= 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 are
$$
1,\quad15,\quad20,\quad12,\quad12.
$$
A 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 number
$$
1+\text{a nonempty proper subset sum}
$$
divides $60$. Hence the only possible normal-subgroup orders are $1$ and $60$. This proves the <Simplicity of the alternating group A5>.
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