Orbits of $G$-action on Cartesian product of sets

Question

We know that G acts transitively on the left-cosets of $H \leq G$, where the action is given by $g.(aH) = (ga) H$. Now define G-action on the Cartesian product of two such sets $[G:H] \times [G:H]$, naturally as $g.(aH, bH) = ((ga)H, (gb)H)$. This action is no longer necessarily transitive. It can have multiple orbits (unless original G-action is two-transitive), yet G-action on each orbit is isomorphic to G-action on $[G:K_o]$ for some $K_o \leq G$.

My question is this: What can we say about this new action? is there a way to infer these subgroups $K_o \quad o \in [O]$ from $H$? Do subgroups appear with some multiplicities?

Answer 1

If $G$ is finite abelian then $G/H\times G/H\simeq n G/H$ where "$\simeq$" denotes the isomorphism of $G$-sets, $n=|G/H|$ and $nX$ denotes the disjoint union of $n$ copies of $X$.

The more general result can be found in T. tom Dieck "Representation groups" in Chapter 5.2:

Problem 2. If $G$ is finite abelian and $K,L\subseteq G$ are subgroups then $$[G/K][G/L]=n[G/K\cap L]$$ where $n=|G||K|^{-1}|L|^{-1}|K\cap L|$.

This is formulated in the language of the Burnside ring $A(G)$ of a group $G$. In the Burnside ring the Cartesian product (with pointwise action) becomes simply a multiplication. And addition is the disjoint union. Therefore your question boils down to studing $A(G)$, at least in finite case.

Things become harder when $G$ is finite non-abelian. Recall that the Burnside ring embeds into a product of integers via

$$\varphi:A(G)\to \bigoplus_{H\in Con(G)}\mathbb{Z}$$ $$\varphi([X])_H=|X^H|$$

where $X^H=\{x\in X\ |\ hx=x\text{ for all }h\in H\}$ and $Con(G)$ are all subgroups of $G$ up to conjugacy ($\varphi$ does not depend on the choice). The homomorphism is also know as the characteristic embedding.

Now I'm not sure whether all subrings of $\mathbb{Z}^m$ are realizable as a Burnside ring of some finite group $G$ via the characteristic embedding. But many are (I couldn't find a counterexample). Why is that important? Well, because the image of $\varphi$ is (freely) generated by $\varphi([G/H])$. And so the answer to your question boils down to:

Having a subring $R\subseteq \mathbb{Z}^n$ with free generators $\{e_1,\ldots, e_n\}$, where $e_1=(1,\ldots, 1)$ how can I decompose $e_i^2$ as a sum of those vectors?

Example. Let $G=S_3$. Then $G$ has only $4$ subgroups up to conjugacy: $G$, trivial $E$ and two cyclic subgroups $C_2, C_3$ of order $2$ and $3$ respectively. And the image of $\varphi$ is generated by

$$\varphi([G/G])=e_1=(1,1,1,1)$$ $$\varphi([G/C_3])=e_2=(2,0,2,0)$$ $$\varphi([G/C_2])=e_3=(3,1,0,0)$$ $$\varphi([G/E])=e_4=(6,0,0,0)$$

With that we easily calculate:

$$e_1^2=e_1\text{ (always)}$$ $$e_2^2=2e_2$$ $$e_3^2=e_3+e_4$$ $$e_4^2=6e_4$$

There probably is some algorithm for that but I doubt there is a closed formula. Anyway I strongly suggest you read the T. tom Dieck's book.

And for infinite $G$ this gets infinitely harder. :)

Side note: This also shows that $G/H\times G/H$ is very rarely transitive. The only way for it to happen is when every coordinate of $\varphi([G/H])$ is either $0$ or $1$, meaning that for any subgroup $K\subseteq G$ there is at most one $g\in G$ such that $kgH=gH$ for all $k\in K$.