So I promised you symmetries…(part 3)

Here’s my introduction to groups and also part 1 and part 2 of this topic.

In this post I’ll introduce the symmetric group, sometimes called the full symmetric group. There are two reasons for doing these groups at this time:

  1. These groups include Dn as a subgroup, just as Dn included Cn, and so we are gradually expanding our horizons.
  2. We will be able to work with these group elements (relatively) easily, which will help with our ‘hands on’ approach.

Definition. For any natural number n, the symmetric group on n letters, denoted by Sn, consists of all bijections from {1,…,n} to {1,…,n}. We combine these elements by composition of the bijections.

Such a definition may not be very instructive. In order to make these groups easier to work with, we will see these group elements as permutations, and introduce simple notation to complement this idea. Let’s have a picture of what we’re doing, by considering our example of a square again.


We’ve found 8 symmetries which permute the vertices in different ways. But if we forget about the lines joining these 4 vertices, there are quite a few permutations we’ve not yet written down. First, we’ve not included any permutation which fixes A, but moves B, C, and D. Let’s consider a more appropriate picture,

symmetric groups

lay these out in a more sensible way

symmetric groups2editted

and finally, since they will grant us more freedom, let’s use numbers as labels.

symmetric groups3

Laying the points out in this way means we can picture the different permutations: I often draw arrows between the points showing where each point {1,…,n} is sent.

Before continuing: write down all of the possible permutations involving the numbers 1,2,3, and 4. How many are there? How many would there be for n numbers?

Hopefully you’ve had a play. This will tell you that there are 24 permutations of the numbers 1 to 4. This generalises to there being n! permutations of the numbers 1 to n. Right! Rather than talk about sends 1 to 2 and 2 to 1 and fixes 3 and 4, it would be nice to also have some proper notation. We’ll then see how to combine the elements of the group using this notation.

Cycles and cycle type. I’ll start by giving some examples. If you want to write down the permutation which swaps 1 and 2 and fixes the numbers 3 and 4, you’d write

(1 2)(3)(4)

and since saying the numbers you are going to fix is really redundant information we could write this as (1  2). Below is a picture to associate with such a permutation.

symmetric groups4

We call a permutation which only moves two points a transposition and say this because it transposes these two points. Similarly a cycle of length r, for example (1 2 … r), is called an r-cycle.

Before continuing: how many transpositions are there in S4? What about Sn?

Important: the cycle (1 2) really means ‘swap the numbers that are in the 1st and 2nd position’, which is a vital distinction. I’ll give some examples below. We may combine these permutations by doing the left most permutation and then moving the the second left most permutation (so that we are applying the permutations from left to right).

Before continuing: write out all of the possible permutations in S4 using cycle notation.

The reason for the important distinction is so that composition behaves properly for us to get a group. I’ll try to make this clear with some examples. We’ll introduce some slightly confusing notation which we will then be able to do away with. Let [1, 2, 3, 4, 5] be an ordered set and permutations of this be written as above. If a=(1 2) and b=(3 4), we have

[1, 2, 3, 4, 5](1 2) = [2, 1, 3, 4, 5]

[1, 2, 3, 4, 5](3 4) = [1, 2, 4, 3, 5]

[1, 2, 3, 4, 5](1 2)(3 4) = [2, 1, 3, 4, 5](3 4) = [2, 1, 4, 3, 5]

so that ab=ba. Setting c=(2 3) we have

[1, 2, 3, 4, 5](1 2)(2 3) = [2, 1, 3, 4, 5](2 3) = [2, 3, 1, 4, 5] – notice that (2 3) here has swapped the numbers which are in the 2nd and 3rd position, NOT the numbers 2 and 3.

[1, 2, 3, 4, 5](2 3)(1 2) = [1, 3, 2, 4, 5](1 2) = [3, 1, 2, 4, 5]

and so ac ≠ ca.

Remark. The work above (and hopefully I have always been consistent in previous posts) has been with right actions. This means that the group elements appear on the right of the set that they permute and so we read our permutation from left to right: abc means that we apply a then b and then c. In exactly the same way we could have used left group actions but, for any given question or proof, you should be consistent with the action you’ve used.

Definition. Let g be a permutation. We say that the support of g, denoted supp(g), is the set of all points moved by g.

It is generally clearer to write elements in disjoint cycle notation, which means that the cycles have disjoint support e.g. (1 5 3)(4 7) is in disjoint cycle notation but (1 5 3)(3 4) is not since both cycles permute the third entry. It is always possible to write a permutation in disjoint cycle notation…have a think why, but I’ll explain this in my next post.

How do we combine cycles? I’ll do this with the example of (1 2)(2 3), seen above.

[1, 2, 3, 4, 5](1 2)(2 3) = [2, 3, 1, 4, 5]

and so now we must look for where each of the points in {1, 2, 3, 4, 5} have been sent. Starting with 1 -though this is arbitrary- we see that it has been sent to the third position. This means our permutation starts (1 3 …). Now 3 has been sent to the second position and so we have (1 3 2…). Finally 2 has been sent to the first position and so we see that the cycle should be (1 3 2). The points 4 and 5 are not in the support of this permutation and so we have that (1 2)(2 3) = (1 3 2). There is a shortcut to doing this, so see if you can spot it (but I’ll show you next time).

Before continuing: check that the elements you wrote out for S4 are in disjoint cycle notation,  work out the order of each of these elements, and then write out the Cayley table for S4.

Are these new groups? As we’ve done before, it’s worth thinking whether we have seen these groups before. Firstly, from your Cayley tables, it’s worth noting, for n≥3, that Sn is not commutative. Question: what orders can elements have in Sn?

Spoilers! First of all we will see that these are a new family of groups. Computing the possible orders of elements in Sn is sufficient to do this. We note that an r-cycle has order r. Thus Sn clearly has elements with orders 1, 2, …, n. For m ≥ 4 this is enough to say that we have a different family from Cm and Dm. Picking cycles a and b with disjoint support we also obtain that |ab| (the order of ab) is |a||b|. Notice however that it will not always be possible to satisfy the disjoint support condition for all 1,…,n.

In order to get to grips with this new notation and multiplication, you’ll need plenty of practice. You can use the calculator available here to check you answers. Note that * is used for multiplying elements ($ provides conjugation, something we’ll see later).

Thanks for reading (and don’t forget to work on that multiplication!)

Next time:

  • an easier way to combine cycles
  • how we can always write an element in disjoint cycle notation
  • investigating when two elements (written in disjoint cycle notation) commute

A challenge: write the permutation (1 5 3)(4 7) in as many ways as possible.

So I promised you symmetries…(part 3)

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