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



Maximal order of elements in SL(n,q)
Yes. It is shown in the paper 


*Darafsheh, M.R.*, [**Order of elements in the groups related to the general linear group.**](http://dx.doi.org/10.1016/j.ffa.2004.12.003), Finite Fields Appl. 11, No. 4, 738-747 (2005). [ZBL1147.20043](https://zbmath.org/?q=an:1147.20043).


(*Theorem 1*)
that the maximal order is 
$$\frac{q^n -1}{q-1},$$ *except* in the case $SL(2, p),$ where the maximum is $2p.$


