 OK, so we can start. So we were talking about the CELO theorems. So we briefly call the statement. We had a P-CELO group. So if G is a group of order, say, P to the m times R, where P and R are relatively prime, a P-CELO subgroup is a subgroup H in G, whose order is the maximal possible power of P to the m. And so the first CELO theorem says that such subgroups always exist. So this is the first CELO theorem, which says G has a P-CELO subgroup. OK. And so I will still, because we're using it a few times today, recall the corollary, the main first corollary, which is Cauchy's theorem, which says that if a prime P divides the order of the group, then the group contains an element of order P. So then we had made a few first applications. And now I wanted to come to the following, which I had just stated but not proved the last time. So this is the following theorem. So every group, actually, I maybe have to be slightly careful about the statement the last time, so let P be a prime, which is at least three. And then every group is looking of order to P is cyclic or diheter. So it means it's either equal to the cyclic group with two P elements, or it's the diheter group Dp. And so I just, again, recall the statement the diheter group Dn is a group of order 2n with the following two condition, which is generated by elements Ap, such that A to the n is equal to 1, B to the b squared is equal to 1, and A, so Bab or Bab to the minus 1, whatever you want, is equal to A to the minus 1. And you can easily see that these conditions allow you to write down all the elements in the group. And so it's determined uniquely by this up to isomorphism. So let's try to prove this. So we will meet the use Cauchy's theorem. So this thing has two P elements. So in particular, P is a divisor. So let G be our group of order 2P. So by Cauchy's theorem, there is an element of order P. So G contains an element, which I call A, of order P. And also B is the divisor, so also 2 is the divisor of 2P, obviously, and an element B of order 2. So we look at the subgroup generated by this element of order P. So this has P elements. So as our group has 2P elements, we have that H is a normal subgroup. No, H has index 2 in G. And it is a subgroup. So by an exercise, we find that it is a normal subgroup. So we have found this wonderful normal subgroup. So therefore, if we write BAB, this is because B squared is equal to 1. This is the same as BAB to the minus 1. So this is the conjugation of A by some element in G, actually by B. So it follows that this must also lie in H. So also this thing must lie in H, because H is a normal subgroup. So this is equal to A to the I for some I, because it lies in the subgroup H, which is generated by A. So let's see what we can say about this I. So we can, for instance, do it twice. So we take B squared A B squared. This means we apply B. So obviously B squared is equal. So A is obviously equal to B squared A B squared, because B squared is equal to 1. On the other hand, we can also view this as B, BAB, B. So this is equal to B, A to the I, B. And you can easily see as B squared is equal to 0, we can write this also equal as BAB to the I. You just write this I times next to each other, always between two factors, you have B squared, which is equal to 1. And so this is the same. So this is A to the I to the I, this is A to the I squared. What? What? A to the I. Yes. So if you write this, BAB times BAB. This is BAB, BAB, and this is equal to 1. And so you just know. So in other words, we have, so it follows, that A to the I squared minus 1, so if you multiply it I squared minus 1 with itself, this is equal to 1. We just divide by A on both sides. Now we know H is a cyclic group of order P generated by A. So we know that A to the, A to some number is equal to 1, if and only if P divides that number. So A is order P plus, we have, it follows that P divides I squared minus 1, which I can write as I plus 1 times I minus 1. And as P is a prime number, if it divides the product of two integers, it has to divide one of these two integers. So it follows, I may be right now, divides like this. P divides I plus 1, or P divides I minus 1. Now I want to look at these two cases. We will find that in one case, the group is cyclic. In the other one, it's the hedger. So let's see. So first look, so if P divides I minus 1, what do we have? We have, and it means that A to the I minus 1 is equal to 1. So it means, if we multiply by A, we get A to the I is equal to A. But that just means, A to the I was BAB. So BAB, which is the same as BAB to the minus 1, is equal to A. So that means BA is equal to AB. So A and B commute. And as our group is generated, so let me see. So A and B commute. Anyway, so let's say not go too fast. So you, I mean, it follows. So yeah, let me. So if A and B commute, you can say that obviously B is not in the subgroup generated by A. So it means that A and B generate G. Because if you take, so this contains already P elements. If you also add to it the element B and all the products