# The conjugacy problem in Out(Fm) when the polynomial restrictions are non-growing

Gabriel Bartlett

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Gabriel Bartlett. The conjugacy problem in Out(Fm) when the polynomial restrictions are non-growing. 2025. hal-05154752v3

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# The conjugacy problem in Out($F_m$) when the polynomial restrictions are non-growing

Gabriel Bartlett

August 8, 2025

## Abstract

We prove that the conjugacy problem in Out($F_m$) is solvable for the class of outer automorphisms whose restrictions to their polynomial subgroups are of finite order. To do this, we first investigate the structure of suspensions of free groups by automorphisms whose outer class is of finite order. We then apply a reduction of our main result to certain problems on groups of this form.

## Contents

<table><tr><td>1 Preliminaries</td><td>3</td></tr><tr><td>1.1 Mapping tori and Reduction</td><td>3</td></tr><tr><td>1.2 Realization of an automorphism and Center</td><td>6</td></tr><tr><td>2 Congruences effectively separating torsion</td><td>7</td></tr><tr><td>2.1 Congruences of the quotient by the center</td><td>7</td></tr><tr><td>2.2 Congruences of the entire suspension</td><td>10</td></tr><tr><td>2.3 Digression: Conjugacy separability of non-growing suspensions</td><td>12</td></tr><tr><td>3 Mixed Whitehead problem</td><td>12</td></tr><tr><td>4 Conjugacy problem</td><td>14</td></tr></table>

## Introduction

### Context

In 1911, Dehn introduced three fundamental problems in group theory, including the conjugacy problem. A group $G$ has a solvable conjugacy problem if there exists an algorithm which takes two elements of $G$ and responds as to whether or not they are conjugate in $G$. An early instance occurs when one classifies conjugacy classes of general linear groups.
In many classes of groups, explicit solutions to this problem have been found. For instance, the problem is solved for free groups simply by considering the corresponding cyclically reduced words, and in general for hyperbolic groups and many groups with hyperbolic geometry. In 2006, Bogopolski, Martino, Maslakova, and Ventura solved the problem for free-by-cyclic groups (see [BMMV06]).

While the problem is particularly interesting for automorphism groups, it is remarkable that it is not solved for groups of outer automorphisms of free groups. However some progress has been made toward a full solution. The first observation to make surrounding this particular instance of the conjugacy problem is a strong link with the isomorphism problem among mapping tori. Indeed, the mapping torus constructed from an automorphism is isomorphic to another by a fibre- and orientation-preserving isomorphism if and only if the outer automorphisms are conjugate. In [Sel95], the problem was solved for irreducible automorphisms, and in [Dah16] for atoroidal automorphisms. In [DT21], Dahmani and Touikan give a reduction of this problem to a number of sub-problems concerning mapping tori. The conjugacy problem in $Out(F_m)$ is therefore reduced to the study of groups of the form $F_m \times_\varphi \langle t \rangle$, and their subgroups, for so-called polynomially growing automorphisms $\varphi$. In [DT24], Dahmani and Touikan apply this reduction to solve the problem for outer automorphisms whose polynomial restrictions are unipotent and linearly growing. In [FH24], Feighn and Handel solve it for unipotent polynomially growing (UPG) automorphisms, using a completely independent strategy. Furthermore, in [DFMT25], the authors solve the conjugacy problem for all automorphisms in $Out(F_3)$.

## Main result

In this paper, we apply Dahmani and Touikan's reduction (see Theorem 1.6) to a certain class $\mathcal{A}$ of outer automorphisms. This class consists of automorphisms whose restrictions to their polynomial subgroups are of finite order.

We wish to prove the following theorem (see Theorem 4.1).

**Theorem A.** Let $F$ be a finitely generated free group. There is an algorithm that decides whether any two outer automorphisms of $F$, whose restrictions to their polynomial subgroups are of finite order, are conjugate in $Out(F)$.

Here, a subgroup is polynomial if its conjugacy class is stable by some power of the automorphism and each element grows polynomially by iterative applications of the automorphism in question (see subsection 1.1).

The automorphisms in Theorem A may have a very complicated exponentially growing part. For instance, start with a (very complicated) graph $X$. Now append edges $e_1, \dots, e_k$ with one vertex $v_i$ in $X$. To the other end of each $e_i$, glue a graph $X_i$. Let $Y$ be the total graph (it looks like a bunch of flowers $X_i$ planted in a pot $X$). Produce an automorphism of the fundamental group of $Y$ by, on the one hand, an atoroidal homotopy equivalence of $X$ fixing each $v_i$, and on the other hand, a symmetry of each $X_i$. This would produce an example.
To prove Theorem A, we will need to study specific problems in $F_k \times_{\psi} \langle t \rangle$, with $[\psi]$ of finite order. Among them is the problem of having congruences that (effectively) separate torsion. This is the case for a group $G$ if there is a (computable) subgroup $H < G$, that is characteristic and of finite index, such that $\text{Out}(G) \to \text{Out}(G/H)$ has torsion-free kernel (see Definition 1.2). We must also prove that these groups have a solvable fibre- and orientation-preserving mixed Whitehead problem. That is to say, given any two tuples of conjugacy classes of tuples of $F_k \times_{\psi} \langle t \rangle$, we wish to decide whether there exists a fibre- and orientation-preserving outer automorphism sending one to the other. These conditions will need to be treated separately.

We will therefore devote Section 2 to proving that free-by-cyclic groups with finite order monodromy have congruences that effectively separate torsion. In a short digression, we then prove that such groups are also conjugacy separable, though this is not used to prove Theorem A. We will then prove in Section 3 that the same class of groups has a solvable fibre- and orientation-preserving mixed Whitehead problem. Finally, we will use these results to prove in Section 4 that $\mathcal{A}$ satisfies the conditions of Theorem 1.6 and therefore that the conjugacy problem in $\text{Out}(F_m)$ for outer automorphisms in $\mathcal{A}$ is solvable, which will finish the proof of Theorem A.

# Acknowledgments

I thank my doctoral supervisor, François Dahmani, for his time, guidance and insight throughout the development of this proof. This research was supported in part by the International Centre for Theoretical Sciences (ICTS) for the program - Geometry in Groups (code: ICTS/GiG2024/07).

# 1 Preliminaries

## 1.1 Mapping tori and Reduction

Given a group $G$, let $\mathcal{Z}(G)$ denote its center and $\text{Aut}(G)$ its automorphism group. For $g_0 \in G$, we will denote by $\text{ad}_{g_0} \in \text{Aut}(G)$ the inner automorphism $g \mapsto g_0^{-1}gg_0$ and by $\text{Inn}(G)$ the normal subgroup of $\text{Aut}(G)$ made up of all inner automorphisms of $G$. We will also consider the group of outer automorphisms of $G$, $\text{Out}(G) = \text{Aut}(G)/\text{Inn}(G)$. A subgroup $H$ of $G$ is said to be *characteristic* if it is stable by all automorphisms of $G$.

