1. Introduction

IJMNTA

International Journal of Modern Nonlinear Theory and Application

2167-9479

Scientific Research Publishing

10.4236/ijmnta.2022.112002

IJMNTA-118072

Articles

Engineering Physics&Mathematics

Evaluate All the Order of Every Element in the Higher Order of Group for Addition and Multiplication Composition

Md.

Abdul Mannan

¹^*Nazmun

Nahar

¹Halima

Akter

¹Momtaz

Begum

²Md.

Amanat Ullah

³Shabnam

Mustari

⁴

Department of Computer Science & Engineering, Prime University, Dhaka, Bangladesh

Department of Mathematics, Uttara University, Dhaka, Bangladesh

Department of Mechanical Engineering, Sonargaon University, Dhaka, Bangladesh

Department of Computer Science & Engineering, Uttara University, Dhaka, Bangladesh

28062022

110211307, March 202225, June 2022 28, June 2022

2014

This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/

This paper aims at treating a study on the order of every element for addition and multiplication composition in the higher order of groups for different algebraic structures as groups; order of a group and order of element of a group in real numbers. Here we discuss the higher order of groups in different types of order which will give us practical knowledge to see the applications of the addition and multiplication composition. If G is a finite group, n is a positive integer and a &isins; G , then the order of the products na. When G is a finite group, every element must have finite order. However, the converse is false: there are infinite groups where each element has finite order. For example, in the group of all roots of unity in C^× each element has finite order. Finally, we find out the order of every element of a group in different types of higher order of group.

Order of Element of a Group Addition Composition Multiplication Composition Modulo

1. Introduction

A group is a particular type of an algebraic system. Here, we propose to study the groups of the order of an element of a group, order of group and the integral powers of an element of a group etc. There are a number of common conventions regarding group notation. The group notation is o or *. We will frequently omit the symbol for the group operation but we will also often write the operation as · or + when it represents multiplication or addition in a ring, and write 1 or 0 for the corresponding identity elements respectively. It’s addition +, multiplication × or (.) is used as binary operation. If the group operation is denoted as a multiplication, then an element a ∈ G is said to be order n if n is the least positive integer such that a n = e or O ( a ) ≤ n i.e., if a n = e and a r ≠ e ∀ r ∈ N s.t. r < n . The order of a is denoted by O ( a ) . If a n ≠ e for any n ∈ N , then a is said to be of zero order or infinite order. It is currently a feature of finite group theory that many theorems on finite groups [1] can only be proved by reducing them to a check that some property holds for all finite simple groups. Then the classification of finite simple groups (see i.e. [2] [3] [4]) comes into play and one has to be able to handle the three different families of simple groups with appropriate techniques. A further extension of Thompson’s characterization was obtained much later in [5], using the full force of the classification of finite simple groups. Thus, each element of a finite group appears exactly once in each row of the table. It follows that each row (respectively column) is a rearrangement of the elements of the group. Next, we discuss the extension of the associative property to products with any number of factors. More specifically, we will prove the so-called generalized associative law which states that in a set with associative operation, a product of factors is unchanged regardless of how parentheses are inserted as long as the factors and their order of appearance in the product are unchanged, as we use additison and multiplication related theorem of group of different orders. As a result we find out the order of group of higher order group. But here we discuss the order of group of higher order group as 50. Finally, we find out the order of every element of a group in different types of the higher order groups ( [6] - [15]).

2. Characterization of the Order of an Element of a Group

We begin this section related to definition with the following characterization of the order of an element of a group

Definition-1 (Multiplication Composition) ( [16] [17]):

Let e be the identity element in (G,). An element a ∈ G is said to be order n if n ∈ Z + such that a n = e or O ( a ) ≤ n . i.e., if a n = e and a r ≠ e ∀ r ∈ N s.t. r < n . The order of a is denoted by O ( a ) . If a n ≠ e for any n ∈ N , then a is said to be of zero order or infinite order.

Definition-1 (Addition Composition) ( [18] [19]):

Let e be the identity element in (G, +). An element a ∈ G is said to be order n if n ∈ Z + such that n a = e or O ( a ) ≤ n . i.e., if n a = e and a r ≠ e ∀ r ∈ N s.t. 0 < r < n . The order of a is denoted by O ( a ) . If n a ≠ e for any n ∈ N , then a is said to be of zero order or infinite order.

Example:

1) In multiplicative group Q 0 , O ( a ) = 1 , O ( − 1 ) = 2 and O ( a ) = ∞ ∀ a ∈ Q 0 s.t. a ≠ ± 1 .

2) In (Z, +), O ( 0 ) = 1 and all other elements have infinite order.

Torsion free group [20]: A group G is called a torsion free group if the identity element e is the only element of finite order.

Example: (Z, +), (Q, +), (R, +).

Torsion group or periodic group [21]: A group G is said to be a torsion group or periodic group if every element of G is of finite order.

Example: The multiplication group is { 1 , − 1 , i , − i } .

Mixed Group [22]: A group G is said to be a mixed group if ∃ at least two elements a , b ∈ G s.t.

1) o ( a ) is finite, a ≠ e 2) o ( b ) = ∞ .

Example: The multiplicative group ( Q 0 , . ) , where Q 0 ≠ 0 is a mixed group.

For o ( − 1 ) = 2 , e ≠ − 1 , 0 ( a ) = ∞ if a ≠ ± 1 and a ∈ Q 0 .

3. Properties of the Order of an Element of a Group

We begin this section of the following theorem related properties of the order of an element of a group.

3.1. Theorem [<xref ref-type="bibr" rid="scirp.118072-ref23">23</xref>]

Show that the order of every element of a finite group is finite.

Proof: Let G be a finite group with multiplication composition.

Let a ∈ G be an arbitrary element.

Now we will prove that O ( a ) is finite.

By closure property, all the elements a² = a ∙ a, a³ = a ∙ a ∙ a, ∙∙∙ etc. belong to G

i.e. a, a², a³, a⁴, a⁵, a⁶, a⁷, … etc. belong to G.

But all these elements are not distinct. Since G is finite.

Let e be the identity in G, then a 0 = e .

Let us suppose that:

a m = a n where m > n ⇒ a m a − n = a n a − n = a 0 = e ⇒ a m − n = e ⇒ a p = e , where p = m − n > 0 , as m > n

also m and n are finite and hence p is a finite positive integer.

Now p is a positive integer s.t. a p = e .

This proves that:

o ( a ) ≤ p = finite number i . e . o ( a ) ≤ a finite number ⇒ o ( a ) is finite .

Remark: The order of any element of a finite group can never exceed the order of the group.

3.2. Theorem [<xref ref-type="bibr" rid="scirp.118072-ref24">24</xref>]

If an element a of a group G is of order n, then a m = e if and only if n is a divisor of m, i.e. m = nq.

Proof: Let a ∈ G be arbitrary element, s.t. o ( a ) = n and so a n = e .

Suppose that a m = e . Now we will prove that n is a divisor of m:

o ( a ) = n , a m = e ⇒ o ( a ) ≤ m ⇒ n ≤ m

if n = m, then clearly n is a divisor of m,

if m > n, then by division algorithm,

∃ q , r ∈ Z s .t . m = n q + r , where 0 ≤ r < n (1)

now a m = a n q + r = ( a n ) q a r = e q a r = e a r = a r ⇒ a m = a r ⇒ e = a r ⇒ a r = e

hence

a r = e , where 0 ≤ r < n (2)

if 0 < r < n, then a r = e is not possible as o ( a ) = n .

if r = 0, then a r = e is possible. So that r = 0.

From (2), then we get, m = nq.

Conversely, Let m = nq. Now we will prove that a m = e .

Now a m = a n q = ( a n ) q = e q = e ⇒ a m = e .

Remark: This theorem can also be expressed in the following ways.

Let a ∈ G be arbitrary element.

1) If o ( a ) = n , then a m = e ⇔ m = n q , q ∈ Z .

2) If a ∈ G is of order n, then there exist an integer m for which a m = e if m is a multiple of n.

4. Results and Discussion

Here, we discuss the result of order of every element for addition and multiplication composition in the higher order of group. As usual we can use addition and multiplication related theorem to evaluate order of group of different orders such as order 2 , 3 , 4 , 5 , ⋯ , 20 etc., i.e. whose order not so high (Not Higher Order Groups). As a result we use addition and multiplication related theorem to evaluate order of group of higher order group as like as 40 , 41 , 42 , 43 , ⋯ , 49 etc. Finally, here we discuss the order of group of higher order group as 50.

4.1. ( [<xref ref-type="bibr" rid="scirp.118072-ref25">25</xref>] - [<xref ref-type="bibr" rid="scirp.118072-ref31">31</xref>]) Find Order of Every Element of the Group {0, 1, 2, 3, …, 49} the Composition Being Addition Modulo 50

Solution: Let (G, +₅₀) is a group. Where G = { 0 , 1 , 2 , 3 , ⋯ , 49 } and +₅₀ denotes addition modulo 50. Here e = 0.

In addition notation,

na = e and n is a least positive integer,

⇒ O ( a ) = n ∴ O ( 0 ) = 1

for O ( e ) = 1 for identity element of every group.

To determine O (1):

1 ( 1 ) = 1 , 2 ( 1 ) = 1 + 50 1 = 2 , 3 ( 1 ) = 1 + 50 2 ( 1 ) = 1 + 50 2 = 3 , 4 ( 1 ) = 1 + 50 3 ( 1 ) = 1 + 50 3 = 4 , 5 ( 1 ) = 1 + 50 4 ( 1 ) = 1 + 50 4 = 5 , 6 ( 1 ) = 1 + 50 5 ( 1 ) = 1 + 50 5 = 6 , 7 ( 1 ) = 1 + 50 6 ( 1 ) = 1 + 50 6 = 7 ,

8 ( 1 ) = 1 + 50 7 ( 1 ) = 1 + 50 2 = 8 , 9 ( 1 ) = 1 + 50 8 ( 1 ) = 1 + 50 8 = 9 , 10 ( 1 ) = 1 + 50 9 ( 1 ) = 1 + 50 9 = 10 , 11 ( 1 ) = 1 + 50 10 ( 1 ) = 1 + 50 10 = 11 , ⋯ , 50 ( 1 ) = 1 + 50 49 ( 1 ) = 1 + 50 49 = 0 = e

Thus 50(1) = e and n (a) ≠ e for n < 50. ∴ O ( 1 ) = 50 .

