Countable groups are outer automorphisms of many HNN extensions

Following on from a previous post on theorems true for all groups, I'd like to record some elements of Logan's proof that HNN extensions of triangle groups can have outer automorphism group isomorphic to any countable group you'd like

A \emph{triangle group} is a group with a presentation of the form \[T_{i, j, k}:=\langle a, b; a^i, b^j, (ab)^k\rangle\].
If $i=j=k$ we shall write $T_i:=T_{i, i, i}$ for the corresponding \emph{equilateral triangle group}. The main goal will be to prove

Theorem 1: Fix an equilateral triangle group $T_{i}$ with $i\geq6$. For every countable group presentation $\mathcal{P}$ there exists an automorphism-induced HNN-extension $T_{\mathcal{P}}$ of $T_i$ such that $\operatorname{Out}(T_{\mathcal{P}})\cong Q$ and $\operatorname{Aut}(T_{\mathcal{P}})\cong T_{\mathcal{P}}\rtimes Q$, where $Q:=\pi_1(\mathcal{P})$.

One of the advantages of this construction is that explicit examples (I suppose I really mean explicit presentations) are easy to construct, as compared to previous results involving small cancellation theory and so on.

Outer automorphisms and Serre's property FA

Let $H$ be a group, $K\lneq H$ a proper subgroup and $\phi\in\operatorname{Aut}(H)$ an automorphism.\footnote{If $H=K$ then $G=H\rtimes_{\phi}\mathbb{Z}$ is the mapping torus of $\phi$. Here the outer automorphism group is very different because $\phi$ extends to an inner automorphism.}
Then the \emph{automorphism-induced HNN-extension of $H$ with associated subgroup $K$ and with associated automorphism $\phi$} is the group with relative presentation
\[
H\ast_{(K, \phi)}=\langle H, t; tkt^{-1}=\phi(k), k\in K\rangle.
\]
The groups $T_{\mathcal{P}}$ in Theorem 1 are automorphism-induced HNN-extensions. The fact that they are automorphism-induced is crucial to our paper. For example, the following fact particular to automorphism-induced HNN-extensions, combined with theorem 10 below, shows that in fact the construction in Theorem 1 is functorial: if $K<K_0< H$ then $H\ast_{(K, \phi)}\twoheadrightarrow H\ast_{(K_0, \phi)}$.

Automorphism-induced HNN-extensions are ``nice'', in the sense that they are an easy class of groups to work with and possess nicer properties than general HNN-extensions. In general, an HNN-extension may exhibit a variety of pathological properties.
The simplest examples of HNN-extensions are Baumslag--Solitar groups (HNN-extensions of the infinite cyclic group), and these also provide the standard examples of badly-behaved HNN-extensions. For example, Baumslag--Solitar groups can have non-finitely generated outer automorphism group, aren't soluble, aren't residually finitely etc.

The construction underlying Theorem 1 is surprising because it shows that automorphism-induced HNN-extensions can still be ``wild'', even when the base group is well-behaved.

As another example of the tractability of automorphism-induced HNN-extensions, we note that the notation $H\ast_{(K, \phi)}$ emphasises that the second associated subgroup $\phi(K)$ can, in practice, be ignored when studying these groups. This in demonstrated in Theorem 2 and Lemma 3, where the descriptions of $\operatorname{Out}(H\ast_{(K, \phi)})$ mention $K$ but not $\phi(K)$.

Describing \boldmath{$\operatorname{Out}(G)$

For $G=H\ast_{(K, \phi)}$, Theorem 2 now describes $\operatorname{Out}(G)$ subject to certain conditions.
One of the conditions required by Theorem 2 is that the group $H$ has Serre's property FA, that is, every action of $H$ on any tree has a global fixed point. For finitely generated groups, this is equivalent to not mapping onto the infinite cyclic group and not splitting non-trivially as an HNN-extension or free product with amalgamation, so is in some sense a natural property and has led to lots of fruitful research in many directions. For instance, all finite groups have this property, which leads one to ask whether all finite groups have the following generalisation of FA: do all actions on finite contractible complexes of dimension $n \geq 2$ have a fixed point? It was known that in general the answer is no when $n \geq 3$, but finally it was shown that the answer is yes when $n=2$.

