 In this video, we're going to prove that the dihedral group is in fact generated by the symmetries R and S. So remember, R is the rotational symmetry of the regular in-gon that we're going to turn the in-gon counterclockwise by one click. Precisely, we're going to spin around the center of the in-gon by the angle 2 pi over n. And then the map S is going to be reflection across the horizontal axis, which if we coincide our in-gon with the nth roots of unity in the complex plane, this is going to be reflection across the real axis, a.k.a. complex conjugation. Now, before we prove the general theorem, we're going to first prove a lemma. And because after all, that's what lemmas are in mathematical, you know, theory, mathematical proofs here. Lemmas are those results which, although by themselves might not be super useful, they can be used to prove important theorems and propositions and the such. So what we're going to prove here in this lemma is that if we take the product R to the k, S inverse, this is the same thing as R to the negative k. So that is, if we sandwich R to the k between two S's here, that's the same thing as switching the direction. Remember, R coincides to a counterclockwise rotation. So its inverse is actually going to represent a clockwise rotation. Now, this is the equation we're going to prove, but the equation we actually want is this one right here, which to be aware, if we times both sides of the equation by S, the S's cancel here and we get S R to the k is equal to R to the negative kS. And before proving this statement, I want to show you why this equation is kind of useful right here. Now, dn is an example of a non-ability in group. The elements in this group do not necessarily commute. What we instead get is a commutation relation. What I mean by that here is that although the maps S, you know, R and S are not equal, their products are not equal to each other. R, S, and S are different symmetries. But what we can argue is the following. If ever you have an S that's to the left of an R, you can move the R past the S. It's just because the operation is non-abelian, it doesn't commute. Here's a price we have to pay a toll, so to speak, to go down this toll road, right? And so the cost of moving an R past an S is that you have to take the inverse of R. So every time an S passes by an R, you'll invert it. You'll switch it from counterclockwise to a clockwise rotation. That's what this equation right here is telling us. That S R to the k is equal to R to the negative k times S. There's this toll of inversion for moving an R past an S. And in terms of being a non-commutative group here, it's actually a fairly cheap toll, right? Forabelian groups, you think of these are groups which have freeways, right? There's no toll for taking the road. But there is a price that has to be paid for if you want to commute in a non-commutative group here. All right, so let's get back to the identity we want to prove. S R to the k S inverse is equal to R to the negative k. So let's start with the left-hand side of this expression and prove that it's equal to the right-hand side. So we have S R to the k times S inverse. Now remember that reflection across any line is going to be an order two symmetry. So that is the order of S is equal to two right here. And so particularly S squared is equal to one. The significance of that, of course, is if you move one of the S to the other side, we see that S inverse is its own identity. Excuse me, it's its own inverse, that's what I meant to say. So the element R S R to the k S inverse can be rewritten as S R to the k S. So what we see here is that we're looking at the map where we're going to reflect, rotate, and then reflect it back again. So we have to show that S R to the k S is equal to R to the negative k there. And we're going to make a geometric argument behind this because after all these are symmetries of the regular in-gone. So in terms of the geometric transformation, the map S R to the k S and we're going to be moving, of course, right to left because our element would be over here on the right-hand side. So the S on the left is going to act on it first. Admittedly, this is a palindrome. So if you did read it left to right, you'd be right for the wrong reasons or for the left reasons or whatever. S to the S times R to the k times S. This means reflect across the real axis. You're then going to rotate by the angle 2 pi k over N counterclockwise and then you're going to reflect back. And I want you to convince yourself that the net effect on the polygon, this is just going to be rotation. It's the same rotation 2 pi k over N, but this will be a clockwise rotation which of course rotating clockwise by k clicks means you're taking R to the negative 1 and to the kth power which of course is equal to R to negative k. And so this then establishes the equation we're trying to do. So this is just a geometric example of such a thing. And so what I'm going to do is I'm actually going to kind of think of a picture right here. We'll just draw it real quick just to try to convince ourselves. And so imagine we have our unit circle. And let's think of what happens to, let's say let's take the number 1 and see what happens to it. So if we go 1, 2, 3, 4, k clicks over here. So we have our point. So this is where this is where one would rotate