Return to Answer

Let $R = \mathbb{Z}_N[X^{\pm 1}]$ be the Laurent polynomials ring over $\mathbb{Z}_N = \mathbb{Z}/N \mathbb{Z}$ , let $U$ be the unit group of $R$ and let $\theta$ be the ring automorphism of $R$ induced by the map $X \mapsto X^{-1}$. Set $C = \langle \theta \rangle \simeq \mathbb{Z}_2$. Note that $C$ identifies with a subgroup of $\operatorname{Aut}(U)$ via restriction to $U$.

Claim. We have $\operatorname{Aut}(L_N) \simeq R \rtimes (U \rtimes C)$ where $U$ acts on the additive group of $R$ via multiplication and $C$ acts on $U$ via restriction.

To make the above isomorphism concrete, we introduce the following notation.

For $x,y$ two elements of a group $G$, we set $x^y : = y^{-1} x y$.

The subgroup $\Phi:= \langle \phi_f \, \vert \, f \in R\rangle$ is an Abelian normal subgroup of $\operatorname{Aut}(L_N)$ which is isomorphic to the additive group of $R$. The subgroup $\Psi: = \langle \psi_u \, \vert \, u \in U\rangle$ is an Abelian subgroup of $\operatorname{Aut}(L_N)$ which is isomorphic to $U$ and $\langle \Psi, \tau\rangle \simeq U \rtimes C$. Indeed, we have $$(1)\quad \phi_f \circ \phi_g = \phi_{f + g}, \,\, \psi_u \circ \psi_v = \psi_{uv}$$ and $$(2)\quad\phi_f^{\psi_u} = \phi_{u^{-1}f},\,\, \phi_f^{\tau} = \phi_{-X \cdot \theta(f)}, \,\, \psi_u^{\tau} = \psi_{\theta(u)}$$ for every $f,g \in R, u,v \in U$.

Proof of the Claim. Observe first that $L_N$ identifies with $R \rtimes \langle t \rangle$ (to see this, use the map $a_0^ft^k \mapsto (f, t^k)$) where $t$ acts on $R$ via multiplication by $X$, i.e., $f^t := X \cdot f$ for $f \in R$. It is easy to verify that $\phi_f$ is a well-defined automorphism of $L_N$ which leaves $R$ point-wise invariant. Besides, as already noted by
HenrikRüping in a comment, the normal subgroup $R$ is actually a characteristic subgroup of $L_N$. Indeed the subgroup $R$ in the previous decomposition identifies with the preimage of the torsion subgroup of $(L_N)_{ab} \simeq \mathbb{Z}_N \times \mathbb{Z}$, the abelianization of $L_N$. Let $\varphi \in \operatorname{Aut}(L_N)$. Let us prove that $\varphi = \phi_f \circ \psi_u \circ \tau^{e}$ for some $f \in R, u \in U$ and $e \in \{0,1\}$. There is $f \in R, \epsilon \in \{-1, 1\}$ such that $\varphi(t) = a_0^ft^{\epsilon}$. Replacing $\varphi$ with $\varphi \circ \tau$ if need be, we can assume, without loss of generality, that $\varphi(t) = a_0^ft$. Replacing $\varphi$ with $\phi_{-f} \circ \varphi$, if needed, we can assume, w.l.o.g., that $f = 0$, that is, $\varphi(t) = t$. Since $\varphi$ preserves $R$, there is $u \in R$ such that $\varphi(a_0) = a_0^u$. Expressing the fact that $\varphi$ is surjective, i.e., that $t$ and $a_0^{u}$ generates $L_N$, is easily seen to be equivalent to the fact that $u$ generates $R$ as an $R$-module. Therefore $u \in U$ and $\varphi = \psi_u$. We have thus demonstrated that $\operatorname{Aut}(L_N) = \Phi \cdot \langle\Psi, \tau\rangle$. The semi-direct product decompositions now follow from the relations (2).

Let $R = \mathbb{Z}_N[X^{\pm 1}]$ be the Laurent polynomials ring over $\mathbb{Z}_N = \mathbb{Z}/N \mathbb{Z}$ , let $U$ be the unit group of $R$ and let $\theta$ be the ring automorphism of $R$ induced by the map $X \mapsto X^{-1}$. Set $C = \langle \theta \rangle \simeq \mathbb{Z}_2$. Note that $C$ identifies with a subgroup of $\operatorname{Aut}(U)$ via restriction to $U$.

