# Contracting non-adjacent points in the icosahedron [closed]

Are there $$2$$ non-adjacent points in the icosahedron graph $$G$$ such that contracting them leaves the Hadwiger number unchanged?

## closed as off-topic by Jan-Christoph Schlage-Puchta, Alexey Ustinov, YCor, Pace Nielsen, Ivan IzmestievApr 2 at 10:25

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• Up to symmetry, there's clearly only two graphs obtained by contracting. Why not simply do it? – verret Mar 28 at 6:00

The icosahedron graph is distance transitive, meaning that for any two pairs $$(a,b)$$ and $$(c,d)$$ of vertices of the icosahedral graph such that $$\text{dist}(a,b) = \text{dist}(c,d)$$, there is an automorphism $$\sigma$$ such that $$\sigma(a) = c$$ and $$\sigma(b) = d$$. Since the icosahedral graph has diameter 3, there are only two choices of pairs of non-adjacent vertices up to automorphisms: those pairs at distance 2 and those at distance 3. Note that the icosahedral graph is planar, and thus it does not have $$K_5$$ as a minor. This means that its Hadwiger number is at most 4 (and it is actually equal to this according to Sage). By computation in Sage, you can quickly show that merging two vertices at distance 2 or at distance 3 results in graphs with $$K_5$$ as a minor, thus the Hadwiger number does increase.