# Why don't Archimedean solids give finite subgroups of $SO(3,\Bbb R)$?

I know that the Platonic solids correspond to finite subgroups of $SO(3,\Bbb R)$. For example, the tetrahedron corresponds to a subgroup isomorphic to $A_4$. The cube and octahedron to one isomorphic to $S_4$, and the dodeca and icosahedron to one isomorphic to $A_5$.

But is there any easy way to see why the Archimedean solids (two or more regular polygons meeting in identical vertices, meaning for any two vertices there is an isometry taking one to the other) don't give rise to finite subgroups?

For example, symmetric truncation preserves the symmetries -- so the truncated cube, truncated octahedron, and cuboctahedron all have rotational symmetry group $S_4$.