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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

**Physics Forums | Science Articles, Homework Help, Discussion**