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Unit II-Lesson 3- Dihedral and Multiplicative Groups (PALMA).pdf

The document discusses dihedral groups Dn, which are the symmetry groups of regular polygons with n sides. It specifically examines D4, the symmetry group of a square with 8 elements - the 4 rotations (r0, r1, r2, r3) and 4 reflections (s0, s1, s2, s3). It then shows that D4 forms a group by verifying the group properties of closure, associativity, identity, and inverse elements. The document will next explore the symmetry group D5 of a pentagon.

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MTH 601 ABSTRACT ALGEBRA Dihedral Groups and Multiplicative Group Jayson C. Palma (Reporter) PhD Math Student
a dihedral group is the group of symmetries of a regular polygon, which includes rotations and reflections. denoted by Dn
Dn = 2n elements D4 = 2(4) = 8 elements D8 = 2(8) = 16 elements
We will explore D4, the eight symmetries of a square. A B C D A B C D r0 = A B C D A B C D
We will explore D4, the eight symmetries of a square. A B C D D A B C r1 = A B C D B C D A
We will explore D4, the eight symmetries of a square. A B C D C D A B r2 = A B C D C D A B
We will explore D4, the eight symmetries of a square. A B C D B C D A r3 = A B C D D A B C
We will explore D4, the eight symmetries of a square. A B C D D B s0 = A B C D A D C B A C
We will explore D4, the eight symmetries of a square. A C s1 = A B C D C B A D C A D B B D
We will explore D4, the eight symmetries of a square. A B C D C A s2 = A B C D D C B A D B
We will explore D4, the eight symmetries of a square. A B C D A C s3 = A B C D B A D C B D
r0 = A B C D A B C D r1 = A B C D B C D A r2 = A B C D C D A B r3 = A B C D D A B C s0 = A B C D A D C B s1 = A B C D C B A D s2 = A B C D D C B A s3 = A B C D B A D C
r0 r1 r2 r3 s0 s1 s2 s3 r0 r1 r2 r3 s0 s1 s2 s3
r0r0 = A B C D A B C D A B C D A B C D = A B C D A B C D A B C D A B C D A B C D = = r0
r0 r1 r2 r3 s0 s1 s2 s3 r0 r1 r2 r3 s0 s1 s2 s3 r0
r0r1 = A B C D A B C D A B C D B C D A = A B C D B C D A B C D A A B C D B C D A = = r1
r0 r1 r2 r3 s0 s1 s2 s3 r0 r1 r2 r3 s0 s1 s2 s3 r1
r0r2 = A B C D A B C D A B C D C D A B = A B C D C D A B C D A B A B C D C D A B = = r2
r0 r1 r2 r3 s0 s1 s2 s3 r0 r1 r2 r3 s0 s1 s2 s3 r2
r3s2 = A B C D D A B C A B C D D C B A = A B C D D C B A C B A D A B C D C B A D = = s1
r0 r1 r2 r3 s0 s1 s2 s3 r0 r1 r2 r3 s0 s1 s2 s3 s1
s2s1 = A B C D D C B A A B C D C B A D = A B C D C B A D B C D A A B C D B C D A = = r1
r0 r1 r2 r3 s0 s1 s2 s3 r0 r1 r2 r3 s0 s1 s2 s3 r1
r0 r1 r2 r3 s0 s1 s2 s3 r0 r0 r1 r2 r3 s0 s1 s2 s3 r1 r1 r2 r3 r0 s3 s2 s0 s1 r2 r2 r3 r0 r1 s1 s0 s3 s2 r3 r3 r0 r1 r2 s2 s3 s1 s0 s0 s0 s2 s1 s3 r0 r2 r1 r3 s1 s1 s3 s0 s2 r2 r0 r3 r1 s2 s2 s1 s3 s0 r3 r1 r0 r2 s3 s3 s0 s2 s1 r1 r3 r2 r0
r0 r1 r2 r3 s0 s1 s2 s3 r0 r0 r1 r2 r3 s0 s1 s2 s3 r1 r1 r2 r3 r0 s3 s2 s0 s1 r2 r2 r3 r0 r1 s1 s0 s3 s2 r3 r3 r0 r1 r2 s2 s3 s1 s0 s0 s0 s2 s1 s3 r0 r2 r1 r3 s1 s1 s3 s0 s2 r2 r0 r3 r1 s2 s2 s1 s3 s0 r3 r1 r0 r2 s3 s3 s0 s2 s1 r1 r3 r2 r0 Closure s2s1 = r1 r3s0 = s2
r0 r1 r2 r3 s0 s1 s2 s3 r0 r0 r1 r2 r3 s0 s1 s2 s3 r1 r1 r2 r3 r0 s3 s2 s0 s1 r2 r2 r3 r0 r1 s1 s0 s3 s2 r3 r3 r0 r1 r2 s2 s3 s1 s0 s0 s0 s2 s1 s3 r0 r2 r1 r3 s1 s1 s3 s0 s2 r2 r0 r3 r1 s2 s2 s1 s3 s0 r3 r1 r0 r2 s3 s3 s0 s2 s1 r1 r3 r2 r0 Associativity (r2s1)s3 = r2(s1s3) s0 s3 = r2 r1
