I know that $A_5$ is simple. But how do I tell which even conjugacy classes of $S_5$ split into two conjugacy classes of alternating permutations. When we know that $A_5$ is normal and normal subgroup is a disjoint union of conjugacy classes.
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See the list of Related questions in the right margin. In addition to the answers you received here they contain a lot of useful information. I picked the duplicate target largely because it was a highly voted question, and covers all the alternating groups. – Jyrki Lahtonen Nov 05 '15 at 16:39
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This is answered in this question. Take $G=S_5$ and $H=A_5$. Then, if $\sigma\in A_5$ and $C_{S_5}(\sigma)\leq A_5$ we have $|\mathrm{Conj}_{A_5}(\sigma)|=\frac{1}{2}|\mathrm{Conj}_{S_5}(\sigma)|$, so $\mathrm{Conj}_{S_5}(\sigma)$ splits into two $A_5$-conjugacy classes. If $C_{S_5}(\sigma)\nleq A_5$, then $|\mathrm{Conj}_{S_5}(\sigma)|=|\mathrm{Conj}_{A_5}(\sigma)|$, so $\mathrm{Conj}_{S_5}(\sigma)=\mathrm{Conj}_{A_5}(\sigma)$.
Put another way, the $S_5$-conjugacy class of $\sigma\in A_5$ equals the $A_5$-conjugacy class if, and only if $\sigma$ commutes with an odd permutation. If $\sigma$ does not commute with an odd permutation, then the $S_5$-conjugacy class splits into two $A_5$-conjugacy classes of equal size.
This works for any $n>2$, not just $n=5$.
David Hill
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