Equivalent definitions of symmetry group of regular n-gon (dihedral group)

Question

Let $P_n$ be a fixed regular convex $n$-gon in the plane. For a metric space $M$ we denote by $\text{Isom}(M)$ the set of distance-preserving maps $M \to M$. How can I show that $$ D_n := \left\{\, f \in \text{Isom}(\mathbb{R}^2) : f(P_n) = P_n \right\} \cong \text{Isom}(P_n) $$ as groups where $\mathbb{R}^2$ is given the euclidean and $P_n$ the induced metric? Also is there a general criteria or classification for subsets like $P_n \subseteq \mathbb{R}^2$ for which these two groups are isomorphic?

Answer 1

I assume that you aren't talking about abstract isomorphism, but rather that each isometry of $\mathbf{P}_n$ extends uniquely to an isometry of $\mathbf{R}^2$. The answer is contained in the following proposition.

Let $S \subseteq T \subseteq \mathbf{R}^2$, and let $f \colon S \to \mathbf{R}^2$ be a distance preserving map. Assume that $S$ is not contained in a line. Then $f$ extends uniquely to a distance preserving map from $T$ into the plane.

Here is an outline of the proof.

For uniqueness, start by taking $A, B, C \in S$ that are not aligned. Show that their images $A', B', C',$ are not aligned. Now if you take any point $M$ in $T$, and call its image $M'$, the distances $A'M'$, $B'M'$ and $C'M'$ are uniquely determined. Show that there is at most one point $M'$ in the plane for which $A'M'$, $B'M'$ and $C'M'$ take the prescribed values.

For existence, first assume that $S = \{A, B, C\}$, $T = \mathbf{R}^2$, let $A'B'C'$ be any triangle with the same side lengths as $ABC$, and let $f$ be the mapping taking $A$, $B$ and $C$ to $A'$, $B'$ and $C'$, respectively. Then first let $\phi_1$ be the reflection taking $A$ to $A'$. Write $B_1 = \phi_1(B)$. Let $\phi_2$ be the reflection with respect to the bisector of the angle $B'A'B_1$. Then since $\phi_2$ takes ray $A'B_1$ to ray $A'B'$, and the distances $A'B_1$ and $A'B'$ are equal, we have $\phi_2(B_1) = B'$. Thus $\phi_2 \circ \phi_1$ takes $A$ to $A'$ and $B$ to $B'$. Finally, let $C_2 = \phi_2 \circ \phi_1 (C)$. We have $A'C_2 = A'C'$, and $B'C_2 = B'C'$. This shows that the points $C_2$ and $C'$ are either identical or symmetric about the line $A'B'$. In the former case, $\phi_2 \circ \phi_1$ is the desired extension. In the latter, use $\phi_3 \circ \phi_2 \circ \phi_1$, where $\phi_3$ is reflection with respect to $A'B'$.

For the general case, let $f \colon S \to \mathbf{R}^2$ be a distance-preserving map. Let $A, B, C$ be three non-aligned points in $S$, and let $f_0$ denote the restriction of $f$ to $S_0 = \{A, B, C\}$. Then by the existence part above, $f_0$ has an extension $f_1$ to all of $\mathbf{R}^2$. Since $f_1|_S$ and $f$ both extend $f_0$ from $S_0$ to $S$, we must have $f_1|_S = f$ by the uniqueness part. Therefore $f_1$ extends $f$ to the entire plane.