alternating group has index 2 in the symmetric group, the

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 $\pi :S_n\setminus A_n\to A_n$ by $\pi (\sigma )=\pi \sigma$, where $\pi \sigma$ is the product of $\pi$ and $\sigma$.

One-to-one:

$$\pi (\sigma )=\pi (\delta )⟹\sigma =\delta$$

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

$$\pi (\alpha )=\lambda .$$

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