 In this video, we are going to prove the very infamous Lagrangian theorem. Lagrangian theorem actually has a lot of different versions you could say, because when one gives a name to a theorem, the idea is that this is sort of a big deal. We want to be able to reference this in the future. And so you could be like, theorem 6.2.1 in this lecture series math 42.20 for, you know, we could try to reference it internally to the textbook or the course we're in. But what happens when the term is over, the semester's over, you get a different book or whatever, no one's going to be referencing those numbers all the time. A theorem that's important needs a name. And oftentimes, it's named after a mathematician who either originally proved it, or more often than not, who originally published it. You know, it's not always the case that the name is the right name, right? And Fair Mog gets named, lots of theorems are named after Fair Mog, for which I don't know if he can be proven to have proved. I don't know if he proved any of them, who knows. He certainly did not prove his so-called last theorem. But of course, he was right that the proof would not fit inside the margin. So I'll give Fair Mog a little bit of credit in that regard, I suppose. But Lagrange's theorem is a very important theorem for group theory, particularly it's useful in finite group theory. And so what I'm trying to say is, when referencing a theorem, if it has a name, it's a big deal, you should know it. But also, when it comes to a theorem with a name, there oftentimes will be many corollaries that follow said theorem. And each and every one of those theorems might actually be referenced as Lagrange's theorem. So in this video alone, we're gonna improve Lagrange's theorem as we have stated here on the screen. And we're also gonna prove three corollaries to Lagrange's theorem very quickly. And in the future, if I ever need to use any one of those four results, I will refer to it as Lagrange's theorem, not the corollary of Lagrange's theorem. And this is because Lagrange's theorem is not necessarily a unique statement in group theory, it's sort of an idea of theorems. And so anything connected to what you see now on the screen is often referred to as Lagrange's theorem. So what we are gonna do are the version of Lagrange's theorem we are gonna prove is that imagine we have a finite group, which we have a finite group G and we have a subgroup H, we are gonna prove that the order of G is equal to the order of H times the index of H inside of G. Now I wanna point out to you that the word finite here could be dropped and this statement would still be true. And we'd have to adapt the proof though and start talking about some infinite cardinals and things like that, which although we can do, it goes a little bit beyond the scope of this undergraduate sequence in abstract algebra. So we're gonna prove it just for finite groups for which case we've basically already proven it. The proof goes by the following. G is partitioned by left cosets. We've argued before that cosets form a partition of a group. Left cosets or right cosets doesn't matter which one you wanna use. The G is partitioned by the left cosets. So that's the first part. So what that means is that G is gonna look like a disjoint union of all these cosets, like so. So it's gonna be a union of all of these cosets. And so we wanna find some, we don't wanna say wanna repeat cosets. So what we wanna do is basically the following. You often get something called a transversal. A transversal is gonna be here a set of representatives, representatives of all the distinct cosets, okay? So for example, if we looked at a previous example that we've done here, T for example, the group Z6 with the subgroup H, which is the cyclic subgroup genera by three. We saw that there were three cosets in that situation. There was H, there's one plus H and there's two plus H. So as a transversal of these cosets, we could take the set zero, one and two. So that's an example of a coset, a transversal of the coset, excuse me. If we take the group S3 and we take H to be the alternating subgroup, then we have like H and one, two, three we could take as a transversal. We often like to take the identity as a transverse inside the transversal. So one and one, two, three would be such an example. And then in contrast, if we take the subgroup this time to be K, then we have three cosets. And so we could take as a transversal the identity one, two, three and one, three, two. Those would all be examples of transversals we could do. All right? So that's what one means by a transversal. So G is a union of all of these partitions for which if we take G from elements of a transversal, then there's gonna be no overlap whatsoever because notice that G, H, intersect, G prime, H is gonna be the empty set so long as G and G inverse disagree with each other. Assume we take elements from the same transversal. Of course, if G and G prime represent the same coset they'll actually be equal. So that's the thing is when you intersect cosets together one of two things happens. You're either gonna get the