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!