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.

For more readable pdf version: ConjugatePermutationsInA_n

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