Exercise 14.2.7

Proof.

(a)

Consider the two maps $$\phi \{\begin{array}{ccc}\hfill H\hfill & \hfill \to \hfill & \hfill G^n\hfill \\ \hfill \varphi \hfill & \hfill \mapsto \hfill & \hfill (\varphi (1),\dots ,\varphi (n)),\hfill \end{array}\psi \{\begin{array}{ccc}\hfill G^n\hfill & \hfill \to \hfill & \hfill H\hfill \\ \hfill (x_1,\dots ,x_n)\hfill & \hfill \mapsto \hfill & \hfill \xi \{\begin{array}{ccc}\hfill \{1,\dots ,n\}\hfill & \hfill \text{→}\hfill & \hfill G\hfill \\ \hfill i\hfill & \hfill \mapsto \hfill & \hfill x_i.\hfill \end{array}\hfill \end{array}$$

Then $\psi \circ \phi =1_H$ and $\phi \circ \psi =1_{G^n}$, therefore $\phi$ is bijective.

Moreover, for all $(\varphi ,\chi )\in H$,

$$\begin{array}{llll}\hfill \phi (\mathit{\varphi \chi })& =((\mathit{\varphi \chi })(1),\dots ,(\mathit{\varphi \chi })(n))& \hfill & \\ \hfill & =(\varphi (1)\chi (1),\dots ,\varphi (n)\chi (n))& \hfill & \\ \hfill & =(\varphi (1),\dots ,\varphi (n))(\chi (1),\dots ,\chi (n))& \hfill & \\ \hfill & =\phi (\varphi )\phi (\chi )).& \hfill & \end{array}$$

Therefore $H\simeq G^n$ via $\phi$.

(b,c)

If we define ${\varphi }^{\tau }$, for $\tau \in S_n$ and $\varphi \in H$, by $({\varphi }^{\tau })(i)=\varphi (\tau (i)),i=1,\dots ,n$, we obtain a right action: if $\tau ,{\tau }^{\prime }\in S_n$, for all $i\in \{1,\dots ,n\}$, $$({({\varphi }^{\tau })}^{{\tau }^{\prime }})(i)=({\varphi }^{\tau })({\tau }^{\prime }(i))=\varphi (\tau ({\tau }^{\prime }(i)))=\varphi ((\tau {\tau }^{\prime })(i))={\varphi }^{\tau {\tau }^{\prime }}(i)),$$

thus ${({\varphi }^{\tau })}^{{\tau }^{\prime }}={\varphi }^{\tau {\tau }^{\prime }}$. To obtain a left action, we must define, as in part (c),

$$(\tau \cdot \varphi )(i)=\varphi ({\tau }^{-1}(i)),i=1,\dots ,n.$$

Then

$$({\tau }^{\prime }\cdot (\tau \cdot \varphi ))(i)=(\tau \cdot \varphi )({\tau {}^{\prime }}^{-1}(i))=\varphi ({\tau }^{-1}({\tau {}^{\prime }}^{-1}(i)))=\varphi {({\tau }^{\prime }\tau )}^{-1}(i)=(({\tau }^{\prime }\tau )\cdot \varphi )(i),$$

so that ${\tau }^{\prime }\cdot (\tau \cdot \varphi )=({\tau }^{\prime }\tau )\cdot \varphi )$ (and $e\cdot \tau =\tau$).

This is a proof of part (c), and this explains the recurrent and stressful injonction from D.A.Cox “Be sure you understand why the inverse is necessary”.

Using this action for part (b), we define $(\tau ,\varphi )$ for $\tau \in S_n,\varphi \in H=G^{\{1,\cdots ,n\}}$,by

$$(\tau ,\varphi )=(\tau ;\varphi (1),\dots ,\varphi (n)),$$

so that

$$(\tau ,\varphi )=(\tau ;g_1,\dots ,g_n)⟺\varphi (1)=g_1,\dots ,\varphi (n)=g_n.$$

If $(\tau ,\varphi )=(\tau ;g_1,\dots ,g_n),({\tau }^{\prime },{\varphi }^{\prime })=({\tau }^{\prime };g_1^{\prime },\dots ,g_n^{\prime })$, then

$$\begin{array}{llll}\hfill (\tau ,\varphi )({\tau }^{\prime },{\varphi }^{\prime })& =(\tau ;g_1,\dots ,g_n)({\tau }^{\prime };g_1^{\prime },\dots ,g_n^{\prime })& \hfill & \\ \hfill & =(\tau {\tau }^{\prime };g_{{\tau }^{\prime }(1)}{g}_1^{\prime },\dots ,g_{{\tau }^{\prime }(n)}{g}_n^{\prime })& \hfill & \\ \hfill & =(\tau {\tau }^{\prime };\varphi ({\tau }^{\prime }(1)){\varphi }^{\prime }(1),\dots ,\varphi ({\tau }^{\prime }(n)){\varphi }^{\prime }(n))& \hfill & \\ \hfill & =(\tau {\tau }^{\prime };({({\tau }^{\prime })}^{-1}\cdot \varphi )(1){\varphi }^{\prime }(1),\dots ,({({\tau }^{\prime })}^{-1}\cdot \varphi )(n){\varphi }^{\prime }(n))& \hfill & \\ \hfill & =(\tau {\tau }^{\prime },({({\tau }^{\prime })}^{-1}\cdot \varphi ){\varphi }^{\prime }).& \hfill & \end{array}$$

