 In this video, we're going to prove the first Sieloff theorem, which is similar to Cauchy's theorem, both in its statement and in its proof. We really can think of the first Sieloff theorem as a generalization of Cauchy's theorem. So recall that Cauchy's theorem says that if you have a finite group and you have a prime that divides the order of that group, then you have a subgroup, which is necessarily cyclic, mind you, but you have a subgroup of G whose order is p, that prime that divides it. Sieloff's theorem is going to generalize that. So if p divides the order of the group, in fact, if p to the r divides the order of the group, where r is some integer power of p there, then G has a subgroup of order p to the r. So Cauchy's theorem is just the case where you have p to the first, right? So that's the situation where r equals 1. Sieloff's first theorem generalizes that. So if p squared divides the order of the group, you have a subgroup of order p squared. If p cubed divides the order of the group, you have a subgroup of order p cubed. If p to the fourth divides the order of the group, then there's a subgroup of order p to the four. That's what the first Sieloff theorem gives us. We're generalizing Cauchy's theorem. And the proof of the first Sieloff theorem is actually almost identical to the proof of Cauchy's theorem we just saw. There is one critical difference that we will see. And at that moment, we actually will need to use Cauchy's theorem, which is why we couldn't just do it all in once. We had to do Cauchy's theorem first then the first Sieloff theorem. All right. So how's the proof go? Like Cauchy's theorem, it's going to be proof by induction on the order of G. Of course, if you have the trivial subgroup, this statement we're trying to prove is vacuously true. So we move on to a non-trivial subgroup. If the order of the group is a prime, then the group is necessarily cyclic. And this statement is true for cyclic groups, because for a cyclic group, you have a unique subgroup for every divisor of the order of the group. In particular, you'll have a subgroup whose order is a power of prime. You get all of them, but in particular, you have the powers of the prime. So then we're going to proceed by induction. The induction hypothesis will be that all groups whose order is less than the order of G assume the result holds for them. We then are going to rely upon the class equation. So if the order of the group, the order of the group is going to equal the order of the center of the group plus the sizes of each of the non-trivial-consciously classes for that group. So remember that by group actions here, if you have the conjugation action working on the group, then the centralizers will be the isotropy subgroups of the conjugation action. The index of the centralizer then gives you the size of that orbit, the size of the congisy class. And then the center of the group is the collection of those elements which commute with everything. And therefore, those are the elements whose congisy classes just themselves, they're trivial. So this, the class equation gives us a partition of the group with respect to the conjugation action. Now, the elements x1, x2, all the way up to xk, these are representatives of the k non-trivial congisy classes. In particular, these are elements which are non-central. Therefore, their centralizers will be proper subgroups because there's at least something that doesn't commute with it. So not everything commutes with it. That proper statement will be important when we get to induction. So the first assumption is suppose there's one of these centralizer indices that's not divisible by p. Well, apply Lagrange's theorem. Lagrange's theorem says we can factor the order of g using the order of the centralizer and the index of that same centralizer. We know by assumption that p to the r divides the order of g. So by Euclid's lemma, these p's have to divide the other side. But by assumption, these p's can't divide the index. So all r of the p's are gonna have to divide the order of the centralizer. Now, since the centralizer, like we said, it's a proper subgroup, the induction hypothesis applies. And since p to the r divides it, that means we get a subgroup of order p to the r. That subgroup of the centralizer is a subgroup of the whole group. And therefore, we have the subgroup that we desire. Let's move on to the next one. The next one, what if p divides all of these indices? Well, by assumption, p to the r divides the order of g. So in particular, p by itself, because r is at least one, p will divide the order of g. So p divides all the indices, it divides the order of g. So by the class equation, p is gonna divide the order of the center. This is where Koshy's theorem is gonna come into play here. Since p divides