False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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FALSE: HNN normal form is canonical without choosing transversals

Statement

The HNN normal form of an element is canonical even before any transversal choice is made.

Facts & Assumptions

Given: An HNN extension with chosen associated subgroup $2 Z \leq Z$ and identity isomorphism.

[L1]

HNN normal form is defined relative to explicit transversal data. (The transversal data used for HNN normal forms)

[L2]

Uniqueness holds only after those transversals are fixed. (Normal forms in an HNN extension are unique relative to chosen transversals)

Refutation

technique · direct

1.1L1givenalgebra

Let $A = ⟨ a ⟩ ≅ Z$ , let $C_{-} = C_{+} = ⟨ a^{2} ⟩$ , and let $ϕ = id$ . With transversal $S = {e, a}$ , the element $t a$ is already in normal form. With transversal $S^{'} = {e, a^{- 1}}$ for the same odd coset, write $a = a^{2} a^{- 1}$ and use the relation $t a^{2} = a^{2} t$ to obtain the different normal form $a^{2} t a^{- 1}$ .

2.1L1L2step 1.1algebra∎

Both written words represent the same group element, but they are distinct as normal forms until the transversal choice is fixed. So the uniqueness clause of [L2] is relative, not canonical without data, and the statement is false.

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

Roger C. Lyndon and Paul E. Schupp, Combinatorial Group Theory (standard reference, not scraped)