This is in symmetric groups $S_n$. I don't know how to approach this problem. I guess I have to use the definition of $$\mathrm{sgn}(\sigma)=\prod_{1 \leq i < j \leq n} \frac{\sigma(j)-\sigma(i)}{j-i}$$

I know I have to use this on the permutation $(i \ j)$ which represents a transposition of the $i$th and $j$th elements of the permutation, but I don't know how to approach this.

  • $\begingroup$ Do an example for, say, $n=5$. $\endgroup$ – arctic tern Apr 9 '17 at 19:42
  • $\begingroup$ I know it will give me $-1$ but how can I do it for $n = n$? $\endgroup$ – The Bosco Apr 9 '17 at 19:45
  • $\begingroup$ Count the number of inversions. $\endgroup$ – Bernard Apr 9 '17 at 19:51
  • $\begingroup$ I know it is only 1 inversion, but I have to use that formula because if I use the definition that an inversion changes the parity I will not be proving anything. $\endgroup$ – The Bosco Apr 9 '17 at 19:59
  • $\begingroup$ You misunderstand me the number of inversions is the number of pairs $(i,j),\enspace i<j\; such that $\sigma(i)>\sigma(j)$. It's what you can read by examining each factor of your formula. $\endgroup$ – Bernard Apr 9 '17 at 21:20

Suppose the transposition $\tau$ permutes $i$ and $j=i+k$ ($k\ge 1$). So the list of elements between $1$ and $n$: $$1,\dots, i-1, i,i+1, \dots , i+k-1,j=i+k,i+k+1,\dots,n$$ becomes $$1,\dots, i-1, j=i+k ,i+1, \dots , i+k-1,i,i+k+1,\dots,n.$$ Therefore $k$ pairs with $j$: $(i,j), (i+1,j),\dots ,(i+k-1,j)$ undergo an inversion.

Also the $k-1$ pairs with $i$: $(i,i+1),\dots, (i, i+k-1)$ undergo an inversion (the pair $(i,j)$ has already been counted in the first series).

In all we therefore have $2k-1$ inversions — an odd number.


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