A transitive action on more than one point is doubly transitive exactly when it has rank two

For the forward direction, suppose the action is doubly transitive. Then every $\beta \mathrm{\ne }\alpha$ can be sent to every other $\gamma \mathrm{\ne }\alpha$ by some element fixing $\alpha$, because [L2] applies to the ordered pairs $(\alpha ,\beta )$ and $(\alpha ,\gamma )$. Since $\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }>1$, the complement $\mathrm{\Omega }\setminus \{\alpha \}$ is nonempty, so the two $G_{\alpha }$-orbits are exactly $\{\alpha \}$ and $\mathrm{\Omega }\setminus \{\alpha \}$. Hence the rank is two by [L1].

For the converse direction, suppose the rank is two. Then [L1] says the only $G_{\alpha }$-orbits are $\{\alpha \}$ and $\mathrm{\Omega }\setminus \{\alpha \}$, so $G_{\alpha }$ is transitive on the complement of $\alpha$. Given ordered pairs $(x,y)$ and $(x^{\mathrm{\prime }},y^{\mathrm{\prime }})$ with $x\mathrm{\ne }y$ and $x^{\mathrm{\prime }}\mathrm{\ne }{y}^{\mathrm{\prime }}$, choose $g\in G$ with $g\cdot x=x^{\mathrm{\prime }}$ by transitivity. The stabilizer satisfies $G_{x^{\mathrm{\prime }}}=g{G}_x{g}^{-1}$, and because the action is transitive the rank-two hypothesis at $\alpha$ implies the same two-suborbit description at $x$. Hence $G_{x^{\mathrm{\prime }}}$ is transitive on $\mathrm{\Omega }\setminus \{x^{\mathrm{\prime }}\}$, so some $h\in G_{x^{\mathrm{\prime }}}$ sends $g\cdot y$ to $y^{\mathrm{\prime }}$. Then $hg$ sends $(x,y)$ to $(x^{\mathrm{\prime }},y^{\mathrm{\prime }})$. Therefore the action is doubly transitive by [L2].

The two directions of steps 1.1 and 1.2 prove the equivalence.