All the proofs I have seen for this theorem uses the same argument: First prove that the identity permutation has even parity. Then let a be (one of) the first elements to appear in a transposition representation of a permutation in S(adsbygoogle = window.adsbygoogle || []).push({}); _{n}. Then identify all the other transpositions in the representation that also feature a and play with the order and such using two defined "moves" that "move" a in a specified direction. Since only a finite number of transpositions feature a, we can eventually cancel two, leaving us with a representation of 2 less transpositions. The theorem follows by an induction argument.

I think this is a somewhat ugly and clumsy proof, probably my least favorite in my algebra book. Does anyone know if there exists another, shorter proof, which does not invoke this "moving" argument?

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# Proving that the parity of a permutation is well defined.

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