 And so we have been talking about groups. So we just write some of the words that we had so far. We have talked about groups. Then we talked about subgroups. And so if we have subgroups, we have the decomposition of the group into cosets. And we had looked at normal subgroups. And if we have a normal subgroup, we had seen that if we take the quotient, so the set of cosets, then we get that the quotient g mod h is a quotient group, is a group. So now we want to talk about the next kind of concept with a group homomorphisms. So it's a common thing in mathematics. You have some set with a structure, like a group, or a vector space, or topological space. And then the way one of the main things you use to study such sets of structure is the homomorphism between them, so the maps between such sets with a structure which are compatible with the structure. And so for a group, the structure is that you have this product. So a homomorphism of groups will be one such that the product goes to the product. So that's very simple. So definition. So we take two groups. So then a group homomorphism from a to b, so f e from g h is a map which sends the product to the product. So with phi of a, phi of the product a times b is equal to phi of a times phi of b. So here the product is in g, and here the product is in h. So this is very simple. So for instance, I mean there are many examples of group homomorphisms. So for instance, if we have the map, the multiplication by k from z to z, which sends, so n t l z is z we take with addition. So we just write like this. So this is a group homomorphism. Because obviously this says that if we take, so the map is an integer n is sent to k times n, and k for k some integer. And then it's clear that if I take, I mean it's a homomorphism because if I take n plus m times k, everybody knows, but this with the flow that this is k times n plus k times n. So this is a simple example of a homomorphism. So now let us, so there are some simple remarks about this. So first, a group homomorphism sends the neutral element to the neutral element and the inverse element to the inverse element. So let z phi from g to h be a group homomorphism. And say e, the neutral element of g and e prime, the neutral element of h. Then we have the phi of e, z to e prime. And phi of a to the minus, so phi of a to the minus 1 is equal to phi of a to the minus 1. For all a in a, for all a in g, well, that's really, maybe I do not explain this because it follows directly. So if you want this in exercise, but it's kind of tricky. And then another remark is that if phi from g to h and psi from h to l are group homomorphisms, then their composition is a group homomorphism. And again, this is direct from the definition, but maybe if you take, so certainly you can take the composition. And then if you take psi composed is phi of a for any a in g. This obviously is psi of phi a. And now if I do this for a product, well, then I first, it's psi is a group homomorphism. So phi is a group homomorphism, so it pulls out like this. And then we do the same again. OK, so these are completely straightforward things. Then if we have a group homomorphism, we have two kind of subgroups, one in g and one in h, which come with it. One is the kernel and one is the image. So the kernel consists of all the elements which map to the one in h. And the image is obviously the image. And so you can look at it. So definition. So we take group homomorphism again, group homomorphism. And then we have, so the kernel of phi is the curl of phi, which is the set of all elements which are mapped to the neutral element here. So this is the set of all a in g, such that phi of a is equal to 1. It's 1, I denote the neutral element in h. And the other thing that we have is the image. The image is just what the image is. So phi, the image of f, is equal to the set. So phi is the set of all phi of a, where a is an element in g. So this is the subset of g and this is the subset of h. And now these are obviously not just subgroups, but not just subsets, but they are subgroups. And in fact, we will see that the kernel is even the normal subgroup. It is quite simple. So all this lemma, the answer integration, phi from g to h is the group homomorphism. Then the image of phi is the subgroup of h. And the kernel of phi is a normal subgroup of g. Now in fact, we will see in a moment that in some sense, the normal subgroups or it is true that the normal subgroups are precisely the kernels of homomorphisms of g to some other group. And somehow the second statement is more important than the first. But in any case, they are both quite simple. It's just checking the definitions. Well, so to be a subgroup, we have to see that the product of any two elements and the inverse of any element lies in it. So if h1 equal to phi of A and h2 equal to phi of E are elements in the image of phi, we have to see to show that h1, h2, and say h1 for the minus 1 are in the image of the inverse of any element and the part of any two elements. Well, and this kind of trivial, no, because h1, h2 is equal to phi of A times phi of E equal to phi of AB. And so it's in the image after by definition and h1 to the minus 1 is equal to