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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 $K$?

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Surely all the dihedral groups also can be viewed as orthogonal transformations in $\mathbf{R}^3$? –  Jyrki Lahtonen Sep 3 '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. –  user8186 Sep 3 '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 3 '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 3 '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 3 '11 at 18:11

2 Answers 2

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.

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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.

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