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 $even$.
Best Answer
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.