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Let $G$ be finitely generated soluble by finite group (i.e $G$ has soluble normal subgroup $K$ such that $G/K$ is finite), such that each proper quotient of $G$ is finite by nilpotent by using induction on the derived length of $K$, i wont to prove that $G$ is abelien by (finite by nilpotent).

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  • $\begingroup$ This doesn't make much sense. "We can assume that" only makes sense if you are trying to prove something. So what are you trying to prove? Also $G$ is a quotient of $G$, so $G$ itself is finite by nilpotent by hypothesis. $\endgroup$ – Derek Holt Mar 23 '15 at 14:30
  • $\begingroup$ Now you have "I want to prove that we can prove that...". $\endgroup$ – Derek Holt Mar 23 '15 at 14:32
  • $\begingroup$ sorry ,is it sense know $\endgroup$ – amel Mar 23 '15 at 14:34
  • $\begingroup$ Yes, but it seems very easy. If $K=K^{(1)} > K^{(2)} > \cdots > K^{(k)}=1$ is the derived series of $K$, then $K^{(k-1)}$ is abelian and $G/K^{(k-1)}$ is a proper quotient, and hence is finite by nilpotent. $\endgroup$ – Derek Holt Mar 23 '15 at 14:37
  • $\begingroup$ yes i do this ,but if $G$ be finitely generated soluble by finite group using induction on the derived length of K is G is abelien by (finite by nilpotent)???? $\endgroup$ – amel Mar 23 '15 at 14:40

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