Proving the number of even and odd permutations of a subgroup $H

Question

Let $H<S_{n}$ and suppose $H$ is not contained in $A_{n}$. Write $H$ as $$H=E\cup O$$ where $E$ and $O$ represent the sets of even and odd elements, respectively. Let $E=\{{\alpha_1,...,\alpha_n}\}$ and $O=\{{\beta_1,...,\beta_m}\}$. Since $O$ is non-empty, it contains at least one element, say $\beta_1$. The same goes for $E$ because $e\in E$. Our goal is to prove $m=n$. The products $\beta_1\alpha_i$, $i=1,...n$, are all distinct elements in $O$, and it follows that $O$ has at least $n$ elements, i.e. $m\geq n$. In the case of strict inequality, $m>n$, add at least one extra odd element ($\beta_{n+1}$) to the list $$\beta_1\alpha_1,...,\beta_1\alpha_n,\beta_{n+1}$$

This is where I find myself stuck. Any help appreciated. Can this route work?

Answer 1

Notice that $\beta_1^{-1}\beta_{n+1}$ is a product of two odd permutations in $H$, thus it is even; so $\beta_1^{-1}\beta_{n+1}\in E$, which implies $\beta_1^{-1}\beta_{n+1}=\alpha_k$ for some $k\leq n$, and thus $\beta_{n+1}=\beta_1\alpha_k$.

So there is actually no extra odd permutation, which shows $m=n$.

Answer 2

Just to provide a proof of the general statement in the title:

Take a subgroup $H < S_n$. If $H \cap A_n \neq \varnothing$, then it has at least one odd permutation $\sigma$. Consider the function (not a homomorphism) $T_\sigma\colon H \to H$ such that $T_\sigma(\tau) = \sigma\tau$. Then note that

1. this is an bijective function. Indeed if $\sigma\tau = \sigma\mu$ then multiplying on the left by $\sigma^{-1}$ gives you $\tau = \sigma$, so $T_\sigma$ is injective. Since $H$ is finite, it's surjective too, and hence a bijection.

2. $T_\sigma$ sends even permutations to odd permutations, and sends odd permutations to even permutations.

So $T_\sigma$ swaps the subset of odd permutations in $H$ with the subset of even permutations in $H$, and does so bijectively, preserving their size. Therefore the size of these two subsets, every permutations and odd permutations, must be the same. So half of the element of $H$ must be even and the other half odd, provided $H \not< A_n$.

Answer 3

I think there’s a much simpler proof than the other answers. Since $H \nleq A_n, S_n=HA_n$. Therefore, $\Bbb Z/2 \Bbb Z \cong S_n/A_n =HA_n/A_n\cong H/(H \cap A_n)$, so $\vert H \setminus A_n \vert = \vert H \cap A_n \vert$. This proves the claim.