Simple proof that there is no isomorphism between any two of $ Aut(\hat{C}) $(Riemann Sphere),$ Aut(H^+) $(upper half plane) and $ Aut(C) $ Referring the groups of automorphisms (holomorphic bijections) of the respective domains.
This is a homework problem. Is a basic course, so sophisticated answers may not be of help (it has a simple solution according to my teacher).  Also, it looks like an Algebra problem, but I’ve been assured that there is a solution within complex analysis, so if anyone can give a non algebraic proof, it will be appreciated. I tried constructing a conformal mapping (between two of the domains) using a supposed isomorphism, also calculate the unit roots (of second degree, and some of higher degree) in each of the groups, and a couple ideas more, but with no luck. 
Thanks in advance for your help (and sorry about my poor English).
Edit:The question was edited to avoid sophisticated algebraic answers.  Please, just use the very basic of algebra in your solution. Is a complex analysis exercise! (of course, if you just want to share a sohpisticated answer that can help other users, welcome.)
 A: $L(C)$ is solvable, the other two aren't. 
${\rm PSL}(2,\mathbb{C})$ contains a subgroup isomorphic to $\mathbb{R}^2$, ${\rm PSL}(2,\mathbb{R})$ doesn't.
A: In $Aut(\mathbb{C})=L(\mathbb{C})$, the product of two elements of order $7$ cannot be of order $3$. In $Aut(H^+) \le Aut(\bar{\mathbb{C}})$ it can. 
In the hyperbolic half plane model ($H^+$) take an equilateral triangle $ABC$ with angles $\frac{2\pi}7$. Let $a$ and $b$ be the rotations about $A$ and $B$, respectively, by angle $2\pi/7$. Then $a^{-1}b$ moves $ABC$ to $CAB$, so $a^{-1}b$ is of order $3$. 

In $Aut(\bar{\mathbb{C}})$ there are two elements of order $2$ whose product also is of order $2$: just take three perpendicular diameters of the sphere and consider the reflections about them. For instance $z\mapsto -z$ and $z\mapsto \frac1z$ are of order 2, and their composition is $\frac{-1}z$.
$Aut(H^+)$ consists of oriented isometries of the half plane model and all elements of order 2 are reflections about points. If you take two reflections about two distinct points, say $A$ and $B$ then their product will be a translation on the line $AB$ which cannot be of order 2.
