 Good afternoon. So we can have the second lecture of algebra. So the last time I gave the definition of a group and of a subgroup and some easy examples. And now I will want to talk about co-sets for a group. It's first thing, so for that I will also co-sets of a subgroup inside a group. So for this, I first want to briefly review, although I expect you know it, equivalence relations. So that we have everything in place. So review of equivalence relations. So equivalence relations occur very often in mathematics. Whenever we want to maybe somehow identify two things, because in some sense they can be identified, then this is due to an equivalence relation. So let us see what it is. First, let's see what the relation is. If A is a set, the relation on A is just a subset of the product of A with itself. It's just a subset contained in A times A. And we would maybe denote the right, say, A is related to B for AB is in R. So this is a kind of such a general concept that it's almost without any use. So now we look at special relations, which are equivalence relations, which allow us in a suitable sense to identify two things if they are related by this relation. So such a relation, so this relation is called an equivalence relation. If it satisfies two properties, the three properties, the following three actions fold. So first, every element is related to itself. So A equivalent to A for all A and A. So this is called useful for reflexivity. The second one is if A is related to B, then B is related to A. So that's a symmetry AB and A. And the third one is the, in some sense, the most important one because it's the one that is most difficult to check and practice, so which is the transitivity. If A is related to B and B is related to C, then it follows that A is related to C. And we say that A and B are equivalent if we have this relation to C, A, and B are equivalent. So as I said, in practice, when you want to show that something is an equivalent relation, you usually find that these first two things are obvious. And the only one where one might have to put some work is the third one. So and then very often in mathematics, it happens that one is looking at certain things, certain mathematical objects. And one finds that there is an equivalent relation on them. And then one will want to identify these objects if they are equivalent, which means that one replaces the set by the set of equivalence classes. You see what the set of elements equivalent to it. And so we want to introduce this. So first, let's look at this example. So for instance, again, we take an element k and z, which is non-zero. Then we say we define A is equivalent to B if k divides A minus B. So this means divides. And then you can easily check this is an equivalent relation. And so now what I want to show. So if you have an equivalent relation into a set, on a set, we find that the set is decomposed into disjoint subsets, the equivalence classes. So each set will be of the form or the elements which are equivalent to one given element. So an equivalent relation determines a decomposition. So say on a set A determines a decomposition of A into disjoint subsets. And these subsets are called the equivalence classes. And so let's see how that goes. So I will actually want to, so that means OK. So I'll actually define it. So let A set an equivalence relation. So the equivalence class of an element A in A is, so this is I noted by A, which is just the set of all B in A such that A, so say B is equivalent to A. So this is the equivalence class. And then the remark is that if we have an equivalent relation, we get a decomposition of A into equivalence classes. So let A be non-empty set, non-empty. And we take the equivalent relation on it. So the distinct equivalence classes with respect to the equivalent relation form a decomposition of A into disjoint sets. So this is obviously quite trivial. So the first thing is that the union of all the equivalence classes, so what is being stated here is that the union of all equivalence classes should be the whole of A. And two such equivalence classes are either equal or disjoint. So we know, after all, by our action, the flexibility that A is equivalent to A. So thus, for every A in A, we have that the class of A contains A as an element. So therefore, A is the union of the A with A in A. And the second statement is that they are disjoint. It's also very simple, so either disjoint or equal. So let's take two elements. So we want to see that they are either disjoint or equal. So let's assume they are not disjoint, so that they have something in common, then we have to show they are equal. Assume the equivalence class of A intersected the equivalence class of B is non-empty. So that is, we can take an element in this intersection. We want to use this to show that they are indeed equal. A is equal to B. I mean, that the equivalence class of A is equal to the equivalence class of B. So we have that, so x lies in A. So it follows that, say, x is equivalent to A, or if you want, A is equivalent to x. And x lies also in B. So we also have x is