# Does “commutator subgroup is in Frattini” guarantee nilpotency?

In nilpotent groups, the commutator subgroup is contained in the Frattini subgroup. On the other hand, if the commutator subgroup of a finite group is contained in Frattini subgroup, can we conclude that $G$ is nilpotent?

• If $G$ is satisfies that condition, then every maximal subgroup is normal, and so if $G$ is finite then $G$ must be nilpotent. But there are infinite non-nilpotent groups that satisfy it, such as the Grigorchuk group. – Derek Holt Sep 17 '15 at 15:00
• Thanks for the clarification. You may post it as answer. I would like to know about Grigorchuk groups. – Groups Sep 18 '15 at 4:26

If $[G,G] \le \Phi(G)$ then every maximal subgroup $M$ of $G$ contains $[G,G]$ and hence is normal in $G$, and then $G/M$ must be abelain and simple, so $|G:M|$ is prime. Conversely, if every maximal subgroup of $G$ is normal of prime index then $[G,G] \le \Phi(G)$, so the conditions are equivalent.

It is well-known that a finite group is nilpotnent if and only if every maximal subgroup is normal, so for finite groups, the condition $[G,G] \le \Phi(G)$ is equivalent to nilpotency.

However, there are non-nilpotent infinite examples, such as the Grigorchuk group, which is an infinite finitely generated $2$-group in which all maximal subgroups have index $2$. In fact $[G,G]=\Phi(G)$ in this example, and has index $8$ in $G$.