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?


  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.

  • 1
    $\begingroup$ There are only two groups with six elements (up to isomorphism), and only one of those two groups is noncommutative. $\endgroup$ Jun 12 '19 at 13:16
  • 1
    $\begingroup$ 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 $S_3$. $\endgroup$
    – Lubin
    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 $\mathbb{C}^\times$.


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