 In this video, we are going to state improve Silov's third theorem about Silov p-subgroups. If g is a finite group and if p is a prime dividing the order of g, well by the first Silov subgroup theorem, first Silov theorem, we know that Silov p-subgroups exist. And by the second Silov subgroup theorem, we know that all Silov p-subgroups are conjugates of each other. In the third theorem, so we've proven so far as we've proven they exist, they're all conjugates of each other. And in the third theorem, we're actually going to count how many Silov p-subgroups can we have. And this is a very, very important result, especially as we start studying simple groups in the not too distant future. Well, the next lecture for this lecture series. It turns out by the third theorem here that the number of Silov p-subgroups of a group is congruent to one mod p, and it has to divide the order of g. For which let me point out to you here that in particular, if you can factor the order of g as p to the r times m, such that p doesn't divide m, meaning that p to the r is the maximum power of p that divides g right there, that's all the powers of p. In that situation, because of these conditions, the number of Silov p-subgroups is congruent to one mod p and divides g. That number divides g, so it means it has to divide this. But if that number is one mod p, that means that it's going to have to divide m right here. So using this number m, we can actually consider how many Silov p-subgroups could the group possibly have. And that consideration has some very interesting applications, which like I said, we will explore those in later videos. So as seen above, g acts on the Silov p-subgroups by conjugation. This is something that's easy to see, I suppose, but in particular by the Silov second theorem, this action is a single orbit. If we look at the action of the Silov p-subgroups, you have some p1, some p2, some p3, all the way down to pk, the number of Silov p-subgroups here. This is one orbit and g can act on this set by conjugation. Therefore, the size of this orbit is the index of the normalizer of any Silov subgroup, and this is something that we get from the fundamental counting principle. So when g acts on the Silov subgroups by conjugation, we know that k, the size of this orbit, is going to equal the index of g with respect to the normalizer of any of these ones. It doesn't matter which one you choose, they're all the same because they're all in there. In particular, indices always divide the order of the group, and so that gives us the very first one, right? The number of subgroups we have because it's a single orbit by the second theorem, that number k, is an index of a group because of the fundamental counting principle, so it divides the order of g. This is a consequence of Lagrange's theorem. So that gives us the second statement right here. So what we have to prove is that this number k is also 1 mod p. Now, to do that, we're going to fix our attention on a single Silov p subgroup, let's call it capital P, and I claim that this group p also acts on the Silov p subgroups by conjugation. Now, it takes the orbit we had before, but it's going to break it up into smaller orbits. One of them is pretty clear, one of them is just going to be p itself because, after all, if I take an element that's in p and then I conjugate p by it, because it's a subgroup, it's closed under multiplication, so we're going to end up with, this is just going to be p, and so there's one element all by itself when we restrict the action just to capital P. I claim that there are no other fixed points in this situation, and to prove that, we're going to use lemma 1515 that we proved in the very first video for lecture seven. Take a look at that one if you don't know what I'm talking about right now, but let me just remind you of what it says, that let p be a Silov p subgroup of g, let x be an element of g whose order is a power of p. If x belongs to the normalizer of that Silov subgroup, then x actually belongs to that Silov subgroup itself. That's what that tells us. So if there's some other Silov subgroup that was all by itself, that means that xqx inverse would equal q, which would mean x belongs to the normalizer of q, but as the order of x is some power of p because x belongs to a Silov subgroup by Lagrange's theorem, since it's a power p, all elements will have a power p order, then that lemma applies and forces that x actually belongs to q, which means q and p were the same Silov subgroup there, and so there's no other fixed points. That's what we're trying to say here. So thus, there's no other fixed points. You're going to have some p2, some p3 going on, and then you can have a bunch of these other ones. There's no other fixed points. There's just the one right here. So if we come and look at the class equation, we often use the class equation for conjugation of a group on itself, but the class equation works for any group action whatsoever, for which remember what the class equation tells us. If we have a group action, take the g set x, it can be partitioned with respect to the orbits. Let's take all of the orbits of fixed points, which a fixed point means that its orbit is just a single element. If we put all of those singletons together, we're going to get this stable set x sub p here, p is the acting group in this situation, and then we add together all of the non-trivial orbits. What's going to happen in that situation? Well, as we observe, there's only one C law p sub group that's fixed by capital P, and that was p itself. So x sub p will have only one. What about these other ones? These other orbits by the fundamental counting principle are going to be, they're going to be indices that look something like this, but it's a little bit different. It's going to look like p dot, well, some other things. It's the normalizer with what's going on here. We talked about this previously, but I actually don't care what that is. If you take an index of a sub group of p that has to divide p, where p's order is p to the r, and so each of these orbits is some power of p. It might not be all of p r, probably not, but these are going to be powers of a prime. So the size of x, the number of C law p sub groups is one plus a bunch of powers of p non-trivial powers of p. So if you reduce that mod p, all of these things disappear and you're left with just one, which then finishes the proof of the third theorem of C law. So to finish this video, I want to provide some examples of this theorem here. So consider you have a sub group of order six, which six, of course, factors as two times three. By C law, third theorem, if we want to consider the possible three groups, the number, sometimes we use a symbol like n sub k for, you know, this is n sub k is the number of C law three sub groups. This is not a standard notation, but it's used from time to time. It's convenient in this situation. So you can think of n sub k. This is the size of C law p of g, like so. So we're asking, you know, what would n three be? Well, what we saw from the, from the third theorem is n three divides six, but n three is also congruent to one mod three. So how many numbers divide six, but are one mod three? So one divides six, two does, but that doesn't work. Three does, but that doesn't work. And then you get six, which doesn't work either. In particular, because you have to divide six, but you have to do one mod three, it has to divide two and be one mod six. So one's the only possibility when it comes to, when it comes to n three, you only get one. So every group