# Number of conjugacy classes are there over $\mathbb{F}_p$

I am struggling with the next exercise of my HW:

How many conjugacy classes are in $GL_3(\mathbb{F}_p)$? And how many in $SL_2(\mathbb{F}_p)$?

It's on the topic of Frobenius normal form of finitely generated modules over $\mathbb{F}_p$.

I'd appreciate any idea.

• The case $\operatorname{GL_3}(\Bbb{F}_p)$

For a given conjugacy class $c$ each element has an identical Jordan form (which lives in $GL_3(\Bbb{F}_p^3))$. If the minimal polynomial has degree $3$ then its Jordan normal form looks like $$\begin{pmatrix}\alpha & 0 & 0\\ 0 & \beta & 0 \\ 0 & 0 & \gamma \end{pmatrix}, \begin{pmatrix}\alpha & 1 & 0\\ 0 & \alpha & 0 \\ 0 & 0 & \beta \end{pmatrix} \text{ or } \begin{pmatrix}\alpha & 1 & 0\\ 0 & \alpha & 1 \\ 0 & 0 & \alpha \end{pmatrix}$$ for three distinct $\alpha, \beta$ and $\gamma$, the Jordan block necessary for the degree to be 3. This accounts for $p \cdotp\cdot(p-1)$ monomials of the form $x^3+ax^2+bx+c$ with $c \neq 0$.

If the polynomial has degree $2$ then each monomial with two different roots stands for two conjugacy classes with representatives $$\begin{pmatrix}\alpha & 0 & 0\\ 0 & \alpha & 0 \\ 0 & 0 & \beta \end{pmatrix} \text{ or } \begin{pmatrix}\alpha & 0 & 0\\ 0 & \beta & 0 \\ 0 & 0 & \beta \end{pmatrix}$$ Since now $\alpha, \beta \in \Bbb F_p$ we have $(p-1)(p-2)/2$ monomials of the form $(x-\alpha)(x-\beta)$ standing for $(p-1)(p-2)$ classes. We have to add $p-1$ classes with monomial a quadratic of degree $2$ with discriminant $0$. These are represented by the matrices $$\begin{pmatrix}\alpha & 1 & 0\\ 0 & \alpha & 0 \\ 0 & 0 & \alpha \end{pmatrix}$$ Finally the monomials of degree $1$ all give rise to scalar matrices, so there are $p-1$ of those classes.

Finally we obtain the result for the number of classes : $p^2(p-1) + (p-1)(p-2)+ (p-1)+ (p-1) = (p-1)p(p+1)$.

• The case $\operatorname{SL_2}(\Bbb{F}_p)$

We work analoguously as in the first case, by degree of monomial.

For degree two we have three possibilities : the first one is when the discriminant of the minimal polynomial is a square. This gives us $(p-3)/2$ matrices of the form $\left(\begin{smallmatrix} \alpha & 0 \\0 & 1/\alpha \end{smallmatrix}\right)$ where $\alpha \in \Bbb F_p \setminus \{-1,0,1\}$. Another case in degree $2$ is when the discriminant is not a square and $\alpha, \beta$ live in $\Bbb F_{p^2}$. There are $(p-1)/2$ of them. Then we have the matrices of the form $\left(\begin{smallmatrix} \alpha & \gamma \\0 & \alpha \end{smallmatrix}\right)$. A small calculation shows that there for each valur of $\alpha = -1,1$ there are two classes deoepending if $\gamma$ is a quadratic residue or not counting for $4$ additional classes, and if we add the two scalar classes $I$ and $-I$ we get the following result:

$(p - 3) / 2 + (p - 1) / 2 + 6 = p + 4$ classes

• The original question was regarding $SL_2(\mathbb{F}_7)$ – niceGuy Feb 27 '17 at 12:25
• @niceGuy: I'll remove that part and replace it later, thanks – Marc Bogaerts Feb 27 '17 at 12:59
• @MarcBogaerts in the $SL_2(7)$ case, how did you count the $(p-1)/2$? Also, why are there 4 additional classes and not 2? In the Jordan normal form, isn't the superdiagonal entry always 1? – Sai Feb 28 '17 at 18:10
• 1) Because if you interchange $\alpha$ and $\beta$ upi stay in the same class. 2) Becayse there are two "jordan-like" non conjugate forms as explained in another comment. Moreover the superdiagonal form can be $-1,-1$. – Marc Bogaerts Feb 28 '17 at 18:34
• @MarcBogaerts 1) But where did the $(p-1)$ come from? Why are there $(p-1)/2$ irreducible monics with constant term 1? 2) Do you mean $\pm 1$? Why? – Sai Feb 28 '17 at 18:43

For the general linear group I suggest that you count the (irreducible) linear polynomials with non-zero constant term, the irreducible quadratics, and finally the irreducible cubics. You then get a conjugacy class for each cubic, $C(f(X)$; a conjugacy class for each pair (linear, irreducible quadratic), $C(X-\alpha)\oplus C(q(X)$; and then you are left dealing with the elements where the characteristic polynomial is the product of linear factors. These last will give you classes for types $C((X-\alpha)^3)$, $C((X-\alpha)^2)\oplus C(X-\beta)$, $C(X-\alpha)\oplus C(X-\beta) \oplus C(X-\gamma)$. (Note, $\alpha=\beta$ is possible.)

For the special group you now need to identify which of these classes is in the group, and then investigate the relative sizes of the centraliser of an element in the special group and the general group to see whether the $GL$-orbit splits into smaller $SL$-orbits.

• This was my approach too, but I end up with a different answer than Mac Bogaerts. My count is $\frac{1}{3}(p^3-p)+\frac{1}{2}(p^2-p)(p-1)+3(p-1)+2\binom{p-1}{2}+\binom{p-1}{3}$, which is not $p^3-p$. Groupprops also says $p^3-p$. Do you see what I counted wrong? – Sai Feb 27 '17 at 14:21
• @Sai I would have got $+2(p-1)(p-2)$ and not $+2\binom{p-1}{2}$ since the cases $\alpha, \alpha,\beta$ and $\beta, \beta, \alpha$ are different. I don't know if that corrects it. – ancientmathematician Feb 27 '17 at 16:02
• It does. Now I see my answer for $SL_2(7)$ is wrong too. Here's my calculation. If Jordan form exists: 2 diagonal forms, 2 Jordan block forms, $(p-3)/2$ forms with distinct eigenvalues. If not, then number of irreducible monics with constant term 1. groupprops says there are 4 Jordan block forms. Why? And what to do with the case without a Jordan normal form? – Sai Feb 28 '17 at 18:04
• @Sai: In SL(2,p) a jordan from doesn't necessarily exists, because the adjoint action isn't always with a matrix of determinant $1$, But the next best thing is that sometimes there is a number different from one. See my answer : there are two "jordan like " blocks, one with a quadratic residue (like 1) and one with a quadratic non-residue. – Marc Bogaerts Feb 28 '17 at 18:28
• @MarcBogaerts I don't follow. In the link, the second row of the table says "Jordan block". Do they mean "Jordan-like"? – Sai Feb 28 '17 at 18:42