Proof. First, note that \(C_0\) is maximal abelian in \(A_0\) or \(B_0\). Indeed, if not, then taking \(a_0 \in A_0 \setminus C_0\) and \(b_0 \in B_0 \setminus C_0\) centralizing \(C_0\), since \(C_0 \neq 1\) and \(G\) is commutative transitive (as it is CSA), we have that \(a,b\) commute. However, this contradicts the normal form theorem, since there can be no relation between elements of \(A_0 \setminus C_0\) and elements of \(B_0 \setminus C_0\).

First, we will show that we can enlarge the splitting to a splitting \(G = A *_C B\) where \(C\) is malnormal in \(A\) and \(B\), and then we will show that in fact \(A = \mathrm{Fix}(\delta)\).

Without loss of generality, suppose that \(C_0\) is maximal abelian in \(B_0\) and let \(C_A\) be the centralizer of \(C_0\) in \(A_0\) (which is the maximal abelian subgroup of \(A_0\) containing \(C_0\)). We can assume that \(C_A \neq C_0\), since otherwise \(C_0\) is already malnormal in both \(A_0\) and \(B_0\) and we are done. Let \(A = A_0, B = \langle B_0, C_A \rangle\) and \(C = C_A\). We claim that \(G = A *_C B\), that \(C\) is maximal abelian in both \(A\) and \(B\) and \(\delta\) still acts as a separating Dehn twist of this splitting with respect to \(c_0\).

\(C\) is maximal abelian in \(A\): Since \(C\) is the centralizer of \(C_0\) in \(A\), we have that \(C\) is maximal abelian in \(A\).

\(C\) is maximal abelian in \(B\): Suppose \(g \in B\) is in the maximal abelian subgroup of \(B\) containing \(C\). Then \(g\) commutes with some \(c \in C \setminus C_0\). We can write \(g = b_1 \cdots a_n\) for some \(a_i \in C_A \setminus B_0 \subset A_0 \setminus B_0 = A_0 \setminus C_0\) and \(b_i \in B_0 \setminus C_A \subset B_0 \setminus C_0\), with all letters non-trivial except possibly \(b_1\) or \(a_n\). We have \(cgc^{-1}g^{-1} = 1\), which yields:

\[c (b_1 a_1 \cdots a_n) c^{-1} (b_1 a_1 \cdots b_n a_n)^{-1} = 1 \iff c b_1 a_1 \cdots b_n a_n c^{-1}a_n^{-1} b_n^{-1} \cdots a_1^{-1} b_1^{-1} = 1\]

Since \(a_n \in C\), this becomes:

\[c b_1 a_1 \cdots b_n c^{-1} b_n^{-1} \cdots a_1^{-1} b_1^{-1} = 1\]

Since \(c \in C \setminus C_0\), we have \(c \in A_0 \setminus C_0\). If \(n \geq 1\) and \(g \notin C\), then \(b_n \notin C_A\), so \(b_n \in B_0 \setminus C_0\), We thus have an alternating word in \(A_0 \setminus B_0\) and \(B_0 \setminus C_0\) being the trivial element, contradicting the normal form theorem for \(G = A_0 *_{C_0} B_0\). Therefore, \(n = 1\) and \(g = a_1 \in C_A\). We conclude that \(C\) is maximal abelian, and hence malnormal, in \(B\).

\(A \cap B = C\): Note that \(C \subseteq A,B\) and hence \(C \subseteq A \cap B\). Let \(g \in A \cap B\). Since \(g \in B = \langle B_0, C_A \rangle\), we can write \(g = a_1 b_1 \cdots a_n b_n\) for some \(a_i \in C_A \setminus B_0 \subset A_0 \setminus B_0 = A_0 \setminus C_0\) and \(b_i \in B_0 \setminus C_A \subset B_0 \setminus C_0\). Suppose for contradiction that \(g \notin C\), so there is at least one letter \(b_i\). We then have:

\[1 = a_1 b_1 \cdots a_n b_n g^{-1}\]

Since \(g \in A = A_0\), this is an alternating non-trivial word in \(A_0 \setminus C_0\) and \(B_0 \setminus C_0\), contradicting the normal form theorem for \(G = A_0 *_{C_0} B_0\). Therefore, we must have \(g \in C_A = C\).

\(G = A *_C B\): Since \(G = \langle A_0, B_0 \rangle\) and since \(A \geq A_0, B \geq B_0\) we have \(G = \langle A, B\rangle\). Since we showed above that \(A \cap B = C\), it remains to show the normal form theorem for \(A,B,C\). Suppose that:

\[c a_1 b_1 \cdots a_n b_n = 1\]

for \(c \in C, a_i \in A \setminus C, b_i \in B \setminus C\).

Note that \(A \setminus C \subset A_0 \setminus C_0\), so each \(a_i \in A_0 \setminus C_0\).

We can write each \(b_i\) as \(b_i = b_{i1}a_{i1} \cdots b_{im_i} a_{im_i}\) for \(b_{ij} \in B_0 \setminus C_A\) and \(a_{ij} \in C_A \setminus B_0\). Substituting these expressions above, we obtain:

\[c a_1 (b_{11}a_{11} \cdots b_{1m_1} a_{1m_1})a_2 \cdots a_n (b_{n1}a_{n1} \cdots b_{nm_n} a_{nm_n}) = 1\]

Note that \(B_0 \setminus C_A \subset B_0 \setminus C_0\) and \(C_A \setminus B_0 \subset A_0 \setminus C_0\), so each \(b_{ij} \in B_0 \setminus C_0\) and \(a_{ij} \in A_0 \setminus C_0\). Also, note that \(ca_1 \notin C_0\), since otherwise \(a_1 \in C\). Thus, we obtain a non-empty word in \(A_0 \setminus C_0\) and \(B_0 \setminus C_0\) equal to the identity in \(G\), contradicting the normal form theorem for \(G = A_0 *_{C_0} B_0\). We conclude that the word the \(c a_1 b_1 \cdots a_n b_n\) must be empty, and so the normal form theorem holds for \(A,B,C\) and thus that \(G = A *_C B\).

\(\delta\) still acts as a separating Dehn twist of \(G = A *_C B\): We still have \(\delta\) being the identity on \(A = A_0\). We need to show that \(\delta\) acts by conjugation via \(c_0\) on \(B\). We know this is true on \(B_0\). On \(C_A\), \(\delta\) fixes \(C_A\) pointwise since \(C_A \subset A\). Also, since \(C_A\) is abelian and contains \(c_0\), we have that \(\delta\) acts as conjugation by \(c_0\) on \(C_A\) (which is trivial). Therefore, \(\delta\) acts as conjugation by \(c_0\) on \(B\) as well. We conclude that \(\delta\) acts as a separating Dehn twist of \(G = A *_C B\).

Now we show that \(A = \text{Fix}(\delta)\). Indeed, by definition \(A \subseteq \text{Fix}(\delta)\). Let \(h \in \text{Fix}(\delta)\). As \(h \in A_0 *_{C_0} B_0\), it has a normal form, say, \(h = c_h a_1b_1 \cdots a_nb_n\) where \(c_h \in C_0\), each \(a_i \in A_0 \setminus C_0\) and each \(b_i \in B_0 \setminus C_0\). Now \[c_h a_1(b_1)^{c_0} \cdots a_n(b_n)^{c_0} = \delta (h) = h = c_h a_1b_1 \cdots a_nb_n \implies (b_1)^{c_0} \cdots a_n(b_n)^{c_0} b_n^{-1}a_n^{-1} \cdots b_1^{-1} = 1\]

Then there must be cancellations, in particular \((b_n)^{c_0} b_n^{-1} \in C_0\) (otherwise it will contradict the fact that every reduced alternating word in \(A_0 *_{C_0} B_0\) is non-trivial). This implies that \(b_n c_0^{-1} b_n^{-1} \in C_0 \implies C_0^{b_n} \cap C_0\neq 1\). But \(C_0\) is maximal abelian in \(B_0\), hence malnormal in \(B_0\). Thus, we obtain \(b_n \in C_0\). Then \((b_n)^{c_0} b_n^{-1} = 1\) will reduce \((b_1)^{c_0} \cdots a_n(b_n)^{c_0} b_n^{-1}a_n^{-1} \cdots b_1^{-1} = 1\) to \((b_1)^{c_0} \cdots a_{n-1}(b_{n-1})^{c_0} b_{n-1}^{-1}a_{n-1}^{-1} \cdots b_1^{-1} = 1\). By the same argument as before, we can show that \(b_{n-1} \in C_0\). Continuing the same way we get each \(b_i\) is in \(C_0\) implies hence \(h \in A\). Hence \(\text{Fix} (\delta) = A\).

We will next show that \(C\) is maximal abelian in \(G\). Let \(C_G\) be the maximal abelian subgroup of \(G\) containing \(C\). We will show that \(C_G \subseteq C\). Let \(g \in C_G\). Then \(g\) has a unique normal form, say, \(g = c' a_1b_1 \cdots a_n b_n\), for \(c' \in C_0\), \(a_i \in A_0 \setminus C_0\) and \(b_i \in B_0 \setminus C_0\). Since \(g \in C_G\), we have that \(g\) commutes with \(c_0\), and hence \(c_0 g c_0^{-1} = g\). This yields:

\[c' a_1^{c_0}b_1^{c_0} \cdots a_n^{c_0} b_n^{c_0} = c' a_1b_1 \cdots a_n b_n \implies a_1b_1 \cdots a_n b_n (b_n^{-1})^{c_0}(a_n^{-1})^{c_0} \cdots (b_1^{-1})^{c_0} (a_1^{-1})^{c_0} = 1\]

By using the same method for proving \(\text{Fix} (\delta) = A\) above (which boiled down to centralizers of \(C_0\) being maximal abelian and hence malnormal), we can deduce that each \(a_i\) must be in the centralizer \(C_A\) of \(C_0\) in \(A_0\) and each \(b_i\) must be in the centralizer of \(C_0\) in \(B_0\), which is \(C_0\) since \(C_0\) is maximal abelian in \(B_0\). This implies that \(g \in A_0\). This proves the claim.

