# Why all nilpotent and finitely generated groups are Max?

An exercise asks to prove that if a group $$G$$ is nilpotent and finitely generated then it satisfy Max condition, in other words all non empty and totally ordered, with respect of inclusion $$\subseteq$$, subsets of this set $$\{H:H\text{ is a subgroup of }G\}$$ have a maximum.

A suggestion says to use this result:

Thm.(Mal'cev) If $$Z(G)$$ is torsion free then also $$Z_{i+1}(G)/Z_i(G)$$ is it for every integer $$n\geq 0$$.

• Please make your question as self-contained as possible. Right now it is very hard to understand what you are asking. – Alex Provost Mar 6 '19 at 16:25
• You could prove it by induction on the class of $G$, using the facts that it is true in the case $n=1$ when $G$ is abelian, and all subgroups of a finitely generated nilpotent group are nilpotent. I don't see immediately how the theorem of Mal'cev helps. – Derek Holt Mar 6 '19 at 17:02
• In order to use induction I should prove a normal finitely generated subgroup $N$, so it's Max and also $G/N$ is Max so $G$ Max. All its subgroups are nilpotent – Jihlbert Mar 6 '19 at 17:27

You need to prove that every subgroup is f.g. Proof by induction on nilpotency length $$c$$. If $$c=0$$, then $$G=1$$ and we're done. Suppose $$c\ge 1$$.
Let $$Z$$ be the center of $$G$$. Since $$G/Z$$ has nilpotency length $$\le c-1$$, by induction it has max, and is an iterated extension of finitely generated abelian groups, and hence is finitely presented group. Since $$G$$ is finitely presented, this implies that $$Z$$ is finitely generated as normal subgroup, and hence that $$Z$$ is finitely generated. Since $$G/Z$$ and $$Z$$ satisfy max, so does $$G$$.
I used the lemma that if $$G$$ is f.g., and $$N$$ is a normal subgroup with $$G/N$$ f.p., then $$N$$ is f.g. as normal subgroup. This reduces to the case when $$G$$ is free, and in this case this amounts to the result that being finitely presented does not depend on the choice of generating subset.