Number Theory Final Requirement

1

GENERATORS AND ORDERS OF ℤn

VICTORIA SAFFIRE A. LILLO

Submitted to the Faculty of the Graduate Studies
Don Mariano Marcos Memorial State University

Mid – La Union Campus
City of San Fernando, La Union

In Partial Fulfillment of the
Requirements for the

Subject Number Theory

Master of Arts in Technology Education
(Mathematics)

May 14, 2022

2

I. INTRODUCTION:

Finding patterns is extremely important. To find patterns among
problems, we look for things that are the same (or very similar) in each
problem. Patterns exist among numbers and different types of numbers.
We need to look for both. It is worth to spend a little thought in getting at
the root reason why mathematics, because of its very abstractness, must
always remain one of the most important topics for thought. Let us try to
make clear to ourselves why explanations of the order of events necessarily
tend to become mathematical.

Algebra is a broad section of mathematics and abstract algebra is
one of algebra’s sub areas which studies algebraic structures singly. In
abstract algebra, a group is a set of elements defined with an operation
that integrates any two of its elements to form a third element satisfying
four axioms. These axioms to be satisfied by a group together with the
operation are: closure, associativity, identity and invertibility are called
group axioms. The integers as a number set together with the addition
operation is a familiar example of a group and denoted with (ℤ, +).

This study has been conducted with the aim to present the simplest
possible way of determining the generators of groups of integers ℤ when
asked for the list of generators and orders of groups. The conclusions that
have been reached are based on the patterns observed using the set of

3

integers modulo 2 to 11. The various information to support the discussion
and conclusion have been collected from online sources, it’s been
organized and presented in the simplest way possible in order for the
readers to easily digest what is the content all about. The format of the
discussion is mainly how the researcher understood and analyzed the
cyclic group specifically the set of integers modulo n (ℤ ), how to generate
its subgroups, how to determine the number of generators and to enlist
the number of generators. Because of the few examples used in the course
of study, suggestions to further prove the claims are embedded in the
recommendations.

4

II. DISCUSSION:

Definition of Cyclic Groups
A group is cyclic if it’s generated by a single element = ⟨ ⟩. In

working with groups you typically use additive notation (+) and
multiplicative notation (×). This is done even if the elements of the group
are not numbers and the group operation is not numerical instead,
something like geometric transformation or function composition. When
using the additive notation, the identity element is denoted by 0. When
using multiplicative notation the identity element is 1.

Let be a group with operation ×.
Pick, ∈ .
The group generated by is the smallest subgroup of containing
.
⟨x⟩ = {… , −4, −3, −2, −1, 1, , 2, 3, 4, … }
It contains , its inverse, identity element 1, all powers of and
powers of its inverse. We call it a group generated by . If = ⟨x⟩ for some
, then we call a cyclic group.
Let be a group with operation +.
Pick, ∈ .
The group generated by is the smallest subgroup of containing
.
⟨y⟩ = {… , −3y, −2 , − , 0, , 2 , 3 , … }

5

It contains y, its inverse, identity element 0, all positive and negative
multiples of y. If = ⟨y⟩ for some y, then we call a cyclic group.

There are types of cyclic groups the infinite and the finite.

Infinite Cyclic Group
The ℤ, + or integers ℤ under addition is an infinite cyclic group.
Claim: ℤ = ⟨1⟩ or the set of integers generated by 1.
⟨1⟩ = {… , −4, −3, −2, −1, 0, 1, 2, 3,4 … }
It contains 1, identity element 0, the additive inverse of 1 which is

-1, multiples of 1 and -1. It covers all the integers, that is why, ℤ is a cyclic
group and it is infinite.

Finite Cyclic Group
Let = integers under addition. This a finite group of n

elements.
Elements: {0, 1, 2, … , − 1}
is cyclic: = ⟨1⟩

{ . . ., -2, -1, 0, 1, 2, . . .,n–1, n, n + 1, n + 2, . . ., 2n–1, 2n, 2n + 1,…}

{ . . ., n-2, n-1, 0, 1, 2, . . .,n–1, 0, 1, 2, . . ., n–1, 0, 1,…}

≡ 0 ( ) − 1 ≡ − 1 ( )

+ 1 ≡ 1 ( ) − 2 ≡ − 2 ( )

+ 2 ≡ 2 ( ) − 3 ≡ − 3 ( )

6

+ 3 ≡ 3 ( ) − 4 ≡ − 4 ( )

..

..

..

The group generated by 1 repeat itself, it cycle through the number

0 through n – 1 over and over, this is why its cyclic.

The ℤ/ ℤ, + or the integers under addition is a finite cyclic

group.

Generators of Cyclic Groups

A cyclic group of order by Euler’s phi function or sometimes called

Euler’s totient function has

( ) = (1 − 1 ) (1 − 1 ) … (1 − 1 )
1 2

generators where 1, 2, . . . , are prime factors. The order of the group is

its size.

Example 1: The set of integers modulo 2 or in symbols ℤ2. The cyclic group
of order 2 is defined as the unique group of order 2.
Solution: Since ℤ2 = {0,1}, n = 2.

1. Find the factors of 2, 2 = 2.1
2. Use the Euler’s totient function to find the number of
generators:

(2) = 2 (1 − 1) = 2 (1) = 1

22

7

∴There is only 1 generator in ℤ2.

3. From ℤ2 = {0,1}, list the subgroups of ℤ2.
⟨1⟩ = 1 ≡ 1 ( 2)
= 1 + 1 = 2 ≡ 0 ( 2)
⟨1⟩ = {0,1} = ℤ2

Note: ⟨1⟩ is a generator for all ℤ .
Since ⟨1⟩ is equal to ℤ3, therefore, the generator of ℤ2 is ⟨ ⟩.

