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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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    @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?*2011-09-04

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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.

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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.