Let $m \in \mathbb{N}^*$. Consider $F_m$, the free group with $m$ generators. As in [Lev09], we recall the following. Let $\Phi \in \text{Out}(F_m)$ and $\varphi \in \Phi$. An element $g \in F_m$ is said to grow at most *polynomially* (resp. at least *exponentially*) under $\varphi$, if the lengths of the reduced words $(\varphi^n(g))_n$ grow at most polynomially (resp. at least exponentially). The same can be said of the conjugacy class $[g]$ under $\Phi$, using the length of the cyclically reduced words. This is independent of the choice of basis. It is a well-known fact (see [Lev09, Theorem 6.2]) that an element $g \in F_m$ grows either polynomially or exponentially under $\varphi$. We say that $\varphi$ (reps. $\Phi$)
is *polynomially growing* if every $g \in F_m$ (resp. $[g]$) grows polynomially. If $P$ is a subgroup of $F_m$, it is *polynomial* if there exist $k \in \mathbb{N}^*$ and $\alpha \in \Phi^k$ such that $\alpha(P) = P$ and $\alpha|_P$ is polynomially growing.

**Proposition 1.1.** [Lev09, Proposition 1.4] Let $\Phi \in \mathrm{Out}(F_m)$.

1. For $g \in F_m$, $[g]$ grows polynomially for $\Phi$ if and only if it is contained in a polynomial subgroup.

2. Every non-trivial polynomial subgroup is contained in a unique maximal one.

3. For $\varphi \in \Phi$ and for $g \neq 1$ polynomial for $\Phi$, the maximal polynomial subgroup $P(g)$ containing $g$ is the set of all polynomially growing elements under $\varphi$.

4. Maximal polynomial subgroups have finite rank, there are finitely many conjugacy classes of them, and they are malnormal (for any two $P$ and $Q$ and for $g \in F_m$, if $P \cap g^{-1}Qg$ is non-trivial, then $P = Q$ and $g \in P$).

For $\varphi \in \mathrm{Aut}(F_m)$, we define the mapping torus of $\varphi$,

$$ \mathbb{T}_{\varphi} = F_m \rtimes_{\varphi} \langle t_{\varphi} \rangle = \langle x_1, \dots, x_m, t_{\varphi} \mid \forall 1 \le i \le m, t_{\varphi}^{-1} x_i t_{\varphi} = \varphi(x_i) \rangle $$

We will sometimes omit the $\varphi$ from $t_\varphi$ if there is no ambiguity. Note that we can consider $F_m$ and $\langle t \rangle \cong \mathbb{Z}$ as subgroups of $F_m \rtimes_\varphi \langle t \rangle$, the former being also distinguished, and all elements of $F_m \rtimes_\varphi \langle t \rangle$ can be written $t^l f$ with $l \in \mathbb{Z}$ and $f \in F_m$, due to the relations $t^{-1}x_i t = \varphi(x_i)$.

We say that $F_m$ is the *fibre* of $\mathbb{T}_\varphi$, the coset $tF_m$ determines the positive orientation and that the automorphism $\varphi$ is the *monodromy*. Let $\alpha$ and $\beta$ be automorphisms of $F_m$. An isomorphism $\psi: F_m \rtimes_\alpha \langle t_\alpha \rangle \to F_m \rtimes_\beta \langle t_\beta \rangle$ is said to be *fibre-preserving* if $\psi(F_m) = F_m$. It is *orientation-preserving* if $\psi(t_\alpha) \in t_\beta F_m$.

Let $\mathrm{Aut}_{fo}(\mathbb{T}_\varphi)$ (resp. $\mathrm{Out}_{fo}(\mathbb{T}_\varphi)$) denote the group of automorphisms (resp. outer automorphisms) that preserve the fibre and orientation. Two outer automorphisms $\Phi_1, \Phi_2 \in \mathrm{Out}(F_m)$ are conjugate if and only if, for some $\varphi_1 \in \Phi_1$ and $\varphi_2 \in \Phi_2$, $F_m \rtimes_{\varphi_1} \langle t_1 \rangle$ and $F_m \rtimes_{\varphi_2} \langle t_2 \rangle$ are isomorphic by a fibre- and orientation-preserving isomorphism (Note that this is namely the case when $[\varphi_1] = [\varphi_2]$). Therefore the conjugacy problem in $\mathrm{Out}(F_m)$ is equivalent to the fibre- and orientation-preserving isomorphism problem in the class of groups of the form $F_m \rtimes_\varphi \langle t \rangle$ with $[\varphi] \in \mathrm{Out}(F_m)$.

We give the following definitions as in [DT21].

**Definition 1.2.** *Say that a group $G$ has congruences separating torsion if it has a characteristic subgroup $G_0$ of finite index such that the quotient $G \to G/G_0$ induces a morphism $\mathrm{Out}(G) \to \mathrm{Out}(G/G_0)$ whose kernel is torsion free. We say that congruences separate torsion effectively in $G$ if we can take $G_0$ to be computable.*
*Remark 1.3.* This means that to prove that a group $G$ has congruences separating torsion, we must find a finite index characteristic subgroup $G_0$ such that any non-trivial finite order outer automorphism of $G$ is still non-trivial on $G/G_0$.

Dahmani and Touikan have proven that free groups of finite rank have congruences effectively separating torsion ([DT21, Corollary 3.2]).

**Definition 1.4.** Let $\mathcal{H}$ be a class of groups. $\mathcal{H}$ is hereditarily algorithmically tractable if the following holds:

* $\mathcal{H}$ is closed under taking finitely generated subgroups and is effectively coherent: finitely generated subgroups are finitely presentable and a presentation is computable,

* the presentations of groups in $\mathcal{H}$ are recursively enumerable,

* $\mathcal{H}$ has a uniform solution to the conjugacy problem,

* $\mathcal{H}$ has a uniform solution to the generation problem.

**Definition 1.5.** Let $\varphi \in \text{Aut}(F_m)$ and let $H$ be a subgroup of $F_m$ whose conjugacy class is preserved by some power of $\varphi$. Let $l > 0$ be the smallest integer such that there exists $a \in F_m$ for which $\varphi^l(H) = \text{ad}_a(H)$. The sub-mapping torus is the subgroup $\langle H, t^l a^{-1} \rangle$ of $F_m \rtimes_{\varphi} \langle t \rangle$. It does not depend on the choice of $a$ and is isomorphic to $H \rtimes_{\text{ad}_{a^{-1}} \circ \varphi^l} \langle t^l a^{-1} \rangle$.

We will now state the reduction theorem that will enable us to prove Theorem A.

**Theorem 1.6.** [*DT21, Theorem A*] Let $\mathcal{O}\mathcal{A}$ be a class of outer automorphisms of $F_m$. Let $\mathcal{P}$ be the collection of sub-mapping tori constructed from the maximal polynomial subgroups under each $\Phi \in \mathcal{O}\mathcal{A}$. Let $\mathcal{P}'$ be the class of all groups in $\mathcal{P}$ and their finitely generated subgroups. Assume that $\mathcal{P}$ and $\mathcal{P}'$ satisfy the following:

1. $\mathcal{P}'$ is a hereditarily algorithmically tractable class of groups

2. the fibre- and orientation-preserving isomorphism problem in $\mathcal{P}$ is solvable

3. the groups of $\mathcal{P}'$ have congruences that effectively separate torsion

4. the fibre- and orientation-preserving mixed Whitehead problem is solvable in every group of $\mathcal{P}$.

