 In this video we're going to apply Burnside's counting theorem to solve some combinatorial problems using group actions. Now before we do that it's important to never underestimate a theorem which counts something. Combinatorics is a very powerful discipline in mathematics. Counting things can be very very useful so we should always always pay attention to things that count things like Burnside's theorem, the fundamental counting theorem, when it comes to these group actions. Burnside's theorem is a very important combinatorial tool especially in the scenarios that we're going to do in this video because it can often be difficult to count the orbits of a generic group action. Now before we get to our combinatorial example let me present one more example of a group action that is I want to present a way of constructing a new group action from an old group action because these are the type of things we're going to count in these combinatorial problems. So imagine we have a group G that acts upon a set X, it's just some generic group action, I don't care what it is it's just we have a group action. Then if you take any other set Y, it might have nothing to do with X, I don't care, but if you take any set Y you can form the set of functions which is commonly denoted as Y to the X. So Y to the X will be the set of all functions of the form F takes as its domain X and as its codomain Y. So Y to the X is the set of functions. The set of functions Y to the X is itself a G set in a natural way. The set of functions inherits the G set structure from X and you do it in the following way. Well, since Y to the X is a collection of functions and G is a group, we have to make sense on what does it mean for a group element to act upon a function. It should produce a new function. So what is the function G dot F? So the image of the function F when it's acted upon by G. It should be a function that takes as its domain X and then lands in Y. But remember the domain of the function F is a G set so it makes sense to act on X by the element G and that's exactly what we do. So to define the function, I'm going to tell you what the rule is because a function is rule that assigns everything to its domain, something in the codomain. So the function given by the rule G dot F will be if you take an arbitrary element of the domain X, G dot F of X will then be F of G dot X. So in some regard, you're sticking the G inside of that. So the function that's acted upon by F will be that function which then always acts upon its input by that element G. This is a lot easier to do if you have a left action. If you have a right action, you can do the same thing, but you have to do it on the right and it's got to be like an inverse that's common when you use right actions. We always do left actions, of course, in this video. So we can construct an action on functions by just pushing the action on to the input. That's exactly the type of action we're going to see in just a second. So now let's get to applications of Burnside's theorem. How can we use Burnside's theorem to help us count combinatorial objects? Well, imagine for the moment a square such as square is visualized right here. Imagine we have the task where we want to color the vertices of a square. So like we want to color this one, we want to color this one, we want to color this vertex, and maybe we want to color this vertex where repetition could be allowed. We could allow color this a different vertex, the same color. So on the screen a moment ago with my circles, I did yellow, yellow, green and blue. That would be a coloring of this square. So imagine we're considering how many different ways we could color a square. Now for simplicity sake, let's just focus on two colors because if you all have one color, there's only one way to do it. You color all the corners the same color. Alright, we don't worry about that. But let's say we want to label or we want to place two colors on our square and we'll just make it simple to say black and white. How many ways can we do it? Well, if you have your square, basically you take the first one, there's two options, you could do black or white, then you take the next one, you could do black or white, you could do the next one black or white two options there two options there. So in essence, you have two options, two options, two options, two options. So the counting principle in play here is that if these decisions are made independent of each other, which they are, the coloring one vertex doesn't affect the other, you're going to get two times two times two times two, which is equal to 16. There are 16 ways to color that square. But that's not exactly true, is it? I mean, if the if the square is sort of stuck to the ground, you can't move it, then sure, there's gonna be 16 ways to color it. But what if the square is allowed to move? What if it can move freely in the plane or move freely in the space? You could rotate and flip it kind of like the Dihegel group does to a square. What about that situation? Well, I want you to look at these four squares for a moment. Here's a square which has a black corner, and then three white corners. Here's one where this of course is going to be in the northwest corner is black. But what if the northeast corner is black, all the others are white? Or what about the southeast is black, all the others are white? Or what about the southwest corner is black and everything else is white? These would be considered four different colorings by the strategy we were employing a moment ago. But do we really want to consider these different colorings? Like imagine this is some toy that a kid plays with. You start spinning