$G$ acts faithfully on $Omega$, $Aleq G$, $A$ transitive on $Omega$. Then $|C_G(A)|$ is a divisor of $|Omega|$.

Question

$G$ acts faithfully on $Omega$, $Aleq G$, $A$ 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 $Gamma:Omegatimes G to Omega$ be the action. Then the restriction $Theta$ of $Gamma$ to $Omegatimes A$ is an action and we are told it is transitive. Let $|Omega|=n$. $Gamma$ induces a homomorphism $Gamma':Gto S_n$ that is one to one. Also $Theta$ induces a homomorphism $Theta':Ato S_n$ 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 $alphainOmega$ and $A_alpha$ is the stabilizer of $alpha$ in $A$ then $|Omega|=|alpha^A|=  |A|/|A_alpha|=|A|/|G_alphacap A|$ (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?

David A. Craven · Answer

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 $gin C_G(A)$ stabilize a point $xin Omega$. Since $A$ is transitive on $Omega$, for any $yin Omega$ there exists $ain A$ 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.

$G$ acts faithfully on $Omega$, $Aleq G$, $A$ transitive on $Omega$. Then $|C_G(A)|$ is a divisor of $|Omega|$.

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