Each element $\begin{pmatrix} a & b \\ c & d\\ \end{pmatrix}$ of $GL_2(\Bbb C)$ gives rise to a Möbius transformation $z \rightarrow \frac{az+b}{cz+d}$.
The first part of the question was to show that these transformations form a group under composition of functions. I managed to do this, but I got stuck on the second part. We have to show that this group is isomorphic to the quotient of $GL_2(\Bbb C)$ by its centre.
I wanted to use the first Isomorphism Theorem. But since we want to show that it's an isomorphism to $GL_2(\Bbb C) / centre$, we would have to find an $f : GL_2(\Bbb C) \rightarrow \text{Möbius group}$ where the Kernel = Centre.
I don't really know how I could come up with an $f$ that satisfies this. So far I only tried $f: \begin{pmatrix} a & b \\ c & d\\ \end{pmatrix} \rightarrow \frac{az+b}{cz+d}$ but then you don't get a homomorphism.
Tips on how to find a suitable $f$ are very welcome!
Hint 1: two Möbius transformations are "equal" (ie, they should be in the same equivalence class) if $A=\lambda A'$ ($A$ and $A'$ the matrices)
Hint 2: the center of $GL_2(\mathbb{C})$ are the scalar matrices $\lambda I_2$ with $\lambda\neq0$