So I promised you symmetries…

Hello again and welcome! Following from my first post about group theory I figured I aught to give you a better overview by giving examples which will form a foundation for further study.

Your first family, cyclic groups. Picture a square. Actually, have it for free: rotations1 We can rotate this and keep it the same…more specifically it becomes


and then we get back to where we started. Ok. So at the moment we have these rotations by 90°. It makes sense to say that a rotation by 360° is the same as a rotation by 0°. Why? Well let’s assume that we didn’t do this. Then …,-450°, -360°, -270°, -180°, -90°, 0°, 90°, 180°, 270°, 360°, 450°,… would all be different. Also, to combine two rotations, we’d just add the numbers together. Try dividing all of the values of our rotations by 90. Look familiar? Before continuing: think about this for a bit.

So, does it look familiar? If not, think about the structure of these rotations. Angles which are a multiple of 90° form our elements. We combine these by adding them together. So the structure of this group is actually just the same as our first example, the whole numbers with addition! Also notice that no ‘squareness’ is captured by this group so let’s see what happens if we consider a rotation by 360° as the same as a rotation by 0°. This means that we only really have 4 different rotations. I’ll go through these below but before continuing: see if you can work out what they are and how they are combined.

An answer for what the group of rotations of the square should look like could consist of the four rotations; 0°, 90°, 180°, 270° (I say could because -270°, -180°, -90°, 0° or even -180°, -90°, 0°, 90° could be equally justified by the condition that 360° is the same as 0°). How can we combine these? If you haven’t had a go yet, try writing out the answers we obtain from combining these elements. Notice that, as with our first example of the whole numbers, we have that the order doesn’t matter: rotating by x° and then y° would give the same answer as rotating by y° and then x°. There is a very useful way of presenting this information, devised by Arthur Cayley and so known as a Cayley table. Let me show you the Cayley table for the group of rotations of the square and then I’ll explain what it all means.

90° 180° 270°
90° 180° 270°
90° 90° 180° 270°
180° 180° 270° 90°
270° 270° 90° 180°

So how do we read the information from the table? I’ve used colours to try and help here. We want to know how to combine any two elements from our group. The geometric visualisation here is to compose two rotations (do one followed by the other). As mentioned above, the order we compose our rotations doesn’t matter. But this might not always be the case (it won’t always be the case) and so we’re going to be careful about the order we do things.

Definition: two elements x and y commute with respect to ∗ if x∗y=y∗x. If all the elements of a group commute, we say it is commutative or (more commonly) abelian.

Ok, so first comes an element chosen from those in red. We will compose this (denoted by ∗ for this example) with an element in blue. The result can then be read off in purple e.g.

  • 180° ∗ 90°270°
  • ∗ 180° = 180°
  • 270°90° =

So why is a Cayley table useful? Well because firstly, the one above provides the information of 16 equations. But also the information is more accessible. What do I mean? Well, we can see that 0° is the identity element since it has no effect on anything it’s multiplied by. As a convention we usually list the identity element first. Also, it’s easy to see what the inverse of each element is since you just need to scan a pink row or column until you find the identity element.

There are two subtleties worth highlighting now. First, the angles written down really represent rotations by these numbers, and so writing rd where d refers to the amount of degrees clockwise rotated would be clearer. Second, these rotations can be thought of as an abstract set of rules, but we’ve been thinking of them through how they act on the square. So just assuming people are aware that the rotation in red is done first may not be clear (and using colours in this way doesn’t seem very clear, rigorous, and what if you have hundreds of elements!)…one way to get round this would be to let ◊ denote our square and then write, for example

((◊)r180°) ∗ r90° = (◊)r270°

which again could be annoying to find a symbol for the shape we are acting on each time. Using a capital letter could remedy this. This leaves us with

((X)r180°) ∗ r90° = (X)r270°

and is hopefully now completely clear. But what does this have to do with the heading (quite a while ago now) of cyclic? We’ve done one example of the rotations of the square. Using any of the notations above (though I would recommend the final notation as it is the clearest) let’s see if you can generalise. Before continuing: write out the group of rotations of the triangle and the pentagon.

