# Under which hypotheses do the dihedral groups specify regular polygons in plane?

For any regular polygon in the plane, we have its associated dihedral group, and my question concerns the other direction.

Say we have some path-connected subset of $\mathbb{R}^2$, what are the hypotheses under which its group of symmetries is a dihedral group iff it is a regular polygon?

My work: I first thought that path-connectedness might be a sufficient hypothesis, but I soon thought up some obvious counterexamples; however, all of my examples are nonconvex, so I suspect that maybe convexity might be sufficient, but I'm scared to attempt a proof because I feel I'd have to do a lot of hand-waving. (this is wrong, see the prof. Myerson's answer)

Another case where I haven't been able to produce a counterexample is a bounded convex, path-connected set, such that it has only $n$ vertices (more analytic/rigorous characterisation of this property may be non-differentiability at those points). Any counterexamples/proofs would be much appreciated!

One of your conjectures seems to be on target:

Proposition: If $S$ is a piece-wise linear convex closed contour with $n$ vertices (i.e. a shape that consists of $n$ straight lines joined in a closed convex shape end to end), then the symmetry group of $S$ contains $n$ rotations if and only if $S$ is a regular $n$-gon.

Corollary: If $S$ is a piece-wise linear convex closed contour with $n$ vertices, then its symmetry group is $D_{2n}$ if and only if it is a regular $n$-gon.

Proof of the Corollary: If $S$ is a regular $n$-gon, then clearly the symmetry group is dihedral. Conversely, if the symmetry group is dihedral, then by the classification of finite subgroups of $GL_2(\mathbb{R})$, the cyclic subgroup of order $n$ is what you think it is: rotations of the plane. By the above proposition, $S$ must be a regular $n$-gon.

Proof of the Proposition: This is clear: if there are $n$-rotations that preserve $S$, then all angles must be equal and all side lengths must be equal, since any angle is mapped to any angle by a rotation, and similarly for sides.

• Thanks Alex! (required characters) – user5501 Feb 29 '12 at 22:07

Convexity is not enough. Take a regular $n$-sided polygon, and cut off a tiny triangle from each vertex. You get a $2n$-sided equiangular convex polygon whose sides alternate in length, and it has exactly the same symmetries as the original $n$-sided polygon. Or replace each edge on the $n$-gon with some curve with some symmetry in such a way as to keep the shape convex. There are lot's of ways to fiddle with a polygon without affecting its symmetries.

• Thanks Gerry, that was helpful. Seeing that there are a lot of, as you say, "ways to fiddle with a polygon without affecting its symmetries", do you think it is a reasonable prospect to expect an answer to the main question? – user5501 Feb 27 '12 at 10:03