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 $|N|=24$.
Then I'm searching for elements of $N$ of order $2$: i.e. $g\in N$ s.t. $g^2=1$. So let's look at the minimal polynomial of such an element, call it $f$. It will divide $X^2-1$, so $f$ must be $X-1,\;X+1$ or $X^2-1$.
If $f(X)=X-1$ then $g=1$ so it wouldn't have order $2$. Hence $f(X)\neq X-1$.
My teacher wrote that $f(X)=X+1$ because if $f(X)=X^2-1$ then $\det(g)=-1$, from which we have a contradiction because $g\notin N$. But I can't understand why!
I thought that $G$ is a subset of the ring $\mathcal{M}_2(\mathbb F_3)$, which is not a integer domain, hence I can't argue as follows: $g-1\neq0,\;g^2-1=0\Rightarrow g+1=0$. But no more.
Can someone help me to understand why $f(X)=X^2-1\;\Longrightarrow\det(g)=-1$?
Thanks a lot!
If its minimal polynomial of an element in $SL_2(F_3)$ is $X^2-1$, then that polynomial is also its characteristic polynomial and its independent term is its determinant: this is absurd, as the element is in $SL_2(F_3)$.