I am aware that questions on this topic are around on this site, but they all seem to require information about the group that is not available to me in this problem.
Consider the special linear group $G = SL(2,\mathbb{F_{3}})$, i.e. the group of $2 \times 2$ matrices of determinant $1$ over the field of order $3$, $\mathbb{F_{3}}$.
I want to find the number of Sylow 3-subgroups of $G$. I have so far shown that $|G|=24$ and that the centraliser of $G$, $Z(G) \neq \{1_{G}\}$. By Sylow's theorem I know that $n_{3} \in \{1,4\}$ as well as $n_{2} \in \{1,3\}$. Now if I knew how many Sylow 2-subgroups of $G$ there are, I think this problem becomes fairly easy. But I don't have this information and am stuck.
Any help would be appreciated!
There is a property who says that the number of elementes by order 3 is $n_3(3-1)$.
If $n_3$=1 then you have just 2 elements by order 3. Let consider the next matrices :
$\begin{pmatrix} \hat{1} & \hat{1} \\ \hat{0} & \hat{1} \end{pmatrix} \quad, \begin{pmatrix} \hat{1} & \hat{0} \\ \hat{1} & \hat{1} \end{pmatrix} \quad, \begin{pmatrix} \hat{2} & \hat{2} \\ \hat{1} & \hat{0} \end{pmatrix} \quad $
They are in G and have order 3. So, $n_3$=4.