When a group action is transitive, is it for all elements of the group acting on the set?

Question

I know that a group action is transitive when there is one orbit. Say that $G$ is a group acting on the set $A$. The identity element of $G$ will clearly create $|A|$-many orbits. But the other elements will create each their own set of orbits. Will all of these elements of $G$ give the same total number of orbits?

Answer 1

Elements of $G$ do not have orbits. Elements of $A$ have orbits (and orbits are subsets of $A$). If a group $G$ is acting on a set $A$ and $a \in A$, then we denote the orbit of $a$ by $\operatorname{orb}_{G}(a)$ and define $$ \operatorname{orb}_{G}(a) : = \{ g \cdot a : g \in G \} \subseteq A. $$

For any $a \in A$ $$ |\operatorname{orb}_{G}(a)| = (G : \operatorname{stab}_{G}(a) ), $$ where $\operatorname{stab}_{G}(a) \leq G$ is the stabilizer of $a$, i.e. $$ \operatorname{stab}_{G}(a) : = \{ g \in G : g \cdot a = a \} \subseteq G. $$

So orbits can have varying cardinalities.

EDIT:

With the examples you are considering, I believe you are looking at the specific example of a subgroup $H \leq G$ acting on $G$ via left multiplication: for any $h \in H$ and $g \in G$, we define $$ h \cdot g : = hg. $$ The orbit of $g \in G$ under the action of $H$ is then $$ \operatorname{orb}_{H}(g) = \{ hg : h \in H \}, $$ which is precisely the right coset $Hg$. All of this is to say that the orbit of $g \in G$ under the action of left multiplication by $H$ is the right coset $Hg$. It is worth mentioning a couple of things:

• The cosets $Hg$ partition $G$ and are all of the same cardinality. In particular, if $|H| < |G|$, then this action is not transitive.
• When $H$ acts by right translation, the orbits are the left cosets of $H$ in $G$.
• When $H$ is a normal subgroup of $G$, then the left and right cosets of $H$ in $G$ are the same and there is no distinction.
• This is a special group action that will not work all of the time. In particular, you are relying on the fact that the set you are acting on, $G$, is actually a group and you know how to multiply elements of the group $H$ and the set $G$. For instance, $S_{n}$ acts on the set $\{ 1 , \dots , n \}$, but there is no canonical multiplication of an element of $S_{n}$ and an element of $\{1 , \dots , n \}$.

With the examples you are discussing in the comments, it seems that you are looking at the example where $H$ is the cyclic subgroup generated by some element.

Answer 2

The right answer to the question is "Have you tried looking at any concrete examples?" I would start with some obvious ones like the symmetric group $S_3$ acting on $\{1, 2, 3\}$, the alternating group $A_4$ acting on $\{1, 2, 3, 4\}$, and the cyclic group $C_6$ acting on $\{1, 2, 3, 4, 5, 6\}$ by the operation "add 1 modulo 6". Look at all the elements in these groups, and consider the orbits into which they divide the set. The second example may be particularly instructive.

(It is amazing how much energy is devoted in comments to pointless pedantry about whether or not an element divides the set into orbits, given that (1) everyone complaining about this phrasing should be capable of understanding what the OP means via the unimportant transition from an element to the cyclic group it generates, and (2) this point is irrelevant to the question being asked.)