ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01

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

Example

The presentation

G = ⟨ x_{1}, \dots, x_{7} ∣ x_{1} x_{2} x_{3} x_{4} x_{5} x_{6} x_{7} ⟩

is a strict $C^{'} (1 / 6)$ presentation with no proper-power relator, so $G$ is torsion-free.

Facts & Assumptions

Given: The displayed one-relator presentation.

[L1]

A $C^{'} (1 / 6)$ presentation with no proper-power relators defines a torsion-free group (A C prime(1/6) presentation with no proper-power relators defines a torsion-free group).

Verification

technique · direct

1.1

The symmetrised relator set has no nontrivial pieces, because distinct cyclic conjugates start with distinct letters. So the $C^{'} (1 / 6)$ condition holds vacuously. The relator is not a proper power because its letters are all distinct in cyclic order.

given

2.1

Apply [L1] to the presentation from step 1.1. Therefore the group $G$ is torsion-free.

L1step 1.1∎

Depends on

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

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

3 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

GAP SmallCancellation manual, Chapter 1: Small Cancellation Theory — the classical conditions (standard reference, not scraped)
Jay Williams, Universal Countable Borel Quasi-Orders (standard reference, not scraped)
Nicholas Touikan, An Introduction to Combinatorial and Geometric Group Theory, Section 3.5 (standard reference, not scraped)
Clara Löh, Geometric Group Theory: An Introduction, Section 7.4.1 (standard reference, not scraped)