 OK, so good afternoon. So everything, do we have any questions or comments? So last time, we were talking about group homomorphisms. So I don't say much about it. You know, it's the obvious thing. You have a, it's a map fee from one group to another, which is compatible with the product. So if I take the image of the product, it's the product of the images in the second group. Now, and then we have to introduce the image. So phi of g is the image of g. And we have the kernel. This is the set of all elements which are mapped to the unit element in the second group. So I could write this as d to the minus 1 of 1. And we know that this homomorphism is injective, if and only if the kernel consists only of the identity element of g. Then we had looked in particular if, say, h. And so one thing is that the image is a subgroup of g, subgroup of h. And the kernel is a normal subgroup, the source g. At any rate, we can also look at the case if we have a group g and n, a normal subgroup. We can form the quotient group, which is a group. And then we actually have the natural map, g. The natural projection gt from g to g mod n, which sends an element to the corresponding coset. And this is a group homomorphism. And finally, we had the homomorphism theorem, which says that if I have a subjective morphism between groups, g from g h with subjective homomorphism, then we can factor it through the quotient group. So let's say by the kernel, so let k be the kernel of phi. Then there exists a morphism phi bar from g mod k to h morphism, such that this map factors via the quotient map, so such that phi is equal to phi bar composed with pi. And here we have said the subjective morphism, so actually this is an isomorphism. So up to isomorphism, every subjective morphism is just dividing by a normal subgroup. Okay, so that was more or less the last thing we did. Now we come to a new topic, which is group operations. So I think I mentioned in the first lecture that originally groups were always viewed as groups of symmetry, so groups are often related to symmetry. And so that means you have some object and the group will be the set of, will somehow be all the things you can do with it without changing that thing. So instance, if you, a typical example would be you have a square and you ask yourself what are the kind of reflections and rotations in the plane that you can do so that the square is mapped into itself. So it would be this set of reflections and rotation would turn out to be a group. So groups are often, so somehow symmetry groups. So for instance, we can look at the reflections and rotations of a square. So this kind of, this means somehow that every element in the group will somehow move the points in the square. And so one can express this in a different way that we have a group operation. That the group operates on a set. So that means that we have a group, if G is a group and X is a set. So somehow an operation will be the way how you can, if you take an element X here, you can apply to it in some way an element of the group to get another element in X. In such a way that it's somehow compatible with being a group. So the only structure you have for a group is that you can multiply elements. So it should somehow be, so what you want is if you have G in G and X in X, you want to somehow associate it to that GX in G times X in X. So the operation of G on X. And so let me, as I said, compatible with the fact that G is a group. So let me write this down properly. So definition, let G be a group, X, a non-empty set. So do I actually, yeah, an operation or one says also action of G on X is a map which looks like multiplication. So it is a map. So I also noted by that dot like the product in the group but it's different. So dot from G times X to X, which sends a pair of a group element and an element in this set X to something that we call G times X. Such that somehow some reasonable properties are fulfilled. The first one is that the neutral element in the group shouldn't do anything. So I have such that first one times X is equal to X. For all X in X, and the second statement is that it should be compatible with the product in the group. So that means if we take the product of two elements in the group and apply it to X, this is the same as if we first apply one and then reply the other. So note that this is sometimes called maybe associative law, but it's not really because here, this is the product in the group. And here we have twice applied the action. But we want it to be compatible so that the product acts as twice applying one element after the other. Okay, so also we usually say, so if we have an action of G on X, we say G acts. So let's look at maybe some examples. So as usual there are stupid examples. So the most stupid one is the action of the group which doesn't do anything at all. So we have the trivial action, we have G is a group and X is a set. And we define G times X for G and G and X in X to be X for all X in X and G in G. So this means that every element in G acts by not acting at all by just sending every element to itself. So this is obviously in action because nothing happens and these conditions are somehow void. Then another case is the symmetric group, which acts in a natural way on the set one to N. Because that's the way how it is actually defined as a symmetric group. So SN acts on the set one N in the obvious way. The symmetric group consists of all bijections of the set one N to itself. And so we let it act on the set by applying the corresponding bijection to the