Let $X$ be a smooth irreducible subvariety of an algebraic group over a field, assume $X$ is invariant under $n$-th power map for every integer $n$ ($n=0$ means the identity element is in $X$). Must $X$ be a subgroup?


The motivation for this question is the intuition that if a cone in a linear space is smooth, then it shall be a linear subspace.



Smooth irreducible subvarieties in an algebraic group that are stable under power maps
This is false in nonabelian unipotent groups. For these groups, the exponential map is algebraic, and an isomorphism. The image of any linear subspace under this map will be smooth, irreducible, and invariant under the $n$th power map. But it will not be a subgroup unless the subspace is closed under the Lie bracket. For instance, the set of matrices of the form


$$1+ \begin{pmatrix} 0 & a & 0 \\ 0 & 0 & b \\ 0 & 0 & 0 \end{pmatrix}+ \begin{pmatrix} 0 & a & 0 \\ 0 & 0 & b \\ 0 & 0 & 0 \end{pmatrix}^2/2 + \dots = \begin{pmatrix} 1 & a & ab/2 \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix} $$ is smooth, irreducible, and closed under $n$th powers since $$\begin{pmatrix} 1 & a & ab/2 \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix}^n =\begin{pmatrix} 1 & na & n^2 ab/2 \\ 0 & 1 & nb \\ 0 & 0 & 1 \end{pmatrix} ,$$ but isn't a subgroup.


Since most reductive groups contain nonabelian unipotent subgroups, this will not be true for reductive groups either.


