 So now we can start. So we were talking about the symmetric group. Oops. We had introduced the cycle cycles in the symmetric group. So this was just to have an element, say, A1 to AR, where this is just a set of elements of set of numbers from 1 to n, such that sigma of ai will be equal to ai plus 1. And if i is smaller than r, and it's equal to A1 for i equal to r. So that really means the permutation permutes these elements cyclically. And obviously sigma of n is equal to n of k is equal to k if k is not an element in the cycle. We had then seen that every element in Sn has a cycle decomposition. So we had this theorem. Every sigma in Sn is the product. So sigma is equal to sigma 1 times sigma s of disjoint cycles. So disjoint cycles just means that the supports are disjoint. So the sets of elements which are actually permuted by them. And we also had seen that this product is essentially unique. And it's called the cycle decomposition of sigma. And we want to see that if we want to now use this to describe the conjugacy classes of Sn. So the claim is that what we want to show is that there's a one-one correspondence between the conjugacy classes for elements in Sn. So a conjugacy class is just an equivalence class under the operation of being conjugated. So where a is equivalent to g, a, g to the minus 1, or this case in Sn. So we want to have one-one correspondence between the conjugacy class in Sn and the partitions of the number n. So I have to remind you what the partition is. So the partition of the number n is a tuple, say n1 to say nr of positive integers. So that means that ni is bigger than 0, which are somehow ordered. So with n1 is bigger equal to n2, bigger equal and so on, bigger equal to nr, and the sum of the ni is equal to n. So for instance, the partitions of 4 are you have 4, 3, 1, 2, 2, 1, 1, and 1, 1, 1, 1. You can easily see that these are all. So in this case, there are five different partitions of 4. And now we want to associate to every conjugacy class of elements in Sn the partition and want to see that this is the one-one correspondence. So let sigma in Sn be a permutation. And let sigma equal to sigma 1 times sigma r be its cycle decomposition. So the cycle decomposition really means that these are. That sigma is a product of these disjoint cycles and that the union of the supports is the whole of the set 1n. So every integer from 1 to n occurs in one of these cycles. Some of these cycles might be one cycle, which then means they are the n-cells identity, but that's OK. So we can look at the so let l of sigma i be the length of the cycle sigma i. So that means sigma i is a l of sigma i cycle. So for instance, this thing would be an r cycle. The length of this thing is r. So obviously, this composition is well defined up to reordering them because it's the sigma i commute. So by reordering, we can assume that the length of sigma 1 is bigger equal to that of sigma 2 and so on. We can just decide in which order we want to enumerate them. And then the tuple, say, l of sigma 1, l of sigma 2, and so on, l of sigma r is called the cycle type of sigma. Now I claim that this thing is a partition of the number n. So clearly, these numbers, l of sigma i, are positive integers because they are the length of the cycle of the cycle. So it's some positive number. So the l of sigma i are positive integers. And we have ordered them in such a way that they are descending in this way. And we have also clearly that l of sigma 1 is bigger equal to l of sigma r by our definition. So the only thing we have to see is that the sum of them is n. But we have that sigma is equal to sigma 1 times sigma r is a product of disjoint cycles. So every integer from 1 to n occurs only in one of them. And the support, the union of the supports of the sigma i is the set 1n. This is what we wanted for a cycle decomposition. So that means that if we take all the numbers, so every number from 1 to n occurs precisely once among all the numbers occurring in the cycles. So that means the length of the cycle of a cycle is how many numbers occur in it. So every number from 1 to n occurs precisely in one cycle. So the sum of the length of the cycle is the total number of numbers that occur in all the cycles. So this number is n. So the sum i equals 1 to r, l of sigma i is equal to n. So this is a partition of n. OK. And now we want to see. So we can look at maybe some example to make sure we understand what we're doing here. So if we, for instance, take, again, the element sigma to be the element in S6. So 1, 2, 3, 4, 5, 6. And then what is it here? 6, 2, 5, 3, 4, 1. So then we can make the cycle decomposition. So 1 goes to 6, and 6 goes to 1, and 2 goes to 2, and 3 goes to 5, 5 goes to 4, and 4 goes to 3. So this is the cycle