 Okay. Good afternoon. So I'm, as you already know, I'm Götze and I'm giving you the course in algebra. So my room is 110 and one of our postdocs who's called Tarek Abdel-Gadir. I'm not sure. So he said that if his room is 122, so if you have questions that you do not want to ask to me or whatever or anyway, if you just have, you can also talk to him. He would like also to see you anyway. I'm usually in my room whenever you have questions about what I say you can come either to me or to him. And okay. So the topic of this course is algebra. And we will be somehow have three parts. We'll be dealing with groups, rings, and fields. In the case of the fields, we'll do Galois theory also. So this course essentially starts from zero. So, I mean, we start with the definition of a group. So it's quite elementary. Maybe for some of you it will be boring. I mean, if it's really boring for everybody, we can kind of change it. But on the other hand, if it's only boring for some of you and some others actually are learning something, we keep it the way it is. And so I will just start by the definition of group. If you have any problems, if I'm going to fast or you don't understand something, you tell me. If you cannot read my handwriting, you also tell me, but I can not permanently improve it. You know, you will also have to learn to read it. Okay. So let's start. So what's the group? By the way, you have my notes. I have prepared some notes. I will actually be following the notes very closely. I've just prepared them before starting this course. They are so new that they are also still full of misprints, as you will have maybe noticed. But you know, anyway, I will, I hopefully will say the correct things in the lecture. So a group will be a set, will be with a binary operation. So somehow you can multiply two elements, A, B. You can multiply them and get an element A times B. So somehow you know, you, so it somehow is set in which you can multiply the elements and some obvious actions should hold. There should be the associative law. So it doesn't play a role how you put brackets for this operation. And you have a neutral element and every element has an inverse. So let me write this down properly. So definition. So we take a non-empty set, set G with the binary operation. So this is multiplication from G times G to G, which sends a pair of elements AB to something which we want to call A times B. So just, you know, binary operation just means that for two elements we get one element is called the group. If the following actions are true, well, first we have the associative law. So we can, so in, it's written like this. So if you have A times B, so for, this is always for all A, B, C elements in G. If we take A times B and we multiply this by C, this is the same as if we multiply B with C and on the other side, multiply it by A. So this just means, you know, we have this operation. So A times, so we can, it just means we first do this operation. We multiply them, get an element of G, we multiply with this, and then we do this with the binary operation. Yeah, this you, it is in principle, so binary is, with the binary operation. So I will try to maybe write bigger. So this is the first thing, and I mean, you know, it's kind of a standard thing for any kind of operation that you need is a stativity. It's the first thing that you always require. Second one, we have a neutral element. So that means there exists an element E in G such that we have that E times A is equal to A times E is equal to A for all A in G. So it's an element for which the multiplication, it doesn't do anything. And the last one is that every element has an inverse so that if you multiply the element by the inverse, you get the neutral element. So for every element for all A in G, we have an element A to the minus one in G such that A times A to the minus one is equal to A to the minus one times A is equal to E. Okay, so these are the simple actions. And it somehow captures some part of what we used to if we kind of multiply elements in Z or in the complex numbers, the real numbers or the rational numbers, or if we add integers. And so we will, we can by this definition, we can simplify slightly the notations. For instance, as the bracketing doesn't play any role, we can also forget the brackets. So we write also ADC for, so first we write AB for A times B. So we usually will kind of suppress the operation and we can also write ADC for ADC, which is the same as ADC. After all, it doesn't play a role. It's verified. So we will start. Now I want to, as an exercise in these definitions, which are kind of straightforward definitions, so just abstractly use the definitions. We want to prove some elementary properties of groups, but really elementary. So this should seem completely trivial to you. So we have remark. Let's say first the neutral element is unique. So as I said, this is not really, these are not particularly great results, but the point is that we have here our actions and we want to argue only the actions and you have to get familiar with these abstract arguments if you are not already. So let E and E prime be