For $\delta\in\operatorname{Aut}(H)$ we write $\widehat{\delta}$ to mean the outer automorphism $\delta\operatorname{Inn}(H)\in\operatorname{Out}(H)$. For $g\in H$ we write $\gamma_g$ for the inner automorphism of $H$ acting as conjugation by $g$, so $\gamma_g(h)=g^{-1}hg$ for all $h\in H$.

We define the subgroup $A_{(K, \phi)}$ of $\operatorname{Aut}(H)$ as follows:
\[A_{(K, \phi)}:=\{\delta\in\operatorname{Aut}(H): \delta(K)=K, \:\exists\: a\in H \textnormal{ s.t. } \delta\phi(k)=\phi\delta\gamma_a(k) \:\forall\: k\in K\}\]


Theorem 2: Let $G=H\ast_{(K, \phi)}$ be an automorphism-induced $HNN$-extension. Assume that:

1. $H$ has Serre's property FA;
2. there does not exist any $b\in H$ such that $\phi(K)\lneq \gamma_b(K)$ and there does not exist any $c\in H$ such that $\gamma_c(K)\lneq \phi(K)$; and
3. $Z(H)=1$.

Then there exists a short exact sequence:
\[1\rightarrow C_H(K)\rtimes\frac{N_H(K)}{K} \rightarrow \operatorname{Out}^0(G)\rightarrow \frac{A_{(K, \phi)}\operatorname{Inn}(H)}{\operatorname{Inn}(H)}\rightarrow 1\]
where either $\operatorname{Out}^0(G)=\operatorname{Out}(G)$ or there exists some $\delta\in\operatorname{Aut}(H)$ and some $a\in H$ such that $\delta(K)=\phi(K)$, $\delta^2\gamma_a(K)=K$ and $\delta\phi^{-1}(k)=\phi\delta\gamma_a(k)$ for all $k\in K$, whence $\operatorname{Out}^0(G)$ has index two in $\operatorname{Out}(G)$.


The assumption that $Z(H)=1$ means that the short exact sequence here is easily digestible, but there is a generalisation which doesn't have this assumption.

Lemma 3 now refines the description of $\operatorname{Out}(G)$ given by Theorem 2.
%These conditions imply that $\operatorname{Out}(G)\cong N_H(K)/K$ and that $\operatorname{Aut}(G)= \operatorname{Inn}(G)\rtimes\operatorname{Out}(G)$. The fact that the automorphism group splits in this way also follows from our previous paper.
We shall often use $\operatorname{Out}^0(G)$ to mean an index-one or-two subgroup of $\operatorname{Out}(G)$. Then by $\operatorname{Aut}^0(G)$ we shall mean the full pre-image of $\operatorname{Out}^0(G)$ in $\operatorname{Aut}(G)$ under the natural map (see Lemmas 3 and 9, and Theorem 10).

It follows with some care that


Lemma 3: Suppose $H$, $K$ and $\phi$ are such that the following hold:

1. $H$ has Serre's property FA;
2. $C_H(K)$ is trivial;
3. $\delta(K)\cap K=1$ for all automorphisms $\delta\not\in \operatorname{Inn}(H)$; and 
4. $\phi\not\in\operatorname{Inn}(H)$.

Let $G=H\ast_{(K, \phi)}$. Then there exists an index-one or-two subgroup $\operatorname{Out}^0(G)$ of $\operatorname{Out}(G)$ such that $\operatorname{Out}^0(G)\cong N_H(K)/K$.
Moreover, $\operatorname{Aut}^0(G)= \operatorname{Inn}(G)\rtimes\operatorname{Out}^0(G)$, and so $\operatorname{Aut}^0(G)\cong G\rtimes N_H(K)/K$. Suppose, in addition, that the following holds:

  5. $\widehat{\phi}$ has odd or infinite order in $\operatorname{Out}(H)$.

Then $\operatorname{Out}^0(G)=\operatorname{Out}(G)$ and $\operatorname{Aut}^0(G)=\operatorname{Aut}(G)$.

