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If $F$ is a finite field of $q$ elements and $G$ is the group of all automorphisms $\sigma$ of the field of rational functions $F(x)$ such that $\sigma$ is given by $$\sigma(x) = \frac{ax + b}{cx + d}$$ with $a, b, c, d\in F$ and $ad - bc \neq 0$, then how does one prove that the order of $G$ is $q^3 - q$?

I understand that the order of $G$ is bounded by $q^4$, and I understand the general strategy of "counting" the number of quadruples $(a, b, c, d) \in F^4$ such that $ad - bc \neq 0$, and I have tried working with some examples, namely $q = 2, 3$, but I am blanking on how to get a general proof/computation for the general case.

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Start by determining the number of matrices $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ with $ad-bc\neq 0$: there are $q^2-1$ choices for the first column (any non-zero vector in $F^2$ will do), and then $q^2-q$ choices for the second column, as it can be any vector not in the span of the first column.

So there are $(q^2-1)(q^2-q)$ matrices with $ad-bc\neq 0$, but any two matrices that differ by a non-zero scalar define the same transformation $\sigma$, and conversely if two matrices define the same transformation then they differ by a non-zero scalar. So dividing by $q-1$, it follows that there are $$\frac{(q^2-1)(q^2-q)}{q-1}=q(q^2-1)=q^3-q$$ different transformations $\sigma$.