equivalent to B. So by symmetry and transitivity, we have A is equivalent to x, and x is equivalent to B. So it follows that by transitivity, that A is equivalent to B. So now, assume y is an element in A. So that means y is equivalent to A. Then by transitivity, using this, we also have that y is equivalent to B. So in other words, y is an element of the class of B. So that means every element of the equivalence class of A is an element of the equivalent class of B. So we have shown that A is contained in B. But obviously, the situation is completely symmetric. We could have exchanged the role of A and B. We would have gotten the corresponding result that B is contained in A. So therefore, A, so by symmetry, we have A. And so we have shown that if we have two equivalence classes, if they have a non-empty intersection, then they are equal. And that's what we wanted to show. OK, so that was simple. Now, I should maybe say just a few more words. So if I don't know what I want to call this definition. But anyway, so let again, we have an equivalence relation on a set A. And so if we have an element A, in some equivalence class B, then we say that A is a representative of B. Actually, in this case, as we have seen, we know that A is equal to B. But anyway, that's not relevant. So what very often happens in kind of the daily life of a mathematician is that you have some nice set on which you have some nice structure. And you have an equivalence relation. And you want to somehow have the structure go to the equivalence classes. So the way you usually do this is that you define it in terms of the representatives. So for instance, you want to take the sum of two equivalence classes with respect to some group structures. And you do this by taking the equivalence class of the sum. And this kind of definition will make sense, if and only if it's independent of the choice of the representative. And so often we want to define something for the equivalence classes by making the definition, I say it in a rather vague way, but I think it's pretty obvious what I mean by making the definition in terms of representatives. So this will make sense, will lead to a valid definition. And one would usually say, one says it is well defined. If this definition is independent of the choice of representatives, and if it's not independent, then this will not be a valid definition. So we can look at a very stupid case, which is related to the previous one. So we had before this equivalence relation, which we had so on. So we have example on z. We had this equivalence relation. So we have chosen element k, say in z, which is non-zero. And we have an equivalence relation, which we had just seen, that n is equivalent to m, if and only if k divides n minus m. This was an equivalence relation, as I had claimed before. So we want to, so I call z modulo kz, the set of equivalence classes for this relation. So we have a set whose elements are the equivalence classes. And we want to, so equivalence classes, according to this, are no n for some n in z. So now we want to make this into a group with somehow using the addition on z. So define in addition on z mod kz by, if I add n plus m, so if I took to add the class, the equivalence class of n to the equivalence class of m, is supposed to be defined as the equivalence class of n plus m. So this ippa-ora does not necessarily make sense because we have defined the sum of these two equivalence class in terms of the representatives. I've written n plus m. But the claim is this is well-defined. Because what do we have to see? We have to see that if I have another representative here and another representative here, I get a representative here of the same class. So in other words, we have to see if n is equivalent to n prime and m is equivalent to m prime, then it follows that n plus m is equivalent to n prime plus m prime. That means that it's well-defined. Well, and that is actually obvious. Because what does it say? So in other words, if k divides n minus n prime and k divides m minus m prime, we want to see that k divides n plus m minus n prime plus m prime. But you know, that's clear. You know, this difference is l times k. This is r times k. And this is l plus r times k. So this is clear. So this is actually well-defined. And one can check with this definition. z mod kz is in a BN group. I mean, if you want this in exercise, so the neutral element B0 and the inverse. So I have the notation here. I take the additive notation for this group. So minus n is equal to the class of minus n. You can check that this is true. Anyways, this kind of follows easily from the corresponding property for z. OK. Now, we want to look at, we have defined what the group is and what the subgroup is. And now I want to see that a subgroup in a group defines us an equivalence relation on the bigger group. And the equivalence classes are the so-called cosets. And we'll study these afterwards and see what we can do with it. So I'm talking about cosets. So as I said, so if h subset g is a subgroup, then this defines an equivalence relation on g. And the equivalence