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113 HW 5_Math 113

# 113 HW 5_Math 113 - Mathematics 113 Homework 5 Curtis...

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Mathematics 113: Homework # 5 Curtis Tekell Dr. Poirier October 11, 2010 Exercise 1. Let G be a group and let S be a set with a binary operation × . Let f : G S be a surjective map of sets such that f ( ab ) = f ( a ) × f ( b ) for all a, b G . Then S is a group under × and f is a group homomorphism. Proof. Let a, b, c S , and suppose x, y, z G with f ( x ) = a , f ( y ) = b and f ( z ) = c . As the operation on G is associative, we wee that [ a × b ] × c = [ f ( x ) × f ( y )] × f ( z ) = f ( x ) × [ f ( y ) × f ( z )] = a × [ b × c ] , and × is associative. Let 1 S = f (1 G ). Then 1 S a = f (1 G x ) = f ( x ) = a = f ( x ) = f ( x 1 G ) = a 1 S , and 1 S is an identity for S . Additionally, af ( x - 1 ) = f ( xx - 1 ) = f (1 G ) = 1 S = f (1 G ) = f ( x - 1 x ) = f ( x - 1 ) a and a - 1 = f ( x - 1 ), thus all elements of S are invertable. It follows that S is a group. Exercise 2. The canonical epimorphism π : G G/H is in fact an epimorphism with Ker ( π ) = H Proof. Given gH G/H , π ( g ) = gH and π is surjective. π ( ab ) = abH = aHbH = π ( a ) π ( b ) shows that π is a homomorphism. Finally, Ker ( π ) = { g G : gH = H } = H .

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113 HW 5_Math 113 - Mathematics 113 Homework 5 Curtis...

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