> [!NOTE] Lemma (Parity of the inverse of a permutation) > Let $S_{n}$ denote the [[Symmetric Groups of Finite Degree|nth symmetric group]]. Let $\rho\in S_{n}$ be a [[Permutation of Finite Degree|permutation of n letters]]. Then $\text{sgn }(\rho^{-1})=\text{sgn }(\rho)$ where $\text{sgn }(\rho)$ denotes the [[Parity of a Permutation of n letters|sign]] of $\rho.$ **Proof**: Suppose $\rho = \tau_{1}\tau_{2}\dots \tau_{n-1} \tau_{n}$ where $\tau_{i}$ are [[Cyclic Permutation of n Letters|2-cycles]] and $n\in \mathbb{N}^{+}.$ Then by [[Inverse of the Product of Group Elements|socks-shoes property]], $\rho^{-1}=\tau_{n}\tau_{n-1}\dots \tau_{2}\tau_{1}$ since a $2$-cycle is Thus the parity of $\rho^{-1}$ is the same as the parity of $n$ which is the same as the parity of $\rho.$