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$G$ is a surface group of genus $g\geq 2$ (the fundamental group of closed orientable surface of genus g). $F$ is a free group of rank $2g$ with basis $\{x_1,\dots,x_{2g}\}$. $\phi$ is a surjective homomorphism from $F$ to $G$ such that $G$ is generated by $\phi(x_1),\dots,\phi(x_{2g})$.

  1. What can we say about the kernel of $\phi$?

  2. Is it always the normal closure of an element $r$ in $F$?

  3. Can we write $r$ explicitly if it is?

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    $\begingroup$ How can you have an endomorphism from $F$ to $G$? You must mean either "homomorphism" or "epimorphism". $\endgroup$ – Jim Belk Nov 20 '13 at 4:54
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    $\begingroup$ As an aside, I don't at all understand the votes to close this question. It seems like a good question to me. $\endgroup$ – Jim Belk Nov 20 '13 at 5:00
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    $\begingroup$ This is an intersting question, possibly more suitable for MathOverflow. So why are there three votes to close? $\endgroup$ – Derek Holt Nov 20 '13 at 8:48
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    $\begingroup$ @DerekHolt I disagree that this is more suitable for MathOverflow. It is essentially just an application of Seifert-van Kampen (in fact, I just checked and this is the example Wikipedia uses! See also here). On the other hand, if you interpret 2 as saying "Is it always the normal closure of an element $r$..." then things become interesting (I don't think this is the interpretation the OP was going for). $\endgroup$ – user1729 Nov 20 '13 at 9:48
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    $\begingroup$ Interestingly, the alternative interpretation has a positive answer, which is due to a recent paper of Lars Louder. He proves that the generating tuple $(\phi(x_1), \ldots, \phi(x_{2g})$ is Nielsen equivalent to the "standard" generating tuple $(y_1, \ldots, y_{2g})$, from which this result follows. $\endgroup$ – user1729 Nov 20 '13 at 9:51
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As I said in the comments above, this has a positive solution due to a recent preprint of Lars Louder, which can be found here. The paper proves that surface groups have a "single Nielsen equivalence class of generating $2g$-tuples". I'll explain what this means, and then I will explain why this solves the problem. (In the comments below, Lars Louder has pointed out that the quoted result is due to Zieschang for genus $g\neq 3$, and that his own results are much more general.)

Let $(x_1, x_2, \ldots, x_n)$ be a generating tuple for a group $G$. Then clearly all permutations of this tuple are generating tuples for $G$, as are $(x_1^{-1}, x_2, \ldots, x_n)$ and $(x_1x_2, x_2, \ldots, x_n)$. These alterations correspond to Nielsen transformations, or equivalently, to automorphisms of the free group $F(x_1, \ldots, x_n)$. Indeed the Nielsen transformations I described above generate all Nielsen transformations, that is, if you repeat the above alterations then you obtain all of the "obvious" generating tuples for $G$, for example if $n=3$ then $(x_1^{-1}, x_2x_1x_3, x_2x_1)$ will be obtained, and so on. If two generating $n$-tuples can be obtained one from the other in this way then they are said to be Nielsen equivalent. They partition the generating tuples into Nielsen equivalence classes.

Fix $G=\langle y_1, z_1, \ldots, y_g, z_g; [y_1, z_1]\cdots [y_g, z_g]\rangle$. What Lars Louder have proven is that if $\mathcal{L}_1=(x_1, x_2, \ldots, x_{2g})$ generates $G$ (note that each $x_i$ denotes a word over $y_{\ast}$ and $z_{\ast}$) then $\mathcal{L}$ can be obtained from the standard generating tuple $\mathcal{L}_0=(y_1, z_1, \ldots, y_g, z_g)$. It is then an easy exercise to see that when you go from the generating tuple $\mathcal{L}_0$ to $\mathcal{L}_1$ via your Nielsen transformation $\phi$, so $\phi(y_1)=x_1$, $\phi(z_1)=x_2$, etc. you obtain the following presentation of $G$. $$\langle y_1, z_1, \ldots, y_{g}, z_g; [x_1, x_2]\cdots [x_{2g-1}, x_{2g}]\rangle$$ Recall that each $x_i$ denotes a word over $y_{\ast}$ and $z_{\ast}$, so this presentation makes sense.

There are some old results in a similar vein from the 70s. For example, Steve Pride proved that one-ended two-generator, one-relator groups with torsion have a single Nielsen equivalence class of generating pairs (and this solves the isomorphism problem for such groups), while at around the same time Brunner proved that the Baumslag-Solitar group $BS(2, 3)=\langle a, t; t^{-1}a^3t=a^3\rangle$ has infinitely many Nielsen equivalence classes of generating pairs. In fact, Brunner proves something stronger. Two generating tuples $(x_1, \ldots, x_n)$ and $(y_1, \ldots, y_n)$ of a group $G$ are said to lie in the same T-system is there exists an automorphism $\phi$ of $G$ such that $(\phi(x_1), \ldots, \phi(x_n))$ is Nielsen equivalent to $(y_1, \ldots, y_n)$. Brunner proves that $BS(2, 3)$ has infinitely many T-systems. (The "T" stands for transitivity, if I recall correctly.)

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  • $\begingroup$ Good answer! In Lyndon & Schupp's book "combinatorials group theory" page 93, they stated $BS(2,3)=<a,t;t^{-1}a^2t=a^3>$ has no presentation with two generators $t, a^4$ and only a single relator. Do you know how many T-systems the Fuchsian type group has? $\endgroup$ – J.C. Wu Nov 21 '13 at 11:07
  • $\begingroup$ Lyndon and Schupp's example is -essentially- the first step in Brunner's proof. In fact, if you turn to p96 they briefly discuss Brunner's work. What do you mean by "the Fuchsian type group"? (I suspect my answer will be "no, I do not know" though.) $\endgroup$ – user1729 Nov 21 '13 at 11:23
  • $\begingroup$ Thank you let me turn to page 96. I see Ziechang had established the analogous result for Fuchsian type group (line 2). $\endgroup$ – J.C. Wu Nov 21 '13 at 11:33
  • $\begingroup$ Careful though - the analogous result is not saying that these groups have one Nielsen equivalence class. Rather, it is saying that automorphisms of the group when viewed in the given presentation keep the given generating tuple in a fixed Nielsen equivalence class. $\endgroup$ – user1729 Nov 21 '13 at 11:55
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    $\begingroup$ I don't have any reputation points so I can't comment on user1729's answer, but I'd like to note that, as stated, the theorem he attributes to me is originally due to Zieschang, but only for $g\neq 3$. The theorem I proved is stronger in that there is no restriction on the genus, orientability, or the rank of the free group, provided there is a surjective map $F\to G$, i.e., a generating tuple $(y_1,\dotsc,y_n)$ is equivalent to $(x_1,\dotsc,x_{2g},1,\dotsc,1)$ (in the oriented case). $\endgroup$ – user129933 Feb 19 '14 at 11:13

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