neutral elements. So that means if I multiply anything by E, I get back the other thing I multiply by. So we have this property. So then while we can say E is equal to E times E prime because after all, E prime is also neutral element and this is equal to E prime because E is a neutral element. And so it's equal to E prime. Okay. So the neutral element is unique. Any two neutral elements are equal. And then, so this was the first one. Second one is, so the inverse is unique. That's somehow similar. So let N prime the inverses of A in element in G. Well, so we can write B prime, same. We want it there as B times A, E. E is our neutral element. So this is B prime times A, D because E is inverse to A and this is equal to D prime. So this was like this. Now we can use the associativity. So this B prime A times B. This is E equal to B. Okay. So it's really straightforward. Then we've just used the actions. The third statement would be, so it's enough to be an inverse. Actually it's also true for a neutral element. It's enough to be that one of these two identities holds. So that it is a left inverse thing or left identity. But now we have to say it for identities. So let B in G be an element with B times A is equal to the identity element. Then it is the inverse. Where this one can do in different ways. But I can also just, I mean, I would just write like this if I take the inverse. So I'm actually, then I can write this as E times A to the minus one. And E I can write as B A and as A is the inverse. This is equal to B. Okay. So these are all very simple things. And then maybe I don't do the other ones which are similar. I mean, they are even simpler. So maybe fourth one is that if I take the inverse of A to the minus one, this is A. Well, this is actually almost total logical. We know that A times A to the minus one is equal to A to the minus one times A equal to E, which says precisely that A is the inverse of A to the minus one. And then finally, as one uses it often, if I take the inverse of the product, this is the product of the inverses, but the other way around. Well, and that's in some sense clear. If I take A B times E to the minus one, A to the minus one, then this is, you know, A B E to the minus one, A to the minus one. So I first can cancel these. This is A, A to the minus one, and this is E. So they are inverse to each other. Obviously, okay. So much for that. So this was very simple. So there's another property which is also useful, which is sometimes called the cancellation property. So that is, if I have an entity in group, so let's again A, B, C, the elements in G, and assume that A, B is equal to A, C, then it follows. So if, then it follows that B is equal to C. So you can always cancel equal factors. And in the same way, if B A is equal to C A, and also for that B is equal to C. And this is obvious. You just multiply with the inverse of A on both sides. Here you multiply with A to the minus one on this side, get B equal to C. So if I take A to the minus one, A, B, then before they were equal, now I multiply. This will be equal to A to the minus one, A, C. And I can forget the brackets as usual. And then this is E. So this is equal to B. And this is equal to C. And the same way for the other equation. So there's another property that we are very used to with the groups that we know is that many groups are commutative. So it doesn't play a role in which order the elements are when we multiply them. This is true for the integers. This is true for the rational numbers and so on. So I will, as although you all know it, I will still kind of write it down. So definition, a group G, G is called commutative if, well, if I take A times B is equal to B times A for all A and B in G. And I want to also introduce notation for that. So the kind of standard, it is most of the time when you have a commutative group, not always, but most of the time you write it in additive notation. So you write plus instead of times. And so often for commutative groups, so write the operation as plus. So this then says we write, we write A plus B instead of A times B. And we write minus A instead of A to the minus one. And we can write A minus B instead of A times B to the minus one and we write zero for the neutral element. So this is most of the time when we have a commutative group we will do that, but not always. So now I want to give you a few examples which you essentially all know to see that groups are all over the place. And so the first example, I mean, anyway, first are there any questions until now? I mean, until now I've just written here these definitions and have some simple computations. So if there's anything which is not clear at this point, you should really tell me because then I have to change my approach. But if it should feel kind of trivial to you, but you know, you shouldn't be extremely super bored, but you know, okay. Any comments? No. Okay. So can you now essentially read what I write or is it very difficult? It's okay. What? No, no. What? Yes. I mean, well, I mean, what does it mean? We have a binary operation which satisfies certain properties and then we call, we