Exercise 14.2.6

Proof.

(a)

Let $G$ be any group and let $A\subset S_n$ be a permutation group. Then set $$A\wr G=\{(\tau ;g_1,\dots ,g_n)|\tau \in A,g_1,\dots ,g_n\in G\},$$

with an operation on this set defined by

$$(\tau ;g_1,\dots ,g_n)({\tau }^{\prime };g_1^{\prime },\dots ,g_n^{\prime })=(\tau {\tau }^{\prime };g_{{\tau }^{\prime }(1)}{g}_1^{\prime },\dots ,g_{{\tau }^{\prime }(n)}{g}_n^{\prime })\in A\wr G.$$

We write $e$ the identity of $G$, and $\text{()}$ the identity of $S_n$.

$\text{∙}$

Let $\lambda =(\tau ;g_1,\dots ,g_n),{\lambda }^{\prime }=({\tau }^{\prime };g_1^{\prime },\dots ,g_n^{\prime }),{\lambda }^{″}=({\tau }^{″};g_1^{″},\dots ,g_n^{″})$ be elements of $A\wr G$. Then $$\begin{array}{llll}\hfill \lambda ({\lambda }^{\prime }{\lambda }^{″})& =(\tau ;g_1,\dots ,g_n)({\tau }^{\prime }{\tau }^{″};g_{{\tau }^{″}(1)}{g}_1^{″},\dots ,g_{{\tau }^{″}(n)}{g}_n^{″})& \hfill & \\ \hfill & =(\tau {\tau }^{\prime }{\tau }^{″};g_{({\tau }^{\prime }{\tau }^{″})(1)}{g}_{{\tau }^{″}(1)}^{″}{g}_1^{″},\dots ,g_{({\tau }^{\prime }{\tau }^{″})(n)}{g}_{{\tau }^{″}(n)}^{″}{g}_n^{″})& \hfill & \\ \hfill (\lambda {\lambda }^{\prime }){\lambda }^{″}& =(\tau {\tau }^{\prime };g_{{\tau }^{\prime }(1)}{g}_1^{\prime },\dots ,g_{{\tau }^{\prime }(n)}{g}_n^{\prime })({\tau }^{″};g_1^{″},\dots ,g_n^{″})& \hfill & \\ \hfill & =(\tau {\tau }^{\prime };h_1,\dots ,h_n)({\tau }^{″};g_1^{″},\dots ,g_n^{″})(\text{where }{h}_k=g_{{\tau }^{\prime }(k)}{g}_k^{\prime })& \hfill & \\ \hfill & =(\tau {\tau }^{\prime }{\tau }^{″};h_{{\tau }^{″}(1)}{g}_1^{″},\dots ,h_{{\tau }^{″}(n)}{g}_n^{″})& \hfill & \\ \hfill & =(\tau {\tau }^{\prime }{\tau }^{″};g_{{\tau }^{\prime }({\tau }^{″}(1))}{g}_{{\tau }^{″}(1)}^{″}{g}_1^{″},\cdots ,g_{{\tau }^{\prime }({\tau }^{″}(n))}{g}_{{\tau }^{″}(n)}^{″}{g}_n^{″})& \hfill & \\ \hfill & =(\tau {\tau }^{\prime }{\tau }^{″};g_{({\tau }^{\prime }{\tau }^{″})(1)}{g}_{{\tau }^{″}(1)}^{″}{g}_1^{″},\dots ,g_{({\tau }^{\prime }{\tau }^{″})(n)}{g}_{{\tau }^{″}(n)}^{″}{g}_n^{″})& \hfill & \end{array}$$

thus $\lambda ({\lambda }^{\prime }{\lambda }^{″})=(\lambda {\lambda }^{\prime }){\lambda }^{″}$, and the law is associative.

$\text{∙}$

Write $𝜀=(();e,\dots ,e)=(\iota ;e_1,\dots ,e_n)$, where $\iota =()$, and $e_k=e,k=1,\dots ,n$. Then $$\begin{array}{llll}\hfill \mathit{𝜀\lambda }& =(\iota ;e_1,\dots ,e_n)(\tau ;g_1,\dots ,g_n)& \hfill & \\ \hfill & =(\tau ;e_{{\tau }^{\prime }(1)}{g}_1,\dots ,e_{{\tau }^{\prime }(n)}{g}_n)& \hfill & \\ \hfill & =(\tau ;g_1,\dots ,g_n)=\lambda (\text{since }{e}_{{\tau }^{\prime }(k)}=e)& \hfill & \\ \hfill \mathit{\lambda 𝜀}& =(\tau ;g_{\iota (1)}{e}_1,\dots ,g_{\iota (n)}{e}_n)& \hfill & \\ \hfill & =(\tau ;g_1,\dots ,g_n)=\lambda (\text{since }\iota (k)=k,e_k=e).& \hfill & \end{array}$$

Therefore $𝜀=(();e,\dots ,e)$ is the identity of $A\wr G$.

