Does faithfull action of abelian group always has regular orbit? Good afternoon. I recall that a regular orbit is the orbit of the size equal to the acting group. My question is motivated by the following post. I don't understand why the faithfull action of abelian group always has regular orbit? (The last part of solution in the mentioned post). I will be gratefull for hints and expalanations.
 A: Let me correct some definitions.
The action of $G$ on $X$ is faithful if the only element of $G$ that fixes all points of $X$ is the identity element, $1\in G$. 
The action of $G$ on $X$ is regular if the action is transitive and the only element of $G$ that fixes at least one point of $X$ is $1\in G$. 
It is not true that a faithful action of an abelian group is regular. For example, is $\sigma = (1\;2)(3\;4\;5)$, then $G=\langle \sigma\rangle$ acts faithfully but not regularly on $X = \{1, 2, 3, 4, 5\}$.
What is true is that, if $G$ is abelian, any faithful transitive action is regular. For this you must show that if $G$ acts transitively on $X$ and $g\in G$ has at least one fixed point in $X$, then it fixes all elements of $X$. The argument goes like this: if $gx_0=x_0$ for some $x_0\in X$, then for any $h\in G$ we have $g(hx_0)=h(gx_0)=hx_0$, so $hx_0$ is a fixed point of $g$. Thus $g$ sixes all points of $Gx_0=X$. Thus it is reasonable to say that a faithful orbit of an abelian group is regular.
A: More of a long comment actually:
This is not true. Consider the subgroup $\{(1 2), (3 4), (1 2)(3 4)\} \subset S_4$, which is isomorphic to the Klein four group. It is abelian and acts faithfully on the set $\{1, 2, 3, 4\}$ of four elements but does not have a regular orbit. I think the other post explicitly uses that the group under consideration is not just abelian but also cyclic, but we have to think a bit more about if and how that would give us the desired property.
UPDATE: It seems that the 'other post' I am referring to has since been deleted. Is that correct? Otherwise my post doesn't make much sense.
