Does non-abelian $\Phi(G)$ imply, that $p^5 | |G|$ for some prime $p$?

Question

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.

Answer 1

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 $\Phi(G)$.