On Conjugacy Classes of Alternating Group $A_n$ In Dummit & Foote, page 131

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

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*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.


*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 ?
 A: 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 $A_4$.
A: 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 ...