Therefore, \(\text{Fix}(\delta) = A_0 = A\) and \(C = C_A = C_G\) is maximal abelian in \(G\) and hence maximal abelian in both \(A\) and \(B\). ◻

Proof. Consider a path \([gv, u]\cup[u, v]\cup[v, hu] = ge \cup e \cup he\) for any \(g \in \Gamma_u\setminus\Gamma_e\) and \(h \in \Gamma_v\setminus\Gamma_e\), where \(e = [u, v] = [\tilde{A},\tilde{B}]\) is the fundamental domain in the Bass-Serre tree \(T\). We will prove that the three edge stabilizers intersect trivially. Note that this will be sufficient to prove the \(2-\)acylindricity of the action of \(\Gamma_\delta\) on \(T\) since any geodesic segment of length 3 is a \(\Gamma\)-translate of the segment \(ge \cup e \cup he\). \(\mathrm{Stab}(ge) \cap \mathrm{Stab}(e) = g\mathrm{Stab}(e)g^{-1} \cap \mathrm{Stab}(e) = g(C \oplus \langle t \rangle)g^{-1} \cap (C \oplus \langle t \rangle) = (gCg^{-1} \cap C) \oplus \langle t \rangle = \langle t \rangle\), because \(C\) is malnormal in \(A\) and \(B\). Similarly, \(\mathrm{Stab}(he) \cap \mathrm{Stab}(e) = \langle t_1 \rangle\). As \(t_1 = c_0t\), \(\langle t_1 \rangle \cap \langle t \rangle = 1\). ◻

Proof. We will first construct a \(\Gamma\)-equivariant injection between the vertex sets \(V(T)\) and \(V(S)\) of \(T\) and \(S\).

We first show that \(\tilde{A}\) is elliptic on \(S\).

By Lemma 5, it follows that for each \(g \in \Gamma\), the stabilizer of the vertex \(g \tilde{A}\) (respectively, \(g \tilde{B}\)) fixes the unique vertex \(g v_{\tilde{A}}\) (respectively, \(g v_{\tilde{B}}\)) of \(S\). This yields a \(\Gamma\)-equivariant map \(f: V(T) \to V(S)\) defined by a sending a vertex \(v\) of \(T\) to the unique of \(S\) stabilized by \(\mathrm{Stab}_{\Gamma}(v)\)

We now show that \(f\) is injective. It suffices to show that the stabilizer \(\Gamma_{v_{\tilde{A}}}\) of \(v_{\tilde{A}}\) has unique fixed vertex \(\tilde{A}\). It will follow similarly that \(\Gamma_{v_{\tilde{B}}}\) fixes the unique vertex \(\tilde{B}\) of \(T\).

We have that the stabilizer \(\Gamma_{v_{\tilde{A}}} \cong H \rtimes \mathbb{Z}\) with \(H\) non-abelian. Indeed, we have \(\tilde{A} < \Gamma_{v_{\tilde{A}}}\) by Lemma 5, and \(A\) is a free product of free abelian groups with at least 2 free factors, hence \(A\) contains the free group \(F_2\), and thus so does \(\tilde{A}\). Therefore, \(H\) cannot be abelian, since otherwise \(H \rtimes \mathbb{Z}\) would be solvable and contain \(F_2\), a contradiction. Since \(H\) is non-abelian, we have that \(\Gamma_{v_{\tilde{A}}} \cong H \oplus \mathbb{Z}\), and so arguing as in Lemma 5, we obtain that \(\Gamma_{v_{\tilde{A}}}\) fixes a unique vertex of \(T\). This vertex must be \(\tilde{A}\), since otherwise \(\tilde{A}\) would fix more than one vertex, thus it would fix an edge of \(T\), and hence would be abelian. Similarly, it follows that \(gv_{\tilde{A}}\) fixes the unique vertex \(g\tilde{A}\) of \(T\), and similarly for \(g v_{\tilde{B}}\). This gives a \(\Gamma\)-equivariant left inverse \(h\) of \(f\) defined by \(h(gv_{\tilde{A}}) = \tilde{A}\) and \(h(gv_{\tilde{B}}) = \tilde{B}\). Therefore, \(f\) is injective, and hence a \(\Gamma\)-equivariant bijection onto its image.

We now show that \(f\) extends to a map of trees. To do this, we need to show that \([v_{\tilde{A}}, v_{\tilde{B}}]\) is an edge of \(S\).

Hence, \(f\) is an isomorphism of \(\Gamma\)-trees onto its image \(f(T)\). But since \(S\) is minimal, it follows that \(f(T) = S\). Hence, \(f\) is a \(\Gamma\)-equivariant isomorphism from \(T\) to \(S\). ◻

Proof. Suppose first that there exists such an isomorphism \(f\). Write \(F=f|_{G}\in \mathop{\mathrm{Aut}}(G)\), then for all \(\gamma \in G\), \[F^{-1} \circ \phi \circ F (\gamma) = F^{-1}(\phi(F(\gamma)) =f^{-1}(\phi(F(\gamma))= f^{-1}(tF(\gamma)t^{-1})= f^{-1}(t) \gamma f(t^{-1}) =s\gamma s^{-1} = \psi(\gamma)\]

Now suppose that \(\psi\) and \(\phi\) are conjugate in \(\mathop{\mathrm{Aut}}(G)\) by an automorphism \(F \in \mathop{\mathrm{Aut}}(G)\), i.e.that \(\psi = F \phi F^{-1}\). Define a map \(f: G \rtimes_{\phi} \langle t \rangle \to G \rtimes_{\psi} \langle s \rangle\) by putting \(f \vert_G = F\) and \(f(t) = s\). We claim that \(f\) extends to an isomorphism \(G \rtimes_{\phi} \langle t \rangle \to G \rtimes_{\psi} \langle s \rangle\). Indeed, to first check that \(f\) is a homomorphism we need to check that the relation

\[tgt^{-1} = \phi(g)\]

is preserved by \(f\), i.e.that \(f(t)f(g)f(t)^{-1} = f(\phi(g))\). We have:

\[\begin{align} f(t)f(g)f(t)^{-1} &= s F(g) s^{-1} \\ &= \psi(F(g)) \\ &= F(\phi(g)) \text{, since \psi = F\phi F^{-1}}\\ &= f(\phi(g)) \end{align}\] Therefore, \(f\) extends to a homomorphism.

We have that \(f\) is invertible since \(f^{-1}\) is given by \(f^{-1} \vert_G = F^{-1}\) and \(f^{-1}(s) = t\) (that \(f^{-1}\) is a homomorphism follows exactly as the argument to show that \(f\) is a homomorphism). Therefore, \(f\) is an isomorphism such that \(f(G) \subseteq G\) and \(f(t)=s\). ◻

Proof. Let \(\gamma\) be such that \(\tilde{A} \subseteq \tilde{A}^{\gamma}\). Then \(\tilde{A}\) fixes the vertices \(\tilde{A}\) and \(\gamma \tilde{A}\) in the Bass–Serre tree \(T\) of the induced splitting of \(\Gamma\). If \(\gamma \notin \tilde{A}\), then \(\gamma \tilde{A} \neq \tilde{A}\), and so \(d_T(\tilde{A}, \gamma \tilde{A}) \geq 2\). Then the geodesic path in \(T\) from \(\tilde{A}\) to \(\gamma \tilde{A}\) contains an edge of \(T\), which implies that \(\mathrm{Stab}_{\Gamma}([\tilde{A}, \gamma \tilde{A}])\) is abelian (since edge stabilizers of \(T\) are abelian). However, \(\mathrm{Stab}_{\Gamma}([\tilde{A}, \gamma \tilde{A}]) = \tilde{A} \cap \tilde{A}^{\gamma} = \tilde{A}\). Therefore, we obtain that \(\tilde{A}\) is abelian, a contradiction. Therefore, we must have that \(\gamma \in \tilde{A}\).

The proof for \(\tilde{B}\) is analogous, since \(\tilde{B}\) is non-abelian. ◻

Proof. As in 3.1, denote \(\tilde{A} = A \oplus \langle t \rangle\), \(\tilde{B} = B \oplus \langle t_1 \rangle\), \(\tilde{C} = C \oplus \langle t \rangle = C \oplus \langle t_1 \rangle\), etc.

Suppose first that \(\delta\) and \(\delta'\) are conjugate in \(\mathop{\mathrm{Aut}}(G)\). Then by Lemma 7, since \(\delta, \delta'\) are conjugate in \(\mathop{\mathrm{Aut}}(G)\), there exists an isomorphism \(f: \Gamma_{\delta} \to \Gamma_{\delta'}\) inducing an automorphism of \(G\) and mapping \(t\) to \(t'\). Let \(T\) and \(T'\) be the Bass-Serre trees corresponding to the given splittings of \(\Gamma_{\delta}\) and \(\Gamma_{\delta'}\).

The isomorphism \(f\) induces an action \(\cdot_f\) of \(\Gamma_{\delta}\) on \(T'\) via, for all \(g \in \Gamma_{\delta}\) and all \(x \in T'\), \(g \cdot_f x = f(g) \cdot x\), where the \(\cdot\) on the right hand side is the action of \(\Gamma_{\delta'}\) on \(T'\). Since \(T'\) is 2-acylindrical with respect to the \(\Gamma_{\delta'}\)-action, it is 2-acylindrical with respect to the \(\Gamma_{\delta}\)-action. The vertex stabilizers of \(T'\) with respect to the \(\Gamma_{\delta}\) action are isomorphic to \(f^{-1}(A' \oplus \mathbb{Z}) \cong f^{-1}(A') \oplus \mathbb{Z}\) or \(f^{-1}(B' \oplus \mathbb{Z}) \cong f^{-1}(B') \oplus \mathbb{Z}\). Since \(f\) is an isomorphism preserving \(G\), we have that \(f^{-1}(A') < G\) and \(f^{-1}(B') < G\) and both \(f^{-1}(A')\) and \(f^{-1}(B')\) are non-abelian since \(A',B'\) are non-abelian. Also, the edge stabilizers of \(T'\) with respect to the \(\Gamma_{\delta}\) action are of isomorphic to \(f^{-1}(C' \oplus \mathbb{Z})\), which is abelian. Therefore, \(T'\) satisfies the conditions of Lemma 4 as a \(\Gamma_{\delta}\)-tree and hence \(T\) and \(T'\) are isomorphic as \(\Gamma_{\delta}\)-trees.

