> [!NOTE] Theorem ($A_{n}$ is a subgroup of $S_{n}$) > Let $n\in\mathbb{N}^{+}.$ The [[nth Alternating Group|nth alternating group]] is a [[Subgroup|subgroup]] of the [[Symmetric Groups of Finite Degree|nth symmetric group]]. **Proof**: We check that it passes the [[Two-Step Subgroup Test|two-step subgroup test]]. Firstly, the identity is a product of zero $2$-cycles and so it is even. Next, take $\alpha,\beta\in A_{n},$ then we can write each as an even number of $2$-cycles. Therefore the product $\alpha\beta$ can be written as an even number of $2$-cycles hence $\alpha\beta\in A_{n}.$ Finally we can write $\alpha=a_{1}a_{2}\dots a_{m}$where the $a_{i}$ are $2$-cycles and $m$ is even. Using [[Inverse of the Product of Group Elements|socks-shoes property]], $\alpha^{-1} = (a_{1}a_{2}\dots a_{m})^{-1} = a_{m}^{-1} a_{m-1}^{-1}\dots a_{1}^{-1} = a_{m}a_{m-1}\dots a_{1} $Hence $\alpha^{-1}$ is even and $\alpha^{-1}\in A_{n}.$