 In this final video for lecture six, I want to very quickly define the notion of a normalizer. Normalizer is very similar to the idea of a centralizer, but they're very important when we talk about Seeloff theorems and we consider simple groups, which remember simple groups don't have normal subgroups. So what is a normalizer? Suppose we have a subgroup H inside of some other group G. We define its normalizer, commonly denoted as N of H. This is going to be the set of all elements of the group such that the left coset coincides with the right coset. And another way of phrasing it is that we want all the elements of the group such that if you conjugate the subgroup by G, you get back H again. Now I want you to recall that a group acts on its set of subgroups by conjugation. Kind of like we're seeing right here. You can conjugate a subgroup and get another subgroup. So the collection of subgroups is itself a G set. The group acts on it by conjugation. Now under this action, the normalizer H or N of H right there is none other than the isotropy subgroup of H with respect to this group action. So in particular, H sub N is a subgroup of G because isotropy subgroups are always subgroups. Now this relates to the centralizer because a group acts on itself by conjugation. And in that situation, the isotropy subgroups are the centralizers. But the group also acts on its collection of subgroups by conjugation. It's a similar action, but the sets are different. G is acting the same way, but the sets are different. You have the elements versus the subgroups. And in this situation, when you're acting on the subgroups, you get the isotropy subgroups to be the normalizer. So that's the relation. That's the similarity between centralizers and normalizers. They're both isotropy subgroups of a conjugation action. But the set that G is conjugating is different sets. There's the elements of the group versus the subgroups of the group. I should also mention that when it comes to the normalizer, the subgroup H is normal inside of the normalizer. Because the normalizer, you take all the elements so that the left and right cosets are the same thing. Or another way, you're taking all the elements of the group for which when conjugating on H, you're closed and still inside of H. So the normalizer is a group for which H is normal inside of it. And in fact, the normalizer is the maximal subgroup of G for which H is normal inside of it. So a very quick corollary here is that the normalizer of H is equal to G if and only if H is normal inside of G. In particular, H is always normal inside of its normalizer. I also want you to be aware that the elements in the normalizer, those are elements of the group which commute with the set H. So they commute with the set H like this, but they don't necessarily commute with elements of H. So you should want to be very cautious about that. This commutation is only at the set level. It's not on the element level. The set G times H is equal to H times G, not elements though. Individually, things can be a little bit more shuffled around there. In particular, if I were to take any element in H, notice that HH is equal to H, which is equal to HH, like so. So what this tells us is that the normalizer of H always contains H itself. So if H is a non-trivial subgroup, then N of H will be non-trivial. But of course, if H is the trivial subgroup, it's normal. So it's normalizers all of G. And so in particular, what I'm trying to say is normalizers are never trivial. They're always a non-trivial subgroup. But also, if you take an element from the center of the group, if you take some element of the center of the group, then you're going to have that, well, Z, H, Z, this equals Z, Z inverse H here. This equals H for anything. So in particular, the central elements commute with every element. Therefore, they commute with every set. If you take the subgroup H, this is going to equal Z, Z inverse H, which is just H, like so. So this tells us that the center is always a subgroup of the normalizer as well. So these are some important observations on normalizers. The normalizer of a subgroup always contains the subgroup. It also always contains the center of the group. There could be other things in there as well, but those are two subgroups that are guaranteed to be inside of a normalizer. Now, if you take, for example, your group S3, this is a centerless group. Its center is actually just the identity. So this last statement doesn't really do us a lot of good in that situation, but still consider it. Let's take the subgroup of the two-cycle 1, 2. What is its normalizer? Well, H is not a normal subgroup, so the normalizer can't be all of S3. It has to be something smaller, but the only subgroup smaller than S3 that contains H is H itself. So it turns out the normalizer of a transposition in S3 is just itself. You can't actually get any bigger than that. And so in some essence, it's very hard to normalize these things, because when you normalize it, you didn't gain anything else. Typically, normalizers grow, but not in this situation. Now let's switch gears a little bit. What if we look at the subgroup K? K is still inside of S3, of course. If you take the alternating group, that is the cyclic subgroup generated by 1, 2, 3, it is a normal subgroup. K is normal inside of A3, so its normalizer would be all of S3. So those are some sort of simple examples. Now let's switch gears and look at the dihedral group D4. I am focusing on non-Abelian groups right now, because in Abelian groups, every subgroup is normal. Therefore, every normalizer is the whole group. There's not a lot of interest to go on there. But if you look at some non-Abelian groups like D4, let's look at a non-normal subgroup, the subgroup generated by S, a single reflection there. Well, it's not normal, the subgroup, so its normalizer can't be everything. It contains itself, so the normalizer has to be a subgroup of D4 that's divisible by 2, but not everything. D4, of course, has order 8. So that tells you the possible orders of the normalizer of H is 2 and 4. If it were 2, it would just be H. So is there something else that could normalize H? It's possible, and it turns out there actually is something else. Consider the element R squared. R squared is central, right? The center of this group, ZD4, this is actually the identity in R squared. And like I said, the normalizer always contains the center. So this means that the normalizer has to contain the subgroup itself, it has to contain the center, and then it'll also contain anything we can produce from them, so like S times R squared. In particular, we're looking for the subgroup generated by S and generated by the central element R squared. So the normalizer in this situation would have order 4. And so if you conjugate H by any element of this group, you get back H, because that's what a normalizer does. And so as we further study Seelof p-subgroups, the tool of a normalizer will be very, very useful, and that's what we're introducing at the end here of lecture 6. We're going to use it in all of the subsequent lectures we cover in our study of Seelof theory. Thanks for watching, everyone. Give this video and any other of these videos a like. If you learned something, subscribe to the channel to get some updates about future math videos like this one. And as always, post your comments below. If you have any questions, I'm glad to answer them. Bye, everyone.