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 $\operatorname{SL} \left({n, K}\right)$ be the special linear group of order $n$ over $K$.

Then $\operatorname{SL} \left({n, K}\right)$ is a subgroup of the general linear group $\operatorname{GL} \left({n, K}\right)$.

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

So $\operatorname{SL} \left({n, K}\right)$ is a subset of $\operatorname{GL} \left({n, K}\right)$.

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

Let $\mathbf A$ and $\mathbf B$ be elements of $\operatorname{SL} \left({n, K}\right)$.

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

From Determinant of Inverse Matrix:
 * $\det \left({\mathbf A^{-1}}\right) = \dfrac 1 {\det \left({\mathbf A}\right)}$

and so:
 * $\det \left({\mathbf A^{-1}}\right) = 1$

So $\mathbf A^{-1} \in \operatorname{SL} \left({n, K}\right)$.

Also, from Determinant of Matrix Product:


 * $\det \left({\mathbf A \mathbf B}\right) = \det \left({\mathbf A}\right) \det \left({\mathbf B}\right) = 1$

Hence the result from the Two-Step Subgroup Test.