Let $F$ be a nonabelian free profinite group, $H \leq_c F$ finitely generated with $[F:H] = \infty$. Must there be some $\{1\} \neq N \lhd_c F$ such that $N \cap H = \{1\}$?
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2$\begingroup$ You first need to assume that $H$ has infinite index. $\endgroup$– YCorCommented Nov 2, 2014 at 21:27
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$\begingroup$ @YCor I just forgot it, edited now. $\endgroup$– PabloCommented Nov 3, 2014 at 5:33
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