 This lecture is part of an online mathematics course on group theory, and will mostly be about groups of prime power order. So we start with the following problem. We've classified all groups of order at most eight. So what we want to do now is to classify groups G of order nine. And let's first suppose that G is abelian. Well, we know the orders of the elements must divide nine, so there must be one, three, or nine. So suppose G has an element of order nine. Well, this implies G is cyclic, and so isomorphic to Z over nine Z. If G has no element of order nine, then that means every element is ordered three or one. So if you write G additively, this means G is a vector space over the field of three elements. So G is isomorphic to two-dimensional vector space, which is just Z modulo three Z times Z modulo three Z. So abelian groups of order nine are quite easy to classify. You notice, by the way, that this works for groups of order P squared for any P. So let's just assume that G is actually that G is of order P squared for P prime. So what we want to do is to show that G is abelian, in which case it will be either cyclic of order P squared, or it will be product of two cyclic groups of order P. So we're going to prove this in two steps. The first one is every group of order P to the N, this is where N is greater than or and P is prime, has non-trivial center. So we're actually only going to use, in this lecture, we're only going to use this for groups of order P squared, but later on this will be useful for higher powers of P. So what does the center mean? Well, the center is the elements G that commute with everything else. So GH equals HG for all H. Non-trivial means more than one element. Obviously, the identity element is in the center of any group, but that's not terribly interesting. So far, most of the groups we've had so far have had a non-trivial center. If you look at the group S3 of order 6, the center is trivial. So for groups that are not of prime power order, this one is order 6, for example, in general, that they don't necessarily have anything in the center apart from the identity element. And the proof of this, what we're going to do is we look at the joint action of G on G. So you remember this is the action where G acts on an element H by taking it to GHG to the minus one. And the orbits are the same as conjugacy classes. And the orbits of size, the center is just the orbits of size one, because saying an orbit is of size one just says that G of H is equal to H for all H, which just says that H commutes with G. And all orbits have size, well, the size of an orbit is the order of G divided by the order of the subgroup fixing an element of that orbit. So you remember the order of G is the size of an orbit times the subgroup fixing an element of the order. So this is some power of P because we're assuming G has order of power of P. So let's look at the order of G. This is equal to the number of orbits of size one plus the sum over all orbits of size greater than one times the size of the orbit. And now let's look at this. Well, this term here is divisible by P because we assumed N was greater than zero. Of course, if the group is order P to the one, then it doesn't have a non-trivial center. This bit here is divisible by P because each term is a power of P. So this is divisible by P. Well, it must be at least one because we've got the identity element in it. So it's at least P in size. So we've shown that every group of order P to the N has a non-trivial center. The center must have at least P elements in it. So that's the first result we need. The second result we need says that if G modulo the center of G is cyclic, then G is equal to the center of G. So this notation here just stands for the center of G. And the center of G is kind of obviously a normal subgroup because if we take G, H, G to the minus one for something in some element H, and here it's equal to H. So the center of G is a normal subgroup. And since it's normal, we can form a quotient group. And what I'm saying is that this quotient group can only be cyclic if G is equal to its center in which case this group is trivial. Of course, there are plenty of cases where G modulo its center is not cyclic. So for example, suppose G is the quaternion group of order eight, then you remember the center of G consists of two elements plus or minus one. And G modulo the center of G is just a product of two groups of order two. So here G modulo the center of G is not cyclic and it's not trivial, it's a product of other groups. Anyway, this follows fairly easily. So suppose G modulo ZG is cyclic. Pick some element G in G such that the image generates Z modulo GZ, which we can do, this is more or less the definition of saying that this group is cyclic, that there is something generated by it. Then every element of G is of the form G to the N times Z, where Z is in the center of G and N is an integer. And this follows because if we've got any element, its image is in here is equal to the image of some power of G. So if we just divide by that power of G, we're left with something in the center. And you notice that G to the M, Z1 and G to the N, Z2 commute. That's because Z1 commutes with everything and Z2 commutes with everything because it's in the center and G to the M commutes with G to the N because they're both powers of G. So in other words, everything in G commutes with everything else in G. So G is equal to the center of G and the quotient must in fact be just one element. So we've got these two results, one of which says the center of a P group is non-trivial. A P group is just a group of order of power of P. And the other says the quotient of a group by the center can't be a non-trivial cyclic group. So now let's take G to be of order P squared or 3 squared if you prefer. And look at the order of the center of G. The order of the center of G divides the order of G, which is P squared. So the center of G has order one P or P squared. And this is not possible by our first theorem as the center of G must have ordered bigger than one because G is a P group. And this is not possible by our second theorem because it would imply that G modulo Z of G has ordered P. So G over Z of G is cyclic. And if it is cyclic, it would have to be trivial. So this would be trivial, but it is order P, which is impossible. So the center of G must have ordered P squared. So the center of G is equal to G. So G is abelian. So summary, any group of order P squared where P is a prime is isomorphic to Z modulo P squared Z or Z over PZ times Z over PZ. I'll just finish by pointing out an important consequence of this. The consequence is that all P groups are nilpotent. I'll explain what nilpotent means in a moment. So a P group means order a power of P for some P prime. Nilpotent means we can kill the group by repeatedly, repeatedly killing the center. So what this means is suppose I take a group G0. Then I put G1 equals G0 modulo the center of G0. So I've killed off the center of G0. Then I put G2 equals G1 over the center of G1. And then I put G3 equals G2 over the center of G2 and so on. And if eventually Gn is trivial, so if Gn is trivial for some n, the group is called, so the group G0 is called nilpotent. The name nilpotent seems a bit odd since there doesn't seem to be anything that's nilpotent anywhere in sight here. But it actually comes from the theory of Lie algebras. It turns out that for Lie groups, if the Lie algebra of the Lie group consists of nilpotent matrices, that means matrices such that some power is zero, then the Lie group is nilpotent. But the word nilpotent doesn't really fit very well with finite groups, so we're sort of stuck with it. Notice, by the way, that if you kill the center of a group, the resulting group might still have a non-trivial center. For instance, we saw this with the quaternions. If we take G0 to be the quaternions, then G1 is the quaternions modulo, the group plus or minus 1. So even though we've killed the center of G0, G1 might still have a center. There are things, there are elements of the quaternions that don't commute with each other, but do commute once if you've killed off the element minus 1. So in this case, G2 would be the trivial group. So next lecture, we will be covering groups of, classifying groups of order 10 using Cauchy's theorem.