r0 r1 r2 r3 s0 s1 s2 s3 r0 r0 r1 r2 r3 s0 s1 s2 s3 r1 r1 r2 r3 r0 s3 s2 s0 s1 r2 r2 r3 r0 r1 s1 s0 s3 s2 r3 r3 r0 r1 r2 s2 s3 s1 s0 s0 s0 s2 s1 s3 r0 r2 r1 r3 s1 s1 s3 s0 s2 r2 r0 r3 r1 s2 s2 s1 s3 s0 r3 r1 r0 r2 s3 s3 s0 s2 s1 r1 r3 r2 r0 Associativity (r2s1)s3 = r2(s1s3) s0 s3 = r2 r1 r3 = r3
r0 r1 r2 r3 s0 s1 s2 s3 r0 r0 r1 r2 r3 s0 s1 s2 s3 r1 r1 r2 r3 r0 s3 s2 s0 s1 r2 r2 r3 r0 r1 s1 s0 s3 s2 r3 r3 r0 r1 r2 s2 s3 s1 s0 s0 s0 s2 s1 s3 r0 r2 r1 r3 s1 s1 s3 s0 s2 r2 r0 r3 r1 s2 s2 s1 s3 s0 r3 r1 r0 r2 s3 s3 s0 s2 s1 r1 r3 r2 r0 Associativity (s0s3)r2 = s0(s3r2) r3 r2 = s0s2
r0 r1 r2 r3 s0 s1 s2 s3 r0 r0 r1 r2 r3 s0 s1 s2 s3 r1 r1 r2 r3 r0 s3 s2 s0 s1 r2 r2 r3 r0 r1 s1 s0 s3 s2 r3 r3 r0 r1 r2 s2 s3 s1 s0 s0 s0 s2 s1 s3 r0 r2 r1 r3 s1 s1 s3 s0 s2 r2 r0 r3 r1 s2 s2 s1 s3 s0 r3 r1 r0 r2 s3 s3 s0 s2 s1 r1 r3 r2 r0 Associativity (s0s3)r2 = s0(s3r2) r3 r2 = s0s2 r1 = r1
r0 r1 r2 r3 s0 s1 s2 s3 r0 r0 r1 r2 r3 s0 s1 s2 s3 r1 r1 r2 r3 r0 s3 s2 s0 s1 r2 r2 r3 r0 r1 s1 s0 s3 s2 r3 r3 r0 r1 r2 s2 s3 s1 s0 s0 s0 s2 s1 s3 r0 r2 r1 r3 s1 s1 s3 s0 s2 r2 r0 r3 r1 s2 s2 s1 s3 s0 r3 r1 r0 r2 s3 s3 s0 s2 s1 r1 r3 r2 r0 Identity r3e = r3 r3 r0 = r3 e = r0
r0 r1 r2 r3 s0 s1 s2 s3 r0 r0 r1 r2 r3 s0 s1 s2 s3 r1 r1 r2 r3 r0 s3 s2 s0 s1 r2 r2 r3 r0 r1 s1 s0 s3 s2 r3 r3 r0 r1 r2 s2 s3 s1 s0 s0 s0 s2 s1 s3 r0 r2 r1 r3 s1 s1 s3 s0 s2 r2 r0 r3 r1 s2 s2 s1 s3 s0 r3 r1 r0 r2 s3 s3 s0 s2 s1 r1 r3 r2 r0 Identity s2e = s2 s2 r0 = s2 e = r0
r0 r1 r2 r3 s0 s1 s2 s3 r0 r0 r1 r2 r3 s0 s1 s2 s3 r1 r1 r2 r3 r0 s3 s2 s0 s1 r2 r2 r3 r0 r1 s1 s0 s3 s2 r3 r3 r0 r1 r2 s2 s3 s1 s0 s0 s0 s2 s1 s3 r0 r2 r1 r3 s1 s1 s3 s0 s2 r2 r0 r3 r1 s2 s2 s1 s3 s0 r3 r1 r0 r2 s3 s3 s0 s2 s1 r1 r3 r2 r0 Inverse s2 a- = e s2 s2= r0
r0 r1 r2 r3 s0 s1 s2 s3 r0 r0 r1 r2 r3 s0 s1 s2 s3 r1 r1 r2 r3 r0 s3 s2 s0 s1 r2 r2 r3 r0 r1 s1 s0 s3 s2 r3 r3 r0 r1 r2 s2 s3 s1 s0 s0 s0 s2 s1 s3 r0 r2 r1 r3 s1 s1 s3 s0 s2 r2 r0 r3 r1 s2 s2 s1 s3 s0 r3 r1 r0 r2 s3 s3 s0 s2 s1 r1 r3 r2 r0 Inverse s0 a- = e s0s0 = r0
We will explore D5, the ten symmetries of a pentagon. 1 2 3 4 1 2 3 4 r0 = 1 2 3 4 5 1 2 3 4 5 5 5
We will explore D5, the ten symmetries of a pentagon. 1 2 3 4 5 1 2 3 r1 = 1 2 3 4 5 2 3 4 5 5 4 1
We will explore D5, the ten symmetries of a pentagon. 1 2 3 4 4 5 1 2 r2 = 1 2 3 4 5 3 4 5 1 5 3 2
We will explore D5, the ten symmetries of a pentagon. 1 2 3 4 3 4 5 1 r3 = 1 2 3 4 5 4 5 1 2 5 2 3
We will explore D5, the ten symmetries of a pentagon. 1 2 3 4 2 3 4 5 r4 = 1 2 3 4 5 5 1 2 3 5 1 4
We will explore D5, the ten symmetries of a pentagon. 2 3 4 5 4 3 s0 = 1 2 3 4 5 1 5 4 3 5 2 2 1 1
We will explore D5, the ten symmetries of a pentagon. 1 3 4 1 5 s1 = 1 2 3 4 5 3 2 1 5 5 4 4 3 2 2
We will explore D5, the ten symmetries of a pentagon. 1 2 4 2 s2 = 1 2 3 4 5 5 4 3 2 5 1 1 5 4 3 3
We will explore D5, the ten symmetries of a pentagon. 1 2 3 s3 = 1 2 3 4 5 2 1 5 4 5 3 3 2 1 5 4 4
We will explore D5, the ten symmetries of a pentagon. 1 2 3 4 s4 = 1 2 3 4 5 4 3 2 1 5 4 3 2 1 5 5
r0 = 1 2 3 4 5 1 2 3 4 5 r1 = 1 2 3 4 5 2 3 4 5 1 r2 = 1 2 3 4 5 3 4 5 1 2 r3 = 1 2 3 4 5 4 5 1 2 3 r4 = 1 2 3 4 5 5 1 2 3 4 s0 = 1 2 3 4 5 1 5 4 3 2 s1 = 1 2 3 4 5 3 2 1 5 4 s2 = 1 2 3 4 5 5 4 3 2 1 s3 = 1 2 3 4 5 2 1 5 4 3 s4 = 1 2 3 4 5 4 3 2 1 5