Malcharacteristic subgroups

 A malnormal subgroup intersects its conjugates trivially apart from in the obvious place in the obvious way. More formally, a subgroup $M\leq H$ is malnormal in $H$ if the following implication holds, where $h^g:=g^{-1}hg$ and $M^g:=\{h^g; h\in M\}$.
\[M^g\cap M\neq1\Rightarrow g\in M\]
A malcharacteristic subgroup is a subgroup that intersects its automorphic orbit trivially apart from in the obvious place in the obvious way.
Definition: A subgroup $M\leq H$ is malcharacteristic in $H$ if for all $\delta\in\operatorname{Aut}(H)$ the following implication holds, where $\operatorname{Inn}(M):=\{\gamma_h; h\in M\}$.
\[\delta(M)\cap M\neq1\Rightarrow\delta\in\operatorname{Inn}(M)\]
Malcharacteristic subgroups may also be characterised as follows.

Lemma 4: Let $M$ be a subgroup of $H$. The following are equivalent.

1. $M$ is malcharacteristic in $H$.
2. $\operatorname{Inn}(M)$ is malnormal in $\operatorname{Inn}(H)$.
3. $M$ is malnormal in $H$ and the following implication holds: \[\delta(M)\cap M\neq1\Rightarrow\delta\in\operatorname{Inn}(H).\]

The proof of Lemma 4 is routine and so omitted. We record two further observations whose proofs are routine and so omitted.

Lemma 5: If $M$ is malcharacteristic in $H$ and $A$ is malnormal in $M$ then $A$ is malcharacteristic in $H$.

This is dual to the well-known result that if $N$ is normal in $H$ and $C$ is characteristic in $N$ then $C$ is normal in $H$.


Lemma 6: If $M$ is malnormal in $H$ and $K\neq1$ is a normal subgroup of $M$ then $N_H(K)=M$.
Note that we may replace the word ``malnormal'' in Lemma 6 with the word ``malcharacteristic''.

Free, malcharacteristic subgroups

Lemma 7 constructs free, malcharacteristic subgroups from a single ``seed'' subgroup $M$. This lemma is central to Theorem 10. We use $\infty$ to denote the first infinite cardinal $|\mathbb{N}|$.


Lemma 7: If a group $H$ contains a malcharacteristic subgroup $M$ which is free of rank $m>1$ (possibly $m=\infty$) then for each $n\in\mathbb{N}\cup\{\infty\}$ the group $H$ contains a malcharacteristic subgroup $M_n$ which is free of rank $n$.

Proof: Suppose $M=\langle \mathbf{x}\rangle$ is a malcharacteristic subgroup of $H$ which is free on the set $\mathbf{x}$, where $|\mathbf{x}|=m$. Let $x_1, x_2\in \mathbf{x}$ and consider $A=\langle x_1, x_2\rangle$ (clearly $A$ is free of rank two). Now, $A$ is malnormal in $M$ and so $A$ is malcharacteristic in $H$, by Lemma 5. Therefore, again by Lemma 5, it is sufficient to prove that $A\cong F(x_1, x_2)$, the free group of rank two, contains malnormal subgroups of arbitrary rank (possibly countably infinite).

To see that $F(x_1, x_2)$ contains malcharacteristic subgroups of arbitrary rank $n\in\mathbb{N}\cup\{\infty\}$, we will quote the following two facts:
1. for each such $n$ there exists a $C^{\prime}(1/6)$ small cancellation set $\mathbf{s}_n\subset F(x_1, x_2)$ which has cardinality $n$ and where no element of $\mathbf{s}_n$ is a proper power.
2. If $n<\infty$ then $M_n:=\langle\mathbf{s}_n\rangle$ is a malnormal subgroup of $F(x_1, x_2)$ which is free of rank $n$.

If $n=\infty$ then denote by $y_i$ for $i\in\mathbb{N}$ the elements of $\mathbf{s}_{\infty}$, so $\mathbf{s}_{\infty}=\{y_1, y_2, \ldots\}$, write $M_i=\langle y_1, \ldots, y_i\rangle$, and write $M_{\infty}:=\langle\mathbf{s}_{\infty}\rangle$. As $M_{\infty}$ is the union of the chain of subgroups $M_1<M_2<\ldots$,
any identity of the form $U_0(\mathbf{s}_{\infty})=1$ or $W(x_1, x_2)^{-1}U_1(\mathbf{s}_{\infty})W(x_1, x_2)=U_2(\mathbf{s}_{\infty})$ which holds in $M_{\infty}$ also holds in $M_i$ for some $i\in\mathbb{N}$.

