 In our previous video, we introduced the notion of left and right cosets. And we provide some examples for groups of order six and argued that although sometimes the left and right cosets are equal to each other, in general, they're not equal to each other. Now, there was a lot of the calculations I kind of sped through and told that the view work can kind of fill in some of the details. And the reason I was able to do some of those calculations so quick is not because I have a computer of a brain or because I'm the most brilliant man alive, no. It really comes down to that I could predict what some of the latter results were gonna be because of the former ones, because it turns out that cosets form a partition on the group. That is the cosets are the equivalence classes for a very specific equivalence relation. That's we're gonna prove in this video right now. Suppose we have a subgroup H inside of G. These can be finite groups or infinite groups. It doesn't matter. We're gonna define a relation on the elements of G. We're gonna call it X twiddle Y. And this is gonna happen only if X inverse Y belongs to the subgroup H. So I'm gonna record that right here. So as we go down the page, you're always gonna remember that X inverse Y being an H is equivalent to X being related to Y. And I claim that this relationship is an equivalence relationship. If it's an equivalence relationship, it then has equivalence classes. Those equivalence classes are going to be the left cosets of H. Therefore, the left cosets form a partition of the group because equivalence relationships correspond to partitions. So let's first see that it's an equivalence relationship. We have to show three things. First, we have to show that it's reflective, the reflexive property. Then we have to show that it's symmetric, the relationship. And then lastly, we have to show that it's transitive. That's what our goal is in this proof. So the reflective property. So imagine that X is inside the group. We need to show that X is related to itself. Well, if we take X inverse X, that's equal to the identity. And since H is a subgroup, we see that the identity is contained inside of H. Therefore, X will be related to itself and we see that the relationship is reflexive. So that's the first thing on our checklist right here. Great. Then the next things, we have to show the symmetric property. When showing the symmetric property, we assume one relation. We have to prove the other relationship. So assume that X is related to Y. Well, what that means is that X inverse times Y is contained inside of H. Now, since H is a subgroup, H is closed under inversion. So if we take the element X inverse Y, its inverse will be contained inside of H. So we get X inverse Y inverse will be inside of H because it's a subgroup. But by the Shusot principle, X inverse Y inverse is equal to Y inverse of X, which notice how things got switched around. Y inverse X belongs to H, which tells us that Y is related to X. This shows us the symmetric principle for which two axioms down, one more to go. Let's talk about transitivity right here. So to show transitivity, we're gonna make two assumptions. So we're gonna assume that X is related to Y and Y is related to Z. We then have to show that X is related to Z. So how are we gonna do that? Well, by the first one, if X is related to Y, that means X inverse Y is part of H. And because Y is related to Z, that says that Y inverse Z is related to H right here. Now finally, because H is a subgroup, H is closed under multiplication. So the product of any two things in H will be inside of H. In particular, X inverse Y times Y inverse Z to elements in H, their product is gonna be in H. But if we simplify the product there, the Y and Y inverse cancel, you use associativity as well, you end up with X inverse Z belonging to H. Therefore, X is related to Z and thus we can conclude that the relationship is transitive. So we have all of our properties now, reflexive, symmetric and transitive. So we can then conclude that this relationship tilde is an equivalence relationship. Now, before we go on and finish the proof here, I wanna point out some important things here. Look at these green boxes that are still on the screen. In order to prove the reflexive property, we use the fact that H has the identity. To show the symmetric property, we use the fact that H is closed under inversion. And to show transitivity, we show that H is closed under multiplication. You know what? Those are the three axioms of the subgroup. Closure under multiplication, identities and inverses. We used all three of those axioms and each of the axioms of an equivalence relationship was based upon one of the three axioms of the subgroup. It's always as if the two things were in this one-to-one correspondence. This is not a coincidence. The axioms chosen for equivalence relationship subgroups and groups themselves, there is this nice trinity going on on these axioms and that the axioms of the subgroup, exactly what was necessary to prove that this is an equivalence relationship. Pretty neat how those things correspond like that. So the last thing to show is that now that we have an equivalence relationship, that equivalence relationship has equivalence classes. We wanna argue that the equivalence classes coincide with the left cosets. So we have to show that the equivalence class represented by G is equal to the coset, left coset GH. So let's take an element X inside of GH. Well, by definition of a coset, there has to be some element little H so that X is equal to G times H. Now let's consider that for a moment. Let's take G inverse times X. Well, since X is equal to GH, you can cancel the Gs and so this is equal to H, which clearly belongs to the subgroup. Therefore, if G inverse X is inside of H, this tells us that G is related to X. And as this is an equivalence relationship, that means X is contained in the same equivalence class that contains G. But as X was an arbitrary element of the left coset, this shows that the entire coset GH is contained inside of this equivalence class G. All right? Now, if we were arguing with only finite sets, we could then perhaps make an argument that the two finite sets have the same size. Therefore, they're equal to each other. But I want an argument that also works for infinite sets as well. So we have to kind of reverse this direction. But be aware that the reverse direction is essentially the same argument. If you take X inside of the equivalence class containing G right here, this means that G is related to X, which that means that G inverse X is inside of H, which means there exists some H inside of H that G inverse X equals H. And multiply on the left by G, we're gonna get that X equals GH, and therefore this belongs to the left coset. And that gives us the other containment. And so since we have containment in both directions, we invoke equality right there. And as the congruence class G was also chosen arbitrarily, this shows that every congruence class with respect to this equivalence relationship is in fact a left coset. So the left cosets partition a group. Now I want you to, I also want to be pointed out here that we prove this for left cosets, but there's nothing particularly special about left cosets in this situation. We could also define an equivalence relationship on G in the following way. We say that X is related to Y if and only if we have that X, Y inverse is in H, right? So notice that we put the inverse on the right-hand side this time, we put it on the right factor as opposed to the left factor. In this situation, by almost the exact same proof, you change the appropriate parts, mutatis mutandis, a Latin phrase, which means exactly that, just change the appropriate parts. We would then see that therefore this twiddle here is it's an equivalence relation where the equivalence classes say of G are gonna equal the right cosets. Therefore the right cosets likewise partition the group. And so this is what we see right here that the right cosets correspond to the equivalence relationship where we put the inverse on the right factor. On the other hand, the left cosets were generated by the equivalence relationship where we put the inverse on the left factor. So just kind of summarizing here, left in X inverse Y being part of H, this coincides with the left cosets and then X Y inverse being in H, this coincides with the right cosets. And these are the equivalence relationships associated to these things. And so both left and right cosets are their partitions, their equivalence classes right there. And the distinction between left and right is really just kind of a matter of preference. For the most part, we're gonna be doing left cosets in this lecture series, but one could equally do things with right cosets and it wouldn't make much of a difference whatsoever. In fact, a lot of people in group theory really like doing right cosets, especially in representation theory. Again, that's not a requirement though, but for this class, so we are gonna focus on left cosets.