# Stabilizer of a doubly transitive is maximal?

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.

• I think you wrote $H$ instead of $Y$ in many places in line $2$. Verify! – Bhaskar Vashishth Oct 8 '14 at 19:58
• why $G_x$ acts transitively on $X$\{$x$}. Are you using $2$-transitivity here? But Subgroup of a doubly transitive need not be doubly transitive? – Bhaskar Vashishth Oct 8 '14 at 20:05
• Ok got it. Thanks alot – Bhaskar Vashishth Oct 8 '14 at 20:10
• Can you explain why $Y\neq X$ ? – learning_math Aug 8 '16 at 5:24
• $Y=X$ would imply $H=G$. – Derek Holt Aug 8 '16 at 8:29

Here is a direct proof:

Recall that the action of $G$ being doubly transitive means that for two any ordered pairs of distinct elements $(x_0,x_1),(y_0,y_1)\in X\times X$ there is some $g\in G$ such that $(gx_0,gx_1)=(y_0,y_1)$. Note that double-transitivity implies transitivity.

We need to show two things: (1) given any $x\in X$, $G_x\neq G$, and (2), if $H$ is a subgroup of $G$ with the property $G_x\subseteq H\subsetneq G$ then $H=G_x$. To show (1), let $x,x'\in X$ be distinct elements. Since the $G$-action is also transitive, there is some $g\in G$ for which $gx=x'$, hence $g\notin G_x$, hence $G\supsetneq G_x$. To show (2), let $x\in X$ and let $H$ be a proper subgroup of $G$ with $G_x\subsetneq H$; we'll derive a contradiction. Let $h\in H\backslash G_x$ and $g_0\in G\backslash H$, so that the elements $x,hx, g_0x$ are all distinct because neither $h$ nor $g_0$ stabilize $x$. Applying double-transitivity to the pairs $(x,hx)$ and $(x,g_0x)$, there must be some $g\in G$ such that $gx=x$ and $ghx=g_0x$, thus $g,g_0^{-1}gh\in G_x$. Since $G_x\subseteq H$, it follows that each of $g,h,g,g_0^{-1}gh$ must all lie in $H$. But then $$g_0=gh(h^{-1}g^{-1}g_0)=gh(g_0^{-1}gh)^{-1}\in H,$$ which contradicts the fact that $g_0\in G\backslash H$.