Therefore, as each $M_i$ with $i\in\mathbb{N}$ is free and malnormal in $F(x_1, x_2)$ it follows that $M_{\infty}$ is a malnormal subgroup of $F(x_1, x_2)$ which is free of countably-infinite rank. $\blacksquare$

Lemma 8: Let $i, j, k\geq6$. The subgroup $M=\langle x, y\rangle$, $x$ and $y$ as below with $\rho\gg\max(i,j,k)$, is a malcharacteristic subgroup of $T_{i, j, k}=\langle a, b; a^i, b^j, (ab)^k\rangle$ which is free of rank two.
\begin{align*}x&:=(ab^{-1})^{3}(a^2b^{-1})^{3}(ab^{-1})^3(a^2b^{-1})^4\ldots (ab^{-1})^3(a^2b^{-1})^{\rho+2}\\y&:=(ab^{-1})^3(a^2b^{-1})^{\rho+3}(ab^{-1})^3(a^2b^{-1})^{\rho+4}\ldots (ab^{-1})^{3}(a^2b^{-1})^{2\rho+2}\end{align*}

The proof of this is omitted, since it is the long technical heart of Logan's paper. The interested reader is invited to consult the paper itself but I have nothing to add here

The construction underlying Theorem 1

Recall that by $\operatorname{Aut}^0(G)$ we mean the full pre-image of $\operatorname{Out}^0(G)$ in $\operatorname{Aut}(G)$ under the natural map.

Lemma 9: Suppose that $H$ and $\phi\in\operatorname{Aut}(H)$ are such that Conditions 1 and 4 of Lemma 3 hold. Suppose that $M_n$ is a malcharacteristic subgroup of $H$ which is free of rank $n>1$, and $K\neq1$ is a normal subgroup of $M_n$.

Let $G=H\ast_{(K, \phi)}$. Then there exists an index-one or-two subgroup $\operatorname{Out}^0(G)$ of $\operatorname{Out}(G)$ such that $\operatorname{Out}^0(G)\cong M_n/K$. Moreover, $\operatorname{Aut}^0(G)= \operatorname{Inn}(G)\rtimes\operatorname{Out}^0(G)$, and so $\operatorname{Aut}^0(G)\cong G\rtimes M_n/K$.

Suppose, in addition, that $\phi$ is such that Condition 5 of Lemma 3 holds. Then $\operatorname{Out}^0(G)=\operatorname{Out}(G)$ and $\operatorname{Aut}^0(G)=\operatorname{Aut}(G)$.

Proof: By Lemma 6, $N_H(K)=M_{n}$. Therefore, to prove the result it is sufficient to prove that $H$, $K$ and $\phi$ satisfy Conditions 2 and 3 of Lemma 3.
The subgroup $M_n$ is malcharacteristic in $H$, so 3 holds. We now prove that $C_H(K)$ is trivial, so 2 holds. Suppose that $g\in H$ is such that $[k, g]=1$ for all $k\in K$. Then $g\in M_n$, by malnormality of $M_n$, and so $C_H(K)\leq M_n$. As $M_n$ is free we have that $K$ is cyclic. Now, non-trivial normal subgroups of non-cyclic free groups can never be cyclic. Thus, as $M_n$ is free of rank $n>1$ we conclude that $C_H(K)$ is trivial, as required.

The final, additional condition that $\widehat{\phi}$ has odd or infinite order in $\operatorname{Out}(H)$ is precisely Condition 5 of Lemma 3. Hence, $\operatorname{Out}^0(H\ast_{(K, \phi)})=\operatorname{Out}(H\ast_{(K, \phi)})$ and $\operatorname{Aut}^0(H\ast_{(K, \phi)})=\operatorname{Aut}(H\ast_{(K, \phi)})$ as required.
\end{proof}

If $\mathbf{r}\subseteq F(\mathbf{x})$ then we write $\langle\langle\mathbf{r}\rangle\rangle$ for the normal closure of the set $\mathbf{r}$ in $F(\mathbf{x})$, so $F(\mathbf{x})/\langle\langle\mathbf{r}\rangle\rangle=\pi_1(\langle\mathbf{x}; \mathbf{r}\rangle)$.

