Timeline for Explicit element in free group which is killed by every solvable quotient

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May 7, 2012 at 10:57	comment	added	Lior Bary-Soroker		One can reduce the question to non-finitely generated free profinite groups, since the commutator subgroup of the free profinite group on 2 generators is free of countable rank.
May 7, 2012 at 10:46	comment	added	YCor		I agree this kernel is nontrivial. But writing down an explicit nontrivial element in this kernel is certainly doable, but doesn't seem too immediate. [I really mean explicit, not only finding a sequence for which some limit points are nontrivial elements in the kernel.]
May 7, 2012 at 9:48	comment	added	Lior Bary-Soroker		You are right, it is in the kernel on the pronilpotent completion. To check the the kernel $K$ onto the maximal prosolvable quotient is non trivial, just take $U$ open normal such that $\widehat{F_2}^{profinite}/U \cong A_5$. This will assure you that $KU=\widehat{F_2}^{profinite}$, hence by iso-2, $K/K\cap U=A_5$. To find an explicit element in $K$ shouldn't be too difficult, I think.
May 6, 2012 at 22:18	comment	added	YCor		This question is interesting. However, in your example I don't see how (with which choice of exponents) you can arrange this word to be converging to a nontrivial element. [by the way, any limit point of a sequence as you construct is in the kernel of the map to the pronilpotent, not prosolvable, completion].
May 6, 2012 at 21:20	history	answered	Lior Bary-Soroker	CC BY-SA 3.0