By Blackwood O.
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2 The Fitting subgroup 51 Proof: Define C = CG F G and Z = Z F G . Clearly Z S. Suppose, for a contradiction, that Z is strictly contained in S. Since S is normal in G, this implies that S/Z contains a minimal normal subgroup M/Z of G/Z. Since S/Z is soluble, M/Z is abelian and so M Z. Since M C, we find that M M F G M = 1, and so M is nilpotent (of class at most 2). 3. The fact that M C now implies that M Z. This contradicts the fact that M/Z is a minimal normal subgroup of G/Z, and so the theorem follows.
2 Let G be a soluble group of order p1 1 p2 2 · · · pk k . Then (i) G has a Sylow system; (ii) any two Sylow systems are conjugate in G; (iii) if H G and Q1 Qk is a Sylow system for H then there is a Sylow k . Pk for G such that Qi = H ∩ Pi for all i ∈ 1 2 system P1 pk \ pi . We say Proof: For each i ∈ 1 2 k , define i = p1 p2 that a subgroup H is a Hall pi -complement if H is a Hall i -subgroup. Let be the set of all Sylow systems in G and let i be the set of all Hall pi -complements in G for 1 i k.
Proof: Let be fixed. We prove the theorem by induction (on n − r). When r = n, the group G is trivial and so the result holds. Assume, as an inductive hypothesis, that any permutation group on with more than r orbits satisfies the theorem. Let G have r orbits on , and let lie in a non-trivial orbit of G. We may write = 1 ∪ 2 ∪ · · · ∪ k , where the sets i are orbits of the stabiliser G of in G. 3 Permutations and primitivity 53 . Since is a non-trivial orbit, k > 1. For i ∈ 1 2 k−1 , that k = let gi ∈ G be such that gi ∈ i .