By Lemma 5, we have that \(\tilde{A}\) fixes a unique vertex \(\gamma \tilde{A}'\) of \(T'\) for some \(\gamma \in \Gamma'\). We can assume that \(\gamma \in G\) since we can absorb the \(t'\)-component of \(\gamma\) into \(\tilde{A}'\). We then have \(f(\tilde{A}) \subseteq \mathrm{Stab}_{\Gamma'}(\gamma \tilde{A}') = (\tilde{A}')^{\gamma}\). Similarly, we have \(f^{-1}(\tilde{A}') \subseteq \tilde{A}^{\lambda}\) for some \(\lambda \in G\).

We will show that \(f(\tilde{A}) = (\tilde{A}')^{\gamma}\). From the inclusions \(f(\tilde{A}) \subseteq (\tilde{A}')^{\gamma}\) and \(f^{-1}(\tilde{A}') \subseteq \tilde{A}^{\lambda}\), we have \(f(\tilde{A}) \subseteq (\tilde{A}')^{\gamma} \subseteq f(\tilde{A}^{\lambda})^{\gamma} = f(\tilde{A}^{f^{-1}(\gamma)\lambda})\). Hence, we obtain \(\tilde{A} \subseteq \tilde{A}^{f^{-1}(\gamma) \lambda }\), which implies by Lemma 8 that \(f^{-1}(\gamma) \lambda \in \tilde{A}\) and hence that \(\tilde{A}^{f^{-1}(\gamma) \lambda } = \tilde{A}\). Therefore, we have \(f(\tilde{A}) \subseteq (\tilde{A}')^{\gamma} \subseteq f(\tilde{A}^{f^{-1}(\gamma)\lambda}) = f(\tilde{A})\). Thus, \(f(\tilde{A}) = (\tilde{A}')^{\gamma}\).

We similarly obtain that \(f(\tilde{B)} = (\tilde{B}')^{\eta}\) for some \(\eta \in G\). By Lemma 5, we obtain that \([\gamma \tilde{A}', \eta \tilde{B}']\) is an edge in \(T'\) (because the isomorphism \(T \to T'\) maps \(\tilde{A}\) to \(\gamma \tilde{A}\) and \(\tilde{B}\) to \(\eta \tilde{B}\), and \([\tilde{A}, \tilde{B}]\) is an edge in \(T\), hence \([\gamma \tilde{A}', \eta \tilde{B}']\) is an edge in \(T'\)). Thus, we have \(\eta \tilde{B}' = \alpha \gamma \tilde{B}'\) for some \(\alpha \in \mathrm{Stab}_{\Gamma'}(\gamma \tilde{A}') = (\tilde{A}')^{\gamma}\), so we can assume \(\eta = \alpha\gamma\), and in fact that \(\eta = \gamma\) by replacing \(\gamma\) with \(\alpha^{-1} \gamma\), since \((\tilde{A'})^{\alpha \gamma} = (\tilde{A}')^{\gamma}\). We then have \(f(\tilde{C}) = f(\tilde{A} \cap \tilde{B}) = f(\tilde{A}) \cap f(\tilde{B}) = (\tilde{A}')^{\gamma} \cap (\tilde{B}')^{\gamma} = (\tilde{C}')^{\gamma}\).

Let \(\phi = \mathrm{Ad}(\gamma^{-1})f\). Since \(f \vert_{\tilde{A}}\) is an isomorphism from \(\tilde{A}\) to \((\tilde{A}')^{\gamma}\), we have that \(\phi \vert_{\tilde{A}}\) is an isomorphism from \(\tilde{A}\) to \(\tilde{A}'\). Similarly, \(\phi \vert_{\tilde{B}}\) is an isomorphism from \(\tilde{B}\) to \(\tilde{B}'\) and \(\phi \vert_{\tilde{A}}(\tilde{C}) = \tilde{C}' = \phi \vert_{\tilde{B}}(\tilde{C})\). Furthermore, \(\phi \vert_{\tilde{A}}, \phi \vert_{\tilde{B}}\) agree on \(\tilde{C}\) since they are restrictions of the same map \(\phi\). Next, since \(\phi \vert_{\tilde{A}}\) is an isomorphism from \(\tilde{A}\) to \(\tilde{A}'\) and since \(\langle t \rangle = Z(\tilde{A})\) (respectively, \(\langle t' \rangle = Z(\tilde{A}')\)), we can assume up to changing the sign of \(\phi\) on \(\langle t \rangle\) that \(\phi(t) = t'\). Therefore, \(\phi\) is an isomorphism \(\Gamma \to \Gamma'\) satisfying \(\phi(G) = G\) and \(\phi(t) = t'\), which yields that \(\delta' = \delta^{\phi}\). We then have \(\phi(A) \subseteq \tilde{A}' \cap G = A'\) and similarly \(\phi^{-1}(A') \subseteq \tilde{A} \cap G = A\), so \(\phi(A) = A'\). Similarly, \(\phi(B) = B'\) and \(\phi(C)=C'\).

We now claim that \(\phi(c_0) = c_0'\). Indeed, since \(\delta' = \delta^{\phi}\), we have that \(\delta'\) acts on \(\phi(B) = B'\) as conjugation by \(\phi(c_0)\). Also, by definition, \(\delta'\) acts on \(B'\) as conjugation by \(c_0'\). Thus, \((c_0')^{-1}\phi(c_0) \in C_G(B') = 1\) since \(B'\) is a non-abelian subgroup of \(G\). Thus, we obtain \(\phi(c_0) = c_0'\).

Now let \(\phi_A = \phi \vert_A, \phi_B = \phi \vert_B\). We then have that \(\phi_A : A \to A'\) and \(\phi_B : B \to B'\) are isomorphisms that map \(C\) to \(C'\), \(c_0\) to \(c_0'\) and agree on \(C\), and hence are the required isomorphisms.

Conversely, suppose there exist such an isomorphisms \(\phi_A,\phi_B\). Since \(\phi_A, \phi_B\) agree on \(C\) and map \(C\) to \(C'\), the free product isomorphism \(\phi_A * \phi_B\) descends to an isomorphism \(\Phi : G \to G\) preserving the splittings \(G = A *_C B = A' *_{C'} B'\). Since \(\phi_A(c_0) = \phi_B(c_0) = c_0'\), mapping \(t\) to \(t'\) extends \(\Phi\) to an isomorphism \(\tilde{\Phi} : \Gamma_{\delta} \to \Gamma_{\delta'}\). By Lemma 7, we have that \(\delta\) and \(\delta'\) are conjugate in \(\mathop{\mathrm{Aut}}(G)\). ◻

Proof. As shown in the beginning of Section 3.1, given that \(\delta\) is a Dehn twist, there exists a splitting of its mapping torus as an amalgamated free product over a free abelian subgroup:

\[\Gamma_{\delta} = (A \oplus \langle t \rangle) *_{C \oplus \langle t \rangle = C \oplus \langle t_1 \rangle} (B \oplus \langle t_1 \rangle)\]

Thus, starting from the standard presentation \(\Gamma_{\delta} = \langle S, t \vert R, tst^{-1} = \delta(s), \text{ for each } s \in S \rangle\), there is a finite sequence of Tietze transformations that will take this given presentation to the presentation associated with the above splitting of \(\Gamma_{\delta}\) as an amalgamated free product.

Thus, we obtain an algorithm to compute the splitting as follows. We enumerate all presentations of \(\Gamma_{\delta}\) obtained from the standard presentation by applying finitely many Tietze transformation, checking at each step whether the current presentation is equal to the presentation associated with the above splitting of \(\Gamma_{\delta}\). Since \(\Gamma_{\delta}\) has a presentation associated with the above splitting (since we know that \(\delta\) is a Dehn twist), the algorithm will terminate at this presentation in finite time. ◻

Proof. Consider a geodesic segment of length 3 in the associated Bass–Serre tree: \((gD, D, hD)\) (it suffices to consider this particular segment because the action of \(G\) on edges in the Bass–Serre tree is transitive). We have to show that \(D^g \cap D \cap D^h = 1\). We either have that \(g,h\) are rotations about the vertices \(e^{-1}A, A\), respectively or \(g,h\) are translations respectively taking the vertices \(A\) to \(e^{-1}A\) and \(e^{-1}A\) to \(A\), so that \(h \in eA \setminus D\) and \(g \in e^{-1}A \setminus D\).

First, suppose that \(g,h\) are both elliptic. Then \(g \in A^{e^{-1}} = \langle a^{e^{-1}}, d \rangle * \langle d^{e^{-1}} \rangle\) and \(h \in A\). We then have \(D \cap D^g\) is either equal to \(D\) or is trivial, being non-trivial if and only if \(g \in \langle a^{e^{-1}}, d \rangle\). Similarly, \(D \cap D^h\) is either \(D\) or is trivial, being non-trivial if and only if \(h \in D\) since \(D\) is malnormal in \(A\). Therefore, \(D^g \cap D \cap D^h \neq 1\) if and only if \(g \in \langle a^{e^{-1}}, d \rangle\) and \(h \in D\). But then since \(h \in D\), we have \(hD = D\), so the segment \((gD, D, hD)\) has length at most 2, a contradiction. Therefore, we can assume that at least one of \(g,h\) is hyperbolic.

However, if \(k \in G\) is hyperbolic, then if \(D^k \cap D \neq 1\), letting \(B\) denote the maximal abelian subgroup of \(G\) containing \(D\), we would obtain \(B^k \cap B \neq 1\), which implies that \(k \in B\) since \(B\) is malnormal (as \(G\) is CSA). In this case, we have that \(B = \langle a,b \rangle^{e^{-1}} < A^{e^{-1}}\), so \(k \in A^{e^{-1}}\) and hence is elliptic, a contradiction. Therefore, if \(g\) or \(h\) is hyperbolic, we must have \(D^g \cap D = 1\) or \(D^h \cap D = 1\).

