Show that the Möbius group is isomorphic to the quotient of $GL_2(\Bbb C)$ by its centre.

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!

• You should use $$f: \pmatrix{a&b\cr c&d\cr}\mapsto [z\mapsto \frac{az+b}{cz+d}].$$ The kernel consists of those matrices $A$ such that $f(A)=f(I_2)$. – Jyrki Lahtonen Dec 28 '17 at 11:49
• @JyrkiLahtonen I never saw a function notated like this. I don't know how I could show that this is a homomorphism. Could you please explain that aswel? – user423841 Dec 28 '17 at 11:57
• You can use whatever notation you want. The point is that $\dfrac{az+b}{cz+d}$ is a complex number or infinity, but a Möbius transformation is a function that sends a complex number to another – Jyrki Lahtonen Dec 28 '17 at 12:00
• The operations that you need to prove to be compatible with $f$ are 1) the product of matrices, 2) the composition of two Möbius transformations. – Jyrki Lahtonen Dec 28 '17 at 12:03

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 $\lambda\neq0$