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Let $X=(x_1, \ldots, x_n)$ be an $n$-tuple of elements of a given group $G$. Then two $n$-tuples $X$ and $Y$ are Nielsen equivalent if there exists an automorphism of the free group on $n$-generators, $\phi\in \operatorname{Aut}(F_n)$, such that $X\phi=(x_1\phi, \ldots, x_n\phi)=Y$. Also, $X$ and $Y$ are said to lie in the same $T$-system if there exists an automorphism of $G$, $\psi\in \operatorname{Aut}(G)$, such that $X\psi$ is Nielsen equivalent to $Y$.

It is (relatively) well-known that both "Nielsen equivalence" and "lying in the same $T$-system" are equivalence relations.

I am interested in what is known about generating $n$-tuples of one-relator groups, so groups of the form $\langle x_1, \ldots, x_n; R^m \rangle$ where $m \geq 1$ and $R \neq S^i$ for any words $S\in F(X)$ and $i > 1$ (so $R$ is not a proper power of any element of the free group).

Now, I know that if $n = 2$ and $m > 1$ then there is only one Nielsen equivalence class. This is in a paper published in 1977 by S. J. Pride (The isomorphism problem for two-generator one-relator groups is solvable, Trans. Amer. Math. Soc. 227 (1977) 109-139) which uses this fact to solve the isomorphism problem for such groups. There is a recent paper published in 2005 by I. Kapovich and P. Schupp (Genericity, the Arzhantseva-Ol'shanskii method and the isomorphism problem for one-relator groups, Math. Ann. 331 (2005) 1-19) about what happens for almost all finite $n$ and $m > 1$.

However, I was wondering what happens if $m = 1$. My question is the following,

Let $G=\langle X; R^m\rangle$, $|X|=2$, $m=1$. What can be said about the number of Nielsen equivalence classes in the $T$-system of $X$? Are there finitely many, or infinitely many? What if $|X|>2$?

Note that it is well-known that if $m > 1$ then these groups are residually finite (whether or not the 185 page proof of this result is true or not is beside the point!) and so generating them should be easy. However, if $m = 1$ then you might end up with non-Hopfian groups, never mind non-residually finite! So, the epimorphism which is surjective but not injective in your group will (if I remember correctly) necessarily move your generating tuple out of its $T$-system. By a result of A. M. Brunner (Transitivity-systems of certain one-relator groups, Springer Lecture Notes in Math. 372 (1974) 131-140), this is certainly what happens with the Baumslag-Solitar group $BS(2, 3) = \langle a,b \, \vert \, ab^2a^{-1} = b^2 \rangle$ and the map $a\mapsto a$, $b\mapsto b^2$. Indeed, this map takes your generating pair to a new $T$-system every time! That is, the generating pairs $(a, b^{2^n})$ lie in different $T$-systems for all $n$. Clearly this means there are infinitely many Nielsen equivalence classes, and so infinitely many $T$-systems. This is why I want to know about what happens in your given $T$-system!

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$X\subset G$ or $X\subset F_n$? – Boris Novikov Nov 8 '11 at 14:57
2  
Perhaps you already know this, but Gerhard Rosenberger has done a lot of work in this area, so you might find something of use among his papers. – James Nov 8 '11 at 16:14
    
@Boris Novikov: $X$ is an $n$-tuple of words in the free group $F_n$. – ADL Nov 8 '11 at 17:40
    
@ADL: About the epimorphism which "necessarily moves your generating tuple out of its $T$-system", I am interested in a reference/proof sketch. – Luc Guyot Jun 15 at 22:57
    
How can it be beside the point of something that its proof be correct or not? :-| – Mariano Suárez-Alvarez Jun 16 at 19:12

The solvable Baumslag-Solitar group $BS(1, n) = \langle a, b \, \vert \, aba^{-1} = b^n \rangle$ with $n \in \mathbb{Z} \setminus \{0\}$, has only one $T$-system of generating pairs and any number of Nielsen classes, including infinity, can be achieved for a suitable choice of $n$. This preprint presents of proof of my claim, see Corollary E.

