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Prove that sign$(\sigma \tau)$ = sign$(\sigma)$sign$(\tau)$ for any permutations $\sigma, \tau \in S_n$.

I think the two thing's I'm trying to show are:

  • If sign$(\sigma)$ = sign$(\tau) = \pm 1 \implies$ sign$(\sigma \tau)$ = $1$
  • Wlog, if sign$(\sigma) = 1$, sign$(\tau) = - 1 \implies$ sign$(\sigma \tau)$ = $-1$

but I'm not sure how to start this. Can someone give me a hint?

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sign is defined as 1 or -1 depending on whether the number of transpositions you can write a permutation in is even or odd.

if $\sigma$ is written as $k$ transpositions and $\tau$ is written as $t$ then $\sigma \tau$ is $k+t$ transpositions.

This proves that $\text{sgn}(\sigma)\text{sgn}(\tau)=\text{sgn}(\sigma\tau)$ because even + even = even, even + odd = odd, odd + odd = even

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    $\begingroup$ Note that the decomposition of permutation isn't unique you have to prove that the parity of number of transpositions in a decomposition of permutation is unique $\endgroup$ – user63181 Apr 21 '13 at 12:53
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    $\begingroup$ Agree with @Sami. This answer simply reduces the question to another. $\endgroup$ – Jyrki Lahtonen Apr 21 '13 at 16:34
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We define the signature of the permutation $\sigma$: $$\epsilon(\sigma)=(-1)^N$$ where $N$ is the number of inversion.

Let $\sigma$ and $\tau$ two permutations with $P$ and $Q$ inversions respectively. The inversions of the composition $\tau \sigma$ are:

  • the inversions $\{i,j\}$ of $\sigma$ s.t $\{\sigma(i),\sigma(j)\}$ is not an inversion for $\tau$

  • The pairs $\{i,j\}$ which is not inversions of $\sigma$ s.t $\{\sigma(i),\sigma(j)\}$ is an inversion of $\tau$.

If we add the numbers of inversions of $\sigma$ and $\tau$ we have twice the number $R$ of inversions $\{i,j\}$ of $\sigma$ s.t. $\{\sigma(i),\sigma(j)\}$ is also an inversion of $\tau$. So the total number of inversions of $\tau\sigma$ is $N=P+Q-2R$. The signature of $\tau\sigma$ is $$\epsilon(\tau\sigma)=(-1)^{P+Q-2R}=(-1)^P(-1)^Q=\epsilon(\tau)\epsilon(\sigma)$$

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  • Let $\pi\in S_n$ and set $\Delta=\prod_{1\leq i\leq j\leq n}(x_i-x_j)$.
  • Set $\Delta^{\pi}=\prod_{1\leq i\leq j\leq n}(x_{(i)\pi}-x_{(j)\pi})$.
  • Show that $\Delta^{\pi}=\text{Sgn}(\pi)\Delta$.
  • Show that for $\pi,\phi\in S_n$, $\text{Sgn}(\pi\phi)=\text{Sgn}(\pi)\text{Sgn}(\phi)$ by showing that $(\Delta^{\pi})^{\phi}=\Delta^{{\pi}{\phi}}$.

  • Note that $\Delta^{{\pi}{\phi}}=\text{Sgn}(\pi\phi)\Delta$ and $(\Delta^{\pi})^{\phi}=\text{Sgn}(\pi)\text{Sgn}(\phi)\Delta$.

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  • $\begingroup$ Yes! $\;+1^+\;\large \ddot\smile$ $\endgroup$ – amWhy Apr 22 '13 at 0:36
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Given that you are asking this question, you may have not seen the following. But typically, $sgn$ is defined to be a homomorphism from any symmetric group to the multiplicative group $\{1,-1\}$, with the alternating group as kernel (i.e. the even permutations map to 1). So your formula is immediate by the definition of a homomorphism. But again, my guess is that you have not been given this characterization of $sgn$.

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Alternatively you can show that $\operatorname{sgn}$ factors through the group of $n \times n$ permutation matrices using the determinant map. Since $\det$ is a homomorphism we get the result.

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