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Modern Algebra: Stabilizers and Conjugacy Classes of Dodec

  1. Sep 9, 2015 #1
    My professor was proving that the Dodecahedron is isomorphic to ##A_5## and in the process utilized the stabilizer (which one can intuit ) of an edge, vertex or face to determine the conjugacy class (which is hard to intuit) of elements of the same order. This seems like a valuable skill but I couldn't follow his argument. Can anyone explain?
  2. jcsd
  3. Sep 9, 2015 #2


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    Do you have some group or group action associated to the Dodecahedron, or are you referring to its symmetries, or maybe something else?
  4. Sep 9, 2015 #3
    We were studying the Icosahedral (or dodecahedral which ever you prefer) group or equivalently the rotational symmetries of the Icosahedron.

    I should elaborate on my question with a little more specifics using an example. In one step he claimed all elements of order 3 are conjugate and to prove it he noted that the stabilizers ##\{1,\rho, \rho^{-1}\}## of two opposite vertices are the same. Realizing this it seemed immediate to him that all the elements or order 3 were conjugate. I on the other hand didn't see how he made that jump.
  5. Sep 12, 2015 #4


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    what is obvious is that all stabilizer groups of all vertices are conjugate, since they are all in the same orbit. And it seems visible that conjugating a counterclockwise rotation about one vertex by a rotation to the opposite vertex, gives you the counterclockwise ratation about that opposite vertex. This however, by your observation, equals the clockwise rotation about the original vertex, so R is also conjugate to R^-1. Since all the order three subgroups are conjugate, and the two elements of order three in each one are conjugate to each other, all elements of order three are conjugate. how's that?
  6. Sep 13, 2015 #5
    It's clear now. Thanks!
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