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.