# Action of $\text{PSL}(2,q)$ on the projective line: How to know if $(az+b)/(cz+d)$ belongs to $\text{PSL}(2,q$)?

The special linear group $$\text{PSL(2,\mathbb{F}) }$$ over the finite field $$\mathbb{F}$$ acts on the projective line $$\mathbb{F}\cup \{\infty \}$$ by the way $$\text{z \to }\frac{a\cdot z+b}{c\cdot z+d}$$

However, how to know if this transformation belongs to $$\text{PSL(2,\mathbb{F})}$$ in the case $$-\mathbf{1}$$ is a square in $$\mathbb{F}$$?

So, let $$-\mathbf{1}=\alpha ^2$$ and let, for instance, $$\text{z \to }\frac{1}{z}=\frac{\alpha }{\text{\alpha z}}$$. Although it is the same transformation in both cases, the first case represent the matrix with determinant $$-\mathbf{1}$$, and the second with determinant $$\mathbf{1}$$. Does this example belong to $$\text{PSL(2,\mathbb{F}) }$$ or not?

It does belong to $$\operatorname{PSL}(2,\mathbb F)$$. The matrices $$M=\begin{pmatrix}a&b\\ c&d\end{pmatrix}\in \operatorname{SL}(2,\mathbb F)$$ and $$\alpha M\in\operatorname{GL}(2,\mathbb F)$$ induce the same element of $$\operatorname{PGL}(2,\mathbb F)\geq\operatorname{PSL}(2,\mathbb F)$$, and it happens to be in $$\operatorname{PSL}(2,\mathbb F)$$. It's enough that there is one matrix in $$\operatorname{SL}(2,\mathbb F)$$ which induces the given transformation. Not every matrix in $$\operatorname{GL}(2,\mathbb F)$$ which induces the same transformation has to be in $$\operatorname{SL}(2,\mathbb F)$$.