Exponent of Sylow $p$-subgroup of classical groups over a field of characteristic $p$ - MathOverflow

Exponent of Sylow $p$-subgroup of classical groups over a field of characteristic $p$

2012-12-12T03:21:06Z

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 $GF(q)$.

Answer by Jim Humphreys for Exponent of Sylow $p$-subgroup of classical groups over a field of characteristic $p$

2012-12-13T17:14:14Z

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.)

Answer by Nick Gill for Exponent of Sylow $p$-subgroup of classical groups over a field of characteristic $p$

2012-12-14T10:18:12Z

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!)