Let group $G$ acts transitively on finite sets $\Omega$ and $\Psi$, with $1<|\Omega|<|\Psi|$. If $G$ is simple, show that $|\Psi|$ cannot be prime. 
Let group $G$ acts transitively on finite sets $\Omega$ and $\Psi$, with $1<|\Omega|<|\Psi|$. If $G$ is simple, show that $|\Psi|$ cannot be prime.


My attempt:
First let's consider when $|G|<\infty$.
Since $G$ acts transitively on finite sets $\Omega$ and $\Psi$, we have $$ |\Omega|=\frac{|G|}{|G_\omega|}\quad\text{and}\quad |\Psi|=\frac{|G|}{|G_\psi|} $$
where $\omega\in \Omega$ and $ \psi\in \Psi $.
Now suppose the converse, $|\Psi|=p$ is prime then clearly $p$ cannot be the smallest prime dividing $|G|$, i.e. we must have another prime $q<p$ with $q\ \vert\ |G|$.
I think we should construct a normal subgroup of $G$ to get a contradiction but I failed... Can someone give me a hint? Thank you.
 A: Let $ \phi $ be a transitive action of $G$. If $G$ is simple then the only normal subgroups of the group are $G$ or $ \{ 1 \} $. The kernel of a group action $ \ker ( \phi ) $ is a normal subgroup (this is the subgroup of $G$ that fixes every element of a set being acted on). So the kernel is either $ \{ 1 \} $ or $ G $.
If the kernel of the action is $G$, then $G$ fixes every element, so the action is not transitive - because the sets have more than one element. Hence $ \ker ( \phi ) =  \{ 1 \} $.
Suppose $ g, h \in G $ such that $ x^{g} = x ^{h} $ for all $x \in \Omega $. Then $x ^{ gh^{-1} } = x $. So $ gh^{-1} $ fixes every element and $gh^{-1} = 1$ and that $g = h$. This means that every element of $G$ gives a different permutation of $ \Omega $. The same reasoning shows that every element of $G$ gives a different permutation of $ \Psi $. I.e. $G$ acts faithfully on these sets.
There are only finitely many different permutations of $ \Omega $ and $ \Psi $ because they are finite sets. Thus there must only be finitely many elements of $G$.
Let $n = | \Omega | $ and $ m = | \Psi |$. The fact that elements of $G$ are permutations of $ \Omega $ and $ \Psi $ and $G$ acts faithfully on both sets means that $G$ can be embedded in both $S_{n} $ and $S_{m} $.
So $| G | $ divides $|S_{n} | =  n! $ and also divides $|S_{m} | = m! $. By the orbit stabiliser theorem, $m = | \Psi |$ divides $|G| $ because $ \Psi $ is a single orbit. So $m$ divides $ n! $. If $m$ is prime then $m $ does not divide $ n! $ because $ n < m $. Hence $ | \Psi | $ is not prime.
