 So, yeah, so last time we talked about group operations, so that, so we have a group G acting on a set X, so this is a map from G times X to X, so G pair of G and X is mapped to something called G times X, which satisfies two actions, namely that if we kind of twice act with two elements in this way, we compose the action twice, it's the same as if we multiply the elements in the group and act by that, and then the identity element should act as the identity, so this is for all X and all G and H. And we had introduced two important concepts connected with group operations. One is the stabilizer and the other one is the orbit. So, the stabilizer of an element will just be the set of all elements in the group which map it to itself. So, this is GX, which is the set of all G and G, such that G times X is equal to X, and we also have the orbit, which is just all the elements of X. So, the orbit of an element X and X will be the set of all elements of X, which we can reach by acting by elements of G on it. So, this is G of X, which is the set of all G times X is G and G. So, this is a subgroup of G and this is a subset of X. And we had seen that if we have a group operation, then we get a decomposition of X into this joint. So, then the orbits, the different orbits form a decomposition of X into this joint sets because they are actually equivalence classes of an equivalence relation. And we finished by the orbit stabilizer theorem which said that where we have a group G acting on a set X and we take a system of representatives for the orbits. So, in other words, this is a set which contains one element out of each orbit. And then we find that the number, so I want GX on X and in this case the X is finite. Then the number of elements of X is equal, this is kind of trivial, sum over all these representatives of the number of elements in the orbit of X. This is just the statement that X is the union, the disjoint union of the orbits. And we can rewrite this as the sum over all X and R of the index of G in GX because one finds that the number of elements in the orbits is precisely the index of GX in G. So, this is actually a very simple result but it's useful and therefore it's a theorem. So, now we want to look at the special case of an action, namely the action of a group by itself by conjugation, group G on itself. So, first what does it mean, conjugation? So, we have G a group and we see they say that G and H in G are conjugated if there is an element A in G such that say H is equal to AGA to the minus 1. And you can easily see that this is an equivalence relation. Okay, this is a trivial exercise. And so we can look at the equivalence classes which are called conjugacy classes of elements are called conjugacy classes of G. So, okay, we want to say something about these conjugacy classes and we want to see that this actually is an operation of G on itself. So, let's first define something more. So, for an element X in the group, the centralizer of this element consists of all elements commuting with it. So, let X in G. So, the centralizer of X is just, so I write it with Z of X to be the set of all G and G such that GX is equal to XG. So, these are all elements which commute with our given element X. Have some simple remark. First, the centralizer of any element X is a subgroup of G. That's very easy. Then, I can look at the center. In the exercise, I use the different notations C of G, but I want it now with the Z. So, the center of G is, I mean, we know it's a normal subgroup of G and this will always be is a subgroup of the centralizer of any element X. But that's obvious by definition. I mean, the center consists of all elements in the group which commute with all elements in the group. So, in particular, that's a subset of all those which commute with a given element X. And it's also obviously a subgroup. And finally, we have an element X in G is in the center if its centralizer is the whole of G. Again, this is trivial because the elements in the center are the elements in the group which commute with all elements in the group. And that's precisely what's being said here. So, this is really a remark. There's nothing to prove. Everything is directly from the definition. So, it turns out that the conjugacy classes are somehow very important invariant of the group. And if one wants to understand the group, one has to understand the conjugacy class. We will not really do that. Maybe in the next semester, you will also hear something about representation theory of groups. And you find, for instance, there's a close relation to the representations of a group and the conjugacy classes of the group. So, but now we want to describe this and we want to describe the conjugacy action of a group on itself so that the conjugacy classes will actually be the orbits. So, the definition. The conjugation or action of G on itself is just the obvious thing. It's just