Orbits of the action of $G$ on $\Omega × \Omega$ I've read that orbits are the equivalence classes of the equivalence relation induced by the existence of $g$ $\in$ $G$ such that for every $x$, $y$ $\in$ $\Omega$, $x^g = y$ (here, exponentiation is used to denote group action).
Now, if $G$ is transitive on $\Omega$, then the action of $G$ on $\Omega$ will induce just one orbit. In my understanding, this will also imply that for all $x, y \in \Omega$, $x \sim y$?
Furthermore, if we now let $G$ act on $\Omega × \Omega$, using the natural action $(x,y)^g = (x^g, y^g)$, will the definition of orbit mentioned above still apply? $G$ is not not necessarily transitive on $\Omega × \Omega$, so we now expect to see distinct orbits. How do we now determine which elements of $\Omega × \Omega$ are related?
Thank you very much. 
 A: As some of the comments mention, it's true that if $G$ acts transitively on $\Omega$, then, by definition, the action has a single orbit, namely, $\Omega$ itself.
One can say at least some things about the natural action $(x, y)^g := (x^g, y^g)$ of $G$ on $\Omega \times \Omega$. For example, for any element of the diagonal $\Delta := \{(x, x) : x \in \Omega\} \subset \Omega \times \Omega$, we have $(x, x)^g = (x^g, x^g) \in \Delta$, so $\Delta$ is a union of orbits, and hence so is $(\Omega \times \Omega) - \Delta$. Thus (if $|\Omega| > 1$) the action of $G$ on $\Omega \times \Omega$ is never transitive.
Can you show using the definition of the action on $\Omega \times \Omega$ that, since $G$ acts transitively on $\Omega$, $\Delta$ is a single orbit?
On the other hand, the decomposition of $(\Omega \times \Omega) - \Delta$ into orbits depends on the nature of the original action, and various behaviors are possible. We acan nalyze the $G$-orbit structure on $\Omega \times \Omega$ just like any other group actions. For any $(x, y) \in \Omega \times \Omega$, $(x, y), (x', y')$ are in the same orbit iff there is a $g \in G$ such that $(x', y') = (x, y)^g$, or, unwinding the definition, such that $x' = x^g$ and $y' = y^g$.
Two "extremal" behaviors are exhibited in the following examples; working out the orbit structure of $G$ on $\Omega \times \Omega$ for both would give you some sense of the possible behaviors:


*

*$G = S_{\Omega}$, the usual permutation action. For concreteness, you might like to take $\Omega = \{1, \ldots, n\}$.

*$\Omega = G$, the left regular action (i.e., $h^g := gh$).



 1. In this case, $(\Omega \times \Omega) \setminus \Delta$ is a single orbit, and so we say that the action of $G$ on $\Omega$ is doubly transitive.

 2. In this case, $(g, h) \sim (g', h')$ implies that there is a $k$ such that $g' = kg, h' = kh$, and so $g'g^{-1} = h' h^{-1}$ (and the converse holds too). Thus the orbits are the sets $H_h := \{(g, hg) : g \in G\}$. In this case, only the identity $e_G \in G$ fixes any element of $\Omega$, so we say that the action of $G$ on $\Omega$ is free.

