# Möbius transformations and groups

I have two questions regarding Möbius transformations and groups. In my notes there are two statements, which I can't prove/understand why they hold.

1. The subgroup of Möbius Transformations which maps the set $$\{z_1,z_2,z_3\}$$ to itself is isomorphic to $$S_3$$, the permutation group of three elements.

It is clear that the size of this subgroup and $$S_3$$ is the same, but what group isomorphism should I use between them?

Also,

1. The subgroup of Möbius transformations for which $$f(z_1)=z_1$$ and $$f(z_2)=z_2$$ is isomorphic to $$\Bbb{C}^*$$ where this is the group of $$\Bbb{C}\setminus \{0\}$$, under $$\times$$.

This has really confused me, I'm just not sure how I should be constructing these isomorphisms.

Any help appreciated, thanks.

• There are only two groups with six elements (up to isomorphism), and only one of those two groups is noncommutative. Jun 12 '19 at 13:16
• I truly don’t understand what was troubling you about #1. Each such Möb induces a permutation of your three points. If there are six such Möbs, you have six permutations of three things. That’s an $S_3$. Jun 13 '19 at 3:36

1. First we show that the group $$G$$ of Mobius transformations sending $$\{z_1,z_2,z_3\}$$ to $$\{z_1,z_2,z_3\}$$ is isomorphic to $$S_3$$. Given a bijection $$\sigma:\{1,2,3\} \to \{1,2,3\}$$ define a Mobius transformation $$\varphi_\sigma$$ which sends $$z_1$$ to $$z_{\sigma(1)}$$, $$z_2$$ to $$z_{\sigma(2)}$$ and $$z_3$$ to $$z_{\sigma(3)}$$. There is a unique such transformation as, for any triple of three distinct point on $$\hat{\mathbb{C}} = \mathbb{C} \cup \{\infty\}$$ there is a unique Mobius transformation taking these to any other triple of distinct points in $$\hat{\mathbb{C}}$$. For reference see Rudin's Real and Complex Analysis p.g. 280. I claim $$\sigma \mapsto \varphi_\sigma$$ is an isomorphism of groups. It's straightforward to check that this is a group homomorphism and that it's injective. To see it's surjective, given $$\varphi: \hat{\mathbb{C}} \to \hat{\mathbb{C}}$$ a Mobius transformation sending $$\{z_1,z_2,z_3\}$$ to itself, define $$\sigma:\{1,2,3\} \to \{1,2,3\}$$ by $$\sigma(i) = j$$ if $$\varphi(i) = j$$. Then $$\varphi = \varphi_\sigma$$. This addresses question 1.
2. Now we consider question 2. Consider the case when $$z_1 = 0$$ and $$z_2 = \infty$$. If $$\varphi(z) = \frac{az + b}{cz + d}$$ then $$\varphi(0) = 0$$ and $$\varphi(\infty) = \infty$$ immediately give $$b = 0$$ and $$c = 0$$. Thus $$\varphi(z) = \frac{a}{d}z$$. If you send $$\lambda \mapsto \varphi_\lambda$$ where $$\varphi_\lambda(z) = \lambda z$$ then this will be an isomorphism of groups from $$\mathbb{C}^\times$$ to the group of Mobius transformations fixing $$0$$ and $$\infty$$. To get the general case let $$f$$ be a Mobius transformation sending $$z_1$$ to $$0$$ and $$z_2$$ to $$\infty$$. Then $$\varphi \mapsto f \circ \varphi \circ f^{-1}$$ gives a group isomorphism from the group of Mobius transformation fixing $$z_1$$ and $$z_2$$ to the group of Mobius transformations fixing $$0$$ and $$\infty$$. Thus the group of Mobius transformations fixing $$z_1$$ and $$z_2$$ is isomorphic to $$\mathbb{C}^\times$$.