College Algebra Exam Review 132

# College Algebra Exam Review 132 - As in the proof of the...

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142 2. BASIC THEORY OF GROUPS for a homomorphism ' W G ! G to “factor through” a quotient map ± W G ! G=N , that is to be written as a composition ' D Q ' ı ± . Proposition 2.7.14. (Factorization Theorem) Let ' W G ! G be a surjec- tive homomorphism of groups with kernel K . Let N ± K be a subgroup that is normal in G , and let ± W G ! G=N denote the quotient map. Then there is a surjective homomorphism Q ' W G=N ! G such that Q ' ı ± D ' . (See the following diagram.) The kernel of Q ' is K=N ± G=N . G ' q q q q q G ± q q q q q ± ± ± ± ± ± q q q q q q q Q ' G=N Proof. Let us remark that the conclusion follows from Proposition 2.7.13 and the homomorphism theorem. The map Q ' is G=N ²! .G=N/=.K=N/ Š G=K Š G: However, it is more transparent to prove the result from scratch, following the model of the homomorphism theorem.
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Unformatted text preview: As in the proof of the homomorphism theorem, there is only one way to deﬁne Q ' consistent with the requirement that Q ' ı ± D ' , namely Q '.aN/ D '.a/ . It is necessary to check that this is well deﬁned 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 deﬁned. 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 deﬁnes a homomorphism of G=N onto G=K with kernel K=N ....
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