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.
Asked
Active
Viewed 362 times
1
-
1Why are assuming that $G$ is infinite? It is unnecessary. – Derek Holt Feb 25 '15 at 15:24
-
yes of course ,just because i always work with infinite group – amel Feb 25 '15 at 16:03
1 Answers
1
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?
Dietrich Burde
- 140,055