denote it either by times or by plus. But you know, it's not, you know, you can obviously even on the integers you could define a binary operation which has nothing to do with the addition or the multiplication. You know, it's just that we call it like that. We call, in this case, we call it sum, but it doesn't have to mean the sum. You know, it just is a notation that we put. I will give some examples although in that case it actually is the sum. Anyway, so first we give some examples. So we first can take, for instance, the first example is the trivial group. So a group should be a non-empty set. So in particular, the smallest non-empty set consists of one element. We could, for instance, take the element which consists of an element which we call one or E and this becomes a group by defining that E times E is defined to be equal to E and then you can check that this is a group. Obviously, E is the neutral element and it's also its own inverse. Okay? And so this is kind of the most stupid example of a group. So we can take, for instance, the integers set with the usual addition z plus. So really an integer is an integer and you just add them the usual way. This will be in a b. Everybody knows that this is the case, that you can add integers and the order which you add them doesn't play a role. Then, on the other hand, for instance, if we take z together with the multiplication, this is not a group. So again, we take the standard multiplication on the integers. This is not a group because not every element has an inverse. For instance, two has no, so you find that the neutral, so it has a neutral element. Obviously, it's one, but the inverse, so two, for instance, has no inverse. Actually, that was even a bit worse. I take maybe z without zero. It's not a group, otherwise it's even more wrong. I mean, also z obviously is not a group, but in that case you don't even have a neutral element. Then maybe we forget about this. So let me have a slightly more complicated example. So let k be some integer, say positive integer. So by this I mean that I want to define a group structure on the set of all integers from zero to k. So let me write although it's not standard. So I just write zk for the set of the element zero one until k minus one is set with k elements. And we want to make this into a group. And in different guises, we will kind of encounter this many times. So I do it like this. So if we have an integer, so if n in z is an integer, I denote by, I mean that's maybe not standard anyway, but I say n underlined the rest of division by k. So the rest of dividing n by k. So that is, we write n equal to d times k plus n bar, where d is an integer and n is an element of this set. And you learn in high school that you can always do this and this is the rest of this division. So now we want to make this. So then we define a group structure. So we define an addition on the set. So an operation which we call addition on zk, which I denote is like a different way. So I say maybe plus. So I say n. So if you have two elements here, n, which are elements of zk. So these are numbers from zero to k minus one. I can add them, I say, and I denote the addition like this. This is done by taking the usual sum as integers and then taking the rest under division by k. So if I have two numbers, which lie between zero and k minus one, if I take the sum, it doesn't necessarily lie between zero and k minus one, but I can, if it's too large, I can subtract k from it and I get an integer in here. And this I define to be the sum. And I claim this is an nabian group. This makes this thing into an nabian group. So that k with this operation plus is an nabian group. Well, this is kind of trivial. So we have to check the action. So we, for instance, about the sensitivity. So what we have to prove is that if we take such a triple sum m plus n plus l, where these elements zk, this is the same as doing it the other round. We have just to look at the definition. So this is we take n plus m, the usual sum, and the rest under division by k. And then we sum this. So maybe for the moment I write plus l, which is by definition n plus m plus k plus l. And so, you know, but this just means we, you know, we sum these elements and we subtract the corresponding multiple of k until we get in the elements at k. So this is just the same as the sum of all three rest under division by k. And this now is symmetric. So if we can kind of perform this operation backwards, this is equal to n plus m plus l equal to, well, that's what I wanted. I mean, you just make these steps on the other side backwards and you get this. Because this is already completely symmetric and you know that in the, so this would be first, if you write like this, it would be first like this. But you know the sensitivity holds in z. So you can also, so maybe you write it like this. This is equal to n plus m plus l. Because here we are in the integers, in the integers the addition is additive and then you go backwards and you get n plus m plus m. And the commutativity is also clear. You know, you know that if you take the sum n plus m, this is n plus m, the