element. So if sigma is an element in SN times some number K, which is an element in one N, is defined to be sigma of K. And now again it is clear that this is an operation. I just say because obviously the neutral element of SN is the identity map. And the identity map maps every element to itself. So that's precisely what's required there. And the product in the symmetric group is defined as the composition of maps. And so it just says if I take tau times sigma times K applied to K, this is tau applied to sigma of K, which is just there. So this is again by definition, this is clear. Now as I said the group structure sigma tau has been defined to be sigma composed with tau, so the group structure is this. And so therefore obviously this is the case. So then we have some other action. For instance, we have some actions of the group on itself. So G is a group. Then I have for instance the left translation. Left translation is a map from G times G to G, which is just actually the product in the group itself. So G, H is mapped to G times H. Now this is the product in the group. I claim that just multiplying in the group is also an operation. This is completely clear because if I multiply an element in the group with one, it's a neutral element, I map it to itself. And then as now the group operation is really just a multiplication, the second statement just becomes the associative law. So one remark that we can make is that if I have an operation of a group on a set, then every element in the group defines a bijection of the set to itself by just kind of applying the element in the group to the elements in the set. So this is followed, so I make this a definition. So let G act on a non-empty set X. So then we have the map. So for G, we have the multiplication by G, which is mg from X to X, which sends an element X and X to G times X. And the claim is that this is a bijection. So mg. Well, that is kind of obvious because you can easily see if you apply m of G to the minus 1, this will be the inverse map. So let's spell this out. I mean, first I want to see. So obviously, note we have m of 1 is equal to the identity map of X, which sends every element X to itself because that's what this action says, that the element 1 in the group sends every element to itself. And so therefore, if I leave the action for a moment, if I take any element G, and I form mg composed with, say, mg to the minus 1 of the inverse element in the group, and we apply this to any element X, all X and X, what do we have? Well, this is G to the minus 1 times X. So the operation in this way, which according to our action was this. So this is 1, and so this is equal to X. So we see therefore that this composition is the identity. And obviously, in the same way, we have that mg to the minus 1 composed with mg is the identity on X. So we see that this map mg is a bijection because we can see what its inverse map is. It's just the map associated to the inverse element. So mg is a bijection, so in other words, we have that mg can be viewed as an element of the symmetric group of X, the set of all bijections of X to itself. Now we want to use this to prove so-called Cayley's theorem, which is actually not very difficult, but it's a useful thing. So we have theorem, it's supposed to be due to Cayley's theorem. Maybe it's not in complete generality. So I say every finite group is a subgroup. Maybe I call the group G is a subgroup of some symmetric group. Yeah, you're right. It's isomorphic to a subgroup. Yeah, that's certainly true. Thank you. Every finite group G is isomorphic to a subgroup of a symmetric group SN, so a symmetric group in N letters. In fact, G is isomorphic to a subgroup of a symmetric group with as many letters as G has elements. So in some sense, maybe it's not completely obvious, but we have essentially already seen it. So we have to somehow see how every element in the group will act as a bijection of the group to itself. Well, it details by the group acting on itself by left translation. So let's do that. So after all, if we take the symmetric group on G, so just a set of bijections of G to itself, this is, as we know, isomorphic to the symmetric group in N. So maybe just like this, put to SN for N equal to the number of elements of G. So we have to show that G is isomorphic to a subgroup with bijections of G to itself. And so for this, so first I make a tiny lemma. So this is the statement is a bit more general. So if let G act on a set, on a non-empty set X, and we have seen in this definition that then this, so for then the map, we have this map M from G to the symmetric group on X, so the set of bijection of X to itself, which sends any element G to MG. We have here, if a group G acts on a non-empty set, we have for every element G and G, we can multiply by it in this form. You know, just apply G to X for all X. And this gives a bijection of X to itself, so an element in Sx. So we have this map here. And the claim is, and this is a group homomorphism. This is also kind of obvious, but I want to state. Yes? Yeah, no, I don't think it's really necessary. I just, you know, it's isomorphic to a symmetric group I think it is true that if G is, it's true for every group, obviously the finite or non-finite, that G is isomorphic to a subgroup of S of G. But here I want it to be a subset of a finite, isomorphic to a subgroup of a finite symmetric group, then obviously it has to be finite, but you're right. So the proof also will show that, so more generally. So for any G, G is isomorphic to a subgroup of S of G. But I'm