Normal subgroups in $GL_2(\mathbb F_3)$

Question

I'm searching for all the normal subgroups of $G:=GL_2(\mathbb F_3)$.

Till now I found $N:=SL_2(\mathbb F_3)$ subgroup of index $2$, $Q_8$ subgroup of index $6$ and the center $Z:=Z(G)=\{\pm\mathbb I_2\}$ of index $24$.

First question: are there other normal subgroups?

My attempt: let $M\unlhd G$. Then we have to separate two cases: $Z\le M$ and $Z\nleq M$. In the first case $M$ must be one of the subgroup I've already written. In the latter case, first we note that $N, M\unlhd G\Rightarrow M\cap N\unlhd G\Rightarrow M\cap N\unlhd N$. But $Z=\{\pm\mathbb I\}$ and we know that $g=-\mathbb I$ is the only element of order $2$ in $N$. Hence the condition $Z\nleq M$ means that $M$ doesn't contain any element of order $2$ and so neither $M\cap N$ can contain any element of order $2$. But $|N|=3\cdot2^3$ hence by Cauchy we have that $M\cap N$ must have oder $3$, i.e. $M\cap N\in\operatorname{Syl}_3(N)$. So we would have only one $3$-Sylow in $N$ and this is absurd since we know that $n_3(N)=4$.

Thus we conclude that a normal subgroup in $G$ must necessarely be in $\{Z,Q_8,N\}$.

Am I right? Is this correct?

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The following problem is already solved, so you can not look at that.

EDIT 2: we know that $\overline N:=N/Z\simeq A_4$; then $G/Z:=\overline G$ acts on $\overline N$ with conjugation: $$ \gamma:G/Z\longrightarrow\operatorname{Aut}(N/Z)\\xZ\longmapsto\gamma_{xZ}:nZ\mapsto n^xZ $$ and $\ker(\gamma)=C_{\overline G}(\overline N)=:\overline C$. I have to prove that $|\overline C|=1$.

Now $\;\overline C\cap\overline N=Z(\overline N)=Z(A_4)=1$. I know that $\overline N$ contains $8$ element of order $3$ (which corresponds to the $8$ element of order $3$ in $N$ and the $8$ element of order $6$ in $N$) so I deduced that $\overline N$ can't contain any element of order $3$ (but in order to make this precise as it should be, I'd have to prove that $G$ doesn't contain any other element of order $3$ or $6$). Hence $|\overline C|\in\{1,2,4,8\}$ (in fact $|\overline G|=3\cdot2^3$).

Now I'm able to show that $|\overline C|=2$ leads to a contradiction (but it's long and I won't write it).

Second question: how can I show that $|\overline C|\neq4,8$?

Moreover it's clear that $|\overline G:\overline N|=2$ but I can't know how to use it.

Thank you all

Answer 1

You have found all the normal subgroups of $G = \operatorname{GL}_2(\mathbb{F}_3)$.

Here is an outline for one possible proof that they are the only ones.

Now $|G| = 2^4 \cdot 3$.

1. $G$ has $4$ Sylow $3$-subgroups, hence there are no normal subgroups of order $3$.

2. A subgroup of order $6$ is the normalizer of a $3$-Sylow, hence there are no normal subgroups of order $6$.

3. $G$ has $3$ Sylow $2$-subgroups, so there are no normal subgroups of order $2^4$.

4. Using the conjugation action on $2$-Sylows we see that $G$ has a unique normal subgroup $Q$ of order $2^3$.

5. If $N$ is a normal subgroup of order $2^2$, then $G/N \cong A_4$. But $Q/N$ is central in $G/N$ and $A_4$ has trivial center, so this is absurd.

6. A normal subgroup of order $2$ is central, so $Z(G)$ is the unique normal subgroup of order $2$.

7. If $N$ is a normal subgroup of order $12$, then $N \cong A_4$. But then $N$ has a characteristic subgroup of order $2^2$, which would also have to be normal in $G$.

8. If $N$ is a normal subgroup of order $24$, then $A^2 \in N$ for all $A \in G$. Conclude that $\pm I \in N$. Use the characteristic polynomial to show that $\operatorname{SL}_2(\mathbb{F}_3) \leq N$ and thus $N = \operatorname{SL}_2(\mathbb{F}_3)$.