Finite Permutation Groups

1 Multiply transitive groups Theorem 1.1. Let Ω be a finite set and G ≤ Sym(Ω) be 2–transitive. Let N E G be a minimal normal subgroup. Then one of the following holds: (a) N is regular and elementary abelian. (b) N is primitive, simple and not abelian. Proof. First we show that N is unique. Suppose that M is another minimal normal subgroup of G, so N ∩M = {e} and therefore [N,M ] = {e}. Since non–trivial normal subgroups of the primitive group G are transitive, we obtain that both M and N are transitive. Together with Lemma ??? and M ≤ CG(N) and N ≤ CG(M) we get that both M and N are regular. By Theorem ???, regular normal subgroups of finite 2–transitive groups are abelian. Hence is abelian. But transitive abelian groups are regular, which forces M = = N . If N is regular, then we are done by Theorem ???.