r0 r1 r2 r3 r4 s0 s1 s2 s3 s4 r0 r1 r2 r3 r4 s0 s1 s2 s3 s4
r4s1 = 1 2 3 4 5 5 1 2 3 4 = 2 1 5 4 = = s3 1 2 3 4 5 3 2 1 5 4 1 2 3 4 5 3 2 1 5 4 3 1 2 3 4 5 2 1 5 4 3
r0 r1 r2 r3 r4 s0 s1 s2 s3 s4 r0 r1 r2 r3 r4 s3 s0 s1 s2 s3 s4
s1s0 = 1 2 3 4 5 3 2 1 5 4 = 3 4 5 1 = = r2 1 2 3 4 5 1 5 4 3 2 1 2 3 4 5 1 5 4 3 2 2 1 2 3 4 5 3 4 5 1 2
r0 r1 r2 r3 r4 s0 s1 s2 s3 s4 r0 r1 r2 r3 r4 s0 s1 r2 s2 s3 s4
r0 r1 r2 r3 r4 s0 s1 s2 s3 s4 r0 r0 r1 r2 r3 r4 s0 s1 s2 s3 s4 r1 r1 r2 r3 r4 r0 s3 s4 s0 s1 s2 r2 r2 r3 r4 r0 r1 s1 s2 s3 s4 s0 r3 r3 r4 r0 r1 r2 s4 s0 s1 s2 s3 r4 r4 r0 r1 r2 r3 s2 s3 s4 s0 s1 s0 s0 s2 s4 s1 s3 r0 r3 r1 r4 r2 s1 s1 s3 s0 s2 s4 r2 r0 r3 r1 r4 s2 s2 s4 s1 s3 s0 r4 r2 r0 r3 r1 s3 s3 s0 s2 s4 s1 r1 r4 r2 r0 r3 s4 s4 s1 s3 s0 s2 r3 r1 r4 r2 r0
r0 r1 r2 r3 r4 s0 s1 s2 s3 s4 r0 r0 r1 r2 r3 r4 s0 s1 s2 s3 s4 r1 r1 r2 r3 r4 r0 s3 s4 s0 s1 s2 r2 r2 r3 r4 r0 r1 s1 s2 s3 s4 s0 r3 r3 r4 r0 r1 r2 s4 s0 s1 s2 s3 r4 r4 r0 r1 r2 r3 s2 s3 s4 s0 s1 s0 s0 s2 s4 s1 s3 r0 r3 r1 r4 r2 s1 s1 s3 s0 s2 s4 r2 r0 r3 r1 r4 s2 s2 s4 s1 s3 s0 r4 r2 r0 r3 r1 s3 s3 s0 s2 s4 s1 r1 r4 r2 r0 r3 s4 s4 s1 s3 s0 s2 r3 r1 r4 r2 r0 Closure s3r2 = s2 r4s3 = s0
r0 r1 r2 r3 r4 s0 s1 s2 s3 s4 r0 r0 r1 r2 r3 r4 s0 s1 s2 s3 s4 r1 r1 r2 r3 r4 r0 s3 s4 s0 s1 s2 r2 r2 r3 r4 r0 r1 s1 s2 s3 s4 s0 r3 r3 r4 r0 r1 r2 s4 s0 s1 s2 s3 r4 r4 r0 r1 r2 r3 s2 s3 s4 s0 s1 s0 s0 s2 s4 s1 s3 r0 r3 r1 r4 r2 s1 s1 s3 s0 s2 s4 r2 r0 r3 r1 r4 s2 s2 s4 s1 s3 s0 r4 r2 r0 r3 r1 s3 s3 s0 s2 s4 s1 r1 r4 r2 r0 r3 s4 s4 s1 s3 s0 s2 r3 r1 r4 r2 r0 Associativity (s0r4)s3 = s0(r4s3) s3 = s0 s3 s0
r0 r1 r2 r3 r4 s0 s1 s2 s3 s4 r0 r0 r1 r2 r3 r4 s0 s1 s2 s3 s4 r1 r1 r2 r3 r4 r0 s3 s4 s0 s1 s2 r2 r2 r3 r4 r0 r1 s1 s2 s3 s4 s0 r3 r3 r4 r0 r1 r2 s4 s0 s1 s2 s3 r4 r4 r0 r1 r2 r3 s2 s3 s4 s0 s1 s0 s0 s2 s4 s1 s3 r0 r3 r1 r4 r2 s1 s1 s3 s0 s2 s4 r2 r0 r3 r1 r4 s2 s2 s4 s1 s3 s0 r4 r2 r0 r3 r1 s3 s3 s0 s2 s4 s1 r1 r4 r2 r0 r3 s4 s4 s1 s3 s0 s2 r3 r1 r4 r2 r0 Associativity (s0r4)s3 = s0(r4s3) s3 = s0 s3 s0 r0 = r0
r0 r1 r2 r3 r4 s0 s1 s2 s3 s4 r0 r0 r1 r2 r3 r4 s0 s1 s2 s3 s4 r1 r1 r2 r3 r4 r0 s3 s4 s0 s1 s2 r2 r2 r3 r4 r0 r1 s1 s2 s3 s4 s0 r3 r3 r4 r0 r1 r2 s4 s0 s1 s2 s3 r4 r4 r0 r1 r2 r3 s2 s3 s4 s0 s1 s0 s0 s2 s4 s1 s3 r0 r3 r1 r4 r2 s1 s1 s3 s0 s2 s4 r2 r0 r3 r1 r4 s2 s2 s4 s1 s3 s0 r4 r2 r0 r3 r1 s3 s3 s0 s2 s4 s1 r1 r4 r2 r0 r3 s4 s4 s1 s3 s0 s2 r3 r1 r4 r2 r0 Identity s2 = s2 s2 = s2 r0 e
r0 r1 r2 r3 r4 s0 s1 s2 s3 s4 r0 r0 r1 r2 r3 r4 s0 s1 s2 s3 s4 r1 r1 r2 r3 r4 r0 s3 s4 s0 s1 s2 r2 r2 r3 r4 r0 r1 s1 s2 s3 s4 s0 r3 r3 r4 r0 r1 r2 s4 s0 s1 s2 s3 r4 r4 r0 r1 r2 r3 s2 s3 s4 s0 s1 s0 s0 s2 s4 s1 s3 r0 r3 r1 r4 r2 s1 s1 s3 s0 s2 s4 r2 r0 r3 r1 r4 s2 s2 s4 s1 s3 s0 r4 r2 r0 r3 r1 s3 s3 s0 s2 s4 s1 r1 r4 r2 r0 r3 s4 s4 s1 s3 s0 s2 r3 r1 r4 r2 r0 Inverse r4 = e r4 = r0 r1 a-
