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Unformatted text preview: As in the proof of the homomorphism theorem, there is only one way to dene Q ' consistent with the requirement that Q ' D ' , namely Q '.aN/ D '.a/ . It is necessary to check that this is well dened and a homomorphism. But if aN D bN , then b 1 a 2 N K D ker .'/ , so '.b 1 a/ D e , or '.a/ D '.b/ . This shows that the map Q ' is well dened. The homomorphism property follows as in the proof of the homomorphism theorem. n Corollary 2.7.15. Let N K G be subgroups with both N and K normal in G . Then xN 7! xK denes a homomorphism of G=N onto G=K with kernel K=N ....
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- Fall '08