HW4_soln - Homework#4 Solutions p 67#8 In U(14 we have 32...

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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 3 = 5 = { 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 ) .
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Proof. Taking x = a in the lemma of the preceding problem immediately gives C ( a ) C ( a k ) . We must establish the reverse inclusion. Since ( k, n ) = 1, we know that there is an m Z so that mk mod n = 1. Since | a | = n
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