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 $A_n$.

Solution 1:

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:

1. 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.

2. 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.