defined by G times H is equal to, say, G H G to the minus 1. So, the element G acts on the element H by this conjugation. So, by definition, the orbits are precisely the conjugacy classes. These equivalence classes are precisely the conjugacy classes. So, the conjugacy class of an element G is the same as the orbit because, you know, after all, this is precisely how it was defined. And we note also that the stabilizer of an element, say, X in G with respect to this conjugation action is precisely the centralize of X, you know, because by definition, the centralize of X is equal to the set of all G and G such that G X G to the minus 1 is equal to X and this is obviously equal by multiplying by both sides by G and multiplying on this side by G to the set of all G and G such that G X is equal to X G, which is... So, now I said it precisely wrong. So, this is the stabilizer is, by definition, this and this is the centralizer of X. So, this statement is also obvious. So, now we just basically, we just want to reformulate the orbit stabilizer theorem for this conjugation action and then it gets a new name. It's called the class equation. So, it's just another way to account, in some sense, the elements in the group. But it's actually just a special case of the orbit stabilizer theorem, but still it's a useful result. Therefore, it gets its own name. So, theorem of class equation. So, we take G, a final group. And so, we take R to be a set. So, a subset of G, which contains precisely one element from every controversy class. So, such that every element in G, minus the centralizer of G, every element in G, which does not lie in the center of G, is conjugated to precisely one element in R. Precisely one element in R. Okay, then the number of elements in G is equal to the number of elements in the center of G, plus the sum over all elements in R of the index of the centralizer of X in G. So, in some sense, note that, in some sense, I have... Here, it's formulated in some sense in an unnecessarily complicated way, which you will see in a moment. But this is usually the way one wants to use it. So, that's why we write it like this. So, note we can say it also as follows. So, maybe I can prove it and prove it, and then we can see. So, we take a system of representatives of the orbits of the conjugation action of G on itself. Then, maybe I'll read this. Then we can apply the orbit stabilizer theorem and that says that... So, by the orbit stabilizer theorem, we have that the number of elements in G is the sum over all X in R The index... So, R prime was our system of representatives. The index of the stabilizer, with respect to the conjugation action, of this X in G. And we remember that we had just seen that this stabilizer was actually the centralizer of X. So, we get this formula. Still with a prime. And now we have to... We see that if our element X lies in the center of G, then it commutes with all other elements. So, its centralizer is the whole of G. So, this index is equal to 1. And so, we put just... So, it means that on other words, we also have that if it lies in the center of G, then it's also clear that the orbit G of X is just equal to X, because, after all, by conjugation, it commutes with every element G in G. So, if I make G X G to the minus 1, then this is equal to X. So, it means that the center... So, therefore, we can just put minus the center of G. So, if you have an element in the center, then it is automatically equal to its own conjugacy class. So, in R prime, all the elements of the center are represented. And we just remove these. And in that case, if we have an element in the center, then this index is equal to 1, because these two are equal. And so, this formula is just equal to the number of elements in the center of G plus the sum over all X in R, the index of G C X. So, we see that this is just a reformulation of the orbit stabilizer theorem, where we also kind of have singled out the elements in the center, because there the orbit contains only one element. Okay. So, now we want to use this to show that any group which has a number of elements, the square of a prime number is a b. So, we know... So, let p be a prime number. We know, we proved it some time, that if we take... So, if the number of elements in a group is equal to p, then p is cyclic. In particular, it is what? I mean, it wouldn't make much sense otherwise. Then G is cyclic. So, now we want to see what happens for the next case when we have a square. So, now we want to show that if the number of elements in G is p squared, then G is a billion or commutative. And for this, we want to use this result. So, the first statement is... So, maybe... Do I give the definition? Yeah. So, first I want to give the definition. So, if I have a group... So, p is still a prime number. A group p, G, is called a p-group. If the number of elements in the group is a power of p, if there