Finding a normal subgroup in $G := GL_2( \mathbb F_7)$

Question

For now, i know Sylow Groups and the theorem of structure for abelian groups.

$G := GL_2( \mathbb F_7)$, $|G|=2^5 \cdot 3^2 \cdot 7 $

I m trying to show that:

There exists a normal sub group in $G$ of order $2^5 \cdot 3 \cdot 7$.

I have no clue at all. Any idea?

Answer 1

Define $\varphi:G\to \mathbb{F_7}^\times$ by $\varphi(A)=detA$. This is a homomorphism, its image is $\mathbb{F_7}^\times\cong\mathbb{Z_6}$. Hence by an isomorphism theorem $G/Ker\varphi\cong\mathbb{Z_6}$. That means $G/Ker\varphi$ has a normal subgroup of index $3$. And now from the correspondence theorem we get that $G$ must have a normal subgroup of index $3$. (a subgroup which contains $Ker\varphi$, but it isn't important for your exercise)

Answer 2

First consider $det:GL_2(7)\rightarrow \mathbb{F}_7^\times$, where $det$ is just the determinate of the matrix so for $\begin{bmatrix}a&b\\ c&d\end{bmatrix}$ it would be $ad-bc$.

The kernel of $det$ is $SL_2(7)$ which contains matrices of determinant $1$. The map $det$ is a surjection onto $\mathbb{F}_7^\times$ which has order $6$ so we conclude that $|GL_2(7):SL_2(7)|=6$. Clearly we need to find an overgroup of $SL_2(7)$ to find a subgroup of index $3$ of $GL_2(7)$.

Indeed we can, if we consider the group $G$ consisting of matrices with determinant equal to $1$ or $-1\mod 7$. This is a proper subgroup (exercise) and is of index $3$.

Hint for exercise:

For the exercise consider what happens to the determinant when you multiply elements of determinant $1$ or $-1$.

It remains to show that $G$ is normal in $GL_2(7)$. We will do do this using the determinant map, let $h\in GL_2(7)$ and $g\in G$, then $det(hgh^{-1})=det(g)=1$ or $-1$ and so we conclude $g\in G$. In particular $G$ is normal.

Answer 3

Consider the set $H$ of elements $\pi \in GL_2(\mathbb{F}_7)$ s.t. det$(\pi) \in \{1,6\}$ [note that $\{1,6\}$ is an index-3 subgroup of $(\mathbb{F}_7)^{\times}$. This is a group, is normal, and has index 3 in $GL_2(\mathbb{F}_7)$.

We first show that $H$ is a group:

1. Let $\pi \in H$. Then $\pi^{-1} \in GL_2(\mathbb{F}_7)$ and det($\pi^{-1})$ must be det$^{-1}(\pi)$, so det($\pi^{-1})$ must be in $\{1,6\}$ as det$(\pi)$ is. So $\pi^{-1}$ is in $H$.

2. Likewise, if $\pi$ and $\pi'$ are in $H$, then $\pi\pi'$ is in $GL_2(\mathbb{F}_7)$. Furthermore, det$(\pi\pi')$ $=$ det$(\pi)$det$(\pi')$ which is in $\{1,6\}$ as both det$(\pi)$ and det$(\pi')$ are in $\{1,6\}$.

So $H$ is a group. We now show that $H$ is normal. Letting $\sigma$ be any element in $GL_2(\mathbb{F}_7)$. Then $\sigma^{-1}\pi\sigma$ is also in $GL_2(\mathbb{F}_7)$. Furthermore, det$(\sigma^{-1}\pi\sigma)$ $=$ det$(\pi) \in \{1,6\}$ so $\sigma^{-1}\pi\sigma$ is also in $H$. Thus $H$ is normal.

Can you check that $H$ has index 3 in $GL_2(\mathbb{F}_7)$?