satisfies q(s(gˉ))=gˉ. If hˉ=gˉq(t) with tSS1, then s(gˉ)1s(hˉ)t1N, so s(gˉ)1s(hˉ)SM+1. Chaining along a shortest quotient expression gives dG(s(gˉ),s(hˉ))(M+1)dG/N(gˉ,hˉ), so s:G/NG is Lipschitz.

L1L3L6L8step 1.1choose
3.1

For every gG, the elements s(q(g)) and g lie in the same coset, so s(q(g))1gN and therefore dG(s(q(g)),g)M. Thus qs=idG/N and sq is at bounded distance from idG. Therefore q is a quasi-isometry with quasi-inverse s.

L5step 2.1

Depends on

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