of order six has a unique C law three subgroup. I want you to consider the two groups of order six. You have z six and s three. Um, z six has a unique subgroup looking at that real quick. It has a unique subgroup of order three. It's isomorphic to z three. It has a unique subgroup of order two. And then you have the trivial subgroup. It has only one subgroup of order three, just like we predicted. What about s three? Well, you have the alternating group a three. You're going to have, um, you're going to have three subgroups that are isomorphic to z two. If we look at the lattice of all these subgroups, but in particular, you have a subgroup, a unique subgroup of order three. Now I should mention that as a consequence of C law of second theorem that if you have a unique C law of subgroup, it has to be normal. Um, because it's going to have the conjugate of a C law of P subgroup is a C law of P subgroup. Um, if there's only one of them, it's going to have to equal itself. And I guess you don't need C law of second theorem. That's just true in general that if you have a unique subgroup of a given order, it'll be normal. And so if there's only one C law of P subgroup, then it's the only one of that order. So it's going to be normal. That's a very important thing. So every group of order six has a normal subgroup of order three, but in retrospect, that's not too surprising. Order three means your index two. And we've seen before that every subgroup of index two has to be normal. So this is the C law theory sort of overkill in the situation, but it's starting to show to you why that could be useful. Well, what about two groups? Okay, how many C law of two groups could you have? Well, by the C by C law of third theorem, we have that in two has to be a number which divides six and is one mod two, which if you're one mod two, that actually means you'd have to divide three and be one mod two, but that gives you one and three. So there are some possibilities here. If you look at Z six, it has a unique subgroup of order two, but S three has three subgroups of order two. These were the two possibilities that C law of third theorem allowed for us. And all of these possibilities were in fact realized. Let me clean up my screen a little bit. Let's come down here and talk about groups of order 12 for a moment. What type of C law of subgroups can you have with a group of order 12? Well, if you look at n three for a moment, we need a number which divides four, but is one mod three. One is always going to work in that situation. You could always get at least one, you always get at least one, but you also could get four two divides four, but it's two mod three, four divides four and is one mod three. So that's a possibility. You could get one or four. What about in two? Well, n two, it's going to be just like we saw before. N two, if you take away the two part, you have to be one mod two and you have to divide 12, which means you're going to divide three in this situation. You get one and three. So these are the possibilities we could get. We could get a unique subgroup of order three and order four. We could get maybe a non-trivial, you could get three subgroups of order two, order four actually maybe, or maybe we could get four and one, or maybe we could get four and three. These are these, these four possibilities here. Not all of them are realizable because of reasons we'll talk about in later lectures, but I do want to show you some examples of this. If you look at Z 12 for a moment, Z 12, it has as its lattice of subgroups, we got Z six, we got Z four, we got Z three, we got Z two, and we got one. Like so in particular, it has a unique subgroup of order three and has a unique subgroup of order four. These are the C law three and two subgroups respectively. So for Z 12, we get that in two is one and we get that in three is one. That's a possibility. What if we switch over to the alternating group? What does it look like? Well, in terms of its groups, A four, you have the client for group, which has order four. It contains three subgroups of order two. So these are going to be cyclic subgroups of order two. And then you also have these subgroups of order three floating around. It'll have four of those. So just drawing the haza diagram here, like so. And so if you start to count what are the C law subgroups, you have your C law two subgroup right there. There's only one of them. It's normal. But you have three, excuse me, you have four C law three subgroups, just like we said before. So that is a possibility. We have one and one. We then have four and one when it comes to A four. And then if we look at another group of order 12, take D six, for example, D six, I erase my numbers there in three could equal one and four and two could equal one and three. What happens with D six? I'm not going to draw the entire lattice. I mean, I guess I could. It's not so bad. We did D six. D six, of course, is going to contain a cyclic subgroup of order six, which gives us Z two. And it gives us Z three. And this all sits above the identity, like so. You also have these, you're going to have a bunch of the reflections there. So you should have like six of those over here. And I can draw some more over here. Z two, Z two, and Z two, like so. So what else can you get? If you brought this Z two with the Z three together, you're going to form a dihedral group, a D three, and that actually would get all of these. Same thing over here, this dihedral group, boom, boom, boom, and boom. If you were to bring together various of these Z twos, you could bring them together and form a Klein four group. You get something like this. And then you're going to get several of these. And this is kind of where it gets messy. And this was my hesitancy to draw this. But if you bring these together, you get all of these Klein four groups. And then we'll do one more over here. Bring that one together, bring that one together, bring that one together. And so I guess what I'm trying to say here is in this picture, you're going to get a unique subgroup of order three like so. That's going to give you this right here. But you're going to get three different Klein four groups, which are going to be your Sealoff two subgroups. So we do in fact get this. So we see all the possibilities now. We can get a subgroup, which gets one in one, we can get a subgroup, we can get a group that gets one in three, we can get one that is four in one. It turns out that four in three is not possible. And that's mostly because it counts too many elements. Because if you have, if you have four, four Sealoff three subgroups, that's those are going to be four times, you know, this Z three right here. These elements are going to contain some element, it's inverse, it's square in the identity. So if you just count these ones, you're going to get eight elements right there. So eight elements of order three, right? Now we've counted these eight elements of order three. If we take those away from everything else, 12 minus eight, you end up with four elements left. And you have to have at least one Sealoff two subgroup, which has order four. So this right here is the Sealoff two subgroup. These four elements, it's a client four group, or maybe it's a sickly group of order four, it doesn't really matter. You have to count it for all of them. So it turns out the option of four and three isn't possible. But this argument that we did hear, these possible considerations, these, when used correctly, can be very powerful tools in group theory. And we'll talk about those more in the next video.