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# feb23 - Math 4124 Wednesday February 23 February 23...

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Math 4124 Wednesday, February 23 February 23, Ungraded Homework Exercise 3.3.4 on page 101 Let C be a normal subgroup of the group A and let D be a normal subgroup of the group B . Prove that ( C × D ) ( A × B ) and ( A × B ) / ( C × D ) = ( A / C ) × ( B / C ) . Define θ : A × B ( A / C ) × ( B / D ) by θ ( a , b ) = ( aC , bD ) . Let a , a 1 A and b , b 1 B . Then θ ( ( a , b )( a 1 , b 1 ) ) = θ ( aa 1 , bb 1 ) = ( aa 1 C , bb 1 D ) = ( aCa 1 C , bDb 1 D ) = ( aC , bD )( a 1 C , b 1 D ) = θ ( a , b ) θ ( a 1 , b 1 ) Thus θ is a homomorphism. θ is onto because given ( aC , bD ) ( A / C ) × ( B / D ) , θ ( a , b ) = ( aC , bD ) . ker θ = { ( c , d ) | θ ( c , d ) = e } , which is { ( c , d ) | ( cC , dD ) = ( C , D ) } . Now cC = C if and only if c C , and dD = D if and only if d D . We deduce that ker θ = C × D . The result now follows from the fundamental homomorphism theorem (Theorem 16 on page 97). Prove that C × / { 1 , - 1 } = C × . Let G = C × . Of course, G is the nonzero complex numbers under multiplication. Define θ : C C by θ ( x ) = x 2 for x G . Then θ is a mapping, because if x 6 = 0, then x 2 6 = 0 and so certainly θ maps G to G . Next θ is a homomorphism, because if x , y G , then θ ( xy ) = ( xy )

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