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Unformatted text preview: 7.3.12 . If G is a &nite group of order n , and p is the least prime such that p j n , show that any subgroup of index p is normal in G . Let H be a subgroup of index p in G . Consider the action of G on the set S of left cosets of H given by a & xH = axH as in Exercise 7.3.2. Let ’ be the corresponding homomorphism from G to Sym S , that is, ’ ( a ) ( xH ) = axH . Note that ker ’ ¡ H because if a 2 ker ’ , then aH = H . Now G= ker ’ is isomorphic to a subgroup of Sym S , by the fundamental homomorphism theorem, so j G= ker ’ j divides j Sym S j = p ! . Also j G= ker ’ j divides j G j = n . As p is the smallest prime divisor of n , it follows that j G= ker ’ j is either 1 or p . In the &rst case, ker ’ = G , which is impossible because ker ’ ¡ H . So j G= ker ’ j = p , the index of H in G . But ker ’ ¡ H , so ker ’ = H , whence H is normal because it is the kernel of a homomorphism. 7.4.1 . Let G be a &nite abelian group, and let p be a prime divisor of j G j ....
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This note was uploaded on 03/24/2012 for the course MATH 203 taught by Professor Ankit during the Spring '12 term at Evergreen.
 Spring '12
 ankit
 Algebra, Sets

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