4·1. One-ended subgroups

One can deduce from Theorem 1·1, that given a finite group A and a one-ended finitely presented group B, every automorphism $A \wr B \to A \wr B$ sends conjugates of B to conjugates of B. This property has strong consequences on the structure of automorphisms of wreath products, which we record in the next section. First of all, let us justify our observation.

Proof. Let $\varphi \;:\; A \wr B \to A \wr B$ be an automorphism. Lemma 4·1 implies that $\varphi(B)$ is contained in a conjugate of B, say $gBg^{-1}$ . Similarly, $\varphi^{-1}(B)$ must be contained in some conjugate $hBh^{-1}$ . We have

\begin{align*}\varphi(B) \subset gBg^{-1} = g \varphi ( \varphi^{-1}(B)) g^{-1}\subset g \varphi(h) \varphi(B) \varphi(h)^{-1}g^{-1},\end{align*}

But B (and a fortiori $\varphi(B)$ ) is almost malnormal in $A \wr B$ (see for instance [ Reference Genevois and TesseraGT24 , fact 4·30]), so the inclusions above must be equalities since B is infinite, hence $\varphi(B)=gBg^{-1}$ .

Of course, Lemma 4·1 does not hold if one does not assume that the subgroup under consideration is finitely presented (since otherwise the statement would apply to the whole wreath product $A \wr B$ when A is non-trivial and B infinite). Nevertheless, Corollary 4·2 turns out to hold without assuming finite presentability of B: such statement is actually a consequence of [ Reference Mohammadi HassanabadiHas78 , theorem 2·4] (based on an earlier result of Houghton [ Reference HoughtonHou75 ]). For the sake of clarity we shall provide the reader with a short proof of this fact. In [ Reference Mohammadi HassanabadiHas78 ], the result holds more generally for certain permutational wreath products. We shall state it and prove it under that generality, as the argument is the same, and the result is of independent interest.

Recall that, given two groups A, B and a set X on which B acts, we define

\begin{align*}A \mathrm{Wr}_X B \;:\!=\; \left( \prod\limits_X A \right) \rtimes B \text{ and } A \wr_X B\;:\!=\; \left( \bigoplus\limits_X A \right) \rtimes B,\end{align*}

where B acts on the direct sum and the direct product by permuting the coordinates according to its action on X, i.e. $b \cdot h(x) = h(b^{-1}x)$ for all $b \in B$ , $x \in X$ , and $h \in \prod_XA$ or $\bigoplus_X A$ . Following [ Reference CornulierCor13 ], also recall that, given a group G acting on a set S, a subset $M \subset S$ is commensurated if the symmetric difference $gM \triangle M$ is finite for every $g \in G$ . For instance, if G is finitely generated, then G must be multi-ended if the action on itself by left-multiplication commensurates a subset that is neither finite nor cofinite (see e.g. [ Reference Scott and WallSW79 , page 173]).

Claim 4·4. An element $g \in \prod_{B/H} A$ conjugates the map $\sigma$ to the trivial section if and only if

\begin{align*}f(a)(xH) = g(xH) g(a^{-1}xH)^{-1}\end{align*}

for all $a \in B$ and $xH \in B/H$ .

Observe that

\begin{align*}(g (g^{-1})^a,a) = (g,1) ( 1,a) (g,1)^{-1} \text{ and } \sigma(a)= (\,f(a),a)\end{align*}

for all $a \in B$ and $xH \in B/H$ , so the claim follows.

Thus, our theorem will follow from the construction of an element $g \in \prod_{B/H} A$ satisfying Claim 4·4. First, we need to record a couple of elementary properties satisfied by f.

Because $(\,f(1),1)=\sigma(1)= 1$ , the first equality is clear. Next, given $a,b \in B$ , we know that

\begin{align*}\sigma(a)\sigma(b)= (\,f(a),a) (\,f(b),b)= (\,f(a) f(b)^a, ab)\end{align*}

must coincide with

\begin{align*}\sigma(ab)= (\,f(ab),ab),\end{align*}

so our second equality follows.

As a consequence of Claim 4·5, we deduce that:

By assumption, every morphism $H \to A$ is trivial, so it follows from the claim that $f(h)(H)=1$ for every $h \in H$ . As a consequence, the map

\begin{align*}g \;:\; \left\{ \begin{array}{c@{\quad}c@{\quad}c} B/H & \to & A \\[5pt] bH & \mapsto & f(b)(bH) \end{array} \right.\end{align*}

is well-defined. Indeed, for all $b \in B$ and $h \in H$ , it follows from Claims 4·5 and 4·6 that

\begin{align*}f(bh)(bH)= f(b)(bH) f(h)(H) = f(b)(bH).\end{align*}

Next, observe that, given $a \in B$ and $x \in B$ , applying Claim 4·5 with $b=a^{-1}x$ yields

\begin{align*}f(a)(xH) =f(x)(xH) f(a^{-1}x)(a^{-1}xH)^{-1} = g(xH) g(a^{-1}xH)^{-1}.\end{align*}

Thus, Claim 4·4 applies and allows us to conclude the proof of the first assertion of our theorem.

