Let $F$ be a finitely generated free group. Consider $R\subset F$ a finite set of relations and denote with $G$ the quotient group of $F$ by the normal closure of $R$.

Now suppose we are given another set of generator $g_1,\ldots,g_k$ of $G$.

The question is the following: is it possible to find $y_1,\ldots,y_k\in F$ such that

  1. the image of $y_i$ in G is $g_i$ for every $1\leq i\leq k$,

  2. the set $\{y_1,\ldots,y_k,R\}$ generates $F$ ?

If it is not the case, is it possible if we suppose that $G$ is free?

I'm able to answer the question only in one case: if $R=\{r\}$ and $G$ is free, then $r$ is an element of a base of $F$ so the answer is yes. I learned the proof of this fact in Lyndon-Schupp's book and it uses a kind of Freiheitssatz, so it doesn't help too much for the general case.


Suppose that $G=\{1\}$ and its generating set is empty. Then you are asking if an arbitrary finite subset $R\subset F$ whose normal closure is $F$, in fact generates $F$: This is clearly false. Namely, start with an arbitrary finite generating set $S$ of $F$. Now, apply conjugation by suitable elements $g_j$ of $F$ to the generators $s_j\in S, j=1,...,n$, so that the conjugate elements $s_j'=g_j s_j g_j^{-1}$ have axes (in the standard Cayley graph of $F$) which are sufficiently far from each other (I think, disjoint is enough). Then the ping-pong argument will show that the set $R=\{s_j', j=1,...,n\}$ generates an infinite index subgroup of $F$. On the other hand, the normal closures of $R$ and $S$ are the same.

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    $\begingroup$ uhm, you are right, the question is quite stupid. $\endgroup$ – Alessandro Carderi Dec 11 '13 at 18:14
  • $\begingroup$ @AlessandroCarderi yes it's a stupid question. You should get banned. Like me. $\endgroup$ – StudySmarterNotHarder Aug 21 '19 at 4:57

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