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 $a^2=g$.)

9$\begingroup$ Using Jordandecomposition it is easy to show that $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ has no square root in $\mathrm{SL}(2,\mathbb{C})$. $\endgroup$ – SHP Aug 16 '17 at 13:13

1$\begingroup$ The expression "complex reductive" may cause some confusion here. The term "reductive" was apparently introduced by Borel and Tits in their 1965 treatment of algebraic groups, where the unipotent radical is required to be trivial. But Jordan decomposition is not welldefined in an arbitrary real or complex Lie group. $\endgroup$ – Jim Humphreys Aug 16 '17 at 13:32

$\begingroup$ @JimHumphreys I'm interested in connected complex linear algebraic groups for which every algebraic representation is isomorphic to a direct sum of irreducible subrepresentations. $\endgroup$ – Pete Aug 16 '17 at 13:56

1$\begingroup$ @Pete: This just means reductive algebraic group over an algebraically closed field of characteristic 0 (but the complex numbers are unnecessary for this). Over any algebraically closed field, "(connected) reductive" translates structurally into the almostdirect product of a semisimple (connected) algebraic group and a central torus. But in prime characteristic the representation theory of a semisimple group is more complicated. $\endgroup$ – Jim Humphreys Aug 16 '17 at 15:27
This question was raised, both in characteristic 0 and in arbitrary characteristic, in backtoback 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.
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 $x \in \mathfrak {g}=\operatorname{Lie}(G)$.

2$\begingroup$ Just stating the obvious for completeness sake: more generally, same argument above shows the existence of $n$th roots in compact connected Lie groups, $n\in\mathbb{N}$. $\endgroup$ – M.G. Aug 16 '17 at 15:57