For a transitive and faithful group action the stabilizers are not normal

Question

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.

Answer 1

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.

Answer 2

By transitivity, all the stabilizers are conjugate. If any of them was normal, they would be all equal, hence one, say $H$. But then the kernel (= intersection of all the stabilizers) would be again $H$, which is not trivial because $|H|=\frac{|G|}{|X|}>1$, by assumption. Contradiction.

Answer 3

Here's a nonelementary way of proving the statement, using character theory.

Let $x\in X$ and suppose that $G_x\unlhd G$, or equivalently, $\text{Core}_G(G_x)=G_x$. If $G$ acts transitively on $X$ then the induced character $(1_{G_x})^G$ is the permutation character of the action. This is, $$(1_{G_x})^G(g)=|\{x\in X: gx=x\}|, \text{ for each }g\in G.$$ As the action is faithful, the only $g\in G$ such that $gx=x$ for all $x\in X$ is $g=1$. Hence, $\ker(1_{G_x})^G=1$. But, $$1=\ker(1_{G_x})^G=\bigcap_{g\in G}(\ker 1_{G_x})^g=\bigcap_{g\in G}(G_x)^g=\text{Core}_G(G_x)=G_x,$$ which is a contradiction to $1\subsetneq G_x$, as you proved.

Answer 4

Here's another approach. As it is mentioned, $1<G_x$, for every $x\in X$. I'll introduced a conjugate of an element in $G_x$, such that this conjugate does not belong to $G_x$.

Let $1\ne h\in G_x$. As the action is faithful there exists $y\in X$, such that $hy\ne y$, and also there exists $g\in G$ such that $gx=y$ (which implies $g^{-1}y=x$), since the action is transitive. Now we can see that $$(g^{-1}hg)x=g^{-1}hy\ne g^{-1}y=x$$.