I have an assignment for my linear Algebra II course which prompts me to disprove the following statement. Let $ A, B \in M_{n}\mathbb{(R)} $ be two matrices. $AB \in O(n) \implies A, B \in O(n) $. It's pretty clear that it isn't true as soon as you realize that $AA^{-1} = \mathbb{I_{D}} \in O(n)$ but $A$ isn't necessarily orthogonal. Nonetheless, another line of reasoning gave me a surprising result.
By applying the definition of orthogonal matrices, and got: $AB \in O(n) \iff (AB)^{T}AB = \mathbb{I_{D}} $ and $ (AB)^T = (AB)^{-1} $ which implies that $A$ and $ B$ are invertible. Let $C = BB^T $ and consider:
$\begin{gather} (AB)(AB)^{T} &= \mathbb{I_{D}} \\ \iff(AB)B^{T}A^{T} &= \mathbb{I_{D}} \\ \iff A(BB^{T})A^{T} &= \mathbb{I_{D}} \\ \iff ACA^{T} &= \mathbb{I_{D}} \\ \end{gather}$
Since $ A$ is invertible, it follows that $A^T$ is invertible as well. Let $ D^{-1} = A^T$. Therefore, we get that:
$\begin{gather} ACD^{-1} &= \mathbb{I_{D}} \\ \end{gather}$
Which implies that $ C $ is similar to the identity matrix, and therefore that $B \in O(n) \implies A \in O(n)$. Can anyone point me where I'm wrong ? Probably made a wrong assumption without seeing it.