College Algebra Exam Review 146

# College Algebra Exam Review 146 - a.n ˛ a.n 00;aa a 00 We...

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156 3. PRODUCTS OF GROUPS of A and N . The semidirect product N Ì ˛ A has the following features: It contains (isomorphic copies of) A and N as subgroups, with N normal; the intersection of these subgroups is the identity, and the product of these subgroups is N Ì ˛ A ; and we have the commutation relation an D ˛ a .n/a for a 2 A and n 2 N . The construction is straightforward. As a set, N Ì ˛ A is N ± A , but now the product is deﬁned by .n;a/.n 0 ;a 0 / D .n˛ a .n 0 /;aa 0 / . Proposition 3.2.3. Let N and A be groups, and ˛ W A ! Aut .N/ a homo- morphism of A into the automorphism group of N . The Cartesian product N ± A is a group under the multiplication .n;a/.n 0 ;a 0 / D .n˛ a .n 0 /;aa 0 / . This group is denoted N Ì ˛ A . Q N D f .n;e/ W n 2 N g and Q A D f .e;a/ W a 2 A g are subgroups of N Ì ˛ A , with Q N Š N and Q A Š A , and Q N is normal in N Ì ˛ A . We have .e;a/.n;e/ D a .n/;e/.e;a/ D a .n/;a/ for all n 2 N and a 2 A . Proof. We ﬁrst have to check the associativity of the product on N Ì ˛ A . Let .n;a/ , .n 0 ;a 0 / , and .n 00 ;a 00 / be elements of N Ì ˛ A . We compute ..n;a/.n 0 ;a 0 //.n 00 ;a 00 / D .n˛ a .n 0 /;aa 0 /.n 00 ;a 00 / D .n˛ a .n 0 aa 0 .n 00 /;aa 0 a 00 /: On the other hand, .n;a/. .n 0 ;a 0 /.n 00 ;a 00 // D .n;a/.n 0 ˛ a 0 .n 00 /;a 0 a 00 / D .n˛
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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 ﬁrst 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 Ì ˛ A . Finally, we have to ﬁnd the inverse of an element of N Ì ˛ 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...
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