From now on, we assume that C lies in $A \wr_{B/H} B$ . Notice that B commensurates every preimage $g^{-1}(S)$ for $S \subset A$ . Indeed, given an $a \in B$ , it follows from Claim 4·4 that

\begin{align*}g^{-1}(S) \Delta a g^{-1}(S) = g^{-1}(S) \Delta g_a^{-1}(S) \subset \mathrm{supp}(\,f(a)),\end{align*}

for every $S \subset A$ , where $g_a$ denotes the map $g(a^{-1} \!\cdot\!)$ . The supports of the f(a) being finite, we indeed deduce that B commensurates each $g^{-1}(S)$ in $B/H$ . By assumption, each $g^{-1}(S)$ must be either finite or cofinite. We distinguish three cases. First, if g has finite support, then we are done. Next, if there exists some $m \in A$ such that $g^{-1}(m)$ is cofinite, then it follows from Claim 4·4 that the map $B/H \to A$ defined by $bH \mapsto g(bH)m^{-1}$ , which has finite support, conjugates C in B. Finally, assume that g has infinite support and that $g^{-1}(m)$ is not cofinite for every $m \in A$ . Then g has a cofinite support and it takes each value of A only finitely many times. Therefore, we can find a partition $A= S_1 \sqcup S_2$ with $g^{-1}(S_1)$ and $g^{-1}(S_2)$ both infinite (and disjoint), contradicting the fact that these two subsets must be finite or cofinite.

4·2. Consequences on automorphisms

Fix two groups F, H. In our definitions of automorphisms below, we refer to an element of $F \wr H$ as a pair (c, p) where $c \;:\; H \to F$ is a finitely supported map and where $p \in H$ . Consequently, $(c_1,p_1) \cdot (c_2,p_2)= (c_1(\cdot) c_2(p_1^{-1} \cdot), p_1p_2)$ for all $(c_1,p_1),(c_2,p_2) \in F \wr H$ .

Automorphisms of H. Given an automorphism $\varphi \in \mathrm{Aut}(H)$ , the map

\begin{align*}\overline{\varphi} \;:\; \left\{ \begin{array}{c@{\quad}c@{\quad}c} F \wr H & \to & F \wr H \\[5pt] (c,p) & \mapsto & \left( c \circ \varphi^{-1}, \varphi(p) \right) \end{array} \right.\end{align*}

defines an automorphism of $F \wr H$ . Moreover, the map $\varphi \mapsto \overline{\varphi}$ defines a monomorphism $\mathrm{Aut}(H) \hookrightarrow \mathrm{Aut}(F \wr H)$ . In the sequel, we will identify $\mathrm{Aut}(H)$ with its image in $\mathrm{Aut}(F \wr H)$ when this does not cause ambiguity.

H-equivariant automorphisms of $\bigoplus_H F$ . For every automorphism $\varsigma \in \mathrm{Aut}(\bigoplus_H F)$ that is equivariant with respect to the action of H on $\bigoplus_H F$ by coordinate-shifting, i.e. $\varsigma(h \cdot c)= h \cdot \varsigma(c)$ for all $h \in H$ and $c \in \bigoplus_H F$ , the map

\begin{align*}\overline{\varsigma} \;:\; \left\{ \begin{array}{c@{\quad}c@{\quad}c} F \wr H & \to & F \wr H \\[5pt] (c,p) & \mapsto & (\varsigma(c),p) \end{array} \right.\end{align*}

defines an automorphism of $F \wr H$ . We denote by $\mathrm{Aut}_H(\bigoplus_H F)$ the subgroup of $\mathrm{Aut}(F \wr H)$ consisting of all these automorphisms.

Verifying that these maps define automorphisms is a direct and straightforward computation. Notice that, for every automorphism $\varphi \in \mathrm{Aut}(F)$ ,

\begin{align*}\widehat{\varphi} \;:\; (c,h) \mapsto (\varphi \circ c,h), \ (c,h) \in F \wr H\end{align*}

yields an example of an H-equivariant automorphism of $\bigoplus_H F$ . When there is no possible confusion, we identify $\mathrm{Aut}(F)$ with its image in $\mathrm{Aut}(F \wr H)$ .

Our first observation is that these automorphisms, together with the inner automorphisms, are sufficient to generate the automorphism group in many cases.

Proposition 4·8. Let F be a finite group and H a one-ended finitely generated group. Then

\begin{align*}\mathrm{Aut}(F \wr H) = \mathrm{Inn}(F \wr H) \cdot \left(\mathrm{Aut}_H \left( \bigoplus\limits_H F \right)\rtimes \mathrm{Aut}(H)\right).\end{align*}

In particular, $\mathrm{Aut}(F \wr H)$ is generated by the inner automorphisms, the automorphisms of H, and the H-equivariant automorphisms of $\bigoplus_H F$ .

