Evaluate All the Order of Every Element in the Higher Order of Group for Addition and Multiplication Composition ()
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
is said to be order n if n is the least positive integer such that
or
i.e., if
and
s.t.
. The order of a is denoted by
. If
for any
, 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
is said to be order n if
such that
or
. i.e., if
and
s.t.
. The order of a is denoted by
. If
for any
, 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
is said to be order n if
such that
or
. i.e., if
and
s.t.
. The order of a is denoted by
. If
for any
, then a is said to be of zero order or infinite order.
Example:
1) In multiplicative group
,
,
and
s.t.
.
2) In (Z, +),
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
.
Mixed Group [22]: A group G is said to be a mixed group if
at least two elements
s.t.
1)
is finite,
2)
.
Example: The multiplicative group
, where
is a mixed group.
For
,
,
if
and
.
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 [23]
Show that the order of every element of a finite group is finite.
Proof: Let G be a finite group with multiplication composition.
Let
be an arbitrary element.
Now we will prove that
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
.
Let us suppose that:
also m and n are finite and hence p is a finite positive integer.
Now p is a positive integer s.t.
.
This proves that:
.
Remark: The order of any element of a finite group can never exceed the order of the group.
3.2. Theorem [24]
If an element a of a group G is of order n, then
if and only if n is a divisor of m, i.e. m = nq.
Proof: Let
be arbitrary element, s.t.
and so
.
Suppose that
. Now we will prove that n is a divisor of m:
if n = m, then clearly n is a divisor of m,
if m > n, then by division algorithm,
(1)
now
hence
(2)
if 0 < r < n, then
is not possible as
.
if r = 0, then
is possible. So that r = 0.
From (2), then we get, m = nq.
Conversely, Let m = nq. Now we will prove that
.
Now
.
Remark: This theorem can also be expressed in the following ways.
Let
be arbitrary element.
1) If
, then
.
2) If
is of order n, then there exist an integer m for which
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
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
etc. Finally, here we discuss the order of group of higher order group as 50.
4.1. ( [25] - [31]) 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
and +50 denotes addition modulo 50. Here e = 0.
In addition notation,
na = e and n is a least positive integer,
for
for identity element of every group.
To determine O (1):
Thus 50(1) = e and n (a) ≠ e for n < 50.
.
To determine O (2):
Thus 25(2) = e and n (a) ≠ e for n < 50.
.
To determine O (3):
Thus 50(3) = e and n (a) ≠ e for n < 50.
.
To determine O (4):
Thus 25(4) = e and n (a) ≠ e for n < 50.
.
To determine O (5):
Thus 10(5) = e and n (a) ≠e for n < 50.
.
To determine O (6):
Thus 25(6) = e and n (a) ≠ e for n < 50.
.
To determine O (7):
Thus 50(7) = e and n (a) ≠ e for n < 50.
.
To determine O(8):
Thus 25(8) = e and n (a) ≠ e for n < 50.
.
To determine O (9):
Thus 50(9) = e and n (a) ≠ e for n < 50.
.
To determine O (10):
Thus 5(10) = e and n (a) ≠ e for n < 50.
.
To determine O (11):
Thus 50(11) = e and n (a) ≠ e for n < 50.
.
To determine O (12):
Thus 25(12) = e and n (a) ≠ e for n < 50.
.
To determine O (13):
Thus 50(13) = e and n (a) ≠ e for n < 50.
.
To determine O (14):
Thus 25(14) = e and n (a) ≠ e for n < 50.
.
To determine O (15):
Thus 10(15) = e and n (a) ≠ e for n < 50.
.
To determine O (16):
Thus 25(16) = e and n (a) ≠ e for n < 50.
.
To determine O (17):
Thus 50(17) = e and n (a) ≠ e for n < 50.
.
To determine O (18):
Thus 25(18) = e and n (a) ≠ e for n < 50.
.
