On some subgroups of $GL_2(\mathbb F_3)$ 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!
Then:
$|G|=(3^2-1)(3^2-3)=48=3\cdot2^{4}$
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: 
$\bullet$ How can I determine $Z(N)$ and $Z(G)$? I read that $Z(G)=\{\pm\bf{1}\}$, but I can understand why (without computations) there aren't other elements in the center.
$\bullet$ 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.
$\bullet$ 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!
 A: Centers!
Finding centers of matrix groups can be done with linear algebra (and almost no computation). The only time I do any computation is when I claim $\operatorname{SL}_n(F)$ has an element with a one-dimensional eigenspace, but that is theoretically just a known property of Jordan blocks.
Eigenspaces are submodules for the centralizer
Let $G \leq \operatorname{GL}_n(F)$ be a matrix group over a field $F$, and let $z,g \in G$ commute, that is, $zg=gz$ or $G \in C_G(z)$. If $F$ is large enough, then $z$ has eigenvalues. Let $v$ be an eigenvector of $z$ with $zv = \lambda v$. Consider $z(gv) = (zg)v = (gz)v = g(zv) = g (\lambda v ) = \lambda (gv)$. This just says that $gv$ is also an eigenvector of $z$ with the same eigenvalue $\lambda$. In other words, each eigenspace $V_{\lambda,g} = \{ v : zv = \lambda v \}$ is a $C_G(z)$-submodule of $V$.
Big groups have elements with one-dimensional eigenspaces
Now if $G$ is big enough, then it has some element $g$ with a one-dimensional eigenspace. For instance $\operatorname{SL}_n(F)$ has a Jordan block: a matrix with constant diagonals, the main diagonal and the one right above it having 1s, and the other diagonals being 0s. If $F$ is big enough, then you could also use a diagonal matrix with one diagonal entry unique (and the others all being 1s, say), but if $F=\mathbb{Z}/2\mathbb{Z}$ there isn't actually a second entry for the diagonal matrix.
So big groups only have scalar matrices in the center
So if $z \in Z(G)$, then $z \in C_G(g)$, so $zv \in V_{1,g}$ whenever $v \in V_{1,g}$. Since $\dim(V_{1,g})=1$, $zv = \lambda v$ for some $\lambda \in F^\times$. In other words, $v \in V_{\lambda,z}$. Now $V_{\lambda,z}$ is a $G$-module (for $G=\operatorname{GL}_n(F)$ or $G=\operatorname{SL}_n(F)$), but $V$ doesn't really have many $G$-submodules, since $G$ acts transitively on the nonzero vectors. Hence $V_{\lambda,z} = V$, and $z$ acts as the scalar $\lambda$ on all of $V$. In other words, $z = \lambda I_n$.
Slightly confusing thing for your field
Over a field like $F=\mathbb{Z}/3\mathbb{Z}$, there are actually only two invertible scalars: $\pm 1$, but in general $Z(\operatorname{GL}_n(F)) \cong F^\times$ can have more than just $\pm1$.
This might be confusing, since $Z(\operatorname{SL}_n(F))$ only contains the scalars with determinant 1, that is, with $\lambda^n=1$. For $n=2$, that also gives you $Z(\operatorname{SL}_n(F))=\pm1$, but for every field, not just $\mathbb{Z}/3\mathbb{Z}$.
