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A group theoretic puzzle in solid geometry

  1. Feb 15, 2005 #1

    mathwonk

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    imagine a regular dodecahedron, and its group of rotations. this group is "simple" of order 60, hence isomorphic to the group A5, of even permutations of 5 elements.

    i.e. each face is preserved by 5 rotations, yet can be translated to 12 different faces, hence there are 60 motions in all.

    can one see how the rotations of a dodecahedron do indeed permute faithfully some collection of 5 objects associated to that solid?

    Note that these rotations certainly permute the 12 faces, the 20 vertices, and the 30 edges, hence also the 6 axes joining centers of opposite faces, and the 15 axes joining opposite edges, and the 10 axes joining opposite pairs of vertices, but what collection of 5 objects is permuted?

    It helps to look at a picture.
     
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  3. Feb 15, 2005 #2

    AKG

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    I'm not exactly sure what you're asking. What do you mean when you say that "these rotations certainly permute the 12 faces?" Do you mean that given any two faces, there is some rotation (of the 60) that rotates one of the faces to the other? Couldn't we just arbitrarily choose any 5 objects related to the dodecahedron to answer your question? I guess these 5 objects should be all of the objects of that type (i.e. we wouldn't just choose any 5 corners because there are 20 corners in all).

    EDIT: Originally, I guessed that we would take those 10 axes joining the 20 opposite pairs of corners, and take the axes joining midpoints of opposite axes, but all those axes share the same midpoint (the center of the dodecahedron), so that doesn't work, but perhaps there is something in the idea of pairing those 10 axes to get 5 pairs of axes, attaching some "meaningful" object to each pair, and then arguing that those 5 objects are preserved under rotations. This is assuming I understand what you're asking.
     
    Last edited: Feb 15, 2005
  4. Feb 15, 2005 #3

    AKG

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    For everyone:

    http://mathworld.wolfram.com/Dodecahedron.html

    You can even rotate the images around to get a better idea of what's going on, by clicking on them and dragging.
     
  5. Feb 15, 2005 #4

    mathwonk

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    you have to find a set of 5 things such that every rotation takes one of them to another one, or the same one. i.e. such that you get a group homomorphism from the group of rotations nito the group of epormutations of the 5 objects.

    for example every rotation peserving the dodecahedron sends faces to faces, so gives a permutation of the 12 faces. Indeed you could consider the pairs of opposite faces but this only represents your group in the permutation group on 6 objects.

    you cannot choose a subset of the vertices because the group of all rotations will not map that subset into itself.
     
  6. Feb 18, 2005 #5
    Dodecahedron Problem

    Since the shape has 30 edges, and this number is divisable by 5, should I be aiming to construct a set of object symetrical to one another that consists of 6 edges? or is that going to send me down the garden path?
    And would twelve of the rotations map one of the constructed 5 objects onto itself?

    Just a stab in the dark, but I'm inclined to select 3 sets of opposite edges, where the sets of three edges map to each other with 4 basic rotations, that is two rotations around one of the six axis and two rotations around another one of the axis, and if an edge has vertices (v1,v2) it moves to like :
    (v1,v2)-> (v2,v3)-> (v3,v4) -> (v4,v5)->(v5,v6), where vi does not equal vj, and (v3,v4) is an opposite edge.
    Ill post later when I find out I'm probbaly wrong!!!
     
    Last edited: Feb 18, 2005
  7. Feb 18, 2005 #6

    mathwonk

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    that is exactly correct! In fact if you look hard, you can see how to embed a cube for each choice of three opposite pairs of edges. such that the 6 faces of the cube correspond to the 6 edges.
     
  8. Feb 19, 2005 #7
    I still can't see how you can imbed a cube. How does an edge correspond to the face of a cube? An interesting problem though to get me thinking about groups again.
     
  9. Feb 19, 2005 #8

    mathwonk

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    i think the face of the cube lies right under the edge of the dodecahedron, and the edge of the dodecahedron bisects the face of the cube.
     
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