To determine O (19):
Thus 50(19) = e and n (a) ≠ e for n < 50.
.
To determine O (20):
Thus 10(20) = e and n (a) ≠ e for n < 50.
.
To determine O (21):
Thus 50(21) = e and n (a) ≠ e for n < 50.
.
To determine O (22):
Thus 25(22) = e and n (a) ≠ e for n < 50.
.
To determine O (23):
Thus 50(23) = e and n (a) ≠ e for n < 50.
.
To determine O (24):
Thus 25(24) = e and n (a) ≠ e for n < 50.
.
To determine O (25):
Thus 2(25) =e and n (a) ≠ e for n < 50.
.
To determine O (26):
Thus 25(26) = e and n (a) ≠ e for n < 50.
.
To determine O (27):
Thus 50 (27) = e and n (a) ≠ e for n < 50.
.
To determine O (28):
Thus 25(28) = e and n (a) ≠ e for n < 50.
.
To determine O (29):
Thus 50(29) = e and n (a) ≠ e for n < 50.
.
To determine O (30):
Thus 10(30) = e and n (a) ≠ e for n < 50.
.
To determine O (31):
Thus 50(31) = e and n (a) ≠ e for n < 50.
.
To determine O (32):
Thus 25(32) = e and n (a) ≠ e for n < 50.
.
To determine O (33):
Thus 50(33) = e and n (a) ≠ e for n < 50.
.
To determine O (34):
Thus 25(34) = e and n (a) ≠ e for n < 50.
.
To determine O (35):
Thus 10(35) = e and n (a) ≠ e for n < 50.
.
To determine O (36):
Thus 25(36) = e and n (a) ≠ e for n < 50.
.
To determine O (37):
Thus 50(37) = e and n (a) ≠ e for n < 50.
.
To determine O (38):
Thus 25(38) = e and n (a) ≠ e for n < 50.
.
To determine O (39):
Thus 50(39) = e and n (a) ≠ e for n < 50.
.
To determine O (40):
Thus 10(40) = e and n (a) ≠ e for n < 50.
.
To determine O (41):
Thus 50(41) = e and n (a) ≠ e for n < 50.
.
To determine O (42):
Thus 25(42) = e and n (a) ≠ e forn < 50.
.
To determine O (43):
Thus 50(43) = e and n (a) ≠ e for n < 50.
.
To determine O (44):
Thus 25(44) = e and n (a) ≠ e for n < 50.
.
To determine O (45):
Thus 10(45) = e and n (a) ≠ e for n < 50.
.
To determine O (46):
Thus 25(46) = e and n (a) ≠ e forn < 50.
.
To determine O (47):
Thus 50(47) = e and n (a) ≠ e for n < 50.
.
To determine O(48):
Thus 25(48) = e and n (a) ≠ e for n < 50.
.
To determine O (49):
Thus 50(49) = e and n(a) ≠ e for n < 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. ( [32] - [37]) Find the Order of Every Element in the Multiplication Group
Solution:
The identity element of the given group is
.
