Why the orthogonal group $G=O(3)$ does not contain a normal subgroup, a cyclic group of order 2, $N=\mathbb{Z}/2\mathbb{Z}$?
It looks that any $g \in G$ satisfies $$ g N g^{-1}=N $$ where $N=\mathbb{Z}/2\mathbb{Z}$ is generated by $-diagonal(1,1,1)$ of rank-3 matrix.
So why the orthogonal group $G=O(3)$ does not contain a normal subgroup $N=\mathbb{Z}/2\mathbb{Z}$, but only an $SO(3)$?
You are quite right that $N = \{ \pm I \}$ is a cyclic subgroup of order two that is normal in $O(3)$. In fact, it is the center of $O(3)$. We also have the internal direct product decomposition $O(3) = SO(3) \times \{ \pm I \}$, or more generally $O(n) = SO(n) \times \{ \pm I \}$ for any odd $n$.