 In this video, I want to state improve the first isomorphism theorem of group theory. And in the next videos for lecture 34, we're actually going to state improve the second and third isomorphism theorem. So take a look for those. So before we do that, I first wanted to find what it means for a function to be natural. Now, to be more precise, I should be doing this in the context of homological algebra, which I don't want to do that in this video, because that kind of goes beyond the scope of this first semester course in group theory here. But we can't define at least what it means for a homomorphism from a group to a quotient group, what it means to be natural in that context. So let n be a normal subgroup of g. So n is just any normal subgroup of g. Well, since n is normal, it makes sense to talk about the quotient group g mod n. That would be well-defined. And so then what we're going to do is we're going to construct a homomorphism from g to a quotient group. All right. And in the literature, this is commonly referred to as eta, the Greek letter eta right there. And so really we don't know anything about the group g except that it has a normal subgroup. Because it has a normal subgroup, we can form a factor group. But in terms of how do we how do we define any formula between these things? And it turns out it's very natural, right? And this is kind of the name, the idea behind the name here. It's natural to send an element to the coset that contains it. Because, you know, if we don't know any of the structure of, if we don't know any of the structure of the group, how in the world would we decide where things go? So the natural choice is to send g, little g to the coset that contains little g. Now this is in fact a homomorphism. It's going to be well-defined. And so notice what happens here is if you take eta of gh, and I guess I shouldn't say, there's really nothing to worry about being well-defined because g itself, it's just an element. We're sending it to its coset. So to show that this is a natural map, we just have to show that it's homomorphic, right? So eta of gh, what does that mean? Well, you're going to, that'll just be gh of n, right? So it sends it to the coset that contains gn. Well, since this is a normal subgroup, we've seen previously that gh of n will become g times n and h times n, which is just eta of g and eta of h. So this is a homomorphism called the natural map or sometimes called the canonical map, right? And the natural map gets its name because it really is the only natural way to send an element of g into g mod n without knowing any more about the group structure. That's what I was trying to say earlier. Note also that the kernel of the natural map, right? So if we look at the kernel of this map, the kernel of eta, this is actually equal to n itself, right? Because the things that map to n, like if, for example, x was nn when this happens, if you take eta of x, this would be x of n, x times n, excuse me, which is just equal to n, which is the identity inside g mod n, like so. So the natural map is a map whose kernel is the normal subgroup n, but n itself is sort of just chosen arbitrarily, just an arbitrary normal subgroup. Thus, every kernel of a homomorphism is a normal subgroup. We saw that in the previous video, but also every normal subgroup is the kernel of a homomorphism. So in this way, normal subgroups and kernels of homomorphisms represent the same collection of subsets of a group. Alright, so let's get to the main idea for this video, the so-called first isomorphism theorem for group theory. Some things I want to mention before we dive into the statement and proof of the first isomorphism theorem. First of all, some textbooks refer to this as the fundamental homomorphism theorem. Because, well, whenever you use the word fundamental to describe a theorem, it's meant to be a big, stinking deal, right? It's important, right? And I would absolutely say that the first isomorphism is an extremely critical, extremely important theorem in the development of group theory, absolutely. I should also mention that I keep on saying the first isomorphism theorem of group theory for two reasons. Well, the name first suggests there's probably some other ones. And as I already mentioned in this video, we'll talk about the second and third isomorphism theorems later, not in this video. But the first isomorphism theorem of group theory is that as we talk about group homomorphisms, we have this property stated right here. But as you talk about rings and modules and other algebraic sets, it turns out that there's a corresponding first, second, and third isomorphism theorems for rings, modules, etc. And it's going to be the exact same statement, the exact same proof, where you just change the appropriate parts as you go from groups to rings to modules to vector spaces to Boolean algebras, whatever you want. And that's actually because this is sort of like our first look into something called universal algebra. As you work between, as you move between algebraic categories, what things remain true. And the isomorphism theorems actually are true statements between all of them. And that's because the proofs are of a very natural state. All right, so we'll talk, we'll talk about that thing here. So the first isomorphism