# Can we prove that the identity permutation is even this way?

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 $even$.

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