Let $F$ be a free pro-$p$ group (for a prime number $p$) on a finite set $X$, $\Phi$ the abstract subgroup generated by $X$, $\{1\} \neq N \lhd_c F$. Is it possible that $\Phi \cap N = \{1\}$?
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1$\begingroup$ For $\#(X)\ge 2$, yes, it's an exercise: pick any pro-$p$-group $G$ with a dense free subgroup $\Phi$ (e.g. a congruence subgroup in $SL_2(\mathbf{Z}_p)$), then if $N$ is the kernel of the natural map $F\to G$, then $N\cap\Phi=\{1\}$. $\endgroup$– YCorCommented Sep 21, 2014 at 11:14
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$\begingroup$ @YCor: In your example, the group $\Phi$ is finitely generated, right? $\endgroup$– PabloCommented Sep 21, 2014 at 11:16
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1$\begingroup$ $\Phi$ is free over $X$ which is any finite set of cardinal $\ge 2$. $\endgroup$– YCorCommented Sep 21, 2014 at 11:17
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$\begingroup$ @YCor: Do you think that there is a chance to find another basis $Y$ for $F$ such that the abstract subgroup it generates intersects $N$ nontrivially? $\endgroup$– PabloCommented Sep 21, 2014 at 11:19
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1$\begingroup$ Yes: for instance any element in the kernel of the canonical morphism from the free profinite group over $X$ to the free pro-$p$-group over $X$. $\endgroup$– YCorCommented Sep 21, 2014 at 17:24
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