Thread: The Size of GL_2(Z_7)

1. The Size of GL_2(Z_7)

Hello everyone. I am asked to find the size of the group $G=GL_2(\mathbb{Z}_7)$ and would like to verify my work.

This group acts on the vector space $\mathbb{Z}_7\times \mathbb{Z}_7$ with basis $\{(1,0), (0,1)\}$. From this, we have the orbit-stabilizer theorem: if $x\in \mathbb{Z}_7\times \mathbb{Z}_7$, then $|G|=|Gx|\cdot |G_x|$, that is, the size of $G$ is the size of the orbit of $x$ times the size of the stabilizer of $x$. Take $x=(1,0)$. If $A\in G$, $Ax$ can possibly be any nonzero vector. There are 48 of these. Therefore, $|Gx|=48$. Then, if $Ax=x$, we must have $A(0,1)$ equal to another vector not a multiple of $x$, since $A$ is invertible. In other words, $A(0,1)$ cannot be $(0,0), (1,0), (2,0),\ldots,(6,0)$. There are 42 of these, so $|G_x|=42$.

We conclude that $|G|=48\cdot 42=2016$.

2. Originally Posted by roninpro
Hello everyone. I am asked to find the size of the group $G=GL_2(\mathbb{Z}_7)$ and would like to verify my work.

This group acts on the vector space $\mathbb{Z}_7\times \mathbb{Z}_7$ with basis $\{(1,0), (0,1)\}$. From this, we have the orbit-stabilizer theorem: if $x\in \mathbb{Z}_7\times \mathbb{Z}_7$, then $|G|=|Gx|\cdot |G_x|$, that is, the size of $G$ is the size of the orbit of $x$ times the size of the stabilizer of $x$. Take $x=(1,0)$. If $A\in G$, $Ax$ can possibly be any nonzero vector. There are 48 of these. Therefore, $|Gx|=48$. Then, if $Ax=x$, we must have $A(0,1)$ equal to another vector not a multiple of $x$, since $A$ is invertible. In other words, $A(0,1)$ cannot be $(0,0), (1,0), (2,0),\ldots,(6,0)$. There are 42 of these, so $|G_x|=42$.

We conclude that $|G|=48\cdot 42=2016$.

Personally I don't see why you can't just count the possibilities based on the columns. In general $\displaystyle \text{GL}_n\left(\mathbb{F}_p\right)=\prod_{j=0}^{ n-1}\left(p^n-p^j\right)$. So, $\left|\text{GL}_2\left(\mathbb{Z}_7\right)\right|= \left(7^2-1\right)\left(7^2-7\right)=2016$.