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Suppose $G$ is a finite group, such that $\Phi(G)$ is non-abelian. Does there always exist such prime $p$, that $p^5 | |G|$? Here $\Phi$ stands for Frattini subgroup.

Using the same method, as the one used in the answer to “If $|G|=p^3q^2$ then $\Phi(G)$ is cyclic for primes $p\neq q$.”, we can reduce this question to the following ones:

Suppose $G$ is a finite group, such that $\Phi(G) \cong D_4$. Is it always true, that $32 | |G|$?

Suppose $G$ is a finite group, such that $\Phi(G) \cong Q_8$. Is it always true, that $32 | |G|$?

Suppose $G$ is a finite group, such that $\Phi(G) \cong (C_p \times C_p) \rtimes C_p$ for some odd prime $p$. Is it always true, that $p^5 | |G|$?

Suppose $G$ is a finite group, such that $\Phi(G) \cong C_{p^2} \rtimes C_p$ for some odd prime $p$. Is it always true, that $p^5 | |G|$?

The answer to the main question is positive iff the answer is positive in all those four particular cases.

A problem, similar to those “reduced ones” was solved for $C_p \times C_p$ here: A question about Frattini subgroup of specific form However, the solution seems to rely on the structure of $Aut(C_p \times C_p)$ and thus the method does not seem to be directly applicable in our case (though, probably, something similar may be...).

And I do not know how to proceed further.

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    $\begingroup$ Note that there is no finite group such that $\Phi(G) \cong Q_8$. $\endgroup$
    – the_fox
    Feb 21, 2019 at 12:44
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    $\begingroup$ An obvious note is that the answer is yes if $G$ is a $p$-group for prime $p$ (since for $G$ of order dividing $p^4$, the Frattini subgroup has order dividing $p^2). Of course you know this, but the reader (like me) might not. $\endgroup$
    – YCor
    Feb 21, 2019 at 13:18
  • $\begingroup$ The case with $\Phi(G) \cong D_4$: math.stackexchange.com/questions/3143850/… $\endgroup$ May 2, 2019 at 8:55
  • $\begingroup$ Having read @Derek Holt on your previous question, I do wonder if it helps here (when $p$ is odd) to look at the action of $G$ on $V:=\Phi(G)/Z(\Phi(G))$? Of course the automorphism group isn't the full linear group as we have to respect the structure of the commutator map $(x,y)\mapsto [x,y]$ on $V\times V$. $\endgroup$ May 2, 2019 at 9:18

1 Answer 1

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It is proved in Lemma 1 of

W.M. Hill and D.B. Parker, The nilpotence class of the Frattini subgroup, Israel Journal of Math, vol 15, 211-215 (1973)

that a nonabelian group $N$ of order $p^3$ (for any prime $p$) cannot occur as a normal subgroup and contained in the Frattini subgroup of any finite group.

So in particular there is no finite group $G$ with $\Phi(G)$ nonabelian of order $p^3$.

The proof is short, but it uses results proved in Huppert's Endliche Gruppen.

It's easy to see that $D_8$ (that's my preferred notation for the dihedral group of order $8$) cannot occur as $\Phi(G)$, because the centralizer in $G$ of its characteristic cyclic subgroup of order $4$ would have index $2$ in $G$ and so would be maximal but would not contain $\Phi(G)$.

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