$\textbf{Definition}$ Let $G$ be a group acting on a set $X$. Let $|X|\geq 2$. We say that the action of $G$ on $X$ is doubly transitive iff for any $y_1,y_2\in X$ such that $y_1\neq y_2$ and $x_1,x_2\in X$ such that $x_1\neq x_2$ there is $g\in G$ such that $gx_1=y_1$ and $gx_2=y_2$.

For $x\in X$ let $\text{Stab}_G(x)$ be the stabilizer of $x$.

Prove that the following are equivalent.

  1. action of $G$ on $X$ is doubly transitive.

  2. The action of $G$ is transitive on $X$ and the action of $\text{Stab}_G(x)$ on $X-\lbrace x\rbrace$ is transitive for all $x\in X$.

I am able to prove that (1) $\Rightarrow$ (2). How to prove (2) $\Rightarrow$ (1) ?


If $G$ satisfies 2., and all four of $x_1,x_2, y_1,y_2\in X$ are distinct, then let $g_1$ be an element in $\operatorname{Stab}_G(x_1)$ with $g_1x_2=y_2$ and $g_2\in\operatorname{Stab}_G(y_2)$ with $g_2x_1=y_1$. Then look at $g=g_2g_1$.

If, for instance, $x_1=y_2$, you need to revise the above, but the idea still works.

Edit: Even simpler, and without having to care for special cases: let $g_1\in G$ be any element with $g_1x_1=y_1$, and let $g_2\in \operatorname{Stab}_G(y_1)$ be such that $g_2(g_1x_2)=y_2$. Define $g$ as above.

  • $\begingroup$ There is a second part to this question. Assume that $|X|\geq 3$. Prove that, if $\text{Stab}_G(x)$ acts transitively on $X-\lbrace x\rbrace$ for all $x\in X$, then $G$ acts doubly transitively on $X$. Any idea ? $\endgroup$ – learning_math Aug 7 '16 at 12:09
  • 1
    $\begingroup$ Yes. Let $x$ be an element other than $x_1$ and $y_1$ (which exists since there are at least three elements in $X$), and take the element you need from $\operatorname{Stab}_G(x)$. $\endgroup$ – Arthur Aug 7 '16 at 12:12

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