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lagrange's theorem for finite groups and verification stage by stage

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Yes, you are right, you have to elaborate more on what you already wrote before. This is the same
instruction that I posted before.
Discuss at least five different theorems for the topic you choose. You must indicate the following:
1. The mathematician who created the theorem.
2. The idea behind the theorem.
3. An example you would use to show students, if you are teaching a lesson on this topic.
• The paper must be 7-10 pages long (double spaced, 12 size Times New Roman font) excluding
references page or any appendix.
• Includes at least three references.
• Uses the MLA Format for this paper.
Starting from the statement of the Lagrange’s theorem for finite groups, we consider a set and an
operation to construct a finite group.
Then we consider a non – empty subset H of that group and verify either the composition table or the
necessary and sufficient condition to be a subgroup and claim H is a subgroup of the group.
Then, we construct the left cosets and pick up the distinct left cosets.
We show that the union of these distinct left cosets forms the entire group.
Further, every left coset consists of equal number of elements.
So, the number of distinct left cosets is nothing but the quotient when the order of the subgroup
divides the order of the group.
We verify all these points with the help of the following example and in the end, we accept that the
Lagrange’s theorem for finite groups is true.
For,
Consider
[ ] [ ] [ ] [ ] [ ] [ ]
{ }
6
0 , 1 , 2 , 3 , 4 , 5=¢
is the set of residue classes modulo 6.
That means, in view of the division algorithm on integers, when an integer is divided by 6 leaves a
remainder 0 or 1 or 2 or 3 or 4 or 5 only. No more or less than these.
So, let us identify an integer regardless of whether it is negative or positive with respect to its
remainder.
For instance, the integer – 245 can be written as
-245 = - 246 + 1 = 6(-41) + 1
Dividend = divisor (quotient) + remainder.
We wrote like this to see the remainder is in the given list of 0, 1, 2, 3, 4, and 5.
If we write – 245 = -240 – 5 = 6(-40) – 5, then the remainder will be negative which is wrong.
So, the division of a negative integer must be performed as shown in the first case.
This way, all the integers are arranged into 6 classes by virtue of their remainder when divided by 6.
So, the classes formed are called the residue classes.
In the present case, our set
6
¢
is having these residue classes as elements.
So, we have to define a suitable operation in view of the residue classes.
We call the operations suitable to residue classes as ‘addition modulo 6’ denoted by
6
+
and
‘multiplication modulo 6’ denoted by
6
×
.
The performance of these operations is also like the integer operation itself.
For example,
[ ] [ ] [ ] [ ]
6
3 4 7 mod 6 1+ = =
[ ] [ ] [ ] [ ]
6
2 4 6mod 6 0+ = =
[ ] [ ] [ ] [ ]
6
2 4 8mod 6 2× = =
Now, we shall take up addition modulo 6 as the operation and show that the set of residue classes
modulo 6 under this operation forms a group.
We fill the composition table which satisfies all the properties of a group.

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From this table, it is easy to follow that [0] is the identity with respect to addition modulo 6.
[1] is the additive inverse of [5] and vice versa
[2] is the additive inverse of [4] and vice versa
[3] is the additive inverse of itself.
Associative law and closure law are readily verified from the composition table.
Thus,
( )
6 6
, +¢
is a group.
We can further verify that this is an abelian group and thus, this is an example for a finite abelian
group.
We shall take up
[ ] [ ]
{ }
0 , 3H =
Clearly, the elements of H are the elements of
6
¢
.
So, H is a non empty subset of the residue class group.
The order of a group is nothing but the number of elements of the group. It is denoted by O (G).
So,
( )
6
6O =¢
Also, we show that H is a subgroup of
( )
6 6
, +¢
It is enough to show the necessary and sufficient condition for H to be a subgroup.
But, we show the composition table as above under the same operation of the group to show that H is
a subgroup of
( )
6 6
, +¢
[ ] [ ]
[ ] [ ] [ ]
[ ] [ ] [ ]
6
0 3
0 0 3
3 3 0
+
See that the closure law under addition and inverses are readily available in the table.
The identity [0] is present in H.
The inverse of [0] is [0] itself and the inverse of [3] is [3] itself.
From this, inverses exist in H.
Therefore, H = {[0], [3]} is a subgroup of
( )
6 6
, +¢
.
See that the order of H is O (H) = 2.
Since 2 divides 6, we can write
( ) ( )
6
|O H O ¢
which verifies the Lagrange’s theorem for finite
groups.
Further, we dealt with the left cosets of H in G in the proof of the theorem.
Now, the left cosets of H in
6
¢
are
[ ] [ ] [ ] [ ]
{ }
[ ] [ ] [ ] [ ]
{ }
[ ] [ ]
{ }
[ ] [ ] [ ] [ ]
{ }
[ ] [ ] [ ] [ ]
{ }
[ ] [ ]
{ }
[ ] [ ] [ ] [ ]
{ }
[ ] [ ] [ ] [ ]
{ }
[ ] [ ]
{ }
[ ] [ ] [ ] [ ]
{ }
[ ] [ ] [ ] [ ]
{ }
[ ] [ ]
{ }
[ ] [ ] [ ] [ ]
{ }
[ ] [ ] [ ] [ ]
{ }
[ ] [ ]
{ }
6 6
6 6
6 6
6 6
6 6
1 1 0 , 3 1 0 , 1 3 1 , 4
2 2 0 , 3 2 0 , 2 3 2 , 5
3 3 0 , 3 3 0 , 3 3 3 , 0
4 4 0 , 3 4 0 , 4 3 4 , 1
5 5 0 , 3 5 0 , 5 3 5 , 2
H
H
H
H
H
+ = + = + + =
+ = + = + + =
+ = + = + + =
+ = + = + + =
+ = + = + + =
We have constructed the left cosets of 5 elements with H and it is easy to see that the distinct left
cosets are
[ ] [ ]
{ }
[ ] [ ]
{ }
[ ] [ ]
{ }
0 , 3 , 1 , 4 , 2 , 5
only.
See that by considering the union of all these left cosets, we get the entire
( )
6 6
, +¢
[ ] [ ] [ ] [ ] [ ] [ ]
{ }
0 , 1 , 2 , 3 , 4 , 5=

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Also, the number of distinct left cosets = 3 =
( )
( )
6
O
O H
¢
which we expressed in the course of proof of
the Lagrange’s theorem.

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