# 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!!

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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.
you are supposing that the action is on a finite set $X$ of cardinality $n$ so that you identified $Aut(X)$ with $S_n$ why? and where did you use abelian here? –  palio Jul 11 '11 at 19:55