Example 2: The set of integers modulo 3 or in symbols ℤ3. The cyclic group
of order 3 is defined as the unique group of order 3.
Solution: Since ℤ3 = {0,1,2}, n = 3.

1. Find the factors of 3, 3 = 3.1
2. Use the Euler’s totient function to find the number of
generators:

(3) = 3 (1 − 1) = 3 (2) = 2

33

∴There are 2 generators in ℤ3.

3. From ℤ3 = {0,1,2}, list the subgroups of ℤ3.
⟨1⟩ = 1 ≡ 1 ( 3)
= 1 + 1 = 2 ≡ 2 ( 3)
= 1 + 1 + 1 = 3 ≡ 0 ( 3)
⟨1⟩ = {0,1,2} = ℤ3

Since ⟨1⟩ is a generator for all ℤ . Therefore, ⟨1⟩ = ℤ3.

8

⟨2⟩ = 2 ≡ 2 ( 3)
= 2 + 2 = 4 ≡ 1 ( 3)
= 2 + 2 + 2 = 6 ≡ 0 ( 3)

⟨2⟩ = {0,1,2} = ℤ3
Since ⟨1⟩ and ⟨2⟩ are equal to ℤ3, therefore, the generators of ℤ3 are
⟨ ⟩ and ⟨ ⟩.

Example 3: The set of integers modulo 4 or in symbols ℤ4. The cyclic group
of order 4 is defined as the unique group of order 4.

Solution: Since ℤ4 = {0,1,2,3}, n = 4.
1. Find the factors of 4, 4 = 2.2
2. Use the Euler’s phi function to find the number of generators:

(4) = 4 (1 − 1) = 4 (1) = 2

22

∴There are 2 generators in ℤ4.

3. From ℤ4 = {0,1,2,3}, list the subgroups of ℤ4.
⟨1⟩ = 1 ≡ 1 ( 4)
= 1 + 1 = 2 ≡ 2 ( 4)
= 1 + 1 + 1 = 3 ≡ 3 ( 4)
= 1 + 1 + 1 + 1 = 4 ≡ 0 ( 4)
⟨1⟩ = {0,1,2,3} = ℤ4

Since ⟨1⟩ is a generator for all ℤ . Therefore, ⟨1⟩ = ℤ4.
⟨2⟩ = 2 ≡ 2 ( 4)

9

= 2 + 2 = 4 ≡ 0 ( 4)
⟨2⟩ = {0,2} ≠ ℤ4
⟨3⟩ = 3 ≡ 3 ( 4)

= 3 + 3 = 6 ≡ 2 ( 4)
= 3 + 3 + 3 = 9 ≡ 1 ( 4)
= 3 + 3 + 3 + 3 = 12 ≡ 0 ( 4)
⟨3⟩ = {0,1,2,3} = ℤ4
Since ⟨1⟩ and ⟨3⟩ are equal to ℤ4, therefore, the generators of ℤ4 are
⟨ ⟩ and ⟨ ⟩.

Example 4: The set of integers modulo 5 or in symbols ℤ5. The cyclic group
of order 5 is defined as the unique group of order 5.
Solution: Since ℤ5 = {0,1,2,3,4}, n = 5.

1. Find the factors of 5, 5 = 5.1
2. Use the Euler’s totient function to find the number of
generators:

(5) = 5 (1 − 1) = 5 (4) = 4

55

∴There are 4 generators in ℤ3.

3. From ℤ5 = {0,1,2,3,4}, list the subgroups of ℤ5.
Since ⟨1⟩ is a generator for all ℤ , ⟨1⟩ = ℤ5.

⟨2⟩ = 2 ≡ 2 ( 5)
= 2 + 2 = 4 ≡ 4 ( 5)

10

= 2 + 2 + 2 = 6 ≡ 1 ( 5)
= 2 + 2 + 2 + 2 = 8 ≡ 3 ( 5)
= 2 + 2 + 2 + 2 + 2 = 10 ≡ 0 ( 5)
⟨2⟩ = {0,1,2,3,4} = ℤ5
⟨3⟩ = 3 ≡ 3 ( 5)
= 3 + 3 = 6 ≡ 1 ( 5)
= 3 + 3 + 3 = 9 ≡ 4 ( 5)
= 3 + 3 + 3 + 3 = 12 ≡ 2 ( 5)
= 3 + 3 + 3 + 3 + 3 = 15 ≡ 0 ( 5)
⟨3⟩ = {0,1,2,3,4} = ℤ5
⟨4⟩ = 4 ≡ 4 ( 5)
= 4 + 4 = 8 ≡ 3 ( 5)
= 4 + 4 + 4 = 12 ≡ 2 ( 5)
= 4 + 4 + 4 + 4 = 16 ≡ 1 ( 5)
= 4 + 4 + 4 + 4 + 4 = 20 ≡ 0 ( 5)
⟨4⟩ = {0,1,2,3,4} = ℤ5
Since ⟨1⟩,⟨2⟩, ⟨3⟩ and ⟨4⟩ are equal to ℤ5, therefore, the generators of
ℤ5 are ⟨ ⟩, ⟨ ⟩, ⟨ ⟩ and ⟨ ⟩.

Example 5: The set of integers modulo 6 or in symbols ℤ6. The cyclic group
of order 6 is defined as the unique group of order 6.
Solution: Since ℤ6 = {0,1,2,3,4,5}, n = 6.

1. Find the factors of 6, 6 = 2.3

11

2. Use the Euler’s totient function to find the number of
generators:

(6) = 6 (1 − 1) (1 − 1) = 6 (1) (2) = 2

2 3 23

∴There are 2 generators in ℤ6.