To determine O (2):

1 ( 2 ) = 2 , 2 ( 2 ) = 2 + 50 2 = 4 , 3 ( 2 ) = 2 + 50 2 ( 2 ) = 2 + 50 4 = 6 , 4 ( 2 ) = 2 + 50 3 ( 2 ) = 2 + 50 6 = 8 , 5 ( 2 ) = 2 + 50 4 ( 2 ) = 2 + 50 8 = 10 , 6 ( 2 ) = 2 + 50 5 ( 2 ) = 2 + 50 10 = 12 , 7 ( 2 ) = 2 + 50 6 ( 2 ) = 2 + 50 12 = 14 , 8 ( 2 ) = 2 + 50 7 ( 2 ) = 2 + 50 14 = 16 , 9 ( 2 ) = 2 + 50 8 ( 2 ) = 2 + 50 16 = 18 , 10 ( 2 ) = 2 + 50 9 ( 2 ) = 2 + 50 18 = 20 , ⋯ , 25 ( 2 ) = 2 + 50 24 ( 2 ) = 2 + 50 48 = 0 = e

Thus 25(2) = e and n (a) ≠ e for n < 50. ∴ O ( 2 ) = 25 .

To determine O (3):

1 ( 3 ) = 3 , 2 ( 3 ) = 3 + 50 3 = 6 , 3 ( 3 ) = 3 + 50 2 ( 3 ) = 3 + 50 6 = 9 , 4 ( 3 ) = 3 + 50 3 ( 3 ) = 3 + 50 9 = 12 , 5 ( 3 ) = 3 + 50 4 ( 3 ) = 3 + 50 12 = 15 , 6 ( 3 ) = 3 + 50 5 ( 3 ) = 3 + 50 15 = 18 , 7 ( 3 ) = 3 + 50 6 ( 3 ) = 3 + 50 18 = 21 , 8 ( 3 ) = 3 + 50 7 ( 3 ) = 3 + 50 21 = 24 , ⋯ , 16 ( 3 ) = 3 + 50 15 ( 3 ) = 3 + 50 45 = 48 , 17 ( 3 ) = 3 + 50 16 ( 3 ) = 3 + 50 48 = 1 , 18 ( 3 ) = 3 + 50 17 ( 3 ) = 3 + 50 1 = 4 , ⋯

26 ( 3 ) = 3 + 50 25 ( 3 ) = 3 + 50 25 = 28 , 27 ( 3 ) = 3 + 50 26 ( 3 ) = 3 + 50 28 = 31 , 28 ( 3 ) = 3 + 50 27 ( 3 ) = 3 + 50 31 = 34 , ⋯ , 38 ( 3 ) = 3 + 50 37 ( 3 ) = 3 + 50 11 = 14 , 39 ( 3 ) = 3 + 50 38 ( 3 ) = 3 + 50 14 = 17 , 40 ( 3 ) = 3 + 50 39 ( 3 ) = 3 + 50 17 = 20 , ⋯ , 48 ( 3 ) = 3 + 50 47 ( 3 ) = 3 + 50 41 = 44 , 49 ( 3 ) = 3 + 50 48 ( 3 ) = 3 + 50 44 = 47 , 50 ( 3 ) = 3 + 50 49 ( 3 ) = 3 + 50 47 = 0 = e

Thus 50(3) = e and n (a) ≠ e for n < 50. ∴ O ( 3 ) = 50 .

To determine O (4):

1 ( 4 ) = 4 , 2 ( 4 ) = 4 + 50 4 = 8 , 3 ( 4 ) = 4 + 50 2 ( 4 ) = 4 + 50 8 = 12 , 4 ( 4 ) = 4 + 50 3 ( 4 ) = 4 + 50 12 = 16 , 5 ( 4 ) = 4 + 50 4 ( 4 ) = 4 + 50 16 = 20 , 6 ( 4 ) = 4 + 50 5 ( 4 ) = 4 + 50 20 = 24 , 7 ( 4 ) = 4 + 50 6 ( 4 ) = 4 + 50 24 = 28 , 8 ( 4 ) = 4 + 50 7 ( 4 ) = 4 + 50 28 = 32 , ⋯ , 16 ( 4 ) = 4 + 50 15 ( 4 ) = 4 + 50 10 = 14 ,

17 ( 4 ) = 4 + 50 16 ( 4 ) = 4 + 50 14 = 18 , 18 ( 4 ) = 4 + 50 17 ( 4 ) = 4 + 50 18 = 22 , ⋯ , 23 ( 4 ) = 4 + 50 22 ( 4 ) = 4 + 50 38 = 42 , 24 ( 4 ) = 4 + 50 23 ( 4 ) = 4 + 50 42 = 46 , 25 ( 4 ) = 4 + 50 24 ( 4 ) = 4 + 50 46 = 0 = e

Thus 25(4) = e and n (a) ≠ e for n < 50. ∴ O ( 4 ) = 25 .

To determine O (5):

1 ( 5 ) = 5 , 2 ( 5 ) = 5 + 50 5 = 10 , 3 ( 5 ) = 5 + 50 2 ( 5 ) = 5 + 50 10 = 15 , 4 ( 5 ) = 5 + 50 3 ( 5 ) = 5 + 50 15 = 20 , 5 ( 5 ) = 5 + 50 4 ( 5 ) = 5 + 50 20 = 25 , 6 ( 5 ) = 5 + 50 5 ( 5 ) = 5 + 50 25 = 30 , 7 ( 5 ) = 5 + 50 6 ( 5 ) = 5 + 50 30 = 35 , 8 ( 5 ) = 5 + 50 7 ( 5 ) = 5 + 50 35 = 40 , 9 ( 5 ) = 5 + 50 8 ( 5 ) = 5 + 50 40 = 45 , 10 ( 5 ) = 5 + 50 9 ( 5 ) = 5 + 50 45 = 0 = e

Thus 10(5) = e and n (a) ≠e for n < 50. ∴ O ( 5 ) = 10 .

To determine O (6):

1 ( 6 ) = 6 , 2 ( 6 ) = 6 + 50 6 = 12 , 3 ( 6 ) = 6 + 50 2 ( 6 ) = 6 + 50 12 = 18 , 4 ( 6 ) = 6 + 50 3 ( 6 ) = 6 + 50 18 = 24 , 5 ( 6 ) = 6 + 50 4 ( 6 ) = 6 + 50 24 = 30 , 6 ( 6 ) = 6 + 50 5 ( 6 ) = 6 + 50 30 = 36 , 7 ( 6 ) = 6 + 50 6 ( 6 ) = 6 + 50 36 = 42 , 8 ( 6 ) = 6 + 50 7 ( 6 ) = 6 + 50 42 = 48 , 9 ( 6 ) = 6 + 50 8 ( 6 ) = 6 + 50 48 = 4 , ⋯ ,

16 ( 6 ) = 6 + 50 15 ( 6 ) = 6 + 50 40 = 46 , 17 ( 6 ) = 6 + 50 16 ( 6 ) = 6 + 50 46 = 2 , 18 ( 6 ) = 6 + 50 17 ( 6 ) = 6 + 50 2 = 8 , ⋯ , 23 ( 6 ) = 6 + 50 22 ( 6 ) = 6 + 50 32 = 38 , 24 ( 6 ) = 6 + 50 23 ( 6 ) = 6 + 50 38 = 44 , 25 ( 6 ) = 6 + 50 24 ( 6 ) = 6 + 50 44 = 0 = e

Thus 25(6) = e and n (a) ≠ e for n < 50. ∴ O ( 6 ) = 25 .

To determine O (7):

1 ( 7 ) = 7 , 2 ( 7 ) = 7 + 50 7 = 14 , 3 ( 7 ) = 7 + 50 2 ( 7 ) = 7 + 50 14 = 21 , 4 ( 7 ) = 7 + 50 3 ( 7 ) = 7 + 50 21 = 28 , 5 ( 7 ) = 7 + 50 4 ( 7 ) = 7 + 50 28 = 35 , 6 ( 7 ) = 7 + 50 5 ( 7 ) = 7 + 50 35 = 42 , 7 ( 7 ) = 7 + 50 6 ( 7 ) = 7 + 50 42 = 49 , 8 ( 7 ) = 7 + 50 7 ( 7 ) = 7 + 50 49 = 6 , ⋯ , 16 ( 7 ) = 7 + 50 15 ( 7 ) = 7 + 50 5 = 12 , 17 ( 7 ) = 7 + 50 16 ( 7 ) = 7 + 50 12 = 19 , 18 ( 7 ) = 7 + 50 17 ( 7 ) = 7 + 50 19 = 26 , ⋯ ,

26 ( 7 ) = 7 + 50 25 ( 7 ) = 7 + 50 25 = 32 , 27 ( 7 ) = 7 + 50 26 ( 7 ) = 7 + 50 32 = 39 , 28 ( 7 ) = 7 + 50 27 ( 7 ) = 7 + 50 39 = 46 , ⋯ , 38 ( 7 ) = 7 + 50 37 ( 7 ) = 7 + 50 9 = 16 , 39 ( 7 ) = 7 + 50 38 ( 7 ) = 7 + 50 16 = 23 , 40 ( 7 ) = 7 + 50 39 ( 7 ) = 7 + 50 23 = 30 , ⋯ , 48 ( 7 ) = 7 + 50 47 ( 7 ) = 7 + 50 29 = 36 , 49 ( 7 ) = 7 + 50 48 ( 7 ) = 7 + 50 36 = 43 , 50 ( 7 ) = 7 + 50 49 ( 7 ) = 7 + 50 43 = 0 = e

Thus 50(7) = e and n (a) ≠ e for n < 50. ∴ O ( 7 ) = 50 .

