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?

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

  • 1
    $\begingroup$ There are no other proper nontrivial normal subgroups, so you have found them all. That is not exactly easy to prove, but you should at least see how far you can get. Can you rule out any specific orders as possible orders of normal subgroups for example? $\endgroup$ – Derek Holt Jun 5 '14 at 15:08
  • $\begingroup$ You already solved your second question in math.stackexchange.com/questions/820704 $\endgroup$ – Derek Holt Jun 5 '14 at 15:59
  • 1
    $\begingroup$ You proved there that $C_{S_4}(A_4)=1$, so $A_4$ has trivial centre. You cannot have $|\overline{C}|=4$ or $8$, since that would imply that $\overline{C} \cap A_4 \le Z(A_4)$ was nontrivial. $\endgroup$ – Derek Holt Jun 5 '14 at 16:45
  • $\begingroup$ No because if I use the result in my link, I should suppose to know that $\overline G\simeq S_4$... but I can't because this is what I (really) want to prove (the second question is my only problem I have in proving that $\overline G\simeq S_4$)! $\endgroup$ – Joe Jun 5 '14 at 16:47
  • $\begingroup$ Yeah now it's clear! $\endgroup$ – Joe Jun 5 '14 at 16:48

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 $N = \operatorname{SL}_2(\mathbb{F}_3)$.

  • $\begingroup$ Thank you so much spin, your answer is very useful. But, don't you think we can show that there are no other normal subgroups using what I wrote in my Edit 1? $\endgroup$ – Joe Jun 6 '14 at 9:03
  • $\begingroup$ @Joe: I don't know, I have not taken a look at that. I wrote this answer before those edits. $\endgroup$ – spin Jun 6 '14 at 13:42
  • $\begingroup$ However thank you! You gave me precious informations! $\endgroup$ – Joe Jun 6 '14 at 15:13

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