Are there any known results of the following type?
For $G=SO(3)$, does it contain a free subgroup $H=F_2$, the free group generated by two letters, such that $\forall g\in G, g^2\in H$ implies $g^2=e,$ or $g\in H$?
Rk: see a related question here
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What do you really mean? – mesel Mar 03 '14 at 14:54
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If you do not require $G$ to be free group,you can set $G=HxZ_2$. – mesel Mar 03 '14 at 15:03
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@mesel, you are right, I should directly say $G=SO(3)$ and ask whether such free subgroup exists or not. – ougao Mar 03 '14 at 15:14
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First of all, the orthogonal group SO(3) does contain free subgroups of rank 2. (Read about Banach-Tarski paradox to find why.) However, none of them has the property you want, since you can take roots of any order of any element in the special orthogonal group (as it has surjective exponential map, see here). Now, apply this property to a free generator of your free group embedded in SO(3).
Moishe Kohan
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That's because an element of the free basis does not admit nontrivial roots in a free group. – Moishe Kohan Mar 05 '14 at 02:50