Proving the number of even and odd permutations of a subgroup $H<S_{n}$ are equal, provided $H$ is not contained in $A_{n}$

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?

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 $m=n$.