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A cyclically reduced sequence has no pinch in an HNN extension either internally or across its cyclic junction. If its stable-letter length is positive, every positive power remains reduced, so Britton's lemma gives infinite order. Repeated cyclic conjugation and pinch removal shows that any finite-order element of an HNN extension is conjugate into its base group.

$\boxed{BS(3,4)\text{ is not residually finite}.}$ We exhibit a nonidentity element killed by every finite quotient.

In any finite image, let $r$ be the order of the image of $a$. The relation conjugating $a^3$ to $a^4$ givesThus $\gcd (r,3)=\gcd (r,4)=1$. Since $3$ is invertible modulo $r$, the relation implies that the image of $ba{b}^{-1}$ is a power of the image of $a$. Therefore every finite image kills the group commutatorusing $[s,t]=s^{-1}{t}^{-1}st$.

View $BS(3,4)$ as an HNN extension of $⟨a⟩\cong \mathbb{Z}$, with associated subgroups $⟨a^3⟩$ and $⟨a^4⟩$. A pinch in an HNN extension would be $b{a}^{3j}{b}^{-1}$ or $b^{-1}{a}^{4j}b$. The word $w$ has none: its intervening exponents are $-1,1,1$, incompatible with the required divisibilities $3,4,3$. By Britton's lemma, $w\ne 1$. Hence finite quotients fail to separate this nonidentity element, proving the conclusion.

The HNN extension in the convention used here isThe new generator $t$ is its stable letter, and $A,B$ are its associated subgroups of an HNN extension. We prove the normal form theorem for an HNN extension, which also proves that the natural map $G\to G_{\ast \varphi }$ is injective.

Set $C_{+}=A$, $C_{-}=B$, ${\psi }_{+}=\varphi$, and ${\psi }_{-}={\varphi }^{-1}$. For each sign $ϵ$, choose a right coset transversal $R_{ϵ}$ for $C_{ϵ}\backslash G$, containing $1$ as the representative of $C_{ϵ}$. Thus every $g\in G$ has a unique decomposition $g=cr$, with $c\in C_{ϵ}$ and $r\in R_{ϵ}$. A normal form iswhere $g_0\in G$ is unrestricted and the case $k=0$ is just $g_0$.

Here is a constructive proof. Let $\mathcal{N}$ be the set of these formal sequences. Left multiplication by $h\in G$ changes only $g_0$ to $h{g}_0$. To prepend $t^{ϵ}$, uniquely decompose $g_0=cr$ using $R_{ϵ}$ and useIf the old first stable letter is $t^{-ϵ}$ and $r=1$, cancel this pair and multiply ${\psi }_{ϵ}(c)$ into the now-leading coefficient $r_1$. Otherwise retain the new first stable letter. Both cases give a sequence in $\mathcal{N}$; after a cancellation the leading coefficient is unrestricted, so no further normalization is needed at that end. Prepending the letters of a word from right to left proves existence.

These operations define permutations of $\mathcal{N}$. The base-group operations satisfy $L_h{L}_{h^{\prime }}=L_{h{h}^{\prime }}$. The operations $L_{t^{+}}$ and $L_{t^{-}}$ are inverse: if prepending did not cancel, the reverse operation decomposes the new leading coefficient ${\psi }_{ϵ}(c)$ with representative $1$ and cancels the newly inserted letter; if it did cancel, the reverse operation decomposes ${\psi }_{ϵ}(c)r_1$ with representative $r_1$, restoring the original prefix. The normal-form restriction rules out an unwanted second cancellation in the latter case.

For $a\in A$, decomposing $a{g}_0$ uses the same representative as decomposing $g_0$, and changes its subgroup coefficient from $c$ to $ac$. Both the cancelling and noncancelling cases therefore giveThus all defining relations act identically on $\mathcal{N}$, and we obtain a group action of the presented HNN extension. Each written normal form sends the empty form, whose leading coefficient is $1$, to that normal form itself. Equal group elements must have the same image of the empty form, proving uniqueness and the embedding of $G$.

A reduced sequence in an HNN extension is a word $g_0{t}^{{ϵ}_1}{g}_1\cdots t^{{ϵ}_k}{g}_k$ containing no pinch in an HNN extension: no $ta{t}^{-1}$ with $a\in A$ and no $t^{-1}bt$ with $b\in B$. Equivalently, whenever ${ϵ}_{i+1}=-{ϵ}_i$, one requires $g_i\notin C_{{ϵ}_i}$.

Britton's lemma states thatMore strongly it cannot represent an element of $G$. To prove this from the normal form theorem for an HNN extension, normalize the coefficients from right to left. Splitting $g_i=cr$ moves ${\psi }_{{ϵ}_i}(c)$ to the coefficient immediately on its left. If the preceding stable letter has the opposite sign, this transported factor belongs to the subgroup relevant to that inverse pair; multiplying by it cannot turn a coefficient outside that subgroup into one inside it. If the signs agree, cancellation is impossible anyway. Hence no stable letter disappears during normalization of a reduced sequence. Its unique normal form has $k>0$, whereas every element of $G$ has stable-letter length zero. This proves Britton's lemma. The same proof works with finitely many stable letters, with pinches requiring the same letter and its inverse.

A reduced sequence is a word $g_0{t}^{{ϵ}_1}{g}_1\cdots t^{{ϵ}_k}{g}_k$ with no pinch in an HNN extension. If $k>0$, Britton's lemma makes it nontrivial and prevents it from representing a base-group element. Reduced sequences need not be unique; uniqueness requires the fixed coset representatives in the normal form theorem for an HNN extension.