Simplicity of $A_n$ I have seen two proofs of the simplicity of $A_n,~ n \geq 5$ (Dummit & Foote, Hungerford). But, neither of them are such that they 'stick' to the head (at least my head). In a sense, I still do not have the feeling that I know why they are simple and why should it be 5 and not any other number (perhaps this is only because 3-cycles become conjugate in $A_n$ after $n$ becomes greater than 4).

What is the most illuminating proof of the simplicity of $A_n,~ n \geq 5$ that you know?


The one I like most goes roughly as follows (my reference is in French [Daniel Perrin, Cours d'algèbre], but maybe it's the one in Jacobson's Basic Algebra I) : 


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*you prove that A(5) is simple by considering the cardinal of the conjugacy classes and seeing that nothing can be a nontrivial normal subgroup (because no nontrivial union of conjugacy classes including {id} has a cardinal dividing 60). Actually, you don't have to know precisely the conjugacy classes in A(5).

*Then, you consider a normal subgroup N in A(n), n > 5 which is strictly larger than {id} and you prove (*) that it contains an element fixing at least n-5 points. The fact that A(5) is simple then gives that N contains every even permutation fixing the same n - 5 points. In particular, it contains a 3-cycle, and therefore contains all of A(n).
To prove (*), you consider a commutator [x,y], where x is nontrivial in your normal subgroup and y is a 3-cycle: by the very definition, it is the product of the 3-cycle and the conjugate of its inverse. So it's the product of two 3-cycles and has at the very least n-6 fixed points. But it's easy to see that you can chose the 3-cycle so that the commutator has n-5 fixed points (it is enough that the two 3-cycles have overlapping supports).
I like this proof because it keeps the "magical computation" part to a minimum, that simply amounts to the fact that you have automatically knowledge about a commutator if you have knowledge about one of his factors.