phi of A to the minus 1. And we have seen, well, actually, I didn't say it, but it's trivial and you can check it. It's also in the image. So let's see for the kernel. That's not much more difficult. So we have to see if two elements lie in the kernel. Then the product lies there and the inverse lies there. That's to be a subgroup. So first we check it's a subgroup. So let A and P elements in the kernel phi. So that means that there are elements in G and we have phi of A is equal to phi of B is equal to 1. Then obviously, phi of AB is equal to phi of A times phi of B which is equal to 1 times 1 equal to 1. And if I was saying with the inverse, phi of A to the minus 1, by what we have seen, that's phi of A to the minus 1. And A maps to 1. So it's 1 to the minus 1, which is 1. So this is all very, I mean, in the future, I will not always do such three things, but it's still beginning. So anyway, one can see I'm just using the definitions. The last one is it's a normal subgroup. So what does that mean? So it means that if we assume, so let H be an element in the kernel of P and A be any element in G, then we have to see that if we conjugate by it, so if I write A to C, A H A to the minus 1 is in the kernel. Well, we just write it down. So we take phi of A H A to the minus 1. This is by the fact that it's a homomorphism, phi of A, phi of H, phi of A to the minus 1 is in the kernel. So this is 1. We know that this is the inverse of phi of A. So this is phi of A times phi to the A of minus 1 is equal to 1. OK, so we see, so anyway, so it follows directly from the definitions that the image of phi is a subgroup of H and the kernel of phi is a normal subgroup of G. As usual, when we have homomorphisms, we have isomorphisms. So a different definition, so homomorphism phi from G to H is called an isomorphism if it is projective. So it's easy to check that if phi is an isomorphism, then its inverse map is also homomorphism that follows. So if phi is an isomorphism, then phi to the minus 1, so just the inverse map, is also a homomorphism therefore an isomorphism. Phi to the minus 1 is the map, so it's with the composition, which is the identity. And it's also easy to see that, so we can say G and H are called isomorphic if there is an isomorphism from G to H. I may be right, denote this G is isomorphic to H. And it's straightforward that this is an equivalent selection, because obviously G is isomorphic itself by the identity. If G is isomorphic to H via phi, then H is isomorphic to G via the inverse map of phi. And the composition of isomorphisms is easily seen to be an isomorphism. I mean, it's the composition of homomorphisms and then if both of them are bijective, the composition is perfect. So that's true. And so if one wants, one could say that one aim of group theory could be to describe all isomorphism classes of groups. It's a somehow impossible task. But that could be one thing that one could be trying to do. So just as a remark, so if phi from G to H is a homomorphism of groups, then we know that the image of phi is a subgroup of H. We just said it a minute ago. And so one can always view phi as a subjective group homomorphism phi from G to the image of phi. So therefore, when we can, in some sense, by replacing H by its image, always assume that our group homomorphism is subjective. And so in order, so if the morphism is injective, then it will be an isomorphism onto its image. And so it's somehow important to know what is the criterion for a homomorphism, a subjective homomorphism, to be an isomorphism. And we can see this in terms of the kernel of the map. So the homomorphism will be injective if and only if its kernel consists only of the neutral element. So lemma get phi from G to H, a group homomorphism. Then phi is injective if and only if the kernel of phi is equal to 1, where 1 is the neutral element of G. So in particular, if I have a subjective group homomorphism from G to H, it will be an isomorphism if and only if its kernel consists only of 1. So this is all, I think today we don't really have any real proof. So but we can look at this elementary argument to kind of practice. So we assume phi is injective, and we want to show the kernel is equal to 1. So we denote, say, by E, the neutral element of H. We have already denoted by 1 the neutral element of G. So obviously, we know that f of 1 is equal to E. That's true for any phi of 1 is equal to E. That's true for any group homomorphism. So thus it follows that the kernel of phi contains 1. We have to see the other inclusion, that every element which is mapped to the neutral element is 1. So f is injective, but we have assumed, thus it follows. So if it's still called phi, phi is injective, thus phi of A is different from phi of 1, which is E, for A different from 1. In other words, the kernel is precisely 1. So this is kind of stupid. And now the other direction, which is so to speak the non-trivial direction, but it's not very non-trivial. So we assume the kernel is 1. So let's take two elements which are mapped to the same element. We have to see that A is equal to B. So say we multiply by phi of B to the minus 1. So it follows that phi of A B to the minus 1 is equal to phi