# 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.

## 2 Answers

Given an element of $O(n)$, we may represent it in suitable bases as a diagonal block matrix $\mathrm{diag}(R_1,\dotsc,R_n,\pm 1,\dotsc,\pm 1)$, where $R_i$ are $2\times 2$ rotation matrices. If it lies in $SO(n)$, the number of $-1$s is even. Thus, it suffices to find roots of the $R_i$ (which is easy, we just scale the rotation angle) and of $\mathrm{diag}(1,1)=I$ (which is trivial) and $M=\mathrm{diag}(-1,-1)$. But $M$ is again a rotation matrix with angle $\pi$.

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 $h^p=g$.