On the normalizer of an iterated wreath product

TL;DR

This paper characterizes the normalizer of an iterated wreath product of a group G within the symmetric group on G^n, revealing it as a semi-direct product involving automorphisms of G.

Contribution

It provides a precise description of the normalizer of iterated wreath products in symmetric groups, including a recursive description of automorphism actions.

Findings

Normalizer equals a semi-direct product involving automorphisms

Explicit recursive description of automorphism action

Applicable to both finite and infinite groups

Abstract

Given a group $G$ and $n\geq 0$, let $W(G,n)$ be the associated iterated wreath product -- unrestricted when $G$ is infinite -- viewed as a permutation group on $G^n$. We prove that the normalizer of $W(G,n)$ in the symmetric group $S(G^n)$ is equal to $M_n\ltimes W(G,n)$, where $M_n$ is isomorphic to~$\mathrm{Aut}(G)^n$. The action of $\mathrm{Aut}(G)^n$ on $W(G,n)$ is recursively described.

Equations

$$S(X_y{)}^k=S(X_{yk}),h_y^k=h_{yk},H_y^k=H_{yk},h\in H,y\in Y,k\in K.$$

$$Z\alpha =Z=M(kf)=M(h_y^{\alpha })=M(h_y)\alpha \subseteq X_y\alpha ,$$

$$x_y^{\prime }\alpha =x_y{h}_y\alpha =x_y\alpha f\in X_{y^{\prime }}.$$

$$x_k\gamma \Gamma =x_{k\gamma },x\in X,k\in K,\gamma \in \mathrm{Aut}(K),$$

$$h_k^{\gamma \Gamma }=h_{k\gamma },h\in H,k\in K,\gamma \in \mathrm{Aut}(K),$$

$$k{\Lambda }^{\gamma \Gamma }=k\gamma \Lambda ,k\in K,\gamma \in \mathrm{Aut}(K).$$

$$(x{m}^k{)}_k$$

$$=(xm{)}_1(k\Lambda )b^k=(xm{)}_k{b}^k=(xm{)}_k{b}_k^k=(xm{)}_k{h}_k^k=(xm{h}^k{)}_k.$$

$$x_k\beta =x_k{\beta }_k=x_k{m}_k^k=x_k{m}_k=(xm{)}_k=x_k{m}^{\ast },k\in K,$$

Full text

The author was partially supported by NSERC grant RGPIN-2020-04062

1. Introduction

Given a group $G$ and $n\geq 0$, we let $W(G,n)$ stand for the associated iterated standard wreath product, taken in the unrestricted sense when $G$ is infinite, where $G$ is viewed as a permutation group on itself via the regular representation and, correspondingly, $W(G,n)$ is viewed as a permutation group on $G^{n}$.

If $p$ is a prime number, $W(C_{p},n)$ is a Sylow $p$-subgroup of the symmetric group $S_{p^{n}}$ [K], and the normalizer of $W(C_{p},n)$ in $S_{p^{n}}$ is equal to $M_{n}\ltimes W(C_{p},n)$, where $M_{n}$ is isomorphic to $C_{p-1}^{n}$ [BJ, Appendix 13]. As it turns out, $C_{p}$ is but the simplest group for which the latter holds.

Theorem. The normalizer of $W(G,n)$ in the symmetric group $S(G^{n})$ is equal to $M_{n}\ltimes W(G,n)$, where $M_{n}$ is isomorphic to $\mathrm{Aut}(G)^{n}$.

The action of $\mathrm{Aut}(G)^{n}$ on $W(G,n)$ is recursively described in Section 3.

The problem of computing normalizers in the general context of finite permutation groups is addressed in [CCR] from an algorithmic standpoint (see also references therein). The normalizers of certain finite subgroups of the so-called homogeneous symmetric groups are determined in [GK]. The automorphism group of $W(C_{p},n)$ for $p$ odd is studied in [B] and [L], and the the automorphism group of the normalizer $W(C_{p},n)$ in $S_{p^{n}}$ is described in [BL]. A variation of $W(C_{p},n)$ with first term $C_{p^{e}}$, as well as the corresponding automorphism group, is investigated in [R]. The irreducible representations of the iterated wreath product of finite cyclic groups are elucidated in [IW].

