This is a lesson about dihedral and multiplicative groups

1. MTH 601 ABSTRACT ALGEBRA
Dihedral Groups and Multiplicative Group
Jayson C. Palma (Reporter)
PhD Math Student

2. a dihedral group is the group of symmetries of a
regular polygon, which includes rotations and
reflections.
denoted by Dn

3. Dn = 2n elements
D4 = 2(4)
= 8 elements
D8 = 2(8)
= 16 elements

4. 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

5. 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

6. 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

7. 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

8. 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

9. 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

10. 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

11. 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

12. 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

13. r0 r1 r2 r3 s0 s1 s2 s3
r0
r1
r2
r3
s0
s1
s2
s3

14. 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

15. r0 r1 r2 r3 s0 s1 s2 s3
r0
r1
r2
r3
s0
s1
s2
s3
r0

16. 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

17. r0 r1 r2 r3 s0 s1 s2 s3
r0
r1
r2
r3
s0
s1
s2
s3
r1

18. 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

19. r0 r1 r2 r3 s0 s1 s2 s3
r0
r1
r2
r3
s0
s1
s2
s3
r2

20. 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

21. r0 r1 r2 r3 s0 s1 s2 s3
r0
r1
r2
r3
s0
s1
s2
s3
s1

22. 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

23. r0 r1 r2 r3 s0 s1 s2 s3
r0
r1
r2
r3
s0
s1
s2
s3
r1

24. 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

25. 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

26. 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

27. 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

28. 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

29. 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

30. 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

31. 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

32. 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

33. 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

34. 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

35. 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

36. 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

37. 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

38. 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

39. 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

40. 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

41. 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

42. 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

43. 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

44. 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

45. r0 r1 r2 r3 r4
s0 s1 s2 s3 s4
r0
r1
r2
r3
r4
s0
s1
s2
s3
s4

46. 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

47. r0 r1 r2 r3 r4
s0 s1 s2 s3 s4
r0
r1
r2
r3
r4 s3
s0
s1
s2
s3
s4

48. 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

49. r0 r1 r2 r3 r4
s0 s1 s2 s3 s4
r0
r1
r2
r3
r4
s0
s1 r2
s2
s3
s4

50. 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

51. 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

52. 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

53. 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

54. 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

55. 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-

56. 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.

57. 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.

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