a group, the center, by Koshy's theorem, there's an element little z inside of the center whose order is p. Let's take capital Z to be the cyclic subgroup generated by little z. Now, this right here is a very important observation, not just for this proof, but in general. Whenever you take a subgroup of the center of a group, it's always a normal subgroup in the entire group. To prove this, I want you to remember that a subgroup is normal, if and only if it's closed under conjugation. If I take an element inside the center, call it x, and I take an element of the group and you conjugate a central element by any elements in the group, since it's central, it'll commute. x and g inverse will commute, g and g inverse cancel out, you just get back x. If you take any central element and you conjugate it, you get back x. That was what we saw earlier with the class equation. Because the conjugation action is just the identity on central elements, any subset of the center is closed under conjugation. If you take a subgroup of the center, that's a subgroup that's closed under conjugation, so it's a normal subgroup. Subgroups of the center are always normal in g. This is a very common trick. We take a single element of the center that has the property we want, such as order p, and then that subgroup by itself will be normal. What do we do with normal subgroups? We mod them out. I want you to consider now the quotient group of g mod z. What's going on there? Let's consider the order of h. The order of h will be the order of g divided by the order of z. We don't know a whole lot about g, but we do know that the order of z is p. The order of g divided by p will be something strictly smaller than the order of g. The inductive hypothesis applies to the group g mod z. Now, because we took away a factor of p, we knew that p to the r divided g, the order of g, so then we can infer that there's a subgroup of h whose order is p to the r minus 1, so we lost one of those p factors because p to the r minus 1 divides the order of g divided by p because of the same right here. That's okay though. Take that subgroup. If we lift it back up to g, so we take a subgroup of h, it lifts to a subgroup of g by the correspondence theorem. That subgroup we just lifted back up will have order p to the r, which then gave us the subgroup of p to the r that we were looking for. Therefore, the result follows by induction. Now that we've proven the Seelof first theorem, I'm actually ready to define what is a Seelof p subgroup. Given a finite group g, let p be some prime divisor of the group. We say that a subgroup is a Seelof p subgroup, typically we'll call this capital P. It's a Seelof p subgroup of g if it's a maximal p subgroup. What does maximal mean in this situation? Maximal means that if q is a p subgroup, and if p is a subgroup of q, then we actually have that p equals q. So it's maximal in the sense that it's maximal on the lattice of p subgroups. Any p subgroup that contains a Seelof p subgroup is actually the p-loft subgroup itself. You can't have a larger p subgroup than that. Then the collection of Seelof p subgroups will denote Syl subp of g. I should mention that the Seelof subgroups of any group are extremely important. These are an extremely important topic for finite group theory. Now that we have Seelof's first theory, imagine that, suppose that p to the r divides the order of g, but p to the r plus 1 does not divide the order of g. Consider that situation. So p to the r is the maximal number of primes that divide the order of g. By the first Seelof theory, there exists a subgroup, there exists a subgroup p such that the order of p equals p to the r. Now if there was a p subgroup that contained capital P, its order would have to be strictly bigger. It would have to be p to the r plus 1, but that's not possible by Lagrange's theorem. Therefore, this group p is necessarily maximal and a k a, it's a Seelof subgroup. So the first Seelof theorem phrase slightly different says Seelof p subgroups exist. So the first theorem really is the existence of Seelof subgroups. When the prime in consideration is clear from context, it's often omitted. You just say a Seelof subgroup there. And so the first theorem tells us that Seelof subgroups exist. The way we phrase is actually a little bit stronger. We're saying there are subgroups of order p to the r, p to the r minus 1, p to the r minus 2, p to the r minus 3, p to the r minus 4, all the way down until we get to p. So we have all these different ones. But there's a little bit of a trade-off that we got from that, that way of phrasing the first Seelof theory, the theorem there, is that we're saying Seelof subgroups are maximal. If you have the maximum order, that necessarily will make you maximal. But that doesn't mean that