Question

Let m>1 be an odd natural number, x a m-cycle in Am, the alternating group in m letters, C the conjugacy class of x in Am.

Questiom: How can I describe the elements in the set { j | x^j in C} in terms of m?

For instance, if C' is the conjugacy class of x in Sm, the symmetric group in m letters, then { j | x^j in C} = { j | (j,m)=1 }, where (j,m) = Greatest common divisor of j and m. But in Am, C' splits in two conjugacy classes of Am of the same size: C and the conjugacy class of (1 2)x(1 2) in Am.

Thank you in advance. Fernando.

Answer 1

The set is the quadratic residues when $m$ is prime, but usually not when $m$ is composite. For example, $(0,1,2,3,4,5,6,7,8)$ is conjugate to $(0,2,4,6,8,1,3,5,7)$ in $A_9$ even though $2$ is not a square mod $9$, so there is no additional condition beyond $(j,9)=1$.

For $m$ odd, the sign of the permutation on $\mathbb Z/ m\mathbb Z$ of multiplication by $j$ is the Jacobi symbol $\big(\frac jm\big)$. (This perspective on the Jacobi symbol is natural from one of Gauss's proofs of quadratic reciprocity, but it's also theorem 1 here. Also see Zolotarev's lemma.) Since there are two conjugacy classes of $m$-cycles in $A_m$, $\big(\frac jm\big)=+1$ iff $x$ is conjugate to $x^j$ in $A_m$.

Answer 2

Thank you, Douglas.

With the notation giving above and that giving in the paper of Marek Szyjewski (that you refered me), the following statements are equivalent: 1) x^j in C, 2) sgn(lambda_ j )=1, 3) J(j,m)=1, J the Jacobi symbol.

1) <=> 2) is easy (I have chequed). 2) <=> 3) is Theor. 1 of the paper of Marek Szyjewski. This is an unplubished article yet. I had no time to chequed all of it; I have only chequed Case 1, but I guess that Case 2 and 3 are correct.(?)

I am interested in the case m=3 p, with p>3 prime. I need to prove that there exist j, with j mod 3 =2, such that x^j in C. This amounts to prove that there exists j, 0< j < m, such that: -) ( j,m)=1, -) j mod 3 =2 (i.e. J( j,3)= -1), -) J( j,p)= -1, because J( j,m)=J( j,3) J( j,p).

Do you have any clue for that?

Thank you in advance.