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Here's the permutation:

$\pi\sigma= \left( \begin{array}{ccc} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9\\ 5 & 6 & 2 & 9 & 1 & 7 & 8 & 3 & 4 \ \end{array} \right)$

I re-wrote this a product of disjoint cylces $$(1 \;\; 5)(2\;\;6\;\;7\;\;8\;\;3)(4\;\;9)$$ and found the $\mathbb{ord}(\pi\sigma)=10$ (I don't know if this helps). The issue I'm having is finding what the sign the permutation is. Here's what I've done

Work

Considering that we have $3$ orbits, I think we can deduce that the length of our permutation is $6$ (if this computation is correct --- $l(\pi\sigma)=9-|\mathbb{orb}(\pi\sigma)|=9-3=6$). Now, there is a theorem that states that the sign of a permutation, say $\mathbb{sgn}(\chi)$ where $\chi$ is our permutation, is $(-1)^{l(\chi)-1}$. Then for our problem, we have $$\mathbb{sgn}(\pi\sigma)=(-1)^{l(\pi\sigma)-1}=(-1)^{6-1}=(-1)^5=-1$$ $$\implies \mathbb{sgn}(\pi\sigma)=-1$$ and so $\pi\sigma$ is an odd permutation.

My question: was this correct? And if correct, could I have deduced this in a different way without the use of orbits?

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  • $\begingroup$ Perhaps you could give a reference to your definition of length: my definition of length (in Bruhat order) implies $\text{sgn}(\sigma) = (-1)^{l(\sigma)}$ $\endgroup$ – zcn Aug 2 '14 at 18:25
  • $\begingroup$ Number, Groups, and Codes by J.F. Humphreys. Theorem 4.2.10 on page 167 $\endgroup$ – Savage Henry Aug 4 '14 at 18:57
  • $\begingroup$ Thank you for sourcing - it seems that our definitions of length are the same $\endgroup$ – zcn Aug 5 '14 at 0:10
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$\DeclareMathOperator{\sgn}{\text{sgn}}$Sign is in fact a homomorphism $\sgn : S_n \to \{\pm 1\} \cong \mathbb{Z}/2\mathbb{Z}$, with kernel $A_n$. Thus if $\sigma = \tau_1 \ldots \tau_r$, then $\sgn(\sigma) = \sgn(\tau_1)\ldots\sgn(\tau_r)$. It then suffices to determine the sign of an $k$-cycle, which is just $(-1)^{k-1}$.

In your case, $\sgn((1 \;\; 5)(2\;\;6\;\;7\;\;8\;\;3)(4\;\;9)) = (-1)^{2-1}(-1)^{5-1}(-1)^{2-1} = 1$, not $-1$.

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  • $\begingroup$ Thank you. I re-read the provided theorems and everything you stated makes complete sense to me now. $\endgroup$ – Savage Henry Aug 2 '14 at 19:57

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