Starters book in Algebra continued

Now lets discuss about the Second chapter named as SUBGROUPS . As mentioned before I am following the sequence of chapters from Herstein.

IMPORTANT IDEAS:

i) First go through the definition very well. You will see that H is a subgroup of G when H is a group under the same operation of G, and H is a subset of G. That's all.
You only need to remember that H is a subgroup of G iff H is closed under the same operation of G and has an inverse of every element in H.
Now does any question pop out in your mind?.................. If yes then you are on the right track in Group theory but if NO then let me tell you the question,
At the very beginning of this discussion I wrote "H is a subgroup of G when H is a group under the same operation of G, and H is a subset of G." So where the hell are associative and identity property?
You must be thinking now that hmmmmm huh!!!????Thinking

Lets pause and think that if elements of G are associative why wont be H's elements.
Is the identity same as G's? Have a look at yourself.

ii) One of the most important ideas of this chapter is understanding the "COSETS"
Cosets are nothing but collection of elements of the form $latex ha$ or $latex ah$ where $latex h\in H, a\in G$

Other useful links:-

Abstract Algebra | Starters handbook for College Math

Hello, this is a discussion page for the college students who are in various prestigious colleges throughout India, and are keen to pursue Mathematics. Abstract Algebra plays a pivotal role in college mathematics, and it mainly focuses on three things
GROUPS, RINGS, and FIELDS.
Though Field is not in the course of some colleges, eventually it will be very helpful.
Now this discussion will be on some of the popular topics from Groups, Rings and Fields(to some extent).

AT A GLANCE WHAT IS IMPORTANT IN GROUPS
I am going to discuss chapterwise so in this post I give the first chapter of Group (Taking Herstein as reference)

Topics to study, and Problems that helps to build concepts (Chapter 2)

Definition of a group (Abelian, Cyclic imp) I.N. Herstein (Sec 2.3 prob 4,8,11 are imp for starters)

Hint of the problems
4) (Sol hint) Just take three consecutive integers like k, k+1, k+2 and try to change
[ab = a^kab/a^k], from where you will actually get [ba^k=a^kb]
Similarly try to find the same relation using k+2. you will get it.

8) (Sol hint) The group is finite this is your hint and use the closure property of the group. to give you a little more hint closure always doesn't happen like ab it is also a.a and a.a.a.a like this. Try to use that.

11) (Sol hint)
 This problems tests you on the "uniqueness of the inverse", property of a group. Remember the group is of even order containing identity as an element. So without it there are odd no of elements in that group and all the elements has a unique inverse. So what now!! DO IT YOURSELF!!

SOME IMPORTANT THEOREMS IN THIS CHAPTER (I am taking Herstein as my reference Chap 2)

Uniqueness and existence of inverse and identity in a group (With proof). (Always watch the operation closely)
Very carefully understand the left and right multiplication. (One of the reasons why group theory become more absurd sometimes)

If you become comfortable then try problems like 14, 18, 19 (Chap 2)

14) (Sol hint) You are already given that your operation is product, the finite set you have abides associativity and closure property of a group.
Now use the cancellation property to establish the identity and inverse property.
First try using the fact that the group is finite then use the closure property to show the existence of identity. (Problem 8 will be helpful) From there using cancellation you can get the inverse property also.

18) (Sol Hint) Don't use the given hint, you now know something about groups and matrices too, so why not put the noncommutative criteria of matrices under multiplication in good use??

19) You now can solve it can't you?? Go on then. Good Luck.

As you are all starting as reference you can use books like:
1) Contemporary Abstract Algebra (J. Gallian Cengage)
2) Abstract Algebra (Dummit Foote)
3) Topics in Algebra (I.N. Herstein)
4)Higher Algebra by S.K.Mapa (This is a very basic book if you are not comfortable with the previous mentioned books then build your basic from this one)

NBHM 2015 some algebra questions' solution

Here is a post that contains some questions of algebra and their solution from NBHM 2015 Examination. Go through it and try them.

QUESTION [NBHM(January)(2015) 1.1]

Solve the following equation, given that it's roots are in arithmetic progression:
[x^3-9x^2+28x-30 =0]

DISCUSSION: Now every root in the above mentioned equation is in arithmetic progression.
So let the roots be a, a+d, a+2d.

