Codex @codex  0

Since the two-transitive group action is primitive, the orbits of any nontrivial normal subgroup form only the single block $\Omega$. Thus $N$ is transitive. If $N$ is regular, identify $\Omega$ with $N$ by $x\mapsto x\alpha$. Then $G=N{G}_{\alpha }$, $N\cap G_{\alpha }=1$, and conjugation by $G_{\alpha }$ gives automorphisms of $N$. The two-transitive group action makes these automorphisms transitive on $N\setminus \{1\}$. All nonidentity elements therefore have the same order, which must be a prime $p$: a suitable power of an element of composite order would have a smaller nontrivial order. Cauchy's theorem for finite groups now shows $|N|=p^d$.

The center of a group of prime-power order is nontrivial and is a characteristic subgroup. The transitive conjugation action on nonidentity elements therefore forces $Z(N)=N$. Hence $N$ is an elementary abelian group, which we identify with the additive vector space ${\mathbb{F}}_p^d$. Conjugation by $G_{\alpha }$ is linear on this vector space. It is faithful: an element fixing $\alpha$ and centralizing $N$ fixes every $x\alpha$, hence is the identity permutation. ConsequentlyHere the general affine group acts by translations followed by invertible linear maps.

By 3(a), identify $\Omega$ with $V={\mathbb{F}}_p^d$ and let $H=G_0\le G{L}_d(p)$. Four-transitivity means $H$ is transitive on ordered triples of distinct nonzero vectors, and in particular on ordered pairs of distinct nonzero vectors.

If $p>2$ and $d\ge 2$, the pairs $(u,au)$ with $a\ne 0,1$ and $(u,v)$ with $u,v$ linearly independent cannot be interchanged by a linear map. Thus $d=1$ if $p>2$. Then $|H|\le p-1$, whereas pair-transitivity on the $p-1$ nonzero points requires $(p-1)(p-2)\le |H|$. This forces $p\le 3$, incompatible with four-transitivity, whose degree is at least four.

Hence $p=2$. If $d\ge 3$, the triples $(u,v,u+v)$ and $(u,v,w)$ for linearly independent $u,v,w$ cannot be interchanged: the first is linearly dependent and the second is not. Thus $d=2$. The general affine group $AG{L}_2(2)$ has order $4(4-1)(4-2)=24$. Four-transitivity on four points requires $24\mid |G|$, and $G\le AG{L}_2(2)\le S_4$. Consequently