Thus, in all cases, we obtain that \(D^g \cap D \cap D^h = 1\), so the splitting is 2-acylindrical. ◻

Proof. Denote \(\tilde{A} = A \oplus \langle t \rangle\) and \(\tilde{D} = D \oplus \langle t \rangle\). Consider \(g = e^{-1}ae\) and \(h = a\). Then \(g \in \mathrm{Stab}_{\Gamma}(e^{-1}\tilde{A}) \setminus \mathrm{Stab}_{\Gamma}(\tilde{D})\) and \(h \in \mathrm{Stab}_{\Gamma}(\tilde{A}) \setminus \mathrm{Stab}_{\Gamma}(\tilde{D})\), so that \((g\tilde{D}, \tilde{D}, h\tilde{D})\) is a geodesic edge path of length 3 in the associated Bass–Serre tree. We have:

\[\begin{align} \tilde{D} \cap \tilde{D}^g &= (D \oplus \langle t \rangle) \cap (D^g \oplus \langle t^g \rangle) \\ &= (D \oplus \langle t \rangle) \cap (D^g \oplus \langle t^{e^{-1}ae} \rangle) \\ &= (D \oplus \langle t \rangle) \cap (D^g \oplus \langle (tb^{-1})^{e^{-1}a} \rangle) \text{, since t^e = tw_2^{-1} = tb^{-1}}\\ &= (D \oplus \langle t \rangle) \cap (D^g \oplus \langle t^{e^{-1}a}(b^{-1})^{e^{-1}a} \rangle) \\ &=(D \oplus \langle t \rangle) \cap (D^g \oplus \langle (td)d^{-1}\rangle) \text{, since t^{e^{-1}} = tw_1 = td and b^{e^{-1}} = d} \\ &= (D \oplus \langle t \rangle) \cap (D^g \oplus \langle t\rangle) \\ &\supseteq \langle t \rangle \end{align}\]

Similarly, we have \(\langle t \rangle \cap \langle t \rangle^h = \langle t \rangle \cap \langle t \rangle^a = \langle t \rangle\) since \(t\) centralizes \(A\). Thus, \(\tilde{D} \cap \tilde{D}^h \supseteq \langle t \rangle\). We conclude that \(\tilde{D}^g \cap \tilde{D} \cap \tilde{D}^h \supseteq \langle t \rangle \neq 1\), so the above splitting of \(\Gamma\) is not 2-acylindrical.

In fact, the splitting is not \(k\)-acylindrical for any \(k\). Indeed, consider the sequence of group elements \(g_1 = a^{e^{-1}}, g_2 = a^{e^{-1}}a, g_3 = a^{e^{-1}}aa^{e^{-1}}, \ldots\) and \(h_1 = a, h_2 = aa^{e^{-1}}, h_3=aa^{e^{-1}}a, \ldots\) (the \(g_i,h_i\) are an alternating product of \(a\) and \(a^{e^{-1}}\) where we alternate multiplying on the right by \(a\) and \(a^{e^{-1}}\)). Then the sequence of edges \(\gamma = (\ldots,g_2 \tilde{D}, g_1 \tilde{D}, \tilde{D}, h_1 \tilde{D}, h_2 \tilde{D}, \ldots)\) is a geodesic. Indeed, it’s enough to prove that \((\tilde{D}, h_1 \tilde{D}, h_2 \tilde{D}, \ldots)\) is a geodesic, since \((\ldots,g_2 \tilde{D}, g_1 \tilde{D}, \tilde{D}) = \tilde{D} \cup a^{e^{-1}}(\tilde{D}, h_1 \tilde{D}, h_2 \tilde{D}, \ldots)\), and \(\tilde{D}\) and \(a^{e^{-1}}(\tilde{D}, h_1 \tilde{D}, h_2 \tilde{D}, \ldots)\) meet at a vertex, so their concatenation is a geodesic. We proceed by induction on \(n\).

For \(n=1\), since \(a \notin \tilde{D}\), we have that \((\tilde{D}, h_1\tilde{D}) = (\tilde{D}, a\tilde{D})\) is a geodesic.

Suppose that \((\tilde{D}, h_1 \tilde{D}, h_2 \tilde{D}, \ldots, h_n \tilde{D})\) is a geodesic. Then \((h_n \tilde{D}, h_{n+1} \tilde{D}) = h_n(\tilde{D}, a\tilde{D})\) or \((h_n \tilde{D}, h_{n+1} \tilde{D}) = h_n(\tilde{D}, a^{e^{-1}}\tilde{D})\), which are geodesics in each case. Since \((\tilde{D}, h_1 \tilde{D}, h_2 \tilde{D}, \ldots, h_{n+1} \tilde{D}) = (\tilde{D}, h_1 \tilde{D}, h_2 \tilde{D}, \ldots, h_n \tilde{D}) \cup (h_n \tilde{D}, h_{n+1}\tilde{D})\) and since \((\tilde{D}, h_1 \tilde{D}, h_2 \tilde{D}, \ldots, h_n \tilde{D}), (h_n \tilde{D}, h_{n+1}\tilde{D})\) meet at a single vertex (since we do not have \(h_{n+1}\tilde{D} = h_{n-1} \tilde{D}\), as otherwise \(a a^{e^{-1}}\tilde{D} = \tilde{D}\) or \(a^{e^{-1}} a \tilde{D} = \tilde{D}\), a contradiction since neither \(aa^{e^{-1}}\) nor \(a^{e^{-1}}a\) are in \(\tilde{D}\)), we have that \((\tilde{D}, h_1 \tilde{D}, h_2 \tilde{D}, \ldots, h_{n+1} \tilde{D})\) is a geodesic.

However, since \(a\) and \(a^{e^{-1}}\) both commute with \(t\), we have that \(t \in P\mathrm{Stab}_{\Gamma}(\gamma)\). Hence, the splitting is not \(k\)-acylindrical for any \(k\). ◻

Proof. Let \(b \in B\). We first show that there exist \(b_1,\ldots,b_n \in B\) such that either \(b_i \in A \cup Ae\) for all \(i\) or \(b_i \in A^{e^{-1}} \cup e^{-1}A\) for all \(i\) and \(b = b_1 b_2 \cdots b_n\).

In the tree \(T_G\), connect the edges \(D\) and \(bD\) by a geodesic path \(\gamma\) (where \(D\) here refers to the edge in the fundamental domain of \(T_G\) having vertices \(e^{-1}A\) and \(A\)). Write \(\gamma = (g_0D = D, g_1D, \ldots, g_nD = bD)\) (\(g_0 = 1\) and \(g_n = b\)). Since the extremal edges of \(\gamma\) are \(D\) and \(bD\), we have that \(P\mathrm{Stab}(\gamma) = D \cap D^b = D\) since \(b \in B\). Thus, for all \(i\), we have \(D \cap D^{g_i} = P\mathrm{Stab}([D, g_i D]) \geq P\mathrm{Stab}(\gamma) = D\), so that \(g_i \in B\) for all \(i\). Set \(b_i = g_{i-1}^{-1}g_{i}\) for all \(i = 1,\ldots,n\). Note that \(b_i \in B\) for all \(i\) since \(g_i \in B\) for all \(i\). Since \(g_{i-1}D\) and \(g_iD\) are adjacent edges in \(T_G\), we have that \(g_{i-1}^{-1}g_iD\) is adjacent to \(D\). Therefore, either \(b_i = g_{i-1}^{-1}g_i \in A \cup Ae\) or \(b_i = g_{i-1}^{-1}g_i \in A^{e^{-1}} \cup e^{-1}A\). Since the edges \(g_iD\) are at increasing distance from \(D\) and since \(\gamma\) is a geodesic edge path, we must have either \(b_i = g_{i-1}^{-1}g_i \in A \cup Ae\) or \(b_i = g_{i-1}^{-1}g_i \in A^{e^{-1}} \cup e^{-1}A\) for all \(i\).

Given such \(b_i\), we have \(\delta(b_i) = b_i d_i\) for \(d_i \in \{1, w_1, w_1^{-1}\}\), since if \(b_i \in A \cup Ae\), then \(\delta(b_i) = b_i\) (if \(b_i \in A\)) or \(\delta(b_i) = b_i w_1\) (if \(b_i \in Ae\)), and if \(b_i \in A^{e^{-1}} \cup e^{-1}A\), then \(\delta(b_i) = b_i\) (if \(b_i \in A^{e^{-1}}\)) or \(\delta(b_i) = b_i w_1^{-1}\) (if \(b_i \in e^{-1}A\)). Therefore, if \(b \in B\), then we have that \(\delta(b) = b w_1^m\) for some \(m \in \mathbb{Z}\). ◻

Proof of Theorem 18. Since edge stabilizers of \(T\) are abelian, it is clear that \(\sim\) is reflexive and symmetric, so it remains to prove that it is transitive. Recall that edge stabilizers of \(T\) are of the form \(\tilde{D}^g = D^g \oplus \langle t^g \rangle\) for \(g \in \Gamma\). Note that \(\sim\) descends to a conjugation invariant relation on edge stabilizers of \(T\), so it suffices to prove that if \(\langle \tilde{D}^g, \tilde{D}\rangle\) and \(\langle \tilde{D}^h, \tilde{D}\rangle\) are abelian then so is \(\langle \tilde{D}^{g^{-1}h}, \tilde{D}\rangle\).

We begin by finding a necessary and sufficient condition on \(g\) to ensure that \(\langle \tilde{D}^g, \tilde{D}\rangle\) is abelian.

For \(\langle \tilde{D}^g, \tilde{D}\rangle\) to be abelian, we need that each element of \(D^g\) as well as \(t^g\) commute with each element of \(\tilde{D}\). Writing \(g = g_0 t^n\) for some \(n \in \mathbb{Z}\) and \(g_0 \in G\), we see that \(\tilde{D}^g = \tilde{D}^{g_0}\), so we can assume that \(g \in G\).

