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?

  • $\begingroup$ 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? $\endgroup$ – WE Tutorial School Oct 1 at 13:34
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    $\begingroup$ Note that in particular the permutation $\sigma$ has at most one cycle of length $1$ - ie fixes at most one of the $n$ points. $\endgroup$ – Mark Bennet Oct 1 at 13:34
  • $\begingroup$ Are you talking about the centralizer in $S_n\supset A_n$? $\endgroup$ – WE Tutorial School Oct 1 at 13:35
  • $\begingroup$ Yeah, i'm talking about the centralizer in $S_n$ $\endgroup$ – Saber98 Oct 1 at 13:42
  • $\begingroup$ 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? $\endgroup$ – Saber98 Oct 1 at 16:32

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, $C_{S_n}(\sigma)\subset A_n$.

  • $\begingroup$ 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? $\endgroup$ – Saber98 Oct 2 at 10:28
  • $\begingroup$ No, you can have any $\alpha_i$. Take $\sigma=(12345)(678)$ and $\rho=(12345)^3(678)^2$. Then, $\rho$ commutes with $\sigma$. $\endgroup$ – David Hill Oct 2 at 14:48

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