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As is well-known, all finite subgroups of $SO(3)$, except for cyclic and dihedral groups, are isomorphic to one of:

  • $A_4$
  • $S_4$
  • $A_5$

The classical proof of this fact uses the geometry of regular polyhedra, their symmetries and rotations.

Are there any algebraic proofs?

(I mean any proofs that take $SO(3)$ as a group of matrices or operators, but not as a group of rotations of three-dimensional space.)

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    $\begingroup$ There is no subgroup of $SO(3)$ isomorphic to $S_5$. There are subgroups isomorphic to $A_5$, however. $\endgroup$ – Servaes Sep 9 '13 at 0:06
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You can find a proof of this using the Riemann-Hurwiz formula in McKean & Moll's book Elliptic Curves: Function Theory, Geometry, Arithmetic. (Some might argue that this is a geometric proof, but the Riemann-Hurwiz formula really belongs to algebraic geometry. So it is an algebro-geometric proof!)

In essence, the proof looks like this: there is an embedding of $SO(3)$ into $\Gamma = PSL_2(\mathbf C)$, given by considering rotations of the sphere as automorphisms of the Riemann sphere, which are described by Möbius transformations. The problem then boils down to classifying finite subgroups of $\Gamma$. If $G$ is such a subgroup, then $\mathbf P^1/\Gamma$ can be made into a compact Riemann surface isomorphic to $\mathbf P^1$. The class formula for $G$ is then encoded into the ramification data of the quotient map $\mathbf P^1 \to \mathbf P^1/G$. Using the Riemann-Hurwiz formula, one obtains a finite set of possibilities for the class equation. Then one realizes each possibility explicitly.

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