So I'm looking to prove that the number of even permutations is equal to the number of odd permutations in $S_n$

I want to prove this using the fact that the map:

$\alpha: S_n \to C_2$ defined as $(\sigma \in S_n) \mapsto sign(\sigma)$

is a surjective homomorphism


  • $S_n$ is the Symmetric group of degree n (i.e. the set of permutations on $X_n = \{1,2,...,n\}$)
  • $C_2 = \{-1,1\}$
  • $sign(\sigma) = 1 \iff \sigma$ is an even permutation (is product of even # of transpositions)
  • $sign(\sigma) = -1 \iff \sigma$ is an odd permutation (is product of odd # of transpositions)

I understand that this means that $Ker(\alpha) = \{\sigma \in S_n | sign(\sigma) = 1\}$ (that is, that kernel of $\alpha$ is the set of even permutations) but how does this tell me that $|Ker(\alpha)| = |S_n - Ker(\alpha)|$?

This question is motivated by this Stack Q here: Prove that in $S_n$ there are an equal number of even and odd permutations.

  • $\begingroup$ $K=\ker(\alpha)$ has index $2$ and so $S_n= K \cup \sigma K$ for some $\sigma$. These are two parts of equal size. $\endgroup$
    – lhf
    Commented Feb 14, 2022 at 9:47

1 Answer 1


First, assume $n\geq2$, because the result is false for $n=0$ and $n=1$.

Then use the fact that, quite generally, in any group, any subgroup has the same cardinality as any of its cosets, a bijection being given by multiplication (on the left or on the right, depending on whether it's a left or right coset) by any fixed element of the coset.

To apply this in your specific situation, let $\pi$ be any odd permutation of $\{1,2,\dots,n\}$, for example the transposition $(1\,2)$. (This is where you use that $n\geq2$.) Then the function $\sigma\mapsto\pi\sigma$ is a bijection from the even permutations to the odd permutations.

  • 1
    $\begingroup$ Oh I think I see, so I know I have the $Ker(\alpha)$ which is the set of all even permutations. If I multiply that set (either on the left or on the right) by an odd permutation (so something simple like (1 2) ) then I have a left (or right) coset $(1 2)Ker(\alpha)$ (or $Ker(\alpha)(1 2)$) whose elements must be all the odd permutations but whose cardinality (by properties of cosets) will be equal to the cardinality of $Ker(\alpha)$ $\endgroup$ Commented Feb 14, 2022 at 2:08

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