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

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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

Facts used

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


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.