Is it true that if $G$ is a group acting $2$-transitively on a set $X$ , then if $x\in X$, then $G_x$ (stabilizer) is maximal in $G$.
I think it must be true as a conclusion of $2$ theorems, as following-
$\textbf{1)}$ Every doubly transitive $G$-set is primitive.
$\textbf{2)}$ Let $X$ be a transitive $G$-set, then $X$ is primitive iff for each $x\in X$ , $G_x$ is a maximal subgroup.
But when I try to prove it directly using $2$-transitive property, to show $G_x$ is maximal, I don't see how to? Any help/hints?
Suppose that $G_x < H < G$ (with strict inequalities) and let $Y = x^H$ be the orbit of $x$ under $H$. Since $G_x<H$, $Y \ne \{x\}$ and, since $H<G$, $Y \ne X$. So there exists $y \in Y \setminus \{x\}$ and $z \in X \setminus Y$.
Since $G_x < H$ and $G_x$ acts transitively on $X \setminus \{x\}$, there exists $h \in H$ with $y^h=z$, which contradicts the fact that $y$ and $z$ are in different orbits of $H$. So $G_x$ is indeed maximal in $G$.
All I have really done here is to glue together the proofs of the two results you mentioned.