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!

• Could you explain why you don't want to compute anything? Computations are pretty good at showing you, among many other things, what the center is and how many Sylow subgroups there are. – Siddharth Venkatesh May 29 '14 at 1:49
• For the third problem, you can note that $G = Z(G)N$ (or just note that it is $HN$, where $H = \{\pm 1\}.)$ Hence, any normal subgroup of $N$ is also a normal subgroup of $G$. – Siddharth Venkatesh May 29 '14 at 1:52
• @SiddharthVenkatesh:Thanks a lot! However these questions are part of very very long exercise that our Group Theory teacher did in class. However he runs a lot, so it's really hard to write down everything he says, hence now I'm facing a lot of troubles. The request he gave wasn't "no computation at all", but "the less computation you do, the better is the work". In general "less computations=smarter"... however, if you have a solution with some computations, write it down, I'd be happy to make you enjoy at least a +1! – Joe May 29 '14 at 2:03
• Sorry but... if $Z(G)=\{\pm\bf1\}$ then it should be $Z(G)\le N$, hence $Z(G)N=N$. Where did I wrong? – Joe May 29 '14 at 2:18
• Oops. I thought we had odd dimension. You're right. – Siddharth Venkatesh May 29 '14 at 4:29

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 $\mathbb{Z}/3\mathbb{Z}$.

• thanks a lot! But can you help me with the other two questions? – Joe May 29 '14 at 11:32