This preview shows page 1. Sign up to view the full content.
Unformatted text preview: a .n a .n 00 //;aa a 00 /: We have n a .n a .n 00 // D n a .n / a . a .n 00 // D n a .n / aa .n 00 /; where the rst equality uses that a is an automorphism of N , and the second uses that is a homomorphism of A into Aut .N/ . This proves associativity of the product. It is easy to check that .e;e/ serves as the identity of N A . Finally, we have to nd the inverse of an element of N A . If .n ;a / is to be the inverse of .n;a/ , we must have aa D e and n a .n / D e . Thus a D a 1 and n D 1 a .n 1 / D a 1 .n 1 / . Thus our candidate for .n;a/ 1 is . a 1 .n 1 /;a 1 / . Now we have to check this candidate by multiplying by .n;a/ on either side. This is left as an exercise. n...
View
Full
Document
 Fall '08
 EVERAGE
 Algebra

Click to edit the document details