Showing that a transitive abelian permutation group is necessarily regular I am trying to show that a transitive, abelian permutation group acting on a set $X$ is necessarily regular, given this hint: 'Given $g \in G$, consider the set $X^g:=\{x \in X\,|\,gx=x\}$.  Show that if $G$ is transitive and abelian, then the only possibilities for $X^g$ are $\varnothing$ or $X$.'

I know that $X^g=X$ iff $g$ is the identity of $G$.  And...that's about all I've got.  I don't even see how, once the above is shown to be true, the result follows.

I would really appreciate some hints to help me in the right direction.

This is from Isaacs' Algebra: A Graduate course.  He introduces permutation groups first, but group actions haven't entered the scene yet.  I am alright with a hint in terms of actions though.

Thanks!
Hint: Let $1\not=g \in G$ and suppose $gx=x$ for some $x\in X$.  Then take another element $y\in X$.  Using the fact that $G$ is transitive, we have that some other element $h\in G$ such that $hy=x$.  Now use the fact that $G$ is abelian to derive a contradiction.
Let $gx=y$, $hx=y$, where $g,h$-elements of the group $G$; $x,y$-elements of the set $A$. $G$ acts on $A$
Let $t$ some element of $A$. As $G$-transitive,there exist $f$ in $G$ that
$t=fx$ As $G$ is abelian we have:
$$gt=g(fx)=(gf)x=(fg)x=f(gx)=fy=f(hx)=(fh)x=(hf)x=h(fx)=ht$$
So for any $t$ in $A$ we have $gt=ht$, so $g=t$, then $G$-regular