 In this video, we're going to prove the third isomorphism theorem for groups. There's an analogous third isomorphism theorem for other algebraic categories, like rings and modules and the like. We're going to prove this in the context of the third isomorphism for groups. Now, it turns out that the third isomorphism theorem is going to be very similar to the second isomorphism theorem in the nature of its proof. I guess we should probably give the statement of this theorem first. Let G be a group where G could be any group, finite, infinite, doesn't matter. We're going to take two normal subgroups of G, so both H and N are normal subgroups of G. We're also going to have the stipulation that N is a subgroup of H, okay? Those are our assumptions. H and N are normal inside of G and N is also inside of H right there. So then what the third isomorphism theorem tells us, that if we take the quotient group G mod N and we mod it out by H mod N, which is a subgroup of G mod N, that's isomorphic to G mod H right there. And so this one kind of right here is like, when you look at it, like of a college algebra student, we're gonna walk in the room right now and be like, let's see, G divided by N divided by H divided by N. Okay, you cancel out the Ns and you get G over H. That makes sense, right? If you were to look at it like, I'm gonna take five sevenths. I'm gonna divide that by three sevenths. Well, we're used to, I'm gonna multiply by the reciprocal, right? And I get seven over three, then those cancel, I get five thirds, right? For arithmetic, that's exactly justifiable. It turns out we can do the same thing for groups in perhaps a less obvious way. So that's the statement of the theorem. Make sure that these groups actually are well-defined before we proceed here. So note that by assumption H is a normal subgroup of G, that's what we were assuming earlier. And as such, for all H inside of H and for all G inside of G, we have that G H G inverse is the side of H. So the fact that H is a normal subgroup means it's closed under conjugation. You can conjugate an element of H by any element of G and you'll stay inside of H. So if we were to restrict our attention, that is to say, since H contains all G conjugates of elements of H, then consider the following. What if we were to take G N times H N times G N inverse? So what if I look at a conjugate and view this as an element of the factor group G mod N, right? So H N, this is a typical element of the factor group H mod N, and then G N is a typical element of G mod N right here, which since these are normal subgroups, this is going to look like G H G inverse times N, which as G H G inverse is the side of H, this would be inside of H mod N. So what I'm saying here is that because H is normal in G, we're going to see that H mod N is normal inside of G mod N, right? That is if you take a normal subgroup and you mod it out by some other normal subgroup, you still have a normal subgroup. And that's of course why we need that N is inside of H, otherwise we might not be able to mod H out by N. In the next video, we talk about the correspondence theorem, we can talk about what happens when this assumption is dropped, but this is a legitimate, H mod N is a legitimate normal subgroup of G mod N. So this group makes sense, it's well-defined, and since H was normal in G, there's nothing wrong with the definition of G mod H right there. All right, so let's get to the proof. The proof of the third isomorphism theorem is very similar to the second isomorphism theorem in so much that we are going to use the first isomorphism theorem. When you want to prove that two things are isomorphic using the first isomorphism theorem, what you're going to do is you're going to first establish a homomorphism. So you want to come up with a surjective homomorphism, that's your goal here. So how are you going to do this? Well, we're going to establish a rule that takes something from G mod N to G mod H, right? So you'll notice that you're only going to grab, so you want to send the numerator, you want to send the numerator of the factor group to the other group right there. So that's what we're doing here. So G mod N is going to map to G mod H. And then after you do that, once you have this surjective homomorphism, what we're then going to do is argue that the kernel of this map, the kernel of phi, is in fact the denominator H mod N. And then by the first isomorphism theorem, we can see that these two groups are isomorphic. So what's the homomorphism we're going to use? So G mod N mapping to G mod H, what can we do? Well, a typical element of G mod N is just going to be some element G times N. A typical element of G mod H is going to be an element from G times the subgroup H right there. And so that's actually going to the rule, right? We're going to take the coset GN and we're going to send it to coset GH. So we're going to use the same representative and move it to this new coset, all right? But as this map is dependent upon the representative, we're concerned whether it's well-defined, all right? So what if we use different representatives here? So suppose H is another element of GN. Of course, G is in there, but what if H is a different one? And maybe I shouldn't have used H because that means the suggestions inside of H right there. We don't exactly know that. But anyway, suppose that H is something inside GN. That means there's some little N inside of capital N such that H equals G times N right here. We have that factorization, okay? Since N is a subgroup of H, we know that little N is inside of H because it was inside of N. So it's inside the bigger set as well. So when we do this, we're going to take phi of HN, that then goes to H big H, which then when you factor the little H, you're going to get GN big H. Well, because N is inside of H, we get that NH is just the same thing as H. Those cosets are the same. And so GNH is just the same thing as GH, which GH is the image of GN under phi. So this map is in fact well-defined. That's kind of a hard part. It wasn't even that hard though. The homomorphic part is going to be almost just the exact same thing. If we take two different cosets, I guess they could be the same cosets. We just take two