Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • 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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FALSE: any two infinite finitely generated groups are quasi-isometric

Statement refuted

any two infinite finitely generated groups are quasi-isometric.

Facts & Assumptions

Given: The proposed claim together with the witness named in the Statement refuted.

[F1]

A finitely generated group is quasi-isometric to a metric space when its word metric for some, equivalently every, finite generating set is (The quasi-isometry type of a finitely generated group).

[L1]

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).

[L2]

Balls of a word metric are finite if and only if the generating set is finite (Balls of a word metric are finite if and only if the generating set is finite).

[L3]

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).

[L4]

B(x,r) is the open ball, Bˉ(x,r) the closed ball and S(x,r) the sphere of centre x and radius r. The radius is always a strictly positive real; a ball of radius 0 or of negative radius is never written in this library. (Open ball, closed ball and sphere in a metric space).

[L5]

A free group on a set X is a group F(X) together with a map i:XF(X) such that, for every group G and every function u:XG, there is a unique group homomorphism u^:F(X)G satisfying (Free group on a set of generators).

[L6]

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

[L7]

A map is (L,C)-coarse Lipschitz when dY(f(x),f(x))LdX(x,x)+C, and an (L,C)-quasi-isometric embedding satisfies in addition L1dX(x,x)CdY(f(x),f(x)) (Coarse Lipschitz maps and quasi-isometric embeddings).

[L8]

In a free group with respect to a free basis, word length is reduced-word length (With respect to a free basis, the word length of an element is the length of its reduced word).

[L9]

For every real number there is a larger natural number (Every complete ordered field is Archimedean).

Refutation

technique · contradiction
1.1

The claim asserts a single quasi-isometry class for all infinite finitely generated groups.

F1assume-contra
1.2

For each integer n1, the open ball of radius n+1 in the integers has 2n+1 elements. In the free group F2=a,b, the 2n positive words of length n in the letters a,b are distinct reduced words, so the corresponding ball has at least 2n elements.

L1L2L4L5L6L8
2.1

Suppose there were a quasi-isometry f:F2Z. By [L3] choose a coarse Lipschitz quasi-inverse g:ZF2, coarse-Lipschitz constants A,B0 for f, and a bound R0 with dF2(g(f(x)),x)R for every xF2. If f(x)=f(x), then dF2(x,x)dF2(x,g(f(x)))+dF2(g(f(x)),x)2R, so every fibre f1(y) lies in the open ball BF2(g(y),2R+1). By [L1], left multiplication by g(y)1 bijects that ball with BF2(e,2R+1), so every fibre has at most K:=BF2(e,2R+1)< elements by [L2]. Moreover, if xBF2(e,n+1) then dZ(f(x),f(e))AdF2(x,e)+B<A(n+1)+B, so f(BF2(e,n+1)) has at most 2A(n+1)+B+1 elements. Hence 2nBF2(e,n+1)K(2A(n+1)+B+1). But 2nn2 for n4 by induction, while 2A(n+1)+B+12An+2A+2B+3. By [L9] choose a natural n4 so large that n>4KA and n2>2K(2A+2B+3); then n2>K(2An+2A+2B+3), contradicting the displayed bound. Thus F2 and Z are not quasi-isometric.

L1L2L3L7L8L9step 1.1step 1.2discharge-contradiction

Depends on

Used by

Nothing in the library uses this result yet.

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Sources