 So, just as a warm up we have gotten into permutation of objects which I call it as a symmetric group. Since we had a weak break just as a warm up let me just go through the slide which you have already seen, but just for continuity ok. So, permutation of three objects is the symmetric group step ok. So, is the symmetric group. Group elements can be written in this format which we seen it right. One format is of this type and then we also could see that if you interchange columns the element is the same ok you do not get a new element everybody is with me. And then we also understood how to do because you can exchange the columns and it is the same element you can do the multiplication so that the final result after pi 2 becomes the initial one and then you write it ok. So, then you get the final result which is showing you the group multiplication property right. We did these things elaborately in the last lecture and I also said how to see the inverse you can go in the reverse direction looking at how the if a goes to b then you see that b goes to a will be the reverse operation on every column then you will get the inverse is that right ok. So, then a couple of things pi 6 was having this property that pi 6 cube will be identity similarly pi 5 cube will also be identity. So, essentially your symmetric group of degree 3 has elements we have seen this earlier as writing it in letters where a and b were generators of this abstract group has some meaning by comparing it with the permutation of 3 objects ok. So, such comparison of one to one on two correspondence is what we call it as an isomorphism between one group and another group ok. So, this is isomorphic to the permutation elements is that clear? An order of a permutation group of degree n will be a symmetric group involving permutation of n objects ok. So, this is a symmetric group what is the order? Order is going to be factorial n and we say that it is a symmetric group of degree n and subgroups you can start forming subgroups out of this n factorial elements you can take some subset which satisfies the 4 axioms of the group which is the subgroup those subgroups are called permutation groups and what Kale's theorem says is that any finite group like the way we wrote an abstract group with 2 generators it gives you a finite number of element finite order group you can show them that they are isomorphic to some subgroups of the symmetric group of degree n ok. For example, we saw that the permutation group sorry the abstract group where you had a and b as generators with a b equal to b a square you had 6 elements you can show it to be exactly same as your symmetric group of order degree 3 right this is what we have seen is that clear? So, you had a group identity a b a b a b squared and what else p squared and this is isomorphic to identity pi 1 pi 2 sorry this I think is pi 2 this is pi 1 and this is nothing, but your symmetric group of order 3. So, this was generated with a and b and this one is the permutation of 3 objects and these 2 are isomorphic to each other ok. So, I am just saying that a finite group of some degree of some order you can always find it to be isomorphic either to the symmetric group of degree n or subgroups of symmetric group of degree n that is all I am trying to say. Like for example, if you just take a group H 1 which is e b b squared you can show this to be isomorphic to e pi pi 5 and pi 6 which is a subgroup of this symmetric group you all agree. So, any finite group will always be isomorphic to subgroups of the symmetric group of degree n and that these subgroups are sometimes called as a permutation groups ok. So, this is trivially a permutation group the total set the subsets which satisfy group axioms are called the permutation ok. So, that is the Kelly's theorem and any permutation element this also we saw that you can equivalently be represented as a product of disjoint permutation cycles. So, if you remember these elements this one can be written as 1, 2, 3 which is disjoint cycles which you know these are one cycles right and this one was 1, 2 exchange, 3 was not touched and so on right. This one is I do not know which one was 1, 3 and which one was 2, 3 is this 2, 3. Anyway one of them is 2, 3 and the other one is this one is 1, 2, 3 this is a 3 cycle and this one is 1, 3, 2. So, this many times people do not write these one cycles they just write only the non-trivial cycles, but it you can see that any element will be a product of. So, no 2 elements here will have an overlap with the next one. So, if it is 1, 2, 1, 2 is a non-trivial cycle 3 has no overlap with it. So, that is what I mean by saying that any permutation element can be written as a product of disjoint permutation cycles. So, this is 3, 1 cycles this is 1, 2 cycle and 1, 1 cycle. So, in this class all these 3 all of them go into it all the 3 belong to the cycle structure where one of them is a 2 cycle and another one is a 1 cycle. This is an identity element which is trivial which is 3, 1 cycles and then these 2, 1, 3 cycle. So, you can break every permutation element each element you can call it as a permutation element into a cycle structure and typically the cycle structure will have this cycle structure is respected by 3 elements and this cycle structure is respected by 2 element they are 2 distinct elements which share the same cycle structure. So, that is the any permutation element can be equivalently represented as a product of disjoint permutation cycles. And this also I have already discussed that if you had a non-trivial permutation group of 7 objects you can write this disjoint cycle by looking at it looking at that element 1 goes to 5, 5 goes to 6, 6 goes to 1. So, 1, 5, 6 is 1 permutation element sorry