of A phi of B to the minus 1 is equal to 1. Because if I multiply this side by phi to the minus 1, I get 1. And so as the map is injective, as the kernel is equal to 1, we get that A B to the minus 1 is in the kernel. So it follows that A B to the minus 1 is equal to 1. So that means that B to the minus 1 is the inverse of A, which is the same as saying that A is equal to B. OK, so the map was injected. OK, just let's look at one example of an isomorphism. Which I need as an exercise. So let's say f from m to n, the bijection of sets. Then say, whatever, I write f star from the symmetric, so the set of permutations of the set m. So the bijection of m to itself is equal to the same thing for m is an isomorphism. So which map? So this means, so if I have such a map, so assume an element here I call sigma. This is mapped to what is it? f to the minus 1, composed of sigma, composed of f. So we have a map from n to itself. Let me see whether it's correct. So I want to apply it to an element in m. So I actually don't do that. So I try again. So I take an element in here. So a map from, I don't know, from, so let's just see. I mean, maybe I write the diagram and seem to be a bit stupid. So we have, I mean, we have here this f goes from m to n. And assume we have here this map, I think it's just this map sigma from m to itself. And we want to find here we have f again n. And so we just want to, instead of doing we associate to this, as soon as sigma is here. So to the sigma, you apply this thing where we first apply, no, it was correct. So we first take an element of n, apply f to the minus 1. We go here, we apply sigma, and we apply f again. And that's what it is. So whenever we have a projection from m to itself, we give us another projection here by doing this kind of conjugation. This gives us, so this claim is an isomorphism. It's easy to check. It's a homomorphism because you can check it's a homomorphism because if you have the composition of sigma and tau, you can put f to the minus 1 in the middle and you have this. And it's an isomorphism because obviously you can, if you do the same with f to the minus 1, this will give you the inverse map. This is an isomorphism inverse. This sigma goes to f of minus 1 in sigma. So in particular, if m has finally many elements, then we get that the symmetric group, so the set of permutations of m is isomorphic to the symmetric group in as many letters as m. So the permutations of the set 1 to m, 1 to the number of elements of m. OK, so these are simple examples. I didn't do all the details, but you can check it. So finally, we want to see that if we have a, so we want to come to the fact that the normal subgroups are precisely the kernels of group homomorphisms. And then you can factorize through the kernel of homomorphism, so we'll see that in a moment. So let me write this down. So first, this is very easy. Lemma, let G be a group. Or maybe I will first kind of say, so let G be a group. And say n in G be a normal subgroup. Then we know that G mod H is a group, no? G mod n is a group with the obvious map. And so we have we'll call pi from G to G mod n, which is the map that we had before we associate to an element G here, the map Gn. This we call the canonical protection. And it's kind of clear from the definition that this will be a group homomorphism, because we have precisely defined the group structure here in such a way that it's a group homomorphism. So this is, we have the lemma, so G is a group and is a normal subgroup, the natural map from G to G mod n is a subjective group homomorphism with kernel. And so in particular, we see that the normal subgroups are precisely the kernels of group homomorphism. Because we know that the kernel of a group homomorphism is a normal subgroup. And on the other hand, this says that if we have a normal subgroup, we find a group homomorphism of which it is the kernel. But there's not very much to say here. It's basically all by definition. So first, by definition, we know that if A and B are elements of G, then pi of A B is equal to, so I'll just put it like this, pi of A times pi of B is equal to A n times B n. This was how it was defined, so pi of A is A n of B n. And the multiplication was defined precisely in the way so that this is A B n, which by definition is equal to pi of A B. So this says it's a group homomorphism. We have precisely defined the structure of the group on the quotient group so that this becomes a group homomorphism. And then so what's the kernel? We have that A is in the kernel of pi, if and only if its image, if A times n is equal to n. A times n consists of all, so this means in particular that A, so that means that A times n is an element in n for all n in n. So for instance, it follows that A times 1 is an element in n, so A is an n. And conversely, if A and n, then certainly A times n, because n is a subgroup, A times n is an n for all n in n. OK, so this is very simple. So thus, we find that the kernel of pi is indeed equal to n, because it consists for these A such that A times n is equal to n. OK, so now we come to something which is not more difficult, but still I call it the theorem. So the homomorphism theorem, which says