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 $G$!