# 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 $\Phi(G) \cong Q_8$. Feb 21, 2019 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 $p^2). Of course you know this, but the reader (like me) might not. – YCor Feb 21, 2019 at 13:18 • The case with$\Phi(G) \cong D_4$: math.stackexchange.com/questions/3143850/… May 2, 2019 at 8:55 • Having read @Derek Holt on your previous question, I do wonder if it helps here (when$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\$. May 2, 2019 at 9:18

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 $$p^3$$.
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 $$\Phi(G)$$.