 In this video, I want to state and prove the fundamental counting principle of group actions. We are hesitant to put adjectives like fundamental into a theorem's name, unless it's actually deserving of that, fundamental. This is a very important property for group actions. This is one of the reasons, not the only reason, of course, but it's one of the reasons why we care about group actions. One should never underestimate a theorem that counts something. So imagine we have a group that's acting on a set X. If you take an element X inside of the set X, then the following statement's gonna be true. The cardinality of the orbit of X is equal to the index of the stabilizer of X inside of group G here. So remember what this notation means here. O sub X, this is the orbit of X. This is the collection, like let me write it over here. So the orbit X here, this is the set of all elements Y inside of X such that Y equals G dot X. So this is the set of everything that is the image of X with respect to some element acting upon it. This is the orbit. The stabilizer GX right here, remember what is this? This is all the elements G inside of X such that G dot X equals X. So if we take the index, so this is the number of cosets, of the stabilizer, this gives you the cardinality of the orbit. Now I'm gonna prove this theorem for the infinite case. I should say I'm not gonna assume anything about cardinality. This can be true for infinite or finite. If you assume things for finite, you can get a slightly easier result here, but it's true for the general. So I want to give us the strongest version of this argument here. To prove that these two things are equal, what I'm gonna do, since after all the index G colon GX counts the number of subsets of GX inside of G, what I need to do is then provide a bijection between the set OX within the set of left cosets of GX. Because after all, the index is the cardinality of this set right here. So if I can find a bijection between OX and G mod GX, then this bijection will prove that their cardinalities are equal, proving the fundamental counting principle here. So let's then describe what is that bijection gonna be. If I take an arbitrary element of the orbit, like we said over here, every element of the orbit can be factored as some G dot X. You get the statement right here. So in fact, I could actually replace the symbol Y with just the symbol G dot X. G dot X represents an arbitrary element of OX so long as I don't specify what the element G is. If little G is allowed to be arbitrary, then G dot X is an arbitrary element of the orbit. That's gonna be important in terms of our construction of the bijection. All right, now let's get to that. We define a function from OX to G mod X. Now be aware, we don't know that GX is a normal subgroup or not. So I'm not claiming that G mod GX is a quotient group. I'm just taking the set of left cosets which may or may not be a group, doesn't matter. OX also is a subset of a set, right? The set X does not have any algebraic structure except that a group acts upon it. So it has, we don't know if there's any internal algebra to the set X. We only have this external action from a group. As such, when we take the subset OX, we don't claim there's any algebraic structure here. We're not claiming that phi is gonna be a niceomorphism or a homomorphism of any kind. We are just trying to construct a bijection. Now the rule is that we're gonna take the element G.X and I send it to the coset represented by G. Now there's a lot of problems here that we have to check. Is this thing even well-defined? Because after all, you could have the case that G.X is equal to H.X for two different group elements. That's exactly what we're considering right here. Now we then need to show that G.X and H.X, even if they're equal, but H and G are different. We just show they represent the same coset. That's all that we need. So what we're gonna do is we're gonna dot both sides on the left by H inverse, okay? So that gets us here. The left-hand side, not much to say, but on the right-hand side, there's some simplification, right? Because if you have H.H.X, by compatibility, we can then factor, or we can re-associate the right-hand side. So we get H inverse H, that's a product, dot X. Well, H inverse H is gonna be the identity which does nothing to X. So what this tells us is that H inverse G dot X is the same thing as X. So therefore, H inverse G stabilizes X. H inverse X, excuse me, H inverse G, that's a typo, belongs to G sub X. It stabilizes it, like so. So if H inverse G belongs to GX, that means that G and H represent the same coset of GX. So therefore, this map is in fact well-defined. It doesn't matter which representation of the element you choose. G or H, you'll get the same coset with respect to the stabilizer. All right, so now we have a function. Is it a bijection? Is it one-to-one? That's our next target here. To show that it's one-to-one, we're gonna consider two elements that have the same image. So F of G dot X is equal to F of H dot X. We're not assuming that G dot X and H dot X are the same thing anymore, but we're gonna prove that they are. But honestly, proving that this thing as well is injective is actually very similar to proving it's well-defined. We're basically gonna provide the exact same argument backwards. So if F of G dot X is equal to F of H dot X, that means that the cosets G GX is equal to H GX. For which if the two cosets are the same, that means the quotient H inverse G belongs to G of X. Again, that's what it means for cosets to be equal to each other. But if H inverse G belongs to the stabilizer, that means H inverse G dot X is equal to X, for which then we can dot both sides by H inverse. Excuse me, not H inverse, just by H. We dot both sides by H. On the right-hand side, we'll just get H dot X. On the left-hand side, we use some compatibility there. We're gonna get H H inverse dot G dot X, right? H H inverse gives you the identity which doesn't do anything to the group action. So it simplifies just to be GH. So G, excuse me, GX. So G dot X is equal to H dot X. This shows us that it's one to one. You actually see this a lot, that when you define a function and it's because of representations, it might be well-defined. You often see that the argument that proves it's well-defined if reversed will show that it's one to one. That's certainly not a universal principle, but it's one that shows up enough that it's worth paying attention to. So now our function is injective. Is it surjective? I actually claim that surjectivity is pretty obvious. If you take an arbitrary element of G mod GX, it's a coset that'll look like little G, GX. Who maps onto that will clearly the element G dot X will map onto it. So surjectivity is immediate and therefore this function is a bijection. Since it's a bijection, this then tells us that the cardinality of OX is equal to the cardinality of G mod GX, which then by definition, this is the index. The index here, thus finishing the proof. Now we did this proof for arbitrary groups. We made no assumptions about infinite or finite or anything like that. But if you restrict this to the special case where G is a finite group, then the index of GX inside of G can then be written as the quotient of integers, the order of G divided by the order of GX. And so if this is equal to the cardinality of OX, then you can clear the denominators in this equation and you end up with the order of G is equal to the cardinality of the orbit times the order of the stabilizer. And this counting principle can be very, very useful. So in particular, since orbits have cardinality equal to the index of a subgroup by Lagrange's theorem, we get, and that's what we see on the screen right now, we get that the size of an orbit divides the order of a group. That's a very, very important thing. The cardinality of a orbit always divides the cardinality of the group. And so this fundamental counting principle is essentially, at least when we restrict it to the finite case, is essentially the group action version of Lagrange's theorem, for which in the previous lecture series, Math 42-20, we use Lagrange's theorem a lot. It's very important to finite group theory. So therefore when you have finite group actions, the fundamental counting principle serves essentially the same role as Lagrange's theorem, which is justifying why it's so fundamental. Unlike Lagrange's theorem, though we don't necessarily attach it to anyone's name individually. So this brings us to the end of lecture two for our lecture series. 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