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Collins' conjugacy theorem for cyclically reduced HNN words
Statement
Let and be cyclically Britton-reduced HNN words of positive stable-letter length. Then and are conjugate in the HNN extension if and only if some cyclic permutation of is conjugate to 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 and of positive stable-letter length.
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)
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
Suppose first that and are conjugate. Then [L2] gives a cyclic permutation of that is conjugate to 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.
Conversely, if some cyclic permutation of is conjugate to by an element of the base group, then [L1] shows that is conjugate to . Composing with the given base-group conjugacy yields that is conjugate to .
The two directions prove the equivalence.
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Sources
- Roger C. Lyndon and Paul E. Schupp, Combinatorial Group Theory (standard reference, not scraped)