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Need help with the following :

Let G be a non-planar graph. Suppose further that G\e is planar for every edge e of G. Show that at most 6 vertices of G have degree 3 or greater.

So far, I know that from Kuratowski's Theorem G contains a subgraph H such that H is a subdivision of K5 or K3,3. Then I proved by contradiction that since G/e is planar we must have that G=H, namely that G itself is a subdivision of K5 or K3,3. Then I know that the number of edges of G - number of vertices of G is either 5 or 3 and I'm thinking maybe the Handshaking Lemma can help me here but I don't see how. Maybe I can prove the result by contradiction but I don't know which facts to use.

Thanks you.

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In a subdivision of a graph, all nodes added by the subdivision must have degree 2; hence, nodes in $H$ with degree 3 or greater must be in $K_5$ or $K_{3,3}$. In both graphs there are clearly at most 6, since they have at most 6 nodes total.

  • 0
    Is a node a different name for a vertex? – 2017-02-07
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    @TeddyRoze, yes, it is. – 2017-02-07
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    How did I not see that ? Great answer, thank you very much ! – 2017-02-07