Fool's quest

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.

For more readable pdf version: ConjugatePermutationsInA_n

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