Prove that this sum equals 0 We have $n\geq3$ and $S_{n}$ the set of permutations of length n. For every $\sigma\in S_{n}$, we have $\epsilon(\sigma)$ the sign of permutation $\sigma$ and $Fix(\sigma)$ the set of fixed points of $\sigma$. Prove that $$\sum_{\sigma\in{S_{n}}} \epsilon(\sigma)\mid{Fix(\sigma)}\mid=0$$.
 A: Reverse the order of summation: the sum can be written as $$\sum_{\sigma \in S_n} \sum_{i\in [n], \sigma(i) = i} \varepsilon(\sigma) = 
\sum_{\sigma \in S_n, i \in [n], \sigma(i) = i} \varepsilon(\sigma) = \sum_{i\in [n]} \sum_{\sigma\in S_n, \sigma(i) = i} \varepsilon(\sigma)$$
The latter inner sum is 0, since the permutations in $S_n$ that fix a given $i$ can be sign-preservingly matched with the permutations in $S_{n-1}$.
A: With the notation from Wikpedia on random
permutations
we find that the sign $\sigma(\pi)$ of a permutation $\pi$ is given by
$$\sigma(\pi) = \prod_{c\in\pi} (-1)^{|c|-1}$$
where the product is over all cycles $c$ of $\pi.$
Using combinatorial classes as in Analytic Combinatorics by Flajolet
and Sedgewick, we have the following class $\mathcal{P}$  of permutations
with fixed points marked:
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\mathcal{P} = \textsc{SET}(
-\mathcal{Z} + \mathcal{V} \mathcal{Z}
+ \textsc{CYC}_{=1}(\mathcal{Z})
+ \mathcal{U} \times \textsc{CYC}_{=2}(\mathcal{Z}) 
+ \mathcal{U}^2 \times \textsc{CYC}_{=3}(\mathcal{Z})  
\\ + \mathcal{U}^3 \times \textsc{CYC}_{=4}(\mathcal{Z})  
+ \cdots).$$
Translating to generating functions we obtain
$$G(z,u,v) =
\exp\left(-z+vz+\sum_{k\ge 1} u^{k-1} \frac{z^k}{k}\right)
\\ = \exp\left(-z+vz+\frac{1}{u} \log\frac{1}{1-uz}\right)
\\ = \exp(-z+vz) \left(\frac{1}{1-uz}\right)^{1/u}.$$
We thus have
$$n! [z^n] G(z, -1, v) = n! [z^n] \exp(-z+vz) (1+z)
= \sum_{\pi\in S_n} \sigma(\pi) v^{\nu(\pi)}$$
where $\nu(\pi)$ is the number of fixed points.
It follows that the desired statistic is given by
$$n! [z^n] \left.\frac{\partial}{\partial v}
\exp(-z+vz) (1+z)\right|_{v=1}
\\ = n! [z^n] \left.\exp(-z+vz) z (1+z)\right|_{v=1}
= n! [z^n] (z+z^2).$$
This is one when $n=1$ and two when $n=2$ and zero otherwise.
A: Your sum is the sum of the values of the character of the representation of $S_n$ obtained as a tensor product of the tautological permutation representation $V$ and the sign representation $\mathrm{sgn}$. Since $V\otimes\mathrm{sgn}$ does not have trivial summands, that sum is zero.