For $x,y$ two elements of a group $G$, we set $x^y : = y^{-1} x y$.

Claim. We have $\operatorname{Aut}(L_N) \simeq R \rtimes (U \rtimes C)$ where $U$ acts on the additive group of $R$ via multiplication, $C$ acts on $U$ via restriction of the previous action and the action of $C$ on the additive group of $R$ is specified by: $f^{\theta} = - \frac{\theta(f)}{X}$ for $f \in R$.

To make the above isomorphism concrete, we introduce the following notation.

It is not difficult to check that all the above maps actually induce automorphisms of $L_N$; details can be added on demand.

The subgroup $\Phi:= \langle \phi_f \, \vert \, f \in R\rangle$ is an Abelian normal subgroup of $\operatorname{Aut}(L_N)$ which is isomorphic to the additive group of $R$. The subgroup $\Psi: = \langle \psi_u \, \vert \, u \in U\rangle$ is an Abelian subgroup of $\operatorname{Aut}(L_N)$ which is isomorphic to $U$ and $\langle \Psi, \tau\rangle \simeq U \rtimes C$. Indeed, we have $$(1)\quad \phi_f \circ \phi_g = \phi_{f + g}, \,\, \psi_u \circ \psi_v = \psi_{uv}$$ and $$(2)\quad\phi_f^{\psi_u} = \phi_{u^{-1}f},\,\, \phi_f^{\tau} = \phi_{-\frac{\theta(f)}{X}}, \,\, \psi_u^{\tau} = \psi_{\theta(u)}$$ for every $f,g \in R$ and every $u,v \in U$.

Proof of the Claim. Observe first that $L_N$ identifies with $R \rtimes \langle t \rangle$ (to see this, use the map $a_0^ft^k \mapsto (f, t^k)$) where $t$ acts on $R$ via multiplication by $X$, i.e., $f^t := X \cdot f$ for $f \in R$. It is easy to verify that $\phi_f$ is a well-defined automorphism of $L_N$ which leaves $R$ point-wise invariant. Besides, as already noted by
HenrikRüping in a comment, the normal subgroup $R$ is actually a characteristic subgroup of $L_N$. Indeed the subgroup $R$ in the previous decomposition identifies with the preimage of the torsion subgroup of $(L_N)_{ab} \simeq \mathbb{Z}_N \times \mathbb{Z}$, the abelianization of $L_N$. Let $\varphi \in \operatorname{Aut}(L_N)$. Let us prove that $\varphi = \phi_f \circ \psi_u \circ \tau^{e}$ for some $f \in R, u \in U$ and $e \in \{0,1\}$. As $\varphi$ preserves $R$, it induces an automorphism of $L_N/R \simeq \langle t \rangle$. Hence there is $f \in R, \epsilon \in \{-1, 1\}$ such that $\varphi(t) = a_0^ft^{\epsilon}$. Replacing $\varphi$ with $\varphi \circ \tau$ if needed, we can assume, without loss of generality, that $\varphi(t) = a_0^ft$. Replacing $\varphi$ with $\phi_{-f} \circ \varphi$, if needed, we can assume, w.l.o.g., that $f = 0$, that is, $\varphi(t) = t$. Since $\varphi$ preserves $R$, there is $u \in R$ such that $\varphi(a_0) = a_0^u$. Expressing the fact that $\varphi$ is surjective, i.e., that $t$ and $a_0^{u}$ generates $L_N$, is easily seen to be equivalent to the fact that $u$ generates $R$ as an $R$-module. Therefore $u \in U$ and $\varphi = \psi_u$. We have thus demonstrated that $\operatorname{Aut}(L_N) = \Phi \cdot \langle\Psi, \tau\rangle$. The semi-direct product decompositions now follow from the relations (2).

