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Unformatted text preview: Algebra , Assignment 7 Solutions 1. Let G be a finite group and let a,b ∈ G with a ∼ b . [Here ∼ denotes conjugacy.] Show o ( a ) = o ( b ). Solution: So b = g- 1 ag so if a n = e then b n = g- 1 a n g = g- 1 g = e . Also a = gbg- 1 so if b n = e then a n = gb n g- 1 = gg- 1 = e . As the powers giving e are the same the minimal powers (the orders) are the same. 2. Let G denote the set of linear functions f ( x ) = mx + b on the real line with m negationslash = 0. Denote such a function by ( m,b ). Define a product f * g as the function h ( x ) = g ( f ( x )). (Check assignment 1 and the solutions for earlier work on this group.) Recall C ( f ) ,N ( f ) denote the conjugate class and the normalizer of f . Solution Note: We’ll write g ( x ) = cx + d below and look at when fg = gf . We have g- 1 ( x ) = ( x − d ) /c . (a) Describe C ( f ) and N ( f ) when m negationslash = 1 and b negationslash = 0. Solution: For g ∈ N ( f ) we want gf = fg so m ( cx + d ) + b = c ( mx + b ) + d so that md + b = cb + d or ( m − 1) d = b ( c − 1). Note m − 1 negationslash = 0 and b negationslash = 0. So for each c negationslash = 1 we have d = b ( c − 1) / ( m − 1). This can also be written − b m − 1 = − d c − 1 which has a nice interpretation. Set β =- b m- 1 so that f ( β ) = β ....
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This note was uploaded on 04/01/2011 for the course MATH 101 taught by Professor Josephs during the Fall '08 term at NYU.
- Fall '08