$\DeclareMathOperator\ss{ss}$Let $G$ be a reductive group over a field $F$ of characteristic 0. (Here not necessarily $F=\overline{F}$.) Consider the square map
$$
G(F)\longrightarrow G(F), \quad g\mapsto g^2.
$$


I know this map is not surjective in general even when $F=\overline{F}$. But how big is the image? Is the image Zariski open?


Also what if one restricts to semisimple elements? Namely, let $G(F)^{\ss}$ be the set of semisimple elements. Is the map
$$
G(F)^{\ss}\longrightarrow G(F)^{\ss}, \quad g\mapsto g^2,
$$
surjective, say, when $F=\overline{F}$?



Image of square map on reductive group
The answers to both questions are Yes.


Let $F$ be a field of characteristic 0 and ${\overline F}$ be a fixed algebraic closure of $F$.
Let $G$ be a connected reductive $F$-group.
By abuse of notation we identify $G$ with the set of ${\overline F}$-points $G({\overline F})$.
Let $G\_s$ denote the subset of semisimple elements in $G$.
For a natural number $n>1$ we consider the map
$$\varphi\_n\colon G\to G,\quad g\mapsto g^n.$$



> 
> **Proposition 1.** $\varphi\_n(G\_s)=G\_s$.
> 
> 
> 


*Proof.*  

Let $s\in G\_s$. It is known that $s\in T$ for some maximal torus $T\subseteq G$.
For an element $s\in T$, it is clear that there exists an element $s'\in T\subseteq G\_s$
such that $(s')^n=s$, which proves the proposition.



> 
> **Theorem 2.** The set $\varphi\_n(G(F))$ is Zariski-dense in $G$.
> 
> 
> 


*Proof.* Let $U\subseteq G$ be a nonempty open subset.
Consider the open subset $U\_n:=\varphi\_n^{-1}(U)\subseteq G$.
We show that $U\_n$ is nonempty.
Indeed, consider the subset of regular semisimple elements $G\_r\subseteq G\_s\subseteq G$.
It is known that $G\_r$ is open and nonempty.
Thus $U\cap G\_r$ is open and nonempty.
Since $U\cap G\_r\subseteq G\_s$, by Proposition 1 $\varphi\_n^{-1}(U\cap G\_r)$ is nonempty.
Thus $U\_n=\varphi\_n^{-1}(U)$ is indeed nonempty.


By Corollary 18.3 in Borel's book (2nd edition) the set $G(F)$ is Zariski-dense in $G$.
Thus there exists an $F$-point $g\in G(F)\cap U\_n$. Then $g^n=\varphi\_n(g)\in \varphi\_n(G(F))\cap U$.
Thus $\varphi\_n(G(F))$ is Zariski-dense in $G$, as required.


