Suppose $n\geq 3$ and let $k$ be $n$ or $n-1$, whichever is odd. Show that the set of $k$-cycles in $A_n$ is not a conjugacy class in $A_n$. Question: Suppose $n\geq 3$ and let $k$ be $n$ or $n-1$, whichever is odd.  Show that the set of $k$-cycles in $A_n$ is not a conjugacy class in $A_n$.
My thoughts: Only cycles of the same length can be in the same conjugacy class.  So I wanted to show that all the cycles in the cycle decomposition would all have different odd lengths, but I wasn't getting anywhere.  So I wanted to try and prove the contradiction or contrapositive by playing with $|A_n:C_{A_n}(\sigma)|$, where $\sigma$ is one of the $k$ cycles, but that wasn't getting me anywhere either.  Any help is greatly appreciated!  Thank you.
 A: Fix positive integer $n \geqslant 3$. I will answer the question by completely classifying all conjugacy classes in $A_n$.
Lemma 1:
Let $B$ be a nonempty set. Fix $b \in B$. We make $S_n$ act on $B$ and $A_n$ act on $\text{Orb}_{S_n}(b)$. Then:

*

*The action of $A_n$ on $\text{Orb}_{S_n}(b)$ is transitive if and only if $\text{Stab}_{S_n}(b)$ has an odd permutation.


*If the action of $A_n$ on $\text{Orb}_{S_n}(b)$ is not transitive, then there is an odd permutation $\tau \in S_n$ such that $\text{Orb}_{S_n}(b)$ is a disjoint union of $\text{Orb}_{A_n}(b)$ and $\text{Orb}_{A_n}(\tau b)$. In this case, $ |\text{Orb}_{A_n}(b) | =  |\text{Orb}_{A_n}(\tau b)|$.
Proof:
Fix an arbitrary odd permutation $\tau \in S_n$. Since:
\begin{equation*}
        \text{Orb}_{S_n}(b) = \{\sigma b | \, \sigma \in S_n\} = \{\sigma b |, \sigma \in A_n \cup \tau^{-1} A_n\} = \{\sigma b |, \sigma \in A_n\} \cup \{\sigma \circ \tau b |, \sigma \in  A_n\} = \text{Orb}_{A_n}(b) \cup \text{Orb}_{A_n}( \tau b)
\end{equation*}
We see that we have at most $2$ orbits. The action is transitive if and only if there is a $\sigma \in A_n$ such that $\sigma b = \tau b$ if and only if $\sigma^{-1} \tau \in \text{Stab}_{S_n}(b)$, which proves 1. Assume from now on that the action is not transitive. If $x \in \text{Orb}_{A_n}(b) \cap \text{Orb}_{A_n}( \tau b)$, then there exists $\sigma_1$ and $\sigma_2 \in A_n$ such that $x = \sigma_1 b = \sigma_2 \tau b$ so that $\tau^{-1}\circ \sigma_2^{-1} \circ \sigma_1 \in \text{Stab}_{S_n}(b)$, which is a contradiction. This proves the union to be disjoint. Finally, note that the function $f: \text{Orb}_{A_n}(b) \to \text{Orb}_{A_n}(\tau b)$ defined by $f(x)=\tau x$ is a bijection.////
If $S_n$ acts on itself by conjugation, then in the above lemma $B=S_n$. For each $\sigma \in S_n$, depending on whether $\sigma$ commutes with an odd permutation, $\text{Orb}_{S_n}(\sigma)$ either equals to $\text{Orb}_{A_n}(\sigma)$ or equals to a disjoint union of $\text{Orb}_{A_n}(\sigma)$ and $\text{Orb}_{A_n}(\tau \circ \sigma)$ for an arbitrary odd permutation $\tau$.
Lemma2:
$\sigma \in S_n$ does not commute with any odd permutation if and only if the cycle type of $\sigma$ has only distinct odd integers.
Proof:
Suppose $\sigma$ does not commute with any odd permutation. Since it commutes with its own cycles, it cannot have any cycle of even length. If $\sigma$'s cycle type has two same odd integers $2r+1$, then a part of $\sigma$ is expressible as $(i_1 \cdots i_{2r+1})(j_1 \cdots j_{2r+1})$, which commutes with $(i_1 \, j_1)\cdots (i_{2r+1} \, j_{2r+1})$. If the cycle type of $\sigma$ has only distinct integers, then it commutes only with permutations in $\langle \sigma \rangle$. Therefore, $\sigma$ commutes only with even permutations if its cycle type has only distinct odd integers. ////
The above two lemmas combined with the paragraph in the middle completely characterizes the conjugacy classes in $A_n$ because you know that two permutations have the same cycle type in $S_n$ if and only if they belong to the same conjugacy class. To answer the question more directly:
If $n$ is odd, then $k=n$ and the cycle type is one single odd integer. If $n$ is even, then $k=n-1$ and the cycle type is $(1,n-1)$, which is a collection of distinct odd integers. Either case the above two lemmas apply and the result holds.
