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I need to show this graph is not planar.

enter image description here

I've attempted to find a subgraph that is a subdivision of $K_5$ or $K_{3,3}$ but haven't been successful yet.

Carlos Andres Henao Acevedo
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    Every vertex is of degree $3$ and the graph is bipartite, so I'd suggest looking for a $K_{3,3}.$ – saulspatz Jun 20 '20 at 02:28
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    This is the Heawood graph. You can see a subdivision of $K_{3,3}$ in it here if you can’t make use of the bipartite nature to find one on your own. – Brian M. Scott Jun 20 '20 at 02:30

2 Answers2

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Is this a subdivision of $K_{3,3}$?

enter image description here

Carlos Andres Henao Acevedo
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Since the graph has a Hamilton cycle, there is a much easier way to prove it has a Hamilton cycle than searching for subdivision of $K_{3,3}.$

enter image description here

Now matter how the graph is drawn in the plane, there will be a Hamilton cycle, which will divid the plane into an interior and an exterior. The chords ID and HM will intersect if both are drawn in the interior, and they will also intersect if both are drawn in the exterior, so to get a plane drawing, one must be drawn in the interior and one in the exterior. However, chord FK intersects both these chords. We have to draw at least two of these chords in the same region of the plane (by the pigeonhole principle), and they will intersect.

This is a simple case of Tutte's conflict graph theorem.

saulspatz
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