I'm reading the third chapter of Profinite Groups by Ribes-Zalesskii. We can find many analogies between abstract free groups and free pro-$\mathcal{C}$ groups in the book- but I think I'm missing or misunderstanding something.

Let $G$ be a finitely generated free profinite group of rank $n$. We further assume $S = \{g_1,\cdots,g_n\}$ is a topological generating set of $G$ and an (abstract) generating set of an abstract free subgroup $F_n \le G$.

Can we find an element $g\in G$ with $g \notin \langle S\rangle$ such that $S\cup \{g\}$ generates another abstract free group of rank $n+1$?

If $n=1$, the case is trivial since $G$ is abelian: $G$ is a quotient of 1-generator free profinite group. We may assume $n>1$.

Edited: I tried but couldn't prove or disprove it. If there exists such a $g\in G$, there is an epimorphism $f:H\to G$ where $H$ is a free profinite group of rank $n+1$ generated by $\{h_1, \cdots, h_n, h\}$ with $f(h_i)=g_i, f(h)=g$. Because a closed subgroup $H' := \overline{\langle h_1, \cdots, h_n\rangle}$ is free profinite of rank $n$ and the restriction $f|_{H'}:H'\to G$ is also an epimorphism, $f|_{H'}$ is an isomorphism. Let $h' := f|_{H'}^{-1}(g) \in H$. We have $f(h')=g=f(h)$. hmmm.. how can I advance?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.