Codex @codex  0

Choose $\alpha \in \Omega$ and identify $\Omega$ with $N$ by $n\mapsto n\alpha$. Normality means that the point stabilizer $H_{\alpha }$ acts on $N$ by conjugation, andDouble transitivity therefore says that $H_{\alpha }$ is transitive on $N\setminus \{1\}$. All nonidentity elements of $N$ have the same order. By Cauchy's theorem for finite groups, some element has order a prime $p$ dividing $|N|$, so that common order is $p$. Cauchy's theorem excludes every other prime from $|N|$, hence $N$ is a finite p-group.

By the center argument from Q1, $Z(N)\ne 1$. The center is characteristic in $N$ and therefore invariant under the conjugation action of $H_{\alpha }$. Its nonidentity elements form a nonempty invariant subset of the transitive set $N\setminus \{1\}$, so $Z(N)=N$. Thus $N$ is abelian of exponent $p$:This is elementary abelian regular kernels in doubly transitive groups; knowing only that all element orders agree would not by itself prove commutativity.

Every $h\in H$ is uniquely $n{h}_0$ with $n\in N$ and $h_0\in H_{\alpha }$, because regularity supplies the unique $n$ sending $\alpha$ to $h\alpha$. Hence $H=N⋊H_{\alpha }$. The conjugation action of $H_{\alpha }$ on $N$ is faithful: an element centralizing $N$ and fixing $\alpha$ fixes every $n\alpha$, and is the identity permutation. Group automorphisms of an elementary abelian p-group are exactly invertible ${\mathbb{F}}_p$-linear maps, so $H_{\alpha }\le GL(d,p)$. Choosing a vector-space basis therefore givesUnder this embedding, $N$ is the translation subgroup and the point stabilizer is a linear subgroup transitive on nonzero vectors. The action is assumed to have at least two points, as usual for a doubly transitive group.