HW4_soln - Homework #4 Solutions p 67, #8. In U (14) we...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ) ....
View Full Document

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.

Page1 / 4

HW4_soln - Homework #4 Solutions p 67, #8. In U (14) we...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online