To determine O(8):

1 ( 8 ) = 8 , 2 ( 8 ) = 8 + 50 8 = 16 , 3 ( 8 ) = 8 + 50 2 ( 8 ) = 8 + 50 16 = 24 , 4 ( 8 ) = 8 + 50 3 ( 8 ) = 8 + 50 24 = 32 , 5 ( 8 ) = 8 + 50 4 ( 8 ) = 8 + 50 32 = 40 , 6 ( 8 ) = 8 + 50 5 ( 8 ) = 8 + 50 40 = 48 , 7 ( 8 ) = 8 + 50 6 ( 8 ) = 8 + 50 48 = 6 , ⋯ , 16 ( 8 ) = 8 + 50 15 ( 8 ) = 8 + 50 20 = 28 , 17 ( 8 ) = 8 + 50 16 ( 8 ) = 8 + 50 28 = 36 , 18 ( 8 ) = 8 + 50 17 ( 8 ) = 8 + 50 36 = 44 , 19 ( 8 ) = 8 + 50 18 ( 8 ) = 8 + 50 44 = 2 , ⋯ , 24 ( 8 ) = 8 + 50 23 ( 8 ) = 8 + 50 34 = 42 , 25 ( 8 ) = 8 + 50 24 ( 8 ) = 8 + 50 42 = 0 = e

Thus 25(8) = e and n (a) ≠ e for n < 50. ∴ O ( 8 ) = 25 .

To determine O (9):

1 ( 9 ) = 9 , 2 ( 9 ) = 9 + 50 9 = 18 , 3 ( 9 ) = 9 + 50 2 ( 9 ) = 9 + 50 18 = 27 , 4 ( 9 ) = 9 + 50 3 ( 9 ) = 9 + 50 27 = 36 , 5 ( 9 ) = 9 + 50 4 ( 9 ) = 9 + 50 36 = 45 , 6 ( 9 ) = 9 + 50 5 ( 9 ) = 9 + 50 45 = 4 , ⋯ , 16 ( 9 ) = 9 + 50 15 ( 9 ) = 9 + 50 35 = 44 , 17 ( 9 ) = 9 + 50 16 ( 9 ) = 9 + 50 44 = 3 , 18 ( 9 ) = 9 + 50 17 ( 9 ) = 9 + 50 3 = 12 , ⋯ , 26 ( 9 ) = 9 + 50 25 ( 9 ) = 9 + 50 25 = 34 , 27 ( 9 ) = 9 + 50 26 ( 9 ) = 9 + 50 34 = 43 ,

28 ( 9 ) = 9 + 50 27 ( 9 ) = 9 + 50 43 = 2 , ⋯ , 38 ( 9 ) = 9 + 50 37 ( 9 ) = 9 + 50 33 = 42 , 39 ( 9 ) = 9 + 50 38 ( 9 ) = 9 + 50 42 = 1 , 40 ( 9 ) = 9 + 50 39 ( 9 ) = 9 + 50 1 = 10 , ⋯ , 48 ( 9 ) = 9 + 50 47 ( 9 ) = 9 + 50 23 = 32 , 49 ( 9 ) = 9 + 50 48 ( 9 ) = 9 + 50 32 = 41 , 50 ( 9 ) = 9 + 50 49 ( 9 ) = 9 + 50 41 = 0 = e

Thus 50(9) = e and n (a) ≠ e for n < 50. ∴ O ( 9 ) = 50 .

To determine O (10):

1 ( 10 ) = 10 , 2 ( 10 ) = 10 + 50 10 = 20 , 3 ( 10 ) = 10 + 50 2 ( 10 ) = 10 + 50 20 = 30 , 4 ( 10 ) = 10 + 50 3 ( 10 ) = 10 + 50 30 = 40 , 5 ( 10 ) = 10 + 50 4 ( 10 ) = 10 + 50 40 = 0 = e ,

Thus 5(10) = e and n (a) ≠ e for n < 50. ∴ O ( 10 ) = 5 .

To determine O (11):

1 ( 11 ) = 11 , 2 ( 11 ) = 11 + 50 11 = 22 , 3 ( 11 ) = 11 + 50 2 ( 11 ) = 11 + 50 22 = 33 , 4 ( 11 ) = 11 + 50 3 ( 11 ) = 11 + 50 33 = 44 , 5 ( 11 ) = 11 + 50 4 ( 11 ) = 11 + 50 44 = 5 , 6 ( 11 ) = 11 + 50 5 ( 11 ) = 11 + 50 5 = 16 , ⋯ , 16 ( 11 ) = 11 + 50 15 ( 11 ) = 11 + 50 15 = 26 , 17 ( 11 ) = 11 + 50 16 ( 11 ) = 11 + 50 26 = 37 , 18 ( 11 ) = 11 + 50 17 ( 11 ) = 11 + 50 37 = 48 , ⋯ , 26 ( 11 ) = 11 + 50 25 ( 11 ) = 11 + 50 25 = 36 , 27 ( 11 ) = 11 + 50 26 ( 11 ) = 11 + 50 36 = 47 ,

28 ( 11 ) = 11 + 50 27 ( 11 ) = 11 + 50 47 = 8 , ⋯ , 38 ( 11 ) = 11 + 50 37 ( 11 ) = 11 + 50 7 = 18 , 39 ( 11 ) = 11 + 50 38 ( 11 ) = 11 + 50 18 = 29 , 40 ( 11 ) = 11 + 50 39 ( 11 ) = 11 + 50 29 = 40 , ⋯ , 48 ( 11 ) = 11 + 50 47 ( 11 ) = 11 + 50 17 = 28 , 49 ( 11 ) = 11 + 50 48 ( 11 ) = 11 + 50 28 = 39 , 50 ( 11 ) = 11 + 50 49 ( 11 ) = 11 + 50 39 = 0 = e

Thus 50(11) = e and n (a) ≠ e for n < 50. ∴ O ( 11 ) = 50 .

To determine O (12):

1 ( 12 ) = 12 , 2 ( 12 ) = 12 + 50 12 = 24 , 3 ( 12 ) = 12 + 50 2 ( 12 ) = 12 + 50 24 = 36 , 4 ( 12 ) = 12 + 50 3 ( 12 ) = 12 + 50 36 = 48 , 5 ( 12 ) = 12 + 50 4 ( 12 ) = 12 + 50 48 = 10 , ⋯ , 16 ( 12 ) = 12 + 50 15 ( 12 ) = 12 + 50 30 = 42 , 17 ( 12 ) = 12 + 50 16 ( 12 ) = 12 + 50 42 = 4 , 18 ( 12 ) = 12 + 50 17 ( 12 ) = 12 + 50 4 = 16 , ⋯ , 24 ( 12 ) = 12 + 50 23 ( 12 ) = 12 + 50 26 = 38 , 25 ( 12 ) = 12 + 50 24 ( 12 ) = 12 + 50 38 = 0 = e

Thus 25(12) = e and n (a) ≠ e for n < 50. ∴ O ( 12 ) = 25 .

To determine O (13):

1 ( 13 ) = 13 , 2 ( 13 ) = 13 + 50 13 = 26 , 3 ( 13 ) = 13 + 50 2 ( 13 ) = 13 + 50 26 = 39 , 4 ( 13 ) = 13 + 50 3 ( 13 ) = 13 + 50 39 = 2 , 5 ( 13 ) = 13 + 50 4 ( 13 ) = 13 + 50 2 = 15 , ⋯ , 16 ( 13 ) = 13 + 50 15 ( 13 ) = 13 + 50 45 = 8 , 17 ( 13 ) = 13 + 50 16 ( 13 ) = 13 + 50 8 = 21 , ⋯ , 26 ( 13 ) = 13 + 50 25 ( 13 ) = 13 + 50 25 = 38 , 27 ( 13 ) = 13 + 50 26 ( 13 ) = 13 + 50 38 = 1 ,

28 ( 13 ) = 13 + 50 27 ( 13 ) = 13 + 50 1 = 14 , ⋯ , 38 ( 13 ) = 13 + 50 37 ( 13 ) = 13 + 50 31 = 44 , 39 ( 13 ) = 13 + 50 38 ( 13 ) = 13 + 50 44 = 7 , ⋯ , 48 ( 13 ) = 13 + 50 47 ( 13 ) = 13 + 50 11 = 24 , 49 ( 13 ) = 13 + 50 48 ( 13 ) = 13 + 50 24 = 37 , 50 ( 13 ) = 13 + 50 49 ( 13 ) = 13 + 50 37 = 0 = e

Thus 50(13) = e and n (a) ≠ e for n < 50. ∴ O ( 13 ) = 50 .

To determine O (14):

1 ( 14 ) = 14 , 2 ( 14 ) = 14 + 50 14 = 28 , 3 ( 14 ) = 14 + 50 2 ( 14 ) = 14 + 50 28 = 42 , 4 ( 14 ) = 14 + 50 3 ( 14 ) = 14 + 50 42 = 6 , ⋯ , 16 ( 14 ) = 14 + 50 15 ( 14 ) = 14 + 50 10 = 24 , 17 ( 14 ) = 14 + 50 16 ( 14 ) = 14 + 50 24 = 38 , ⋯ , 18 ( 14 ) = 14 + 50 17 ( 14 ) = 14 + 50 38 = 2 , 24 ( 14 ) = 14 + 50 23 ( 14 ) = 14 + 50 22 = 36 , 25 ( 14 ) = 14 + 50 24 ( 14 ) = 14 + 50 36 = 0 = e

Thus 25(14) = e and n (a) ≠ e for n < 50. ∴ O ( 14 ) = 25 .

