 This video is part of an online mathematics course on group theory, and today we will be covering dihedral groups. So you remember dihedral groups, often denoted by D2n, are groups of order 2n, which are the symmetries of a regular n-gon. They have two n elements because you can rotate it. There are n possible rotations and n possible reflections. So let's write out the first few. So D6 will just be the group of symmetries of a triangle. And you remember this is the same as the symmetric group S3, which are all permutations of the three points. D8 is symmetries of a square, which we made use of in the lecture on rooks. Then we have D10, which would be the symmetries of a regular pentagon, D12, symmetries of a hexagon, and so on. And the group Dn is generated by two elements, a and b, such that a to the n equals 1, and b squared equals 1, and b, a, b to the minus 1 is equal to a inverse. So a is order n, b is order 2, and if you conjugate a by b, you turn a into its inverse. And these, the element a is just a rotation. So for D10, for example, if we have a pentagon, the element a will just rotate it like this, whereas the element b might do something like this. So you might just reflect it along a vertical line. So here's the element b, and you see if we have a line here, the element b is just a reflection in the line. And these two relations are obvious, and this one you have to maybe think about for a few seconds, but you can see that this also holds. Well, it's sometimes useful to have D4 as well. So what's D4? Well, D4 is generated by elements such that a squared equals 1, b squared equals 1, and b, a, b to the minus 1 is a to the minus 1, which is a. So b and a commute, and D4 is just z over 2z times z over 2z. You can think of this as being a regular polygon of two sides, which gives you a very, very long thin rectangle. It's sort of infinitely thin or something. But it's also sometimes useful to have D2. Well, D2 would say a is equal to 1, and b squared is equal to 1. So D2 is just c over 2z, just the group of order 2. And I don't really know how to draw something with just one side. So I'll leave it at that. So D2 and actually makes sense for n greater than or equal to 1. So now let's try and find what are the conjugacy classes of dihedral groups? Well, let's first of all look at say D2n for n even. So let's look at the conjugacy classes of say the group of symmetries of a hexagon. And we can get some reflections. So we could get a reflection in this line here or in this line here or in this line here. So we get three reflections. Well, we also get three more reflections because we can also reflect in these lines here. And these three reflections are all conjugated essentially because the group acts transitively on these green lines. So acts transitively on the reflections corresponding to and these ones are also conjugate. But these green reflections aren't the same as the red reflections. So we get plus three reflections. And then we get rotations. So let's think what what sort of rotations we get. Well, we get the identity rotation, which does nothing. And then we have two rotations by one sixth of a revolution. So we can either do that or that. And then we can get two rotations by two sixth of a revolution. So we can either do that or that. And finally, we get one more rotation by 180 degrees. So we sort of rotate all the way around there. So the conjugacy classes, we get one plus two plus two plus one rotations. And we notice this one is order two. And these ones, the various orders three and six and so on. And if you look at this, we see we get three different conjugacy classes of involutions. You get one involution that's a rotation by 180 degrees and two more involutions. These are reflections. I should have said involution is the same as element of order two. They are particularly important for various reasons. So here we have the involutions. So we've got three classes of involutions. And we also got one element in the center because this element here, rotation by 180 degrees is in the center. So the center has order two. And something very similar happens for D to N whenever N is even. We again get three classes of involutions, one of which is an element of order two in the center. And then we get the identity elements and a bunch of pairs of rotations. Now let's look at what happens for D to N for N odd. This ends up being a bit different. So let's do a pentagon and let's look at reflections. Well, we can reflect in this line or this line or this line or this line or yes, I missed out one. So we get five reflections, which are all order two. And this time they all form one conjugacy class. You can see any two of these lines. There's a symmetry of the pentagon that takes one to the other. And you see there's a difference between the case when the polygon has an odd or an even number of sides, because if it has an even number of sides, one of these reflections can either go through two vertices or it can go through two edges. Whereas if it has an odd number of sides and something going through a vertex ends up going through an edge on the other side. So there's only one conjugacy class of reflections instead of two. Now let's look at rotations. Well, again, we can have an identity rotation of order one. And we can have rotations by fifth of a revolution of order two. And we can have rotations of two fifths of a revolution. Sorry, order five, not order two, again of order five. So we get two pairs of in this case it happens to be order five. And we notice there's no element in the center apart from the identity. So the center is trivial, meaning that it only contains one. So when there were an even number of sides, we had this rotation by 180 degrees, which commutes with everything. But for an odd number of sides, there isn't a rotation by 180 degrees. So there's no center. So this is a sort of difference between regular polygons with an odd or an even number of sides, they have different numbers of involutions. And they have the ones with an even number of sides have a center and the ones with an odd number of sides don't accept for D two. So D two is a kind of slightly bizarre special case. So apart from D two, there's one conjugacy class of involutions. In fact, even for D two, there's one conjugacy class come to think of it. And the identity element and then various conjugacy classes of order two consisting of rotations, these rotations may have various orders. They all have order five in this case, because five happens to be prime. If five wasn't prime, then you could get various devices of it. So there's another funny property of dihedral groups of even order. Suppose we look at the group of rotations and reflections of the hexagon. Now inside the hexagon, I can draw a triangle. And we notice that all