Prove that for $n\ge3$, every nontrivial normal subgroup of $S_n$ either contains a $3$-cycle or a product of two disjoint $2$-cycles.
This question cannot directly use the advanced fact that $A_n$ is simple group for $n\ge5$ and classification of normal subgroup of $S_n$ ($n\ge 3$) like here.
My attempts:
A normal subgroup $N$ of $G$ is union of conjugacy classes, here. The conjugacy class of $S_n$ should have same cycle type, here.
Conjugacy class itself is not closed in group multiplication. I want to prove that a normal subgroup generated by any conjugacy class of $S_n$ will contain a $3$-cycle or a product of two disjoint $2$-cycles.
That is to construct a $3$-cycle or a product of two disjoint $2$-cycles just by multiplying group element with same cycle type. However I don't have idea how to proceed, or is there other easy way to prove original problem?
