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 $n \geq 2$


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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.