College Algebra Exam Review 128

College Algebra Exam Review 128 - aN D bN , b 1 a 2 N D ker...

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138 2. BASIC THEORY OF GROUPS Homomorphism Theorems The features that we have noticed in the several examples are quite general: Theorem 2.7.6. (Homomorphism theorem). Let ' W G ±! G be a surjec- tive homomorphism with kernel N . Let ± W G ±! G=N be the quotient homomorphism. There is a group isomorphism Q ' W G=N ±! G satisfying Q ' ı ± D ' . (See the following diagram.) G ' q q q q q G ± q q q q q ± ± ± ± ± ± q q q q q q q = Q ' G=N Proof. There is only one possible way to define Q ' so that it will satisfy Q ' ı ± D ' , namely Q '.aN/ D '.a/ . It is necessary to check that Q ' is well-defined, i.e., that Q '.aN/ does not depend on the choice of the representative of the coset aN . Suppose that aN D bN ; we have to check that '.a/ D '.b/ . But
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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 sub-groups and (surjective) homomorphisms are two sides of one coin: Given...
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