# 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!

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$O(2)=\mathbb{Z}/(2)\rtimes SO(2)$. $\mathbb{Z}/(2)$ conjugation is inversion: reflect rotate reflect=rotate inverse. – yoyo Mar 27 '11 at 19:15

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.

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Typo: "...the center of SO(n)xZ_2 is..." (not SO(n)). Thanks for the answer though! – gnometorule Nov 4 '12 at 23:39
@gnometorule: thanks, I just edited it. – Eric O. Korman Nov 5 '12 at 3:29