Then the conjugacy problem in $\text{Out}(F_m)$ for automorphisms in $\mathcal{O}\mathcal{A}$ is solvable.

Let us denote by $\mathcal{A}$ the set of outer automorphisms whose polynomially growing parts are of finite order. More specifically, it is the set of outer automorphisms $\Phi \in \text{Out}(F_m)$ such that for $\varphi \in \Phi$, and for $P$ a polynomial subgroup for $\varphi$, there exist $k \in \mathbb{N}^*$ and $\gamma_P \in F_m$ such that $\text{ad}_{\gamma_P} \circ \varphi^k$ restricts to identity on $\mathcal{P}$.
Remark 1.7. Let $\Phi \in \mathcal{A}$, $\varphi \in \Phi$, and take $P$ a non-trivial polynomial subgroup for $\varphi$. Consider the sub-mapping torus constructed from $P$. It can be written $P \rtimes_{\text{ad}_{a^{-1}} \circ \varphi^l} \langle t^l a^{-1} \rangle$, for certain $a$ and $l$ as in the definition. Now $P$ is free, and $[\text{ad}_{a^{-1}} \circ \varphi^l] \in \text{Out}(P)$ is of finite order. Indeed, $l$ is the smallest integer such that $\varphi^l(P)$ is a conjugate of $P$ in $F_m$, therefore $l \mid k$. Now $(\text{ad}_{a^{-1}} \circ \varphi^l)^{\frac{k}{l}} = \text{ad}_b \circ \varphi^k$ with $b = a^{-1}\varphi^l(a^{-1}) \dots \varphi^{(\frac{k}{l}-1)^l}(a^{-1})$. Now $\text{ad}_{\gamma_P} \circ \varphi^k$ restricts to identity on $P$, so on $P$,

$$ \text{ad}_b \circ \varphi^k = \text{ad}_b \circ \varphi^k \circ (\text{ad}_{\gamma_P} \circ \varphi^k)^{-1} = \text{ad}_{b\gamma_P^{-1}} $$

And since $P$ is non-trivial and is preserved by conjugation by $b\gamma_P^{-1}$ in $F_m$, then $b\gamma_P^{-1} \in P$. Therefore, in $\text{Out}(P)$, $[\text{ad}_{a^{-1}} \circ \varphi^l]^l = [\text{ad}_{b\gamma_P^{-1}}] = [\text{id}_P]$. Therefore, all sub-mapping tori constructed from $\Phi \in \mathcal{A}$ and a finitely generated polynomial subgroup, are of the form $F_M \rtimes_{\alpha} \langle t_{\alpha} \rangle$, with $[\alpha] \in \text{Out}(F_M)$ of finite order (and $M$ not necessarily equal to $m$). In particular, this is the case for maximal polynomial subgroups, since they are finitely generated, by Theorem 1.1.

## 1.2 Realization of an automorphism and Center

Let $X$ be a connected graph, $v_0 \in X^{(0)}$ a vertex, and let $\sigma : X \to X$ be an isometry. Then we may construct an outer automorphism of the fundamental group $\pi_1(X, v_0)$ induced by $\sigma$. Indeed, the isometry $\sigma$ induces an isomorphism $\sigma_* : \pi_1(X, v_0) \to \pi_1(X, \sigma(v_0))$. We then choose a path $c$ from $v_0$ to $\sigma(v_0)$ in $X$. Concatenating paths provides us with an isomorphism

$$ \alpha_c : \begin{cases} \pi_1(X, \sigma(v_0)) & \to \pi_1(X, v_0) \\ x & \mapsto cx\bar{c} \end{cases} $$

where $\bar{c}$ is the path $c$ taken in the other direction. Then, by composition, we get an automorphism $\varphi_c = \alpha_c \circ \sigma_* \in \text{Aut}(\pi_1(X, v_0))$. In fact, through this process we can construct the entire outer class $\Phi$ of $\varphi_c$ by considering all the possible paths $c$. Furthermore, this construction yields a group homomorphism

$$ \Omega_{X,v_0} : \begin{cases} \text{Isom}(X) & \to \text{Out}(\pi_1(X, v_0)) \\ \sigma & \mapsto \Phi_{\sigma} \end{cases} $$

We recall the following realization theorem due to Culler in [Cul84], stated previously under a more geometric form in [Zim81, Satz 3.1(a)] by Zimmerman.

**Theorem 1.8 (Realization theorem).** [Cul84, Theorem 2.1] Let $\Phi \in \text{Out}(F_m)$ be of finite order. There exists a connected graph $X$ and an isometry $\sigma : X \to X$, such that there is an isomorphism $\psi : \pi_1(X, v_0) \cong F_m$, with $v_0 \in X$ some vertex, and such that $\sigma$ induces $\Phi$.

In other words, for $\alpha \in \Omega_{X,v_0}(\sigma)$, $[\psi \circ \alpha \circ \psi^{-1}] = \Phi$.
**Remark 1.9.** If $\Phi_1, \Phi_2 \in \text{Out}(F_m)$ are conjugate, they are realized by the same isometry. Assume $\Phi_1 = \Theta\Phi_2\Theta^{-1}$ and that $\Phi_2$ is realized as in the theorem. Consider $\psi' = \theta \circ \psi : \pi_1(X, v_0) \to F_m$ with $\theta \in \Theta$. Then

$$ \Phi_1 = \Theta[\psi \circ \alpha \circ \psi^{-1}]\Theta^{-1} = [\psi' \circ \alpha \circ \psi'^{-1}] $$

We recall the following proposition, well-known to experts. (See, for instance, [BP24].)

**Proposition 1.10.** Let $\Phi \in \text{Out}(F_m)$ be of finite order $k$ and let $\varphi \in \Phi$. Then $\mathcal{Z}(F_m \times_{\varphi} \langle t \rangle)$ is infinite cyclic. In fact, $\mathcal{Z}(F_m \times_{\varphi} \langle t \rangle) = \langle t^k f_0 \rangle$, where $f_0$ is the element in $F_m$ such that $\varphi^k = \text{ad}_{f_0^{-1}}$. Furthermore, $t$ and $f_0$ commute.

**Remark 1.11.** The converse is true. In fact if $\Phi$ is not of finite order, the center is trivial. (See, for instance, [BP24, Lemma 3.4])

# 2 Congruences effectively separating torsion

This section is dedicated to proving the following theorem.

**Theorem 2.1.** Let $m \ge 1$. Let $\Phi \in \text{Out}(F_m)$ be of finite order, and let $\varphi \in \Phi$. Then $F_m \times_{\varphi} \langle t \rangle$ has congruences effectively separating torsion.

**Remark 2.2.** The case of $m = 1$ can be done explicitly. Indeed, if $m = 1$ we are dealing either with $Z \times Z$, for which [DT24, Theorem 1.5] applies, or $Z \times Z = \langle f, t | t^{-1}ft = f^{-1} \rangle$. Now $\text{Out}(Z \times Z)$ is finite, of order 4, generated

by the classes of $\alpha : \begin{cases} f \mapsto f \\ t \mapsto t^{-1} \end{cases}$ and $\beta : \begin{cases} f \mapsto f \\ t \mapsto tf \end{cases}$. Therefore $[\alpha]$ is the only finite order outer automorphism. We may consider the subgroup $\langle f, t^4 \rangle$, which is characteristic and of finite index, and $[\alpha]$ descends to a non-trivial automorphism of the quotient, $Z/4Z$.

