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Is there a free subgroup of rank 3 in $SO_3$?

How is this possible?

Here are some facts:

  1. $F_2 \subset F_3$

  2. $F_2 \not\cong F_3$

Fact 2 implies that $F_3$ is isomorphic to a proper subgroup of $F_2$, i.e. one in which there are exist some relations between the generators of $F_2$.

But I'm really struggling to come up with anything else.

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  • Does it help if I mention that $F_2$ in fact contains a subgroup isomorphic to the free group on $\aleph_0$ generators? Use the universal property of $F_3$ to show that you can construct a suitable homomorphism $F_3 \to F_2$, and then use the characterisation of $F_2$ in terms of words to show that the homomorphism is injective. – Zhen Lin Dec 09 '11 at 22:05
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    In fact $F_2$ contains the free group on countably many generators. See the linked question. It is not true that there exist any relations between the generators of $F_2$ in a proper subgroup of $F_2$, and I don't really understand how this could be (since $F_2$ is, by definition, free). Perhaps you are confusing subgroups with quotient groups. – Qiaochu Yuan Dec 09 '11 at 22:05

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