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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.

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

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Hint:

$$sgn(\sigma)\cdot sgn(\tau)=sgn(\sigma \tau)$$

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  • $\begingroup$ So, you just say that sigma(j)^-1 - sigma(i)^-1 = sigma(j) -sigma(i). And why is that? $\endgroup$ – Kath Mar 27 at 11:02
  • $\begingroup$ It is much harder to see in the first formula , the second hint is a lot better (and the first leads to the second hint anyway). I will edit the first hint out. $\endgroup$ – idle mathematician Mar 27 at 11:05

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