sol7 - Algebra Assignment 7 Solutions 1 Let G be a nite...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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 . Then g N ( f ) if and only if g ( β ) = β . For the centralizer in general g - 1 fg ( x ) = g ( f ( g - 1 ( x ))) = c bracketleftbigg m parenleftbigg x d c
Image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern