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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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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
    
@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

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