Boone hnn tower and auxiliary subgroups

Since every ${\varphi }_i$ is an isomorphism of actual subgroups of $G_0$, the successive construction in [F1] forms $G_2$ with embedded bases. Its relations need only be imposed on the free bases, because conjugation and ${\varphi }_i$ preserve products and inverses. The tape relation $r_i^{-1}(sx)r_i=s{x}^{-1}$ is equivalent to $sx{r}_i=r_{i}s{x}^{-1}$ and then $r_{i}s=sx{r}_{i}x$. Thus the finite presentation of $G_2$ is exactly the tape/state/rule portion of $\mathcal{B}$.

The automorphism used in [F1] fixes $H$ and carries $⟨q_{a(i)}⟩\ast T_1$ onto $A_i$. A reduced free-product word with a state syllable cannot belong to $H$, so $A_i\cap H=T_1$. Similarly $B_i\cap H=T_{-1}$. If an element of $T_{\eta }$ equals $x^m$, applying $\rho$ gives identity, and the injectivity on $T_{\eta }$ gives that element equal to $1$. The infinite order of $x$ then gives $m=0$. Hence $A_i\cap ⟨x⟩=B_i\cap ⟨x⟩=\{1\}$.

Let a nonempty freely reduced word on $x,r_i$ be given, consolidating consecutive $x$ letters into powers. A possible rule pinch has the form $r_i^{ϵ}{x}^m{r}_i^{-ϵ}$. If $m\mathrm{\ne }0$, step 1.2 excludes subgroup membership. If $m=0$, the displayed pair would cancel freely, contrary to the chosen spelling. Thus a word with rule letters is nonidentity by multiple-letter Britton in [F1]; without rule letters it is a nonzero power of the embedded $x$. This proves freeness of $C$ on $x,r_i$, including the case of zero rule letters.

The identity map of the actual subgroup $C\le G_2$ is an isomorphism, so adjoining $t$ with that edge map is an HNN extension. Choose representatives by [A1]; [F2]–[F3] embed $G_2$ in $G_3$. Commutation with each $x,r_i$ is equivalent to commutation with every product and inverse, hence with all of $C$.

By [F4], $q$ is a specified generator in the embedded state group, so $q^{-1}tq\in G_3$ is defined. Let $D$ be the subgroup generated by the finite list $x,r_i,q^{-1}tq$. Its identity map is an isomorphism regardless of relations among that list. Adjoining $k$ with this edge map therefore embeds $G_3$ by [F2]–[F3], again using [A1]. Requiring commutation with the displayed generators is equivalent to commutation with $D$. By [F5], these are exactly the remaining defining relations of $\mathcal{B}$; both presentations are the same free-group quotient. This proves the tower and every asserted embedding.