exists a number n bigger than 0, such that the number of elements in G is equal to p to the n. We'll see that... When we do the Seelow theorems, that they also have something to do with p-groups contained in given groups. So, now the first statement we want to make that the center of a p-group has at least... The number of elements in the center of a p-group is at least p. For position, let G be a p-group. So, the number of elements in the group is some power of p. Then, the number of elements in the center of G is at least p. Okay. So, we want to prove that. So, we assume that the number of elements in G is equal to p to the n for some n bigger than 0. And we know that the center... I mean, we know the center of G is a subgroup of G. So, we know that the number of elements in the subgroup always will divide the number of elements in the group for finite groups. So, we know that the number of elements in C of G divides the number of elements in G. But obviously, the divisors of p to the n are just the powers of p. So, 1 p p squared under p to the n minus 1. Thus, the number of elements in G is a power of p. So, it would be p to the m where m is bigger equal to 0. But it could also be that m is equal to 0 so that the center has only one element, in which case we would not have proven our proposition. So, we only have to see... Therefore, it is enough to see that the center of G is not equal to 1. That there is at least one non-chival element because we know that the number of elements in the center is a power of p. So, it must be positive power of p. So, it's at least p. So, we assume by indirect argument that we assume that the center of G is equal to 1. So, in particular, the center of G has precisely one element. So, then we apply the class equation. So, what does it say? It says that the number of elements in G, which we happen to know is p to the n, is 1... 1 is the number of elements in the center plus the sum over all x and r of the index of the centralizer of x in G, where r was a system of representatives for the non-chival conjugacy classes. So, it means that... So, x are a system of representatives for the conjugacy classes with more than one element. Or, in other words, for the conjugacy classes of elements which do not lie in the center of x. So, here, the sum is precisely over those elements for which the centralizer is not equal to the whole of G. The centralizer is a subgroup. So, maybe I can also write this. So, this is equal to the set of all x. So, representatives for... No, anyway. So, I'll just say it again. So, if I have an element here, we take precisely the representatives of conjugacy classes for which the centralizer, for which the conjugacy class is more than one element, so where the element does not lie in the center, which is equivalent to the fact that the centralizer of x is not equal to G. So, for x and r, we have that the centralizer of x is not equal to G. But the centralizer of x is a subgroup. So, but z of x, in this case, is a subgroup of G. So, we know that the number of elements in a subgroup is a power of p. And so, we take p to the n, divide by a smaller power of p. So, the number of elements here will be... This index is a positive power of p. So, we find that for all x and r, we have that this index is divisible by p. So, to say it again, we know that this is a subgroup of G, which is not equal to G. So, its number of elements is a power of p, which is smaller than this n. And so, the quotient is a positive power of p. This is divisible by p. But then, we see our contradiction. We have that p to the n is 1 plus something divisible by p. That contradiction, p to the n is equal to 1 plus a number divisible by p. Okay? And so, with this, we find that indeed it's not true that the center consists only of one element. And so, therefore, it is a positive power of p and therefore it's bigger or equal to p. And now, we want to prove this statement, which I announced here, that if I have a group such that the number of elements is p squared for prime number, then it's a bn. So, it's a proposition. Every group of order p squared, where p is a prime, is a bn. You know, for instance, that all groups of order 4 are bn. On the other hand, we know that the symmetric group in three letters, which has six elements, is non-a bn. So, anyway. So, we take such a group. So, we have to show... So, we want to show the group is commutative. So, we can reformulate this in a different way. So, we have to... So, g is a bn. If, for all elements, x in g, we have that the centralizer of x is equal to the whole of g. This is just by definition. The centralizer of x is the set of all elements in g which commute with x. And so, if for every element in x, the set of all elements which commute with x is everything, it means every element commutes with every element. So, let's see. So, we want to show that. So, obviously, if x... So, now, if