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Unformatted text preview: aN D bN , b 1 a 2 N D ker .'/ , e D '.b 1 a/ D '.b/ 1 '.a/ , '.b/ D '.a/: The same computation shows that Q ' is injective. In fact, Q '.aN/ D Q '.bN/ ) '.a/ D '.b/ ) aN D bN: The surjectivity of Q ' follows from that of ' , since ' D Q ' . Finally, Q ' is a homomorphism because Q '.aNbN/ D Q '.abN/ D '.ab/ D '.a/'.b/ D Q '.aN/ Q '.bN/: n A slightly different proof is suggested in Exercise 2.7.1 . The two theorems (Theorems 2.7.1 and 2.7.6 ) say that normal subgroups and (surjective) homomorphisms are two sides of one coin: Given...
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 Fall '08
 EVERAGE
 Algebra

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