Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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FALSE: Greendlinger's lemma holds for every finite presentation

Statement

Greendlinger's lemma holds for every finite presentation.

Facts & Assumptions

Given: The presentation a,baba1b1 of Z2.

[L1]

Greendlinger's conclusion on this page is proved only for reduced C(1/6) diagrams (In a reduced C prime(1/6) null diagram, some face contributes more than half of its boundary to the outer boundary).

[F1]

A group presentation is the quotient by the normal closure of its defining relators (Group presentation by generators and relations).

Refutation

technique · direct
1.1

The relator aba1b1 and its cyclic conjugates have long overlaps, so the presentation does not satisfy the C(1/6) hypothesis required by [L1].

L1F1given
2.1

In the square grid van Kampen diagrams for commutator powers in Z2, every face can meet the outer boundary in exactly two of its four edges, never in more than half. Thus the characteristic Greendlinger conclusion fails for these null words.

step 1.1given
3.1

Therefore Greendlinger's lemma does not extend to arbitrary finite presentations. The statement is false.

step 1.1step 2.1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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