General Linear Group to Determinant is Homomorphism

Theorem
Let $\operatorname{GL} \left({n, \R}\right)$ be the general linear group over the field of real numbers.

Let $\R \setminus \left\{{0}\right\}$ denote the multiplicative group of real numbers.

Let $\det: \operatorname{GL} \left({n, \R}\right) \to \R \setminus \left\{{0}\right\}$ be the mapping:
 * $\mathbf A \mapsto \det \left({\mathbf A}\right)$

where $\det \left({\mathbf A}\right)$ is the determinant of $\mathbf A$.

Then $\det$ is a group homomorphism.

Its kernel is the special linear group $\operatorname{SL} \left({n, \R}\right)$.

Proof
From Determinant of Matrix Product:
 * $\det \left({\mathbf A \mathbf B}\right) = \det \left({\mathbf A}\right) \det \left({\mathbf B}\right)$

which is seen to be a group homomorphism by definition.

The special linear group $\operatorname{SL} \left({n, \R}\right)$ is the subset of $\operatorname{GL} \left({n, \R}\right)$ such that:
 * $\forall \mathbf A \in \operatorname{SL} \left({n, \R}\right): \det \left({\mathbf A}\right) = 1$

From Real Multiplication Identity is One:
 * $1$ is the identity of the multiplicative group of real numbers.

It follows by definition that $\operatorname{SL} \left({n, \R}\right)$ is the kernel of the $\det$ mapping.