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 So this binary operation is associative. (Identity) Take e = - 1 and x R , we have, e x - 1) + x 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 . 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, ± ab cd ² 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 frst case gives, ± 10 01 ² , ± 11 ² , ± ² , and the second, ± ² , ± ² , ± ² . So | G | =6 . b. We’re looking For a matrix ± ab cd ² , such that ± ²± [0] [1] [1] [0] ² = ± [1] [0] [0] [1] ² , which means ± ba dc ² = ± [1] [0] [0] [1] ² . So ± ² = ± [0] [1] [1] [0] ² .
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This note was uploaded on 11/02/2011 for the course MATH 235 taught by Professor Goren during the Fall '07 term at McGill.

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

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