= Solution
The face action gives \b[$|G|=60$], as calculated above. Every <normal subgroup> contains the identity and is a union of whole <conjugacy classes>. Thus its possible order is $1$ plus a subset sum of $15,20,12,12$. The distinct resulting numbers are
$$
1,13,16,21,25,28,33,36,40,45,48,60.
$$
By <Lagrange's theorem for finite groups>, a subgroup order divides $60$. Only $1$ and $60$ in this list do so. Consequently \b[the <normal subgroups> are $\{1\}$ and $G$], and the <rotational symmetry group of a regular icosahedron> is a <simple group>.
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