We have $2$ congruent, regular $n$-gons in the plane. Compute the size of the symmetry group of this configuration.

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Suppose we have two congruent regular n-gons in the plane. The symmetry group of this configuration is the number of ways to pick up both polygons and put them down so that they cover the same points. Find the size of this group using the orbit-stabilizer theorem.

Conceptually, I believe that I understand what I am being asked:

Two $n$-gons are congruent if they are of the same shape and size but represented in terms of some reflections/rotations of each other. So, we have two copies of the same $n$-gon. The symmetry group of a regular $n$-gon is the dihedral group of order $2n$. If I need to place down two copies of the same $n$-gon such that they cover the same points in the $x$–$y$ plane, then I need to choose coordinates $(x,y)$ where $x, y \in D_{2n}$. In other words, I am being asked to compute the size of $D_{2n}\times D_{2n}$, which is simply $(2n)^2 = 4n^2$.

My issues lie with my understanding of group actions on sets and the orbit-stabilizer theorem. I have attempted the problem below. I am very confused about how to apply relevant definitions to the problem at hand (specifically, how to apply these definitions to both $n$-gons). Please let me know if there are any mistakes/misconceptions in my solution.

Defintions

  • We define $\pi: G \to \operatorname{Sym}(X)$ as the action of a group $G$ on some set $X$ under a homomorphism $\pi$. The group $G$ is represented in a fashion specified by the homomorphism $\pi$ as permutations of the set $X$.

  • The orbit of $x \in X$ under the group action $\pi$ is defined as $\operatorname{Orb}_{\pi}(x) = \{y: \pi(g)(x) = y$ for some $ g \in G\}$

  • The stabilizer of $x \in X$ in $G$ is $G_x = \{g \in G: \pi(g)(x)=x\}$. In other words, $G_x$ is the subset of $G$, given $\pi$, which maps $x$ to itself.

  • The orbit stabilizer theorem states that: $|G| = |\operatorname{Orb}_{\pi}(x)||G_x|$

Solution

  • Let $X$ be the Cartesian product of the two sets of vertices of both $n$-gons. Then, we defined the group action of $G$ under $\pi$ on $X$ as the set of permutations of $X$ such that the two sets of vertices cover the same points on the $x,y$-plane. For clarity, let $V_1$ and $V_2$ represent both sets of vertices. A vertex of $V_1$ can only correspond to one vertex of $V_2$ given that they are permutations of each other.

  • Any vertex of $V_1$ can be taken to any other vertex of $V_1$ via a rotation. The same is true for $V_2$. This implies that there are $n$ transformations we can apply to each vertex of an $n$-gon such that the vertex changes positions. So, there are $n(n) = n^2$ ways to combine all possible positions of both sets of vertices. This corresponds to the size of $\operatorname{Orb}_\pi(x)$. In other words, we should have that $|\operatorname{Orb}_\pi(x)| = n^2$

  • Now, if we consider what transformations fix each of the vertices of an $n$-gon, we arrive at the identiy rotation and a reflexion about a line between the origin and some vertex of the $n$-gon. So, there are $2$ transformations that we can apply which fix the vertex of an $n$-gon. Combining these transformations between the two $n$-gons yields 4 possible ways to fix $(v_1\in V_1,v_2\in V_2) \in X$. This corresponds to the size of $G_x$. In other words, we should have that $|G_x|=4$.

$\implies$ Applying the orbit-stabilizer theorem, we get that: $|G|=4n^2$.

Best Answer

First a few remarks:

  1. The group $D_{2n}$ describes the ways you can pick up a regular $n$-gon and place it back on the plane so that it covers the same points. Picking up two congruent regular $n$-gons and placing them back, there $2$ ways to choose where to place them, and then $|D_{2n}|=2n$ ways to orient them in that place. So you get more than $4n^2$ ways to place the two $n$-gons!
  2. Defining an action as a group homomorphism $\pi:\ G\ \longrightarrow\ \operatorname{Sym}(X)$, an element $g\in G$ acts on $x\in X$ by mapping $x$ to $\big(\pi(g)\big)(x)$. It is common to omit the homomorphism $\pi$ from the notation, as it is usually clear from the context, and to simply write $g\cdot x$ or even $gx$. I'll do that from here on.
  3. A cleaner, more common way to define the orbit is $$\operatorname{Orb}_{\pi}(x)=\{gx:\ g\in G\}.$$ The orbit is usually denoted $Gx$, with the homomorphism $\pi$ implicit from the context.

Now for your solution:

  1. The question concerns two congruent regular $n$-gons in the plane, and the symmetries obtained by picking them up and putting them back to cover the same region. Together the two $n$-gons have $2n$ vertices, but of course not every permutation of the vertices corresponds to an allowed symmetry. This should also be clear because you have already computed the total number of symmetries; it isn't $(2n)!$.

    Now you start off by taking for $X$ the Cartesian product of the sets of vertices of the $n$-gons, and for $G$ the group of all symmetries of this set. This Cartesian product has $n\times n=n^2$ elements, and $(n^2)!=\text{huge}$ permutations. That can't possibly be right. Stick to the original geometric setup!

Reading further, I cannot make sense of any of your arguments, so I'll just give you a rough framework for how to approach the question:

  • Take for $X$ a finite set that captures everything that goes on in the geometric picture. The $n$-gons are fully determined by their vertices, for example. The group acting on this set is simply the group given in the question; the group of symmetries of the two $n$-gons. There is no need to describe this group in any alternative way.
  • Now to apply the orbit-stabilizer theorem, consider the orbit and stabilizer of a vertex. First the easy part: What is the size of its orbit? That is, how many vertices can this vertex be mapped to?
  • Then what is the stabilizer of a vertex: How many symmetries of the two $n$-gons are there that keep one given vertex in place? How many symmetries for the $n$-gon containing that vertex? How many for the other one?

Now you have the sizes of the orbit and the stabilizer of a vertex, so you can use the orbit-stabilizer theorem to determine the size of the group of symmetries.

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