cosets inside of G mod N. So let's say like GN and HN. Again, I'm not supposing H is inside of capital R. I'm not saying little H is inside of capital H there. But if we take phi of GN times HN, that product, well, the product of cosets will be GHN for which its image under phi will be GH times capital H. But as H is a normal subgroup, we can factor that coset as GH times HH. In which case the image or GH is the image of GN and HH is the image of GH right there. That's clearly a typo. That should be HH right there. Sorry about that. And in which case there you go. It's homomorphic. Is it surjective? It's kind of obvious, right? Because like we kind of said earlier, right? A typical element of G mod N just looks like GN. A typical element of G mod H looks like GH, right? So if you're like, how am I going to land on side of GH here? We'll just take GN and land on it. So the surjectivity is kind of clear. It's kind of like the fun thing about this proof. It's like at the very beginning there's like, well, what map can we come up with? This is kind of like the only one you could do. You have to follow your nose. But then everything just falls through with the proof. Everything works out perfectly. It's just so natural here. All right, so what about the kernel? What can we say about the kernel here? Now we have to check to use the first isomorphism theorem and get this last statement right here. We have to show that the kernel of phi is equal to H mod N. Because if that's true, then the first isomorphism theorem would show us that H mod N mod the kernel of phi is isomorphic to the image of phi, which of course, if the kernel is H mod N, we get that. And since the map is surjective, like we said earlier, since the map is surjective, we get that the image here is going to be the whole co-domain, which is G mod H. So if we can prove the kernel is H mod N, we're done. But this is also fairly immediate, right? Because the kernel is the thing that maps to the identity. What is the identity of G mod H? It's going to look like just G. So if some element GN, which is going to map to GH, if that's the identity, that suggests that G would have to be inside of H, right? Which if little G belongs to H, that would suggest that GH belongs to H mod N, like so. In which case, then it follows by the first isomorphism theorem. So consider quotients of the infinite cyclic group Z. So if we take the group Z mod MNZ and then mod that out by MZ mod out MNZ. So I want you to check to see these things make sense, right? This cyclic group, like all cyclic groups, they're a billion, so everything is normal. We have that MNZ is a subgroup of Z. But we also have that MNZ is a subgroup of M, which is a subgroup of Z, right? Because multiples of MN are multiples of M necessarily. So therefore, this is a legitimate factor group. But by the third isomorphism theorem, these things cancel out. And this should be Z mod MZ, which, as we've seen before, is isomorphic to ZM. This is the cyclic group of order M. So to give some understanding of what this group is even about and how we can interpret this in a different way, well, if you take Z mod MNZ, that really is just the group ZMN, that is the cyclic group of order M times N. Well, what if we mod this out by the unique subgroup generated by M, right? Because after all, that's what this group is. It's a subgroup generated by M. Well, since the subgroup has order MN, the GCD calculation is going to show that this subgroup generated by M is actually isomorphic to ZN. It's a cyclic subgroup of order N. And so this we can write as ZMN divided by ZN. Since cyclic groups have unique subgroups given their order, I could replace the cyclic subgroup generated by M with just the cyclic subgroup of order N. And there's no ambiguity in doing that thing right there. And so kind of what you would expect with the third isomorphism there, it's like, oh, the N's cancel out and we end up with this thing being ZM in the end. So that's kind of how you want to think about it. If you take a cyclic group and you divide it and you mod it out by one of its cyclic subgroups, you're going to get a cyclic subgroup of the corresponding order where you take the quotient of the orders. It's very, very nice. If you take a cyclic subgroup of order 36 and you mod out by a cyclic subgroup of order four, you're going to get a cyclic subgroup of order nine. That's what we get from the third isomorphism there. It's pretty simple when you say it that way. Let's take a look at our good old friend S4 again. Let's take the group G to be S4. We're going to take a normal subgroup A4 to be H. And we're going to take the Klein 4 group to be another normal subgroup, which sits inside of H. So this would be though. So we have S4, A4, and D4 right here. So if I take G mod H, for this example, that means S4 mod A4, and a symmetric group mod out, the alternating group is going to be the cyclic group of order two, so we get Z2. That's pretty nice. I mean, there's only one group of order two, so it's got to be Z2. If we take G mod N, that is if we take the symmetric group S4 mod out the Klein 4 group, when we prove by the second isomorphism theorem that this is isomorphic to S3, all right? And so then if we take H mod N, that is the alternating group A4 mod out V4, well A4 is a group of order 12, V4 is a group of order four, so when you mod it out, you're going to have to get a group of order three, which that's a prime number, this is only going to be, the only thing that could be Z3, which Z3 is also the same thing as A3, right? It's the cyclic subgroup of order three inside of S3. That's what we're trying to get right here. So we get the following calculations right here, okay? So now let's see what this says, what is the third isomorphism theorem say about this? If I take G mod N, mod out H mod N, what does that mean? So if you take S4 mod V4 and you mod out A4 mod V4, the idea is we should be able to cancel out the V4s and we end up with S3 over A3, which is Z2, which is the same thing as A4 mod, sorry, S4 mod A4. So we would expect this because that's what the third isomorphism theorem is trying to tell us here. And so it's a fun little example to demonstrate what this theorem does for us.