 Welcome back to our lecture series Math 4230 abstract Edward II for students at Southern Utah University. As usual, I'll be a professor today, Dr. Andrew Misseldine. In lecture six, we introduced the notion of a Seelof P subgroup and we proved the first Seelof theorem, which basically says that Seelof P subgroups exist. We actually prove that for every power of a prime that divides the order of the group, there is a subgroup of that order that exists. In particular, those who have the maximum number of P as their exponent there, those will necessarily be a Seelof P subgroup, although there could potentially be other ones, but we'll prove very shortly in the second law that that thing is not the case. The goal of lecture seven is to prove the second and third theorems of Seelof about Seelof P subgroups and such. In this video, there's two lemmas I want to present. That'll be very useful for both proving the remaining two Seelof theorems, but also just our general study of Seelof P subgroups and P groups in general. Let me present these two lemmas in this video, and then the next video we will, of course, prove the second Seelof theorem. Our first lemma is about a Seelof P subgroup, which remember, a Seelof P subgroup means it's a maximal P subgroup of G. Suppose P is a Seelof P subgroup, and suppose that little x is an element whose order is a power of P. If x belongs to the normalizer of P, which remember, what does the normalizer mean? The normalizer is the collection of elements in the group such that x P x inverse is equal to P. And this is a statement of the set, not of the elements. If you conjugate the subgroup, you get back the exact same subgroup. Now, if x is in the normalizer of the Seelof P subgroup, then it turns out that x actually belongs to the Seelof P subgroup itself. So x has an order of a power of a prime. So if a P element normalizes a Seelof P subgroup, then it actually belongs to that Seelof P subgroup. Okay, so how we prove this? Well, note, when it comes to normalizers, normalizers always have the subgroup they're normalizing as a normal subgroup. That's the whole point of normalizing. And in particular, the normalizer is the maximal subgroup for which that is the normalizer in P of P is the maximal subgroup for which P is normal inside of it. In particular, since P is a normal subgroup of its normalizer, you can quotient it out and look at the quotient group in P mod P. All right, so consider this quotient group for a moment and then take this element x, take the coset that contains x, it would be x P and take the cyclic subgroup generated by x P inside of this quotient group, call that cyclic subgroup K. And so this is a cyclic subgroup of NP mod P. All right, now since the element x is a power of P, its order is something, I don't know what it is, P to the A. We have that x to the P will also be, its order will also be a power of P, right? Because if I take x P and I raise it to the P to the A power, this is equal to x to the P to the A times P, which would then give you the identity times P, which is just P, like so. Now I'm not saying that this element's order is P to the A, but by Lagrange's theorem, if P to the A sends this to the identity, then the order of the element must divide P to the A. But as it's a power of a prime, the only things that divide a power of a prime are powers of that same prime. You know, possibly smaller powers, but it's a power of that prime, so that's important here. So the reason we go to the quotient group is we construct this element here, great. But then we're gonna lift it back up using the correspondence theorem, which says, the correspondence theorem says there's a unique subgroup of the normalizer NP, such that H will contain P as a subgroup. And that's because P was the group we quotient out. The correspondence theorem always will give us a subgroup that contains P, but also we have that H mod P corresponds with K. So H mod P gives us K. So the correspondence theorem gives us this theorem H, it exists, but what we care about H here is that the order of H will be the order of P times the order of K, all right? By assumption, since the order of XP was a power of P, the cyclic subgroup generated by XP will also have order of power of P. So K is a P group, all right? So now look at H, we have a P group times a P group, that is the order of P is a power of P. The order of K is a power of P. So H has as its order a power of P, it is a P group. But whoa, whoa, whoa, wait a second. H is a subgroup that contains P. H is a P subgroup. P is a Seeloff P subgroup, which means it's a maximal P subgroup. Whenever you have this maximal condition, it means since H contains P, but P is maximal as a P subgroup, it must be by the maximality here that H and P are one in the same thing. So if H and P are the one in the same thing, when I take H mod P, this is actually P mod P, which is gonna have to give you the trivial