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3.3: Dihedral Groups (Group of Symmetries)

( \newcommand{\kernel}{\mathrm{null}\,}\)

 Dihedral groups are an essential class of abstract algebra groups that arise naturally in geometry and other areas of mathematics.

Definition: Dihedral Groups

 Let n (\geq 2) \in \mathbb{Z} Then the Dihedral group D_n is defined by   D_n=<r,s | s^2 =e, r^n=e, srs=r^{-1}>=\left\{e, r,r^2,r^3, \cdots, r^{n-1}, s, rs,\cdots, r^{n-1}s \right\} and, |D_n|= 2n.

Alternatively, for  n (\geq 3) \in \mathbb{Z}, a dihedral group is a group of symmetries of a regular polygon with n sides. The dihedral group of order 2n, denoted by D_n, is the group of all possible rotations and reflections of the regular n sided polygon. in this case r=(1,2,3,\cdots, n) represents a rotation of (360/n) degrees clockwise about the center of the polygon, and s=(1,n-1)(2,n-2)(3,n-3)\cdots  represents the rotation of 180 degrees about a line through a vertex n and the center of the figure. If n is odd, s fixes only one vertex n and if If n=2m is even, s fixes  vertices n and m.

clipboard_e1777b790230b720635d939227030f0d5.png
Figure \PageIndex{1}: Reflections of polygons. (Copyright; Tikz Library)

Thus the group D_n consists of 2n elements, which can be depicted as follows:

  • n rotations, denoted by R_0, R_{360/n}, R_{(360)(2)/n} ..., R_{(n-1)360/n}, where R_{i (360)/n} represents a rotation of (360i/n) degrees clockwise about the center of the polygon, for i=1,\cdots, n..
  • n reflections, denoted by F_0, F_1, F_2, ..., F_{(n-1)}, where F_i represents a reflection across a line passing through the center of the polygon and one of its vertices.

All rigid motions on n-gon: 

a. To replace the first vertex, we have n choices.

b. the second vertex can be replaced by k^{th} vertex by k+1 or k-1.

The group operation in D_n,  is the composition of symmetries.

 

Recall from chapter 2, section 2.1 Example 11 

Example \PageIndex{1}

Consider the equilateral triangle below. What rigid motions can we make to it that will result in the triangle occupying the same space? What happens if we combine rigid motions? Keep in mind that "doing nothing" is also a rigid motion.

alt

Symmetries of equilateral triangles: 

Screen Shot 2023-06-29 at 10.23.13 AM.png 

Let r  be a rotation then there are three rotations  I, r, r^2. Let s be a reflection, then there are  s, rs, r^2s  elements.  In this case, r^3=s^2=I and sr=r^2s. Thus D_3=\{I,r,r^2,s, rs, r^2s \}.

 

Theorem \PageIndex{1}

Screen Shot 2023-07-06 at 1.54.53 PM.pngDihedral groups  D_n are non-abelian, for integers n ≥ 3.

Proof

Let a,b \in D_3.

We will show that ab \ne ba, \; \forall a,b \in D_n, n \ge 3.

D_3=\{e,(AB),(AC),(BC),(ABC),(ACB)\}. 

Since D_3 is isomorphic to S_3, D_3is non-abelian.

Since D_3 \le D_n, \forall \; n\ge 3, D_n is non-abelian for n \ge 3.

The Dihedral group D_3 is isomorphic to S_3.

Example \PageIndex{2}

Consider the possible rotations and reflections of a  4-sided regular polygon -Square, D_4.

Screen Shot 2023-07-06 at 10.21.02 AM.png

 

Cayley Table for D_4

 

R_0

R_{90}

R_{180}

R_{270}

H

V

D

D’

R_0

R_0

R_{90}

R_{180}

R_{270}

H

V

D

D’

R_{90}

R_{90}

R_{180}

R_{270}

R_0

D’

D

H

V

R_{180}

R_{180}

R_{270}

R_0

R_{90}

V

H

D’

D

R_{270}

R_{270}

R_0

R_{90}

R_{180}

D

D’

V

H

H

H

D

V

D’

R_0

R_{180}

R_{90}

R_{270}

V

V

D’

H

D

R_{180}

R_0

R_{270}

R_{90}

D

D

V

D’

H

R_{270}

R_{90}

R_0

R_{180}

D’

D’

H

D

V

R_{90}

R_{270}

R_{180}

R_0

 

Example \PageIndex{5}

D_5 dihedral group of order 10, is the group of symmetries of a regular pentagon. It is composed of 10 elements, which can be represented as rotations and reflections of the pentagon.