Dihedral Group Symbol Our Thoughts D1 Shell Petroleum uses the symbol to the left. This shell shape has no rotations (other than the identity) and has only one mirror line (vertical). Therefore, like Mickey Mouse, the figure is said to be bilaterally symmetric and it fits into the category D1. D2 An example of D2 that is easily spotted is the logo for the Columbia Broadcasting System (CBS). The "eye" shape within the circle prevents the figure from being able to rotate by any rotation other than a 1/2 turn. Additionally, the figure has only two ways in which it can be reflected onto itself. D3 The luxury car, Mercedes-Benz, uses a symbol with three rotations and 3 mirror lines. Therefore, the emblem is an example of D3. If we were to convert this figure into a peace sign, however, we would lose 2 of the rotations and two of the reflection lines. This would leave a D1 figure.
D4 The symbol for Purina is a great example of a finite figure of the category D4. It is easy to see that there are four mirror reflections of the figure (one vertical, one horizontal, and two diagonal) as well as four rotations. In other words, rotating the figure four times gives the original figure (the identity). D5 The symbol for Chrysler is a great example of a finite figure of the category D5. In other words, the symbol has five rotations and five axes of reflection. D8 This finite figure is a dihedral group of order 8 due to its eight reflections and eight rotations. The symmetries are created by two squares placed on top of each other and offset by 90 degrees.
In modular arithmetic, the integers coprime (relatively prime) to n from the set {0, 1, …, n-1} of n non-negative integers form a group under multiplication modulo n, called the multiplicative group of integers modulo n. denoted by U(n)
U(8) = {1, 3, 5, 7} Binary operation: Multiplication modulo 8 Lets construct the group table for U(8) 1 3 5 7 1 3 5 7 1 3 5 7 3 1 7 5 5 7 1 3 7 5 3 1 Group Properties Closure * 5 * 5 = 1 7 * 3 = 5
U(8) = {1, 3, 5, 7} Binary operation: Multiplication modulo 8 Lets construct the group table for U(8) 1 3 5 7 1 3 5 7 1 3 5 7 3 1 7 5 5 7 1 3 7 5 3 1 Group Properties Associativity * (3 * 5) * 1 = 3 * (5 * 1) 7 * 1 = 3 * 5
U(8) = {1, 3, 5, 7} Binary operation: Multiplication modulo 8 Lets construct the group table for U(8) 1 3 5 7 1 3 5 7 1 3 5 7 3 1 7 5 5 7 1 3 7 5 3 1 Group Properties Associativity * (3 * 5) * 1 = 3 * (5 * 1) 7 * 1 = 3 * 5 7 = 7
U(8) = {1, 3, 5, 7} Binary operation: Multiplication modulo 8 Lets construct the group table for U(8) 1 3 5 7 1 3 5 7 1 3 5 7 3 1 7 5 5 7 1 3 7 5 3 1 Group Properties Identity * 3 * e = 3 3 *1 = 3
U(8) = {1, 3, 5, 7} Binary operation: Multiplication modulo 8 Lets construct the group table for U(8) 1 3 5 7 1 3 5 7 1 3 5 7 3 1 7 5 5 7 1 3 7 5 3 1 Group Properties Identity * 7 * e = 7 7 *1 = 7
U(8) = {1, 3, 5, 7} Binary operation: Multiplication modulo 8 Lets construct the group table for U(8) 1 3 5 7 1 3 5 7 1 3 5 7 3 1 7 5 5 7 1 3 7 5 3 1 Group Properties Inverse * 7 * a- = e 7 *7 = 1
U(8) = {1, 3, 5, 7} Binary operation: Multiplication modulo 8 Lets construct the group table for U(8) 1 3 5 7 1 3 5 7 1 3 5 7 3 1 7 5 5 7 1 3 7 5 3 1 Group Properties Inverse * 5 * a- = e 5 *5 = 1
Unit II-Lesson 3- Dihedral and Multiplicative Groups (PALMA).pdf

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Unit II-Lesson 3- Dihedral and Multiplicative Groups (PALMA).pdf

  • 1.
    MTH 601 ABSTRACTALGEBRA Dihedral Groups and Multiplicative Group Jayson C. Palma (Reporter) PhD Math Student
  • 2.
    a dihedral groupis the group of symmetries of a regular polygon, which includes rotations and reflections. denoted by Dn
  • 3.
