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Math Help - Conjugacy Class Equation

  1. #1
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    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
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  2. #2
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    Look here, the last post.
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  3. #3
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    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?
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  4. #4
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    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)|.
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  5. #5
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    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
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