 Solution

An element of the general linear group over a finite field is an invertible $2\times 2$ matrix. Its columns must form an ordered basis of ${\mathbb{F}}_p^2$. There are $p^2-1$ choices for the first column. Its linear span contains $p$ vectors, so there are $p^2-p$ choices for a second column outside that linear span. Thus the order of a general linear group over a finite field isThe determinant is a surjective group homomorphism $G{L}_2({\mathbb{F}}_p)\to {\mathbb{F}}_p^{\times }$: for each $a\ne 0$, the diagonal matrix $\mathrm{diag}(a,1)$ has determinant $a$. Its kernel of a group homomorphism is the special linear group over a finite field. The first isomorphism theorem gives $|G{L}_2|/|S{L}_2|=p-1$, hence