## Conjugacy classes in A_n

We know that two permutations in ${S_n}$ are conjugate if and only if their decompositions consist of the same cycle type. And a conjugacy class in ${S_n}$ of even permutations is either equal to a single conjugacy class in ${A_n}$, or splits into two conjugacy classes in ${A_n}$. So two even permutations of the same cycle type may not be conjugate in ${A_n}$. In this article we introduce a simple and practicable criterion for determining whether two even permutations are conjugate in ${A_n}$.

For convience, we assume that all permutations here have already been decomposed into disjoint cycles.

Let ${a}$ and ${b}$ be two conjugate even permutations in ${S_n}$, then we can easily compute a permutation ${\tau \in S_n}$ such that ${\tau a\tau^{-1} = b}$. Let ${\sigma}$ be another permutation (${\sigma}$ may equal to ${\tau}$) such that ${\sigma a\sigma^{-1} = b}$, then ${\tau^{-1} \sigma a\sigma^{-1} \tau = a}$, which means ${\tau^{-1} \sigma \in Stab_{S_n}(a)}$. Then ${\sigma \in \tau Stab_{S_n}(a)}$, which means that any ${\sigma}$ satisfy ${\sigma a\sigma^{-1} = b}$ if and only if ${\sigma \in \tau Stab_{S_n}(a)}$.

We observe that if ${Stab_{S_n}(a)}$ contains of both odd and even permutations, then there exists an element ${\pi}$ of ${Stab_{S_n}(a)}$ such that ${\sigma = \tau \pi}$ is even, which implies ${a}$ and ${b}$ are conjugate in ${A_n}$. And since ${(1)}$ is in ${Stab_{S_n}(a)}$, if all elements of ${Stab_{S_n}(a)}$ have the same parity, then ${Stab_{S_n}(a)}$ consists of only even permutations, hence ${\tau}$ and ${\sigma}$ have the same parity. So in this case ${a}$ and ${b}$ are cocnjugate if and only if ${\tau}$ is even.

First we claim that:
Theorem 1. If the number of distinct integers in ${a}$ is less than ${n-1}$, then ${a}$ and ${b}$ are conjugate in ${A_n}$.

Proof: Since ${a}$ and ${b}$ are conjugate in ${S_n}$, then there exists an permutation ${\tau \in S_n}$ such that ${\tau a\tau^{-1} = a}$. By hypothesis, we have at least two distinct integers not in ${a}$, say ${p}$ and ${q}$. then ${\sigma = \tau(p, q)}$ has different parity with ${\tau}$ and ${\sigma \in Stab_{S_n}(a)}$. Hence proves our theorem. $\Box$

Lemma 2. If the number of distinct integers in ${a}$ is greater than or equal to ${n-1}$, then all permutations in ${Stab_{S_n}(a)}$ are even if and only if ${a}$ consists of cycles of distinct odd length.

Proof: Let ${a}$ consists of cycles of distinct odd length, say ${a = (s_0, s_1, \ldots, s_{s-1})(t_0, t_1, \dots, t_{t-1})\cdots(q_0, q_1, \ldots, q_{q-1})}$, where ${s, t, \ldots, q}$ are distinct odd integers. If ${\sigma \in Stab_{S_n}(a)}$, namely ${\sigma a\sigma^{-1} = a}$, then ${\sigma (s_0, s_1, \ldots, s_{s-1}) \sigma^{-1} = (s_0, s_1, \ldots, s_{s-1})}$. This means the effect of ${\sigma}$ acting on ${(s_0, s_1, \ldots, s_{s-1})}$ is “pushing forward” each integer ${x}$ steps in the cycle, namely ${s_i\mapsto s_{i+x}}$, where ${0\le x \le s-1}$, and all subscripts are taken modulo ${s}$.

When only considering ${(s_0, s_1, \ldots, s_{s-1})}$, we may assume without loss of generality that ${\sigma}$ consists of integers in ${(s_0, s_1, \ldots, s_{s-1})}$. Then ${\sigma}$ consists of cycles of the same odd length (e.g. when ${x\mid s}$, ${\sigma = (s_0, s_x, \ldots, s_{(k-1)x})(s_1, s_{1+x}, \ldots, s_{1+(k-1)x})\cdots(s_{x-1}, s_{2x-1},\ldots, s_{kx-1})}$), which means ${\sigma}$ is even. The effect of ${\sigma}$ acting on other cycles follows in the same fashion. So ${\sigma}$ consists of multiple such permutations and no other cycles of length greater than ${1}$ can be added to this permutation since there is at most one integer unused. Hence ${\sigma}$ is even.

For the converse, it is equivalent to say that if ${a}$ doesn’t consist of cycles of distinct odd length, then there exists at least one odd permutation in ${Stab_{S_n}(a)}$. There are two cases.

${\bullet}$ If ${(p_0, p_1, \ldots, p_{p-1})}$ is a cycle of even length in ${a}$, then ${\sigma = (p_0, p_1, \ldots, p_{p-1})}$ is an odd permutation such that ${\sigma a\sigma^{-1} = a}$.

${\bullet}$ If ${(e_0, e_1, \ldots, e_{m-1})}$ and ${(f_0, f_1, \ldots, f_{m-1})}$ are two cycles of the same odd length in ${a}$, then ${\sigma = (e_0, f_0)(e_1, f_1)\cdots(e_{m-1}, f_{m-1})}$ is the desired odd permutation such that ${\sigma a\sigma^{-1} = a}$. $\Box$

Hence we’ve proved:
Theorem 3. Let ${a}$ and ${b}$ be two conjugate even permutations in ${S_n}$ and ${\tau}$ any permutation such that ${\tau a\tau^{-1} = b}$. Then ${a}$ and ${b}$ aren’t conjugate in ${A_n}$ if and only if the number of distinct integers in ${a}$ is greater than or equal to ${n-1}$, ${\tau}$ is odd, and if ${a}$ consists of cycles of distinct odd length.