that if we have a surjective group homomorphism, then we have some, then we can divide by its kernel and get an isomorphism. So we can factor out the kernel. This theorem, I mean, it's a big, much to call a theorem, but it's often called homomorphism theorem. Sometimes first isomorphism theorem or second isomorphism I never remember, but in Germany it's called homomorphism theorem, which says that to let phi from G to H be a surjective group homomorphism with kernel k, then there exists an isomorphism, say phi bar, from G mod k to H such that if I take this map, we first apply, so if we take phi, this is equal to applying phi bar after the natural projection pi. So in other words, we have this diagram. You can either take phi directly to go to H or we can take the natural projection to G mod k and apply this map phi bar here and this diagram commutes. So we get the same whether we go use this map or the composition of these two. So in some sense, this statement says that if you want up to this isomorphism here, a surjective group homomorphism is the same as the up to composing going isomorphism. It's the same as not just dividing by a normal subgroup. So we can always make a surjective group homomorphism into an isomorphism by dividing by the kernel. So let's see how this goes. It's actually quite simple. So we see that there exists such a map. In fact, it's also unique, but this which satisfies this. And so the only reasonable thing how we can try to prove it is to use this as a definition for phi bar and then see whether it works. Maybe so just to claim it. So in particular, we have that h is isomorphic to g model of the kernel of phi. So let's prove it. We want that phi is equal to phi bar composed with pi. We want to find the phi bar with this property. So the reasonable thing is to use this as a definition. So what does it mean? So that means for all elements a in g, we should have that phi of a is equal to phi bar of t of a, which is just phi bar of a plus a times k of this. So the natural projection is that. So that means the reasonable thing to do is to say that we define phi bar of a k to be phi of a. So this formula forces us to make this definition because we have required this to be true. And so we make this definition. And now obviously we have to check whether that makes sense. So the first thing we have to check whether it's well-defined. Because we have defined here phi bar on an equivalence class in terms of the representative a. So it must be well-defined. It should be independent of a. So first, phi bar is well-defined. So what does it mean? So in other words, if a k is equal to b k, then we need, we have to show that phi of a is equal to phi of b. But that's kind of trivial, no? Because so what do we have? So that a k is equal to b k means that there is an element of k such that b is equal to a times this element. So a large k equal to b large k means there exists an element. So a small k in large k such that a is equal to a times k. So then what is phi of b is equal to phi of a times k. It's equal to phi of a times phi of k because it's a group homomorphism. And so this is phi of a times 1 equal to phi of a. So that's fine. So we have indeed seen that if a k is equal to b k, then phi of a is equal to phi of b, so it's well-defined. And the second statement is that it's a group homomorphism. But that's basically trivial, too. Because we have made our definitions in this way. So phi is homomorphism, phi bar is homomorphism. How does it go? Well, if I take phi bar of a k, say, times b k, then according to our definition, so first this is equal to phi bar of a b times k because that's how the group structure goes. And this we had said is phi of a b equal to phi of a times phi of a. And according to our definition, phi of a is equal to phi bar of a k. And that shows it's a group homomorphism. And it's clear that phi bar is more subjective because the image, after all, is the same as the image of phi. Because we send phi of a k to phi of a. So we get the whole image phi bar of a k. So the image is the same. And we also want to show that phi bar is injective. So in other words, we need to show that the kernel of phi bar is equal to the unit element in this quotient that's 1 times k, so to the neutral element. But by definition, we have that element a is in the kernel of phi, while if and only if phi bar, so a plus a k is in the kernel of phi, if and only if phi of a is equal to 1. Because after all, that was the phi bar of a k is equal to phi of a. And so this is if and only if a is an element of k. And this is equivalent to a k is equal. OK, so the kernel is just this. So we get this isomorphism. And by definition, by our definition, we know that this holds. Because we have precisely said phi bar, we have phi is equal to phi bar. OK, so are there any questions or comments until now? So let's do some simple example. Let say g be a cyclic group of order k. So the number of elements is k. And then g is isomorphic to z mod k times k is some positive integer. So it's quite clear. We take a generator of g. So that means it's an element such that we get g by just taking all powers of a. So that means that g is equal to the set a to