Groups Lecture 17

in which we explore conjugacy classes in S_n and A_n, and prove that A_5 is simple.

Having proved last time that the conjugacy classes in S_n are determined by cycle type, we listed all sizes of the ccls of S_4, and also the sizes of their centralisers, using Orbit-Stabiliser. Knowing the ccls and their sizes, we could have a look at what normal subgroups we could find, because normal subgroups must be unions of ccls. In S_4 we found four such, and also determined the corresponding quotients. Here our “groups of order 6” result came in useful. I left you to try the same for S_5.

We then looked at ccls in A_n. It is not quite the same there, as some ccls can split. We saw, using Orbit-Stabiliser for both S_n and A_n, that the ccl of \sigma\in A_n splits in A_n if and only if \sigma commutes with no odd permutation in S_n. So we just have two possible situations for any \sigma\in A_n: since C_{A_n}(\sigma)=C_{S_n}(\sigma)\cap A_n, we either have |C_{S_n}(\sigma)|=2|C_{A_n}(\sigma)| and the same ccls, or we have C_{S_n}(\sigma)=C_{A_n}(\sigma) and the ccl splits. We explored this in A_4 and A_5, and in both cases found only one ccl which split. We used the sizes of the ccls also to prove that A_5 is simple: the only unions of ccls (including e) which give a size dividing |A_5|=60 are \{e\} and all of A_5.

Understanding today’s lecture

Getting your head round the splitting ccls can take a few tries. I remember my lecturer explaining it with “A_n has index 2 in S_n, and C_{A_n}(\sigma) has index 1 or 2 in C_{S_n}(\sigma), so the only possibilities are these.” I didn’t really understand it with the index, it helped me to write down both Orbit-Stabiliser equations explicitly, and then I saw what was going on. So if you don’t understand my explanation, see if you can reformulate it in a way that you like better. Or look in a book, or talk to your colleagues, or your supervisor, or come and ask me :-).

Preparing for Lecture 18

Next time we will finish the promised “groups up to order 8” by looking at the missing order 8. We will see that there are quite a few, which is in fact because 8 is a power of 2. Apart from the obvious products of cyclic groups, we also have a dihedral one and a “new” one, called the Quaternions. We will look at those in a bit more detail. We will then start on matrix groups. We’ve seen a few already throughout, so you could look back to find matrix groups in your lecture notes as a preparation.

Going a little deeper

We saw in the examples of A_4 and A_5 that only one of the ccls split, that of the largest possible cycle (which is actually in A_n). You could investigate what happens in general for A_n. Will it always be just the largest possible cycle? Or could there be others as well? I think I started investigating this once, but I can’t quite remember how far I got, so I’d be interested to hear your answers! (Maybe not on the blog, so you don’t spoil it for others who want to try. Or at least with sufficient spoiler alert.)


2 thoughts on “Groups Lecture 17

  1. Spoiler: (partial) Solution to when conjugacy classes split in A_n.

    I’m pretty sure that in A_8 the (5)(3) conjugacy class splits, whereas (7)(1) is the largest cycle, since I can’t find any way that an odd permutation can commute with it. More generally, I think that a conjugacy class splits if and only if it is made up of distinct odd cycles. The ‘only if’ is easy to prove (if g contains an even cycle then it will commute with this cycle, and if there are two odd cycles of the same length you can swap all of the elements in the two cycles). The ‘if’ seems intuitive, but I do not have a rigorous proof.


    • Good start! Can you find the size of the centraliser of a 5,3 cycle in S_8? I’ve just calculated it, and that will help you to prove what the actual centraliser is and whether it contains any odd elements.

      You are right about the even length cycles, and the odd length of same length. (Typo in your answer: I think you mean “even length” and “odd length” in the bracket sentence. Remember that even length cycles are odd and odd length cycles are even, so this distinction really matters!)

      Additional question: will the “longest length even cycle” always have a splitting ccl? You’ve shown that it is not always the only one, but is it always one? That might be a first step towards the “odd length cycles of different length” proof.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s