I'm having a bit trouble proving this statment
Given a permutation $ \sigma \in A_n $ such that its cycle decomposition contains only odd and different length cycles, prove that every member in its centralizer is even.
In other words..
If $ \sigma \in A_n $ : $ \sigma = c_1c_2...cs$ where $c_1,c_2,...,c_s$ are disjoint permutation of odd length such that $length(c_i) \neq length(c_j)$ if $ i \neq j $, then every member of C($\sigma$) is even.
Any idea?
Couple of preliminary observations:
First note that $\rho\in C_{S_n}(\sigma)$ if, and only if $\rho\sigma\rho^{-1}=\sigma$. Next, if $\sigma=(a_1,\ldots,a_k)$, then $\rho\sigma\rho^{-1}=(\rho(a_1),\ldots,\rho(a_k))$. Now, if $\sigma=\sigma_1\sigma_2\cdots\sigma_k$ is a product of disjoint cycles of length $\ell_1\geq\cdots\geq \ell_k$, then $$\rho\sigma\rho^{-1}=\rho\sigma_1\rho^{-1}\rho\sigma_2\rho^{-1}\cdots\rho\sigma_k\rho^{-1}$$ is again a product of disjoint cycles of length $\ell_1\geq\cdots\geq\ell_k$. In particular, $\rho\sigma\rho^{-1}=\sigma$ if, and only if, $\rho$ either cyclically permutes the entries of each $\sigma_i$, or $\rho$ permutes cycles of the same length (i.e. $\rho\sigma_i\rho^{-1}=\sigma_j$ with $\ell_i=\ell_j$).
Now, under the assumption that $\sigma=\sigma_1\cdots\sigma_k\in A_n$ with $\ell_1>\cdots>\ell_k$ and all $\ell_i$ odd, it follows that $\rho\sigma\rho^{-1}=\sigma$ if, and only if, $\rho\sigma_i\rho^{-1}=\sigma_i$ for all $i$, so $\rho$ cyclically permutes the entries in each $\sigma_i$. It follows that $\rho\in\langle \sigma_1,\ldots,\sigma_k\rangle$, so $C_{S_n}(\sigma)=\langle \sigma_1,\ldots,\sigma_k\rangle$. But, since $\ell_i$ is odd, $\sigma_i\in A_n$ for each $i$. Hence, $C_{S_n}(\sigma)\subset A_n$.