For $g\in\operatorname{SL}(2,q)$, do we have $\operatorname{tr}(g)=\operatorname{tr}(g^{-1})?$

Question

The Question:

For $g\in\operatorname{SL}(2,q)$ and $q=p^r$ for a prime $p$ with $r\in\Bbb N$, do we have $\operatorname{tr}(g)=\operatorname{tr}(g^{-1})?$

Here $\operatorname{tr}(h)$ is the trace of the matrix $h$.

Thoughts:

I think so. According to a preprint, it holds for $\operatorname{SL}(2,\Bbb K)$, for $\Bbb K=\Bbb{R,C}$.

My guess is that it should fall nicely from $\det(g)=1$ but I don't see how just yet.

Does diagonalising $g$ work, as we are over a finite field? I suppose the answer would be something along those lines.

I feel like I'm missing something obvious.

Answer 1

Since $\det g = 1$, the inverse of $g$ is the adjugate of $g$. So, regardless of the ground field $\mathbb{K}$, for a $2 \times 2$ matrix in $\operatorname{SL}(2, \mathbb{K})$ $$ g = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, $$ we have $$ g^{-1} = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}. $$

It's clear that either matrix has trace $a + d$.