Difference between regular and transitive action in terms of stabilizers in a group

Question

A group action on a set is said to be transitive if it has only one orbit. Now, a group action on the same set is said to be regular if all the stabilizers of the elements of the set are trivial.

My doubt is, if there is only one orbit, then, by orbit-stabilizer, all stabilizers must be trivial right? What am I missing here? Any hints? Thanks beforehand.

Answer 1

To add on to the answer by MathFail, the reason this doesn't contradict orbit stabilizer is that orbit-stabilizer (for a finite group) tells us that $|G_x| \cdot |G \cdot x| = |G|$. If our group acts transitively on a set $X$, it tells us that $|G \cdot x| = |X|$, and so, plugging that into orbit-stabilizer gives that $|X| \cdot |G_x| = |G|$. If $|X|$ is smaller than $|G|$, we can have $|G_x| > 1$! Note, however, that this implies for a finite group $G$ acting on a set $X$ with $|G| = |X|$, we have that the action is regular iff it is transitive.

With this in mind, is it possible to come up with an example of a group action that is regular but not transitive?

Answer 2

No, it is false. Let $G=\langle g\rangle$ and $|G|=4$, and set $X=\{a, b\}$. Define

$$ea=a, eb=b, ga=b, gb=a$$

hence $G$ is transitively acting on $X$, but the stablizer

$$G_a=\{e, g^2\}=G_b$$

is non-trivial.

Answer 3

if there is only one orbit, then, by orbit-stabilizer, all stabilizers must be trivial right?

Not really. Rather, the orbit-stabilizer tells you that all the stabilizers have order $\frac{|G|}{|X|}$, which is not necessarily equal to $1$.