We prove that the alternating group $A_n$ has index 2 in the symmetric group $S_n$ i.e., $A_n$ has the same cardinality as its complement $S_n\setminus A_n$ The proof is function-theoretic. Its idea is similar to the proof in the parent topic, but the focus is less on algebraic aspect.

Let $\pi\in S_n\setminus A_n$ Define $\pi:S_n \setminus A_n\rightarrow A_n$ by $\pi(\sigma)=\pi\sigma$ where $\pi\sigma$ is the product of $\pi$ and $\sigma$

One-to-one: \begin{equation*} \pi(\sigma)=\pi(\delta) \Longrightarrow \sigma=\delta \end{equation*}since $\pi^{-1}$ exists and $\pi^{-1}\pi\sigma=\pi^{-1}\pi\delta$

Onto: Given $\alpha\in A_n$ there exists an element in $S_n\setminus A_n$ namely $\lambda=\pi^{-1}\alpha$ such that \begin{equation*} \pi(\alpha)=\lambda. \end{equation*}(The element $\lambda$ is in $S_n\setminus A_n$ because $\pi^{-1}$ is and the product of an odd permutation and an even permutation is odd.)

The function $\pi:S_n \setminus A_n\rightarrow A_n$ is, therefore, a one-to-one correspondence, so both sets $S_n \setminus A_n$ and $A_n$ have the same cardinality.