TITLE: Why isn't $SO(3)\times SO(3)$ isomorphic to $SO(4)$?
QUESTION [15 upvotes]: Why is $SO(3)\times SO(3)$ isomorphic to $SO(4)$?

REPLY [30 votes]: This isn't quite true: $SO(3) \times SO(3)$ is isomorphic to $SO(4) / \mathbb{Z}_2$, where $\mathbb{Z}_2 = \{1,-1\}$.  (Topologically speaking, $SO(4)$ is a double cover of $SO(3)\times SO(3)$.)
The simple explanation for this is the following:

$SO(3)$ is isomorphic to $U(\mathbb{H})/\mathbb{Z}_2$, where $U(\mathbb{H})$ is the group of unit quaternions and $\mathbb{Z}_2 = \{1,-1\}$.  Specifically, the action of $U(\mathbb{H})$ on $\mathbb{R}^3$ is by conjugation, where $\mathbb{R}^3$ is identified with the set of quaternions of the form $ai+bj+ck$ for $a,b,c\in\mathbb{R}$.  (See the Wikipedia article on quaternions and spatial rotation for more information on this action.)
$SO(4)$ is isomorphic to $\bigl(U(\mathbb{H})\times U(\mathbb{H})\bigr)/\mathbb{Z}_2$, where $\mathbb{Z}_2 = \{(1,1),(-1,-1)\}$.  In particular, any rotation of $\mathbb{R}^4$ can be defined by an equation of the form
$$
R(x) \;=\; axb
$$
where $a$ and $b$ are quaternions and the input vector $x\in\mathbb{R}^4$ is interpreted as a quaternion.

One consequence of this is that the spin group $\text{Spin}(3)$ is isomorphic to $U(\mathbb{H})$, while $\text{Spin}(4)$ is isomorphic to $U(\mathbb{H}) \times U(\mathbb{H})$.  Thus,
$$
\text{Spin}(4) \;\cong\; \text{Spin}(3) \times \text{Spin}(3).
$$
The statement you gave is also true on the level of Lie algebras, i.e.
$$
\mathfrak{so}(4) \;\cong\; \mathfrak{so}(3) \times \mathfrak{so}(3).
$$

REPLY [5 votes]: On the level of Lie algebras we have that ${\mathfrak {so}}(n)$ are just antisymmetric matrices $n \times n$. It turns out that the six-dimensional space of such $4\times 4$ matrices decomposes into two three-dimensional subspaces that are each closed under taking commutators and each of them satisfies precisely the commutation relations of $\mathfrak{so}(3)$.
Because exponentiation defines an isomorphism between a neighborhood of the identity and a Lie algebra, we have that the two groups are locally isomorphic. It only remains to check global properties, like simple-connectedness, number of components, etc., to be sure the groups are really isomorphic (and not just an universal cover of each other, say, as in the case of $SO(n)$ and $Spin(n)$.)