rest of this, but the addition in the integers is commutative. So this is m plus m. And obviously the neutral element for this addition is the element 0. Because if we add 0 to element in z, you get back the same element. It doesn't change when we take the rest under division by k. And the inverse of some element l, k minus 1, is okay. If l is equal to 0, we take 0. And if l, and otherwise we take the inverse of another l, which is different from 0, otherwise the inverse of l is k minus l. Because if we take l plus k minus l, then obviously this is the rest under division by k of l plus k minus l, which is the rest under division by k of k, which is 0. Okay, so this is a kind of, it's a very simple example, but somehow we have, you will see that this can be viewed as an example of taking a quotient group of a group by a normal subgroup. The moment we don't need that to know that. So another, I give one more example, if I, so a bit more abstract, which you know from linear algebra. So let, I don't know which number I'm now at, but anyway, five. Let v be a vector space. Then I can call, say the, maybe the group of automorphisms of v is a set of all, say, v from v to v, where v is a linear map. So vector space, say, r vector space or something does matter. v is a linear map. So phi is linear and phi is projective. So then v is from a group, so out of v is a group. Yeah, well, I can take any vector, any field, but you know, whatever, I could say the field, say v is an r vector space, and then this would be r linear, but you know, whatever, no, any field. So out of v is a group. Well, you know, you know that, so the group operation is a composition of maps. So I can say that, so phi times c would be phi composed with c. I know they are both projective maps. I can compose them and I get another projective linear map. And so, so we know it's a standard fact that one learns early in the one's life that composition of map is always associative. So this is associative from the definition just by remembering what it means to compose map. And we see that if, so if I take the identity of v from v to v, which sends any vector to itself, is a neutral element. Because if I compose with the identity, I get back the same map. And you know, if phi is an automorphism of v, it is a projective map. So it has a projective linear map. So it has a projective linear inverse. So we have phi to the minus one v, the inverse map. And then it is also the inverse in the group. So phi to the minus one is the inverse element in, out of v. Because by definition, to be the inverse map precisely means that both compositions of phi and phi to the minus one are the identity. phi composed with phi to the minus one is equal to phi to the minus one composed with phi is equal to the identity of v. That's what it means to be the inverse map. And that's the same definition as what it means to be the inverse element in the group. Okay. A similar thing that we can look at is just, you know, the symmetric group on a set, which is a more general statement. So if, so six, let m be any set. We can just call, so the group s of m is just defined to be the set of permutations of m is just a set of all bijections of m to the set. And by the same token as before, I mean the same argument will actually see that this is a group by composition. So that means, so again, phi times c is equal to phi composed with c and the composition of bijective maps is a bijective maps. And we have again the identity of m is a neutral element and the inverse map to a given bijection is the inverse element. Okay. So this is, I mean somehow okay. So we have many groups. This is not, so traditionally when groups were introduced, they were always viewed as groups of permutations of some set. So they were always thought of being that. That's how groups were first considered. I mean, this was what the group was supposed to be. And so in that sense, this kind of goes back to the standard things. And actually, this comes up also in the context of Galois theory, which we will do at the very end of our course. In fact, groups were invented to do Galois theory, which is, you know, somehow. Anyway, so we have here. So we look at a special case of this when we have a kind of very simple set. So if m, we can look at the symmetric group on just the set of elements one to n. So on this finite man is called the symmetric group n. And I will usually denote it, I think, by, I don't know what my notation is. Maybe I will just call it sm. Okay. Usually we have some kind of Gothic s or something, but as I cannot write that, I write like this sm. This is, so it's a special case of this where the set m of which we take the permutations. So all the bijective maps to itself is a finite set, one to n. And here one can be very explicit. So again, the group operation is just a composition like this. And we want to make some computations in this. So what, yes? Yeah, okay. No, it has nothing to do with the order of the group. Actually, the order of the group is much bigger. Actually, I'm slightly wondering whether I want to call it the symmetric group of order n after your comment. So