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 $SL \left({n, K}\right)$.

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

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

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

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

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

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

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

Hence the result from the Two-Step Subgroup Test.