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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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10  
Sure: a free group on two generators contains subgroups of arbitrary (up to countable) rank. –  t.b. Aug 4 '11 at 16:17
    
To expand on Theo's comment - every subgroup of a free group is free. So take three elements $a$, $b$ and $c$, such that $a\not\in\langle b, c\rangle$, etc, and you'll be done. So, for example, take $x$, $y^{-1}xy$ and $y^{-2}xy^2$. Also, $[F_2, F_2]\cong F_{\infty}$. –  user1729 Aug 4 '11 at 16:21
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@Swlabr: It's not quite that easy. For example, if you start with a free group in $x$ and $y$, and you take $b=x^2$, $c=y^2$, then $\langle b,c\rangle$ is free of rank $2$, $a=x\notin\langle b,c\rangle$, but $\langle a,b,c\rangle = \langle a,c\rangle$ is still free of rank $2$. Simply not being in the subgroup is not enough. –  Arturo Magidin Aug 4 '11 at 16:50
    
Sorry, should have said that $\langle a, b\rangle$ etc. is non-cyclic. I think that is sufficient... –  user1729 Aug 4 '11 at 17:38

2 Answers 2

up vote 8 down vote accepted

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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If you haven't seen this kind of topological argument here before, you should prove the Shreier index formula: an index k subgroup of F_n generators is free on k(n-1)+1 generators. –  Noah Snyder Aug 4 '11 at 17:29
    
@Qiaochu, could you please explain how your result allows to embed $F_3$ in $SO_3$?? –  janmarqz Jan 10 at 22:50

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.

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see detail for $F_3\hookrightarrow F_2$ in [juanmarqz.wordpress.com/2014/01/08/… à la Schreier. –  janmarqz Jan 10 at 22:52

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