Home > representation theory > Conjugacy classes in the finite general and special linear groups

## Conjugacy classes in the finite general and special linear groups

Now that I’m finally done with school for the summer, I’d like to get back into the routine of blogging regularly. If you were following last summer: I never completed my project of understanding the Weil representation, so I probably won’t be continuing that series of posts. I may be helping some people complete that project this summer, in which case I can hopefully link to some further information eventually.

This week I’m going to give a detailed description of the conjugacy classes in $\text{GL}_2(\mathbb{F}_q)$ and $\text{SL}_2(\mathbb{F}_q)$, where $\mathbb{F}_q$ is the finite field with $q$ elements. This is relevant to representation theory because the conjugacy classes in a finite group correspond bijectively to irreducible representations, and in particular we will find out how many irreducible representations these groups have. A quick Google search reveals that it is easy to find the final answers, but somewhat harder to find a careful explanation, which is what I will attempt now.

First, the general linear group: for any $A \in \text{GL}_2(\mathbb{F}_q)$, consider the $\mathbb{F}_q[t]$-module $\mathbb{F}_q^2$ where $t$ acts by $A$. Two matrices are conjugate if and only if the corresponding modules are isomorphic, and it is easy to analyze these isomorphism classes using the structure theorem for principal ideal domains. Note that since we are counting invertible matrices, we need only consider polynomials with nonzero constant term.

• The nonzero scalar matrices are precisely the center of $\text{GL}_2(\mathbb{F}_q)$, so these account for $q-1$ conjugacy classes with one element each.
• For each $\lambda,\mu \in \mathbb{F}_q^{\times}$ such that $\lambda \neq \mu$, there is the semisimple conjugacy class of matrices with minimal polynomial $(x-\lambda)(x-\mu)$: the centralizer of such a matrix is a split maximal torus $\mathbb{F}_q^{\times} \times \mathbb{F}_q^{\times} \subset \text{GL}_2(\mathbb{F}_q)$, so each of these $\frac12 (q-1)(q-2)$ conjugacy classes has $q^2+q$ elements.
•  For each $\lambda \in \mathbb{F}_q^{\times}$ there is a conjugacy class of matrices with minimal polynomial $(x-\lambda)^2$ which are not semisimple, and hence conjugate to a Jordan block. If we write a Jordan block as $\lambda I + N$, where $N$ is the nilpotent matrix defined by $Ne_1 = 0$ and $Ne_2 = e_1$, it is easy to see that the centralizer consists of matrices of the form $a I + bN$ where $a,b \in \mathbb{F}_q$ and $a \neq 0$. Thus each of these $q-1$ conjugacy classes has $q^2 - 1$ elements.
•  Finally, there are the matrices which have no eigenvalue in $\mathbb{F}_q$, and therefore have a conjugate pair of eigenvalues $\alpha,\alpha^q \in \mathbb{F}_{q^2} \setminus \mathbb{F}_q$. Such matrices are semisimple because $\mathbb{F}_q$ is perfect, so their conjugacy class is determined by their eigenvalues, and in particular we see that there are $\frac12 (q^2 - q)$ conjugacy classes of these matrices. If $A \in \text{GL}_2(\mathbb{F}_q)$ has eigenvalue $\alpha \in \mathbb{F}_{q^2} \setminus \mathbb{F}_q$, then the subalgebra $\mathbb{F}_q[A] \subset \text{Mat}_2(\mathbb{F}_q)$ is isomorphic to $\mathbb{F}_{q^2}$. If we use the basis $1,\alpha$ to identify $\mathbb{F}_{q^2}$ with $\mathbb{F}_q \oplus \mathbb{F}_q$, we get an isomorphism $\text{Mat}_2(\mathbb{F}_q) \cong \text{End}_{\mathbb{F}_q}(\mathbb{F}_{q^2})$, and here $\mathbb{F}_q[A]$ corresponds to $\text{End}_{\mathbb{F}_{q^2}}(\mathbb{F}_{q^2}) \cong \mathbb{F}_{q^2}$. The centralizer of this subalgebra is $\mathbb{F}_{q^2}^{\times}$, so we see that the centralizer of $A$ in $\text{GL}_2(\mathbb{F}_q)$ is isomorphic to the non-split torus $\mathbb{F}_{q^2}^{\times}$ and in particular the conjugacy class of $A$ has $q^2 - q$ elements.

Note that the total number of conjugacy classes of $\text{GL}_2(\mathbb{F}_q)$ is

$q-1 + \frac12 (q-1)(q-2) + q-1 + \frac12 (q^2 - q) = q^2 - 1.$

As for $\text{SL}_2(\mathbb{F}_q)$, we first find the $\text{GL}_2(\mathbb{F}_q)$-conjugacy classes in $\text{SL}_2(\mathbb{F}_q)$ and then determine how they split into $\text{SL}_2(\mathbb{F}_q)$-conjugacy classes. Unfortunately, we must now keep track of whether $q$ is even or odd.

