 So let's finally see where this is going, cosets lead us to Lagrange's theorem. And if this is the first video that you see, you're not going to understand much. Start with the first one, I'll link to it down below. I'll put it up in the corner there to start with a video on cosets. Before we get to Lagrange's theorem, let's consider partitioning of sets. Remember if I have this finite set and I can, it's almost like a Venn diagram, partitioning means I can subdivide it into subsets such as if this was a set S and this was my subset S1 and make it Si and Sj etc. That if I take the intersection between these, those are always the empty set, they are disjoint. So if I can divide a set into these disjoint subsets, I have partitioned that whole set. So just remember partitioning. And what we are suggesting here with Lagrange's theorem, so let's just say Lagrange's theorem, Lagrange, Lagrange's theorem is that we can take the set that makes up a group and we can partition it into a subgroup. So let's imagine that's the whole group. I'm going to suggest that we can have this, that's the whole group G, the set that makes up the group G under some binary operation. I can partition it into a subgroup and all the cosets and all its cosets. Now it's going to be either all its left cosets or all its right cosets. Remember the two situations where we have what, when the left and right cosets are equal, there's videos for that. Remember that would be when we have a normal subgroup and when we have an abelian, when obviously we have an abelian group where we can do, when the commutivity applies. So the Lagrange's theorem, that's not what the Lagrange's theorem says, but what we are claiming here is that we can partition the set that makes up a group. Now let's just consider just the left ones for the moment. I'm going to have AH and that is going to be all these, remember such as H is an element of H, that's my left coset. Now imagine we are just going to, we've seen what happens when A is an element of H, but now let's just consider A is not an element of H. So A is going to be not an element of H and we've shown that then it's not, you know, if it's not then it's not going to be there. So I have all these left cosets there. And I'm suggesting that there's another element B and that it's, if I have an element B, then it's not going to be an element of this coset. It's going to be this element. Now if that is so, just think about it logically, that means that I have these two cosets and their intersection is going to equal the empty set. They're going to be this joint. So I'm stating that there's going to be this other element B to make another coset and that these cosets are this joint. That means I'm going to have this specific thing here if I can show that to be true. If I can show that there's an element B that's not in, you know, it's not one of these. Remember this is going to be one element. B might be one of the elements. I might have A O H, A not O, it's binary operation with H being B. But imagine then B equals B is not in there. I can find a B that's not in there. That means that I'm going to be able to create another subset, another coset and that coset is completely this joint from this coset. So how can we prove that? We can prove that by suggesting that this empty, this empty, this is not an empty set. That the intersection is not empty. That there is some overlap there. And if there's some overlap that would mean, you know, I'm suggesting here that B is an element. B is an element of this left coset. That there is such an element. And if that is so, you know, then this is not the empty set. In other words, if I take this left coset B, it's going to equal one of these that are in there. But I've got to make it some other H so that I'm not simply just sticking to one of the Hs. Remember if we had Cayley's, if we had Cayley's table and it's H means I've got to cycle through all of the Hs. Not cycle through, but it's A times the first H, A times then by times, I mean binary operation, the second H, the third H, the fourth H, whatever. And I don't want these to be exactly equal. I want to be totally arbitrary about this. Now, if this is so, I'm stating now to the contrary that this is not the empty set. So I can find this B that's an element of A H. And that means B is going to equal to H inverse. If I take H inverse on both sides, by associativity I can do these two. But remember this and this, they are both elements of H. So what I've just written here, this is an element of H. So what I've written here is B is by contrast, I said that it's not. It is an element of H. And we said that it's not an element of H. So what we've shown here is by contradiction that there will be this B with forming a left coset that will be quite distinct from, will have a non-empty set, will be disjoint from this coset. And so on and so on and so on, I can do and I can find another one C arbitrarily to show that it would be disjoint from this one, which is disjoint from this one. In other words, I'm going to find this very specific partitioning of the set G that makes up the group G. By the subgroup H and by all the cosets. It's easy, it's easy to prove by contradiction here. Now think about this though for a moment now. We know that if the cardinality or how many elements there are in the subgroup H, that is going to equal how many. And we saw that because remember if we had this coset, it'll have exactly the same number of elements, we've proven that, it'll have the same as that. And I've just shown that these must, there are some distinct ones. They are totally disjoint but they must all have, I'm going to call them AI, they must all have the same cardinality. And if they are disjoint and I add all of them up, imagine that there are H with all of these. Imagine that there are M of them. There are M or let's just imagine that the cardinality or the number of elements in there is M. So there's got to be M times N of these where N is how many there are. 1, 2, 3, N is an element, N is an element of the natural numbers. So imagine this one was 5, the net one must be 5, the net one must be 5, the net one must be 5, that one must be 5. So it's just so many multiples, N multiples of 5. And that brings us to the Granger's theorem only for finite sets. We can't do this for infinite sets. For finite sets is that the number of elements I've just made up the whole of G, that must be some multiple of the cardinality of the set of the subgroup, the set that makes up the subgroup. A better way to write this is that this divides G, the cardinality of G. Okay, it must divide. So if I have that H, if I have that, say for instance, G is 7, you know, this is a lovely thing. So say for instance G is 8. I cannot make up a subgroup H that has 3. There is 3 elements in it. There's so many subgroups I can immediately throw out. The subgroup must have the identity element. All the properties of groups must be there, but there's an easier way to get rid of possible subgroups is if the cardinality of that subgroup does not divide the cardinality of the group from which it is taken, then it's gone. And it's so clear and easy to see from this logical argument. This is just a simple fact. So I can really take out many of these. They're beautiful, so Lagrange's theorem is very powerful for us in that way. And another thing to think about this is what if G is, the cardinality of G is a prime number. Now we've just seen, so the only thing that can happen here is it must be either equal to, H must equal G or H must just have a single element and that is the G's identity element. So those are the two trivial subgroups of G and that's also quite nice, a nice thing to immediately see. So if there are a prime number of elements inside of the set that makes up the group, then I can only have the two trivial subgroups. The two trivial subgroups. And there's one more beautiful thing that comes from this and that is the real power behind this, the real power and it's an isomorphism. If I take X an element of G and I make up the cyclic group of G, I make up a cyclic group with an element of G and I let G, the cardinality of G be a prime number. So that is going to equal X to the power 0, that is just the identity element of G and then X just itself, then X binary operation X, then X binary operation X, binary operation X, etc. And it means I've got to stop some way because it can only have this subset, can only have this cyclic subgroup that I'm making, the cyclic subgroup that I'm making, can only have either, because this one has a prime number of elements, that means this cyclic subgroup can only have either the one and it's only this one or it must have the same elements of this. So if that's a prime number and let's make that prime number P, it means I can only go up to X P minus 1. I can only go up to X P minus 1 because together with this one it will give me P elements and if I go to X to the power P, that will automatically be X to the power 0 again. And that means the cyclic or groups with prime number elements in it, they are isomorphic and if that prime number says P, that is isomorphic to the cyclic group with that number of elements. And that's a very easy group to understand the cyclic group, very easy to understand group and with finite number of elements and that is isomorphic to this group, these set of groups with prime numbered elements inside of them. And that means it's very easy to understand all of them because they are isomorphic to the cyclic group with a prime number of elements. So all in all we've combined everything that we've learned, start with the first video on cosets, you won't just understand what's going on here, but see how logically we can just let our minds flow and all this flows logically from there, which gives us this very nice idea here, the Granger's theorem, let's remember it like this, and which gives us this isomorphism with a cyclic group with prime elements if we just consider groups with a prime number of elements, the set that makes up the group with prime number of elements. Fantastic stuff you've got to agree.