How do I write a group action in terms of the action of a subgroup? Suppose $H < K < G$ are finite groups and $G$ acts primitively by (right) multiplication on the set $\Gamma = G/K$ of (right) cosets of $K$ in $G$, and $K$ acts primitively on the set $\Delta = K/H$ of cosets of $H$ in $K$.  Let's assume that $H$ is core-free in $K$ and $K$ is core-free in $G$, so these are, indeed, permutation groups.
I suspect there is a natural way to write the action of $G$ on the set $\Omega = G/H$ in terms of (a composition of) the actions $G$ on $\Gamma$ and $K$ on $\Delta$.  In other words, I think the action of $G$ on $\Omega$ could be viewed as (1) permuting the cosets of $H$ which lie within a single $K$-coset, followed by (2) permuting the cosets of $K$.  So it seems there is a wreath product underlying this, but I'm having trouble writing it down.  
Here's what I have so far.  Perhaps someone who knows more group theory can tell me what's right or wrong with it:
For each $g\in G$, we have the action $g : Hy \mapsto Hyg$.  Suppose $g = kx$, for some $k \in K$ and some $x \in G$, where $x$ is a $K$-coset representative.  Then the action of $g$ "factors through" the action of $k$ as follows:
$$g: Hy \mapsto Hyk \mapsto Hykx$$
Now, since we assumed these are permutation groups, we have $K\hookrightarrow Sym(\Delta)$ and $G\hookrightarrow Sym(\Gamma)$.  Let $\mathcal K$ and $\mathcal G$ be the images of $K$ and $G$ under these embeddings.
Is the action of $G$ on the set $\Omega=G/H$ somehow related to the wreath product
$$\mathcal K^{\Gamma} \rtimes \mathcal G \; ?$$
(Incidentally, I'm fairly certain I don't need the primitivity assumptions, but in my application I happen to know that $H$ is maximal in $K$ and $K$ is maximal in $G$.)

Update: Professor Holt's answer below is perfectly clear, but I recently came across this nice article by Cheryl Praeger describing in detail exactly what I had in mind.
 A: Yes, you are correct that there is an embedding into the wreath product. This follows directly from the following general result on imprimitive permutation groups.
Let $G$ act imprimitively on $\Omega$ with block system $\Gamma$. Let $\alpha \in \Gamma$ and let $K$ be the restriction to $\alpha$ of the stabilizer in $G$ of the block $\alpha$. Let $\rho: G \to G^\Gamma$ be the natural homomorphism from $G$ to the induced action on the block system. Then $G$ embeds into the wreath
product of $K$ with $\rho(G)$.
This result seems very believable. The proof is not very hard but it also not totally easy. I think it might be in Peter Cameron's book on Permutation Groups, but I don't have that to hand.
Let me see if I can reconstruct it. The wreath product in question is the semidirect product of the group $K^\Gamma$ of functions $f:\Gamma \to K$ with $\rho(G)$, where the action of $\rho(G)$ on $K^\Gamma$ is given by $f^g(\beta) = f(\beta^{g^{-1}})$ (where I've just written $f^g$ rather than $f^{\rho(g)}$).
We have $\Gamma = \{ \alpha^t : t \in T \}$, where $T$ is a right transversal of the block stabilizer $G_\alpha$ in $G$. Now we can define our embedding $\phi$ from $G$ to the wreath product.
We put $\phi(g) = f_g \rho(g)$, where $f_g \in K^\Gamma$ is defined by $f_g(\alpha^t) = k|_\alpha$, where $tg = kt'$ with $k \in G_\alpha$ and $t' \in T$.
So we have to show that this is a homomorphism, which reduces to showing that, for $g,h \in G$, $f_{gh} = f_g f_h^{g^{-1}}$.
For $t \in T$, we can write $tgh = kt'h = klt''$, where $k|_\alpha = f_g(\alpha^t)$ and
$l|_\alpha = f_h(\alpha^{t'}) = f_h(\alpha^{tg}) = f_h^{g^{-1}}(\alpha^t),$
and also $f_{gh}(\alpha^t) = (kl)|_\alpha$, which proves what we want!
I guess we should also need to prove that $\phi$ is a monomorphism, but I'll leave that to the reader! An element in $\ker \phi$ is in $\ker \rho$ and can be shown to restrict to the identity on all blocks, so it must be the identity permutation.