We know that
, 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
To determine o(a3)
Here, o(a50) = e, l.c.m of 50 and 3 is 150
To determine o(a4)
Here, o(a50) = e, l.c.m of 50 and 4 is 100
To determine o(a5)
Here, o(a50) = e, l.c.m of 50 and 5 is 50
To determine o(a6)
Here, o(a50) = e, l.c.m of 50 and 6 is 150
To determine o(a7)
Here, o(a50) = e, l.c.m of 50 and 7 is 350
To determine o(a8)
Here, o(a50) = e, l.c.m of 50 and 8 is 200
To determine o(a9)
Here, o(a^50 )=e, l.c.m of 50 and 9 is 450
To determine o(a10)
Here, o(a50) = e, l.c.m of 50 and 10 is 50
To determine o(a11)
Here, o(a50) = e, l.c.m of 50 and 11 is 550
To determine o(a12)
Here, o(a50) = e, l.c.m of 50 and 12 is 300
To determine o(a13)
Here, o(a50) = e, l.c.m of 50 and 13 is 650
To determine o(a14)
Here, o(a50) = e, l.c.m of 50 and 14 is 350
To determine o(a15)
Here, o(a50) = e, l.c.m of 50 and 15 is 150
To determine o(a16)
Here, o(a50) = e, l.c.m of 50 and 16 is 400
To determine o(a17)
Here, o(a50) = e, l.c.m of 50 and 17 is 850
To determine o(a19)
Here, o(a50) = e, l.c.m of 50 and 18 is 450
To determine o(a19)
Here, o(a50) = e, l.c.m of 50 and 19 is 950
To determine o(a20)
Here, o(a50) = e, l.c.m of 50 and 20 is 20
To determine o(a21)
Here, o(a50) = e, l.c.m of 50 and 21 is 1050
To determine o(a22)
Here, o(a50) = e, l.c.m of 50 and 22 is 550
To determine o(a23)
Here, o(a50) = e, l.c.m of 50 and 23 is 1150
To determine o(a24)
Here, o(a50) = e, l.c.m of 50 and 24 is 600
To determine o(a25)
Here, o(a50) = e, l.c.m of 50 and 25 is 50
To determine o(a26)
Here, o(a50) = e, l.c.m of 50 and 26 is 650
To determine o(a27)
Here, o(a50) = e, l.c.m of 50 and 27 is 1350
To determine o(a28)
Here, o(a50) = e, l.c.m of 50 and 28 is 700
To determine o(a29)
Here, o(a50) = e, l.c.m of 50 and 29 is 1450
To determine o(a30)
Here, o(a50) = e, l.c.m of 50 and 30 is 150
To determine o(a31)
Here, o(a50) = e, l.c.m of 50 and 31 is 1550
To determine o(a32)
Here, o(a50) = e, l.c.m of 50 and 32 is 800
To determine o(a33)
Here, o(a50) = e, l.c.m of 50 and 33 is 1650
To determine o(a34)
Here, o(a50) = e, l.c.m of 50 and 34 is 850
To determine o(a35)
Here, o(a50) = e, l.c.m of 50 and 35 is 350
To determine o(a36)
Here, o(a50) = e, l.c.m of 50 and 36 is 900
To determine o(a37)
Here, o(a50) = e, l.c.m of 50 and 31 is 1850
To determine o(a38)
Here, o(a50) = e, l.c.m of 50 and 38 is 950
To determine o(a39)
Here, o(a50) = e, l.c.m of 50 and 39 is 1950
To determine o(a40)
Here, o(a50) = e, l.c.m of 50 and 40 is 200
To determine o(a41)
Here, o(a50) = e, l.c.m of 50 and 41 is 2050
To determine o(a42)
Here, o(a50) = e, l.c.m of 50 and 42 is 1050
To determine o(a43)
Here, o(a50) = e, l.c.m of 50 and 43 is 2150
To determine o(a44)
Here, o(a50) = e, l.c.m of 50 and 44 is 1100
To determine o(a45)
Here, o(a50) = e, l.c.m of 50 and 45 is 450
To determine o(a46)
Here, o(a50) = e, l.c.m of 50 and 46 is 1150
To determine o(a47)
Here, o(a50) = e, l.c.m of 50 and 47 is 2350
To determine o(a48)
Here, o(a50) = e, l.c.m of 50 and 48 is 1200
To determine o(a49)
Here, o(a50) = e, l.c.m of 50 and 49 is 2450
To determine o(a50)
Here, o(a50) = e, l.c.m of 50 and 50 is 50
Remarkable:
1) The addition modulo m, denoted by +m on the set Z as follows:
,
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
in which an = e = identity element, first find out least common multiple (i.e. (l.c.m)) = λ) of m and n. Then
.
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.