 Solution

The conjugacy classes in $A_5$ consist of the identity, the $15$ double transpositions, the $20$ three-cycles, and two classes of $12$ five-cycles. Thus their sizes areA normal subgroup is a union of conjugacy classes containing the identity, and its order must divide $60$ by Lagrange's theorem. Checking sums of a subset of $15,20,12,12$ shows that no numberdivides $60$. Hence the only possible normal-subgroup orders are $1$ and $60$. This proves the Simplicity of the alternating group A5.