 discussed what is a group and you need a set not only a set and a group operation and it should satisfy the four properties closure it should have an identity element for every element in the set there should be an inverse element and associativity property has to be satisfied. And if on top of it if you combine using the group operation the order does not matter whether you write and a group operation B as B group operation A then we call it abelian which is commutative property. And then we also went on to looking at subsets of this above set subsets of this above set right which is called as a subgroup which is denoted by this subset should also satisfy this is called subgroup if it satisfies all the satisfying all the four properties ok. And then I also said that you can draw a multiplication table involving the elements right. So, a simple thing which we took was E then we did this rotation by 90 degrees which is which let me call it as a rotation by 180 degrees can be called as a squared rotation by 270 can be called as a cube ok. So, this is sometimes denoted as c 4 which is nothing, but rotation by 2 pi by 4 ok. So, you can write a multiplication table for this and you can see that this multiplication table you just multiply E with E A with E and so on you can write those elements. And whatever entries you are writing here will all be some kind of a permutation. So, some kind of a permutation and this is what we call it as a multiplication table. So, in this particular case order of the group order of c 4 c 4 is what I call it as this group of rotation by pi by 2 and this order means the number of elements the number of elements is 4 here 1 2 3 4 which is belonging to the column or the number of elements on the row of the multiplication table is what decides for you the order of the group ok. And we also went on to talk about we did some few examples I specifically said if you take 2 cross 2 matrices with complex entries and if you put the group operation as multiplication then this commutative property may not be satisfied. If it is matrix multiplication 2 cross 2 matrices for each one of them then that multiplication A times B for all elements in the set will not satisfy B times A. Some elements can satisfy but that is not the criteria it should be satisfied for all A and B belonging to G ok. If that is satisfied then it is an abelian group. So, a set of 2 cross 2 matrices with complex entries with the group operation which is multiplication matrix multiplication it is going to not satisfy that property. So, it is a non abelian group ok. So, order of the group is what I said is denoted by mod of that group. So, this we denoted by this is the short and notation which we will keep following in future and then subgroup as I said is a subset which satisfies all the 4 axioms of the group and then we went on to generate us. In this particular case which is rotation by pi by 2 you do see that with a simple rotation by pi by 2 you can generate other elements which belongs to the set. So, we call this A is a generator of C 4 especially the C denotes the cyclic group and this 4 should be interpreted as rotation by 2 pi by 4 ok. So, C n will be. So, then here you require that the order of the generator turns out to be identity element ok. So, C n is a cyclic group generated by A such that A power n is identity ok. So, order of C n will be n is that ok and you can see that cyclic group is generated by one generator ok and it is also an abelian group C n is an abelian group ok because the definition of a generator is that all the elements in that group the set has to be powers of a generator ok. So, any element in C n will be some A to the power of m and you can write it in whichever order you want. So, suppose I write A n 1 is 1 element A n 2 is another element this does not really matter because it is A n 1 plus n 2 which is same as writing A n 2 A n 1. So, with a single generator ok there is only cyclic group has only one generator with a single generator you cannot achieve any other group other than the abelian. You can write all possible powers of that generator for that group ok. So, this is cyclic group is this clear. So, this is for cyclic group now we look at for general groups ok. So, suppose you take some abstract group G and say that A and B are generators. So, what we mean is that any arbitrary element of G should be re-writeable as A to the n 1, B to the n 2, A to the m 1, B to the m 2 you know dot or dot. So, this is what I call it as a arbitrary word, word generated using powers of each of the generators. This is what you mean by saying it is a generator. Just like I said the 26 alphabets is a generator of all words in the group you can create arbitrary words using those alphabets. Some could repeat itself the group operation there is concatenation, but it is definitely not a group because we do not find inverse of them. This is an analogy all possible words. So, you can take powers of one generator multiply with another power of another generator it does not stop here can keep continuing like this any number of times. So, if you do that that is what I call it as an arbitrary word which you generate using powers of each of the generators then those are the elements of this group. So, this is what I call it as G for example, is this. So, it is going to be order of that finite group can be infinite actually here. So, order of G may be very large ok. So, let me call it infinite it is countable, but infinite. If suppose I put an additional constraint ok. So, constraints like a squared is identity, b squared is identity and I also add a b equal to b a then many of these words may not be independent ok. It will give back the same elements