 In this video, we're gonna prove two very important properties, counting properties about left cosets and also by analog right cosets. And so let's talk about those. Let G be a group with H being a subgroup and let G be a specific element of that group. Then I claim that the order of the group H is equal to the cardinality of any of its cosets. Since the representative G is chosen arbitrarily, we're gonna show that these two sets have the exact same cardinality. And this is gonna be true for finite sets, which will be extremely important, as this will lead to Lagrange's theorem. But it's also gonna be true for infinite sets as well, that if H is countable, then all of its cosets have to be countable as well. And now how do you show that two sets have the same cardinality? The basic idea from combinatorics is to provide a bijection between the two sets. So I'm gonna show that there's a one-to-one onto map from H to GH. We're gonna call that map phi. And we're gonna define it by the following rule. Phi of H is gonna equal G times H. Now this is a map we've actually ran into before. This is the map that we would call left translation. Translation, that is we multiplication on the left by the element G. Now we saw this before when we proved that the, when we proved that the order of the alternating group is equal to the order, well, we proved it's into N factorial over S. The basic argument was the following. It was hinged upon the idea that the number of even partitions was this permutations, excuse me, was equal to the number of odd permutations. And we showed there was a bijection from even permutations to odd permutations. And we did it using right translation. We took an even permutation sigma and we mapped it to sigma times one, two, which would be odd transposition. That was right translation right there. And we argued that right translation is bijection. And so although we talked about right translation by one, two, there's nothing special about the element one, two. Right translation is always gonna be a bijection and so is left translation. The side that you multiply by doesn't matter. So even though we've proven that before, we're gonna actually provide the general proof right now. It's gonna be exactly the same thing. Mutatis mutandus changed the appropriate parts from that previous video. So we had to first show that it's injective. So assume that phi of H equals phi of H prime for two potentially different elements of H, right? We don't actually know. Well, if phi of H equals phi of H prime, that means that GH equals GH prime, okay? Cause that's what the map does. But then because we have a G on the left of both sides, we can cancel the G on the left by the cancellation law and we see that H equals H prime. And so the map is in fact injective. So notice here to get injectivity, we just have that cancellation. So this argument actually would be applicable for any quasi group, a group which basically, not binary operation that has the cancellation property. We don't necessarily need associativity right there or inverses or anything, we just need cancellation to show that left or right translation is injective. Well, why is it surjective? Well, to show surjectivity, suppose we have an arbitrary element of the co-domain, so X. Well, if X is on side the left coset, that means there's some element little H so that X is equal to G times little H. And now we have our candidate right there. Well, how are we gonna map onto that? Well, phi of H is gonna equal GH, which is gonna equal X. And so that shows surjectivity. And so to show that this thing is surjective in general, we just need the existence of this little element right here. And this is gonna, again, for a quasi group because it's Cayley table forms a Latin square, we're gonna see that for a quasi group, translation left or right is always gonna be a bijection. But sticking to the realm of groups right now, we've now shown that left translation, we see that left translation is in fact a bijection. It's both one to one and onto. Therefore, the domain of left translation is gonna be having the same size as the co-domain of left translation. So that proves that these two cosets have the same size. And as the cosets, as the coset GH was arbitrary, this shows that all cosets have the exact same size. Their card value will be exactly the order of the group H. And I should also mention that in particular, it's not just that all the left cosets have the same size, but this argument could have easily been made into an argument about right translation. Right translation by the element G is also a bijection. And that therefore shows that the order of H is equal to the order of HG, the right coset. And so in particular, all right cosets will have the same size. It's gonna be the order of H. And they're gonna be the same size as the left cosets. Now, we're not saying that GH is equal to HG. Left and right cosets are not the same thing in general. But what we are saying is that each and every coset will have the exact same size. And that size is the order of H. Doesn't matter if you're left or right, all cosets will have the same size. All right, that's a very important counting principle. We should never underestimate a theorem that counts something. All cosets will have the exact same size with respect to H. Of course, if you switch the subgroup H, then the size of the cosets will change between on the new subgroup H right there. But another thing we can also say is that all given any subgroup H inside of G, the number of left cosets is equal to the number of right cosets. We know the cosets have the same size, but we also have the same number. Now, if we restrict ourselves just to finite groups, this would be a very simple argument. It basically would follow up from Lagrange's theorem, which we'll prove a little bit later. But what we wanna do is actually prove this in the most general setting possible. We actually wanna prove this for infinite groups as well. And also to avoid some circular reasoning with Lagrange's theorem. We're gonna prove this right now. But I want you to be aware that this theorem right here, these two theorems is essentially the foundation for which Lagrange's theorem comes from. The way we are presenting this, Lagrange's theorem is gonna be an immediate corollary, essentially, of these two theorems we're proving right now. So if we wanna prove the number of left cosets is