College Algebra Exam Review 139

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/15/2011 for the course MAC 1105 taught by Professor Everage during the Fall '08 term at FSU.

Ask a homework question - tutors are online