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Unformatted text preview: Math 3124 Thursday, November 17 Eighth Homework Solutions 1. (a) ( AB ) = det ( AB ) = det ( A ) det ( B ) = ( A ) ( B ) , so is a homomorphism. It is onto because given x Z # 5 , we have x 1 = x , and the foregoing matrix is in GL 2 ( Z 5 ) . (b) A ker if and only if det ( A ) = [ 1 ] . (c) Since K = ker and is onto (from above), this follows immediately from the Fundamental Homomorphism Theorem. (d) Since the property of being cyclic is invariant under isomorphism, it will be suffi cient to determine whether Z # 5 is cyclic. The answer is yes, because Z # 5 = h [ 2 ] i . 2. Define : G H by ( x ) = x 4 . Since Z # 509 is abelian, we have ( xy ) = ( xy ) 4 = x 4 y 4 = ( x ) ( y ) and so is a homomorphism. It is onto by the definition of H . Therefore by the Fundamental Homomorphism Theorem G / ker = H . We need to determine ker . Now [ 208 ] ker because [ 208 ] 4 = [ 1 ] 2 = [ 1 ] . Also [ 208 ] 2 = [ 1 ] 6 = [ 1...
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 Fall '08
 PARRY
 Math, Algebra

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