Is $O(4)\cong SO(4)\times C_2$?

Question

My claim is $O(4)\ncong SO(4)\times C_2$, since there is no "$-I$" role as in the proof that $O(3)\cong SO(3)\times C_2$, as $\det(-I)=1$ in $\mathbb{R}^4$. However, I cannot proceed like $O(2)$ case that both $SO(2)$ and $C_2$ are abelian while $O(2)$ is not. Are there any properties I missed?

Answer 1

Hint. Show $Z(O(4))=Z(SO(4))$ in order to prove $Z(O(4))\not\cong Z(SO(4)\times C_2)$.