    Dn = 2nelements D4 = 2(4) = 8 elements D8 = 2(8) = 16 elements
  • 4.
    We will exploreD4, the eight symmetries of a square. A B C D A B C D r0 = A B C D A B C D
  • 5.
    We will exploreD4, the eight symmetries of a square. A B C D D A B C r1 = A B C D B C D A
  • 6.
    We will exploreD4, the eight symmetries of a square. A B C D C D A B r2 = A B C D C D A B
  • 7.
    We will exploreD4, the eight symmetries of a square. A B C D B C D A r3 = A B C D D A B C
  • 8.
    We will exploreD4, the eight symmetries of a square. A B C D D B s0 = A B C D A D C B A C
  • 9.
    We will exploreD4, the eight symmetries of a square. A C s1 = A B C D C B A D C A D B B D
  • 10.
    We will exploreD4, the eight symmetries of a square. A B C D C A s2 = A B C D D C B A D B
  • 11.
    We will exploreD4, the eight symmetries of a square. A B C D A C s3 = A B C D B A D C B D
  • 12.
    r0 = A BC D A B C D r1 = A B C D B C D A r2 = A B C D C D A B r3 = A B C D D A B C s0 = A B C D A D C B s1 = A B C D C B A D s2 = A B C D D C B A s3 = A B C D B A D C
  • 13.
    r0 r1 r2r3 s0 s1 s2 s3 r0 r1 r2 r3 s0 s1 s2 s3
  • 14.
    r0r0 = A BC D A B C D A B C D A B C D = A B C D A B C D A B C D A B C D A B C D = = r0
  • 15.
    r0 r1 r2r3 s0 s1 s2 s3 r0 r1 r2 r3 s0 s1 s2 s3 r0
  • 16.
    r0r1 = A BC D A B C D A B C D B C D A = A B C D B C D A B C D A A B C D B C D A = = r1
  • 17.
    r0 r1 r2r3 s0 s1 s2 s3 r0 r1 r2 r3 s0 s1 s2 s3 r1
  • 18.
    r0r2 = A BC D A B C D A B C D C D A B = A B C D C D A B C D A B A B C D C D A B = = r2
  • 19.
    r0 r1 r2r3 s0 s1 s2 s3 r0 r1 r2 r3 s0 s1 s2 s3 r2
  • 20.
    r3s2 = A BC D D A B C A B C D D C B A = A B C D D C B A C B A D A B C D C B A D = = s1
  • 21.
    r0 r1 r2r3 s0 s1 s2 s3 r0 r1 r2 r3 s0 s1 s2 s3 s1
  • 22.
    s2s1 = A BC D D C B A A B C D C B A D = A B C D C B A D B C D A A B C D B C D A = = r1
  • 23.
    r0 r1 r2r3 s0 s1 s2 s3 r0 r1 r2 r3 s0 s1 s2 s3 r1
  • 24.
    r0 r1 r2r3 s0 s1 s2 s3 r0 r0 r1 r2 r3 s0 s1 s2 s3 r1 r1 r2 r3 r0 s3 s2 s0 s1 r2 r2 r3 r0 r1 s1 s0 s3 s2 r3 r3 r0 r1 r2 s2 s3 s1 s0 s0 s0 s2 s1 s3 r0 r2 r1 r3 s1 s1 s3 s0 s2 r2 r0 r3 r1 s2 s2 s1 s3 s0 r3 r1 r0 r2 s3 s3 s0 s2 s1 r1 r3 r2 r0
  • 25.
    r0 r1 r2r3 s0 s1 s2 s3 r0 r0 r1 r2 r3 s0 s1 s2 s3 r1 r1 r2 r3 r0 s3 s2 s0 s1 r2 r2 r3 r0 r1 s1 s0 s3 s2 r3 r3 r0 r1 r2 s2 s3 s1 s0 s0 s0 s2 s1 s3 r0 r2 r1 r3 s1 s1 s3 s0 s2 r2 r0 r3 r1 s2 s2 s1 s3 s0 r3 r1 r0 r2 s3 s3 s0 s2 s1 r1 r3 r2 r0 Closure s2s1 = r1 r3s0 = s2
  • 26.
    r0 r1 r2r3 s0 s1 s2 s3 r0 r0 r1 r2 r3 s0 s1 s2 s3 r1 r1 r2 r3 r0 s3 s2 s0 s1 r2 r2 r3 r0 r1 s1 s0 s3 s2 r3 r3 r0 r1 r2 s2 s3 s1 s0 s0 s0 s2 s1 s3 r0 r2 r1 r3 s1 s1 s3 s0 s2 r2 r0 r3 r1 s2 s2 s1 s3 s0 r3 r1 r0 r2 s3 s3 s0 s2 s1 r1 r3 r2 r0 Associativity (r2s1)s3 = r2(s1s3) s0 s3 = r2 r1
  • 27.
    r0 r1 r2r3 s0 s1 s2 s3 r0 r0 r1 r2 r3 s0 s1 s2 s3 r1 r1 r2 r3 r0 s3 s2 s0 s1 r2 r2 r3 r0 r1 s1 s0 s3 s2 r3 r3 r0 r1 r2 s2 s3 s1 s0 s0 s0 s2 s1 s3 r0 r2 r1 r3 s1 s1 s3 s0 s2 r2 r0 r3 r1 s2 s2 s1 s3 s0 r3 r1 r0 r2 s3 s3 s0 s2 s1 r1 r3 r2 r0 Associativity (r2s1)s3 = r2(s1s3) s0 s3 = r2 r1 r3 = r3
  • 28.
