 In this video we're gonna prove the second isomorphism theorem for groups. Analogous theorems can be stated for other algebraic categories like rings and modules, vector spaces and the like, but this video will focus on the second isomorphism theorem for groups. So let G be a group, this could be finite infinity group, doesn't matter. Let H be a subgroup of G and let N be a normal subgroup of G, okay? And so then the second isomorphism theorem under those assumption tells us that if you take the subgroup HN and you mod it out by N, this will be isomorphic to the group H modded out by H intersect N. So before we dive into the proof of this thing, let's make sure that this statement even makes sense because if we go about trying to prove nonsense, well, that's gonna be a big waste of time. Now note that by assumption we have that N is a normal subgroup of G, okay? So since N is a normal subgroup of G, this means that N contains all of the G conjugates of elements of N. So N here is closed under conjugation. In particular, N contains all of the H conjugates of elements from N. So let me elaborate a little bit more about that. So because N is normal in G, we know that for all say N inside of N and for all G inside of G, woops, we'll have that G and G inverse is inside of N. That's what I mean by it's closed under conjugation. But G, right, has H as a subset. It's a subgroup, but in particular, it's a subset. So if it's true that for all little G and G, it's closed under conjugation, I could restrict my set to something smaller, right? H will contain fewer elements than G. So if I restrict the set that I'm gonna conjugate by, then I'll still be closed under conjugation because we're a normal subgroup. So if N is closed under G conjugates, then N will also be closed under H conjugates. So that's an important observation to make right here. Also, well, I guess the conclusion we make from that is that notice N will be a normal subgroup of H times N. So we proved previously that HN is gonna be a subgroup of G. So first of all, HN is a group because H is a subgroup and N is a normal subgroup, okay? So it makes sense to talk about N being a subgroup of that. Now clearly, N, which is just equal to E times N, this is gonna be a subset of HN because H contains the identity. So the fact that N is inside of HN makes sense. As N is a group inside of another group, it's a subgroup, that's what that means. So the fact to say that N is a subgroup of HN follows from the fact that HN is a group. But the reason we can say it's a normal subgroup is this conversation up here that we're closed under conjugations of H. But in fact, I said it's closed under conjugations of H, we could also be like, oh, it's closed under conjugations of HN. If I take some element X, which is inside of HN, then you're gonna get these conjugations right there. So that's what I'm trying to say here, that is we restrict the group and it's still closed under conjugation in that regard. So summarizing what we're just saying there, we can say that N is a normal subgroup of HN. The reason this is significant is that this quotient actually makes sense. N has to be a normal subgroup of the numerator there, which it is, in fact, the case. So on the left-hand side, we have a legitimate statement. That's a legitimate quotient group. What about the other side? Since HN mod N is well-defined, can we say the same thing for H mod out H intersect N? Sure, we can. Well, from what we've also seen previously, we've seen that H intersect N clearly, if you take a set intersect another set, that's gonna be a subset of one of those things. So like H intersect N is a subset of N. But since the intersection of subgroups is itself a subgroup, we get that H intersect N will be a subgroup of N. And then by the same argument as we said before, H will be closed under H conjugates because that just comes from closure and multiplication in H. And as N is a normal subgroup, it'll be closed under H conjugates as well. So H intersect N will be closed under H conjugates. So in fact, H intersect N is a normal subgroup of H. We've proven this explicitly previously. So this tells us that in fact H mod H intersect N is also a well-defined quotient group. So that whole discussion is just to make sense that these two groups are well-defined quotient groups. And it turns out that the same quotient group up to isomorphism. All right, so how are we gonna prove this? So it turns out when one wants to prove that two sub, or two groups are isomorphic, there's essentially two strategies to do that. I may be saying there's three strategies, one proves this. So the first strategy is to actually design an explicit function, a specific isomorphism between the two. It's like, here's the formula for the isomorphism that shows the two groups are isomorphic. That's a nice strategy. Another one that's a little bit more difficult and this is the one I'm adding to my list of two that became three, is if you do have some type of classification theorem, kind of like the fundamental theorem of finite abelian groups, there could be like a certain sufficient collection of invariance that when they agree, they determine the group. And so you could determine that two groups are isomorphic without an explicit isomorphism