Action of product of groups gives bijection between $X/(G\times H)$ and $(X/G)/H$.

Question

Let $X$ be a set and suppose we have $G$ and $H$ are groups with actions on $X$ and $G$ and $H$ commute to each other. This gives us action of $G\times H$ on $X$. I want to show $X/(G\times H)$ is bijective to $(X/G)/H$ where $X/G$ means equivalence classes under left action. My idea is to show the action of $G\times H$ on $X$ gives rise to action of $H$ on $X/G$, but how can I prove this? Also I am not sure how to produce the bijection after I got that.

Answer 1

Let $p:X\to X/G$ and $q:X/G\to(X/G)/H$ be the natural maps induced by this action of $G\times H$ on $X.$

$$\begin{align}q(p(x))=q(p(y))&\iff\exists h\in H\quad Gx=hGy\\&\iff\exists h\in H\quad Gx=Ghy\\&\iff\exists h\in H\quad x\in Ghy\\&\iff x\in(G\times H)y \end{align}$$

Answer 2

This works just as well for extensions, not just products: If $G$ acts on a set $X$ and $N \leq G$ is any subgroup of $G$, then we have a quotient map $p_N\colon X \to N\backslash X$ from $X$ to the orbits of the $N$-action on $X$.

Now if $N$ is a normal subgroup of $G$, then for any $g \in G, x\in X$ we have $$ \tag{$\dagger$} g(N.x) = gN.x = (gNg^{-1}).gx = N.(gx) $$ so that $N.x \mapsto gN.x$ defines an action of $G$ on the set of orbits of $N$ on $X$. This action clearly then descends to an action of $G/N$ on $N\backslash X$ given by $gN.(N.x) = gN.x$, ($\forall g\in G, x\in X$).

If $p_{G/N}\colon N\backslash X \to G\backslash N \backslash X$ denotes the quotient map with respect to the $G$ (or $G/N$) action on $N\backslash X$, and $q\colon X\to G\backslash X$ the quotient map for the action of $G$ on $X$, then $q = p_{G/N}\circ p_N$.

Indeed $p_{G/N}\circ p_N(x) = p_{G/N}\circ p_N(y)$ if and only if there is some $g\in G$ such that $gN(Nx) = Ny$. But using $(\dagger)$ we see that $gN(Nx) = g.(Nx) = N(gx)$, hence this is equivalent to $N.gx = N.y$, which in turn is equivalent to the existence of some $n \in N$ such that $y=n.(gx)$, that is $y \in G.x$ or $q(x)=q(y)$.