 In this video, we're going to prove our final theorem of lecture 34, which is sometimes entitled the fourth isomorphism theorem. More often than not, I feel like it's called the correspondence theorem, which that's what the label we're going to take here. And the correspondence theorem, this is the correspondence theorem for the sake of groups in other algebraic categories like rings and modules and such. There's also a corresponding correspondence theorem. What the correspondence theorem says is that it essentially homomorphisms preserve haze diagrams. So the statement is actually sort of a mouthful itself. So let n be a normal subgroup of g. Then there exists a one-to-one correspondence, which is where it gets its name. There's a one-to-one correspondence between the set of subgroups of g. You might call this the lattice of subgroups of g containing n and the set of subgroups of g mod n. So just so you're aware of this set, we don't consider every subgroup of g. We only consider those subgroups that contain n. So if you take the set of subgroups of g containing n and you take the set of subgroups of g mod n, then there's a one-to-one correspondence between those, the bijection. And the bijection is simply you take the subgroup h, which h is a subgroup of g, but it also contains n. And you're going to map this over to h mod n, which h mod n is going to be a subgroup of g mod n and clearly it will contain n mod n, which n mod n is just the trivial subgroup of g mod n. And so that's the bijection. So furthermore, this correspondence restricts to a one-to-one correspondence between the set of normal subgroups of g containing n and the set of normal subgroups of g mod n. So this correspondence, there's a one-to-one correspondence between the subgroups of g containing n and the subgroups of g mod n. And when you restrict your attention to normal subgroups, that's still a one-to-one correspondence. That is, normal subgroups will map to normal subgroups here. All right, so how are we going to accomplish this? So this map where you send the subgroup h to h mod n, this is essentially just the natural map, the quotient map, eta, which eta will send g to g mod n. So remember what this thing does is you have an element g, which will map it over to its coset g n. That's the natural map. So this is a homomorphism. So if h is a subgroup of g, wherein is a normal subgroup of that, I want to remember here that we've proven previously that the homomorphic image of a subgroup is a subgroup. So phi of h is, in fact, going to be a subgroup of g mod n, which is the co-domain in that situation. But what is g mod h? g mod h, because it contains n, this is just going to be h mod n. That's what we want to try to convince ourselves of, that why is this correspondence legitimate here? So h mod n is a subgroup of g mod n, and eta, since it's a homomorphism, in fact, every subgroup of g mod n is of this form, of this form right here, since eta is surjective. Therefore, the natural map establishes a correspondence between the subgroups, the subgroups of g containing n and the subgroups of g mod n, namely this one right here. So this is really the kicker, that because n is a subgroup of h, then phi of h is actually identical to just h mod n, like so. Let's see, so furthermore, we want to show that this is one-to-one correspondence. So this says that, okay, we can send h over to h mod n, the eta does exactly that, but that just, it's a function relationship. Is it in fact a bijection like we claim it is? Well, let's establish a inverse function. So furthermore, h is the pre-image of h mod n. How do we know that, right? So we know that if you have a homomorphism, pre-images are subgroups. So the pre-image of a subgroup h mod n is going to be a subgroup of g, okay? It's going to be a subgroup that does contain n. So n is going to be contained inside of there. And h maps onto that thing here. And so when you do these things back to back, right? So we see that h is going to map over to h mod n, which then maps over back to h when you do the inverse here. And so because these, because this correspondence has an inverse and necessarily has to be bijective, so it's a one-to-one correspondence. And so finally, if h mod n is normal inside of g mod n, we have seen previously that pre-images will send normal subgroups to normal subgroups. So the pre-image h mod n has to be normal in g, but that pre-image is h itself. So if h mod n is normal, then h will be normal. In the other direction though, right? This is the one we don't prove in general. If h is a normal subgroup of g, then h mod n, which is the image eta of h, I claim this will be normal inside of g mod n. This follows from the fact that eta here is going to be a surjective map. So maybe we see a little bit more detail on what's going on there. Why is that? So if we take g n and we times this by h n and we times that by g n inverse, this is going to equal g h g inverse n, which this element right here, since n is normal, g h g inverse will be inside of h. So this thing is going to live inside of h n, which h n is the image of eta h right there. These things are the same thing. So in fact, we do get normality. And so the correspondence theorem then tells us that normal subgroups will correspond to normal subgroups. I want to provide an example of this and we're going to look at a picture to help illustrate this. So the groups are going to play here. We're going to take the symmetric group S4 and we're going to consider the normal subgroup V4. So this is our g and this is our n going on in this situation here. So there are six subgroups of S4 which contain V4. That is S4 itself, V4 itself, the alternating group A4 will contain V4 as well. The other three groups are going to be groups which are dihedral groups. That is these are groups that are isomorphic to D4, the symmetries of the square. So one of them is going to be the subgroup generated by 1, 2, 3, 4 and 1, 2, 3, 4. So there's a four cycle and then a 2, 2 cycle. Another dihedral group will be generated by 1, 3, 4, 2, the four cycle and 1, 3, 4, 2, the 2, 2 cycle. You may be able to see the pattern we're doing here. And then the third dihedral group will be generated by the four cycle 1, 4, 2, 3 and then the 2, 2 cycle 1, 4, 2, 3. So those three groups are all going to be isomorphic to each other. They're isomorphic to D4 and they all contain the Klein 4 group. The first three groups are normal while the dihedral groups are non-normal. So S4 of course is normal inside of itself. V4 is normal inside of S4 and so is the alternating group. So these green groups you can see right here are normal subgroups that contain V4. The Ds are not. The Haase diagram for S4 and for S4 mod V4 which is isomorphic to S3 are illustrated below. So you can see here these are the subgroups of S4 that contain V4. This right here is the entire Haase diagram for S4 mod V4 which S4 mod V4 is S3 right here. And so you're going to see the correspondence here. S4 mod V4 is going to be S3. A4 mod V4 like we saw before that's equal to A3 which is just a cyclic group of order 3. V4 mod V4 of course is just the identity and then these other groups when you mod out the Klein 4 group you're basically going to get something that looks like a 2 cycle. You get something like the cyclic subgroup generated by 1, 2. The cyclic subgroup generated by 1, 3. And then this would be isomorphic to I should say these are isomorphic to this is isomorphic to the cyclic subgroup generated by 1, 4. Like something like that. So this is the usual Haase diagram for S3. So we see the two pictures of the same. That's what the correspondence theorem is telling us. The correspondence theorem tells us that these Haase diagrams, what's going on there, these Haase diagrams are going to be isomorphic to each other. But in particular if we restrict our attention to the normal subgroups, in this case the Haase diagram of normal subgroups also is isomorphic to each other. And so this is what's guaranteed by that correspondence theorem. The corresponding Haase diagrams are the same.