0
$\begingroup$

I am trying to prove a theorem for my Algebra Homework and I am stuck in a specific problem, I would be much grateful if someone can help me.

If $\sigma \in A_n$ and the cycle type of $\sigma$ consists of cycles with distinct odd lenght, how can I prove (without using Group Actions) that the set of permutations that commutes with $\sigma$ is the group generated by the cycles in the cycle decomposition of $\sigma?$ Or, at least that the centralizer $C_{S_n}(\sigma)$ only admits even permutations.

The cycle decomposition in this problem includes the 1-cycles.

Thank you.

$\endgroup$
  • $\begingroup$ Do you require $\sigma$ had length at least $n-1$? Otherwise you could have say $\sigma=(1,2,3)\in A_5$ and $(1,2,3)(4,5)=(4,5)(1,2,3)$ $\endgroup$ – Robert Chamberlain Oct 15 '16 at 12:01
  • $\begingroup$ @TheRob, in this case $\sigma=(123)$ does not have cycle decomposition with all cycles with distinct odd length $\endgroup$ – bttmbrcelo Oct 15 '16 at 14:14
  • $\begingroup$ @user26857 the cycle $(123)=(123)(4)(5)$ it contains 2 cycles with length 1, it is not distinct length. Here we are considering the cycle decomposition that shows all letters (because we included 1 cycles)... so, there is no cycles which is disjoint from the all ones in the decomposition. $\endgroup$ – bttmbrcelo Oct 15 '16 at 15:27
  • $\begingroup$ On one side you consider "the cycle decomposition that shows all letters (because we included 1 cycles)", but on the other side say "the cycle (123)=(123)(4)(5) contains 2 cycles with length 1, it is not distinct length." How to make these agree? $\endgroup$ – user26857 Oct 15 '16 at 19:25
  • $\begingroup$ Does not is? I dont understand why $\endgroup$ – bttmbrcelo Oct 15 '16 at 19:44

Your Answer

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

Browse other questions tagged or ask your own question.