Results 1 to 5 of 5

Math Help - Conjugacy Class Equation

  1. #1
    Newbie
    Joined
    Apr 2008
    Posts
    3

    Conjugacy Class Equation

    Hello there,

    Can someone please explain to me the quickest method of finding out the Class equation of Dihedral groups, D_8, D_10, D_12 etc.

    I am sure there is a quicker way than doing all the calculations, if someone can please tell me in ENGLISH what it is [since I can't understand the mathematical way]

    I had initially thought there was a pattern, but it didnt work for D_8, since I thought if you have D_2n, then the class equation is
    {1}, {x^k,x^m}, {x^w, x^v},....,{xy, ...., x^n-1 y}

    where k+m = n, and w+v = n

    This however was not the case for D_8, since {xy, ...., x^n-1 y} was split into 2, so if someone can please explain.

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Look here, the last post.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2008
    Posts
    3
    I dont understand this part:

    The second step that I would do is choose an element not from the ones chosen, say x. Now we will use the theorem which will make computations a little easier. The number of elements conjugate with x has to be [G:\text{C}(x)]. Now 1,x,x^2,x^3 \in \text{C}(x) thus it has at least four elements, by Lagrange's theorem it can therefore have 4\mbox{ or } 8 elements it cannot be 8 since x is not in the center thus \text{C}(x) = \{1,x,x^2,x^3\} thus [G:\text{C}(x)] = 2 which means the number of elements conjugate with x is just two. Now yxy^{-1} = x^{3}yy^{-1} = x^3 thus x^3 is conjugate to x, we can stop here by the theorem thus \{x,x^3\} is another conjugacy class.

    The third step that I would do is choose an element not from the ones chosen, say y. Now we will use the theorem which will make computations a little easier. The number of elements conjugate with y has to be [G:\text{C}(y)]. Now 1,x^2,y\text{C}(y) thus it has at least three elements it cannot have all eight thus by Lagrange's theorem |\text{C}(y)| = 4 which implies that [G:\text{C}(y)] = 2. Now xyx^{-1} = x^2y thus x^2y is conjugate to y, we can stop here by the theorem thus \{y,x^2y\} is another conjugacy class.
    Can you please simplify the above if possible?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by dcclanuk View Post
    Can you please simplify the above if possible?
    The centralizer of an element x, denoted by x, is the subgroup [tex]\{ a\in G|ax = xa\}. Think of it as a set of all elements which commute with x, we denote it by \text{C}(x).

    The theorem (for finite groups) says that the number of elements in the same conjugacy class as x is equal to [G:\text{C}(x)].

    Remember what [G:\text{C}(x)] means. It means the number of cosets of \text{C}(x). It is in fact equal to |G|/|\text{C}(x)|.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Apr 2008
    Posts
    3
    Quote Originally Posted by ThePerfectHacker View Post
    The centralizer of an element x, denoted by x, is the subgroup [tex]\{ a\in G|ax = xa\}. Think of it as a set of all elements which commute with x, we denote it by \text{C}(x).

    The theorem (for finite groups) says that the number of elements in the same conjugacy class as x is equal to [G:\text{C}(x)].

    Remember what [G:\text{C}(x)] means. It means the number of cosets of \text{C}(x). It is in fact equal to |G|/|\text{C}(x)|.
    Sorry to sound dumb, but I don't understand how u still carried out the above calculations
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. conjugacy class
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: January 27th 2011, 07:38 AM
  2. conjugacy class
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 1st 2010, 08:09 PM
  3. Conjugacy class
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: October 19th 2009, 08:44 PM
  4. conjugacy class and class equation
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 1st 2009, 07:52 PM
  5. conjugacy class
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: November 2nd 2008, 10:55 AM

Search Tags


/mathhelpforum @mathhelpforum