Determinant as a group homomorphism

Example

Let $n$ be a positive integer and let $\mathrm{GL}_n(\mathbb{R})$ be the group of invertible $n \times n$ real matrices under multiplication. The determinant is a map $\det \colon \mathrm{GL}_n(\mathbb{R}) \to \mathbb{R}^\times$ to the group of nonzero real numbers under multiplication. It is a group homomorphism, because $\det(AB) = \det(A) \det(B)$ for all $A, B$.

Remarks

The kernel of $\det$ is the special linear group $\mathrm{SL}_n(\mathbb{R})$ of matrices of determinant $1$, so this group is normal in $\mathrm{GL}_n(\mathbb{R})$. The determinant is surjective onto $\mathbb{R}^\times$, so $\mathrm{GL}_n(\mathbb{R}) / \mathrm{SL}_n(\mathbb{R}) \cong \mathbb{R}^\times$. The same holds over each field $F$, with $\mathrm{GL}_n(F)$, $\mathrm{SL}_n(F)$ and $F^\times$.

Explanations

L1 human

The determinant turns a product of matrices into a product of numbers. That is exactly what a homomorphism does. The matrices that it sends to $1$ form an important subgroup, the special linear group.