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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!

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2 Answers 2

up vote 1 down vote accepted

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.

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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.

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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

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