 So, permutation cycles. So, let us consider one particular element. So, now let us look at this particular element which I have put it on the screen ok. So, I have written an element first of all looking at that element you should be able to tell me. So, this will be an element of a symmetric group of degree this is one element in the symmetric group of degree 7 and you will see that 1 goes to 5 and 5 goes to 6, 6 goes to 1 ok. This is what you see. It can be compactly written as 1 goes to 5, 5 goes to 6, 6 goes to 1. It is an equivalent description rather than writing in a long hand way. Once I put it in parenthesis 1, 5, 6 it means 1 goes to 5, 5 goes to 6, 6 goes to 1 ok. And then let us see the other one 2 goes to 7, 7 goes to 2 right sorry this is 1 and this is 2. So, 2 goes to 7, 7 goes to 2. This is a better notation which takes care of that element. And what else? 3 remains 3, 4 remains 4 that is it something else 1, 2, 3, 4, 5, 6, 7 yeah 7 elements. This is what I call it as a the same element written in cycle structure. So, I am just writing the same element in a cycle structure. But these objects did not change there is no need to mention it at all. Whatever will change is what we mention in the cycle structure. The rest which I do not mention are going to remain as unchanged that is what is the meaning. So, this is what is a cycle structure. So, let us write the cycle structure for these elements. This you can write it as 1, 2 and 3. Anyway it is equivalent to not mentioning it you can just leave it as if it is some kind of a call it as an identity element. What about this? 1, 2, this one will be 1, 1, 3, this one, this one is 2, 3 cycle structure. What about this? 1, 2, 3, this one, 1 goes to 3, 3 goes to 2, 2 goes to 1. Are you all with me? If you understand the simple example others you will follow ok. So, this is the cycle structure for an arbitrary element. I have just taken an order degree 7 symmetric group just to say that looking at this you could write the cycle structure. The advantage is that the elements in the cycle structure are also in disjoint subsets. You cannot have one element common in the, if it is common then everything will get into one higher cycle right. They are all disjoint and you can play around when you do the multiplication little better. That is why the cycle structure becomes much more handleable if you go to higher degree groups ok. So, that is why I am trying to push you to this ok. So, I will explain that you can write it as a cycle structure. So, you can write this one and cycle decomposition is useful for multiplications of 2 permutation elements. Because if you see here if I write it as a 1, 2 cycle structure, if I do it twice 1 becomes 2, 2 becomes 1 again you know you can see that 1, 2 square is identity. Similarly, 1, 3 square is identity and you can see that 1, 2, 3 you can cyclically permute it and when you do it third time you will come back to the identity. You know there are a lot of advantages of seeing in the cycle structure. So, order really does not matter whether I call 1, 5, 6 permutation followed by 2, 7 or 2, 7. So, this will be same as you could have done 2, 7 and 1, 5, 6. So, these kinds of you know commuting elements are easy ok. So, this is what I show here and also when you do 1, 2, 3 when you write it this order also does not matter the initial starting point could have been 2, 2, 3, 1 that is same as 1, 2, 3 ok. So, these also I have said in the slide here that 1, 2, 3 is same as 2, 3, 1 which is same as 3, 1, 2 really does not matter. There are these technical definition any 2 cycle we call it as a transposition and then whenever I want to write an inverse of a transposition it is the opposite 1, 2, 1 go you know it goes to this and then by this equivalence you can show that pi 2 is same as pi 2 inverse right. So, transposition or order 2 elements transposition is on a 2 cycle, cycle with 2 elements and transposition will always be an order 2 element. It is 1 if you do it again you will get back at it and inverse of the transposition is the same element is this clear ok. Inverse of 1, 2, 3 is 1, 3, 2 can you check that also you could probably do it here if you take the inverse here do the inverse here. So, what does this mean in the inverse I already taught you that you have to see 2 goes to 1, 2 goes to 1, 3 goes to 2 and then 1 goes to 3. What is this? I did the reverse and you see that this is nothing, but which is nothing, but the phi 