Show that the order of the group \(\mathrm{GL}_n(\mathbb{F}_q)\) of invertible \(n\times n\) matrices over the field \(\mathbb{F}_q\) of \(q\) elements is \((q^n-1)(q^n-q)\cdots(q^n-q^{n-1})\).
Solution
An invertible \(n\times n\) matrix over \(\FF_q\) is the same thing as an ordered basis of the vector space \(\FF_q^n\), namely its ordered list of columns.
1. The first column can be any nonzero vector, giving \(q^n-1\) choices.
There are \(q^n\) vectors in \(\FF_q^n\), and only the zero vector is forbidden.
2. After choosing \(k\) linearly independent columns, the next column has \(q^n-q^k\) choices.
The span of \(k\) independent vectors is a \(k\)-dimensional subspace, hence has \(q^k\) elements. The next column must lie outside that span.
3. Therefore\[ |\GL_n(\FF_q)|=(q^n-1)(q^n-q)(q^n-q^2)\cdots(q^n-q^{n-1}). \]
Choose the columns successively and multiply the numbers of choices from 1 and 2.