It remains to prove the theorem for $m \ge 2$.

## 2.1 Congruences of the quotient by the center

We will first prove the following proposition, which will later allow us to prove Theorem 2.1.

**Proposition 2.3.** Let $m \ge 2$. Let $\Phi \in \text{Out}(F_m)$ be of finite order $k > 1$, and let $\varphi \in \Phi$. Then

$$ Q \stackrel{\text{def}}{=} F_m \times_{\varphi} \langle t \rangle / \mathcal{Z}(F_m \times_{\varphi} \langle t \rangle) $$

has congruences separating torsion effectively.

We will prove this result in the following two steps.

**Proposition 2.4.** If $Q$ is as in Proposition 2.3, then it is virtually free, has no non-trivial finite normal subgroups and its center is trivial.
**Proposition 2.5.** Let $V$ be a virtually free group with trivial center.

Then Out($V$) has finitely many conjugacy classes of finite order elements.

Furthermore, there is an effective procedure to compute a finite list in Aut($V$) containing a representative of each conjugacy class of finite order elements of Out($V$).

Finally, if $V$ also has no non-trivial finite normal subgroups, then it has congruences separating torsion effectively.

Proposition 2.3 immediately follows from these.

*Proof of Proposition 2.4.* Recall that if $\varphi^k = \text{ad}_{f_0^{-1}}$, then $x = t^k f_0$ is such that $\mathcal{Z}(F_m \rtimes_{\varphi} \langle t \rangle) = \langle x \rangle$. Note that since $\langle x \rangle \cap F_m = \{1\}$, then $F_m \hookrightarrow Q$, ie. $Q$ contains a copy of $F_m$. Furthermore, $F_m$ is of finite index $k$ in $Q$ as $Q/F_m \cong \mathbb{Z}/k\mathbb{Z}$ ($Q = \bigsqcup_{i=0}^{k-1} t^i \bar{F}_m$). Hence, $Q$ is virtually free.

Let us show that $Q$ has no non-trivial finite normal subgroups. Assume for contradiction that such a subgroup $K < Q$ exists. Note that since $K$ is finite, it has no elements of infinite order, and hence $K \cap F_m = \{1\}$. Therefore $K \hookrightarrow Q/F_m \cong \mathbb{Z}/k\mathbb{Z}$. So $K$ is finite cyclic, generated by some element $\tau = \bar{t}^a \bar{f}_1$ of $Q$, with $f_1 \in F_m$ and $a \in \llbracket 1, k-1 \rrbracket$.

Let us prove that $K$ is in fact central. Take $y \in F_m \rtimes_{\varphi} \langle t \rangle$ and consider $\bar{y}^{-1}\tau\bar{y}$. In $F_m \rtimes_{\varphi} \langle t \rangle$, $y^{-1}(t^a f_1)y$ can be written $t^a f$ for some $f \in F_m$ since, conjugates in $F_m \rtimes_{\varphi} \langle t \rangle$ must have the same exponent of $t$. Now by denoting $f' = f_1^{-1}f$, we get $t^a f = (t^a f_1)f'$, so that $\bar{y}^{-1}\tau\bar{y} = \bar{t}^a \bar{f} = \tau\bar{f}'$. On the other hand, since $K$ is normal, $\bar{y}^{-1}\tau\bar{y} \in K$, and since $K$ is also cyclic generated by $\tau$, $\bar{y}^{-1}\tau\bar{y} = \tau^l$, for some $l \in \mathbb{Z}$. Therefore, $\bar{y}^{-1}\tau\bar{y} = \tau^l = \tau\bar{f}'$. Hence $\tau^{l-1} = \bar{f}' \in F_m \cap K$. Now $\tau \neq 1$, since we took $K$ to be non-trivial, so $l=1$. Therefore, for all $\bar{y} \in Q$, $\bar{y}^{-1}\tau\bar{y} = \tau$. So $K \subset \mathcal{Z}(Q)$. Furthermore, note that $\mathcal{Z}(Q)$ is also finite cyclic since $m \ge 2$ and therefore, as for $K$, $F_m \cap \mathcal{Z}(Q) = \{1\}$. We may then assume that $K = \mathcal{Z}(Q)$.

Now since $\tau = \bar{t}^a f_1$ is central in $Q = F_m \rtimes_{\varphi} \langle t \rangle / \langle t^k f_0 \rangle$, for all $f \in F_m$ there exists $w \in \mathbb{Z}$ such that $(t^a f_1)^{-1} f(t^a f_1) = f(t^k f_0)^w \in F_m$. Also, for all $f \in F_m$, $(t^a f_1)^{-1} f(t^a f_1) \in F_m$. Thus $w=0$ and $t^a f_1$ commutes with all $F_m \hookrightarrow F_m \rtimes_{\varphi} \langle t \rangle$. Recall that $\Phi \in \text{Out}(F_m)$ is of finite order $k > 1$. Now we have just shown that, for $f \in F_m \hookrightarrow F_m \rtimes_{\varphi} \langle t \rangle$, $f_1^{-1}(t^{-a} f t^a) f_1 = f$. In other words, in Aut($F_m$), $\text{ad}_{f_1} \circ \text{ad}_{t^a} = \text{id}_{F_m}$. Therefore $\varphi^a = \text{ad}_{t^a} = \text{ad}_{f_1^{-1}} \in \text{Inn}(F_m)$. So $k \mid a$, which is impossible, as $1 \le a \le k-1$. We have therefore reached a contradiction, having assumed that $Q$ had non-trivial finite normal subgroups. Hence $Q$ does not have any non-trivial finite normal subgroups. We have also just proven that $\mathcal{Z}(Q)$ is trivial. $\square$

Let us recall two notions that will be important. Firstly, a group $G$ is *conjugacy separable* if every element $g \in G$ is *conjugacy distinguished*, ie. for all $h \in G$ such that $h$ and $g$ are not conjugate in $G$, there is a normal subgroup $N \trianglelefteq G$ of finite index, such that the images $\bar{h}$ and $\bar{g}$ in $G/N$ are still not conjugate. We will use Dyer's theorem that virtually free groups are conjugacy separable [Dye79].
The second concerns pointwise inner automorphisms. If $G$ is a group, an automorphism $\psi$ is pointwise inner if, for all $g \in G$, $\psi(g)$ is conjugate to $g$. (We also say that the outer automorphism $[\psi]$ is pointwise inner.) In many cases, such automorphisms are inner. This was first observed by Grossman in [Gro74], but we will use the comprehensive result of Minasyan and Osin [MO10, Corollary 1.2] that in a (relatively) hyperbolic group without non-trivial finite normal subgroups, all pointwise inner automorphisms are inner.

The relevance of these two notions is given by the following proposition, which we will use to prove Proposition 2.5.

