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Step 2

In order to find a free group of rotations of 3D space, i.e. that behaves just like (or "is isomorphic to") the free group F, two orthogonal axes are taken (e.g. the x and z axes). Then, A is taken to be a rotation of θ = arccos (1 3) {\textstyle \theta =\arccos \left({\frac {1}{3}}\right)} about the x axis, and B to be a rotation of θ {\displaystyle \theta } about the z axis (there are many other suitable pairs of irrational multiples of π that could be used here as well).

The group of rotations generated by A and B will be called H. Let ω {\displaystyle \omega } be an element of H that starts with a positive rotation about the z axis, that is, an element of the form ω = … b k 3 a k 2 b k 1 {\displaystyle \omega =\ldots b^{k_{3}}a^{k_{2}}b^{k_{1}}} with k 1 > 0, k 2, k 3, …, k n ≠ 0, n ≥ 1 {\displaystyle k_{1}>0,\ k_{2},k_{3},\ldots,k_{n}\neq 0,\ n\geq 1}. It can be shown by induction that ω {\displaystyle \omega } maps the point (1, 0, 0) {\displaystyle (1,0,0)} to (k 3 N, l 2 3 N, m 3 N) {\textstyle \left({\frac {k}{3^{N}}},{\frac {l{\sqrt {2}}}{3^{N}}},{\frac {m}{3^{N}}}\right)}, for some k, l, m ∈ Z, N ∈ N {\displaystyle k,l,m\in \mathbb {Z},N\in \mathbb {N} }. Analyzing k, l {\displaystyle k,l} and m {\displaystyle m} modulo 3, one can show that l ≠ 0 {\displaystyle l\neq 0}. The same argument repeated (by symmetry of the problem) is valid when ω {\displaystyle \omega } starts with a negative rotation about the z axis, or a rotation about the x axis. This shows that if ω {\displaystyle \omega } is given by a non-trivial word in A and B, then ω ≠ e {\displaystyle \omega \neq e}. Therefore, the group H is a free group, isomorphic to F.

The two rotations behave just like the elements a and b in the group F: there is now a paradoxical decomposition of H.

This step cannot be performed in two dimensions since it involves rotations in three dimensions. If two rotations are taken about the same axis, the resulting group is the abelian circle group and does not have the property required in step 1.

An alternative arithmetic proof of the existence of free groups in some special orthogonal groups using integral quaternions leads to paradoxical decompositions of the rotation group.