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The group of orthogonal transformations of $\mathbb{R}^3$ is direct product of the group of rotations and the group $Z=\langle x\mapsto -x\rangle\cong \mathbb{Z}/2$. The finite subgroups of group of rotations are well known: rotational symmetries of regular polyhedrons.

These subgroups, along with their direct product with $Z$ will give some (families) of finite groups of orthogonal transformations.

It looks that some finite groups of orthogonal transformations need not be of this form (see Shafarevich- Algebra I; I couldn't understand arguments there).

How do we cover all finite subgroups of orthogonal transformations?

A more general question in this context may be: How to get subgroups of $H\times K$ knowing subgroups of $H$ and $K$?

  • Surely all the dihedral groups also can be viewed as orthogonal transformations in $\mathbf{R}^3$? – Jyrki Lahtonen Sep 03 '11 at 11:14
  • @Jyrki: yes; take regular polygon in $XY$ plane centered at origin; rotational symmetries form dihedral group- reflections in $D_n$ are obtained by rotation about an axis through origin in $XY$ plane, and rotations of $D_n$ are simply rotations about $Z$ axis. The group $D_n$ is in fact a *group of rotatios in $\mathbb{R}^3$, not just orthogonal transformations*. –  Sep 03 '11 at 11:40
  • I'm confused: I think every finite subgroup of $O_n$ is either a finite rotation group or the direct product of a finite rotation group and $Z \cong \mathbb{Z}/2$. The antipodal map $\rho: x \mapsto -x$ is in the center of $O_3$. Given any subgroup $H$ of $O_3$, either it contains $\rho$ or it doesn't. If it doesn't, it's a rotation group. If it does, let $K$ be its rotation subgroup; then $H = K \cup \rho K$. But $\rho$ commutes with everything, thus $\rho K = K \rho$ and $H$ is a direct product of $K$ and $Z$. What's wrong with this? – Ben Blum-Smith Sep 03 '11 at 14:48
  • Edit: in 2nd to last line, the statement $\rho K = K \rho$ isn't really what I meant. The important point is that $\rho$ commutes with each individual element of $K$. – Ben Blum-Smith Sep 03 '11 at 14:55
  • group: I was just pointing out that a finite subgroup of $O(3)$ doesn't need to be a symmetry group of a Platonic polyhedron. @Ben: you undoubtedly know this, but your explanation seems to leave out the possibility that $H\cap SO(3)$ may be of index $2$ in $H$ even if $\rho\notin H$. For example, the symmetries of the regular tetrahedron form the group $S_4\le O(3)$. This copy of $S_4$ does not contain the antipodal map, but does not consist of rotations only, as the reflections have determinant $=-1.$ The tetrahedron is not stable under $\rho$ anyway. Here $S_4\cap SO(3)=A_4$. – Jyrki Lahtonen Sep 03 '11 at 18:11
  • @Jyrki - oh, yes, now that you mention it I was totally neglecting this possibility. (Symmetries of an object that has some orientation-reversing symmetry but not antipodal symmetry.) So in this case $H$ can be a semidirect product of $K$ and $Z$... – Ben Blum-Smith Sep 03 '11 at 19:11
  • @Jyrki: Yes, a finite subgroup subgroup of $O_3$ doesn't need to be a symmetry group of a platonic polyhedron. $O_3\cong SO(3) \times \mathbb{Z}/2$. So if $H$ is a finite subgroup of $SO(3)$ and $K=\mathbb{Z}/2$ or ${1}$, then $H \times K$ is a finite subgroup of $O_3$. But, there are some finite subgroups of $O_3$ which are not of this form. How do get such groups? In general, how do we obtain subgroups of direct product of two groups, if we know subgroups of component groups? –  Sep 04 '11 at 07:55

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The well-known Goursat's lemma is a classification of subdirect products of two groups. If all the subgroups of the factors are known (as it is in your case), then the lemma actually gives you a description of all the subgroups of the direct product. This question was also discussed on MO.

Community
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darko
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For an enumeration of all finite subgroups of orthogonal transformations in dimensions 1 through 4, see the book "On Quaternions and Octonions" by Conway and Smith, Chapters 1-4. The enumeration for dimension $i$ is in Chapter $i$.

The enumeration of subgroups of a direct product plays a role, and is discussed in Chapter 4, Section 4.3.

Ted
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