In Dummit & Foote, page 131

Let $K$ be a conjugacy class and suppose that $K$ is subset of $A_n$ .

  1. Show that if $\sigma$ belongs to $S_n$ then , $\sigma$ does not commute with any odd permutation if and only if the cycle type of consists of distinct odd integers.

  2. Deduce that $K$ is a union of two -conjugacy classes in $A_n$ if and only if the cycle type of an element of $K$ consists of distinct odd integers.

[Hint: Assume first that $\sigma$ belongs to $S_n$ does not commute with any odd permutation. Observe that $\sigma$ commutes with each cycle in its own cycle decomposition, so that each cycle must have odd length. If two cycles have the same odd length , find a product of transpositions which interchanges them and commutes with $\sigma$ . Conversely, if the cycle type of $\sigma$ consists of distinct integers, prove that $\sigma$ commutes only with the subgroup generated by the cycles in its cycle decomposition.]

the exercise number 21 , i uesed the hints to prove that the cycles of cycle decomposition of $\sigma $ must be of odd length , but i don't know how to prove that the cycle type is distinct .

the text said find transposition which interchanges them and commute $\sigma$ ,

my question is , interchange what ?!!!

what does this mean ?

can anyone help ?

  • 7
    $\begingroup$ Please state the question for those who don't have Dummit&Foote at hand. $\endgroup$
    – Ludolila
    Feb 17, 2013 at 21:53
  • $\begingroup$ @TaraB $\sigma $ is an elemeny of $S_n$ in the exercise which i mentioned in my question . it's also an element in $A_n$ $\endgroup$ Feb 17, 2013 at 21:54
  • $\begingroup$ my question is , what is the question ?!!! $\endgroup$ Feb 17, 2013 at 21:54
  • 1
    $\begingroup$ @gnometorule , check it back plz :) $\endgroup$ Feb 17, 2013 at 22:02
  • $\begingroup$ @JyrkiLahtonen , i added the hall of question , plz check it back :) $\endgroup$ Feb 17, 2013 at 22:04

2 Answers 2


Hint for part 1: The product of two 3-cycles $\sigma=(123)(456)$ commutes with $\tau=(14)(25)(36)$. This is because $\tau (123)\tau^{-1}=(456)$ and $\tau(456)\tau^{-1}=(123)$. The permutation $\tau$ is odd, because it has an odd number of transpositions. Generalize to conclude that if a permutation has two cycles of an equal odd length then it commutes with an odd permutation.

Hint for part 2: Remember that the size of the conjugacy class of an element $x$ in a group $G$ is $|G|/|C_G(x)|$. Here we have both $S_n$ and $A_n$ assuming the role of $G$. But obviously $C_{A_n}(x)=C_{S_n}(x)\cap A_n$, and $C_{S_n}(x)$ consists either of only even permutations (when intersecting with $A_n$ won't change anything) or ...


Hint: sometimes it's hard to understand the general idea, and it might be a good idea to have an example in mind. Think of $A_4$. The cycle types of elements in $A_4$ are: $1111$(identity), $22,13$. The only cycle type that consists of distinct odd integers is $13$. That is, $K$ is a union of two conjugacy classes in $A_4$ iff $K$ has an element of the form $(- - -)$. Now, you can (easily) check that the elements $(123),(132),(124),(142),(234),(243),(134),(143)$ form two conjugacy classes in $A_4$.


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.