decomposition. So the lengths of the cycle are 3, 2, and 1. So the cycle type is this partition 3, 2, 1 of the number 6. So now we want to see what the cycle type has to do with the conjugacy class. So the claim is that two partitions are conjugated to each other if and only if they have the same cycle type. Lemma, two permutations, sigma and tau in Sn, are conjugated if and only if sigma and tau have the same cycle type. OK. So we want to see that. So we have to see what does mean to be. So we take our element sigma, the element is Bs. So this is the cycle decomposition of sigma. So now we want to take an element conjugated to it and see that it has the same cycle type. So let, maybe I call this in pi. Let tau be an element in Sn. And so we want to conjugate sigma with tau to see what happens to the cycle type. So if we form tau, sigma, tau to the minus 1, still conjugated element, what does it do? So we want to find out what this is. So we have this cycle decomposition. So we write sigma i equal tau, so ai, a1i, until a, say, m of i, i. OK. So I think it has a double index, no? So the i-th cycle goes from a1 to ami, but it depends on this. And then we put, so for all i from 1 to r and j from 1 to m of i, we put bi, b, that was maybe, bjii to be tau of agi. And we put, say, bi equal to b1i until a, b, m of i, i. So we make a new cycle by replacing the ai's, the aij's, by the bij's. And now what do we have? So if we take tau, sigma, tau, so tau, say, sigma, tau to the minus 1, and we apply this to some element, bij, what is it? First we apply tau to the minus 1 to this. The tau to the minus 1 of this is aij, no? Because tau of ai of r, here it is, yeah. So it was the other way around, but anyway, ji. So the tau of aij of aji is bji. So the tau to the minus 1 of bji is aji. So this is tau of sigma of aij aji. Now we know what sigma does to this, because sigma is the product of this disjoint cycle, sigma i. So applied to this element, it's just sigma i applied to this, which sends this one to the next one in the same cycle. So this is tau of, so there are basically two possibilities. This is tau of aji plus 1i. If i is smaller than mi, and it is the first one, so a1i, if j is equal to mi, no? That's how it is. So you just have this cycle, you turn it around, you go one further. This one to the next, and so on, the last one comes back. And this tau applied to this. But now tau of this is the corresponding thing with b. So this is bj plus 1, i, j is smaller than mi, and it's b1i if j is equal to mi. So that means this thing permutes just the bijs, the bjis in the same way as here the ajis are promoted. So it follows that tau, sigma tau to the minus 1 is equal to, say, pi 1 times pi r, where pi i is equal to precisely this thing. We're going to write it on the next line, where, as I already wrote here, pi i is equal to b1i until bm i. So we see that what happens is actually very simple. If you have a cycle and you conjugate it with a certain element, then what happens is just that you get, again, a cycle of the same length where the elements in the cycle are replaced by the images under the element by which you conjugate. So in particular, we see that these have the same lengths as before. And so clearly, we have that the cycle type pi equal to pi 1 to pi r is equal to that of sigma. So we see that if two permutations are conjugated, they have the same cycle type. And now for the converse, it's actually not very much. And basically, we trace the steps. So assume, we basically just have to see that all the steps can be reversed. So assume that sigma and pi have the same type of type. So we can write sigma equal to sigma 1 times sigma r pi equal to pi 1 times pi r, where the length of the sigma i is equal to the length of the pi i for all i. So that means I can write sigma i equal to, say, a 1 i until a, say, mi i. And pi i, I can write as b 1 i until b mi i for just some elements a 1 i. So for this, just some elements in one n, and the same here. So as these have the same number of elements, and the union of all these, I mean, the union of the supports of them is up to the set 1n. We can find the bijection from 1n to itself by just sending each aji to the corresponding bji. So let tau from 1n to 1n. So I send aji, where in this decomposition, to the corresponding bji. We know that these are disjoint subsets of 1n, and the union is the whole of 1n. So this is a bijection of the set 1n to itself. And by what we have proven here, we know precisely, so by the above, we know if we take tau sigma tau to the minus 1, then this is equal to pi. Because we have precisely seen that this is, we are precisely back in the situation, and we know that if we conjugate in this way, we get that the cycle with the aji is replaced by the