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Dihedral Groups

Another special type of permutation group is the dihedral group. Recall the symmetry group of an equilateral triangle in . Such groups consist of the rigid motions of a regular n-sided polygon or n-gon.

Dihedral Groups

Another special type of permutation group is the dihedral group. Recall the symmetry group of an equilateral triangle in . Such groups consist of the rigid motions of a regular \(n\)-sided polygon or \(n\)-gon. For \(n = 3, 4, \ldots\), we define the nth dihedral group to be the group of rigid motions of a regular \(n\)-gon. We will denote this group by \(D_n\). \(D_n\) the dihedral group We can number the vertices of a regular \(n\)-gon by \(1, 2, \ldots, n\) (). Notice that there are exactly \(n\) choices to replace the first vertex. If we replace the first vertex by \(k\), then the second vertex must be replaced either by vertex \(k+1\) or by vertex \(k-1\); hence, there are \(2n\) possible rigid motions of the \(n\)-gon. We summarize these results in the following theorem.

Theorem

The group \(D_n\), \(n \geq 3\), consists of all products of the two elements \(r\) and \(s\), where \(r\) has order \(n\) and \(s\) has order \(2\), and these two elements satisfy the relation \(srs = r^{-1}\).

Proof

The possible motions of a regular \(n\)-gon are either reflections or rotations (). There are exactly \(n\) possible rotations: \[\begin{aligned}\end{aligned}\]. We will denote the rotation \(360^{\circ} /n\) by \(r\). The rotation \(r\) generates all of the other rotations. That is, \[\begin{aligned}\end{aligned}\].

Label the \(n\) reflections \(s_1, s_2, \ldots, s_n\), 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 \(s_1 = s_{n/2 + 1}, s_2 = s_{n/2 + 2}, \ldots, s_{n/2} = s_n\). 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 (). In either case, the order of each \(s_k\) is two. Let \(s = s_1\). Then \(s^2 = 1\) and \(r^n = 1\). Since any rigid motion \(t\) of the \(n\)-gon replaces the first vertex by the vertex \(k\), the second vertex must be replaced by either \(k+1\) or by \(k-1\). If the second vertex is replaced by \(k+1\), then \(t = r^k\). If the second vertex is replaced by \(k-1\), then \(t = r^k s\).Since we are in an abstract group, we will adopt the convention that group elements are multiplied left to right. Hence, \(r\) and \(s\) generate \(D_n\). That is, \(D_n\) consists of all finite products of \(r\) and \(s\), \[\begin{aligned}\end{aligned}\]. We will leave the proof that \(srs = r^{-1}\) as an exercise.

Example

The group of rigid motions of a square, \(D_4\), consists of eight elements. With the vertices numbered \(1\), \(2\), \(3\), \(4\) (), the rotations are \[\begin{aligned}r & = (1 \, 2 \, 3 \, 4) \\ r^2 & = (1 \, 3)(2 \, 4) \\ r^3 & = (1 \, 4 \, 3 \, 2) \\ r^4 & = (1)\end{aligned}\] and the reflections are \[\begin{aligned}s_1 & = (2 \, 4) \\ s_2 & = (1 \, 3)\end{aligned}\]. The order of \(D_4\) is \(8\). The remaining two elements are \[\begin{aligned}r s_1 & = (1 \, 2)(3 \, 4) \\ r^3 s_1 & = (1 \, 4)(2 \, 3)\end{aligned}\].

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. By rigid motion, we mean a rotation with the axis of rotation about opposing faces, edges, or vertices. 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 \(6 \cdot 4 = 24\). We have just proved the following proposition.

Theorem

The group of rigid motions of a cube is \(S_4\).

Proof

From , we already know that the motion group of the cube has \(24\) elements, the same number of elements as there are in \(S_4\). There are exactly four diagonals in the cube. If we label these diagonals \(1\), \(2\), \(3\), and \(4\), we must show that the motion group of the cube will give us any permutation of the diagonals (). 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 (). There are six such axes, giving all transpositions in \(S_4\). Since every element in \(S_4\) is the product of a finite number of transpositions, the motion group of a cube must be \(S_4\).

A permutation group is a very concrete representation of a group, and Sage support for permutations groups is very good making Sage a natural place for beginners to learn about group theory.

Symbols used here

^\circ

degrees
1/360 of a full turn. 180° = π radians.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

x \in A,\ A \subseteq B

element of, subset
x belongs to A; every element of A is in B.

\blacksquare\ \text{or}\ \square

end of proof (halmos)
Marks the point where the statement has been established.

a \equiv b \pmod n

congruent modulo n
n divides a − b; a and b have the same remainder.

a \mid b,\ \gcd(a,b)

divides, greatest common divisor
b is a multiple of a; the largest number dividing both.

(G, \cdot),\ e,\ g^{-1}

group, identity, inverse
A set with an operation; the do-nothing element; the element that undoes g.

G \cong H,\ G / N

isomorphic, quotient group
Same structure; the group of cosets of a normal subgroup N.

\mathbb{Z}/n\mathbb{Z},\ \mathbb{Z}_n

integers modulo n
The remainders 0…n−1 with clock arithmetic.

\operatorname{Hom}(A, B),\ f \circ g

arrows from A to B, composition
The set of morphisms; do g then f.

Questions people ask

What is a group, in plain words?

A set with one operation that is associative, has an identity, and lets every element be undone. Symmetries of any object form a group — that is where the idea came from.

What is the difference between a ring and a field?

A ring has addition and multiplication that behave like the integers (you cannot always divide); a field is a ring where every non-zero element has a reciprocal, like the rationals or the reals.

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Parts of this page are adapted from Judson, Abstract Algebra: Theory and Applications (GFDL 1.3). Condensed and re-explained here; errors are ours.

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