 In this final video for lecture 32, I actually wanted to prove some properties of homomorphisms, that is, these are functions which preserve the group operations. We go from one group to another. In this situation, we'll call it the map phi. It'll be a map from G to H right here. Now, if the function preserves the multiplication, right? So we see that phi of A times B is equal to phi of A times phi of B. That's what we mean by this homomorphic property, preserves the operation. What else can we guarantee that it's preserved, right? What type of group structure is preserved? So the first principle that we're gonna prove here is that if E is the identity of the group G, then phi of E will actually be the identity of the group H. So homomorphisms always preserve the identity. And I should mention that every isomorphism actually is a homomorphism as well. Isomorphism is just bijective homomorphisms. So this also tells us that an isomorphism will send an identity to an identity. Homomorphisms will send inverses to inverses, right? So phi of G inverse is equal to phi of G and then inverse, right? So this will send an inverse to an inverse. Isomorphisms would have to do the same thing. If we have a subgroup K inside of G, then phi of K will be a subgroup of G, okay? And so a homomorphism sends subgroups to subgroups. Now if K was a normal subgroup in G, then phi of K will be a normal subgroup in the image of phi, which is often just called phi of G right there. Now I can't claim that phi of K will be normal inside of H, but it will be normal inside of the image of the homomorphism. And then the last one, it turns out that pre-images work a lot better than images when it comes to basically everything, but particularly homomorphisms. If you have a subgroup of the co-domain, then it's pre-image, which remember what that is. The pre-image, we're looking for all the elements G inside of G, such that phi of G lives inside of K, right? So the pre-image is all the elements inside of the domain that map into this set K. And it could be empty, right? That's a possibility. Now I mean, not for subgroups of course, because the identity will map to the identity, which is inside of K. But these things, K could be this very large set, but the pre-image could be trivial for all we know. But the point is, if you take a subgroup of the co-domain, then the pre-image will be a subgroup of the domain. And if K is normal in H, then the pre-image will be normal in the domain, which is pretty impressive. So let's take a look at these results right here. Let's prove property A that identities are preserved. So let's suppose that E prime is the identity of H, okay? And we think that phi of E is the identity of H. That's what we're trying to prove right here. So we have to show, in fact, that E prime is equal to phi of E. That's what we're trying to do right now. So let's start off with the element phi of E for a moment. Well, since you're in the group H, right? This thing belongs to H. We could times that by the identity phi, or E prime there, times phi of E. So that clearly would just give us back phi of E, right? Because E prime is the identity of H. Now, on the other hand, if I look inside of the function, right? E is the same thing as E squared, because it's the identity of G, E times E would be E. And then by the homomorphic properties, it becomes phi of E times phi of E. So compare these two statements right here. You have E prime times phi of E. You have phi of E times phi of E. Since we're in a group, we can cancel on the right, and we get that E prime is equal to phi of E. That then proves that we want. So phi of E is equal to the identity of H. The identity always maps to the identity. Let's move on to property B, which says that homomorphisms preserve inverses, excuse me. So what we have to show is we have to show that phi of G inverse is equal to phi of G, and then that's raised to the inverse, right? So we're looking for the inverse of phi of G. So what we need to do, kind of backing up for a second, is we want to show that phi of G times phi of G inverse is equal to the identity E prime right there. If we can do that, then since inverses are unique, if this element walks like an inverse and quacks like an inverse, then it must be the inverse. So we're gonna multiply phi of G times phi of G inverse, like we see right here. But since we have a homomorphism, we can bring the two elements inside together so we get phi of G times G inverse, which as this is an element, this is here in G, in capital G, G times G inverse will become the identity, which is a phi of E, which we mentioned a moment ago, phi of E is equal to E prime, the identity of H. And so this tells us by the uniqueness of inverses that phi of G inverse is the inverse of phi of G. This proves property B. Now we move on to property C. Remember, we wanna show that the image of a subgroup is a subgroup. So let's take a subgroup K inside of G, all right? Well, phi of K, by definition, this is gonna be a set of all phi of X's where X was inside of K. That's just the definition of this thing. So we have to show that this thing is a subgroup. So it has to be closed under multiplication, it has to contain the identity, it has to be closed under inverses. Let's, well, okay, K contains the identity because itself is a subgroup. And remember that E prime, the identity of H is equal to phi of E. But phi of E is inside of phi of K because E was inside of K, so therefore we get that's inside of there, right? So the identity is inside of phi of K. So we got the identity, great. Let's show closure under multiplication. If we take two arbitrary elements A and B inside of phi of K, what does it need to be inside the image? That means there are some elements X and Y inside of K such that A equals phi of X and B equals phi of Y. So when we multiply these together A times B, well, this would be phi of X times phi of Y by the homomorphic property phi of X times phi of Y is equal to phi of X, Y, which this right here is an