corresponding cycle with the bji. So it's anyway good to know that this conjugation has this very simple effect on the cycle. So as a corollary, we now can count the conjugacy classes of Sn. So the number of conjugacy classes of Sn is equal to the number of partitions of n. And basically, we have shown it proof. So we have seen that the map, which associates to a partition, so the map from the conjugacy classes in Sn to the partitions of the number n, which associates to a conjugacy class of an element sigma, the cycle type of sigma. So we have seen that, so what have we seen? We have seen that the cycle type of a partition depends only on the conjugacy class. So that means this map is well defined as a map from conjugacy classes to partitions. And we have seen that, in fact, two elements are conjugated to each other, so in the same conjugacy class, if and only if the partition is the same. So that means this map is well defined and injective. So we have to see it's surjective. And that is very simple. We just have, for every partition, we have to write down an element in the symmetric group, which has that cycle type, but that's kind of trivial. No, we just have to. So for instance, so let's say if p equal to n1 to nR is a partition of n, then we can take sigma. So we just have to have something where the lengths of the cycle are given by this. So it means the sum of d and i is equal to n. And we just take, so say, 1 until n1 as the first cycle. Then we take n1 plus 1 until n1 plus n2. So the length of this will be n2. And you can kind of see how it goes on. In the end, we have n1 plus nR minus 1 plus 1 until the sum of all d and i, which is equal to n. So this is a cycle decomposition of some permutation. Because these are just disjoint cycles. And we see that the length of the corresponding cycles is here n1, here n2, and so on. Finally, here it's nR. So this has cycle type. So we have this projection. So this was as much as I wanted to say about this cycle type and the conjugacy classes in SN. Now finally, to finish with the symmetric group, I want to talk about the sign of a permutation. So this is something that you might be even familiar with from linear algebra. Sometimes one of the definitions of the determinant is that you take the product over all ways how you can take one element out of every row and column. And you multiply by the sign of the permutation and you sum them all up. So it already occurs there. Now I want to introduce it here. So there's one n invariant. So a number you can associate to a partition of a partition is its sign. And in fact, the way how one usually understands it is it is minus 1 to the m, where m is equal to the number of transpositions, which one uses in, so the partition is called sigma, used to write sigma. So if I write sigma as a product of transpositions, there will be many different ways how to do it. But the claim is that whenever you write it, the number of transpositions you need is either even or odd. And this does not depend on how you write it. And minus 1 to the number of transpositions you need is the sign of the partition. Now, so this is what it is for some reason. It's not now the easiest to work with for me at this moment. So I give another definition, and then I show it's equal to this one. Because obviously here the problem is that whether this definition is well-defined or not. You could expect that sometimes you need an even number, sometimes a not number. But this is not the case. So definition, I give a much more complicated definition, which is a bit crazy somehow. So let n be equal to 2 be some positive integer. And so for sigma in Sn, the sign of sigma is, so it's called, I denote by epsilon of sigma, and I write down some crazy formula. So this is the sum over all i, j, sigma of i, minus sigma of j divided by i minus j. So these are all integers. So you can certainly do this. And the sum means it's the sum is over all pairs of integers i comma j with i and j, just integers from 1 to n, and i bigger than j. So I take this sum. So if you look at it in the first moment, you look at it, you might even wonder whether this is a sign. This is supposed to be plus or minus 1. So in the first moment, you might even not be sure whether this is plus or minus 1. It looks like a rational number. But we'll see the moment is not the case. So now let's study it. So the first claim is that epsilon of sigma is equal to minus 1 to the m, where m is not precisely this, but something similar, where with m equal to the number of pairs. Let me write it equal to the number of pairs j with, so again, ij pairs in 1n with i bigger than j and sigma of i smaller than sigma of j. So this is the number of elements where kind of the order of the elements is reversed by the