solutions2 - Spring 2010 MATH 4320 Solutions to Prelim 2...

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Spring 2010 MATH 4320: Solutions to Prelim 2 Instructor: Yuri Berest Problem 1. ( a ) This is straightforward: for example, we have π 1 [( x 1 , x 2 ) · ( y 1 , y 2 ))] = π 1 [( x 1 y 1 , x 2 y 2 )] = x 1 y 1 = π 1 [( x 1 , x 2 )] π 1 [( y 1 , y 2 )] , and similarly for π 2 . Both π 1 and π 2 are surjective, because given any x 1 H 1 and any x 2 H 2 , we have π 1 [( x 1 , x 2 )] = x 1 and π 2 [( x 1 , x 2 )] = x 2 . Finally, Ker( π 1 ) = ( e 1 , H 2 ) and Ker( π 2 ) = ( H 1 , e 2 ) . ( b ) Given any group G with two homomorphisms f 1 : G H 1 and f 2 : G H 2 , we define ϕ : G H 1 × H 2 by ϕ ( x ) := ( f 1 ( x ) , f 2 ( x )) for x G . Then, we see at once that ( π 1 ϕ )( x ) = π 1 [ ϕ ( x )] = π 1 [( f 1 ( x ) , f 2 ( x ))] = f 1 ( x ) and ( π 2 ϕ )( x ) = π 2 [ ϕ ( x )] = π 2 [( f 1 ( x ) , f 2 ( x ))] = f 2 ( x ) for all x G . In other words, f 1 = π 1 ϕ and f 2 = π 2 ϕ as required. Now, suppose there is another homomorphism, say ψ : G H 1 × H 2 , satisfying f 1 = π 1 ψ and f 2 = π 2 ψ . By definition of H 1 × H 2 , the elements ψ ( x ) in the image of ψ can be written as ψ ( x ) = ( g 1 ( x ) , g 2 ( x )) , where g 1 ( x ) H 1 and g 2 ( x ) H 2 . Since f 1 = π 1 ψ and f 2 = π 2 ψ , we have f 1 ( x ) = ( π 1 ψ )( x ) = π 1 [ ψ ( x )] = π 1 [( g 1 ( x ) , g 2 ( x ))] = g 1 ( x ) and f 2 ( x ) = ( π 2 ψ )( x ) = π 2 [ ψ ( x )] = π 2 [( g 1 ( x ) , g 2 ( x ))] = g 2 ( x ) for all x G . Thus, g 1 ( x ) = f 1 ( x ) and g 2 ( x ) = f 2 ( x ) , and therefore ψ ( x ) = ϕ ( x ) for all x G . This proves that ψ = ϕ , which means the uniqueness of ϕ . Problem 2. ( a ) For each h H and for all x 1 and x 2 in N , we have f ( h )( x 1 x 2 ) := h ( x 1 x 2 ) h - 1 = h x 1 ( h - 1 h ) x 2 h - 1 = ( h x 1 h - 1 ) ( h x 2 h - 1 ) . So
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This note was uploaded on 06/07/2010 for the course MATH 201 taught by Professor Crissinger during the Spring '08 term at University of Delaware.

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solutions2 - Spring 2010 MATH 4320 Solutions to Prelim 2...

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