around your square. If you started off by coloring the northwestern corner black, they want you spin it around. It's really the same square because once you pull that square out of the box, each and every one of those squares is a possible orientation. So the problem we had earlier is when we started coloring and we got two times two times two times two is we believed we could tell the corners apart. Like there was some labeling we thought they were distinguishable. So like oh, the first vertex could be black or white. The second vertex could be black or white. The third and the fourth could be black or white. But what happens when we can't tell the vertices apart? Well, when you start rotating the square, when you start flipping the square, you can't tell apart who was who was the original vertex one because it got moved around. So because of these equivalent orientations, it becomes a lot harder to count the number of square colorings we can have. So this is what Burnside's theorem comes into play because as I'm talking about moving the square, this is a D4 action on the square. The D4, the dihedral group there, it rotates the square, it reflects the square. And so if you give a coloring to the square, any rotation of that coloring would be an equivalent coloring. Any reflection of that coloring would be an equivalent coloring. So as D4 acts on the square with its colors, that creates equivalent colorings and we don't want to count those colorings as different ones. So our 16 number has actually over counted the numbers because two different colorings are really the same, but we counted them as if they're different. Now D4 does act on the square, but the G set that we're looking for is not going to be the square. It's going to be colorings of the square. So imagine we have our set Y right here for which Y is the square. Okay, so we have 1234 as the square right there. So the vertices of the square. We're not trying to count, we're not trying to count the orientations of the square. We're trying to count the colorings. A coloring of the square is a function from the set 1234 to the set BW where we can think of B is black and W is white. So as we think of this function, we have some function here F where you have Y over here and then black and white over here. We have to make a choice on these vertices. It's like, oh, the first vertex is black. The second one is white. The second one is white. The fourth one is black. Every function is a coloring. It's an assignment of the vertices to a color. So the colorings are what we're trying to look at the equivalent colorings. So the set that we actually care about is the set of functions from the set 1234 to the colors black and white. So we're looking at this function set right here. What are the possible colorings? But when you think of your function, you have two options for this first, two options for the second, two options for the third, two options for the fourth. So this number 16 that we got earlier is significant, but we were counting the wrong thing. 16 is counting the number of colorings of the square, but we're trying to count the number of equivalent colorings. So two different equivalent colorings are considered the same. And so we're counting the orbits of the coloring under the orient, the rotations and reflections of the square. So Burnside's theorem is relevant here with D4 acting on this set X because it's a function set. It's a function set on a G set on a D4 set. So we can do that. So let's look at the stabilizers for a moment. I should say the stable sets. Who is stabilized by the identity? That's an easy one. Every function, every coloring with its orientation is fixed and therefore you're going to get 16. The stable set of the identity is always the entire thing. So we get 16 right there. So let's think about rotations for a moment. Who gets fixed by a rotation? So if you have a square over here and you rotate it, well, if this color was white, when you rotate it, then this color has to be white, okay? And so this one will move over and become this one over here. But this has to be the same thing we started with. So if this one was white, when we rotate it, it will move down here and that would then be white. So this one has to be white. Similarly, if this one was white, because it starts and stops at the same spot, that the square, this will move over here and be white. And this one also be over here coming white. So basically, when you rotate the square, the only way you can have the same square when you're done is if all of the colors are the same. So they're all white or they're all black. I did the example with white, but we can do it all with all black. So there's two options. There's the all white square and there's the all black square. By similar reasoning, if you rotate three times, that's the same thing as rotating counterclockwise once. And the only thing that will be left stable by that action would be the all black or the all white square. So you only get two options right there. Okay, what about R2? If we rotate twice, that one gives a little bit more flexibility here, because if this one was white and we rotate it twice, then that means this one's going to have to be white, but then when you rotate it back, that's going to be white again. So these two colors, these two corners have to be the same color. Okay, but this one right here, it could potentially be a different color because when you rotate and come over here, this one's going to have to agree with it, but then this will rotate back and it'll have to agree with it as well. So these ones on a different diagonal actually could disagree with each other. They don't have to