The maximal order of an element of $\mathrm{GL}(n,\mathbb{F}_q)$ is $q^n-1$, where the characteristic of $\mathbb{F}_q$ is odd $p$.  See [here][1] for a nice proof that uses the Cayley-Hamilton Theorem.  

However, for $\mathrm{SL}(2,\mathbb{F}_q)$ the maximal order is $2p$ if $q=p$ and $q+1$ otherwise.

To see this, note that over $\mathbb{F}_{q^2}$ any such matrix $A$ is conjugate to an upper-triangular matrix.  If it is not diagonal then its eigenvalues are repeated and $\pm 1$; thus its order is bounded by $2p$. If it is diagonal then the eigenvalues are in $\mathbb{F}_q$ or properly in the degree two extension.  If the former then the order divides $|\mathbb{F}_q*|=q-1$, and if the latter then $x\mapsto x^q$ generates the Galois group of the extension showing the eigenvalue $a$ satisfies $a^q=a^{-1}\implies a^{q+1}=1 \implies |a|{\large \mid} q+1$.

I did some searching on MO and other places online but did not find a generalization of this for $n\geq 3$.

> Is there a nice formula for the maximal order of an element in $\mathrm{SL}(n,\mathbb{F}_q)$?  If so, what is the proof or reference?

**Remark:** I would also be interested in the answer to the same question for other finite groups of Lie type, as well as the even characteristic case.



  [1]: https://mathoverflow.net/questions/109483/maximal-order-of-elements-in-gln-p