with it, you get two P elements. So G is commutative. Because the two elements will generate what? I didn't get the word. You have to say speak louder. I don't know what you said. So I mean, maybe it was a bit fast. But so we have H is a subgroup which has P elements. B does not lie in the subgroup. And so therefore, B does not lie in H. Because B has order 2 and this has order P. And so it doesn't lie in that. So therefore if we, so the subgroup generated by A and B must have order at least 2 times P. And so we find that it's generated by them. And these two elements commute as we have seen. So G is commutative. Because you just write any word in A and B and then you can they commute. So we have a commutative. So it means in particular that so thus we also have the group subgenerated by B is a normal subgroup. And we have that intersection of these two subgroups consists just of the unit element. Because all elements different from the unit element here have order P. And all elements different from the unit element here have order 2. And this cannot be the case for one element. And also the group is generated. And so we can, and by the argument I just gave, we have that if we take all products of elements in A and B, this gives the whole group. So we can apply this theorem we had before, which tells us that, so by previous theorem, which tells us under what conditions you get a product, you have that this is the product of these two groups. We have G is equal to A times B. But this is isomorphic to a cyclic group of order P. And this is isomorphic to a cyclic order group of order 2. And then if we want, we can apply a previous theorem, which tells us if the two such numbers are relatively prime, in this case P and 2, then the product of the cyclic groups is equal to the cyclic group of the product order. This is isomorphic to Z, module 2P. So we find in this case it's a cyclic group. And here we have used this case. We have the other case that P divides I plus 1. So well, we just look what we have. So that means what? So then that just means that A to the I plus 1 is equal to 1. So in other words, if I divide by A, A to the I is equal to A to the minus 1. So thus we find that BAB is equal to A to the minus 1. We already know that A to the P is equal to 1, that B squared is equal to 1. And we have assumed that the group has 2P elements. And again, by the same argument as before, the group is generated by these two. So we see that this is precisely the condition that characterizes the Tahitian group. So the group is the Tahitian group into P elements. So it's Dp. So it says the Tahitian group Dp has 2P elements. OK, so yeah. Well, in some sense we can, I think we can already say it from the beginning. Because we have a group of order 2P. We have an element regardless of the commutativity or anything like that. So we have that A is a subgroup of order P. And we have B an element in the group, which is not in the subgroup. So it means that kind of the other coset with respect to this is so it really implies, so we have A as P elements. And we have the group, and we have B is not in A. So we could make, we can just look at the cosets for A. So we have that G is equal to A plus B, so union B, A. Now these are the two cosets for the set. So it means that actually every element can be written as an element as a power of A plus B times a power of A. So in particular, the group is generated by A and B. This is independent of commutativity or anything. It's just by the fact that this thing has indexed 2 in the group, and we have found one element which does not lie in the group. So I didn't maybe explain this so properly, but it's a very simple fact, doesn't have anything to do with the rest. Is it clear? Yeah, that's true. So that's what I, why do we know it's a cyclic group? Yeah, yeah, no, that's what I deduced. But you say you can see it more directly. Yeah, okay, yeah, maybe you're right. But I don't quite see. Yeah, yeah. So if you just take, so you just say if you take the powers of A, B, you can get all these elements. Yeah, okay, no, that is in some sense true. If you just look at all the products and you see what you get, you find all the elements. That is true, yeah, okay. Okay, so I needn't make such a complicated story, but okay. Okay, so now we want to come to the second C-lo theorem, which is a bit more complicated to state, but so it says maybe something about how the different P-groups and how the different P-C-lo subgroups are related to each other in some way by conjugation. And so the second C-lo theorem says in particular, so one kind of consequence of it, which is easier to remember, that all C-lo subgroups are conjugated to each other. But the precise formulation is it says somewhat more, which makes it also a bit more complicated. So this is the second C-lo theorem. So what does it say? Let's say if we take any subgroup, so yeah, of G, whose order is divisible by P. So let H be a P-C-lo subgroup of G, then there is a group which is conjugate to H such that the