the cycle and then 2 and 7 forms another cycle the rest of the 1 cycles typically people do not mention, but if you want you can mention it. So, cycle decomposition is generally helpful if you want to do multiplication because you know you can just concentrate on objects which are only subsets not on all the n objects when you do the multiplication that way it is useful. So, if you have no symbol in common you can write either the 3 cycle followed by a 2 cycle or a 2 cycle followed by 3 cycle basically the order does not matter it because you can do the exchange between 4 and 5 or permutation between 4 and 5th object 1 to 3 you can do a cycle permutation, but those 2 are independent. So, you can do it in whichever order you want. So, these are advantages if you write it in the cycle structure that you can use these properties. And this is that column exchange which I said is equivalent to just doing the cyclic way of writing it both are equivalent in the cycle structure the way of writing this is same as exchanging the columns. And whenever you have a 2 cycle 2 cycle is nothing, but exchange of object 1 and object 2 or any 2 objects. So, like this here it is exchange of 1 to 2, 2 to 1 it is exchange of 1 to 3, 3 to 1. So, it is always an exchange of 2 objects it is called as a transposition inverse of the transposition is the same element. So, inverse of a 3 cycle this also I have told you how to do it you have to instead of going in a cyclic manner you go in the anticlockwise manner. So, you say that the 1 goes to 3, 3 goes to 2, 2 goes to 1. So, you write the inverse of the 3 cycle by reversing the direction of exchange ok. So, 1, 2, 3 if you do 1 goes to 2, 2 goes to 3, 3 goes to 1 that is different, but instead you can do the other way round 1 goes to 3, 3 goes to 2 in the reverse direction then this will be the inverse of inverse of the select ok. So, that is why these 2 are inverses of each other any questions. So, this also you saw any n cycle whenever you have a 3 cycle I could try to write it as a 1, 2, 3 can be written as a product of transposition. So, in any n cycle you could try and write it as a product of transposition or any k cycle you can write it as a product of transposition. So, what is the meaning of this? So, 1 goes to 2, 2 goes to 3, 3 goes to 4, 4 goes to 5, 5 goes to 1. So, if you write this in the element fashion 1, 2, 3, 4, 5. So, this is the meaning of the cycle structure and you can rewrite this. So, this is a 5 cycle you can rewrite this as 1, 2, 1, 3, 1, 4, 1, 5 you miss anything. So, it involves 4 transpositions similarly this one involves 2 transpositions and so on. Now, I am giving it as a statement to you we verified this last time I want you to verify for some cases please verify it, but this will be generalizable just like mathematical induction you first verify for a couple of things and you can generalize and write a expression. So, that is all I have done, but you have to check it out ok. So, I have tried to recap some more examples sigma and pi and also we did pi sigma which gave you a 6 cycle structure and then sigma pi is another 6 cycle structure and interestingly the order of multiplying 2 different elements gave you the same cycle structure irrespective of whatever, but the elements are different ok. So, as an element this is very different from this element, but both have the same cycle structure. So, it is a 6 cycle structure. So, this is also a 6 cycle structure that is important to note here ok. So, this I have already said any k cycle can be broken up into a product of transpositions transpositions are nothing, but 2 cycles. If you have odd number of transposition or even number of transposition the corresponding element is called as a even permutation or an odd permutation. So, this one is called even permutation and these are called what about this? This is also even permutation. If you take an odd permutation cycle and multiply with an even permutation cycle, what do you expect? Now, there could be an odd permutation. You take 2 even permutation if you multiply it will always be even. So, does that tell you something? Can a subset of only odd permutations satisfy group properties or can a subset of even permutation satisfy group property? How different? Even permutation. So, even permutation elements that is what happened here, what is this? This is nothing, but this element was 1, 3, 2 and 1, 2, 3 right. It is a 3 cycle involves even permutation and you find that it satisfies the group properties. So, this is a subgroup which is generated by the 3 cycle and it is going to be only even permutation elements. Is that clear? So, now coming to abstract definition with all these warm up which we have done. So, whenever I talk about permutation elements, I look at it as if there is a set of elements you know I want repetitions. So, basically let me write it. So, 1 cycle I want 1 cycle, 1 cycle I 2, 2 cycle and so on. You get what I am saying? So, you will have objects which are let me call it as A 1, A 2, A I 1 and then I could have B 1 what is a better way of writing this. So, I am basically saying this 1 will be or let me call it C 1, then B 2, C 2. So, there will be an I 2, 2 cycles and so on. So, all these things which I have given as examples falls into this picture. This identity element means I 2 is 0, I 2 is 0, I 2 is 0. So, I k and all is 0, only I 1 will be 3, clear? Similarly, if I look at this it corresponds to I 2 equal to 1 and I 1 equal to 1, 1, 2 cycle and 1, 1 cycle. So, this will be the most general cycle structure for any symmetric group of order degree n, where I can write it in a cycle structure, but there is a constraint. What is the constraint? Here if you see if the group is of degree 3, the total number