I am trying to find representatives for all conjugacy classes of elements of order 15 in $A_{11}$.
It is not hard to see that $(12345)(678)$ and $(12345)(678)(9 10 11)$ are the representatives for the conjugacy classes of order $15$ in $S_{11}$.
In the second case, the same element is a representative for $A_{11}$ as this conjugacy classes does not split in $A_{11}$ because we have two cycles of the same length. However, in the first case, this conjugacy class splits in $A_{11}$, and I am having a hard time finding representatives of each of these conjugacy classes. Any thoughts?
Best Answer
Let $\sigma\in S_n$ be an even permutation. The size of its conjugcay class in $S_n$ is, by the orbit-stabilizer theorem, equal to $$|[\sigma]_{S_n}|=|S_n|/|C_{S_n}(\sigma)|.\qquad(*)$$ Similarly, in $A_n$ we get $$ |[\sigma]_{A_n}|=|A_n|/|C_{A_n}(\sigma)|.\qquad(**) $$ Here $|A_n|=|S_n|/2$ and $C_{A_n}(\sigma)=C_{S_n}(\sigma)\cap A_n$. In other words, $C_{A_n}(\sigma)$ consists of the even permutations in $C_{S_n}(\sigma)$. It follows that either $C_{A_n}(\sigma)=C_{S_n}(\sigma)$, or $C_{A_n}(\sigma)$ is an index two subgroup of $C_{S_n}(\sigma)$. The conjugacy class splits in two in the former case (see equations $(*)$ and $(**)$). We also see that the latter case (=no splitting) occurs, when there are odd permutations in $C_{S_n}(\sigma)$.
The 2-cycle $(9;10)$ is in the centralizer $C_{S_{11}}((12345)(678))$. This means that the conjugacy class of $\sigma=(12345)(678)$ does not split in $A_{11}$.