Finite 2-transitive groups have affine or almost simple socle type

Choose a minimal normal subgroup $N\mathrm{\ne }1$, possible by finiteness. It is transitive by [L1]. If $N$ is regular, identify $\mathrm{\Omega }$ with $N$ by $u\mapsto u\alpha$. Conjugation by $G_{\alpha }$ is transitive on $N\setminus \{1\}$, so all nonidentity elements have the same order, necessarily a prime $p$ by taking powers. By Cauchy's theorem $N$ is a $p$-group; its nontrivial characteristic center equals $N$ by [L3]. Thus $N$ is elementary abelian. A permutation centralizing a regular group is determined by its value at $\alpha$; for abelian $N$ it is the corresponding translation. A second minimal normal subgroup would therefore lie in $N$ by [L2], which is impossible. The affine conclusion follows from [L6].

Assume henceforth that $N$ is nonregular. Since $N_{\alpha }⊴G_{\alpha }$, all its orbits on $\mathrm{\Omega }\setminus \{\alpha \}$ have a common size $m$. Transitivity of $N$ gives the same size at every point. Here $m>1$, for otherwise $N_{\alpha }$ fixes every point and is trivial.

We establish a finite permutation fact: a faithful transitive group $U$ whose nontrivial suborbits all have size $m>1$ is primitive or has trivial two-point stabilizers. Suppose $B$ is a block of size $1<k<n$, with $b\in B$ and $i\mathrm{\notin }B$. Since $U_b$ preserves $B$, $m\mid (k-1)$, so $\gcd (k,m)=1$. The union of the translates of $B$ under $U_i$ is a union both of blocks of size $k$ and of suborbits of size $m$; its size is at most $km$, so equals $km$. Thus the setwise stabilizer $U_{i,B}$ has index $m$ in $U_i$.

For every $b\in B$, $U_{i,b}\le U_{i,B}$ and both have index $m$, so they agree. In particular this group fixes $B$ pointwise. If $b^{\mathrm{\prime }}\in B\setminus \{b\}$, then $U_{i,b}\le U_{(B)}\le U_{b,b^{\mathrm{\prime }}},$ where $U_{(B)}$ is the pointwise stabilizer. All two-point stabilizers have order $\mathrm{\mid }U\mathrm{\mid }\mathrm{/}(nm)$, so both inclusions are equalities. Applying the same argument to the block $C$ containing $i$ gives $U_{(B)}=U_{(C)}$. This holds for every block, so their common pointwise stabilizer fixes all points and is trivial. Both within-block and between-block two-point stabilizers are therefore trivial.

If $N$ were imprimitive, steps 3.1–4.1 would give trivial two-point stabilizers. Put $h=\mathrm{\mid }{N}_{\alpha }\mathrm{\mid }>1$. Counting the elements fixing exactly one point shows that the set $D$ of fixed-point-free elements of $N$ has size $\mathrm{\mid }D\mathrm{\mid }=nh-1-n(h-1)=n-1.$ The set $D$ is conjugation invariant under $G$. The number $a$ of its elements carrying one point to a different specified point is independent of the ordered pair by $2$-transitivity. Counting these incidences gives $n(n-1)=an(n-1)$, hence $a=1$. Conjugating the ordered pair $(\alpha ,d\alpha )$ to $(\alpha ,e\alpha )$ for $d,e\in D$ now shows that $d,e$ are conjugate in $G$.

The action of $N_{\alpha }$ on the other points is free, so $h\mid (n-1)$. For each prime $p\mid n$, Cauchy's theorem gives an element of order $p$ in $N$, necessarily in $D$ since $p\nmid h$. All elements of $D$ have the same order, so $n$ is a power of a single prime $p$. A Sylow $p$-subgroup $P$ of $N$ has order $n$, because $p\nmid h$. All its nonidentity elements are fixed-point-free. Thus $P=D\cup \{1\}$ is a nontrivial proper $G$-normal subgroup of $N$, contradicting minimality. Hence $N$ is primitive.

Suppose $N$ is not simple and choose a nontrivial proper minimal normal subgroup $M$ of $N$. It is not $G$-normal, so it has a distinct $G$-conjugate $M^{\mathrm{\prime }}$. By [L2] for primitive $N$, these are its only two minimal normal subgroups, and they commute and are regular. Thus $M{M}^{\mathrm{\prime }}$ is $G$-normal, whence $N=M{M}^{\mathrm{\prime }}$; also $M\cap M^{\mathrm{\prime }}=1$. The group $M$ is nonabelian: otherwise its permutation centralizer would be $M$, forcing the commuting regular subgroup $M^{\mathrm{\prime }}$ to equal $M$.

Let $H=N_G(M)$. Its index in $G$ is two, it contains $N$, and it is transitive. Consequently $G=H{G}_{\alpha }$ and $[G_{\alpha }:H_{\alpha }]=2$. Its normal subgroup $H_{\alpha }⊴G_{\alpha }$ has either one orbit or two equal-sized orbits on $\mathrm{\Omega }\setminus \{\alpha \}$. Under the regular identification with $M$, these are automorphism orbits on $M\setminus \{1\}$. Thus there are at most two nonidentity element orders in $M$.

If $\mathrm{\mid }M\mathrm{\mid }$ has only one prime divisor, its nontrivial center makes it abelian by [L3] and [L4], contradicting step 7.1. Otherwise Cauchy's theorem shows that there are exactly two prime divisors $p,q$, and the two equal-sized orbits consist of elements of orders $p$ and $q$. No other nonidentity order is possible.

For an element $x$ of order $p$, the centralizer $C_M(x)$ has order a power of $p$: if $q$ divided its order, Cauchy's theorem would give a commuting element $y$ of order $q$, and $xy$ would have order $pq$, an impossibility. By [L5], every $M$-conjugacy class of elements of order $p$ has size divisible by $q$. The total number of such elements, $(\mathrm{\mid }M\mathrm{\mid }-1)\mathrm{/}2$, is therefore divisible by $q$, contrary to $q\mid \mathrm{\mid }M\mathrm{\mid }$. This contradiction proves that $N$ is simple. It is nonabelian, since a faithful transitive abelian group is regular.

Since $N$ is nonregular, [L2] excludes a second minimal normal subgroup of $G$. If its normal centralizer $C_G(N)$ were nontrivial, it would contain a minimal normal subgroup of $G$ and thus contain $N$, contrary to nonabelian simplicity. So conjugation embeds $G$ in $\mathrm{Aut}(N)$. The subgroup $N$ maps to its inner automorphism group, isomorphic to $N$ because $Z(N)=1$. This is the almost simple alternative. Together with step 1.1 it proves the claim.