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I've been trying to attack this problem in many ways, but Couldn't figure out the right answer .

The question is, That a finitely generated subgroup of SO3 when all the elements are of finite order, can not be dense . This is a preceding question to proving that these subgroups are finite (so this statement is forbidden) .

It can be proven alternatively that this subgroup must be closed, but this is also a tough argument .

Raz
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The two questions also can be treated in a different order. We can first solve the second question. Indeed, all finitely generated torsion subgroups of $SO(3)$ are finite, see here:

Finitely generated torsion subgroup of $SO(3,\mathbb{R})$ is finite

Hence such subgroups cannot be dense.

Dietrich Burde
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  • Thanks for the quick answer . Yes this is true, but it kinda loses the whole point of the ex., proving this independently of Schur's Theorem, out of topological (or other) properties – Raz Feb 11 '18 at 18:07
  • Do you have the whole text of this exercise? What are the basic results of the lecture before this exercise? – Dietrich Burde Feb 11 '18 at 19:12
  • No primary results : It is a classical theorem of I. Schur that a finitely generated subgroup of GLn(C) all of whose elements are of finite order (called a periodic, or torsion, group) is a finite group. Challenge (). Prove, independently of Schur's theorem, that a periodic subgroup of SO3(R) cannot be dense. – Raz Feb 11 '18 at 21:13
  • Then use Selberg's theorem. This is very elementary. – Dietrich Burde Feb 11 '18 at 22:56