I'm currently working through Dobson, Malnič, Marušič's Symmetry in Graphs (2022), and am trying to prove the following proposition (pp. 208):
Exercise 5.2.1 If $G \leqslant \text{Sym}(\Omega)$ is a regular permutation group, then $G=G^{(2)}$, i.e. $G$ is 2-closed.
Given that $G^{(2)}=\bigcap_{i=1}^{|\Omega|-1}\text{Aut}(\Gamma_i)$, where $\Gamma_i$ are the orbital digraphs of $G$, my approach to prove 5.2.1 has been to determine the structure of an orbital digraph of a regular permutation group (a union of disjoint directed cycles, according to my working). However I'm unsure if this is a right approach, as the notion of 2-closure is fairly new to me.
Any advice on how to proceed would be appreciated.
For $\alpha,\beta \in\Omega$, let $\Gamma$ be the orbital digraph containing the directed edge $\alpha \to \beta$.
Since the stabilizer $G_\alpha$ of $\alpha$ in $G$ is trivial, this is the only edge in $\Gamma$ with source $\alpha$. So any element in ${\rm Aut}(\Gamma)$ that fixes $\alpha$ must fix $\beta$. So no element in $G^{(2)}$ fixes both $\alpha$ and $\beta$.
Since $G \le G^{(2)}$, $G^{(2)}$ is transitive with trivial point stabilizers, and hence $G = G^{(2)}$.