Conjugacy between cyclically Britton-reduced HNN words reduces to base-group conjugacy after cyclic permutation

We first record the boundary calculation used below. Let $p=a_0{t}^{{\epsilon }_1}{a}_1\cdots t^{{\epsilon }_n}{a}_n$ be cyclically Britton-reduced with $n>0$, and write a positive-length Britton-reduced conjugator as $y=b{t}^{\delta }z$, where $\mathrm{\mid }z{\mathrm{\mid }}_t=\mathrm{\mid }y{\mathrm{\mid }}_t-1$. At the two interfaces with $p$ in $y^{-1}py$, only $t^{\delta }$ and $t^{-\delta }$ from this first syllable of $y$ can participate. A pin at the $y^{-1},p$ interface has the literal form $t^{-\delta }(b^{-1}{a}_0)t^{{\epsilon }_1}$ with ${\epsilon }_1=\delta$; a pin at the $p,y$ interface has the literal form $t^{{\epsilon }_n}(a_{n}b)t^{\delta }$ with ${\epsilon }_n=-\delta$. In either case the relation $tc{t}^{-1}=\varphi (c)$ or its inverse removes the displayed pair. The untouched copy of $t^{\delta }$ becomes the opposite end syllable, and a direct substitution of the same relation shows that the remaining central word is a base-group conjugate of the corresponding literal cyclic permutation $p_1$ of $p$. Absorb that base conjugacy into $z$ to obtain $y^{-1}py=y_1^{-1}{p}_1{y}_1$ with $\mathrm{\mid }{y}_1{\mathrm{\mid }}_t=\mathrm{\mid }y{\mathrm{\mid }}_t-1$. The two signs cover the two associated subgroups, and [L1] and [L2] show that $p_1$ is again cyclically Britton-reduced. If neither interface is a pin, the displayed word is Britton-reduced, because its three factors already are, and it has stable-letter length $\mathrm{\mid }p{\mathrm{\mid }}_t+2\mathrm{\mid }y{\mathrm{\mid }}_t$.

Let $n=\mathrm{\mid }u{\mathrm{\mid }}_t$ and $r=\mathrm{\mid }v{\mathrm{\mid }}_t$. Among all equations $y^{-1}py=q$, with $p$ a cyclic permutation of $u$, $q$ a cyclic permutation of $v$, and $y$ Britton-reduced, choose one minimizing $\mathrm{\mid }y{\mathrm{\mid }}_t$; the original conjugacy supplies such an equation. If $\mathrm{\mid }y{\mathrm{\mid }}_t>0$ and a boundary pin occurs, step 1.1 shortens $y$ while rotating $p$, contrary to the choice. If no pin occurs, step 1.1 gives a Britton-reduced representative of $q$ of length $n+2\mathrm{\mid }y{\mathrm{\mid }}_t$. Normalizing a Britton-reduced word by [L3] only transfers associated-subgroup coefficients and performs no stable-letter cancellation, so uniqueness in [L3] gives $r=n+2\mathrm{\mid }y{\mathrm{\mid }}_t$. Apply the same alternatives to $p=yq{y}^{-1}$. A boundary pin now shortens the same chosen conjugator while rotating $q$, again contradicting minimality; no boundary pin gives $n=r+2\mathrm{\mid }y{\mathrm{\mid }}_t$, incompatible with the preceding equality. Thus $\mathrm{\mid }y{\mathrm{\mid }}_t=0$, and [L3] also gives $n=r$.

Now keep $v$ fixed. Among all equations $x^{-1}{u}^{\mathrm{\prime }}x=v$, with $u^{\mathrm{\prime }}$ a cyclic permutation of $u$ and $x$ Britton-reduced, choose one minimizing $\mathrm{\mid }x{\mathrm{\mid }}_t$. Such equations exist by the hypothesis. If $\mathrm{\mid }x{\mathrm{\mid }}_t>0$ and a boundary pin occurs, step 1.1 produces $x_1^{-1}{u}^{\mathrm{\prime }\mathrm{\prime }}{x}_1=v$ with $u^{\mathrm{\prime }\mathrm{\prime }}$ a cyclic permutation of $u$ and $\mathrm{\mid }{x}_1{\mathrm{\mid }}_t=\mathrm{\mid }x{\mathrm{\mid }}_t-1$, contradicting the choice.

If instead no boundary pin occurs, step 1.1 makes $x^{-1}{u}^{\mathrm{\prime }}x$ a Britton-reduced word of length $n+2\mathrm{\mid }x{\mathrm{\mid }}_t$. It represents $v$, whose Britton-reduced length is $n$ by step 2.1. As in step 2.1, [L3] makes these lengths equal, a contradiction. Hence $\mathrm{\mid }x{\mathrm{\mid }}_t=0$, so $x\in A$, and the chosen cyclic permutation $u^{\mathrm{\prime }}$ is conjugate to $v$ by a base-group element as required.