Show that for transitive subgroup $A$, its centralizer divides $|\Omega| = |A : A_{\alpha}|$ Let $G$ be a group acting faithful on a finite set $\Omega$ and suppose the subgroup $A \le G$ acts transitive on $\Omega$. Then $|C_G(A)|$ divides $|\Omega|$, and if in additon $A$ is abelian, then $C_G(A) = A$.
How to solve this?
What I got: If some $x \in C_G(A)$ fixes a point $\alpha$, then as for each $a \in A$ we have $\alpha^{ax} = \alpha^{xa} = \alpha^a$ by transitivity it fixes all points, hence as the action is faithful we have $x = 1$. This shows that 
$$
 C_G(A) \cap G_{\alpha} = 1
$$
for all $\alpha \in \Omega$ and hence $C_G(A) \subseteq G \setminus \bigcup_{g\in G} G_{\alpha^g}$. But I do not see how to show that the order of the centralizer divides $|\Omega| = |G : G_{\alpha}| = |A : A_{\alpha}|$.
The rest is clear, as by Frattini $G = AG_{\alpha}$ and if $A$ is abelian $A \le C_G(A)$ and hence $G = C_G(A)G_{\alpha}$. Using $C_G(A) \cap G_{\alpha} = 1$ and hence $A \cap G_{\alpha} = 1$ we find 
$$
 |A||G_{\alpha}| = |G| = |C_G(A)||G_{\alpha}|
$$
so that $|A| = |C_G(A)|$ which gives $A = C_G(A)$.
So just the fact that the order of the centralizer divides $|\Omega|$ is missing. If $C_G(A)G_{\alpha}$ would be a group, then we could reason that as its order divides $|G| = |G:G_{\alpha}||G_{\alpha}|$ and $|C_G(A)G_{\alpha}| = |C_G(A)||G_{\alpha}|$ that $|C_G(A)|$ must divide $|G : G_{\alpha}|$, but in general we do not know that $C_G(A)G_{\alpha}$ is a group? 

When you showed that $C_G(A)\cap G_{\alpha}=1$, noticed that you showed it for any $\alpha \in \Omega$. Thus, orbit of $\alpha$ has exactly $|C_G(A)|$ elements under the action of $C_G(A)$ on $\Omega$. It directly showed that $|C_G(A)|$ divides $|\Omega|$.
