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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 at 12:44
  • $\begingroup$ 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 at 13:18

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