---
tag: 0JQ1KNQW2K
kind: statement
role: theorem
name: Order of the general linear group over a finite field
msc: [20G40]
uses: [0SEX9F9J5Z]
sources:
  - url: https://en.wikipedia.org/wiki/General_linear_group
    locator: Over finite fields
  - url: https://en.wikipedia.org/wiki/General_linear_group
    locator: History
---

## Statement

Let $q$ be a prime power, let $\mathbb{F}_q$ be the finite field with $q$ elements, and let $n \geq 1$.
The group $\mathrm{GL}_n(\mathbb{F}_q)$ of the invertible $n \times n$ matrices over $\mathbb{F}_q$ has [[order|0B85S2V2DR]] $\prod_{k=0}^{n-1} (q^n - q^k) = (q^n - 1)(q^n - q) \cdots (q^n - q^{n-1})$.

## Remarks

A matrix is invertible exactly when its columns are linearly independent.
There are $q^n - 1$ choices for the first column. Column $k+1$ has $q^n - q^k$ choices, because it must avoid the $q^k$ vectors in the span of the first $k$ columns.
For example, $|\mathrm{GL}_2(\mathbb{F}_2)| = 3 \cdot 2 = 6$, and $\mathrm{GL}_2(\mathbb{F}_2) \cong S_3$.
Also, $|\mathrm{GL}_3(\mathbb{F}_2)| = 7 \cdot 6 \cdot 4 = 168$.

## History

Évariste Galois computed the order of the general linear group over a prime field in 1832, in his last letter to Chevalier.

## Explanations

### L1 human

To build an invertible matrix, choose its columns one after the other.
Each new column must not lie in the span of the columns that you already chose.
Counting the choices at each step gives the product formula.