empty set or you're gonna get that they're actually equal to each other depending on whether it's the same coset or not. Cosets do not overlap at all unless they overlap entirely. That's what it means for the cosets to form a partition. So I'm gonna get this stuff off the screen right now. So G is a destroyed union of all of the cosets. So you're gonna have something like the following. G is gonna look like G1, H, union G2, H, union all the way down to say G, T, H, where G1 or GI here is just an element of this transversal T. But each and every one of these cosets has the same size. Each of these cosets G1, H is gonna equal GJ, H all of these cosets have the same cardinality. And so because of that, we're gonna see that the order of G, it's gonna look like, well, it's the sum of the size of all these cosets. But as each of these cosets has the same thing, this is just gonna give us some number times the order of H. But what's the number there? Well, the number turned out to be how many cosets do we have? And that's exactly what the index says right here. So we make this argument using finite groups right here, but essentially this argument becomes the same even with infinite sets. I want to be aware that there's really nothing in this proof that actually requires that things be finite. The reason that people mostly say finite here is that oftentimes this equation is rewritten in the following way that the index of a group is equal to the order of G divided by the order of H. And that's where things can get a little bit funky if you use infinite numbers. But this proof is actually perfectly acceptable for an infinite group right here because this could be made into an infinite sum because it came from an infinite union, which again, we don't have to worry about that too much. In particular though, if the order of G can be factored as H times the index of H, then that means that the order of H divides the order of G and that's the critical takeaway from the Granges theorem. The order of a subgroup must divide the order of G. Now let's look at those corollaries I promised you here. Suppose that G is some finite group. Again, this word isn't really necessary in this case, but it mostly just comes down to the division. So we'll keep it infinite here. Let little G be the element of that. Then that means that the order of an element divides the order of G. In particular, if you take an element to the order of the group, you're always gonna get back the identity. Now I'm not saying the order of little G is the order of the whole group, but if you raise an element to any multiple of its order, you'll get the identity. And as the order of little G divides the order of capital G, that means capital G is a multiple of little G. And so the argument is a quick, immediate consequences of the Granges theorem we just saw a moment ago. Let's say the order of capital G is N and let's say the order of little G is M and let's form the cyclic subgroup generated by little G. Well, the order of the cyclic subgroup will equal the order of the element itself. And by Lagrange's theorem, the order of a subgroup divides the order of the whole group. So M divides N. So therefore there's some integer K such that KM equals N. And therefore if you take G to the N, which is the order of capital G, this is G to the KM, which is equal to G to the M to the K, which is E to the K, which is the identity. It's that simple. The order of an element is the order of a cyclic subgroup and therefore it will divide the order of the group. All right. Suppose, for example, that G has prime order. Then if a group has prime order, then I actually claim that it's cyclic. And in fact, every non-identity element is a generator of that group. So groups of prime order are always gonna be cyclic. This is gonna be our first classification theorem. That is, can we classify groups up to certain types? And every group of prime order has to be cyclic. It's a immediate consequence of Lagrange's theorem. So take an element of G and let's say that its order is N. Well, by Lagrange's theorem, the order of an element must divide P. But P is a prime number, so there's only two options. N is either one or it's P. Well, if N is equal to one, then the order of G, if that's one, that means it's the identity. The cyclic subgroup generated by the identity, of course, is just the trivial subgroup itself. All right, well, what if the order is P? Well, if the order of G is P, well, P with the order of the whole group, capital G, and therefore that means the cyclic subgroup generated G has to be the whole group and therefore G is cyclic, like we just argued a moment ago. So for groups of prime order, they're only cyclic and anyone who's not the identity must generate the entire group. This divisibility condition is extremely powerful for finite groups. And so let's do our last corollary. We're proving these corollaries in like 30 seconds flat. Suppose we have two groups, two subgroups, H and K. These are finite, these are subgroups inside of G right here. And let's say we have a chain of subgroups. So K is inside of H and H is inside of