Let $G$ be a classical group of dimension $n$ over $GF(q)$ where $q=p^f$ is a prime power, and $P$ be a Sylow $p$-subgroup of $G$. What is the maximal order of elements, i.e. the exponent, of $P$?

For $G=GL(n,q)$, it can be easily seen that the exponent of $P$ is the least power of $p$ greater than or equal to $n$. This gives a upper bound for exponents of classical groups of dimension $n$ over $GF(q)$.

  • $\begingroup$ According to the theorem of Borel and Tits discussed in mathoverflow.net/questions/104201/…, if $G$ is a connected reductive group over a perfect field $k$ of characteristic $p > 0$, any finite $p$-subgroup of $G(k)$ is contained in $U(k)$ where $U$ is the ($k$-split) unipotent radical of a minimal parabolic $k$-subgroup $P$ of $G$. All such $P$ are $G(k)$-conjugate, so such $U(k)$ are the Sylow $p$-subgroups of $G(k)$ when $k$ is finite. The general context for your question is to compute the exponent of $U(k)$. Try more to settle GL$_n$ by yourself. $\endgroup$
    – user29720
    Dec 12, 2012 at 5:28
  • 1
    $\begingroup$ Although root system considerations give an upper bound on $m$ for which $p^m$ kills the group, with the bound on $m$ depending only on the Killing-Cartan "type" of the classical group (not on the size of the finite field or its characteristic), the exact exponent may be sensitive to peculiar features of small finite fields or small characteristic (e.g., surprising commutation of some root groups in symplectic and odd special orthogonal groups in characteristic 2). $\endgroup$
    – user29720
    Dec 12, 2012 at 13:26
  • $\begingroup$ What is the exact formulation of the upper bound? $\endgroup$ Dec 12, 2012 at 13:38
  • $\begingroup$ I recommend Gorenstein, Lyons, Solomon vol 3 which gives a full description of the Sylow p-subgroups of the classical groups. (I don't have a copy in front of me so can't be more precise.) Off the top of my head, I think they all contain an element with one Jordan block, hence will have the same exponent as a Sylow-$p$ of $GL_n(q)$... But I could be wrong... $\endgroup$
    – Nick Gill
    Dec 12, 2012 at 16:05
  • 2
    $\begingroup$ "largest power of $p$ greater than or equal to $n$"? $\endgroup$
    – Someone
    Dec 13, 2012 at 10:07

3 Answers 3


The question is reasonable (even in the full generality of finite groups of Lie type in the defining characteristic $p$). However, the answer requires case-by-case study, as in related questions about classifying unipotent classes and centralizers of unipotent elements. Most of the information needed starts out in the ambient algebraic group $G$ and leads to fairly uniform results for the finite groups when the prime is "good" (meaning odd for symplectic or orthogonal groups). When studying unipotents, the isogeny class of the group doesn't matter, or whether you consider general linear or special linear groups. In good characteristic it's mostly enought to look at nilotents in the Lie algebra of $G$, but bad primes require some extra study within the group setting.

Probably the most straightforward way to think about the exponent of a Sylow $p$-subgroup (coming from a maximal unipotent subgroup of the algebraic group) is to look at a smallest possible faithful representation of $G$, say of dimension $n$. Since the $n$th power of each nilpotent matrix vanishes, a rough upper bound on the possible orders of unipotents in $G$ is given by the least (not largest) power of $p$ which is $\geq n$.

What about a precise determination of the exponent? Since there are only finitely many unipotent classes in $G$ (even for bad primes), you mainly need to locate the largest Jordan block size among those occurring for any unipotent matrix in a class. If this size is $d$, then the exponent is the least power of $p$ which is $\geq d$. For classical groups, at least when $p$ is odd, the unipotent classes in $G$ and then a finite subgroup of Lie type are fairly easy to obtain from those of the ambient general linear group. There is a lot of published work in this direction, going back to a paper by G.E. Wall. Useful books include the 1985 Simple Groups of Lie Type by Roger Carter and a recent AMS book Unipotent and Nilpotent Classes in Simple Algebraic Groups and Lie Algebras by Martin Liebeck and Gary Seitz.

Probably the regular unipotent elements have the largest orders, though in the symplectic and orthogonal groups they need not involve just a single large Jordan block as they would in the special linear group. Anyway, each type and each small prime need some attention here. For finite or algebraic groups of exceptional types, all of this is more delicate in terms of working with a convenient matrix representation (the adjoint representation being the smallest for type $E_8$). So much attention in the literature is directed at these groups: see for example the influential paper by Ross Lawther: Jordan block sizes of unipotent elements in exceptional algebraic groups, Comm. Algebra 23 (1995), no. 11, 4125–4156.

Final remark: In the study of finite simple groups, the exponent is not usually a major player. For example, it doesn't show up directly in the character table. The main themes have been the classification of conjugacy classes and the determination of corresponding centralizers in $G$ and the finite groups.