every maximal p subgroup has the maximum order, right? Could it be possible that like p cube divides the order of g? But could we have a Seelof subgroup of order p squared? Because it's maximal in the sense that no other p subgroup contains it. There's no subgroup of order p cube that contains this subgroup of p squared. Is that a possibility? Well, by the second Seelof theorem, which we'll prove in the next lecture, we'll actually find out that's not a possibility. But we want to consider just some examples of Seelof subgroups at the moment. So consider an Abelian group like Z12. If you have order 12, that factors as 2 squared times 3. Z12 does have a subgroup of order 3. Notice 3 to the first is the biggest power of 3 divides 12. If you take the subgroup generated by the element 4, that'll give you a subgroup of order 3. And in fact, that's the only subgroup of order 3. This group would have a unique, has a unique Seelof 3 subgroup. Because actually there's no other three subgroups of the group at all. What about 2 on the other hand? Well, the group Z4 does have a subgroup of order 4. It's the group generated by 3, the cyclic subgroup generated by 3. And it'll be a Seelof subgroup because it has order 4. What other two subgroups does the group have? Well, there's also the group generated by 6, which will be isomorphic to Z2. But I should mention that this is actually contained inside of the subgroup generated by 3. So it's a 2 group, but it's not a maximal 2 group, and therefore it's not a Seelof 2 group. So it turns out this is the only Seelof 2 group. So the cyclic group of order 12 has a unique Seelof 3 subgroup and a unique Seelof 2 subgroup. Let's look at a non-Abelian example or two. S3 is a non-Abelian group. It's the symmetric group on 3 letters. Its order is 6. So its prime divisors are 2 and 3. It has a unique subgroup of order 3. That's just the alternating group A3. That's the only 3 group in the entire group. It actually has 2, excuse me, it has 3 2 subgroups. If I take the cyclic subgroup generated by any of the 2 cycles, that gives you a 2 group. They're each isomorphic to Z2. It's a 2 group and they're going to be maximal because they have the maximum order of a 2 group. And so you actually have 3 distinct Seelof 2 subgroups. A group can have distinct Seelof subgroups. If we come to D4, D4, the dihedral group, its order is 8, which is 2 cubed. D4 is itself a 2 group because its order is a power of 2 and therefore it is a maximal 2 group because it's the largest subgroup possible. So if your group is itself a P group, then it's its own Seelof P group. So the group itself is its Seelof 2 subgroup, which of course is unique in that situation. You could say the same thing about Q8, the Quatorian group of order 8, because it's a 2 group, it's its own Seelof 2 subgroup. Let's come down to A4. This is a very interesting non-Abelian group to consider. Its order is 12, so just like Z12 we did earlier, right? It has order 4 and order 2 for the prime powers there, 2 squared and 3. A4 has a unique Seelof 2 subgroup and that's actually going to be the Klein 4 group, which you see right here. You have the identity in all the 2 2 cycles. Now the other 2 subgroups of A4 consist of just taking the identity in one of the 2 2 cycles, but those all live inside of the Klein 4 group. So the Klein 4 group is the only maximal Seelof, the only maximal 2 subgroup, so it makes it a Seelof 2 subgroup. It's unique, but in A4 you have lots of, you have lots of 3 subgroups. One of them, you just pick your favorite 3 cycle, like 1 2 3, then the cyclic subgroup generated by it gives you the identity, it gives you 1 2 3, and it gives you 1 3 2, like so. That gives you a 3 group and because 3 is the largest power of 3 that divides A4, there's no larger 3 subgroup than this one right here. So this makes it a maximal one, this makes it a Seelof one, but this is the one we get if you generate it using 1 2 3. If you use any of the other 3 cycles other than 1 2 3 in its inverse, you can get a different one. And so it turns out that A4 actually has 4 Seelof 3 subgroups. So this is something we've seen so far that we can have multiple Seelof subgroups or we can have unique ones. Now some interesting patterns here is that when the Seelof subgroup was unique, it actually was normal. And when it was distinct, they're not normal, but even when they're distinct, they're still isomorphic. All of the Seelof 3 subgroups of A4 here are cyclic subgroups of order 3, they're all Z3. So that's sort of an interesting thing. When it's unique, it's normal, if and only if, and when they're different, they're still isomorphic to each other. And this is, we're going to prove these observations with the second and third Seelof theorems, which we'll do in the course in the next lecture.