classes are the so-called cosets. And we will maybe now go to the property definition. So we have this thing. We have g is a group. h is a subgroup. So we define an equivalence relation, which I now denote a little bit different on g as follows. We see that a is congruent to b. So a and p are elements in g. a is called congruent to b, modulo h. If and only if, what is it? If I take that there exists an element h and h, such that, which we want it, a is equal to b times h. So whenever I can obtain a from b by multiplying by h, so obviously this is an equivalence relation or the other round, then I call them equivalent. So I say a is congruent to b, mod h. So this is an equivalent relation. Well, that's quite easy to see again. So first, obviously, a is equivalent to a itself, mod h. We can take our element h here, take h to be 1. And a is equal to a times 1. And we also see this is the first property, this is the reflexivity. And then we have the symmetry, which is also trivial. So if a is equal to b times h, then and h is an h, I can multiply by h to the minus 1. And I get that b is equal to h to b to a times h to the minus 1. And if h is an h, then as h is a subgroup, this is also an h. So we also get that b is congruent to a mod h. So the first two properties are clear. And in this case, also the third one is not really more difficult. So transitivity. So assume that a is congruent to b mod h. So a is equal to bh for some h in h. And b is congruent to c mod h. So b is congruent to c times h prime for some h prime in h. Well, then obviously, a is equal to, I mean, we just put it in, c h prime h. And h prime h is a part of two elements. And h is an element of h. So this is in equivalence relation. So yeah, that's all not very exciting, but OK. So the equivalence classes are called the cosets. So I'll write it more correctly, finishing at h b sub of g. Then so for every element a in g, we call the set, say ah, which is just somehow what looks like it's written here. Every product of an element of a with an element of h. So a times h, where h is an element in h, is called the right, a left coset. I find it always, I mean, it's how it's called. Although I always find it strange because the h lies on the right, but that's the way it's called. So it's called a left coset of h in g. So you can see by the definition we gave before, that the coset of a is precisely equal to the equivalence class of this equivalence relation. So by the above, so with respect to the above equivalence relation, we have that ah is the equivalence class of a. And it's clear by definition that if I take the equivalence class of the coset of one, this is just h. So we have the decomposition of g into equivalence classes. And one of these, so which are these cosets, one of them is actually equal to h. And now the next thing we will see that all the equivalence classes have the same number of elements in a simple sense. They are all bijective to each other. So they are all as big as the sub-comp h. So remark, so let ah and ph be two cosets. So this is called the left coset, but I will only consider left coset. So from now on, they are just called cosets. So let ab be two cosets of h and g. Then I claim they are bijective to each other. So we have a map from ah to ph, which sends just for any h and h, the element h small h to b small h is a bijection. h is a bijection. I mean, by definition, if that's very simple, I can write down what this map is. This map is just, in a suitable sense, the multiplication. So how is it? It's b. It's the multiplication from this side by bA to the minus 1. So because you can see this map is given by multiplying any element here with b times A to the minus 1. And so this will certainly send any element ah to the corresponding bh. And it is a bijection because we can obviously give its inverse is the multiplication well by the inverse element, so by ab to the minus 1. So if I have any element in ah, first I multiply it by this. I get the ah is sent to bh. And then I apply this, it's sent back to ah. So this is the bijection in both directions. So we find that any two cosets are bijective to each other. So in particular, if the group is finite, or if the cosets are finite, or if only the cosets are finite, they have the same number of elements. So ah, and so this allows us to make some kind of simple counting argument. It's all not very good. So the definition, let G be a group, H a subgroup. So we say the index of H in G is ah, no. I don't want to say that now, subgroup. So then we denote, so the set, so the set. Of cosets of H in G is denoted G mod H. We call the quotient set of G by H. And so it contains, so by definition, G mod H. This set of all ah with A and G. And the order of this group of the set, so the number of elements, so the set, the number G mod H, so the index of H in G is ah, the number of elements of the set with a usual convention that we write it's infinite if this is not a finite set. So with the, and so is, so I should say it is denoted ah, G H, and defined to be the number of elements in this quotient set. And so when