**Proposition 2.6.** [DT21 Proposition 3.1] Let $G$ be a finitely presented group such that the following holds:

1. $G$ is conjugacy separable,

2. $G$ has no non-trivial, pointwise inner, finite-order, outer automorphisms,

3. we are given a finite list $\{\alpha_1, \dots, \alpha_k\} \subset \mathrm{Aut}(G)$ containing a representative of the conjugacy class of every finite-order element of $\mathrm{Out}(G)$.

Then $G$ has congruences separating torsion effectively.

Recall that the Dunwoody-Stallings deformation space of a group $G$ is a graph whose vertices are equivalence classes of certain essential $G$-trees on which $G$ acts such that all edge stabilizers are finite. If $G$ is hyperbolic, a finite fundamental domain for this action can be computed, as well as the stabilizers of each of its points, by [DG11, Theorem 7.1, Lemma 7.5]. Let us now prove Proposition 2.5

*Proof of Proposition 2.5.* We apply a variation on Culler’s proof of Theorem 1.8 ([Cul84, Theorem 2.1]). Let $H$ be a finite subgroup of $\mathrm{Out}(V)$. This proof relies on constructing an extension $\Gamma$ of $V$ by $H$, in the sense that we wish to obtain the following short exact sequence:

$$1 \to V \to \Gamma \to H \to 1$$

corresponding to the inclusion homomorphism $H \hookrightarrow \mathrm{Out}(V)$. Recall that, by Eilenberg-MacLane theory [EM47] on cohomological obstructions (see also its rendition in [Bro82, Ch. 4]) since $V$ has trivial center, such an extension exists (and is even unique up to equivalence). Now $V$ is of finite index in $\Gamma$, and $V$ is virtually free. Therefore $\Gamma$ is also virtually free. Recall that by Karrass, Pietrowski and Solitar’s corollary of Stalling’s theorem [KPS73, Theorem 1] there exists $\mathcal{X}$ a finite graph of finite groups, and $v_0$ a vertex, such that $\Gamma = \pi_1(\mathcal{X}, v_0)$. Now, since $V$ is a subgroup of $\Gamma$, and $\Gamma$ acts on $T$ the Bass-Serre tree of $\mathcal{X}$, then so does $V$. Now $V$ does not necessarily act freely on $T$ since $V$ is only virtually free. So $V \setminus T$ is a graph of groups, on which $V \setminus \Gamma \cong H$ acts and $\pi_1(V \setminus T) \cong V$.

In other words, we have found a tree in the Stallings-Dunwoody space $\mathcal{D}$ of $V$ on which $H$ acts, that is the Bass-Serre tree of the graph of groups $V \setminus T$. Thus
$H$ fixes a point in $\mathcal{D}$. Now, since $V$ is virtually free, it is hyperbolic. Therefore, we may apply the results of [DG11] recalled above and compute $(\tau_i)_{1\le i\le N}$ a fundamental domain in $\mathcal{D}$ for the action of $G$. Therefore a conjugate of $H$ fixes some $\tau_i$. Hence, there exists $H'$, conjugate to the subgroup $H$, that is in the stabilizer in $\text{Out}(V)$ of $\tau_i$, denoted by $\text{Out}_{\tau_i}(V)$.

Now, $\text{Out}_{\tau_i}(V)$ is finite. Indeed, consider $\Psi \in \text{Out}_{\tau_i}(V)$. Then $\Psi$ is an outer automorphism of $V$ that preserves $\tau_i$ and therefore acts $V$-equivariantly on the underlying tree $T_i$. Therefore $\Psi$ is an automorphism of the graph of groups $V\setminus\tau_i$, in the sense defined in [DG11, Definition 2.19]. Now, $V\setminus\tau_i$ is a finite graph of finite groups ([Bas93, Theorem 8.4]), and $\Psi$ is determined by a tuple $(F, (\varphi_e), (\varphi_v), (\gamma_e))$ of

- a graph isomorphism $F$ of the underlying graph of $V\setminus\tau_i$,

- for each edge $e$ in $V\setminus\tau_i$, an isomorphism $\varphi_e$ between the edge groups of $e$ and $F(e)$,

- for each vertex $v$ in $V\setminus\tau_i$, an isomorphism $\varphi_v$ between the vertex groups of $v$ and $F(v)$,

- for each edge $e$ in $V\setminus\tau_i$, an element $\gamma_e$ in the vertex group of $F(t(e))$, with $t(e)$ the terminal end of edge $e$,

with certain extra conditions (see [DG11, Definition 2.19]). For each of these items, there are only finitely many possibilities, since the graph is finite and every vertex and edge group as well. Therefore, there are only finitely many possibilities for $\Psi$. Hence $\text{Out}_{\tau_i}(V)$ is finite.

We have therefore proven that, if $\Phi \in \text{Out}(V)$ is of finite order (here we take the finite subgroup $H$ we considered above to be $\langle\Phi\rangle$), $\Phi$ has a conjugate in one of the $(\text{Out}_{\tau_i}(V))_{i\in\llbracket 1,N\rrbracket}$. Since each of these is finite, there are only finitely many conjugacy classes of finite order elements of $\text{Out}(V)$.

Furthermore, since there is an effective procedure to compute $(\tau_i)_{i\in\llbracket 1,N\rrbracket}$, we may explicitly construct all the graph of group automorphisms of each of the $(\tau_i)_{i\in\llbracket 1,N\rrbracket}$. We may then take a representative in $\text{Aut}(V)$ of each of these, which will provide us with a finite list in $\text{Aut}(V)$ containing a representative of the conjugacy class of every finite-order element of $\text{Out}(V)$.

Let us now suppose that $V$ has no finite normal subgroups. Since $V$ has trivial center, by [MO10, Corollary 1.2] all pointwise inner automorphisms are inner. Furthermore, since $V$ is virtually free, by [Dye79] it is conjugacy separable. Therefore we may apply Proposition 2.6 and conclude. $\square$

## 2.2 Congruences of the entire suspension

We may now prove Theorem 2.1 that states that for $m \ge 1$, $\Phi \in \text{Out}(F_m)$ of finite order, and $\varphi \in \Phi$, $F_m \rtimes_{\varphi} \langle t \rangle$ has congruences separating torsion effectively.
*Proof of Theorem 2.1.* Recall, we have already proven the case where $m = 1$ in Remark 2.2. Let us now consider $m \ge 2$.

Recall that if $k$ is the order of $\Phi$ and $\phi^k = \text{ad}_{f_0^{-1}}$, then $x = t^k f_0$ is such that $\mathcal{Z}(\mathbb{T}_\phi) = \langle x \rangle$. Note that, since the center of $F_m \hookrightarrow \mathbb{T}_\phi$ is trivial, $x \notin F_m$. Let $Q = \mathbb{T}_\phi / \langle x \rangle$. The following sequence is exact:

$$1 \to \langle x \rangle \to F_m \rtimes_{\phi} \langle t \rangle \to Q \to 1$$

We will show that there exists a finite quotient of $F_m \rtimes_{\phi} \langle t \rangle$ on which any non-trivial, finite order outer automorphism of $F_m \rtimes_{\phi} \langle t \rangle$ induces an outer automorphism that is still non-trivial. Let $\Psi \in \text{Out}(F_m \rtimes_{\phi} \langle t \rangle)$ be non-trivial of finite order. Note that $\langle x \rangle = \mathcal{Z}(F \rtimes_{\phi} \langle t \rangle)$ is characteristic. Hence we may consider $\Psi_Q$ the outer automorphism induced on the quotient.

