The proof has two parts: first show $\mathcal{M}$ is a [group](/page/Group) under composition, then construct the isomorphism with $\mathrm{GL}_2(\mathbb{C})/Z$. **Step 1: $\mathcal{M}$ is a group under composition.** - *Identity:* $f(z) = z = \frac{1 \cdot z + 0}{0 \cdot z + 1} \in \mathcal{M}$. - *Closure:* If $f_1(z) = \frac{a_1 z + b_1}{c_1 z + d_1}$ and $f_2(z) = \frac{a_2 z + b_2}{c_2 z + d_2}$, then $f_2 \circ f_1(z) = \frac{(a_1 a_2 + b_2 c_1)z + (a_2 b_1 + b_2 d_1)}{(c_2 a_1 + d_2 c_1)z + (c_2 b_1 + d_1 d_2)}$, whose determinant is $(a_1 d_1 - b_1 c_1)(a_2 d_2 - b_2 c_2) \neq 0$. - *Inverses:* For $f(z) = \frac{az+b}{cz+d}$, the map $f^*(z) = \frac{dz - b}{-cz + a}$ satisfies $f \circ f^* = f^* \circ f = \mathrm{id}$, with determinant $ad - bc \neq 0$. - *Associativity:* Inherited from composition of [functions](/page/Function). **Step 2: Construct the surjective homomorphism.** Define $\Phi : \mathrm{GL}_2(\mathbb{C}) \to \mathcal{M}$ by $\Phi\begin{pmatrix} a & b \\ c & d \end{pmatrix} = f(z) = \frac{az + b}{cz + d}$. The closure computation in Step 1 shows that the composition of Möbius maps corresponds to matrix multiplication, so $\Phi$ is a homomorphism. Surjectivity is immediate: every $f \in \mathcal{M}$ is $\Phi$ of its coefficient matrix. **Step 3: Identify the kernel.** $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \ker(\Phi)$ iff $\frac{az + b}{cz + d} = z$ for all $z \in \mathbb{C}_\infty$. Evaluating at $z = \infty$ gives $c = 0$; at $z = 0$ gives $b = 0$; at $z = 1$ gives $a = d$. So $\ker(\Phi) = \left\{\begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix} : \lambda \in \mathbb{C} \setminus \{0\}\right\} = Z$. **Step 4: Apply the First Isomorphism Theorem.** By the [First Isomorphism Theorem](/theorems/791), $\mathrm{GL}_2(\mathbb{C}) / Z \cong \mathcal{M}$.