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

Statement
Suppose $$G$$ is a fact about::slender nilpotent group, i.e., a group that is both a fact about::nilpotent group and a fact about::slender group (every subgroup is finitely generated, or equivalently, every ascending chain of subgroups stabilizes after a finite length). Then, $$G$$ is a fact about::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) uses::Nilpotent implies every maximal subgroup is normal
 * 2) uses::Nilpotence is quotient-closed

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

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

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 $$G$$: 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: 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) $$MN = G$$: $$MN$$ 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 $$G$$: 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) $$\Phi(G) \le Z(G)$$: $$\Phi(G)$$ 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) $$[G,G] \le \Phi(G)$$ (Given data used: $$G$$ nilpotent) : 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.