Orthogonal Group is Subgroup of General Linear Group

Theorem
Let $k$ be a field.

Let $\operatorname O \left({n, k}\right)$ be the $n$th orthogonal group on $k$.

Let $\operatorname{GL} \left({n, k}\right)$ be the $n$th general linear group on $k$.

Then $\operatorname O \left({n, k}\right)$ is a subgroup of $\operatorname{GL} \left({n, k}\right)$.

Proof
From Unit Matrix is Orthogonal, the unit matrix $\mathbf I_n$ is orthogonal.

Let $\mathbf A, \mathbf B \in \operatorname O \left({n, k}\right)$.

Then, by definition, $\mathbf A$ and $\mathbf B$ are orthogonal.

Then by Inverse of Orthogonal Matrix is Orthogonal:
 * $\mathbf B^{-1}$ is a orthogonal matrix.

By Product of Orthogonal Matrices is Orthogonal Matrix:
 * $\mathbf A \mathbf B^{-1}$ is a orthogonal matrix.

Thus by definition of orthogonal group:
 * $\mathbf A \mathbf B^{-1} \in \operatorname O \left({n, k}\right)$

Hence the result by One-Step Subgroup Test.