we have such a... And so, let's take an element in x in g. So, if x is an element in the center of g, then obviously the centralizer of x is equal to g, essentially by definition. On the other hand, if x does not lie in the center of g, that actually will finally not be able to happen if g is a bn, but if we assume we have an element x which does not lie in the center of g, then it means that if I take the centralizer of x, this certainly contains x because obviously x commutes with itself, and it also contains the center of g because the elements in the center of g commute with every element. But x does not lie in the center of g. So, this contains p elements. So, it means it contains these elements. All the elements in the center plus also x itself. But we know the number of elements of the center of g is bigger equal to p, and so the number of elements in the centralizer of x is strictly bigger than the number of elements in the center of g because we have also the element x. So, that means the number of elements in z of x is bigger equal to p plus 1. But the centralizer of x is a subgroup of g, and so g has p squared elements, and we have a subgroup which has at least p plus 1 elements. As the number of elements of a subgroup must be divisor, there is no divisor which is bigger equal to p plus 1 of p squared except for p squared itself. So, it follows that as number of elements of c of x is a divisor of p squared, so number of elements of g equal to p squared, it follows that z of x is actually equal to g. So, we have found that for every element x in g, the center is equal to g. So, it means g is commutative. I mean, if you think this to the end, obviously we find that this case actually does not occur because the group is commutative. We have proven it's commutative, so it x is always in the center of g because all elements are, but we are making an indirect argument, so that's okay. Okay, so this was this result. So, now I want to, are there any questions about this? Comments? Time? So, now I want to look at something a bit more special. I want to look at the symmetric group again, and I want in particular to understand the conjugacy classes of the symmetric group. So, I will, we have talked about the symmetric group in n letters before, and now I want to look at it slightly more carefully. So, talk about the symmetric group S n. So, we want to describe this group more precisely. We will introduce new notations for elements in the symmetric group in terms of so-called cycles, and then we want to see that the conjugacy classes of the symmetric group can be described in terms of these cycles, the length of the cycles. So, we call that S n, this was the set of all bijections from the set 1 n to itself with the, as a group operation, the composition. So, sigma tau was equal to sigma composed with tau. And there's a, okay. We have introduced the notation. So, for sigma, so we write sigma as some kind of a matrix where we write on the top the numbers 1, 2 and so on until n, and in the bottom we write the images. So, sigma of 1, and so this is certainly a way how one can describe all permutations. Now we want to introduce a new notation in terms of so-called cycles, which many of you might also know. So, you can just say it definition. So, say let we take some elements a1 to ar, the distinct elements of our set 1 to n. So, we take, we pick out r numbers from 1 to n, different numbers. And then the cycle a1 to ar is the following permutation. Well, it's what one might think if one understands the word cycle. It means that it should be something which maps a1 to a2, a2 to a3 and so on, ar minus 1 to ar, and ar back to a1. And all other elements which are not among a1 to ar should stay fixed. So, this is the map. Permutation sigma in Sn, which is given by sigma of ai is equal to ai plus 1, if 4i equals 1 to ar minus 1. And sigma of ar should be, well maybe just say that, whatever. And sigma of ar, so a1 for i equals r. So, sigma of ar is equal to a1. So, you really have like, if you would kind of put them all on a circle, you kind of turn it once around. And obviously, and sigma of b is equal to b for all b which do not belong to the set a1 to ar. So, this is the cycle. So, we call r, the number r, is called the length of the cycle, of the cycle a1 to ar. And we say that if I have a cycle of length r, I also call it r cycle. And in particular, the two cycles are somewhat more important than they are called transpositions, two cycle transposition. So, if you have a two cycle, it just means you have two elements a1, a2, and the corresponding permutation kind of exchanges them at least all other ones fixed. And you may be called, so we call a1 to ar, the support of the cycle. And we call some cycles disjoint if their supports are disjoint. So, cycles a1 to ar, b1 to bs are called disjoint. Well, if the corresponding sets of elements in one n are disjoint, so