Algebra 1 HW 2 S

Algebra 1 HW 2 S - Solutions to Assignment 2 October 2...

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Solutions to Assignment 2 October 2, 2011 (1) (Associativity) Let x, y, z R , x ( y z ) = x ( y + z + 1) = x + ( y + z + 1) + 1 = x + y + z + 2 ( x y ) z = ( x + y + 1) z = x + y + 1 + z + 1 = x + y + z + 2 So this binary operation is associative. (Identity) Take e = - 1 and x R , we have, e x = ( - 1) + x + 1 = x and x e = x + ( - 1) + 1 = x. So -1 is the idenitity for this operation. (Inverses) For each x R , x ( - x - 2) = x + ( - x - 2) + 1 = - 1 and, ( - x - 2) x = ( - x - 2) + x + 1 = - 1 . So - x - 2 is the inverse of x . So ( R , ) is a group. (2) a. GL 2 ( Z / 2 Z ) consists of matrices with entries in Z / 2 Z of determinant nonzero. But there are only two elements in Z / 2 Z , and to have determinant nonzero, means to have determinant [1] so we have, a b c d GL 2 ( Z / 2 Z ) iff ad - bd = 1 ( Z / 2 Z ) 1
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So, either ad=1 and bc=0, or ad=0 and bc=1. The first case gives, 1 0 0 1 , 1 1 0 1 , 1 0 1 1 , and the second, 0 1 1 0 , 1 1 1 0 , 0 1 1 1 . So | G | = 6 . b. We’re looking for a matrix a b c d , such that a b c d [0] [1] [1] [0] = [1] [0] [0] [1] , which means b a d c = [1] [0] [0] [1] . So a b c d = [0] [1] [1] [0] .
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