Why is the orthogonal group $\operatorname{O}(2n,\mathbb R)$ not the direct product of $\operatorname{SO}(2n, \mathbb R)$ and $\mathbb Z_2$? We know that when $n$ is odd, $\operatorname{O}_n(\mathbb R) \simeq \operatorname{SO}_n (\mathbb R) \times \mathbb Z_2$.
However, this seems not true when $n$ is even. But I have no idea how to prove something is not a direct product.
I have tried to verify some basic properties of direct product. For example, $\operatorname{SO}_n(\mathbb R)$ is a normal subgroup of $\operatorname{O}_n(\mathbb R)$, whenever $n$ is odd or even. But they are not helpful.
So, is this statement true and how to prove it?
Thank you!
 A: Look at the centers:  the center of $\operatorname{O}(n)$ is $\pm \operatorname{Id}$.  When $n$ is even, this is also the center of $\operatorname{SO}(n)$.  Therefore for even $n$ the center of $\operatorname{SO}(n) \times \mathbb Z_2$ is $\{\pm \operatorname{Id} \} \times \mathbb Z_2$, which is bigger than the center of $\operatorname{O}(n)$.
EDIT:  This works for $n \ge 3$.  For $n=2$, $\operatorname{O}(2)$ is non-abelian while $\operatorname{SO}(2) \times \mathbb Z_2$ is.
A: In this answer, we for fun generalize and write out the construction in more detail.


*

*An element in the orthogonal group$^1$ $O(n,\mathbb{F})$ has determinant $\pm 1$. The center is $$Z(O(n,\mathbb{F}))~=~\{\pm \mathbb{1}\}~\cong~\mathbb{Z}_2,\tag{1}$$
$$Z(SO(n,\mathbb{F}))~=~\left\{\begin{array}{c}\{ \mathbb{1}\}\text{ if $n$ odd},\cr
\{\pm \mathbb{1}\}\text{ if $n$ even}.\end{array}\right.\tag{2}$$
Here $n\in\mathbb{N}$.

*$O(n,\mathbb{F})$ has 2 distinct components
$$O(n,\mathbb{F})~=~SO(n,\mathbb{F}) ~\sqcup~ P\cdot SO(n,\mathbb{F}),\tag{3}$$
where $P\in O(n,\mathbb{F})$ is a fixed element with $\det P=-1$.

*There is always a group isomorphism from the semidirect product
$$ \mathbb{Z}_2 ~\ltimes~ SO(n,\mathbb{F})~~\stackrel{\Phi}{\cong}~~O(n,\mathbb{F}) \tag{4}$$
given by 
$$ ((-1)^p, M)~~\stackrel{\Phi}{\mapsto}~~ P^p\cdot M, \qquad p~\in~\{0,1\},\qquad M~\in~SO(n,\mathbb{F}) . \tag{5}$$
The factor $(-1)^p$ is the determinant. Explicitly, the semidirect product reads
$$ ((-1)^{p_1}, M_1)\cdot ((-1)^{p_2}, M_2)~=~((-1)^{p_1+p_2}, P^{-p_2}\cdot M_1\cdot P^{p_2}\cdot M_2). \tag{6}$$

*On one hand, if we choose the fixed element $P$ to belong to the center (1), the semidirect product (6) becomes a direct product. This is precisely possible if $n$ is odd. 

*On the other hand, for $n$ even, then $O(n,\mathbb{F})$ and the direct product $\mathbb{Z}_2 \times SO(n,\mathbb{F})$ have different centers, so they cannot be isomorphic, cf. answer by Eric O. Korman.
--
$^1$ Here $\mathbb{F}$ is a field with characteristic different from 2.
