Multiply transitive group action

Remarks

A $1$-transitive action is a transitive action. A $k$-transitive action is $j$-transitive for every $j \le k$. For $\abs{\Omega} \ge 2$, an action is $2$-transitive if and only if it is transitive and the stabiliser $G_\omega$ of one point $\omega$ is transitive on $\Omega \setminus \set{\omega}$.

Example: the symmetric group $S_n$ is $n$-transitive on $\set{1, \dots, n}$, since any bijection between two orderings of distinct points extends to a permutation. For $n \ge 3$, the alternating group $A_n$ is $(n-2)$-transitive. Indeed, let $\sigma \in S_n$ do the job, and let $\tau$ swap the two points not among $y_1, \dots, y_{n-2}$. Then $\tau \sigma$ also does the job, and one of $\sigma$ and $\tau \sigma$ is even. For a finite field $\F_q$, the maps $x \mapsto ax + b$ with $a \ne 0$ act sharply $2$-transitively on $\F_q$.

Non-example: the eight symmetries of a square, acting on its corners $1, 2, 3, 4$ in cyclic order. The action is transitive, but no symmetry sends the pair $(1, 2)$ of neighbouring corners to the pair $(1, 3)$ of opposite corners, so it is not $2$-transitive.

Each $2$-transitive action is primitive. Camille Jordan showed that for $k \ge 4$ the only finite sharply $k$-transitive groups are $S_k$, $S_{k+1}$, $A_{k+2}$, and the Mathieu groups $M_{11}$ for $k = 4$ and $M_{12}$ for $k = 5$.