 In this video, we're gonna prove the following theorem. Let H and K both be subgroups of some group G. And suppose that the orders of the subgroups H and K are both finite. It does not actually have to be true that G is itself a finite group, but H and K, we're gonna have to be finite groups in this situation. Then the cardinality of the set H times K, which is not itself necessarily a subgroup. So I'm not gonna call it the order, but the cardinality of the set H times K will be the product of the orders of H and K divided by the order of their intersection. For which if you clear the denominators, you could write this as the cardinality of HK times the cardinality, well in this case the order of H intersect K. This is equal to the order of H times the order of K. And this is actually the identity that we are gonna prove in just a second right here. And admittedly, when you get rid of the division, you could make this into a statement about infinite groups for which I'm not gonna worry about whether this proof is valid for infinite or not. Infinite cardinals is sort of a different type of arithmetic I don't wanna worry about at this moment in our lecture series, but be aware that the identity we're looking for follows immediately from the identity we're gonna prove right here, okay? And so the way we're gonna do this, the following, we're gonna consider a map from the direct product of H times K to this forbenius product H by K, right? And the key observation here, well there's a couple of observations that one could say is key, but notice that when you take the order of H cross K, this is equal to the order of H times the order of K. So that connects you to the right-hand side right here. And so what we're gonna then show is that, well, this map goes over to HK for which you're gonna get the co-domain, right? But it turns out that this isn't gonna be a one-to-one map necessarily. This is gonna be an H intersect K to one map. That is, when we look at the pre-images of elements of this map feed, every pre-image is gonna have the same size and that size is gonna be the order of H intersect K. And so by combining these principles you take, since the pre-images partition the domain, we're gonna get this factorization going on from there. So that's the strategy we're gonna employ right here. So the map feed is defined by the following very simple rule. That feed of H comma K will equal the product H times K. Great, that's gonna be, that's the map there. It's gonna be well-defined and surjective is also fairly obvious because after all the set HK is by definition the product of every possible H times every possible K, right? So HK definitely belongs to the set H times K and every element can be written in that fashion. We wouldn't expect this thing to be injective though and like I was hinting towards above this thing is likely not gonna be injective. It is of course when the intersection of H and K is trivial but it turns out the pre-images will actually be the same size as H intersect K. So let's explore that for a little bit. So take an arbitrary element X inside of the intersection. We know there's always the identity. There could potentially be other things because H intersect K will be a subgroup of both H and K and hence a subgroup of G as well. So consider the following situation. You have this just element X sitting in your back pocket right there. So if you look at phi of H times K, well, phi of H and K, excuse me, that'll be the product H times K. Well, let's insert an inconspicuous identity element sitting between H and K. Well, that clearly, that doesn't change anything. But E, the identity is of course just the product of any element times its inverse, great. And by associativity we can write this as HX times X inverse K, right? But remember, okay, X belongs to H intersect K. So X is inside of H, which means since H is in H and X is in H, this thing belongs to H. But we also know that X inverse, well, since X is in the intersection, X is in K, K is a subgroup. So its inverse will be in K, right? And then since K is in K and X inverse is in K, their product is in K. And so we see that HX, HX is in H, X inverse K is in K. So therefore, HX comma X inverse K, this order pair, we can see as an element of H cross K, for which it's in the domain of fee, this is actually equal to the same thing. So fee of HK is equal to fee HX comma H inverse K. And so assuming X is a non-identity element, this would show us that this thing is not gonna be one-to-one, right? That the images of these two ordered pairs is the same under this map fee, right? And now we wanna show that whenever two ordered pairs have the same image with respect to fee, we can uniquely identify an element X inside of H intersect K, right? So let's go in that direction. Let's suppose that the image of two ordered pairs H and K and H prime, K prime are the same, okay? So that would mean the product HK, H times K and H prime times K prime is equal to each other if their images are the same, right? And so then, well, we could times both sides of the equation, let's play this over here for a second. We got HK and then H inverse, excuse me, H prime, K prime, right? So the first thing to do is we could multiply both sides on the left by H inverse, H prime inverse, I should say, and we could also multiply on the right by K inverse. So you'll see that K, K inverse cancels out on the left and then H prime inverse, H prime cancels out on the right. And so you end up with the product H prime inverse H, admittedly that right there is in H. On the right, you're gonna get K prime K inverse, which is an element in K, but because of this equality, we actually, I mean, it's very cool, right? These things are equal to each other. And so therefore this element right here, which is equal to this element, it's in H, it's in K. So we get that, if you call that element X for a moment, X is equal to H prime inverse H and K prime K inverse. X here is an element that lives in H, it lives in K, so it lives inside of H intersect K. And so then playing around with this product, this element again here, HK, well, I can write this as H inverse, or H prime H prime inverse H, right? So I can just write H that way, because this right here is just the identity again. And then let's multiply it by an identity element on the right, K prime inverse K prime, right? So that's the same element, right? They're equal to each other, but then you see that with a little bit of tweaking things, so we'll just not do anything to the first one yet, but if you do something in the second one, notice just by the inverse, the shoe sock principle that K prime, K inverse, inverse, if you were to distribute that around, you twist things around, you end up with K, K prime inverse. So that's just equality still. But this right here is X, and this right here is X inverse. So notice that HK, this is equal to H prime X times X inverse K prime, okay? So given this element, I could factor out this element X, which belongs to the intersection of H and K. And so because of that, we then can uniquely determine the pre-image here. If you have an element HK, the pre-image will look like things of this form, H, X, X inverse K, where X can range over the intersection of H intersect K that the pre-image, the pre-image, there's this natural identification, there's a bijection really with the set H intersect K. Or it's a little bit easier to describe it the other way around where you have your element X, you're gonna map it to the order pair HX, H inverse K, like so. And so you're gonna, basically the previous argument is showing that this map right here is a bijection. And so since this is a bijection, this shows that fee inverse of HK, the size of this set is always equal to the size of H intersect K right here. That this is true for any pre-image, right? This is true for any order pair H and K, you get that. And so therefore since H cross K is partitioned by the pre-image as a fee and each pre-image has the cardinality H, the order of H intersect K, we then get, well, we then get the observation we made earlier, right, this one, we get that the domain will have the same cardinality as the co-domain times these pre-images are, like we said earlier, we can divide, we get this identity right here, that the co-domain will look like the domain divided by these pre-images that all have the same size. But then as the order of H cross K is equal to the order of H times the order of K, we've then established the identity we have right here. And so this is a counting argument that the size of the product of two subgroups, we can compute this formula for these finite groups. And the very first time I saw this identity, it was from my very first abstract algebra class for which we were using the text abstract algebra by John Frally, for which he's very famous for making the popular statement, well, popular, never underestimate a theorem that counts something. And as personally an algebraic commentatorist myself, I take that very, very seriously. One should never underestimate a theorem that counts something. Counting formulas can be very powerful as we will see in the next video of this lecture.