> [!NOTE] Theorem ($A_{n}$ is a subgroup of $S_{n}$) > Let $n\in\mathbb{N}^{+}.$ The [[nth Alternating Group|nth alternating group]] is a [[Normal Subgroup|normal subgroup]] of the [[Symmetric Groups of Finite Degree|nth symmetric group]]: that is for all $\sigma\in S_{n},$ we have $\sigma A_{n}=A_{n} \sigma.$ ###### Proof Suppose $\sigma$ is [[Parity of a Permutation of n letters|even]]. Then $\sigma A_{n}=A_{n}=A_{n}\sigma$: clearly $\sigma A_{n} \subset A_{n}.$ note that $\sigma^{-1}$ is even by [[Parity of the inverse of a permutation|parity of inverse of permutation]]. Let $\tau\in A_{n}$ then $\tau=\sigma \sigma^{-1} \tau$ and $\sigma^{-1}\tau\in A_{n}$ since it is even. Thus $\tau\in \sigma A_{n}$ and so $\sigma A_{n}=A_{n}.$ Similarly, $A_{n}\sigma=A_{n}.$ Now suppose $\sigma$ is odd. Then $\sigma A_{n}=S_{n}\setminus A_{n}=A_{n}\sigma$ by a similar argument. $\square$ Note this is effectively the same proof used in [[Subgroup of Index 2 is Normal]]. ###### Proof We know that $[S_{n}:A_{n}]=2$ (this can be seen a consequence of [[Sign of Permutation of n Letters is a Homomorphism|first isomorphism theorem]]). It follows from [[Subgroup of Index 2 is Normal]] zxxthat $A_{n}$ is normal.