Square root in complex reductive groups Let $G$ be a connected complex reductive linear algebraic group. Does every $g\in G$ have a square root? (That is, some $a\in G$ such that $a^2=g$.)
 A: As the comment shows the answer is negative in general. Perhaps it is worth to mention that for connected compact Lie groups the answer is yes, because its
exponential map is surjective. In general, if the exponential map of a Lie group $G$ is surjective, then every group element $g \in G$ has a square root, i.e. an element $a \in G$ with $a^2 = g$, since $\exp(x)$ has $\exp(x/2)$ as a square root for any $x \in \mathfrak {g}=\operatorname{Lie}(G)$.
A: This question was raised, both in characteristic 0 and in arbitrary characteristic, in back-to-back 2003 papers by R. Steinberg here and P. Chatterjee here.   I recall reviewing these papers together for Math Reviews.   The answer is similar for all power maps (not just squares), but the adjective "reductive" is replaced in both papers by the more precise "semisimple".  (It's best to treat algebraic tori separately.)     
Even in characteristic 0 the answer to your question is sometimes negative, but these two papers do give precise criteria for all power maps.  
