I seen the equation $\sum_\sigma(\operatorname{sgn} \sigma)^2=k!$, where $\sigma\in S_k$ ($S_k$ is the symmetric group). How can I prove this equation?
2026-04-28 08:27:58.1777364878
Summation of sgn function
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$\text{sgn}\sigma \in \{1,-1\}$, then $(\text{sgn}\sigma)^2 = 1$ for all $\sigma \in S_k$. Hence: $$ \sum_{\sigma}(\text{sgn}\sigma)^2 = \sum_{\sigma} 1 = k! $$