Exam1-Solns

# Exam1-Solns - Exam 1 Solutions 1. (a) Not a group. 0 has no...

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Exam 1 Solutions 1. (a) Not a group. 0 has no inverse. (b) Not a group. It’s not associative. x/ ( y/z ) = xz/y , ( x/y ) /z = x/yz . (c) Is a group. Closure : Let x, y, G . x * y = x + y + xy = - 1 ( x + 1)( y + 1) = 0 x = - 1 or y = - 1. Since - 1 6∈ G , we have closure. Associative : x * ( y * z ) = x * ( y + z + yz ) = x + y + z + yz + xy + xz + xyz . ( x * y ) * z = ( x + y + xy ) * z = x + y + z + xy + xz + yz + xyz . Thus associativity holds. Identity : x * 0 = x + 0 + 0 = x and 0 * x = 0 + x + 0 = x . Thus 0 is the identity. Inverses : x - 1 = - x 1+ x G (since - 1 6∈ G ). x * ( - x 1+ x ) = x + - x 1+ x + - x 2 1+ x = 0. By commutativity of addition and multiplication of reals, we get - x 1+ x * x = 0 as well. 2. e B and f ( e ) = e so e f - 1 ( B ) 6 = Ø. Let x, y f - 1 ( B ) f ( x ) , f ( y ) B . Thus f ( x ) f ( y ) - 1 B since B is a group. Hence f ( xy - 1 ) B since f is a homomorphism. So xy - 1 f - 1 ( B ) and f - 1 ( B ) G
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## This note was uploaded on 04/19/2011 for the course MATH 21373 taught by Professor Johnson during the Spring '11 term at Carnegie Mellon.

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