subgroup there. In particular, this is just the element P, right? It's the trivial subgroup. This tells us that K, which is H mod P, is the trivial subgroup, which implies if K was the trivial subgroup, it's the subgroup generated by XP. This has to be just P itself, the coset. So XP is actually equal to P, which that implies that X belongs to the Seeloff P subgroup P itself. So proving the first lemma there. Now let's look at a second lemma that's gonna be necessary as we study P groups and such. This lemma doesn't actually say anything about Seeloff P subgroups, but it's about normalizers, which normalizers are a very important tool we use to study Seeloff subgroups. So imagine we have two subgroups H and K of some group G. Then the number of distinct H conjugates of K. Let's stop and look at that for a moment. So you have a subgroup K. If you look at any conjugate of a subgroup, this is also a subgroup. It's actually an isomorphic subgroup, but you'll get another subgroup. Now what if I conjugate K by some element of H? So I don't conjugate by all elements of the group. I only look at the conjugates with respect to the subgroup H. That's what we mean by H conjugates of K. We look at all the possible subgroups we can form by conjugating the subgroup K by some elements of the subgroup H. The number of H conjugates of K is gonna equal the index of H with respect to the intersection of the normalizer of K with H right there. So how are we gonna get this? Well, we're trying to say the number of H conjugates of K is equal to an index where here we have a group action going on. We have that if you look at the conjugates of K, H acts on them by conjugation. And so there's a group action going on here. The group G naturally acts on the set of its subgroups which include K by conjugation. If we restrict G down to H, H will also act on the set of subgroups by conjugation. It just turns out that by restricting the subgroup down to a subgroup, you might not be able to connect things that were connected before. So in terms of the partition, the orbits associated to the action, by restricting the group, you actually might have more orbits, but that's okay. G conjugates all of its subgroups. H will also act on all of the subgroups of G by conjugation including K right here. So we need to compute the size of the orbit of K with respect to this H conjugation action. And this is where of course the fundamental counting principle comes into play. The orbit of K with respect to the H action will be H index HK where HK is the stabilizer, the isotropy subgroup of K with respect to the conjugation action. So what we need to do to prove this theorem is we're looking at these indices here. We have to prove that the isotropy subgroup of H acting on K is in fact equal to the normalizer of K intersect H. So we wanna prove these two groups, these two subgroups are the same thing. So we just wanna prove their subsets of each other. One direction is pretty easy. Clearly, HK is a subgroup of H. We prove that stabilizers are always subgroups of the group that's acting. In this situation, H is acting on K in its conjugates and therefore the stabilizer will be a subgroup. So that's pretty easy. So we get that H is a subgroup, excuse me, HK is a subgroup of H. We wanna then show that HK is a subgroup of the normalizer. Well, how are we gonna do that? Well, HK, the stabilizer of K with respect to the H action is a subgroup of the stabilizer of K with respect to the G action. Why is that? Well, everything that stabilizes K in H is an element of G that also will stabilize K. The action didn't change, it's just the number of elements in play might change. Now it could be that G has more elements that stabilize K that H doesn't. So it could be bigger, but HK is a subgroup of GK. But GK is just the normalizer of K. That's how we define the normalizer by this conjugation action here. And so therefore HK is a subgroup of the normalizer. And since it's a subgroup of both the normalizer of K and H, it's gonna be a subgroup of their intersection. That gives us the first direction. We got that one. So we have to go the other direction. So this one's a little bit harder, but it's not so bad. So let's take an element that belongs to the intersection of the normalizer of K and it's intersection with H there. So in particular, X belongs to H. So X can act on K by conjugation. Now, if X is inside the normalizer of K, that means it stabilizes K. That means that XKX inverse is equal to X. So because of that, the element X stabilizes K, but it belongs to H. So it's inside the stabilizer of K with respect to the H action. That's telling us here that since X is inside of HK, we didn't finish this equality, which then finishes the result. So we prove these two very important lemmas. These lemmas have general applications in group theory, but in particular, we will apply them to the Sealoff theorems in the next video.