not so sure how useful that practically is because, you know, if G is infinite, then S of G will be very large. So then, but you're right that this is not, I just stated it in this simple way. But there's no need for this assumption. But obviously I want here just a finite symmetric group. Okay, so we want to see that that's kind of, so you want to see that it's a homomorphism that's basically obvious. It's again basically the definition, you know, so this is the remark that, so, you know, just by definition we have, if X is an element in X, then G and H are elements in G. Then by definition, if I take MG, if I take MGH of X, this is GH times X, which is G times H times X. On the other hand, H times X is MH of X. And, you know, multiplying by this is, so this is MG of MH of X. And by definition of the, I mean, the group structure here on the symmetric group is, after all, just the composition. So that means that MGH is equal to MG times MH. Okay, so this is actually, but I still call it lemma. And now we want to use the lemma with respect to the left translation action of G on itself. So we have the left translation, G times GG, G comma H, maps to G times H is the left translation action of G on G. Then with respect to this, I have the map M. So for this, we have the group homomorphism M, which we have just been here, from G to S of G, okay, which was the map which sends an element G in G to the left modification action by G, which is a map from G to itself. So with this, so the image of this map will be, so M of G is the subgroup of S of G. And in order to see that G is isomorphic to a subgroup of S of G, we want to show that this map is injective, subgroup H plus of G to show G is isomorphic to H. We have to show, say that the kernel of M is equal to the element one in the group G. And obviously this proof will work without any assumption on whether G is finite or infinity, as you noticed. So we take an element G in the kernel of M. So that means that the left multiplication by G is the identity of G on to itself. So M G is equal to the identity of G. So in other words, for all H in G, we have G times H is equal to H. Now we are in a group, this is actually the multiplication in the group, so we can multiply by the inverse of H, and so we get G is equal to 1. So this was not very difficult. Okay, and so this proofs that the kernel consists just of the element one. And so G is isomorphic to a subgroup of a symmetric group. And if G is finite, it's isomorphic to a subgroup of a finite symmetric group. This actually, I think, does even play, I mean, also practical role. For instance, sometimes the easiest way to compute in a group is actually to represent it as a subgroup of a symmetric group and two computations in a symmetric group. And also in computer algebra programs, there are several ways how you can represent a group. I mean, if it's a non-commutative group, you can either represent it as a subgroup, as some group of matrices, or you can represent it as a subgroup of some symmetric group. And then actually the computer algebra programs use that to do computations in groups and find out whether certain things can happen or not. So it's really of practical importance. There's one drawback, obviously, with this representation in them as a subgroup of the symmetric group. It's kind of not very efficient. Because according to this, we have seen that G is isomorphic to a subgroup of Sn, where N is equal to the number of elements of G. So G is isomorphic to a subgroup. I mean, if G has N elements, it's isomorphic to a subgroup of a symmetric group which has N factorial elements. So it's not very, but still, in some cases, it's still the best you can do. OK. So now I want to get to some more concrete. So let's see. So now I want to look at the example of a group, which is kind of basically defined as a subgroup of a symmetric group which I hinted in at the beginning. So the group of symmetries of a regular N-gon, which is called the dihedral group. So first given informal definition so that you can imagine what it is. And then I just define it as a subgroup of a symmetric group. So the dihedral group is the group of symmetries. So maybe N must at least be, I don't know, this positive integer to make it meaningful. Maybe we should have at least three. But anyway, it doesn't matter. The N is the group of reflections and rotations of a regular N-gon. So we now have this regular N-gon. I don't know what I want here, so this is a bit big. Maybe I make it just four. And then there are two obvious things. I mean symmetries of the plane that you can do in order to map it to itself. So maybe the assume the vertices are number 1, 2, 3, 4. Either you can rotate it, maybe, for instance, in this case, by 90 degrees. And it maps the thing itself. Or you can take, for instance, any, you can take this dividing line. And you can kind of reflect it along this line. This also is a map that maps this thing to itself. And so I can do this more generally for any substance. So number, the edges, so we can view this as a subgroup of the symmetric group on N letters, where N is the number. For instance, if we make this rotation by 90 degrees, then this will send, so here, rotation by 90 degrees, will send, will be the element in SN, which sends 1 to 2, 2 to 3, 3 to 4, and 4 to 1. Because I just turn it around. And if I make this reflection here, so this reflection, well, this will be what? So 1 and 2 are interchanged, and 3 and 4 interchanged. So this is 1, 2, 2, 1, 3, 4, 4, 3. So for these particular two elements, I