Theorem 10.20.
The dihedral group, \(D_n\text{,}\) is a subgroup of \(S_n\) of order \(2n\text{.}\)
Section 10.2 Dihedral Groups
Another special type of permutation group is the dihedral group. Recall the symmetry group of an equilateral triangle in Chapter 8. Such groups consist of the rigid motions of a regular \(n\)-sided polygon or \(n\)-gon. For \(n = 3, 4, \ldots\text{,}\) we define the nth dihedral group to be the group of rigid motions of a regular \(n\)-gon. We will denote this group by \(D_n\text{.}\) We can number the vertices of a regular \(n\)-gon by \(1, 2, \ldots, n\) (Figure 10.19). Notice that there are exactly \(n\) choices to replace the first vertex. If we replace the first vertex by \(k\text{,}\) then the second vertex must be replaced either by vertex \(k+1\) or by vertex \(k-1\text{;}\) hence, there are \(2n\) possible rigid motions of the \(n\)-gon. We summarize these results in the following theorem.
Theorem 10.23.
The group \(D_n\text{,}\) \(n \geq 3\text{,}\) consists of all products of the two elements \(r\) and \(s\text{,}\) satisfying the relations
\begin{align*}
r^n & = 1\\
s^2 & = 1\\
srs & = r^{-1}.
\end{align*}
Proof.
The possible motions of a regular \(n\)-gon are either reflections or rotations (Figure 10.21). There are exactly \(n\) possible rotations:
\begin{equation*}
\identity, \frac{360^{\circ} }{n}, 2 \cdot \frac{360^{\circ} }{n}, \ldots, (n-1) \cdot \frac{360^{\circ} }{n}.
\end{equation*}
We will denote the rotation \(360^{\circ} /n\) by \(r\text{.}\) The rotation \(r\) generates all of the other rotations. That is,
\begin{equation*}
r^k = k \cdot \frac{360^{\circ} }{n}.
\end{equation*}
Label the \(n\) reflections \(s_1, s_2, \ldots, s_n\text{,}\) where \(s_k\) is the reflection that leaves vertex \(k\) fixed. There are two cases of reflections, depending on whether \(n\) is even or odd. If there are an even number of vertices, then two vertices are left fixed by a reflection, and \(s_1 = s_{n/2 + 1}, s_2 = s_{n/2 + 2}, \ldots, s_{n/2} = s_n\text{.}\) If there are an odd number of vertices, then only a single vertex is left fixed by a reflection and \(s_1, s_2, \ldots, s_n\) are distinct (Figure 10.22). In either case, the order of each \(s_k\) is two. Let \(s = s_1\text{.}\) Then \(s^2 = 1\) and \(r^n = 1\text{.}\) Since any rigid motion \(t\) of the \(n\)-gon replaces the first vertex by the vertex \(k\text{,}\) the second vertex must be replaced by either \(k+1\) or by \(k-1\text{.}\) If the second vertex is replaced by \(k+1\text{,}\) then \(t = r^k\text{.}\) If the second vertex is replaced by \(k-1\text{,}\) then \(t = s r^k\text{.}\) Hence, \(r\) and \(s\) generate \(D_n\text{.}\) That is, \(D_n\) consists of all finite products of \(r\) and \(s\text{,}\)
\begin{equation*}
D_n = \{1, r, r^2, \ldots, r^{n-1}, s, sr, sr^2, \ldots, sr^{n-1}\}.
\end{equation*}
We will leave the proof that \(srs = r^{-1}\) as an exercise.
Example 10.24.
The group of rigid motions of a square, \(D_4\text{,}\) consists of eight elements. With the vertices numbered \(1\text{,}\) \(2\text{,}\) \(3\text{,}\) \(4\) (Figure 10.25), the rotations are
\begin{align*}
r & = (1234)\\
r^2 & = (13)(24)\\
r^3 & = (1432)\\
r^4 & = (1)
\end{align*}
and the reflections are
\begin{align*}
s_1 & = (24)\\
s_2 & = (13).
\end{align*}
The order of \(D_4\) is \(8\text{.}\) The remaining two elements are
\begin{align*}
r s_1 & = (12)(34)\\
r^3 s_1 & = (14)(23).
\end{align*}
Subsection The Motion Group of a Cube
We can investigate the groups of rigid motions of geometric objects other than a regular \(n\)-sided polygon to obtain interesting examples of permutation groups. Let us consider the group of rigid motions of a cube. One of the first questions that we can ask about this group is “what is its order?” A cube has \(6\) sides. If a particular side is facing upward, then there are four possible rotations of the cube that will preserve the upward-facing side. Hence, the order of the group is \(6 \cdot 4 = 24\text{.}\) We have just proved the following proposition.
Proposition 10.27.
The group of rigid motions of a cube contains \(24\) elements.
Theorem 10.28.
The group of rigid motions of a cube is \(S_4\text{.}\)
Proof.
From Proposition 10.27, we already know that the motion group of the cube has \(24\) elements, the same number of elements as there are in \(S_4\text{.}\) There are exactly four diagonals in the cube. If we label these diagonals \(1\text{,}\) \(2\text{,}\) \(3\text{,}\) and \(4\text{,}\) we must show that the motion group of the cube will give us any permutation of the diagonals (Figure 10.26). If we can obtain all of these permutations, then \(S_4\) and the group of rigid motions of the cube must be the same. To obtain a transposition we can rotate the cube \(180^{\circ}\) about the axis joining the midpoints of opposite edges (Figure 10.29). There are six such axes, giving all transpositions in \(S_4\text{.}\) Since every element in \(S_4\) is the product of a finite number of transpositions, the motion group of a cube must be \(S_4\text{.}\)
