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There are known free subgroups of rank 2 in the set of rotations about the origin in $\mathbb{R}^3$, $SO_3$. For instance, the rotations by angle $\arccos \frac {1}{3}$ about the $z$- and $x$-axis generate such a free subgroup.

Are there free subgroups of rank 3 (or higher) in $SO_3$?

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    Sorry, should have said that $\langle a, b\rangle$ etc. is non-cyclic. I think that is sufficient...2011-08-04

2 Answers 2

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Here's a short topological proof that $F_2$ contains the free group on countably many generators. The key is that

the classifying space $S^1 \vee S^1$ of $F_2$ has a covering space which is homotopic to a wedge of countably many circles

and this space has fundamental group free on countably many generators by Seifert-van Kampen. The relevant covering space consists of a circle attached to every integer point on $\mathbb{R}$, where the covering map sends the edges between consecutive integers to one loop $y$ in $S^1 \vee S^1$ and sends the circles to the other loop $x$. The fundamental group of this covering space injects into $F_2$, and in fact it is freely generated by elements of the form $y^{-n} x y^n$, as can be seen from the contraction which takes all of $\mathbb{R}$ to a point.

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    @Qiaochu, could you please explain how your result allows to embed $F_3$ in $SO_3$??2014-01-10
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Yes. As soon as you've got a rank 2 free group ($F_{2}$), you've got any higher (countable) rank free group, since the free group of rank 2 contains (free) subgroups of all countable ranks. For instance, the derived subgroup of $F_{2}$ is free of countably infinite rank.