Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01
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FALSE: a presentation with no proper-power relators is automatically torsion-free

Statement

A presentation with no proper-power relators is automatically torsion-free.

Facts & Assumptions

Given: The presentation G=a,bab2,a2b.

[L1]

The torsion-free conclusion on this page needs both the C(1/6) hypothesis and the no-proper-power hypothesis (A C prime(1/6) presentation with no proper-power relators defines a torsion-free group).

Refutation

technique · direct
1.1

Neither relator ab2 nor a2b is a proper power: each is cyclically reduced of length 3 and is not a repetition of a shorter cyclic word.

given
2.1

From ab2=1 one gets a=b2, and substituting this into a2b=1 gives b4b=1, hence b3=1 in the sense of [F1]. Moreover, if C3=tt3=1, then the assignment at, bt satisfies both relators, so it induces a surjective homomorphism GC3. Therefore the image of b is nontrivial and G contains a nontrivial torsion element.

F1step 1.1algebra
3.1

Therefore the absence of proper-power relators alone does not force torsion-freeness. By [L1], the missing small-cancellation hypothesis is load-bearing.

L1step 2.1

Depends on

Used by

Dependency tree · two levels

14 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