be the same color. So basically what I'm saying here is when you rotate this thing, the corners that are opposite of each other, they will have to be the same color. There's two options there, but then these ones, they will have to be the same color too, and it could be the same color as this one, but it doesn't have to be. So you get two options for this color and two options for this color. So that gives you two squared options, which then turns out to be four. So that'll take care of the rotations. What about reflections now? Well, if we think of reflections here, let's do s. s is the horizontal reflection, so we reflect like that. Well, with the horizontal reflection, whatever this color on this corner has to be, it has to be this color too, because those two are just going to swap each other. But then this one, it could be a different color. It'll come down here and agree with this color. So those colors have to agree with each other for this permutation. So in other words, we're saying that this side of the square has to have the same color, which there's two options there. And this side of the square is going to have to have the same color too, which it could be a different color, but you'll have two options there. That gives you two squared, which is equal to four. Alright? Jumping down to r squared s, we get a very similar picture there, because in this situation, you have now a horizontal line. And by similar reasoning, these two vertices will have to be the same color, which gives you two options. And these two vertices will have to be the same color, which gives us two options. Because we're looking for what is a square that's not changed when you do these rotations. There's reflections in this situation. So like if these are black and these are white, when you reflect it, you won't be able to tell the difference on the coloring. It doesn't give you a different coloring. So those are stable. Those are stabilized by r squared s. So again, you're going to get two squared, which is equal to four. You'll notice I skipped rs, because that is a reflection, but it's a diagonal reflection that actually has a different effect on things. Because notice that when you look at this element right here, this corner, whatever its coloring is, let's say it's black, when you reflect it over, if this is the same coloring, this would have to also be black. Okay? Basically, everyone who's in the same orbit of the D4 action on the square has to have the same color. But what about this one right here? This one could be white because these two colors doesn't affect it. But this one could also be white or black. It doesn't matter what the other colorings are. These, if we look at the orbits of the vertices by the D4 action, these two color, these two vertices have to have the same color. There's two options there. But this one, because it's left fixed by the reflection, it could be its own color. You have two options. But also this corner down here could also be a different color. It doesn't have to coincide with the other vertices. So you actually get two times, two times two options, which is equal to eight. And so then if we do this whole exercise one more time with our cube desk, that's the diagonal, that's the reflection across this diagonal, I should say. Same thing. This could be any color. This could be any color. And let me use a different color to emphasize. It doesn't have to be the same color. But then these ones, they do have to be the same. So you get two, two, two, two times two times two is eight. And that then gives you the size of that right there. All right. So let me bring up my notes here. So then what we then see is we counted up all these stable sets. We then get that by Burnside's theorem, the number of orbits we equal one over the group, the group's order, this is the diagonal group, it's order eight. And then you add up all of these stable set sizes. So we had sixteen plus two plus four plus two plus four plus eight plus four plus eight. For which, while some of these numbers don't look like they're divisible by eight, notice if you take two plus four plus two, that's an eight. Four plus four is eight. You actually get sixteen plus eight plus eight plus eight. All of these are actually divisible by eight, which of course has to be the case. K is an integer, but yet you're dividing it by an integer. That means this has to add up to be something that's divisible by eight. It'll happen eventually. So sixteen plus eight plus eight plus eight plus eight is divisible by eight. You get two plus one plus one plus one plus one, which adds up to be six. This tells us, by Burnside's theorem, there are six colorings of the square. And I can give them to you real quick. There's the all-white coloring. There's the all-black coloring. We talked about those. We also saw the example where one of the corners is black and the other corners are white. While there are ways you could rotate that one, those aren't considered different squares. There's also its dual where you have a white corner and three black corners. But then what about the other two? There should be two more. Well, these other ones are going to be when you have two blacks and two whites. So what if like both the tops are black and both of the bottoms are white? That's one. And then the very last one is what if the colors are kitty corner of each other? What if the blacks are diagonal from each other and the whites are diagonal from each other? So up to equivalents, these six squares are the only squares that you can get by coloring, right? Every other square will be equivalent to them up to rotation and reflections, which is what we're trying to consider here. Before we do another example, I want to point out a very important observation here. Looking at the sizes of all these stable sets, what's in common with all of them? We end up with 16, 2, 4, 2, 4, 8, 4, 8. These are all powers of two. And notice where they came from, right? 