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Could only prepare ground for this in my last post here, with a reduced group and set size.

Kindly give hints, as sizes of subsets that can think of are too small: $X_3.$ Though read of tetrahedron too, but not clear where the fourth vertex will lie. Also,the subsets have to be disjoint; so if $X_4$ is taken then $X_3$ is removed.

jiten
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    The group of symmetries in particular acts by isometries, so if ${v_1,v_2}$ is in the same orbit as ${v_1',v_2'}$ then $d(v_1,v_2)=d(v_1',v_2')$. Thus the level sets of $d$ partition $X$ into $G$-invariant subsets. Now can you check whether $d(v_1,v_2)=d(v_1',v_2')$ implies there is $g\in G$ sending ${v_1,v_2}$ to ${v_1',v_2'}$? If not, you may have to subdivide your partition further. – David Sheard Aug 03 '22 at 13:28
  • @DavidSheard Please note that by now Orbits not covered. But, next week has it. So can read, though not expected. But, request more details/elaboration. – jiten Aug 03 '22 at 13:30
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    OK, I'm not using any special properties of orbits like the Orbit Stabiliser Theorem (although if you had met this it might help), I'm just using the terminology. By "${v_1,v_2}$ is in the same orbit as ${v'_1,v'_2}$" I merely mean "there exists $g\in G$ such that $g\star{v_1,v_2}={v'_1,v'_2}$. – David Sheard Aug 03 '22 at 13:35
  • @DavidSheard Sorry, it was easy. Also, what is meant by: "level sets of $d$ ". – jiten Aug 03 '22 at 13:37
  • @DavidSheard If not totally incorrect, checking for $2.60.66 $ actions is infeasible. Or, you imply some shortcut. – jiten Aug 03 '22 at 13:44
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    "Level set" is just a fancy way to say $\left{{v_1,v_2}\mid d(v_1,v_2)=c\right}$ for some fixed $c\in\mathbb{R}^{\ge0}$. Ie the set of all pairs of vertices which are some fixed distance apart. – David Sheard Aug 03 '22 at 13:48
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    I don't know what you mean? Isn't there just a single action in your question: that of the symmetry group of the icosahedron on the icosahedron? – David Sheard Aug 03 '22 at 13:51
  • @DavidSheard There is a single action. But the verb of 'applying a symmetry to both vertices in a pair simultaneously', occurs for each symmetry (which are $2.60$) and each pair (which are $66$). Please tell where wrong. – jiten Aug 03 '22 at 21:21

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I shall expand on the hints I provided in the comments. First note that any symmetry of the icosahedron preserves distances, and so if the pair of vertices $\{u,v\}$ get transformed into $$g\star\{u,v\}=\{g\star u,g\star v\}=:\{u',v'\},$$ then $d(u,v)=d(u',v')$.

Suppose therefore that we have a partition $X=X_1\sqcup\cdots\sqcup X_k$ such that $G$ acts transitively on each $X_i$. Then the observation above implies that if $\{u,v\},\{u',v'\}\in X_i$ for some $i$, then $d(u,v)=d(u',v')$. I claim that taking exactly the partition of $X$ coming from $d$ is sufficient.

In order to prove this, let $X_i=\{\{u,v\}\mid d(u,v)=c\}$ for some fixed $c$. Let $\{u,v\},\{u',v'\}\in X_i$. Now, a key observation is that while $G$ may not act transitively on $X$, it does act transitively on the set of vertices of the dodecahedron. Therefore we can find $g\in G$ such that $g\star u'=u$.

If we define $w=g\star v'$ then we can reduce to the case we have $\{u,v\},\{u,w\}\in X_i$, and want to show that there is $h\in G$ such that $h\star\{u,v\}=\{u,w\}$. Such an $h$ must lie in the stabiliser of $u$, $\textrm{Stab}_G(u)$. By considering a picture of an icosahedron, t is easy to convince yourself (by putting $u$ at the top say), then $\textrm{Stab}_G(u)$ acts transitively on the set of all vertices a fixed distance from $u$.

Thus, the required $h$ exists, and the element $gh$ maps $\{u,v\}\mapsto\{u,w\}$ as required.

enter image description here

(Image by DTR (CC BY-SA 3.0))

This picture also shows that there are three possible distances for a pair of points to be apart, which I'm calling $d_1, d_2,d_3$. Let $X_1, X_2,X_3$ be the corresponding sets.

Finally we must compute the cardinalities of these sets. $|X_1|$ is the number of edges in the icosahedon, ie $\frac{20\times 3}{2}=30$. $|X_3|$ is the number of antipodal vertices, ie $\frac{20\times 3}{5\times 2}=6$. This leaves $|X_2|=|X|-|X_1|-|X_3|=30$.

David Sheard
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  • Why it is that only three types of edges are there? – jiten Aug 04 '22 at 11:03
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    Look at the picture. Once the vertex $u$ is fixed every vertex is either one of the dark blue vertices, one of the purple vertices, or the red vertex. All vertices of the same colour are the same distance from $u$, so there are only three possible distances. – David Sheard Aug 04 '22 at 12:03
  • What property was used in getting number of edges as $= $ number of triangles $\cdot\frac 32.$ Also, why $X_1$ is for a different quantity (edges), while $X_2$ is about vertices. – jiten Aug 04 '22 at 14:27
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    Each of the 20 triangles contributes 3 edges, but this double counts the edges since when the triangles are glued together two edges get identified, hence $20\times 3/2$. $X_1$ corresponds the the dark blue vertices which are exactly one edge away from $u$. Thus the pairs of vertices in $X_1$ correspond the the edges in the icosahedron. Meanwhile $X_3$ corresponds to the red vertex which is antipodal to $u$. They are just different ways to describe the relationships between pairs of vertices in the picture. The number of antipodal pairs of vertices is half the number of vertices. – David Sheard Aug 04 '22 at 14:37