Prove that exactly half of the permutations in $S_n$ have a negative signature

Question

If $S_n=\lbrace \alpha : \mathbb{Z}_n \rightarrow \mathbb{Z}_n \mid \alpha \text{ is one-to-one and onto}\rbrace$ is the symmetric group on $n$ letters, that is, $S_n $ consists of all permutations of $n$ objects. Define $Sgn:S_n \rightarrow \lbrace 1, −1\rbrace$ by $Sgn(\alpha)=(−1)^{\text{number of inversions in }\alpha}$. Show that $\sum\limits_{\alpha \in S_n}Sgn(\alpha)=0$, that is, prove that exactly half of the permutations in $S_n$ have a negative $Sgn$.

What I tried: I know that when n = 4,

$Sgn(\alpha_1)=1$

$Sgn(\alpha_2)= −1$

$Sgn(\alpha_3)=−1$

$Sgn(\alpha_4)=1$

$Sgn(\alpha_5)=1$

$Sgn(\alpha_6)=−1$

And clearly half of those $Sgn$'s are negative, but I'm not sure how to expand it to a general case.

Answer 1

Show that $sgn$ is a homomorphism from $S_n$ to $\{\pm 1\}$. Show that this is surjective. Why does this imply your result?

p.s: Let me know if you want more details. It's probably better for you to work it out yourself.

Answer 2

For $n > 1$, fix a transposition $\sigma \in S_n$ (actually, any odd permutation will do); can you show that the map $L_{\sigma} : S_n \to S_n$ defined by $L_{\sigma} : \tau \mapsto \sigma\tau$

1. is bijective, and
2. maps $A_n$ onto its complement?

Here, $A_n$ denotes the set of even permutations in $S_n$, i.e., those of positive sign.