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Let G infinite group.

if G is supersoluble group then $[G,G]$ is nilpotent.

a group G is called supersolvable (or supersoluble) if it has an invariant normal series whose factors are all cyclic. Since a normal series has finite length by definition.

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    $\begingroup$ Why are assuming that $G$ is infinite? It is unnecessary. $\endgroup$ – Derek Holt Feb 25 '15 at 15:24
  • $\begingroup$ yes of course ,just because i always work with infinite group $\endgroup$ – amel Feb 25 '15 at 16:03
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This is Theorem $4.20$ in Keith Conrad's notes Subgroup Series II, where a detailed proof is given.

Rekark: One has to be careful not to use a wrong argument, which might come to mind: since a supersolvable group is clearly solvable, it should have nilpotent commutator subgroup. This is false, with counterexample $G=S_4$, see If a finite group $G$ is solvable, is $[G,G]$ nilpotent?

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