Polygon in $\mathbb{R}^2$ whose symmetry group is $\mathbb{Z}/3\mathbb{Z}$.

Question

Question:

Describe a polygon in $\mathbb{R}^2$ whose symmetry group is $\mathbb{Z}/3\mathbb{Z}$. What is the fewest number of vertices that such a polygon can have?

My attempts:

Since the symmetry of a polygon relies on permutating its vertices I searched for some isomorphic group whose elements are permutations. I found $\mathbb{Z}/3\mathbb{Z} \cong A_3$, $A_3 = \{1, (1,2,3), (1,3,2)\}$ (even permutations of $S_3$) so I thought I needed some triangle with just rotational symmetries, however, there is no such triangle. Since $A_3$ only permutes 3 vertices, I don't know if it is useful.

Searching for polygons with only rotational symmetries I found this one:

Whose symmetry group is $\mathbb{Z}/3\mathbb{Z}$, I have also read this: Minimal symmetry group of polygon in $\mathbb{R}^2$. However, I still don't know how to come up with such a shape and prove that 6 is the fewest number of vertices it can have.

Thanks in advance.

Answer 1

Any symmetry of a polygon must permute its vertices. Reflections have order $2$, so they are not allowed. Among rotations, the only symmetries with order $3$ are rotations by $120^{\circ}$. Therefore, the only symmetries of this polygon are $120^{\circ}$ rotations. If there is a vertex that is fixed by this symmetry, then it must be the center of rotation. Since every vertex has exactly $2$ edges, there cannot be any $120^{\circ}$ rotation symmetry around that vertex. Therefore, all vertices are part of equilateral triangles around the center. If our polygon has $3$ vertices, then it is an equilateral triangle, which has reflectional symmetry. Since the number of vertices must be a multiple of $3$ because every vertex is part of a $3$-cycle of vertices, $6$ is the minimum number of vertices.

Another way to see this is that any symmetry of a polygon permutes its edges. Since there is a symmetry of order $3$ and no edge can be mapped to itself, the number of edges is a multiple of $3.$ The number of edges cannot be $3,$ because then the polygon then has additional symmetries. Therefore, the minimum number of edges is $6$. This argument avoids the special case of a vertex being the center of symmetry, so it is easier to understand.