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?


1 Answer 1


They do, they just don't give any symmetry groups we didn't already get from a Platonic Solid.

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

In general, you'd just need to convince yourself that whatever operation we perform on a Platonic Solid to yield an Archimedean Solid -- truncation, expansion, and "snubbification" -- preserves symmetries exactly; none missing, nothing new.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .