 In this video, I want to prove the second Silov theorem. You'll recall that the first Silov theorem told us that Silov p-subgroups exist. In fact, we actually get p-groups of various sizes. They're guaranteed to exist by the first theorem. But in particular, the Silov first theorem says that Silov p-subgroups exist. The second theorem says that different Silov p-subgroups are really not that different. Imagine we have some finite group g and p is some prime divisor dividing the order of g. Then we get that all Silov p-subgroups are conjugates of each other. So in particular, if p and q are two distinct Silov p-subgroups, there exists some element of the group g so that gpg inverse is equal to q. All of the Silov subgroups are conjugates of each other. So before proving this, I want to make some statements. If this is true, which it is, by Silov second theorem, all Silov p-subgroups are conjugates to each other. And as conjugation is an inner automorphism on g, conjugation induces a group isomorphism between Silov p-subgroups. In particular, you're gonna get p and q are isomorphic to each other. They're the same group, different elements, but they have the same group structure. In particular, this implies that all of the Silov p-subgroups have the same order because the order of one will then be the other. The order of p is equal to the order of q. Now, if p to the r divides the order of g and we have that p to the r plus one does not divide the order of g, the Silov first theorem guarantees there exists a subgroup p of g whose order is p to the r. So there's someone who has this and it has to be a Silov subgroup because it's maximal. And therefore this then would tell us that every Silov p-subgroup has an order p to the r. So the only Silov p-subgroups are those who have the maximal possible order. And so that's what we wanna get. So the consequences of the Silov second theorem are very, very important. So let's prove that all Silov p-subgroups are conjugates of each other. So like we said a moment ago, let's suppose that the order of g is p to the r times m where p doesn't divide m, that's to say that r is the maximal power of p that divides g. So by the Silov first theorem, there exists a subgroup p whose order is p to the r. We were just talking about that. And because it's the largest possible p-subgroup, it has to be a maximal p-subgroup. So that makes it a Silov p-subgroup. So let's consider the orbit of this Silov subgroup p with regard to conjugation by the whole group g. So let's say that orbit looks like this. There's some p1, p2 up to pk. And we can assume that p1 is the subgroup p. Now, these are all conjugates of p. And so these are all groups whose order is gonna have to be the same so they're all p to the r. So all of these groups are necessarily Silov p-subgroups. What we wanna show is that every Silov p-subgroup is inside of this orbit. That is there's only one orbit with regard to g conjugation. Now to do this, we're gonna use the fundamental counting principle of group actions. That is the size of the orbit k is the same thing as the index of g with the normalizer of this Silov p-subgroup p. So let's play around using Lagrange's theorem for a moment. We know that the order of g by assumption was p to the r times m, where p doesn't divide m because p to the r was the maximum power of p. Well, by Lagrange's theorem, we can factor g as the order of the normalizer of p times the index of the normalizer of p. Now, by the fundamental counting principle, this index is equal to k. And likewise, because p is a subgroup of its normalizer, that means the order of p, which is p to the r, it has to divide into the p. Again, by Lagrange's theorem. So since p to the r divides this, that means there's no p's left that can divide k. So p does not divide the index of this right here. So that's a very important observation there. So p doesn't divide k. All right, so now let's suppose we take an arbitrary Silov p-subgroup. Let's call it q for a moment. We wanna argue that q belongs to the orbit of p. There's something we can conjugate p by to get q. So how do we do that? Well, consider conjugation, consider conjugation by p on the set o. So what I mean by this is that we take o here, which has now disappeared, right? O, we had p, we had p2, we had p3, all the way up to pk. Well, since g acts on this set by conjugation, we could restrict it to some subgroup q. q acts on this set by conjugation. In particular, this might partition into smaller sets. How many would we have? The number of q conjugates we would get for the subgroup p is gonna equal the index of q with the intersection between the normalizer p and q. This was given by limit 15.110, which we actually proved in the previous video. Take that out, check that out if you didn't see it already. So the number of smaller conjugates is gonna be given by this number. Now, of course, as q is a p group, all of the indices right here have to be powers of p because the index, any index of q divides the order of q. q is a p group, so its order is some power of p. So these indices are likewise some power of p. So as q conjugates the Seeloff subgroups in the orbit here, we can potentially get smaller orbits. You have some subgroups there, you have some of the subgroups here, and then we just have a bunch of these smaller orbits, right? But since q is a p group, all of these orbits, all these smaller orbits would have to have sizes of power of primes. So you have some p to the e1, some p to the e2, all the way down to some p to the eS or something like that. So we broke up into something smaller. We all have the powers of a prime. So what this tells us is that k can be written as p to the e1 plus p to the e2, all the way down to p to the eS, what we called it. So k is a sum of powers of p. Now, if each of these powers is at least one, then that sum will be divisible by p, which means k is divisible by p, but we just said a moment ago, it's not on the screen anymore, that p can't divide k, so it's a problem. So at least one of these powers of a prime has to be one. That is one of these e's have to be zero. We have to have some p to the zero showing up here in order for that to be a one here. Basically, you're saying, I'm saying one of these things has to be a singleton all by itself. And so without the loss of generality, let's assume it's just p, because again, all of these CLOF subgroups do have the same order of p to the r. So in that situation, we have that q, the index of q with npi, which we can call it one for all I care. It doesn't matter, one of them is, npi intersect q is equal to one. This would imply that the two groups are actually one and the same thing. So this tells us that q equals npi intersect q, like so. So they have to equal each other. So in particular, if we take any x inside of q, it's gonna belong to this normalizer. It normalizes pi. So xpi x inverse is equal to pi. But lemma 1519, which we also proved in the previous lemma says that if an element of p prime power normalizes a CLOF subgroup, then it has to belong to that CLOF subgroup itself. So x belongs to pi. But as x was an arbitrary element of q, this shows that every element of q belongs to pi. But q is a maximal p subgroup, it's a CLOF subgroup. Therefore, if a p group contains a CLOF subgroup, they actually are the same group. So q equals pi. But pi was a conjugate of p. And so as pi is, since the subgroup q was an arbitrary CLOF p subgroup, this says that every CLOF p subgroup is a conjugates to each other. This then proves the second CLOF theorem, and then we gain all the important implications of that theorem.