maybe one could also say the symmetric group, sometimes I would call it symmetric group on n letters. So somehow you have n letters and you permute them. Maybe that's a better name. Okay. So we introduced the following notation for such a permutation. So if I have an element, so for you write an element f from, so I've got it in sn, so it's a bijection from the set one n to itself. In sn, we write this as, so we write first all the elements one to n, one, two, three, and so on until n. And below we write their image. So f of one, f of two, f of n. Okay. So we have some notation how to write this thing. Later we will also find other notations. So let's look at, so we can look at some examples. For instance, s two, all the bijections from the set one, two to itself, they're not very many. There's, it consists of two elements. There's either the identity element, so the identity element, which is just, so it consists of one. So the identity, which I, in this notation would be one, two, one, two. And there's only one more element, which is permuting the two. So one, two, two, one. So this has two elements. We can also look at s three. Again, we start with the identity element. So, so every element goes to itself. And then we have to look at all possibilities what we can do. We can first permute these two. So this is, there's left second to it's one, two, three, one, three, two. Then we can make this go to two, one, two, and skip the order of the other ones. So two, three, then we can exchange these two. Then we have exhausted everything we can do if one goes to two. And then we can get one goes to three. We can keep the order of the other one. So we have two, three, which goes to one, two, or we can exchange the order of the other ones. And you can easily convince yourself that these are all elements. Maybe it could be, and so how does one, and then this is our, I mean, actually maybe in the, in this case, we can see that the group is not commutative. So, so we've seen that this is a group because the special case of this group, but it's not commutative. So s three is not commutative. And we can see this by seeing what happens if we multiply two elements. So for instance, we can look at these two elements. One, two, three goes to one, three, two. And we multiply it by one, two, three goes to two, one, three. Yeah. And so what does this give to us? So we have to remember that the operation was composition. So in this case, we have here one goes to two, and then two goes to three. So one goes to three. Here two goes to one, and then one goes to itself. And here three goes to three, and then three goes to two. So this is how, what this gives. And we can also do it the other way around. Do the same thing. So one goes to one, and then one goes to two. You can already see it's not commutative. So two goes to three, and then three goes to itself. And finally three goes to two, and then two goes to one. So we see it's not commutative. What? Yeah. Yeah, that's true. So you don't, I mean, I like, okay. Now that's correct, yeah. But I mean, I, it's not, okay. You have to, some books on algebra do this, and some don't. I mean, it's not like everybody doesn't do it. I'm kind of going against the, so obviously in all other fields of mathematics, which is not algebra, one does it in this way. So in algebra, one sometimes also does the other way around. But you know, I kind of, I prefer to do it the way, which is kind of general. But you know, obviously, you know, if you, I mean, if the multiplication is defined the other round, we'll get different results. But it's kind of equivalent. No, it's not a big difference. And normally the composition of maps is always done in this way, although it is somehow slightly, you know, you know, it feels maybe slightly wrong. Yeah, but this again is only done in group theory and not in any other field. So I mean, somehow I don't, you know, I don't like the idea that, I mean, in some sense, I've, you know, maybe it's for some purposes slightly easier, but I think, you know, there's a standard notation in mathematics and we use that. I mean, it's not, even if in group theory, sometimes one does a different thing. But you know, you're free to do it the other round. What? It's confusing for you or? Yeah. Well, but the different, there's no real big difference. So, and otherwise it's confusing for me. I mean, I cannot, you know, and so, I mean, I hope you can manage, you know, I mean, it's just, and I mean, in, yeah, I don't know, in the books, well, I don't know. I mean, okay, no, no, but you're right. It is very often, so in group theory you find that, but it's kind of strange. I find it, you know, that there's one small group of mathematicians who does it different from everybody else is somehow also wrong. So, yeah, so if you want, so an exercise could be, I mean, and this is not homework, but I'm just saying if it should be easy to check that if you look at the number of elements, so actually, so I, so the number of elements of Sn is equal to n factorial, so for, so n factorial as you, I hope, for knows is just the product of all numbers up to n. So, maybe