Proof. Let $\varphi \in \mathrm{Aut}(F \wr H)$ be an automorphism. As a consequence of Corollary 4·7, $\varphi(H)$ is a conjugate of H. Up to composing $\varphi$ with an inner automorphism, we assume that $\varphi(H)=H$ . Up to composing $\varphi$ with an automorphism of H, we also assume that $\varphi$ fixes H pointwise. Because $\bigoplus_HF$ is characteristic [ Reference NeumannNeu64 , theorem 9·12], it follows that there exists an automorphism $\varsigma \;:\; \bigoplus_H F \to \bigoplus_H F$ such that $\varphi(\,f,p)=(\varsigma(\,f),p)$ for every $(\,f,p) \in F \wr H$ . Necessarily, $\varsigma$ must be H-equivariant, i.e. $\varphi$ is induced by an H-equivariant automorphism of $\bigoplus_H F$ .

Thus, we have proved the decomposition

\begin{align*}\mathrm{Aut}(F \wr H) = \mathrm{Inn}(F \wr H) \cdot \mathrm{Aut}(H) \cdot \mathrm{Aut}_H \left( \bigoplus\limits_H F \right).\end{align*}

As $\mathrm{Aut}(H)$ intersects trivially and normalises $\mathrm{Aut}_H\left( \bigoplus_H F \right)$ , the desired conclusion follows.

In view of Proposition 4·8, a natural problem is to investigate the structure of the group of H-equivariant automorphisms of $\bigoplus_H F$ . In the sequel, we focus on the case when F is cyclic. Even restricting to this case, we shall see that understanding H-equivariant automorphisms may be quite tricky.

Units of F [H]. If F is cyclic, identifying $\bigoplus_H F$ with the group ring F[H] allows us to endow $\bigoplus_H F$ with a product $\cdot $ such that $(\bigoplus_H F, + , \cdot )$ is a ring. For every $u \in \bigoplus_H F$ (i.e. left- and right-invertible element), the map

\begin{align*}\overline{u} \;:\; \left\{ \begin{array}{c@{\quad}c@{\quad}c} F \wr H & \to & F \wr H \\[5pt] (c,p) & \mapsto & (c \cdot u, p) \end{array} \right.\end{align*}

defines an automorphism of $F \wr H$ . Observe that two distinct units induce two distinct automorphisms of $F \wr H$ . We denote by $F[H]^\times$ the subgroup of $\mathrm{Aut}(F \wr H)$ consisting of all these automorphisms.

Verifying that $\overline{u}$ defines an automorphism of $F \wr H$ is a straightforward computation once we have observed that the action of H on $\bigoplus_H F$ by coordinate-shifting coincides with the action of H by left multiplication in the ring F[H]. This observation also implies that $F[H]^\times \subset \mathrm{Aut}_H(\bigoplus_HF)$ . The reverse inclusion turns out to hold as well:

Proof. Let $\sigma \in \mathrm{Aut}( \bigoplus_H F)$ be an H-equivariant automorphism. Being H-equivariant amounts to saying that $\sigma(h\cdot c) = h \cdot \sigma(c)$ for all $h \in H$ and $c \in \bigoplus_H F$ . Because $\sigma$ is automatically F-linear, as F is cyclic, it follows that $\sigma(c_1 \cdot c_2) =c_1 \cdot \sigma(c_2)$ for all $c_1,c_2 \in \bigoplus_H F$ . In other words, $\sigma$ is a morphism of $(\bigoplus_H F)$ -module. Consequently,

\begin{align*}\sigma(c)= \sigma(c \cdot 1) = c \cdot \sigma(1) \text{ for every } c \in \bigoplus_H F.\end{align*}

The same observation applies to the inverse $\sigma^{-1}$ of $\sigma$ , i.e. $\sigma^{-1}(c)=c \cdot \sigma^{-1}(1)$ for every $c \in \bigoplus_H F$ . From

\begin{align*}1 = \sigma \circ \sigma^{-1}(1)= \sigma^{-1}(1) \sigma(1) \text{ and } 1 = \sigma^{-1} \circ \sigma (1) = \sigma(1) \sigma^{-1}(1),\end{align*}

we deduce that $\sigma^{-1}(1)$ is the left- and right-inverse of $\sigma(1)$ . In particular, $\sigma(1)$ is a unit.

As a consequence of Lemma 4·9, understanding the H-equivariant automorphisms of $F \wr H$ is as hard as understanding the units of the group ring F[H]. The latter problem is well-known to be quite difficult. Indeed, the so-called Kaplansky unit conjecture, which states that the group ring F[H] has only trivial units when F is a field and H torsion-free, remained an open problem for a long time and has been disproved only recently [ Reference GardamGar21 ]. Therefore, Lemma 4·9 suggests that the structure of $\mathrm{Aut}_H(\bigoplus_H F)$ , and a fortiori of $\mathrm{Aut}(F \wr H)$ , may be highly non-trivial and goes beyond the scope of our current understanding in full generality.

Before turning to the proof of Corollary 1·8 from the introduction, let us record the following elementary observation:

Proof. Let $g \in \bigoplus_BA$ be an element normalising B, i.e. $gBg^{-1}=B$ . For every $b \in B$ , observe that $gbg^{-1} \cdot b^{-1} \in gBg^{-1} \cdot B=B$ and that

\begin{align*}g \cdot bg^{-1}b^{-1} \in \bigoplus_BA \cdot \bigoplus_BA = \bigoplus_BA,\end{align*}

hence $gbg^{-1}b^{-1}=1$ . In other words, g commutes with all the elements of B. Because B acts on $\bigoplus_BA$ by permuting the coordinates, it follows that all the coordinates of g must be identical. When thought of as a map $B \to A$ , this amounts to saying that g is constant.

