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Mar 30, 2016 at 20:52 comment added Uri Bader @Andreas a conceptual (though not terribly elementary) way to see that $SO(3)$ contains a free group is by the following argument. First note that it is enough to show that $SU(2)$ contains one. To see that $SU(2)$ has one, by a Baire category argument, it is enough to see that for each word in $F_2$ there is a pair in $SU(2)$ which does not satisfy it. As this is an algebraic question, it is enough to show it for the complexification, $SL_2(\mathbb{C})$. But this is obvious, as it contains $SL_2(\mathbb{Z})$, which contains a free group...
Nov 28, 2010 at 13:26 comment added S. Carnahan Two such elements are given in this blog post: sbseminar.wordpress.com/2007/09/17/…
Nov 28, 2010 at 13:26 comment added Andreas Thom Right, the only non-trivial part of showing that there exists at least one pair of rotations, which generate a free group, was proved first by Felix Hausdorff.
Nov 28, 2010 at 13:25 vote accept Marcin Kotowski
Nov 28, 2010 at 13:22 history answered Qiaochu Yuan CC BY-SA 2.5