# 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.

• Actually you want a proof for this: PSL$(2,\mathbb C)$ is a simple group. Try google!
– user26857
Mar 3 '13 at 12:38
• What is $\phi$? Jun 24 '13 at 19:11
• I do not understand the votes to close - is it merely because this is an old, yet unanswered question? Personally, I think it should be left open and that someone should, you know, answer it... Jun 24 '13 at 19:36
• @user1729 I completely agree. Jun 25 '13 at 18:12
• Look at this Apr 5 '18 at 23:29

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 $$-xy+y \neq 0$$, $$N=M$$.

• Great answer. It is hard to find a proof of the fact that $PSL_2(\mathcal{C})$ is simple, and this proof is easy to understand. I think that there is a typo in the line "This implies that $S_1 \circ S_4\circ S_3\circ S_2\in N$...", the second element should be $S_2\circ S_1\circ S_4\circ S_3$ or am I wrong? Dec 27 '19 at 4:46
• You are correct, thank you! Dec 28 '19 at 5:45