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Thread: [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
    Attached Thumbnails Attached Thumbnails [SOLVED] Dihedral Group D8-d8.jpg  
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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 $\displaystyle r$. Now $\displaystyle C(r)$ contains $\displaystyle \{e,r,...,r^7\}$. Thus, $\displaystyle |C(r)|\geq 8$. By Lagrange's theorem this means $\displaystyle |C(r)| = 8 \text{ or }16$ since $\displaystyle r\not \in \text{Z}(D_8)$ it means $\displaystyle |C(r)|=8$. Thus, $\displaystyle |G:C(r)|=2$ which means the number of elements in the conjugacy class of $\displaystyle r$ is $\displaystyle 2$. Conjugating $\displaystyle r$ by $\displaystyle s$ we get $\displaystyle srs^{-1} = srs=r^{-1}s^2=r^7$. Thus, $\displaystyle \{r,r^7\}$ is another conjugacy class. Repeating the same argument with $\displaystyle r^2$ we find that $\displaystyle |C(r^2)| = 8$ (since $\displaystyle C(r^2)$ at least contains $\displaystyle \{e,r,...r^7\}$. Thus, the number of element in the conjugacy class of $\displaystyle r^2$ is $\displaystyle 2$. Conjugating $\displaystyle r^2$ by $\displaystyle s$ (it does not have to be $\displaystyle s$ in fact it can be any element) we get $\displaystyle sr^2s^{-1} = r^{-2}s^2 = r^6$. Thus, $\displaystyle \{r^2,r^6\}$ is another conjugacy class. Reapting the same argument we find $\displaystyle \{r^3,r^5\}$ as another conjugacy class. But we are still left with $\displaystyle r^4$. However, $\displaystyle r^4$ commutes with $\displaystyle s$. Therefore, $\displaystyle r^4 \in \text{Z}(D_8)$ and it is in its own conjugacy class.

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

    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: $\displaystyle \left< r\right>,\left< r^2,s\right>$, i.e. a copy of $\displaystyle C_8$ and $\displaystyle D_4$.

    This is Mine 12th Post!!!
    Last edited by ThePerfectHacker; Jul 19th 2008 at 07:17 PM.
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