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Let $F$ be a free profinite group of rank $\aleph_0$, and let $d \in \mathbb{N}$.

Let $N_d \lhd_c F$ be the intersection of all open normal subgroups $L \lhd_o F$ for which $F/L$ can be generated by at most $d$ elements.

Can it be that $N_d = \{1\}$?

The equivalent question in finite groups asks whether every finite group is an image of a finite subdirect product of groups generated by at most $d$ elements:

Can it be that for every finite group $G$ there exists some $r \in \mathbb{N}$ and a finite group $H$ with normal subgroups $H_1, \dots, H_r \lhd H$ such that:

  1. $ \bigcap_{i=1}^{r} H_i = \{1\}$

  2. For each $1 \leq i \leq r$, $H/H_i$ can be generated by at most $d$ elements.

  3. $G$ Is a homomorphic image of $H$.

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    $\begingroup$ The answer to the analogous question in the discrete setting is 'yes'. For every finite subset $X\subseteq F_\infty$, there is a 2-generator finite group $Q$ and an epimorphism $q:F_\infty\to Q$ so that $q$ is injective on $X$. The profinite version may follow, but it's not completely clear to me right now. $\endgroup$ – HJRW Dec 3 '14 at 20:34
  • $\begingroup$ @HJRW Yes, for discrete groups this is true because simple groups are 2-generated but do not satisfy any law. For profinite groups the answer seems to be negative however (maybe I will write an answer when I will have the details). The reason for this lies in the following compactness argument which works for profinite groups but fails for abstract groups : If the intersection is trivial, then for any open subgroup $U \lhd_o F$ there is a FINITE intersection which is contained in $U$ (this also explains the equivalence with the formulation in finite groups). $\endgroup$ – Pablo Dec 3 '14 at 22:25
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    $\begingroup$ You don't need to appeal to the classification for the discrete case. It follows because $F_{n+1}$ is (fully) residually $F_n$, for any $n>1$. $\endgroup$ – HJRW Dec 4 '14 at 6:55
  • $\begingroup$ Yes, I believe I was mistaken in my thinking. Comment deleted. $\endgroup$ – Geoff Robinson Dec 4 '14 at 10:28
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    $\begingroup$ The one in the title would seem to indicate you just need to separate a word from 1 in a not necessarily onto homomorphism to a 2-generated finite group, which you can do since symmetric groups are 2-generated. $\endgroup$ – Benjamin Steinberg Dec 5 '14 at 2:44

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