3. From ℤ6 = {0,1,2,3,4,5}, list the subgroups of ℤ6.
Since ⟨1⟩ is a generator for all ℤ , ⟨1⟩ = ℤ6.

⟨2⟩ = 2 ≡ 2 ( 6)
= 2 + 2 = 4 ≡ 4 ( 6)
= 2 + 2 + 2 = 6 ≡ 0 ( 6)

⟨2⟩ = {0,2,4} ≠ ℤ6
⟨3⟩ = 3 ≡ 3 ( 6)

= 3 + 3 = 6 ≡ 0 ( 6)
⟨3⟩ = {0,3} ≠ ℤ6
⟨4⟩ = 4 ≡ 4 ( 6)

= 4 + 4 = 8 ≡ 2 ( 6)
= 4 + 4 + 4 = 12 ≡ 0 ( 6)
⟨4⟩ = {0,2,4} ≠ ℤ6
⟨5⟩ = 5 ≡ 5 ( 6)
= 5 + 5 = 10 ≡ 4( 6)
= 5 + 5 + 5 = 15 ≡ 3 ( 6)
= 5 + 5 + 5 + 5 = 20 ≡ 2 ( 6)
= 5 + 5 + 5 + 5 + 5 = 25 ≡ 1 ( 6)

12

= 5 + 5 + 5 + 5 + 5 + 5 = 30 ≡ 0 ( 6)
⟨5⟩ = {0,1,2,3,4,5} = ℤ6
Since ⟨1⟩ and ⟨4⟩ are equal to ℤ6, therefore, the generators of ℤ6 are
⟨ ⟩ and ⟨ ⟩.

Example 6: The set of integers modulo 6 or in symbols ℤ7. The cyclic group
of order 6 is defined as the unique group of order 7.
Solution: Since ℤ7 = {0,1,2,3,4,5,6}, n = 7.

1. Find the factors of 7, 7 = 7.1
2. Use the Euler’s totient function to find the number of
generators:

(7) = 7 (1 − 1) = 7 (6) = 6

77

∴There are 6 generators in ℤ6.

3. From ℤ7 = {0,1,2,3,4,5,6}, list the subgroups of ℤ7.
Since ⟨1⟩ is a generator for all ℤ , ⟨1⟩ = ℤ7.

⟨2⟩ = 2 ≡ 2 ( 7)
= 2 + 2 = 4 ≡ 4 ( 7)
= 2 + 2 + 2 = 6 ≡ 6 ( 7)
= 2 + 2 + 2 + 2 = 8 ≡ 1 ( 7)
= 2 + 2 + 2 + 2 + 2 = 10 ≡ 3 ( 7)
= 2 + 2 + 2 + 2 + 2 + 2 = 12 ≡ 5 ( 7)
= 2 + 2 + 2 + 2 + 2 + 2 + 2 = 14 ≡ 0 ( 7)

13

⟨2⟩ = {0,1,2,3,4,5,6} = ℤ7
⟨3⟩ = 3 ≡ 3 ( 7)

= 3 + 3 = 6 ≡ 6 ( 7)
= 3 + 3 + 3 = 9 ≡ 2 ( 7)
= 3 + 3 + 3 + 3 = 12 ≡ 5 ( 7)
= 3 + 3 + 3 + 3 + 3 = 15 ≡ 1 ( 7)
= 3 + 3 + 3 + 3 + 3 + 3 = 18 ≡ 4 ( 7)
= 3 + 3 + 3 + 3 + 3 + 3 + 3 = 21 ≡ 0 ( 7)
⟨3⟩ = {0,1,2,3,4,5,6} = ℤ7
⟨4⟩ = 4 ≡ 4 ( 7)
= 4 + 4 = 8 ≡ 1 ( 7)
= 4 + 4 + 4 = 12 ≡ 5 ( 7)
= 4 + 4 + 4 + 4 = 16 ≡ 2 ( 7)
= 4 + 4 + 4 + 4 + 4 = 20 ≡ 6 ( 7)
= 4 + 4 + 4 + 4 + 4 + 4 = 24 ≡ 3 ( 7)
= 4 + 4 + 4 + 4 + 4 + 4 + 4 = 28 ≡ 0 ( 7)
⟨4⟩ = {0,1,2,3,4,5,6} = ℤ7
⟨5⟩ = 5 ≡ 5 ( 7)
= 5 + 5 = 10 ≡ 3 ( 7)
= 5 + 5 + 5 = 15 ≡ 1 ( 7)
= 5 + 5 + 5 + 5 = 20 ≡ 6 ( 7)
= 5 + 5 + 5 + 5 + 5 = 25 ≡ 4 ( 7)
= 5 + 5 + 5 + 5 + 5 + 5 = 30 ≡ 2 ( 7)

14

= 5 + 5 + 5 + 5 + 5 + 5 + 5 = 35 ≡ 0 ( 7)
⟨5⟩ = {0,1,2,3,4,5,6} = ℤ7
⟨6⟩ = 6 ≡ 6 ( 7)

= 6 + 6 = 12 ≡ 5 ( 7)
= 6 + 6 + 6 = 18 ≡ 4 ( 7)
= 6 + 6 + 6 + 6 = 24 ≡ 3 ( 7)
= 6 + 6 + 6 + 6 + 6 = 30 ≡ 2 ( 7)
= 6 + 6 + 6 + 6 + 6 + 6 = 36 ≡ 1 ( 7)
= 6 + 6 + 6 + 6 + 6 + 6 + 6 = 42 ≡ 0 ( 7)
⟨6⟩ = {0,1,2,3,4,5,6} = ℤ7
Since ⟨1⟩,⟨2⟩, ⟨3⟩ , ⟨4⟩, ⟨5⟩ and ⟨6⟩ are equal to ℤ7, therefore, the
generators of ℤ7 are ⟨ ⟩, ⟨ ⟩, ⟨ ⟩ , ⟨ ⟩, ⟨ ⟩ and ⟨ ⟩.

