Two permutations have different parity when each element in one is a function of an element in the other Consider two permutations $\rho$ and $\sigma$ of the set $\{1,2,...,n\}$ where $n$ is even. Suppose $\rho = ( \rho_1, \rho_2,..., \rho_n)$ and $\sigma_{n-i}=n-\rho_i + 1$ for $i=1,2,...,n-1$ and $\sigma_i = n - \rho_i + 1$ for $i=n$.
Notice that $\rho_i = \sigma_i$ if and only if $\rho_i  + \rho_{n-i} = n+1$. Also notice that $\rho_n$ and $\rho_{n/2}$ must always be different from $\sigma_n$ and $\sigma_{n/2}$, respectively.
I am fairly certain that these permutations have different parity, but how can I show this? Any help would be appreciated.
 A: Let $\alpha$ be the permutation given by $\alpha(i) = n-i$ for $i < n$ and $\alpha(n) = n$. And let $\beta$ be the permutation given by $\beta(i) = n - i + 1$. Then $\sigma = \beta\rho\alpha$. That is $\sigma(i) = \beta(\rho(\alpha(i)))$. Since $\text{sgn}(xy) = \text{sgn}(x)\text{sgn}(y)$ we get $\text{sgn}(\sigma) = \text{sgn}(\beta)\text{sgn}(\rho)\text{sgn}(\alpha)$. So $\sigma$ and $\rho$ will have opposite parity if $\alpha$ and $\beta$ have different parity. And this is the case: $\alpha$ can be written as a product of the $\frac{n}{2}-1$ transpositions $(1\text{ }(n-1)), (2\text{ }(n-2)), \ldots, (\frac{n}{2}-1 \text{ }\frac{n}{2}+1)$. And similarly $\beta$ can be written as the product of the $\frac{n}{2}$ transpositions $(1\text{ } n), (2\text{ }(n-1)), \ldots, (\frac{n}{2} \text{ }\frac{n}{2}+1)$. That means that $\text{sgn}(\alpha) = (-1)^{\frac{n}{2}-1}$ and $\text{sgn}(\beta) = (-1)^{\frac{n}{2}}$. Hence $\text{sgn}(\alpha)\text{sgn}(\beta) = -1$. And so $\text{sgn}(\sigma) = \text{sgn}(\alpha)\text{sgn}(\rho)\text{sgn}(\beta) = -\text{sgn}(\rho),$ showing that $\sigma$ and $\rho$ have opposite parity.
