Slender nilpotent and every proper subgroup is abelian implies Frattini-in-center

Statement

Suppose $G$ is a Slender nilpotent group (?), i.e., a group that is both a Nilpotent group (?) and a Slender group (?) (every subgroup is finitely generated, or equivalently, every ascending chain of subgroups stabilizes after a finite length). Then, $G$ is a Frattini-in-center group (?), i.e., the commutator subgroup of $G$ is contained in the Frattini subgroup of $G$, which in turn is contained in the center of $G$.

Related facts

• Finite non-abelian and every proper subgroup is abelian implies not simple
• Finite non-abelian and every proper subgroup is abelian implies metabelian
• Classification of finite non-abelian groups in which every proper subgroup is abelian
• Schmidt-Iwasawa theorem

Facts used

1. Nilpotent implies every maximal subgroup is normal
2. Nilpotence is quotient-closed

Proof

Given: A slender nilpotent group $G$ such that every proper subgroup of $G$ is abelian.

To prove: $[G,G]\leq \Phi (G)\leq Z(G)$.

Proof: If $G$ is abelian, we are done, so we assume that $G$ is non-abelian.

1. Every non-identity element of $G$ is contained in a maximal subgroup of $G$: [SHOW MORE]
   Consider the cyclic subgroup generated by that element. This is nontrivial and not the whole group, because $G$ is non-abelian. If the subgroup is not maximal, consider a bigger proper subgroup of $G$ containing it. Keep proceeding this way until the subgroup becomes maximal. If the process does not stop, we have an infinite ascending chain of subgroups, contradicting the slenderness assumption.
2. $G$ has at least two maximal subgroups $M$ and $N$, both of which are normal in it: [SHOW MORE]
   Pick any non-identity element $g\in G$. By step (1), there is a maximal subgroup $M$ of $G$ containing $g$. Since $M$ is proper, there exists $h\in G\setminus M$. By step (1), there exists a maximal subgroup $N$ of $G$ containing $h$. By fact (1), both $M$ and $N$ are normal in $G$.
3. $MN=G$: [SHOW MORE]
   $MN$ is a subgroup since both $M$ and $N$ are normal, and it equals $G$ since both are maximal and distinct.
4. $M\cap N$ is contained in the center of $G$: [SHOW MORE]
   By assumption, both $M$ and $N$ are abelian, so $C_{G}(M\cap N)$ contains both $M$ and $N$, so it must contain $MN=G$.
5. $\Phi (G)\leq Z(G)$: [SHOW MORE]
   $\Phi (G)$ is the intersection of all maximal subgroups, so it is contained in $M\cap N$. So, by the previous step, it is contained in $Z(G)$.
6. $[G,G]\leq \Phi (G)$ (Given data used: $G$ nilpotent) : [SHOW MORE]
   By fact (1), every maximal subgroup is normal (hence maximal normal). The quotient is nilpotent by fact (2), so it is a simple nilpotent group, and hence must be abelian. In particular, $[G,G]$ is contained in every maximal subgroup. Thus, $[G,G]$ is contained in the intersection $\Phi (G)$.

The last two steps complete the proof.