$G$ acts faithfully on $\Omega$, $A\leq G$, $A$ transitive on $\Omega$. Then $|C_G(A)|$ is a divisor of $|\Omega|$. If in addition $A$ is abelian then $C_G(A)=A$. $G$ and $\Omega$ are finite.
Let $\Gamma:\Omega\times G \to \Omega$ be the action. Then the restriction $\Theta$ of $\Gamma$ to $\Omega\times A$ is an action and we are told it is transitive. Let $|\Omega|=n$. $\Gamma$ induces a homomorphism $\Gamma':G\to S_n$ that is one to one. Also $\Theta$ induces a homomorphism $\Theta':A\to S_n$ which is the restriction of $\Gamma'$ to $A$. Therefor as $\Gamma'$ is one to one so is $\Theta'$, that is $\Theta$ is faithful.
The image of $G$ under $\Gamma'$ is isomorphic to $G$ and a subgroup of $S_n$. Hence $|G|$ divides $n!$ (1). Also, if $\alpha\in\Omega$ and $A_\alpha$ is the stabilizer of $\alpha$ in $A$ then $|\Omega|=|\alpha^A|= |A|/|A_\alpha|=|A|/|G_\alpha\cap A|$ (2) because of the transitivity of $A$. Aside from this I don't find any further equations to help me prove the statement. Could somebody give me a hint?
First, we need that $C_G(A)$ acts semi-regularly on $\Omega$, i.e., no non-identity element of $C_G(A)$ stabilizes a point on $\Omega$. To see this, let $g\in C_G(A)$ stabilize a point $x\in \Omega$. Since $A$ is transitive on $\Omega$, for any $y\in \Omega$ there exists $a\in A$ such that $xa=y$. But then $g^a=a$, but as is standard, if $g$ stabilizes $x$ then $g^a$ stabilizes $xa=y$. Thus $g$ stabilizes $\Omega$, and $g=1$. This proves that $C_G(A)$ acts semi-regularly on $\Omega$.
Each orbit has length $|C_G(A)|$, by the orbit-stabilizer theorem, and so $|C_G(A)|$ divides $|\Omega|$, as needed.