theorem, what does it state? It says if psi is a group homomorphism from G to H, we get where G and H are groups. And let's say that the kernel of psi is the normal subgroup N inside of G. Then, so we have a group, we just have a group homomorphism. That's really all the assumption says the fact that it's kernel is called N is just because we want to call it something for simplicity. So we have a group homomorphism, then there exists a unique homomorphism fee that goes from G mod N to H, such that psi will factor as fee composed with eta, where eta is the natural map. And so a special case of this homomorphism factorization tells us that in particular the map fee, which goes from G mod N to the image of psi, is an isomorphism. And this is often how people think of the first isomorphism theorem, that's why I get its name. The first isomorphism theorem tells us that G mod N is isomorphic to phi of psi, right? Which remember, since N is the kernel of psi, often people, in other words, people say the following. A group mod out its kernel is isomorphic to the image of that same map, right? So G mod the kernel of homomorphism is isomorphic to the image of the homomorphism, which is a very, very important thing. We'll see some examples of this in just a second. Now, I want to explain a little bit what this picture here on the screen is describing. So this diagram is what's referred to in the literature as a commutative diagram. A commutative diagram. These things are ubiquitous in homological algebra. And the reason they get their name is because all the paths from G to H are equal to each other. So it doesn't matter how you go there. So if you want to go from G to H, you could go there through psi. You could also take the detour through G mod H. And it's the fact that it doesn't matter which path you take. Following the arrows, it's all the same. So that's the example of a commutative diagram. The dashed line right here, this dashed line for phi implies that phi is uniquely constructed from psi. So the idea is when you start this process, you have just just have the group homomorphism psi. Well, you always have in play the natural map. And so the idea is with psi, you can uniquely construct phi. Hence the dashed line there. So it's constructed but in a unique manner. There's only one phi that'll make this diagram commute. All right. So let's get to the proof right here. So how are we going to construct this map phi that I'm talking about right here? So we're given psi. We have eta. What can we do? So we have to construct this map phi that goes from G mod N to H. And it turns out there's really only one natural way to get from G mod N to H. Because we really know nothing about G and we know nothing about H except for a few things. We know that G is a group. We know that H is a group. And we know that there's a homomorphism from G to H. But that homomorphism could be the trivial map for all we know, right? Psi could just send everything to the identity. That might not mean much, right? But the only way we know is that G homomorphically maps into H, okay? And because G is a group and psi is a homomorphism, it does have a kernel, right? That kernels will be a normal subgroup. So the factor group is well-defined. We have that. So we don't really know much about it other than those sort of like those things that have to be true about groups. And so we don't really have a lot of options. And so the really only one option we can do to define the map psi, excuse me, to define the map phi, is to define phi of GN to be psi of G, right? So if we take a little element G inside of big G, right? We know that old map over here to H as psi of G. We have that. Well, what happens to, if we want this diagram to be commutative, what happens to little G? Well, as we move it over here, we're going to go to eta of G, which is just its coset GN. So to finish this, we basically have to send GN to psi of G. That's kind of like the only option if we want this thing to be commutative. When I was in graduate school in a homological class, I had a professor who would always refer to this idea as you have to follow your nose. And that's to say that you can't like the fog is so thick in front of you that you can see nothing except for the nose on your face. And in homological algebra, you do these so-called diagram chases that we're doing right now. You just have to kind of follow what's in front of you. You can't see where you're going, but there's really only one path to follow. So you just follow your nose here. And that's what we're doing. So phi of GN has to map to psi of G. But the concern here is we're defining a map on a coset based upon its representative. Is this thing going to be well-defined? That should be our concern whenever we define functions on quotient groups based upon the representative of the coset and not the coset itself. So let's first check to see if this thing is well-defined. Let's take an element h, which is inside of the same coset GN. And so that means there's going to exist some element N inside of the kernel so that h equals GN. So we want to consider what happens in this situation. So if we take phi of hN, this will map to psi of h based upon the rule that we have right here, which since psi, well, excuse me, since we have the formula h