2. Unrestricted wreath product of permutation groups

Given permutation groups $H$ and $K$ acting on sets $X$ and $Y$ respectively, the unrestricted wreath product $H\wr K$ of $H$ and $K$ is defined as a permutation group on $Z=X\times Y$ as follows.

For $y\in Y$ we set $x_{y}=(x,y)$, where $x\in X$, and let $X_{y}=\{x_{y}\,|\,x\in X\}$, so that ${\mathcal{P}}=\{X_{y}\,|\,y\in Y\}$ is partition of $Z$. Given $h\in H$ and $y\in Y$ we define $h_{y}\in S(X_{y})$ by $x_{y}h_{y}=(xh)_{y}$ for all $x\in X$ and let $H_{y}=\{h_{y}\,|\,h\in H\}$, whence the maps $h\mapsto h_{y}$ and $x\mapsto x_{y}$ yield a similarity from $H$ to $H_{y}$.

We naturally view each $S(X_{y})$ as a subgroup of $S(Z)$. Let $S_{Y}(Z)$ be the subgroup of $S(Z)$ of all permutations $f$ of $Z$ such that $X_{y}f=X_{y}$ for all $y\in Y$, and let $B$ be the subgroup of all $f\in S_{Y}(Z)$ such that the restriction $f_{y}$ of $f$ to $X_{y}$ is in $H_{y}$ for each $y\in Y$.

We have an embedding $\Lambda:K\to S(Z)$, where $k\Lambda$ is given by $x_{y}\mapsto x_{yk}$ for all $x\in X$, $y\in Y$, and $k\in K$. If no risk of confusion is possible we will identify $K$ with $K\Lambda$. Thus

[TABLE]

It follows that $K$ normalizes $B$ with $B\cap K=1$, and we set $W=H\wr K=K\ltimes B$ and $N=N_{S(Z)}W$. The group $B$ is known as the base group of $W$ and is isomorphic to the Cartesian product of the $H_{y}$, $y\in Y$. Note that if $Y$ is finite then $B$ is the internal direct product of its subgroups $H_{y}$, $y\in Y$. Observe also that if $|X|=1$ then $W$ is similar to $K$ (this case arises in the inductive proof of our main result when passing from $n=0$ to $n=1$).

Lemma A. If $H$ and $K$ are transitive then so is $W$.

Proof. Given $x,x^{\prime}\in X$ and $y,y^{\prime}$ there are $h\in H$ and $k\in K$ such that $xh=x^{\prime}$ and $yk=y^{\prime}$, so $x_{y}h_{y}k=x^{\prime}_{y^{\prime}}$.

Lemma B. If $K$ is semiregular, then conjugation by any element of $N$ sends every $H_{y}$ back into $B$.

Proof. Suppose, if possible, that there is $\alpha\in N$ such that for some $y\in Y$ and $h\in H$ we have $h_{y}^{\alpha}\notin B$. Then $h_{y}^{\alpha}=kf$, where $1\neq k\in K$ and $f\in B$. For $\gamma\in S(Z)$ let $M(\gamma)$ be the subset of $Z$ of points moved by $\gamma$. Thus $M(kf)=Z$ and $M(h_{y})\subseteq X_{y}$, so

[TABLE]

which implies $Y=\{y\}$ and a fortiori $K=1$, a contradiction.

Lemma C. If $K$ is semiregular and $H$ is transitive, then the partition ${\mathcal{P}}$ of $Z$ is $N$-stable. Thus, there is a group homomorphism $\Omega:N\to S({\mathcal{P}})$, say $\alpha\mapsto\widetilde{\alpha}$, with kernel $N\cap S_{Y}(Z)$.

Proof. Let $\alpha\in N$ and $y\in Y$. Take any $x\in X$. Then $x_{y}\alpha\in X_{y^{\prime}}$ for a unique $y^{\prime}\in Y$. Given any other $x^{\prime}\in X$, we have $x^{\prime}=xh$ for some $h\in H$, so that $x^{\prime}_{y}=x_{y}h_{y}$. By Lemma B, $h_{y}^{\alpha}=f\in B$, so

[TABLE]

Thus $X_{y}\alpha\subseteq X_{y^{\prime}}$ and therefore $X_{y}\subseteq X_{y^{\prime}}\alpha^{-1}$. But $\alpha^{-1}\in N$ and $X_{y^{\prime}}\alpha^{-1}\cap X_{y}\neq\emptyset$, so by above $X_{y^{\prime}}\alpha^{-1}\subseteq X_{y}$. Hence $X_{y^{\prime}}\alpha^{-1}=X_{y}$ and therefore $X_{y}\alpha=X_{y^{\prime}}$.

