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?

  • 4
    $\begingroup$ Elements don't have orbits. Subgroups do. So the element $g\in G$ will have a cyclic subgroup associated with it, and that subgroup might or might not be transitive. But we can't say how many orbits you'll get. $\endgroup$ Jul 22, 2022 at 19:47
  • 2
    $\begingroup$ I never said that elements of $G$ have orbits. I said that $g∈G$ produces a certain number of orbits. My question is, if $g1,g2∈G$, will $g1,g2$ create the same quantity of distinct orbits of the elements of $A$? $\endgroup$
    – user1080674
    Jul 24, 2022 at 14:00
  • 3
    $\begingroup$ Elements $g$ don’t produce orbits, either. $\endgroup$ Jul 24, 2022 at 14:06
  • $\begingroup$ It partitions $A$ into a set of orbits of the same cardinality. $\endgroup$
    – user1080674
    Jul 24, 2022 at 14:19
  • 1
    $\begingroup$ Essentially, you have invented a concept of “orbits which are ‘produced’ by $g$,” without definition, and then invented properties of it. For example the dihedral group $D_8$ acts on the the set $\{1,2,3,4\}$ transitively, and yet some non-identity elements of $D_8$ fix some. For example, the reflection $(1,3)$ fixes $2$ and $4,$ and the only possible definition of the “orbits” of $(1,3)$ are the three orbits, $\{1,3\} ,\{2\},\{4\},$ two of which are size $1$ and one of which is size $2.$ $\endgroup$ Jul 24, 2022 at 14:37

2 Answers 2


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.


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.

  • $\begingroup$ I never said that elements of $G$ have orbits. I said that $g \in G$ produces a certain number of orbits. My question is, if $g_1, g_2 \in G$, will $g_1, g_2$ create the same quantity of distinct orbits of the elements of $A$? $\endgroup$
    – user1080674
    Jul 24, 2022 at 13:58
  • $\begingroup$ @NinaDumectra I do not understand your question. Explain how $g_{1}, g_{2}$ produce orbits. Can you write down, in set theoretic language, “the orbit produced by $g_{1}$?” $\endgroup$
    – Oiler
    Jul 24, 2022 at 14:10
  • $\begingroup$ Take the identity element of $S_3$ acting on $S_3$. It creates 6 orbits each of size 1 of all the elements of $S_3$. If you take the element $(12)$ and act on $S_3$, it partitions $S_3$ into 3 distinct orbits of size 2. For a transitive group action, will every element of $G$ always partition the set $A$ into 1 orbit? $\endgroup$
    – user1080674
    Jul 24, 2022 at 14:20
  • 1
    $\begingroup$ @NinaDumectra You need to specify the group action; I am assuming you are talking about $S_{3}$ acting on itself via left multiplication. I still am not 100% certain what you are asking, but I believe the answer to your question is no and I think your very own example with $S_{3}$ shows why. A group acting on itself via left multiplication is always transitive and for $S_{3}$, the element $(1 2)$ does not partition the set $A$ into 1 orbit. $\endgroup$
    – Oiler
    Jul 24, 2022 at 20:17
  • 1
    $\begingroup$ Perhaps the terminological issues here are larger than the actual mathematical ones... $\endgroup$ Jul 24, 2022 at 21:56

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.)


You must log in to answer this question.