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.

  • $\begingroup$ Yes to your first question. It is not possible to say without more information (about the group, about $\Omega$, and about the action) what the orbits of the second action will be like. In general we can say say $(x_1, y_1) \sim (x_2, y_2)$ if there is a $g \in G$ such that $x_2 = x_1^g$ and $y_2 = x_2^g$. $\endgroup$ Feb 12 '18 at 15:30
  • $\begingroup$ If G is transitive on $\Omega$ then there will be only one orbit. Not sure in the latter case, more information is needed. $\endgroup$ Feb 12 '18 at 15:30
  • $\begingroup$ Has the expression "doubly-transitive action" been mentioned where you are reading/hearing this stuff? $\endgroup$ Feb 12 '18 at 15:31
  • $\begingroup$ @AbhiramNatarajan Then, in that case, will $(x_1, y_1)$ and $(x_2, y_2)$ be in the same orbit? $\endgroup$ Feb 13 '18 at 2:38
  • $\begingroup$ @ClémentGuérin $G$ is doubly-transitive if $\Omega$ has 2 orbits, if I'm not mistaken? $\endgroup$ Feb 13 '18 at 2:42

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:

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

  • $\begingroup$ Hi, does the choice of $g$ have anything to do with which orbit an element belongs to? Say, for $S_3$ acting on $T= \{1,2,3\}$ x $\{1,2,3\}$, the identity permutation would produce the diagonal orbit which would contain all elements of $T$? $\endgroup$ Feb 13 '18 at 14:18
  • $\begingroup$ I don't think either part of the question makes sense as written. The orbit of $\omega \in \Omega$ is the set $\{\omega^g : g \in G\}$ of the all of the elements of $\Omega$ one can "reach" via the action of $G$. The orbit decomposition depends only on the action itself, there's no choice of an element $g \in G$ in the picture. $\endgroup$ Feb 13 '18 at 15:08
  • $\begingroup$ The action of $S_3$ on $\Omega := \{1, 2, 3\}$ by permutations is transitive, so we know that this action has a single orbit, $\Omega$. This means that the induced action on $\Omega \times \Omega$ acts transitively on the diagonal $\Delta=\{(1,1),(2,2),(3,3)\}$, because if $g$ maps $x$ to $x'$, then it also maps $(x,x)$ to $(x',x')$. $\endgroup$ Feb 13 '18 at 15:11
  • $\begingroup$ Oh, okay. So, in this case, the $S_3$ has 2 orbits on $T$, the diagonal $\Delta$ and $T$ \ $\Delta$, since these are the two "unique" sets produced by applying each $g \in G$ to the elements of $T$? $\endgroup$ Feb 13 '18 at 23:07
  • $\begingroup$ That's correct. In particular, $\Delta \setminus T$ is an an orbit because for any pairs $(a, b)$, $(c, d)$, where $a$ and $b$ are distinct and so are $c$ and $d$, there is a permutation mapping $(a, b) \mapsto (c, d)$, i.e., mapping $a \mapsto c$ and $b \mapsto d$. $\endgroup$ Feb 14 '18 at 15:29

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