If we have two congruent regular n-gons in the plane, and the symmetry group is the numbers of ways to pick up both of the polygons and put them down so that they cover the same points, what would the size of this group be? I can see that I am supposed to use the result that for a group $G$, we have that $|G| = |G_x||Orb(x)|$ for $x\in G$. I would think that the size of the orbit of any $x$ in $G$ would just be $n$ (I am not sure if that is right). Also for the stabilizer of any $x$, that would be the set of elements that when conjugated with $x$, give you $x$ back, and in this case the only element that I can think that would do that would be the identity, so then would $|G_x|=1$?
2026-03-30 04:56:40.1774846600
Figuring out size of a symmetry group
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