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College Algebra Exam Review 131

# College Algebra Exam Review 131 - W G ² G= K Then ı W G...

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2.7. QUOTIENT GROUPS AND HOMOMORPHISM THEOREMS 141 For a subgroup A of G containing N , ' 1 .'.A// is a subgroup of G which a priori contains A . If x is in that subgroup, then there is an a 2 A such that '.x/ D '.a/ . This is equivalent to a 1 x 2 ker .'/ D N . Hence, x 2 aN aA D A . This shows that ' 1 .'.A// D A , which completes the proof of part (a). Let B D ' 1 . B/ . For part (b), we have to show that B is normal in G if, and only if, B is normal in G . Suppose B is normal in G . Let g 2 G and x 2 B . Then '.gxg 1 / D '.g/'.x/'.g/ 1 2 B; because '.x/ 2 B , and B is normal in G . But this means that gxg 1 2 ' 1 . B/ D B , and thus B is normal in G . Conversely, suppose B is normal in G . For N g 2 G and N x 2 B , there exist g 2 G and x 2 B such that '.g/ D N g and '.x/ D N x . Therefore, N g N x N g 1 D '.gxg 1 /: But gxg 1 2 B , by normality of B , so N g N x N g 1 2 '.B/ D B . Therefore, B is normal in G . n Proposition 2.7.13. Let ' W G ! G be a surjective homomorphism with kernel N . Let K be a normal subgroup of G and let K D ' 1 . K/ . Then G=K Š G= K . Equivalently, G=K Š .G=N /=.K=N / . Proof. Write for the quotient homomorphism
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Unformatted text preview: W G ²! G= K . Then ı ' W G ²! G= K is a surjective homomorphism, because it is a com-position of surjective homomorphisms. The kernel of ı ' is the set of x 2 G such that '.x/ 2 ker . / D K ; that is, ker . ı '/ D ' ± 1 . K/ D K . According to the homomorphism theorem, Theorem 2.7.6 , G= K Š G= ker . ı '/ D G=K: More explicitly, the isomorphism G=K ²! G= K is xK 7! ı '.x/ D '.x/ K: Using the homomorphism theorem again, we can identify G with G=N . This identiﬁcation carries K to the image of K in G=N , namely K=N . Therefore, .G=N/=.K=N/ Š G= K Š G=K: n The following is a very useful generalization of the homomorphism theorem. We call it the Factorization Theorem, because it gives a condition...
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