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So, as per one of my previous questions, I'm working through some problems in Armstrong's book Groups and Symmetry. The first two thirds of the question I've managed to grind through (after much spatially-related struggle) but here I seem to be--quite literally--drawing a blank. Can anyone guide me toward a geometric object whose (full) symmetry group corresponds to $A_4\times\mathbb Z_2$? I know that $A_4$ corresponds to the proper rotations of a tetrahedron, but its full symmetry group $S_4$ isn't isomorphic to the product group in question here. Any help you can offer me is much appreciated!

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up vote 12 down vote accepted

(Pictures from Wikipedia.)

An object in three dimensions with symmetry group isomorphic to $A_4\times\mathbb{Z}_2$ is said to have pyritohedral symmetry. This is the symmetry of a volleyball:

enter image description here

Equivalently, it is the symmetry of a cube whose faces have been subdivided in a certain way:

enter image description here

Polyhedra with this symmetry group include the pyritohedron, which is a certain type of irregular dodecahedron:

enter image description here

Pyritohedral symmetry is so-named because it is the point symmetry group for one of the crystal structures of the mineral pyrite.

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For people like me who are more accustomed to Schoenflies notation: this corresponds to point group $T_h$. This gives a buckyball derivative as an example of a molecule belonging to this point group. – J. M. Dec 1 '11 at 4:12
Additionally, as I've just realized, if you consider a tetrahedron with its inscribed dual tetrahedron and subsequently expand the interior dual until its edges intersect the exterior one, that solid possesses the desired symmetry as well. Pictured here...… – AsinglePANCAKE Dec 1 '11 at 18:25
@AsinglePANCAKE: No that is not an example, as its full symmetry group is that of a cube (its convex hull), which is $S_4\times\mathbb{Z}/2$ – Marc van Leeuwen Dec 2 '11 at 10:48
@MarcvanLeeuwen:Oops, you're right! You don't happen to be related to my good friend Latisha Chaniqua Yolanda van Leeuwen?? – AsinglePANCAKE Dec 5 '11 at 2:34
@AsinglePANCAKE: No, not as far as I know. But "van Leeuwen" is a very common name in some places. – Marc van Leeuwen Dec 5 '11 at 10:46

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