We say that $\langle\mathbf{y}; \mathbf{s}\rangle$ is a \emph{quotient presentation} of $\langle\mathbf{x}; \mathbf{r}\rangle$ if $\mathbf{x}=\mathbf{y}$ and $\langle\langle\mathbf{r}\rangle\rangle\lneq \langle\langle\mathbf{s}\rangle\rangle$.
We now have the main result of this section.


Theorem 10: Suppose that the group $H$ has

1. Serre's property FA;
2. non-trivial outer automorphism group; and 
3. a malcharacteristic subgroup which is free of rank two.

Then:

 (i) For every countable group presentation $\mathcal{P}$ there exists an automorphism-induced HNN-extension $H_{\mathcal{P}}$ of $H$ such that $\operatorname{Out}^0(H_{\mathcal{P}})\cong \pi_1(\mathcal{P})$, where $\operatorname{Out}^0(H_{\mathcal{P}})$ is an index-one or-two subgroup of $\operatorname{Out}(H_{\mathcal{P}})$. Moreover, $\operatorname{Aut}^0(H_{\mathcal{P}})= \operatorname{Inn}(H_{\mathcal{P}})\rtimes\operatorname{Out}^0(H_{\mathcal{P}})$, and so $\operatorname{Aut}^0(H_{\mathcal{P}})\cong H_{\mathcal{P}}\rtimes \pi_1(\mathcal{P})$.

   (ii) For $\mathcal{P}_1$ and $\mathcal{P}_2$ countable group presentations, if $\mathcal{P}_2$ is a quotient presentation of $\mathcal{P}_1$ then there is a surjection $H_{\mathcal{P}_1}\twoheadrightarrow H_{\mathcal{P}_2}$.

Suppose, in addition, that the group $H$ has
 4. a non-inner automorphism $\phi$ such that $\widehat{\phi}\in\operatorname{Out}(H)$ has odd or infinite order.

Then (iii) $\operatorname{Out}^0(H_{\mathcal{P}})=\operatorname{Out}(H_{\mathcal{P}})$ and $\operatorname{Aut}^0(H_{\mathcal{P}})=\operatorname{Aut}(H_{\mathcal{P}})$.

In the construction of Theorem 10 we simply specify the subgroups $M_n$ and $K$ in Lemma 9.

Proof: By Lemma 7, and as $H$ contains a malcharacteristic subgroup which is free of rank two, the group $H$ contains malcharacteristic subgroups $M_n$ which are free of rank $n$ for any given $n\in\mathbb{N}\cup\{\infty\}$. Fix for each $n\in\mathbb{N}\cup\{\infty\}$ such a subgroup $M_n$.

Let $\mathcal{P}=\langle \mathbf{x}; \mathbf{r}\rangle$ be a given presentation. Let $H_{\mathcal{P}}$ be the automorphism-induced HNN-extension $H\ast_{(K_{\mathcal{P}}, \phi)}$ of $H$ where $\phi\in\operatorname{Aut}(H)$ is a non-inner automorphism (with $\widehat{\phi}$ of odd or infinite order if such an automorphism exists) and where $K_{\mathcal{P}}$ is as follows:
Consider the presentation $\widehat{\mathcal{P}}=\langle {\mathbf{x}}, p, q; {\mathbf{r}}, p, q\rangle$, $p, q\not\in \mathbf{x}^{\pm1}$, and we shall write $\widehat{\mathbf{x}}=\mathbf{x}\sqcup\{p, q\}$ and $\widehat{\mathbf{r}}=\mathbf{r}\sqcup\{p, q\}$ (so $\widehat{\mathcal{P}}=\langle\widehat{\mathbf{x}}; \widehat{\mathbf{r}}\rangle$). Note that $\pi_1(\mathcal{P})\cong\pi_1(\widehat{\mathcal{P}})$, that $|\widehat{\mathbf{x}}|>1$, and that $\widehat{\mathbf{r}}$ is non-empty. Choose $M_{|\widehat{\mathbf{x}}|}$ to be malcharacteristic in $H$ of rank $|\widehat{\mathbf{x}}|$ and take $K_{\mathcal{P}}$ to be the normal subgroup of $M_{|\widehat{\mathbf{x}}|}$ associated with $\langle\langle\widehat{\mathbf{r}}\rangle\rangle$, the normal closure in $F(\widehat{\mathbf{x}})$ of $\widehat{\mathbf{r}}$.

