Exercise 1.4.7 (Order of $\mathrm{GL}_2(\mathbb{F}_p)$)

Question

Let $p$ be a prime. Prove that the order of $\mathrm{GL}_2(\mathbb{F}_p)$ is $p^4 - p^3 - p^2 + p$ (do not just quote the order formula in this section). [Substract the number of $2 	imes 2$ matrices which are {\bf not} invertible from the total number of $2 	imes 2$ matrices over $\mathbb{F}_p$. You may use the fact that a $2 	imes 2$ matrix is not invertible if and only if one row is a multiple of the other.]

Richard Ganaye · Accepted Answer

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\begin{proof} The total number of $2 	imes 2$ matrices over $\mathbb{F}_p$ is $N = p^4$ (we choose the $4$ coefficients of the matrix in $\F_p$).

Now we count the number of matrices $M = \begin{pmatrix} a & b \ c &d \end{pmatrix} \in \mathrm{GL}_2(\mathbb{F}_p)$ which are {\bf not} invertible.

\begin{itemize}
\item If the first row is $(0,0)$, then $M = \begin{pmatrix} 0 & 0 \ c &d \end{pmatrix}$ is not invertible. There are $p^2$ choices for $c$ and $d$, so there are $N_1 = p^2$ such matrices

\item If the first row is $(a,b)
e (0,0)$, there are $p^2 - 1$ choices for $(a,b)$.

For each of theses choices, $M$ is not invertible if and only if there is some $\lambda \in \F_p$ so that $(c,d) = \lambda (a,b)$. There are $p$ choices for $\lambda$, so there are $N_2 = (p^2 - 1) p$ matrices $M$ not invertible in this case.

\end{itemize}
So the number of invertible matrices is
$$|\mathrm{GL}_2(\mathbb{F}_p)| = N - (N_1+N_2)  =p^4 - [p^2 + (p^2 - 1) p] = p^4 - p^3 - p^2 +p.$$
\end{proof}

Note: Alternatively,  the number of invertible matrices is the product of the number of choices for a nonzero first row $(a, b)$, that is $p^2 - 1$, by the number of choices of a  second row $(c,d)$ non collinear to $(a,b)$, among $p^2 - p$, so that
$$|\mathrm{GL}_2(\mathbb{F}_p)| = (p^2 - 1)(p^2 - p) = p^4 - p^3 - p^2 + p.$$

Exercise 1.4.7 (Order of $\mathrm{GL}_2(\mathbb{F}_p)$)

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Exercise 1.4.7 (Order of $\mathrm{GL}_2(\mathbb{F}_p)$)

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