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

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## 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 $G$.

## 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 $G$: [SHOW MORE]
2. $G$ has at least two maximal subgroups $M$ and $N$, both of which are normal in it: [SHOW MORE]
3. $MN = G$: [SHOW MORE]
4. $M \cap N$ is contained in the center of $G$: [SHOW MORE]
5. $\Phi(G) \le Z(G)$: [SHOW MORE]
6. $[G,G] \le \Phi(G)$ (Given data used: $G$ nilpotent) : [SHOW MORE]

The last two steps complete the proof.