Abstract
Suppose the normalizer N of a subgroup A of a simple group G is a Frobenius group with kernel A, and the intersection of A with any other conjugate subgroup of G is trivial, and suppose, if A is elementary Abelian, that ¦a¦> 2n+1, where n=¦N:A¦. It is proved that if A has a complement B in G, then G acts doubly transitively on the set of right cosets of G modulo B, the subgroup B is maximal in G, and ¦B¦ is divisible by ¦a¦−1. The proof makes essential use of the coherence of a certain set of irreducible characters of N.
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