Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
3  
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
2  
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 2

up vote 4 down vote accepted

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

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.