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19 results · all verified · 3 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 16 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Cayley Graphs, Word Metrics and Quasi-Isometry — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-26Open item page →

The Cayley graph of Z for the generating set {1} is a line and its word metric is mn

Example

The Cayley graph of Z for the generating set {1} is a line and its word metric is mn.

Facts & Assumptions

Given: The objects and hypotheses in the Example.

[F1]

The Cayley graph of a group G with respect to a subset S has vertex set G and edge set {{g,gs}:gG, s(SS1){e}} (The Cayley graph of a group with respect to a subset).

[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]

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).

[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).

[L5]

A free abelian group on a set X is an abelian group A(X) together with a map i:XA(X) such that, for every abelian group B and every function u:XB, there is a unique group homomorphism u^:A(X)B satisfying (Free abelian group on a set).

Verification

technique · direct
1.1

With S={1} the symmetrised set is {1,1} and the edges join n to n±1, so the Cayley graph is a two-way infinite path.

F1L5
2.1

The word length of n is n, since n is a product of n copies of 1 or of 1 and no shorter expression exists.

L1L2L3step 1.1
3.1

So the word metric is d(m,n)=mn, the metric induced from the real line.

L2L4step 2.1
ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The word metrics of Z for {1} and for {2,3} differ at 1 and are bilipschitz equivalent

Example

The word metrics of Z for {1} and for {2,3} differ at 1 and are bilipschitz equivalent.

Facts & Assumptions

Given: The objects and hypotheses in the Example.

[F1]

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).

[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]

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

[L3]

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).

[L4]
  • 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).

Verification

technique · direct
1.1

For S={1} the length of 1 is one, while for S={2,3} the element 1 is not a one-letter word in SS1={±2,±3} and satisfies 1=3+(2), so its length is two. Thus the two metrics already differ at the pair (0,1).

F1L1algebra
2.1

The comparison theorem gives constants: every member of one symmetrised set has length at most three in the other, so the identity is bilipschitz with constant three.

L1L2L3L4step 1.1
ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-26Open item page →

The Cayley graph of Zn for the standard basis is the integer lattice, and its word metric is the sum of coordinate differences

Example

The Cayley graph of Zn for the standard basis is the integer lattice, and its word metric is the sum of coordinate differences.

Facts & Assumptions

Given: The objects and hypotheses in the Example.

[F1]

The Cayley graph of a group G with respect to a subset S has vertex set G and edge set {{g,gs}:gG, s(SS1){e}} (The Cayley graph of a group with respect to a subset).

[L1]

A free abelian group on a set X is an abelian group A(X) together with a map i:XA(X) such that, for every abelian group B and every function u:XB, there is a unique group homomorphism u^:A(X)B satisfying (Free abelian group on a set).

[L2]

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).

[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]

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).

[L5]

d1(x,y):=k<nxkyk,d2(x,y):= k<n(xkyk)2 ,d(x,y):=max{xkyk:k<n}. (Rn as the set of functions nR, and d1, d2, d are metrics on it).

Verification

technique · direct
1.1

With the standard basis as generating set, the neighbours of a tuple are those differing by one in a single coordinate, so the Cayley graph is the integer lattice.

F1L1
2.1

The word length of a tuple is the sum of the absolute values of its coordinates: that many steps suffice, and each step changes the sum by at most one.

L2L3L4step 1.1
3.1

So the word metric is the restriction of the 1 metric, and the inclusion into that normed space is a quasi-isometry.

L3L5step 2.1
ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-26Open item page →

The Cayley graph of the free group on two generators is the tree in which every vertex has four neighbours

Example

The Cayley graph of the free group on two generators is the tree in which every vertex has four neighbours.

Facts & Assumptions

Given: The objects and hypotheses in the Example.

[F1]

The Cayley graph of a group G with respect to a subset S has vertex set G and edge set {{g,gs}:gG, s(SS1){e}} (The Cayley graph of a group with respect to a subset).

[L1]

The Cayley graph of a free group with respect to a free basis is a tree (The Cayley graph of a free group with respect to a free basis is a tree).

[L2]

Every vertex of a Cayley graph has the same degree, and the graph is locally finite exactly when the symmetrised generating set is finite (Cayley-graph neighbourhoods are equipotent, and local finiteness is equivalent to finiteness of the symmetrised subset).

