Orthogonal Group is Subgroup of General Linear Group
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Theorem
Let $k$ be a field.
Let $\map {\operatorname O} {n, k}$ be the $n$th orthogonal group on $k$.
Let $\GL {n, k}$ be the $n$th general linear group on $k$.
Then $\map {\operatorname O} {n, k}$ is a subgroup of $\GL {n, k}$.
Proof
From Unit Matrix is Orthogonal, the unit matrix $\mathbf I_n$ is orthogonal.
Let $\mathbf A, \mathbf B \in \map {\operatorname O} {n, k}$.
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:
- $\mathbf A \mathbf B^{-1} \in \map {\operatorname O} {n, k}$
Hence the result by One-Step Subgroup Test.
$\blacksquare$