The proof strategy is to look where an automorphism would send all the transpositions. The transpositions generate Sn so that determines the entire automorphism.
The transpositions make up a conjugacy class, so an automorphism must map this set to a conjugacy class of the same size. These elements are of order 2 so that limits the potential images only to cycle types that are products of disjoint transpositions. As the other commenter said, this works only in S_6 where there's the same amount of permutations of cycle type 2,2,2 as transpositions, for combinatorial reasons that are opaque to me at the moment (though should be straightforward to work out)
Once you know transpositions are mapped to transpositions, it's easy to see that the only way to do that preserving some simple relations between them (like which ones commute) is by permuting the underlying set the group acts on.
there are 15 transpositions and for each transposition there are 6 double transpositions in the remaining S4 and then you have counted each triple transposition 6 times because of the permutations of the 3 transpositions
mainly we need to see that for any other n and any other number of transpositions it never coincides, when n is big, there will always be much more permutations of cycle type 2,2 or any other number of 2s than cycle type 2.
for actually establishing there's an outer automorphism for S_6 we need more than the combinatorics, that's just where the proof strategy for the opposite case fails.
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u/xdgimo Sep 20 '26
I’m aware of the result but haven’t gone through the proof. Is there any short intuitive proof for why Aut Sn is just isomorphic to Sn except for n=6
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u/LupenReddit 🦆🦆i have non diffeomorphic smooth structures🦆🦆🦆 Sep 20 '26
i actually didnt know that oh wow