In order to check your answers, I’ll introduce the whole family of groups. The rotations of the square (a 4-gon) provided us with a group with 4 distinct elements. One of these was the ‘do nothing’ rotation which we denoted by r0° . In just the same way, every cyclic group can be thought of as the rotations of some n-gon, and is denoted Cn (or sometimes Zn or 0a2067f5n). They also have a key property, one that is often used to define them (rather than taking the geometric route of symmetries as we have done above). So what is it you ask? Well, let’s think about how much we can get by continually applying one symmetry. Let’s think back to the rotations of the square. There were four of these. But we got to each one by rotating by 90° again and again (see the second picture of this post). What about the triangle? Well, we can get all of the rotations by rotating by 120° again and again. Can you see what the pattern will be?

 Number of sides Angle of smallest rotation
3 120°
4 90°
5 72°

We could also set n=2 and so look at the rotational symmetries of a line. This would consist of the ‘do nothing’ symmetry r and the half turn r180°. Thus another definition of cyclic groups is that all of their elements can be made by just applying one element again and again. We say that such an element generates the group and any element that does this we call a generator. What group does r generate? Well, we call this the trivial group (because it has the most trivial structure possible for a group) and it fits with the other cyclic groups by being called C1. We have that r120°  and  r240°  are generators of C3 and that r90°  and  r270° are generators of C4. Can you find any pattern of which 2 generators we will have for Cfor any n≥3?

Definition: the smallest number of times an element x must be multiplied by itself to get to the identity is called the order of the element, e.g. the order of r180° is 2. The order of the identity element is always defined to be 1.

Note that the order of an element is another property that can be deduced reasonably easily from the Cayley table of a group. Before continuing: compute the order of all of the elements of Cn for n=3, 4, and 5. What order must we have for the smallest rotation in Cn? What must the order of any generator be?

But what about the circle? You may know that the circle has infinite rotational symmetry, and this can be visualised geometrically by the fact that rotations by irrational numbers, for example by \sqrt{2}°, send the circle to itself. Any positive power we raise this rotation to will not get us back to where we started (can you prove this?). This means that a rotation by \sqrt{2}° does not generate the group of rotations of a circle since it cannot produce r180° : if it did then there would be a power which would then be power equal to r180°∗r180°, contradicting the above (which said no positive power will get us back to where we started). Another way to see that the group of rotations of the circle cannot be cyclic is to observe that, if r is in the group, then so is r where y=x/2. The upshot of all of this is that the group of rotations of the circle has a much more complex structure than any of the groups we’ve seen so far.

Let’s just think about the groups we have seen so far. For any whole number n≥1, we have a cyclic group with n elements. This means we have an infinite family of groups. Although these describe the rotational symmetries of a n-gon, we also have Cayley tables for each of these groups. Now, if we found another group which had the same structure of Cayley table (i.e. the only difference between the tables is the letters we have used for the elements or the ordering of the columns/rows) then we could conclude that this other group is a finite cyclic group, and so not new at all! So every finite cyclic group is really the same as a Cn for some n≥1. This notion of same is captured by something called an isomorphism, but talking about isomorphisms is probably best left for another time!

So can we have a cyclic group which is infinite? Let’s consider r where x=\sqrt{2} (the element of the group of rotations of the circle that we saw above). We’ve already remarked that if m is a whole number satisfying


then m must be 0. Here comes a (possibly counter intuitive) convention: when given a generator, it generates all whole number powers of itself (positive, negative and 0). In the finite cases this did not come into play since taking large enough powers of the element we always have inverses for each of the elements we produce and so only positive powers were necessary. But for any element of infinite order (such as r) without this convention we will not have generated a group, since we will not have inverses. Hence this convention makes sense when working with groups. We will look at more general structures where we do not use this convention (worlds without inverses, what?!) but will leave this for after we have covered groups in sufficient detail. Getting back on track, r generates an infinite cyclic group. Does this seem like quite a strange concept?

Question: thinking back to our first ever example, the whole numbers with addition, what does the number 1 generate? If you’re desperate to know if you’ve got it right, or just want spoilers before next time, here’s the answer (the 2nd paragraph of this page).

So I promised you symmetries…

4 thoughts on “So I promised you symmetries…

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