Example 7: The set of integers modulo 8 or in symbols ℤ8. The cyclic group
of order 8 is defined as the unique group of order 8.
Solution: Since ℤ8 = {0,1,2,3,4,5,6,7}, n = 8.

1. Find the factors of 8, 8 = 2.2.2
2. Use the Euler’s totient function to find the number of
generators:

(8) = 8 (1 − 1) = 8 (1) = 4

22

∴There are 4 generators in ℤ8.

3. From ℤ8 = {0,1,2,3,4,5,6,7}, list the subgroups of ℤ8.

15

Since ⟨1⟩ is a generator for all ℤ , ⟨1⟩ = ℤ8.
⟨2⟩ = 2 ≡ 2 ( 8)
= 2 + 2 = 4 ≡ 4 ( 8)
= 2 + 2 + 2 = 6 ≡ 6 ( 8)
= 2 + 2 + 2 + 2 = 8 ≡ 0 ( 8)
⟨2⟩ = {0,2,4,6} ≠ ℤ8
⟨3⟩ = 3 ≡ 3 ( 8)
= 3 + 3 = 6 ≡ 6 ( 8)
= 3 + 3 + 3 = 9 ≡ 1 ( 8)
= 3 + 3 + 3 + 3 = 12 ≡ 4 ( 8)
= 3 + 3 + 3 + 3 + 3 = 15 ≡ 7 ( 8)
= 3 + 3 + 3 + 3 + 3 + 3 = 18 ≡ 2 ( 8)
= 3 + 3 + 3 + 3 + 3 + 3 + 3 = 21 ≡ 5 ( 8)
= 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 = 24 ≡ 0 ( 8)
⟨3⟩ = {0,1,2,3,4,5,6,7} = ℤ8
⟨4⟩ = 4 ≡ 4 ( 8)
= 4 + 4 = 8 ≡ 0 ( 8)
⟨4⟩ = {0,4} ≠ ℤ8

⟨5⟩ = 5 ≡ 5 ( 8)
= 5 + 5 = 10 ≡ 2 ( 8)
= 5 + 5 + 5 = 15 ≡ 7 ( 8)
= 5 + 5 + 5 + 5 = 20 ≡ 4 ( 8)

16

= 5 + 5 + 5 + 5 + 5 = 25 ≡ 1 ( 8)
= 5 + 5 + 5 + 5 + 5 + 5 = 30 ≡ 6 ( 8)
= 5 + 5 + 5 + 5 + 5 + 5 + 5 = 35 ≡ 3 ( 8)
= 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 = 40 ≡ 0 ( 8)
⟨5⟩ = {0,1,2,3,4,5,6,7} = ℤ8
⟨6⟩ = 6 ≡ 6 ( 8)
= 6 + 6 = 12 ≡ 4 ( 8)
= 6 + 6 + 6 = 18 ≡ 2 ( 8)
= 6 + 6 + 6 + 6 = 24 ≡ 0 ( 8)
⟨6⟩ = {0,2,4,6} ≠ ℤ8
⟨7⟩ = 7 ≡ 7 ( 8)
= 7 + 7 = 14 ≡ 6 ( 8)
= 7 + 7 + 7 = 21 ≡ 5 ( 8)
= 7 + 7 + 7 + 7 = 28 ≡ 4 ( 8)
= 7 + 7 + 7 + 7 + 7 = 35 ≡ 3 ( 8)
= 7 + 7 + 7 + 7 + 7 + 7 = 42 ≡ 2 ( 8)
= 7 + 7 + 7 + 7 + 7 + 7 + 7 = 49 ≡ 1 ( 8)
= 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 = 56 ≡ 0 ( 8)
⟨7⟩ = {0,1,2,3,4,5,6,7} = ℤ8
Since ⟨1⟩, ⟨3⟩,⟨5⟩, and ⟨7⟩ are equal to ℤ8, therefore, the generators of
ℤ8 are ⟨ ⟩, ⟨ ⟩, ⟨ ⟩ , and ⟨ ⟩.

17

Example 8: The set of integers modulo 9 or in symbols ℤ9. The cyclic group
of order 9 is defined as the unique group of order 9.
Solution: Since ℤ9 = {0,1,2,3,4,5,6,7,8}, n = 9.

1. Find the factors of 9, 9 = 3.3
2. Use the Euler’s totient function to find the number of
generators:

(9) = 9 (1 − 1) = 9 (2) = 6

33

∴There are 6 generators in ℤ9.

3. From ℤ9 = {0,1,2,3,4,5,6,7,8}, list the subgroups of ℤ9.
Since ⟨1⟩ is a generator for all ℤ , ⟨1⟩ = ℤ9.

⟨2⟩ = 2 ≡ 2 ( 9)
= 2 + 2 = 4 ≡ 4 ( 9)
= 2 + 2 + 2 = 6 ≡ 6 ( 9)
= 2 + 2 + 2 + 2 = 8 ≡ 8 ( 9)
= 2 + 2 + 2 + 2 + 2 = 10 ≡ 1 ( 9)
= 2 + 2 + 2 + 2 + 2 + 2 = 12 ≡ 3 ( 9)
= 2 + 2 + 2 + 2 + 2 + 2 + 2 = 14 ≡ 5 ( 9)
= 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 = 16 ≡ 7 ( 9)
= 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 = 18 ≡ 0 ( 9)

⟨2⟩ = {0,1,2,3,4,5,6,7,8} = ℤ9
⟨3⟩ = 3 ≡ 3 ( 9)

= 3 + 3 = 6 ≡ 6 ( 9)

18

= 3 + 3 + 3 = 9 ≡ 0 ( 9)
⟨3⟩ = {0,3,6} ≠ ℤ9
⟨4⟩ = 4 ≡ 4 ( 9)