• The center of $\text{SL}_2(\mathbb{F}_q)$ is trivial if $q$ is even or $\{ \pm I \}$ if $q$ is odd. Hence this accounts for one conjugacy class if $q$ is even or two if $q$ is odd, with one element each in either case.
• For each $\lambda \in \mathbb{F}_q^{\times}$ with $\lambda \neq \pm 1$, there is the semisimple conjugacy class of matrices with minimal polynomial $(x-\lambda)(x-\lambda^{-1})$. If $q$ is even then there are $\frac12 (q-2)$ of these conjugacy classes, and if $q$ is odd then there are $\frac12 (q-3)$. We already saw that the stabilizer of such a matrix in $\text{GL}_2(\mathbb{F}_q)$ is a split maximal torus, so each conjugacy class has $q^2+1$ elements.
• There are matrices with minimal polynomial $(x \pm 1)^2$ which are not semisimple, and hence conjugate to a Jordan block. If $q$ is even then there is only one such$\text{GL}_2(\mathbb{F}_q)$-conjugacy class, and if $q$ is odd then there are two. We saw that the stabilizer in $\text{GL}_2(\mathbb{F}_q)$ of such a matrix has $q(q-1)$ elements, so these $\text{GL}_2(\mathbb{F}_q)$-conjugacy classes contain $q^2-1$ matrices each.
• The conjugacy classes of matrices which have no eigenvalue in $\mathbb{F}_q$ are parameterized by conjugate pairs $\alpha,\alpha^q \in \mathbb{F}_{q^2} \setminus \mathbb{F}_q$ where $\alpha^{q+1} = 1$. The latter equation has $q+1$ solutions in the cyclic group $\mathbb{F}_{q^2}^{\times}$, and if $q$ is even only one of those solutions comes from $\mathbb{F}_q$, while if $q$ is odd then two do. Thus there are $\frac12 q$ such conjugacy classes if $q$ is even and $\frac12 (q-1)$ if $q$ is odd. As we saw, the stabilizers in $\text{GL}_2(\mathbb{F}_q)$ of these matrices are non-split maximal tori, so each of these conjugacy classes has $q^2-q$ elements.

So we have described the  $\text{GL}_2(\mathbb{F}_q)$-conjugacy classes in  $\text{SL}_2(\mathbb{F}_q)$, but it remains to see how these split as $\text{SL}_2(\mathbb{F}_q)$-conjugacy classes. We will show momentarily that semisimple $\text{GL}_2(\mathbb{F}_q)$-conjugacy classes in $\text{SL}_2(\mathbb{F}_q)$ do not split further as $\text{SL}_2(\mathbb{F}_q)$-conjugacy classes, and here the only non-semisimple matrices are conjugate to one of the Jordan blocks $\pm I + N$ (where $N$ is the nilpotent matrix mentioned earlier). Let’s write $G = \text{GL}_2(\mathbb{F}_q)$ and $H = \text{SL}_2(\mathbb{F}_q)$ for the moment to improve the notation. Now if $X$ is the $G$-conjugacy class in $H$ of $A \in H$, then $X \cong G/C_G(A)$ as $G$-sets and in particular as $H$-sets. In particular we get a bijection $X/H \cong G/HC_G(A) \cong \mathbb{F}_q^{\times}/\text{det} C_G(A)$. We saw earlier that if $A$ is a Jordan block then $C_G(A)$ consists of matrices of the form $aI + bN$ with $a,b \in \mathbb{F}_q$ and $a \neq 0$, so $\text{det} C_G(A) = \mathbb{F}_q^{\times 2}$ is the subgroup of squares. Thus if $q$ is odd then the two $G$-conjugacy classes of $\pm I + N$ split into two $H$-conjugacy classes with $\frac12 (q^2-1)$ elements each, and if $q$ is even then the $G$-conjugacy class of $I + N$ does not split further as an $H$-conjugacy class. We see now that if $q$ is odd then $\text{SL}_2(\mathbb{F}_q)$ has

$2 + \frac12 (q-3) + 4 + \frac12 (q-1) = q+4$

conjugacy classes, and if $q$ is even then the number is

$1 + \frac12 (q-2) + 1 + \frac12 q = q+1.$

It remains to show that if $X \subset H$ is a semisimple $G$-conjugacy class, then $X$ does not split further as an $H$-conjugacy class. This is true for $G = \text{GL}_n(F)$ and $H = \text{SL}_n(F)$ where $n \geq 1$ is arbitrary and $F$ is any field with the property that the norm map $N_{E/F} : E^{\times} \to F^{\times}$ is surjective for any finite extension $E/F$. Even more generally, suppose $R$ is a finite-dimensional commutative semisimple algebra over such a field $F$, and $M$ a finite-dimensional $R$-module. Then we have the determinant map $\text{Aut}_F(M) \to F^{\times}$, and we claim the subgroup $\text{Aut}_R(M)$ surjects onto $F^{\times}$. Now $R \cong E_1 \times \cdots \times E_r$ by the semisimplicity hypothesis, where each $E_i$ is a finite field extension of $F$, so $M \cong M_1 \times \cdots \times M_r$ where each $M_i$ is an $E_i$-vector space and $R$ acts diagonally. Thus the automorphism group splits as well:

$\text{Aut}_R(M) \cong \text{Aut}_{E_1}(M_1) \times \cdots \times \text{Aut}_{E_r}(M_r)$.

It is enough to show that $\text{det} : \text{Aut}_{E_i}(M_i) \to F^{\times}$ is surjective for some $1 \leq i \leq r$, so we have reduced to the case that $R = E$ is a finite field extension of $F$. But now we can see from the definitions that the determinant $\text{Aut}_E(M) \to F^{\times}$ factors into the determinant $\text{Aut}_E(M) \to E^{\times}$ followed by the norm $E^{\times} \to F^{\times}$, and the latter is surjective by assumption. Applying this to the case when $A \in H$ is semisimple, $R = F[A] \subset \text{Mat}_n(F)$, and $M = F^n$, we have $\text{Aut}_R(M) = C_G(A)$ and the claim follows.