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I'm having a hard time counting. I need to count the number of elements for the multiplicative group of invertible $2\times 2$ matrices $GL_{2}\left( \mathbb{F}_{p}\right) $ with elements from the field $\mathbb{F}_{p}$ where $p$ is any prime number. David S. Dummit and Richar M. Foote's Abstract Algebra (3rd edition) has a solution on page 413 but that approach is by counting the possible number of basis of vector spaces but I wanted to see if there was a direct way. The book says that $\left\vert GL_{n}\left( \mathbb{F}_{p}\right) \right\vert =\left( p^{n}-1\right) \left( p^{n}-p\right) ...\left( p^{n}-p^{n-1}\right) $ so my case reduces to $p^{4}-p^{3}-p^{2}+p$. For any matrix $\left( \begin{array}{cc} a & b \\ c & d% \end{array}% \right) $ with $a,b,c,d\in \mathbb{F}_{p}$, I understand that $p^{4}$ is the total number of 2x2 matrices and that we need to subtract the possible combinations for $ad=bc$ (non-invertible elements) but I can't account for the rest. In particular, that plus sign is absolutely baffling. Any hint would be appreciated. I'd then take this up for the general $n$x$n$ case but first I need to dispose the 2x2 case.

P.S. I tried looking for a pre-existing answer and a related answer is present here but it's by the same approach as the book and I'd like to have an approach without resorting to linear independence. This answer, too, was a case for $p=3$ but I couldn't see the pattern.

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Coward
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    For $n\gt 2$, the determinant is difficult to work with, while going row by row making sure the rows are linearly independent works smoothly. – André Nicolas Aug 18 '14 at 05:57
  • @AndréNicolas, yes the details will get messy for the generalised determinant formula.. I'll see what I can do. Thanks for the heads-up! – Coward Aug 19 '14 at 11:28

1 Answers1

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How many quadruples $(a,b,c,d)$ are there in which $ad=bc$?

Case 1, $a=0$. Then $d$ is arbitrary and $b$ or $c$, or both, must be zero. There are $p$ choices for $d$ and $2p-1$ for $b$ and $c$.

Case 2, $a\ne0$. Then $b,c$ are arbitrary and $d=bc/a$. There are $p-1$ choices for $a$ and $p^2$ for $b$ and $c$.

Total, $$p(2p-1)+(p-1)p^2=p^3+p^2-p\ ,$$ and as you have already pointed out you must now subtract this from $p^4$.

BTW for the general case I doubt that you will be able to avoid the argument involving linear independence.

David
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