To determine O (15):

1 ( 15 ) = 15 , 2 ( 15 ) = 15 + 50 15 = 30 , 3 ( 15 ) = 15 + 50 2 ( 15 ) = 15 + 50 30 = 45 , 4 ( 15 ) = 15 + 50 3 ( 15 ) = 15 + 50 45 = 10 , ⋯ , 8 ( 15 ) = 15 + 50 7 ( 15 ) = 15 + 50 5 = 20 , 9 ( 15 ) = 15 + 50 8 ( 15 ) = 15 + 50 20 = 35 , 10 ( 15 ) = 15 + 50 9 ( 15 ) = 15 + 50 5 = 0 = e

Thus 10(15) = e and n (a) ≠ e for n < 50. ∴ O ( 15 ) = 10 .

To determine O (16):

1 ( 16 ) = 16 , 2 ( 16 ) = 16 + 50 16 = 32 , 3 ( 16 ) = 16 + 50 2 ( 16 ) = 16 + 50 32 = 48 , 4 ( 16 ) = 16 + 50 3 ( 16 ) = 16 + 50 48 = 14 , ⋯ , 16 ( 16 ) = 16 + 50 15 ( 16 ) = 16 + 50 40 = 6 , 17 ( 16 ) = 16 + 50 16 ( 16 ) = 16 + 50 6 = 22 , ⋯ , 24 ( 16 ) = 16 + 50 23 ( 16 ) = 16 + 50 18 = 34 , 25 ( 16 ) = 16 + 50 24 ( 16 ) = 16 + 50 34 = 0 = e

Thus 25(16) = e and n (a) ≠ e for n < 50. ∴ O ( 16 ) = 25 .

To determine O (17):

1 ( 17 ) = 17 , 2 ( 17 ) = 17 + 50 17 = 34 , 3 ( 17 ) = 17 + 50 2 ( 17 ) = 17 + 50 34 = 1 , 4 ( 17 ) = 17 + 50 3 ( 17 ) = 17 + 50 1 = 18 , ⋯ , 16 ( 17 ) = 17 + 50 15 ( 17 ) = 17 + 50 5 = 22 , 17 ( 17 ) = 17 + 50 16 ( 17 ) = 17 + 50 22 = 39 , ⋯ , 26 ( 17 ) = 17 + 50 25 ( 17 ) = 17 + 50 25 = 42 , 27 ( 17 ) = 17 + 50 26 ( 17 ) = 17 + 50 42 = 9 , ⋯ , 38 ( 17 ) = 17 + 50 37 ( 17 ) = 17 + 50 29 = 46 , 39 ( 17 ) = 17 + 50 38 ( 17 ) = 17 + 50 46 = 13 , ⋯ , 49 ( 17 ) = 17 + 50 48 ( 17 ) = 17 + 50 16 = 33 , 50 ( 17 ) = 17 + 50 49 ( 17 ) = 17 + 50 33 = 0 = e

Thus 50(17) = e and n (a) ≠ e for n < 50. ∴ O ( 17 ) = 50 .

To determine O (18):

1 ( 18 ) = 18 , 2 ( 18 ) = 18 + 50 18 = 36 , 3 ( 18 ) = 18 + 50 2 ( 18 ) = 18 + 50 36 = 4 , 4 ( 18 ) = 18 + 50 3 ( 18 ) = 18 + 50 4 = 22 , ⋯ , 16 ( 18 ) = 18 + 50 15 ( 18 ) = 18 + 50 20 = 38 , 17 ( 18 ) = 18 + 50 16 ( 18 ) = 18 + 50 38 = 6 , ⋯ , 24 ( 18 ) = 18 + 50 23 ( 18 ) = 18 + 50 14 = 32 , 25 ( 18 ) = 18 + 50 24 ( 18 ) = 18 + 50 32 = 0 = e

Thus 25(18) = e and n (a) ≠ e for n < 50. ∴ O ( 18 ) = 25 .

To determine O (19):

1 ( 19 ) = 19 , 2 ( 19 ) = 19 + 50 19 = 38 , 3 ( 19 ) = 19 + 50 2 ( 19 ) = 19 + 50 38 = 7 , ⋯ , 16 ( 19 ) = 19 + 50 15 ( 19 ) = 19 + 50 35 = 4 , 17 ( 19 ) = 19 + 50 16 ( 19 ) = 19 + 50 4 = 23 , ⋯ , 26 ( 19 ) = 19 + 50 25 ( 19 ) = 19 + 50 25 = 44 , 27 ( 19 ) = 19 + 50 26 ( 19 ) = 19 + 50 44 = 13 , ⋯ , 38 ( 19 ) = 19 + 50 37 ( 19 ) = 19 + 50 3 = 22 , 39 ( 19 ) = 19 + 50 38 ( 19 ) = 19 + 50 22 = 41 , ⋯ , 49 ( 19 ) = 19 + 50 48 ( 19 ) = 19 + 50 12 = 31 , 50 ( 19 ) = 19 + 50 49 ( 19 ) = 19 + 50 31 = 0 = e

Thus 50(19) = e and n (a) ≠ e for n < 50. ∴ O ( 19 ) = 50 .

To determine O (20):

1 ( 20 ) = 20 , 2 ( 20 ) = 20 + 50 20 = 40 , 3 ( 20 ) = 20 + 50 2 ( 20 ) = 20 + 50 40 = 10 , ⋯ , 8 ( 20 ) = 20 + 50 7 ( 20 ) = 20 + 50 40 = 10 , 9 ( 20 ) = 20 + 50 8 ( 20 ) = 20 + 50 10 = 30 , 10 ( 20 ) = 20 + 50 9 ( 20 ) = 20 + 50 30 = 0 = e

Thus 10(20) = e and n (a) ≠ e for n < 50. ∴ O ( 20 ) = 10 .

To determine O (21):

1 ( 21 ) = 21 , 2 ( 21 ) = 21 + 50 21 = 42 , 3 ( 21 ) = 21 + 50 2 ( 21 ) = 21 + 50 42 = 13 , ⋯ , 16 ( 21 ) = 21 + 50 15 ( 21 ) = 21 + 50 15 = 36 , 17 ( 21 ) = 21 + 50 16 ( 21 ) = 21 + 50 36 = 7 , ⋯ , 26 ( 21 ) = 21 + 50 25 ( 21 ) = 21 + 50 25 = 46 , 27 ( 21 ) = 21 + 50 26 ( 21 ) = 21 + 50 46 = 17 , ⋯ , 38 ( 21 ) = 21 + 50 37 ( 21 ) = 21 + 50 27 = 48 , 39 ( 21 ) = 21 + 50 38 ( 21 ) = 21 + 50 48 = 19 , ⋯ , 49 ( 21 ) = 21 + 50 48 ( 21 ) = 21 + 50 8 = 29 , 50 ( 21 ) = 21 + 50 49 ( 21 ) = 21 + 50 29 = 0 = e

Thus 50(21) = e and n (a) ≠ e for n < 50. ∴ O ( 21 ) = 50 .

To determine O (22):

1 ( 22 ) = 22 , 2 ( 22 ) = 22 + 50 22 = 44 , 3 ( 22 ) = 22 + 50 2 ( 22 ) = 22 + 50 44 = 16 , ⋯ , 16 ( 22 ) = 22 + 50 15 ( 22 ) = 22 + 50 30 = 2 , 17 ( 22 ) = 22 + 50 16 ( 22 ) = 22 + 50 2 = 24 , ⋯ , 24 ( 22 ) = 22 + 50 23 ( 22 ) = 22 + 50 6 = 28 , 25 ( 22 ) = 22 + 50 24 ( 22 ) = 22 + 50 28 = 0 = e

Thus 25(22) = e and n (a) ≠ e for n < 50. ∴ O ( 22 ) = 25 .

To determine O (23):

1 ( 23 ) = 23 , 2 ( 23 ) = 23 + 50 23 = 46 , 3 ( 23 ) = 23 + 50 2 ( 23 ) = 23 + 50 46 = 19 , ⋯ , 16 ( 23 ) = 23 + 50 15 ( 23 ) = 23 + 50 45 = 18 , 17 ( 23 ) = 23 + 50 16 ( 23 ) = 23 + 50 18 = 41 , ⋯ , 26 ( 23 ) = 23 + 50 25 ( 23 ) = 23 + 50 25 = 48 , 27 ( 23 ) = 23 + 50 26 ( 23 ) = 23 + 50 48 = 21 , ⋯ , 38 ( 23 ) = 23 + 50 37 ( 23 ) = 23 + 50 1 = 24 , 39 ( 23 ) = 23 + 50 38 ( 23 ) = 23 + 50 24 = 47 , ⋯ , 49 ( 23 ) = 23 + 50 48 ( 23 ) = 23 + 50 4 = 27 , 50 ( 23 ) = 23 + 50 49 ( 23 ) = 23 + 50 27 = 0 = e

Thus 50(23) = e and n (a) ≠ e for n < 50. ∴ O ( 23 ) = 50 .

To determine O (24):

1 ( 24 ) = 24 , 2 ( 24 ) = 24 + 50 24 = 48 , 3 ( 24 ) = 24 + 50 2 ( 24 ) = 24 + 50 48 = 22 , ⋯ , 16 ( 24 ) = 24 + 50 15 ( 24 ) = 24 + 50 10 = 34 , 17 ( 24 ) = 24 + 50 16 ( 24 ) = 24 + 50 34 = 8 , ⋯ , 24 ( 24 ) = 24 + 50 23 ( 24 ) = 24 + 50 2 = 26 , 25 ( 24 ) = 24 + 50 24 ( 24 ) = 24 + 50 26 = 0 = e

Thus 25(24) = e and n (a) ≠ e for n < 50. ∴ O ( 24 ) = 25 .

To determine O (25):

1 ( 25 ) = 25 , 2 ( 25 ) = 25 + 50 25 = 0 = e

Thus 2(25) =e and n (a) ≠ e for n < 50. ∴ O ( 25 ) = 2 .

To determine O (26):

1 ( 26 ) = 26 , 2 ( 26 ) = 26 + 50 26 = 2 , ⋯ , 16 ( 26 ) = 26 + 50 15 ( 26 ) = 26 + 50 40 = 16 , 17 ( 26 ) = 26 + 50 16 ( 26 ) = 26 + 50 16 = 42 , ⋯ , 24 ( 26 ) = 26 + 50 23 ( 26 ) = 26 + 50 48 = 24 , 25 ( 26 ) = 26 + 50 24 ( 26 ) = 26 + 50 24 = 0 = e

Thus 25(26) = e and n (a) ≠ e for n < 50. ∴ O ( 26 ) = 25 .

