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Math Help - [SOLVED] Dihedral Group D8

  1. #1
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    [SOLVED] Dihedral Group D8

    Me again, How do I find the conjugacy classes of the attached? If someone could provide the answer and show me how its done so i can follow it, that would be great
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  2. #2
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    The number of elements of a conjugacy class of an element is equal to the index of the centralizer of the element.

    Look at the elements which we did not use yet, for example r. Now C(r) contains \{e,r,...,r^7\}. Thus, |C(r)|\geq 8. By Lagrange's theorem this means |C(r)| = 8 \text{ or }16 since r\not \in \text{Z}(D_8) it means |C(r)|=8. Thus, |G:C(r)|=2 which means the number of elements in the conjugacy class of r is 2. Conjugating r by s we get srs^{-1} = srs=r^{-1}s^2=r^7. Thus, \{r,r^7\} is another conjugacy class. Repeating the same argument with r^2 we find that |C(r^2)| = 8 (since C(r^2) at least contains \{e,r,...r^7\}. Thus, the number of element in the conjugacy class of r^2 is 2. Conjugating r^2 by s (it does not have to be s in fact it can be any element) we get sr^2s^{-1} = r^{-2}s^2 = r^6. Thus, \{r^2,r^6\} is another conjugacy class. Reapting the same argument we find \{r^3,r^5\} as another conjugacy class. But we are still left with r^4. However, r^4 commutes with s. Therefore, r^4 \in \text{Z}(D_8) and it is in its own conjugacy class.

    The conjugacy classes are: \{e\}, \{r^4\}, \{s,r^2s,r^4s,r^6s\}, \{rs,r^3s,r^5s,r^7s\}, \{r,r^7\},\{r^2,r^6\},\{r^3,r^5\}.

    To find normal subgroups of order 8 it is sufficient to find subgroups of order 8 because since they are index two they are automatically normal. It actually has three subgroups of order 8. But the two more obvious ones are: \left< r\right>,\left< r^2,s\right>, i.e. a copy of C_8 and D_4.

    This is Mine 12th Post!!!
    Last edited by ThePerfectHacker; July 19th 2008 at 08:17 PM.
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