What's the maximum order of an element in $SL_2(\mathbb{Z} /p\mathbb{Z})$ for $p>2$ prime?

Question

I know the answer is $2p$ as I've checked it for $p=3,5$ and $71$. The characteristic polynomial of a matrix $A\in$ $SL_2(\mathbb{Z} /p\mathbb{Z})$ is $P_A(x)=x^2-tr(A)x+1$, so if this polynomial has a solution, the matrix can be diagonalized. I've studied all of the cases when a matrix is diagonalizable:

1. $A$ has double eigenvalues $1$ or $-1$ of multiplicity $2$, so it's similar to a matrix $\left(\begin{matrix} \pm 1 & 0 \\ 0 & \pm 1 \end{matrix}\right)$, which have respectively orders $1$ or $2$.

2. $A$ two distinct eigenvalues $\omega$ and $\omega ^{-1}$, with $\omega \in (\mathbb{Z}/p\mathbb{Z})^*$, so it's similar to $\left(\begin{matrix} \omega & 0 \\ 0 & \omega ^{-1} \end{matrix}\right)$, with order $p-1$.

3. $A$ has a double eigenvalue $1$ of multiplicity $1$, so it's similar to $\left(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}\right)$, of order $p$.

4. $A$ has a double eigenvalue $-1$ of multiplicity $1$, so it's similar to $\left(\begin{matrix} -1 & 1 \\ 0 & -1 \end{matrix}\right)$, of order $2p$.

This covers every case except when $P_A(x)$ has no solution. So, I'd just need to prove that, in that case, the order is $\le 2p$.

Answer 1

@reuns answered your question in a comment to an answer that has now been deleted, so I will just repeat the argument.

If the characteristic polynomial of an element $g \in {\rm SL}(2,p)$ is irreducible over ${\mathbb F}_p$ then it has two roots over ${\mathbb F}_{p^2}$ that are Galois conjugates, so they are $a$ and $a^p$ for some $a \in {\mathbb F}_{p^2}$.

Now, since the determinant of $g$ is $1$, we have $a^{p+1}=1$, and hence the order of $g$ divides $p+1$.

This completes the proof that the largest order of an element is $2p$ when $p$ is odd.