Proof of Corollary 1·8. It follows from Proposition 4·8 and Lemma 4·9 that

\begin{align*}\mathrm{Aut}(A \wr B ) = \mathrm{Inn}(A \wr B ) \cdot A [B]^\times \cdot \mathrm{Aut}(B).\end{align*}

Because a conjugation by an element of $A \wr B$ decomposes as a product of a conjugation by an element of $\bigoplus_BA$ with a conjugation by an element of B, and that the latter also belongs to $\mathrm{Aut}(B)$ , we have

\begin{align*}\mathrm{Aut}(A \wr B ) = \bigoplus\limits_B A \cdot \mathrm{Aut}(B) \cdot A[B]^\times \cdot \mathrm{Aut}(B)= \bigoplus\limits_{B} A \cdot A [B]^\times \cdot \mathrm{Aut}(B),\end{align*}

where the second equality follows from the fact that $\mathrm{Aut}(B)$ normalises $A[B]^\times$ , which implies $\mathrm{Aut}(B) \cdot A[B]^\times = A[B]^\times \cdot \mathrm{Aut}(B)$ . Clearly, $\bigoplus_{B} A$ and $A [B]^\times$ intersect trivially and commute, so $\bigoplus_{B} A \cdot A [B]^\times = \bigoplus_{B} A \oplus A [B]^\times$ . Also, observe that $\mathrm{Aut}(B)$ intersects trivially $\bigoplus_{B} A \cdot A [B]^\times$ . Indeed, if $\bar{\varphi} = \iota(g) \circ \bar{u}$ for some automorphism $\varphi \in \mathrm{Aut}(B)$ , some unit $u \in A[B]^\times$ , and some element $g \in \bigoplus_BA$ , then $\varphi(p)=gpg^{-1}$ for every $p \in B$ . But we deduce from $B= \bar{\varphi}(B)=gBg^{-1}$ and from Fact 4·10 that $g \in B$ , and in fact that $g=1$ since we already know that g belongs to $\bigoplus_BA$ . Therefore, $\varphi$ must be the identity, justifying that $\mathrm{Aut}(B)$ indeed intersects trivially $\bigoplus_{B} A \cdot A [B]^\times$ . Moreover, $\mathrm{Aut}(B)$ clearly normalises both $\bigoplus_{B} A$ and $A [B]^\times$ . The desired decomposition of $\mathrm{Aut}(A \wr B)$ follows.

Outside the one-ended case, the elementary automorphisms introduced so far are not sufficient to generate the automorphism group of the corresponding lamplighter group. It seems difficult to exhibit a structure that is both completely general and really informative. Nevertheless, one can push further.

Partial conjugations. Assume that F is an abelian group and that H splits as a free product $H_1 \ast H_2$ . Given an $a \in \bigoplus_H F$ , the map

\begin{align*}\kappa \;:\; s \mapsto \left\{ \begin{array}{c@{\quad}l} s & \text{if $s \in F \cup H_1$} \\[5pt] asa^{-1} & \text{if $s \in H_2$} \end{array} \right.\end{align*}

induces an automorphism of $F \wr H$ .

Here, $\kappa$ is only defined on generators of $F \wr H$ , and it is naturally extended to words of generators. In order to justify that $\kappa$ induces a well-defined morphism $F \wr H \to F \wr H$ , it suffices to show that a relator of $F,H_1,H_2$ or from the relative presentation

\begin{align*}\left\langle F, H \mid \left[ hfh^{-1}, f \right] =1 \text{ where } f \in F, h \in H \backslash \{1\} \right\rangle\end{align*}

has always trivial image under $\kappa$ . Clearly, $\kappa$ sends a relator of $F,H_1,H_2$ to the identity element, so the desired conclusion follows from the observation that $\kappa$ fixes $\bigoplus_H F$ pointwise. Indeed, let $f \in F$ and $h \in H$ be two elements. Decompose h as an alternating product $a_1b_1 \cdots a_nb_n$ where $a_1,\ldots, a_n \in H_1$ and $b_1, \ldots, b_n \in H_2$ . Observe that

\begin{align*}\begin{array}{l@{\quad}c@{\quad}l} \kappa \left( a_1b_1 \cdot f \cdot b_1^{-1}a_1^{-1} \right) & = & a_1 ab_1a^{-1} \cdot f \cdot ab_1^{-1} a^{-1}a_1^{-1}\\[5pt] \\[5pt] & = & a_1 ab_1 \cdot f \cdot b_1^{-1} a^{-1}a_1^{-1} = a_1b_1 \cdot f \cdot b_1^{-1}a_1, \end{array} \end{align*}

where the penultimate equality is justified by the fact that F is abelian and where the last equality is justified by the fact that $b_1fb_1^{-1}$ commutes with a since $\bigoplus_HF$ is abelian. By iterating, one shows that $\kappa \left( hfh^{-1} \right) = hfh^{-1}$ , as desired. That the kernel of $\kappa$ is trivial is a direct consequence of the facts that $\kappa$ induces the identity on H and restricts to the identity on $\bigoplus_HF$ . Finally, the surjectivity of $\kappa$ follows from the easy observation that $F, H_1,H_2$ all lie in the image of $\kappa$ .