For \(D^g\) to centralize \(\tilde{D}\), we need \(D^g\) to centralize \(D\) and \(t\). We have that \(D^g\) centralizes \(D\) if and only if \(g \in B\), since if \(D^g\) centralizes \(D\), then for any \(d \in D \setminus 1\) we have \(d^g \in B\) and so \(1 \neq d \in B \cap B^{g^{-1}}\), giving that \(g \in B\) by malnormality of \(B\) in \(G\) (since \(G\) is CSA). Conversely, if \(g \in B\), then \(D^g = D\) which centralizes \(D\) and \(t\). Therefore, \(D^g\) centralizes \(\tilde{D}\) if and only if \(g \in B\). We now check that for each \(g \in B\), \(t^g\) also centralizes \(\tilde{D}\). We have \(t^g = gtg^{-1} = g(g^{-1})^t t\). If \(g \in B\), by Lemma 13, we have that \(g^t = \delta(g) = g w_1^n\) for some \(n \in \mathbb{Z}\). Thus, \(g (g^{-1})^t = g (gw_1^n)^{-1} = w_1^{-n}\), so that \(t^g = w_1^{-n} t\), which centralizes both \(D\) and \(t\).

Therefore, for \(g \in G\), we have that \(\tilde{D}^g\) centralizes \(\tilde{D}\) if and only if \(g \in B\). Thus, for \(g \in \Gamma\), \(\tilde{D}^g\) centralizes \(\tilde{D}\) if and only if \(g \in \tilde{B} = B \rtimes \langle t \rangle\) (this shows, in fact, that \(\langle \tilde{D}^g, \tilde{D} \rangle\) being abelian implies that \(\tilde{D}^g=\tilde{D}\)).

If \(\tilde{D}^g\) and \(\tilde{D}^h\) both centralize \(\tilde{D}\), then \(g,h \in \tilde{B}\) and hence \(g^{-1}h \in \tilde{B}\), therefore, \(\tilde{D}^{g^{-1}h}\) centralizes \(\tilde{D}\) and so \(\langle \tilde{D}^{g^{-1}h}, \tilde{D} \rangle\) is abelian. We conclude that \(\sim\) is an equivalence relation. ◻

Proof. By Lemma 13, for each \(b \in B\) we have that \(\delta(b) = bw^m\) for some \(m \in \mathbb{Z}\). Thus, for \(n \neq 0\), we have \(\delta^n(b) = bw^{mn}\). If \(\delta^n(b) = b\), then \(bw^{mn} = b\), which implies \(w^{mn} = 1\), so that \(m=n=0\) since \(w\) has infinite order. Thus, \(\delta(b) = b\). ◻

Proof. Let \(G = {A_0}*_{D_0^e=C_0}\) be some HNN extension on which \(\delta\) acts as a non-separating Dehn twist (i.e.such that \(\delta\vert_{A_0} = \mathrm{id}\), \(D_0\) is free abelian and \(\delta(e) = ew_1\) for some \(w_1 \in D\)). Denote \(A = \mathrm{Fix}(\delta)\) and let \(D\) (respectively, \(C\)) be the maximal abelian subgroup of \(D_0\) (respectively, \(C_0\)) in \(A\). Denote \(w_2 = w_1^e\). We will prove that \(G = A*_{D^e=C}\).

First, note that \(D^e < A\) since if \(d \in D\), then \(\delta(ede^{-1}) = ew_1 d w_1^{-1}e^{-1} = ede^{-1}\) since \(d\) commutes with \(w_1\). Therefore, since \(C_0 = D_0^e\) in \(A_0\), we have that \(C = D^e\) also holds in \(A\).

Consider the graph \(\mathcal{G}\) with vertices left cosets \(gA\) of \(A\) in \(G\) and an edge joining the cosets \(ge^{-1}A\) and \(gA\) for each \(g \in G\). We will prove that \(\mathcal{G}\) is a tree. This will then imply that \(G = A*_{D^e=C}\) since \(G\) acts by isometries on \(\mathcal{G}\) with one orbit of vertices and one orbit of edges, and the stabilizer of the vertex \(A\) is \(A\) while the stabilizer of the fundamental edge \([e^{-1}A, A]\) is \(A^{e^{-1}} \cap A\). Note that \(A^{e^{-1}} \cap A = D\) since \(D < A^{e^{-1}} \cap A\) by above, and conversely if \(a \in A\) with \(a^e \in A\) then \(eae^{-1}=\delta(eae^{-1}) = ew_1 a w_1^{-1}e^{-1}\), so \([a,w_1]=1\) and hence \(w_1 \in D^a \cap D\), so that \(a \in D\) by malnormality of \(D\) in \(A\) (since \(D\) is maximal abelian in \(A\) and \(A\) is CSA).

Let \(T\) be the Bass–Serre tree of the splitting \(G = {A_0}*_{D_0^e=C_0}\). Then in \(T\) we have induced subgraphs corresponding to the cosets of \(A\) in \(G\), since \(A_0 < A\).

Note that the subgraphs of \(T\) corresponding to the cosets \(hA\) are convex, hence subtrees of \(T\). Indeed, note that \(\delta\) is an isometry of \(T\) since given an edge \([gA_0, ghe^{\pm 1}A_0]\) for \(h \in A_0\) and \(g \in G\), applying \(\delta\) to the endpoints of this edge yields the vertices \(\delta(gA_0) = \delta(g)A_0\) and

\[\delta(gheA_0) = \delta(g) \delta(h) ew_1 A_0=\delta(g) \delta(h) e A_0\] or \[\delta(ghe^{-1}A_0) = \delta(g) \delta(h) w_1^{-1}e^{-1} A_0=\delta(g) \delta(h) e^{-1}w_2^{-1} A_0=\delta(g) \delta(h) e^{-1} A_0\], and \(\delta(h) \in A_0\) so that \([\delta(gA_0), \delta(ghe^{\pm1}A_0)]\) is an edge. Then if \(gA_0 \subseteq A\), we have \(\delta([A_0, gA_0]) = [A_0, gA_0]\), so that \([A_0, gA_0] \subset A\), so that \(A\) (and hence also coset subgraphs \(hA\) in \(T\)) is convex.

We have that the cosets \(A\) and \(hA\) are adjacent in \(\mathcal{G}\) if and only if the subgraphs \(A\) and \(hA\) are at distance 1 in \(T\) (i.e.are joined by an edge in \(T\)). Indeed, suppose that \(A\) and \(hA\) are adjacent in \(\mathcal{G}\), so that \(hA = g e^{\pm1}A\) for some \(g \in A\). Then there exists \(a \in A\) and \(g \in A\) such that \(h = g e^{\pm 1} a\). We can decompose \(a = a_0 g_0\) for some \(a_0 \in A_0\) and \(g_0 \in A\). Then we obtain \(h = g e^{\pm 1} a_0 g_0\), so that \(h g_0^{-1} = g e^{\pm 1}a_0\), so that \(hg_0^{-1}A_0 = g e^{\pm 1}A_0\), so that the cosets \(gA_0 \subset A\) and \(hg_0 A_0 \subset hA\) are adjacent in \(T\). Conversely, if \(A\) and \(hA\) are adjacent in \(T\), then there exists \(g_1,g_2 \in A\) such that \(g_1 A_0\) and \(hg_2 A_0\) are adjacent in \(T\), so that there exists \(g \in G\) such that \(hg_2A_0 = g_1 ge^{\pm 1}A_0\), so that \(hA = g_1 g e^{\pm 1}A\), so that \(hA\) is adjacent to \(g_1A=A\) in \(\mathcal{G}\).

We then have that \(\mathcal{G}\) is connected since given a coset \(hA\), we can join any pair of minimal distance vertices of the subgraphs \(A\) and \(hA\) by a geodesic \(\gamma\) in \(T\) and then cover the vertices of \(\gamma\) by pairwise distinct cosets \(h_1 A, \ldots, h_nA\) with \(h_1 A=A\) and \(h_nA=hA\). Since the cosets \(h_i A\) cover the vertices of \(\gamma\), we have that minimal distance cosets \(h_iA\) and \(h_jA\) are joined by an edge, hence are adjacent in \(\mathcal{G}\). Thus, \(h_1A, \ldots, h_n A\) is an edge path in \(\mathcal{G}\) from \(A\) to \(hA\).

We have that \(\mathcal{G}\) is simply connected since if \(A=h_1 A, \ldots, h_nA=A\) is a simple closed path in \(\mathcal{G}\), then connecting the minimal distance vertices \(v_i, v_{i+1}\) of cosets \(h_iA, h_{i+1}A\) by edges in \(T\) and connecting \(v_{i-1}, v_i\) by the geodesic \([v_{i-1},v_i] \subset h_iA\), since \(h_i A\) is convex, we obtain a simple closed path in \(T\), contradicting that \(T\) is a tree.

Therefore, \(\mathcal{G}\) is a tree, and we conclude that \(G = A*_{D^e=C}\). ◻

Proof. Recall that the tree of cylinders \(T_C\) of the associated splitting \(\Gamma = \tilde{A}_{\tilde{D}^e = \tilde{C}}\) has two orbits of vertices (the orbit of the white vertex \(\tilde{A}\) and the orbit of the black vertex corresponding to the cylinder containing \(\tilde{D}\), which we identify with its stabilizer \(\tilde{B} = B \rtimes \langle t \rangle\), where \(B\) is the maximal abelian subgroup of \(G\) containing \(D\)), and at most two orbits of edges (the orbits of the edges \(E=[e^{-1}\tilde{A}, \tilde{B}]\) and \(F=[\tilde{B}, \tilde{A}]\)).

Note that \(B \cap A^{e^{-1}} = D = B \cap A\). Indeed, since \(\delta\) fixes \(B \cap A^{e^{-1}}\) pointwise, as if \(e^{-1}ae \in B\) for \(a \in A\) then \(\delta(e^{-1}ae) = w_1^{-1}e^{-1}aew_1 = e^{-1}ae\) since \(e^{-1}ae \in B\). Therefore, since \(A = \mathrm{Fix}(\delta)\), we have \(B \cap A^{e^{-1}} < A\). Thus, \(B \cap A^{e^{-1}} < A \cap A^{e^{-1}}=D\). Conversely, \(D < B \cap A^{e^{-1}}\), so we obtain \(B \cap A^{e^{-1}} = D\). Similarly, if \(a \in B \cap A\), then \(\delta(eae^{-1}) = ew_1 a w_1^{-1}e^{-1} = eae^{-1}\), so \((B \cap A)^{e} < A\), which implies \(B \cap A < A^{e^{-1}}\), which yields \(B \cap A < A \cap A^{e^{-1}}=D\), so that \(B \cap A = D\) (as conversely \(D < B\) and \(D < A\)). Therefore, the stabilizers of the edges \(E\) and \(F\) are respectively, \(\tilde{B} \cap \tilde{A}^{e^{-1}} = \tilde{D}\) and \(\tilde{B} \cap \tilde{A} = \tilde{D}\).

