Suppose $\Omega$ is a finite set with $|\Omega| \geq 5$. Let $G$ act faithfully on $\Omega$ such that $G$ is 4-transitive on $\Omega$. Let $N$ be a normal, nontrivial, nonregular subgroup of $G$. I have three questions:
(Let $\omega \in \Omega$)
a. Show that the point stabilizer $N_{\omega}$ is a normal subgroup of $G_{\omega}$.
b. Show that $G_{\omega}$ is 2-transitive on $\Omega - \{ {\omega} \}$.
c. Show that $N$ is 2-transitive on $\Omega$.
I've already done (a.). What I did was to get a $g \in G_{\omega}, n \in N_{\omega}$. Then, since $N$ is normal, $g^{-1}ng \in N$. And since $g$ and $n$ fix $\omega$, $g^{-1}ng \in N_{\omega}$. This means that $g^{-1}N_{\omega}g \subseteq N_{\omega}$. Therefore, $N_{\omega}$ is a normal subgroup of $G_{\omega}$.
However, I'm stuck at (b.). How do I show that given any $a, b, c, d \in \Omega - \{ \omega \}$, $a$ distinct from $b$, $c$ distinct from $d$, there is a $g \in G_{\omega}$ such that $(a, b)^g = (a^g, b^g) = (c, d)$? And how would I use this to prove the last statement?
You are given that $G$ is $4$-transitive on $\Omega$, and the triples $(a,b,\omega)$ and $(c,d,\omega)$ each consist of three distinct elements (given, as stated, that $a\neq b$ and $c\neq d$). Therefore, there is an element $g$ in $G$ for which $(a,b,\omega)^{g} = (c,d,\omega)$. In particular, $\omega^{g} = \omega$, so that $g$ belongs to $G_{\omega}$.