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Symmetry of the square

  1. Apr 21, 2007 #1
    Hi all,

    I was thinking of symmetries today and this kept bugging me. I wonder if anyone can help me with this.

    Consider the symmetries of a triangle. You can do 3 rotations and 3 reflections to get them all. Number the vertices, create an ordered set with 3 elements that contain the numbers of the vertices. Starting with the topmost and working clockwise you'd get (1,2,3), (2,3,1), etc. By doing all the symmetries you can get all possible combinations of three numbers.
    Consider the symmetries of a square. You can do 4 rotations and 4 reflections to get them all. Do the same procedure as above, that is, number the vertices and construct the ordered sets. Now you don't end up with all possible combinations of 4 numbers. For example: number the top left hand side vertex 1, the top right hand side 2, the bottom right 3 and the bottom left 4. I cannot think of a way of getting the set (1,2,4,3). Can anyone?
    Last edited: Apr 21, 2007
  2. jcsd
  3. Apr 21, 2007 #2


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    Any symmetry is going to preserve which vertices are adjacent to which others. So, while in a triangle, all edges are adjacent, so you can't use this to rule out any lists like (1,3,2), etc. as representing actual symmetries, for the square you can rule out (1,2,4,3) since this would make 2 adjacent to 4, which it isn't in the original square.
  4. Apr 22, 2007 #3
    Yeah, I realised that I was ignoring the edges when re-thinking about this a few minutes later... I'm sorry for posting such a silly question :)
  5. Apr 22, 2007 #4

    Chris Hillman

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    By the way, Siberius, the symmetry group of the square is the eight element dihedral group often denoted [itex]D_4[/itex] (confusingly, some authors denote this group [itex]D_8[/itex] but the former notation is definitely more common and IMO preferable).

    Another point to remember: when you used reflections in computing the symmetry groups of the triangle and the cube, these are (as linear transformations) "improper" (they are isometries, but they don't have determinant one). The "proper" symmetry group correspond to an index two subgroup of the guys you found. Can you figure out the symmetry group of the tetrahedron and the three-cube? As permutation groups acting on the vertices, or alternatively, on the edges. (Would the group acting on the vertices be isomorphic to the group acting on the edges?)

    I can't resist pointing out that you can generate the two groups you found (allowing "improper" isometries) using only reflections. This is related to the beautiful method of Coxeter for classifying such reflection groups, which turns out to be the same as Dynkin's method of classifying the simple complex Lie algebras. (Keyword: Coxeter-Dynkin diagrams.) Reflection groups have a very beautiful theory which is the subject of many books.

    A very readable and fun high school textbook:

    Groups and their Graphs, by Israel Grossman and Wilhelm Magnus, Mathematical Association of America, 1964.

    Two very readable and fun undergraduate textbooks:

    Permutation Groups by Peter J. Cameron, Cambridge University Press, 1999.

    Groups and Geometry by Peter M. Neumann, Gabrielle A. Stoy, and the late Edward C. Thompson, Oxford University Press, 1994.

    Kleinian geometry is also fun:

    Transformation Geometry : an Introduction to Symmetry, by George E. Martin, Springer, 1982.
    Last edited: Apr 22, 2007
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