$\text{∙}$

Set $\mu =({\tau }^{-1};h_1,\dots ,h_n)=({\tau }^{-1};g_{{\tau }^{-1}(1)}^{-1},\dots ,g_{{\tau }^{-1}(n)}^{-1})$, with $h_k=g_{{\tau }^{-1}(k)}^{-1},k=1,\dots ,n$. Then $$\begin{array}{llll}\hfill \mathit{\lambda \mu }& =(\tau ;g_1,\dots ,g_n)({\tau }^{-1};h_1,\dots ,h_n)& \hfill & \\ \hfill & =(();g_{{\tau }^{-1}(1)}{h}_1,\dots ,g_{{\tau }^{-1}(n)}{h}_n)& \hfill & \\ \hfill & =(();g_{{\tau }^{-1}(1)}{g}_{{\tau }^{-1}(1)}^{-1},\dots ,g_{{\tau }^{-1}(n)}{g}_{{\tau }^{-1}(n)}^{-1})& \hfill & \\ \hfill & =(();e,\dots ,e)=𝜀& \hfill & \\ \hfill \mathit{\mu \lambda }& =({\tau }^{-1};h_1,\dots ,h_n)(\tau ;g_1,\dots ,g_n)& \hfill & \\ \hfill & =(();h_{\tau (1)}{g}_1,\dots ,h_{\tau (n)}{g}_n& \hfill & \\ \hfill & =(();g_{{\tau }^{-1}(\tau (1))}^{-1}{g}_1,\dots ,g_{{\tau }^{-1}(\tau (n))}^{-1}{g}_n)& \hfill & \\ \hfill & =(();g_1^{-1}{g}_1,\dots ,g_n^{-1}{g}_n)=(();e,\dots ,e)=𝜀.& \hfill & \end{array}$$

Therefore every element in $A\wr G$ is invertible.

$A\wr G$ is a group under the multiplication defined in the Mathematical Notes.

(b)

For the group $A\wr G$ of part (a), where $A\subset S_n$ and $G$ is a group, we show the following lemma:

Lemma. The map

$$\phi \{\begin{array}{cc}\hfill A\wr G\hfill & \hfill \to A\hfill \\ \hfill (\tau ;g_1,\dots ,g_n)\hfill & \hfill \mapsto \tau \hfill \end{array}$$

is a group homomorphism that is surjective and whose kernel is isomorphic to $G^n$.

Let $\lambda =(\tau ;g_1,\dots ,g_n),{\lambda }^{\prime }=({\tau }^{\prime };g_1^{\prime },\dots ,g_n^{\prime })$ be any elements of $A\wr G$. By definition, $\lambda {\lambda }^{\prime }=(\tau {\tau }^{\prime };g_{{\tau }^{\prime }(1)}{g}_1^{\prime },\dots ,g_{{\tau }^{\prime }(n)}{g}_n^{\prime }),$ so that

$$\phi (\lambda {\lambda }^{\prime })=\tau {\tau }^{\prime }=\phi (\lambda )\phi ({\lambda }^{\prime }).$$

$\phi$ is a group homormphism.

If $\tau$ is any element of $A$, then $\phi (\tau ;e,\dots ,e)=\tau$, where $(\tau ;e,\dots ,e)\in A\wr G$. Therefore $\phi$ is surjective.

Moreover $(\tau ;g_1,\dots ,g_n)\in \ker <!--\; FUNCTION\; APPLICATION\; -->\phi$ if and only if $\tau =()$, therefore

$$\ker <!--\; FUNCTION\; APPLICATION\; -->\phi =\{\iota ;g_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,g_n)|(g_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,g_n)\in G^n\},\text{where }\iota =().$$

Consider

$$\psi \{\begin{array}{ccc}\hfill \ker <!--\; FUNCTION\; APPLICATION\; -->\phi \hfill & \hfill \to \hfill & \hfill G^n\hfill \\ \hfill (\iota ;g_1,\dots ,g_n)\hfill & \hfill \mapsto \hfill & \hfill (g_1,\dots ,g_n)\hfill \end{array}$$

Then $\psi$ is bijective (with inverse map $(g_1,\dots ,g_n)\mapsto (\iota ,g_1,\dots ,g_n)$). We verify that $\psi$ is a group homomorphism: if $\lambda =(\iota ;g_1,\dots ,g_n),{\lambda }^{\prime }=(\iota ;g_1^{\prime },\dots ,g_n^{\prime })$ are elements of $\ker <!--\; FUNCTION\; APPLICATION\; -->\phi$, then

$$\begin{array}{llll}\hfill \psi (\lambda {\lambda }^{\prime })& =\psi ((\iota ;g_1,\dots ,g_n)(\iota ;g_1^{\prime },\dots ,g_n^{\prime }))& \hfill & \\ \hfill & =\psi (\iota ;g_{\iota (1)}{g}_1^{\prime },\dots ,g_{\iota (n)}{g}_n^{\prime })& \hfill & \\ \hfill & =\psi (\iota ;g_1{g}_1^{\prime },\dots ,g_n{g}_n^{\prime })(\text{since }\iota (k)=k)& \hfill & \\ \hfill & =(g_1{g}_1^{\prime },\dots ,g_n{g}_n^{\prime })& \hfill & \\ \hfill & =(g_1\dots ,g_n)(g_1^{\prime },\dots ,g_n^{\prime })& \hfill & \\ \hfill & =\psi (\lambda )\psi ({\lambda }^{\prime }).& \hfill & \end{array}$$

So $\psi$ is an group isomorphism, and $\ker <!--\; FUNCTION\; APPLICATION\; -->\phi \simeq G^n$.

(c)

By part (b), since $\phi$ is a surjective homomorphism, $$(A\wr G)∕\ker <!--\; FUNCTION\; APPLICATION\; -->\phi \simeq A,$$

and $\ker <!--\; FUNCTION\; APPLICATION\; -->\phi \simeq G^n$. Therefore

$$|A|=|A\wr G|∕|\ker <!--\; FUNCTION\; APPLICATION\; -->\phi |=|A\wr G|∕|G{\text{|}}^n,$$

which proves

$$|A\wr G|=|A||G{\text{|}}^n.$$