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Let $n \ge 3$. Number the vertices of a regular n-gon by the numbers $1,2,...,n$. Each symmetry of the n-gon corresponds to a permutation of its vertices and hence to an element of $S_n$. The subgroup of $S_n$ of symmetries of the n-gon is the subgroup generated by the rotation $\sigma = (1,2,3...n)$ and the reflection $p= (1n)(2n-1)(3n-2)...$ is denoted by $D_n$. I need to show that $|D_n|=2n$ and list the elements. So i know that $n$ has rotational symmetries and $n$ reflection symmetries. So $<x,a | a^n = x^2 =1, xax^-1 = a^-1>$. I dont know where to go from here. I also need to show that $D_6$ is isomorphic to $<R^2, U^2>_G$.

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We can show $D_{n}$ has $2n$ elements as follows. You already know $D_n$ contains $n$ rotations (generated by $\sigma$) and $n$ reflections (including $p$) and hence at least $2n$ elements. If an element of the symmetry group $D_n$ maps vertex 1 to vertex $i$, then it must map vertex 2 to a vertex adjacent to $i$, i.e. to either $i-1$ or $i+1$. Once the map for vertex 2 is determined, the map for all the other remaining vertices is automatically determined, for eg, if 2 was mapped to $i-1$, then 3 must be mapped to $i-2$ (because 3 must be mapped to a vertex adjacent to $i-1$, i.e. to $i$ or $i-2$, but $i$ is already the image of vertex 1). Thus, the total number of symmetries in $D_n$ is $n$ (the number of ways to map vertex 1) times 2 (the number of ways to map the vertex 2), i.e $2n$.

In general, one way to show there are no others symmetries is to use the Orbit-Stabilizer Lemma, which states that the size of a symmetry group $G \le S_n$ equals the number of elements in the stabilizer subgroup $G_x$ times the size of the orbit of $x$ under the action of $G$, in short $|G|=|G_x| ~|x^G|$. In the case above, the number of symmetries that fix the vertex 1 is 2, i.e. $|G_1|=2$, and $G$ takes 1 to every other vertex, whence the orbit of 1 is the set of all $n$ vertices, so $|1^G|=n$. Thus, $|G|=2n$ by the orbit-stabilizer lemma.

The elements of $D_n$ can be listed in the form $\{p^i \sigma^j: i=0,1, j=0,1,\ldots,n-1\}$.

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