Is it true that any finite graph has a $K_n$ minor, where $n$ is a minimal vertex degree?

  • $\begingroup$ Not if the graph is infinite (e.g. $n=3$ and the infinite binary tree). $\endgroup$ – M. Winter Jan 31 at 14:37


The edge-graph of the icosahedron is regular of degree five, but does not have a $K_5$ minor because it is planar (Kuratowski's theorem).

  • $\begingroup$ This is a great (counter)example. $\endgroup$ – Steven Stadnicki Jan 31 at 17:00
  • $\begingroup$ Also the smallest if I am not mistaken. No $K_4$-minor means the graph has a vertex of degree 2. $\endgroup$ – Gordon Royle Feb 1 at 7:11

More generally, it is a classic result (independently due to Kostochka and Thomason) that minimum degree $O(n \sqrt{\log n})$ suffices to force a $K_n$ minor, and conversely, there are graphs with minimum degree $\Omega(n\sqrt{\log n})$ that do not contain a $K_n$ minor. Moreover, we know precisely what the constant term should be. See here to access the paper by Thomason.

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    $\begingroup$ Where can I read about this in detail? $\endgroup$ – Eschatum Verus Feb 1 at 3:00
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    $\begingroup$ I added a link to the corresponding paper by Thomason. Note that Thomason's result is stated for average degree, but every graph with average degree $d$ contains a subgraph of minimum degree $\frac{d}{2}$. $\endgroup$ – Tony Huynh Feb 1 at 5:22

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