intersection of it with K is a P-C-lo subgroup of K. So then there is a conjugate subgroup H prime equal to say G H G to the minus 1 of G such that K intersected H prime is a P-C-lo subgroup of K. So in particular it somehow says how you can get P-C-lo subgroups of subgroups by just taking a P- C-lo subgroup of the big group, conjugating it and taking the intersection. So we want to give a couple of corollaries. So first, let K G be a subgroup, which is a P-group. So that means the number of elements is power of P. Then K is contained in a P-C-lo subgroup of G. So largely, obviously. So you will always be contained in a group with the maximum P power of elements. And the second one is the one I said before. So the P-C-lo subgroups of G are all conjugate. I mean, you should notice that if I have a P-C-lo subgroup, if I take a conjugate of it, it will also be a subgroup. And as a conjugation is a bijection, it also has P to the M element. So it's also P-C-lo subgroup. So all the conjugates of P-C-lo subgroups are P-C-lo subgroups. And here we have the converse that the P-C-lo subgroup. You can also get all the P-C-lo subgroups as conjugates of just one of them. What does maximum mean? It's not contained in. So it's the biggest possible P-group. But there can be bigger subgroups, I think. But that's kind of trivial anyway. Because this P-C-lo group has the P power of elements. So there can be no other P-group which is bigger. Because the maximal number which divides the number of elements in the group is P to the M. So the maximality follows just from the number of elements. But it is true. It's also maximal. But it's a little bit like a statement that everything is contained in. It's maximal in the stronger sense that everything is contained in one of them. So that's kind of OK. So now let's first prove this corollary. So I just write also down that as a conjugation from H to G H, G to the minus 1 is a bijection, we have that a conjugate of a P-C-lo subgroup is a P-C-lo subgroup. So that I said before. So let's start with proof of 1. So if K is a P-subgroup of G, so it's a P-group, that means the number of elements is the power of P, a subgroup of key, and it's a subgroup of G. Then, obviously, just by definition, K is a P-C-lo subgroup of itself. It contains the maximum P-power of elements. That's kind of a P-group. It's a P-C-lo subgroup of itself, kind of tautologically. So we can apply the theorem to this wonderful situation. So thus by the theorem, so the second zero theorem, if H is a C-lo subgroup of G, then there is this conjugate and so on. So then there is a conjugate H prime equal to G H G to the minus 1, such that if I take the intersection K intersected H prime, then this is a P-C-lo subgroup of K, actually the unique P-C-lo subgroup of K, because obviously the only P- subgroup of K, the only P-C-lo subgroup of K itself is K itself, because test of all the elements. So I could also actually be unique. So this must therefore be equal to K, which is nothing else that K is a subgroup of H prime. And as H prime is a C-lo subgroup, we find that K is contained in it, which is, so H prime is a P-C-lo subgroup. And so we find, kind of by this trivial application, that this first one is true. Now let's look at the second. So we were right here containing K. So let's look at the second. So what is the statement? The P-C-lo subgroups are all conjugates. So let's take two P-C-lo subgroups of G. So then there exists a conjugate, say H prime equal to GH, G to the minus 1 of H, which contains K. So this is by part 1. K is a P-group. And we have said that every P-group is contained in, actually by the proof of part 1. In the proof of part 1, we have just taken, we've seen that not only it is contained in a P-C-lo subgroup, but for any given P-C-lo subgroup, it's contained in a conjugate of that, and it's proof. So H contains K, but H has P to the m elements, and K has P to the m elements, so they're equal. So H prime contains K. H prime contains K, and the number of elements in H prime is equal to the number of elements in K, which is P to the m. So they're equal. So we find that, indeed, all P-C-lo subgroups are conjugate. So this is a slightly complicated statement, but that's how it is. And maybe the corollary, which just says some special cases of it, is slightly easier to understand. I mean, keep in mind. I want to directly go to the third C-lo theorem and then give another application. So the first C-lo theorem says that there are always P-C-lo subgroups. The second one says in particular that they are all conjugate, and the third one says something about how many P-C-lo subgroups there are. Third C-lo theorem tells you not quite completely how many there are, but at least gives you some idea. It tells you something about how many P-C-lo subgroups there are. So the statement is the following. So again, I repeat the usual assumptions that we have that G was a finite group of order P to the m times r, where p does not divide r. Yes? So let S be the number of P-C-lo subgroups. And we can