Return to Answer

This is an old question, but I wanted to write a proof that $A_5$ is simple via symmetries of the icosahedron, using as little group theory as possible. I don't think that it can lead to a proof of the unsolvavility of the quintic without the usual group theory (permutations, normal subgroups, quotients, solvable groups), and, of course, field theory.

Regarding the question Q2, there is an interesting (though not quite elementary) application: The fact that the icosahedron group $G$ is simple is the first step in the construction of Poincaré's homology sphere, a 3-manifold $M$ that has the same homology of the sphere $S^3$, but is not homeomorphic to $S^3$, because its fundamental group is nontrivial. It is said that this construction allowed Poincaré to reject a preliminar wrong version of his famous conjecture, which was originally stated in terms of homology (a 3-manifold with the homology of the sphere is homeomorphic to a sphere, as happens in dimension 2).

The homology group $H_1(M)$ is obtained by abelianization of the fundamental group $\Pi_1(M)=2G$, so we must prove that the commutator subgroup $[2G,2G]$ equals $2G$. Observe that $\pi([2G,2G])=[G,G]=G$, where the first equality holds because $\pi$ maps $2G$ onto $G$, and the second one holds because $G$ is nonabelian and simple, so the nontrivial normal subgroup $[G,G]$ must be equal to the whole group. So $\pi$ restricts to a surjective morphism $[2G,2G]\to G$ and there are two possiblities:

Since the angle-of-rotation distance in $SO(\mathbb R^3)$ is just the double of the geodesic distance in $S^3$ (which is the angle of vectors), to verify what I claim we just have to prove that $2G$ is $\frac{2\pi}{12}$-dense in $S^3$. This, if true, can be checked numerically with error of less than 1 degree. But it would be more interesting to have a picture of the Voronoi-cell decomposition of $S^3$ with respect to the discrete set $2G$. I wonder if this leads to the other usual presentation of the Poincaré sphere (as a quotient of the dodecahedron, identifying oposite faces by a $\frac{2\pi}{10}$ right-handed twist).

This is an old question, but I wanted to write a proof that $A_5$ is simple via symmetries of the icosahedron, using as little group theory as possible. I don't think that it can lead to a proof of the unsolvability of the quintic without the usual group theory (permutations, normal subgroups, quotients, solvable groups), and, of course, field theory.

Regarding the question Q2, there is an interesting (though not quite elementary) application: The fact that the icosahedron group $G$ is simple is a step in the construction of Poincaré's homology sphere, a 3-manifold $M$ that has the same homology of the sphere $S^3$, but is not homeomorphic to $S^3$, because its fundamental group is nontrivial. A construction of this kind allowed Poincaré to reject a preliminary wrong version of his famous conjecture, which was originally stated in terms of homology (a 3-manifold with the homology of the sphere is homeomorphic to a sphere, as happens in dimension 2).

The homology group $H_1(M)$ is obtained by abelianization of the fundamental group $\Pi_1(M)=2G$, so we must prove that the commutator subgroup $[2G,2G]$ equals $2G$. Observe that $\pi([2G,2G])=[G,G]=G$, where the first equality holds because $\pi$ maps $2G$ onto $G$, and the second one holds because $G$ is nonabelian and simple, so the nontrivial normal subgroup $[G,G]$ must be equal to the whole group. So $\pi$ restricts to a surjective morphism $[2G,2G]\to G$ and there are two possibilities:

Since the angle-of-rotation distance in $SO(\mathbb R^3)$ is just the double of the geodesic distance in $S^3$ (which is the angle of vectors), to verify what I claim we just have to prove that $2G$ is $\frac{2\pi}{12}$-dense in $S^3$. This, if true, can be checked numerically with error of less than 1 degree. But it would be more interesting to have a picture of the Voronoi-cell decomposition of $S^3$ with respect to the discrete set $2G$. I wonder if this leads to the other usual presentation of the Poincaré sphere (as a quotient of the dodecahedron, identifying opposite faces by a $\frac{2\pi}{10}$ right-handed twist).

Let $G$ be its group of rotations. It has order $2*30=60$, because if $e$ is an oriented edge, then for any oriented edge $e'$, there is exactly one rotation that carries $e$ to $e'$. Any nontrivial rotation $g$ is of one of three types, according to whether its axis contains a vertex, an interior point of an edge (which must be the midpoint) or an interior point of a face (which must be the center). Its order is 5, 2 or 3, respectively.

Let $G$ be its group of rotations. It has order $2*30=60$, because if $e$ is an oriented edge, then for any oriented edge $e'$, there is exactly one rotation that carries $e$ to $e'$. Any nontrivial rotation $g$ is of one of three types, according to whether its axis contains a vertex, an interior point of an edge (which must be the midpoint) or an interior point of a face (which must be the center). Its order is 5, 2 or 3, respectively.