Special Linear Group is Subgroup of General Linear Group

Theorem
Let $K$ be a field whose zero is $0_K$ and unity is $1_K$.

Let $\SL {n, K}$ be the special linear group of order $n$ over $K$.

Then $\SL {n, K}$ is a subgroup of the general linear group $\GL {n, K}$.

Proof
Because the determinants of the elements of $\SL {n, K}$ are not $0_K$, then are invertible.

So $\SL {n, K}$ is a subset of $\GL {n, K}$.

Now we need to show that $\SL {n, K}$ is a subgroup of $\GL {n, K}$.

Let $\mathbf A$ and $\mathbf B$ be elements of $\SL {n, K}$.

As $\mathbf A$ is invertible we have that it has an inverse $\mathbf A^{-1} \in \GL {n, K}$.

From Determinant of Inverse Matrix:
 * $\det \paren {\mathbf A^{-1} } = \dfrac 1 {\det \paren {\mathbf A} }$

and so:
 * $\det \paren {\mathbf A^{-1} } = 1$

So $\mathbf A^{-1} \in \SL {n, K}$.

Also, from Determinant of Matrix Product:


 * $\det \paren {\mathbf A \mathbf B} = \det \paren {\mathbf A} \det \paren {\mathbf B} = 1$

Hence the result from the Two-Step Subgroup Test.