 In our previous video for lecture four, we introduced the idea of the class equation, which gives us a way of measuring the order of finite group using its congessy classes. I mentioned that this class equation is very, very important. So in this video, I want to study the idea of p-groups and then use the class equation to help us better understand the structure of a p-group. What's a p-group? Well, in this situation, p is a prime number, it can be two, it can be three, five, any prime number, it doesn't matter. We say that a finite group is a p-group if the order of the group is p to the n. That is, the order of the group is a power of prime, or another way of saying that is that only one prime divides the order of the group. Some examples of that, we've seen many, right? Examples of two groups would be like z4, the climb four group, d4, q8. The cyclic group of order four, as the name suggests, has order four. The climb four group also in a group of order four. These are examples of a billion two groups because their orders are two to a power. d4 is a non-abelian group of order eight. The quaternion group is also a non-abelian group of order eight. These are all examples of two groups. Examples of three groups would be things like z9 and z3 cross z3. So this is the elementary belian group of order nine. These are three groups. And we can do this, of course, for various groups, right? We can take cyclic groups and elementary belian groups. These are all examples of p-groups for their respective orders. It's easy to come up with abelian groups because we have the fundamental theorem of finite abelian groups. If we just take direct products of cyclic p-groups, that'll give us an abelian p-group. Non-abelian p-groups are a little bit more difficult to come across here. I shouldn't say that. I mean, they're quite ubiquitous, but at this level of our development, it's difficult to describe some of these. We'll get there eventually, of course. But I should give you some non-examples, right? Like s3 is not a p-group. The order of s3 is six, which actually means that both two and three divided. It's not a power of a prime. Many groups are not p-groups. But p-groups, in some respect, are the atomic building blocks of other groups in some essence. I won't say too much about that. But in some regard, they're the simplest of all groups. Not simple in the group theory sense that, I mean, p-groups will have normal sub-groups. We'll actually, we'll kind of get to that in just a second. But from a conceptually point of view, they're very, very elementary in comparison to other groups, right? Oftentimes, we try to improve things about groups. We go to p-groups first, because if we can prove it for p-groups, then we can climb the hierarchy of groups, more complicated groups, more complicated groups. p-groups are some of the most simple type of groups. There are those elementary type of groups. But like I said a moment ago, p-groups are not simple in the group theory sense. p-groups always have non-trivial normal sub-groups. So if g is a p-group, it'll have a non-trivial center. That is, there's some element of the group that commutes with everything other than the identity, right? I wanna give you some examples, right? d4, its center is actually one and r squared. If we take the quaternion group, q8, its center, its center of course is the identity, but you also get negative one. These are two examples of non-apelian two groups, but their centers are non-trivial. This happens for every p-group there is, and this is a direct consequence of the class equation. So remember the class equation, for any finite group, the order of that group will equal the order of its center plus a bunch of indices, and particularly this will be the indices of centralizers, which this coincides with the order of congenital classes, right? Well, since these are p-groups, we can specialize this, right? We know that p divides the order of g. In fact, p is the only prime that divides the order of g. If we unravel these indices, as these are finite groups, right, the index here, g index sum centralizer, where xi is just some element, doesn't matter what it is really. By Lagrange's theorem, since these are finite groups, this is gonna be the order of g divided by the order of the centralizer here, for which order of g is some power of p, so it's p the n. Now, you take away some of it, because c of xi here is a subgroup of g. By Lagrange's theorem, its order divides order of g, and that's the important thing to know about p-groups, is that by Lagrange's theorem, the subgroup of a p-group is also a p-group. So if you take these centralizers, they're also p-groups themselves, and the p-group divided by a p-group is still gonna be a p-group. So for p-groups, quotients are also p-groups. This is, you start to see why it's nice and simple to study p-groups, right? Subgroups and quotients of p-groups are themselves p-groups. Therefore, we know that p must divide this index because p divides this thing right here, all right? The centralizer can't be everything. It's not the whole, this isn't gonna be one, because if g was equal to the centralizer, that actually means xi would belong to the center. So these things over here are those who are not central elements. So therefore, this quotient is not one, but it's gonna be some power of p, so p divides it. So p divides this thing, p divides this individual index, so it divides all the sums. So if we rewrite this class equation, we get that the order of z, I'm just gonna write that z for a moment, this is gonna be order of g minus the sum of all these things, g index cxi. So g divides the right-hand side here, so p divides the right-hand side, so p has to divide the center as well. So p divides the center, which means the center can't equal one because p doesn't divide that. So there has to be something in the center other than the identity. Now, it might not be much. I mean, like with these two examples right here, the center has order two, but that's the point. These are two groups, two has to divide that. So these are small centers, but there is still something in there. And once the center is non-trivial, that can give you a lot of information. Now, of course, for an Abelian group, the center is the whole group, so that doesn't say much, but this is very crucial for non-Abelian p groups because the center is non-trivial, there's something in there, there's some central element. And in particular, the center is a normal subgroup. So you can use that in various ways. I wanna give you an application of this. So this next corollary, the result