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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.

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Can't you just define the submaps as restrictions and back out to G? – Greg L Feb 20 '12 at 6:38
@Greg I'm not sure what you mean by "the submaps" and "back out." Could you elaborate? – William DeMeo Feb 20 '12 at 7:07
up vote 3 down vote accepted

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.

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Awesome! Thank you very much, Professor Holt. It took me a while to digest your answer, since I'm still getting used to the terminology. (I'm more accustomed to "congruences" than "block systems", but group theory language can sometimes make things simpler than universal algebra language, so I should be bilingual!) Thank you again. – William DeMeo Feb 20 '12 at 9:13
Sorry, one follow up question. What is the kernel of $\rho$? Isn't it the subgroup of elements of $G$ that leave the blocks of $\Gamma$ where they are (but may permute the elements within the blocks). Isn't that just $K$? I hope not, because I'm assuming $K$ is core-free in $G$! – William DeMeo Feb 20 '12 at 9:51
Yes, as you say, the kernel of $\rho$ is the subgroup that fixes each individual block, and may permute the blocks among themselves. It is not (usually) equal to $K$, which is the stabilizer of a single block, but to the core of $K$ in $G$. I know you were assuming that $K$ is core-free in $G$, but you didn't really need to assume that! – Derek Holt Feb 20 '12 at 10:05
Of course! ker $\rho = \bigcap_{g\in G} K^g$. I am so dense! In my application $K$ happens to be core-free, and I wanted to be sure that this doesn't prevent me from applying your answer. But, assuming I understand it correctly, your representation $\rho$ is just the action of $G$ on the cosets of $K$. So core-free $K$ means this action is faithful. (By the way, I noticed in Dixon and Mortimer there is a universal wreath product embedding theorem, but it assumes $K$ is a normal subgroup.) Thank you again for all your help! – William DeMeo Feb 20 '12 at 10:42

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