Infinite group G is polycyclic then subgroup of fitting is nilpotent.

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.

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