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I saw in a proof that for each $n > 1$, the symmetric group $S_n$ satisfies $$\sum_{g\in S_n} \varepsilon(g) =0,$$ where $\varepsilon$ is the signature. Is that true?

I checked it is true for $S_2$ and $S_3$ but false for $S_4$ (I found this sum equals $-6$ by counting the conjugacy classes of $S_4$).

Thanks for your help.

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    $\begingroup$ Hint: $\varepsilon:S_n\to\{+1,-1\}$ is a homomorphism. $\endgroup$ – Thomas Andrews Apr 15 at 15:04
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    $\begingroup$ Alternative hint: The sum is the determinant of the $n \times n$-matrix $\begin{pmatrix} 1 & 1 & \cdots & 1 \\ 1 & 1 & \cdots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \cdots & 1 \end{pmatrix}$. $\endgroup$ – darij grinberg Apr 15 at 17:52
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It's true. Let $x$ be any transposition, then $$\begin{align} \sum\limits_{g \in S_n} \varepsilon(g) &= \sum\limits_{g \in S_n} \varepsilon(xg) \\ &= \sum\limits_{g \in S_n} -\varepsilon(g) \\ &= -\sum\limits_{g \in S_n} \varepsilon(g). \end{align}$$

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    $\begingroup$ This is very elegant! $\endgroup$ – Shaun Apr 15 at 17:10
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If signature means sign, then the sum is zero because exactly half the elements of $S_n$ have sign $+1$ and half the elements have sign $-1$.

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