the n, n, and z. And so we know I can maybe define a map phi from z to g, which sends an element n in z to a to the n. Then we have seen in the first lecture that I have a to the n plus m is equal to a to the n times a m. So that means this is a group of morphism. And this statement here says it's a subjective group of morphism. And what is the kernel? We have actually seen before that the order of g in this case is equal to the order of a, which is equal to the minimum over all k, a to the n is equal to 1. So we see that k is an element in the kernel of phi, and there's no smaller n bigger than 0 that is in the kernel of phi. So k is the minimal positive integer of this property. On the other hand, we see that if k is in the kernel, then also all multiples of k are in the kernel. Because if I have a to the k is equal to 1, then a to the 2k will be 1 times 1 and so on. And on the other hand, if this was not equal, I would find a smaller element. And it's easy to see. So you can find a smaller positive element in kernel of phi. But just if n is in the kernel of phi, then you see from this that a to the k is also equal to 1. So you can multiply by any power of a to the k. So it follows that n minus d times k is in the kernel of phi for all d and z. And so we can do division with rest. So if n is not a multiple of k, you find that there exists a d such that n minus dk is bigger than 0 and smaller than k. And this is a contradiction to our choice of k. OK, and so we have that the kernel is k times z. And so in other words, we find that we have a subjective homomorphism from z to the cyclic subgroup generated by a to the g, whose kernel is k times z. So g, yes? You cannot see it. Otherwise, OK, so it was not a mathematical question. OK, because otherwise, we have this. So thus it follows that g is isomorphic to z mod kz. Because we have a subjective homomorphism from z to g, whose kernel is kz. OK, now we want to say, well, it's not very, want to say a little bit about automorphisms. It's not very, so the g is a group. So first, an endomorphism of g is a homomorphism of g to z. It is an isomorphism if it's an isomorphism. Now we'll see in a moment that the automorphisms of a group form by themselves a group. And we describe some of these automorphisms. So I will actually denote out of g to be, in the moment, the set of automorphisms of g. Oh, yeah, that would be a bit tautological, but anyway. In fact, it wouldn't make sense. Yeah, isomorphism. So now, first I have to remark, we want to see that the automorphisms of g are a group. So I claim that out is a group. It is a subgroup of just the bijections of g towards z. So subgroup of s of g. Remember that s of g was just a set of all maps which are bijective. So in order to check this, we have to see that the composition, so the group structure here of the group of permutations of g is just bi-composition. So we have to see that the composition of two automorphisms and automorphisms, this is obvious. We know that the composition of isomorphism is an isomorphism, and it goes from g to d, it goes from g to g, and that's it. And the inverse of an automorphism is an automorphism, again, that's obvious. OK, so this is clear. We have already proven it. So now we want to look at a special case of automorphisms which are the so-called inner automorphisms. So we get somehow, to every element in the group, we can associate an automorphism of the group by kind of what one could call a conjugation by that element in the group. So this is the following statement. So let g be a group, and a g. So for every, actually, so first we note that the map, which I call tau a, from g to g, which sends an element b to a, b, a to the minus 1, is an automorphism of g. So first, it's obviously a homomorphism, because if we have tau a of b times c, this is no, yes. So if you have two elements b of c of g, then it should be compatible with product. If you have two elements b of c of g, then it should be compatible with product. So this, by definition, is a, b, c, a to the minus 1. And then we can multiply suitably with 1, a, a to the minus 1. And so this is tau a of b times tau a of c. So this shows that this is a homomorphism. And it is bijective, because we can immediately see what the inverse is. So if we take tau a to the minus 1, then this is, by the same statement, this is also a homomorphism of g to itself. And it's clear that tau a, that this will be equal to the identity and conversely. I mean, obviously, you just write it down and you see me here, this will be the case. So this is indeed an automorphism. And then, so an automorphism is called an inner automorphism. So the p from g to g is called an inner automorphism if it is of this form. So if phi is equal to tau a for some a in g. So I want to briefly describe what the inner automorphism is. So one can maybe see, for instance, if, for example, if g is a vegan, is commutative, then the inner automorphisms are just the identity. So I maybe should say I denote by in of g the set of inner automorphisms. Now we want to show that if g is commutative, then obviously the set of inner automorphisms of g consists just of the identity of g. Because as g is commutative, we