Show that transposition is an odd permutation 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.

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.
