If $q=p^a$ , where $p$ is a prime number, then I would like to know the number of conjugacy classes related to elements of order $p$ and $2$ in the simple group $PSL(2,q)$ .
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2$\begingroup$ I'd guess that a direct route can be taken by applying the rational canonical form on $GL(2,q)$, moding out the center, and the semidirect product $PGL(2,q) = PSL(2,q) \rtimes F_q^\times / (F_q^\times)^2. $ $\endgroup$– Marc PalmCommented Sep 20, 2013 at 11:28
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3$\begingroup$ This is a standard question but not at all research-level. It's been understood since the time of Frobenius and Dickson, and has become standard in group theory textbooks. Already in 1968 it was just part of Exercise 21 at the end of Chapter 2 in Gorenstein's Finite Groups. All it requires is a mixture of elementary group theory and matrix theory, starting with $2 \times 2$ matrices. More suitable for Stack Exchange. (Also, the groups are not actually simple for all $q$.) $\endgroup$– Jim HumphreysCommented Sep 20, 2013 at 13:07
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1$\begingroup$ @Jim: I agree that this is more suitable for Stack Exchange. But just to clarify, that exercise in Gorenstein follows an 11-page exposition of Dickson's results on such subgroups. Although Dickson's work was elementary, it was by no means trivial -- e.g., Gorenstein called it "brilliant". $\endgroup$– Michael ZieveCommented Sep 20, 2013 at 13:23
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$\begingroup$ @Michael: The result needed here doesn't depend on the detailed work of Dickson, however. The steps actually involved are quite elementary. I only mentioned Gorenstein to emphasize that these classes have been around for a long time in the literature. $\endgroup$– Jim HumphreysCommented Sep 20, 2013 at 15:28
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2 Answers
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There is one conjugacy class of elements of order 2, and if $p$ is odd then there are two conjugacy classes of elements of order $p$. This goes back to Dickson's 1901 book on Linear Groups.
Added later: for order $p$ elements this can be seen as follows. All order-$p$ elements of $PSL(2,q)$ are conjugate to elements of any prescribed Sylow $p$-subgroup. One such Sylow $p$-subgroup $S$ of $PSL(2,q)$ consists of the upper-triangular matrices with $1$'s on the diagonal. Now consider the action of $PSL(2,q)$ on the projective line $\mathbb{P}^1(\mathbb{F}_q)$. Each nonidentity element of $S$ has a unique fixed point, namely $\infty$. Thus, any element of $PSL(2,q)$ which conjugates one nonidentity element of $S$ to another must fix $\infty$, and hence must be upper-triangular. Finally, one easily checks that $$ \left(\begin{matrix} a & b \\ 0 & a^{-1} \end{matrix}\right) \left(\begin{matrix} 1 & c \\ 0 & 1 \end{matrix}\right) \left(\begin{matrix} a & b \\ 0 & a^{-1} \end{matrix}\right)^{-1} = \left(\begin{matrix} 1 & a^2 c \\ 0 & 1 \end{matrix}\right). $$ It follows that the conjugacy classes of order-$p$ elements are in bijection with $\mathbb{F}_p^\times/(\mathbb{F}_p^\times)^2$, so there are two such classes if $p$ is odd and one if $p$ is even.
For order $2$ I don't know a proof from first principles that is as short as the one above; the quickest proof I know is the one given in the proof of Lemma A.3 of my paper with Bob Guralnick titled "Polynomials with PSL(2) monodromy".
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1$\begingroup$ Wonderful! Thanks a million for your help. $\endgroup$– TinaCommented Sep 20, 2013 at 10:21
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To amplify my comments (in community-wiki style:
1) Note that it's OK when discussing just unipotent elements (here those of order $p$) to work instead with the matrix group $G:=\mathrm{SL}(2,q)$, since $\mathrm{PSL}(2,q)$ is just the quotient by the center $\{\pm I\}$ which consists of semisimple elements (and is trivial for $p=2$). Also, the methods are basically the same for all $q$ with $p$ fixed.
2) Usually your question is part of a more general computation of conjugacy classes : size of each and total number of classes. This may be easier to organize, since the total number of elements i $|G|=q(q-1)(q+1)$. In any case, the tally of classes is most often part of the search for ordinary characters. (For a short exposition, see my old paper in Amer. Math. Monthly 82 (1975), 21--39.) Though I don't have most textbooks at hand, I'd also suggest looking at the arguments on page 230 of L. Dornhoff Group Representation Theory, Part A, Dekker, 1971. This is now out-of-print but has useful short chapters on many topics including this standard example.
3) Maybe it's worth emphasizing that the difference between $p=2$ and other primes for this purpose is that all nonzero elements of $\mathbb{F}_q$ are squares in the first case but only half the elements in the second case. Thus a unipotent matrix in $G$ is conjugate for odd $p$ to just one of the matrices $$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \quad \begin{pmatrix} 1 & \nu \\ 0 & 1 \end{pmatrix}$$ with $\nu$ a fixed nonsquare. There are $q^2$ unipotent matrices in all (special case of a general result of Steinberg), half in each class for $p$ odd. The usual canonical form theory just has to be adapted a bit over a finite field.
4) If you also want to look at elements of order 2 when $p$ is odd, you have to work a little more. Again it's probably easiest to determine (as in Dornhoff) all classes of semisimple, unipotent, or mixed elements, along with the sizes of their classes. It's not clear to me what motivates the format of your question, however.
ADDED: To be more explicit about elements of order 2 when $p$ is odd, just look at the standard list of classes in $G$ and note there is a unique one containing elements of order 4 (hence order 2 in the projective group). Such elements are semisimple, but may be diagonalizable over $\mathbb{F}_q$ or not depending on which of $q-1$ or $q+1$ is divisible by 4.