 In the previous video, we showed that every symmetry in the dihedral group Dn has a so-called normal form. Symmetry sigma can be factored uniquely as r to the k times s to the l, where r is the principal rotational symmetry. You rotate counterclockwise by the angle two pi over n, and s is gonna be the horizontal reflective symmetry across the real axis. So we think of our in-gone as the complex in-sroots of unity. So we have this normal forms. And I should mention that every element in the dihedral group is uniquely determined by its normal form. So for example, if you have two elements, say sigma and rho, and sigma has a normal form r to the k, s to the l, if rho has that same normal form r to the k, s to the l, well, that actually means these two things are equal to each other. So rho and sigma are the same element. So two elements have the same normal form, are actually the same element. But let's say that you have two different normal forms for the same element. So sigma looks like r to the k, s to the l, and r to the k prime times s to the l prime. Well, by working this equation right here, multiplying on the left by the inverse of r to the k prime and on the right by the inverse of s to the l, you're gonna see that you get r to the negative k prime, r to the k is equal to s to the l prime times s to the negative l. So we have a, so if you, when you take r to any power, write this composite, it's gonna turn out to be some type of rotational symmetry. Which one it is? I don't know. And then right here, when you multiply these things together, you're gonna get some type of reflective symmetry. I guess I should say it's a little bit more simple. When you take s to some power, you're either gonna get s or you're gonna get the identity. The identity is not a rotation. So the old, excuse me, s is not a rotation. The only way that then a power of s can coincide with a rotational symmetry is if it was the identity, like so. So if it's the identity, that means that the powers of s combined together to get some even number, we have to have it that s to the l prime is actually equal to, equal to s to the l. So that s to the negative l had to be the inverse of s to the l prime. So this would tell us that l prime is congruent to l mod two. So the same thing's happening over here that the only way that r to the negative k prime times r to k is equal to the identity is that, in fact, r to the k and r to the k prime were actually the same element. That doesn't mean the integers are the same, but they are gonna be congruent to each other mod n where n is the order of r in the situation. And it turns out this argument right here gives us a way of counting the number of elements in our dihedral group, dn right here. So what's going on? So dn, we can claim, actually has order two to the n. And this comes from the normal forms here. So every element in dn can be written in the form r to the k times s to the l where k and l can be uniquely chosen between, so that k must be between zero and n minus one. So there's n options for k because the order of r is n. And then likewise l has to be chosen between zero and one. So you're gonna get two options because the order of s is two. So if you have n options for k and two options for l, you're gonna get n times too many options total following from the fundamental multiplicative principle of counting. So then then the order of the dihedral group is two to the n, that's great. Now I do wanna mention that there are some group theories out there that when they talk about the dihedral group, they don't use the notation dn, they use the notation d2n. And the reason they're doing that is they're using the subscript as describing the order of the group. So it's like, oh yeah, if you take the dihedral group of order eight, you might compare it with other groups of order eight, like the cyclic group of order eight or the quaternion group of order eight. So the subscript is describing the order. As opposed to the notation we use here, where the n describes the polygon that we're describing here, the symmetry groups of said polygon. So we would say that the order of dn is equal to two n, but they would say that the order of d2n is equal to two n, which seems clear I guess. But then you get into sort of confusing situations like what does d8 represent? Are we talking about the dihedral group of order eight or the symmetry group of the regular octagon? Those two groups are not the same because the symmetry group of the octagon is actually 16. And so when you are looking at the dihedral group in any mathematical publication, you do have to make sure which notation is the author using, dn or d2n. dn is the notation we're gonna use for our series here that coincides with Judson's notation in the textbook. It's also the notation I prefer because again, as an algebraic commentatorist, as a representation theorist to me, it makes sense to be describing the set that the dihedral group is acting on as opposed to just the order of the group. But both of them are commonly used and I don't even know which one is used more. It might be even be used equally. So you have to be aware that dihedral group could be denoted as dn or d2n. The two n notation comes from the fact that we're talking about the dihedral group of order two to the n. So let's consider some of the dihedral groups we've talked about already. So the symmetries of the equilateral triangle d3, for example, we could orient the vertices of the triangle to be complex roots of unity and I'm just gonna call them a one, two, three. So rotational symmetry, R, it would send one to two. Two would go to three and then three goes to one. So that gives us the traditional three cycle, one, two, three. For the square, I wanna mention to you that if you take this rotation R, you're gonna go one to two, two to three, three to four, and then four to one. And so you get the four cycle, one, two, three, four. So in general for the dihedral group, R is gonna look like one, two, three, all the way up to n. If you label the vertices in this counterclockwise rotation, like we see here on the screen right now. What about S? Well, S is gonna be reflection across the horizontal line. Whoops. So in this case, you're gonna swap two and three for the triangle. CCS looks like the vertex two, three. Well, R squared would be to rotate twice. So one goes to three. Three goes to two, and then two goes to one. So you get one, three, two. It's also a three cycle. This of course also coincides with R to the negative one. It's a counterclockwise rotation. How do you get RS? Well, RS would look like one, two, three, which is R times by S, which is two, three. Notice that one goes to two. Two goes to three, which three goes to one. So you get this two cycle, one, two, and then three will remain fixed. Three goes to two and two goes to three. So RS is the reflection across the line that leaves three fixed. And the idea is you had S, you had like the line associated to S right here. You're gonna rotate a half angle, and that's where you're gonna get your next symmetry right here. And that's where you get RS. And then R squared S is the other one. It's gonna be reflection, reflection. Let's see. Oh, that's a typo there. Sorry about that. That should be the other one. That should be one, three. So reflection through two right here. The issue of course being, you rotate this thing twice and you can compute that. If you take one, three, two, and you times it by two, three, let's see, one goes to three. Three goes to two, which two goes to one. So one goes to three here. And then what happens to two, two goes to three, three goes to two. So we get the following. Sorry about the typo. One, three is what you get right there for the other symmetry. For the square, if we take reflection across the horizontal here, two goes to four, four goes to two. So we get this thing right here. If we do a double rotation, one goes to three, three goes to one. And then two goes to four and four goes to two. So we get this guy right here. Triple rotation is just gonna be clockwise rotation. One goes to four, four goes to three, three goes to two, two goes to one, like we see right here. The other ones, I'll let you compute them. What would RS look like? RS, we'll do this one together. So you're gonna get one, two, three, four times two, four. So one goes to two, two goes to four, which goes to one. So you're gonna get one, two as a two cycle. What happens to three? Three goes to four, and then four goes to two, which goes to three. So you get this two, two cycle, one, two, three, four. So with symmetries of the square, you'll notice the following. You have that the horizontal reflection and the vertical reflection, these ones are perpendicular to each other. That corresponds to S and R squared S. And then the diagonal ones, these reflections right here, that'll coincide with RS and R cubed S. And so that kind of discusses the symmetry groups for these regular indons, which is the dihedral group. Some other symmetry groups that we could also discuss would be like three-dimensional objects, symmetries of three-dimensional objects. Like let's take the platonic solids, for example. We could take the tetrahedron. The tetrahedron will have as its symmetry group, the alternating group A4. The cube will have as its symmetry group S4, the symmetric group there. And that's something we can talk about at some other time, right? Actually, I think I leave these as homework assignments for my students here. The octahedron, it'll actually have the same symmetry group as its dual platonic solid, which is the cube. And then there's also the symmetry groups for like the icosahedron and the dodecahedron, which ones are a little bit more complicated so we are not gonna talk about those in this series. But it'll be left as a homework exercise to prove the symmetry groups for the tetrahedron and the cube are exactly A4 and S4.