# Prove that the number of even and odd permutations in $S_n$ is equal using maps

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

where:

• $$S_n$$ 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.

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

First, assume $$n\geq2$$, because the result is false for $$n=0$$ and $$n=1$$.
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.
• 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 $Ker(\alpha)$ Commented Feb 14, 2022 at 2:08