Any help with the question which follows will be greatly appreciated. I'm working through Dixon and Mortimer's Permutation Groups and have a question regarding a particular semi-direct product related to wreath products. First I'll give a (slightly expanded) account of what Dixon and Mortimer say, and then my attempt to untangle their claims.
Let $\Omega$ be a set and $\Sigma$ a partition of $\Omega$ into equally sized subsets. We define the automorphism group of $\Sigma$ to be $$ G:=\mathrm{Aut}(\Sigma) := \{x \in \mathrm{Sym}(\Omega) : \forall \Delta \subseteq \Omega,\ \Delta^x \in \Sigma \iff \Delta \in \Sigma\} $$ where the notation $\Delta^x = \{\alpha^x : \alpha \in \Delta\}$ (recalling that $\mathrm{Sym}(\Omega)$ acts on $\Omega$). Then $G$ has a natural action on $\Sigma$ given by $\Delta \mapsto \Delta^x$ for every $x \in G$ and every $\Delta \in \Sigma$. Let $B$ denote the kernel of this action: $$ B = \{x \in G : \Delta^x = \Delta, \ \forall \Delta \in \Sigma\} $$ Begin with the observation that we have an isomorphism $$ B \cong \prod_{\Delta \in \Sigma} \mathrm{Sym}(\Delta) $$ given by the map $x \mapsto (x|_{\Delta})_{\Delta \in \Sigma}$ (each $x \in \mathrm{Sym}(\Omega)$ and fixes each $\Delta \in \Sigma$, thus restricts to a permutation on each such $\Delta$).
Next Dixon and Mortimer observe that $\mathrm{Sym}(\Sigma)$ acts on $B$ "by permuting the components of the elements of $B$ in a natural way." Given this action, it makes sense to consider the semi-direct product $$ B \rtimes \mathrm{Sym}(\Sigma) $$ and the authors claim: "it's easy to see that" $$ G \cong B \rtimes \mathrm{Sym}(\Sigma). $$
So here's my attempt at unpacking a bit of what they claim. Before we try to write-down the action of $\mathrm{Sym}(\Sigma)$ on $B$, we fix some notation. Every $\Delta \in \Sigma$ has the same cardinality, so we can choose for each pair $\Delta, \Delta^\prime \in \Sigma$ a particular bijection $$ \phi_{\Delta}^{\Delta^\prime} : \Delta \to \Delta^\prime $$ with the properties that $$ \phi_{\Delta^\prime}^{\Delta} = (\phi_{\Delta}^{\Delta^\prime})^{-1} $$ and $$ \phi_{\Delta^\prime}^{\Delta^{\prime \prime}} \circ \phi_{\Delta}^{\Delta^\prime} = \phi_{\Delta}^{\Delta^{\prime \prime}} $$ for every $\Delta, \Delta^\prime , \Delta^{\prime \prime} \in \Sigma$ (composition is left to right, so in the above composite $\phi_{\Delta^\prime}^{\Delta^{\prime \prime}}$ is first).
Now, for the action of $\mathrm{Sym}(\Sigma)$ on $B$, represent an element of $B$ by its image $$ (x_\Delta)_{\Delta \in \Sigma} \in \prod_{\Delta \in \Sigma} \mathrm{Sym}(\Delta). $$ Then given $\sigma \in \mathrm{Sym}(\Sigma)$, we define $$ (x_\Delta)_{\Delta \in \Sigma} ^{\sigma} := (x_{(\Delta)\sigma}^\Delta)_{\Delta\in \Sigma} $$ where $$ x_{(\Delta)\sigma}^{\Delta} = \phi_{\Delta}^{(\Delta)\sigma} \circ x_{(\Delta)\sigma} \circ \phi_{(\Delta)\sigma}^{\Delta} \in \mathrm{Sym}(\Delta) $$ Note that we apply elements in Symmetric groups on the right so that the image of $\Delta \in \Sigma$ under the permutation $\sigma \in \mathrm{Sym}(\Sigma)$ is denoted $(\Delta)\sigma$.
Question 1. Does this seem like the correct action?
Now we want to actually prove that there's an isomorphism $B \rtimes \mathrm{Sym} (\Sigma)\cong G$, so I began by trying to write down a map $\Psi : B \rtimes \mathrm{Sym} (\Sigma) \to G$ as follows: send
$$ ((x_\Delta)_{\Delta\in \Sigma} , \ \sigma) \in B \rtimes \mathrm{Sym} (\Sigma) $$ to the map $x : \Omega \to \Omega$ defined by $$ x : \alpha \mapsto (\alpha)x_{(\Delta)\sigma}^{\Delta} $$ where $\alpha \in \Delta$.
Thus far I've been able to prove that $\Psi$ is a well defined set map (by which I mean that the image actually lands in $G$). I've been struggling proving the homomorphism property however.
Question 2. Does this seem like the correct way to define the homomorphism (eventual isomorphism) from $B \rtimes \mathrm{Sym} (\Sigma) \to G$? If it's not, do you know the correct map to define? If it is, do you know how to prove it's a homomorphism?
Question 3. Any obvious ways to clean up this unruly notation?
Any guidance is appreciated!
I just want to note. You said that the bloc of $\Sigma$ are all of the same size. And $B=\prod_{\Delta\in \Sigma}Sym(\Delta)$, then $$B\simeq Sym(|\Delta|)^{|\Sigma|}$$ for some $\Delta \in \Sigma$.
Then $$G\simeq Sym(|\Delta|)^{|\Sigma|}\rtimes Sym(\Sigma)=Sym(\Delta)\wr Sym(\Sigma)$$.