Let $R = \mathbb{Z}_N[X^{\pm 1}]$ be the Laurent polynomials ring over $\mathbb{Z}_N = \mathbb{Z}/N \mathbb{Z}$ , let $U$ be unit group of $R$ and let $\theta$ be the ring automorphism of $R$ induced by the map $X \mapsto X^{-1}$. Set $C = \langle \theta \rangle \simeq \mathbb{Z}_2$. Note that $C$ identifies with a subgroup of $\operatorname{Aut}(U)$ via restriction to $U$.

Claim. We have $\operatorname{Aut}(L_N) \simeq R \rtimes (U \rtimes C)$ where $U$ acts on the additive group of $R$ via multiplication and $C$ acts on $U$ via restriction.

Note 1. We actually have an explicit description of $U$ thanks to [2, Exercise 3.17] (see also [4, Proposition 4.8]).

Note 2. The remarks and the proof below are not original. They only consist in a specialisation of [2], which can already be done with the results of [1] in the specific case of $L_N$.

To make the above isomorphism concrete, we introduce the following notation.

For $x, y$ two elements of a group $G$, we set $x^y : = y^{-1} x y$.

For $f = c_{-d}X^{-d} + \cdots + c_d X^d \in R$, we define $$a_0^f := (a_0^{c_{-d}})^{t^{-d}} \cdots (a_0^{c_d})^{t^d}$$ and let $\phi_f \in \operatorname{Aut}(L_N)$ be defined by $$\phi_f(a_0) = a_0,\, \phi_f(t) = a_0^ft.$$

For $u \in U$, let $\psi_u \in \operatorname{Aut}(L_N)$ be defined by $$ \psi_u(a_0) = a_0^u,\, \psi_u(t) = t. $$

Finally, let $\tau \in \operatorname{Aut}(L_N)$ be defined by $$\tau(a_0) = a_0,\, \tau(t) = t^{-1}.$$

The subgroup $\Phi:= \langle \phi_f \, \vert \, f \in R\rangle$ is an Abelian normal subgroup of $\operatorname{Aut}(L_N)$ which is isomorphic to the additive group of $R$. The subgroup $\Psi: = \langle \psi_u \, \vert \, u \in U\rangle$ is an Abelian subgroup of $\operatorname{Aut}(L_N)$ which is isomorphic to $U$ and $\langle \Psi, \tau\rangle \simeq U \rtimes C$. Indeed, we have $$(1)\quad \phi_f \circ \phi_g = \phi_{f + g}, \,\, \psi_u \circ \psi_v = \psi_{uv}$$ and $$(2)\quad\phi_f^{\psi_u} = \phi_{u^{-1}f},\,\, \phi_f^{\tau} = \phi_{-X \cdot \theta(f)}, \,\, \psi_u^{\tau} = \psi_{\theta(u)}$$ for every $f,g \in R, u,v \in U$.

We are now in position to prove the claim.

Proof of the Claim. Observe first that $L_N$ identifies with $R \rtimes \langle t \rangle$ (to see this, use the map $a_0^ft^k \mapsto (f, t^k)$) where $t$ acts on $R$ via multiplication by $X$, i.e., $f^t := X \cdot f$ for $f \in R$. It is easy to verify that $\phi_f$ is a well-defined automorphism of $L_N$ which leaves $R$ point-wise invariant. Besides, as already noted by
HenrikRüping in a comment, the normal subgroup $R$ is actually a characteristic subgroup of $L_N$. Indeed the subgroup $R$ in the previous decomposition identifies with the preimage of the torsion subgroup of $(L_N)_{ab} \simeq \mathbb{Z}_N \times \mathbb{Z}$, the abelianization of $L_N$. Let $\varphi \in \operatorname{Aut}(L_N)$. Let us prove that $\varphi = \phi_f \circ \psi_u \circ \tau^{e}$ for some $f \in R, u \in U$ and $e \in \{0,1\}$. There is $f \in R, \epsilon \in \{-1, 1\}$ such that $\varphi(t) = a_0^ft^{\epsilon}$. Replacing $\varphi$ with $\varphi \circ \tau$ if need be, we can assume, without loss of generality, that $\varphi(t) = a_0^ft$. Replacing $\varphi$ with $\phi_{-f} \circ \varphi$, if needed, we can assume, w.l.o.g., that $f = 0$, that is, $\varphi(t) = t$. Since $\varphi$ preserves $R$, there is $u \in R$ such that $\varphi(a_0) = a_0^u$. Expressing the fact that $\varphi$ is surjective, i.e., that $t$ and $a_0^{u}$ generates $L_N$, is easily seen to be equivalent to the fact that $u$ generates $R$ as an $R$-module. Therefore $u \in U$ and $\varphi = \psi_u$. We have thus demonstrated that $\operatorname{Aut}(L_N) = \Phi \cdot \langle\Psi, \tau\rangle$. The semi-direct product decompositions now follow from the relations (2).