Transvections. Assume that F is abelian and that H splits as a free product $H' \ast Z$ , where Z is cyclic, say generated by $z \in Z$ of order $r \in \mathbb{N} \cup \{\infty\}$ . Given an $a \in \bigoplus_H F$ such that $(az)^r=1$ if $r\lt \infty$ , the map

\begin{align*}\kappa \;:\; s \mapsto \left\{ \begin{array}{c@{\quad}l} s & \text{if $s \in F \cup H'$} \\[5pt] as & \text{if $s=z$}, \end{array} \right.\end{align*}

induces an automorphism of $F \wr H$ .

Again, $\kappa$ is only defined on generators and is naturally extended to words of generators. In order to justify that $\kappa$ induces a well-defined automorphism of $F \wr H$ , it suffices to reproduce the previous argument almost word for word, starting from the relative presentation

\begin{align*}\left\langle F,H',z \mid z^r=1, \ \left[hfh^{-1},f \right]=1 \text{ where } f \in F, h \in \langle H',z \rangle \backslash\{1\} \right\rangle.\end{align*}

When r is finite, the elements $a \in \bigoplus_H F$ satisfying $(az)^r=1$ are obtained in the following way. Decompose $\mathrm{supp}(a)$ as a disjoint union $S_1 \sqcup \cdots \sqcup S_k$ of $\langle z \rangle$ -orbits. Write a as $a_1 \cdots a_k$ such that $\mathrm{supp}(a_i)=S_i$ for every $1 \leq i \leq k$ . Observe that

\begin{align*}\begin{array}{l@{\quad}c@{\quad}l} (az)^r & = & a \cdot zaz^{-1} \cdots z^{r-1}az^{-r+1} \\[5pt] & = & \left( a_1 \cdot za_1z^{-1} \cdots z^{r-1}a_1z^{-r+1} \right) \cdots \left( a_k \cdot za_kz^{-1} \cdots z^{r-1}a_kz^{-r+1} \right). \end{array}\end{align*}

Because z has order r, each $S_i$ has size r and z acts on it through a cyclic shifting, hence

\begin{align*}a_i \cdot za_iz^{-1} \cdots z^{r-1}a_i z^{-r+1} = \prod\limits_{s \in S_i} s.\end{align*}

Thus, in order to construct a, we took a collection of $\langle z \rangle$ -invariant subsets $S_1, \ldots, S_k \subset H$ , and, for each $1 \leq i \leq k$ , we coloured the lamps in $S_i$ such that the product of all the colours is trivial.

We emphasise that the free product decompositions used in order to define partial conjugations and transvections may be trivial. For instance, $\mathbb{Z} \wr \mathbb{Z}$ has transvections (which turn out to generate the outer automorphism group up to finite index, see Proposition 4·13 below).

Thanks to these new families of automorphisms, we are now able to generalise Proposition 4·8 to some multi-ended groups.

Proposition 4·11. Let F be a finite cyclic group and H a finitely generated group that decomposes as a free product of one-ended and cyclic groups. Then

\begin{align*}\mathrm{Aut}(F \wr H) = \left(\mathrm{PConj}(F \wr H) \cdot \mathrm{Trans}(F \wr H) \cdot F[H]^\times\right)\rtimes \mathrm{Aut}(H) .\end{align*}

We emphasize that, in this statement, the partial conjugations and the transvections are taken with respect to a fixed free product decomposition.

Proof of Proposition 4·11. Decompose H as a free product $H_1 \ast \cdots \ast H_n$ of one-ended and cyclic groups, and let $\varphi\in \mathrm{Aut}(F \wr H)$ be an automorphism. Because $\bigoplus_H F$ is characteristic, up to composing $\varphi$ with an automorphism of H we assume that $\varphi$ induces the identity on H.

Assume that there exists some $1 \leq i \leq n$ such that $H_i$ is one-ended. As a consequence of Theorem 1·1, $\varphi(H_i)$ must lie in a conjugate of H. So there exist $1 \leq j \leq n$ and $g \in F \wr H$ such that $\varphi(H_i) \subset gH_jg^{-1}$ . Since we know that the projection of $\varphi(H_i)$ on H is $H_i$ , necessarily $j=i$ and $g \in (\bigoplus_H F) H_i$ . Therefore, $\varphi(H_i) \subset a H_i a^{-1}$ for some $a \in \bigoplus_H F$ . By composing $\varphi$ with a partial conjugation, we can assume that $\varphi(H_i) \subset H_i$ . Because $\varphi$ induces the identity on H, it follows that $\varphi$ fixes $H_i$ pointwise.

