Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Dear all,

Consider the finite simple Ree groups $G=^2\hspace{-1mm}G_2(3^{2n+1})$ where $n$ is a positive integer. I would like to know the orders of conjugacy class representatives of $G$ and from there to know the number of conjugacy classes of $3$-singular elements of $G$ (an element is 3-singular if its order is divisible by 3). This is probably well-known but I couldn't find it somewhere in the literature.

Thank you in advance!

share|improve this question
2  
Have you looked at the papers of H.N. Ward? –  Geoff Robinson May 28 '13 at 17:05
    
@Geoff Robinson: I have looked the paper: Ward, Harold N. On Ree's series of simple groups. Trans. Amer. Math. Soc. 121 1966, 62–89. It has a lot of information on the Ree groups but I still couldn't figure out the information needed. –  Hung Nguyen May 28 '13 at 18:33
    
See Jim Humphreys's answer, which gives more detail. –  Geoff Robinson May 28 '13 at 18:59

2 Answers 2

up vote 3 down vote accepted

To amplify Geoff's brief comment, the most standard source for details about (most of) the ordinary characters of a Ree group of type $G_2$ (specified by an odd power of $3$ at least $27$ which we call $q$ here but is sometimes written $q^2$) is the old paper by H.N. Ward based on his thesis. This is now freely available online here.

Note that Ward leaves aside the more complicated Ree groups of type $F_4$ in characteristic 2. Instead he starts with certain known properties of the $G_2$ Ree groups (which define for him a group of "Ree type") and then deduces a great deal about the group including its simplicity.

From Ward's organization of the ordinary irreducible characters you can obtain the number of these (which I guess is $q+8$), equal to the number of conjugacy classes. On the other hand, Steinberg showed that the number of semisimple ($3$-regular) classes is just $q$. This correlates with the fact that the $3$-modular irreducible representations are obtained by restricting certain irreducible rational representations of the ambient algebraic group, whose highest weights are easily specified. (His original results are in Nagoya J. Math. in 1963, but are also included in his 1967-68 Yale lectures here.)

While Steinberg's method is probably optimal, I'm not sure whether there are better methods by now to enumerate all conjugacy classes than the somewhat indirect ones used by Ward. (At any rate, it would be interesting to derive the ordinary characters systematically using the later methods of Deligne-Lusztig.)

share|improve this answer
    
@Jim Humphreys: Thank you very much! Can you point out the reference for the Steinberg's result that you mentioned? –  Hung Nguyen May 28 '13 at 19:09
    
See my edited version above. There is also an exposition in my LMS Lecture Note Ser. 326 Modular Representations of Finite Groups of Lie Type (Cambridge, 2006), Chapter 20. –  Jim Humphreys May 28 '13 at 20:42

I think there is enough information in Chapter III of Ward's paper to determine the classes of 3-singular elements, and the paragraph numbers below refer to that).

Let $P \in {\rm Syl}_3(G)$. Then $|P|=q^3$, with upper and lower central series of $P$ both $1 < C < P_1 < P$ (using Ward's notation), where $|C|=q$, $P_1$ is elementary abelian of order $q^2$, and $P/P_1$ is elementary abelian.

The normalizer of $P$ is a semidirect product of $P$ with a cyclic group of order $q-1$, which acts fixed-point-freely on $C$ and on $P/P_1$, but acts as a cycle of order $(q-1)/2$ on the middle layer $P_1/P$. The element of order 2 in this cyclic group has centralizer $C_2 \times L_2(q)$.

The $q-1$ elements in $C \setminus \{1\}$ form a single class. The $q^2-q$ elements of $P_1 \setminus C$ (which have order 3) split into two classes (paragraph 7). These elements centralize an element of order 2, and there are also two classes of elements of order 6. Finally, the $q^3-q^2$ elements in $P \setminus P_1$ split into three classes of elements of order 9 (paragraphs 9,10,11).

share|improve this answer
    
@Derek Holt: Thank you very much. This helps a lot. –  Hung Nguyen May 28 '13 at 20:58

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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