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