We suppose for the remainder of this section that $K$ is regular. Thus, we may assume without loss that $Y=K$ under the regular representation.

Lemma D. The map $\Gamma:\mathrm{Aut}(K)\to N$ given by

[TABLE]

is a group monomorphism satisfying

[TABLE]

[TABLE]

Proof. This is a routine calculation.

We suppose for the remainder of this section that $H$ is transitive.

Lemma E. The image of $\Omega:N\to S({\mathcal{P}})$ is $\widetilde{\mathrm{Aut}(K)\Gamma}\ltimes\widetilde{\Lambda K}$.

Proof. As $B\subseteq\ker\Omega$, we see that $\widetilde{N}$ is contained in the normalizer of the regular subgroup $\widetilde{\Lambda K}$. It is well-known [H, Ch. 6] that this normalizer is the semidirect product of $\widetilde{\Lambda K}$ by the group of permutations of ${\mathcal{P}}$ associated to the automorphisms of $K$. This group is precisely $\widetilde{\mathrm{Aut}(K)\Gamma}$.

If no risk of confusion is possible, we will write $\mathrm{Aut}(K)\ltimes K$ instead of $\mathrm{Aut}(K)\Gamma\ltimes\Lambda K$.

Lemma F. We have $N=(\mathrm{Aut}(K)\ltimes K)(N\cap S_{K}(Z))$.

Proof. This is a consequence of Lemmas C and E.

We assume in what follows that $N_{S(X)}H=M\ltimes H$ for some subgroup $M$ of $S(X)$. To each $m\in M$ we associate $m^{*}\in N\cap S_{K}(Z)$ defined by $x_{k}m^{*}=(xm)_{k}$ for all $x\in X$ and $k\in K$. It is clear that the map $m\mapsto m^{*}$ yields a group monomorphism $M\to N\cap S_{K}(Z)$ whose image will be denoted by $M^{*}$.

Lemma G. We have $N=M^{*}\mathrm{Aut}(K)\ltimes W$, where $\mathrm{Aut}(K)\cap M^{*}=1$, $[M^{*},K]=1$, and $[M^{*},\mathrm{Aut}(K)]=1$.

Proof. Recall that for $f\in S_{K}(Z)$, $h\in H$, and $k\in K$, we write $f_{k}$ for the restriction of $f$ to $X_{k}$, and $h_{k}\in H_{k}$ for the map $x_{k}\mapsto(xh)_{k}$. Given $m\in M$ and $k\in K$, we let $m_{k}\in S(X_{k})$ be given by $x_{k}m_{k}=(xm)_{k}$ for all $x\in X$, and set $M_{k}=\{m_{k}\,|\,m\in M\}$. Thus $N_{S(X_{k})}H_{k}=M_{k}\ltimes H_{k}$.

We claim that $N\cap S_{K}(Z)=M^{*}B$. Clearly $M^{*}B\subseteq N\cap S_{K}(Z)$. Let $\alpha\in N\cap S_{K}(Z)$. Given any $k\in K$, we have $H_{k}^{\alpha}\subseteq W\cap S(X_{k})=H_{k}$, whence $\alpha_{k}\in M_{k}\ltimes H_{k}$. Thus, there is $b\in B$ such that $\beta=\alpha b$ satisfies $\beta_{k}\in M_{k}$ for all $k\in K$. We assert that $\beta=m^{*}$ for some $m\in M$, so that $\alpha=m^{*}b^{-1}\in M^{*}B$, whence $N\cap S_{K}(Z)\subseteq M^{*}B$, thereby proving the claim. To see the assertion, let $k\in K$. As $\beta_{k}\in M_{k}$, there exists $m^{k}\in M$ (where the superscript indicates dependence on $k$) such that $\beta_{k}=m^{k}_{k}$, that is, $x_{k}\beta_{k}=(xm^{k})_{k}$ for all $x\in X$. Setting $m=m^{1}$, we proceed to show that $\beta=m^{*}$. As $k\Lambda\in W$ and $\beta\in N$, we have $(k\Lambda)^{\beta}\in W$, so there exist $\ell^{k}\in K$ and $b^{k}\in B$ (where the superscripts indicate dependence on $k$) such that $(k\Lambda_{k})^{\beta}=(\ell^{k}\Lambda)b^{k}$. Here $\beta,k\Lambda,\ell^{k}\Lambda,b^{k}\in N$, with $\beta,b^{k}\in S_{K}(Z)$, so making use of the homomorphism from Lemma C, we infer $\ell^{k}=k$. On the other hand, as $b^{k}\in B$, the very definition of $B$ ensures the existence of $h^{k}\in H$ (where the superscript indicates dependence on $k$) such that $b^{k}_{k}=h^{k}_{k}$, that is, $(xb^{k})_{k}=(xh^{k})_{k}$ for all $x\in X$. Thus for every $x\in X$, we have

