# Thread: The Automorphisms of the Symmetric Group

1. ## The Automorphisms of the Symmetric Group

I was assigned in my algebra class to find the automorphisms of S_n. For S_2 it was trivial - it is just the identity transformation. After a lot of work I was also able to show that for Aut(S_3) is isomorhpic to S_3 and Aut(S_4) is isomorphic to S_4. I basically used the fact that the center of S_n is trivial and hence Inn(S_n) is isomorpic to S_n. Since Inn(S_n) is a normal subgroup of Aut(S_n), this showed that Aut(S_n) has at least n! elements. I then showed that Aut(S_3) and Aut(S_4) have at most 3! and 4! elements by usining some propeties of their generators, multiplication tables, etc... It was quite tedius.

Now, my intuition told me that then Aut(S_n) is isomorhpic to S_n except for n = 2. Unfortunatly I found out via wikipedia that this is not true, that n = 6 is an exception.

Now more searching on the web showed this proof that looked pretty easy to understand. However, upon reading it, I just don't understand it. He uses the term "similiar elements" which I have no clue what he means. Also, how does he get this part of the eqaution - n!/(2^k * k! * (n-2k)!). And how can he say that n*(n-1)/2 = n!/(2^k * k! * (n-2k)!) implies k = 1 for all n except 2 and 6? I used mathematica to prove it to me for the first thousand integers but I can't prove it, and that does not sit well with me.

If anybody could elucidate the proof or provide an alternate I would be very much so in their debt! Thanks.

2. Segal uses "similar" to mean conjugate. A "class of similar elements" means a conjugacy class.

The gist of Segal's argument is that an automorphism of a symmetric group must take a transposition to an element of order 2, but there is no a priori reason to suppose that it will take a transposition to a transposition (and in fact the outer automorphism of S_6 takes a transposition to a product of three transpositions).

So suppose that an automorphism of S_n takes a transposition to a product of k disjoint transpositions. The conjugacy class of the transposition contains ${n\choose2}=\frac{n(n-1)}2$ elements. The conjugacy class of the product of k disjoint transpositions contains $\frac{n!}{2^kk!(n-2k)!}$ elements (i.e. the number of ways of choosing k pairs from n objects).

Segal skips the calculation which shows that these numbers can only be equal for k=1, unless n=6. Here's how I did it.

If $\frac{n(n-1)}2 = \frac{n!}{2^kk!(n-2k)!}$ then $(n-2)! = 2^{k-1}k!(n-2k)!$. Divide both sides by (n-2k)!(2k-2)! and you get

${n-2\choose2k-2} = \frac{2^{k-1}k!}{(2k-2)!}$. Now look at the possible values of k. We're not interested in k=1, so start with k=2. The equation then says ${n-2\choose2} = 2$, which has no solution for n. If k=3 then we get ${n-2\choose4} = 1$, which has the solution n=6. If $k\geqslant4$ then $\frac{2^{k-1}k!}{(2k-2)!}<1$, so the equation has no solutions.

The rest of Segal's proof is ingenious but not too hard to follow. What he does not do is to provide an example of an outer automorphism of S_6. I found a very nice geometrical construction of that here. (I also came across this account of a recent series of lectures that J-P Serre has been giving at Harvard, on the numbers 2 to 8 seen from a (very!) advanced viewpoint. The lecture on 6 includes some comments on $\mathrm{Aut}(S_6)$.)

3. Opalg:

I know that every permuation can be made into a product of disjoint cycles. And I know that every permutation can be made into a product of transpositions. But it seems in the proof, that a permuation can be made into a product of DISJOINT transpositions? Why is this the case.

Thanks again.

4. Originally Posted by mpetnuch
Opalg:

I know that every permuation can be made into a product of disjoint cycles. And I know that every permutation can be made into a product of transpositions. But it seems in the proof, that a permuation can be made into a product of DISJOINT transpositions? Why is this the case.
This is because the permutation in question is the image of a transposition (under an automorphism of S_n). An automorphism must take a transposition to an element of order 2. The only elements of order 2 in S_n are those whose disjoint cycle decomposition consists of (disjoint) transpositions.

5. Originally Posted by Opalg
The rest of Segal's proof is ingenious but not too hard to follow. What he does not do is to provide an example of an outer automorphism of S_6. I found a very nice geometrical construction of that here. (I also came across this account of a recent series of lectures that J-P Serre has been giving at Harvard, on the numbers 2 to 8 seen from a (very!) advanced viewpoint. The lecture on 6 includes some comments on $\mathrm{Aut}(S_6)$.)
Nice subject

I need to understand the symbol that Segal used to complete his proof . Could you write it in another manner ?

6. Originally Posted by leonardo1000
I need to understand the symbol that Segal used to complete his proof . Could you write it in another manner ?
Do you mean the notation $t = \begin{pmatrix}1&2&\ldots&r&\ldots&n\\ a_2&b_2&\ldots&b_r&\ldots&b_n\end{pmatrix}$ ? That just denotes the permutation that takes 1 to a_2, and r to b_r for 2≤r≤n.

7. I got it finally, but I read the pages you gave it here to understand how to get the outer automorphism , but I couldn't understand it how? since there are a lots of things I didn't get it in the class , could u clarify it here please ?