 Welcome back to our lecture series Math 4230, abstract algebra 2 for students at Southern Utah University. As usual, be a professor today, Dr. Andrew Misseldine. Lecture 6 represents the start of our coverage of chapter 15, which is going to be about the Seeloff theorems and their applications towards simple groups. In section 15.1 of Tom Judson's textbook, we, following his table contents, I should say, we're going to develop and prove the three Seeloff theorems. In this lecture, we're going to prove the first Seeloff theorem, and in the next lecture, lecture 7 will prove Seeloff theorems 2 and 3. But before we get to see the first Seeloff theorem, I actually want to prove in this video something called Cauchy's theorem. Now, in abstract algebra 1, we saw Lagrange's theorem, which told us that the order of any subgroup divides the order of the whole group, at least for finite groups, right? Infinite groups, the arithmetic, it's a little bit more messy. So just delving into finite groups, the order of a subgroup always divided the order of the group. So if H is a subgroup of B, then we have that the order of H divides the order of G, something like that. It's a very, very useful statement. But this is Lagrange's theorem. It's tempting to hope, or at least it's hopeful, that you can have a converse to Lagrange's theorem. That is, for every divisor of the group, there's a subgroup that obtains that order. But that's actually not the case, right? You have groups like, say, A4, the alternating group there. Its order is 12, and it does have a subgroup of order one, the trivial subgroup. You do have a subgroup of order two, because you could take a subgroup generated, a cyclic subgroup generated by a 2-2 cycle. It does have a subgroup generated, I should say a subgroup order of order three, because it's generated by a three cycle. You do have a subgroup of order four, because if you take the identity in the three 2-2 cycles, it forms a client four group, and you go from there, you also got a subgroup of order 12, because it's the whole group itself. But what about six? Six divides 12, but A4 does not have a subgroup of order six. Try and try as you might, you can't get one. And in fact, we can prove that there is no subgroup of order six. We're not gonna do that in this video. But this is sort of like the simplest counter example, at least one of the simplest ones, if there's one smaller, but I'm pretty sure this is gonna be our simplest counter example of the converse of the Granges theorem, that there do exist groups where you don't have a subgroup for some divisor of the order of the finite group. And that might seem a little too bad, but that's the reality of it. Now, while we don't get the full converse of the Granges theorems, we do get partial converses. For example, when it comes to cyclic groups, for cyclic groups, you actually have unique subgroup for every divisor of the group. That's pretty easy to form. And then also when you get abelian groups, we have a lot of structure on abelian groups. I mean, by the fundamental theorem of finite abelian groups, we can write every abelian group as a direct product of cyclic groups. And as such, we can basically get a subgroup of any divisor order we want. So it's actually pretty easy to get a subgroup of a fixed size for abelian groups, but not abelian groups like A4, things get problematic. So this is where the Seeloff theorems come into play in particular Cauchy's theorem as well. Because these give us some partial converse to Lagrange's theorem. It gives us some conditions which will guarantee the existence of subgroups of specified orders. But instead of a general factor, we can only sort of guarantee it for prime powered or prime powered ordered subgroups. So that's the issue with six. Six is the only divisor of 12 other than 12 itself. That's not a power or a prime. And so that's why we don't get six, but two, three, four, they're all powers of primes. By combining the Seeloff theory here, we can actually guarantee that there's groups, subgroups of that order. So with that, let's start with Cauchy's theorem. This is extremely useful theorem here. Cauchy's theorem says that if G is a finite group and P is a prime, such that P divides the order of G, then there is an element of G whose order is P. So every prime divisor of the order of the group, that group will have an element whose order is that prime divisor. And the proof of Cauchy's theorem is gonna follow by induction. And we're gonna play induction on the order of G, like so. So if the order of G is equal to one, it's the trivial group. This is vacuously true in that situation. So let's then go up. If the order of G is a prime number, we've seen previously that the only groups whose order is a prime is a cyclic group. And like I mentioned, with cyclic groups, we've studied those in abstract order one. Cyclic groups have a unique subgroup for every order. In particular, if you take a prime divisor, it's sub, there will be a subgroup whose order is P and it'll be generated by an element of order P, right? So it's pretty trivial when you're prime, all right? So then let's assume we maybe have composite order. Let's suppose that the result holds for all groups whose order is less than G. So that's the induction hypothesis right now. So considering the order of G, every group less than G, its order, if its order is smaller, which it does because it's smaller, it satisfies this property. So the basis of Cauchy's theorem is induction plus the class equation. Remember the class equation, it tells us that the order of the group can be decomposed as the order of the center of the group. The center of the group is all of the elements of the group that commute with everything. And then we add to the order of the center a bunch of indices. These are indices of centralizer subgroups for which you'll remember that