By Lemma 9, this construction proves Point 1 of the theorem. Also by Lemma 9, if there exists $\widehat{\phi}\in\operatorname{Out}(H)$ which has odd or infinite order then $\operatorname{Out}^0(H_{\mathcal{P}})=\operatorname{Out}(H_{\mathcal{P}})$ and $\operatorname{Aut}^0(H_{\mathcal{P}})=\operatorname{Aut}(H_{\mathcal{P}})$.

To see that this construction proves Point 2 of the theorem, note that as $\mathcal{P}_2=\langle\mathbf{x}; \mathbf{s}\rangle$ is a quotient presentation of $\mathcal{P}_2=\langle \mathbf{x}; \mathbf{r}\rangle$ we have that $\langle\langle \mathbf{r}, p, q\rangle\rangle\lneq\langle\langle \mathbf{r}, \mathbf{s}, p, q\rangle\rangle\leq F(\mathbf{x}, p, q)$. Therefore, adding the relators $tUt^{-1}\phi(U)^{-1}$ to $H_{\mathcal{Q}}$, where $U\in M_{|\widehat{\mathbf{x}}|}$ corresponds to an element of $\langle\langle\mathbf{r}, \mathbf{s}, p, q\rangle\rangle\setminus\langle\langle\mathbf{r}, p, q\rangle\rangle$, induces the required group homomorphism $H_{\mathcal{P}_1}\twoheadrightarrow H_{\mathcal{P}_2}$. $\blacksquare$

Note that because the subgroup $K_{\mathcal{P}}$ in the proof of Theorem 10 is free, the presentation of the HNN-extension $H_{\mathcal{P}}=H\ast_{(K_{\mathcal{P}}, \phi)}$ is aspherical, and so minimal by a theorem of Chiswell-Collins-Huebschmann. Thus, for $\mathcal{P}=\langle \mathbf{x}; \mathbf{r}\rangle$, the group $H_{\mathcal{P}}$ in the construction is finitely presented if and only if the set $\mathbf{x}$ and the group $\pi_1(\mathcal{P})\cong\operatorname{Out}(H_{\mathcal{P}})$ are both finite.
o obtain the groups $T_{\mathcal{P}}$ from Theorem 1, take in Theorem 10 the group $H$ to be $T_i$, the automorphism $\phi$ to be $a\mapsto b$, $b\mapsto b^{-1}a^{-1}$, and the malcharacteristic subgroup in Condition 2 to be the subgroup $M:=\langle x, y\rangle$ stated in the section on malcharacteristic subgroups, above.

 Theorem 10 proves Theorem 1 modulo proving that the subgroup $M=\langle x, y\rangle$ of $T_{i, j, k}$, $i, j, k\geq6$, is malcharacteristic and free of rank two (which we quoted as Lemma 8); which is Condition 3 of Theorem 10 holds for these triangle groups. This is because the groups $T_{i, j, k}$ are well-known to possess Conditions 1 and 2, while the groups $T_i$ additionally possess Condition 4 (the map $\phi: a\mapsto b$, $b\mapsto b^{-1}a^{-1}$ is non-inner of order three).

