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

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

Let and be two conjugate even permutations in , then we can easily compute a permutation such that . Let be another permutation ( may equal to ) such that , then , which means . Then , which means that any satisfy if and only if .

We observe that if contains of both odd and even permutations, then there exists an element of such that is even, which implies and are conjugate in . And since is in , if all elements of have the same parity, then consists of only even permutations, hence and have the same parity. So in this case and are cocnjugate if and only if is even.

First we claim that:

**Theorem 1.** * If the number of distinct integers in is less than , then and are conjugate in . *

*Proof:* Since and are conjugate in , then there exists an permutation such that . By hypothesis, we have at least two distinct integers not in , say and . then has different parity with and . Hence proves our theorem.

**Lemma 2.** * If the number of distinct integers in is greater than or equal to , then all permutations in are even if and only if consists of cycles of distinct odd length. *

*Proof:* Let consists of cycles of distinct odd length, say , where are distinct odd integers. If , namely , then . This means the effect of acting on is “pushing forward” each integer steps in the cycle, namely , where , and all subscripts are taken modulo .

When only considering , we may assume without loss of generality that consists of integers in . Then consists of cycles of the same odd length (e.g. when , ), which means is even. The effect of acting on other cycles follows in the same fashion. So consists of multiple such permutations and no other cycles of length greater than can be added to this permutation since there is at most one integer unused. Hence is even.

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

If is a cycle of even length in , then is an odd permutation such that .

If and are two cycles of the same odd length in , then is the desired odd permutation such that .

Hence we’ve proved:

**Theorem 3.** * Let and be two conjugate even permutations in and any permutation such that . Then and aren’t conjugate in if and only if the number of distinct integers in is greater than or equal to , is odd, and if consists of cycles of distinct odd length. *

For more readable pdf version: ConjugatePermutationsInA_n