To determine O (27):

1 ( 27 ) = 27 , 2 ( 27 ) = 27 + 50 27 = 4 , ⋯ , 16 ( 27 ) = 27 + 50 15 ( 27 ) = 27 + 50 5 = 32 , 17 ( 27 ) = 27 + 50 16 ( 27 ) = 27 + 50 32 = 9 , ⋯ , 26 ( 27 ) = 27 + 50 25 ( 27 ) = 27 + 50 25 = 2 , 27 ( 27 ) = 27 + 50 26 ( 27 ) = 27 + 50 2 = 29 , ⋯ , 38 ( 27 ) = 27 + 50 37 ( 27 ) = 27 + 50 49 = 26 , 39 ( 27 ) = 27 + 50 38 ( 27 ) = 27 + 50 26 = 3 , ⋯ , 49 ( 27 ) = 27 + 50 48 ( 27 ) = 27 + 50 46 = 23 , 50 ( 27 ) = 27 + 50 49 ( 27 ) = 27 + 50 23 = 0 = e

Thus 50 (27) = e and n (a) ≠ e for n < 50. ∴ O ( 27 ) = 50 .

To determine O (28):

1 ( 28 ) = 28 , 2 ( 28 ) = 28 + 50 28 = 6 , ⋯ , 16 ( 28 ) = 28 + 50 15 ( 28 ) = 28 + 50 20 = 48 , 17 ( 28 ) = 28 + 50 16 ( 28 ) = 28 + 50 48 = 26 , ⋯ , 24 ( 28 ) = 28 + 50 23 ( 28 ) = 28 + 50 44 = 22 , 25 ( 28 ) = 28 + 50 24 ( 28 ) = 28 + 50 22 = 0 = e

Thus 25(28) = e and n (a) ≠ e for n < 50. ∴ O ( 28 ) = 25 .

To determine O (29):

1 ( 29 ) = 29 , 2 ( 29 ) = 29 + 50 29 = 8 , ⋯ , 16 ( 29 ) = 29 + 50 15 ( 29 ) = 29 + 50 35 = 14 , 17 ( 29 ) = 29 + 50 16 ( 29 ) = 29 + 50 14 = 43 , ⋯ , 26 ( 29 ) = 29 + 50 25 ( 29 ) = 29 + 50 25 = 4 , 27 ( 29 ) = 29 + 50 26 ( 29 ) = 29 + 50 4 = 33 , ⋯ , 38 ( 29 ) = 29 + 50 37 ( 29 ) = 29 + 50 23 = 2 , 39 ( 29 ) = 29 + 50 38 ( 29 ) = 29 + 50 2 = 31 , ⋯ , 49 ( 29 ) = 29 + 50 48 ( 29 ) = 29 + 50 42 = 21 , 50 ( 29 ) = 29 + 50 49 ( 29 ) = 29 + 50 21 = 0 = e

Thus 50(29) = e and n (a) ≠ e for n < 50. ∴ O ( 29 ) = 50 .

To determine O (30):

1 ( 30 ) = 30 , 2 ( 30 ) = 30 + 50 30 = 10 , 3 ( 30 ) = 30 + 50 2 ( 30 ) = 30 + 50 10 = 40 , ⋯ , 8 ( 30 ) = 30 + 50 7 ( 30 ) = 30 + 50 10 = 40 , 9 ( 30 ) = 30 + 50 8 ( 30 ) = 30 + 50 40 = 20 , 10 ( 30 ) = 30 + 50 9 ( 30 ) = 30 + 50 20 = 0 = e

Thus 10(30) = e and n (a) ≠ e for n < 50. ∴ O ( 30 ) = 10 .

To determine O (31):

1 ( 31 ) = 31 , 2 ( 31 ) = 31 + 50 31 = 12 , ⋯ , 16 ( 31 ) = 31 + 50 15 ( 31 ) = 31 + 50 15 = 46 , 17 ( 31 ) = 31 + 50 16 ( 31 ) = 31 + 50 46 = 27 , ⋯ , 26 ( 31 ) = 31 + 50 25 ( 31 ) = 31 + 50 25 = 6 , 27 ( 31 ) = 31 + 50 26 ( 31 ) = 31 + 50 6 = 37 , ⋯ , 38 ( 31 ) = 31 + 50 37 ( 31 ) = 31 + 50 47 = 28 , 39 ( 31 ) = 31 + 50 38 ( 31 ) = 31 + 50 28 = 9 , ⋯ , 49 ( 31 ) = 31 + 50 48 ( 31 ) = 31 + 50 38 = 19 , 50 ( 31 ) = 31 + 50 49 ( 31 ) = 31 + 50 19 = 0 = e

Thus 50(31) = e and n (a) ≠ e for n < 50. ∴ O ( 31 ) = 50 .

To determine O (32):

1 ( 32 ) = 32 , 2 ( 32 ) = 32 + 50 32 = 14 , ⋯ , 16 ( 32 ) = 32 + 50 15 ( 32 ) = 32 + 50 30 = 12 , 17 ( 32 ) = 32 + 50 16 ( 32 ) = 32 + 50 12 = 44 , ⋯ , 24 ( 32 ) = 32 + 50 23 ( 32 ) = 32 + 50 36 = 18 , 25 ( 32 ) = 32 + 50 24 ( 32 ) = 32 + 50 18 = 0 = e

Thus 25(32) = e and n (a) ≠ e for n < 50. ∴ O ( 32 ) = 25 .

To determine O (33):

1 ( 33 ) = 33 , 2 ( 33 ) = 33 + 50 33 = 16 , ⋯ , 16 ( 33 ) = 33 + 50 15 ( 33 ) = 33 + 50 45 = 28 , 17 ( 33 ) = 33 + 50 16 ( 33 ) = 33 + 50 28 = 11 , ⋯ , 26 ( 33 ) = 33 + 50 25 ( 33 ) = 33 + 50 25 = 8 , 27 ( 33 ) = 33 + 50 26 ( 33 ) = 33 + 50 8 = 41 , ⋯ , 38 ( 33 ) = 33 + 50 37 ( 33 ) = 33 + 50 21 = 4 , 39 ( 33 ) = 33 + 50 38 ( 33 ) = 33 + 50 4 = 37 , ⋯ , 49 ( 33 ) = 33 + 50 48 ( 33 ) = 33 + 50 34 = 17 , 50 ( 33 ) = 33 + 50 49 ( 33 ) = 33 + 50 17 = 0 = e

Thus 50(33) = e and n (a) ≠ e for n < 50. ∴ O ( 33 ) = 50 .

To determine O (34):

1 ( 34 ) = 34 , 2 ( 34 ) = 34 + 50 34 = 18 , ⋯ , 16 ( 34 ) = 34 + 50 15 ( 34 ) = 34 + 50 10 = 44 , 17 ( 34 ) = 34 + 50 16 ( 34 ) = 34 + 50 44 = 28 , ⋯ , 24 ( 34 ) = 34 + 50 23 ( 34 ) = 34 + 50 32 = 16 , 25 ( 34 ) = 34 + 50 24 ( 34 ) = 34 + 50 16 = 0 = e

Thus 25(34) = e and n (a) ≠ e for n < 50. ∴ O ( 34 ) = 25 .

To determine O (35):

1 ( 35 ) = 35 , 2 ( 35 ) = 35 + 50 35 = 20 , 3 ( 35 ) = 35 + 50 2 ( 35 ) = 35 + 50 20 = 5 , ⋯ , 8 ( 35 ) = 35 + 50 7 ( 35 ) = 35 + 50 45 = 30 , 9 ( 35 ) = 35 + 50 8 ( 35 ) = 35 + 50 30 = 15 , 10 ( 35 ) = 35 + 50 9 ( 35 ) = 35 + 50 15 = 0 = e

Thus 10(35) = e and n (a) ≠ e for n < 50. ∴ O ( 35 ) = 10 .

To determine O (36):

1 ( 36 ) = 36 , 2 ( 36 ) = 36 + 50 36 = 22 , ⋯ , 16 ( 36 ) = 36 + 50 15 ( 36 ) = 36 + 50 40 = 36 , 17 ( 36 ) = 36 + 50 16 ( 36 ) = 36 + 50 36 = 12 , 24 ( 36 ) = 36 + 50 23 ( 36 ) = 36 + 50 28 = 14 , 25 ( 36 ) = 36 + 50 24 ( 36 ) = 36 + 50 14 = 0 = e

Thus 25(36) = e and n (a) ≠ e for n < 50. ∴ O ( 36 ) = 25 .

To determine O (37):

1 ( 37 ) = 37 , 2 ( 37 ) = 37 + 50 37 = 24 , ⋯ , 16 ( 37 ) = 37 + 50 15 ( 37 ) = 37 + 50 5 = 42 , 17 ( 37 ) = 37 + 50 16 ( 37 ) = 37 + 50 42 = 29 , ⋯ , 26 ( 37 ) = 37 + 50 25 ( 37 ) = 37 + 50 25 = 12 , 27 ( 37 ) = 37 + 50 26 ( 37 ) = 37 + 50 12 = 49 , ⋯ , 38 ( 37 ) = 37 + 50 37 ( 37 ) = 37 + 50 = 19 , 39 ( 37 ) = 37 + 50 38 ( 37 ) = 37 + 50 19 = 6 , ⋯ , 49 ( 37 ) = 37 + 50 48 ( 37 ) = 37 + 50 26 = 13 , 50 ( 37 ) = 37 + 50 49 ( 37 ) = 37 + 50 13 = 0 = e

Thus 50(37) = e and n (a) ≠ e for n < 50. ∴ O ( 37 ) = 50 .

