Problem

Let \(p\) be a prime. Prove that the order of \(GL_2(\mathbb{F}_p)\) is \(p^4 - p^3 - p^2 + p\).

Solution

Note about the notation in this section of the text that

$$ \mathbb{F}_p = \mathbb{Z}/p\mathbb{Z} $$

Therefore, by definition of \(GL_n(F)\),

$$ GL_2(\mathbb{F}_p) = \{A \mid A \text{ is a } 2 \times 2 \text{ matrix with entries from } \mathbb{Z}/p\mathbb{Z} \text{ such that } det(A) \ne 0\} $$

Let \(A = \begin{bmatrix}a & b\\ c & d\end{bmatrix}\) be an arbitrary element of \(GL_2(\mathbb{F}_p)\). Recall that the determinant of \(A\) is defined as

$$ \mid A \mid = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc. $$

First, notice that for \(ad - bc\) to be non-zero, \(\begin{pmatrix}a \\ c\end{pmatrix}\) must be a non-zero vector. There are \(p^2 - 1\) choices for that.

Then, once \(\begin{pmatrix}a \\ c\end{pmatrix}\) is fixed, to the same goal of making \(ad - bc\) non-zero, \(\begin{pmatrix}b\\d\end{pmatrix}\) can be anything but a multiple of \(\begin{pmatrix}a\\c\end{pmatrix}\). There are \(p^2 - p\) choices for that.

Therefore,

$$ GL_2(\mathbb{F}_p) = (p^2 - 1)(p^2 - p) = p^4 - p^3 - p^2 + p. $$

$$\tag*{$\blacksquare$}$$