16 was 2 to the fourth. 4 was 2 squared, 8 was 2 cubed. We also had a 2, which of course is just 2 to the first. They're all powers of two. Is that a coincidence? The answer, of course, is no. Each of them has a form. Each of the stable sets had the form 2 to the A. And where did where do these numbers come from? Where does the base come from? Where does the power come from? The base the base is easy to explain here. The base is the number of colors. Because whenever there was a decision that had to be made, that decision was either black or white. Every decision had two options you could appeal to. And then the letter A, it represented how many choices had to be made. That depended on the orbit structure that was going on with our square. And that comes directly from the dihedral action on the square. If we wrote our elements of the dihedral group as permutations on the set 1, 2, 3, 4, it would look something like this. The identity fixes everything. 1, 2, 3, 4 is what R is. R squared is 1, 3 and 2, 4. And then we can keep on going with each of every one of these. We could think of the dihedral group as a permutation group acting on the set 1, 2, 3, 4. Now I want you to look at the orbit structures here. The identity has four different cycles in its decomposition. We typically skip one cycles because they just leave the elements fixed. But for the sake of counting, I want to include all of the one cycles. So the identity has four one cycles. Remember what the size of X1 was? It was 2 to the fourth, 4 being the number of orbits, the number of cycles in this decomposition here. Then what about R? If you remember from the previous slide, R, when we looked at XR, it had just 2, which is 2 to the first. Your base 2, because you had two colors, two options. But then the exponent was 1 because you had one cycle in the permutation action there. What about the other ones? Well, for XR squared, we had a 2 squared. 2 came from the two colors. That's the base. And then the exponent, there's two cycles in the decomposition. So that's how all of these are going to work. So the how many for this one? You had 2 to the first, one cycle. For this one, you had 2 squared. 2 cycles. This one, R, S, you actually had 2 cubed. Where do you get that? You get 1, 2, 3 cycles in the permutation representation. And therefore, you get 2 to the third. The number of cycles tells you the number of options you had. And you had 2 options for each. That's where the 2 to the 2 cubed came from. So with this perspective about the permutation cycles and the number of colors, this idea of coloring up to equivalence becomes a whole lot easier. Let's look at another example of this. I want to redo this exact same example that we did a moment ago, but this time let's color the squares vertices with four different colors. If I have four different colors, how many possible ways could we color the square up to equivalence? Well, we're trying to compute k. What are the number of orbits of this group action up to equivalence? That's what orbits are measuring here. Well, the group is still the dihedral group. It still has order a. So you have this one eighth in front. So then look at a stable set. Let's see if I can get this to fit on the same screen. Voila! We have all of the permutation decompositions of the dihedral group right here. So the first one, you have now four colors, but you still have four cycles. So you get four to the four. That's going to be 256. For the next one, you have four colors, one cycle, in which case you're going to get four to the first. R squared has two cycles, so you get four squared. R cubed has one cycle, so you get four to the first. S has two cycles, so you get four squared. R s has three cycles, so you get four cubed, which is 64. You get four squared for R squared s has two cycles, and then r cubed s also has three cycles, so you get four cubed right there. So with that perspective, it's like, voila! I can pull this thing out really, really quickly, because I know the cycle structures for this permutation action. We have to take 1 eighth of 256 plus 8 plus 16 plus 16 plus 64 plus 16 plus 64. That simplifies just to be 55. So while the two colored example I probably could have done it by hand, this one's going to be a lot harder because there are 55 distinct colorings, and that's not even considering that some colorings are equivalent. These are 55 distinct colorings up to equivalents. Burnside's theorem is a very effective counting technique when we have to count functions of a permutation action. So to finish this lecture, let me give you one more example. We're going to switch things up a little bit. This time we're going to ask ourselves how many ways can we label a four-sided die? So like if you're really into Dungeon and Dragons and other role-playing games, six-sided die are not enough for you. You want your dice to have four sides and eight sides and 20 sides. You know, you take all the platonic solids there. So if we think of like a tetrahedron, right? Something like this. So let's color the vertices of the tetrahedron. So you have like there's one, two, three, four vertices. So what if we want to, we're not coloring this one, I'm sorry, we're just going to label this one. So we could label it like that. Could I label it something different? Like could I label like we'll do this one one, this one two, this one three, this one