we just saw, it's not a direct consequence of the class equation, but it's a corollary of the theorem we just proved. Why is having a non-trivial center useful, okay? I claim that if you're a group of order p squared, you are necessarily Abelian. Now, by the fundamental theorem of finite Abelian groups, that means you either are zp squared of the cyclic group or you're zp cross zp, that is, you're the elementary Abelian group of order p squared. Those are the only two groups of order p squared up to isomorphism. All right, but that now follow if it's Abelian. Why is a group of order p squared Abelian? Well, by the previous theorem, since this is a p group, right, your order's p squared, the center of that group is non-trivial. So it can't be, the order of the center can't be one. But by Lagrange's theorem, since the center is a subgroup, its order must divide p squared. The only other options are p and p squared. Now, in the case that the center has order p squared, that actually means that zg is equal to g and thus g is an Abelian group, so we're done. So there's only other case to consider is this one. Now, I want you to be aware that this is the hardest part of the argument, but it's actually sort of vacuous, right? Because if a group is Abelian, its center has to be everything. So this is not a possibility, but we're still gonna get, we're still gonna get Abelian here. Even though this is sort of a vacuous case, it's sort of a weird logical head scratcher, but just trust me, it's gonna be okay. Let's suppose that the center has order p. Now, what I'm gonna do is I'm gonna take the g, I'm gonna take g and mod out by its center. The center is always a normal subgroup. And since the order of g is p squared, if you take a group of order p and divide it in this situation, the order here is gonna be p squared divided by p. So this is p. So this is the order of the quotient group, g mod out zg, okay? And so the quotient group g mod zg has order p, but we've also previously seen in the math 42.20 abstract out of one that if a group has order of a prime, then it must be a cyclic group. So g mod zg is isomorphic to zp. So it's a cyclic group. The quotient is a cyclic group. Now, also in math 42.20, there was a homework exercise that was assigned that said that if a group mod out its center is cyclic, then it actually turns out the original group was a b-link in the first place. So that will give us what we want here. Now, this was a homework exercise. This, so I actually wanted to provide the solution to it. And it's a question I gave in the first semester. So if you're watching this video, I'm assuming you finished that semester. So you're not gonna get a simple answer to that question. But let's actually look at the proof of that one. If g mod zg is cyclic, then the group is abelian. So for simplicity, let's just call the center of the group z. So I don't have to put all this extra notation when it's clear z is the center here. And so we'll assume that g mod z is cyclic. That means there's a single element that generates the entire group. Now g mod z, it's a quotient group. So the elements of g mod z are cosets. Az is an element of that, it's some coset. And let's say that generates the entire thing. So there's some element of the group so that the coset az generates the quotient group. Now I wanna show that g is abelian. Take two arbitrary elements of the group g. So because the quotient group is cyclic, the coset gz is some power of az. So some power az to the n will give you gz. But as this is coset multiplication, this is the same thing as an to the z. Similarly, there's some power m so that az is equal to amz. So we're gonna use that to help us out here. So gz and az can both be written as some power of az. Since gz is equal to a to the nz, there has to exist some element of the center called z so that g equals an times z. Similarly, h has to be factored as a to the m times z prime where z prime also belongs to the center. So consider the following equation. If you take g times h, well, this is gonna equal anz times amz prime. Redoing the parentheses, let's look at just the product z a to the m. Since z is central, this becomes az a to the mz. Redoing parentheses again, we have a to the n a to the m. We have z times z prime. Well, this first expression here, it's just the exponents of a here. By exponential rules, this becomes a to the n plus m like so, but a to the n plus m is also equal to a to the m times a to the n. In particular, if you take the powers, two different powers of the same element, these things commute with each other. And as z and z prime are central elements, they commute with each other. So z times z prime becomes z prime times z. Redoing parentheses, we now get a to the n times z prime. z prime is again central, so it commutes with a to the n and redoing the parentheses one last time. You get a to the m times z prime, a to the n. a to the mz prime is h, a to the n times z is g. And so we see that gh commutes and becomes hg. That then finishes this homework exercise, this claim here, for which noticing what's going on here. If you'd have a group, you might not buy it center, you guys, if that's cyclic, that quotient group, then the group itself is a billion. That finishes our claim. So coming back to the original statement here, g modded center is a group of order p, which necessarily is isomorphic zp, so it's cyclic. So that means that g was originally a billion itself. That gives us what we want. And that finishes the proof. Now, like I said, that second case is sort of silly because we ended up with what we wanted, but that's impossible given the assumption. So technically that second case actually is vacuous. It's not possible, it leads to contradiction. But nonetheless, people often skip over because the contradiction you get was actually the thing you wanted in the first place. So we kind of skip over those details nonetheless. So I wanted to show you some theory behind p groups, right? There's only two p groups of order p squared, and they're the two abelian ones. And this is all a consequence of the class equation, which has to do with the sizes of these consciousness classes. It's a pretty impressive result. And so that brings us to the end, of course, to lecture four about the class equation. Thanks for watching. If you learned anything, please hit the like button for these videos. If you wanna see more like this in the future, please subscribe to the channel. And as always, if you have any questions, feel free to post them in the comments below, and I will gladly answer them at my soonest convenience.