permutation. This what? Well, you're right. This would not make any, I think I certainly want the product there. What? What? OK, let me see. No. Yeah, if I sum, I get 0. So if the product is OK, this term will never be 0, because i is different from j. And this is the permutation, so they always are different. So this was just a misprint, which, however, I did not. OK, this is certainly the product. I think now in the proof, it also is everywhere the product. And then the second statement is that this epsilon, so the map which associates to permutation its sign, is actually a group homomorphism. So secondly, the map sigma epsilon from SN to the set consisting of the elements 1 and minus 1 together with the product. So this is a subgroup of the, whatever, the rational numbers with multiplication. So 1 times 1 is 1, minus 1 times minus 1 is 1, and 1 times minus 1 is minus 1. So this map is a group homomorphism, so where these are really the integers 1 and minus 1 with the multiplication. Well, it's obviously subjective. But I mean, that's if n is at least 2. But I mean, that's kind of trivial. We'll see that in a, but it just means if you have a, for instance, if you just have a trans, according to this story, if you just replace two things, if you place 1 and minus, if you have two elements, say, you always have the transposition 1, 2, and this transposition will have, according to the part 1, the sine minus 1. And obviously, the identity, according to the same story, has always the sine plus 1. So it's a trivially subjective. So this is a group homomorphism. And third is that the statement that I used as a definition. So if sigma equal to tau 1 times tau k is a decomposition as a product of transpositions with tau i transpositions, then it follows that epsilon of sigma is equal to minus 1 to the k. So which was what I first said would be the definition. So let's prove it. So we have to deal with this idiotic formula. So let's see. Let's just write down, as before, product i bigger than 0 sigma of i minus sigma of j. So we are afterwards supposed to divide it by this. So we want to compare it with this. So let's look at what this is. Well, clearly, I can write this as a product of all i bigger than j, where sigma i is also bigger than sigma of j of the same thing times the product where now sigma i is smaller than sigma of j, sigma i minus sigma j. That's not very difficult. But now, if you exchange the role, if you here take the product the other way around, you get each time a sine minus 1. So this is equal, if I want product i bigger than j, sigma i bigger than j. OK. So anyway, so I can replace this by sigma j minus sigma i. And I get the sine minus. So if I do this here, I get minus 1 to the number of elements here. So I should maybe say that m was the number of pairs. So I claim this is minus 1 to the m times the product over all sigma i bigger than sigma j of sigma i minus sigma j. Because what do we have? So this is the product over all. So what I mean is that if we get this, I think I'm confused. So if i is bigger, so here we have this. In the second product, we can exchange the role of i and j. If we exchange the role of i and j, then we get now that j is bigger than i is smaller than j. Sigma i becomes bigger than sigma of j. And now we have exchanged i and j, so we get the sine minus. And so we get minus 1, if we do this precisely, m times, where m is the number of elements which occur here. m is the number of pairs ij, where i is bigger than j, and sigma of i is smaller than sigma of j. So these are precisely those which occur in the second product. They are precisely m factors. Is it clear, or should I start again saying it? You make some kind of, you look slightly unhappy? Should I explain it again? Or is it clear? What? It's fine. I hope it's fine for everybody. Anyway, so we have this. So this will be this. And now, so now what do we have here? What? What? Yeah. OK, maybe I write a little bit more. I write back what I had before. Sigma of i minus sigma of j. So first, I should notice that these are precisely m factors. Because m was precisely the number of pairs ij, such that i is bigger than j, and sigma of i is smaller than sigma of j. These are precisely those which are here. And now, in this last product, we want to exchange i and j, so i with j. So what does it mean? It means now that first, we get that now i is smaller than j, and sigma of i is bigger than sigma of j. So just the last factor looks like this. And here, we have sigma of i minus sigma of j, which precisely changes sign. So we still should write sigma of i minus sigma of j, which is the same as minus m. Because now this i, so we have exchanged the role of i and j. So now what was i before is now j and vice versa. So i is smaller than