Elements living in the conjugacy class and in the centralizer of an $m$-cycle in $A_m$

Question

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

Question: How can I describe the elements in the set $\{ j \mid x^j \in C\}$ in terms of $m$?

For instance, if $C^\prime$ is the conjugacy class of $x$ in $S_m$, the symmetric group in $m$ letters, then $$ \{ j \mid x^j \in C\} = \{ j \mid (j,m)=1 \}, $$ where $(j,m)$ is the greatest common divisor of $j$ and $m$. But in $A_m$, $C^\prime$ splits in two conjugacy classes of $A_m$ of the same size: $C$ and the conjugacy class of $(1 2)\times(1 2)$ in $A_m$.

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 $A_m$.