[L3]

A cycle is a closed walk of length at least three with distinct vertices apart from its endpoints; a forest is a simple graph with no cycle and a tree is a connected forest (Cycles, trees and forests in a simple graph on an arbitrary vertex set).

[L4]

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).

[L5]

The subset B is a free basis of F if (F,i) is a free group on the set B in the sense of. (A free basis of a group).

Verification

technique · direct
1.1

A two-element free basis generates and the general theorem makes the Cayley graph a tree.

F1L1L3L4L5
2.1

The symmetrised set has four elements and none is the identity, so every vertex has degree four.

F1L2step 1.1
ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

The dihedral group of order eight has Cayley graphs that are a cycle of length eight and a cube

Example

The dihedral group of order eight has Cayley graphs that are a cycle of length eight and a cube.

Facts & Assumptions

Given: The objects and hypotheses in the Example.

[F1]

The Cayley graph of a group G with respect to a subset S has vertex set G and edge set {{g,gs}:gG, s(SS1){e}} (The Cayley graph of a group with respect to a subset).

[L1]

Every vertex of a Cayley graph has the same degree, and the graph is locally finite exactly when the symmetrised generating set is finite (Cayley-graph neighbourhoods are equipotent, and local finiteness is equivalent to finiteness of the symmetrised subset).

[L2]

For D=Dih(C4)=r,s, one has r4=s2=1, srs1=r1, and every element is uniquely ri or ris for 0i<4 ( Dih(Cn)=CnC2 with inversion action has order 2n and the dihedral relations).

[L4]

The degree of v is degG(v):=NG(v), equivalently the number of edges incident with v. A graph is r-regular when every vertex has degree r; it is cubic when it is 3-regular. (Adjacency, incidence, open and closed neighbourhoods, vertex degree, minimum degree and maximum degree).

Verification

technique · direct
1.1

For the generating set {r,s}, the four vertices 1,r,r2,r3 form a 4-cycle under right multiplication by r±1, and the four vertices s,rs,r2s,r3s form another. Right multiplication by s joins ri to ris for each i. Thus the graph is two 4-cycles joined at corresponding vertices, which is the cube.

F1L1L2L3L4
2.1

For the generating set {s,rs} both generators are involutions and they generate because (rs)s=r. Alternating them gives the eight-cycle 1,s,r3,r3s,r2,r2s,r,rs,1, whose consecutive vertices differ by right multiplication by s or rs. These are all eight group elements, and every vertex has only the two displayed neighbours, so this Cayley graph is C8.

F1L1L2L4step 1.1
ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-26Open item page →

The inclusion of Z in R is a quasi-isometry that is neither surjective nor a bilipschitz equivalence

Example

The inclusion of Z in R is a quasi-isometry that is neither surjective nor a bilipschitz equivalence.

Facts & Assumptions

Given: The objects and hypotheses in the Example.

[F1]

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).

[L1]

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).

[L2]

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).

[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]

It is written x and called the integer part, or floor, of x. (Integer part: for every real x there is exactly one integer m with mx<m+1).

Verification

technique · direct
1.1

The inclusion ι:ZR preserves distances exactly, so it is a quasi-isometric embedding with constants one and zero.

F1L3L5
2.1

Let g:RZ be the integer-part map. Then g(n)=n for every integer n, while every real x satisfies xg(x)<1; so gι=idZ and ιg is at bounded distance from idR. Therefore ι is a quasi-isometry.

L1L4step 1.1
3.1

It is not a bilipschitz equivalence because it is not surjective, and a bilipschitz equivalence must in particular be bijective.

L2step 2.1
ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-26Open item page →

The subgroup 2Z×Z has index two in Z2 and its inclusion is a quasi-isometry

Example

The subgroup 2Z×Z has index two in Z2 and its inclusion is a quasi-isometry.

Facts & Assumptions

Given: The objects and hypotheses in the Example.

[L1]

A subgroup of finite index in a finitely generated group is finitely generated and its inclusion is a quasi-isometry (A subgroup of finite index in a finitely generated group is finitely generated, and its inclusion is a quasi-isometry).

[F1]
[L2]

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).

[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]

A free abelian group on a set X is an abelian group A(X) together with a map i:XA(X) such that, for every abelian group B and every function u:XB, there is a unique group homomorphism u^:A(X)B satisfying (Free abelian group on a set).

