How to prove this permutation summation is zero. The function $\mu_j$ characterises the position of the $j^\text{th}$-particle, where
$$\mu_j = \left(j - \frac{N+1}{2}\right)d,$$
and $N$ is the total number of particles. I want calculate this summation:
$$\sum_\sigma \left[\sum_{j=1}^N \mu_j \mu_{\sigma(j)}\right],$$
where $\sigma$ ranges over elements of the permutation group on $N$ objects. Implementation in Mathematica finds that this evaluates to zero. How can I use the properties of the permutation group to analytically prove this. 
 A: As @DanielFisher in his comment pointed, lets change the order of summation:
$$\sum_{j=1}^n \mu_j \left[\sum_\sigma \mu_{\sigma(j)}\right]$$
For each $j$ there are exactly $(N-1)!$ permutations $\sigma$, such that $\sigma(j)=i$. Our summation can be than changed to:
$$\sum_{j=1}^N \mu_j \left[\sum_{i=1}^N \mu_i\cdot (N-1)!\right] = (N-1)!\sum_{j=1}^N \mu_j \left[\sum_{i=1}^N \mu_i \right]$$
Substitute $\mu_j = \left(j-\frac{N+1}{2}\right)$:
$$(N-1)!\sum_{j=1}^N \left(j-\frac{N+1}{2}\right) \left[\sum_{i=1}^N \left(i-\frac{N+1}{2}\right) \right]=\\
\\= (N-1)!\sum_{j=1}^N \sum_{i=1}^N \left(j-\frac{N+1}{2}\right)  \left(i-\frac{N+1}{2}\right) =\\
=(N-1)!\sum_{j=1}^N \sum_{i=1}^N ij - i\frac{N+1}{2}-j\frac{N+1}{2}+\left(\frac{N+1}{2}\right)^2=
(N-1)!\left(\sum_{j=1}^N \sum_{i=1}^N ij - 2\sum_{j=1}^N \sum_{i=1}^Ni\frac{N+1}{2}+\sum_{j=1}^N \sum_{i=1}^N\left(\frac{N+1}{2}\right)^2\right)=\\
=(N-1)!\left(\left(\frac{N(N+1)}{2}\right)^2 -2 N \frac{N+1}{2} \frac{N(N+1)}{2}+ N^2\left(\frac{N+1}{2}\right)^2 \right) = \\
=(N-1)!\left(2\left(\frac{N(N+1)}{2}\right)^2-2\left(\frac{N(N+1)}{2}\right)^2\right) = 0$$
