Is there a polygon in $\mathbb{R}^2$ whose symmetry group is isomorphic to $\mathbb{Z}\backslash 3\mathbb{Z}$?
I believe I found such a polygon, it’s an equilateral triangle with $3$ smaller equilateral triangles cut out on each side. My idea was to maintain the rotational symmetries, and eliminate the reflectional symmetries. This shape has $12$ vertices.
I am wondering if there is a polygon with less than $12$ vertices that has symmetry group isomorphic to $\mathbb{Z}\backslash 3\mathbb{Z}$.
Any help would be appreciated.
Best Answer
A (skew) three-pointed ninja star has six vertices:
This is minimal. Suppose we have a polygon $P$ with $\mathbb{Z}_3$ symmetry. Any orbit other than the origin has three vertices. If there were only one triangular orbit, the polygon would either be an equilateral triangle or (if we include the center) three equal line segments at $120^\circ$ from each other, and in either case the full symmetry group is dihedral. Thus there are at least two triangular orbits, for a minimum of six vertices.