Let $n$ be a positive integer, $p$ a (positive rational) prime, and $\mathbb K$ an algebraically closed field. If ${\rm char}(\mathbb K) = 0$ then ${\rm GL}_n(\mathbb K)$ is divisible (see <a href="http://mathoverflow.net/questions/124810/all-and-the-only-algebraically-closed-fields-s-t-any-regular-n-by-n-matrix-has-a">here</a>). But what about ${\rm SL}_n(\mathbb K)$? Again, the answer is negative if $\mathbb K$ has finite characteristic, and again we may assume that (i) our matrices are in Jordan normal form and (ii) $\mathbb K$ is the complex field, thanks to the Lefschetz principle and the fact that, by Laplace's formula, the determinant of an $n$-by-$n$ "formal matrix" can be expressed as a wff in the first-order language $\mathcal L = (+, \cdot, -, 0, 1)$ of (the theory of) rings. Thus, let us consider a Jordan matrix of size $n \times n$, say $J = {\rm diag}(J_1, \ldots, J_m)$, where $J_i$ is a Jordan block of size $k_i \times k_i$, with $k_1 + \cdots + k_m = n$. If $\det(J) = 1$ and $p \nmid \gcd(k_1, \ldots, k_m)$, it is not difficult to prove that there exists $A \in {\rm SL}_n(\mathbb C)$ such that $A^p = J$. In particular, ${\rm SL}_2(\mathbb C)$ is divisible.

>> **Question 1.** What about the other cases? 

The problem should be well-known and I _feel_ that the answer is in the negative, but so far I couldn't either get a reference or find a counterexample by myself.

>> **Question 2.** Does anyone know where to find either/both of them? 

Thanks in advance for any help.