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I want to show that a polygon $P$ is regular, i.e. all the angles are equal, iff $\Sigma(P)$, the symmetry group, acts transitively on Vert$(P)$.

S_Alex
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  • There's a difference between the "symmetric" group and the "symmetry" group. The first one doesn't act on the polygon (unless it is a triangle), the second one by definition does. – Matt Samuel Jun 10 '20 at 20:08
  • I mean symmetry group – S_Alex Jun 10 '20 at 20:09
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    Perhaps I don't fully understand your set up, but a rectangle has a symmetry group that is transitive on its vertices, even if the rectangle is not a square (i.e. not a regular polygon). – hardmath Jun 10 '20 at 20:14
  • You're right. I modified the question. By regularity I just mean that the angles are equal. Then both directions are not clear for me. – S_Alex Jun 10 '20 at 20:50

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The term equiangular polygon (Wikipedia) is preferred to describe the condition that all (interior) angles of a polygon are equal. To be regular a polygon is conventionally defined to be both equiangular and equilateral (all sides of equal length).

It is easy enough to show that if the symmetry group of a polygon is vertex transitive then all interior angles are equal. Indeed if some symmetry of the polygon maps vertex $V$ to vertex $V'$, then that map establishes the equality of the angles at $V,V'$.

The other direction is not true. An equiangular polygon need not have a vertex transitive symmetry group. Indeed something of the sort was already shown by the earlier Question Are all equiangular odd polygons also equilateral? The example described there of an equiangular pentagon which is not equilateral perturbs a regular pentagon so that instead of having all sides equal, one gets an equiangular pentagon with bilateral symmetry but not a vertex transitive symmetry group.

Indeed one might as easily have an equiangular pentagon with all five edge lengths different, so that the symmetry group is trivial and thus vertex transitivity is lacking. Wikimedia contributor J Hokkanen posted an image of such a pentagon (licensed CC A-S-A 3.0):

equiangular pentagon without any equal sides

It is possible to introduce additional assumptions in order to draw the conclusion of vertex transitive symmetry (or even regularity) from equiangularity. The Wikipedia article linked above might suggest some ideas of this kind, and if you are interested I can elaborate.

hardmath
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