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Good morning,

I am having a hard time trying to describe the commutator subgroup of a surface group. Namely, if $S$ is a compact orientable surface and $G$ its fundamental subgroup, let's recall that $G$ is given by the following presentation $ \langle a_1, b_1, ..., a_g, b_g \ | \ \prod_{1\leq i \leq g}{[a_i,bi]} =1 \rangle$. Its commutator subgroup is an infinite rank free group. Can one give a set of generator of this subgroup ? Or even better, could one make such a set free ?

I would be grateful for any hint or reference !

Selim

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Just a guess for the genus 2 surface: Take a simple loop $c$ which separates $S$ in two one-holed tori. Then $c$ is clearly in the commutator subgroup of $G$. Now, consider the orbit of $c$ under the entire group $Aut(G)$. The result, I think, is a generating set for the commutator subgroup of $G$. –  studiosus Nov 22 '13 at 10:53
    
What you say is right and can be extended in the general case, since $G$ is a free group, $Aut(G)$ acts transitively on free set of generator. However, I'm searching for an explicit description of a set of generator, which could be found using your method if one is able to describe $Aut(G)$ or the orbit of an element through its action. –  Selim Ghazouani Nov 22 '13 at 11:01
    
See this question math.stackexchange.com/questions/184176/… for the answer in the case of commutator subgroup of the free group. The same method should work for surface groups. –  studiosus Nov 22 '13 at 12:27
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@SelimGhazouani, $\;G\;$ as given by you can't be a free group since it is defined by means of non-trivial relations. –  DonAntonio Nov 22 '13 at 22:44
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