Residual finiteness

Let $\mathbb{H}_{\operatorname{Fin}}$ be the class of groups $H_{\mathcal{P}}$ where $\pi_1(\mathcal{P})$ is finite.
Note that these presentations $\mathcal{P}$ are countable presentations of finite groups. In particular, if the input presentation $\mathcal{P}=\langle \mathbf{x}; \mathbf{r}\rangle$ in Theorem 11 has infinite generating set, so $|\mathbf{x}|=\infty$, then the presentations $\mathcal{P}_g=\langle \mathbf{x}; \mathbf{r}, \mathbf{s}\rangle$ in the following proof are presentations of finite groups but with infinite generating set $\mathbf{x}$. It follows that if $|\mathbf{x}|=\infty$ then the associated subgroups $K_{\mathcal{P}_g}$ cannot be finitely generated.
If $U$ and $V$ are words then we write $U\leq V$ to mean that $U$ is a subword of $V$.



Theorem 11: If $\pi_1(\mathcal{P})$ is residually finite then $H_{\mathcal{P}}$ is residually-$\mathbb{H}_{\operatorname{Fin}}$.


Proof: Recall that $H_{\mathcal{P}}$ is an HNN-extension $H\ast_{(K_{\mathcal{P}}, \phi)}$. Let $g\in H\ast_{(K_{\mathcal{P}}, \phi)}\setminus \{1\}$ be arbitrary. By Britton's lemma, $g$ is represented by a word $W=h_0t^{\epsilon_1}h_1\cdots t^{\epsilon_m}h_m$ which does not contain any subwords of the form $tkt^{-1}$ or $t^{-1}\phi(k)t$ for any $k\in K_{\mathcal{P}}$.
For those $h_i$ in $N_H(K_{\mathcal{P}})$ such that $th_it^{-1}\leq W$, write $\overline{h}_i:=h_iK_{\mathcal{P}}\in M_n/K_{\mathcal{P}}$, and for those $h_i$ in $N_H(\phi(K_{\mathcal{P}}))$ such that $t^{-1}h_it\leq W$, write $\overline{h}_i:=\phi^{-1}(h_i)K_{\mathcal{P}}\in M_n/K_{\mathcal{P}}$.

As $M_n/K_{\mathcal{P}}\cong \pi_1(\mathcal{P})$ is residually finite there exists a finite group $P_g$ and a homomorphism $\overline{\sigma}_g: M_n/K_{\mathcal{P}}\rightarrow P_g$ such that $\overline{\sigma}_g(\overline{h}_i)\neq1$ for each $\overline{h}_i$ (if there are no $\overline{h}_i$, so for example if every $h_i\not\in N_H(K_{\mathcal{P}})\cup N_H(\phi(K_{\mathcal{P}}))$, then we may take $P_g$ to be trivial).
Now, there exists a quotient presentation $\mathcal{P}_g=\langle\mathbf{x}; \mathbf{r}, \mathbf{s}\rangle$ of $\mathcal{P}=\langle \mathbf{x}; \mathbf{r}\rangle$ which corresponds to the map $\overline{\sigma}_g$. Consider the group $T_{\mathcal{P}_g}=H\ast_{(K_{\mathcal{P}_g}, \phi)}$ from Theorem 1. Then, the map $\sigma_g: T_{\mathcal{P}}\twoheadrightarrow T_{\mathcal{P}_g}$ from Theorem 10.\ref{point:mainconstruction3} is obtained by adding the relators $ tUt^{-1}\phi(U)^{-1}$, where $U$ corresponds to an element of $\langle\langle\mathbf{r}, \mathbf{s}, p, q\rangle\rangle\setminus\langle\langle\mathbf{r}, p, q\rangle\rangle\subseteq F(\mathbf{x}, p, q)$. In particular, this map naturally corresponds to the map $\overline{\sigma}_g$, and so if $h_i\in N_H(K_{\mathcal{P}})$ and $th_it^{-1}\leq W$ then $h_i\not\in K_{\mathcal{P}_g}$, and similarly if $h_i\in N_H(\phi(K_{\mathcal{P}}))$ and $t^{-1}h_it\leq W$ then $h_i\not\in \phi(K_{\mathcal{P}_g})$.
Note also that $N_H(K_{\mathcal{P}})=N_H(K_{\mathcal{P}_g})$.

