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This must have been known to the ancients, but I am having some trouble finding the references: what can be said (especially geometrically) about the normal closure of an element in a surface group? Especially an elliptic element (in an orbifold group)...

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@Igor: what do you want to know about these subgroups? They are typically free products of cyclic groups generated by the conjugates of the element. –  Mark Sapir Apr 9 '11 at 21:49
    
@Mark: what you say is already (a) interesting and (b) not obvious to me. Why is that true? For a torsion-free surface group, since the abelianization is 1-dimensional, the normal closure is of infinite index, hence infinitely generated free, but it is not clear what the canonical generators are (are all the $a g a^{-1}$ the free generators?) and with torsion that is less clear. Plus, if you look at the quotient by the normal closure, do you get something interesting? –  Igor Rivin Apr 10 '11 at 0:35
    
@Igor: By a theorem of Delzant for every element $g$ of a torsion-free hyperbolic group $G$ the normal closure of some power $g^n$ is freely generated by conjugates $hg^nh^{-1}$. I think that in the case of surface groups $n$ should be 1. Every torsion-free hyperbolic group is SQ-universal by Olshanskii. The proof implies that every 1-related group embeds into the quotient you described. In fact if you take "any random" free subgroup $F$ of the surface group $S$, and every element $r$ in $F$, then $F/\ll r\gg$ naturally embeds into $S/\ll r\gg$. –  Mark Sapir Apr 10 '11 at 1:22
    
@Igor: Delzant, Thomas(F-STRAS-I) Sous-groupes distingués et quotients des groupes hyperboliques. (French) [Distinguished subgroups and quotients of hyperbolic groups] Duke Math. J. 83 (1996), no. 3, 661–682. - the paper by Delzant. –  Mark Sapir Apr 10 '11 at 2:09
    
@Mark: Thanks for the reference! I looked briefly at the paper, and if I understand correctly, the family of elements $h g^n h^{-1}$ always satisfies the small cancelation condition for $n$ large enough. That's not quite satisfying (since one can always get this kind of result by ping-pong in groups which are themselves very far from free (eg, lattices in higher rank lie groups). If you can actually get $n=1,$ that's a different matter entirely.... –  Igor Rivin Apr 10 '11 at 2:42
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1 Answer

This doesn't say much about the elliptic case, but there is Hempel's theorem that the normal closure of an element $\alpha$ contains a (power of a) simple curve $\beta$ if and only $\alpha$ is a (power of a) simple curve. Furthermore, any such $\beta$ is either a power of $\alpha$ or a commutator of $\alpha$ with a simple loop $\gamma$ intersecting $\alpha$ in one point.

You should be able to glean some structural information about the normal closure from his paper.

John Hempel, One-relator surface groups, Mathematical Proceedings of the Cambridge Philosophical Society (1990), 108: 467-474

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@Richard: thanks, I saw that paper (and there is a sequel by Jim Howie), but I am not quite sure yet what to make of it... –  Igor Rivin Apr 10 '11 at 16:38
    
Sure thing. If you're interested in this kind of topological statement, there is also a theorem of Maskit that the covering corresponding to $\langle \langle \alpha \rangle \rangle$ is planar. –  Richard Kent Apr 10 '11 at 18:22
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