say something about this number. We can say two things. Namely, then S, this number of P-C-lo subgroup, divides the number r, so the rest here. And S is congruent 1 modulo p. So it's S can be. So it has these two conditions. It's a divisor of r, and it has the property that if you divide, take the rest of division by p, you get 1. So this is a rather strong restriction. So in particular, if the numbers P and r are somehow not too large, very often this will be almost impossible to fulfill, or it will force S to be 1 or something. So let's give an application of this where one actually precisely uses that. And this is another characterization of many classification of many groups. So the statement is, let G be a group of order, so G, which is equal to P times Q. Where P and Q are primes, P is bigger than Q, and P and Q are primes. So we have two prime numbers, P and Q, one is bigger than the other, and in addition, we have, so if Q does not divide the number P minus 1, then G is cyclic. So if we have this kind of, so for quite a number of possible number of elements of the group G, we can completely say what the group is. Up to isomorphism. So this is true, for instance, if the order of group is equal to 15, 33, 35, and hope 35 is correct, and so on. So there are many cases. OK, so let's try to prove this. And this actually is an application mostly of the third zero theorem. So we take NP to be the number of P zero subgroups. So we have these two primes, NP equal to the number of P zero subgroups of G. We want to find out what this is. We know that by the third zero theorem, we have that NP is congruent to one modulo P. And so yeah, OK, I wrote the number symbol, but number of, I mean, sometimes one just had this to be number, but maybe I shouldn't, at least not without warning you. So NP is congruent one mod P, and it's supposed to be a divisor of Q, but P is bigger than Q. So if it's congruent to one mod P and a divisor of Q, it must be one. Because if it would be already P plus one, it cannot be a divisor of Q, because it would be bigger than Q. And in P divides, I write it now like this, Q. So as P is bigger than Q, it follows that NP is equal to one. So there's only one P zero subgroup. But we know that all P zero subgroups are conjugate to each other. And also if you take the conjugation of a P zero subgroup by some element, so if you take an applied element to a P zero subgroup, by a conjugation you get another P, you get a P zero subgroup. So it follows. So if G is an element in G, then G H G to the minus one is it so? So let H be the unique P zero subgroup. Then if G is in G, then we have G H G to the minus one is also P zero subgroup, but there's only one. So this must be equal to H. So it means that H is a normal subgroup. So we see that if there's only one P zero subgroup, it always has to be a normal subgroup. So this is actually we have done by applying the second. Well, not really. And now let's look at the number N Q to the number of Q zero subgroups. We will want to show that this is also, this number is also one by a similar argument. So we know that N Q is equal, is congruent to one mod Q. And is it correct? Yes? And P divides the number. No, and N Q divides P. But now, so N Q has to divide P. So P has only two divisors, namely one and P. So it follows is equal to one, or N Q is equal to P. Only two possibilities. So we want to see that N Q is equal to one, so we want to see that this is not possible. So if N Q is equal to P, we know that N Q is congruent to one mod P. So then P is congruent to one mod P, mod Q, obviously. Sorry for the misprints. And so it means that Q divides P minus one, which is precisely what we have excluded here. We have excluded precisely because we didn't want this to happen. So thus it follows that N Q is equal to one. So this is a contradiction. And so N Q is equal to one. And so now the number of Q zero subgroups is also equal to one. By the argument we have made before, it means that the unique Q zero subgroup is a normal subgroup. So K of G is a normal subgroup. So note that these are H is isomorphic to Z mod PZ, because it is a group with P elements and every group with P elements is cyclic. And K is isomorphic to Z mod QZ. In particular, every element in this thing, which is different from one, has order P. Every element in this one, which is different from one, has order Q. So we find that H intersected K consists only of the one element. And so we can apply a previous theorem, although maybe we have to explain a little bit. We had this theorem which said that if we have a group and we have two normal subgroups whose intersection is one, and if you take the product of any of them, it's the whole of the group, you get that the group is the product of these two normal subgroups. But if you look at the proof, we had shown that the map from the product to the group is injective, is a homomorphism, it is injective. And we used the fact and surjective. So for the injectivity, we used only this. And for the fact that it was well-defined, only