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I/J)~ = 2:l~p~j

B) Same with lI aE !. If M is an R-module, HomR(M, R) is dual to M and its denoted by MD. 4 If M is isomorphic to R n where R is commutative, then its dual MD is isomorphic to Rn as an R-module. 1 SYLOW'S THEOREMS. Let G be a finite group. Let p be a prime. Assume that #G = pb· m where (p, m) = 1. 1 If pal#G, then there is a subgroup H C G with pa = #H. Proof. Let X be the set of subsets L of G with pa elements. Then #X = (pb . m)(pb . m - 1) ... (pbm - pa + 1) (pa)(pa - 1) ... :... because the power of p dividing pb.

Let F' = {e E Elrp(e) = e for all rp in Aut(E/F)}. Clearly F' is a subfield of E containing F. Define the field extension FeE is Galois if E is finite over F and F' = F. Thus Galois extensions are very symmetric. Here Aut(E / F) is called the Galois group of the Galois field extension FeE. 2 R c C is a Galois extension. 1 FeE is Galois if and only if E is a splitting field of a separable polynomial in F[X]. Let E / F be Galois. 1 Aut(E/F) is finite. Proof. Let E = F( e1, ... , ed). Let rp E A ut( E / F).

### Absolutely summing operators from the disc algebra

by Ronald

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Categories: Algebra