= 4 + 4 = 8 ≡ 8 ( 9)
= 4 + 4 + 4 = 12 ≡ 3 ( 9)
= 4 + 4 + 4 + 4 = 16 ≡ 7 ( 9)
= 4 + 4 + 4 + 4 + 4 = 20 ≡ 2 ( 9)
= 4 + 4 + 4 + 4 + 4 + 4 = 24 ≡ 6 ( 9)
= 4 + 4 + 4 + 4 + 4 + 4 + 4 = 28 ≡ 1 ( 9)
= 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 = 32 ≡ 5 ( 9)
= 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 = 36 ≡ 0 ( 9)
⟨4⟩ = {0,1,2,3,4,5,6,7,8} = ℤ9
⟨5⟩ = 5 ≡ 5 ( 9)
= 5 + 5 = 10 ≡ 1 ( 9)
= 5 + 5 + 5 = 15 ≡ 6 ( 9)
= 5 + 5 + 5 + 5 = 20 ≡ 2 ( 9)
= 5 + 5 + 5 + 5 + 5 = 25 ≡ 7 ( 9)
= 5 + 5 + 5 + 5 + 5 + 5 = 30 ≡ 3 ( 9)
= 5 + 5 + 5 + 5 + 5 + 5 + 5 = 35 ≡ 8 ( 9)
= 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 = 40 ≡ 4 ( 9)
= 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 = 45 ≡ 0 ( 9)
⟨5⟩ = {0,1,2,3,4,5,6,7,8} = ℤ9

19

⟨6⟩ = 6 ≡ 6 ( 9)
= 6 + 6 = 12 ≡ 3 ( 9)
= 6 + 6 + 6 = 18 ≡ 0 ( 9)

⟨6⟩ = {0,3,6} ≠ ℤ9
⟨7⟩ = 7 ≡ 7 ( 9)

= 7 + 7 = 14 ≡ 5 ( 9)
= 7 + 7 + 7 = 21 ≡ 3 ( 9)
= 7 + 7 + 7 + 7 = 28 ≡ 1 ( 9)
= 7 + 7 + 7 + 7 + 7 = 35 ≡ 8 ( 9)
= 7 + 7 + 7 + 7 + 7 + 7 = 42 ≡ 6 ( 9)
= 7 + 7 + 7 + 7 + 7 + 7 + 7 = 49 ≡ 4 ( 9)
= 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 = 56 ≡ 2 ( 9)
= 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 = 63 ≡ 0 ( 9)
⟨7⟩ = {0,1,2,3,4,5,6,7,8} = ℤ9
⟨8⟩ = 8 ≡ 8 ( 9)
= 8 + 8 = 16 ≡ 7 ( 9)
= 8 + 8 + 8 = 24 ≡ 6 ( 9)
= 8 + 8 + 8 + 8 = 32 ≡ 5 ( 9)
= 8 + 8 + 8 + 8 + 8 = 40 ≡ 4 ( 9)
= 8 + 8 + 8 + 8 + 8 + 8 = 48 ≡ 3 ( 9)
= 8 + 8 + 8 + 8 + 8 + 8 + 8 = 56 ≡ 2 ( 9)
= 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 = 64 ≡ 1 ( 9)
= 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 = 72 ≡ 0 ( 9)

20

⟨8⟩ = {0,1,2,3,4,5,6,7,8} = ℤ9
Since ⟨1⟩, ⟨2⟩,⟨4⟩,⟨5⟩,⟨7⟩, and ⟨8⟩ are equal to ℤ9, therefore, the
generators of ℤ9 are ⟨ ⟩, ⟨ ⟩,⟨ ⟩,⟨ ⟩,⟨ ⟩, and ⟨ ⟩.
Example 9: The set of integers modulo 10 or in symbols ℤ10. The cyclic
group of order 10 is defined as the unique group of order 10.

Solution: Since ℤ10 = {0,1,2,3,4,5,6,7,8,9}, n = 10.
1. Find the factors of 10, 10 = 5.2

2. Use the Euler’s totient function to find the number of

generators:

(10) = 10 (1 − 1) (1 − 1) = 10 (4) (1) = 4
52 52

∴There are 4 generators in ℤ10.

3. From ℤ10 = {0,1,2,3,4,5,6,7,8,9}, list the subgroups of ℤ10.
Since ⟨1⟩ is a generator for all ℤ , ⟨1⟩ = ℤ10.

⟨2⟩ = 2 ≡ 2 ( 10)

= 2 + 2 = 4 ≡ 4 ( 10)

= 2 + 2 + 2 = 6 ≡ 6 ( 10)

= 2 + 2 + 2 + 2 = 8 ≡ 8 ( 10)

= 2 + 2 + 2 + 2 + 2 = 10 ≡ 0 ( 10)

⟨2⟩ = {0,2,4,6,8} ≠ ℤ10
⟨3⟩ = 3 ≡ 3 ( 10)

= 3 + 3 = 6 ≡ 6 ( 10)

= 3 + 3 + 3 = 9 ≡ 9 ( 10)

