Finite groups of Lie type include those obtained as rational points of a connected simple algebrraic group over a finite field $k = \mathbb{F}_q$ of characteristic $p$: these are split or quasi-split. There are also several families obtained less directly from algebraic groups: the Suzuki groups $^2\!B_2(q)$ and the Ree groups $^2\!G_2(q),\: ^2\!F_4(q)$, where $q = p^{2n+1}$ with $n \geq 1$ is an odd power of (respectively) $2,3,2$. [Caution: Sometimes $q$ is written here as $q^2$ to emphasize the similarity of group orders to those of untwisted groups.]

Counting the total number of conjugacy classes in such a group is a natural problem, relative to the determination of ordinary characters. Case-by-case study has been done for many of the families, especially the exceptional types; but it's unclear how much can be expressed uniformly. (Older results are surveyed in Chapter 8 of my 1995 AMS book on conjugacy classes.) It helps to assume the ambient algebraic group is simply connected, in which case Steinberg proved for split and quasi-split types that the number of semisimple classes (whose elements have order prime to $p$) is $p^r$ with $r$ the Lie rank. For Suzuki or Ree groups, $r$ is the rank of the BN-pair: 1, 1, 2.

For Suzuki groups the total number of classes is $q + 3$ (Deriziotis), while for Ree groups of type $G_2$ the number is $q+8$ (Ward). The latter groups came up recently here.

How many conjugacy classes do the Ree groups of type $F_4$ have?

This should be computable from known data, though not easily. Judging from all special cases I'm aware of, the answer should be given by a polynomial $q^2 + aq + b$, where $a, b \in \mathbb{Z}$ are independent of $q$. [Note however that for most other families of groups of Lie type, there are extra complications related to isogeny type, bad primes, and such.]

The underlying question, of course, is whether one can predict a priori what the polynomials will look like for all groups (coming from simply connected algebraic groups) starting with the highest degree term $q^r$.


1 Answer 1


According to

Fleischmann, Peter; Janiszczak, Ingo. On the computation of conjugacy classes of Chevalley groups. Appl. Alg. in Eng., Comm. and Comp. 1996, 7(3), 221--234

the class number of ${\rm F}_4(q)$ is

  • $q^4 + 2q^3 + 6q^2 + 10q + 19$ if $q = 2^n$,

  • $q^4 + 2q^3 + 7q^2 + 15q + 30$ if $q = 3^n$, and

  • $q^4 + 2q^3 + 7q^2 + 15q + 31$ if $q = p^n$ where $p > 3$

(see page 233).

According to Frank Lübeck's database on finite groups of Lie type, the class number of $^2{\rm F}_4(q^2)$ is $q^4+4q^2+17$. In that database you also find class numbers for many other types of finite groups of Lie type.

  • $\begingroup$ @Stefan: I should have emphasized that the Ree group of type $F_4$ is not the original Chevalley group studied by Fleischmann-Janiszczak but rather a proper subgroup of it: fixed points of a special Frobenius morphism involving a graph symmetry which interchanges long and short roots $\endgroup$ Jun 1, 2013 at 16:31
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    $\begingroup$ @Jim: I have added the class number formula for $^2{\rm F}_4(q^2)$ -- is that what you are looking for? $\endgroup$
    – Stefan Kohl
    Jun 1, 2013 at 17:21
  • $\begingroup$ @Stefan: Yes, the added information is very helpful. The older published results on $F_4(q)$ were included in my book, but I haven't yet checked out everything Lubeck and others have computed in recent years. (Note that Lubeck is using the alternative convention which involves $q^2$ rather than $q$, so in my notation above the class number would read $q^2 +4q + 17$.) Meanwhile I'm still curious to understand a priori the format of such results, which is intuitively consistent with Jordan decomposition but hard to make rigorous. $\endgroup$ Jun 1, 2013 at 17:42

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