equals GN, psi of h will be equal psi of GN. But psi is a homomorphism. It's a group homomorphism. So we can factor as psi of G and psi of N. But what do we know about N? N was the kernel of psi. So the image of N, whatever N turned out to be, is going to be trivial. It's just going to be identity. And if you multiply identity times anything, you just get back that element. So we're going to see that psi of h is equal to psi of G. But psi of h is just phi of hN and psi of G is just phi of GN. So this does tell us in fact that phi is a well-defined map. It's independent of the representative. But is it a homomorphism, right? So phi here is the only map that makes this thing work. That makes this diagram commute. But we want it to be a homomorphism. We've established that phi is a function. It's well-defined. But is it a homomorphism? Well, to see the homomorphic property, we see the following. All right. So note here that it's also a homomorphism. If we take two cosets of G mod N, so we'll take GN and HN, like so. And so multiply them together by coset multiplication, phi of GN times HN becomes phi of GNH, which that means that'll map to psi of GH, which as psi is a group homomorphism, you're going to get psi of G times psi of H, which then psi of G is equal to phi of GN and psi of H is equal to phi of HN, like so. And so this gives us that this thing is homomorphic, right? Phi of GN times HN is equal to phi of GN times phi of HN. So this is a homomorphism from the factor group to H. Notice also that by construction, psi of G, which is equal to phi of GN, that's how we defined it above right here, this is good. Well, G of N, G time then, excuse me, is eta of N, which then by function composition, we get exactly this, right? Psi of G is the same thing as phi composed with eta of G. And as the element G was arbitrary, this tells us that psi is equal to phi composed with eta. And so this then proves, whoops, coming back to our picture right here, this then proves the main statement that given any group of homomorphism psi right here, there exists a unique homomorphism phi such that this diagram commutes, sometimes it does a little circle here to show that the diagram is commutative, psi equals phi composed with eta. Now I want to prove this special case of this, right, about the isomorphism. So coming back down into our proof here. Let's suppose, oh, I guess I take that back. What I want to do next is I want to prove that this map phi is unique, all right? So we've constructed a homomorphism, but we want to prove that it's unique. That phi is the only, is the only map that makes this diagram commute. So let's suppose we have another map, let's call it chi. chi goes from g mod n to h. And suppose that, suppose that psi is equal to chi composed with eta. Now we don't have, we don't have cancellation when it comes to functions, right? Like if this was a group where we'd be like, oh, okay, let the chi composed with eta is equal to phi composed with eta. Well, I'm just going to cancel eta and get that chi equals phi. Well, we do that like in a permutation group, right? Because the permutations are bijections, right? We can cancel them. Eta is not an injective map. It would only be injected if the kernel was trivial, which would only happen if it's, if basically, if, what I'm trying to say here is that only in the trivial case would we be able to cancel. So we don't have cancellation here. So we'd have to prove something else in order to get that phi and chi are the same thing. So what we're going to do is we're going to look at the following right here. Take phi of gn, right? By construction, phi of gn is equal to psi of g. Well, psi of g, since that equals chi of eta, well, we can use that factorization here, chi of eta of g. Well, eta of g is just gn right here. So we get that phi of gn is equal to chi of gn, where gn is just an arbitrary coset inside of g mod n. Well, since they agree on arbitrary elements of the domain, that means the two functions are actually the same. So we get that phi equals chi. So phi is the unique map that makes the above diagram commute. All right. So now let's get to this last statement about isomorphism here. So finally, we show that, so I want to show here that, because after what's our goal at the moment, we want to show that g mod n is isomorphic to the image of psi. Like so. And so how are we going to do that? Well, in order to do this, that doesn't look like a psi, try that again. In order to do that, we actually want to argue that phi is going to be an injective map. It's going to be injective. The reason why that's important is that we want to show that from g mod n to this thing right here as we go into h, right? This map phi might not be surjective. We have no idea. But even though it might not be surjective, it can be surjective. You've kind of restrict the code domain. What do we mean by that? Because if it's not surjective, it means there could be things in h that we're missing. But there's nothing in the image of psi that we're missing because by definition, the things in the image are the things that something landed on, right? So you can't miss anything that you hit. You can only miss the things that you miss. And so that's why we kind of replace h instead with the image of psi. So that will automatically