We now prove that $\sigma_g(g)\neq1$, which proves the result. Suppose $th_it^{-1}\leq W$. If $h_i\in N_H(K_{\mathcal{P}})$ then $h_i\not\in K_{\mathcal{P}_g}$ by the above. On the other hand, if $h_i\not\in N_H(K_{\mathcal{P}})$ then $h_i\not\in K_{\mathcal{P}_g}\leq N_H(K_{\mathcal{P}})$. Hence, $W$ does not contain any subword of the form $tkt^{-1}$ for any $k\in K_{\mathcal{P}_g}$. Similarly, $W$ does not contain any subword of the form $t^{-1}\phi(k)t$ for any $k\in K_{\mathcal{P}_g}$. That $\sigma_g(g)\neq1$ now follows from Britton's Lemma. $\blacksquare$

Theorem 12: Let $\mathbb{T}_{\operatorname{Fin}}$ be the class of groups $T_{\mathcal{Q}}$ where $\pi_1(\mathcal{Q})$ is finite. If $\pi_1(\mathcal{P})$ is residually finite then $T_{\mathcal{P}}$ is residually-$\mathbb{T}_{\operatorname{Fin}}$.

Theorem 12 follows from Theorem 11. Bumagin--Wise asked if every countable group $Q$ can be realised as the outer automorphism group of a finitely generated, residually finite group $G_Q$. We now prove Corollary 13 (which is a corollary of Theorem 12). Corollary 13 answers this question of Bumagin--Wise for all finitely generated, residually finite groups $Q$ by taking $G_Q:=T_{\mathcal{P}}$ for $\mathcal{P}$ a presentation of $Q$ with finite generating set.


Corollary 13: If the presentation $\mathcal{P}$ has a finite generating set and the group $\pi_1(\mathcal{P})$ is residually finite then the group $T_{\mathcal{P}}$ is residually finite.


We will need the following result of Scott:

Theorem 14: Triangle groups are LERF, i.e. all finitely generated subgroups are separable/closed in the profinite topology.

Proof of Corollary 13: We write $h:=T_i$. By Theorem 12 it is sufficient to prove that if $\mathcal{P}=\langle \mathbf{x}; \mathbf{r}\rangle$ has finite generating set $\mathbf{x}$ and $\pi_1(\mathcal{P})$ is finite then $T_{\mathcal{P}}$ is residually finite. Under these conditions the associated subgroup $K_{\mathcal{P}}$ of the group $T_{\mathcal{P}}=H\ast_{(K_{\mathcal{P}}, \phi)}$ has finite index in the subgroup $M_{|\mathbf{x}|}$. As $M_{|\mathbf{x}|}$ is a free group of finite rank ${|\mathbf{x}|}$, the subgroup $K_{\mathcal{P}}$ is finitely generated.

Now, triangle groups are LERF. Hence, $K_{\mathcal{P}}$ is separable in the triangle group $H$. Therefore, the automorphism-induced HNN-extension $T_{\mathcal{P}}$ is residually finite by a result of Baumslag and Tretkoff. $\blacksquare$


In order to extend Corollary 13 to all countable residually finite groups it would be necessary to prove that if $\mathcal{P}=\langle \mathbf{x}; \mathbf{r}\rangle$ is a presentation of a finite group where $|\mathbf{x}|=\infty$ then the group $T_{\mathcal{P}}$ is residually finite. (In Corollary 13 we explicitly use the fact that $\mathbf{x}$ is a finite set.)


Corollary 14: Let $\mathbb{T}_{\operatorname{Fin}}^{\infty}$ be the class of groups $T_{\mathcal{P}}$ where $\pi_1(\mathcal{P})$ is finite but where $\mathcal{P}$ has infinite generating set.
Suppose that every element of $\mathbb{T}_{\operatorname{Fin}}^{\infty}$ is residually finite.
Then for every countable, residually finite group $Q$ there exists a finitely generated, residually finite group $G_Q$ such that $\operatorname{Out}(G_Q)\cong Q$.

Note that Corollary 14 represents the best application of Theorem 1 in this direction, in the sense that if $T_{\mathcal{P}}$ is residually finite then $Q:=\pi_1(\mathcal{P})$ is residually finite. This is because $Q$ embeds in $\operatorname{Aut}(T_{\mathcal{P}})$, so if $Q$ is not residually finite then $T_{\mathcal{P}}$ is not residually finite.