Special Orthogonal Group is Subgroup of Orthogonal Group

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Let $k$ be a field.

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

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

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


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

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

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:

$\mathbf A \mathbf B^{-1} \in \operatorname {SO} \left({n, k}\right)$

Hence the result by One-Step Subgroup Test.