On wikipedia:Möbius transformation in the linked paragraph it is shown that $$ PGL(2, \mathbb C) \cong \mbox{Aut}(\hat{\mathbb C}) \quad \mbox{and} \quad PSL(2, \mathbb C) \cong \mbox{Aut}(\hat{\mathbb C}) $$ hence we would have $PGL(2, \mathbb C) \cong PSL(2,\mathbb C)$. I am new to this stuff, but this makes me wondering. As my algebra book does not mentions anything like that for the general groups $PSL(2, \mathbb F), PGL(2,\mathbb F)$ it discusses.
As I see it, the map from $SL(2, \mathbb F)$ (i.e. the matrices with determinant one) give exactly those transformations where $ad - bc = 1$, and on the side of the mappings these are all those mappings where $ad - bc $ is a square (i.e $=u^2$ for some $u \in \mathbb F$, as by multiplying the nominator and denominator by $1/u$ we could switch freely between maps where this is a square and where this is one). So for example over $\mathbb F = \mathbb R$ all mappings with negative determinant are not in the image of $SL(2, \mathbb R)$, but all with positive determinant as each positive number is a square. In $\mathbb C$ every number is a square, so both group are indeed the same. For the finite field $\mathbb F_5 = \{ 0,1,2,4,5\}$ the only squares $\ne 0$ are $1$ and $4$, so only the mappings with this determinant are in $PSL(2, \mathbb F_5)$
Are my objections all right? Is there anything more to say about these relations?