6 l ok. So, that is why I said very you just do the inverse of the 3 cycle and you will see that phi 5 and phi 6 are inverses of each other ok. Is this clear? So, I have tried to take you on a warm up how to do multiplication group multiplication in the notation of symmetric group elements of this type. I have also said now you can also equivalently write it in the cycle structure fashion. Any 2 cycle is what is called transposition it is an order 2 element. This element is it is its own inverse and the other elements you can start finding if you want to find the inverse of any arbitrary element you do the reverse and write down the element I will give you the inverse and you can check whether the inverse lies in that set 1, 2, 3 and 1, 3, 2 lies in that set. So, 1, 2, 3 inverse is 1, 3, 2 ok. So, this much I have given you a clarity. The next one is suppose I have this was a simple exam. I could have had a cycle which had 4 or 5 elements. This is a 3 cycle, this is a 2 cycle. So, this particular element is a product of a 3 cycle and a 2 cycle. You could have 4 cycle and a 3 cycle and then the question is can what do we mean by a 3 cycle? It is actually a composition of transpositions. Any 3 cycle will be a composition of 2 transposition, any n cycle will be a composition of n transpositions ok. You can play around with transposition. Transposition is a fundamental object and you can play around with these transposition to generate all possible n cycles ok. So, this is what I am trying to say here. Every n cycle can be written as a product of transpositions. So, I have given you a k cycle 1 to k. So, the first check you have to do is do it for 1, 2, 3 whether it is satisfied. Can you check that out for this case? So, what have I written? 1 to k is 1, 2, 1, 3, 1 k. So, let us write it. So, let us do it for 1, 2, 3. So, let us write it in the long hand. So, I am just verifying the statement which I am writing that k cycle can be broken up into k transpositions and then just verifying that 1, 2, 3 is given by 1, 2 multiplying 1, 3 ok. So, let us take another element of this S 7, the symmetric group of degree 7. Now, the cycle structure of this element sigma which I have shown has a 2 cycle under 4 cycle. Others are all 1 is not changed. You can see that 3 goes to 6, 6 goes to 3. So, that is why it is a 2 cycle. And then you have a 4 going to 5, 5 going to 7, 7 goes to 2, 2 goes to 4. So, this is 2, 4, 5, 7. Is that clear? Notation. Now, the next thing is do multiplication. Pi, I anyway gave you. Let us multiply pi with sigma. What will be the cycle structure of pi with sigma? That is one question you can ask. Explicitly do it. Please do this and verify. You will see that you get a cycle structure with 6 elements. And if you do sigma pi, it is some other element. It is not the same element. Why it is not the same element? Permutations are not abelian group. If you do pi with sigma, the answer need not match with sigma with pi. You agree right. But what is seen here is something you know little bit I want you to think. So, what is happening? You do pi sigma, you find a cycle structure with 6 elements or a 6 cycle. You do sigma pi, you again find only a 6 cycle. You are not finding a cycle which is different. What does that tell you? You can have pi with a different cycle structure, sigma with a different cycle structure and you know pi sigma is not same as sigma pi. Interestingly, if I find a 6 cycle structure for pi sigma product, I find the same 6 cycle structure, element is different. This pi phi and pi 6 are 3 cycles, but they are different elements. Elements could be different, but the cycle structure remains the same. That is very something interesting and that leads to making one more comment. So, what we find is pi sigma has a 6 cycle structure. So, let me 1, 7, 4, 5, 3, 6. Sigma pi is 1, 5, 2, 4, 6, 3. These two are not same, pi sigma is not equal to sigma pi. But you can see that pi sigma pi inverse can give you a new element. So, pi sigma pi inverse can give you a new element in general and this element and this element are called what? We did this in the beginning of the lecture, conjugate elements. They are conjugate elements. What you can show is that if this has a 6 cycle, the conjugation, conjugation will give you the same 6 cycle. It will be a different element, but the cycle structure remains the same. This is what we can show. So, now, I have given you a hint on the conjugacy classes. What is the hint? In