Draw the Cayley Graph of the symmetry group of the triangular Prism.

I am having a difficult time with this question. So far I know that the symmetry group has order 12, and also the symmetry group is D_3h. I just cannot seem to wrap my head around how to draw the cayley graph for this. Any help would be great. Thanks in advance.

  • $\begingroup$ By "triangular prism" you must mean "regular tetrahedron". To get a Cayley graph you also need to choose generators - does the homework question specify what those should be? Finally, what do you know about the elements of the group? Are you able to list them in some way? $\endgroup$ – Alon Amit Sep 28 '11 at 4:43
  • $\begingroup$ No it is not the regular tetrahedron. It is a pentahedron, so the two trinagular faces are parallel. Yes, the first thing to do is to get the generators chosen. I know there are 4, and no I do not know all of the elements. That is partially my problem. I can draw the triangular prism and figure out the reflections and rotations. I just seem to work better with having permutations as elements when I draw cayley graphs. I am unable to do this at this point with this problem. $\endgroup$ – Betty Sep 28 '11 at 4:52
  • $\begingroup$ What is "D_3h"? $\endgroup$ – Arturo Magidin Sep 28 '11 at 5:23
  • $\begingroup$ Is the "triangular prism" like a triangular peg? Two (equilateral?) triangles, one on top and one on the bottom, and three rectangular sides? $\endgroup$ – Arturo Magidin Sep 28 '11 at 5:31
  • 1
    $\begingroup$ @Arturo: Yes, a prism over S is { (x,y,z) : (x,y) in S, 0 ≤ z ≤ 1 } or so. D_3h is literally the symmetry group of this prism. I think the D_3 means the dihedral group of order 6, acting planewise in its natural action as the symmetry group of the triangular cross-sections. The sub h means an additional generator is added, probably a horizontal reflection, meaning, it takes one triangular base to the opposite triangular base. $\endgroup$ – Jack Schmidt Sep 28 '11 at 14:53

Why do you have four generators? I would think three is natural: rotation of 120 degrees, reflection about a plane parallel to the triangle, reflection about a plane orthogonal to the triangle.

Now write down the 12 elements of the group that you have in some organized way and draw edges between them when multiplication with a generator from the right gets you from one element to the other. Label this edge with the corresponding generator. That's your Caley graph. (Each element should have three outgoing edges and three incoming edges.)


You say you work better when you have permutations as elements. Well, we can work with permutations. Draw the prism and name its vertices 1 through 6 in some way. Now figure out the reflections and rotations; each one maps the set of vertices onto itself, so you get a permutation, and that permutation uniquely defines the symmetry of the prism. From this point onwards you may forget about the geometry and just work with the permutations.

Now, the generators: I'm not sure what you mean by "the first thing to do is get the generators chosen. I know there are 4..." - a group can be generated by more than one set of generators, and you can choose any set you like I guess if the problem doesn't specify one (you can point out to the prof, TA or author that "the Cayley graph of a group" is not a well defined concept). Now that you have permutations to work with, you may find it easier to pick some generating set and then draw the edges from each group element to its product by each generator.

  • $\begingroup$ The triangular prism would have 6 vertices not 8. $\endgroup$ – Joseph Malkevitch Sep 28 '11 at 14:37
  • $\begingroup$ Thank you Alon and Stefan. I will update the finished cayley graph when I am finished. $\endgroup$ – Betty Sep 28 '11 at 16:37
  • $\begingroup$ Of course :-) it was too late at night. $\endgroup$ – Alon Amit Sep 28 '11 at 23:14

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