# 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?

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To compute the size of the conjugacy class it suffices, by the orbit-stabilizer theorem, to compute the size of the centralizer. Can you do that? – Qiaochu Yuan Mar 1 '12 at 22:01
I suppose if I figure out the first one, I can do the second one. The centralizer of $\begin{pmatrix} -1 & 0 \\ 0 & 1\end{pmatrix}$ is all $g \in PGL_{2}(\mathbb{F}_{q})$ such that $\begin{pmatrix} -1 & 0 \\ 0 & 1\end{pmatrix}g\begin{pmatrix} -1 & 0 \\ 0 & 1\end{pmatrix}^{-1} = g$. Writing $g = \begin{pmatrix} a & b \\ c & d\end{pmatrix}$, this forces $b = c = 0$ and hence $g$ is a diagonal matrix. Since $g \in PGL_{2}(\mathbb{F}_{q})$, we must have its determinant nonzero, so $a, d \neq 0$. And this implies that the centralizer has size $(q-1)^{2}$? – 387492 Mar 1 '12 at 22:09
Wait, so if the centralizer has size $(q - 1)^{2}$, as $|PGL_{2}(\mathbb{F}_{q})| = (q - 1)q(q + 1)$, wouldn't I have a conjugacy class of size $q(q + 1)/(q - 1)$? I feel like something is wrong here. – 387492 Mar 1 '12 at 22:14
You are in $PGL$ and not in $GL$, so you are counting elements in the centralizer wrong. – Mariano Suárez-Alvarez Mar 1 '12 at 22:21
Wait, so in $PGL$, $\begin{pmatrix} a & 0\\ 0 & d\end{pmatrix} \equiv \begin{pmatrix} 1 & 0\\ 0 & da^{-1}\end{pmatrix}$. There are $q - 1$ such $da^{-1}$, but this still gives me the wrong count for the centralizer. – 387492 Mar 1 '12 at 23:15

## 2 Answers

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 $\dfrac{q(q-1)(q+1)}{2(q-1)}=\dfrac{q(q+1)}{2}$.

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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 ${\rm GL}(2,q)$).

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