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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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