Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedverified 2026-09-23 (gpt-6-sol)
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In a C prime(1/6) group, every nontrivial torsion element is conjugate to a power of a relator root

Statement

Let G=⟨X∣R⟩ be a symmetrised C′(1/6) presentation. Every nontrivial torsion element of G is conjugate to a power of a root of some defining relator.

Facts & Assumptions

Given: A nontrivial torsion element g∈G.

[F2]

The published local theorem C prime one sixth torsion elements come from relator roots proves that a nonidentity finite-order element in the symmetrised C′(1/6) convention of Sc toolkit symmetrised relators and pieces is conjugate to a power of a root of a cyclic conjugate of a defining relator. This convention uses nonempty cyclically reduced words and permits proper powers and infinite relator sets.

Proof

technique · direct
1.1F2given

Because g is a nontrivial torsion element in the stated symmetrised C′(1/6) presentation, [F2] applies. It supplies a cyclic conjugate r′=vu of a defining word r=uv and a word root c of r′ such that g is conjugate to a power of the image of c. In the symmetrised relator set, r′ is itself a defining relator; equivalently r′ and r are conjugate in the free group. Thus this is a power of a root of a defining relator as claimed.

2.1F1step 1.1∎

Powers are interpreted as in [F1], so step 1.1 is exactly the claimed conclusion.

Depends on

Used by

Dependency tree · two levels

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Sources