    r0 r1 r2r3 s0 s1 s2 s3 r0 r0 r1 r2 r3 s0 s1 s2 s3 r1 r1 r2 r3 r0 s3 s2 s0 s1 r2 r2 r3 r0 r1 s1 s0 s3 s2 r3 r3 r0 r1 r2 s2 s3 s1 s0 s0 s0 s2 s1 s3 r0 r2 r1 r3 s1 s1 s3 s0 s2 r2 r0 r3 r1 s2 s2 s1 s3 s0 r3 r1 r0 r2 s3 s3 s0 s2 s1 r1 r3 r2 r0 Associativity (s0s3)r2 = s0(s3r2) r3 r2 = s0s2
  • 29.
    r0 r1 r2r3 s0 s1 s2 s3 r0 r0 r1 r2 r3 s0 s1 s2 s3 r1 r1 r2 r3 r0 s3 s2 s0 s1 r2 r2 r3 r0 r1 s1 s0 s3 s2 r3 r3 r0 r1 r2 s2 s3 s1 s0 s0 s0 s2 s1 s3 r0 r2 r1 r3 s1 s1 s3 s0 s2 r2 r0 r3 r1 s2 s2 s1 s3 s0 r3 r1 r0 r2 s3 s3 s0 s2 s1 r1 r3 r2 r0 Associativity (s0s3)r2 = s0(s3r2) r3 r2 = s0s2 r1 = r1
  • 30.
    r0 r1 r2r3 s0 s1 s2 s3 r0 r0 r1 r2 r3 s0 s1 s2 s3 r1 r1 r2 r3 r0 s3 s2 s0 s1 r2 r2 r3 r0 r1 s1 s0 s3 s2 r3 r3 r0 r1 r2 s2 s3 s1 s0 s0 s0 s2 s1 s3 r0 r2 r1 r3 s1 s1 s3 s0 s2 r2 r0 r3 r1 s2 s2 s1 s3 s0 r3 r1 r0 r2 s3 s3 s0 s2 s1 r1 r3 r2 r0 Identity r3e = r3 r3 r0 = r3 e = r0
  • 31.
    r0 r1 r2r3 s0 s1 s2 s3 r0 r0 r1 r2 r3 s0 s1 s2 s3 r1 r1 r2 r3 r0 s3 s2 s0 s1 r2 r2 r3 r0 r1 s1 s0 s3 s2 r3 r3 r0 r1 r2 s2 s3 s1 s0 s0 s0 s2 s1 s3 r0 r2 r1 r3 s1 s1 s3 s0 s2 r2 r0 r3 r1 s2 s2 s1 s3 s0 r3 r1 r0 r2 s3 s3 s0 s2 s1 r1 r3 r2 r0 Identity s2e = s2 s2 r0 = s2 e = r0
  • 32.
    r0 r1 r2r3 s0 s1 s2 s3 r0 r0 r1 r2 r3 s0 s1 s2 s3 r1 r1 r2 r3 r0 s3 s2 s0 s1 r2 r2 r3 r0 r1 s1 s0 s3 s2 r3 r3 r0 r1 r2 s2 s3 s1 s0 s0 s0 s2 s1 s3 r0 r2 r1 r3 s1 s1 s3 s0 s2 r2 r0 r3 r1 s2 s2 s1 s3 s0 r3 r1 r0 r2 s3 s3 s0 s2 s1 r1 r3 r2 r0 Inverse s2 a- = e s2 s2= r0
  • 33.
    r0 r1 r2r3 s0 s1 s2 s3 r0 r0 r1 r2 r3 s0 s1 s2 s3 r1 r1 r2 r3 r0 s3 s2 s0 s1 r2 r2 r3 r0 r1 s1 s0 s3 s2 r3 r3 r0 r1 r2 s2 s3 s1 s0 s0 s0 s2 s1 s3 r0 r2 r1 r3 s1 s1 s3 s0 s2 r2 r0 r3 r1 s2 s2 s1 s3 s0 r3 r1 r0 r2 s3 s3 s0 s2 s1 r1 r3 r2 r0 Inverse s0 a- = e s0s0 = r0
  • 34.
    We will exploreD5, the ten symmetries of a pentagon. 1 2 3 4 1 2 3 4 r0 = 1 2 3 4 5 1 2 3 4 5 5 5
  • 35.
    We will exploreD5, the ten symmetries of a pentagon. 1 2 3 4 5 1 2 3 r1 = 1 2 3 4 5 2 3 4 5 5 4 1
  • 36.
    We will exploreD5, the ten symmetries of a pentagon. 1 2 3 4 4 5 1 2 r2 = 1 2 3 4 5 3 4 5 1 5 3 2
  • 37.
    We will exploreD5, the ten symmetries of a pentagon. 1 2 3 4 3 4 5 1 r3 = 1 2 3 4 5 4 5 1 2 5 2 3
  • 38.
    We will exploreD5, the ten symmetries of a pentagon. 1 2 3 4 2 3 4 5 r4 = 1 2 3 4 5 5 1 2 3 5 1 4
  • 39.
    We will exploreD5, the ten symmetries of a pentagon. 2 3 4 5 4 3 s0 = 1 2 3 4 5 1 5 4 3 5 2 2 1 1
  • 40.
    We will exploreD5, the ten symmetries of a pentagon. 1 3 4 1 5 s1 = 1 2 3 4 5 3 2 1 5 5 4 4 3 2 2
  • 41.
    We will exploreD5, the ten symmetries of a pentagon. 1 2 4 2 s2 = 1 2 3 4 5 5 4 3 2 5 1 1 5 4 3 3
  • 42.
    We will exploreD5, the ten symmetries of a pentagon. 1 2 3 s3 = 1 2 3 4 5 2 1 5 4 5 3 3 2 1 5 4 4
  • 43.