four. But if it's a four-sided die, the idea is you're going to be rolling this, you're going to be casting this die as you spin it around. It's going to move around. It's still the same die even though if we change our orientation, what labels actually give us a different labeling of this thing? That's an interesting question, but group theory provides a very slick way of counting the ways we can label the four vertices of a tetrahedron of a four-sided die here. And the important observation is to remember the symmetry group of the tetrahedron. If we look at just rotational symmetries, because as we roll our four-sided die, we can't reflect it through space. We're just rolling it. That's rotational symmetry. The rotational symmetry group of the tetrahedron was a four. This is something we talked about in the first semester of abstract algebra, which of course with a four remember you have the identity. You're going to have three two two cycles. So you have the as we listed these things, you have the identity, you're going to have the two cycles, two two cycles, one two three four, you have one three two four, and we have one four two three. We're also going to have eight three cycles. So we have like one two three and we have one three two. Okay, we're going to have one three four and we're going to have one four three. Let's see is there another one. We have one two four and we have one four two. And then lastly, we're going to have two three four and we're going to have two four three. So these are, these are eight three cycles. Okay, now the the three cycles actually show up in two different consciously classes and I don't claim, I don't claim that I have any orientation with those consciously classes all right now, but it turns out if you know the conjugates, you can actually simplify these things a lot more because conjugates in a permutation group will have to have the same cycle structure, but nonetheless it's really just the cycle structure we know. So we have one element which really that one element is going to look like one two three four okay. We're going to have these two two cycles which have two cycles each as the name suggests and then for the three cycles, well there are three cycles but there's a fixed element so we should really should think of it as like one two three and then four as a separate thing. So they have two orbits each, each of the three cycles. But what's left fixed right? How many things are left fixed going on with this with this right here? So with the tetrahedron, the idea leaves everything fixed. So you get the whole thing, you get all possible labels here which are going to get four factorial of this. This is a little bit different than the colorings we did before because we can't repeat colors anymore. A label this time is a permutation. We have to label this one one, this one two, this one three, this one four. So there's four factorial ways you can color or you could label the dye. Okay. Now let's ask ourselves what's left fixed by a two two cycle. So if you have a two two cycle like if you take one two three four you're going to swap those things around. You're going to swap those things around. The issue here is that with a with a two two cycle everything that's moved around but we can distinguish the vertices because they're labeled. This isn't like the coloring did before. When you do a two two cycle nothing is left fixed. Everything got moved around. Okay. So nothing's left fixed. So we get that gives us a different labeling. So it turns out if you take your three two two cycles each of their stable sets are empty. It has size zero. So that's actually just going to be ignored in the sum. The same thing is also going to happen with the three cycles. Okay. There's eight of them but each of their stable sets is empty. If we look at the stable set of one two three this is actually the empty set. There is no tetrahedron labeled where if we swap one two and three it would be the same labeling because one two and three moved. You would have to swap. It's not like the colors we did earlier where you swapped a black with a black. No. You're swapping a one with a two. That's noticeable. And as such these labels are you see them move the stable set is going to be empty here as well. So by Burnside's theorem the number of colorings I should say the number of labeling here is going to be one over the order of the group which is 12. That's a four. You have four factorial plus zero plus zero. So you get four factorial which is 24 over 12. That just gives you two. So it turns out there are only two labelings of a four-sided die. One of them is on the screen right now. Can you come up with the other one that is different? That's actually not equivalent to the one on the board right now. It's an interesting question. I actually would challenge you also to do the same question with a six-sided die. How many ways can you label a six-sided die so that when you roll them they're actually different labels. You have you I'll give you of course the hint there are six sides to the six-sided die obviously you have to label one through six. But what's the symmetry group there? We proved previously that the symmetry group was it's isomorphic to S4. Use that to try to identify what are the number of labels you can put on a six-sided die. Thanks for watching this video and this lecture everyone. I hope you learned something. Give us a like if you did. If you want to see more videos like this please subscribe to the channel and as always if you have any questions or comments or answers to questions I gave to you in the video put those comments below and I'll answer them at my soonest convenience.