We know that a permutation is $even$ if it can be written as a product of $even$ number of transpositions.

I read so many proofs regarding the parity of the identity permutation but found them all too long and sometimes hard. I tried this:

Let $B$ be an even permutation, then $B^{-1}$ is also even. Same if odd ($i.e.$ both a permutation and its inverse are products of the same number of transpositions).

The identity permutation $id$ = $B$$B^{-1}$, so if $B$ is a product of $r$ transpositions, then $B^{-1}$ is also a product of $r$ transpositions. Therefore the identity is a product of $2r$ transpositions and hence $even$.

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    $\begingroup$ One could alternatively prove the proposition by pointing out that $0$ is an even number. $\endgroup$ – Arthur Apr 30 '17 at 16:55
  • $\begingroup$ @CatalinZara: Yes I corrected it. Is my proof correct? $\endgroup$ – Nour Apr 30 '17 at 17:17
  • $\begingroup$ @Arthur: Mmmm, why? didn't get it. $\endgroup$ – Nour Apr 30 '17 at 17:18
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    $\begingroup$ The identity premutation may be written as the product of no transpositions. Zero is an even number, therefore the identity permutation is even. $\endgroup$ – Arthur Apr 30 '17 at 17:20
  • $\begingroup$ @Arthur. Oh right! But is my proof acceptable? $\endgroup$ – Nour Apr 30 '17 at 17:25

In your proof, you are assuming that a permutation cannot be both even and odd (for otherwise, you could express B as a product of an even number of transpositions, and B^(-1) as a product of an odd number of transpositions, and then your proof falls flat). To prove that a permutation cannot be BOTH even and odd, you need the fact that identity can only be expressed as a product of an even number of permutations. So you do need an alternative approach to proving this result. That's why most proofs are lengthy.


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