I should say in this context, as this also already came up, so if G is a group, the order of G is, well, it's just the number of elements of G, so it's just this number, but then one sometimes also denotes it as in a different way is order of G, which is just G, the number of elements of G. So, we write the order of G is infinite, so equal to infinity if G is infinite, so if it's not a finite group, once the integers have this, and so here the claim would be that the order of Sn is n factorial, well, maybe I don't do this example. Okay, so one more piece of notation, so it's kind of useful to have some kind of notation like one uses it for the integers that one can look at powers of elements in the group and things like that, and so we just introduce this, so if G is a group and A an element in the group, in the group, then we, and we have an integer, so first we want to define A to the n for an integer n in Z, this is just supposed to be an element in G, so if n is bigger than zero, so a positive integer, we let A to the n just be the product of n times element A, A times, times A n times, okay? If n is equal to zero, A to the zero, we define to be the unit element E, E is the neutral element, so this is our definition, and if n is, say, bigger than zero and we take A to the mean of minus n, so we take a negative power, so if n is smaller than zero, we say that A to the n is defined to be the inverse of A to the power n, well, maybe minus n, okay? So this defines, gives a definition for A to the n for all elements in Z, so this is no problem, and we obviously, in order for this to be making any kind of sense, we have to know that kind of the standard rules that you know for, you know, the usual numbers when taking power still apply, so this I formulate as an exercise, so show that for any element A, we have A to the, so how shall I put it? A to the n times A to the m is equal to the A to the n times m for all n and m integers, and if we take A to the minus n, this is the same as A to the minus 1 to the n, or maybe I want to do the other round, this was a definition, so this is A to the n to the minus 1, okay? And this is basically, or again, a simple exercise in the definitions, you know how to add and multiply integers, so these are just integers and then you have to see that you, that this always holds, and this again for all n and z, so we can write this power kind of notation, okay? So much for the, you know, the very basic things about groups, now we want to, are there any questions under here? Now we want to introduce subgroups, so these are, you know, subgroups, subsets of a group which, you know, if I restrict the group operation of the bigger group to the smaller set, the smaller set becomes also a group, so let's do that. So just to make our life a bit easier, I will, from now on, we usually denote by one the neutral element of our group, so not always, sometimes we might denote it by zero, or sometimes we might still denote it by E, but whenever I write one, it will be the neutral element of our group, and in fact to just say, so if this were already had, if our, if G is a billion, and we use the additive notation, then I already had said that we still denote the neutral element as zero, no? So here if you write a plus b for the group operation, then we denote the neutral element as zero. So let's do the subgroup, so definition, so as I said it's a subset of a group, such that with the operation of the group it becomes a group itself, so let G be a group, H subset G is subset, so H is a subgroup of G, well first I just say what I just said, so if first for any two elements a and b in H, we have that a times b is also an element in H, so here we have the operation in G, and the operation in G has the property that it will send, and we want that it has the property that if you have two elements in this subgroup, the product also lies there, and the second one is that with this operation H is a group, and with this operation, so with the restriction modification from G times G to H times H, H is a group. Well strictly speaking it follows from this, but because we know that the group is non-empty, but obviously I can say that it should be a non-empty subset. What? Which condition? No, no I'm actually asking this, I'm making, in the moment this is maybe not the most useful definition, but it is the most natural definition, a subgroup of a group is a subset which with the operation of the group is a group, of the MN group is a group, this is the definition, and then obviously that's what I'm saying here, so I'm saying the product of any two elements must be there, and then if I take this as operation it is a group, that's the requirement, so that's a definition, but you are right that this is usually not how one states it, because what one usually states is a criterion to be a group, and that I can also write, so we have the following criterion, so lemma, so let G be a group, and H, so a non-empty subset H of G is a group, is a subgroup, well if we have this property and also another one, so if for all, so if we have that A times B is in H for all A, B in H, and A to