 Welcome back to our lecture series, Math 4230, Abstract Algebra 2 for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Missildine. In lecture five, we actually want to use what we've been learning about group actions and apply it towards, well, an application. Specifically, an application towards commonatorics. You will recall that commonatorics is the mathematics of counting thing. We want to enumerate things. And it turns out, symmetry is a big part of commonatorics. Algebra and commonatorics, they play nice with each other very well. And in this lecture five, we're going to talk about Burnside's counting theorem, how we can use group actions to help us count certain commonatorial structures. Now, before we present Burnside's theorem with its proof, I want to remind the viewer of a very important observation here. So imagine we have a group which is acting upon a set. If we take some element little g inside the group and some element little x inside the set. I want you to remember that, well, first, the symbol g sub x, this is the stabilizer of little x here. This is the set of all elements g dot x, where g can range over g here, right? These are all the elements of the group that stabilize, that fix the element x. In particular, this is a subgroup of g here. What I want to remind us here is that if we look at the stabilizer of gx, this is actually the conjugate subgroup of the isotropy subgroup gx, where you actually conjugate it by the element g. So stabilizers of elements in the same orbit are actually isomorphic to each other. Now note, if you have a subgroup, if you take a conjugate of that subgroup, you're going to be isomorphic to that subgroup. Because after all, conjugation is an inner automorphism. So the stabilizer of x is isomorphic to the stabilizer of gx. Now x and gx are going to belong to the same orbit of the g action. So we say that x and gx are g equivalent. So whenever you have two g equivalent elements they're in the same orbit, their stabilizers will be isomorphic in particular for us as we will be counting things as their stabilizers are isomorphic. It means that their stabilizers will have the same order. They're the same size. And that's going to be important as we work with these, when we work with these isotropy subgroups presenting Burnside's theorem. So here is the theorem of Burnside that we're talking about in this lecture. Imagine we have a finite group g that's acting upon some set x. Let little k denote the number of g orbits of x. So that's how many orbits there are in this group action. Then what our theorem is going to tell us is that the number of orbits of the x. Because the orbits form a partition of x, they all combine together to give us x here. But how many orbits are there? This is going to equal one over the order of g. And then you add together the size of all the stable sets where you range over the elements of little g right here. So as a reminder, what are these things meant? We talked about the stabilizer just a moment ago. What is the stable set x of g? This is the set of all elements of the form where you have g dot x equals x right here. But this time g is fixed and you allow the elements x to range here. So this will be a subset of the g set x. And so we gather all of those elements which are left fixed that are stabilized by the element g right here. So we can compute the number of orbits of the group action by looking at the sizes of these stable sets. How do we do this? Well, this equation comes about by analyzing this equation right here. Consider the equation g dot x equals x. And I want you to treat this as an equation with two variables where the elements of the group g are allowed to vary and the elements x of the g set are also allowed to vary. So with an equation with two variables there, how many solutions are there to this equation? Well, it depends on how you count them. If you consider this with respect to stable sets, which again gx here, we want all of the elements that satisfy the relationship g dot x equals x as x ranges over capital X right here. So if you have a fixed g in mind, like so if this variable is nailed down, it's like we keep g fixed, then you want to grab all of the x's that work with that g. Well, that's exactly the set x sub g. So if you allow then g to vary because we have two variables, while g is fixed, you'll just grab x sub g. But then you allow g to vary over all the possible elements of the group. That gives you all the possible stable sets. And so the solutions to the equation gx, if we think of those as ordered pairs, those can be identified with elements of this set here. So this counts all of those solutions. We take the sum as g ranges over g of the sizes of the stable sets xg. But what if we change our perspective? What if we think about the stabilizers, which we said earlier, the stabilizers g sub x. This is the collection of all the things that satisfied the relationship g dot x equals x, where g ranges over g here, right? So if x is now fixed, so if we fix x and we're like what g's work here, then g sub x will grab all those possible g's so we get all these different stabilizers. Then if you allow little x to range over all of the possible x's inside the g set, that'll also capture all of the same solutions. So if we're counting solutions to this equation, the stable sets count the solution by enumerating all the possible stable sets. But also, if we enumerate all the possible stabilizers, that will also give us all of the solutions. Now in combinatorics, this is what we call a combinatorial proof. That is, we counted the same set in two different ways. What's the set? We're counting the solutions to the above equation of two variables, the solutions in two variables there. We have then two different numbers that count the same set. Well, if they count the same set, that means they actually equal each other. And so we get this very important equation that the sum