CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-01

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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 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).

[F1]

Group powers are the powers from Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ .

Proof

technique · direct

1.1

Let $g$ be a torsion element. By [L1], $g$ is conjugate to $v^{k}$ , where some defining relator has the form $v^{m}$ .

L1given

2.1

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 $v^{k}$ is trivial by [F1]. Hence $g = 1$ .

F1step 1.1algebra

3.1

Since every torsion element is trivial, the group is torsion-free.

step 2.1∎

Depends on

Used by

Dependency tree · two levels

13 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)