Is it true that any finite graph has a $K_n$ minor, where $n$ is a minimal vertex degree?
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$\begingroup$ Not if the graph is infinite (e.g. $n=3$ and the infinite binary tree). $\endgroup$ – M. Winter Jan 31 '20 at 14:37
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No.
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).
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$\begingroup$ 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 '20 at 7:11
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More generally, it is a classic result (independently due to Kostochka and Thomason) that minimum degree $(\alpha+o(1))n \sqrt{\log n}$ suffices to force a $K_n$ minor, where $\alpha$ is an explicit constant. Conversely, there are random graphs with minimum degree $\Omega(n\sqrt{\log n})$ that do not contain a $K_n$ minor. See here to access the paper by Thomason.
answered Jan 31 '20 at 16:36
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3$\begingroup$ 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 '20 at 5:22