(d)

Consider the map $$\phi \{\begin{array}{ccc}\hfill A\wr G\hfill & \hfill \to \hfill & \hfill H⋊A\hfill \\ \hfill (\tau ,\varphi )\hfill & \hfill \mapsto \hfill & \hfill (\tau \cdot \varphi ,\tau ).\hfill \end{array}$$

If $\psi :H⋊A\to A\wr G$ is defined by $\psi (\chi ,\tau )=(\tau ,{\tau }^{-1}\cdot \chi )$, then, for all $\tau \in S_n,\varphi ,\chi \in H$,

$$\begin{array}{llll}\hfill (\psi \circ \phi )(\tau ,\varphi )& =\psi (\tau \cdot \varphi ,\tau )=(\tau ,{\tau }^{-1}\cdot (\tau \cdot \varphi )=(\tau ,\varphi ),& \hfill & \\ \hfill (\phi \circ \psi )(\chi ,\tau )& =\phi (\tau ,{\tau }^{-1}\cdot \chi )=(\tau \cdot ({\tau }^{-1}\cdot \chi ),\tau )=(\chi ,\tau ).& \hfill & \end{array}$$

Thus $\psi \circ \phi =1_{A\wr G},\phi \circ \psi =1_{H⋊A}$. This proves that $\phi$ is bijective.

Recall that the binary operation in $H⋊A$ is defined by (6.9):

$$(\varphi ,\tau )({\varphi }^{\prime },{\tau }^{\prime })=(\varphi (\tau \cdot {\varphi }^{\prime }),\tau {\tau }^{\prime }).$$

We verify that $\phi$ is a group homomorphism. Note first that, for $\tau \in S_n,\mathit{\varphi \chi }\in H$,

$$\tau \cdot (\mathit{\varphi \chi })=(\tau \cdot \varphi )(\tau \cdot \chi ).$$

Indeed, for all $i\in \{1,\dots ,n\}$,

$$\begin{array}{llll}\hfill (\tau \cdot (\mathit{\varphi \chi }))(i)& =(\mathit{\varphi \chi })({\tau }^{-1}(i))& \hfill & \\ \hfill & =\varphi ({\tau }^{-1}(i))\chi ({\tau }^{-1}(i))& \hfill & \\ \hfill & =(\tau \cdot \varphi )(i)(\tau \cdot \chi )(i)& \hfill & \\ \hfill & =((\tau \cdot \varphi )(\tau \cdot \chi ))(i).& \hfill & \end{array}$$

Using this rule, we obtain

$$\begin{array}{llll}\hfill \phi ((\tau ,\varphi )({\tau }^{\prime },{\varphi }^{\prime }))& =\phi (\tau {\tau }^{\prime };({({\tau }^{\prime })}^{-1}\cdot \varphi ){\varphi }^{\prime })& \hfill & \\ \hfill & =((\tau {\tau }^{\prime })\cdot ({({\tau }^{\prime })}^{-1}\cdot \varphi ){\varphi }^{\prime }),\tau {\tau }^{\prime })& \hfill & \\ \hfill & =((\tau {\tau }^{\prime })\cdot ({({\tau }^{\prime })}^{-1}\cdot \varphi )((\tau {\tau }^{\prime })\cdot {\varphi }^{\prime }),\tau {\tau }^{\prime })& \hfill & \\ \hfill & =((\tau \cdot \varphi )((\tau {\tau }^{\prime })\cdot {\varphi }^{\prime }),\tau {\tau }^{\prime }),& \hfill & \\ \hfill & & \hfill \end{array}$$

and using the binary operation in $H⋊A$,

$$\begin{array}{llll}\hfill \phi (\tau ,\varphi )\phi ({\tau }^{\prime },{\varphi }^{\prime })& =(\tau \cdot \varphi ,\tau )(({\tau }^{\prime }\cdot {\varphi }^{\prime },{\tau }^{\prime })& \hfill & \\ \hfill & =((\tau \cdot \varphi )(\tau \cdot ({\tau }^{\prime }\cdot {\varphi }^{\prime })),\tau {\tau }^{\prime })& \hfill & \\ \hfill & =((\tau \cdot \varphi )((\tau {\tau }^{\prime })\cdot {\varphi }^{\prime }),\tau {\tau }^{\prime }),& \hfill & \end{array}$$

thus $\phi ((\tau ,\varphi )({\tau }^{\prime },{\varphi }^{\prime }))=\phi (\tau ,\varphi )\phi ({\tau }^{\prime },{\varphi }^{\prime })$. We have proved that $\phi$ is a group isomorphism, so

$$A\wr G\simeq H⋊A=G^{\{1,\dots ,n\}}⋊A.$$

Wreath products can be represented by semidirect products.