Let G be a infinite polycyclic group i.e soluble and satisfy max. Show that the subgroup of fitting of G is nilpotent. he subgroup of fitting of G is subgroup of G generated by normal nilpotent subgroup of G.

  • $\begingroup$ the subgroup of generated by normal nilpotent subgroup $\endgroup$ – amel Jan 20 '15 at 9:52
  • $\begingroup$ This is a question in advanced group theory. I'd expect some real effort/work has already been done here. $\endgroup$ – Timbuc Jan 20 '15 at 9:53
  • $\begingroup$ $$Fit(G):=\langle\; N\lhd G\;:\;\;N\;\;\text{is nilpotent}\;\rangle$$ $\endgroup$ – Timbuc Jan 20 '15 at 9:54
  • $\begingroup$ Do you know how to show this when there are only finitely many normal nilpotent subgroups? $\endgroup$ – Tobias Kildetoft Jan 20 '15 at 10:00
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    $\begingroup$ Then let $N_i$ for $i\in I$ be the normal nilpotent subgroups of $G$ (let's say that $I$ contains the natural numbers for simplicity). Consider the chain $N_1\leq N_1N_2\leq \cdots$ $\endgroup$ – Tobias Kildetoft Jan 20 '15 at 10:33

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