Three cases arise.

**First case** Suppose that there exists $\psi \in \Psi$ such that the restriction $\psi|_{\langle x \rangle}$ is non-trivial. Therefore, $\psi(x) = x^{-1}$. Indeed, as $\mathcal{Z}(F_m \rtimes_{\phi} \langle t \rangle)$ is characteristic, the restriction $\psi|_{\langle x \rangle}$ is an automorphism of $\mathcal{Z}(F_m \rtimes_{\phi} \langle t \rangle) \cong \mathbb{Z}$. Since there are only two such automorphism, identity and inversion, the restriction must be the latter. Now consider $H = \langle x^3 \rangle \subset F_m \rtimes_{\phi} \langle t \rangle$. More precisely, since $x = t^k f_0$, we get $F_m \rtimes_{\phi} \langle t \rangle / H \cong (F_m / \langle f_0^3 \rangle) \rtimes_{\phi} (\mathbb{Z} / 3k\mathbb{Z})$ as $tf_0 = f_0t$. Denote with a bar the induced elements and homomorphisms in $F_m \rtimes_{\phi} \langle t \rangle / H$. Since $\bar{x}$ is central, its conjugacy class is trivial, and since $\bar{x} \neq \bar{\psi}(\bar{x})$ in $F_m \rtimes_{\phi} \langle t \rangle / H$, they cannot be conjugates. Thus $\psi$ is still non-inner on the quotient.

**Second Case** Suppose that $\Psi_Q$ is non-trivial. By Proposition 2.3, $Q$ has congruences separating torsion effectively, so there is a computable finite index characteristic subgroup $L < Q$ such that $\Psi_Q$ is still non-trivial on the quotient $Q/L$.

**Third and final case** Suppose that $\Psi_Q$ is trivial and that for all $\psi \in \Psi$, $\psi|_{\langle x \rangle}$ is trivial. Take $\psi \in \Psi$ such that $\psi_Q$ is trivial. Then for all $y \in F_m \rtimes_{\phi} \langle t \rangle$, there exists $r \in \mathbb{Z}$ such that $\psi(y) = yx^r$. And since $\psi$ itself is non-trivial, there exists $y \in F_m \rtimes_{\phi} \langle t \rangle$ such that the integer $r$ is non-zero. However, we can write $y$ as $t^l f$ with $f \in F_m$ and $l \in \mathbb{Z}$. Hence, since $x = t^k f_0$ and since $t$ and $f_0$ commute by Proposition 1.10,

$$\psi(y) = t^l f x^r = t^l x^r f = t^l (t^k f_0)^r f = t^{l+rk} f_0^r f$$

Furthermore, for all $n \in \mathbb{N}$, $\psi^n(y) = t^{l+nrk} f_0^{nr} f$, since $\psi|_{\langle x \rangle}$ is trivial. Now, for all $n \in \mathbb{N}^*$, $nrk$ is a non-zero integer. Hence $\psi^n(y)$ and $y$ cannot be conjugate in $F_m \rtimes_{\phi} \langle t \rangle$, since they have different exponents in $t$. This however is impossible, since we took $\Psi$ to be of finite order. Therefore this case does not arise.
**In conclusion** Set $H = x^3 \cap (\langle x \rangle L)$ a computable finite index, characteristic subgroup of $F_m \times_{\varphi} \langle t \rangle$. Then $\Psi$ is non-trivial on $F_m \times_{\varphi} \langle t \rangle / H$, and since $H$ does not depend on the non-trivial finite order outer automorphism $\Psi$ that we considered, $F_m \times_{\varphi} \langle t \rangle$ does indeed have congruences separating torsion effectively. □

## 2.3 Digression: Conjugacy separability of non-growing suspensions

As a side result that will not be used in what follows, we may note the following property on the conjugacy separability of non-growing suspensions.

**Proposition 2.7.** Let $m \ge 1$ and let $[\varphi] \in \text{Out}(F_m)$ be of finite order. Then $F_m \times_{\varphi} \langle t \rangle$ is conjugacy separable.

*Proof.* Recall that if $\varphi^k = \text{ad}_{f_0^{-1}}$, then $x = t^k f_0$ is such that $\mathcal{Z}(F_m \times_{\varphi} \langle t \rangle) = \langle x \rangle$. Take two elements of $F_m \times_{\varphi} \langle t \rangle$ that are non conjugate, say $a = t^p f$ and $b = t^q g$.

Suppose first that $p = q$. Let $Q = F_m \times_{\varphi} \langle t \rangle / \mathcal{Z}(F_m \times_{\varphi} \langle t \rangle)$ and consider the images $\bar{a}$ and $\bar{b}$ in $Q$. If $\bar{a}$ and $\bar{b}$ are conjugate in $Q$, then there is $\bar{y} \in Q$ such that $\bar{b} = \bar{y}^{-1}\bar{a}\bar{y}$. Then by lifting the relation to $F_m \times_{\varphi} \langle t \rangle$, there exists $r \in \mathbb{Z}$ such that $\bar{b} = \bar{y}^{-1}\bar{a}\bar{x}^r$. Now, since $p = q$, then $r = 0$. So $a$ and $b$ are conjugate and we reach a contradiction. Therefore, $\bar{a}$ and $\bar{b}$ are not conjugate in $Q$. Now, since $Q$ is virtually free, by [Dye79] it is conjugacy separable. So there is a finite index, normal subgroup $\bar{N} \subsetneq Q$ such that $\bar{a}$ and $\bar{b}$ are still not conjugate on the quotient $Q/\bar{N}$. Then $N' = \langle \mathcal{Z}(F_m \times_{\varphi} \langle t \rangle), N \rangle$, with $N$ a lift of $\bar{N}$, is a finite index normal subgroup such that $a$ and $b$ are still not conjugate on the quotient $F_m \times_{\varphi} \langle t \rangle / N'$.

Now suppose that $p \neq q$. Then instead of taking the quotient by $\langle x \rangle$ entirely, we may consider the quotient by a subgroup $\langle x^r \rangle$ with $r = pq + 1$ for instance. Therefore $\bar{a}$ and $\bar{b}$ are still not conjugate on the quotient as their powers in $\bar{t}$ are different. Hence, we may conclude by the same reasoning, since the quotient is also virtually free. □

# 3 Mixed Whitehead problem

Take $\Phi \in \text{Out}(F_m)$ to be of finite order $k$ and let $\varphi \in \Phi$. We wish to prove that the mixed Whitehead problem on $\text{Out}_{fo}(F_m \times_{\varphi} \langle t \rangle)$ is solvable in the sense recalled in Theorem 3.4 below. For this we will need another result of Dahmani and Guirardel.

**Theorem 3.1.** [*DG11, Theorem 8.1*] Let $G$ be a hyperbolic group, $N \in \mathbb{N}$ and let $S_1, \dots, S_N, S'_1, \dots, S'_N \subset G$ be tuples. There is an effective procedure to determine if there exists $\Psi \in \text{Out}(G)$ such that, for $i \in [1, N]$, $\Psi([S_i]) \in [S'_i]$.

