Alphabeta Math
CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-11
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In ⟨a,b∣aba−1b−1⟩, the trivial word a2b2a−2b−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⟩, the word a2b2a−2b−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⟩, the expanded eight-letter word w=a2b2a−2b−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=hg, then (gh)n=gnhn (Exponent laws in a group: gm+n=gmgn and (gm)n=gmn for all m,n∈Z, and (gh)n=gnhn when g and h 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=e gives ab=ba; hence [L2] gives a2b2=(ab)2, and a2b2a−2b−2=a2b2(a2b2)−1=e.

L1L2F1
1.2

The eight-letter word a,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 aabb, abba−1, bba−1a−1, ba−1a−1b−1, and a−1a−1b−1b−1; none is the relator aba−1b−1 or its inverse bab−1a−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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