if and only if I take the set a1 to ar, then this does not intersect set b1 to bs. We don't have any common elements. So, we can first, we have an obvious remark, cycles are after all, these are permutations. And it is, I claim it is obvious to see that disjoint cycles commute. So, that means if a1 to ar and b1 to bs are disjoint cycles, so this set of these elements does not intersect, there's no common elements here are disjoint cycles. Then a1 to ar times b1 to bs is equal to the other way round. Now, that's kind of obvious because what these things do to elements has nothing to do with each other. So, if I apply this to any element k, then if k is among b1 to bs, then I will apply this part to it and this will not do anything. And this will do the same here and the same if it is one of the a's. And if it doesn't belong to any of these, it will anyway be the identity. So, in any case, they do the same thing. Now, we want to show that every element in the symmetric group can be decomposed, I mean written as a product of disjoint cycles, essentially in a unique way. Every element, say sigma in Sn is the product sigma equal to sigma1 times sigma whatever, s of disjoint cycles. And so if we require that, so disjoint means pairwise disjoint. So, no two of them have any element in common. No, whatever. So, if we require that if I take the union of the supports of the cycles, that this is equal to the whole of the set 1n, then this decomposition is unique. If we require that the union i equals 1 to s, the support. That the union of the supports of the cycles is 1n, then this decomposition is unique up to obviously reordering. Obviously, as the disjoint cycles commute, I can also order them, but I claim that the number of cycles and precisely which cycles they are is unique. It's determined by sigma. So, this second part I will not actually prove. So, I kind of make this an exercise, although it essentially follows from what I prove. But, and so this decomposition, so in this, so under this assumption, so the, maybe I don't need that now, it's coming later. Where is it? Yeah. So, we will call, so, definition. So, if we have such a decomposition into pair voice disjoint cycles, such that the union of the supports is everything, we call this the cycle decomposition of sigma. So, in case, so, composition sigma equal to sigma 1 to sigma n into disjoint cycles with whose union is 1n. So, such that the union of the supports is called the cycle decomposition. So, by the buff, it is unique. So, we can look at an example if I have it somewhere, well, whatever, so we can look at, for instance, let's look at, so, cycle decomposition of say the element 1, 6. What is it? 2, 2, 3, 4, 5, 6, 5, 3, 4, 1. So, how do we do this? So, we can say, we start with the element 1, it is sent to 6, element 6 is sent to 1. So, we have a cycle, so we have, yes, one cycle is 1, 6. So, if you take the element 2, it's mapped to itself. So, that's also a cycle. Now, we take the element 3, it's mapped to 5, 5 is mapped to 4, and 4 is mapped to 3, so back to 3. And that would be our cycle decomposition. So, one thing that one should obviously notice is that if you have a 1 cycle, it's just the map which sends one element to itself, and all other elements to itself, it's just the identity. So, 1 cycle is just the identity. And so, this is why I had to assume that the support is everything in order to get uniqueness. Otherwise, I would have had to assume that it doesn't contain any 1 cycles. That would also be possible. And then, you can also see that this notation, the notation is not precisely unique. So, if you have this element, for instance, 3, 4, 5, so, or whatever, if you have the element 1, 2, 3, this is the map which sends 1 to 2, and 2 to 3, and 3 to 1. But I can also write it as 2, 3, 1, because it's the same thing. 1 is map to 2, 2 is map to 3, and 3 to 1. So, the decomposition into cycles is unique, but the notation is not unique. You can always kind of rotate them and get the same element. Okay, so now I want to prove this theorem. Maybe I can first prove the theorem. So, as I said, I wanted to only prove the existence, but the uniqueness is actually quite easy. So, prove. So, well, we will find that... So, how does one do this? What are these cycles? So, what you actually find is that, you know, if you take the element sigma, it has the property that it sends an element in the cycle to the next element in the cycle that permutes them. So, you find that under the operation of the group generated by sigma, the supports of the cycle will be precisely the orbits. And so that's, therefore, the way how we want to do it. So, let H equal to the cyclic subgroup generated by sigma. So, if I take the map... So, we have this map of multiplication of elements in one N by elements in H. So, we have the multiplication from H times one N to one N, which is just... We take an element in