for the surjectivity, we used that every element in the group is a product of one element of one of the other. But here, we have that the number of elements in the product of the group is P times Q. So we have an injective map from H times K to our, an injective homomorphism from H times K to our group G, which must be bijection because they have the same number of elements. So this means they are isomorphic. So I should maybe have stated that when I did the previous theorem, as you also kind of hinted in some sense, but anyway. So it follows by previous theorem and its proof that G is equal to H times K. So if you have this thing, if you have two normal subgroups of a group with this property, then G is the product. So that means that is equal to Zp Z mod P times Z mod Q, which as P and Q are two different primes, is isomorphic to Z mod PQ PQ Z. So this is the last application I want to make. So what you can see is that at least if the group is not too big, this fact, these zero theorems can actually tell you quite a lot about the group so that in many cases you can actually find out precisely what the group is. But one has to be slightly, if you look at it, in all our cases we somehow always had assumed that the number of elements in the group has very few prime factors. And it kind of gets not so nice anymore. And you cannot say so much or so easily when you have, for instance, some high power of a prime. For instance, we do not know what the groups of order 8 are. It's not so difficult, but from what we have said, we don't know it, or 16, or 12, whatever. OK. So now, obviously, I will have to prove these theorems, the zero theorems. So I've made all these applications. So somehow, you might want to be sure that the theorems are actually true. So let's start with the proof of the first zero theorem. So we start with the elementary lemma, which is not really about group theory, but just about integers, about counting things. So the number of subsets of order p to the m, so with p to the m elements, of a set with, say, n equal to p to the m times r elements, where p does not divide. r is given as follows. This has nothing to, actually, this is independent of this assumption, so n choose p to the m, which is n times n minus 1 until n times p to the m plus 1 divided by p to the m times p to the m minus 1, minus 1, and so on. So the typical element will be p to the k minus, p to the m minus k, for some k and so on, until we find the n with 1. So everybody knows these binomial coefficients, and you know, I hope that they can be defined as the number of subsets, so n choose m is equal to the number of subsets of m elements of a set of n elements, or the number of subsets of m elements of the numbers from 1 to n. So this is more or less how it comes up, if you look at the binomial formula, it comes precisely up by counting these. So this is standard. So this is, I'm not going to prove, because I consider it as standard, but the new thing is that furthermore, p does not divide m, this number n. So the binomial coefficient is not divisible by p. OK, so let's prove this. As I said, the first one is elementary, and I will not end well known. So that this number of subsets is this binomial coefficient is well known. And sometimes one does it in school. So we have to show only this last thing, that p does not divide n. He does not divide n. So here we just have to look at these factors. What I'm claiming is that whenever some power of p divides a factor here, it divides with the same power the corresponding factor here. So for instance, here, obviously p to the m divides this one, but p to the m also divides this, and it's the highest power of p which divides it. So there's no power of p which occurs here. And you can see that it goes on like this. So let me write it. So the claim is whenever, or maybe this is the note, whenever p to the l divides n minus k, then also p to the l divides p to the m minus k, and vice versa. This is just because n is equal to p to the m times r. So the same power of p which divides this, if you take the highest power of p which divides n or n minus anything, namely because this is if and only if, because this is the maximum power of p that divides k. So the maximum power of p which divides this difference is the same in this range as when p k goes from 1 to p to the m minus 1. The maximum power of k which divides n minus k, and the maximum power of p which divides p to the m minus k, is precisely the maximum power of p which divides k. Because k is a number from 1 to p to the m, and so it will always, this is already the highest power of m of p which is possible, so this is this. So just say it again. So the maximum power, maximum, and I have written it before, but it's a slightly strange way, the maximum power of p which divides n minus k. So for k, k is an element from 0 to p to the m minus 1. So the maximum power which divides this is equal to the maximum power of p which divides k. And this is also equal to the maximum power of p which divides p to