2-Transitive permutation groups by Mazurov V. D.

By Mazurov V. D.

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Proof Let a be any root of f{X) in an extension field E of F. Since f{X) is irreducible, f{X) is the unique monic polynomial in F[X] of lowest degree with / ( a ) = 0 (and indeed, f(X) = m^(X)). L Hence every root of f(X) in a splitting field is simple and f(X) is separable. Now if fiX) is a nonconstant polynomial and char F = 0, then / ' ( Z ) ^ 0. If charF = p, then f\X) 7^ 0 unless the exponent of every power of X appearing in f{X) is a multiple of p, proving the proposition. n Recall that an algebraic extension E of F is separable if ma(X) e F[X] is a separable polynomial for every a G E.

14, B/F is normal if and only if B/F is Galois. 8 (2). (4) As we shall see from the proof, in case B is a Galois extension of F, the quotient map Gal(E/F) -> Gal(B/F) is simply given by restriction, a \-^ a \B. o Proof. (1) For each subgroup H and G, let BH=Fix(H). This gives a map r : {subgroups of G} -^ {fields intermediate between F and E}. We show r is a one-to-one correspondence. 6 (2). r is onto: Let F c B c E and let H = {a eG\ a | B = id} = Gal(E/B) c Gal(E/F). 14. But F c B so f(X) e B[X].

0. Note that 5 > 1, since if 5 = 1, 0 = Picfiiei) = )Si6i implies ^i = 0 as €i 7^ 0. , 6„}. Thus we may finally assume that jSi ^ F. Then we have (*r) iSior/(6i) + '" + Ps-icri{es-i) + ati^s) = 0 , / = 1 , . . , n. Since jSi ^ F = Fix(E), there is a OTJ^ G S with crk(Pi) i=- P\. Since S is a group, for any oi € S there is a or^ G S with oi = 0^0j. fe) = 0, for / = 1 , . . , n. Subtracting (**/) from (*j) gives (* * */) (A - crk(Pi))ai(ei) -h • • • + (Ps-i - or,fe-i))a,fe_i) = 0, for / = 1 , .

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