there is no point suppose I do a to the power of 4, a to the power of 4 is a squared squared a squared is identity. So, it is not going to give you anything more. In fact, that will put in restrictions of giving you independent elements of the group to be if I put this condition then the group will involve elements in the set as e then a b and a b. What else can you have? Anything else which I have missed out if I write a squared it is identity you will get back b. If I put b squared you will get back identity again. So, this is the maximal set which you can get for arbitrary words subjected to this condition. So, this is given if you give this condition then it is not going to be infinite the order of this group is going to be 4. So, let us look at some more examples and then it will be clear. So, there was a couple of questions on generators. So, I thought let me stress on this. So, the claim group it is called in the group theory literature as claimed 4 group which is denoted by v. So, it is made of e a b a b. So, this is the multiplication table a and b are generators satisfying a squared equal to b squared equal to e and a b equal to b a. So, this is what is the multiplication table. The next thing I want to give is this is one group I have tried to give with some condition for two generators. I can keep varying this condition and start generating different groups ok. So, the next one which I want you to actually try it out which is really useful is. So, group g generated again by a by two generators such that a squared equal to b cube equal to identity. So, this is different from the claim 4 group which you have been seen ok. Also want to have a b equal to squared a. Now, you start cooking up writing all possible words and see how many will be independent elements. I leave it you to try it out ok. So, the set g will have e a definitely a squared is identities it does not mean anything. You can have b b squared and then you can play around closure properties a with b a with b squared what else b squared a this is independent. I have given you a condition that b squared a is a b. So, it is already there in the set. So, this is not there anything else a squared b what do you want? It will truncate if you give conditions and you can determine the order of the group. What is the order of the group? Let us write is b a independent. So, this implies as a b squared using these conditions you can show that b a is same as a b square. So, just check it out. So, this is also not there. So, order of g is order of g is 1, 2, 3, 4, 5, 6 is 6. Is this abelian? Trivially this condition which I have written clearly shows that it cannot be abelian. So, g is this non abelian. So, non abelian group what else? What are its subgroups? Can we find out its subgroups? A forms a group subgroup. So, let us write. Let me call h 1 as e and a h 2 is e b b square. Is that right? These two I can trivially write. Order of an element is the order of the subgroup, order of a is 2 by this definition. A squared equal to identity means order of the generator a is 2, b cube equal to identity means order of the generator b is 3. So, using that I can write cyclic groups. One cyclic group which is its abelian generated by a, another cyclic group generated by b. Anything else? A b. E a b, good. Is that a group? So, if you take a squared b squared, a squared is identity then you will get a no a b a b right. So, let us write it and see a b a b. So, b a is nothing but a b squared, a is a b squared, b is also a order 2 element. So, this is also a subgroup and anything else that is it. What about a b squared? Is that ok? Order 2 is easy to check, order 2 groups right. So, a b squared, a b squared, b squared a is a b this one is a b. So, this is also identity. So, you have another order 2 subgroup which is e and a b. So, it is interesting to play around you know I am not saying that given an instant value with so many elements or so many generators with certain properties you can actually sit down and see whether you can generate the groups, subgroups. So, clearly you do see that there is h 1, h 2, h 3, h 4 the common intersection is only identity element, but then there are all these other elements which are there ok. So, just to have completeness I just put the multiplication table which you can also rewrite instead of this 4 cross 4 it will become a 6 cross 6 ok. So, you have to write using these properties which is what I have put it in here on the 6 cross 6 matrix. So, which is what I have put it here on the 6 cross 6 matrix. So, e a b, b squared, a b, a b squared on the column and again on the row and then I have multiplied and written down explicitly. Even here you can check whether e and a b and similarly e and a b what happens you can see that it is between e, a b, a b, e. So, you can see that there is a subgroup with e and a b fine. You can figure it out by looking at the subgroups by looking at the multiplication table also you can see which one is the subgroups ok. So, just to summarize keeping this example. So, let me call this example as some kind of a non-Abelian group let me call this with some notation as O 3 we will come to it what it is. This is this will be known as symmetric group of degree 3. We will explain why it is called degree 3, why this 3 is coming on all right. Now, you take it as an abstract group G generated using through two generators one is order 2 generator another one is order 3 generators satisfying this additional constraint and you have generated a 6 element group ok and that group later on we will see we will call it to be symmetric group of degree 3 we will see why we call it and it is clearly non-Abelian because of this property is this clear.