Sharply t-transitive groups

Sharply transitive groups are the smallest possible transitive groups, given $t \in \mathbb N$ a $t$-transitive group $G$ is called sharply t-transitive if $G_{\alpha_1 \alpha_2 \ldots \alpha_t} = 1$ for distinct $\alpha$s. Equivalently, there is exactly one group element that takes $(\alpha_1,\cdots,\alpha_t)$ to any other triple of distinct symbols.

Theorem If $G$ is a $t$-transitive group of degree $n$ (i.e. it acts on $n$ symbols) then it is sharply $t$-transitive iff $|G| = n (n-1)\cdots(n-t-1)$.
proof: For $t=1$ this is the regular action. For $t+1$ any $G_\alpha$ will be $t$-transitive and $|G|=n|G_\alpha|$ since the orbit of $\alpha$ is the whole of $\Omega$, conversely every stabilizer $G_\alpha$ will have the same order - and there are $n$ of them so each $|G_\alpha| = |G|/n$ is $t-1$-transitive so $G$ is $t$-transitive.

Example $S_n$ is sharply $n$-transitive in the natural actions, it is also $n-1$-sharply transitive. Note this isn't a contradiction from the group order result because $|G|=|G|\cdot 1$. Intuitively what's happening is if we choose exactly where $n-1$ symbols go, then it's already decided where the last one must go.

Proposition If $G$ is sharply $2$-transitive of degree $n$ then $G$ has a regular characteristic subgroup which is an elementary abelian $p$-group for some prime $p$ (so $n$ is power of $p$).
proof: Given $G$, let $K$ be the union of the fixed point free elements and 1. We will show it is a group. Since for any distinct $\alpha,\beta \in \Omega$, $G_\alpha \cap G_\beta = G_{\alpha \beta} = 1$ so $G$ is the disjoint union $K \sqcup \bigsqcup_{\alpha \in \Omega} G_\alpha^\#$, $|G_\alpha^\#| = \frac{|G|}{n}-1$ (using $n = |G|/|G_\alpha|$ from the orbit-stab. theorem) so $|K|=n$. Take $\alpha,\beta \in \Omega$ distinct and choose $k \in K^\#$ such that $\alpha k \not = \alpha$. As $G$ is $2$-transitive there exists $g \in G_\alpha$ with $(\alpha h) g = \beta$ then $h^g$ has the same order and fixedpoints as $h$ i.e. none, therefore its in $K$, and $\alpha k^g = \beta$ so $K$ is a transitive "set". For any $\alpha \in \Omega$ the map $K \to \Omega$ given by $k \mapsto \alpha k$ is surjective and hence bijective. Now take take distinct $x,y \in K$ we know $\alpha x \not = \alpha y$ so $\alpha x y^{-1} \not = \alpha$ for all $\alpha$, so $x y^{-1} \in K$. This proves $K$ a subgroup!
Given $k \in K^\#$ all its cycles on $\Omega$ must all have the same length since otherwise some non-identity power of $k$ would have fixed points. So the order of $k$ divides $n$. Similarly for elements of $G_\alpha^\#$ the same argument tells us all its cycles on $\Omega \setminus \{\alpha\}$ have the same length, so its order divides $n-1$. Hence $K$ consists of the elements of $G$ of order dividing $n$ - a property preserved by conjugation - therefore it's a characteristic subgroup. By the corollary from the section on regular normal subgroups it is an elementary abelian $p$-group.