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[This is exam review] I have the following graph:

enter image description here

Question: I need to prove it is not planar by showing a subdivision of $K_5$ or $K_{3,3}$ exists in the graph (to use Kuratowski’s Theorem). I understand the concept of a subdivision (ie. adding some vertices between two vertices to make new edges), but I have no idea how to apply it to get a subdivision of $K_5$ or $K_{3,3}$.

The definition of a subdivision from my textbook: An edge subdivision of a graph G is obtained by applying the following operation, independently, to each edge of G: replace the edge by a path of length 1 or more; if the path has length $m > 1$, then there are $m − 1$ new vertices and $m − 1$ new edges created; if the path has length $m = 1$, then the edge is unchanged.

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A subdivision of $K_5$ will have $5$ vertices of degree $4$, which you don't have, so you're looking for a subdivision of $K_{3,3}$. Get rid of the vertices of degree $2$, replacing the edges $ef,fg$ with $eg$ and $ha,ab$ with $hb$. Finally, remove the edge $bd$. You now have a $K_{3,3}$ with $b,d,g$ on one side and $c,e,h$ on the other side. If we call your original graph $G$, then the subgraph $G-bd$ is a subdivision of this $K_{3,3}$.

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