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Unformatted text preview: Homework #4 Solutions p 67, #8. In U (14) we have 3 2 mod 14 = 9 3 3 mod 14 = 27 mod 14 = 13 3 4 mod 14 = 3 · 13 mod 14 = 39 mod 14 = 11 3 5 mod 14 = 3 · 11 mod 14 = 33 mod 14 = 5 3 6 mod 15 = 3 · 5 mod 14 = 15 mod 14 = 1 and 5 2 mod 14 = 25 mod 14 = 11 5 3 mod 14 = 5 · 11 mod 14 = 55 mod 14 = 13 5 4 mod 14 = 5 · 13 mod 14 = 65 mod 14 = 9 5 5 mod 14 = 5 · 9 mod 14 = 45 mod 14 = 3 5 6 mod 15 = 5 · 3 mod 14 = 15 mod 14 = 1 . Hence h 3 i = h 5 i = { 1 , 3 , 5 , 9 , 11 , 13 } = U (14). p 68, # 16. Lemma 1. Let G , x ∈ G and k ∈ Z . Then C ( x ) ≤ C ( x k ) . Proof. If y ∈ C ( x ) then x = yxy 1 so that x k = ( yxy 1 ) k = yx k y 1 (the latter equality was proven in class) and hence y ∈ C ( x k ). If we apply the lemma with x = a , k = 1 we have C ( a ) ≤ C ( a 1 ) while if we take x = a 1 we get C ( a 1 ) ≤ C (( a 1 ) 1 ) = C ( a ) . proving that C ( a ) = C ( a 1 ). p 68, # 24. We will prove the following more general fact. Proposition 1. Let G be a group, a ∈ G and suppose  a  = n . If ( k,n ) = 1 then C ( a ) = C ( a k ) . Proof. Taking x = a in the lemma of the preceding problem immediately gives C ( a ) ≤ C ( a k ) ....
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This note was uploaded on 02/29/2012 for the course MATH 3362 taught by Professor Ryandaileda during the Fall '10 term at Trinity University.
 Fall '10
 RyanDaileda

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