Multiple-transitivity

Consider actions of rank 2 (meaning that there are two suborbits of $G_\alpha$), $\Omega \setminus \{\alpha\}$ forms a single $G_\alpha$ orbit (because one its other orbit is $\alpha G_\alpha = \{\alpha\}$). $G$ acts on $\Omega^2$ non-transitively because $g(\beta,\beta) = (\gamma,\gamma)$ but if we define the diagonal $Delta = \{(\beta,\beta)\in \Omega\}$ this is a single orbit due to transitivity and so $\Omega^2 \setminus \Delta$ is the interesting part:

Lemma The action of $G$ on $\Omega$ is of rank 2 iff $\Omega^2 \setminus \Delta$ is a single orbit.
proof: If the action has rank 2 then let $(\beta_1,\beta_2), (\gamma_1,\gamma_2)$ lie off the diagonal and pick $x,y \in G$ such that $\alpha x = \beta_1$, $\alpha y = \gamma_1$ then neither of $\beta_2 x^{-1}$ and $\gamma y^{-1}$ are equal to $\alpha$ (otherwise $\beta_1 = \beta_2$ or $\gamma_1 = \gamma_2$) so there exists $h \in G_\alpha$ which maps one to the other $\beta_2 x^{-1} h = \gamma_2 y^{-1}$. Let $g = x^{-1} h y$ and compute $$(\beta_1,\beta_2) g = (\gamma_1,\gamma_2).$$
In the other direction if  $\Omega^2 \setminus \Delta$ is a single orbit the action is at least 2 (since we can fix any one element $\beta$ in the first component and map any other element $\gamma$ not equal to beta to any other element not equal to beta). So suppose the rank were larger than 2, then pick $\beta,\gamma$ in different $G_\alpha$ orbits of $\Omega \setminus \{\alpha\}$ and take $(\alpha,\beta), (\alpha,\gamma) \Omega^2 \setminus \Delta$ there's clearly no way to map from one to the other.

We can generalize this to $\Omega^t$ for any natural $t \le |\Omega|$. Again $\Delta = \{(\alpha,\alpha,\ldots)\}$ is a single orbit, but the interesting part is $\Omega^{(t)} = \{(\alpha_1,\alpha_2,\ldots)|\alpha_i \not = \alpha_j\}$.

Definition The action of $G$ on $\Omega$ is $t$-transitive if the induced action on $\Omega^{(t)}$ is transitive. This is equivalent to saying it can simultaneously map any $t$ distinct points to any other $t$ distinct points.

Lemma In terms of cosets, a group action is 2-transitive iff $G = G_\alpha \cup G_\alpha g G_\alpha$ for any $g \setminus G_\alpha$.
proof:  We saw previous that double cosets correspond to suborbits, write $G_\alpha = G_\alpha 1 G_\alpha$ to see this is the same as rank 2.

Lemma The action of $G$ on $\Omega$ is $t$-transitive iff the action of $G_\alpha$ on $\Omega \setminus \{\alpha\}$ is $(t-1)$-transitive.
proof: write this out.

Corollary If $G$ acts $t$-transitively on $\Omega$ and $|\Omega|=n$ then $|G|$ is divisible by $n(n-1)\cdots(n-t+1)$. WHICH THEOREM DOES THIS DEPEND ON? CHECK OTHER BOOK

Theorem In the natural action $S_n$ is $n$-transitive while $A_n$ is $n-2$ transitive and not $n-1$ transitive.
proof: This is obvious from the fact $S_n$ contains every permutation. For $A_n$ we use induction: it clearly holds for $A_3$. For $n \ge 3$ the stabilizer of any point of $A_n$ is $A_{n-1}$ so by the lemma we complete the induction.