21

= 3 + 3 + 3 + 3 = 12 ≡ 2 ( 10)
= 3 + 3 + 3 + 3 + 3 = 15 ≡ 5 ( 10)
= 3 + 3 + 3 + 3 + 3 + 3 = 18 ≡ 8 ( 10)
= 3 + 3 + 3 + 3 + 3 + 3 + 3 = 21 ≡ 1 ( 10)
= 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 = 24 ≡ 4 ( 10)
= 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 = 27 ≡ 7 ( 10)
= 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 = 30 ≡ 0 ( 10)
⟨3⟩ = {0,1,2,3,4,5,6,78,9, } = ℤ10
⟨4⟩ = 4 ≡ 4 ( 10)
= 4 + 4 = 8 ≡ 8 ( 10)
= 4 + 4 + 4 = 12 ≡ 2 ( 10)
= 4 + 4 + 4 + 4 = 16 ≡ 6 ( 10)
= 4 + 4 + 4 + 4 + 4 = 20 ≡ 0 ( 10)
⟨4⟩ = {0,2,4,6,8} ≠ ℤ10
⟨5⟩ = 5 ≡ 5 ( 10)
= 5 + 5 = 10 ≡ 0 ( 10)
⟨5⟩ = {0,5} ≠ ℤ10
⟨6⟩ = 6 ≡ 6 ( 10)
= 6 + 6 = 12 ≡ 2 ( 10)
= 6 + 6 + 6 = 18 ≡ 8 ( 10)
= 6 + 6 + 6 + 6 = 24 ≡ 4 ( 10)
= 6 + 6 + 6 + 6 + 6 = 30 ≡ 0 ( 10)
⟨6⟩ = {0,2,4,6,8} ≠ ℤ10

22

⟨7⟩ = 7 ≡ 7 ( 10)
= 7 + 7 = 14 ≡ 4 ( 10)
= 7 + 7 + 7 = 21 ≡ 1 ( 10)
= 7 + 7 + 7 + 7 = 28 ≡ 8 ( 10)
= 7 + 7 + 7 + 7 + 7 = 35 ≡ 5 ( 10)
= 7 + 7 + 7 + 7 + 7 + 7 = 42 ≡ 2 ( 10)
= 7 + 7 + 7 + 7 + 7 + 7 + 7 = 49 ≡ 9 ( 10)
= 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 = 56 ≡ 6 ( 10)
= 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 = 63 ≡ 3 ( 10)
= 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 = 70 ≡ 0 ( 10)

⟨7⟩ = {0,1,2,3,4,5,6,7,8,9} = ℤ10
⟨8⟩ = 8 ≡ 8 ( 10)

= 8 + 8 = 16 ≡ 6 ( 10)
= 8 + 8 + 8 = 24 ≡ 4 ( 10)
= 8 + 8 + 8 + 8 = 32 ≡ 2 ( 10)
= 8 + 8 + 8 + 8 + 8 = 40 ≡ 0 ( 10)
⟨8⟩ = {0,2,4,6,8} ≠ ℤ10
⟨9⟩ = 9 ≡ 9 ( 10)
= 9 + 9 = 18 ≡ 8 ( 10)
= 9 + 9 + 9 = 27 ≡ 7 ( 10)
= 9 + 9 + 9 + 9 = 36 ≡ 6 ( 10)
= 9 + 9 + 9 + 9 + 9 = 45 ≡ 5 ( 10)
= 9 + 9 + 9 + 9 + 9 + 9 = 54 ≡ 4 ( 10)

23

= 9 + 9 + 9 + 9 + 9 + 9 + 9 = 63 ≡ 3 ( 10)
= 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 = 72 ≡ 2 ( 10)
= 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 = 81 ≡ 1 ( 10)
= 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 = 90 ≡ 0 ( 10)
⟨9⟩ = {0,1,2,3,4,5,6,7,8,9} = ℤ10
Since ⟨1⟩, ⟨3⟩, ⟨7⟩, and ⟨9⟩ are equal to ℤ10, therefore, the generators
of ℤ10 are ⟨ ⟩, ⟨ ⟩, ⟨ ⟩ , and ⟨ ⟩.

Example 10: The set of integers modulo 11 or in symbols ℤ11. The cyclic
group of order 11 is defined as the unique group of order 11.
Solution: Since ℤ11 = {0,1,2,3,4,5,6,7,8,9,10}, n = 11.

1. Find the factors of 11, 11 = 11.1
2. Use the Euler’s totient function to find the number of
generators:

(11) = 11 (1 − 1 ) = 11 (10) = 10

11 11

∴There are 10 generators in ℤ11.

3. From ℤ11 = {0,1,2,3,4,5,6,7,8,9,10}, list the subgroups of ℤ11.
Since ⟨1⟩ is a generator for all ℤ , ⟨1⟩ = ℤ11.

⟨2⟩ = 2 ≡ 2 ( 11)
= 2 + 2 = 4 ≡ 4 ( 11)
= 2 + 2 + 2 = 6 ≡ 6 ( 11)
= 2 + 2 + 2 + 2 = 8 ≡ 8 ( 11)

24

= 2 + 2 + 2 + 2 + 2 = 10 ≡ 10 ( 11)
= 2 + 2 + 2 + 2 + 2 + 2 = 12 ≡ 1 ( 11)
= 2 + 2 + 2 + 2 + 2 + 2 + 2 = 14 ≡ 3 ( 11)
= 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 = 16 ≡ 5 ( 11)
= 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 = 18 ≡ 7 ( 11)
= 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 = 20 ≡ 9 ( 11)
= 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 = 22 ≡ 0 ( 11)
⟨2⟩ = {0,1,2,3,4,5,6,7,8,9,10} = ℤ11
⟨3⟩ = 3 ≡ 3 ( 11)
= 3 + 3 = 6 ≡ 6 ( 11)
= 3 + 3 + 3 = 9 ≡ 9 ( 11)
= 3 + 3 + 3 + 3 = 12 ≡ 1 ( 11)
= 3 + 3 + 3 + 3 + 3 = 15 ≡ 4 ( 11)
= 3 + 3 + 3 + 3 + 3 + 3 = 18 ≡ 7 ( 11)
= 3 + 3 + 3 + 3 + 3 + 3 + 3 = 21 ≡ 10 ( 11)
= 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 = 24 ≡ 2 ( 11)
= 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 = 27 ≡ 5 ( 11)
= 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 = 30 ≡ 8 ( 11)
= 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 = 33 ≡ 0 ( 11)
⟨3⟩ = {0,1,2,3,4,5,6,78,9,10} = ℤ11
⟨4⟩ = 4 ≡ 4 ( 11)
= 4 + 4 = 8 ≡ 8 ( 11)
= 4 + 4 + 4 = 12 ≡ 1 ( 11)

