Let $G$ be a finite group that acts on a set $X$ faithfully and transitively. Additionally we have $|G| > |X| > 1$. Show that there is no $x\in X$ such that its stabilizer is a normal subgroup of $G$.
I managed to show that we always have $e \subsetneq G_x \subsetneq G$, so the stabilizer is never a trivial normal subgroup. But other than that I'm stuck.
Best Answer
Assume there is $x \in X$ such that $G_x \triangleleft G$. Then for any $g \in G$, $g G_x g^{-1} = G_x$. That is, for any $h \in G_x$ $$ ghg^{-1} x = x $$ This implies that $h$ stablizes $g^{-1}x$. Since $g$ and $h$ are arbitrary, $G_x$ stablizes every element in $G \cdot x$. The transitivity implies that $G \cdot x = X$. So $G_x$ stablizes any $y \in X$. Then the faithfulness implies $G_x = \{e\}$. And you have showed that $G_x = \{e\}$ is impossible.