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 $$\det \left({\mathbf{A}^{-1}}\right) = \frac 1 {\det \left({\mathbf{A}}\right)}$$, 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.