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CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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In ⟨a,b∣aba−1b−1⟩\langle a,b\mid aba^{-1}b^{-1}\rangle, the trivial word a2b2a−2b−2a^2b^2a^{-2}b^{-2} is stuck under free cancellation and delete-only relator rewriting

Statement refuted

Use the same narrow delete-only procedure: freely cancel adjacent inverse pairs, or delete a contiguous occurrence of the displayed relator or its inverse, but never insert a relator and never lengthen the word.

The false claim is that this procedure reduces every identity word to the empty word. In ⟨a,b∣aba−1b−1⟩\langle a,b\mid aba^{-1}b^{-1}\rangle, the word a2b2a−2b−2a^2b^2a^{-2}b^{-2} represents the identity but admits no delete-only step. Any rewriting path from this word to the empty word using relator insertions and deletions must therefore begin by lengthening it.

Facts & Assumptions

Given: The presentation G=⟨a,b∣aba−1b−1⟩G=\langle a,b\mid aba^{-1}b^{-1}\rangle, the expanded eight-letter word w=a2b2a−2b−2w=a^2b^2a^{-2}b^{-2}, and the delete-only procedure just stated.

[L1]

In a presented group, every displayed relator becomes the identity, as do all consequences forced by normality (Group presentation by generators and relations).

[L2]

If gh=hggh=hg, then (gh)n=gnhn(gh)^n=g^nh^n (Exponent laws in a group: gm+n=gmgng^{m+n} = g^{m}g^{n} and (gm)n=gmn(g^{m})^{n} = g^{mn} for all m,n∈Zm, n \in \mathbb{Z}, and (gh)n=gnhn(gh)^{n} = g^{n}h^{n} when gg and hh commute).

[F1]

Group multiplication is associative, has a two-sided identity, and has two-sided inverses (Group and abelian group).

Counterexample

technique · direct
1.1

The relator aba−1b−1=eaba^{-1}b^{-1}=e gives ab=baab=ba; hence [L2] gives a2b2=(ab)2a^2b^2=(ab)^2, and a2b2a−2b−2=a2b2(a2b2)−1=ea^2b^2a^{-2}b^{-2}=a^2b^2(a^2b^2)^{-1}=e.

L1L2F1
1.2

The eight-letter word a,a,b,b,a−1,a−1,b−1,b−1a,a,b,b,a^{-1},a^{-1},b^{-1},b^{-1} has no adjacent inverse pair.

given
1.3

Its five length-four windows are aabbaabb, abba−1abba^{-1}, bba−1a−1bba^{-1}a^{-1}, ba−1a−1b−1ba^{-1}a^{-1}b^{-1}, and a−1a−1b−1b−1a^{-1}a^{-1}b^{-1}b^{-1}; none is the relator aba−1b−1aba^{-1}b^{-1} or its inverse bab−1a−1bab^{-1}a^{-1}.

given
2.1

The nonempty identity word is therefore stuck under the delete-only procedure; since its first move in any nonconstant relator-rewriting path cannot be a cancellation or deletion, such a path must begin with an insertion and increase the length.

step 1.1step 1.2step 1.3∎

Depends on

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