the m minus k. So we see that in this thing, whenever something above is divisible by some power of p, the thing which lies is directly below it is divisible by the same power of p. And so we can just cancel it out, and there's no factor of p left over. And so it's not divisible by it. So the whole thing does p does not divide n. So now let's try to start the proof of the theorem. This is done in a strange trick which is quite remarkable. I don't know how one can come up with it. So we take, so obviously somehow you can want to apply this. So we take s to be the set of all subsets m of g. Such that m has precisely p to the m elements. So m is a subset, not a subgroup of g of order with p to the m elements. We know that the number n we have introduced here is just equal to the number of elements in s. So now we want to, we let g act on it by left multiplication. So g acts on s via left multiplication. So we have per g in g and some m is mapped to g m. Just every element in g, which as you know is a set of all g m in a, obviously left multiplication is a bijection. So it respects the number of elements. So it acts on the set s. And we can, so we have this action. So g can be decomposed into disjoint orbits. So thus, so we can decompose s into disjoint orbits, into the orbits, just like the orbits, with respect to this action. So we have s is a union of disjoint unions, a union of these disjoint orbits. So we have the number of elements and s is the sum of the number of elements in the orbit. So we have number n, which was s, is equal to the sum over all orbits for this action, which the orbit I call o, the number of elements in o, orbits for left multiplication. We know that p does not divide n. That was after the lambda. p does not divide n. So in particular, so if for all orbits, the number of elements in the orbit would be divisible by p, then p would have to divide n. So there must be some orbit such that the number of elements is not divisible by p. So thus, there is an orbit o, which is the orbit of some subset m, such that the number of elements that p does not divide the number of elements in the orbit. I'm confused. OK, so proposition above. Yeah, maybe you can remind me which proposition that is. I say the number of elements in this stabilizer of m is a power of p. Is that obvious to you? It should be, even though it's not in the moment obvious to me. Question is which proposition that is. This is the wrong one. This is not so the wrong one. So maybe as I, in the moment, well, I should at least, what, how much time do I have? We make one attempt to find out what the statement is. So does somebody remember which proposition I'm using here? Can make this an exam question. So who is smarter than me? OK, so yeah, obviously I don't have my notes, so I cannot look for it. So anyway, so I think I will have to leave this for the moment and I will tell you next time why the order of this. It's a bit annoying. Wow, I don't even have the notes. No, I cannot. What? What does the proposition say? No, what does it say? Yeah? OK, so that's the point. Yeah, obviously I should have remembered that. OK, thank you. So the statement is that, so if we have the proposition, which all of you remember very well, was that if G acts on itself by left multiplication, then the order and M is some element. Is a subset of G. Then the order of the stabilizer of the subset divides the number of elements in M. OK? So that was obviously the result we used, because we are looking here at subsets M, whose order is P to the M. So we must have that the order of the stabilizer divides P to the M, so it is a power of P. Thank you. So you managed to help me out. I hope I have still my notes. So now we apply the orbit stabilizer theorem, which says that the number of elements in G, which we know to be P to the M times R, is equal to the number of elements in the stabilizer times the number of elements in the orbit. Now, we know that P does not divide the number of elements in the orbit. So as P does not divide the number of elements in GM, because this was the number N, which we have found not to be divisible by P, it follows that as this thing. So on the other hand, this must be a number of elements. So if the number of elements here is a power of P, it must be the highest possible power of P. It must be P to the M. It follows the number of elements in GM is equal to P to the R. Now, the stabilizer is always a subgroup. So it P to the M, obviously. So it follows that GM is a P0 subgroup of G. It's a bit. It's a very funny proof to construct this group in this very roundabout way. But anyway, it's not supposed to be an easy result. So one has to have a kind of strange shape. It's somehow rather sneaky. I don't know how. Anyway, so it is. Also, you seem to be talking about something completely different all the time. And in the end, this group comes out. OK, so now we will try to prove the second C0 theorem. So maybe I hope you remember the statement, the second C0. So the statement was that if you have a subgroup