Next, assume that there exists some $1 \leq i \leq n$ such that $H_i$ is cyclic. Fix a generator $h_i \in H_i$ . Because $\varphi$ induces the identity on H, there must exist some $a \in \bigoplus_H F$ such that $\varphi(h_i)=ah_i$ . Of course, $h_i$ and $ah_i$ have the same order, since $\varphi$ is an automorphism, so we can assume that $\varphi$ fixes $h_i$ up to composing it with a transvection.

Thus, we have proved that, up to composing $\varphi$ with an automorphism of H followed by a product of pairwise commuting partial conjugations and transvections, we can assume that $\varphi$ fixes H pointwise. Then $\varphi$ is induced by an H-equivariant automorphism of $\bigoplus_H F$ , and the desired conclusion follows from Lemma 4·9.

In other words, we have proved that

\begin{align*}\mathrm{Aut}(F \wr H) = \mathrm{Aut}(H) \cdot \mathrm{PConj}(F \wr H) \cdot \mathrm{Trans}(F \wr H) \cdot F[H]^\times.\end{align*}

Because $\mathrm{Aut}(H)$ normalises $\mathrm{PConj}(F \wr H)$ , $\mathrm{Trans}(F \wr H)$ , and $F[H]^\times$ , and because it intersects trivially $\mathrm{PConj}(F \wr H) \cdot \mathrm{Trans}(F \wr H) \cdot F[H]^\times$ , the desired decomposition follows.

Proposition 4·11 describes the automorphism groups of many interesting lamplighter groups, but we are still far away from a global picture. Let us conclude this section with a few open questions.

In the paper, we have mainly focused on lamps with only finitely many colours. Even though we have seen that the proof of Proposition 4·8 also provides interesting information outside this situation, many cases of interest have not been covered. As a particular case:

We excluded the case $n=1$ from the question because it can be easily solved, due to the fact that $\mathbb{Z}$ satisfies the unit Kaplansky conjecture over $\mathbb{Z}$ . In fact, we can prove more generally that:

Proposition 4·13. Let Z be a cyclic group (finite or infinite). Then

\begin{align*}\mathrm{Aut}(Z \wr \mathbb{Z})= Z[\mathbb{Z}]^\times \cdot \mathrm{Inn}(Z \wr \mathbb{Z}) \cdot \{ \mathrm{Trans}(z), z \in Z\} \cdot \langle \iota \rangle,\end{align*}

where $\iota$ denotes the automorphism that inverts the $\mathbb{Z}$ -factor and where $\mathrm{Trans}(z)$ denotes the transvection associated to z. Moreover:

1. (i) if $Z= \mathbb{Z}$ , then $Z[\mathbb{Z}]^\times = \{ \pm X^n \mid n \in \mathbb{Z}\} = \langle X, -1 \rangle \simeq \mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}$ ;

2. (ii) if $Z= \mathbb{Z}/k\mathbb{Z}$ with $k=p_1^{r_1}\cdots p_n^{r_n}$ the prime decomposition of k, then $Z[\mathbb{Z}]^\times$ consists of elements P such that for each i, there exists exactly one coefficient of P which is prime with $p_i$ . Moreover, $Z[\mathbb{Z}]^\times$ is finitely generated if and only if $r_1= \cdots = r_n=1$ .

As a consequence, $\mathrm{Out}(Z \wr \mathbb{Z})$ is virtually infinite cyclic if $Z= \mathbb{Z}$ , it is virtually $\mathbb{Z}^{n-1}$ if Z has order a product of n distinct primes, and it contains an infinitely generated abelian subgroup of finite index if the order of Z is divisible by a square.

In this statement, we identify $Z[\mathbb{Z}]$ with $Z[X,X^{-1}]$ . Automorphisms of lamplighter groups $(\text{finite cyclic group}) \wr \mathbb{Z}$ have been already considered in the literature; see for instance [ Reference Bachmuth, Baumslag, Dyer and MochizukiBBDM87, Reference Stein, Taback and WongSTW15, Reference Matucci and SilvaMS21 ]. We include a short and elementary proof for completeness.

Proof of Proposition 4·13. Let $\varphi$ be an automorphism of $Z \wr \mathbb{Z}$ . Because $\bigoplus_\mathbb{Z} Z$ is characteristic, $\varphi$ induces an isomorphism $\mathbb{Z} \to \mathbb{Z}$ , which we can assume to be trivial up to composing $\varphi$ with $\iota$ . Consequently, there exists some $a \in \bigoplus_\mathbb{Z} Z$ such that $\varphi(z)=az$ where z denotes a generator of the $\mathbb{Z}$ -factor. Up to composing $\varphi$ with a transvection, we assume that $a=1$ . So $\varphi$ fixes pointwise the $\mathbb{Z}$ -factor. In other words, $\varphi$ has to be a $\mathbb{Z}$ -equivariant automorphism of $\bigoplus_\mathbb{Z} Z$ , which amounts to saying that $\varphi$ belongs to $Z[\mathbb{Z}]^\times$ according to Lemma 4·9. Thus, we have proved that

\begin{align*}\mathrm{Aut}(Z \wr \mathbb{Z})= Z[\mathbb{Z}]^\times \cdot \mathrm{Trans}(Z \wr \mathbb{Z}) \cdot \langle \iota \rangle.\end{align*}

The desired composition of $\mathrm{Aut}(Z \wr \mathbb{Z})$ now follows from the following observation:

Let $\rho \;:\; \bigoplus_\mathbb{Z} Z \to Z$ denote the morphism that computes the sum of the coordinates. We claim that, for every $g \in \bigoplus_\mathbb{Z} Z$ , the transvection $\mathrm{Trans}(g)$ is an inner automorphism if and only if $g \in \mathrm{ker}(\rho)$ . This observation implies Claim 4·14.