ranging over the elements of the group, when you add up the size of all the stable sets, that's equal to the sum where you range over all the elements of the g set, where you add up the orders of the stabilizers. Those two sums are exactly the same thing. Now, let's get back to the observation we made at the very beginning of this video here. If y and x are in the same orbit, so in particular if y is in the same orbit as x, their stabilizers are actually the same, well, they're isomorphic stabilizers, thus they'll be the same order. So when we look at this set here, when we look at g sub x here, though that order is actually gonna show up a couple times. It's gonna show up for every element which is in the same orbit of x. Well, let's try to keep track of that. So instead of taking the entire set x, what if we just look at the orbit of x for a moment? If we just add up those. Well, this stabilizer right here, it'll have the, excuse me, this stabilizer will have the same order as gx. So what you're getting is you're getting a sum of the gx's a bunch of times, where you range over the orbit of x. You just get the same number over and over and over again, so you're gonna get some multiple of the order of gx. But how many things are gonna show up there? Well, how many things were in the orbit? The cardinal of the orbit is gonna count how many times this thing shows up. But when you look at this, we have the order of gx, then we have the size of the orbit, but we know by the fundamental counting principle that the size of the orbit is actually equal to the index of the stabilizer. So we get the index of gx times the order of gx by Lagrange's theorem. That's actually the order of the group itself. So what we see here is that the sum of the orders of the stabilizers when we range over a fixed orbit always adds up to the order of g, right? And this happens for any orbit whatsoever, it doesn't matter. So if we were to range over, if we range over all the elements of x, we're gonna get some number of copies of g. Well, how many copies are we gonna get of g? Well, it's the number of orbits we have. That's what we called k earlier. So then divide both sides of this equation by the order of g. We then end up, so we have this equation right here. If we divide both sides of this by the order of g, we then get Burnside's theorem, which you saw at the top of the screen. It's back there right now. And so Burnside's theorem comes as a consequence of this very nice combinatorial proof. Let's see an example of this. Let's take the set where x is the numbers 1, 2, 3, 4, 5. And let's take g to be the, it's gonna be a permutation group that's isomorphic to the Klein-4 group. G consists of the elements, the identity permutation, the 2-cycle 1, 3, the 2, 2-cycle 1, 3, and 2, 5, and then the 2-cycle 2, 5. So that's a Klein-4 group. It acts on the set 1, 2, 3, 4, 5 in a natural way. It's just the permutation action. What are the orbits of this structure, right? Well, the identity leaves everyone fixed, all right? 1, 3 will send 1 to 3 and 3 back to 1, okay? 1, 3, 2, 4, it'll send 1 to 3 and 3 back to 1, but it'll also send 2 to 5 and 5 back to 2. It leaves 4 fixed. And then lastly, 2, 5, it'll move, it'll swap 2 and 5, it'll leave the other ones fixed. So we see that for this Klein-4 group, the orbits of the group action are exactly these sets you see on the screen. 1 and 3 are together, 2 and 5 are together, and 4 is isolated by itself, okay? And so this gives us a partition of x. The orbits of a group action always partition the set. You'll notice that how many orbits do we have? We have exactly three orbits. So we really count the orbits without Burnside's theorem. This isn't too hard of an example, but alternatively, if we use Burnside's theorem, what are we gonna get? Well, let's first look at the stabilizer of the identity permutation. The identity stabilizes everything. So it sends 1 to 1, 2 to 2, 3 to 3, 4 to 4, 5 to 5. And so the stabilizer of the identity is gonna be the entire G set. That's always the case. So we have that x1 equals x. What if we look at the stable set of the element 1, 3? Well, 1, 3, as the permutation notation suggests, it sends 1 to 3, 3 goes to 1, so those are not left fixed, but it doesn't do anything to 2. It doesn't do anything to 4. It doesn't do anything to 5. So the stable set for 1, 3 is 2, 4, 5, okay? What about the stable set for 1, 3 and 2, 5? Well, the nice thing about permutations, the permutation action here is that the stable sets are easy to calculate. If the number shows up in the cycle decomposition, then it's not stabilized. So the numbers that are missing when we write the permutation notation here, the cycle decomposition, that's the number that's stabilized. So notice 4 is left alone. So the stable set of 1, 3 and 2, 5 is the number 4. And then likewise, who is stabilized by the 2 cycle 2, 5? It's gonna be 1, 3 and 4, because you don't see that in the cycle decomposition here. So we have these four stable sets that we see right here. So if we use Burn Science formula, the group itself has order 4, so we get this coefficient of 1, 4th and front. Then we add together the cardinalities of the four stable sets. We get 5, 3, 1 and 3. And then adding these things together, of course, 5 plus 3 plus 1 plus 3, that's equal to 12. 12 divided by 4 is equal to 3. And that was the number of orbits we saw. So this is a verification, just to check on Burn Science theorem. Now this example was very small. It was much easier to count the orbits by just computing the orbits instead of computing all of these stable sets. I mean, in fact, we had only three orbits to begin with. But as the group actions gets more complicated and more complicated, then the Burn Science approach to computing the number of orbits can actually be much easier to do. And we're actually gonna see that in the next video where we utilize Burn Science theorem to solve some combinatorial problems.