Having looked at cyclic groups, let’s look at dihedral groups. They can be created geometrically by starting with a cyclic group Cn… think of it as rotations of a regular n-gon through multiples of 360°/n… and then imagine that you can also spin the n-gon out of the plane about some axis.

Huh?

Let’s look back at the cyclic group C3. Take an equilateral triangle. Take our group operation r to be a rotation thru 120° (= 360°/3). Then r^2 is a rotation thru 240°, and r^3 is a rotation thru 360° – that is, r^3 = 1, the identity element.

Here a pictures of what r and r^2 do to an equilateral triangle:

(I could have, and probably should have, set things up so that the drawings showed counter-clockwise rotations, but I didn’t. What they’re really showing is the permutations, and they made more sense to me the way I did them, in the previous post.)

I need to get the Abstract Algebra package involved here, so let me do what I must…. I ask for a dihedral group based on a triangle (“3”), and I want it as a set of permutations, so I can apply them to a triangle. Note that the group G is both a set of elements and a binary operation on them; it is not just a set; in this case, the operation is the composition of permutations.

(Don’t be confused by my also having the equilateral triangle as a set, or maybe its vertices as a set, or even just the labels “1”, “2”, “3” as a set. The group is one thing; the object I’m seeing its effect on is another.)

I extract the elements of G… and call that list p for permutations… and I ask for the length of that list:

Let me point out that the group I asked for is of order 6. The notations are not standard – oh, it will almost certainly be “D” but you will see the dihedral group of order 6 denoted either D3 or D6. You just have to find out which notation your book – or any other reference – is using. Or, if you ask someone about D6, you must tell them what you mean by D6 – because it could symbolize the dihedral group of order 6 or the group of order 12. (D3 can only be the dihedral group of order 6.)

Since I have to ask for it as “3”, and since it is symmetries of a 3-gon, I will write it as D3.

OK? My notation is that the dihedral group of order 2n is written Dn, because it applies to a regular n-gon… but in addition to the n rotations, it has n spins.

Here’s my starting point:

The 3rd permutation in G, applied to that triangle, gives me this:

Or I could show it this way:

The point is that I can’t get the “after” image just by rotating the triangle in the plane. I have to spin it out of the plane, about a vertical line thru vertex “3”. We have interchanged labels “1” and “2” while leaving “3” in place. To use a different phrase, I imagine picking up the triangle, turning it over, and putting it back down.

I can rephrase that yet again. This symmetry of the equilateral triangle is not a rotation in the plane, but a rotation in space. The rotations that made up the cyclic group were about the z-axis; now I’m making a rotation about the y-axis.

There’s another way to think of it, too. We could imagine that we have reflected the image across the y-axis. Geometrically I find that less satisfying than spinning an object; but the mathematical description of an inversion is 2-dimensional instead of 3-D.

To be clear, here’s my implicit coordinate system:

Let me try another one….

That’s a spin about a line thru vertex “1” and the center of the triangle. Label “1” is fixed but “2” and “3” have been swapped.

There’s a third spin, of course, about a line thru vertex “2” and the center of the triangle:

The other three elements of the dihedral group are the three rotations thru 0°, 120°, and 240°, i.e. elements of a cyclic group of order 3. The group as a whole, then, should have an identity (order 1), three spins of order 2, and two rotations of order 3.

Let me simply ask for a dihedral group “3” without specifying “permutations”:

We see “Rot” for rotation and “Ref” for reflection – what I’m calling a spin. And there’s a reason for that. It is almost universal to write the elements of a dihedral group as “r” and “s” – and so I use the word “spin”.

We can get the package to do that: RotSym and RefSym can be set to whatever I want…

(I find it strange that it lists r^2 before r.) Anyway, I can ask the package what are the orders of the elements of the group:

As I, at least, expect: one element of order 1… two elements, r and r^2, of order 3… and three elements of order 2 (s, rs, and r^2 s). (Hey, what about things like sr and s r^2? We’ll see – they’re included already.)

We see something interesting when we do not use permutations. It appears that there are two fundamental building blocks (generators) of our group: r and s.

We have our two rotations, r and r^2, and a spin s… but then rs and r^2 s are also of order 2… that is, they are the other two spins. (It doesn’t matter which spin we take to be “s”.)

