 Now we've looked extensively at the centralizer. We looked at it from a computer, from a Mathematica point of view. I showed you, there's some intuition there and then I showed you that it is a group we went through all of those. But now I want to show you something that is the bigger cousin of the centralizer called the normalizer. The normalizer of a set in a group. As per usual we're going to start with a group. It contains a set and a binary operation and I'm going to take some arbitrary subset of the set that makes up the group. We've done this many, many times. I'm going to define the normalizer of a set A in a group G, writing it like that. I'm going to start with the set that makes this up. I'm going to say it's all the elements in G set, all elements of G set such that I have the following. I take A and I really want this to happen for all A element in A. I want it to happen for all that. What do I want to happen to this? I want to conjugate this with G. And that means G, binary operation A, binary operation G inverse. And I'm saying that that has got to be an element of A. So it doesn't equal A. And by the way, that's the other way to set up a centralizer. We just looked at it as commuting. But you can also set it up in this way that this equals little A. Equals that little A. But now I'm relaxing things a bit. I'm saying it's got to be... So let's say this is the set G and I have some small, some part of it. Doesn't matter how big it is and there's an A in there. If I conjugate that A I'm going to land up with another element but that element is also in A. But it is definitely in... It can be little A but it definitely has to be an A. And now I want to claim that this is a group under that same binary operation that makes up the group. So I've got to show closure and I've got to show inverses. So let's look at closure. So the closures say they're better if I have G1 and I have G2 and they are elements of this normalizer. Normalizer of A and G. If that is so, this implies that if I compose them with binary operation there will also be an element of the normalizer of A and G. So all I have is this and that. Now, if this was so, what would we be able to show? We would be able to show that if I take G1 and G2 whatever that binary operation is, whatever that results in and I use it to conjugate, so I take whatever it is again, inverse, that I land up this being an element of A. So I better have this end up an element of A. Now there's something very neat going on here. I just want to show you that. I hope that as you can see in the corner. If I have G1 composed of G2 and I compose that with its inverse, G1 composed of G2's inverse, what do I get? I get the identity element. So I have an element composed of its inverse, I have that. But this inverse, I can rewrite that in a different way. I can take the second one's inverse, compose it with the first one's inverse. I claim that that's the exact same thing. If that was so, if I do this, I better end up with the identity element because by associativity, that gives me the identity element. So it's this, the identity element, that, these two give me the identity element and I'm just left with E. So I'm claiming that I can rewrite this as follows. So going on from here, I have still this and I have A and now I have this new thing, G2 inverse composed of G1 inverse. I'm claiming that those two are the same thing and I'm showing you here that they are exactly the same thing. Now, hang on a minute. If we look at this carefully, what do I have here in the middle? I have an element of my neutralizer, I've been told that it is, with A, with that thing's inverse. So I have exactly that. So this better be, by definition, this is an element of A. Let's call it A star. So I have G1 still and I have G1's inverse still there and I can't replace this with one of the elements in A because this fulfills my definition. I'm told that G2 is an element and it fulfills my definition. Same story, G1 is, fulfills my definition. So this is an element of A. So I have closure that if I have two elements being, two elements being, elements of the neutralizer, their composition is also. We don't have to show associativity because we have that. And now we really have to, the identity element we really don't have to show because, I mean, E, identity element, identity element, A is an element of A. It satisfies that. So apologies for that. I just had to take a call. We were busy with the inverse. So I'm saying if G is an element of the neutralizer of A and G, that implies that G inverse is also an element of the normalizer of A and G. That's what we're saying. And if G inverse is an element, think about it. If G inverse is an element, that means I can plug G inverse in there and G inverse in there. So what I would have in essence is G inverse composed with A, composed with G inverse is inverse. That must be an element of A. That's what's going to happen. But the inverse of the inverse, that's just G. So that's what we're going to have. Now, what can we do with only what we have here? Not