G. Now, of course, by transitivity, K is a subgroup of G, but particular K is a subgroup of H. So then it turns out we can factor, we can factor the index of K as the product of the index of H inside of G and the index of K inside of H, all right? So the index factors that way. Now, this is the first result for which we really are gonna focus on the finite, right? Now the previous result, the corollary, we'll sure if you prime order to the group has to be finite, that's true. On this one right here, if the order of an element dividing the order of G, that doesn't really mean much if G was an infinite group because everything divides infinity. So to speak, one has to be a little bit more careful because you have to start talking about the torsion group, subgroup and things like that. So that one gets a little bit murky. We can make it into an infinite statement, but that's not something we're gonna do right now. And then like I said, the original proof of Lagrange's theorem we did in this video so far, it did not require finite condition whatsoever. Really, the whole business about finite is really gonna come right here. That this proof, I do wanna mention that this result is true for infinite. We can remove the word finite right here and it would still be true, but the proof is gonna fundamentally change. And again, without getting into the infinite cardinals that I predicted, we're just gonna stick with a finite proof right here because the finite proof is pretty nice. It goes like the following. By Lagrange's theorem, the index of K inside of G will be the order of G divided by the order of K. Then as this is just a fraction, a rational number, we can times by a strategic number one. We're gonna times it by the order of H over the order of H, which then we can basically factor this thing. You're gonna get G over H, which is the index of H inside of G. And then you're gonna get H over K, which is the index of K inside of H. And so the factorization follows immediately from the fact that Lagrange's theorem told us that the index of G and H is equal to the order of G divided by the order of H. And so this gives us Lagrange's theorem with its important corollaries. And again, in the future, I might reference any of these corollaries equally as well as Lagrange's theorem. When one talk about Lagrange's theorem, you typically meaning that the order of a subgroup is gonna divide the order of the group. And in those things that are very much related to that principle we refer to as Lagrange's theorem. Before ending this video though, I didn't wanna make one comment about the converse of Lagrange's theorem. So Lagrange's theorem says that the order of a subgroup must divide the order of G. That is the order of a subgroup is a divisor of G. The converse of Lagrange's theorem would be interested in the following. Is every, does every divisor of G have a subgroup of that size? And we saw that for cyclic groups, that thing is true. That every, there's a subgroup for every divisor of the order of a cyclic group. But that's not true in general. If you, for example, take the alternating group A4, it has an order of, well, that should be four factorial divided by two. Just so we're clear, four factorial is 24. And so you're gonna get 12. What are the divisors of 12? You're gonna get one, you're gonna get two, you're gonna get three, you're gonna get four, you're gonna get six, and you're gonna get 12. These are the divisors of 12. Now, of course, A4 does have a subgroup of order of one. It's the trivial subgroup. It has a subgroup of order of 12. It's called the whole group, A4. It does also have, it has a subgroup of order two. For example, you can take the subgroup generated by one, two, and three, four. If you take a two, two cycle, its order will be two. We have a subgroup of order three. For example, if you take the cyclic subgroup generated by a three cycle, which is an even permutation, that'll be three. It does also have a subgroup of order four. This coincides with the Klein four group. You take the subgroup that contains the identity and all of the two two cycles. Let's see, two, four, and then two. We want one, four, and then two, three. So we have the Klein four group. This is a group of order four. But search as you might. It is impossible to construct a group of order six inside of A4. So although six is a divisor, there is no subgroup of A4 of order six. And I'm not gonna supply the argument of this right now. Actually, I think it'd be a great idea for the viewers to try to think about it. Why can't A4 have a subgroup of order six? It's not gonna happen. Basically, the way you'd have to do it is you'd have to combine. You'd have to take a group that contains a three cycle and a two, two cycle. That's kind of like the only hope of constructing a subgroup of order six. But then you can show that any such subgroup actually generates the whole group. And so when you try to get six, you actually get 12. So the converse of the Gauges theorem does not hold. Just because a group order has a divisor does not mean it has a subgroup of set size that does not hold in general.