Now if \(\gamma = (E_1,E_2,E_3)\) is an edge path of length 3 in \(T_C\), then the middle edge of \(\gamma\) is either in the orbit of \(E\) or in the orbit of \(F\). Therefore, we can assume that the middle edge \(E_2\) of \(\gamma\) is \(E\) or \(F\).

First, suppose the middle edge of \(\gamma\) is \(E\). We consider the following cases:

1. \(E_1 = gE\) and \(E_3 = hE\) for some \(g,h \in \Gamma\). Then \(g \in \tilde{A}^{e^{-1}} \setminus \tilde{D} = \tilde{A}^{e^{-1}} \setminus \tilde{B}\), so that \(\tilde{D}^g \cap \tilde{D} = \langle t^g \rangle \cap \langle t \rangle\). Also, \(h \in \tilde{B} \setminus \tilde{D}\), so that \(\tilde{D}^h \cap \tilde{D} = D \oplus (\langle t^h \rangle \cap \langle t \rangle)\). In order for \(\langle t^h \rangle \cap \langle t \rangle \neq 1\), we need \([h,t^n] = 1\) for some \(n \neq 0\), i.e.\(h \in \mathrm{Fix}(\delta^n) \oplus \langle t \rangle = \mathrm{Fix}(\delta) \oplus \langle t \rangle\) by Lemma 14. Thus, we need \(h \in \tilde{A}\) for \(\langle t^h \rangle \cap \langle t \rangle \neq 1\). But \(h \notin \tilde{A}\) since otherwise \(h \in \tilde{B} \cap \tilde{A} = \tilde{D}\). Thus, \(\langle t \rangle^h \cap \langle t \rangle = 1\). Thus, we obtain \(\tilde{D}^g \cap \tilde{D} \leq \langle t \rangle\) while \(\tilde{D}^h \cap \tilde{D} \leq D\), so that \(\mathrm{Stab}(\gamma) = \tilde{D}^g \cap \tilde{D} \cap \tilde{D}^h = 1\).

2. \(E_1 = gE\) and \(E_3 = hF\). As above, we obtain \(\mathrm{Stab}(E_1) \cap \mathrm{Stab}(E_2) \leq \langle t \rangle\). On the other hand, \(h \in \tilde{B} \setminus \tilde{A}\), so that \(\langle t \rangle^h \cap \langle t \rangle = 1\), so that \(\mathrm{Stab}(E_2) \cap \mathrm{Stab}(E_3) \leq D\). Thus, \(\mathrm{Stab}(\gamma) = 1\).

3. \(E_1 = gF\) and \(E_3 = hE\). As above, we obtain \(\mathrm{Stab}(E_3) \cap \mathrm{Stab}(E_2) \leq D\). We have \(g \notin \tilde{B}\) since \(E_1\) does not contain the vertex \(\tilde{B}\). Hence, \(\mathrm{Stab}(E_1) \cap \mathrm{Stab}(E_2) = \tilde{D}^g \cap \tilde{D} \leq \langle t \rangle\). Thus, \(\mathrm{Stab}(\gamma) = 1\)

4. \(E_1 = gF\) and \(E_3 = hF\). Then \(\mathrm{Stab}(E_1) \cap \mathrm{Stab}(E_2) \leq \langle t \rangle\) while \(\mathrm{Stab}(E_3) \cap \mathrm{Stab}(E_2) \leq D\). Thus, \(\mathrm{Stab}(\gamma) = 1\).

Now suppose the middle edge of \(\gamma\) is \(F\). We consider the following cases:

1. \(E_1 = gF\) and \(E_3 = hF\) for some \(g,h \in \Gamma\). Then as above \(\mathrm{Stab}(E_1) \cap \mathrm{Stab}(E_2) \leq D\) while \(\mathrm{Stab}(E_3) \cap \mathrm{Stab}(E_2) \leq \langle t \rangle\), so that \(\mathrm{Stab}(\gamma) = 1\).

2. \(E_1 = gF\) and \(E_3 = hE\). Then \(\mathrm{Stab}(E_1) \cap \mathrm{Stab}(E_2) \leq D\) while \(\mathrm{Stab}(E_3) \cap \mathrm{Stab}(E_2) \leq \langle t \rangle\) since \(h \notin \tilde{B}\). Thus, \(\mathrm{Stab}(\gamma) = 1\).

3. \(E_1 = gE\) and \(E_3 = hF\). Then \(\mathrm{Stab}(E_1) \cap \mathrm{Stab}(E_2) \leq D\) as \(g \in \tilde{B} \setminus \tilde{A}\) while \(\mathrm{Stab}(E_3) \cap \mathrm{Stab}(E_2) \leq \langle t \rangle\) as \(h \in \tilde{A} \setminus \tilde{B}\). Thus, \(\mathrm{Stab}(\gamma) = 1\)

4. \(E_1 = gE\) and \(E_3 = hE\). Then again \(\mathrm{Stab}(E_1) \cap \mathrm{Stab}(E_2) \leq D\) since \(g \in \tilde{B} \setminus \tilde{A}\) and \(\mathrm{Stab}(E_3) \cap \mathrm{Stab}(E_2) \leq \langle t \rangle\) since \(h \notin \tilde{B}\). Thus, \(\mathrm{Stab}(\gamma) = 1\).

Therefore, in every case, we obtain \(\mathrm{Stab}(\gamma) = 1\), so \(T_C\) is 2-acylindrical. ◻

Proof. We will construct a \(\Gamma\)-equivariant bijection from the white vertices of \(T_C\) to the white vertices of \(S\) and then extend this to a \(\Gamma\)-equivariant tree isomorphism \(T_C \to S\).

We first show that \(\tilde{A}\) is elliptic on \(S\).

By Lemma 16, we have that for each white vertex \(u \in V(T_C)\), we have that \(\Gamma_u\) fixes a unique white vertex of \(S\) (since \(\Gamma_u = \tilde{A}^g\) for some \(g \in \Gamma\), so that \(\Gamma_u\) fixes the unique white vertex \(u=gv\) where \(v \in V(T_C)\) with Stab\(_{\Gamma} (v) = \tilde{A}\)). This gives us a \(\Gamma\)-equivariant map from the white vertices of \(T_C\) to the white vertices of \(S\). Similarly, each white vertex stabilizer \(\Gamma_v\) of \(S\) is elliptic on \(T_C\), and it must fix a unique white vertex since otherwise we would obtain \(H \oplus \mathbb{Z}< B \rtimes \langle t \rangle\) for some \(H < G\) with \(Z(H) = 1\), which would imply by taking projections to \(G\) that \(H < B\), contradicting that \(H \neq 1\) and \(Z(H) = 1\) (note that \(Z(A) = 1\) since \(A\) is non-abelian and finitely generated CSA groups are commutative transitive (see [10]), meaning that commutation is a transitive relation among non-trivial elements). Thus, we obtain a \(\Gamma\)-equivariant map from the white vertices of \(S\) to the white vertices of \(T_C\), and this map is inverse to the map from the white vertices of \(T_C\) to the white vertices of \(S\), since \(\tilde{A} \leq \tilde{A}^g\) implies \(\tilde{A} = \tilde{A}^g\) for any \(g \in \Gamma\).

We now extend this map to a tree isomorphism \(T_C \to S\).

Therefore, by Lemma 17\(, [u',v']\) has a single black vertex. Thus, the \(\Gamma\)-equivariant bijection from white vertices of \(T_C\) to white vertices of \(S\) extends to a unique \(\Gamma\)-equivariant bijective tree map \(T_C \to S\).

We conclude that \(T_C\) and \(S\) are isomorphic as \(\Gamma\)-trees. ◻

Proof. Since edge stabilizers of \(T_C\) are abelian and since \(\tilde{A}\) is non-abelian, the proof for \(\tilde{A}\) follows exactly as in the proof of Lemma 8. For \(\tilde{B}\), if \(\tilde{B}\) is non-abelian (i.e.\(\delta\) acts non-trivially on \(B\)), then again the result follows exactly as in the proof of Lemma 8. If \(\tilde{B}\) is abelian (i.e.\(\delta\) acts trivially on \(B\)), then \(B \subseteq A\) and since \(D\) is maximal abelian in \(A\), we obtain that \(D = B\) and so \(\tilde{D} = \tilde{B}\). If \(\gamma \in \Gamma\) is such that \(\tilde{B} \subseteq \tilde{B}^{\gamma}\), then \(\langle \tilde{D}, \tilde{D}^{\gamma}\rangle = \langle \tilde{B}, \tilde{B}^{\gamma}\rangle = \tilde{B}\) which is abelian. It was shown in the proof of Theorem 18 that this implies \(\gamma \in \tilde{D} = \tilde{B}\). ◻

Proof. Let \(\Gamma\), \(\Gamma'\) be the mapping torus corresponding to these two non-separating Dehn twists on \(G\) and \(G'\) respectively, i.e. \(\Gamma = (A*_{D^e = C}) \rtimes_\delta \langle t \rangle\) and \(\Gamma' = (A'*_{D'^{e'} = C'}) \rtimes_{\delta'} \langle t' \rangle\) and let \(T_C\), \(T_C'\) be the tree of cylinders of the associated splitting of the mapping torus \(\Gamma = \tilde{A}*_{\tilde{D}^e = \tilde{C}}\) and \(\Gamma' = \tilde{A'}*_{\tilde{D'}^{e'} = \tilde{C'}}\), respectively. Assume that \(\delta\) and \(\delta'\) are conjugate in \(\mathop{\mathrm{Aut}}(G)\). Then by Lemma 7, there exists an isomorphism, \(f: \Gamma \to \Gamma'\) such that \(f|_G \in \mathop{\mathrm{Aut}}(G)\) and \(f(t)=t'\). The isomorphism \(f\) induces a group action \(._f\) of \(\Gamma\) on the tree of cylinders \(T_C'\): \(g ._f x = f(g).x\), for all \(g \in \Gamma\) and \(x \in T_C'\). This makes \(T_C'\) a \(\Gamma\)-tree that satisfies all the conditions of Lemma 15 as follows.

