proof that S_n is non-abelian for n > 2

In this post, I prove that the symmetric group is non-abelian (snob for: 'not commutative') for .

I'm pretty sure almost every mathematician hates abstract algebra. I could never understand why that is. It's way easier than calculus. I'll admit it's probably not as useful, although it does have uses in physics, chemistry, cryptography and maybe a few other fields. In any case, I feel abstract algebra is one of the best ways to learn how to write proofs, which is why I like it so much.

Anyway, on to the proof.

Consider the transpositions in for , which I'll call , and , which I'll call . Because and , is non-abelian for . QED

That was very short. It's a result you can be pleased with. When I first wrote my proof for this problem, it was much longer because I defined and . These functions were two I picked out at random which I didn't think would commute. Incidentally, the proof using them is very cute. Because my own and are defined for any value of , you'll notice that and . As such, is obviously abelian for . It's easy and fun to see how becomes non-abelian in this case. However, that proof requires two tedious proof-by-induction lemmas for and and so I have opted instead for the shorter version someone recommended to me, even if it isn't as cute.

There is an important lesson to take away from this story. To paraphrase Charles Hoare: "There are two ways of constructing proofs: one way is to make them so simple that there are obviously no deficiencies; the other way is to make them so complicated that there are no obvious deficiencies." As with everything else in nature, the better, plainer choice is usually hard in practice. Think ahead.

(rendering courtesy of servalx02)