orbits of group action on product space and orbits of stablizer are in 1-1 correspondence? How to prove the next question? Thanks
orbits of  group action on product space and orbits of stablizer are in 1-1 correspondence?
Let group $G$ act transitively on a set $X$. Let $x\in X$ and $H=\operatorname{Stab}(x)$. Let $G$ act on $X\times X$ via $g(x_1,x_2)=(gx_1,gx_2)$ for any $g\in G$. Prove that all $G$-orbits in $X\times X$ are in a bijective correspondence with all $H$-orbits in $X$.
 A: Since $G$ is transitive on $X$, the orbits of $G$ on $X \times X$ will be the orbits of elements of the form
$$
(x, y),
$$
for some $y \in X$. 
When are two such elements $(x, y), (x, z)$ in the same $G$-orbit? This happens if and only if there is $g \in G$ such that
$$
(g x, g y) = (x, z),
$$
that is, if and only if there is $g \in H$ such $z = g y$, that is, if and only if $y$ and $z$ are in the same $H$-orbit.
So if 
$$
(x, y_1), (x, y_2), \dots, (x, y_n), 
$$
are representatives of the orbits of $G$ on $X \times X$, then
$$
y_1, y_2, \dots, y_n
$$
will be representatives of the orbits of $H$ on $X$.
A: For $y \in G$, write $\overline{y}_H$ to denote the $H$-orbit of $y$
$$
\overline{y}_H = \{hy : h\in H\}
$$
and write $\overline{(x,y)}_G$ to denote the $G$-orbit of $(x,y) \in G\times G$
$$
\overline{(x,y)}_G = \{(gx,gy) : g\in G\}
$$
Now consider the map $\overline{y}_H \mapsto \overline{(x,y)}_G$ given by
$$
hy \mapsto (hx,hy) = (x,hy)
$$
This is well-defined, and your required bijection (Injectivity is easy, and it is surjective because the action of $G$ on $X$ is transitive)
