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Given a finite group $G$ that is not abelian, nilpotent, or solvable, what is the smallest normal subgroup $H$ in each case such that $G/H$ is abelian, nilpotent, or solvable (respectively)?

In the abelian case it seems clear that the correct subgroup is the commutator subgroup. But what of the other two? Perhaps I'm just shaky with the concepts (of nilpotency and solvability) but I'm having trouble finding the right way to tackle this problem.

Edit: Well I've tried the most obvious thing and it worked out.

For the nilpotent case I took the intersection of all the subgroups in the lower central series, and for the solvable case I took the intersection of all subgroups in the derived/commutator series.

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In many cases the answer's trivial: if $\;G\;$ is simple non-abelian then the minimal normal such sbgp. is $\;G\;$ , of course. Pretty boring. –  DonAntonio Nov 26 '13 at 5:38
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In many cases it isn't, though. One gets some very different smallest normal subgroups of $D_{2n}$ making the quotient nilpotent, depending on the order of $n$. –  Saigyouji Nov 26 '13 at 6:32
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Your answers are correct when $G$ is finite. When $G$ is infinite, it could happen that the lower central series and/or the derived series are non-terminating (this happens in noncyclic free groups, for example), in which case there is no smallest normal subgroup with the required property. For fixed $k>0$, there is always a smallest normal subgroup such that the quotient is nilpotent of class at most $k$ or solvable with derived length at most $k$. –  Derek Holt Nov 26 '13 at 8:27

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