Fix a $g \in \bigoplus_\mathbb{Z} Z$ and let z be a generator of the $\mathbb{Z}$ -factor of $Z \wr \mathbb{Z}$ . Then $\mathrm{Trans}(g)$ is an inner automorphism if and only if there exists $a \in \bigoplus_\mathbb{Z} Z$ such that $g \cdot z = aza^{-1}$ , or equivalently $g=aza^{-1}z^{-1}$ since $aza^{-1} = aza^{-1} z^{-1} \cdot z$ . In other words, we are trying to find an element $a \in \bigoplus_\mathbb{Z} Z$ such that, by adding a to the z-shift or by its inverse, we obtain g. There is only one possibility: the leftmost coordinate of a has to coincide with the leftmost coordinate of g, say $a_i=g_i$ ; next, $a_{i+1}$ has to be $g_{i+1}+g_i$ ; and $a_{i+2}$ has to be $g_{i+2}+g_{i+1}+g_i$ , and so on. A priori, the element a thus defined belong to $\prod_{\mathbb{Z}} Z$ . It belongs to $\bigoplus_{\mathbb{Z}} Z$ if and only if $\rho(g)=0$ . This concludes the proof of Claim 4·14.

Now, we focus on the structure of $Z[\mathbb{Z}]^\times$ , which we identify with $Z[X,X^{-1}]^\times$ . The first item of our statement is a direct consequence of the following observation:

Claim 4·15. If R is an integral domain, then

\begin{align*}R\left[ X,X^{-1} \right]^\times= \{ r X^n \mid r \in R^\times, n \in \mathbb{Z} \} = \langle R^\times, X \rangle \simeq R^\times \oplus \mathbb{Z}.\end{align*}

Finding the units in $R[X,X^{-1}]$ amounts to finding the divisors of the $X^n$ , $n\in \mathbb{N}$ , in R[X]. But, because R is an integral domain, the product of two polynomials such that at least of them is not monomial cannot be monomial since the products of the terms of highest and lowest degrees provide at least two terms. This proves Claim 4·15.

Next, let us prove the second item of our statement. Let $Q\in Z[X,X^{-1}]^\times$ . Then its projection $\bar{Q}$ in $\mathbb Z/p_i\mathbb Z[X,X^{-1}]$ must be invertible as well, which by Claim 4·15 forces Q to be of the required form: indeed, $\bar{Q}$ must be a non-trivial monomial, meaning that exactly one coefficient of Q is prime with $p_i$ . Conversely let $Q=\sum b_kX^{k}$ be such that each $p_i$ divides all but one $b_k$ . If $k\neq \ell$ , we therefore have $b_kb_\ell\in p_1\ldots p_n\mathbb Z$ . Letting $Q'=\sum b_kX^{-k}$ , we deduce that

\[QQ'\in \sum_k b_k^2+p_1\ldots p_nZ[X,X^{-1}].\]

Note that each $p_i$ is prime with $ \sum_k b_k^2$ , which is therefore invertible. On the other hand, since $p_1\ldots p_n$ is nilpotent, so are all the elements in $J=p_1\ldots p_n Z \left[X,X^{-1}\right]$ , and therefore $1+a$ is invertible for all $a\in J$ . This shows that $(\sum_k b_k^2)^{-1}QQ'$ is invertible, and so is Q. Hence we are done.

Let us check that if some $r_i\geq 2$ , then $Z \left[X,X^{-1}\right]^\times$ is not finitely generated. For convenience, we write $p\;:\!=\;p_i$ and $r\;:\!=\;r_i$ . Our description of invertible elements implies that the projection $Z \left[X,X^{-1}\right]\to \mathbb{Z}/p^{r}\mathbb{Z}\left[X,X^{-1}\right]$ induces a surjective morphism between the groups of invertible elements. Hence it is enough to show infinite generation of the group $ \mathbb{Z}/p^{r}\mathbb{Z}\left[X,X^{-1}\right]^\times$ .

In the case where $Z=\mathbb{Z}/p^{r}\mathbb{Z}$ , an element Q is invertible if exactly one coefficient of Q is invertible. In other words,

\[\mathbb{Z}/p^{r}\mathbb{Z}\left[X,X^{-1}\right]^\times=\mathbb{Z}/p^r\mathbb{Z}^\times+p \mathbb{Z}/p^{r}\mathbb{Z}\left[X,X^{-1}\right].\]

Reducing further modulo $p^2$ yields a surjective morphism to

\[G\;:\!=\;\left\langle X,A+ p\mathbb{Z}/p^{2}\mathbb{Z} \left[X,X^{-1}\right] \right\rangle,\]

where A is the image of $\mathbb{Z}/p^r\mathbb{Z}^\times$ in $\mathbb{Z}/p^2\mathbb{Z}^\times$ .

