Centralizer (in $S_n$) of $\sigma \in A_n$ such that its cycle decomposition contains only odd and different length cycles

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?

• I don't understand the question. Every member of $C(\sigma)$ would be an element of $A_n$, right? Therefore, it must be even, no? – WE Tutorial School Oct 1 at 13:34
• Note that in particular the permutation $\sigma$ has at most one cycle of length $1$ - ie fixes at most one of the $n$ points. – Mark Bennet Oct 1 at 13:34
• Are you talking about the centralizer in $S_n\supset A_n$? – WE Tutorial School Oct 1 at 13:35
• Yeah, i'm talking about the centralizer in $S_n$ – Saber98 Oct 1 at 13:42
• A useful idea. Suppose that $\sigma$ has no fixed point; i calculated |C($\sigma$)| = n. It's always true that $\sigma$ commutes with a power of yours. So, how to prove that the order of $\sigma$ is n? – Saber98 Oct 1 at 16:32

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, $$C_{S_n}(\sigma)\subset A_n$$.
• Very nice proof, but i have a doubt. If $\rho$ cyclically permutes the entries in each $σ_i$ then $\rho = σ_1^{\alpha_1}σ_2^{\alpha_2}...σ_k^{\alpha_k}$ such that for every i=1,2...,k $\alpha_i =0$ or $\alpha_i =1$. Right? – Saber98 Oct 2 at 10:28
• No, you can have any $\alpha_i$. Take $\sigma=(12345)(678)$ and $\rho=(12345)^3(678)^2$. Then, $\rho$ commutes with $\sigma$. – David Hill Oct 2 at 14:48