The proof verifies the [group](/page/Group) axioms directly for $\mathcal{M}$ under composition. **Step 1: Closure.** Let $f_1(z) = \frac{a_1 z + b_1}{c_1 z + d_1}$ and $f_2(z) = \frac{a_2 z + b_2}{c_2 z + d_2}$ with $a_1 d_1 - b_1 c_1 \neq 0$ and $a_2 d_2 - b_2 c_2 \neq 0$. Direct computation (excluding special points, then checking them individually) gives: \begin{align*} (f_2 \circ f_1)(z) = \frac{(a_2 a_1 + b_2 c_1)z + (a_2 b_1 + b_2 d_1)}{(c_2 a_1 + d_2 c_1)z + (c_2 b_1 + d_2 d_1)}. \end{align*} The "determinant" of the composition is $(a_1 d_1 - b_1 c_1)(a_2 d_2 - b_2 c_2) \neq 0$, so $f_2 \circ f_1 \in \mathcal{M}$. **Step 2: Associativity.** Composition of [functions](/page/Function) is associative. **Step 3: Identity.** The identity map $I(z) = z = \frac{1 \cdot z + 0}{0 \cdot z + 1}$ is a Möbius map with $ad - bc = 1 \neq 0$. **Step 4: Inverses.** For $f(z) = \frac{az + b}{cz + d}$ with $ad - bc \neq 0$, define $f^*(z) = \frac{dz - b}{-cz + a}$. This is also a Möbius map (its determinant is $da - (-b)(-c) = ad - bc \neq 0$). One verifies $f(f^*(z)) = z = f^*(f(z))$ for all $z \in \mathbb{C}_\infty$ (checking the special points $z = a/c$, $z = -d/c$, and $z = \infty$ individually). Since $\mathcal{M} \subseteq \operatorname{Sym}(\mathbb{C}_\infty)$ (each Möbius map is a bijection of $\mathbb{C}_\infty$), $\mathcal{M}$ is a subgroup of $\operatorname{Sym}(\mathbb{C}_\infty)$.