Verification

technique · direct
1.1

The subgroup of pairs with even first coordinate has index two, with transversal the zero pair and the first basis vector.

F1L4
2.1

The general proposition applies and makes the inclusion a quasi-isometry.

L1L3step 1.1
3.1

Directly, the inclusion doubles the first coordinate of a word expression at worst, so the constants are two and one.

L2L3step 2.1algebra
ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-26Open item page →

The infinite dihedral group is quasi-isometric to Z, and to Z×Z/2

Example

The infinite dihedral group is quasi-isometric to Z, and to Z×Z/2.

Facts & Assumptions

Given: The objects and hypotheses in the Example.

[F1]

The Cayley graph of a group G with respect to a subset S has vertex set G and edge set {{g,gs}:gG, s(SS1){e}} (The Cayley graph of a group with respect to a subset).

[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]

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).

[L3]

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

[L4]

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).

[L5]

The group with presentation (Group presentation by generators and relations).

[L6]

If the evaluation of every rR under u is eH, then there is a unique homomorphism (Von Dyck's theorem: maps of generators that satisfy the relators extend uniquely from a presented group).

[L7]

Let u,vF(X) and put N= ⁣R ⁣F(X). The words u and v represent the same element of XR if and only if (In XR, the words u and v represent the same element if and only if u1v ⁣R ⁣).

Verification

technique · direct
1.1

Take the two presentations of the infinite dihedral group, by two involutions and by an infinite-order element with an inverting involution.

F1L5L6L7
2.1

The Cayley graph for the first is a two-way infinite path, isomorphic to that of the integers with generator one; for the second it is the two-way infinite ladder, isomorphic to that of the integers times a group of order two.

F1L8step 1.1
3.1

Isomorphic Cayley graphs give isometric word metrics, and the comparison theorem transports the identification across generating sets, so all three groups are quasi-isometric.

L1L2L3L4step 2.1
CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-26Open item page →

Taking Z itself as a generating set gives a word metric of diameter one, not bilipschitz equivalent to the standard one

Statement refuted

The comparison theorem for word metrics remains true for arbitrary generating sets.

Facts & Assumptions

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

[F1]

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).

[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]

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

[L4]

Bounded subset. A is bounded if A= or there are x0X and a real r>0 with AB(x0,r). (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

Counterexample

technique · direct
1.1

Taking the whole group of integers as generating set gives every nonzero integer word length one.

F1L1
2.1

So that word metric has diameter one, while the metric for the generating set {1} is unbounded: the distance from 0 to n is n. Its individual balls are finite, as [L2] requires, but their radii are not bounded uniformly.

L1L2L4step 1.1
3.1

The two are therefore not bilipschitz equivalent; the failing hypothesis of the comparison theorem is the finiteness used to take a maximum over the symmetrised set.

L3step 2.1
CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-26Open item page →

A single map exhibiting a quasi-isometry that is discontinuous, non-injective and non-surjective

Statement refuted

A quasi-isometry must be continuous, injective, or surjective.

Facts & Assumptions

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

[F1]

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).

[L1]

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).

[L2]

It is written x and called the integer part, or floor, of x. (Integer part: for every real x there is exactly one integer m with mx<m+1).

Counterexample

technique · constructive
1.1

Let f:RR be f(x)=2x/2, whose image is the even integers, and let ι be the inclusion of that image into R. Then ι is coarse Lipschitz, f(2m)=2m for every even integer 2m, and every real x satisfies xf(x)<2; so f is a quasi-isometry.

F1L1L2L3construct
2.1

It is discontinuous at every even integer, non-injective on each half-open interval [2m,2m+2), and misses every odd integer, so all three failures occur in one map.

L2L4step 1.1discharge-construct
False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26Open item page →

FALSE: the Cayley graph of a group is independent of the chosen generating set

Statement refuted

the Cayley graph of a group is independent of the chosen generating set.

Facts & Assumptions

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

[F1]

The Cayley graph of a group G with respect to a subset S has vertex set G and edge set {{g,gs}:gG, s(SS1){e}} (The Cayley graph of a group with respect to a subset).

[L1]

Every vertex of a Cayley graph has the same degree, and the graph is locally finite exactly when the symmetrised generating set is finite (Cayley-graph neighbourhoods are equipotent, and local finiteness is equivalent to finiteness of the symmetrised subset).