be surjective, right? Every function maps its domain onto its image. So to show that phi actually serves as a group isomorphism from g mod n to the image right here. To show that, we need to show that this function is injective. That's what we're left to do. So to do that, we're going to show that the kernel of phi is trivial. Now, you don't want to get confused here. Now n is the kernel of psi. And now we're trying to show that the kernel of phi is trivial. And so it's just going to be the subgroup n. That's all it's going to be. So let's suppose that phi of g n maps the identity of h. So this is the identity of h right here. Let me apply that psi of g maps to the identity given how we defined phi above. Now, if psi of g equals the identity, that means g is inside the kernel of psi. But n is the kernel of psi. And so that tells us that little g is inside of n. So that g n is actually just in itself. It's the identity. And thus that proves for us that the kernel of phi is just the subgroup n itself. So phi is an injective map. Since you're always surjective onto your image, this tells us that phi is an isomorphism between g mod n and the image of psi. I'm not saying that g mod n is isomorphic to h. I'm saying that g mod n is isomorphic to its image. And so this gives us the first isomorphism theorem. Every quotient group is going to be isomorphic to the image of the associated homomorphism. And so we saw at the beginning of this lecture, at the beginning of this video, excuse me, that homomorphic kernels and normal subgroups are really one of the same thing. What the first isomorphism theorem is telling us in a nutshell is that factor groups and homomorphic images are one of the same thing. Because every factor group is isomorphic to the image of the homomorphism and vice versa. Let's see a few examples, and then that'll be the end of our video here. So I'm going to look at some examples we've seen in previous videos here. So one example we had is if you take the infinite cyclic group z and map it to z n, we're going to use the natural map right here where you just map an integer m to the integer m mod n. So you might reduce it down. Clearly this is going to be a surjective map and the kernel of this map is going to be nz. That is multiples of n will map to zero when you reduce mod n here. So the kernel is going to be nz. So the first isomorphism theorem, let's call this thing right here psi. So this map here is psi. Then what we see is that the kernel of psi is equal to nz. And the image of psi is equal to z mod n since it's an onto map, it's a surjective. So the first isomorphism theorem tells us that z mod nz is isomorphic to zn. But this is actually the simplest proof to argue that z mod nz is isomorphic to the cyclic group of order n because it's just a very quick consequence of the first isomorphism theorem. And honestly in practice the most simple, the most effective and the most common way to prove an isomorphism other than just constructing the isomorphism explicitly. And my opinion is to use the first isomorphism theorem that you show the two groups are isomorphic by taking if you want to show that two groups are isomorphic you basically take a homomorphism from a third group and argue that the quotient of that group is the same as the surjective that is you take an epimorphism and surjective homomorphism and argue through kernels that the two things are isomorphic. We'll see how to do that when we prove the second and third isomorphism theorems later on. And so let's do one more example for this video here. Let's take the map psi which goes from the real numbers under addition to the circle group under multiplication and let's this map psi here should be psi that's a typo psi of theta is going to equal e to the 2 pi i theta like so. This is going to be a surjective map it's going to map on to S1 because every complex number of modules one can be written explicitly in this formula. The kernel psi right here is just going to be the integers because notice how I already stuck in this factor of 2 pi so if I stick any integer in there we're going to get a multiple 2 pi and as this number is equal to cosine of 2 pi theta plus i sine of 2 pi theta. We're going to see that this thing will equal this will equal 1 exactly when we have an integer in for theta. So the kernel of psi is going to be z and this shows us that our the integers is isomorphic to the circle group and there's nothing particularly special about z right here because I kind of insert this 2 pi if you take the real numbers and you mod out by any discrete subgroup you're going to get the circle group we see right here so our mod z is isomorphic to the circle group and then similarly if you take q mod z right so z so S1 right here the circle group this is all of the complex numbers whose modulus is equal to 1. q mod z on that hand so you take the subgroup the subgroup the rational numbers inside of the reals if you take q mod z this will be the isomorphic to the group of all roots of unity that is those complex numbers which some integer power will give you the identity which there are many things in here that don't have that property but you get q mod z which is actually the torsion subgroup of S1 right here but that's a topic we can talk about at the other time