25

= 4 + 4 + 4 + 4 = 16 ≡ 5 ( 11)
= 4 + 4 + 4 + 4 + 4 = 20 ≡ 9 ( 11)
= 4 + 4 + 4 + 4 + 4 + 4 = 24 ≡ 2 ( 11)
= 4 + 4 + 4 + 4 + 4 + 4 + 4 = 28 ≡ 6 ( 11)
= 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 = 32 ≡ 10 ( 11)
= 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 = 36 ≡ 3 ( 11)
= 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 = 40 ≡ 7 ( 11)
= 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 = 44 ≡ 0 ( 11)
⟨4⟩ = {0,1,2,3,4,5,6,7,8,9,10} = ℤ11
⟨5⟩ = 5 ≡ 5 ( 11)
= 5 + 5 = 10 ≡ 10 ( 11)
= 5 + 5 + 5 = 15 ≡ 4 ( 11)
= 5 + 5 + 5 + 5 = 20 ≡ 9 ( 11)
= 5 + 5 + 5 + 5 + 5 = 25 ≡ 3 ( 11)
= 5 + 5 + 5 + 5 + 5 + 5 = 30 ≡ 8 ( 11)
= 5 + 5 + 5 + 5 + 5 + 5 + 5 = 35 ≡ 2 ( 11)
= 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 = 40 ≡ 7 ( 11)
= 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 = 45 ≡ 1 ( 11)
= 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 = 50 ≡ 6 ( 11)
= 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 = 55 ≡ 0 ( 11)
⟨5⟩ = {0,1,2,3,4,5,6,7,8,9,10} = ℤ11

26

⟨6⟩ = 6 ≡ 6 ( 11)
= 6 + 6 = 12 ≡ 1 ( 11)
= 6 + 6 + 6 = 18 ≡ 7 ( 11)
= 6 + 6 + 6 + 6 = 24 ≡ 2 ( 11)
= 6 + 6 + 6 + 6 + 6 = 30 ≡ 8 ( 11)
= 6 + 6 + 6 + 6 + 6 + 6 = 36 ≡ 3 ( 11)
= 6 + 6 + 6 + 6 + 6 + 6 + 6 = 42 ≡ 9 ( 11)
= 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 = 48 ≡ 4 ( 11)
= 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 = 54 ≡ 10 ( 11)
= 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 = 60 ≡ 5 ( 11)
= 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 = 66 ≡ 0 ( 11)

⟨6⟩ = {0,1,2,3,4,5,6,7,8,9,10} = ℤ10
⟨7⟩ = 7 ≡ 7 ( 11)

= 7 + 7 = 14 ≡ 3 ( 11)
= 7 + 7 + 7 = 21 ≡ 11 ( 11)
= 7 + 7 + 7 + 7 = 28 ≡ 6 ( 11)
= 7 + 7 + 7 + 7 + 7 = 35 ≡ 2 ( 11)
= 7 + 7 + 7 + 7 + 7 + 7 = 42 ≡ 9 ( 11)
= 7 + 7 + 7 + 7 + 7 + 7 + 7 = 49 ≡ 5 ( 11)
= 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 = 56 ≡ 1 ( 11)
= 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 = 63 ≡ 8 ( 11)
= 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 = 70 ≡ 4 ( 11)
= 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 = 77 ≡ 0 ( 11)

27

⟨7⟩ = {0,1,2,3,4,5,6,7,8,9,10} = ℤ11
⟨8⟩ = 8 ≡ 8 ( 11)

= 8 + 8 = 16 ≡ 5 ( 11)
= 8 + 8 + 8 = 24 ≡ 2 ( 11)
= 8 + 8 + 8 + 8 = 32 ≡ 10 ( 11)
= 8 + 8 + 8 + 8 + 8 = 40 ≡ 7 ( 11)
= 8 + 8 + 8 + 8 + 8 + 8 = 48 ≡ 4 ( 11)
= 8 + 8 + 8 + 8 + 8 + 8 + 8 = 56 ≡ 1 ( 11)
= 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 = 64 ≡ 9 ( 11)
= 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 = 72 ≡ 6 ( 11)
= 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 = 80 ≡ 3 ( 11)
= 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 = 88 ≡ 0 ( 11)
⟨8⟩ = {0,1,2,3,4,5,6,7,8,9,10} = ℤ11
⟨9⟩ = 9 ≡ 9 ( 11)
= 9 + 9 = 18 ≡ 7 ( 11)
= 9 + 9 + 9 = 27 ≡ 5 ( 11)
= 9 + 9 + 9 + 9 = 36 ≡ 3 ( 11)
= 9 + 9 + 9 + 9 + 9 = 45 ≡ 1 ( 11)
= 9 + 9 + 9 + 9 + 9 + 9 = 54 ≡ 9 ( 11)
= 9 + 9 + 9 + 9 + 9 + 9 + 9 = 63 ≡ 8 ( 11)
= 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 = 72 ≡ 6 ( 11)
= 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 = 81 ≡ 4 ( 11)
= 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 = 90 ≡ 2 ( 11)

28

= 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 = 99 ≡ 0 ( 11)

⟨9⟩ = {0,1,2,3,4,5,6,7,8,9,10} = ℤ11
⟨10⟩ = 10 ≡ 10 ( 11)

= 10 + 10 = 20 ≡ 9 ( 11)

= 10 + 10 + 10 = 30 ≡ 8 ( 11)