K, and if you also have a P0 subgroup of the big group, then you can find a conjugate of this P0 subgroup that the intersection with K is a P0 subgroup of K. So let's try to do that. So let K be a subgroup of G and H, the P0 subgroup of G. We have to show that there exists a conjugate H prime equal to GHG to the minus 1 such that H prime intersected K is a P0 subgroup of K. OK, let's try to do that. So we look at the set of left cosets for H. So let S equal to G mod H be the set of left cosets. Obviously G acts transitively on the set on G mod H by left multiplication. You can just, if you start with H, 1 times H, you can get every G times H by multiplying by some element in G. And so just G times A times H is equal to G H. That's the action. And the stabilizer for this action of the element 1 times H, which is just H, is H itself. If you multiply an element in H with something which is not in H, you get outside H. And if you multiply it with something in H, as H is a subgroup, you stay in H. So the stabilizer of this thing for this action is precisely H. And you can see from this quite easily. So see that the stabilizer, if I take the coset, A H is precisely A H A to the minus 1. Now, you can easily see that if you multiply an element in A H with A H E to the minus 1, you are again in A H. And it's easy to see that it's the only way how you can do that. I mean, anyway, it's easy to see that the stabilizers, if you move the element, the corresponding stabilizers are conjugate to each other. Anyway, so this, if you want is an exercise, but it's kind of obvious. Now we want to restrict this action to the subgroup K of G. So I mean the group that operates is no longer G, but K. So the action is now from K times G mod H to G mod H. K times A H is K. And we want to decompose G mod H into orbits for this action, into K orbits. Before, if we took the G action on this thing, it's transitive, so there's only one G orbit. But now we take a smaller group. There might be more orbits. So as H is supposed to be a P-cellular subgroup, so it follows that the number of elements in H is P to the m. So the number of elements in G mod R, G mod H is therefore the corresponding number of elements in G divided by P to the m. So this is not divisible by P. It's actually, you know that, so as we know, that the number S, number of elements in G mod H, is prime to P. So it's not divisible by P. Because after all, we had that the number of elements in G is equal to P to the m times R. The number of elements in H is equal to P to the m. And by the whatever Lagrange theorem, we have that the number of elements in G mod H is equal to R, which is not divisible by P. So therefore, it's kind of the same trick as before. So therefore, we decompose it into orbits. We have again that the number of elements in S is equal to sum of the elements in the orbits. And so therefore, there must be one orbit such that the number of elements is not divisible by P. K orbit, so K times A H, such that number of elements in K times A H is prime to P, not divisible. So we put H prime equal to A H A to the minus 1. We have seen here that this is precisely the stabilizer in G of this A H. So then H prime is the stabilizer equal to G A H. It's a stabilizer in G. And now instead, we look at the stabilizer for the K action. But a stabilizer is an element which does not move this A H. If we take the stabilizer inside the subgroup, it's precisely the set of all elements in the subgroup which are in the stabilizer here. Because it's still the elements which do not move it. So it follows that the stabilizer, so I could call this for the action of K, A H is H prime intersected K. And again, by the orbit stabilizer theorem, we have that the number of elements in the orbit times the number of elements in. So we have that the number of elements in K times A H times the number of elements in the stabilizer is equal to the number of elements in the group, which acts, which in this case is K. But we know that the number of elements in the orbit is prime to P. P does not divide the number of elements in K of A H in the orbit here. So the highest power of P which divides the number of elements in K will divide the number of elements in the stabilizer. So if we write the number of elements in K equal to, say, P to the S times L, then P to the S divides the number of elements in the stabilizer. On the other hand, we know that H prime is a P group. After all, it is a P subgroup of G. So the number of elements is a power of P. It has a P to the M element. And this stabilizer is a subgroup of H prime. So it means that the number of elements in this thing is divisible by P is a power of P. It's a divisor of P to the M. So we see that this thing is a P group whose number of elements is the maximal possible one for P group, the maximal power of P. So it is a P zero subgroup, K A H, which is equal to H prime intersected K is a P zero subgroup of K. And this was the statement of the second zero theorem. Maybe this time I went slightly over, so I will stop here.