Following the argument given, we reach the case that the minimal normal subgroup N of G is a Frobenius group. Now elementary counting shows that it contains n-1 fixed-point-free elements, where n is the degree. So there are n(n-1) triples (alpha,beta,g), where g is f.p.f. and maps alpha to beta. But there are just n(n-1) pairs (alpha,beta), permuted transitively by G. We conclude:
Now let p be a prime dividing n, and P a Sylow p-subgroup of N. Then P contains a f.p.f. element of order p. So all f.p.f. elements of N have order p, and n is a power of p. Then it follows that P is transitive, and so consists of the identity and all the f.p.f. elements; so P is a regular normal subgroup. This concludes the proof.
The fact that a finite 2-transitive group has a unique minimal normal subgroup is more elementary. By Theorem 4.4, if a primitive group has more than one minimal normal subgroup, then it has just two, and each is the centraliser of the other, so they are non-abelian and regular. But a regular normal subgroup of a finite 2-transitive group is abelian, by Theorem 1.6.
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Peter J. Cameron
2 January 2001.