j, and we have this. So we have precisely m factors. So this part is minus 1 to the m times product i smaller j sigma of i bigger sigma of j sigma of i minus sigma of j. But now, if we put this together, here we have the product, so I've only dealt with the second factor. So this is the product of the i smaller than j sigma of i bigger than j, and here we have the same product with i smaller than j, but the sigma i's are the same. So the condition is now only on the sigma i. So if I now, this was a computation in between, but if we go on here, this is equal to product over all ij such that sigma of i is bigger than sigma of j sigma of i minus sigma of j. So this product means it's the product over all pairs ij such that sigma of i is bigger than sigma of j, because there's no condition on whether i is bigger than j or smaller, because we have used both. But now, sigma is a permutation of the set 1n. So here, if we apply i by sigma of i, we have just reordered the factors. So this is the same. So reordering the same as the product over all i bigger than j i minus j, because I just replace sigma of i by i. If I do this for all i, I get the same factors only in a different order. And so this means if I take the quotient, except that obviously I made one mistake, because I had this factor of minus 1 to the m here, which I dropped. And so the definition was that I'm supposed to take this product and divide it by the product of the i minus j. And what remains is just minus 1 to the m. Maybe you can go through it again also, but OK. So this is 1, then 2. So I want to see it's a group homomorphism. So it's somehow always the same trick, so which takes some time getting used to it, so maybe you can recognize it the second time. So we want to show it's a group homomorphism. So if I take a product of two permutations, then I should get also the product of these signs. So if I take sigma and tau elements in a symmetric group, then we have the product. So we want to compare epsilon of which one we want to tau, sigma tau. We want to compare this to epsilon of sigma times epsilon of tau. In fact, we want to show they're equal. So let's try to compute this. So according to this definition we have given, this is the product over all i bigger than j sigma of tau of i minus sigma of tau of j divided by i minus j. There's nothing wrong with that. So this is just a definition. Now we can certainly multiply this as one often do in analysis. Multiply this with one in a slightly complicated way. So I hope you remember the statement. So we just multiply this with one. So this is product i bigger than j sigma of tau of i minus sigma of tau of j divided by i minus j. So it was by i minus j instead. We multiply it by tau of i minus tau of j. And we multiply by tau of i minus tau of j divided by i minus j. If I want, I can also write product i bigger than j. So we have this. Now this is just we've just multiplied. We have multiplied by this and we have divided by the same. So there's nothing happening. And now let's look at this factor. So I claim, first I write it down and then I give it the reason this is the same as a product over all tau of i bigger tau of j. So it means again, the product over all pairs ij of different elements in 1n such that tau of i is bigger than tau of j of sigma of tau of i minus sigma of tau of j divided by tau of i minus tau of j. Why is that? So if i is smaller than j and tau of i, if i is bigger than j and tau of i is bigger than tau of j, then nothing changes. If i is bigger than j and tau of i is smaller than tau of j, we can replace exchange the order of i and j. So we replace i by j. So then i is smaller than j and tau of i. So that means, so in this, to compare these two. So if i is bigger than j, tau of i and tau of i is bigger than tau of j, we just take the factor here. You'll get the same factor here. Don't change anything. Don't change the factor. And if i is bigger than j and tau of i is smaller than tau of j, we replace i by j and j by i. Then we have now that i is smaller than j. So then the new i is smaller than the new j. And the tau of i will now again be bigger than tau of j. But what is the effect of this? Replace i by j. So now the denominator changes sign because we exchange these two. And the numerator exchanges sign because we flip these two. So then both numerator and denominator change sign. So we get this whole product, we get this product from this product by exchanging i and j in all the factors where this holds. And when we do it, each time we exchange them, both the numerator and the denominator change sign. So the whole thing doesn't change sign. And so they're equal. And then we have the same argument as we had to finish the argument here. If we take the product over all pairs with