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Fix an integer $n>0$. Are there infinite subgroups of $SL_2(\mathbb{C})$ such that every element is $n$-torsion?

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    $\begingroup$ No. Short proof (in $GL_d$ as well): the Zariski closure has the same property; moreover since the identity is isolated among elements of $n$-torsion, it is open and hence by homogeneity all singletons are open. Since the Zariski topology is noetherian, this shows that the group is finite. $\endgroup$
    – YCor
    Aug 16, 2015 at 21:45
  • $\begingroup$ cool sleight-of-hand! this is also helpful to see exactly why characteristic 0 is important: the identity is not isolated among the elements of $p$-torsion in characteristic $p$. $\endgroup$
    – Raju
    Aug 16, 2015 at 22:32

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A theorem of Burnside says that a linear group of finite exponent is finite. So the answer is no.

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  • $\begingroup$ Don't you need them to be finitely generated? It is definitely needed in positive characteristic. $\endgroup$ Aug 16, 2015 at 21:05
  • $\begingroup$ Not in characteristic 0! See Theorem 19A.9 on p. 231 of Graduate Algebra: Noncommutative View (available on google books) $\endgroup$
    – Raju
    Aug 16, 2015 at 22:11
  • $\begingroup$ Raju, maybe I was not clear. My point is that Ehud's answer is imprecise. Either you require that the characteristic is zero or that the group is finitely generated. You cannot remove both conditions. However, in characteristic zero you can generalize in a slightly different way: if you require that the group is finitely generated and torsion (but not necessarily of finite exponent), then it is still finite. $\endgroup$ Aug 17, 2015 at 19:11
  • $\begingroup$ You right, I should have written: a group which is linear over a field of characteristic zero. $\endgroup$
    – Ehud Meir
    Aug 18, 2015 at 0:52

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