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

1. Apr 17, 2012

### espen180

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 Sn. 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?

2. Apr 17, 2012

### DonAntonio

If $\sigma\in S_n$ is a permutation, the DEFINITION of the sign of $\sigma$ is $Syg(\sigma):=\prod_{1\leq i< j\leq n}\frac{i-j}{\sigma(i)-\sigma(j)}$ .

I can't see how the above can't be well defined: the above is 1 if the number of inversions in the order of each pair $(i,j)$ in the image of $\sigma$ is even,

and -1 otherwise. What is there to proof?

DonAntonio

3. Apr 17, 2012

### Office_Shredder

Staff Emeritus
Assuming that is given as the definition of sign, which it isn't always

Wikipedia discusses the equivalence of the two definitions and gives several elegant proofs in my opinion (which equivalently can be thought of as proving that the sign is well defined in the number of transpositions sense)

http://en.wikipedia.org/wiki/Parity_of_a_permutation

4. Apr 17, 2012

### DonAntonio

Well, then it may be what you wrote in the last lines: take one definition and prove that another one is equivalent to it.

This though wasn't what the OP wanted, at least from what I can understand in his question.

DonAntonio

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