have represented them as elements in symmetric group. So one can do more generally. So now, assume we have an N gone. So it looks like this. We have here 1, 2, 3, and it goes on. And here we have N. And then we have, so we have either the rotation by 2 pi i, 2 pi divided by N. Yeah, yeah. But now I just make one step. So I have this rotation by 2 pi divided by N, which is in the same way as here. It sends 1 to 2, 2 to 3, N minus 1 to N, and N to 1. And we can also, so if N is equal to 2M is even, because there's also this difference between even and odd, then we have here 1, 2, and so on. It goes on. And here we have, what is it? So this would be, yeah, I think M, M plus 1, and then it goes on until N. And we have here, we can look at, we can reflect along this line. So reflection along this line, along the line dividing the edge 1, 2, and M, M plus 1. This is, what is it? Well, we can see that it sends always each to the other one. So do I really want it like that? You know, it's maybe not so, I prefer, well, now I've done it like this, not stupid. I'm not even sure I counted it correctly. No, I didn't, it's also counted wrong. So the way I counted it here, if I want here this, I should here have 1, and here should be N, and this is 2. So then it's this, so this is N1, so then, OK? So we have the line which divides these two edges. And this will be just replace any by the other. So 1 is sent to N, 2 is sent to N minus 1, and so on. N is sent to 1. So this is this reflection. And if the number is odd, then it looks a bit different. So if N is equal to 2M plus 1, then we can look at this. We start here with 1, 2, then 3, and then where do we arrive? So here we will have an edge which is opposite to this, which goes from M. Yeah, I don't want it like this. So again, I have here, call this 1, 2, and this will be say N. Then we do it like this, and this will be, if we count correctly, this will be M and M plus 1, I hope. Then it goes on, and here there would be N minus 1, and then we come back. And so we make this reflection. Maybe we should have the angles correct. And so what does that do? Well, in this case, we have done it in such a way that N is mapped to itself, because it's on the reflecting line, so reflect vertex N to the edge, so the middle point of the edge between M and M plus 1. And so what does it do? It will send N to N, and then I find that 1 is sent to N minus 1, 2 to N minus 2, and so on, until, again, N minus 1 is sent to N. In each case, we have described this thing to 1, obviously, yeah. Sorry, thank you anyway. So and I will call, what was I? So this rotation, I will call sigma. I have always fixed N, and this reflection along this line, we will call tau in both cases. Hope it's correct. So what do we now I claim? So now I've described. So then for me, the Tahitian group now, kind of the abstract definition will just be it's a subgroup of the symmetric group in N letters, which is generated by these two elements. So the Tahitian group, the N, is just the subgroup generated by sigma and tau, so the subgroup in the symmetric group in N letters. Now, you can see a few things quite easily. So it's kind of easy, it's kind of clear, that if I take, for instance, what's the order of sigma is obviously equal to N. I mean, from the picture, it's kind of clear that if we apply this N times, then we come back. But also, if you just look at it as permutation, each time it's moved it one more after N steps to come back to have the identity. And if you look at this reflection, you can always see that both for this tau and for this tau, also both in the even and odd case, you send one element to another one, but this other element is sent back to the original element. So if you apply this thing twice, you get the identity. So the order of tau is equal to 2. And slightly more difficult to check, but you just write it down. See if you take tau, sigma, tau, this will be sigma to the minus 1, I hope that's correct. So this is, you can just check this. And the claim is now that this, in some sense, completely describes it. So I will not actually do it. I could call it maybe exercise. So first at Dn, I can just describe all the elements here. This is the set of all sigma to the i tau k, where i, where 0 is smaller equal to i, is smaller than n, and 0 is smaller equal to k, smaller than 2. And all these elements are different. So the first thing is that if you look at this thing, it's just its elements will just be all products of powers of sigma and powers of tau in pleated. So any kind of words in sigma to the k, tau to the l, sigma to the k2, tau to the lk, and so on. You just have some kind of expression like this in powers of sigma and powers of tau. And then one can check that if I apply these two, these, in some sense, three relations, that tau squared is equal to 1, means that you can never, you can always, if you have any power of tau, which appears anywhere which is bigger equal to 2, you can eliminate it. So you have in these words only powers of tau of 1. In the same way, you have power of at most 1. And for the sigma, you have powers of at most n minus 1. And then this relation tells you that you can move the tau is equal to tau to the minus 1. So you can move tau and sigma past each other by doing this. And so you find, if you do this inductively, that all the elements can be described in this way. And then if you just look at how these represent as elements of the symmetric group, you find that all these elements are actually different. And so you have completely described