Results 1 to 3 of 3

Math Help - Conjugacy classes

  1. #1
    Newbie
    Joined
    Feb 2008
    Posts
    20

    Conjugacy classes

    I thought I had the hang of algebra but the questions that I am struggling through at the moment are raising so many problems!
    This one has been driving me mad:
    We are given the fact that if G is a group of order 24, generated by elements x, y, z such that x^2 = y^3 = z^3 = xyz then G is isomorphic to SL2(Z3) with the isomorphism given by:
    x mapped to the 2x2 matrix (0,2;1,0) (i.e 0 2 on the top row and 1 0 on the bottom), y is mapped to the marix (2,0 ; 1,2) and z is mapped to (0,1 ; 2,1). I do not have to prove this and can use it throughout the question.

    I have to compute the conjugacy classes, the centraliser of an element in each conjugacy class and the centre of G.

    The hint given says to start by conjugating a generic matrix (a,b ; c,d) by each generator. I have done this but am now stuck. I know I need to find the elements in the group but I dont want a 24 x 24 table and am not sure how to reduce it by much. Also, am I meant to be working with matrices or the x,y and z? I tried conjugating generators by generators but the inverses of the matrices leave me with matrices with entries like 1/2 or 1/4 and these are not in Z3!
    This is an assessed question so hints would be great or perhaps a similar example. On top of all this I'm not even sure how to tell the difference between the centraliser of an element and the centre of G!
    Thanks!
    Sooz
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by Sooz View Post
    I thought I had the hang of algebra but the questions that I am struggling through at the moment are raising so many problems!
    This one has been driving me mad:
    We are given the fact that if G is a group of order 24, generated by elements x, y, z such that x^2 = y^3 = z^3 = xyz then G is isomorphic to SL2(Z3) with the isomorphism given by:
    x mapped to the 2x2 matrix (0,2;1,0) (i.e 0 2 on the top row and 1 0 on the bottom), y is mapped to the marix (2,0 ; 1,2) and z is mapped to (0,1 ; 2,1). I do not have to prove this and can use it throughout the question.

    I have to compute the conjugacy classes, the centraliser of an element in each conjugacy class and the centre of G.

    The hint given says to start by conjugating a generic matrix (a,b ; c,d) by each generator. I have done this but am now stuck. I know I need to find the elements in the group but I dont want a 24 x 24 table and am not sure how to reduce it by much. Also, am I meant to be working with matrices or the x,y and z? I tried conjugating generators by generators but the inverses of the matrices leave me with matrices with entries like 1/2 or 1/4 and these are not in Z3!
    This is an assessed question so hints would be great or perhaps a similar example. On top of all this I'm not even sure how to tell the difference between the centraliser of an element and the centre of G!
    Thanks!
    This looks like a long problem. Try reading this. The group that is being used in that example is the dihedral group. In case if you never seen it before it is defined as \{1,x,x^2,x^3,y,xy,x^2y,x^3y\} with the properties that x^4 = y^2 = 1 and yx=x^3y. The special linear group that you posted above does not make sense did you mean perhaps x^2=y^3=z^3 = xyz = 1, maybe?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2008
    Posts
    20
    Quote Originally Posted by ThePerfectHacker View Post
    TThe special linear group that you posted above does not make sense did you mean perhaps x^2=y^3=z^3 = xyz = 1, maybe?
    Hi,
    I have asked around and it seems there is no mistake with the question. I don't understand how x^2 + y^3=z^3 = xyz= 1 would be allowed since, for example, x is mapped to (0,2;1,0) which, multiplied by (0,2;1,0) does not equal 1 and this is what x is isomorphic to (unless I have completely misunderstood the question).
    I have tried conjugating a generic matrix by each of the generators and got out 3 matrices in terms of a,b,c and d and I have also conjugated each generator matrix by every other (I sorted out my original problem, it was just me making silly mistakes) and got out a 3x3 table containing various matrices (some of which seem to be the same as others, I'm not even sure if that is a problem!) The trouble is, I am staring at everything I have calculated so far, completely clueless on how to get on with it!
    That example helped a bit but I am still having trouble applying it to my problem where I am working with matrices.
    Any ideas?
    Thanks,
    Sooz
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Conjugacy Classes
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: May 11th 2011, 07:25 PM
  2. conjugacy classes
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 16th 2009, 12:25 PM
  3. conjugacy classes
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: November 5th 2009, 02:13 AM
  4. Conjugacy Classes
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: December 9th 2008, 12:29 PM
  5. Conjugacy Classes in A_n
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 1st 2008, 04:53 PM

Search Tags


/mathhelpforum @mathhelpforum