1. The edge stabilizers, \(\mathrm{Stab}_\Gamma(\tilde{z})\), are of the form \(f^{-1}(\mathrm{Stab}_{\Gamma'}(\tilde{z}))\), for \(\tilde{z} \in E(T_C')\), which is abelian because \(\mathrm{Stab}_{\Gamma'}(\tilde{z})\) is abelian and \(f\) is an isomorphism.

2. White vertex stabilizers of \(T_C'\) in \(\Gamma'\) are of the form \((A' \oplus \langle t' \rangle)^g = A'^g \oplus \langle t'^g \rangle\), for some \(g \in \Gamma'\). Note that \(A'^g\) has trivial center in \(G\). Therefore white vertex stabilizers of \(T_C'\) in \(\Gamma\) look like \(f^{-1}(A'^g \oplus \langle t'^g \rangle) = f^{-1}(A'^g) \oplus f^{-1}(\langle t'^g \rangle) \cong H \oplus \mathbb{Z}\) with \(H < G\) having a trivial center because \(f\) is an isomorphism mapping \(G\) to \(G\).

3. Black vertices in \(T_C'\) have stabilizers in \(\Gamma'\) of the form \((B' \rtimes \langle t' \rangle)^g = B'^g \rtimes \langle t'^g \rangle\), for some \(g \in \Gamma'\), where \(B'\) is the maximal abelian subgroup of \(G\) containing \(D'\). Now black vertex stabilizers of \(T_C'\) in \(\Gamma\) are therefore of the form \(f^{-1}(B'^g \rtimes \langle t'^g \rangle) = f^{-1}(B'^g) \rtimes f^{-1}(\langle t'^g \rangle) \cong K \rtimes \mathbb{Z}\) with \(K < G\) being abelian.

4. Because \(._f\) is an induced action from an isomorphism \(f: \Gamma \to \Gamma'\) with \(f|_G \in \mathop{\mathrm{Aut}}(G)\) and \(f(t)=t'\), and the action of \(\Gamma'\) on \(T_C'\) is \(2\)-acylindrical (due to Lemma 20), the action of \(\Gamma\) on \(T_C'\) is \(2\)-acylindrical as well.

From Lemma 15 it follows that there is a \(\Gamma\)-equivariant isomorphism between the two tree of cylinders \(T_C\) and \(T_C'\).

By Lemma 16, we have that \(\tilde{A}\) fixes a unique white vertex \(\gamma \tilde{A}'\) of \(T_C'\) (where we can assume \(\gamma \in G\)), and \(\tilde{A}'\) also fixes a unique white vertex of \(T_C\). Using the same argument as in the proof of Lemma 9, we obtain \(f(\tilde{A}) = (\tilde{A}')^{\gamma}\), and similarly \(f(\tilde{B}) = (\tilde{B}')^{\eta_1}\), \(f(\tilde{B}^e) = (\tilde{B}')^{\eta_2}\) for some \(\eta_1,\eta_2 \in G\).

Since \([\tilde{A}, \tilde{B}]\) and \([\tilde{A}, e\tilde{B}]\) are edges in \(T_C\) that are, in general, in different \(\Gamma\) orbits, by Lemma 17, we also have that \([\gamma \tilde{A}', \eta_1 \tilde{B}']\) and \([\gamma \tilde{A}', \eta_2 \tilde{B}']\) are edges in \(T_C'\) that are in different \(\Gamma\) orbits. Thus, we have \(\eta_1 \tilde{B}' = \alpha_1 \gamma \tilde{B}'\) and \(\eta_2 \tilde{B}' = \alpha_2 \gamma e' \tilde{B}'\), or \(\eta_1 \tilde{B}' = \alpha_1 \gamma e' \tilde{B}'\) and \(\eta_2 \tilde{B}' = \alpha_2 \gamma \tilde{B}'\) for some \(\alpha_1, \alpha_2 \in (\tilde{A}')^{\gamma}\). Up to replacing \(e\) by \(e^{-1}\) and \(D\) with \(C\), we can assume that the former case occurs. We can assume \(\eta_1 = \gamma\) and \(\eta_2 = \gamma e'\) as in Lemma 9 and so \(f(\tilde{A}) = (\tilde{A}')^{\gamma}\), \(f(\tilde{B}) = (\tilde{B}')^{\gamma}\) and \(f(\tilde{B}^e) = (\tilde{B}')^{\gamma e'}\). Define \(\phi = \mathrm{Ad}(\gamma^{-1}) f\). Then \(\phi(\tilde{A}) = \tilde{A}', \phi(\tilde{B}) = \tilde{B}'\) and \(\phi(\tilde{B}^e) = (\tilde{B}')^{e'}\). These imply that \(\phi(\tilde{D}) = \tilde{D}'\) and \(\phi(\tilde{C}) = \tilde{C}'\) (since \(\tilde{D} = \tilde{A} \cap \tilde{B}\) and \(\tilde{C} = \tilde{A} \cap \tilde{B}^e\), and similarly for \(\tilde{D}'\) and \(\tilde{C}'\)). Since \(\phi (G) = G\), we obtain that \(\phi(A) = A'\), \(\phi(B) = B'\), \(\phi(C) = C'\) and \(\phi(D) = D'\).

Since \(\phi(\tilde{B)} = \tilde{B}'\) and \(\phi(\tilde{B}^e) = (\tilde{B}')^{e'}\), we obtain that \((e')^{-1}\phi(e) \in N_{\Gamma'}(\tilde{B}') \cap G = \tilde{B}' \cap G = B'\) (where \(N_{\Gamma'}(\tilde{B}') = \tilde{B}'\) by Lemma 18), so that \(\phi(e) = e' b'\) for some \(b' \in B'\). Up to re-defining \(e'\) to be \(e'b'\) and re-defining \(w_1'\) to be \((w_1')^{n}\) for some \(n \in \mathbb{Z}\) (which does not change the group \(G\) nor the way \(\delta'\) acts on \(G\), hence does not change \(\Gamma'\)), we can thus assume that \(\phi(e) = e'\). Note also that we can arrange for \(\phi(t) = t'\) since \(\langle t \rangle = Z(\tilde{A})\) (respectively, \(\langle t' \rangle = Z(\tilde{A}'\))) and \(\phi(\tilde{A}) = \tilde{A}'\). Now we have \(\phi(tet^{-1}) = \phi(ew_1)\), and since \(\phi(t) = t'\) and \(\phi(e) = e'\), we obtain \(\phi(w_1) = w_1'\).

Lastly, we check that \(\sigma' \circ \phi \vert_D = \phi \circ \sigma\). For each \(d \in D\), we have \(\sigma(d) = ede^{-1}\), so \(\phi(\sigma(d)) = \phi(ede^{-1}) = e' \phi(d) (e')^{-1}\). On the other hand, we have \(\sigma'(\phi(d)) = e' \phi(d) (e')^{-1}\). Thus, we see that \(\sigma' \circ \phi \vert_D = \phi \circ \sigma\). The map \(\phi \vert_A\) is thus the required isomorphism.

Conversely, let us assume that there is an isomorphism \(\phi: A \to A'\) such that \(\phi(D) \subseteq D'\), \(\phi(C) \subseteq C'\), \(\phi(w_1) = w_1'\) and \(\sigma' \circ \phi|_D = \phi \circ \sigma\). To show that \(\delta\) is conjugate to \(\delta'\) in \(\mathop{\mathrm{Aut}}(G)\), by Lemma 7 it suffices to prove that there exists an isomorphism \(F: \Gamma \to \Gamma'\) such that \(F|_G \in \mathop{\mathrm{Aut}}(G)\) and \(F(t) = t'\). Define \(F: \Gamma \to \Gamma'\) by \(F(t) = t'\), \(F(e) = e'\) and \(F|_A = \phi\). To show that \(F\) is an isomorphism, the only thing that we need to check is if the this map is compatible with the edge maps, i.e. \(\tilde{\sigma} \circ F|_{\tilde{D}} = F \circ \tilde{\sigma}\), where \(\tilde{\sigma}: \tilde{D} \to \tilde{C}\) and \(\tilde{\sigma'}: \tilde{D'} \to \tilde{C'}\) are the edge group isomorphisms associated to the HNN splittings of \(\Gamma\) and \(\Gamma'\) respectively with the following properties: \(\tilde{\sigma}|_{\tilde{D}} = \sigma: D \to C\) and \(\tilde{\sigma'}|_{\tilde{D'}} = \sigma': D' \to C'\) are the edge maps corresponding to the splittings \(G = A*_{D^e = c}\) and \(G = A'*_{D'^{e'} = C'}\) respectively, \(\tilde{\sigma}(t) = tw_2^{-1}\) and \(\tilde{\sigma'}(t') = t'w_2'^{-1}\). Now \(\tilde{\sigma} \circ F|_{\tilde{D}}(t) = \tilde{\sigma'}(t') = t'w_2'^{-1}\) and \(F \circ \tilde{\sigma} (t) = F(tw_2^{-1})=t'w_2'^{-1}\) (as \(F(t) = t'\) and \(F|_A = \phi\) by definition), and note that \(\phi(w_2) = w_2'\) by conditions (3) and (4). This implies that \(\tilde{\sigma} \circ F|_{\tilde{D}}(t) = F \circ \tilde{\sigma}(t)\). Next, let \(d \in D\). Then

\[\begin{align} (\tilde{\sigma'} \circ F|_{\tilde{D}})(d) &= (\tilde{\sigma'} \circ \phi)(d) \\ &= (\sigma' \circ \phi)(d){0.2in} [\phi(D) \subseteq D']\\ &= (\phi \circ \sigma)(d) \\ &= (F \circ \tilde{\sigma})(d){0.2in} [\tilde{\sigma} \vert_D = \sigma \text{ and } \tilde{F} \vert_C = \phi_C]\\ \end{align}\]

This implies that \((\tilde{\sigma} \circ F|_{\tilde{D}})|_D = (F \circ \tilde{\sigma})|_D\).