symmetries of the triangle are contained in the symmetries of the hexagon. So we have D six is contained in D 12 as a subgroup. Well D 12 also contains another group, the group one minus one is contained in D 12. And you can now see D 12 is the product of these two groups. Any symmetry of the hexagon is a symmetry of the triangle possibly multiplied by minus one, because you can, there are really two different triangles and you can switch these two triangles with the element minus one and then do a symmetry of the triangle. So we see D 12 is actually equal to D six times a group of order two. The dihedral group of order 12 splits as a product. And the same thing happens whenever we have a group D 4n with n odd. So the dihedral group of dihedral groups of order D 4, D 12, D 20 and so on all split as a product of two groups for much the same reason. For instance, if we did D 20 that would be, we would have to take a decagon 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 and inside this decagon we would find two pentagons which play the same role as these two triangles. You may ask why doesn't this work for dihedral groups like D 8 or D 16? Well let's see what happens there. Suppose we try and do the same trick for D 8. So we draw two squares. Well now we've got a problem because the element minus one doesn't actually exchange these two squares. In fact the group generated by one and minus one is now actually contained in D 8. So D 16 is not a product of D 8 in this group because this group is contained in D 8. Next we see that any dihedral group is generated by two involutions and we can take these two involutions as follows. Let me just do it for D D 10. One of the involutions is going to be reflection like that and the other is going to be reflection like that. So you can see the product of these two involutions if I call these x and y and x times y is rotation by in this case at the one-fifth of a revolution. So that will give us the two generators for the dihedral group. In fact you can see in the example of this I will just use this small turtle to illustrate. So if I get a pair of mirrors like this then these two mirrors are the two involutions are going to be the two involutions of my dihedral group and you can see if the two mirrors are at right angles then I'm getting the dihedral group of order four. You can see there are four different images of the turtle and if I move them around a bit this is now the dihedral group of order six. You can see there are six small turtles and the angle between them is one-sixth of a revolution and I can go on like this. There we have eight turtles and the angle between these mirrors is one-eighth of a revolution and I can go on to see if I can get 10 turtles. So there's 10 turtles and here we have the dihedral group of order 10 generated by two reflections and obviously you can just go on like that for as long as your patience holds out. Conversely any group generated by two involutions is a dihedral group. Well I have to allow infinite dihedral groups as well here that I'll give an example of them in a moment. So here the group has elements a squared equals one and b squared equals one so both a and b are other inverses. Now we can look at the smallest positive integer such that a, b to the n is equal to one and if we put c equals a, b we find that c to the n equals one and ac, a to the minus one is equal to ba as you can see from an immediate calculation which is equal to c to the minus one. So these elements satisfy the relations for a dihedral group and it's not very difficult to check you actually get the entire dihedral group. You get exactly the dihedral group of order two n from this and in fact if a squared and b squared are both one then you can just write down all elements of the group. There's one a and b and we can't have a squared because that's one and we can't have b squared that's one so we get a, b and ba and if you multiply these by a or b on the left and the right well if you multiply that a and the left you just get b again so then we get a, b, a, b, a, b, a oops overshot b, a, b, a and so on. So it's very easy to write down all elements of the group explicitly and if a, b to the n equals one you can cut this down and show there are only two n of them which must be the dihedral group. So this covers, so groups generated by two elements of order two are very easy to describe they're all dihedral groups. You might think the next case to do is groups generated by an element of order two and an element of order three. Unfortunately these groups turn out to be ridiculously complicated, so complicated you can't possibly classify them. I mean the same with groups generated by any pair of elements of any order other than two and two or I guess one and n. I guess I should just mention the infinite dihedral group if you just miss out this relation completely then you get the infinite dihedral group which you can think of as being a group of symmetries of a regular polygon with an infinite number of sides which just becomes sort of straight line and here there are two reflections because you can either reflect in this line and swap it like that or you can reflect in this line and swap it like that so you can think of this being the group of isometries of the integers. So let's just give an application of this theory of dihedral groups. Suppose g is finite and suppose a and b have order two in g then either a and b are conjugate or a and b both commute with some involution c which seems a bit of a surprising result at first but this is actually quite easy because pick the smallest n with ab to the n equals one which we can do as g is finite so a and b generate the dihedral group d2n and now if n is odd this implies a b a conjugate because we saw that any two involutions in a dihedral group d2n with an order conjugate if n is even you remember the dihedral group d2n has an element c in the center which was ab to the n over two in fact so ab to the n over two has order two and commutes with ab. So what this implies is that if you've got a finite group that either all involutions are conjugates there's only one conjugate because you class it involutions or you can find two different involutions that commute with each other in which case it would have to contain a copy of the Klein four group. By the way there's no obvious analog of this for elements of order three so you can't say that elements of order three are either conjugate or there's a or you can find a product of two groups of order three for instance if you take just a cyclic group of order three then it has two involutions that are not conjugated it doesn't contain a product of two groups of order three. So next lecture we will move on to groups of order 12 and introduce the Silov theorems in order to help study them.