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 $n\geq 0$.

  • 2
    $\begingroup$ Please make your question as self-contained as possible. Right now it is very hard to understand what you are asking. $\endgroup$ – Alex Provost Mar 6 '19 at 16:25
  • $\begingroup$ 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. $\endgroup$ – Derek Holt Mar 6 '19 at 17:02
  • $\begingroup$ 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 $\endgroup$ – 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.

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