A problem about class equation

Question

Let $k$ be a finite field, where $|k|=q$ and $\operatorname{char}k\neq2$, and let $$D=\{A\in\operatorname{SL}(2,k)\mid A \text{ is diagonalizable}\}.$$ Prove that $$|D|=2+\tfrac{1}{2}\cdot(q+1)\cdot q\cdot(q-3).$$ Can anyone help me? Thanks!

Answer 1

1. We know that $|GL(2,k)| = (q^2-1)(q^2-q)$, and $|k^{\ast}| = (q-1)$, so $$ |SL(2,k)| = (q+1)(q^2-q) $$
2. Now, consider $$ \hat{D} := \left\lbrace A_{\lambda} := \begin{pmatrix} \lambda & 0 \\ 0 & \lambda^{-1} \end{pmatrix} : \lambda \in k^{\ast} \right\rbrace $$ and consider the conjugacy class $C(A_{\lambda})$ of $A_{\lambda}$.

a) If $\lambda =1$ or $\lambda = -1$ (which are different since $2\neq 0$), $A_{\lambda}$ is in the center of $SL(2,k)$, and so $$ |C(A_1)| = 1 = |C(A_{-1})| $$ b) For any other $\lambda \in k^{\ast}$, the centralizer of $A_{\lambda}$ is precisely $\hat{D}$, which has cardinality $|k^{\ast}| = (q-1)$. By the Orbit-Stabilizer theorem, $$ |C(A_{\lambda})| = \frac{|SL(2,k)|}{|\hat{D}|} = q(q+1) $$ Each such $\lambda$ occurs in pairs $\{\lambda, \lambda^{-1}\}$, and there are $$ \frac{1}{2}(q-3) $$ such pairs (everything in $k$ excluding $\{0,1,-1\}$). Hence, $$ |D| = 2 + \frac{1}{2}(q-3)q(q+1) $$