the objects, three object cases, you had a cycle structure. There was a 2 cycle structure, you had a 3 cycle structure, right. So, a 3 cycle cannot be conjugate to a 2 cycle, clear? If you have a 2 cycle structure, by conjugation it will go only to a 2 cycle structure, cannot go to a 3 cycle structure. So, that tells you that you will have these 3 which are 1, 2, 2, 3 and 1, 3 belong to a conjugacy class because all the 3 will have a 2 cycle structure. The elements are different, but they have a cycle structure which is 2 cycle structure. These 2 are 1, 2, 3 and 1, 3, 2. There are only 2 non-trivial 2 cycle, 3 cycles and the 2 will belong to 1 conjugacy class. There are 3 transposition, independent transposition, they belong to 1 conjugacy class. Identity element is like 1 cycle and that is always a class by itself, ok. So, this is one way of motivating in terms of the cycle structures. First of all, given a symmetric group of degree 4, suppose you can break it up, right. You can write situations where let us do that, maybe I will stop with that. So, you can have degree 4 identity element. So, I am calling that as degree 4 group, symmetric group, degree 4. So, identity element and then let us say what all you can have? You have to help me, otherwise I will that is the 2 cycles. You can have 1, 2, 3, then 1, 3, 2, 1, 4, 3, you know you can list out all the possibilities, ok. I am not writing the, how many will be there totally? 4 factorial, 24 variables, ok. These cycle structures are all 2 cycle structures. We will all belong to 1 conjugacy class and similarly, 3 cycle structures will belong to 1 conjugacy class and so on. This is also still not at a simpler level. We will do a little more simpler level in the next class, but at least you get the feel of that how many classes will be there is how many partitioning you can do, ok, of an integer 4. So, here it is this means what? There are 4 1 cycles, ok. This means what? There is 1, 2 cycle and 2, 1 cycle. This should add up to 4, this should add up to 4. What does this mean? This is 1 into 3 plus 1, 1 cycle, right. The 1 cycle I will never mention. What else can I have? 1 into 4, yeah. So, I am saying identity element is nothing but 1, 2, 3, 4. 4 1 cycles. So, 1 cycle plus 1 cycle plus 1 cycle plus 1 cycle which you can call it as 4 1 cycles if you want, 4 into 1, maybe that is a better notation, 4 into 1, ok. That is 4. This one is 1, 2 is a 2 cycle, then you have a 3, then you have a 4. So, 1, 2 cycle and 2, 1 cycles, ok. So, maybe this also I can write it as 2 into 1. How about this? It is 1, 3 cycle and 1, 1 cycle. What is this? It is 1, 4 cycle, ok. So, this is the way I am trying to interpret the diagrams and then you can break it up into partitioning of an integer 4 into various ways. Do you know what is the combinatorial formula for that? Maybe you will know it, but maybe next class we could discuss this, is this clear? Yeah, no, it is different, right. Good point. I am not listed everything. So, you can have a 1, 2 and a 3, 4. You can have a 1, 3 and a 2, 4. So, this one will be 2 into 2, 2 cycles, 2 of them and no other 1 cycle. This will also be 4. This is also there, initial formula. So, in fact, you can figure it out at least how many classes are there, how many elements in the class maybe there is a combinatorial, but you can figure out that how many classes are there. So, this is the breaking of an integer 4 into various partitioning. So, let me write that. So, integer n into l 1, 1 cycle, l 2, 2 cycle, l 3, 3 cycle and so on, with the restriction such that someone l 1, is that right? They should add up to how much it can go to, not beyond l n, cannot have a n plus 1 cycle on a degree n object. This is a good constraint. For the 3 objects, it is simple. What are the options? You can have 3, if you want n equal to 3, l 1 is 3. What is that? That is the identity element, 3, 1 cycles. l 2 is 1 that will be your pi 2, pi 3, pi 4. l 3 is 1 satisfies this condition. So, I have broken them into basically the division of integers satisfying this condition. Then once how many such divisions are possible will tell you how many classes are there. So, you have class 1, class 2 and no other possibility here. So, you do the same thing for n equal to 4, just to get some practice. After you do that, you have to also figure out this numbering as 3. Why is number 2 having 3 elements, number 3 having 2 elements? Can you find the combinatorial factors for that also?