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

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.

• Note that there is no finite group such that $\Phi(G) \cong Q_8$. – the_fox Feb 21 at 12:44
• 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. – YCor Feb 21 at 13:18