Consider $G:=GL_2(\mathbb F_3)$. I have to extrapolate as much information about it as I can. Without computations.
First of all: I think someone else has already done this before me, hence if you know where to find some good pdf, please give me the link!
Now:
$|G|=(3^2-1)(3^2-3)=48=3\cdot2^{4}$
We know that $N:=SL_2(\mathbb F_3)\unlhd G$; in fact $N=\ker(\det)$, where $\det:G\rightarrow \mathbb F_3^{\times}$. Being $\det$ surjective, we have that $G/N\simeq\mathbb F_3^{\times}$, from which $|N|=24=3\cdot2^3$.
My problems are:
$\bullet$ Call $n_2(N)$ the number of $2$-Sylows of $N$; we know that $n_2(N)\equiv1 (mod\:2)$ and $n_2(N)|3$ hence $n_2(N)\in\{1,3\}$. But I can't see how to determine $n_2(N)$ exactly. I can't use the fact that $G$ is nilpotent (in which case i know that every $p$-Sylow must be normal, hence it's unique).
$\bullet$ Call $Q$ one of this/these $2$-Sylow(s). How can I prove that it's normal in $N$ and in $G$? Clear that if $n_2(N)=1$, since all the $p$-Sylows are conjugate, then it will be $Q\unlhd N$. Suppose to have proved this. How can we use this to prove that $Q\unlhd G$?
Can someone help me? Thank you all!
It is easier to write down specific Sylow $3$-subgroups of $N$ than Sylow $2$-subgroups, so let's do that, and see where it get us.
One such is $$ \left\{ \left(\begin{array}{rr}1&0\\0&1\end{array}\right), \left(\begin{array}{rr}1&1\\0&1\end{array}\right), \left(\begin{array}{rr}1&-1\\0&1\end{array}\right)\right\},$$ and you get another one by using lower triangular rather than upper triangular matrices. So we have more than one Sylow $3$-subgroup and, hence, by Sylow's Theorem, there must be $4$.
With small groups, it often helps to count numbers of elements of each order. Each of the four Sylow $3$-subgroup has two elements of order $3$, so we have $8$ elements of order $3$ altogether. They all commute with the central element $-I$, and by multiplying each of these $8$ elements of order $3$ by $-I$, we get $8$ elements of order $6$. Since $|{\rm SL}(2,3)| = 24$, we only have $8$ remaining elements. Since no element of order $3$ or $6$ can lie in a Sylow $2$-subgroup, these $8$ remaining elements must form the unique Sylow $2$-subgroup $Q$ of $N$. So $Q \lhd N$.
Try and prove that $Q \lhd G$ yourself.