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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Translated from Matematicheskie Zametki, Vol. 20, No. 2, pp. 177–186, August, 1976.
The author would like to thank V. D. Mazurov for helpful discussions concerning the theorem proved in this note.
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Romanovskii, A.V. Finite groups with Frobenius subgroup. Mathematical Notes of the Academy of Sciences of the USSR 20, 660–665 (1976). https://doi.org/10.1007/BF01155869
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DOI: https://doi.org/10.1007/BF01155869