Hnn Extensions and Brittons Lemma

Because $\alpha$ and $\beta$ are injective by [L1], [L4] shows that both maps identify $C$ with their images $C_{-},C_{+}\le A$ from [L2]. Hence ${\alpha }^{-1}:C_{-}\to C$ exists, and $\varphi =\beta \circ {\alpha }^{-1}$ is a bijective group homomorphism $C_{-}\to C_{+}$. So $\varphi$ is an isomorphism by [L3].

For each $c\in C$, put $x=\alpha (c)\in C_{-}$. Then the defining relator $t\alpha (c)t^{-1}=\beta (c)$ from [L1] becomes $tx{t}^{-1}=\varphi (x)$. Conversely every $x\in C_{-}$ has the form $x=\alpha (c)$ for a unique $c\in C$, so every relator $tx{t}^{-1}=\varphi (x)$ comes from exactly one original relator.

The two presentations therefore have the same generators and the same set of defining relations after the change of notation $x=\alpha (c)$. Hence they present the same group.

By [L1], every pin of the first kind is literally one side of a defining relator of the presentation, so replacing $tc{t}^{-1}$ by $\varphi (c)$ preserves the represented element.

If $d\in C_{+}$, write $d=\varphi (c)$ with $c={\varphi }^{-1}(d)\in C_{-}$. Then step 1.1 applied to $tc{t}^{-1}=\varphi (c)=d$ gives $t^{-1}dt=c={\varphi }^{-1}(d)$ after multiplying by $t^{-1}$ on the left and $t$ on the right. So replacing a pin of the second kind by ${\varphi }^{-1}(d)$ also preserves the represented element.

Starting at the right end of an HNN word, use [L1] to rewrite every coefficient following a $t$ as $cs$ with $c\in C_{-}$ and every coefficient following a $t^{-1}$ as $dr$ with $d\in C_{+}$. Move the subgroup factor to the left by $tc=\varphi (c)t$ or $t^{-1}d={\varphi }^{-1}(d)t^{-1}$. If this creates adjacent inverse stable letters, apply [L2], combine the adjacent base coefficients, and repeat. Each cancellation removes two stable letters; between cancellations the next coset decomposition is unique. The process therefore terminates and yields a transversal normal form.

Let $\mathcal{N}$ be the set of transversal normal forms. For $a\in A$, let ${\lambda }_a$ multiply the initial coefficient by $a$. Define ${\lambda }_t$ by prepending $t$, making the forced decomposition $a_0=cs$ from [L1], and using $tcs=\varphi (c)ts$; define ${\lambda }_{t^{-1}}$ dually from $a_0=dr$ and $t^{-1}dr={\varphi }^{-1}(d)t^{-1}r$. If the new stable letter is inverse to the first old one and $s=e$ or $r=e$, cancel that pair and apply the same front rule again. This recursion terminates because each repetition removes two stable letters.

Uniqueness of the decompositions in [L1] gives ${\lambda }_a{\lambda }_b={\lambda }_{ab}$ and shows directly, in the cancellation and noncancellation cases, that ${\lambda }_t$ and ${\lambda }_{t^{-1}}$ are inverse permutations of $\mathcal{N}$. For $c\in C_{-}$, the front rules reduce the literal pin in $tc{t}^{-1}w$ before touching $w$, so ${\lambda }_t{\lambda }_c{\lambda }_{t^{-1}}={\lambda }_{\varphi (c)}\mathrm{.}$ Thus the factor actions of $A$ and $F(\{t\})$ satisfy every relator in [L3] and descend to an action of the HNN extension on $\mathcal{N}$.

Apply the action from step 1.3 to the length-zero normal word $e_A$. Reading a written normal form from right to left reconstructs it exactly: each terminal coefficient is already in the required transversal, and the nonidentity condition at a sign change prevents cancellation. Hence a normal form sends $e_A$ to that same written normal form. If two normal forms represented one group element, their actions on $e_A$ would agree, so the two written forms would be identical. The identity acts trivially, and therefore its unique normal form is $e_A$.

Choose transversals containing the identity in both associated subgroups. Normalize $w$ by the procedure of [L2]. Because $w$ is Britton-reduced by [L1], no elementary pin reduction is available, so the normalization only replaces each interior coefficient by the corresponding transversal representative in the same coset and leaves the stable-letter length $n$ unchanged.

If $w=1$, the resulting normal form is the identity's normal form from [L2]. Step 1.1 shows that this is possible only when $n=0$, and then uniqueness in [L2] forces the remaining coefficient to be $a_0=e_A$. The contrapositive is exactly the nontriviality clause for Britton-reduced words containing a stable letter.

Regard $a\in A$ as the HNN word of stable-letter length zero. It is Britton-reduced, since it contains no stable letters and hence no pin.

If this word represents the identity in the HNN extension, [L1] forces its unique base coefficient to be $a=e_A$. Therefore distinct elements of $A$ remain distinct in the HNN extension, so the canonical map $A\to G$ is injective.

For any nonzero integer $n$, the word $t^n$ contains a stable letter and has no base coefficient at which a pin could occur, so it is Britton-reduced.