because they get some classification theorem like the fundamental theorem of finite abelian groups there. That's an awesome nice strategy if you have such a classification theorem. Those were sometimes hard to come by. The third strategy, which actually I would think in practice is one of the most common ways to prove two groups is the isomorphic is use some type of universal mapping property. That is a condition such as the first isomorphism theorem that guarantees the existence of an isomorphism. In fact, perhaps like a unique isomorphism like in the first isomorphism theorem. If we can guarantee the existence of a isomorphism theorem of isomorphism then the two groups have to be isomorphic. The main difference here is that the universal mapping property might not make explicit what that is. The isomorphism is sometimes implicit. So what I'm trying to say in a nutshell is you don't wanna underestimate how powerful the first isomorphism theorem is. The second isomorphism theorem we're gonna prove using the first isomorphism theorem. And this is a very important strategy for a budding algebraist. How can you prove that two groups are isomorphic using the first isomorphism theorem? So remember what that says. The first isomorphism theorem tells us that a group if you mod out by the kernel of a homomorphism will be isomorphic to the image of that isomorphism of that homomorphism. So that's what we're gonna do right here. So what we're gonna do is we're gonna establish a homomorphism from H in to H mod K where K here is gonna be H intersected. So really what I'm saying is phi is gonna be a map from H in to H mod out H intersect in. Like so. So basically what I'm doing is I have two factor groups I wanna show are isomorphic to each other. So I'm gonna take the numerator of one of the factor groups and show that the other factor group is a homomorphic image of that numerator. Then we're gonna argue that the denominator of the first factor group its kernel is in fact equal to N. Then we conclude by the first isomorphism theorem what we want. So we're gonna try a homomorphic way of proving these two groups are isomorphic. So what's the map H in? Moving over to H mod K where again K is H intersect in here. Well essentially there's only one thing we really can do. If you take a typical element of H in and it'll look like little H times little N, I'm gonna map this over to the coset HK or again K is H intersect in. Okay, so let's first make clear that this is a well-defined map. So suppose we have two different factorizations of the same element of HN. There could be such an element, right? So let's imagine we have HN which is equal to H prime N prime or H and H prime are inside of H and N and N prime are inside of N. Well, if you take this equation here HN this is equal to H prime N prime. Let's multiply on the left side H prime inverse. H prime inverse here, H prime inverse in which we see that the H primes will cancel on the right-hand side. But let's also take N inverse of both sides so that the N's cancel on the left-hand side. This will show that H prime inverse H is equal to N, excuse me N prime N inverse. Well, the term on the left, this is inside of H because it's a product of two things in H, H is a subgroup. The terms on the right, this is gonna be something inside of N, N is also a subgroup, a normal subgroup mind you, but it's a subgroup. So product of things in N are gonna be an N. So we have something that's in H and in N. So this element, let's call it X just to give it a name. This element actually belongs to H intersect N AKA it's inside of K. So with that in mind, let's consider the image of H prime N prime. Well, by the definition of our map right here, H prime N prime will map to H prime K. But since X belongs to K, right? This right here is just K. Since X belongs to K, you could replace H prime K with H prime XK, right? X represents K just the same way the identity does. And then re-associating this thing we get H prime X, K, right? Essentially, if you have an element inside the group, the group can barf it out without changing the coset whatsoever. So you're gonna get H prime X, which look at X right here, right? X is equal to H prime inverse H. So if I take H prime times X here, you're gonna get H prime inverse times H. Then we're gonna see that H and its inverse will cancel out, H prime and its inverse will cancel out, giving us just an H. So we get HK, which is the image of phi of HN. So this tells us that the map phi is in fact well-defined. It doesn't matter on which factorization of the element you use. So now let's prove it's homomorphic. So if we take the product of two typical elements of H, the capital H, capital N, I'm gonna use HN and H prime N prime. So in this situation, we're not assuming that HN and H prime N prime are equal to each other. There's two arbitrary elements of our group. Well, because we are in the group, this product is gonna equal something in HN. And we saw earlier how this thing works, that you're gonna want to commute this N past this N prime, or H prime somehow. Because N is a normal subgroup, this is allowed, H prime will move to the left, but then the N might switch to a different element of N, capital N. Let's call that new element maybe little N, double