# The number of inversions in a permutation is equal to the number of its inverse permutation.

I need to prove that the sign of a permutation is equal to the sign of the inverse of the permutation. I understand it is true, but how do you proof that the #inversions of $$\sigma$$= #inversions of $$\sigma^{-1}$$? Can anyone help me out? I know that the inversions are not the same, I tried it with an example.

• It is much easier to prove this via the product formula of the sign – idle mathematician Mar 27 at 10:58
• yes the $\-1^{#inversions of \sigma$} right? – Kath Mar 27 at 10:59
• what is the product, $\operatorname{sgn}(\sigma)\operatorname{sgn}(\sigma^{-1})$? – Rylee Lyman Mar 27 at 11:03
• sgn(identity) . – Kath Mar 27 at 11:05

Saying $$\sigma$$ inverts $$i$$ and $$j$$ means that $$i$$ and $$j$$ come in the opposite order to $$\sigma(i)$$ and $$\sigma(j)$$.
Therefore $$\sigma$$ inverts $$i$$ and $$j$$ if and only if $$\sigma^{-1}$$ inverts $$\sigma(i)$$ and $$\sigma(j)$$. This gives a one-to-one correspondence between the inversions of $$\sigma$$ and $$\sigma^{-1}$$.
$$sgn(\sigma)\cdot sgn(\tau)=sgn(\sigma \tau)$$