Refutation

technique · contradiction
1.1

The claim asserts that the isomorphism type of the Cayley graph depends only on the group.

F1assume-contra
2.1

For the integers the generating sets {1} and {2,3} give graphs of degree two and four, so the claim fails.

F1L1step 1.1discharge-contradiction
False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-26Open item page →

FALSE: every word metric is invariant under right translation

Statement refuted

every word metric is invariant under right translation.

Facts & Assumptions

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

[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 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).

[L2]

Right translation by a fixed element displaces every point of a word metric space by exactly the word length of that element (Right translation by a fixed element displaces every point of a word metric space by exactly the word length of that element).

Refutation

technique · contradiction
1.1

The claim asserts dS(gk,hk)=dS(g,h) for all g,h,k.

F1L1assume-contra
2.1

In the free group on two generators with its free basis, take g=e, h=a and k=b: the left sides differ, because dS(b,ab)=b1abS=3 while dS(e,a)=1.

F1L2step 1.1discharge-contradiction
False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-26Open item page →

FALSE: every quasi-isometry is continuous, or bijective

Statement refuted

every quasi-isometry is continuous, or bijective.

Facts & Assumptions

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

[F1]

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).

[L1]

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).

[L2]

It is written x and called the integer part, or floor, of x. (Integer part: for every real x there is exactly one integer m with mx<m+1).

Refutation

technique · contradiction
1.1

The claim asserts that every quasi-isometry is continuous, or that every quasi-isometry is bijective.

F1assume-contra
2.1

The map f(x)=2x/2 is a quasi-isometry of R: the inclusion of the even integers is a coarse Lipschitz quasi-inverse, since f(2m)=2m and every real x satisfies xf(x)<2. But f is not bijective, and it is not continuous at any even integer. So it refutes both readings at once.

F1L1L2L3step 1.1discharge-contradiction
False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-26Open item page →

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
False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-26Open item page →

FALSE: a nontrivial finitely generated group with a word metric is a geodesic metric space

Statement refuted

a nontrivial finitely generated group with a word metric is a geodesic metric space.

Facts & Assumptions

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

[F1]

A geodesic of length L in a metric space is an isometric embedding of the interval [0,L], and the space is geodesic when every two points are the endpoints of one (Geodesics and geodesic metric spaces).

[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]

A group with the word metric of any generating set is a (1,1)-quasi-geodesic space (A group with the word metric of any generating set is a (1,1)-quasi-geodesic space).

Refutation

technique · contradiction
1.1

The claim asserts that every two elements are the endpoints of an isometric embedding of a real interval.

F1assume-contra
2.1

A geodesic between two elements at distance one supplies points at every intermediate real distance, while a word metric takes only integer values.

F1L1step 1.1
3.1

So no nontrivial group with a word metric is geodesic; the correct statement is that it is (1,1)-quasi-geodesic.

L2step 2.1discharge-contradiction
False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-26Open item page →

FALSE: groups with isomorphic Cayley graphs are isomorphic

Statement refuted

groups with isomorphic Cayley graphs are isomorphic.

Facts & Assumptions

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

[F1]

The Cayley graph of a group G with respect to a subset S has vertex set G and edge set {{g,gs}:gG, s(SS1){e}} (The Cayley graph of a group with respect to a subset).

[L1]

On a finite vertex set V, the empty graph has edge set and the complete graph KV has edge set [V]2. When V is an n-element labelled set, these are also denoted Kn and Kn. (Empty and complete graphs, complete bipartite graphs, and the convention that Pn and Cn have n vertices).

[L2]

If G=g is cyclic, then exactly one of the following applies: (Every cyclic group is isomorphic to (Z,+) or to (Z/n,+) for its finite order n1).

[L3]

A graph isomorphism is a bijection φ:VW such that, for all distinct u,vV, (Graph isomorphisms, automorphisms and graph complements).

Refutation

technique · contradiction
1.1

The claim asserts that the isomorphism type of some Cayley graph determines the group.

F1assume-contra
2.1

Taking the whole group as generating set, both the symmetric group on three letters and the cyclic group of order six give the complete graph on six vertices, and those groups are not isomorphic.