The proof is elementary and can be summarized as follows. First identify $G = BS(1, n)$ with $\mathbb{Z}[\frac{1}{n}] \rtimes_n \mathbb{Z}$ where the canonical generator of $\mathbb{Z}$ acts as the multiplication by $n$ on $R = \mathbb{Z}[\frac{1}{n}]$. Then the commutator of a generating pair of $G$ is of the form $(1 - n)u$ for some unit $u$ of $R$ and every unit arises in this way. By Higman's lemma, if two generating pairs $\mathbf{g}_1$ and $\mathbf{g}_2$ of $G$ are Nielsen equivalent then the corresponding units $u_1$ and $u_2$ are related by an element of the form $(-1)^kn^l$ with $k, l \in \mathbb{Z}$. It is not difficult to show that every generating pair of $G$ can be Nielsen reduced to $(a, b')$ for some $b'$ that identifies with a unit of $R$. Now it is immediate to see that if $u_1$ and $u_2$ are related by an element of $(-1)^{\mathbb{Z}}n^{\mathbb{Z}}$ then $\mathbf{g}_1$ and $\mathbf{g}_2$ are Nielsen equivalent. So $G$ has as many Nielsen classes of generating pairs as elements in $R^{\times}/(-1)^{\mathbb{Z}}n^{\mathbb{Z}}$. A careful inspection of $R^{\times}$ shows that the cardinality of the latter is infinite if $n = 6$ and equal to $d \in \mathbb{N} \setminus \{0\}$ if $n = 2^d$. As the group $G$ has only one $T$-system of generating pairs by [Theorem 2.4, 4], it answers OP's question with residually finite instances.

There are some classical examples which can provide insight on OP's question. The fundamental group $\pi_1(\Sigma_g)$ of a closed surface $\Sigma_g$ of genus $g > 0$ is long known to have a single Nielsen equivalence class of generating $2g$-tuples, see [2, 7]. The family of one-relator groups $T(m, n) = \langle a, b \, \vert \, a^m = b^n \rangle$ for $m,n \in \mathbb{Z}$ has been also extensively studied. It follows from [1] and [5] that each $T$-system of generating pairs of $T(m, n)$ contains a unique Nielsen equivalence class (see also [3] for the specific case of the trefoil group). There are infinitely many such $T$-systems if $\vert m \vert,\vert n \vert \ge 2 $ and $\vert m \vert + \vert n \vert > 4$. But all these examples are again residually finite.

A. M. Brunner proved in [4] that for each $n \ge 0$ the generating pair $(a, b^{2^n})$ of the non-Hopfian Baumslag-Solitar group $BS(2, 3) = \langle a, b \, \vert \, ab^2a^{-1} = b^3 \rangle$ lies in a different $T$-system of $BS(2, 3)$ but he left open the question whether each $T$-system has a representative of this form. By [Theorem B, 7], every automorphism of $BS(2, 3)$ is inner except the automorphism induced by $(a, b) \mapsto (a, b^{-1})$. Therefore each $T$-system contains at most two Nielsen equivalence classes and the $T$-system of $(a, b^{2^n})$ contains a single Nielsen equivalence class. More generally, if $p, q > 1$ and neither divides the other, then $\Gamma = \operatorname{Out}(BS(p, q))$ is isomorphic to the dihedral group of order $d = 2\vert p - q \vert$ generated by the images of $(a, b) \mapsto (ab, b)$ and $(a, b) \mapsto (a, b^{-1})$. As a result every $T$-system of generating pairs contains at most $d$ Nielsen equivalence classes and the $T$-system of $(a, b)$ contains a single Nielsen equivalence class. If $p$ properly divides $q$, then $\Gamma$ is not finitely generated [6]. But also in this case, inspection the automorphism group shows that the $T$-system of $(a, b)$ contains a single Nielsen equivalence class.


[1] "Über die gruppen $A^{a}B^{b}=1$", O. Schreier, 1923.
[2] "Über die Nielsensche Kürzungsmethode in freien Produkten mit Amalgam", H. Zieschang, 1970.
[3] "Presentations of the trefoil group", M. J. Dunwwoody, A. Pietrowsky, 1973.
[4] "Transitivity systems of some one-relator groups", A. M. Brunner, 1974.
[5] "Generators of the free product with amalgamation of two infinite cyclic groups", H. Zieschang, 1977.
[6] "Automorphisms and Hopficity of certain Baumslag-Solitar groups", D. J. Collins and F. Levin, 1983.
[7] "Tree actions of automorphism groups", N. D. Gilbert et al., 2000.
[8] "Nielsen equivalence in closed surface groups", L. Louder, 2010, arXiv:1009.0454.

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could you be a little more precise than "anything can happen" in the 1st paragraph? – YCor Jun 14 at 16:03
    
@YCor: thanks for your interest and diligence. By "anything" I meant that any number of Nielsen classes, including infinity, can be achieved for a suitable choice of the parameter $n$. I edited my answer accordingly. The sketched proof should also make this point clearer. – Luc Guyot Jun 14 at 18:08

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