1. [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

2. 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!!!

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class equation of d8

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