 We have learned so many things about groups and here is yet another thing. The end is not in sight. We see what the center of a group is. Now I want to talk to you about a centralizer. Something slightly different, the centralizer. So I'm going to consider once again a very arbitrary group G. It's going to be made up of a set and a binary operation. And I'm going to consider A a subset of this set, an arbitrary subset. I'm just going to take some of these elements. And then I'm going to define the centralizer of A in G. And we write it as such, the centralizer of A in G. Something like that. And we are going to show that this is actually a subgroup of G. And we are going to define it as such. So it's going to be a group and it is a set and it's the set of all elements G. Elements of G set such that G composed with A or G binary operation A equals A, a binary operation G for all A elements in A. So I've got my set G which makes up my group. I take some of the elements out there and I call that subset A. And I take each one of those in any particular order. I'm just taking each one of those A's. And I'm going then through all of the G. Remember which is some of them are in A. But I go through all the G's. So I take my first A and I say is my first G composed with A equals A composed with G. The second one, the third one, the fourth one run through all of G's. Then go to the next A which is in A. And in all of those which we have this commutant. So in other words, if we have this commutant of property, then we call that set this centralizer and it does make a group. It does make a group. Like yesterday I changed those two around on the day before. It doesn't matter. So this is a group I'm saying that C, the centralizer of A and G is a subgroup of G. Now we have to prove the four things. Associativity is included because it is a subset. And you don't actually need to prove the identity element because it just falls out of the inverse being made. So let's just have a look at the two important ones. The closure and the inverse is there. So let's prove that there's closure. So I'm saying if G1 and G2 are elements of this centralizer of A and G, that implies that G1 composed with G2, that binary operation is whatever this element is, is also an element of the centralizer of A and G. So what would that mean if it is? It means if I have G1 and G2 and A, that must equal A and G1 and G2. So that is what we must have. If it fulfills that property, then by all means there is closure because that would fall whatever this element is, it falls in that definition. So let's just have a look at G1 composed with G2, composed with A. Let's write compose this with the identity element. And one way to write the identity element is if I take G2 inverse composed with G2. That is very simply just the identity element. I take G2 and it's inverse in that order. But remember I said that G2 is an element of this. So if it is, it means G2 composed with A equals A composed with G2. So I have G1 composed with A, composed with G2's inverse, composed with G2's inverse. G1 composed with A, composed with G2, composed with G2 inverse, composed with G2. That's what we have there. Now with our associativity we have this and that is just the identity element. So let me go up here, so I have G1 composed with A, composed with, let's say identity element, composed with G2. But G1, remember, is an element. I'm giving that that is an element in there so I can commute those two. So I have A, so that equals that, that equals that. So I have A, composed with G1, composed with G2. So I see that indeed I started off with, I started off with this. G1, composed with G2, composed with A and I show eventually that it does equal A, composed with G1, composed with G2. That fulfills or satisfies my definition of this. So I have shown there is closure. So if G1 and G2 is there, the binary operation between them is also an element. So we have closure. Let's just look at the inverses then. So I'm saying that inverses if G is an element of the center of A and G, that implies that G inverse is an element of the center of A and G. If that is so, what would we have? We would have the G inverse of and A equals A and G inverse. That's what we would have. Now, let's see if we can get there. Let's see if I start with this. And that would be equal to G inverse and A. And I suppose I can put the identity element there. And one way to put the identity element there is once again just G composed with its inverse. That's the identity element, so I've done nothing. Associativity, so I can do those two. But G is an element of that, so I can swap those two around. So I have G inverse and I have G and I have A and I have G inverse. Associativity with those two, that's just the identity element. So I have A composed with G inverse. And I've shown that G inverse of A equals A with G inverse. G inverse composed with A equals A is composed with G inverse. Once again, that satisfies my definition of the centralizer. So we have the fact that if G is in there, its inverse is also in there. We've shown that we have the important properties. And if they are both in there, if they are both in there and G and its inverse, if you compose the two, you get identity elements. Of course, the identity element is one of them. Let's say the identity element falls out. We don't have to prove the identity element every time. So there we have it, the fact that we have this centralizer. And there's something very special then about this set of elements in G. And sometimes we don't always just look at a whole. We just take one of the elements in this group, in the set that makes up the group, and we see what is the centralizer of just that one element. But you can take more than elements in arbitrary element and there will be a centralizer and that centralizer will be a subgroup.