 So this lecture is part of an online mathematics course on group theory, and we'll be covering the C-LOV theorems. So the first question is, how do you pronounce this name? He is a Norwegian mathematician, although it's often pronounced silo in English. I think Norwegians pronounce it something like C-LOV, and my apologies to any Norwegians if I haven't really got it right. So if we've got a group G, a finite group G, then any subgroup H has order dividing G, and we can ask the converse. So suppose a number D divides the order of G, we can ask, does G have a subgroup of order D? And most of the examples of groups we've seen do have subgroups of every order dividing D. However, this isn't true in general. Suppose we take G to be rotations of a tetrahedron, then G has order 12, but G has no subgroup of order 6. You can see this. For example, any group of order 6 has a normal subgroup of order 3. And when we look at the symmetries of a tetrahedron, you can look at all the subgroups of order 3, and you can see that none of them are normalized by some subgroup of order 2. So this just doesn't have subgroups of order 6, even though 6 divides 12. Well, there's one case, a very useful case in which you can show that G has subgroups of order. And this, so the Seelow theorems say the following. Suppose P to the n is the largest power of P dividing the order of the group. Then G has subgroups of order P to the n. Secondly, we can ask how many there are? Well, the number is one mod P and divides the order of G. And thirdly, they're all conjugate. In particular, they're all isomorphic. And fourthly, any subgroup of order power of P is in one of them. So these subgroups are called Seelow P subgroups. So this actually allows you to get hold of the subgroups of a group, because at least for order P to the n for n, the maximum order dividing G, there's a single conjugacy class of subgroups. So what we'll do is prove this theorem. Let's first show existence. So we're trying to show the existence of a subgroup of order P to the n, where P to the n is the largest power dividing the order of G. So first suppose G has a subgroup, let's say a proper subgroup of index, co-prime to P. So if this proper subgroup is H, then P to the n divides the order of H and the order of H is less than the order of G. So H has a subgroup of order P to the n, so G does. So here we're using induction on the order of G. We're assuming that every subgroup of order less than G has P Seelow subgroups. Well that was assuming that G has a subgroup of index co-prime to P. So suppose not. So suppose not. That means all proper subgroups have index divisible by P. And now what we do is we copy a bit of the argument that we used in proving Cauchy's theorem. We look at the order of G, and do you remember this is the order of the center of G plus sum over all conjugacy classes of size greater than one times the size of the conjugacy class. Now the size of the conjugacy class of size greater than one is the index of some proper subgroup of G such as the subgroup fixing an element of it. So all these are divisible by P, and this is divisible by P. So the center of G is divisible by P. So G has an element of order P in the center. So you notice this is almost exactly the same argument that we used to prove part of Cauchy's theorem. Anyway, now what we do is we look at G modulo, the cyclic group of order P where G to the P equals one, G is not equal to one, and G is in the center. So because G is in the center, this is a normal subgroup, so we can form this quotient here. So we now pick the seal of P subgroup S of G over P. So S is order P to the n minus one because the order of this group here is the order of G divided by P. And now look at, we've got a map from G to G over P. And this contains a group S, and we just look at the inverse image. Let's call it S prime. So the inverse image S prime of S has order P to the n, and it's a subgroup, so is seal of P subgroup of G. We've shown that every finite group has seal of P subgroups, which are subgroups of order of the maximal power of a prime dividing G. And so next we want to show that the number of seal of P subgroups is one mod P. Well, suppose S and T are seal of P subgroups with S not equal to T, then S cannot normalize T. Well, what does normalize mean? Well, we say that a subgroup A normalizes B if A, B, A to the minus one is in B for all A in A and B in B. In other words, the conjugate of B by any element of A is just B. So if A was the whole group, then the group normalizes B if and only if B is a normal subgroup. Well, why can't S normalize T? If S normalizes T, then ST is a subgroup because S1 T1 times S2 T2 is equal to S1 times S2 times S2 minus 1 T1 S2 times T2. So the product of two things in ST is an ST because this is in T. But this subgroup has order P to the N, so P to the K for K greater than N because it contains T which has order P to the N. And then it has an additional power of T because of S which is not possible as N is the maximum possible power of P dividing the order of the group. So if we've got two distinct set of subgroups, they cannot normalize each other. Now we look at the orbits of S on the set of P set of subgroups. So let's look at the orbits. There is one orbit with one element, which is just S itself, and all other orbits have P to the K elements for K greater than one. This is because the number of orbits is the order of S divided by the subgroup of S normalizing T, where T is some element in this orbit. And as we said, the subgroup normalizing T can't be the whole of S. So this subgroup here is strictly smaller than S, which means that this number here is divisible by P. So what we've seen is that the orbits of S on the other subgroups, there's one orbit of size one, all other orbits have size divisible by P. So the number of conjugates of S is divisible, sorry, is not divisible by P, is one mod P. Well, now suppose S and T are not conjugates. Well, the number of conjugates of S is one mod P by what is said above, but the number of conjugates of S is also not mod P, because since S is not conjugates of T, the number of conjugates of T on any conjugate of S under the whole group G is zero mod P, because these conjugates are not conjugates of T, so what we said earlier, the number of orbits of T on them is divisible by P. Well, this is a contradiction because the number of conjugates of S can't be one mod P and not mod P, so S must be conjugate to T. So there's only one conjugacy class of seal of P subgroups, so all seal of P subgroups are conjugate and the number is one mod P. So the fourth thing we want to prove, I guess that was the second and third thing sort of combined, is that any subgroup of order P to the A is contained in some P seal of subgroup. And the proof of this, what we do is we pick a maximal one, so if not true, pick a maximal subgroup, let's call it X of order P to the A. So we're picking a maximal subgroup whose order is a power of a prime but not contained in a P seal of subgroup and we're going to get a contradiction. So what we do is we notice the normalizer of X has index divisible by P, the normalizer of X is the things that normalize it, so this is the group of all elements G with GX G to the minus one equals X. So the number of conjugates of X is divisible by P. But by the argument above that we used for seal of P subgroups, the number of conjugates is one mod P, and this is a contradiction, the number of conjugates can't be divisible by P and also one mod P. So X must actually be contained in a P seal of subgroup. So we should give an example. We've said the seal of P subgroups are all conjugate and all isomorphic. That only applies to subgroups divisible by the maximal power of P dividing the order of the group. So suppose we take G to be the dihedral group D4, then G has subgroups isomorphic to Z over 4Z which could just be rotations for example. And it also has subgroups isomorphic to Z over 2Z times Z over 2Z. Because we can for example take the subgroup generated by flipping horizontally and vertically. So you see these two subgroups is both of order 2 squared, but they're not isomorphic, so they're not conjugate. So the part of seal of theorem about subgroups of order P to the N all being conjugate only works when N is the largest power of P to the N is the largest power of P dividing the order of the group. So in the next lecture we're going to use the seal of theorems to classify the groups of order 12.