# 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 $\pi:S_{n}\setminus A_{n}\rightarrow A_{n}$ by $\pi(\sigma)=\pi\sigma$, where $\pi\sigma$ is the product of $\pi$ and $\sigma$.

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

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 $\pi:S_{n}\setminus A_{n}\rightarrow A_{n}$ is, therefore, a one-to-one correspondence, so both sets $S_{n}\setminus A_{n}$ and $A_{n}$ have the same cardinality.

Title alternating group has index 2 in the symmetric group, the AlternatingGroupHasIndex2InTheSymmetricGroupThe 2013-03-22 16:48:49 2013-03-22 16:48:49 yesitis (13730) yesitis (13730) 8 yesitis (13730) Proof msc 20-00