Special Orthogonal Group is Subgroup of Orthogonal Group

Theorem

Let $k$ be a field.

Let $\map {\operatorname O} {n, k}$ be the $n$th orthogonal group on $k$.

Let $\map {\operatorname {SO} } {n, k}$ be the $n$th special orthogonal group on $k$.

Then $\map {\operatorname {SO} } {n, k}$ is a subgroup of $\map {\operatorname O} {n, k}$.

Proof

We have that Unit Matrix is Proper Orthogonal, so $\map {\operatorname {SO} } {n, k}$ is not empty.

Let $\mathbf A, \mathbf B \in \map {\operatorname {SO} } {n, k}$.

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

Then by Inverse of Proper Orthogonal Matrix is Proper Orthogonal:

$\mathbf B^{-1}$ is a proper orthogonal matrix.

By Product of Proper Orthogonal Matrices is Proper Orthogonal Matrix:

$\mathbf A \mathbf B^{-1}$ is a proper orthogonal matrix.

Thus by definition of special orthogonal group:

$\mathbf A \mathbf B^{-1} \in \map {\operatorname {SO} } {n, k}$

Hence the result by One-Step Subgroup Test.

$\blacksquare$