• [1] F. Szechtman. "The group of outer automorphisms of the semidirect product of the additive group of a ring by a group of units", 2004.
• [2] C. Weibel, "The K-book", 2013.
• [3] M. Stein, J. Taback, "Automorphisms of Higher Rank Lamplighter Groups", 2015.
• [4] A. Genevois, R. Tessera, "A note on morphisms to wreath products", 2022.

Let $R = \mathbb{Z}_N[X^{\pm 1}]$ be the Laurent polynomials ring over $\mathbb{Z}_N = \mathbb{Z}/N \mathbb{Z}$ , let $U$ be the unit group of $R$ and let $\theta$ be the ring automorphism of $R$ induced by the map $X \mapsto X^{-1}$. Set $C = \langle \theta \rangle \simeq \mathbb{Z}_2$. Note that $C$ identifies with a subgroup of $\operatorname{Aut}(U)$ via restriction to $U$.

Claim. We have $\operatorname{Aut}(L_N) \simeq R \rtimes (U \rtimes C)$ where $U$ acts on the additive group of $R$ via multiplication and $C$ acts on $U$ via restriction.

Note 1. We actually have an explicit description of $U$ thanks to [2, Exercise 3.17] (see also [4, Proposition 4.8]).

Note 2. The remarks and the proof below are not original. They only consist in a specialisation of [3, Theorem 3.2], which can already be done with the results of [1] in the specific case of $L_N$.

To make the above isomorphism concrete, we introduce the following notation.

For $x,y$ two elements of a group $G$, we set $x^y : = y^{-1} x y$.

For $f = c_{-d}X^{-d} + \cdots + c_d X^d \in R \,(d \ge 0)$, we define $$a_0^f := (a_0^{c_{-d}})^{t^{-d}} \cdots (a_0^{c_d})^{t^d}$$ and let $\phi_f \in \operatorname{Aut}(L_N)$ be defined by $$\phi_f(a_0) = a_0,\, \phi_f(t) = a_0^ft.$$

For $u \in U$, let $\psi_u \in \operatorname{Aut}(L_N)$ be defined by $$ \psi_u(a_0) = a_0^u,\, \psi_u(t) = t. $$

Finally, let $\tau \in \operatorname{Aut}(L_N)$ be defined by $$\tau(a_0) = a_0,\, \tau(t) = t^{-1}.$$

The subgroup $\Phi:= \langle \phi_f \, \vert \, f \in R\rangle$ is an Abelian normal subgroup of $\operatorname{Aut}(L_N)$ which is isomorphic to the additive group of $R$. The subgroup $\Psi: = \langle \psi_u \, \vert \, u \in U\rangle$ is an Abelian subgroup of $\operatorname{Aut}(L_N)$ which is isomorphic to $U$ and $\langle \Psi, \tau\rangle \simeq U \rtimes C$. Indeed, we have $$(1)\quad \phi_f \circ \phi_g = \phi_{f + g}, \,\, \psi_u \circ \psi_v = \psi_{uv}$$ and $$(2)\quad\phi_f^{\psi_u} = \phi_{u^{-1}f},\,\, \phi_f^{\tau} = \phi_{-X \cdot \theta(f)}, \,\, \psi_u^{\tau} = \psi_{\theta(u)}$$ for every $f,g \in R, u,v \in U$.

We are now in position to prove the claim.

