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What is a simple way to prove that for any compact two-dimensional surface $S$ and an element $g$ in $\mathbb \pi_1(S)$ there exists a finite index normal subgroup $\Gamma\subset \pi_1(S)$ such that $g\notin \Gamma$?

In fact, who was first to prove this statement?

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    $\begingroup$ It is a finitely generated subgroup of invertible matrices with coefficients in an algebraic extension of the rationals? $\endgroup$
    – Charlie Frohman
    Commented Jun 9, 2011 at 16:54
  • $\begingroup$ Certainly, the surface groups are lattices in $SL_2({\mathbb C})$. But that is harder to prove than residual finiteness (I think). $\endgroup$
    – user6976
    Commented Jun 9, 2011 at 17:05
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    $\begingroup$ Charlie, thanks for the comment! Unfortunately it is not simple enough for me to understand... You don't suggest to use residual finiteness of linear groups? Also, how do you prove what you wrote? You are more than welcome to develop your comment into a full answer! $\endgroup$
    – aglearner
    Commented Jun 9, 2011 at 17:05
  • $\begingroup$ So the statement is that for Riemann surfaces, the topological group injects into the algebraic fundamental group (the profinite completion of the former). If it is OK to ask questions in comments, I would like to know if this holds more generally for smooth complex algebraic varieties. $\endgroup$
    – Piotr Achinger
    Commented Jun 9, 2011 at 17:31
  • $\begingroup$ For higher-dimensional projective varieties the result is false. In fact, D. Toledo has constructed examples with non-residually finite fundamental group, see archive.numdam.org/article/PMIHES_1993__77__103_0.pdf $\endgroup$
    – Francesco Polizzi
    Commented Jun 9, 2011 at 17:36

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See this text. The proof is just a few lines.

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  • $\begingroup$ Could you possibly clarify, which homomorphism $\pi_1(F)\to S_3$ is meant in this proof? $\endgroup$
    – mathreader
    Commented Dec 29, 2011 at 4:49

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