should add up to be 3, is that right? So, if you see here, you see a constraint that summation over k which runs from 1 to n with an I k multiplied with k has to be n. How many n cycles are possible in a symmetric group of degree n in this cycle structure? So, this is what is important. So, if you had an n cycle, so this is n, then I k has to be 1, ok. So, only then this is satisfied. Is this satisfied for these cases? Also you can check, right. This one is 2, 1 times 2 plus 1 which is 3. This is very important. You have to make sure that it is always a sum adds up to give you. Number of 2 cycles multiplied by 2, this subscript is multiplied here, ok. So, keep this in mind. No, this is only for a element. Any element in the permutation or in the symmetric group of degree n can be broken up into a cycle structure with this constraint. Any particular element if you write, it should satisfy this constraint. The cycle structure should satisfy that condition, good. The next question I am coming to what you are saying is nice that how do you know that there are 3 elements with the same cycle structure? That is what you are asking. That is not visible from this. This is for a element or any element there will be a cycle structure. Only constraint is that the cycle structure what you write should satisfy this condition, good. So, that is what I have shown here. Any permutation element will have ikk cycle where k runs from 1 to n such that this constraint is satisfied. All permutation elements with the above cycle structure, ok, can be shown to be conjugate elements. So, if you see here this element, this element, this element, all of them have i 1 equal to 1 and i 2 equal to 1. So, I am sure you would have checked at some point that pi 2, pi 3 and pi 4 are conjugate to each other. This is same as showing that a, a b, a b squared were conjugate to have you checked this? We have checked this many times, right. So, this is interestingly pi 2, pi 3, pi 4 have a cycle structure 1, 1 cycle and 1, 2 cycle. So, I am just trying to stress the point that conjugate elements of the symmetric group. If an element a is conjugate to another element, you can be sure that these two will be having the same cycle structure. The elements will be different. You see that the element is very different from this element, but as a cycle structure, by cycle structure I mean I just give you the set of integers i 1, i 2, i k. As a cycle structure, they have the same cycle structure, i 1 and i 2 are same, ok. So, then the next question is that how many, how can you say that there will be only three elements with this cycle structure, ok? This is all elements of my permutation symmetric group of degree 3, right. How can I say that there will be only three elements, not more, not less with this particular cycle structure, ok? So, this is i 1 and this is i 2, this is 1 and this is 1 in this notation. Is everyone with me? So, I am not going to prove this for you, but I will leave it you to as a curiosity to sit back and look at how to prove this number of permutation elements, like to get this three you can use this formula. So, also can be derived by combinatorics or of elements with cycle structure. So, now I am going to write cycle structure as i 1, i 2, i k. You understand what it means? i 1 will be the number of one cycles, i 2 will be the number of two cycles and so on. So, in this case, i 1 is 1 and i 2 is 1 and you found the answer to be 3. So, in general all these elements which you find with a particular cycle structure will be conjugate elements and I want to find that number. So, this number is going to be for a permutation group of object n or symmetric group of degree n, it is going to be this times it is going to be i k factorial k to the power of i k. So, let us try this out here. You will have in that case three factorial because this is a degree is 3 and then you have i 1, i 1 is 1, right. It is 1 times 1 to the power of 1 and then i 2 is also 1 times 2 to the power of 1. You agree? Second one is a two cycle, second one is a two cycle. So, you have this. I have explicitly written the product in this case and three cycle is of course, 0. So, i 3 you can put it 0, 0 factorial is 1 and then you can use 3 to the power of 0 which is also 1. So, it does not contribute. So, this turns out to be 3 which is nothing but these three elements. You can also check the second type which is 1, 3 cycle how many elements are there? Can you check? i 3 is 1, i 1 is 0, i 2 is 0 of elements with this cycle structure will be, you know the answer, but it will not but you I just wanted to use the formula and check it out. So, you essentially can determine the conjugacy classes. How many elements are there in the conjugacy classes for which is given by a cycle structure? By conjugacy class I mean a cycle structure and you can find how many elements have that cycle structure and they are conjugate to each other, ok. So, idea is to somehow get to handle these conjugacy classes in a better way. For three objects it is simple, but if suppose I give you 10 objects it becomes much simpler and you need to work only with one candidate of the conjugacy class. Physics does not determine, you know it does not bother about all the elements of the conjugacy class, whether I work with this element the same physics I could get from here and here. So, I can actually work with only one non trivial element of a conjugacy class. I do not need to work with all the elements which is n factorial. So, to break this this methodology will help you to at least determine how many elements are in the conjugacy class which will be like a multiplicative factor and then you work with the candidate of the conjugacy class or candidate of the conjugacy class.