[TABLE]

Hence $m^{k}=mh^{k}$ for all $k\in K$. As $m^{k},m\in M$ and $h^{k}\in H$, with $M\cap H=1$, we deduce $m^{k}=m$ for all $k\in K$. Therefore

[TABLE]

which implies $\beta=m^{*}$, as stated. It now follows from Lemma F that $N=\mathrm{Aut}(K)KM^{*}B$. It is easy to see that $[M^{*},K]=1$ and $[M^{*},\mathrm{Aut}(K)]=1$, whence $N=M^{*}\mathrm{Aut}(K)W$. Here $M^{*}$ normalizes $W$ by construction, and $\mathrm{Aut}(K)$ normalizes $W$ by Lemma D.

Suppose, finally, that $m^{*}\gamma=kb$ for some $m\in M$, $\gamma\in\mathrm{Aut}(K)$, $k\in K$, and $b\in B$. Applying the homomorphism from Lemma C, we see that $k=1$ and $\gamma=1$. Then $m^{*}=b$, so $M\cap H=1$ forces $m=1$ and $b=1$.

3. Proof of the theorem and action of $\mathrm{Aut}(G)^{n}$ on $W(G,n)$

Lemma H. The unrestricted iterated wreath product $W(G,n)$ is a transitive subgroup of $S(G^{n})$.

Proof. We argue by induction on $n$. The result is clear if $n=0$. Suppose $W(G,n)$ is a transitive subgroup of $S(G^{n})$ for some $n\geq 0$. Set $Z=G^{n+1}=X\times Y$, where $X=G^{n}$ and $Y=G$, and take $H=W(G,n)$ and $K=G$. Then $W(G,n+1)$ is a transitive subgroup of $S(Z)$ by Lemma A.

Proof of the theorem. We argue by induction on $n$, the theorem being trivially true when $n=0$. Suppose the result is true for some $n\geq 0$. Adopting the notation of Lemma H, the inductive hypothesis yields $N_{S(X)}W(G,n)=M\ltimes W(G,n)$, where $M\cong\mathrm{Aut}(G)^{n}$. By Lemma H, $W(G,n)$ is transitive, whence Lemma G gives $N=M^{*}\mathrm{Aut}(G)\ltimes W(G,n+1)$, with $M^{*}\cong M$, $M^{*}\cap\mathrm{Aut}(G)=1$, and $[M^{*},\mathrm{Aut}(G)]=1$, so $M^{*}\mathrm{Aut}(G)\cong\mathrm{Aut}(G)^{n+1}$.

Let us recursively indicate the action of $\mathrm{Aut}(G)^{n}$ on $W(G,n)$. If $n=0$ there is nothing to do. Given $n\geq 0$, we have $N_{S(X)}W(G,n)=M\ltimes W(G,n)$ in the notation of Lemma H, where the action of $M\cong\mathrm{Aut}(G)^{n}$ on $W(G,n)$ is assumed to be known. According to Lemma G, we have $N_{S(Z)}W(G,n+1)=(M^{*}\times\mathrm{Aut}(G))\ltimes(G\ltimes B)$ with $[M^{*},G]=1$. The action of $\mathrm{Aut}(G)$ on $G\ltimes B$ is given in Lemma D, while the action of $M^{*}$ on $B$ is described prior to Lemma G.

Acknowledgment. We thank the referee for a careful reading of the manuscript.