To determine O (38):

1 ( 38 ) = 38 , 2 ( 38 ) = 38 + 50 38 = 16 , ⋯ , 16 ( 38 ) = 38 + 50 15 ( 38 ) = 38 + 50 20 = 8 , 17 ( 38 ) = 38 + 50 16 ( 38 ) = 38 + 50 8 = 46 , 24 ( 38 ) = 38 + 50 23 ( 38 ) = 38 + 50 24 = 12 , 25 ( 38 ) = 38 + 50 24 ( 38 ) = 38 + 50 12 = 0 = e

Thus 25(38) = e and n (a) ≠ e for n < 50. ∴ O ( 38 ) = 25 .

To determine O (39):

1 ( 39 ) = 39 , 2 ( 39 ) = 39 + 50 39 = 28 , ⋯ , 16 ( 39 ) = 39 + 50 15 ( 39 ) = 39 + 50 35 = 24 , 17 ( 39 ) = 39 + 50 16 ( 39 ) = 39 + 50 24 = 13 , ⋯ , 26 ( 39 ) = 39 + 50 25 ( 39 ) = 39 + 50 25 = 14 , 27 ( 39 ) = 39 + 50 26 ( 39 ) = 39 + 50 14 = 13 , ⋯ , 38 ( 39 ) = 39 + 50 37 ( 39 ) = 39 + 50 43 = 32 , 39 ( 39 ) = 39 + 50 38 ( 39 ) = 39 + 50 32 = 21 , ⋯ , 49 ( 39 ) = 39 + 50 48 ( 39 ) = 39 + 50 22 = 11 , 50 ( 39 ) = 39 + 50 49 ( 39 ) = 39 + 50 11 = 0 = e

Thus 50(39) = e and n (a) ≠ e for n < 50. ∴ O ( 39 ) = 50 .

To determine O (40):

1 ( 40 ) = 40 , 2 ( 40 ) = 40 + 50 40 = 30 , 3 ( 40 ) = 40 + 50 2 ( 40 ) = 40 + 50 30 = 20 , ⋯ , 8 ( 40 ) = 40 + 50 7 ( 40 ) = 40 + 50 30 = 20 , 9 ( 40 ) = 40 + 50 8 ( 40 ) = 40 + 50 20 = 10 , 10 ( 40 ) = 40 + 50 9 ( 40 ) = 40 + 50 10 = 0 = e

Thus 10(40) = e and n (a) ≠ e for n < 50. ∴ O ( 40 ) = 10 .

To determine O (41):

1 ( 41 ) = 41 , 2 ( 41 ) = 41 + 50 41 = 32 , ⋯ , 16 ( 41 ) = 41 + 50 15 ( 41 ) = 41 + 50 15 = 6 , 17 ( 41 ) = 41 + 50 16 ( 41 ) = 41 + 50 6 = 47 , ⋯ , 26 ( 41 ) = 41 + 50 25 ( 41 ) = 41 + 50 25 = 16 , 27 ( 41 ) = 41 + 50 26 ( 41 ) = 41 + 50 16 = 7 , ⋯ , 38 ( 41 ) = 41 + 50 37 ( 41 ) = 41 + 50 17 = 8 , 39 ( 41 ) = 41 + 50 38 ( 41 ) = 41 + 50 8 = 49 , ⋯ , 49 ( 41 ) = 41 + 50 48 ( 41 ) = 41 + 50 18 = 9 , 50 ( 41 ) = 41 + 50 49 ( 41 ) = 41 + 50 9 = 0 = e

Thus 50(41) = e and n (a) ≠ e for n < 50. ∴ O ( 41 ) = 50 .

To determine O (42):

1 ( 42 ) = 42 , 2 ( 42 ) = 42 + 50 42 = 34 , ⋯ , 16 ( 42 ) = 42 + 50 15 ( 42 ) = 42 + 50 30 = 22 , 17 ( 42 ) = 42 + 50 16 ( 42 ) = 42 + 50 22 = 14 , ⋯ , 24 ( 42 ) = 42 + 50 23 ( 42 ) = 42 + 50 16 = 8 , 25 ( 42 ) = 42 + 50 24 ( 42 ) = 42 + 50 8 = 0 = e

Thus 25(42) = e and n (a) ≠ e forn < 50. ∴ O ( 42 ) = 25 .

To determine O (43):

1 ( 43 ) = 43 , 2 ( 43 ) = 43 + 50 43 = 36 , ⋯ , 16 ( 43 ) = 43 + 50 15 ( 43 ) = 43 + 50 45 = 38 , 17 ( 43 ) = 43 + 50 16 ( 43 ) = 43 + 50 38 = 31 , ⋯ , 26 ( 43 ) = 43 + 50 25 ( 43 ) = 43 + 50 25 = 18 , 27 ( 43 ) = 43 + 50 26 ( 43 ) = 43 + 50 18 = 11 , ⋯ , 38 ( 43 ) = 43 + 50 37 ( 43 ) = 43 + 50 41 = 34 , 39 ( 43 ) = 43 + 50 38 ( 43 ) = 43 + 50 34 = 27 , ⋯ , 49 ( 43 ) = 43 + 50 48 ( 43 ) = 43 + 50 14 = 7 , 50 ( 43 ) = 43 + 50 49 ( 43 ) = 43 + 50 7 = 0 = e

Thus 50(43) = e and n (a) ≠ e for n < 50. ∴ O ( 43 ) = 50 .

To determine O (44):

1 ( 44 ) = 44 , 2 ( 44 ) = 44 + 50 44 = 38 , ⋯ , 16 ( 44 ) = 44 + 50 15 ( 44 ) = 44 + 50 10 = 4 , 17 ( 44 ) = 44 + 50 16 ( 44 ) = 44 + 50 4 = 48 , ⋯ , 24 ( 44 ) = 44 + 50 23 ( 44 ) = 44 + 50 12 = 6 , 25 ( 44 ) = 44 + 50 24 ( 44 ) = 44 + 50 6 = 0 = e

Thus 25(44) = e and n (a) ≠ e for n < 50. ∴ O ( 44 ) = 25 .

To determine O (45):

1 ( 45 ) = 45 , 2 ( 45 ) = 45 + 50 45 = 40 , 3 ( 45 ) = 45 + 50 2 ( 45 ) = 45 + 50 40 = 35 , ⋯ 8 ( 45 ) = 45 + 50 7 ( 45 ) = 45 + 50 15 = 10 , 9 ( 45 ) = 45 + 50 8 ( 45 ) = 45 + 50 10 = 5 , 10 ( 45 ) = 45 + 50 9 ( 45 ) = 45 + 50 5 = 0 = e

Thus 10(45) = e and n (a) ≠ e for n < 50. ∴ O ( 45 ) = 10 .

To determine O (46):

1 ( 46 ) = 46 , 2 ( 46 ) = 46 + 50 46 = 42 , ⋯ , 16 ( 46 ) = 46 + 50 15 ( 46 ) = 46 + 50 40 = 16 , 17 ( 46 ) = 46 + 50 16 ( 46 ) = 46 + 50 16 = 12 , ⋯ , 24 ( 46 ) = 46 + 50 23 ( 46 ) = 46 + 50 8 = 4 , 25 ( 46 ) = 46 + 50 24 ( 46 ) = 46 + 50 4 = 0 = e

Thus 25(46) = e and n (a) ≠ e forn < 50. ∴ O ( 46 ) = 25 .

To determine O (47):

1 ( 47 ) = 47 , 2 ( 47 ) = 47 + 50 47 = 44 , ⋯ , 16 ( 47 ) = 47 + 50 15 ( 47 ) = 47 + 50 5 = 2 , 17 ( 47 ) = 47 + 50 16 ( 47 ) = 47 + 50 2 = 49 , ⋯ , 26 ( 47 ) = 47 + 50 25 ( 47 ) = 47 + 50 25 = 22 , 27 ( 47 ) = 47 + 50 26 ( 47 ) = 47 + 50 22 = 19 , ⋯ , 38 ( 47 ) = 47 + 50 37 ( 47 ) = 47 + 50 39 = 36 ,

39 ( 47 ) = 47 + 50 38 ( 47 ) = 47 + 50 36 = 33 , ⋯ , 49 ( 47 ) = 47 + 50 48 ( 47 ) = 47 + 50 6 = 3 , 50 ( 47 ) = 47 + 50 49 ( 47 ) = 47 + 50 3 = 0 = e

Thus 50(47) = e and n (a) ≠ e for n < 50. ∴ O ( 47 ) = 50 .

To determine O(48):

1 ( 48 ) = 48 , 2 ( 48 ) = 48 + 50 48 = 46 , ⋯ , 16 ( 48 ) = 48 + 50 15 ( 48 ) = 48 + 50 20 = 28 , 17 ( 48 ) = 48 + 50 16 ( 48 ) = 48 + 50 28 = 26 , ⋯ , 24 ( 48 ) = 48 + 50 23 ( 48 ) = 48 + 50 4 = 2 , 25 ( 48 ) = 48 + 50 24 ( 48 ) = 48 + 50 2 = 0 = e

Thus 25(48) = e and n (a) ≠ e for n < 50. ∴ O ( 48 ) = 25 .

To determine O (49):

1 ( 49 ) = 49 , 2 ( 49 ) = 49 + 50 49 = 48 , ⋯ , 16 ( 49 ) = 49 + 50 15 ( 49 ) = 49 + 50 35 = 34 , 17 ( 49 ) = 49 + 50 16 ( 49 ) = 49 + 50 34 = 33 , ⋯ , 26 ( 49 ) = 49 + 50 25 ( 49 ) = 49 + 50 25 = 24 , 27 ( 49 ) = 49 + 50 26 ( 49 ) = 49 + 50 24 = 23 , ⋯ , 38 ( 49 ) = 49 + 50 37 ( 49 ) = 49 + 50 13 = 12 , 39 ( 49 ) = 49 + 50 38 ( 49 ) = 49 + 50 12 = 11 , ⋯ , 49 ( 49 ) = 49 + 50 48 ( 49 ) = 49 + 50 2 = 1 , 50 ( 49 ) = 49 + 50 49 ( 49 ) = 49 + 50 1 = 0 = e

Thus 50(49) = e and n(a) ≠ e for n < 50. ∴ O ( 49 ) = 50 .