= 10 + 10 + 10 + 10 = 40 ≡ 7 ( 11)

= 10 + 10 + 10 + 10 + 10 = 50 ≡ 6 ( 11)

= 10 + 10 + 10 + 10 + 10 + 10 = 60 ≡ 5 ( 11)

= 10 + 10 + 10 + 10 + 10 + 10 + 10 = 70 ≡ 4 ( 11)

= 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 = 80 ≡ 3 ( 11)

= 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 = 90 ≡ 2 ( 11)

= 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 = 100 ≡ 1 ( 11)

= 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 = 110 ≡ 0 ( 11)

⟨10⟩ = {0,1,2,3,4,5,6,7,8,9,10} = ℤ11

Since ⟨1⟩, ⟨2⟩, ⟨3⟩, ⟨4⟩, ⟨5⟩,⟨6⟩, ⟨7⟩,⟨8⟩, ⟨9⟩,⟨10⟩, and ⟨11⟩ are equal to ℤ11,

therefore, the generators of ℤ11 are ⟨ ⟩, ⟨ ⟩,

⟨ ⟩, ⟨ ⟩, ⟨ ⟩,⟨ ⟩, ⟨ ⟩,⟨ ⟩, ⟨ ⟩, and ⟨ ⟩,

29

III. SUMMARY:

The order, the number of generators (using the Euler’s totient

function) and the generator(s) of the ℤ2 to ℤ11 is presented in the following
table.

The set of integers Order ( ) Generator(s)

modulo n(ℤ ), 1 < ≤

11

ℤ2 = {0,1} 2 1 ⟨1⟩
ℤ3 = {0,1,2} 3 2 ⟨1⟩, ⟨2⟩
ℤ4 = {0,1,2,3} 4 2 ⟨1⟩, ⟨3⟩
ℤ5 = {0,1,2,3,4} 5 4 ⟨1⟩, ⟨2⟩, ⟨3⟩, ⟨4⟩
ℤ6 = {0,1,2,3,4,5} 6 2 ⟨1⟩, ⟨5⟩
ℤ7 = {0,1,2,3,4,5,6} 7 6 ⟨1⟩, ⟨2⟩, ⟨3⟩, ⟨4⟩, ⟨5⟩, ⟨6⟩
ℤ8 = {0,1,2,3,4,5,6,7} 8 4 ⟨1⟩, ⟨3⟩, ⟨5⟩, ⟨7⟩
ℤ9 = {0,1,2,3,4,5,6,7,8} 9 6 ⟨1⟩, ⟨2⟩, ⟨4⟩, ⟨5⟩, ⟨7⟩, ⟨8⟩
ℤ10 10 4 ⟨1⟩, ⟨3⟩, ⟨7⟩, ⟨9⟩

= {0,1,2,3,4,5,6,7,8,9}

ℤ11 11 10 ⟨1⟩, ⟨2⟩, ⟨3⟩, ⟨4⟩, ⟨5⟩, ⟨6⟩, ⟨7⟩, ⟨8⟩, ⟨9⟩, ⟨10⟩

= {0,1,2,3,4,5,6,7,8,9,10}

30

The table shows that the order of a certain set of integers modulo n
is determined by the number of elements starting from 0 to n – 1 and it is
equal to the modulo itself. The element ⟨1⟩ is a generator to all ℤ . And the
number of generators in each of the ℤ using the Euler’s totient function
is notable in the column of generators. Anent, for ℤ where is composite,
the generators are the elements which are relatively prime to and that
for ℤ where is prime, every non-zero element is a generator.

IV. CONCLUSION:

Based from the pattern observed in determining the order and
generator(s) of the set of integers modulo n from ℤ2 to ℤ11 , here are the
following conclusions:

1. The order of a group ℤ is the number of elements from 0 to n – 1, thus
the order is the itself.

2. The Euler’s totient function will verify the number of generator(s) in ℤ .

( ) = (1 − 1 ) (1 − 1 ) … (1 − 1 )
1 2

where 1, 2, . . . , are prime factors.
3. The element ⟨1⟩ is a generator to all ℤ .

31

4. In ℤ where n is composite, every non-zero element ∈ ℤ is a generator
if and only if ( , ) = 1. In layman’s term is relatively prime or coprime to
.

5. In ℤ where p is prime, every non-zero element is a generator.

V. RECOMMENDATION:

Due to time constraint, the study covered only few examples
(starting from 2 to 11) in order to generate the conclusions. Therefore, it
is highly recommended to the next researcher who is willing to verify the
results of this study to go beyond and above to prove that the conclusions
are true. It is further suggested to use random in which the value of n
is either composite or prime for the first 100.

32

References:

Abstract Algebra:

https://www.shs-
conferences.org/articles/shsconf/pdf/2016/04/shsconf_erpa2016_0108
2.pdf

Cyclic Group: https://www.youtube.com/watch?v=8A84sA1YuPw

Cyclic Group:
https://en.wikipedia.org/wiki/Cyclic_group#:~:text=A%20cyclic%20grou
p%20is%20a,1%20%3D%20gn%E2%88%921

Definition of Cyclic Group:
https://www.youtube.com/watch?v=hp7bpkNL790

Definition of Group:
https://faculty.cc.gatech.edu/~aboldyre/teaching/Fall05cs6260/numbe
rtheoryhandouts.pdf

Generators: https://www.youtube.com/watch?v=uwUVKwC6E_0

Generator of Cyclic Group:
https://www.youtube.com/watch?v=RMwAmMt_gUw

33

How to find the number of generators of cyclic group/lecture 5:
https://www.youtube.com/watch?v=PvPxrhooAEI
Order of Z: https://www.youtube.com/watch?v=PVsgF27xgPI
Trivial Group e: https://en.wikipedia.org/wiki/Trivial_group