Furthermore, let $\mathcal{P} = ([S_1], \dots, [S_N])$. Denote by $\text{Aut}(G, \mathcal{P})$ the automorphisms of $G$ that preserve each conjugacy class in $\mathcal{P}$. Then $\text{Aut}(G, \mathcal{P})$ is finitely generated and there is an effective procedure that gives a finite generating set.
**Remark 3.2.** Since inner automorphisms preserve all conjugacy classes, the same statement can be made with $Out(G, \mathcal{P})$.

**Remark 3.3.** If there exists such a $\Psi$, sending one tuple $\mathcal{P}$ to another $\mathcal{P}'$, then there is also an effective procedure that produces one. Indeed, we wish to construct $\Psi \in \text{Out}(G)$ such that $\text{Out}(G, \mathcal{P}') = \Psi\text{Out}(G, \mathcal{P})\Psi^{-1}$. Since $\text{Out}(G)$ is finitely generated and there is an effective procedure that computes a finite generating set (see [DG11, Theorem 7.1]), and similarly for $\text{Out}(G, \mathcal{P})$ by Theorem 3.1, then we may simply go through all the words of $\text{Out}(G)$ of increasing length, testing at each step if every generator of $\text{Out}(G, \mathcal{P})$ conjugated by the word preserves every $[S'_i]$. This procedure terminates, since we assumed that such a $\Psi$ exists, and it returns a suitable outer automorphism.

Let us now state and prove the main theorem of this section.

**Theorem 3.4.** Let $m \ge 1$, let $\Phi \in \text{Out}(F_m)$ be of finite order and let $\varphi \in \Phi$. Given two tuples of conjugacy classes of tuples in $F_m \times_{\varphi} \langle t \rangle$, there is an effective procedure to determine if there exists $\Psi \in \text{Out}_{fo}(F_m \times_{\varphi} \langle t \rangle)$ sending one to the other.

*Proof.* First let us note that in the case of $m=1$, if we are dealing with $Z \times Z$, then the problem is solved by elementary linear algebra on $Z$, and if we are dealing with $Z \times Z$ it is trivial since, as stated above in Remark 2.2, $\text{Out}(Z \times Z)$ is finite.

Similarly, one may treat the case $m=2$ explicitly since only a few subcases arise. However the proof is significantly more technical than for $m=1$ and the following covers this case.

Now let $m \ge 2$. Let $\mathcal{P} = ([S_1], \dots, [S_N])$ and $\mathcal{P}' = ([S'_1], \dots, [S'_N])$ be tuples of classes of tuples in $F_m \times_{\varphi} \langle t \rangle$. We begin by checking that for all $i$, the tuples $S_i$ and $S'_i$ have the same length and the exponents in $t$ of each element in $S_i$ matches with the corresponding one in $S'_i$.

As before, we will consider the problem in $Q = F_m \times_{\varphi} \langle t \rangle / \mathcal{Z}(F_m \times_{\varphi} \langle t \rangle)$, then lift the result to $F_m \times_{\varphi} \langle t \rangle$. Recall that if $k$ is the order of $\Phi$ and $\varphi^k = \text{ad}_{f_0}^{-1}$, then $\mathcal{Z}(F_m \times_{\varphi} \langle t \rangle) = \langle t^k f_0 \rangle$. Recall also that $Q$ is virtually free, and therefore hyperbolic. Let $\bar{\mathcal{P}}$ and $\bar{\mathcal{P}}'$ denote the projected tuples in $Q$.

Note that for $\psi \in \text{Aut}_{fo}(F_m \times_{\varphi} \langle t \rangle)$, $\psi(t^k f_0)$ is central, so $\psi(t^k f_0) \in \langle t^k f_0 \rangle$. However $\psi(t^k f_0) \in t^k F_m$, so $\psi(t^k f_0) = t^k f_0$. So there is a natural bijection between $\text{Aut}_{fo}(F_m \times_{\varphi} \langle t \rangle)$ and automorphisms of $Q$ that preserve $F_m$, and send $\bar{t}$ into $\bar{t}F_m$. Therefore, the problem is equivalent to deciding whether there exists $\bar{\psi} \in \text{Aut}(Q)$, sending $\bar{\mathcal{P}}$ to $\bar{\mathcal{P}}'$, that preserves $F_m$, and sends $\bar{t}$ into $\bar{t}F_m$.

Since $Q$ is hyperbolic, by Theorem 3.1, there is an effective procedure to determine if there exists $\bar{\psi} \in \text{Aut}(Q)$ sending $\bar{\mathcal{P}}$ to $\bar{\mathcal{P}}'$. And by Remark 3.3, if such a $\bar{\psi}$ exists, it can be constructed effectively. This however does not entirely solve the problem. Indeed, $\bar{\psi}$ does not necessarily send $F_m$ to $F_m$ nor $\bar{t}$ into $\bar{t}F_m$. If it does not (which can be tested in finitely many steps, by looking at the images of the generators $x_1, \dots, x_m$ and $\bar{t}$), we modify $\bar{\psi}$. For this, we use the second part of Theorem 3.1. There is an effective procedure that gives a finite generating set of $\text{Aut}(Q, \bar{\mathcal{P}}')$, the subgroup of $\text{Aut}(Q)$ that preserves the
conjugacy classes in $\bar{P}'$. Therefore we may compose $\bar{\psi}$ by these automorphisms
without undoing our work. Furthermore, we may also obtain a finite generating
set of $\mathrm{Aut}(Q, \bar{P})$.

Now Aut($Q$) acts on the set of finite index subgroups of $Q$ of index $k$ by
permutation. Since there are only finitely many such subgroups, the group
of permutations is finite. Therefore, we may effectively compute the sub-
group of permutations to which Aut($Q, \bar{P}'$) maps by looking at its genera-
tors, and decide whether there exists an automorphism $\alpha \in \mathrm{Aut}(Q, \bar{P}')$ sending
$\langle \bar{\psi}(x_1), \dots, \bar{\psi}(x_m) \rangle$ to $F_m$. This also produces such an $\alpha$ by taking the word
in the generators of $\mathrm{Aut}(Q, \bar{P}')$ corresponding to a suitable permutation in the
associated subgroup of permutations.

We now need to deal with $\bar{t}$. Consider the stabilizer $\mathcal{S}$ of $F_m$ in the action
of $\mathrm{Aut}(Q, \bar{P}')$ on index $k$ subgroups of $Q$. By [DG11, Lemma 2.1] (itself a
consequence of [MKS04, Theorem 2.7, Corollary 2.7.1]) it is finitely generated
and there is an effective procedure that computes a generating set. Consider
also $\alpha \circ \bar{\psi}(\bar{t}) = \bar{t}^p c$ for some $p \in [0, k-1]$ and $c \in F_m$. Now, $\mathcal{S}$ acts on the finite
set $\{\bar{t}^l F_m \mid l \in [0, k-1]\}$, and as before, it is decidable whether there is an
automorphism $\beta \in \mathcal{S}$ sending $\bar{t}^p F_m$ to $\bar{t} F_m$. In which case $\beta \circ \alpha \circ \bar{\psi} \in \mathrm{Aut}(Q)$
sends $\bar{P}$ to $\bar{P}'$, $F_m$ to $F_m$, and $\bar{t}$ into $\bar{t} F_m$. This automorphism can then be lifted
to an automorphism of $F_m \times_{\varphi} \langle t \rangle$ that solves the problem. $\square$

4 Conjugacy problem

In this section we wish to prove Theorem 4.1, which states that the conjugacy
problem in Out($F_m$), for outer automorphisms whose polynomial parts are of fi-
nite order, is solvable. For this we will use the reduction introduced by Dahmani
and Touikan recalled in Theorem 1.6.

