Sum of signature of elements of $S_n$ is $0$

Question

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.

Answer 1

It's true. Let $x$ be any transposition, then $$\begin{align} \sum\limits_{g \in S_n} \varepsilon(g) &= \sum\limits_{g \in S_n} \varepsilon(xg) \\ &= \sum\limits_{g \in S_n} -\varepsilon(g) \\ &= -\sum\limits_{g \in S_n} \varepsilon(g). \end{align}$$

Answer 2

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 $-1$.