From the chinese remainder theorem to products of transitive G-sets

Note: I am aware of the question Analog to the Chinese Remainder Theorem in groups other than Z_n.

For an abelian group $A$, every transitive $A$-set $M$ is of course isomorphic, as an $A$-set, to a quotient group $A/H$, by picking a point $m\in M$ and letting $H = Stab(m)$. Note that stabiliser groups of different points are conjugate, hence equal.

For a pair of transitive $A$-sets, $N,M$, their product $M\times N$ is an $A$-set by the diagonal action $(m,n) \stackrel{a}{\mapsto} (am,an)$. This is not in general a transitive $A$-set, but is the disjoint union of transitive $A$-sets. An easy result is that

$$Stab(m,n) = Stab(m)\cap Stab(n)$$

The Chinese remainder theorem is precisely the statement

$$\mathbb{Z}/(k)\times\mathbb{Z}/(l) \simeq \mathbb{Z}/((k)\cap (l)) \simeq \mathbb{Z}/(kl)$$

for coprime $k$ and $l$ (and generalised to more than two factors) and so the product of transitive $\mathbb{Z}$-sets is a transitive $\mathbb{Z}$-set. There is also the version where one has to consider the gcd of the factors, and this is when things get a bit more interesting, and break away from the ring-theoretic approach - the disjoint union of rings is not a ring!

Describing the structure of $A/H \times A/K$ for subgroups $H, K \lt A$ is only mildly interesting - it is a disjoint union of a number of copies of isomorphic transitive $A$-sets. This is not what my question is about, but there may be some combinatorial interest in the case of finite $A$. Consider instead a finite nonabelian group $G$ - not necessarily nilpotent! - and a pair of subgroups $H, K \lt G$. Fairly elementary observations show that

$$G/H\cap K \hookrightarrow G/H\times H/K$$

and that generally the orbits look like $G/(H\cap gKg^{-1})$. This seems to me to be an interesting combinatorial/group-theoretic problem, enumerating/classifying the various subgroups $H\cap gKg^{-1} \lt G$, and the number of orbits in the product.

My question is: has anyone done any work on something like this?

Postscript: people know know me might wonder why I was thinking about this. Well, the general problem of determining the structure of the product $G/H\times H/K$ came up thinking about proper-class-sized $G$ with set-sized $G/H$, $G/K$. It quickly became apparent that this would be nontrivial even for finite $G$!

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Alfred Foster rang a few changes on the theme of CRT for general algebras since the 1930's. Alden Pixley might know if your particular variation was considered by Foster or by anyone else. Gerhard "Ask Me About System Design" Paseman, 2011.11.15 –  Gerhard Paseman Nov 16 '11 at 1:27

There has been a good deal of work on this, under the heading of "Burnside rings". The Burnside ring of a finite group $G$ is the Grothendieck ring obtained from the finite $G$-sets with the operations of disjoint union and cartesian product. One of the easy ingredients of the theory is Burnside's notion of the mark of a subgroup $H$ of $G$ in a $G$-set $X$; this is just the number of points in $X$ fixed by the action of all elements of $H$. The mark of any fixed $H$ gives a ring-homomorphism from the Burnside ring of $G$ to $\mathbb Z$, and these homomorphisms, for varying $H$, are jointly monic. These facts let you decompose products of transitive $G$-sets into their transitive parts easily, once you've calculated the table of marks, i.e., the matrix, with rows and columns indexed by (conjugacy classes of) subgroups of $G$, with the $(H,K)$-entry being the mark of $H$ in $G/K$. (For easy calculation, you also want the inverse of this matrix, but that's easy to find because the table of marks is triangular if you order the subgroups by size.)
A more modern point of view is that the mark of $H$ in $G/K$ is just the number of $G$-equivariant maps from $G/H$ to $G/K$. From this point of view, Burnside's results on marks could be regarded as a precursor of the Yoneda Lemma. –  Andreas Blass Nov 16 '11 at 0:09