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The dihedral group

Pick an integer $ n>2$ and let $ R$ be a regular $ n$-gon centered about the origin in the plane. If $ n=3$ then $ R$ is an equilateral triangle, if $ n=4$ then $ R$ is a square, if $ n=5$ then $ R$ is a pentagon, and so on. Let $ G$ denote the set of all linear transformations of the plane 5.3 to itself which preserve the figure $ R$. The binary operation $ \circ : G\times G\rightarrow G$ given by composition of functions gives $ G$ the structure of a group. This group is called the group of symmetries of $ R$. Label the vertices of the $ n$-gon as $ 1$, $ 2$, ..., $ n$. The group $ G$ permutes these vertices amongst themselves, hence each $ g\in G$ may be regarded as a permutation of the set of vertices $ V=\{1,2,...,n\}$. In this way, we may regard $ G$ as a permutation group since it is the subgroup of $ S_n$ generated by the elements of $ G$. The fact that this group has $ 2n$ elements follows from a simple counting argument: Let $ r\in G$ denote the element which rotates $ R$ by $ 2\pi/n$ radians counterclockwise about the center. Let $ L$ be a line of symmetry of $ R$ which bisects the figure into two halves. Let $ s$ denote the element of $ G$ which is reflection about $ L$. There are $ n$ rotations by a multiple of $ 2\pi/n$ radians about the center in $ G$: $ 1,r,r^2,...,r^{n-1}$. There are $ n$ elements of $ G$ which are composed of a reflection about $ L$ and a rotations by a multiple of $ 2\pi/n$ radians about the center: $ s,s\circ r,s\circ r^2,...,s\circ r^{n-1}$. These comprise all the elements of $ G$. One remarkable property of this symmetry group, which we shall use in the example in the next section, is that it is generated by any two distinct reflections in the group:

Lemma 5.9.3   Pick two distinct lines $ L,L'$ of symmetries of $ R$, each of which bisects $ R$ in half, and let $ s$, $ s'$ (resp.) denote the corresponding reflections, regarded as elements in $ S_n$. Then $ G=\langle s,s'\rangle $.

The interested reader is referred to [NST], [R], or [Ar], chapter 5, §3, for a proof. The symmetry group of $ R$ is known as the dihedral group of order $ 2n$, denoted $ D_{2n}$.

Example 5.9.4   Let $ G$ be the symmetry group of the square: i.e., the group of symmetries of the square generated by the rigid motions

\begin{displaymath}
\begin{array}{c}
g_0 = 90 \ {\rm degrees\ clockwise\ rotat...
...,\\
g_4 = {\rm reflection\ about}\ \ell_4,\\
\end{array}
\end{displaymath}

where $ \ell_1, \ell_2, \ell_3$ denote the lines of symmetry in the picture below:
\begin{picture}(400,250)(-100,100)
\put(0,220){\line(1,0){200}}
\put(-10,200...
...){$\ell_3$}
\put(0,120){\line(1,1){200}}
\put(0,100){$\ell_4$}
\end{picture}
The elements of $ G$ are

$\displaystyle 1,g_0,g_0^2,g_0^3,g_1,g_2,g_3,g_4.
$

Let $ X$ be the set of vertices of the square. Then $ G$ acts on $ X$.


next up previous contents index
Next: Example: The two squares Up: Puzzling examples Previous: Puzzling examples   Contents   Index
David Joyner 2002-08-23