the group. In particular, we have that the number of elements of dn is equal to 2n. One should notice that this will be, if n is at least 3, I hope, this is a non-commutative group. Because this relation says that tau, I mean anyway, maybe you can, I hope it's correct with 3. But you can check that these relations tell us that this is a non-commutative group. Basically it's just, this tells you after all that if you, normally this would have to, if it was commutative, then tau, sigma, tau, and you know that tau is equal to tau to the minus 1. So tau, sigma, tau to the minus 1 has to be equal to sigma. But it's equal to sigma to the minus 1. So whenever sigma is not equal to sigma minus 1, which is 2, whenever n is at least 3, this will not be commutative. So that was this one example of a standard group, which often encounters. So now we want to just give a few more standard definitions for group operations. So there's a notion of group operation being transitive, simple transitive. And then we want to have some results about counting elements when, then we want to talk about orbits. And then we want to see how one can decide maybe how many orbits there are and so on. So the definition, we have again, let group G act on a non-empty set X. So the operation is called transitive. Well, if I can move any element of X to any other element of X by means of the operation. So that means for all elements X and Y, for all elements X and Y in X, there exists a group element such that, say, Y is equal to G to apply to X. So the group will somehow send any element of X to any other element of X by just applying the action. And it's called simply transitive if there's precisely one such element for any pair. So it is simply transitive if, so for all X and Y in X, there exists a unique, so that's what this means, G and G, such that Y is equal to G times X. So one of the things that one should notice is that if the action is simply transitive, then G must have the same number of elements as X. So we have a projection from G to X. So if G acts simply transitively on X, then for any element X in X, the map of multiplying by elements of G, so the map from G to X, which sends an element G to G times X, is a projection. And this is actually just a definition. No, because it says simply transitively means that for all X and Y, there exists a unique G which sends such that G times X is equal to Y. And that means precisely that if I fix the X, the map from G to X, which sends X to GX, is a projection. Because for any element in X, there's precisely one G such that maps it there. So now I want to use two other words which are important when one is dealing with group actions, which are the orbits and the stabilizer of an element. So definition. So again, we have GX on X. Then for any element X in X, the orbit of X, which I just denote GX, what is just all the elements that X is sent to by multiplying with elements of G. So this is the set of all G times X with G in G. And any subset, so a subset Y in X, which is of this form, is called an orbit, such that Y is equal to GX for some X in X is called an orbit of the action. So if one compares to this work we had just introduced here, we see that obviously the action is transitive. If and only if for any X in X, we have that the orbit of X is equal to the whole of X. So we can also define. We can also see that the orbits are certain equivalence classes of an equivalence relation on X, which therefore means that X is a disjoint union of orbits for the action. We have an equivalence relation on X by saying that X is equivalent to Y, if and only if. There exists an element G in G, such that Y is equal to G times X. Again, it's straightforward to see that there's an equivalence relation. I think you can certainly see that. And so it is also clear that the orbits of X of the G action are precisely the equivalence classes. In fact, if I take the equivalence class of an element X in X, this is the orbit of X. So one usually calls the set of equivalence classes with respect to this equivalence relation. In other words, the set of orbits is called orbit space and denoted X divided by G. And these are things that, such things one studies quite a lot in algebraic geometry. But OK. So then another thing that we have is the stabilizer. So the stabilizer will consist of all the elements in G which do not move X. So such that G times X is equal to X. And again, our wonderful group action. So the stabilizer of an element X in X is for the moment, unless it needs to confusion, I denoted G lower X. But if you find these two similar, we can decide on a different notation, which is just a set of all elements G in G, such that G times X is equal to X. So this is some subset of G. In fact, it is clear because product of two elements in the stabilizer and the stabilizer in the inverse of an element in the stabilizer will be the stabilizer that GX is a subgroup. So just by definition. We have the following simple observation to relate the number, so the size of an orbit to the number of elements of the stabilizer, so lemma. So we have again a GX on X. And as before, so let X be an element X. There's a bijection. I can take the coset space G modulo GX, so the set of the group G, the cosets with respect to the subgroup GX of G. This set is bijective to the orbit of X. So that's actually not the value. So in particular, if, for instance, the orbit of X is finite, then in particular, for instance, if the number of elements in the orbit is finite, we have it is equal. Then we have the number of