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.

  • $\begingroup$ I think you wrote $H$ instead of $Y$ in many places in line $2$. Verify! $\endgroup$ – Bhaskar Vashishth Oct 8 '14 at 19:58
  • $\begingroup$ 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? $\endgroup$ – Bhaskar Vashishth Oct 8 '14 at 20:05
  • $\begingroup$ Ok got it. Thanks alot $\endgroup$ – Bhaskar Vashishth Oct 8 '14 at 20:10
  • $\begingroup$ Can you explain why $Y\neq X$ ? $\endgroup$ – learning_math Aug 8 '16 at 5:24
  • $\begingroup$ $Y=X$ would imply $H=G$. $\endgroup$ – 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 $g_0\in G\backslash H$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.