 ◻

Proof. Suppose that \(\delta\) was both a separating and non-separating Dehn twist of \(G\). Then by Theorem 14 we can split \(G\) on the one hand as:

\[G = A_1 *_{C_1} B_1\]

with \(C_1\) maximal abelian in \(A_1, B_1\), with \(\delta\) acting as the identity on \(A_1\) and conjugation by some \(c \in C_1\) on \(B_1\). By Theorem 14, we have that \(A_1 = \mathrm{Fix}(\delta)\).

On the other hand, \(G\) also splits as:

\[G = {A_2}*_{D_2^e = C_2}\]

By Lemma 19, we can assume that \(A_2 = \mathrm{Fix}(\delta)\) and that \(D_2, C_2 < A_2\) are maximal abelian. Write \(\delta(e) = ed\) for some \(d \in D_2\). We denote \(A = A_1 =A_2\).

Let \(\Gamma = G \rtimes_{\delta} \langle t \rangle\) denote the mapping torus.

We must have that \(A\) and \(B_1\) are non-abelian. Indeed, suppose that \(A\) is abelian. Then \(C_1 = A\) and so \(G = B_1\) and \(\delta\) is an inner automorphism of \(G\), of the form \(\delta = \mathop{\mathrm{Ad}}(c)\) for some \(c \in A\). Also, from the HNN splitting of \(G\) we obtain \(A = D_2 = C_2\) and \(e\) commutes with \(A\), so \(G = A \oplus \langle e \rangle\). But then \(G\) is abelian and so \(\delta\) is trivial. However, this contradicts that \(\delta\) is non-trivial. Therefore, \(A_1\) cannot be abelian if \(\delta\) is non-trivial.

Now suppose that \(B_1\) is abelian. Then \(C_1 = B_1\) and \(G = A_1\), and so \(\delta\) is trivial, a contradiction. Therefore, we must have that \(A_1\) and \(B_1\) are non-abelian if \(\delta\) is non-trivial and is both separating and non-separating.

Then by Lemma 3, we have a canonical 2-acylindrical splitting \(T\) of \(\Gamma\) as an amalgam.

\[\Gamma = \tilde{A_1} *_{\tilde{C_1}} \tilde{B_1}\]

and by Theorem 20, we have another canonical 2-acylindrical splitting \(T_C\) of of \(\Gamma\) coming from the tree of cylinders construction.

We have that \(T_C\) satisfies the conditions of Lemma 4, and hence \(T\) and \(T_C\) are isomorphic \(\Gamma\)-trees. However, this is a contradiction, as \(T_C/\Gamma\) has solvable vertex group \(\tilde{B_2} = B_2 \rtimes \langle t \rangle\) while the vertex groups of \(T/\Gamma\) contain \(F_2\) (since \(A_1, B_1\) contain \(F_2\), being free products of at least 2 free abelian groups). ◻

Proof. Suppose that \(\delta\) is separating and \(\delta'\) is non-separating. Write associated splittings \(G = A *_C B\) for \(\delta\) and \(G = J*_{H^e=K}\) for \(\delta'\). Let \(f \in \mathop{\mathrm{Aut}}(G)\) be such that \(\delta' = \delta^f\) in \(\mathop{\mathrm{Aut}}(G)\). Then \(\delta\) is also non-separating with respect to the splitting \(G = f^{-1}(J)_{*f^{-1}(H)^{f^{-1}(e)} = f^{-1}(K)}\). Since \(\delta\) is non-trivial, this contradicts Lemma 21. ◻

Proof. We identify \(A \cong \mathbb{Z}^{n+1}\) and \(\mathop{\mathrm{Aut}}(A) = \mathrm{GL}_{n+1}(\mathbb{Z})\) via the choice of basis \((e_1,\ldots,e_{n+1})\).

If \(\gcd(v_1,\ldots,v_n) = \gcd(v_1',\ldots,v_n')\), then by Lemma 20, we have that there exists \(\psi \in \mathrm{GL}_n(\mathbb{Z})\) such that \(\psi(v) = v'\). Consider the block matrix \(M \in \mathrm{GL}_{n+1}(\mathbb{Z})\) with \(\psi\) in the upper left \(n \times n\) block and last row and column \((0,0,\ldots,0,1)\). We then see that:

\[\begin{align} M \delta &= \begin{bmatrix} \psi & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} Id & v \\ 0 & 1 \end{bmatrix} \\ &= \begin{bmatrix} \psi & \psi(v) \\ 0 & 1 \end{bmatrix} \\ &= \begin{bmatrix} \psi & v' \\ 0 & 1 \end{bmatrix} \\ &= \begin{bmatrix} Id & v' \\ 0 & 1 \end{bmatrix} \begin{bmatrix} \psi & 0 \\ 0 & 1 \end{bmatrix} \\ &= \delta' M \end{align}\]

So that \(\delta\) and \(\delta'\) are conjugate by \(M\) in \(\mathrm{GL}_{n+1}(\mathbb{Z})\).

Conversely, suppose that \(\delta' = \delta^M\) for some \(M \in \mathrm{GL}_{n+1}(\mathbb{Z})\). Then for each \(i=1,\ldots,n\) we have:

\[M \delta M^{-1}(e_i) = \delta'(e_i) = e_i\]

so that \(\delta M^{-1}(e_i) = M^{-1}(e_i)\) for all \(i=1,\ldots,n\). We may assume that either both \(\gcd(v_1,\ldots,v_n)\) and \(\gcd(v_1',\ldots,v_n')\) are zero or non-zero, since each of these gcd’s are zero if and only if the corresponding automorphism \(\delta, \delta'\) is trivial, in which case they are conjugate if and only if they are both trivial. Hence, the fixed point set of \(\delta\) and \(\delta'\) is \(\langle e_1, \ldots, e_n \rangle\), and \(M\) preserves this fixed point set. Thus, \(M\) is block diagonal of the form:

\[M = \begin{bmatrix} \psi & x \\ 0 & x_{n+1} \end{bmatrix}\]

with \(x \in \mathbb{Z}^n\) and \(x_{n+1} \in \mathbb{Z}\setminus 0\).

Then:

\[M \delta = \begin{bmatrix} \psi & x \\ 0 & x_{n+1} \end{bmatrix} \begin{bmatrix} Id & v \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} \psi & \psi(v) + x \\ 0 & x_{n+1} \end{bmatrix}\]

and

\[\delta' M = \begin{bmatrix} Id & v' \\ 0 & 1 \end{bmatrix} \begin{bmatrix} \psi & x \\ 0 & x_{n+1} \end{bmatrix} = \begin{bmatrix} \psi & x + v'x_{n+1} \\ 0 & x_{n+1} \end{bmatrix}\]

Since \(M \delta = \delta' M\), we obtain \(\psi(v) + x = x + v'x_{n+1}\), and so \(\psi(v) = v' x_{n+1}\). Since \(\psi\) is an automorphism, this yields \(\gcd(v_1,\ldots,v_n) = \gcd(v_1',\ldots,v_n') x_{n+1}\), so that \(\gcd(v_1',\ldots,v_n')\) divides \(\gcd(v_1,\ldots,v_n)\). Repeating the above argument with the equality \(M^{-1} \delta' = \delta M^{-1}\), we obtain that \(\gcd(v_1,\ldots,v_n)\) divides \(\gcd(v_1',\ldots,v_n')\), so that \(\gcd(v_1,\ldots,v_n)=\gcd(v_1',\ldots,v_n')\), since gcd is non-negative by definition. ◻

Proof. If \(\delta\) is a Dehn twist of a non-trivial splitting of \(G\), then by Sections 3 and Section 4.4, \(\Gamma\) splits as items (3) or (4). By Lemma 21, items (3) and (4) cannot both hold. Also, (1) and (2) cannot hold since Proposition 10 and items (3) and (4) each imply that \(\Gamma\) has finite centre, while (1) and (2) each imply that \(\Gamma\) has infinite centre.

If \(\delta\) is a Dehn twist of a trivial splitting of \(G\), then if \(G\) is non-abelian, this spliting must be a trivial amalgam splitting and \(\delta\) is then an inner automorphism of \(G\). Writing \(\delta = \mathop{\mathrm{Ad}}(g)\) for some \(g \in G\), we then have that \(\Gamma = G \oplus \langle g^{-1}t \rangle\), so item (1) holds. Items (3) and (4) cannot hold since then by Proposition 10, \(\Gamma\) has finite centre \(Z(\Gamma)\), which contradicts that \(gt \in Z(\Gamma)\) and has infinite order. Item (2) cannot hold by Lemma 21. If \(G\) is abelian and \(\delta\) is a non-trivial Dehn twist of a trivial splitting of \(G\) (if \(\delta\) is trivial then it is an inner automorphism and (1) holds), then \(G\) splits as a direct product HNN extension \(G=A \oplus \langle e \rangle\) for some abelian subgroup \(A\) of \(G\) and \(\delta\) is a unimodular automorphism of \(G\), i.e.\(\delta \vert_A = id_A\) and \(\delta(e) = ea_0\) for some \(a_0 \in A\). Therefore, item (2) holds. Item (1) does not hold because if \(G\) is abelian and (1) holds then \(\delta = id_G\) by Lemma 21. Items (3) and (4) also do not hold since they imply that \(\Gamma\) has finite centre by Proposition 10, while (2) implies that \(\Gamma\) has infinite centre. ◻

Proof. By Lemma 7, if \(\delta\) and \(\delta'\) are conjugate, then we have an isomorphism \(f : \Gamma \to \Gamma'\) that maps \(G\) to \(G\) and \(t\) to \(t'\). This isomorphism \(f\) then preserves the decomposition of \(\Gamma\), i.e.\(f\) induces a decomposition of \(\Gamma'\) of the same type in Lemma 21 as \(\Gamma\). ◻