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A transitive set in $\mathbf{R}^n$ is a finite set with a transitive group of symmetries. I want to understand how subsets of a transitive set constrain the group.

Let me start with the example of a tetrahedron. A tetrahedron has symmetry group isomorphic to $S_4$, which is of course nonabelian. However, it is possible to embed the tetrahedron into a larger set, the cube, which has a transitive abelian group of symmetries isomorphic to $C_2^3$. (Of course, the full symmetry group of the cube is $C_2 \times S_4$, which is nonabelian, but my point is there is an abelian subgroup which is still transitive.) So we made things simpler by going bigger.

https://en.wikipedia.org/wiki/Compound_of_two_tetrahedra --> https://en.wikipedia.org/wiki/Cube

[pictures borrowed from Wikipedia]

My specific question is the same with (tetrahedron, abelian) replaced with (icosahedron, soluble).

https://en.wikipedia.org/wiki/Icosahedron --> ???

Is there a finite set in $\mathbf{R}^n$ (with possibly $n > 3$) with a transitive soluble group of symmetries containing the vertices of a regular icosahedron?

I would also appreciate any references to related matters.

EDIT: Achim Krause answered trivially the question I asked but not the question I intended. At the risk of making my question a bit like Douglas Adams's concept of the Universe in that, if ever anyone discovers exactly what it is for and why it is here it will instantly be replaced by something even more bizarre and inexplicable, I now wish to replace "icosahedron" with "snub dodecahedron".

https://en.wikipedia.org/wiki/Snub_dodecahedron

Now the group of rotational symmetries is $A_5$, which acts regularly on the vertices.

The point is I meant to choose a transitive set for which the answer was "obviously" no, but such that it is not clear how to prove that, especially if we allow embedding into possibly higher dimensions.

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    $\begingroup$ Note that in your tetrahedron example there is a subgroup $C_2\times C_2$ of $C_2^3$ which takes the tetrahedron to itself, and acts already transitively on vertices. So we don't really need to go to the cube at all, we're just shifting attention to a transitive subgroup of $S_4$. $\endgroup$
    – Achim Krause
    Commented Mar 12, 2021 at 11:13
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    $\begingroup$ It feels like you should be able to get a negative answer to your question from the classification of finite subgroups of $O(3)$. Basically, there's not many things your solvable group could be: all large enough subgroups of $O(3)$ already fix a common axis, and thus can't act transitively on a set containing the vertices of an icosahedron. The rest is a finite list, of subgroups of symmetry groups of platonic solids. Only the smaller ones of them are solvable, and none of them looks to me like it could act transitively on a superset of the vertices of an icosahedron. $\endgroup$
    – Achim Krause
    Commented Mar 12, 2021 at 11:27
  • $\begingroup$ That is a very good point about the tetrahedron, quite embarrassing. $\endgroup$
    – Sean Eberhard
    Commented Mar 12, 2021 at 11:41
  • $\begingroup$ Any group acting vertex-transitively on the snub dodecahedron needs at least 60 elements. You can check this list to see that there is not much left among the irreducible groups (only $A_5$ and $A_5\times 2$, both not soluble). And I think we agree that reducible groups cannot do it. $\endgroup$
    – M. Winter
    Commented Mar 12, 2021 at 16:07

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The answer to your question is actually "yes", but maybe not in the way you wanted. Indeed, the full (oriented) symmetry group of the icosahedron is isomorphic to $A_5$. The stabilizer of a vertex is cyclic of order $5$, so the set of vertices with action by $A_5$ can be identified with the set of cosets $A_5/C_5$.

The solvable subgroup $A_4\subseteq A_5$ acts freely and transitively on this: It has the right cardinality (12), and it acts freely, since if an $h\in A_4$ stabilizes a coset $gC_5$, it means that $h$ is conjugate to an element of $C_5$. But $A_4$ has not elements of order $5$, so $h=e$.

So you can take the set of vertices to be just the vertices of the icosahedron!

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  • $\begingroup$ Ok, this is also embarrassing, thanks! $\endgroup$
    – Sean Eberhard
    Commented Mar 12, 2021 at 14:00
  • $\begingroup$ I first thought "this can't be". But I just now realized that the generic orbit polytope of the orientation preserving symmetry group of a regular tetrahedron is an icosahedron (the snub tetrahedron). Wow, thank you for reminding me! This was very unexpected. $\endgroup$
    – M. Winter
    Commented Mar 12, 2021 at 14:30
  • $\begingroup$ To visualize this, note the icosahedron acts on five triples of orthogonal golden rectangles; the stabilizer of one of those is A4, which one observes acts transitively (see here) $\endgroup$
    – Beren Gunsolus
    Commented Mar 13, 2021 at 1:46

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