Proof of the Claim. Observe first that $L_N$ identifies with $R \rtimes \langle t \rangle$ (to see this, use the map $a_0^ft^k \mapsto (f, t^k)$) where $t$ acts on $R$ via multiplication by $X$, i.e., $f^t := X \cdot f$ for $f \in R$. It is easy to verify that $\phi_f$ is a well-defined automorphism of $L_N$ which leaves $R$ point-wise invariant. Besides, as already noted by
HenrikRüping in a comment, the normal subgroup $R$ is actually a characteristic subgroup of $L_N$. Indeed the subgroup $R$ in the previous decomposition identifies with the preimage of the torsion subgroup of $(L_N)_{ab} \simeq \mathbb{Z}_N \times \mathbb{Z}$, the abelianization of $L_N$. Let $\varphi \in \operatorname{Aut}(L_N)$. Let us prove that $\varphi = \phi_f \circ \psi_u \circ \tau^{e}$ for some $f \in R, u \in U$ and $e \in \{0,1\}$. There is $f \in R, \epsilon \in \{-1, 1\}$ such that $\varphi(t) = a_0^ft^{\epsilon}$. Replacing $\varphi$ with $\varphi \circ \tau$ if need be, we can assume, without loss of generality, that $\varphi(t) = a_0^ft$. Replacing $\varphi$ with $\phi_{-f} \circ \varphi$, if needed, we can assume, w.l.o.g., that $f = 0$, that is, $\varphi(t) = t$. Since $\varphi$ preserves $R$, there is $u \in R$ such that $\varphi(a_0) = a_0^u$. Expressing the fact that $\varphi$ is surjective, i.e., that $t$ and $a_0^{u}$ generates $L_N$, is easily seen to be equivalent to the fact that $u$ generates $R$ as an $R$-module. Therefore $u \in U$ and $\varphi = \psi_u$. We have thus demonstrated that $\operatorname{Aut}(L_N) = \Phi \cdot \langle\Psi, \tau\rangle$. The semi-direct product decompositions now follow from the relations (2).

• [1] F. Szechtman. "The group of outer automorphisms of the semidirect product of the additive group of a ring by a group of units", 2004.
• [2] C. Weibel, "The K-book", 2013.
• [3] M. Stein, J. Taback, "Automorphisms of Higher Rank Lamplighter Groups", 2015.
• [4] A. Genevois, R. Tessera, "A note on morphisms to wreath products", 2022.

Let $R = \mathbb{Z}_N[X^{\pm 1}]$ be the Laurent polynomials ring over $\mathbb{Z}_N = \mathbb{Z}/N \mathbb{Z}$ , let $U$ be unit group of $R$ and let $\theta$ be the ring automorphism of $R$ induced by the map $X \mapsto X^{-1}$. Set $C = \langle \theta \rangle \simeq \mathbb{Z}_2$. Note that $C$ identifies with a subgroup of $\operatorname{Aut}(U)$ via restriction to $U$.

Claim. We have $\operatorname{Aut}(L_N) \simeq R \rtimes (U \rtimes C)$ where $U$ acts on the additive group of $R$ via multiplication and $C$ acts on $U$ via restriction.

Note 1. We actually have an explicit description of $U$ thanks to [2, Exercise 3.17] (see also [4, Proposition 4.8]).

Note 2. The remarks and the proof below are not original. They only consist in a specialisation of [2], which can already be done with the results of [1] in the specific case of $L_N$.

To make the above isomorphism concrete, we introduce the following notation.

For $x, y$ two elements of a group $G$, we set $x^y : = y^{-1} x y$.

For $f = c_{-d}X^{-d} + \cdots + c_d X^d \in R$, we define $$a_0^f := (a_0^{c_{-d}})^{t^{-d}} \cdots (a_0^{c_d})^{t^d}$$ and let $\phi_f \in \operatorname{Aut}(L_N)$ be defined by $$\phi_f(a_0) = a_0,\, \phi_f(t) = a_0^ft.$$

For $u \in U$, let $\psi_u \in \operatorname{Aut}(L_N)$ be defined by $$ \psi_u(a_0) = a_0^u,\, \psi_u(t) = t. $$

Finally, let $\tau \in \operatorname{Aut}(L_N)$ be defined by $$\tau(a_0) = a_0,\, \tau(t) = t^{-1}.$$