F1L1L2L3step 1.1discharge-contradiction
CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-26Open item page →

The Cayley graphs of Z/2 for {1} and of Z for {1,1} are trees, and neither generating set is free

Statement refuted

Whenever a Cayley graph is a tree, the chosen generating set is a free basis.

Facts & Assumptions

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

[F1]

The Cayley graph of a group G with respect to a subset S has vertex set G and edge set {{g,gs}:gG, s(SS1){e}} (The Cayley graph of a group with respect to a subset).

[L1]

A cycle is a closed walk of length at least three with distinct vertices apart from its endpoints; a forest is a simple graph with no cycle and a tree is a connected forest (Cycles, trees and forests in a simple graph on an arbitrary vertex set).

[L2]

If no product of two members of a generating set is the identity and the Cayley graph is a tree, the set is a free basis (If no product of two members of a generating set is the identity and the Cayley graph is a tree, the set is a free basis).

[L3]

The subset B is a free basis of F if (F,i) is a free group on the set B in the sense of. (A free basis of a group).

[L4]

If G=g is cyclic, then exactly one of the following applies: (Every cyclic group is isomorphic to (Z,+) or to (Z/n,+) for its finite order n1).

Counterexample

technique · constructive
1.1

The Cayley graph of the group of order two for its nonidentity element is a single edge, a tree, and that group is not free.

F1L1L4construct
2.1

The Cayley graph of the integers for the generating set {1,1} is the line, a tree, and that set is not a free basis.

F1L1L4step 1.1
3.1

In both cases a product of two members of the generating set is the identity, which is exactly the hypothesis the converse theorem adds.

L2L3step 1.1step 2.1discharge-construct
ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-26Open item page →

A nonempty metric space of finite diameter has trivial quasi-isometry group

Example

A nonempty metric space of finite diameter has trivial quasi-isometry group.

Facts & Assumptions

Given: The objects and hypotheses in the Example.

[F1]

The quasi-isometry group of a metric space is the set of quasi-isometries of it modulo bounded distance (The quasi-isometry group of a metric space).

[L1]

The quasi-isometry group is a group under composition, and a quasi-isometry induces an isomorphism between the quasi-isometry groups of its source and target (Quasi-isometries modulo bounded distance form a group, and a quasi-isometry induces an isomorphism of these groups).

[L2]

Bounded subset. A is bounded if A= or there are x0X and a real r>0 with AB(x0,r). (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

[L3]

Two maps into a metric space are at bounded distance when the distance between their values is bounded uniformly (Bounded distance between two maps into a metric space).

Verification

technique · direct
1.1

In a space of finite diameter every self-map is at distance at most the diameter from the identity.

F1L2L3
2.1

So there is exactly one bounded-distance class, and the quasi-isometry group is trivial.

F1L1step 1.1
ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-26Open item page →

Scaling maps embed the multiplicative group of nonzero reals into the quasi-isometry group of Z

Example

Scaling maps embed the multiplicative group of nonzero reals into the quasi-isometry group of Z.

Facts & Assumptions

Given: The objects and hypotheses in the Example.

[F1]

The quasi-isometry group of a metric space is the set of quasi-isometries of it modulo bounded distance (The quasi-isometry group of a metric space).

[L1]

The quasi-isometry group is a group under composition, and a quasi-isometry induces an isomorphism between the quasi-isometry groups of its source and target (Quasi-isometries modulo bounded distance form a group, and a quasi-isometry induces an isomorphism of these groups).

[L2]

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).

[L3]

It is written x and called the integer part, or floor, of x. (Integer part: for every real x there is exactly one integer m with mx<m+1).

[L5]

Group isomorphisms, automorphisms and the set Aut(G). (Group isomorphisms, automorphisms and the set Aut(G)).

Verification

technique · direct
1.1

For α0 let qα(n)=αn. The estimate uvuv+1 shows qα is coarse Lipschitz, and q1/α is a quasi-inverse because qα(q1/α(n))n<α+1andq1/α(qα(n))n<1/α+1 for every integer n. So qα is a quasi-isometry of Z.

F1L2L3L4
2.1

Composing the maps for α and β agrees with the map for αβ up to an error of at most α+1, so the assignment α[qα] is a homomorphism on classes.

L3step 1.1
3.1

For αβ the difference αnβn is unbounded, so the homomorphism is injective.

F1L1L5step 2.1

Sources