a centralizer subgroup, what's C of X1 here, the centralizer subgroup, this is all of the elements in the group G that commute with XI right there, X1 in this case. So the centralizers are always subgroups. We take their indices, this is what the class equation gave us because if you do the conjugation action of a group onto itself, then the stabilizers, the isotropy subgroups of that group action are the centralizers. There that is the isotropy of the conjugation action. And then you can apply the class equation for any group action. If you apply it using congisly classes, you get exactly this right here. The center is exactly those elements of the group who are only conjugate to themselves, just as a reminder of this vocabulary and terminology and such. So in particular, those elements that are showing up that are represented over here, the X1, the X2, the XK, these are elements which are non-central. So they are conjugate to multiple elements in the group besides themselves. In particular, since they are non-central elements, their centralizers will be proper subgroups and thus these indices, they're gonna be relevant to consider, okay? So continuing on here, suppose that there is an index G colon C of Xi for some Xi that's not divisible by P. Well, by Lagrange's theorem, the order of G is factorable as the order of the centralizer with then the index of the centralizer. Well, since G by assumption divides the order of G by Euclid's lemma, since it's a prime number, it must divide either this factor or this factor. By assumption, it's not this factor. Therefore, P then divides the order of the centralizer where that centralizer is a proper subgroup. Therefore, we see that we've now produced a subgroup who has an order divisible by prime by P. It's a proper subgroup so it's a strictly smaller order by the inductive hypothesis. The centralizer has an element of order P and as it's a subset of G, that means G has an element of order P. So the result follows by induction in that situation. Well, what's the other possibility? Let's suppose that all of the indices are divisible by P, okay? So the order of G equals the order of Z, the center, and then you have all these centralizer indices, something like that. We could move these all to the other side of the equation. So we get the order of the center is equal to the order of G minus all of these centralizer indices. There's a bunch of those. So the point is this element is divisible by P, this element is divisible by P, all the other elements are divisible by P. So the entire sum is divisible by P, which means that the order of the center must also be divisible by P, okay? And so then we wanna focus here. So we know that P divides the order of the center. Now, we might just wanna say induction at this moment, but we actually don't know that the center is smaller than the group because what if it's an Abelian group? The center is the whole group itself. So the order of the center might not be smaller because the group could be Abelian, but whatever the case, we know that Z of G is an Abelian group. So the fundamental theorem of finite Abelian groups applies to the center. And by that structure theorem, an Abelian group always has an element of order P if, of course, its order is divisible by P that comes from the fundamental theorem. I won't go into the details of that right now. So therefore Z has an element of order P and as it's a subset of G, then G has an element for P and therefore we're done. We couldn't quite cite induction in that situation, but we got really close. If it was a non-Abelian group, then it would have followed immediately by induction, but we have to deal with Abelian groups themselves. So the fundamental theorem of finite Abelian groups takes care of it. And so that then proves Toshi's theorem. Every group whose order is divisible by a prime has an element whose order is a prime, that same prime. And so in particular, let G be a finite group whose prime P divides its order, then G has a subgroup of order P. So we switched from an element of order P to a subgroup of order P. But by Koshy's theorem, we're gonna have some element G inside of G such that the order of G is that prime. Well, then if you take H, then to be the subgroup generated by G, you're gonna get that the order of H is equal to the order of G, which is prime. So if you have an element of order P, then you'll have a subgroup of order P. So Koshy's theorem guarantees that every group has a subgroup order of the primes that divide it. So let's come back to A4, for example, like we saw. A4 is order is 12. Its prime divisors are two and three. So by Koshy's theorem, there is a subgroup of order two and there's a subgroup of order three in particular you're gonna have elements of order two and elements of order three, for which for A4, yeah, the two, two cycles, there's three of them, they have order two and then there's eight, three, three cycles in A4, they have order three. Those are the elements that were guaranteed by Koshy's theorem. I should also mention that Lagrange's theorem says the orders of an element can only divide the group and therefore you can't get any other prime divisors for the orders of elements than those that show up. So two and three are the only prime orders you can have in A4 and they're all realized. That's the limitations of Koshy's theorem though. We got these subgroups of order two and order three. We can't guarantee all subgroup orders like six is not present in A4 like we mentioned before. So six is problematic. Now A4 did have a subgroup of order four, which I said that's a prime power, it's two squared. Why is there a subgroup of order four? It turns out that to guarantee the existence of four for A4, we actually need something stronger than Koshy's theorem, which is gonna be the first Sealoff theorem, which we'll discuss in the next video.