Elements living in the conjugacy class and in the centralizer of an m-cycle in Am 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.
 A: 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 $A_m$. 
A: 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.