equal to the number of right cosets, we wanna prove that these two sets have the same cardinality. Again, we're gonna construct a bijection between these two sets. Our bijection is gonna be the following rule. Phi will send left cosets to right cosets. And the way it's gonna do that is the left coset gh, we're gonna send to the right coset hg inverse. Now you might wonder, well, why didn't we take gh and map it over to hg? Why doesn't that make the natural thing? And although one could do that, it's gonna be a lot harder to prove, it's gonna be a lot harder to prove there's a bijection in this setting. And that's because in general, we can't assume commutivity. So how do you twist things around in a general group? If it's non-Abelian, the inversion map can switch things around. It can switch the order of things. And so that's why we're taking hg inverse here. Now to show that this is a bijection, we have to first of all show it's a function. This wasn't a concern on the previous video, or the previous proof I should say, because we were sending elements to elements. Now we're sending cosets to cosets. gh is mapping to hg inverse. The problem is this map is dependent on the representative. We're sending g to g inverse. What if we picked a different representative? What if we took something like x right here? Is that still gonna map to, because that's gonna wanna map to hx inverse? How do I know that these two sets are the same thing? Is the map dependent upon the representation? And we wanna argue that it's not. Thus it's gonna be a well-defined map. So to show that this map is well-defined, that's the main argument here. Take two different cosets, sorry, two different representatives of the same cosets. Say that ah is equal to bh. So a and b are potentially two different representatives. So we have to then prove that ha inverse is equal to b, or hb inverse. How are we gonna do that? So we assume, we're gonna assume this right here. So we assume ah equals bh. Now we have to then prove this holds. So let's take an arbitrary element of ha inverse. Well, if x belongs to ha inverse, that means there's some element little h so that x is equal to ha inverse. That's what we're gonna use right here. Now this is the reason why we used inverses because although our operation multiplication is non-commutative, inverses can twist things around. If we take the inverse of these elements right here, x inverse, we'll just become x inverse, but then if we take ha inverse inverse by the shoe sock principle, you put your socks on then your shoes, but you first take off your shoes then your socks, ha inverse inverse will switch these things around so that a goes first, h comes second, and you get a, this will look like a inverse inverse, h inverse, but of course the double inverse is just a. This right here, x inverse is equal to a h inverse, which will belong to the left coset represented by a. But by assumption, ah is equal to bh. Therefore, there exists some element k, which belongs to h, so that x inverse equals b times k. But then if we take the inverse of this equation right here, x inverse inverse will bring us back to x, but then we have to compute bk inverse, all right? But by the shoe sock principle again, bk inverse becomes k inverse b inverse. Now since k was inside of h, k inverse is inside of, what did I say? Since k is inside of h, k inverse is inside of h, therefore k inverse b inverse belongs to the coset hb inverse. So this shows us that h, a inverse is a subset of hb inverse. Now again, if we were just finite, we can then invoke that the two things have the same size and we'd be done, but for an infinite situation, we want this to still be true. And so we'd have to then turn this argument around, but there was nothing particularly special about a, changing all the appropriate parts, switching a's to b's and b's to a's, you'll get by a similar argument that hb inverse is a subset of h, a inverse. And so this shows that the two sets are equal to each other. This then argues that the map phi is in fact well defined. It doesn't matter which representative you use. That gives us now we have a function. How do we get an inverse function? How do we show that it's invertible? Well, you can show that it's one to one, you can show that it's onto, that's one way to show that a map is bijective, but another way to show that a map is bijective, well bijective also means that it's invertible, we can just construct an inverse map. So we actually have a candidate for who the inverse is. Phi was a map from left cosets to right cosets. We're gonna come up with a new map called psi, which sends right cosets to left cosets. And it'll be defined by the rule that the right coset hg will be mapped to the left coset g inverse h. And I want you to see what happens when we compose these things together. If we do psi of phi of gh, this is gonna equal psi of hg inverse, which will then equal g inverse inverse of h, which is just gh. So the composition here is the identity map. So these things are inverses of each other. Now, psi by similar reasoning, we could argue is a well-defined function and it's gonna be the inverse of phi. And so since phi has an inverse, it's invertible, that means it's bijective. And because it's bijective, the two sets will have the same size that these two sets right here, left cosets and right cosets have the same size. So summarizing what we've seen so far is that every left coset will have the same size as the order of h and so will every right coset. We've also shown that the number of left cosets is equal to the number of right cosets, even in the case of infinite sets. And so another proposition I do wanna mention here, this is included in Judson's text and this is a pretty important proposition. Let h be the subgroup of g and let x and y be elements of g, then two cosets are equal to each other exactly when one is inside of the other. And this basically just comes from the fact that cosets form a partition. I'm gonna leave this up as an exercise to the reader to prove this one, but this is a consequence of cosets forming our partition relationship. And that's gonna bring us to the end of our lecture 18 in our series. We're gonna talk about some more about cosets in lecture 19, so take a look for that. Also in lecture 19, we'll prove Lagrange's theorem by continuing developing this theory of cosets that we've done in this lecture right here.