# Cayley Graph of the Symmetry Group of the Triangular Prism

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.

• 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? – Alon Amit Sep 28 '11 at 4:43
• 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. – Betty Sep 28 '11 at 4:52
• What is "D_3h"? – Arturo Magidin Sep 28 '11 at 5:23
• Is the "triangular prism" like a triangular peg? Two (equilateral?) triangles, one on top and one on the bottom, and three rectangular sides? – Arturo Magidin Sep 28 '11 at 5:31
• @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. – Jack Schmidt Sep 28 '11 at 14:53