we say this index is infinite, equal to infinity if G mod H is infinite set, OK? So it can certainly happen that G, both G and H, so if G is finite, then certainly the index of any subgroup will also be finite. If G is infinite, it can be that there are some subgroups which are also infinite for which the finite, for which the index is finite. So for instance, example can look at our stupid example that we have kind of always used of Z mod K. So if we have, so let K is an element of Z without 0, and maybe for simplicity I assume that it's actually positive, then we know that K Z is a subgroup of Z. And we can form the quotient set, so the quotient set is Z mod K Z. And you should notice that this is precisely the set before I had defined Z mod K Z in a different way. I had said that Z mod K Z is the quotient of Z by the equivalence relation where we say that two elements and an MI equivalent if the difference is divisible by K. But you can easily check that this is the same statement, that this is the same as the quotient by this subgroup. So note this is the same definition as in the previous example because K Z is just the set of elements in Z which are divisible by K. And so then you find that the two definitions I gave for this are the same. The quotient set is this, and you can check easily that this is a finite set. So Z mod K Z is equal to the equivalence class of 0, 1, until K minus 1. No? Because if we have any element which is, for instance, larger than K, we can do division with rest, and we get an element which is equivalent to it, which is lies in the same class as one of these. And so this quotient set is indeed finite. And so the index, so in other words, we find that the index of K Z in Z is equal to K. As one would maybe also expect. OK, now get our first theorem, which is not very difficult to say the very least. But it's still, it's a theorem because it's useful. And it was maybe even has a name to it, it's a theorem of Lagrange. So let G be a finite group, and H be a subgroup. Then the number of elements of H divides that of G. Then H, so which one can also call, which was also called the order of H, divides G. This, I remind you, means divides. Well, first thing that's still right. So which is the order of G. The order of any subgroup divides the order of the group. And to make it more precisely, more precisely, we have that the order of G is equal to the order of H times the index of H and G. OK, so this is a bit, I've called this theorem, but it's actually quite obvious. But it's called a theorem because it's quite useful. So let's see first how one sees it. So we just use what we have here. So we have that G has a decomposition into equivalence class into cosets. So we know that G has a decomposition into cosets. So we have, and in fact, the number of cosets is G mod H is this index. And they are into G mod H disjoint cosets. So in order to prove that this identity holds, we only need to see that all cosets have H, have as many elements. But we know that already because we have seen that one of the cosets is actually equal to H. And all cosets have the same number of elements. So H is a coset. And all cosets have the same number of elements. So it follows that the number of elements in G is equal to the number of elements in H times index. OK, so this is, as I said, fairly obvious. But I mean, one must say maybe to the defense of Lagrange that he was one of the inventors of group theory. But it's also really useful. I think, again, this was invented for the use in Galois theory. And it's an important little fact. I will give some corollaries to show that this is somehow useful. So the first one is a corollary. So first I define the order of an element. Let A in G in element in a group, where G is a finite group. And so I don't need that. A is an element in a group. G is a group. I'm making a definition of the statement. So the order of G, the order of A is odd of A, which is the smallest positive power of A such that you get 1. So it's equal to the minimum of all n positive integers such that A to the n is equal to 1. If such an n exists, such n exists. And it's infinity otherwise. So in particular, for instance, differently from a very annoying misprint in my notes, the order of 1 is 1. And for every element in the group. And obviously, if G is a finite group, you will find that the order of every element is finite. Let's see. In particular, we get the following statement. Corollary, let G be a finite group. And I take an element A at A in the element in G. Then the order of A divides the order of G. Well, it's not so. So what we actually, I mean, it's kind of clear how one would want to prove that. You want to see that the subgroup generated by A contains precisely the order of A elements. And then, so as it's a subgroup, we use this thing. And so it has to divide the order of the group. So let's see it faster. So let A be the cyclic subgroup, which we introduced the other time. The cyclic subgroup of G generated by A. So it's enough to show that this has as many elements because it's a subgroup, no? Enough to show