originally. Well, if we reflect across the diagonal or across the real axis, one's not going to move. And then we're going to, you know, then we're going to rotate by k steps. So we're over here at that green point page into that. Then if we rotate back, reflect back, excuse me. This point right here then comes down over here. We'll do this one here in blue so it sticks out. And so this blue point would be the final location of the number 1. And notice that's the same thing had we just rotated clockwise by k clicks. Now you might think 1's a little bit exceptional because the reflections didn't seem to bother it as much. But you can take any other point on the unit circle and convince yourself you'll get the exact same thing. So we're just relying on the geometric argument there. Now with that limb approved, we're now ready to present the main theorem for this lecture here. We're calling this theorem 522. The group Dn, the dihedral group, right, is generated by the symmetries R and S. So specifically we're saying that Dn is equal to the cyclic, not the cyclic subgroup because it's not cyclic, but the subgroup generated by R and S. In particular, every symmetry, let's say sigma, can be expressed in the form sigma equals R to the K, S to the L, where K and L are specific integers. And specifically we also know that K is going to, you know, you don't have to choose it to be negative. We can take it to be something non-negative, but it's going to be less than N because N is the order of R here. And likewise, S will be something between 0 and 2, specifically S will be 0 or 1. There's only two options for S, but you're going to get N options for K, although that's a fact we don't need right now. Now from geometry, if we think about the symmetries of just a circle, let's not even worry about the symmetries of an in-gone right now, but let's just think of symmetries of a circle, for example, and can I dare draw a circle? It's going to be okay. Oh boy, that's too hideous. Try this again. That seems acceptable here. Now just from a geometric argument, the symmetries of a circle are going to be two-fold. You have the rotations and the reflections. The symmetry of this circle, we can't have eight. Well, if you just think of the plane itself, the symmetries of the plane have rotations around any point in the plane. You have reflections across any line in the plane. You have translations when you move things in a direction and you have like glide reflections, like footprints in the sand. You have things that kind of move forward, down some line, maybe something like you see right here, be drawing on the screen, a glide reflection. Now things, translations and glide reflections, no point is left fixed. And so for a symmetry of the circle, the center of the circle, which is this very symmetric point, does have to stay fixed. And so the only symmetries that can fix a point are going to be rotations, which the rotation will fix the roto-center of the rotation, right? The center of the rotation, which for the circle would have to be the center of the circle. So we only can allow rotations around its center. And then reflections also leave any point on the axis of reflection would be remained fixed. And so for the circle here, this means we can only take lines that go through the origin, because you have to go through the center of the circle right there. So that's that's sort of a geometric argument that the reflect are the only symmetries of a circle be rotations and reflections. Why does that matter? Well, because the regular in gone is going to be inscribed into a circle. Any symmetry of an in gone is going to be a symmetry of the circle. And so we can only accept symmetries of circles to give us symmetries of our in gone. So the dihedral group, in other words, the dihedral group is going to be a subgroup of the symmetries of the circle group, which is very, very large, right? It's an infinite group, but for DN, it's going to be much, much smaller. All right. So these are the only types of symmetries we have to worry about rotations and reflections. Now, all of the counterclockwise rotational symmetries of the in gone can be thought of as powers of R, because if you take a point like this, this sort of the point number one over here, it has to land on a vertex. We either have to go here, which is one turn, or here, which is two turns, three turns, four turns all the way until we get back to N, right? So one has to land on a vertex. And if we keep track of where one went, we can then identify each of those that we can identify that rotation with some power of R. So we either get R to the first, R squared, R cubed, et cetera. And then counter, or those are counterclockwise rotations. Clockwise rotations can be identified with powers of negative R. So all the rotational symmetries can be identified with R. Great. What about the reflections? This one's going to take a little bit more effort, but we can convince ourselves of it. Imagine we have a reflection across some line L. Now this line is going to have to go through the center of the circle, which coincides with the center of the Ingon. And it'll intersect the Ingon in one of two ways. Either the line goes through a vertex of the Ingon, right? So it goes through an Invertex, and it could maybe go through a vertex on the other side, or it might