0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.