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?
Best Answer
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 $\operatorname{SL}(2,\mathbb F)$.