 Before we go into cosets and where we're really going with cosets, it remains this little mystery about why we're spending so much time on cosets. Remember this little picture from yesterday? We have that A is not an element of H, but I might as well have drawn A being an element of H. I have my left coset, I have my right coset. And the question is, when are these two cosets, is my left coset, when are they equal to each other, my right coset? And of course, if I define this as A in binary operation with H, there's my binary operation. And this I define as H in binary operation with A, such that H is an element of H. You can well imagine that if the group is a billion, then I have commutivity, of course, and then they are equal to each other. Now, there's another specific circumstance in which these two cosets are equal left and right and cosets of an element A of the group G is equal to each other, the left and right cosets, if the subgroup is what is called a normal subgroup. So I just want to spend this little video on normal subgroups. What is a normal subgroup? And if H was a subgroup of G, it's usually written in textbooks as this closed triangle, a smaller than, less than sign, and this line closing it. That's a normal subgroup. And we define, there's a definition of when something is a normal subgroup. And we say that H is stable under conjugation by the elements of G, so that if we have H is an element of H and G is an element of G, we have this fact that H is stable. So G, binary operation with H, binary operation with G inverse is still going to be an element of H. It is stable under conjugation by all the elements inside of G. So H is an element of H, G is an element of G, and we have this conjugation. We say H is stable under conjugation by elements in G because G, binary operation H, binary operation G inverse, those have to be the inverses of each other, which remains an element of H. That is our definition of a normal subgroup. So we are saying, assuming that we have a normal subgroup, so assuming a normal subgroup, can I show that, or I say that this implies that the left coset is equal to the right coset. So how do I do this proof? Very simple. I have two sets there. If I want to prove two sets that are equal to each other, I've got to prove two parts. First of all, I've got to prove that AH is a subgroup of, the left is a subgroup of the right. Now let's just choose an AH that is an element of the left coset because that is the definition of a left coset. I need to find, let's make it H star, which is such that, I want to find that there exists this H star such that A, let's make it as such that we have that this is equal to AH. I need to find that, and if I find that, that means that this is a subgroup inside of that because this will belong to the right coset. And I'm going to let this equal AH A inverse. Why can I do that? Because this one, remember this here is an element of, this is an element of H. And we assumed that this is a normal subgroup. This A belongs to G. And I said that under this special circumstance of a normal subgroup, it remains an element of that. And if I were to rewrite this now, with this being there, I have AH A inverse A, and that is just a binary operation with H, which is that I've shown that this is a subgroup of that. So that's showing the one way, the other way I have to show that the right coset is a subset of the left coset. Again, also what I need to show now is that I have this, remember this is an element of H A, that is by definition. And I need now to find this so that I have, so I need to find this H star such that I have the fact that it's the other way around now. If we just look at it, I needed to show that this was equal to that. I now need to show that something is equal to this one, that I have this H A is equal to being then a subgroup of that. And I just need to, there's just a little caveat in this one. I need to show, I need to use what is called conjugation. So I'm going to let this equal, and I'm going to let this equal this. Now you might say that, you know, that's not right because I had the inverse at the end there, and now I have the inverse in the front. But just think of it like this, we know that A inverse is inverse equals A. So what I really have here, what I really have here is something and it's sort of inverse. So what I have here, see this is let A inverse equal to some element G. So what I really have there is G H and it's inverse, because the inverse of this is just itself. So just see this as G and it's inverse. So that's called conjugation and there's no problem writing it the other way round. And if I have that and I plug that in here, if I plug that in there, I'm going to get exactly what we started off with. So I've shown that this is a subset of that, I've shown that this is a subset of that. So under the assumption that I have a normal subgroup, and we know what a normal subgroup is by definition, we've defined it, we have the fact that the left coset and the right coset are equal to each other. So we have two circumstances. One is if we have an abelian group, then I can just use commutivity. That's a trivial, but I know that if I use this definition of this very special kind of subgroup called a normal subgroup, then it's very easy to show that these two subsets are actually, these two cosets, left and right end cosets are actually equal to each other. So a little detour just to define this normal subgroup and under what circumstances we have cosets being equal to each other.