Conjugacy classes in a matrix group Consider the matrix group $PGL_{2}(\mathbb{F}_{q})$ for $q$ odd. Why is it that $\begin{pmatrix} -1 & 0\\ 0 & 1\end{pmatrix}$ has $q(q + 1)/2$ elements in its conjugacy class while $\begin{pmatrix} 2 & 0\\ 0 & 1\end{pmatrix}$ has $q(q + 1)$ elements?
 A: The centralizer of $\begin{pmatrix} -1 & 0\\ 0 & 1\end{pmatrix}$ in $PGL(2,q)$ is given by all matrices $\begin{pmatrix} a & b\\ c & d\end{pmatrix}$, such that 
$$
\begin{pmatrix} -1 & 0 \\ 0 & 1\end{pmatrix}\begin{pmatrix} a & b\\ c & d\end{pmatrix}\begin{pmatrix} -1 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix} ka & kb\\ kc & kd\end{pmatrix}
$$
 for some constant $k\in\mathbb{F}_q$.
We get that 
$$
\begin{pmatrix} -1 & 0 \\ 0 & 1\end{pmatrix}\begin{pmatrix} a & b\\ c & d\end{pmatrix}\begin{pmatrix} -1 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix} a & -b\\ -c & d\end{pmatrix}
,
$$ which fits our pattern, if either $b=c=0$ or $a=d=0$.  Thus we get $2(q-1)$ total matrices, for a conjugacy class of size $\dfrac{q(q-1)(q+1)}{2(q-1)}=\dfrac{q(q+1)}{2}$.
A: Actually, when $q$ is a power of $3,$ the two matrices are the same, so you need to exclude that case, which I do in the following discussion. You can do an explicit calculation, as has been done by other already, or you can observe that in ${\rm GL}(2,q)$, the first matrix is conjugate to just one scalar multiple of itself (the multiple being $-1,$ of course), while the second matrix is not conjugate to any other scalar multiple of itself (and I'm implictly using the fact that both matrices have the same centralizer in ${\rm GL}(2,q)$).