the minus one is an element in H for all A in H, there's another way of saying, I mean anyway that you would see, okay, so this is, if you want often you find this also as a definition, let's see why this is equivalent, so we have to see that, so the first condition is anyway the same, we have to see that with these two conditions, H will actually be a group with this operation, so first and conversed, so obviously if H in G is a subgroup, then these two conditions hold, because for a group we have an operation on the elements in the group, so the product of any two elements lies in the group, and the inverse of elements also must lie in the group, because it's a group, so then, so maybe I write this one and two, so one and two obviously hold, okay, so conversely, but it's not so difficult, so the first one, so we now want to show it's a group, so what we have to show, I mean first we have to show the laws of that it's a group, so if for A, B, C in the group G, we have the associative law, so thus it also holds in H, because if I have three elements in G, then if I have three elements in H, then in particular elements in G, and so the associative law holds for them, thus it also holds for A, B, C in H, so this is the first extreme, and the second one is we need to have a neutral element, so if, so let A be an element in H, then it follows that A to the minus one is an element in H by this extreme, and we have that A times A to the minus one is equal to the identity element is equal to A to the minus one times A, and this shows two things, so as a, so so it follows that as the as a product of two elements in H, one is an element of H, because this first one here says that the product of any two elements in H lies in H, so and H, one is a neutral element of G, so it's also a neutral element of H, because being a neutral element means if I multiply anything by it, it stays unchanged, so follows that H has a neutral element, and on the other hand this thing we also see here that A to the minus one is the inverse element of A, so and A has an inverse, so we see this is a kind of trivial reformulation, I mean if one wants one can also, I don't, I never found this particularly useful, but you often find this, it's an easy remark that a non-empty subset H subset G is a subgroup, if we can just write this in some sense in one formula, if A times B to the minus one is in H for all A in H, so if you want to go and call this exercise, it's easy to see that this is enough to imply this, you can check that, so now let's look at some examples of subgroups, so for instance, so we have z, we look at z where there is an addition, so integers with standard addition, and so if k is an integer, we can look at, let's say kz to be the set of all integers which are multiples of k, so this is a set of all n times k where n is an integer, maybe I should write it in the same order, doesn't matter as it's commutative, so these are all integers which are divisible by k, so then I claim this is a subgroup and I expect you can, I mean it's kind of clear that, so here we have the additive notation, so the neutral element of z is the element zero, and the element zero is contained in this, and well anyway, so we just have to see that if you take a product of two integers divisible by k, then if you take a sum of two integers which are both divisible by k, then the sum is also divisible by k, and if you take minus an integer which is divisible by k, it's also divisible by k, so if you see anything to write, so if a comma b in are divisible by k, then so is a plus b, and also minus a, and so we precisely, now we have here this additive notation instead of times we have plus, instead of a to the minus one we have minus a, but we see that precisely these two things are fulfilled, okay these are kind of stupid, so maybe if h is a subgroup of a group g, and l is a subgroup, and l is a subgroup of h, then we find that l is also a subgroup of g, this is trivial from the definition, I mean or from this criterion, and it's also so if if say h and l are subgroups of g, then h intersect at l is a subgroup of g, again this follows directly definition if you all buy this criterion, so any the product of any two elements in h lies in h, and the product of any two elements in l lies in h, so if the elements before were both in element in h, then also a product is both in element in h, and similarly for the inverse, anyway you can check it, if it's not obvious to you it's an exercise, then one very special case of a subgroup, so I don't know which number we're at, maybe three, so let g be a group, and a an element in g, then we can look at the cyclic subgroup of g generated by a, so let a the subgroup generated by a, this is just a set of all powers a to the n, where n lies in z, so we take any number of products of a or any number of products of a and a to the minus one, so note we have this if one remembers this exercise that I formulated about these powers, we know that a to the n times a to the m is equal to a to the n plus m, so which means that the product of any two elements in this a again lies in a, and we know that