Now according to the cubic equation's root rule we get that
[a+(a+d)+(a+2d) = 9.......... eq(i)]
[a(a+d)+(a+d)(a+2d)+(a+2d)a = 28..........eq(ii)]
[a(a+d)(a+2d) = 30........... eq(iii)]

These above equations are gonna come in handy.
From eq (i) we clearly get [3a+3d = 9 & that means a+d =3]
So we will use this in our own ways. putting d = 3-a in our eq(iii) we get
[3a(6-a) = 30]
Which eventually will give us a quadratic equation like
[a^2-6a+10 = 0]
Calculate the roots by the same old same old and you have 3+i and 3-i.
Now you see that if a is 3+i and a+d is 3 then clearly a+2d will be 3-i and similarly if you take a = 3-i then a+2d is 3+i and this is your required answer hope you liked it.

QUESTION (1.2) Which of the statement is true?
a. Every group of order 51 is cyclic.
b. Every group of order 151 is cyclic.
c. Every group of order 505 is cyclic.

Discussion: To do this problem I am going to use a result given as a problem in I.N. Herstein (Section 2.9 problem 10, If it's difficult I will discuss the problem latter in another discussion)

It is a simple enough problem if you know the result of the Herstein problem. It tells that if o(G) where G is the group is of the form pq where p and q are distinct primes and p>q.
Then if q does not divide (p-1) then only G is cyclic.
So 51 can be written as [3*17 = 51] and see that 3 does not divide 16 so 51 is clearly cyclic.
Next up is 151.
This is a prime number guys and to prove that as a hint use the Fermat's Theorem which states that
[a^p \equiv a mod(p)] Where p is a prime.So it's obviously a cyclic group.

Last in the plate is 505 and now I give this to you to prove use the same result(theorem what you wish to call it) discussed in this section. For your relief the answer is "no".

A Common but deadly question in Group theory

Let's discuss a Common but deadly question in Group theory.

Question: Is it possible to get an infinite group which has elements of finite order?

Discussion To pursue this discussion which is basically a very good concept for the students who are new in group theory, they must know first about the QUOTIENT GROUPS.

Particularly for this problem I am going to take two very fundamental groups
1) (Q , +) (Rationals under addition)
2) (Z , +) (Integers under addition)
Now can you prove that (2) is a normal subgroup of (1). (Hint: Use the definition of normal subgroups)
So Q/Z under the binary operation will be a quotient group.

Now a little of topic discussion, if you know what will be the elements then you sure know the identity element of Q/Z. DON'T MAKE MISTAKE THAT '0' IS THE IDENTITY OF Q/Z. THE IDENTITY IS Z ITSELF AS THE ELEMENTS OF Q/Z ARE ALL IN FORM OF SETS.

For an example let p/q our usual rational number. Then the elements of Q/Z will be of the form (p/q) + Z
So as I have told you before that the identity is Z so for fun add (p/q) + Z two times
It gives (2p/q) + Z isn't so. Then what will happen if you add it q times? It will be p + Z. Now p is an integer itself. Then p + Z is Z itself. So you get an arbitrary element in Q/Z which has finite order q.(As q is the smallest integer to do so because gcd (p,q) is 1). Hope all of new comers in group theory will understand this.

This kind of question is very important for JAM, TIFR mainly.

Jordan form of a matrix - Problem Discussion

Problem (Artin, chapter 4, 7.1) Determine the Jordan form of a matrix $$ \left[ \begin{array}{ccc} 1 & 1 & 0 \ 0 & 1 & 0 \ 0 & 1 & 1 \end{array} \right] $$

Discussion According to the Jordan form of a matrix, we first determine the characteristic polynomial of the above matrix.

To do that, first we subtract $$ \lambda $$ from each of the diagonal entries of the matrix and then the matrix looks like $$ \left[ \begin{array} {ccc} 1-\lambda & 1& 0\ 0 & 1-\lambda & 0\ 0 & 1 & 1-\lambda \end{array} \right] $$

Now the determinant of this second matrix will give us the desired eigenvalues so the determinant is $$ (1-\lambda)^3 $$

Equating the determinant value = 0 we get that the only eigenvalue of the matrix is 1 and it is a repeated eigenvalue.

So now the Jordan form of the matrix will be of the form $$ \left[ \begin{array} {ccc} 1 & 0 & 0 \ 1 & 1 & 0 \ 0 & 1 & 1 \end{array} \right] $$