I am trying to prove the simplicity of $A_5$ by showing that every non-trivial normal subgroup $H$ contains a 3-cycle, and therefore is all of $A_5$ since the 3-cycles all belong to one conjugacy class and all the 3-cycles generate $A_5$. Now if $H$ contains a 3-cycle already then the proof is done. Otherwise $H$ contains only products of two transpositions or 5-cycles. How can I cook up some general calculation that no matter what I'll always get a 3-cycle from these types of elements?
I've asked a related question here.
If $H$ contains a 5-cycle then it contains all 5-cycles, since all subgroups of order 5 are conjugate. Show that the product of two well-chosen 5-cycles gives a 3-cycle.
If $H$ contains a permutation that is a product of two disjoint 2-cycles, then $H$ contains all such permutations (all are conjugate). Again, pick two and take their product.