ADDED (post-final remark): Looking more closely at known results, I can see that the exponent varies somewhat erratically over powers of $p$ until it seems to stabilize at $p$ when $p \geq h$. Here $h$ is the Coxeter number (not the rank), which in types $A_n, B_n, C_n, D_n$ is respectively $n, 2n, 2n, 2(n-1)$. If this bound on $p$ is optimal, it's reasonable to look for a general proof. It seems consistent with Lawther's tables for exceptional types. (It's also worth comparing with work on elements of prime order in simple complex or compact Lie groups.)

P.S. As Maria points out, there is an exact formulation for all Lie types and all primes in Corollary 0.5 of the 1995 paper by Testerman. I had forgotten about this, but her work and Lawther's have been closely related, and sometimes collaborative as in their two AMS Memoirs no. 674 (1999) and no. 988 (2011). As far as I know, the proof of 0.5 near the end of her paper requires considerable case-by-case work (most of which assumes $p$ is good) together with separate treatment of bad primes. By now there is a lot of related literature, including the recent monograph by Liebeck and Seitz, so I'm not quite sure what the best arguments are. (Also, Maria has corrected my hasty listing of Coxeter numbers: $\mathrm{SL}_{n+1}$ has type $A_n$ and Coxeter number $n+1$.)


Here's a more naive approach than that suggested by Jim in his answer, and Kreck in comments.

One can explicitly write down the Sylow $p$-subgroup as a matrix group using a basis of hyperbolic pairs. If you order this basis correctly, then there will be a Sylow $p$-subgroup of the associated classical group which will be a bunch of upper-triangular matrices. (See Kleidman & Liebeck "The subgroup structure of the finite classical groups" for a full description.)

(Note: I'm avoiding questions about which version of the group you are interested in. Most of the time the natural matrix group version of the classical group is the universal version (the main exceptions involving certain orthogonal groups). In any case, if you avoid bad primes, as Jim describes above, the exponent of a Sylow $p$-subgroup of your group won't depend on which version you're using.)

So, for example, let $e_1,\dots, e_k, f_1,\dots, f_k$ be such a basis for a vector space $V$ over the field of order $q$ where $q$ is odd, where $V$ is equipped with a non-degenerate alternating bilinear form, then the isometry group of the form will be the symplectic group ${\mathrm Sp}_{2k}(q)$ and one can choose a Sylow $p$-subgroup of $G$ such that all elements are upper-triangular. Once you've done this it's easy to observe that $G$ contains an element $$\left(\begin{array}{ccccc} 1 & 1 & & & \\\\ & 1 & 1 & & \\\\ & & \ddots \ddots & & \\\\ & & & 1 & -1 \\\\ & & & & 1 \end{array}\right).$$ (The formatting is a bit screwed. It's supposed to indicate a diagonal of $1$'s. Then on my super-diagonal I have $k$ lots of $1$ to start and $k-1$ lots of $-1$ to finish.) I haven't double-checked, but I believe this is an element of maximal exponent. Its order is the least power of $p$ greater than $k+1$.

It's the same story with orthogonal groups. Again I haven't double-checked that I really am obtaining an element of maximal order, but this should be straight-forward.

In any case, these calculations suggest the following result.

Conjecture: Let $P$ be a Sylow $p$-subgroup of an untwisted classical group $G$ over $\mathbb{F}_q$, and suppose that $p$ is not a bad prime. Then the exponent of $P$ is equal to the least power of $p$ greater than ${\mathrm rk}(G)+1$.

(I'm stating the conjecture conservatively. It's possible that it holds for twisted groups, and for bad primes, but that would just be idle speculation!)

  • $\begingroup$ I meant to say: I have an e-copy of Kleidman & Libeck which I can email should you need it. $\endgroup$
    – Nick Gill
    Dec 14, 2012 at 10:48

For some classical groups (arising from simple algebraic groups), the answer is given in Proposition 0.5 in a paper by Donna Testerman: $A_1$-type overgroups of elements of order $p$ in semisimple algebraic groups and the associated finite groups, J. Algebra, 177 (1995), no. 1, 34--76. I quote this proposition:

Let $G$ be a simple algebraic group defined over an algebraic closed field of arbitrary characteristic $p>0$ and $\sigma$ be a surjective endomorphism of $G$ such that $G_\sigma$ is finite. Then the exponent of a Sylow $p$-subgroup of $G_\sigma$ is the least power of $p$ greater than the height of the highest root in the root system of $G$.

The height of the highest root is equal to $h-1$, where $h$ is the mentioned Coxeter number, and so is $n$, $2n-1$, $2n-1$ and $2n-3$ in $A_n$, $B_n$, $C_n$ and $D_n$ respectively (the Coxeter number in type $A_n$ is $n+1$ not $n$).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.