Let $G:= \Bbb{GL}_2(\Bbb{F}_3)$ be the finite group of order $48$ of invertible matrices over $\Bbb{F}_3$. Determine the center $Z(G)$, representatives $x_1,\dots,x_n$ of the conjugacy class of non central elements and the orders $|C(x_1)|,\dots,|C(x_n)|$ of the respective centralizers, and verify directly the validity of the class equation $$|G| = |Z(G)| + \sum_{i=1}^n[G:C(x_i)].$$
What I've done:
First of all i know that the center is made of the two matrices $E_2$ and $2 E_2$, because there are no other matrices that commutes with every other element of $\Bbb{GL}_2(\Bbb{F}_3)$.
I've shown that the number of conjugacy classes is the number of distinct rational canonical form in $\Bbb{GL}_2(\Bbb{F}_3)$. To prove this fact i've used a theorem from lineare algebra that says that every matrix $A\in \Bbb{GL}_2(\Bbb{F}_3)$ is similar to a matrix in rational canonical form.
The conjugacy class of an element $A \in \Bbb{GL}_2(\Bbb{F}_3)$ is $Cl(A) = \{ PAP^{-1} | P \in \Bbb{GL}_2(\Bbb{F}_3)\}$. Since the rational canonical form of a matrix is unique we know that two matrices $A,B \in \Bbb{GL}_2(\Bbb{F}_3)$ are in the same conjugacy class if and only if they have the same rational canonical form, therefore the number of conjugacy classes is the number of different rational canonical forms.
Now I've counted how many conjugacy classes there are. Namely if $A \in \Bbb{GL}_2(\Bbb{F}_3)$ is in rational canonical form then we have two possibilities: either $A$ is diagonalizable, in which case the rational canonical form would be $A= \lambda E_2,\quad \lambda \in \Bbb{GL}_2(\Bbb{F}_3)\backslash \{0\}$ (We have two possible values of $\lambda$), or $A= \begin{pmatrix} 0 &a\\1&b\end{pmatrix}$ with $a,b \in \Bbb{GL}_2(\Bbb{F}_3), a\neq 0$. The sum of the possible entries are therefore $$2+2\cdot 3 = 8.$$
Therefore there are $6$ representatives, which are not in the center, given by the matrices
\begin{equation*} \underbrace{\quad\begin{pmatrix}0&1\\1&0\end{pmatrix}}_{:=x_1}; \quad \underbrace{\begin{pmatrix}0&1\\1&1\end{pmatrix}}_{:=x_2}; \\ \underbrace{\begin{pmatrix}0&1\\1&2\end{pmatrix}}_{:=x_3}; \quad \underbrace{\begin{pmatrix} 0&2\\1&0\end{pmatrix}}_{:=x_4};\quad \underbrace{\begin{pmatrix}0&2\\1&1\end{pmatrix}}_{:=x_5}; \quad \underbrace{\begin{pmatrix} 0&2\\1&2\end{pmatrix}}_{:=x_6}. \end{equation*}
Edit (some news):
I've found on my script of Algebra the Orbit Stabilizer Theorem that states
\begin{equation}|G| = |Orb(x)| \cdot |Stab(x)|,\qquad \qquad (1)\end{equation}
where $Orb(x)$ are the orbit of the action of $G$ on $X$ and $Stab(x)$ is the stabilizer of $x\in X$. Therefore i defined the action
\begin{align*} G\times G &\to G \\(g,x)&\mapsto g\cdot x := gxg^{-1}. \end{align*}
The orbits of this action are then subsets of the form $$G \cdot x = \{g\cdot x | g\in G \}=\{gxg^{-1}|g\in G\}=Cl(x)$$
And the stabilizer of $x$ is
$$Stab(x) = \{g\in G |g\cdot x = x \iff gxg^{-1}=x \iff gx=xg\}$$
That is all the elements in $G$ that commute with $x$. Now i tried to find all the stabilizer of my representatives, in such a way that i could use $(1)$ to find the order of the orbit and hence the orbit of the conjugacy class (since they are equal).
Question:
Is this the right way to find the size of the conjugacy classes? Is there a smarter way?