The order of every element of the group {0, 1, 2, 3, …, 49}, the composition being addition modulo 50 in the following figure as follows:

4.2. ( [<xref ref-type="bibr" rid="scirp.118072-ref32">32</xref>] - [<xref ref-type="bibr" rid="scirp.118072-ref37">37</xref>]) Find the Order of Every Element in the Multiplication Group G = { a , a 2 , a 3 , a 4 , ⋯ , a 5 0 = e }

Solution:

The identity element of the given group is a 50 = e ⇒ o ( a ) = 50 ∴ o ( a ) = 50 .

We know that o ( a m ) = λ m , where λ = l.c.m of m and n.

To determine o(a²)

Here, o(a⁵⁰) = e, l.c.m of 50 and 2 is 50 ∴ o ( a 2 ) = 50 2 = 25 ⇒ o ( a 2 ) = 25

To determine o(a³)

Here, o(a⁵⁰) = e, l.c.m of 50 and 3 is 150 ∴ o ( a 3 ) = 150 3 = 50 ⇒ o ( a 3 ) = 50

To determine o(a⁴)

Here, o(a⁵⁰) = e, l.c.m of 50 and 4 is 100 ∴ o ( a 4 ) = 100 4 = 25 ⇒ o ( a 4 ) = 25

To determine o(a⁵)

Here, o(a⁵⁰) = e, l.c.m of 50 and 5 is 50 ∴ o ( a 5 ) = 50 5 = 10 ⇒ o ( a 5 ) = 10

To determine o(a⁶)

Here, o(a⁵⁰) = e, l.c.m of 50 and 6 is 150 ∴ o ( a 6 ) = 150 6 = 25 ⇒ o ( a 6 ) = 25

To determine o(a⁷)

Here, o(a⁵⁰) = e, l.c.m of 50 and 7 is 350 ∴ o ( a 7 ) = 350 7 = 50 ⇒ o ( a 7 ) = 50

To determine o(a⁸)

Here, o(a⁵⁰) = e, l.c.m of 50 and 8 is 200 ∴ o ( a 8 ) = 200 8 = 50 ⇒ o ( a 8 ) = 25

To determine o(a⁹)

Here, o(a^50 )=e, l.c.m of 50 and 9 is 450 ∴ o ( a 9 ) = 450 9 = 50 ⇒ o ( a 9 ) = 50

To determine o(a¹⁰)

Here, o(a⁵⁰) = e, l.c.m of 50 and 10 is 50 ∴ o ( a 10 ) = 50 10 = 5 ⇒ o ( a 10 ) = 5

To determine o(a¹¹)

Here, o(a⁵⁰) = e, l.c.m of 50 and 11 is 550 ∴ o ( a 11 ) = 550 11 = 50 ⇒ o ( a 11 ) = 50

To determine o(a¹²)

Here, o(a⁵⁰) = e, l.c.m of 50 and 12 is 300 ∴ o ( a 12 ) = 300 12 = 25 ⇒ o ( a 12 ) = 25

To determine o(a¹³)

Here, o(a⁵⁰) = e, l.c.m of 50 and 13 is 650 ∴ o ( a 13 ) = 650 13 = 50 ⇒ o ( a 13 ) = 50

To determine o(a¹⁴)

Here, o(a⁵⁰) = e, l.c.m of 50 and 14 is 350 ∴ o ( a 14 ) = 350 14 = 25 ⇒ o ( a 14 ) = 25

To determine o(a¹⁵)

Here, o(a⁵⁰) = e, l.c.m of 50 and 15 is 150 ∴ o ( a 15 ) = 150 15 = 10 ⇒ o ( a 15 ) = 10

To determine o(a¹⁶)

Here, o(a⁵⁰) = e, l.c.m of 50 and 16 is 400 ∴ o ( a 16 ) = 400 16 = 25 ⇒ o ( a 16 ) = 25

To determine o(a¹⁷)

Here, o(a⁵⁰) = e, l.c.m of 50 and 17 is 850 ∴ o ( a 17 ) = 850 17 = 50 ⇒ o ( a 17 ) = 50

To determine o(a¹⁹)

Here, o(a⁵⁰) = e, l.c.m of 50 and 18 is 450 ∴ o ( a 18 ) = 450 18 = 25 ⇒ o ( a 18 ) = 25

To determine o(a¹⁹)

Here, o(a⁵⁰) = e, l.c.m of 50 and 19 is 950 ∴ o ( a 19 ) = 950 19 = 50 ⇒ o ( a 19 ) = 50

To determine o(a²⁰)

Here, o(a⁵⁰) = e, l.c.m of 50 and 20 is 20 ∴ o ( a 20 ) = 500 20 = 25 ⇒ o ( a 20 ) = 25

To determine o(a²¹)

Here, o(a⁵⁰) = e, l.c.m of 50 and 21 is 1050 ∴ o ( a 21 ) = 1050 21 = 25 ⇒ o ( a 21 ) = 50

To determine o(a²²)

Here, o(a⁵⁰) = e, l.c.m of 50 and 22 is 550 ∴ o ( a 22 ) = 150 22 = 25 ⇒ o ( a 22 ) = 25

To determine o(a²³)

Here, o(a⁵⁰) = e, l.c.m of 50 and 23 is 1150 ∴ o ( a 23 ) = 1150 23 = 50 ⇒ o ( a 23 ) = 50

To determine o(a²⁴)

Here, o(a⁵⁰) = e, l.c.m of 50 and 24 is 600 ∴ o ( a 24 ) = 600 24 = 25 ⇒ o ( a 24 ) = 25

To determine o(a²⁵)

Here, o(a⁵⁰) = e, l.c.m of 50 and 25 is 50 ∴ o ( a 25 ) = 50 25 = 2 ⇒ o ( a 25 ) = 2

To determine o(a²⁶)

Here, o(a⁵⁰) = e, l.c.m of 50 and 26 is 650 ∴ o ( a 26 ) = 650 26 = 25 ⇒ o ( a 26 ) = 25

To determine o(a²⁷)

Here, o(a⁵⁰) = e, l.c.m of 50 and 27 is 1350 ∴ o ( a 27 ) = 1350 27 = 50 ⇒ o ( a 27 ) = 50

To determine o(a²⁸)

Here, o(a⁵⁰) = e, l.c.m of 50 and 28 is 700 ∴ o ( a 28 ) = 700 28 = 25 ⇒ o ( a 28 ) = 25

To determine o(a²⁹)

Here, o(a⁵⁰) = e, l.c.m of 50 and 29 is 1450 ∴ o ( a 29 ) = 1450 29 = 50 ⇒ o ( a 29 ) = 50

To determine o(a³⁰)

Here, o(a⁵⁰) = e, l.c.m of 50 and 30 is 150 ∴ o ( a 30 ) = 150 30 = 5 ⇒ o ( a 25 ) = 5

To determine o(a³¹)

Here, o(a⁵⁰) = e, l.c.m of 50 and 31 is 1550 ∴ o ( a 31 ) = 1550 31 = 50 ⇒ o ( a 31 ) = 50

To determine o(a³²)

Here, o(a⁵⁰) = e, l.c.m of 50 and 32 is 800 ∴ o ( a 32 ) = 800 32 = 25 ⇒ o ( a 32 ) = 25

To determine o(a³³)

Here, o(a⁵⁰) = e, l.c.m of 50 and 33 is 1650 ∴ o ( a 33 ) = 1650 33 = 50 ⇒ o ( a 33 ) = 50

To determine o(a³⁴)

Here, o(a⁵⁰) = e, l.c.m of 50 and 34 is 850 ∴ o ( a 34 ) = 850 34 = 25 ⇒ o ( a 34 ) = 25

To determine o(a³⁵)

Here, o(a⁵⁰) = e, l.c.m of 50 and 35 is 350 ∴ o ( a 35 ) = 1550 35 = 10 ⇒ o ( a 35 ) = 10

To determine o(a³⁶)

Here, o(a⁵⁰) = e, l.c.m of 50 and 36 is 900 ∴ o ( a 36 ) = 900 36 = 25 ⇒ o ( a 36 ) = 25

To determine o(a³⁷)

Here, o(a⁵⁰) = e, l.c.m of 50 and 31 is 1850 ∴ o ( a 37 ) = 1850 37 = 50 ⇒ o ( a 37 ) = 50

To determine o(a³⁸)

Here, o(a⁵⁰) = e, l.c.m of 50 and 38 is 950 ∴ o ( a 38 ) = 950 38 = 25 ⇒ o ( a 38 ) = 25

To determine o(a³⁹)

Here, o(a⁵⁰) = e, l.c.m of 50 and 39 is 1950 ∴ o ( a 39 ) = 1950 39 = 50 ⇒ o ( a 39 ) = 50

To determine o(a⁴⁰)

Here, o(a⁵⁰) = e, l.c.m of 50 and 40 is 200 ∴ o ( a 40 ) = 200 40 = 5 ⇒ o ( a 40 ) = 5

To determine o(a⁴¹)

Here, o(a⁵⁰) = e, l.c.m of 50 and 41 is 2050 ∴ o ( a 41 ) = 2050 41 = 50 ⇒ o ( a 41 ) = 50

To determine o(a⁴²)

Here, o(a⁵⁰) = e, l.c.m of 50 and 42 is 1050 ∴ o ( a 42 ) = 1050 42 = 25 ⇒ o ( a 42 ) = 5

To determine o(a⁴³)

Here, o(a⁵⁰) = e, l.c.m of 50 and 43 is 2150 ∴ o ( a 43 ) = 2150 43 = 50 ⇒ o ( a 43 ) = 50

To determine o(a⁴⁴)

Here, o(a⁵⁰) = e, l.c.m of 50 and 44 is 1100 ∴ o ( a 44 ) = 1100 44 = 25 ⇒ o ( a 44 ) = 25

To determine o(a⁴⁵)