the subgroup generated by sigma and we apply it to this. So, tau i is mapped to tau of i. So, this is obviously an operation of H on one N. Basically, it's just we have the operation of the symmetric group on the set one N by just applying, you know, the element in the symmetric group to the corresponding set. And now we restrict it to the subgroup H. So, we have an operation. And so, we can look at the orbits. So, we choose some elements, B1 to B. Do you want that? S such that they... such that if I apply H to B1 and so on to H applied to Bs, these are the different orbits of this action. So, we know that... So, this B1... the set B1 to Bs is just a system of representatives for the orbits for this action. And we know that the set one N is equal to the disjoint union of the set. So, now, I mean, I've called here H is just all the powers of this one element sigma. So, let's see. We want to write this. Now, let me see. So, for each element, I equals 1 to S. We choose the minimal power of sigma such that if I apply this power to Bi, I will get Bi back. So, let Mi be equal to the minimum over all k positive integers such that if I take sigma and apply it k times, so sigma to the k of Bi is equal to Bi. This element sigma is an element in the finite group. So, there's some positive power of it, which is the identity element. And so, certainly, there is some minimal positive power which is the identity on this one element Bi. So, then, it is easy to see that if I take this element Bi, sigma of Bi and so on, sigma of Mi minus 1, so the Mi power of sigma applied to Bi, these are distinct elements. They are pairwise distinct. It's like you're doing some exercise. If you would have that, you know, for some smaller number, so if two of these would be equal, then you could, you know, multiply with the corresponding power of, so, sigma to the L of Bi then it follows that if I take, so forth, let's say 0 is more than equal, then you see that you can just multiply with sigma k to the minus k and you would get that Bi is equal to sigma of L minus k to the Bi. And so, we have found a smaller power which is equal to Bi. That's, so we can, so these will be pairwise distinct and how does sigma act on them? So, they are pairwise distinct and we also see if I take the orbit of our element Bi, I claim this is just this element. It's just these elements Bi, sigma of Bi, sigma Mi minus 1 of Bi because, in fact, what you find is that if you take sigma, so in fact, by definition, if you apply sigma to any element here, it gives you the next element and if you take this one and you apply sigma to it, you get this. So, in fact, if I look at this set here, Bi sigma of Bi until sigma Mi minus 1 of Bi, this is a cycle because we see precisely, by definition, if I apply, so let's see, maybe I can explain it slightly better. So, if I, so I get anyway, so, if I, obviously, this is by definition, if I write it like this, it is a cycle, I can write down any distinct elements, I get a cycle, but what I have is that if I take sigma I of, so if, so if X is an element in the set, Bi sigma of Bi sigma Mi minus 1 of Bi, then if I take sigma of X, this will be, as I said, the next one here, cyclically, so this is the same as sigma I of X if sigma I is the cycle because the sigma precisely acts by rotating this by one, putting everyone to the next and the last one to the first and this is precisely what sigma I does by being a cycle. So, thus we get the sigma I for equal to this thing, Bi sigma of Bi until sigma Mi minus 1 of Bi for I equals 1 to S, disjoint cycles, I mean they are disjoint because the elements here are the elements in the orbit and we know the orbits are disjoint and the union of the orbit is the whole of X, so who's such that and the union of the orbits, union of their supports is 1n and now if we take any element in 1n, I claim that it acts by the product of these cycles, so we want to claim that sigma is equal to sigma 1 times sigma S, so this is the cycle decomposition and so we take an element in 1n and so we have to show that these two permutations do the same thing to it, so obviously if you apply sigma to it you get sigma of X and if you apply this we get something else, so let's see what we do get, so let X, so we know that the union of these sets here is the whole of 1n, so it means that X lies in one of them, so then there exists and there is a unique I in, so then there is an I in the set 1 to S such that X is an element in the set, sigma of Pi, sigma Mi minus 1 of Pi, well in fact if we take all these elements the union of all these elements is X so we precisely have that X in fact X is equal to Pi to the power J for some J with J so not J, so now you should protest if I write nonsense, so sigma to the J of Pi for some J which lies between 0 and Mi minus 1, because the elements here are precisely this, Di and then what does sigma do to it so then sigma applied to