Möbius transformations form a simple group 
How to show the group $M$ of Möbius transformations is a simple group?

I know: $SL_2(\mathbb C)/\{+I,-I\}\cong M$ then if $A \lhd M \implies \phi^{-1}(A) \lhd SL_2(\mathbb C)/\{+I,-I\}$.  
So if I can show $SL_2(\mathbb C)/\{+I,-I\}$ is simple probably that answers the question.
 A: You can prove this without having to look at $PSL_2 (\mathbb{C})$. First, consider that any Möbius transformation can be expressed as the composition of dilations, translations and the inversion. In particular, if $T(z) =\dfrac{az+b}{cz+d}$, then $T = S_4 \circ S_3 \circ S_2 \circ S_1$, where $S_1$ and $S_4$ are translations, $S_3$ a dilation and $S_2$ the inversion. Now, let $N \trianglelefteq M$ be non-trivial and suppose $T \in N$ is a translation, i.e., $T(z) = z+b$, with $b \in \mathbb{C^{*}}$ (we don't have to consider $b=0$ because the identity is always there). Let $S(z)=az$, with $a \in \mathbb{C}^{*}$. We have that $S^{-1}\circ T \circ S = z+\dfrac{b}{a} \in N$, so, for any $c \in \mathbb{C^{*}}$, we may choose $a=b/c$ and obtain every translation. In particular, $R(z) = z-1 \in \mathbb{C^{*}}$ and if we take $I(z) = 1/z$ to be the inversion, we can obtain that
$$I\circ R \circ I^{-1} = \dfrac{z}{1-z}$$
$$\left(\dfrac{z}{1-z}\right)\circ R = \dfrac{z-1}{-z}$$
$$(-I)\circ \left(\dfrac{z-1}{-z}\right) \circ (-I) = \dfrac{1}{z+1}$$
$$\left(\dfrac{1}{z+1}\right) \circ R = \dfrac{1}{z} \in N$$
Then, if we take $I \circ S \circ I \circ S^{-1} = a^2 z$, we can obtain every dilation and so $N=M$. Now, suppose we have $S(z)=az \in N$, $a\in \mathbb{C^{*}-\{1\}}$. If we take $T(z) = z+b$, $b \in \mathbb{C^{*}}$, then $ S^{-1} \circ T \circ S \circ T^{-1} = z+ab-b \in N$ $\Rightarrow z+c \in N, \forall c \in \mathbb{C^{*}}$. We have proved the following implications, for $N \trianglelefteq M$:
$(\text{i) }z+b \in N$, for some $b \in \mathbb{C^{*}} \Rightarrow$ $z+c \in N, \forall c \in \mathbb{C^{*}} \Rightarrow 1/z \in N \Rightarrow az \in N, \forall a \in \mathbb{C^{*}} \Rightarrow N = M;$
$(\text{ii) }az \in N$, for some $a \in \mathbb{C^{*}-\{1\}} \Rightarrow$ $z+ab-b \in N \Rightarrow$ $N=M;$
$(\text{iii) }1/z \in N\Rightarrow$ $az \in N, \forall a \in \mathbb{C^{*}} \Rightarrow$ $N=M;$
Finally, it suffices to prove that if we have a general Mobius transformation $T(z) =\dfrac{az+b}{cz+d} \in N$, then we can obtain a dilation, a translation or an inversion. We have that $T = S_4 \circ S_3 \circ S_2 \circ S_1$. This implies that $ S_1 \circ S_4 \circ S_3 \circ S_2 \in N$ and $ S_2 \circ S_1 \circ S_4 \circ S_3 \in N$, so $L = (S_1 \circ S_4 \circ S_3)^2 \in N$. Since $L$ is the composition of translations and dilations, it has the form $L(z)=xz+y $, with $x,y \in \mathbb{C^{*}}$. If $x=1$ and $y=0$, this would imply that $S_1 \circ S_4 \circ S_3 = z \Rightarrow S_4 \circ S_3 = S_1^{-1}$ $\Rightarrow T = S_1^{-1} \circ S_2 \circ S_1 \Rightarrow 1/z \in N$
$ \Rightarrow N=M$. If $x=1$ or $y=0$, we are done. Finally, if $x\neq 1$ and $y \neq 0$, then $$L \circ (z-y) \circ L^{-1} \circ (z+y) = z-xy+y \in N$$ and, since $-xy+y \neq 0$, $N=M$.