the number of elements in this subgroup is equal to the order of A. Because as is the subgroup, we know its number of elements divides the order of G. And so that's it. So well, that's not very, let's see. So you might remember what this was. We have that A is equal to the set of all A to the n, where n is an element in z. So not generate this system. And A to the n wall meant multiplying A with itself n times, or if the n is negative, multiplying inverse of A with itself n times, as you had introduced yesterday. So now we employ, so in z, we have division with rest. So by this number, order of E. So that is, we can write n. So we can write n, n equal to sum number d times order of A plus the rest, where d is an integer. And r is an integer between 0 and this number minus 1. No? Everybody, this you learn, I don't know, maybe in the third class or in the fifth class or something, school. And so what does it mean for this? So if I take A to the n, this I can write as A to the d times the order of A plus r. And I can also write the other round. So this is the same as A to the order of A to the power d times A to the r. Now, A to the order of A is equal to 1. So this is 1. This is equal to A to the r. So we find, so therefore, for each n in z, A to the n is equal to A to the r, for r an integer from 0 to order of A minus 1. So we find that this set here, so we see that it's equal to 1 A A squared A to the r order of A minus 1. So the only thing we have not yet seen is that these elements are really all different. So we have precisely A order of A elements. We want to see that they are different. To see they are all different. Then the number of elements in A is precisely order of A. Well, so if A to the r1 is equal to A to the r2, with, say, 0, smaller equal to r1, smaller than r2, smaller than order of A. Maybe I write like this. Then what do we have? Well, then we have, obviously, we can divide by the other. We get A to the r1 minus r2 is equal to 1. And we also have that r2 minus r1, I want. And we have that, obviously, 0 is smaller equal to r2 minus r1 is smaller than the order of A. No, because we have subtracted the smaller one from the bigger one. And so we know, however, that this order of A is the smallest integer such that we get 1. I mean the smallest positive integer. So it follows that this number must be 0. r2 is equal to r1, in other words. Or we get, say, I want to say r2 minus r1 is equal to 0. So in other words, A to r1 is equal to A. So r1 is equal to r2. So we see, therefore, that A consists of order of A elements of these elements. And all of them are different. So the number of elements in A is order of A. So this then actually shows that the order of any element in a finite group divides the order of the group. It's not very. So another such simple summary is that this basically follows. If we, in particular, if we take any element in a group and we take it in a finite group, we take it as to the power of the order of the group, we get always 1. So also, we're going to have G be a finite group and A an element in G. Then if we take A to the order of G, so the number of elements in G, this is 1. This is something which one uses quite often, but it's obviously quite easy to see. So by the previous corollary, we have that the order of A divides the order of G. So we can write the order of G as sum number of D times the order of A, where D is a positive integer. Well, and so if I take A to the order of G, this will be A to the order of A to the power D. But A to the order of A is already 1. So this is quite easily the case. But this does play a role in some arguments also in anyway. So this was just to have some more. Well, then finally, something about the structure of groups, which is again, is very easy corollary. So obviously, it's difficult to know for a general group what all the subgroups are also for general finite group. But if the number of elements in the group is a prime number, then we know because then there's essentially none. So corollary, so a G be a finite group, such that the order of the group is equal to P with P is a prime number. So I expect you know what a prime number is. So that's a positive integer different from 1, which is only divisible by its seventh one. Then G. So for one thing, I wanted to say that G has basically no. So the only subgroups of G are G and the trivial group. But actually, we can save it more than G is cyclic. So we can precisely see what kind of group it is. So this is very simple. So what does it mean to be cyclic? It means that there is an element in G that G is equal to the cyclic subgroup of G generated by that element. But in this particular case, it's very easy. We just take any element which doesn't happen to be 1. And then the subgroup generated by it will already be G, because there's no room for other subgroups. So for any subgroup H in G, we must have we have the number of elements in H divides the number of elements in G, which is P. But as P is a prime number, it means that the number of elements in H either is 1, which means that H is a trivial subgroup, or it