tau of i bigger than tau of j, tau is a bijection from the set 1n to itself. So we can apply this bijection. So it means that if we replace tau of i by i, we have just permuted the factors. So tau is a bijection. So for each value tau of i, for each tau of i, tau of j, we get the factor. And if we replace tau of i by i for all i, we get also the fact that we get the same factors because tau is just a permutation of 1n. So there's a bijection between the factors in the same way as we did it here. So it means this is equal to the set of all i bigger than j sigma of i minus sigma of j divided by i minus j. So let me say it once more again. I just say that these factors here are just the permutation of these factors. They are permuted by the permutation tau. And then obviously these two things together say that this product, so this together says that epsilon of sigma tau is equal to epsilon of sigma times epsilon of tau. Because here, this is epsilon of tau. And this one we find this epsilon of sigma. So this is part 2. And then the third statement was this one. So if we write sigma as product of transpositions, then epsilon of sigma is minus 1 to the number of transpositions we use. And for this, we only have to see that for transposition that this number is 1. So it is minus 1. So for 3 to prove 3, it is enough to show that epsilon of tau is equal to minus 1 for any transposition. Because as this is a s by 2, this epsilon is a homomorphism. So if I take epsilon of sigma, this is epsilon of sigma 1 times epsilon of tau 1 times epsilon of tau k, each time we get minus 1. So we get minus 1 to the k. OK, so we just see it. So for instance, if tau is equal to 1, 2, so the transposition which exchanges 1 and 2, then we just see, then obviously, epsilon of tau is equal to minus 1. Because there's precisely, if you look at this thing, then all stay the same. So if i bigger than j, then also tau of i is bigger to tau of j for all i, except for 1 and 2. And 1 and 2 are exchanged. So you get 1 factor minus 1, and all the other factors are 1. So 2, 1 is the only pair of i, j with i bigger than j and tau of i smaller than tau of j. And so this gives us minus 1. And so now, on the other hand, we know by the above that if you have a transposition, so now if tau is equal to i, j is a transposition, we know it's conjugated to this one. In fact, if sigma is a permutation in 1n with sigma of 1 is equal to i and sigma of 2 is equal to j, then we know, we have seen this, that if I take tau 1, 2, tau to the minus 1, then this is equal to i, j. So if we apply, again, our epsilon to it, this is a group homomorphism. So it follows that epsilon of, say, this permutation i, j. So epsilon tau, epsilon of 1, 2, epsilon of tau to the minus 1 is equal to epsilon of i, j. So this is some number, plus or minus 1. This is minus 1. This minus 1 times epsilon of tau times epsilon of tau to the minus 1, which is epsilon of tau to the minus 1. So now we are just in the group of 1 minus 1 with multiplication. So this commutative, so we can cancel these two and we get just minus 1. So we see that for any transposition, we get minus 1, and this proves the result. So finally, one can use this to introduce the alternating group. So the permutation sigma in Sn is called even if its sign is 1. So it means it can be written by an even number of transpositions. So the set of even permutations is called the alternating group in n-dettas and denoted En. And note that this is a normal subgroup of Sn. So An is a normal subgroup because it's the kernel of a group homomorphism with An is equal to the kernel of the map epsilon from Sn to 1, set 1 minus 1 with the multiplication. The kernel are those which map tau, the neutral n. OK. So one of the things which, so this group occurs sometimes, one of the things it occurs in is also in Gallo theory, I don't know whether we will mention that later. But the fact whether you can solve the polynomial equation by taking roots and so on, it depends on whether a certain group is solvable, which has something to do with having enough normal subgroups. And it turns out that the fact that you cannot solve all polynomials of degree 5 in terms of taking roots and roots of something is connected to the fact that this group An is simple. That means it has no normal subgroups, but we will see. That's actually not so, is 4n equal 5. So it's just a statement that, for instance, A5 is a simple group. I was thinking of making it an exercise, but it's somewhat difficult. And so maybe I don't. What? I still don't understand you. So what is not commutative? Well, I will have to think about the statement. So you say that, yeah, you could say like that. Yes, if you want. OK, yeah, there are certainly, in principle, well, I don't know precisely