    We will exploreD5, the ten symmetries of a pentagon. 1 2 3 4 s4 = 1 2 3 4 5 4 3 2 1 5 4 3 2 1 5 5
  • 44.
    r0 = 1 23 4 5 1 2 3 4 5 r1 = 1 2 3 4 5 2 3 4 5 1 r2 = 1 2 3 4 5 3 4 5 1 2 r3 = 1 2 3 4 5 4 5 1 2 3 r4 = 1 2 3 4 5 5 1 2 3 4 s0 = 1 2 3 4 5 1 5 4 3 2 s1 = 1 2 3 4 5 3 2 1 5 4 s2 = 1 2 3 4 5 5 4 3 2 1 s3 = 1 2 3 4 5 2 1 5 4 3 s4 = 1 2 3 4 5 4 3 2 1 5
  • 45.
    r0 r1 r2r3 r4 s0 s1 s2 s3 s4 r0 r1 r2 r3 r4 s0 s1 s2 s3 s4
  • 46.
    r4s1 = 1 23 4 5 5 1 2 3 4 = 2 1 5 4 = = s3 1 2 3 4 5 3 2 1 5 4 1 2 3 4 5 3 2 1 5 4 3 1 2 3 4 5 2 1 5 4 3
  • 47.
    r0 r1 r2r3 r4 s0 s1 s2 s3 s4 r0 r1 r2 r3 r4 s3 s0 s1 s2 s3 s4
  • 48.
    s1s0 = 1 23 4 5 3 2 1 5 4 = 3 4 5 1 = = r2 1 2 3 4 5 1 5 4 3 2 1 2 3 4 5 1 5 4 3 2 2 1 2 3 4 5 3 4 5 1 2
  • 49.
    r0 r1 r2r3 r4 s0 s1 s2 s3 s4 r0 r1 r2 r3 r4 s0 s1 r2 s2 s3 s4
  • 50.
    r0 r1 r2r3 r4 s0 s1 s2 s3 s4 r0 r0 r1 r2 r3 r4 s0 s1 s2 s3 s4 r1 r1 r2 r3 r4 r0 s3 s4 s0 s1 s2 r2 r2 r3 r4 r0 r1 s1 s2 s3 s4 s0 r3 r3 r4 r0 r1 r2 s4 s0 s1 s2 s3 r4 r4 r0 r1 r2 r3 s2 s3 s4 s0 s1 s0 s0 s2 s4 s1 s3 r0 r3 r1 r4 r2 s1 s1 s3 s0 s2 s4 r2 r0 r3 r1 r4 s2 s2 s4 s1 s3 s0 r4 r2 r0 r3 r1 s3 s3 s0 s2 s4 s1 r1 r4 r2 r0 r3 s4 s4 s1 s3 s0 s2 r3 r1 r4 r2 r0
  • 51.
    r0 r1 r2r3 r4 s0 s1 s2 s3 s4 r0 r0 r1 r2 r3 r4 s0 s1 s2 s3 s4 r1 r1 r2 r3 r4 r0 s3 s4 s0 s1 s2 r2 r2 r3 r4 r0 r1 s1 s2 s3 s4 s0 r3 r3 r4 r0 r1 r2 s4 s0 s1 s2 s3 r4 r4 r0 r1 r2 r3 s2 s3 s4 s0 s1 s0 s0 s2 s4 s1 s3 r0 r3 r1 r4 r2 s1 s1 s3 s0 s2 s4 r2 r0 r3 r1 r4 s2 s2 s4 s1 s3 s0 r4 r2 r0 r3 r1 s3 s3 s0 s2 s4 s1 r1 r4 r2 r0 r3 s4 s4 s1 s3 s0 s2 r3 r1 r4 r2 r0 Closure s3r2 = s2 r4s3 = s0
  • 52.
    r0 r1 r2r3 r4 s0 s1 s2 s3 s4 r0 r0 r1 r2 r3 r4 s0 s1 s2 s3 s4 r1 r1 r2 r3 r4 r0 s3 s4 s0 s1 s2 r2 r2 r3 r4 r0 r1 s1 s2 s3 s4 s0 r3 r3 r4 r0 r1 r2 s4 s0 s1 s2 s3 r4 r4 r0 r1 r2 r3 s2 s3 s4 s0 s1 s0 s0 s2 s4 s1 s3 r0 r3 r1 r4 r2 s1 s1 s3 s0 s2 s4 r2 r0 r3 r1 r4 s2 s2 s4 s1 s3 s0 r4 r2 r0 r3 r1 s3 s3 s0 s2 s4 s1 r1 r4 r2 r0 r3 s4 s4 s1 s3 s0 s2 r3 r1 r4 r2 r0 Associativity (s0r4)s3 = s0(r4s3) s3 = s0 s3 s0
  • 53.
    r0 r1 r2r3 r4 s0 s1 s2 s3 s4 r0 r0 r1 r2 r3 r4 s0 s1 s2 s3 s4 r1 r1 r2 r3 r4 r0 s3 s4 s0 s1 s2 r2 r2 r3 r4 r0 r1 s1 s2 s3 s4 s0 r3 r3 r4 r0 r1 r2 s4 s0 s1 s2 s3 r4 r4 r0 r1 r2 r3 s2 s3 s4 s0 s1 s0 s0 s2 s4 s1 s3 r0 r3 r1 r4 r2 s1 s1 s3 s0 s2 s4 r2 r0 r3 r1 r4 s2 s2 s4 s1 s3 s0 r4 r2 r0 r3 r1 s3 s3 s0 s2 s4 s1 r1 r4 r2 r0 r3 s4 s4 s1 s3 s0 s2 r3 r1 r4 r2 r0 Associativity (s0r4)s3 = s0(r4s3) s3 = s0 s3 s0 r0 = r0
  • 54.
