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In my project to construct the outer automorphism group for $S_6$ I have come across a need (or desire) to visualize a graph that has 15 vertices, with each vertex having 6 incident edges and 3 incident "triangles." (By my count, that makes 15 total vertices, 45 total edges, and 15 total triangles.) No matter how I try to draw this, I am not successful. I started with a regular 15-sided figure, but this lead nowhere quickly, and I can see clearly that if this graph even exists, it will have to be a concave/self-intersecting polygon, geometrically. Am I barking up the wrong tree, or is there a way to realize this graph?

Edit: In case anybody is wondering, the reason I am interested in this graph is because I am considering all the two-cycles in $S_6$ as "vertices," for which there are $6 \choose 2$ aka 15 total vertices; two vertices are joined by an edge if they commute, ie, if they are disjoint. Three vertices form a "triangle" if they are pairwise disjoint.

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    @DerekHolt, that's exactly how I came up with the construction for the graph! But forgive me if I am being stupid, but I still don't see how that helps me draw the graph2012-10-21

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Start with $15$ disjoint triangles, and identify the vertices in sets of $3$ (belonging to distinct triangles, but making sure that you don't identify all the vertices of one triangle with those of another) so that you end up with $15$ distinct vertices.

FWIW, here are some pictures of the graph as constructed using Derek Holt's comment. The graph is not planar, and I don't know if there is a simpler way to visualize it.

enter image description here

enter image description here

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    @DerekHolt perfect explanation. I was wondering about how $S_6$ acted on the graph, which should have been pretty obvious to me by its construction. Thank you.2012-10-23