prime, like so, for which then here we have something in H, we have something in N, in which case phi will then send it to the coset represented by that element of H, H times H prime, which as this is a coset, it's in fact, this is a coset where K is a normal subgroup of H intersect, excuse me, K is a normal subgroup of H. We get that H prime K is equal to HK times H prime K. You can factor it inside of the quotient group. Now, HK is just the image of phi of HN and H prime K is the image of H prime N prime. So we see in fact that we have a homomorphic map, all right? Then the next thing we need to do, now that we have a homomorphic map to show that it's surjective, this is kind of an obvious statement, right? Because remember phi is supposed to go from HN over to H mod K. So what we're gonna do is notice that you have the element H, like H is a subgroup of HN, right? Because you can just factor everything in little and H is just H times the identity. And so this right here is going to map H, HG, which is just H of course, it'll map down to HK, which if H is an arbitrary element of capital H, then HK will be an arbitrary element of H mod K. So we have that. It's gonna be surjective again, that's a triviality. So then the other thing we had to show, remember the other thing we wanted to show was that the kernel of this map is equal to N. If we can do that, we can invoke the first isomorphism theorem. So let's consider what elements HN will map to K, which is the identity element of H mod K. Well, if HN maps to K, I mean it should be mapping to HK, this would imply that H is inside of K, which remember K is H intersect N. Clearly H is inside of H, but this also tells us that H, little H has to be inside of N, all right? And so this is what then implies for us that the kernel of phi is gonna equal N. Those elements, those elements that live inside, and I mean I should also mention of course that, so if H is inside of K, right? Clearly N is inside of N, like so, this would imply that HN is gonna be inside of N, like we're trying to say here, which is a subgroup of HN. So this is where we get this inference that the kernel is going to be all of N. Again, as we observe right here, if you said N, N will just map to the K as well. And so by the first isomorphism theorem, we have this map phi, which goes from HN to H mod K, for which the kernel, the kernel of phi is equal to N. So the first isomorphism theorem tells us that HN, the kernel of phi, is gonna be isomorphic to the image of phi, which fill in the details there, we get HN, the kernel we've established was N, that's the left-hand side of what the second isomorphism says. And since the map is surjective, since the map is surjective, the image of the homomorphism is the co-domain, which we saw earlier was H mod K, which K was just a substitute. This is H mod H intersect N, like so thus proving the theorem. Let's look at a quick example of this real quick. I said quick twice there. So let's take as a group, the symmetric group on four letters, S4. Let's take the Klein-4 group, which is a normal subgroup in there. And so by the Klein-4 group, of course we mean you have the identity and you have the two two cycles. Well, let's see how those elements right here. That's what we mean by the Klein-4 group. And then let's take the subgroup H to be the subgroup generated by one, two, three and one, two. So you can really think of like, this is S3, this is the symmetric group on three letters sitting inside of the symmetric group of four letters. So H is a subgroup of G and as a normal subgroup. So notice those things have been established, right? H is a subgroup of G because that's really just saying that S3 can be viewed as a subgroup of S4. And we also have the Klein-4 group as a normal subgroup in S4. So we have the assumptions necessary for the second isomorphism theorem. Let's see what it tells us. Let's consider the set H in for a moment. We've learned previously that the set H in, the order of this subgroup is gonna be the order of H times the order of N divided by the intersection H intersect in right there. Well, H is a subgroup of order six, it's S3. We get this one right here and which is the Klein-4 group, it's order four. And then what elements are common to both the Klein-4 group and S3? That turns out just to be one. That is the identity, it's the only thing common. So you get six times four times one are divided by one. That gives you a subgroup of order 24, right? Because the intersection is trivial. Well, as S4 itself is a group of order 24, four factorial, the only subgroup of S4 that says order 24 is S4 itself. So this tells us that HN is equal to S4. So let's consider then these quotients. If I take S4 mod V4, so the symmetric group mod out the Klein-4 group. Well, S4 is the same thing as HN. The Klein-4 group is N. By the second isomorphism theorem, this will be isomorphic to H to mod H intersect N which H was the symmetric group S3 and H intersect N was trivial, right? So if you take a group and mod out the trivial subgroup, you're gonna get S3. Well, you'll get the original group back, which is S3. So this tells us when we put these things together that the symmetric group on four letters mod out the Klein-4 group is in fact isomorphic to S3.