The subgroup $\Phi:= \langle \phi_f \, \vert \, f \in R\rangle$ is an Abelian normal subgroup of $\operatorname{Aut}(L_N)$ which is isomorphic to the additive group of $R$. The subgroup $\Psi: = \langle \psi_u \, \vert \, u \in U\rangle$ is an Abelian subgroup of $\operatorname{Aut}(L_N)$ which is isomorphic to $U$ and $\langle \Psi, \tau\rangle \simeq U \rtimes C$. Indeed, we have $$(1)\quad \phi_f \circ \phi_g = \phi_{f + g}, \,\, \psi_u \circ \psi_v = \psi_{uv}$$ and $$(2)\quad\phi_f^{\psi_u} = \phi_{u^{-1}f},\,\, \phi_f^{\tau} = \phi_{-X \cdot \theta(f)}, \,\, \psi_u^{\tau} = \psi_{\theta(u)}$$ for every $f,g \in R, u,v \in U$.

We are now in position to prove the claim.

Proof of the Claim. Observe first that $L_N$ identifies with $R \rtimes \langle t \rangle$ (to see this, use the map $a_0^ft^k \mapsto (f, t^k)$) where $t$ acts on $R$ via multiplication by $X$, i.e., $f^t := X \cdot f$ for $f \in R$. It is easy to verify that $\phi_f$ is a well-defined automorphism of $L_N$ which leaves $R$ point-wise invariant. Besides, as already noted by
HenrikRüping in a comment, the normal subgroup $R$ is actually a characteristic subgroup of $L_N$. Indeed the subgroup $R$ in the previous decomposition identifies with the preimage of the torsion subgroup of $(L_N)_{ab} \simeq \mathbb{Z}_N \times \mathbb{Z}$, the abelianization of $L_N$. Let $\varphi \in \operatorname{Aut}(L_N)$. Let us prove that $\varphi = \phi_f \circ \psi_u \circ \tau^{e}$ for some $f \in R, u \in U$ and $e \in \{0,1\}$. There is $f \in R, \epsilon \in \{-1, 1\}$ such that $\varphi(t) = a_0^ft^{\epsilon}$. Replacing $\varphi$ with $\varphi \circ \tau$ if need be, we can assume, without loss of generality, that $\varphi(t) = a_0^ft$. Replacing $\varphi$ with $\phi_{-f} \circ \varphi$, if needed, we can assume, w.l.o.g., that $f = 0$., that is, $\varphi(t) = t$. Since $\varphi$ preserves $R$, there is $u \in R$ such that $\varphi(a_0) = a_0^u$. Expressing the fact that $\varphi$ is surjective, i.e., that $t$ and $a_0^{u}$ generates $L_N$, is easily seen to be equivalent to the fact that $u$ generates $R$ as an $R$-module. Therefore $u \in U$ and $\varphi = \psi_u$. We have thus demonstrated that $\operatorname{Aut}(L_N) = \Phi \cdot \Psi$. The semi-direct product decompositions now follow from the relations (2).

• [1] F. Szechtman. "The group of outer automorphisms of the semidirect product of the additive group of a ring by a group of units", 2004.
• [2] C. Weibel, "The K-book", 2013.
• [3] M. Stein, J. Taback, "Automorphisms of Higher Rank Lamplighter Groups", 2015.
• [4] A. Genevois, R. Tessera, "A note on morphisms to wreath products", 2022.

Let $R = \mathbb{Z}_N[X^{\pm 1}]$ be the Laurent polynomials ring over $\mathbb{Z}_N = \mathbb{Z}/N \mathbb{Z}$ , let $U$ be unit group of $R$ and let $\theta$ be the ring automorphism of $R$ induced by the map $X \mapsto X^{-1}$. Set $C = \langle \theta \rangle \simeq \mathbb{Z}_2$. Note that $C$ identifies with a subgroup of $\operatorname{Aut}(U)$ via restriction to $U$.

Claim. We have $\operatorname{Aut}(L_N) \simeq R \rtimes (U \rtimes C)$ where $U$ acts on the additive group of $R$ via multiplication and $C$ acts on $U$ via restriction.

Note 1. We actually have an explicit description of $U$ thanks to [2, Exercise 3.17] (see also [4, Proposition 4.8]).

Note 2. The remarks and the proof below are not original. They only consist in a specialisation of [2], which can already be done with the results of [1] in the specific case of $L_N$.

To make the above isomorphism concrete, we introduce the following notation.

