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 $X$.

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    $\begingroup$ This question is not off-topic (someone voted to close for that reason, strangely). $\endgroup$
    – anon
    Sep 2 '13 at 14:57
  • $\begingroup$ @anon, I agree. $\endgroup$ Sep 2 '13 at 15:05
  • $\begingroup$ sure Thanks. I will remember to click. $\endgroup$
    – Shiquan
    Oct 9 '13 at 13:42

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 $X$.

  • $\begingroup$ Thank you very much Prof. Andreas Caranti! $\endgroup$
    – Shiquan
    Sep 3 '13 at 6:08
  • $\begingroup$ Question to Dr. Caranti: What is the reason for the very first sentence of your solution? Otherwise everything makes perfect sense. Thank you! $\endgroup$
    – user206991
    Jan 10 '15 at 19:23
  • $\begingroup$ @algebrabeginner, I am not sure I understand your question correctly. However, I am setting up locating suitable elements in the orbits, which turn handy later in the proof. $\endgroup$ Jan 11 '15 at 10:11
  • $\begingroup$ @AndreasCaranti what if $G$ acts on $X,Y$ and we seek the orbits of the product action? (No transitivity assumptions.) $\endgroup$
    – Arrow
    May 20 '18 at 12:45
  • $\begingroup$ @Arrow assume everything is finite for simplicity. Let $x_{1}, \dots, x_{n}$ be a set of representatives of the action of $G$ on $X$. Then for $i \ne j$, we have that $(x_{i}, y)$ and $(x_{j}, z)$ are in different orbits. When are $(x_{i}, y)$ and $(x_{j}, z)$ in the same orbit? From now on, it should be more or less like in the previous case. $\endgroup$ May 20 '18 at 13:00

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)


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