    r0 r1 r2r3 r4 s0 s1 s2 s3 s4 r0 r0 r1 r2 r3 r4 s0 s1 s2 s3 s4 r1 r1 r2 r3 r4 r0 s3 s4 s0 s1 s2 r2 r2 r3 r4 r0 r1 s1 s2 s3 s4 s0 r3 r3 r4 r0 r1 r2 s4 s0 s1 s2 s3 r4 r4 r0 r1 r2 r3 s2 s3 s4 s0 s1 s0 s0 s2 s4 s1 s3 r0 r3 r1 r4 r2 s1 s1 s3 s0 s2 s4 r2 r0 r3 r1 r4 s2 s2 s4 s1 s3 s0 r4 r2 r0 r3 r1 s3 s3 s0 s2 s4 s1 r1 r4 r2 r0 r3 s4 s4 s1 s3 s0 s2 r3 r1 r4 r2 r0 Identity s2 = s2 s2 = s2 r0 e
  • 55.
    r0 r1 r2r3 r4 s0 s1 s2 s3 s4 r0 r0 r1 r2 r3 r4 s0 s1 s2 s3 s4 r1 r1 r2 r3 r4 r0 s3 s4 s0 s1 s2 r2 r2 r3 r4 r0 r1 s1 s2 s3 s4 s0 r3 r3 r4 r0 r1 r2 s4 s0 s1 s2 s3 r4 r4 r0 r1 r2 r3 s2 s3 s4 s0 s1 s0 s0 s2 s4 s1 s3 r0 r3 r1 r4 r2 s1 s1 s3 s0 s2 s4 r2 r0 r3 r1 r4 s2 s2 s4 s1 s3 s0 r4 r2 r0 r3 r1 s3 s3 s0 s2 s4 s1 r1 r4 r2 r0 r3 s4 s4 s1 s3 s0 s2 r3 r1 r4 r2 r0 Inverse r4 = e r4 = r0 r1 a-
  • 56.
    Dihedral Group Symbol Our Thoughts D1 ShellPetroleum uses the symbol to the left. This shell shape has no rotations (other than the identity) and has only one mirror line (vertical). Therefore, like Mickey Mouse, the figure is said to be bilaterally symmetric and it fits into the category D1. D2 An example of D2 that is easily spotted is the logo for the Columbia Broadcasting System (CBS). The "eye" shape within the circle prevents the figure from being able to rotate by any rotation other than a 1/2 turn. Additionally, the figure has only two ways in which it can be reflected onto itself. D3 The luxury car, Mercedes-Benz, uses a symbol with three rotations and 3 mirror lines. Therefore, the emblem is an example of D3. If we were to convert this figure into a peace sign, however, we would lose 2 of the rotations and two of the reflection lines. This would leave a D1 figure.
  • 57.
    D4 The symbol forPurina is a great example of a finite figure of the category D4. It is easy to see that there are four mirror reflections of the figure (one vertical, one horizontal, and two diagonal) as well as four rotations. In other words, rotating the figure four times gives the original figure (the identity). D5 The symbol for Chrysler is a great example of a finite figure of the category D5. In other words, the symbol has five rotations and five axes of reflection. D8 This finite figure is a dihedral group of order 8 due to its eight reflections and eight rotations. The symmetries are created by two squares placed on top of each other and offset by 90 degrees.
  • 58.
    In modular arithmetic,the integers coprime (relatively prime) to n from the set {0, 1, …, n-1} of n non-negative integers form a group under multiplication modulo n, called the multiplicative group of integers modulo n. denoted by U(n)
  • 59.
    U(8) = {1,3, 5, 7} Binary operation: Multiplication modulo 8 Lets construct the group table for U(8) 1 3 5 7 1 3 5 7 1 3 5 7 3 1 7 5 5 7 1 3 7 5 3 1 Group Properties Closure * 5 * 5 = 1 7 * 3 = 5
  • 60.
    U(8) = {1,3, 5, 7} Binary operation: Multiplication modulo 8 Lets construct the group table for U(8) 1 3 5 7 1 3 5 7 1 3 5 7 3 1 7 5 5 7 1 3 7 5 3 1 Group Properties Associativity * (3 * 5) * 1 = 3 * (5 * 1) 7 * 1 = 3 * 5
  • 61.
    U(8) = {1,3, 5, 7} Binary operation: Multiplication modulo 8 Lets construct the group table for U(8) 1 3 5 7 1 3 5 7 1 3 5 7 3 1 7 5 5 7 1 3 7 5 3 1 Group Properties Associativity * (3 * 5) * 1 = 3 * (5 * 1) 7 * 1 = 3 * 5 7 = 7
  • 62.
    U(8) = {1,3, 5, 7} Binary operation: Multiplication modulo 8 Lets construct the group table for U(8) 1 3 5 7 1 3 5 7 1 3 5 7 3 1 7 5 5 7 1 3 7 5 3 1 Group Properties Identity * 3 * e = 3 3 *1 = 3
  • 63.
    U(8) = {1,3, 5, 7} Binary operation: Multiplication modulo 8 Lets construct the group table for U(8) 1 3 5 7 1 3 5 7 1 3 5 7 3 1 7 5 5 7 1 3 7 5 3 1 Group Properties Identity * 7 * e = 7 7 *1 = 7
  • 64.
    U(8) = {1,3, 5, 7} Binary operation: Multiplication modulo 8 Lets construct the group table for U(8) 1 3 5 7 1 3 5 7 1 3 5 7 3 1 7 5 5 7 1 3 7 5 3 1 Group Properties Inverse * 7 * a- = e 7 *7 = 1
  • 65.
    U(8) = {1,3, 5, 7} Binary operation: Multiplication modulo 8 Lets construct the group table for U(8) 1 3 5 7 1 3 5 7 1 3 5 7 3 1 7 5 5 7 1 3 7 5 3 1 Group Properties Inverse * 5 * a- = e 5 *5 = 1