what that means now. So because if you, for instance, take the fact is, if you take S3, for instance, as a simplest example. So the, obviously, the even permutations are actually a cyclic group. So they're also commutative. The fact is only that the S3 is actually to isomorphic to the dihedral group in three letters. So it is somehow, it comes from two cyclic groups, but altogether it's not commutative. So it's, I don't know how this answers your question, but you can see it somehow. The way how the two come together is somehow, I mean, sometimes, so in this case, you see that the even permutations are actually commutative. And the quotient is also commutative, but the whole thing isn't. OK, so I took a bit longer than I expect. So now we will want to start to prepare for, in some sense, for the statement and proof of the Silov theorems, and we do this by studying operations of groups on subsets. So if we see operations on subsets. So if we have a, so see if G acts on a set X, on a set S, it will also act on the set of all subsets of S. We want to a little bit investigate this, because we will use it when we try to prove the Silov theorems. And anyway, it's interesting to know. So let's state this. So let G act on a set X, and let U be a subset of S. Then for an element G in G, we can form GU to be the set of all G. Maybe I write time because in operation G times U. So it's not U doesn't have to be G. It's not automatically coset G times U, the set of all G times U, where U is an element in U. So we can associate, we can make an element G in G act on a subset of S, but just taking this set we obtain by applying G to every element of U. So it's easy to see that the map G comma U, maps to G times U, defines an operation of G on the set of subsets of S, of G, on the set of subsets, because you know what you have to see. If you take the identity, it maps every U to itself. If you take the identity element in G, it maps every one to itself. And if you take the product of two elements, it's the same as if you apply first one element and then the other. It's kind of trivial. We can also see we can restrict this action to subsets of S with a fixed number of elements. So I claim if we just take the set of subsets of S, which have five elements, then this operation defines an operation on that set. And this is because as we have learned, the set, the multiplication, the operation, is always by bijection. G times from S to S, which sends an element S to G times S, is a bijection. It follows from the fact that you have an operation that each element operates by bijection. And so if you have a subset with a given number of elements applying G to it, the image will have the same number of elements. And now we want to define the usual things that we have for operations. So we can look at the stabilizer, GU. So if U is a subset of S, no. The stabilizer, GU, is the usual thing. It's the set GU, which is the set of all G and G, with respect to this operation, act as the identity. But what does it mean? It means such that G times U is equal to U. That's the stabilizer. So it means that the set U is mapped to itself. As usual, this is a subgroup of U. And maybe there might be some good. Yeah, well, it couldn't be one of U. So just as a warning, the stabilizer GU of U in this thing is not the set of all elements G and G, which fix all elements of U. So it is not. So GU is not the set of all G in G, such that GU is equal to U for all U in U, which would be the stabilizer of every element. It just stabilizes the set. So each element of the set is sent to some element of the set. But rather, the set of all G in G, such that GU times U, is an element of U for all U in U. Is that a problem? No. OK. Anyway, it's maybe obvious, but I just wanted to indicate. Well, I make one trivial proposition. It's a small remark. So let G act on a set S and let U in S be a subset. Then I'm asked myself, when the stabilizer of U is the whole of G, so then G is equal to GU, if and only if, U is a union of G orbits. OK. So we can see that's quite obviously, and that's basically obvious if you think of what it means. So G, by definition, G is equal to GU, if and only if G times U is an element of U for all U in U. That's essentially the definition. And for all G in G. So the set of all GU's, G times U's, for all G in G is the orbit of U. So that means it's the same as the set of all that GU is contained in U for all U in U. And the statement that, for each element, its orbit is contained in U is the same as saying that U, so for every element, its orbit is contained in U, is the same as saying that U is the union of orbits, namely precisely, the G of U for elements U and U. OK, maybe it's not a good moment to start something new now, so maybe I will stop now.