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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.

Did
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Raton
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  • This might help : http://math.stackexchange.com/questions/1626115/proving-the-number-of-even-and-odd-permutations-of-a-subgroup-hs-n-are-equa/1626142#comment3316035_1626142 (it's not exactly the same though, as it uses some properties of signature, so I'm not sure it is a duplicate) – Arnaud D. Mar 05 '16 at 19:26

2 Answers2

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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.

Arkady
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  • Why does showing that $sgn$ is surjective indicate that there are an equal number of -1 and 1 outputs? Doesn't subjectivity just show that there both values of the codomain appear? – Raton Mar 05 '16 at 19:37
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    Yes, but being a surjective homomorphism is a much stronger condition than being just a surjective function. The preimage of every element in the codomain contains the same number of elements. One way to see this is the first isomorphism theorem which states that $S_n/\text{Ker(sgn)}\cong {\pm 1}$. So, $|S_n|/|\text{Ker(sgn)}|=|{\pm 1}|. So, |S_n|/|\text{Ker(sgn)}|=2$. But $\text{Ker(sgn)}$ is just those permutations with sgn $1$! – Arkady Mar 06 '16 at 10:32
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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.

Travis Willse
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