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 $G$!
 A: 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.)
