# Proving this permutation identity

How can I prove this identity to be correct

$$\sum_\sigma\sum_{k=1}^N\left[c_k(x_{\sigma(k)} - \mu_k)\right] \equiv \sum_\sigma\sum_{k=1}^N\left[c_{\sigma(k)}(x_k - \mu_{\sigma(k)})\right],$$

where σ ranges over elements of the permutation group on N objects and $c_k$ is some constant and $μ_j$ characterises the position of the $j^\text{th}$-particle such that

$$μ_j=\left(j− \frac{N+1}{2}\right)d.$$

I understand that the above follows from the identity that

$$\sum_{k=1}^Nc_{\sigma(k)}\mu_{\sigma(k)} \equiv \sum_{k=1}^Nc_k\mu_k,$$

but I was wondering whether there is a way to prove the identity by use of properties of the permutation group alone.

For any group $G$ the map $\sigma\mapsto\sigma^{-1}$ is a bijection from $G$ to itself. In particular, for any quantity $n_\sigma$, $\sum_{\sigma\in G}n_\sigma=\sum_{\sigma\in G}n_{\sigma^{-1}}$. In your case, $n_{\sigma}=\sum_{k=1}^nc_k(x_{\sigma(k)}-\mu_k)$ gives $\sum\limits_\sigma\sum_{k=1}^nc_k(x_{\sigma(k)}-\mu_k)=\sum\limits_\sigma\sum_{k=1}^nc_k(x_{\sigma^{-1}(k)}-\mu_k)$.
Notice also that each permutation $\sigma$ is by definition a bijection from $\{1,\ldots,k\}$ to itself. In particular, for any quantity $n_k$, $\sum_{k=1}^nn_k=\sum_{k=1}^nn_{\sigma(k)}$. In your case $n_k=c_k(x_{\sigma^{-1}(k)}-\mu_k)$ gives $\sum_{k=1}^nc_k(x_{\sigma^{-1}(k)}-\mu_k)=\sum_{k=1}^nc_{\sigma(k)}(x_k-\mu_{\sigma(k)})$.
Putting these together we obtain $$\sum\limits_\sigma\sum_{k=1}^nc_k(x_{\sigma(k)}-\mu_k)=\sum\limits_\sigma\sum_{k=1}^nc_k(x_{\sigma^{-1}(k)}-\mu_k)=\sum\limits_\sigma\sum_{k=1}^nc_{\sigma(k)}(x_k-\mu_{\sigma(k)})$$