Here, o(a⁵⁰) = e, l.c.m of 50 and 45 is 450 ∴ o ( a 45 ) = 450 45 = 10 ⇒ o ( a 45 ) = 10

To determine o(a⁴⁶)

Here, o(a⁵⁰) = e, l.c.m of 50 and 46 is 1150 ∴ o ( a 46 ) = 1150 46 = 25 ⇒ o ( a 46 ) = 25

To determine o(a⁴⁷)

Here, o(a⁵⁰) = e, l.c.m of 50 and 47 is 2350 ∴ o ( a 47 ) = 2350 47 = 50 ⇒ o ( a 47 ) = 50

To determine o(a⁴⁸)

Here, o(a⁵⁰) = e, l.c.m of 50 and 48 is 1200 ∴ o ( a 48 ) = 1200 48 = 25 ⇒ o ( a 48 ) = 25

To determine o(a⁴⁹)

Here, o(a⁵⁰) = e, l.c.m of 50 and 49 is 2450 ∴ o ( a 49 ) = 2450 49 = 50 ⇒ o ( a 49 ) = 50

To determine o(a⁵⁰)

Here, o(a⁵⁰) = e, l.c.m of 50 and 50 is 50 ∴ o ( a 50 ) = 50 50 = 1 ⇒ o ( a 50 ) = 1

Remarkable:

1) The addition modulo m, denoted by +_m on the set Z as follows:

a + m b = r , 0 ≤ r < m

where r is the least non-negative remainder when ordinary of (a + b) is divided by m;

2) In order to find out the order of an element a m ∈ G in which aⁿ = e = identity element, first find out least common multiple (i.e. (l.c.m)) = λ) of m and n. Then o ( a m ) = λ m .

5. Conclusion

We hope that this work will be useful for group theory related to order of element of a group. Our result is the order of every element of a group in different types of the higher order group. This result has found extensive use in statistics, information theory and geometrics etc. Then all expected results in this paper will help us to understand better solution of complicated order.

Acknowledgements

I would like to thank my respectable teacher Prof. Dr. Nurul Alam Khan for guidance throughout the research process.

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

Cite this paper

Mannan, Md.A., Nahar, N., Akter, H., Begum, M., Ullah, Md.A. and Mustari, S. (2022) Evaluate All the Order of Every Element in the Higher Order of Group for Addition and Multiplication Composition. International Journal of Modern Nonlinear Theory and Application, 11, 11-30. https://doi.org/10.4236/ijmnta.2022.112002

References1

Beltrán, A., Lyons, R., Moretó, A., Navarro, G., Sáez, A. and Tiep, P.H. (2018) Order of Products of Elements in Finite Groups. Journal of the London Mathematical Society, 99, 535-552. https://doi.org/10.1112/jlms.12185

Gow, R. (2000) Commutators in Finite Simple Groups of Lie Type. Bulletin of the London Mathematical Society, 32, 311-315. https://doi.org/10.1112/S0024609300007141

Gorenstein, D. and Lyons, R. (1983) The Local Structure of Finite Groups of Characteristic 2 Type. Memoirs of the American Mathematical Society, Vol. 42, American Mathematical Society, Providence. https://doi.org/10.1090/memo/0276

Gorenstein, D., Lyons, R. and Solomon, R. (1998) The Classification of the Finite Simple Groups, Number 3, Part 3. Mathematical Surveys and Monographs, Vol. 40, American Mathematical Society, Providence. https://doi.org/10.1090/surv/040.3

Robinson, D.J.S. (1982) A Course in the Theory of Groups. Springer-Verlag, New York.

Abdul Mannan, Md., Akter, H. and Mondal, S. (2021) Evaluate All Possible Subgroups of a Group of Order 30 and 42 by Using Sylow’s Theorem. International Journal of Scientific & Engineering Research, 12, 139-153

Hall Jr., M. (1961) On the Number of Sylow Subgroups in a Finite Group. Journal of Algebra, 7, 363-371. https://doi.org/10.1016/0021-8693(67)90076-2

Brauer, R. (1942) On Groups Whose Order Contains a Prime Number to the First Power I. American Journal of Mathematics, 64, 401-420. https://doi.org/10.2307/2371693

Mckay, J. and Wales, D. (1971) The Multipliers of the Simple Groups of Order 604,800 and 50,232,960. Journal of Algebra, 17, 262-273. https://doi.org/10.1016/0021-8693(71)90033-0

Trefethen, S. (2019) Non-Abelian Composition Factors Finite Groups with the CUT-Property. Journal of Algebra, 522, 236-242. https://doi.org/10.1016/j.jalgebra.2018.12.002

Moretó, A. (2021) Multiplicities of Fields of Values of Irreducible Characters of Finite Groups. Proceedings of the American Mathematical Society, 149, 4109-4116. https://doi.org/10.1090/proc/15566

Bächle

,et al. (2019)3 Questions on Cut Groups Advances in Group Theory and Applications 8, 157-160.

Kurdachenko, L.A., Pypka, A.A. and Subbotin, I.Ya. (2019) On the Structure of Groups Whose Non-Normal Subgroups Are Core-Free. Mediterranean Journal of Mathematics, 16, Article No. 136. https://doi.org/10.1007/s00009-019-1427-6

Kurdachenko, L.A., Pypka, A.A. and Subbotin, I.Ya. (2020) On Groups Whose Non-Normal Subgroups Are Either Contranormal or Core-Free. Advances in Group Theory and Applications, 10, 83-125.

Conway, J., Curtis, R., Norton, S., Parker, R. and Wilson, R. (1985) Atlas of Finite Groups. Oxford University, Eynsham.

Kurdachenko, L.A., Longobardi, P. and Maj, M. (2020) Groups with Finitely Many Isomorphism Classes of Non-Normal Subgroups. Advances in Group Theory and Applications, 10, 9-41.

McCann

,et al. (2020)On Products of Cyclic and Non-Abelian Finite p-Groups Advances in Group Theory and Applications 9, 5-37.

Lucchini, A. and Moscatiello, M. (2020) A Probabilistic Version of a Theorem of Laszlo Kovacs and Hyo-Seob Sim. International Journal of Group Theory, 9, 1-6.

Lucchini, A. and Moscatiello, M. (2020) Generation of Finite Groups and Maximal Subgroup Growth. Advances in Group Theory and Applications, 9, 39-49.

Cook, W.J., Hall, J., Klima, V.W. and Murray, C. (2019) Leibniz Algebras with Low-Dimensional maximal Lie Quotients. Involve, 12, 839-853. https://doi.org/10.2140/involve.2019.12.839

Kurdachenko, L.A., Otal, J. and Subbotin, I.Ya. (2019) On Some Properties of the Upper Central Series in Leibniz Algebras. Commentationes Mathematicae Universitatis Carolinae, 60, 161-175. https://doi.org/10.14712/1213-7243.2019.009

Kurdachenko, L.A., Semko, N.N. and Subbotin, I.Ya. (2020) Applying Group Theory Philosophy to Leibniz Algebras: Some New Developments. Advances in Group Theory and Applications, 9, 71-121.

Bächle, A. and Margolis, L. (2019) On the Prime Graph Question for Integral Group Rings of 4-Primary Groups II. Algebras and Representation Theory, 22, 437-457. https://doi.org/10.1007/s10468-018-9776-6

Bächle, A. and Margolis, L. (2019) An Application of Blocks to Torsion Units in Group Rings. Proceedings of the American Mathematical Society, 147, 4221-4231.

Margolis, L. and del Río, á. (2019) Finite Subgroups of Group Rings: A Survey. Advances in Group Theory and Applications, 8, 1-37.

Russo

,et al. (2019)Some Theorems of Fitting Type Advances in Group Theory and Applications 8, 39-47.

de Giovanni, F. and Subbotin, I.Ya. (2019) Some Topics of Classical Group Theory: The Genesis and Current Stage. Advances in Group Theory and Applications, 8, 119-153.

D’Angeli, D., Francoeur, D., Rodaro, E. and Wächter, J.P. (2020) Infinite Automaton Semigroups and Groups Have Infinite Orbits. Journal of Algebra, 553, 119-137. https://doi.org/10.1016/j.jalgebra.2020.02.014

Gillibert, P. (2018) An Automaton Group with Undecidable Order and Engel Problems. Journal of Algebra, 497, 363-392. https://doi.org/10.1016/j.jalgebra.2017.11.049

Cavaleri

, D’Angeli

, Rodaro

A.D.E.

,et al. (2021)Graph Automaton Groups Advances in Group Theory and Applications 11, 75-112.

Casolo, C., Dardano, U. and Rinauro, S. (2018) Groups in Which Each Subgroup Is Commen Surable with a Normal Subgroup. Journal of Algebra, 496, 48-60. https://doi.org/10.1016/j.jalgebra.2017.11.016

Dardano, U. and Rinauro, S. (2019) Groups with Many Subgroups which Are Commensurable with some Normal Subgroup. Advances in Group Theory and Applications, 7, 3-13.

Margolis, L. and Schnabel, O. (2019) The Herzog-Schönheim Conjecture for Small Groups and Harmonic Subgroups. Beiträge zur Algebra und Geometrie, 60, 399-418. https://doi.org/10.1007/s13366-018-0419-1

Chouraqui, F. (2019) The Herzog-Schönheim Conjecture for Finitely Generated Groups. International Journal of Algebra and Computation, 29, 1083-1112. https://doi.org/10.1142/S0218196719500425

Chouraqui, F. (2018) The Space of Coset Partitions of Fn and Herzog-Schönheim Conjecture. arXiv: 1804.11103.

Chouraqui, F. (2020) An Approach to the Herzog-Schönheim Conjecture Using Automata. Developments in Language Theory: 24th International Conference, Tampa, 11-15 May 2020, 55-68. https://doi.org/10.1007/978-3-030-48516-0_5

Chouraqui, F. (2021) Herzog-Schönheim Conjecture, Vanishing Sums of Roots of Unity and Convex Polygons. Communications in Algebra, 49, 4600-4615. https://doi.org/10.1080/00927872.2021.1924766