Algebra 1 HW 3 S

Algebra 1 HW 3 S - ALGEBRA 235 HOMEWORK 3 SOLUTIONS...

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ALGEBRA 235 HOMEWORK 3 SOLUTIONS Question 1 . Let ψ : C × R × be given by ψ ( z )= | z | ,where | x + yi | = ° x 2 + y 2 . (a) Show that ψ is a homomorphism. (b) Calculate ker( ψ ) and im( ψ ). Solution. (a) First, observe that ψ is well-defned, as | z =0i ±andon ly li ± z ° = 0 ±or any z C .W en e e dt oc h e c kt h a t ψ (1) = 1, but this is clear since 1 2 =1 . Next ,let z 1 = x 1 + y 1 i and z 2 = x 2 + y 2 i where z i C × (that is, at least one o± x i ,y i R × ±or i =1 , 2). Then ψ ( z 1 z 2 )= ψ ( x 1 x 2 y 1 y 2 +( x 1 y 2 + y 1 x 2 ) i ) = ° ( x 1 x 2 y 1 y 2 ) 2 +( x 1 y 2 + y 1 x 2 ) 2 = ± x 2 1 x 2 2 + x 2 1 y 2 2 + y 2 1 x 2 2 + y 2 1 y 2 2 = ± ( x 2 1 + y 2 1 )( x 2 2 + y 2 2 ) = ψ ( z 1 ) ψ ( z 2 ) . There±ore, ψ is a homomorphism. (b) By defnition, ker( ψ )= { z C × | ψ ( z )=1 } . In other words, i± z = x + yi ker( ψ ), | z | =1 ,andhence z is an element o± the complex unit circle, S 1 .N o t et h a t z ker( ψ )i ±andon lyi ±( x,y ) is a real solution to x 2 + y 2 = 1. On the other hand, we can write z = e 2 πit ±or some t R . There±ore, ker( ψ )= S 1 = { x + yi | x 2 + y 2 =1 , ( x,y ) R 2 } = { e 2 πit | t R } . Next, we want to compute im( ψ ). Note that since we defne ψ using the positive branch o± the squareroot, | z | > 0 ±or all z C × . Furthermore, since R > 0 C × and | z | = z ±or z R > 0 , it ±ollows that ψ surjects onto R > 0 and im( ψ )= R > 0 . (Here R > 0 := { x R | x> 0 } ). ° Question 2 . Let φ : G H be a homomorphism and let g G have fnite order. (a) Show that the order oF φ ( g ) divides the order oF g . (b) IF φ is an isomorphism, show that | φ ( g ) | = | g | . Solution. (a) Let n = | g | .Th ens in ce φ is a homomorphism, φ ( e G )= e H ,and φ ( g a )= φ ( g ) a ±or all a N .Thu s , e H = φ ( e G )= φ ( g n )= φ
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This note was uploaded on 11/02/2011 for the course MATH 235 taught by Professor Goren during the Fall '07 term at McGill.

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Algebra 1 HW 3 S - ALGEBRA 235 HOMEWORK 3 SOLUTIONS...

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