Let $p$ be a prime number, $F$ a free nonabelian finitely generated pro-$p$ group, $L \lhd_o F$ and $Y$ a basis for $L$ with $y \in Y$. Is there a basis $X$ for $F$ such that $y$ is in the abstract subgroup generated by $X$ ?
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1$\begingroup$ Pablo, I don't know how relevant it is to this specific question, but you should look into the theory of profinite and pro-p trees, as developed by Ribes and Zalesskii. (Not in their published book, but in some papers by them and various co-authors, including me.) This technology makes concrete some of the analogies between profinite (or pro-p) free groups and abstract free groups. $\endgroup$– HJRWCommented Sep 21, 2014 at 19:41
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$\begingroup$ @HJRW: Thanks for this advice, I will look at this. By the way, can you give me some references (just for me to know where to start from)? $\endgroup$– PabloCommented Sep 21, 2014 at 20:01
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1$\begingroup$ I wrote a paper with Zalesskii called 'Profinite properties of graph manifolds', which appeared in Geom. Dedicata. You could look at the references in that. $\endgroup$– HJRWCommented Sep 21, 2014 at 21:37
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1$\begingroup$ It seems the answer is no in general. mathoverflow.net/questions/181463/…: according Barnea's answer, at least for odd $p$ there's a counterexample: namely if $N$ is the intersection of kernels of continuous homomorphisms $F\to\mathrm{GL}_2(\mathbf{Z}_p)$, then $N\neq\{1\}$ (Zubkov's Theorem) and for every basis $X$ of $F$, if $\Phi$ is the subgroup generated by $X$, we have $\Phi\cap N=\{1\}$. Zubkov's theorem is probably also true for $p=2$. $\endgroup$– YCorCommented Sep 22, 2014 at 8:39
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1$\begingroup$ @Pablo: Every non-trivial element of a free pro-$p$ group is a basis element of some open subgroup. So you asking whether every element of a pro-p group (the condition free is not essential) belongs to the abstract subgroup generated by a some smallest generating set. $\endgroup$– Andrei JaikinCommented Oct 18, 2014 at 9:35
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