nov03 - . This shows that [ 14 ] ∈ ker θ . 3. Since [ 14...

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Math 3124 Thursday, November 3 November 3, Ungraded Homework Let C = all cubes in ( Z # 211 , ± ) (that is { x ± x ± x | x Z # 211 } ), and define θ : Z # 211 Z # 211 by θ ( α ) = α 3 . 1. Prove that θ is a homomorphism and that C 6 Z # 211 . 2. Show that [ 14 ] ker θ . 3. Deduce that | C | ± ± 70. 1. Let x , y Z # 211 . Then θ ( x ± y ) = ( x ± y ) 3 = x 3 ± y 3 (because ± is commutative) = θ x ± θ y . Therefore θ is a homomorphism and we can now apply the Fundamen- tal Homomorphism Theorem to deduce that θ ( Z # 211 ) 6 Z # 211 . Since C = θ ( Z # 211 ) , it follows that C 6 Z # 211 . 2. θ [ 14 ] = [ 14 ] 3 = [ 196 ] ± [ 14 ] = [ - 15 ] ± [ 14 ] = [ - 210 ] = [ 1 ]
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Unformatted text preview: . This shows that [ 14 ] ∈ ker θ . 3. Since [ 14 ] 6 = [ 1 ] , the proof of previous result also shows that o ([ 14 ]) = 3. Therefore 3 ± ± | ker θ | by Lagrange’s theorem, say | ker θ | = 3 n where n is a positive integer. By the Fundamental Homomorphism Theorem, | C | = | Z # 211 / ker θ | = 210 / | ker θ | = 70 / n . Therefore | C | n = 70 and we conclude that | C | ± ± 70....
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This note was uploaded on 01/02/2012 for the course MATH 3124 taught by Professor Parry during the Fall '08 term at Virginia Tech.

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