**Theorem 4.1.** Let $m \in \mathbb{N} \setminus \{0\}$ and let $F_m$ be the free group of rank $m$. Let $\mathcal{A}$ be the set of outer automorphisms $\Phi \in \mathrm{Out}(F_m)$ such that for $\varphi \in \Phi$, and for $P$ a polynomial subgroup for $\varphi$, there exist $k \in \mathbb{N}^*$ and $\gamma_P \in F_m$ such that $\mathrm{ad}_{\gamma_P} \circ \varphi^k$ restricts to identity on $P$ (ie. the polynomially growing parts of $\Phi$ are of finite order).

There is an effective procedure that decides whether any two outer automor-
phisms in $\mathcal{A}$ are conjugate in Out($F_m$).

*Proof.* We will prove that the class $\mathcal{A}$ verifies the hypotheses of Theorem 1.6.

Let $\mathcal{P}$ be the collection of sub-mapping tori constructed from the maximal
polynomial subgroups under each $\Phi \in \mathcal{A}$. Then by Remark 1.7, all groups in $\mathcal{P}$
are of the form $F_M \times_{\alpha} \langle t \rangle$, for some $M$ and some $[\alpha] \in \mathrm{Out}(F_M)$ of finite order.

First, let us note that, by Theorem 3.4, the fibre- and orientation-preserving
mixed Whitehead problem is solvable in every group of $\mathcal{P}$.

Let us check that the fibre- and orientation-preserving isomorphism prob-
lem in $\mathcal{P}$ is solvable. Let $\mathbb{T}_1, \mathbb{T}_2 \in \mathcal{P}$. If there exists a fibre- and orientation-
preserving isomorphism between $\mathbb{T}_1$ and $\mathbb{T}_2$, then the respective fibres are iso-
morphic, and therefore have the same rank, say $M$. Hence the underlying
automorphisms, $\alpha_1$ and $\alpha_2$, both belong to Aut($F_M$). Therefore, a fibre- and orientation-preserving isomorphism exists if and only if $\alpha_1$ and $\alpha_2$ are conjugate in Aut($F_M$). Now, $[\alpha_1]$ and $[\alpha_2]$ are of finite order in Out($F_M$), since they arise from elements of $\mathcal{A}$. Therefore, their conjugacy can be tested by checking whether they are realized by the same isometry, using Theorem 1.8 and Remark 1.9.

Let $\mathcal{P}'$ be the class of all groups in $\mathcal{P}$ and their finitely generated subgroups. If $H$ is in $\mathcal{P}'$, it is either a sub-mapping torus of the form $F_M \times_{\varphi} \langle t \rangle$, with $[\varphi]$ of finite order in Out($F_M$), or it is a finitely generated subgroup of such a sub-mapping torus in $\mathcal{P}$. Suppose $H$ is a finitely generated subgroup of some $F_M \times_{\varphi} \langle t \rangle$, with $[\varphi] \in \text{Out}(F_M)$ of finite order. Let $C$ denote the center of $F_M \times_{\varphi} \langle t \rangle$, and recall that $C$ is infinite cyclic by Proposition 1.10.

Let us first suppose that $H \cap C = \{1\}$. Then $H \hookrightarrow (F_M \times_{\varphi} \langle t \rangle)/C$, which is virtually free, as we have seen previously (Proposition 2.3). Now $H$ is torsion free, as it is a subgroup of $F_M \times_{\varphi} \langle t \rangle$, which is torsion free. Therefore $H$ is in fact free and finitely generated.

Let us now suppose that $C_0 \stackrel{\text{def}}{=} H \cap C$ is non-trivial. Then $C_0$ is infinite cyclic. Consider the following short exact sequence.

$$1 \to F_M \to F_M \times_{\varphi} \langle t \rangle \to \mathbb{Z} \to 1$$

And now consider its restriction to $H$.

$$1 \to F_M \cap H \to H \to \mathbb{Z} \to 1$$

Since the term on the right is free, there is a section from $\mathbb{Z}$ to $H$ and therefore $H$ is of the form $H = (F_M \cap H) \times_{\alpha} \mathbb{Z}$. We will now check that $F_M \cap H$ (which is free) is finitely generated. Consider $H_0 := H \cap (F_M \times C_0)$. Since $F_M \times C_0$ is of finite index in $F_M \times_{\varphi} \langle t \rangle$, $H_0$ is of finite index in $H$. Therefore, since $H$ is finitely generated, so is $H_0$. Now $C_0$ is normal in $H$ as it is central, so the image of $H_0$ by the quotient map modulo $C_0$ is $H_0 \cap F_M$. Being the image of a finitely generated group, $H_0 \cap F_M$ is finitely generated. And since $H_0$ is of finite index in $H$, $H_0 \cap F_M$ is of finite index in $H \cap F_M$, and thus $H \cap F_M$ is finitely generated and free, of rank say $M'$. Therefore, we have shown that $H \cong F_{M'} \times_{\alpha} \langle t \rangle$ and has a non-trivial center since $C_0 \subset \mathcal{Z}(H)$. So $[\alpha]$ is of finite order in Out($F_{M'}$), by Remark 1.11.

Hence, we have shown that all groups in $\mathcal{P}'$ are either free of finite rank, or of the form $F_{M'} \times_{\alpha} \langle t_{\alpha} \rangle$, for some $M'$ finite, and with $[\alpha] \in \text{Out}(F_{M'})$ of finite order. Therefore these groups all have congruences that effectively separate torsion, by [DT21, Corollary 3.2] for the former groups, and by Theorem 2.1 for the latter.

Finally, we must prove that $\mathcal{P}'$ is hereditarily algorithmically tractable. It is indeed closed under taking finitely generated subgroups, and is effectively coherent, since the finitely generated subgroups are finitely presentable given their description above, and a presentation is computable. Furthermore, the presentations of groups are recursively enumerable using Tietze transformations (see, for instance, [MKS04, Section 1.5]). Finally, the conjugacy problem is
solvable for groups in $\mathcal{P}'$ by [BMMV06, Theorem 1.1] and so is the generation problem. The class $\mathcal{P}'$ is therefore hereditarily algorithmically tractable.
We may therefore apply Theorem 1.6 to $\mathcal{A}$ and conclude. $\square$

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GABRIEL BARTLETT, UNIVERSITÉ GRENOBLE-ALPES, INSTITUT FOURIER,
38610 GIÈRES, FRANCE
email: gabriel.bartlett@univ-grenoble-alpes.fr