has the same number of elements. It has the number of elements as P, which means that H is G. So there are not so many subgroups. And so now let's take any element with A being an element in G, which is not a neutral element. Then if I take the subgroup, the cyclic subgroup generated by A, this is a subgroup of G. And it contains an element which is different from 1, so which is different from H, different from the trivial subgroup. So by what we have just said, it follows that A, this subgroup, is equal to the whole group. And so G is cyclic. OK, so this was this simple statement. So much for the moment about groups and subgroups and the cosets and the quotient set by subgroup. So are there any questions to this or comments? So have I already told you until now anything that you didn't know before or not? To nobody? Anyway. OK. What? Nothing new. But is it OK or the same? Or I mean, I hope you will soon come to something new. And not for me, for everybody. I mean, if somebody has studied everything, then there will be nothing new. But I think so now we want to talk about normal subgroups and quotient groups. So if G is a group and H is a subgroup, we have the quotient set G mod H. And I mean, the natural thing that you could ask yourself is, is this, again, a group? Is this in a natural way? And it's kind of in a natural way means one wants to do the obvious thing and it should become a group. So what is the obvious thing? I mean, it's kind of clear. So we would want to define the group structure on this quotient set by saying that A H times B H would want to define this as A times B H. Now, obviously, but as I told you, this is, again, a definition in terms of representatives. So in order for this to be a valid definition, we would have to see that it's independent of the representative. So have to see to make this into definition should be independent of representatives. And so we can ask ourselves, so otherwise it's not definition, but just nonsense. So let's see. So let's check. So the point is that, I mean, that we will see, is that this will not always be well-defined. So in general, this will depend on the representatives. And so this is not a definition. But if you put an extra definition, an extra condition on the subgroups, which makes them, and such subgroups we call normal subgroups, then this is well-defined. So we can easily see that we need some condition by looking at some example. So for instance, if we put, say, what we want, A, so the element A equal to the 1. So what we want, so thus we need. So I have to explain. So thus we need A h is equal to B h, that's A prime h. And B h is equal to B prime h. Then it follows that A B h is equivalent to A prime B prime h, or that A. So let's just look at an example. So assume that we take A to be 1, A prime to be h, which is certainly fine, because they are A1 and any element. And this is an arbitrary element in h. And we take just, say, what we do. We take B, any element in G, an arbitrary element, and B prime equal to B. So we look at just one special case to get some to see whether already here we find the condition on h. So we want that, in this case, that A B h, so B h, is equal to A prime B prime h. So this is H B h. And B is an arbitrary element of G, and h is an arbitrary element of H. So in other words, that exists in h prime, in h, such that H B, so this one, is equal, so it's maybe not so easy to distinguish, but anyway, H B is equal to B H prime. Well, if this is the case, it follows that I can multiply by B to the minus 1 on this side. So it follows that H prime is equal to B to the minus 1 H B. So in this case, but remember that B in G was arbitrary, and H in H was arbitrary. So we actually have found the condition. Namely, we have that for all H in H and all B in G, we have that if you want B to the minus 1 H B is again in H, or I can replace B by B to the minus 1, we want that B H B to the minus 1 is an element in H. So in order so that we have a chance, I mean, we haven't proven that this gives us a group structure on the quotient, but in order so that we have the smallest chance that there is such a group structure on the quotient, we need this condition. And which subgroups which satisfy this condition we call normal subgroups. And then we will see that this condition is also sufficient. So definition, let G be a group, H a subgroup. Then H is called a normal subgroup of G if this condition holds. So if, say, A H A to the minus 1 is an element of H for all A in G and all H in H. So we'll see. Let's first see that there are some normal subgroups and there are some groups which are not normal. So example, certainly if G is a billion, G is commutative and all subgroups are normal. This is obvious because if it's commutative, then A commutes with H. So A H A to the minus 1 is just equal to H and H was an H. So this is kind of clear. And let's look at one example of a subgroup, what? Maybe just as a further example, so if we take the trivial subgroup and the whole of G, some kind of the stupid subgroups of G are normal subgroups. In fact, we are not going to use this. There