elements in the orbit is equal to the index of GX in G. Because the index is just precisely the number of cosets. So well, it's not, this is quite simple. It's clear how to define the bijection. We have a, so we define the map. So we want to define this map from this quotient to here. There's only one way how we can attempt to do it, from G mod GX to G of X. So the only thing that we have is that we can apply an element of G to our element X. And this will map us here. So if we have an element here G times GX, so this coset, this should be mapped to G times X. So again, we have defined here a map from such a set of equivalence classes to somewhere in terms of the representative of it, this G. So we have to see it's well defined. So if GX GX is equal to H, so G prime GX, what does it mean? Yeah, okay. Yeah, okay, it was a, it was a theoretical question, but in principle, maybe I should involve you more. So then, yeah, so this is equivalent to say that, I mean, either we can say that, if we can either say what you say, or we can say more close to what? Yeah, yeah, yeah, that's correct, yeah. Okay, so we can say that, say G to the minus 1 times G prime is an element in GX. So we had defined, we had actually given an equivalence, as he said, we had given an equivalence relation on G, whose equivalence classes were the cosets. So which said that these two elements are equivalent, if this satisfies this condition, so we can apply this. And so, well, so then if we take G prime times X, we would want that this is equal to G times X. This is equal to, I'm not quite sure I will get it right. So, yeah, I mean, I somehow, okay, yeah, yeah, okay, okay. So, we can maybe just, so we can say like he says, if we look at this, then this element is in the stabilizer. So it follows according to the story, so it follows that this is equal to X. So we have G to the minus 1, G prime X is equal to X. So we can multiply by G, so we get G prime X is equal to G times X. And so this, and this proves precisely that it's already fine. So, in fact, we can, so we can see also that the map is injective. Because in fact, according to the definition, this is true if and only if. This is the case, and this is if and only if. So it follows that this map is well defined and injective. We have that injectivity one can also see directly, if just turning it around. So if G prime X is equal to G times X, then we have this. And so this is an element in the stabilizer, and then we have this, okay? So therefore, we see that this map is, we get an injective map here. And by definition, the map is subjective. Because we have said here that we sent an element G times 2X to G times X. So we get here all elements of G. So we get all G times X for G and GX. So the map is subjective and injective, so it's a projection. Okay, so we get a projection. Okay, in what time? Where's my, I know it's a bit, well, I can still try. So now we want to use this to prove the so-called orbit stabilizer theorem. Which is somehow, some way how one can count how many orbits there should be. So it's not the very, it's kind of, none of this is very deep, but these are always useful things which one uses all the time. So first, we want to talk about system of representatives, so definition. So if we have that B and equivalence relation on a set X, so we call a subset R in X is called system of representatives if it contains one element in every equivalence class. Precisely one element in every equivalence class. So see, maybe I see precisely one, okay? And then we have the orbit stabilizer theorem. Which tells us how to relate the number of orbits. So theorem to the stabilizers and the number of elements in the orbits. So, what? Well, I don't know, maybe, okay. This is some simple theorem. So this is called orbit stabilizer theorem. It's again, simple observation. So let the group G act on a finite set X. Also non-empty, but whatever. So then the number of elements in X can be written either. So I forgot the system of, yeah. So let R be the system of representatives for the orbits. Then, so that means it contains one element out of each orbit. Or it's a system of representatives for the equivalence relation. But two things are equivalent if they are in the same orbit. So the number of elements of X is equal either to the sum over all equivalence or all representatives over the orbit. And this is equal to the sum again over all representatives in the system of representatives of the index of GX in G. Now this, I call it theorem, but it's actually trivial from what we know. So it's kind of useful because it often happens that we know some of these, and then by this we know the others. So now we know that these orbits are precisely the equivalence classes of an equivalence relation. So it means that X is the disjoint union of the orbits. So it means that X is equal to the disjoint union over all X, small x in R of the orbit of X, orbits are disjoint. And then obviously it follows that the first identity holds by just counting elements. So that's first identity. And the second identity is also clear because we have seen that the orbit is in bijection to the quotient of G by GX. So if the orbit is finite, as in this case because X is finite, then the number of elements in the orbit is equal to the index of G and GX. And second identity follows by the lemma, I don't know, by the previous lemma, which precisely said that number of elements in GX is always equal to the index of G, GX in G. OK, that's all. So thank you very much. Are there further comments, questions? OK.