much. We have to use something else. Let's create this mapping from A to A. I'm going to call it sigma. Under G, we map from A to A. I'm going to take some element A and it's going to map some way. It's going to map to G, A, G inverse. So what we have here is we have the sigma of G maps A to G, A, G inverse. I'm not putting the binary operator in there. I have some space issues here. It would be lovely if this mapping was bijective. If this mapping was bijective. And what we want to show with a bijection is if we have the G inverse, that's going to map A to G inverse A, G. And the way to show that that is a bijection because that would be very nice. That means when I do this, I land up with something that is in A. And it's uniquely in A. There's one here, one there. And we can do this by showing again that this is, and we've done it before, showing that this is a two-sided inverse. And the two-sided inverse means if I take sigma G of the sigma G inverse of A, that I land up with A, and I have that the sigma G inverse of the sigma G of A is also A. Is this true? Well, we have sigma G there. What is the sigma G inverse of A? Well, that is G inverse A, G. And if this is a new thing, and I take this new thing and I plug it into the sigma G, I have that. So remember this is now the new A, which is the new A there. So I have my G. Instead of A, I now have this G inverse A, G. And I have my G inverse in the end and learn, behold, G G inverse, G G inverse, and I have A. The same goes for here. Let's have a look. What is the sigma of G inverse? What is the sigma G of A? The sigma G of A is G, A, G inverse. And the sigma G inverse of this would be G inverse G, A, G inverse G. And that's also A. So by showing that this is a two-sided inverse here, this mapping from A to itself, that this mapping is one to one and on two. And that is really brilliant because what I can replace this with is something to the effect of G, A star, G inverse. G, A star, G inverse. Why is that so? Because A does map, A maps to something to, it's a one to one and on to mapping. So what we have left is G inverse. Instead of A, we have G, A star, G inverse G. And that is just A star, and A star is an element of A. It is just another element of A. So by showing that we can set this up as a one to one and on to mapping here, we show the two-sided inverse. It means I can rewrite A, the element A, just as something else, some other mapping. And that is why, say for instance, this is just A star, this new element is A star, make a bow, and I do this inverse mapping, and I do a mapping of it, G and G inverse. I'm going to, this is this new element, I'm calling this new element A star, so A mapping to A star. If it's one to one and I take this mapping of it, it maps back to one, because we've shown it is one to one and on to. So I can rewrite A as this, G A star, G inverse. And because this was this mapping, I can call this A star, it is still an element of A. I've shown that it's a one to one mapping. It is an element, if I do that to G star, I've got to go back to A because it's one to one and on to mapping. So one to one and on to mapping, shown here by the two sided inverse. I can plug that in for A instead of A and I'm left with A star, which I have chosen to be an element of A. So if G is an element of the neutralizer, it's inverses also. And if they're both there and I compose them, I do get the identity element and I've shown closure already, so that's another way to show the identity element is also there. So we don't have to show the identity element every time. Now, so that's the neutralizer. There's a very special subset A and that is if I turn A into a subgroup of G. If I turn A into a subgroup of G, so I'm not an arbitrary A anymore, I make it specifically, so I add a binary operator and I make it a subgroup. That implies that A would then also be a subgroup of the neutralizer of A and G and even more so, it will be a normal subgroup. Not only a subgroup, but a normal subgroup. Why? Well, because that there, I'm still, that is the definition of a normal subgroup that if I take something, those elements are conjugated and I land back there. Well, that's just a normal subgroup. That's how we defined a normal subgroup. So just two special things about the normalizer. So I think this is easy. This is slightly more difficult, but remember I'm sitting up this A goes to something else. I'm going to call that something else A star. I show by two-sided inverse that this is a one-to-one mapping. I've proven that it is a one-to-one mapping. So if this thing is A star and I conjugate it, it lands back to A. So I can rewrite A as this thing and I'm left with something that is still within A. So I have inverses, I have closure, I have associativity because of that and the inverse falls out. I mean the inverse falls out from there and the inverse falls out of G and G inverse. Now in enclosures in the composition of G and G inverse it gives me E, so E is also an element. So you don't have to prove that every time. So very nicely the normalizer of a group.