-1

In this question, they show that the order of $GL_2(\mathbb{F}_p)$ is $(p^2-1)(p^2-p)$.

For the first column, there are $p^2$ options, and we need to exclude the $0$ column, so there are $p^2-1$. That's clear to me.

For the second column, there are again $p^2$ possible combinations, and that answer says we need to exclude the $p$ different scalar multiples of the first column. That's not clear to me; for example if the first column is $[1 \quad 1]^t$, then there are $p$ different scalar multiples, namely, $[0\quad0]^t,[1\quad1]^t,[2\quad 2]^t,...,[p\quad p]^t$. However, if the first column is $[1\quad p]^t$, what are the $p$ combinations that need to be excluded and why do they need to be excluded?

Kenta S
  • 16,151
  • 15
  • 26
  • 53
Schach21
  • 700

1 Answers1

0

As suggested in your comment, we'll take the first column as $[1\quad 4]^t$ and $p>4$.

We should note that the matrix $\begin{bmatrix}1&1\\4&4\end{bmatrix}$ has determinant zero, so is not in the group $GL_2(\mathbb{F}_p)$. Similarly, if the second column is any multiple of the first column, say $n$ times the first column, then we see that the matrix $\begin{bmatrix}1&n\\4&4n\end{bmatrix}$ also has determinant zero, so is not in the group $GL_2(\mathbb{F}_p)$.

Therefore we are excluding one matrix for each $n\in\mathbb{F}_p$, of which there are $p$ options.

ndhanson3
  • 1,365