Special Linear Group is Subgroup of General Linear Group

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

The set of all order-$n$ square matrices over $K$ whose determinant is $1_K$ is a group under (conventional) matrix multiplication.

The field itself is usually $\R$, $\Q$ or $\C$, but can be any field.

This group is called the Special Linear Group and is denoted $\operatorname{SL} \left({n, K}\right)$, or $\operatorname{SL} \left({n}\right)$ if the field is implicit.

It is a subgroup of the General Linear Group $\operatorname{GL} \left({n, K}\right)$.

Proof
From Inverse of a Matrix, elements of $\operatorname{SL} \left({n, K}\right)$ are invertible as their determinants are not $0_K$.

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

From Determinant of Inverse we have that $\displaystyle \det \left({\mathbf A^{-1}}\right) = \frac 1 {\det \left({\mathbf A}\right)}$, and so $\det \left({\mathbf A^{-1}}\right) = 1$.

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

Hence the result from the Two-Step Subgroup Test.