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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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 Spring '12
 ankit
 Algebra, Sets

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