send b to a b a to the minus 1, which is the same as a a to the minus 1 b, which is b. So every b is sent to itself. OK. So now we want to, so we first want to claim a proposition that the set of inner automorphisms of g is a normal subgroup of all the automorphisms of g. Well, this is actually more remark. We are making remark. So we just check the definition. So we have to see it's a subgroup and it's a normal subgroup. So we just check the definition. So the inner automorphisms are precisely those which are obtained in this way. So if we take, so we have that it's a subgroup, we have to show, so we have that if a b and g, then tau a and tau b are inner automorphisms. So we have to show that their composition is an inner automorphism. And also that tau a to the minus 1 is an inner automorphism. But so far we have already seen that tau a to the minus 1 is equal to tau a to the minus 1. So this is certainly an inner automorphism. We just saw it a minute ago here. And if we, so otherwise we have to see if we take tau a composed with tau b, to apply to some element x, so x is an element of g, then what is it? So it's tau a of b x b to the minus 1 by definition, which is a b x b to the minus 1, a to the minus 1. And we know that this is the inverse of a b. So in other words, this is tau of a times b. So this is also an inner automorphism. And then finally we want to see, so this shows it's a subgroup. Actually, I need to have whatever. And now I want to show it's a normal subgroup. So that means if we conjugate with any element in with any automorphism, it should still, we should still get an inner automorphism. So let phi from g to g, no g, be an automorphism. We have to see that if we take phi tau a phi to the minus 1, this is an inner automorphism. So for all a in g, well, we can just compute what it is. So same as before, we see what it does to any element. So well, so let x be an element in g. So we take phi tau a phi of x, phi to the minus 1 of x. So this is, so first we apply phi to the minus 1 of x. Then we apply tau a to it. And then we apply phi to it. Now phi is a homomorphism. So you can pull out the product goes to the product. So this is phi of a phi of phi to the minus 1 of x. Phi of a to the minus 1, which is the same as phi of a x phi of a minus 1. In other words, we have tau of phi of a. So this is also an inner automorphism. And finally, we want to give an explicit description of the inner automorphism as a quotient of the group itself. So definition, so we have g is a group. So the center of g, which you also have in an exercise, is, I think I use, I don't know if I use the same notation, is c of g, which is the set of all elements in g, such that a g is equal to g a. So I think you are required to prove that that's a normal subgroup of g. Now, unfortunately, I will, from the proof that I give, this will also follow, but maybe you can try to give a different proof. So proposition, we have that c of g is a normal subgroup of g. And the group of inner automorphisms of g is equal to the quotient of g divided by the center. Isomorphic, whatever I am. Well, so what I will do is I, so I just have to give a subjective homomorphism from g to the inner automorphisms whose kernel is c of g. Then this proves at the same time that c of g is a normal subgroup and that this identity holds. So proof, well, and we know what the homomorphism is. After the inner automorphisms are given, by sending an element a in g to tau a. So tau, which sends a to tau a, is hopefully a homomorphism. From g to in, which sends a to tau a. So the claim is, we actually have seen, I just wiped it out, but it's trivial. We have seen it, we have seen that tau of a b is equal to tau of a composed with tau of b. Hope that's correct. So we have that tau is a homomorphism. So now we only want to, we have to show it is, by definition, it is subjective, because the inner automorphisms were precisely those automorphisms which could be obtained as tau a for some a in g. So by definition, it is subjective. So the only thing we have to check is what the kernel is. So the claim is that the kernel of tau is equal to the center of g. And then by the homomorphism theorem, we have this. So assume we have an element a, which lies in the kernel of tau. So by definition, this is if and only if, if I take tau a, this is equal to the identity of g. This is equivalent to saying that for all elements g and g, we have a g a to the minus 1 is equal to g. So tau a is the map which sends g to a g a to the minus 1. And we want this to be the identity. So that's what this, and this is equivalent. We can always multiply on this side by a. That for all g in g, we have that a g is equal to g a. And this, by definition, means that a is in the center. So we see, therefore, that the kernel of tau is the center of g. And therefore, indeed, as the kernel of a group homomorphism, it's a normal subgroup. And the inner automorphisms are g divided by the center of g. So it was just some, it's more than propositions and so on. It was just examples of the definition. And so maybe that's enough for today. Next time, I will talk about actions of groups on sets and what one can see with that. OK, thank you.