go through a midpoint on the other side. Again, my Ingon's not drawn very well. So this one should be going through a vertex, the center of the circle, and through a midpoint on the other side. And that's what's going to happen when you have an odd roots of unity. So like with the fifth roots of unity, your lines are going to go through vertices and then they're going to hit midpoints of the other side. Those are the only reflective symmetries of this Ingon. Now, if you have, for example, an even one, like say a hexagon, right, you can actually have lines of symmetry that go through midpoints of two sides, or they can go through two vertices, just as some examples. Sorry for the crudeness of my drawing here. But those are the possibilities, right? You're either going to be passing through vertices in our midpoints of lines. One of those two things is going to happen. So if row is a symmetry of the Ingon, then L must pass through the origin and can measure, and we can measure and represent L by the angle it makes. So if we kind of think of, oh, it goes through, it makes some angle with the real axis. Let's just call that theta right here. Now, because these lines have to go through either vertices or they have to go through midpoints, that's the thing that happens. This line, L, is either going to go through a vertex or it's going to go through one of the sides at a midpoint. We can measure, we can use that sort of like starting point over here. And so let's see. So in order to be a symmetry, L must pass through a vertex or the midpoint of the edge of the polygon. We mentioned that. So in either case, the angle is going to be some measurement of k times pi over n. So notice that the angle to the first vertex will be 2 pi over n, and then every other vertex will be a multiple of pi over n. The angle to get the midpoint of the first edge, that is going to be pi over n itself. It's half of 2 pi over n. And then to get any other edge, you're going to be taking multiples of pi over n. But some of those multiples will also get you back vertices. So this angle theta, we can say is equal to k times pi over the n, whichever situation we're in. And so let's call R theta the rotation around the origin counterclockwise by this angle theta right here. And so next we want to convince ourselves that the symmetry row is equal to R theta s R theta inverse. Now how do we see this? How do we get something like this? Well, again, this is a geometric argument going into play here. And so I'm going to just redraw the picture so I don't have to move everything up all the time. So imagine we have just our x-axis. That's only what we need right here. And let's take our rotation, oh, sorry, take our line of reflection L right here. And then there's this angle theta. The idea behind this is the following principle. This is the principle of conjugation, which is not complex conjugation. This is the general group conjugation, which will be defined in a previous lecture in this series. But this is kind of like the socks and sandals principle, which is not the same thing as the shoe sock principle. What I mean is the following. Imagine you're walking to your class one day for abstract algebra. And once you get to class, you realize, oh my goodness, I'm wearing socks with sandals. The fashion faux pas we've committed will be never forgivable unless we deal with this very quickly. So how do you take care of this? Well, in order to fix your fashion, you have to take your socks off. But as we've learned before, you can't take your socks off before your shoes or in this case, sandals. So what we have to do is we have to take our sandals off, then we take our socks off, then we put our sandals back on. And so that's what's going on right here. We're going to first take off our sandals, then we take off our socks, then we put our sandals back on, right? And so this sandal socks predicament will help us. And we see this in general when you look at things like GXG inverse. You take your sandals off, then you perform whatever action needs to be done, which in this case is to take off your socks, and then you put your sandals back on. This is what we call in group theory a conjugate. It's somewhat analogous to the complex conjugate we've seen before, at least in terminology, algebraic benefits, someone limited here. But we can apply this to this geometric argument right here. How does one reflect across the line L? Well, I know what S is. S is reflection across the horizontal line, the real axis right there. And so you can relate reflection across L with reflection across the X axis by this socks and sandals principle. So you're going to first rotate the line L so that it coincides with where the real axis used to be. So you take off your sandals. Then you reflect across the horizontal line, which is where L is right now. And then once that's done, you rotate it back to where L was supposed to be. So we were able to take our socks off because we were wearing sandals. So you rotate L to be S, or to be the X axis, the real axis, then you reflect and you rotate it back. The net composition of those three transformations. So we did R theta inverse S, then R theta. That has the same effect as row. But by the previous lemma we've proven five to one, right, that says that we can move a rotation past the reflection. Or in this