# 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!

• $O(2)=\mathbb{Z}/(2)\rtimes SO(2)$. $\mathbb{Z}/(2)$ conjugation is inversion: reflect rotate reflect=rotate inverse.
– yoyo
Commented Mar 27, 2011 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.

• Typo: "...the center of SO(n)xZ_2 is..." (not SO(n)). Thanks for the answer though! Commented Nov 4, 2012 at 23:39
• @gnometorule: thanks, I just edited it. Commented Nov 5, 2012 at 3:29

In this answer, we for fun generalize and write out the construction in more detail.

1. 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 $$n\in\mathbb{N}$$.

2. $$O(n,\mathbb{F})$$ 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 $$\det P=-1$$.

3. 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 $$((-1)^{p_1}, M_1)\cdot ((-1)^{p_2}, M_2)~=~((-1)^{p_1+p_2}, P^{-p_2}\cdot M_1\cdot P^{p_2}\cdot M_2). \tag{6}$$

4. 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.

5. 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.