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Hopefully an easy question: the icosahedral group of order 60 (orientation preserving symmetries of a regular icosahedron) is isomorphic to the alternating group on 5 points. In terms of the icosahedron, what are the 5 "points"?

It would be ideal if the "points" were actually points. The wikipedia article mentions some compounds of inscribed solids, and I think I'd need a physical demonstration to see that. However, according to this question and this question, we should be able to give just a few points, so that the stabilizer of those points has order 12.

However, I am not sure what the points would be. Maybe they are the vertices of an associated tetrahedron. It would be nice if it was easy to describe those vertices, as I certainly don't see any tetrahedra myself, but I can imagine a collection of 4 vertices easily enough.

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Jack Schmidt
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  • in coxeter's "regular polytopes" there is some discussion of the various "compounds" that demonstrate $A_5$ symmetry – yoyo Jan 27 '12 at 18:53
  • @yoyo: i've got a copy here. do you know what page number? – Jack Schmidt Jan 27 '12 at 18:58
  • Maybe it helps to think of the icosahedron as the dual to the dodecahedron. It has the same symmetry, which is obvious 5-fold. See here. – draks ... Jan 27 '12 at 19:01
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    @draks: that is partially what bothers me. A5 has way more than 5-fold symmetry of the dihedral sort. So if I think of the 5 vertices of one of the pentagons as the five "points", then I should be able to leave two of them alone and cycle the other 3, since in A5 there is the permutation (1)(2)(3,4,5). Obviously one can't do that sort of thing to a pentagon though. – Jack Schmidt Jan 27 '12 at 19:03
  • In other words, I wanted the 5 points to be the vertices of a pentagon, but that doesn't work. – Jack Schmidt Jan 27 '12 at 19:04
  • @JackSchmidt section 3.5 pg 46 on the google books version, which is missing pages of course... – yoyo Jan 27 '12 at 19:07
  • The five inscribed tetrahedra the convex hull of whose vertices form the dodecahedron are shown here http://pcjohnson.net/311PROJ/Icosahedron.html – Matthew Towers Jan 27 '12 at 19:25
  • Hehe, mt_'s link is frightening. Someone understands the perm rep, but not me. I think I've found yoyo's reference, but it isn't pictured clearly (in print or in my head). Presumably once I label a physical dodecahedron with Coxeter's labels, everything, including mt_'s link, will make sense. Hopefully I'll have time to draw a simpler picture, since I think the problem is just too many tetrahedra drawn at once. – Jack Schmidt Jan 27 '12 at 19:56
  • What's so nice about this situation is that the rotation group is acting naturally (and fixed-point-freely) on a set of six things (pairs of opposite faces of the dodecahedron). The rotation group is A5, and you are seeing a `strange copy' of A5 in S6, which is of course related to the exceptional automorphism. – Matthew Towers Jan 27 '12 at 22:22

1 Answers1

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Planar icosahedron graph

The icosahedral group $I$ permutes the 5 colors in this coloring of the icosahedron. A rotation of the icosahedron about a vertex gives a 5 cycle permutation of the colors. A rotation of the icosahedron about a triangle face is a 3 cycle permutation of colors. A rotation about an edge gives (ab)(cd) signature.

As @draks says I think the action is more obvious that this coloring is preserved if you think of inscribed tetrahedra in the dodecahedron, but this is the icosahedron equivalent. The corners of those tetrahedra partition the vertices of the dodecahedron, corresponding to this partition of faces in the icosahedron.

Proof $I$ acts on the coloring as $A_5$:

Up to reflection this coloring is unique in having the properties that: each color has 4 faces, no pair of faces sharing an edge or vertex are the same color, and every vertex is incident to all 5 colors. Clearly those properties are preserved by $I$. I don't have a good uniqueness argument other than after you assign the colors around a vertex, you have a single choice you can make before all other colors are forced by the non-adjacency conditions. Therefore $I$ must preserve this coloring.

Once you convince yourself that this coloring is preserved, then the action of the icosahedral group on the set of colors must be $A_5$, since $A_5$ is simple. Any kernel of the action must be trivial.

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