are groups, I mean there are many groups which have the property that the only normal subgroups of the group are 1 and the whole of G. And such subgroups are called simple. And one of the biggest results in group theory of the last century is to give a complete classification of all finite simple groups. And the third one is if that G. So let us just look at a simple example. So G, assume that G is equal to S3. So let's look, so just the permutations of three letters, symmetric group of three letters. So then we take a subgroup H to be a group with two elements. So first, as a subgroup, it has to contain the identity, so which is the identity element. Another one is what is called a transposition. So I exchange 3 and 2, 2 and 3. So this is a subgroup. And why is it there? So you can easily check. So it's straightforward to see that if I take 1, 2, 3, if I take this element and multiply it by itself, then I get the identity. So this will give me the trivial transposition. I mean, you can just look. No, 1 goes to 1, and then it goes again to 1. 2 goes to 3, and then 3 goes to 2. And 3 goes to 2, and 2 goes to 3. So it's this. So this means, first, that it means two things. One thing that this set is closed under multiplication. If I multiply anything by E, I get the thing back. If I multiply this by itself, I get the identity. So this is a subset which is closed under multiplication. And we see that this thing, we have a neutral element which is E, and the inverse of E is E. And the inverse of this element is the element itself. Because if I multiply it by itself, I get the identity. So this H is a subgroup. Inverse of 1, 2, 3, 1, 3, 2 is the same element. And obviously in the same way, we'll do here, obviously in the same way, if I take 1, 2, 3, 2, 1, 3, and I multiply it with itself, this again will be the identity element. So then we can compute. If we look at this element, 1, 2, 3, 2, 1, 3, and I multiplied with this element here in the group, and I multiplied with the inverse of this element, remember that the inverse of this element is the element itself, then we can just compute. So this would be 1 goes to 2, 2 goes to 3. I'm not yet finished, 2, 3. And 3 goes to 3, so I actually was finished. Then 2 goes to 1, 1 goes to 1, and 1 goes to 2. And 3 goes to 3, 3 goes to 2, and 2 goes to 1. And you see, this is an element which is not an element in H. So we have found that indeed, this H is not a normal subgroup. OK, that's what I have. Yeah, and then, well, maybe I can still try. So now we want to actually show that what I know, what was the purpose of defining a normal subgroup, namely that if we divide by it, we still get the quotient set as a group, that this actually is true. So maybe I don't call it this way, I call it a position H in G, a normal subgroup, then the quotient G mod H is a group with the obvious group law with modification A H times B H is equal to AB H. And well, so this is quite simple. We only have to prove that the product is well-defined. In some sense, we essentially already did this when we motivated the definition of a normal subgroup. But let's do it properly. So we take some elements, let AB be the elements in G, so maybe I say A prime, B prime, B elements in G, such that A with A H is equal to A prime H, B H is equal to prime H. So that means that A is equal to A prime H for some small h in H and B is equal to B prime times H prime for some h in H. Well, and then we want to somehow see that AB and A prime, B prime are related. So if I take AB, this is by what we see, A prime H times B prime H prime. Now we can remove the brackets and we can multiply by B prime to the minus 1, which is just 1. Is this what I want? I hope so. Yeah. Now this is, we have an element in H. We have taken the element to the minus 1 times H times this. So this is some element in H. So this is equal to A prime, B prime, H double prime, H prime. So this is a product of two elements in H. So this means that H prime is an element in H. So thus ABH is equal to A prime. OK, so that's quite simple. And the rest, so once it's well defined, the rest of that it's actually group is kind of obvious. So it's clear. So that we take the, so the associativity is clear, follows from that of group G. I can kind of write it out. So if we take A H times BH times CH, well, this is by what we have now ABH times CH. And this is done by multiplying by this. So this is ABC times H. And now you can see, we already don't see the bracket anymore. So we can go backwards on the other side. So after some steps, we get that's AH times BH times CH. And the neutral element clearly is 1 times H, which is equal to H. Because if I take A times H times 1 times H, we get A times H. And the inverse to A times H is A to the minus 1 H. So this is certainly a group. So maybe I should stop now. So I think, OK, so maybe even that was not new. But so I hope you would have some. But is it really? Well, anyway, so for now it's enough. So next time I will work a little bit more with these quotient groups. And I will also then talk about homomorphisms of groups and use them. OK.