Let’s see this. I’ll take s to be the spin about the vertical axis

Our rotation r will still be a 120° CW rotation:

What is rs? First, read that as the composition of functions: s first, followed by r. So “s” applied first gives us

and if we rotate that 120° CW, we get

… which is the spin thru “2”.

(I know. I drew that by asking for the spin thru “2”… but what I really did was just try permutations until I had the one that looked right. I knew what the answer looked like, I just had to find the permutation that gave it to me.)

There’s a contrast here. Geometrically, we have 3 spins and two rotations (three, if we count the rotation thru 0 or 360°, which is the identity). But algebraically, we have

r^3 = 1 and s^2 = 1

and products of the form

r^k s^m, with k = 0, 1, or 2 and s = 0, or 1.

Products of the form “all powers of r first, followed by all powers of s” are all well and good… but what about, for example, s r?

We do the rotation first.

And then we spin about the central vertical line, getting

which is the third spin, i.e. r^2 s. (It’s not the second spin, rs, thru the southwest vertex)

We seem to have

.

That is a crucial result, because it shows that s r ≠ r s, so the dihedral group D3 is not commutative (i.e. not abelian).

In fact, since , we could rewrite that as

.

(If you’re not used to that…. says that r^2 is a right inverse (hence an inverse) for r; and it says that r is a left inverse (hence an inverse) for r^2. Either way, .)

However we write it, that relationship is a crucial piece of the algebraic puzzle. Algebraically (in contrast to geometrically) we define a dihedral group of order 2n by three relationships:

(or or ; it really helps if you remember that s^2 = 1 means .)

Again, the dihedral group Dn of order 2n is not abelian (unless n = 2, in which case we do have r^2 =1, i.e. .)

Let’s look at the multiplication table for D3 (they’re named for the mathematician Arthur Cayley):

Read the horizontal line labeled “s” on the left. Then the column headed by “1” says that s * 1 = s, and the next entry says … then … … and then the last two, and .

How would we confirm those last two? If I remember only the relation , then I write

. QED.

And for the last one,

. QED.

Here, let’s have the dihedral group of order 4: rotations and spins of a square. We can spin it about vertical or horizontal lines through its center, or about the diagonals… we should expect the four rotations 360°/4 (one of which is the identity) that comprise C4, and four spins each of order 2.

Define the group and extract its elements:

Start with a square and labeled vertices:

Spin it about a vertical line through its center. We know we get the image

(We know the answer; I just have to pick the permutation that gives me what the geometry tells me. Why should I work at drawing the answer myself when I can just ask for it?)

About a horizontal line through its center?

Spin about the 1-3 diagonal?

Finally, spin about the 2-4 diagonal?

As before, these spins are not rotations about the z-axis; we have got more symmetries than just the rotations that give us a representation of C4.

Now let’s get the usual representation:

Before we see the orders of the elements, what do you expect? The beginning of the answer is “4 spins of order 2, and the identity of order 1, and a rotation of order 4″… the rest of the answer is that “if r is of order 4, so is r^3, and r^2 is of order 2.”

So I expect five elements of order 2, two of order 4, and the identity of order 1:

And here’s the multiplication table:

Look at the product : we have r s. That’s the defining relation:

.

(I wrote to emphasize that you need to choose the row labeled by “s” and the column labeled by “r^3”.)

You should expect that there are finite dihedral groups of every even order (even of order 2, since I can argue that we have n = 1 and

(We haven’t proved, but it is true, that r^1 = 1 implies r = 1. That is, the only element of order 1 in a group is the identity.)

Also: the dihedral group of order 4 does exist, even though I cannot see how to represent it as symmetries of an arrow. It’s usually called the “Klein 4-group” and we’ll see it when we do direct products. (If you can’t wait, it’s usually described as C2 x C2, the direct product of two cyclic groups of order 2.)

I should also show you how to work with the permutation representations of a group. You might ponder the rows of the multiplication table: each row is a permutation of the labels across the top… and therefore the multiplication table of a group automatically gives us a realization of that group as a group of permutations. That’s a major fact. Although it’s not always feasible to study a specific group as just a group of permutations, in a sense the general abstract definition of a group didn’t get us anything more than if we had defined a group as a self-contained set of permutations.

Next?

Groups of permutations and, I think, the quaternion group.

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