I’m using the formula that the number of conjugacy class is given to be $\frac{1}{|G|}\sum|C_{G}(g)|$, where $C_{G}(g)=\{h \in G ; gh=hg\}$, which is a special result by Burnside’s theorem.
I found that the number of conjugacy class in $D_8$ is 5, so to double check, I listed down all $C_{G}(g)=\{h \in G ; gh=hg\}$.
Let r be a rotation counter clockwise and m be a rotation in the x axis, 1 is the identity.
Well, $|C_G(1)|=|D_8|=16$,
$|C_G(r)|= |C_G(r^2)|= |C_G(r^3)|= |C_G(r^5)|= |C_G(r^6)|= |C_G(r^7)|=|\{1,r,r^2,r^3,r^4,r^5,r^6,r^7\}|=8$,
$|C_G(r^4)|=|\{1,r,r^2,r^3,r^4,r^5,r^6,r^7,m\}|=9$ since $r^4m=mr^4$
Similarly, we can count for the reflections;
$|C_G(m)|=|C_G(r^4m)|=|\{1,m\}|=2$, while the rest of the elements with any reflections only has one element, i.e $|C_G(r^nm)|=|\{1\}|=1,n \neq 0,4$
So the problem comes that my summation is 83, while my $|G|=16$. In this case I won’t get the number of conjugacy class to be 5. Did I do something wrong here? I just merely applied the definition…
Best Answer
$G = \{r,a|r^8 = a^2 = e, ara^{-1} = r^{-1}\}$
$e$ commutes with all elements and is in a conjugacy class all by itself.
$\{e\}$
Rotations fall into conjugacy classes that include their inverses.
$(r^n)(r^m)(r^{-n}) = r^m$
$(ar^n)(r^m)(ar^n)^{-1} = ar^nr^mr^{-n}a = r^{-m}$
$\{r, r^7\},\{r^2, r^6\},\{r^3, r^5\}, \{r^4\}$
Reflections:
$(r^n)(ar^m)(r^{-n}) = r^n (ar^{m-n}) = r^nr^{n-m}a = r^{2n-m}a = ar^{m-2n}$
and $(ar^n)(ar^m)(r^{-n}a) = ar^{2n-m}$
creating congugacy classes of $\{ar, ar^3, ar^5, ar^7\}$ and $\{a, ar^2, ar^4, ar^6\}$
That gives 7 conjugacy classes.
Counting the centralizers.
The identity commutes with everything.
Rotations commute with rotations.
$r^4$ commutes with every reflection, not just one reflection (as suggested above). Which means that $r^4$ commutes with everything.
Every reflection commutes with the identity, $r^4$, itself, and one other reflection.
$(ar^n)(r^{4-n}a) = (r^{4-n}a)(ar^n)=r^4$
$|C_G(e)| = 16\\ |C_G(r)| = 8 \text { times 6}\\ |C_G(r^4)| = 16\\ |C_G(a)| = 4 \text { times 8}$
$\frac {16\times 2 + 8\times 6 + 4\times 8}{16} = 7$