Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.4 Exercise 1.4.1
Prove that $|GL_2(\mathbb{F}_2)| = 6$.
Solution: $GL_2(\mathbb{F}_2$ consists of precisely those $2 \times 2$ matrices over $\mathbb{F}_2$ having a nonzero determinant. Let $A = \left[ {a \atop c} {b \atop d} \right]$ be an arbitrary $2 \times 2$ matrix over $\mathbb{F}_2$. Then $$\mathsf{det}(A) = ad-bc,$$ where the arithmetic is performed in $\mathbb{Z}/(2)$. We need to find all $a, b, c, d \in \mathbb{Z}/(2)$ such that $ad – bc = 1$.
There are two cases: (1) $ad = 0$ and $bc = 1$ and (2) $ad = 1$ and $bc = 0$.
If $ad = 0$ and $bc = 1$, then$ b = c = 1$ and either $a = d = 0$, $a = 0$ and $d = 1$, or $a = 1$ and $d = 0$.
If $ad = 1$ and $bc = 0$, then $a = d = 1$ and either $b = c = 0$, $b = 0$ and $c = 1$, or $b = 1$ and $c = 0$.
This gives the following six matrices: $\left[ {1 \atop 0}{0 \atop 1} \right]$, $\left[ {1 \atop 0}{1 \atop 1} \right]$, $\left[ {1 \atop 1}{0 \atop 1} \right]$, $\left[ {0 \atop 1}{1 \atop 0} \right]$, $\left[ {1 \atop 1}{1 \atop 0} \right]$, and $\left[ {0 \atop 1}{1 \atop 1} \right]$.
