I'm working through Dixon and Mortimer's book Permutation groups. I'm currently working on showing that we can turn a wreath product into a permutation group. I've changed the notation used from the book slightly as I find it easier to use but the reference is the same.
Let $G$ and $H$ act on sets $\Omega$ and $\Psi$ respectively, and let $W=H\wr G = A\rtimes G = H^{\Psi}\rtimes G$. $A$ naturally acts on $\Gamma$ the direct product of $\Psi$ with itself $|\Omega|$ times i.e. $\Gamma = \Psi^{\Omega}$. We can define an action of the wreath product $W$ on $\Gamma$ in the following way: for each $(a,g)\in W$ and $\beta\in\Gamma$ let $\beta^{a,g}:\Omega\rightarrow\Psi$ defined as:
$$\delta\mapsto \left((\delta^{g^{-1}})^{\beta}\right)^{(\delta^{g^{-1}})^{a}}.$$
My issue is showing that this is an action as taken $a,g = 1$, I get, $\delta^{\beta(1_{A},1)}= (\delta^{\beta})^{\delta}$ which doesn't satisfy the definition of a group action. Any help is appreciated!
Best Answer
I don't think you have the correct setup.
Following Dixon and Mortimer, but switching to your notation:
Let $G$ act on $\Omega$, and $H$ act on $\Psi$. We let $A=H^{\Omega}$, and then $H\wr_{\Omega}G = A\rtimes G = H^{\Omega}\rtimes G$. (You have it as $H^{\Psi}$, which is incorrect.)
Since $H$ acts on $\Psi$, then $H^{\Omega}$ acts on $\Psi^{\Omega}$ "coordinatewise". (You have $H^{\Psi}$ acting on $\Psi^{\Omega}$, which is not at all clear). That is, given $f\colon\Omega\to \Psi$ and $h=(h_{\omega})_{\omega\in\Omega}\in H^{\Omega}$, we have $$f^{h}(\omega) = f(\omega)^{h_{\omega}^{-1}}$$ for each $\omega\in\Omega$.
In addition, $G$ acts on $\Psi^{\Omega}$ by acting on the coordinates; that is, given $f\colon\Omega\to\Psi$, and $g\in G$, we have $$(f^g)(\omega) = f(\omega^{g^{-1}}).$$
So now we want to let $A\rtimes G = H^{\Omega}\rtimes G = H\wr_{\Omega} G$ act on $\Psi^{\Omega}$. We define it as follows:
With this definition, what happens when $(a,g)=(1_A,e_G)$? Note that $A=H^{\Omega}$, so $1_A$ is the map that sends every $\omega\in\Omega$ to $e_H\in H$. So we have: $$\begin{align*} \beta^{(1_A,e_G)}(\omega) &= \Bigl( \beta(\omega^{e_G})\Bigr)^{1_A(\omega^{e_G})}\\ &= \Bigl( \beta(\omega)\Bigr)^{1_A(\omega)}\\ &= \Bigl(\beta(\omega)\Bigr)^{e_H}\\ &= \beta(\omega), \end{align*}$$ because by definition $e_H$ acts trivially on $\Psi$. Thus, $\beta^{(1_A,e_G)}=\beta$, as expected.
I suspect you may have mistakenly evaluated $(\delta)^{1_A}$ as $\delta$ (that seems to match your computation?); but as noted above, since $A=H^{\Omega}$, then the identity of $A$ is the map that sends every $\delta$ to $e$, not the identity map of $\Omega$.