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.

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.