All the Conjugacy Classes of the Dihedral Group $D_8$ of Order 8
Problem 54
Determine all the conjugacy classes of the dihedral group
\[D_{8}=\langle r,s \mid r^4=s^2=1, sr=r^{-1}s\rangle\]
of order $8$.
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Hint.
You may directly compute the conjugates of each element
but we are going to use the following theorem to simplify the computations.
Theorem.
The number of conjugates of an element $g$ in a group is the index $|G: C_G(s)|$ of the centralizer of $s$.
Solution.
Let us denote $G=D_8$.
Let $K_x$ be the conjugacy class in $G$ containing the element $x$.
Note that $\langle r \rangle < C_G(r) \lneq G$ and the order $|\langle r \rangle|=4$.
Hence we must have $C_G(r)=\langle r \rangle$.
Thus the element $r$ has $|G:C_G(r)|=2$ conjugates in $G$.
Since $srs^{-1}=r^3$, the conjugacy class $K_r$ containing $r$ is $\{r, r^3\}$.
Since $\langle s \rangle < C_G(s) \lneq G$ and $|\langle s \rangle|=2$, we have either $C_G(s)=\langle s\rangle$ or $|C_G(s)|=4$.
Since $r^2s=sr^2$, we must have $|C_G(s)|=4$ and hence the conjugacy class $K_s$ containing $s$ has $|G:C_G(s)|=2$ elements.
Since $rsr^{-1}=sr^2$, we have $K_s=\{s, sr^2\}$.
We know the center is $Z(G)=\{1,r^2\}$ by Problem Centralizer, normalizer, and center of the dihedral group D8.
Thus $K_1=\{1\}$ and $K_{r^2}=\{r^2\}$.
The remaining elements $sr$ and $sr^3$ should be in the same conjugacy class (otherwise these elements are in the center), thus $K_{sr}=\{sr, sr^3\}$.
In summary, the conjugacy classes of the dihedral group are
\[\{1\}, \{r^2\}, \{r, r^3\}, \{s, sr^2\}, \{sr, sr^3\}.\]
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