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I always tried to understand how the finite reflection groups of $\Bbb R^d$ (of some fixed dimension $d$) relate to the point groups of the same space $\smash{\Bbb R^d}$ (finite subgroup of the orthogonal group $\smash{\mathrm O(\Bbb R^d)}$).

Initially, I was under the impression that each point group is a subgroup of a finite reflection group. This turned out to be wrong, which is obvious in hindsight. Many reflections groups have symmetries in the placement of their mirrors that can be used to enlarge the group.

So let's take these enlarged groups instead. From my geometric understanding, by that I mean the symmetry groups of the uniform polytopes. So I shall call them uniform point groups. Most (or all?) uniform polytopes can be generated from a reflection group, and then it has all the symmetries of this group, but might have more.

Question: Is every point group a subgroup of a uniform point group?

Regardles of the answer to that question, I am open for any statement that sheds light on the placement of reflections groups (or easily derived groups thereof) inside the family of general point groups.


Update Sep. 2019

There seems to exist a counterexample in dimension four, namely, a point group denoted $\pm[I\times C_n]$ that is supposedly not the subgroup of a symmetry group of a uniform polytope. This was mentioned in this answer by Günter Rote. Currently I am not able to verify the claim. So, any hint is welcome.

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    $\begingroup$ $S_n$ are all uniform point groups (of an $(n-1)$-simplex) and contain all finite groups as subgroups. $\endgroup$ – Mikhail Tikhomirov Sep 15 '19 at 16:47
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    $\begingroup$ @MikhailTikhomirov I am interested in the case when a point groups is a subgroup of a reflection/uniform group of the same dimension. Edited the question to make this clearer. $\endgroup$ – M. Winter Sep 15 '19 at 16:48
  • $\begingroup$ @MikhailTikhomirov Any group is isomorphic to a subgroup of the symmetry group of a simplex. But I am not interested in abstract groups, but matrix groups. $\endgroup$ – M. Winter Nov 4 '19 at 13:43
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    $\begingroup$ It would be useful to clarify the meaning of $I\times C_n$. Is it a cyclic group of order $n$ acting as the identity on some $2$-plane? $\endgroup$ – YCor Nov 4 '19 at 17:18
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    $\begingroup$ The reference should be to "Conway and STEIN's book, p. 44, Table 4.1". It comes from the fact that rotations in 4-space can be written as $x\mapsto \bar a\cdot x\cdot b$ for two unit quaternions $a,b$. In 3-space they are $x\mapsto\bar a\cdot x\cdot a$ for a unit quaternion $a$. In this way, every group in $SO_4$ is associated to a subgroup of the (group-theoretic) product of two groups in $SO_3$ (more or less, it's a double-cover). $I\times C_n$ means $I$ is the left group and $C_n$ is the right group. $I$ permutes the 12 great circles in my example. $C_n$ turns each great circle in itself. $\endgroup$ – Günter Rote Nov 6 '19 at 10:19
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Here is a more worked-out and concrete version of the proposed counterexample mentioned in the "Update" of the post.

  1. We start from a symmetric arrangement of 12 great circles $F_1,\ldots,F_{12}$ on the 3-sphere $\mathbb{S}^3$: 12 circles from the Hopf fibration, shown in stereographic projection to $\mathbb{R}^3$. They are the inverse image of the 12 corners of the regular icosahedron under the Hopf map $\mathbb{S}^3\to\mathbb{S}^2$ (fibers of the Hopf fibration).

  2. On each circle we place 70 equidistant points: 840 points on 12 circles.

  3. The supporting hyperplanes of the 3-sphere at these points form a 4-polytope $P$ with 840 equal facets: A perspective view of a facet, a 3-polytope with 40 vertices and 22 sides. (In order to ensure that all faces are equal, the regular 70-gons on the different circles cannot just be placed arbitrarily in Step 2. The points form the orbit of a specially chosen starting point under the group $\pm[I\times C_7]$.)

The facets are thin plates of roughly pentagonal shape centered at points of a circle $F_i$ and lying perpendicular to $F_i$. The plates stack up to form a twisted pentagonal tube that surrounds $F_i$. Such a tube makes one full turn of $360^\circ$ as it winds around the circle. The 12 tubes fill the space around the 3-sphere and enclose it completely.

The symmetry group $G$ of $P$ has 8400 elements: a given plate can be mapped to any of the 840 plates in 10 different ways. There are only rotations (determinant $+1$), no reflections (determinant $-1$). (Because of the special choice of the starting point of the orbit, the group $G$ is larger than the group $\pm[I\times C_7]$ by which the orbit was generated. I guess, with a generic orbit, the symmetry group will reduce to size 840, but the picture will be messier.)

The polytope $P$ is clearly not a uniform polytope: none of its 2-faces is a regular polygon. To definitely answer the question, one would have to argue why $G$ is not a subgroup of the symmetries of a different, uniform, polytope $P'$. One could use the classification of the point groups from the book of Conway and Smith, but maybe there is a more direct argument. The group must contain all rotations of a regular 70-gon.

This example and the programs (in Sage) that were used to produce the images were prepared with the help of my student Laith Rastanawi.

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  • $\begingroup$ Thank you for this elaboration! I think the number 70 plays no particular role here, and we can choose any arbitrarily high number of points on each circle, right? If so, we can make the resulting polytope have more symmetries than any 4-dimensional uniform polytope except for certain duoprisms (14400 might ne the highest number of symmetries, attained for the 120/600-cell). I then believe it might not be too hard to argue that no symmetry group of a duoprism can have the described group as a subgroup. $\endgroup$ – M. Winter Dec 26 '19 at 9:35
  • $\begingroup$ Indeed, 70 can be replaced by any multiple of 10. (10 because the group $I$ (taken as left multiplications) itself already generates 10 points on each circle.) $\endgroup$ – Günter Rote Dec 28 '19 at 5:36
  • $\begingroup$ Here is a proof that $G$ is not contained in the symmetry group of a duoprism. We look at great circles that are kept elementwise fixed by nontrivial elements of $G$: A symmetry that rotates a plate $P$ leaves the whole circle $F_i$ on which $P$ lies pointwise fixed. Hence there are 12 great circles that are left fixed by elements of $G$. On the other hand, there are only two choices of a circle that is pointwise fixed by a nontrivial symmetry of a "large enough'' duoprism (the two "axis circles''). $\endgroup$ – Günter Rote Dec 29 '19 at 6:42

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