For every m, denote $G_m$ be the group spanned by powers of X and elements of $A+ p\mathbb{Z}/p^{2}\mathbb{Z} \left[X,X^{-1}\right] $ of degree at most m in $X,X^{-1}$ . Note that for all $x,y\in \mathbb{Z}/p^2\mathbb{Z}[X,X^{-1}]$ , we have

\[(1+px)(1+py)=1+p(x+y).\]

Hence $G_m$ consists of elements of the form $X^m(a+px)$ , where $a\in A$ and x has degree at most m in $X,X^{-1}$ . In particular $G_m$ is strictly contained in $G_{m+1}$ , and since $G=\cup_m G_m$ , we deduce that G is not finitely generated.

Let us prove that $Z \left[X,X^{-1}\right]^\times$ is finitely generated if $r_i=1$ for all i. Precisely we will prove that a set of generators consists of invertible elements of the form $a_0+a_1X$ , with $a_0,a_1\in Z$ . Let us denote G the group generated by these elements. Assume $P\in Z \left[X,X^{-1}\right]^\times$ . Assume by contradiction that P does not belong to G. On multiplying P by a power of X (which belongs to G), we may assume that $P=a_0+a_1X\ldots+a_mx^m\in Z \left[X\right]$ . We will assume in addition that its degree m is minimal, and that among those, the number of $p_i$ ’s dividing $a_m$ is maximal.

Let i be such that $p_i$ is prime with $a_m$ . On reindexing the $p_i$ ’s, we may assume that $i=n$ . Note that $p_n$ divides $a_0$ . We then consider the element

\[Q=p_1p_2\ldots p_{n-1}X^{-1}+p_n.\]

Since $Q=X^{-1}(p_1p_1p_2\ldots p_{n-1}+p_nX)\in G$ , PQ does not belong to G. Note that the coefficient of degree $-1$ of PQ is a multiple of $p_1\ldots p_n$ hence equals 0. Therefore $PQ\in Z \left[X\right]$ . Besides, the coefficient of degree m of PQ is $c_m=p_nb_m=0$ . By minimality of the degree of P we have $c_m\neq 0$ , and therefore the degree of PQ is exactly m. Since $p_n$ is prime with $b_m$ , we deduce that the number of $p_i$ ’s dividing $c_m$ is strictly larger than that of $b_m$ , contradicting our maximality assumption. Hence $G=Z \left[X,X^{-1}\right]^\times$ and we are done.

Finally, the description of $\mathrm{Out}(Z\wr \mathbb{Z})$ follows from the description of $\mathrm{Aut}(Z\wr \mathbb{Z})$ together with the observation that $Z[\mathbb{Z}]^\times \cap \mathrm{Inn}(Z \wr \mathbb{Z}) = \langle X \rangle$ .

For wreath products of the form $(\text{finite group})\wr ( \text{multi-ended group})$ , we only considered the case where the finite group is abelian, because otherwise the partial conjugations and transvections may not be well-defined. This suggests the following problem.

The difficulty in extending Proposition 4·11 to lamplighters over arbitrary multi-ended groups comes from the torsion, which may introduce sufficient flexibility in order to define new types of automorphisms.

Generalised partial conjugations. Assume that F is abelian and that H splits as a graph of groups with underlying graph $\Gamma$ . Fix a marked vertex $v_0 \in V(\Gamma)$ , an orientation on $\Gamma$ , and, for every edge $e \in E(\Gamma)$ , an element $a_e \in C(H_e) \cap \bigoplus_H F$ . Every path in $\Gamma$ is naturally labelled by the element obtained by reading the oriented edges it passes through. Assume that the label of every cycle is trivial. For every vertex $v \in V(\Gamma)$ , let $a_v \in \bigoplus_H F$ denote the label of a path from $v_0$ to v. (As a consequence of our previous assumption, this label does not depend on the path we chose.) Fix a maximal subtree $T \subset \Gamma$ , let $e_1, \ldots, e_k$ denote the oriented edges outside T, and let $t_1, \ldots, t_k$ denote the corresponding stable letters. The map

\begin{align*}\kappa \;:\; s \mapsto \left\{ \begin{array}{c@{\quad}l} s & \text{if $s \in F$} \\[5pt] a_vsa_v^{-1} & \text{if $s \in H_v$} \\[5pt] a_{o(e_i)}t_i a_{o(e_i)}^{-1} & \text{if $s=t_i$} \end{array} \right.\end{align*}

induces an automorphism of $F \wr H$ .

The proof is in the same spirit as before, but we use Bass-Serre theory in order to find a relative presentation of H. We do not include the details here. Similarly, one can define generalised transvections.

Understanding such automorphisms in full generality turns out to be tricky since they seem to depend on the structures of the finite groups involved and on how they interact together. As a test question, we propose: