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Prove that in $S_n$ there are an equal number of even and odd permutations.

$S_n$ is a group of all possible permutations on a set of $n$ elements. For this problem we can assume $n>1$.

I'm pretty sure I need to prove this by contradiction and show that if the numbers weren't even then $S_n$ wouldn't be a group, but I'm not sure how to go about that.

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The map $\sigma \mapsto (12)\sigma$ is a bijection and it maps even permutations to odd ones and vice-versa.

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The map $\sigma \mapsto \hbox{sign}(\sigma)$ is a surjective homomorphism $S_n \to C_2$ whose kernel is the set of even permutations.

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