Is $O_n(\mathbb{R})\leq GL_{n}(\mathbb{R})$ A Sub Group? Is $O_n(\mathbb{R})\leq GL_{n}(\mathbb{R})$ where $O_n(\mathbb{R})$ are orthonormal matrices.


*

*$I_n^t=I^{-1}_n=I_n$ and therefore: $I_n\in O_n(\mathbb{R})$

*Let $A,B\in O_n(\mathbb{R})$ then we have to prove that $AB^{-1}\in O_n(\mathbb{R})$
We have that if $B\in O_n(\mathbb{R})$ then $B^{-1}=B^{t}\in O_n(\mathbb{R})$ but how can we conclude about  $AB^{-1}\in O_n(\mathbb{R})$?
Can we say that $$(AB^{t})^t=BA^t=BA^{-1}$$
And therefore $AB^{-1}\in O_n(\mathbb{R})$?

Note that the condition of $A$ being invertible and satisfying $A^t=A^{-1}$ is equivalent to $AA^t = I$. With this it is easy to check the properties of being a subgroup.


*

*$II^t = II = I$, hence $I\in O_n$

*Given $A,B\in O_n$ we have $AA^t=BB^t=I$ and hence
$$\begin{align}
(AB^{-1})(AB^{-1})^t&= AB^{-1}(B^{-1})^tA^t\\
&= AB^{-1}(B^t)^{-1}A^t\\
&= A(B^t B)^{-1} A^t\\
&= AA^t\\
&= I.
\end{align}$$
thus $AB^{-1}\in O_n$ as well.