The Cayley table for D_5:

Screenshot 2024-04-08 120102.png

Examples abound of corporations using Dihedral groups, D_n, as their logos. Below are examples of D_1, D_3, and D_5.

Screen Shot 2023-07-06 at 12.48.16 PM.png

 

Example \PageIndex{6}

Consider the Dihedral group D_4D_4=<r,s|r^4=e, s^2=e, srs=r^{-1}>.

Screenshot 2024-04-10 083116.png

  1. Find all the subgroups lattice of D_4, the Dihedral group of order 8.

D_4=\{1,r,r^2,r^3,s,rs,r^2s,r^3s\}. Thus, the subgroup lattice would look like this:

Screen Shot 2023-07-06 at 12.56.43 PM.png

2. Prove that every subgroup of D_4 of odd order is cyclic.

Proof:

We shall show that all odd subgroups of D_4 are cyclic.

The only possible odd subgroup, which is cyclic, is \{e\}.  

Thus, every subgroup of D_4 of odd order is cyclic.◻

Theorem \PageIndex{2}

Let the group D_n=<r,s|r^n=e, s^2=e,srs=r^{-1}>.

  1. Then  r^ks=sr^{-k}.

Proof

Consider r^ks=er^ks=s^2r^ks=ssr^ks=sr^{-k}.

  1. The order of r^k is \frac{n}{\gcd(k,n)}.

Proof

We will first show that r^k=e  if and only if n|k.

Consider e=r^k

Then by the division algorithm, k=na+b where 0 \le b < n.

Thus e=r^k=r^{na+b}=r^{na}r^b=er^b=r^b.

Since the smallest possible integer, m, such that r^m=e is n, b=0.

Conversely, if n|k, then k=ns, s \in \mathbb{Z}.

Thus r^k=r^{ns}=(r^n)^s=e^s=e.

Thus r^k=e iff n|k.

Let b=r^k, \in D_4 since r is a generator of D_n.

We shall show that the smallest integer, m s.t. r^k=e is \frac{n}{k}.

Let d=\gcd(n,k).

Consider e=b^m=r^{km}.

Since this is the smallest integer m such that n|km.  Thus \frac{n}{d}|m(\frac{k}{d}).

Since d is the greatest common divisor of n and k, \frac{n}{d} and \frac{k}{d} are relatively prime.

Hence \frac{n}{d}|m(\frac{k}{d}) means that \frac{n}{d}|m.  The smallest such m is \frac{n}{d}.

Thus |r^k|= \frac{n}{\gcd(n,k)}.◻

Example \PageIndex{7}

Find the centre of D_n.

Solution:

Let D_n=<r,s|r^n=e, s^2=e,srs=r^{-1}>

We shall show that,

Z(D_n) = \{x \in D_n \mid dx = xd, \forall d \in D_n\} = \begin{cases} \{e\}, & \text{if } n \text{ is not divisible by } 2 \\ \{e, r^2\}, & \text{if } n \text{ is divisible by } 2 \end{cases}
 

Let's consider an element x in D_n that commutes with every other element in the group. Therefore, any element in D_n that commutes with every other element is either the identity element e or a rotation r^k where k is a multiple of n/2 (for even n).

  • If x is the identity element e, it commutes with every element and is in the center.
  • If x is a rotation r^k,  where k is a multiple of n/2) (for even n), or x is the identity element e (for odd n), then it commutes with every element, and it is in the center.
  • If x is a rotation r^k,  where k is not a multiple of n/2(for even n), then it commutes with all rotations but not with reflections. Therefore, it cannot commute with every element and is not in the center.

By combining these two statements, we can conclude that the center of D_n consists of the identity element e and the rotation r^{n/2} (if n is divisible by 2), or just the identity element e (if n is not divisible by 2) for even n. For odd n, the center consists only of the identity element e.

 


This page titled 3.3: Dihedral Groups (Group of Symmetries) is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Pamini Thangarajah.

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