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Since each symmetry can be thought of as a permutation of the vertices, the elements of $D_n$ can be thought of as elements of $S_n$. So I'm wondering if there's a systematic way that we can always write elements of $D_n$ in disjoint cycle notation?

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The dihedral group $D_n$ is a subgroup of $S_n$ for all $n\ge 3$, see also here, so we can write the elements in cycle notation. More precisely, $D_n$ is generated by an $n$-cycle $\sigma$ and a $2$-cycle $\tau$ satisfying the conditions $\sigma^n=\tau^2=1$, $\sigma\tau=\tau\sigma^{n-1}$. This gives a systematic way how to write the elements of $D_n$ in cycle notation. For example, for $n=3$, take $\sigma=(123)$ and $\tau=(23)$. Then $D_3=\{id,(12),(13),(23),(123),(132) \}$.

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  • $\begingroup$ Would this be affected by how we define the reflection in $D_n$ (the axis of reflection being the line connecting the edges or vertices) $\endgroup$ – mathchai Dec 15 '15 at 14:01
  • $\begingroup$ No, it would not be affected, because we only speak of a group of permutations which is isomorphic to $D_n$. There are several ways to realize these permutations. $\endgroup$ – Dietrich Burde Dec 15 '15 at 14:06
  • $\begingroup$ @DietrichBurde, you said '$D_n$ is generated by an $n$-cycle $\sigma$ and a $2$-cycle $\tau$'. But, in your example, $D_3$ contains more than one $3$-cycle and more than one $2$-cycle. Am I missing something? $\endgroup$ – Omar Shehab Apr 30 '16 at 5:00
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    $\begingroup$ @OmarShehab "generated" does not mean that the group consists only of the generators. $S_n$ is generated by $(1,2,\ldots .n)$ and $(12)$, but contains all cycles of length $\le n$, so "everything". $\endgroup$ – Dietrich Burde Apr 30 '16 at 7:45

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