Is $SO_n({\mathbb R})$ a divisible group? The title says it all ... Formally, if 
$SO_n(\mathbb R)=\lbrace A\in M_n({\mathbb R}) |AA^{T}=I_n, {\sf det}(A)=1 \rbrace$
and $W\in SO_n(\mathbb R)$, is it true that for every integer $p$,  there is a $V\in SO_n(\mathbb R)$ satisfying $V^p=W$  ?
This is obvious when $n=2$, because rotations in the plane are defined
by an angle which can be divided at will.

Every compact connected Lie group $G$ (and $SO(n)$ is both compact and connected) is divisible since its exponential map $\exp$ is surjective (see here). By surjectivity of $\exp$, every element $g$ of $G$ has the form $\exp(v)$ (with $v\in {\mathfrak g}$, the Lie algebra of $G$) and, thus, for $h=\exp(v/p)$ we obtain $h^p=g$. 
