https://www.dpmms.cam.ac.uk/~wtg10/definition.html

With thanks and regards to Prof. Tim Gowers,

Nalin Pithwa

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# Author: Nalin Pithwa

# Definition of definition : Prof. Tim Gowers, Fields Medalist

# You and your research (you and your studies) by Richard Hamming, AT & T, Bell Labs mathematician.

# Dihedral groups explained by I N Herstein

# Remembering Sir Michael Atiyah’s “Advice to a young mathematician”

# Basic Topology: Esp Order and LUB property

# Equivalence relations and partitions: some core basic theorems

# Cedric Villani, Fields Medalist: Work to live, or live to work?

# Some foundation mathematics

# Precursor to Algebra: Herstein shows a way!

# Riddle of the month: Oct 2017: AMS grad student blog

I am a DSP Engineer and a mathematician working in closely related areas of DSP, Digital Control, Digital Comm and Error Control Coding. I have a passion for both Pure and Applied Mathematics.

https://www.dpmms.cam.ac.uk/~wtg10/definition.html

With thanks and regards to Prof. Tim Gowers,

Nalin Pithwa

Reference: Abstract Algebra, Third Edition, I N Herstein

First consider the following: Let S be the plane, that is, and consider defined by and ; f is the reflection about the y-axis and g is the rotation through 90 degrees in a counterclockwise direction about the origin. We then define , and let * in G be the product of elements in A(S). Clearly, identity mapping;

and

So .

It is a good exercise to verify that and G is a non-abelian group of order 8. This group is called the dihedral group of order 8. [Try to find a formula for that expresses a, b in terms of i, j, s and t.

II) Let S be as in above example and f the mapping in above example. Let and let h be the rotation of the plane about the origin through an angle of in the counterclockwise direction. We then define and define the product * in G via the usual product of mappings. One can verify that identity mapping and . These relations allow us to show that (with some effort) G is a non-abelian group of order 2n. G is called the dihedral group of order 2n.

More later,

Nalin Pithwa

a very gentle, gradual pace of introducing these hardcore concepts…

Suppose R is an equivalence relation on a set S. For each a in S, let [a] denote the set of elements of S to which a is related under R, that is,

We call [a] the equivalence class of a in S under R. The collection of all such equivalence classes is denoted by S/R, that is, . It is called the quotient set of S by R.

The fundamental property of an equivalence relation and its quotient set is contained in the following theorem:

**Theorem I:**

Let R be an equivalence relation on a set S. Then, the quotient set S/R is a partition of S. Specifically,

(i) For each , we have .

(ii) if and only if .

(iii) If , then [a] and [b] are disjoint.

*Proof of (i):*

Since R is reflexive, for every and therefore .

*Proof of (ii):*

Assume: .

we want to show that . That is, we got to prove, (i) and (ii) .

Let ; then, . But, by hypothesis and so, by transitivity, . Accordingly, . Thus, .

To prove that , we observe that implies, by symmetry, that . Then, by a similar argument, we obtain . Consequently, .

Now, assume .

Then by part (i) of this proof that for each , we have . So, also, here ; hence, .

*Proof of (iii):*

Here, we prove the equivalent contrapositive of the statement (iii), that is:

If then .

if then there exists an element with . Hence, and . By symmetry, , and, by transitivity, . Consequently, by proof (ii), .

The **converse of the above theorem is also true. **That is,

**Theorem II:**

Suppose is a partition of set S. Then, there is an equivalence relation on S such that the set of equivalence classes is the same as the partition .

Specifically, for , the equivalence in Theorem I is defined by if a and b belong to the same cell in P.

Thus, we see that there is a one-one correspondence between the equivalence relations on a set S and the partitions of S.

*Proof of Theorem II:*

Let , define if a and b belong to the same cell in P. We need to show that is reflexive, symmetric and transitive.

(i) Let . Since P is a partition there exists some in P such that . Hence, . Thus, is reflexive.

(ii) Symmetry follows from the fact that if , then .

(iii) Suppose and . Then, and . Therefore, . Since P is a partition, . Thus, and so . Thus, is transitive.

Accordingly, is an equivalence relation on S.

Furthermore,

.

Thus, the equivalence classes under are the same as the cells in the partition P.

More later,

Nalin Pithwa.

**Well-Ordering Principle:**

Every non-empty set S of non-negative integers contains a least element; that is, there is some integer a in S such that for all b’s belonging to S.

*Because this principle plays a role in many proofs related to foundations of mathematics, let us use it to show that the set of positive integers has what is known as the Archimedean property.*

**Archimedean property:**

If a and b are any positive integers, then there exists a positive integer n such that .

**Proof:**

*By contradiction:*

Assume that the statement of the theorem is not true so that for some a and b, we have for every positive integer n. Then, the set consists entirely of positive integers. By the Well-Ordering Principle, S will possess a least element, say, . Notice that also lies in S; because S contains all integers of this form. Further, we also have contrary to the choice of as the smallest integer in S. This contradiction arose out of original assumption that the Archimedean property did not hold; hence, the proof. **QED.**

**First Principle of Finite Induction:**

Let S be a set of positive integers with the following properties:

a) the integer 1 belongs to S.

b) Whenever the integer k is in S, the next integer is also in S.

Then, S is the set of all positive integers.

**Second Principle of Finite Induction:**

Let S be a set of positive integers with the following properties:

a) the integer 1 belongs to S.

b) If k is a positive integer such that belong to S, then must also be in S.

Then, S is the set of all positive integers.

So, in lighter vein, we assume a set of positive integers is given just as Kronecker had observed: “God created the natural numbers, all the rest is man-made.”

More later,

Nalin Pithwa.

**Reference: Abstract Algebra, 3rd edition, I. N. Herstein, Prentice Hall International Edition.**

**Problems:**

- Let S be a set having an operation * which assigns an element a*b of S for any . Let us assume that the following two rules hold:

i) If a, b are any objects in S, then

ii) If a, b are any objects in S, then

Show that S can have at most one object.

II. Let S be the set of all integers . For a, b in S, define * by . Verify the following:

(i) unless .

(ii) in general. Under what conditions on a, b, c is ?

(iii) the integer 0 has the property that for every a in S.

(iv) For a in S, .

III) Let S consist oif f the two objects and . We define the operation * on S by subjecting and to the following conditions:

i) i

ii)

iii)

**of verify by explicit calculations that if a, b, c are any elements of S (that is, a, b and c can be any of ** or then

i) is in S

ii)

iii)

iv) There is a particular a in S such that $la=latex a*b=b*a=b$ for all b in S.

,v) Given b in S, then where a is the parituclar element in part “iv” above.

Cheers,

Nalin Pithwa

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