We know that the Lie group (here with real $\mathbb{R}$ coefficients in the matrix representation) --- giving that orthogonal group and spin group are related to the special orthogonal group via $$ 1\to Z/2Z\to Spin(n) \to SO(n) \to 1, $$ $$ 1\to SO(n) \to O(n) \to Z/2Z\to1 $$
It seems that $ O(n)=SO(n) \times Z/2Z,$ if and only if $n=1,3,5, \dots$ is an odd integer. Yes or no?
Do we have $$ Spin(n)=SO(n) \times Z/2Z?$$ for certain $n$? How to show the answer if it is positive or negative?
When $n$ is odd, we do have an isomorphism of Lie groups $SO(n)\times\mathbb{Z}/2\mathbb{Z}\cong O(n)$, with the $\mathbb{Z}/2\mathbb{Z}$ generated by $-I$, the negative of the identity matrix (here we need $n$ to be odd so that $-I\not\in SO(n)$). When $n$ is even, the center of $SO(n)$ consists of just $I$ and $-I$ and so is contained in $SO(n)$, and so we cannot have such an isomorphism.
There is never an isomorphism of Lie groups $Spin(n)=SO(n) \times \mathbb{Z}/2\mathbb{Z}$ for $n>1$ since $Spin(n)$ is connected (it is the connected double cover of $SO(n)$). When $n=1$, $SO(n)$ is trivial and so we do have such an isomorphism.
(If you don't know that $Spin(n)$ is connected for $n>1$, then note that to prove it you just need to check the case $n=2$: you know $Spin(n)$ is a double cover of $SO(n)$, so to check it is a nontrivial double cover, it suffices to check that its restriction to a double cover of $SO(2)\subseteq SO(n)$ is nontrivial.)