every transitive action of an abelian group is regular Why the following is true :

"Every transitive
  action of an abelian group is regular" 

Does this mean that every action of an abelian group is free? because as i understand, a regular action is an action that is transitive and free. but i know this is not true because an abelian group acting trivially is not acting freely!!
 A: In response to comments of the questioner, I expand (it is correct that the statement need not be confined to action on a finite set. The use of the fact that the image 
group is Abelian to show regularity is more subtle).
There is an important word missing. Every faithful transitive action of an Abelian group
is regular. If $\phi:G \to {\rm Sym}(\Omega)$ is a homomorphism and $G\phi$ is a transitive Abelian subgroup of ${\rm Sym}(\Omega)$, then $G\phi$ is regular, since for any $x \in G$, we see that $\langle x \phi \rangle \lhd G\phi$ as $G\phi$ is Abelian, so that $G\phi$ permutes the fixed points of $x\phi$. But since $G\phi$ is transitive, all points of $\Omega$ must be fixed by $x\phi$ and $x \in {\rm ker} \phi$. Hence we have shown that $x\phi$ fixes an element of $\Omega$ if and only if $x \in {\rm ker}\phi$. But since $G\phi$ is transitive by hypothesis, $G\phi$ is regular. 
We have used the fact that $G\phi$ is Abelian (actually just that all its subgroups are normal), though it is not necessary for the proof that $G$ itself be Abelian. However, to say that the action is faithful means that ${\rm ker} \phi =1$, so that $G\phi \cong G$, and so in tht case, if $G\phi$ is Abelian, $G$
must also be Abelian.
