> [!NOTE] Theorem (Normal Form of Dihedral Group Elements) > Let $n\geq 3$. Then [[Dihedral Group|dihedral group]] is given by $D_{2n}=R \cup sR $ where $r\in D_{2n}$ denotes the anticlockwise rotation through $2\pi/n$; $s\in D_{2n}$ denotes the reflection in the line of symmetry through vertex $1$; $R=\langle r\rangle$; and $sR$ is the [[Coset|left coset]] of $R$ containing $s$. > > That is, each symmetry of the $n$-gon can be expressed as $r^ks^l$ where $k$ and $l$ are unique modulo $n$ and $2$ respectively. ###### Proof of Normal Form of Dihedral Group Elements \[MA268\] By definition, the $D_{2n}$ is the group of symmetries of the regular $n$-gon. Label its vertices $1$ through $n$. Let $r\in D_{2n}$ denote the anticlockwise rotation through $2\pi/n$ and $s\in D_{2n}$ denote the reflection in the line of symmetry through vertex $1$. Let $b,c\in D_{2n}$ such that $b(1)=c(1)$ and $b(2)=c(2)$. Let $a=c^{-1}b\in D_{2n}$. Then $a(1)=c^{-1}(b(1))=c^{-1}(c(1))=1$ and similarly $a(2)=2$, that is $a$ fixes vertices $1$ and $2$. In this case, $a$ fixes the whole line segment joining these two vertices. But also $a$ fixes the centre of the $n$-gon. It is clear now that $a$ fixes the whole $n$-gon. Thus $a=\text{id}$ and therefore $b=c$. Now let $a\in D_{2n}$ and $k:=a(1)$. Note that $a$ must map the vertices adjacent to the vertex 1 to the vertices adjacent to vertex $k.$ The vertices adjacent to 1 are 2 and $n.$ The vertices adjacent to $k$ are $k-1$ and $k+1$ (where if $k=n$ we think of vertex $k+1$ as really being vertex 1). Thus, $\text{either}\quad a:\begin{cases}n\mapsto k-1,\\1\mapsto k,\\2\mapsto k+1\end{cases}\quad\text{or}\quad a:\begin{cases}n\mapsto k+1\\1\mapsto k,\\2\mapsto k-1.\end{cases}$ In the first case, let $b=r^k$. Note that $a(1)=k=b(1), \quad \quad a(2)=k+1=b(2),$so $a=b=r^k$. Suppose we're now in the second case. Note that$s:\begin{cases}n\mapsto2\\1\mapsto1\\2\mapsto n.\end{cases}$Thus$as\::\:\begin{cases}n\mapsto k-1,\\1\mapsto k,\\2\mapsto k+1.\end{cases}$Hence $as=r^k$, so $a=r^ks^{-1}=r^ks$. We conclude that $D_{2n}=R \cup sR.$