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Suppose $\psi(n)$ denotes the minimal natural number $k$, such that there exists a finite group $G$, such that $k = \max \{m \in \mathbb{N}| \exists \text{ prime } p, p^m | |G| \}$, and $\Phi(G)$ has nilpotency class exactly $n$. Here $\Phi$ stands for Frattini subgroup. Is there some sort of a closed formula for $\psi(n)$?

What I currently know:

$$\psi(n) \geq n + 3$$ for $n \geq 2$

Proof:

Using the method from the answer to “If $|G|=p^3q^2$ then $\Phi(G)$ is cyclic for primes $p\neq q$.“, the statement becomes reduced to:

If $G$ is a finite group, such, that $\Phi(G)$ is a $p$-group of exact nilpotency class $n \geq 2$, then $p^{n+3}| |G|$. Here $p$ stands for an arbitrary prime number.

Now, suppose, $p^{n+3}$ does not divide $|G|$. Then, $|\Phi(G)| | p^{n + 1}$. That means, that $|\Phi(G)| = p^{n + 1}$, because all groups of order $p^m$, with $m \leq n$, have exact nilpotency class strictly less than $n$. Thus $\Phi(G)$ is a maximal class group. Thus it contains a non-abelian characteristic subgroup of order $p^3$ (which is the second element of its upper central series). And there we receive the contradiction with Lemma 1 from “The nilpotence class of the Frattini subgroup” by W.M. Hill and D.B. Parker, which states:

A non-abelian group of order $p^3$ can non occur as a normal subgroup contained in the Frattini subgroup of any finite group.

EDIT:

Actually that article proves an even more stronger result

$$\psi(n) \geq 2n + 1$$ for $n \geq 2$

(which I did not know before, because I found the full text of the article only today)

However, the question, whether this bound is sharp, or can be bettered, still remains.

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