Hnn Extensions and Brittons Lemma - Examples

Take both associated subgroups to be $A$ and the associated isomorphism to be the identity. Then the defining relation in [L1] becomes $ta{t}^{-1}=a$ for every $a\in A$, so the stable letter commutes with the image of $A$.

The map from the HNN extension to $A\times \mathbb{Z}$ sending $A$ to $A\times \{0\}$ and $t$ to $(e,1)$ satisfies the relation from step 1.1, so [L2] gives a homomorphism. The reverse map sends $(a,n)$ to $a{t}^n$, and the commuting relation makes it a homomorphism inverse to the first one. Hence the HNN extension is $A\times \mathbb{Z}$.

In the base group $A=⟨a⟩\cong \mathbb{Z}$, the subgroups $⟨a^m⟩$ and $⟨a^n⟩$ are isomorphic and the displayed presentation is exactly of the HNN form from [L1].

If $\mathrm{\mid }m\mathrm{\mid }=1$, then $⟨a^m⟩=A$ and [L2] makes $BS(m,n)$ an ascending HNN extension; similarly if $\mathrm{\mid }n\mathrm{\mid }=1$ after reversing the stable letter. In those cases [L3] gives the one-sided normal form. When both $\mathrm{\mid }m\mathrm{\mid }$ and $\mathrm{\mid }n\mathrm{\mid }$ exceed $1$, both associated subgroups are proper, so the extension is not ascending.

The map $\varphi (n)=2n$ is injective, so [L1] gives the presentation $⟨a,t\mid ta{t}^{-1}=a^2⟩$. Its positive associated subgroup is $2\mathbb{Z}\le \mathbb{Z}$.

Under the multiplicative notation $a^k\leftrightarrow k$, the image subgroup $\varphi (\mathbb{Z})=2\mathbb{Z}$ consists exactly of the even exponents. Thus $a^k\mathrm{\notin }\varphi (\mathbb{Z})$ exactly when $k$ is odd. The condition in [L2] therefore specializes to the stated unique forms $t^{-m}{a}^k{t}^n$, with $k$ odd whenever $m,n>0$.

By [L1], the first three letters of $w$ form a pin, so [L2] replaces them by $\varphi (c)$ and gives the shorter word $\varphi (c)t^{-1}dt$.

The remaining stable-letter subword is the second kind of pin, so another application of [L2] gives $\varphi (c){\varphi }^{-1}(d)\in A$. This is the Britton-reduced representative to which [L3] applies.

For every $c\in C$, [L1] gives $t\alpha (c)t^{-1}=\beta (c)$. Hence $t\alpha (C)t^{-1}\subseteq \beta (C)$.

Applying the same relation to $t^{-1}$ shows $\alpha (c)=t^{-1}\beta (c)t$ for every $c\in C$, so $\beta (C)\subseteq t\alpha (C)t^{-1}$. Thus the two subgroups are conjugate exactly as [L2] predicts.

Choose any $c\in C_{-}$. The word $tc{t}^{-1}$ contains a stable letter, and [L1] says it is a pin, so it is not Britton-reduced.

By [L2], the same word reduces to $\varphi (c)\in A$. Thus it is a concrete counterexample to the statement refuted.