College Algebra Exam Review 139

College Algebra Exam Review 139 - 149 3.1 DIRECT PRODUCTS...

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3.1. DIRECT PRODUCTS 149 Proposition 3.1.5. (a) Suppose M and N are normal subgroups of G , and M \ N D f e g . Then for all m 2 M and n 2 N , mn D nm . (b) MN D f mn W m 2 M and n 2 N g is a subgroup and .m;n/ 7! mn is an isomorphism of M ± N onto MN . (c) If MN D G , then G Š M ± N . Proof. We have mn D nm , mnm ± 1 n ± 1 D e . Observe that mnm ± 1 n ± 1 D .mnm ± 1 /n 2 N; since N is normal. Likewise, mnm ± 1 n ± 1 D m.nm ± 1 n ± 1 / 2 M; since M is normal. Since the intersection of M and N is trivial, it follows that mnm ± 1 n ± 1 D e . It is now easy to check that MN is a subgroup, and that .m;n/ 7! mn is a homomorphism of M ± N onto MN . The homomorphism is injective, because if mn D e , then m D n ± 1 2 M \ N D f e g . Assertion (c) is evident. n Notation 3.1.6. When G has subgroups N and M such that N \ M D f e g and NM D G , we will write G D N ± M to convey that G Š N ± M and N and M are subgroups of G . See Remark 3.1.9 . Example 3.1.7. Let’s look at Example 3.1.3 again “from the inside” of Z ab . The subgroup h ŒaŁ i of Z ab generated by ŒaŁ has order b and the subgroup h ŒbŁ i has order a . These two subgroups have trivial intersec- tion because the order of the intersection must divide the orders of both subgroups, and the orders of the subgroups are relatively prime. The two
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