X obviously is equal to sigma to the J plus 1 of Pi which will either be equal to Pi if this was Mi if this J was equal to Mi minus 1 or otherwise it's just the next one so the index is like this so that's what the sigma does and what does sigma, if I take sigma I of Pi of X so if I take sigma J of X for K for J different from I then this is equal to X because the this cycle only permutes the elements which lie in the elements which are here so if it is if our X is none of these elements it is not moved by this so the sigma I sends Pi to sigma of Pi and so on but if I have an element which is none of these here it will just stay fixed and on the other hand if I take sigma I of X this is equal to sigma J of Pi and you know this is just we apply the cycle to it and this precisely says we go one step further in the cycle so this is Pi if J is equal to Mi minus 1 and sigma J plus 1 of Pi if J is different on Mi minus 1 and so we see that and so if I take sigma 1 times sigma S of X then these things commute so X is not so the sigma K which are different from the sigma I will not do anything so this will just be equal to sigma I of X and this is which is as we have seen the same as sigma of X so it's kind of I mean I made it maybe a bit more complicated than it is somehow these different cycles have nothing to do with each other and so our element X will lie in one of these cycles and then only the corresponding cycle X on it the other ones don't do anything and therefore you get the same result so this will prove the result so we see in particular that actually we in the cycle decomposition we find that the supports of the cycles are actually the orbits of the elements under the powers of the thing we want to decompose so as a corollary we find that every element in the symmetric group can be written as a product of transpositions you see where this plus every element sigma NSN is a product of transpositions so we have for this we just have to see that every cycle is a product of transpositions so assume we have because every element NSN is a product of cycles so if every cycle is a product of transposition we are done so we write down any cycle A1 to AR is a cycle and I claim I can write it as a product of transpositions in fact I can just write down what it is so I say I take A1 A2 A1 A3 and it goes on until A1 A2 so I claim this cycle is in this way a product of transpositions so transposition after all was a two cycle well and you know you can just that's kind of clear so here you want to say that this thing maps A1 to A2 so if you take A1 A1 is mapped to A2 then A2 never occurs again so A1 is mapped to A2 now A2 is mapped to A1 here and A1 is A3 but then A3 never occurs again so A2 is mapped to A3 and so on so it's easy to see so this is easy to check so this shows that is a product of transpositions and as every element in the symmetric group is a product of cycles we are done so how much time are basically none let me see so I maybe just give a preview so now we want to use this to describe the conjugacy classes I mean as I said I will only say so we will find that if I have two cycles of the same length then they are conjugated to each other so if you have a if an element in the symmetric group has a cycle decomposition such that the lengths of the cycles are the same for both if there are two elements which have cycle decompositions with the same lengths then they are conjugated to each other and vice versa so that means the conjugacy class of the element will be determined by precisely by the lengths of the cycle in the cycle decomposition of the element and if you look here so such a cycle decomposition is a decomposition into the wet set one you know in particular if you take a support of the cycles this is a decomposition of the set one n into the disjoint subsets so the number of elements in each cycle if you take the tuple of the number of elements in each cycle this is a set of numbers which add up to n and you do this up to the ordering so you find that the conjugacy classes are in one one correspondence to the partitions of the number n so of the ways how you can write the number n as a sum of numbers and so this we will explore the next time and the other time the other thing we will explore the next time is that we also look at the sign of a permutation so if you have a permutation then I had told you here that it is a permutation as a product of transpositions and this writing this permutation as part of permutations of transpositions is not unique there are many ways but you find out that you can if you take minus one to the power the number of transpositions you need to write this thing this is something which only depends on the permutation itself and this is called the sign of the permutation we will prove this and okay that's maybe the beginning and then after that we will start preparing ourselves for proving the Silo theorem