# Group of Invertible Matrices Over a Finite Field and its Stabilizer

## Problem 108

Let $\F_p$ be the finite field of $p$ elements, where $p$ is a prime number.
Let $G_n=\GL_n(\F_p)$ be the group of $n\times n$ invertible matrices with entries in the field $\F_p$. As usual in linear algebra, we may regard the elements of $G_n$ as linear transformations on $\F_p^n$, the $n$-dimensional vector space over $\F_p$. Therefore, $G_n$ acts on $\F_p^n$.

Let $e_n \in \F_p^n$ be the vector $(1,0, \dots,0)$.
(The so-called first standard basis vector in $\F_p^n$.)

Find the size of the $G_n$-orbit of $e_n$, and show that $\Stab_{G_n}(e_n)$ has order $|G_{n-1}|\cdot p^{n-1}$.

Conclude by induction that
$|G_n|=p^{n^2}\prod_{i=1}^{n} \left(1-\frac{1}{p^i} \right).$

## Proof.

Let $\calO$ be the orbit of $e_n$ in $\F_p^n$.
We claim that $\calO=\F_p^n \setminus \{0\}$, hence
$|\calO|=p^n-1.$

To prove the claim, let $a_1 \in \F_p^n$ be a nonzero vector.
Then we can extend this vector to a basis of $\F_p^n$, that is, there is $a_2, \dots, a_n \in \F_p^n$ such that $a_1,\ a_2, \dots, a_n$ is a basis of $\F_P^n$.
Since they are a basis the matrix $A=[a_1 \dots a_n]$ is invertible, that is , $A \in G_n$.
We have
$Ae_n=a_1$ Thus $a_1\in \calO$. It is clear that $0 \not \in \calO$. Thus we proved the claim.

Next we show that
$|\Stab_{G_n}(e_n)|=|G_{n-1}|\cdot p^{n-1}. \tag{*}$ Note that $A \in \Stab_{G_n}(e_n)$ if and only if $A e_n=e_n$.
Thus $A$ is of the form
$\left[\begin{array}{r|r} 1 & A_2 \\ \hline \mathbf{0} & A_1 \end{array} \right],$ where $A_1$ is an $(n-1)\times (n-1)$ matrix, $A_2$ is a $1\times (n-1)$ matrix , and $\mathbf{0}$ is the $(n-1) \times 1$ zero matrix.
Since $A$ is invertible, the matrix $A_1$ must be invertible as well, hence $A_1 \in G_{n-1}$.
The matrix $A_2$ can be anything.
Thus there are $|G_{n-1}|$ choices for $A_1$ and $p^{n-1}$ choices for $A_2$.
In total, there are $|G_{n-1}|p^{n-1}$ possible choices for $A \in \Stab_{G_n}(e_n)$. This proves (*).

Finally we prove that
$|G_n|=p^{n^2}\prod_{i=1}^{n} \left(1-\frac{1}{p^i} \right)$ by induction on $n$.

When $n=1$, we have
$|G_1|=|\F_p\setminus \{0\}|=p-1=p\left(1-\frac{1}{p} \right).$

Now we assume that the formula is true for $n-1$.
By the orbit-stabilizer theorem, we have
$|G_n: \Stab_{G_n}(e_n)|=|\calO|.$ Since $G_n$ is finite, we have
\begin{align*}
|G_n|&=|\Stab_{G_n}(e_n)||\calO|\\
&=(p^n-1)|G_{n-1}|p^{n-1}\\
&=(p^n-1)p^{n-1}\cdot p^{(n-1)^2}\prod_{i=1}^{n-1} \left(1-\frac{1}{p^i} \right) \text{ by the induction hypothesis}\\
&=p^n\left(1-\frac{1}{p^n} \right)p^{n-1}p^{n^2-2n+1} \prod_{i=1}^{n-1} \left(1-\frac{1}{p^i} \right) \\
&=p^{n^2}\prod_{i=1}^{n} \left(1-\frac{1}{p^i} \right).
\end{align*}
Thus the formula is true for $n$ as well.

By induction, the formula is true for any $n$.

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