if we take a to the n to the minus one, this is the same as a to the minus n, so this says that the inverse of any element here lies in a, so contains all products, and here contains inverses, so this means it's a subgroup, so we have a for any element we have a subgroup generated by it, and maybe one can check it's easy to see that a is a billion, so commutative, and I didn't write before what one also says a billion for this for that, a billion actually will be large, so it's just because you know after all the addition in z is a billion, so a to the m times a to the n will be a to the m plus n, which is the same as a to the n plus m, okay, so we get this a billion subgroup of our group, so we call a is called the cyclic subgroup of g generated by a, so example we had this already, if we take the group k times z, this is after all the set of all k times n with n in z we had it before, so this as you can see now we have here the additive notation, so the instead of looking at these powers we take the products, so maybe I should, so this is the cyclic subgroup of z generated by k, so maybe I should kind of also I'm not sure I introduce the notation, so if g, if we have the additive notation for the group, so if we write a plus b for the group operation, then as I said we write zero the neutral element and we write minus a the inverse of a and we would write for n in z we write n times a for what otherwise would have been called instead of a to the n, no, this is instead of a times b, this is instead of one and this is instead of a to the minus one, which does notation, so then you can see that this is the cyclic subgroup generated by k and what time yeah okay, so we will call a group cyclic if it's it looks like that, so definition the group g is called cyclic we write here if it is equal to the cyclic subgroup generated by one of its elements, so if there exists an element a in g such that a the subgroup generated by a is equal to g, so for instance we see that that z is cyclic, so for instance we have it generated, so z with the addition generated by the element one, so any integer can be obtained as n times one for some n that's kind of and we have if we take this group z g, z k which was just the set of elements zero to k minus one for k positive integer and this is also cyclic also generated by one because by definition one plus one is also here equal to two and so on until k minus one and k minus one plus one is equal to zero, so let me see maybe finally I talk about the subgroup generated by several elements, so definition let g be a group and say let u the subset the subset of g, so the subgroup generated by u is what one would call the smallest subgroup of g which contains u, so technically this means is the intersection of all subgroups say h of g which contain u, so obviously g does contain u, so there are such subgroups and we can take the intersection of all of them and we know that the section of two subgroups is again the subgroup and it's easy to see that this also holds for the intersection of any number of subgroups and so we have this we have this subgroup this is called the subgroup generated by u, so this is so in particular for finally many elements if say g1, g k are some elements of g we can consider the subgroup generated by them, so this is just I mean it's denoted maybe like this and it's just defined to be the subgroup generated by the set of these elements, so this is the smallest subgroup of g which contains g1 to gk, you can check as an exercise but it's a kind of standard that if we look if g is equal to z with the addition then if I take the subgroup generated by 4 and 6 this is equal to the subgroup generated by 2 which after all is equal to 2 times hope that's correct yes and you can easily see how you might want to generalize this to any numbers okay so that's maybe enough for for today, so I mean I don't know you can maybe you didn't make very many comments in between so I hope it was clear to you and you can maybe can if it's too fast or too slow you can tell me now or if there's anything that is not clear you can also think until the next time whether you want to ask me some questions so I think tomorrow we meet early no is that correct what at four ah although that's good because I'm actually not a very happy early riser um okay so tomorrow we meet at four and maybe I will try by tomorrow to have some some homeworks which then you have one week to do there should be some elementary things to check with these things so I mean I hope that you know that you find what I told you until now quite elementary and kind of easy maybe it's a bit faster than usual but it's things that you normally would already know and you know basically I've given you some simple definitions and I have given you straightforward applications mostly have a bit many definitions but you're already familiar with them you know with time obviously we it will get slightly more advanced and then you have to you know maybe make a bit more effort but in some sense we do start from zero and everything that is being used kind of is explained here anyway okay thank you for today