image. It's an image of an element, right? Now since K is a subgroup, X, Y is inside of K because X and Y are there individually. K is itself closed under multiplication. So since X, Y is in K, then the image of X, Y is in the image of K. So that shows that phi of K is closed under multiplication. How about inverses? Well, if A inverse, you know, let's take that element, A inverse, since A remember A equals phi of X right here, A inverse will equal phi of X to the inverse, which that is the same thing as we proved earlier, it's just phi of X inverse. As K is a group and X is in K, that means phi or X inverse will be inside of K, which means phi of X inverse will be inside of phi of K. So phi of K is closed under inverses, and since we're closed under identity, the multiplication and inverses, this proves that phi of K is a subgroup of H. That was the first part of C that we wanted to prove. The second part is we want to show, we want to show that phi of K is normal inside of phi of G. Now we're not necessarily saying that's the same thing as H. It might not be normal on H, but we can say that if K was normal inside of G, then the image of K will be normal inside the image of G, which might not be all of H. So to do that, we have to show it's closed under conjugation. So take an arbitrary element of phi of K that looks like phi of X, where X is something inside of K, and then take a typical element of phi of G or the image of phi, where G, little G here is just an element of G. So we take a typical element of phi of G, we take a typical element of phi of K, and we wanna conjugate these things. So we're gonna take phi of G times phi of X times phi of G inverse, all right? Well, by the homomorphic property, this thing is gonna be equal to phi of G times phi of X times phi of G inverse, like so. And then using the homomorphic property again, we can bring this all together, and we get phi of GXG inverse. Now, since K is normal inside of G, that means GXG inverse is inside of K, all right? So the conjugate will be inside of K since it's closed under conjugation. But this is an image element, right? It's the image of something in K, so that means it belongs to something in, this element will belong to phi of K. And so therefore, if you conjugate something in phi of K by something in phi of G or the image of phi, you'll get something back in there. So that proves them property C, like so. The last part, D, we're gonna see here. So D is about pre-images, right? So we wanna show that if K is a subgroup of H now, let's talk about its pre-image. With the pre-image, remember its definitions right here, we're looking for all elements of G so that their images, the image of that element X is inside of K. So what can we say here about this? Notice that phi of E, which we proved earlier, is equal to E prime. E prime is the identity of H. Since K is a subgroup of H, it'll have the identity. So phi of E is inside of K. That means the identity is inside of the pre-image. So it's closed under identity. Okay, let's show that the pre-image is closed under multiplication of a homomorphism. So let's take two elements in the pre-image. X, Y is in the pre-image of K. So what that means is to be in the pre-image that phi of X is inside of K and phi of Y is inside of K. But K is a group, it's a subgroup. So the product of any two elements in K will be in K. So phi of X times phi of Y inside of K, because it's a subgroup. But the homomorphic property tells us that phi of X times phi of Y is equal to phi of X, Y. So phi of X, Y is inside of K, which means that K, Y is in the pre-image of K. So pre-image of K is closed under multiplication, it was closed under identities. We just have to do inverses to show it's a subgroup. So same thing, let X be inside of the pre-image of K. That means phi of X is inside of K. But like we saw earlier, the inverse of phi of K is actually phi of the inverse of X, okay? And so since K is a subgroup, the inverse of an element is inside of K as well. So if phi of X is in K, then the inverse of phi of K is in there. So that tells us that phi of X inverse is in there. Well, by the pre-image then, that tells us that X inverse is in the pre-image of K, for which since X was in there, we see that pre-image of K is closed under inverses. So therefore the pre-image of a subgroup is a subgroup of the domain. All right, so now the last condition about normality. Suppose that K is a normal subgroup of the co-domain. We then wanna show that phi inverse of K is closed under conjugation. So take an element X inside of the pre-image and then take any element inside of the group, the domain. We'll call little G here. So we have to show that GXG inverse is inside of phi inverse of K. This is what we want to establish right now, okay? So what we're gonna do is we're gonna take the element GXG inverse, we're gonna hit it with phi and see what happens. So you get phi of GXG inverse. Well, by the properties, the homomorphic property here, phi of GXG inverse will become phi of G times phi of X times phi of G inverse, which that inverse in the last spot comes out and becomes phi of G to the inverse. That's gonna be inside of K because K is a normal subgroup of H, right? You took something that's inside of K. How do we know that? Well, since X is inside of phi inverse of K, that means phi of X will be inside of K and then you conjugate it by something that's an H, right? Phi of G is inside of H. Since K is normal, this will be inside of K, all right? And so since phi of GXG inverse is inside of K, that means that GXG inverse is inside of phi inverse of K, which then proves that we have a normal subgroup. So in summary, homomorphisms, these are functions that preserve the group operation between groups. Automatically, if it preserves the group operation, you'll preserve the identity, you'll preserve the inverses. The image of a subgroup will be a subgroup. The pre-image of a subgroup will be a subgroup. And the pre-image of a normal subgroup will be a normal subgroup.