For $x, y$ two elements of a group $G$, we set $x^y : = y^{-1} x y$.

For $f = c_{-d}X^{-d} + \cdots + c_d X^d \in R$, we define $$a_0^f := (a_0^{c_{-d}})^{t^{-d}} \cdots (a_0^{c_d})^{t^d}$$ and let $\phi_f \in \operatorname{Aut}(L_N)$ be defined by $$\phi_f(a_0) = a_0,\, \phi_f(t) = a_0^ft.$$

For $u \in U$, let $\psi_u \in \operatorname{Aut}(L_N)$ be defined by $$ \psi_u(a_0) = a_0^u,\, \psi_u(t) = t. $$

Finally, let $\tau \in \operatorname{Aut}(L_N)$ be defined by $$\tau(a_0) = a_0,\, \tau(t) = t^{-1}.$$

The subgroup $\Phi:= \langle \phi_f \, \vert \, f \in R\rangle$ is an Abelian normal subgroup of $\operatorname{Aut}(L_N)$ which is isomorphic to the additive group of $R$. The subgroup $\Psi: = \langle \psi_u \, \vert \, u \in U\rangle$ is an Abelian subgroup of $\operatorname{Aut}(L_N)$ which is isomorphic to $U$ and $\langle \Psi, \tau\rangle \simeq U \rtimes C$. Indeed, we have $$(1)\quad \phi_f \circ \phi_g = \phi_{f + g}, \,\, \psi_u \circ \psi_v = \psi_{uv}$$ and $$(2)\quad\phi_f^{\psi_u} = \phi_{u^{-1}f},\,\, \phi_f^{\tau} = \phi_{-X \cdot \theta(f)}, \,\, \psi_u^{\tau} = \psi_{\theta(u)}$$ for every $f,g \in R, u,v \in U$.

We are now in position to prove the claim.

Proof of the Claim. Observe first that $L_N$ identifies with $R \rtimes \langle t \rangle$ (to see this, use the map $a_0^ft^k \mapsto (f, t^k)$) where $t$ acts on $R$ via multiplication by $X$, i.e., $f^t := X \cdot f$ for $f \in R$. It is easy to verify that $\phi_f$ is a well-defined automorphism of $L_N$ which leaves $R$ point-wise invariant. Besides, as already noted by
HenrikRüping in a comment, the normal subgroup $R$ is actually a characteristic subgroup of $L_N$. Indeed the subgroup $R$ in the previous decomposition identifies with the preimage of the torsion subgroup of $(L_N)_{ab} \simeq \mathbb{Z}_N \times \mathbb{Z}$, the abelianization of $L_N$. Let $\varphi \in \operatorname{Aut}(L_N)$. Let us prove that $\varphi = \phi_f \circ \psi_u \circ \tau^{e}$ for some $f \in R, u \in U$ and $e \in \{0,1\}$. There is $f \in R, \epsilon \in \{-1, 1\}$ such that $\varphi(t) = a_0^ft^{\epsilon}$. Replacing $\varphi$ with $\varphi \circ \tau$ if need be, we can assume, without loss of generality, that $\varphi(t) = a_0^ft$. Replacing $\varphi$ with $\phi_{-f} \circ \varphi$, if needed, we can assume, w.l.o.g., that $f = 0$, that is, $\varphi(t) = t$. Since $\varphi$ preserves $R$, there is $u \in R$ such that $\varphi(a_0) = a_0^u$. Expressing the fact that $\varphi$ is surjective, i.e., that $t$ and $a_0^{u}$ generates $L_N$, is easily seen to be equivalent to the fact that $u$ generates $R$ as an $R$-module. Therefore $u \in U$ and $\varphi = \psi_u$. We have thus demonstrated that $\operatorname{Aut}(L_N) = \Phi \cdot \langle\Psi, \tau\rangle$. The semi-direct product decompositions now follow from the relations (2).

• [1] F. Szechtman. "The group of outer automorphisms of the semidirect product of the additive group of a ring by a group of units", 2004.
• [2] C. Weibel, "The K-book", 2013.
• [3] M. Stein, J. Taback, "Automorphisms of Higher Rank Lamplighter Groups", 2015.
• [4] A. Genevois, R. Tessera, "A note on morphisms to wreath products", 2022.