case, I'm actually going to move the reflection past the invert by the rotation here. Because when we prove that for R, it didn't matter what the rotational angle was. And so it works in this setting as well. So to move S past R inverse is going to give you R row, R row S. So you switch from a clockwise to a counterclockwise rotation with R right here. And so then what happens is when you get an R theta times R theta, R theta squared. But since R is just rotation around the angle K theta over N, if you do that twice, you're going to get rotation about the angle 2 pi K over N. So you get the angle here R 2 theta. But in this situation, R is rotation by 2 pi over N. And we've done that K times. So 2 pi R 2 theta, it will actually be the angle R to the K. And so we've expressed R, R row, excuse me, as R to the KS. And so we see that what we've now accomplished is that any symmetry of the in-gone, whether it's a rotational or reflective symmetry can be written either as a power of R when it's a rotational symmetry. And when it's a reflective symmetry, it can be written as R to the K times S. And so this proves that R and S generate the entire group because every symmetry can be written as some product of R's and S's. Either it looks like R to the K for some K or it looks like R to the KS, depending on whether it's rotational or a symmetry here. And so what do we want to mention next? So each product of R's and S's can be called a word. So you look at something like this, this is what we mean by a word of R and S. It's some product of R's, S's, and their inverses. Now as S is its own inverse, we don't need S inverse. And as R to the N minus 1 is just R inverse, we don't need any R inverses as well. So we think of these as letters in an alphabet, R and S. And what words can we spell with these things? All right? Well, this is what we mean by word. Well, in most general terms, what are the words we can produce using R's and S's, right? Well, like I said, S squared is 1, so we never need to have 2S's in a row. We never need an S to the negative 1 either. And so if we look at a generic word in DN, we're going to be getting something that looks like this. You have some power of R times an S, times some power of R times an S, times some power of R times an S all the way up and we go through this. Now we are allowing the fact that K0 and K to the M could be 0 themselves. So our word could start with an S or it could end with an S, but then the intermediate ones here aren't going to be 0 powers because we can separate specific things like that. So we're going to induct on the number, on the length of these words right here to show that each of these things are going to be things part of our dihedral group right here. So the smallest possible word would just be, and we're actually going to induct on the number of S's inside of this generic word right here. So the smallest possibility is that you have no S's whatsoever. So you get R to the K0. That is just a rotational symmetry. And so that will just be R to the K0. That looks like R to the K0 times S to the 0. Just to be clear here, what we're trying to show right now is that sigma can always be written in the form R to the K times S to the L, the final statement of this proof. And so yeah, if you are just a pure rotation, you can just take the 0th power of S. That's perfectly fine. So then by our inductive hypothesis, if you have M minus 1 mini S's in your word, this can be simplified to look like R to the K times S to the L. And so then let's consider a symmetry which then has one more S in it. Well, basically then you're going to start off with the first part. You're going to start off with the first part right here. This word row, excuse me. Why can't I write that? This word row right here. Then by our inductive hypothesis can be written in this form RKSL. And so then sigma, which then factors as row times S times R to the KM. Well, the row can be written as R to the KSL. And then we have to deal with this S and R to the KM here. But well, we can combine the S's together. This will give you S to the L plus 1 and that might simplify, but I don't really care right now. We can deal with that later if necessary. How do we get rid of, how do we get the Rs together? Well, by lemma 521, which we saw previously in this video, one by one each of these Rs can be moved by an S. Now there are L plus 1 mini S's and there are K sub M mini Rs. And so if we go through that, each R that passes an S will be an inverse. Then it passes the next S, it'll be inverted again, which will go back to the original. You're going to get negative 1 to the L plus 1 for each of the Rs. And there's going to be KM mini Rs. So you can put all those powers together. And so you're going to get R to the K times some power of R. It could be positive, it could be negative, whatever. But these are going to be combined together with the KR's that are already there. And so you're going to get something like R to the K prime times S to the L prime, which in this case L prime would just be L plus 1. And so every word in the dihedral group can be written in this form, R to the K times S to the L. And in fact Rs. So the Rs and S generate the entire group, but even better, every word, every symmetry can be written as some number of rotations followed by a reflection or not a reflection.