# 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 $A_5$.

We started on the proof of what the possible groups of order $8$ are, and we will finish it next time.

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 ask on the forum on the moodle page :-).

Preparing for Lecture 18

Next time we will finish the proof about groups of 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 (of those who are 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.)