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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 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.

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  • $\begingroup$ For completeness: the maximal order of an element of GL(n,Fq) is q^n−1 - the cycle generated by that element is called "Singer cycle" - discussed in many papers and at MO e.g. Geoff Robinson here: mathoverflow.net/a/254509/10446 $\endgroup$
    – Alexander Chervov
    Commented Jun 18, 2017 at 18:02

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Yes. It is shown in the paper

Darafsheh, M.R., Order of elements in the groups related to the general linear group., Finite Fields Appl. 11, No. 4, 738-747 (2005). ZBL1147.20043.

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

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    $\begingroup$ Thank you very much Igor. This question came up yesterday in a meeting with undergraduates in my geometry lab. After thinking about the SL(2,q) case I optimistically guessed the above formula was the answer and that it was a known fact, but I would never have guessed it was a 2005 Theorem. $\endgroup$
    – Sean Lawton
    Commented Jun 17, 2017 at 20:16
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    $\begingroup$ @SeanLawton Just because it was published in 2005 does not mean that it was not known to John Thompson is 1962 :) $\endgroup$
    – Igor Rivin
    Commented Jun 17, 2017 at 20:18
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    $\begingroup$ @AndreasThom Maybe, but that's no reason to take away credit from the guy who published the 2005 paper. Who knows how much of what we all published was known before... $\endgroup$
    – Igor Rivin
    Commented Jun 19, 2017 at 0:28
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    $\begingroup$ @IgorRivin: It is good to have a reference, but I was just thinking that Landau looked at the maximal order of elements in Sym(n) in 1903 because this was a difficult and interesting problem (and the solution relies on the prime number theorem). Those people must have thought about the same problem for linear groups, which is much easier. $\endgroup$
    – Andreas Thom
    Commented Jun 19, 2017 at 7:25
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    $\begingroup$ Do you folks know of any reference for the maximal order of an element in $SL_2(\mathbb{Z}/p^k\mathbb{Z})$, where $p$ is a prime? $\endgroup$
    – Alain Valette
    Commented May 21, 2018 at 17:09

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