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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28
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Collins' conjugacy theorem for cyclically reduced HNN words

Statement

Let u and v be cyclically Britton-reduced HNN words of positive stable-letter length. Then u and v are conjugate in the HNN extension if and only if some cyclic permutation u of u is conjugate to v by an element of the base group.

In particular, conjugate cyclically Britton-reduced words of positive stable-letter length have the same stable-letter length. The length-zero base-group case of Collins' theorem is deferred: it requires a separate chain criterion through the associated subgroups.

Facts & Assumptions

Given: The cyclically Britton-reduced words u and v of positive stable-letter length.

[L1]

Every cyclic permutation of a positive-length cyclically Britton-reduced HNN word is conjugate to the original word and remains cyclically Britton-reduced. (Cyclic permutations of a cyclically Britton-reduced HNN word stay in the same conjugacy class)

[L2]

Conjugacy between positive-length cyclically Britton-reduced words reduces to conjugacy by a base-group element after a cyclic permutation. (Conjugacy between cyclically Britton-reduced HNN words reduces to base-group conjugacy after cyclic permutation)

Proof

technique · direct
1.1

Suppose first that u and v are conjugate. Then [L2] gives a cyclic permutation u of u that is conjugate to v by an element of the base group. Because a cyclic permutation preserves the number of stable letters, conjugate positive-length cyclically Britton-reduced words have the same stable-letter length.

L2given
1.2

Conversely, if some cyclic permutation u of u is conjugate to v by an element of the base group, then [L1] shows that u is conjugate to u. Composing with the given base-group conjugacy yields that u is conjugate to v.

L1given
2.1

The two directions prove the equivalence.

step 1.1step 1.2

Depends on

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Sources