Let, $Mob(\hat{\Bbb C})$ be the group of all Mobius transformations from the extended complex plane to itself i.e. from $\hat{\Bbb C} \to \hat{\Bbb C}$ .
I have been able to prove that (i) $Mob(\hat{\Bbb C}) \cong PSL_{2}(\Bbb C)$ where $PSL_{2}(\Bbb C)$ is defined to be , $PSL_{2}(\Bbb C) := SL_{2}(\Bbb C)/ \{\pm I\}$ and that,
(ii) $SL_{2}(\Bbb C)$ does not have a proper normal subgroup containing its center i.e. $\{\pm I\}$ and thus by correspondence theorem, $PSL_{2}(\Bbb C) = SL_{2}(\Bbb C)/ \{\pm I\}$ is simple.
Combining (i) and (ii) we get that $Mob(\hat{\Bbb C})$ is simple!
My question precisely is,
What is actually the geometry or analysis behind the fact that $Mob(\hat{\Bbb C})$ is simple ?
Assume that $N$ is a normal subgroup of the Mobius group which is not the trivial subgroup. Then $N$ contains some non-identity element $f$. If $f$ is hyperbolic/loxodromic, i.e., if it has two fixed points $z_0$ and $z_1$, then you can pick a third point $z_2$ different from $z_0$ and $z_1$, and conjugate $f$ by a Mobius transformation $h$ which fixes $z_0$, and which maps $z_1$ to $z_2$, to obtain $g = hfh^{-1} \in N$ with fixed points $z_0$ and $z_2$. It is then easy to show that the commutator $k = [f,g] = f^{-1}g^{-1}fg$ fixes $z_0$ with $k'(z_0)=1$, but that it is not the identity, so it must be a parabolic Mobius transformation. This establishes that $N$ contains a parabolic Mobius transformation, and since any two parabolic Mobius transformations are conjugate, $N$ contains all of them. In particular $N$ contains $F(z) = z+1$ and $G(z) = \frac{z}{1+az}$ for any $a \ne 0$. Explicit calculations of the commutator of $F$ and $G$ shows that you get a hyperbolic/loxodromic element with trace $2+a^2$. (This calculation is probably easier with matrices than with explicit Mobius transformations.) This shows that $N$ contains hyperbolic/loxodromic elements with any given trace, so $N$ is the whole Mobius group.