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CorollaryStatement: Literature-sourcedProof: AI-generatedverified 2026-09-24 (gpt-6-sol)
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A C prime(1/6) presentation with no proper-power relators defines a torsion-free group

Statement

A C′(1/6) presentation with no proper-power relators defines a torsion-free group.

Facts & Assumptions

Given: A C′(1/6) presentation in which no defining relator is a proper power.

[L1]

Every nonidentity torsion element is conjugate to a power of a root of a defining relator (In a C prime(1/6) group, every nontrivial torsion element is conjugate to a power of a relator root).

Proof

technique · direct
1.1L1given

The identity has finite order and needs no argument. Suppose for a contradiction that g≠1 is a torsion element. By [L1], g is conjugate to vk, where some defining relator has the form vm.

2.1F1step 1.1algebra

The no-proper-power hypothesis forces m=1. Thus the root word v is itself a defining relator and represents the identity in the presented group, so every power vk is trivial by [F1]. Hence g=1, contradicting step 1.1.

3.1step 2.1∎

Since no nonidentity torsion element exists, the group is torsion-free.

Depends on

Used by

Dependency tree · two levels

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Sources