By [L1], the element represented by $t^n$ is nonidentity for every $n\mathrm{\ne }0$. Hence no nonzero power of $t$ equals $1$, and $t$ has infinite order.

Use $f$ on the base-group generators of a presentation of $A$ and send the stable letter $t$ to $h$. The old relators from $A$ are satisfied because $f$ is a homomorphism, and each new HNN relator from [L1] is satisfied because the hypothesis gives $hf(\alpha (c))h^{-1}=f(\beta (c))$.

Therefore [L2] gives a unique homomorphism $\bar{f}:G\to H$ extending those assignments. By construction it restricts to $f$ on $A$ and sends $t$ to $h$, which is exactly the required universal property.

Choose a right-coset transversal $S$ for $\varphi (A)\mathrm{\backslash }A$ containing $e$, and apply [L2] with the transversal for the negative associated subgroup equal to $\{e\}$. Every coefficient following a letter $t$ is then $e$, so a sign pattern $+,-$ cannot occur: it would create a pin. Thus every transversal normal form has all negative stable letters before all positive stable letters.

Such a normal form has the shape $a_0{t}^{-1}{s}_1\cdots t^{-1}{s}_m{t}^n$, where $s_i\in S$ and the positive-letter coefficients are identities. Repeatedly use $b{t}^{-1}=t^{-1}\varphi (b)$ to move $a_0$ and then the intervening coefficients to the right. This gives one expression $t^{-m}a{t}^n$. If $m,n>0$, normality gives $s_m\mathrm{\ne }e$; the resulting middle coefficient has the form $\varphi (b)s_m$, so it lies outside $\varphi (A)$.

Conversely, start with $t^{-m}a{t}^n$ and recursively decompose the coefficient immediately following the last $t^{-1}$ into its unique form $ds$ with $d\in \varphi (A)$ and $s\in S$, moving $d$ left by $t^{-1}d={\varphi }^{-1}(d)t^{-1}$. This recovers a unique transversal normal form. When $m,n>0$, the condition $a\mathrm{\notin }\varphi (A)$ makes the last representative nonidentity, so no cancellation occurs at the sign change. This reverse construction and the construction of step 2.1 are inverse, and uniqueness in [L2] therefore gives uniqueness of $m,n,a$.

Choose, among all conjugates of the given element, a Britton-reduced representative $w$ of minimal stable-letter length. Such a representative exists because one may first Britton-reduce any conjugate using [L2].

If $w$ were not cyclically Britton-reduced, [L1] would give a pin across the two ends. Conjugating by the initial stable-letter syllable rotates that end-pin into the interior of the word, and [L2] then removes it to produce a conjugate with strictly smaller stable-letter length, contradicting the minimal choice in step 1.1.

Therefore the minimal Britton-reduced representative from step 1.1 has no end-pin and is cyclically Britton-reduced. The nontriviality clause [L3] ensures that the shortening in step 2.1 is genuine whenever a stable letter is present.

Conjugating $w$ by $a_0$ gives $a_0^{-1}w{a}_0=t^{{\epsilon }_1}{a}_1\cdots t^{{\epsilon }_n}(a_n{a}_0)$, so the displayed first cyclic permutation lies in the conjugacy class of $w$.

The end-rotation of the word from step 1.1 is $t^{{\epsilon }_n}(a_n{a}_0)t^{{\epsilon }_1}{a}_1\cdots t^{{\epsilon }_{n-1}}{a}_{n-1}$, which is Britton-reduced by [L1]. That is exactly the criterion saying the word from step 1.1 is cyclically Britton-reduced. Repeating the same conjugation on each successive rotation shows that every cyclic permutation of the stable-letter syllables is conjugate to $w$.

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.

Suppose first that $u$ and $v$ are conjugate. Then [L2] gives a cyclic permutation $u^{\mathrm{\prime }}$ 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.

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

The two directions prove the equivalence.

If a nontrivial element of $A$ became trivial in the HNN extension, the canonical map $A\to G$ would identify it with $e_A$.

That contradicts the injectivity stated in [L1]. Hence the statement is false.

For any $c\in C_{-}$, the word $tc{t}^{-1}$ contains a stable letter and is a pin by [L1].

By the defining relator it equals $\varphi (c)\in A$, so it can even represent the identity when $c=e$. Thus [L2] does not justify the stated claim, and the statement is false.

Let $A=⟨a⟩\cong \mathbb{Z}$, let $C_{-}=C_{+}=⟨a^2⟩$, and let $\varphi =\mathrm{id}$. With transversal $S=\{e,a\}$, the element $ta$ is already in normal form. With transversal $S^{\mathrm{\prime }}=\{e,a^{-1}\}$ 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 $a^{2}t{a}^{-1}$.

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.

Let $A=\mathbb{Z}$ and choose associated subgroups $2\mathbb{Z}$ and $3\mathbb{Z}$ with the isomorphism $2n\mapsto 3n$. By [L2] this defines an HNN extension.

Neither associated subgroup is all of $A$, so this example cannot satisfy the condition in [L1]. Therefore it is an HNN extension that is not ascending, and the statement is false.