The proof verifies the three properties: homomorphism, surjectivity, and identification of the kernel. **Step 1: $\mathrm{GL}_n(\mathbb{R})$ is a [group](/page/Group).** We verify the group axioms under matrix multiplication. Closure: $\det(AB) = \det A \det B \neq 0$, so $AB \in \mathrm{GL}_n(\mathbb{R})$. Associativity: matrix multiplication is associative. Identity: $I_n$ has $\det I_n = 1 \neq 0$. Inverses: $\det A \neq 0$ implies $A$ is invertible, with $\det(A^{-1}) = (\det A)^{-1} \neq 0$. **Step 2: $\det$ is a homomorphism.** For $A, B \in \mathrm{GL}_n(\mathbb{R})$: \begin{align*} \det(A \cdot B) = \det(A) \times \det(B), \end{align*} which is precisely the homomorphism condition from $(\mathrm{GL}_n(\mathbb{R}), \cdot)$ to $(\mathbb{R} \setminus \{0\}, \times)$. **Step 3: $\det$ is surjective.** For any $r \in \mathbb{R} \setminus \{0\}$, the diagonal matrix $A = \operatorname{diag}(r, 1, \ldots, 1)$ satisfies $\det A = r$ and $A \in \mathrm{GL}_n(\mathbb{R})$. **Step 4: Identify the kernel.** $\ker(\det) = \{A \in \mathrm{GL}_n(\mathbb{R}) : \det A = 1\} = \mathrm{SL}_n(\mathbb{R})$. By the [Kernel Is a Normal Subgroup](/theorems/788) theorem, $\mathrm{SL}_n(\mathbb{R}) \unlhd \mathrm{GL}_n(\mathbb{R})$. By the [First Isomorphism Theorem](/theorems/791), $\mathrm{GL}_n(\mathbb{R})/\mathrm{SL}_n(\mathbb{R}) \cong (\mathbb{R} \setminus \{0\}, \times)$.