This question came about when I was naively considering to what extent generating sets for finitely-generated groups are unique.


Let $G$ be a group and let $\phi\_1, \phi\_2 : F\_k \to G$ be two surjective homomorphisms from a free group on $k$ letters $F\_k$. Given a natural number $K \geq k$ let the surjective homomorphisms $\Phi\_1^K, \Phi\_2^K : F\_K \to G$ be the obtained by extending the corresponding maps $\phi\_1$ and $\phi\_2$ to send the $K-k$ other letters in $F\_K$ to the identity.


Does there exist some $K$ such that $\Phi\_1^K \circ \alpha = \Phi\_2^K$ for some automorphism $\alpha$ of $F\_K$?



Stable equivalence of generating sets of a finitely-generated group?
The answer is $K=2k$, which is an exercise. Here is a sketch: Let $f\colon \langle x\_i\rangle\to G$ be a surjective homomorphism, and $h\colon \langle x\_i\rangle\*\langle y\rangle\to G$ the map defined by $h(x\_i)=f(x\_i)$, $h(y)=e$. Let $g\in G$ be arbitrary, and choose $w\in\langle x\_i\rangle$ such that $f(w)=g$. Let $\alpha$ be the automorphism of $\langle x\_i\rangle\*\langle y\rangle$ which maps $y$ to $yw$ and leaves the other generators alone. Then $h(\alpha(y))=g$. Do this $k$ times, one for each element of your second generating tuple, then reverse the process, swapping the roles of the $x$ and $y$.


$K=2k-1$ doesn't suffice, by Kapovich and Weidmann, who in "Nielsen equivalence in a class of random groups" construct small-cancellation examples.


