A question about Frattini subgroup of specific form

Suppose $$p$$ is a prime number and $$G$$ is a finite group, such that $$\Phi(G) = C_p \times C_p$$, where $$\Phi$$ denotes the Frattini subgroup. Is it always true, that $$p^4$$ divides $$|G|$$?

This statement can be easily proved for $$p$$-groups by seeing that no group of order $$p$$, $$p^2$$ or $$p^3$$ (there is a full classification of such groups) possesses a Frattini subgroup of the aforementioned form. Knowing that any finite nilpotent group is a direct product of $$p$$-groups and that the Frattini subgroup of a finite direct product of finite groups is the direct product of their Frattini subgroups, we can reach the same conclusion about finite nilpotent groups. However, I do not know, how to prove this statement in general.

Any help will be appreciated.

• Nice question ! – the_fox Jan 27 at 20:29
• Note that you only need to exclude the case where $|P|=p^3$, $P$ a Sylow $p$-subgroup of $G$, since $\Phi(G)$ does not contain a full Sylow subgroup of $G$. – the_fox Jan 27 at 20:39
• Have you understood a very small possible counterexample, for instance $C_2 \rtimes_\phi (C_3 \times C_3)$ where $\phi$ is conjugation of an element of $C_3 \times C_3$ and the conjugation by the nontrivial element of $C_2$ is inversion? – Eric Towers Jan 27 at 21:28
• @EricTowers That has maximal subgroups of order $6$ and so is not a counterexample. I believe that the conjectured result is true, but I have not quite worked out the details yet. – Derek Holt Jan 27 at 21:32
• @DerekHolt : That rejects the particular example, but not the proposed method for understanding the general failure of counterexamples. – Eric Towers Jan 27 at 21:43

I think the answer is yes, $$p^4$$ divides $$|G|$$. Here is a sketch of how to prove this. This argument seems a bit long and tortuous, and there might be an easier proof. I will just do it for odd $$p$$. A similar but slightly different argument works for $$p=2$$.

Let $$N = \Phi(G) = C_p \times C_p$$, and $$N \le P \in {\rm Syl}_p(G)$$.

Now $$N$$ cannot have a complement in $$G$$, since otherwise that complement would be contained in a maximal subgroup that did not contain $$N$$. So by a theorem of Gaschütz, $$N$$ does not have a complement in $$P$$. So $$N < P$$, and we only have to consider the case when $$|P|=p^3$$. Then, for elements $$g \in P \setminus N$$ must have order $$p^2$$, with $$g^p \in N$$.

Now the conjugation action of $$G$$ on $$N$$ induces a subgroup $$\bar{G} = G/C_G(N)$$ of $${\rm Aut}(N) = {\rm GL}(2,p)$$. If the image $$\bar{P}$$ of $$P$$ in $$\bar{G}$$ is not normal in $$\bar{G}$$, then $$\bar{G}$$ has more than one Sylow $$p$$-subgroup. But any two Sylow $$p$$-subgroups of $${\rm GL}(2,p)$$ generate $${\rm SL}(2,p)$$.

Since we are assuming that $$p$$ is odd, $${\rm SL}(2,p)$$ has a central subgroup $$\bar{T}$$ of order $$2$$ that acts as $$-I_2$$ on $$N$$. Let $$T$$ be the complete inverse image of $$\bar{T}$$ in $$G$$ (so $$|T/C_G(N)|=2$$). Then $$T \lhd G$$. Let $$S \in {\rm Syl}_2(T)$$. Then, by the Frattini Argument, $$G = TN_G(S)$$. So $$p$$ divides $$|N_G(S)|$$, but $$N_G(S) \cap N = 1$$, so a Sylow $$p$$-subgroup of $$N_G(S)$$ has order $$p$$ and complements $$N$$, contrary to what we said above.

So $$\bar{P} \unlhd \bar{G}$$. But then $$M := \langle g^p \mid g \in P \rangle$$ is a normal subgroup of $$G$$ of order $$p$$ contained in $$N$$. The image $$N/M$$ of $$N$$ in $$M$$ has a complement in $$P/M$$, and hence, by Gaschütz's theorem again, $$N/M$$ has a complement $$H/M$$ in $$G/M$$. Then $$|G:H|=p$$ and $$H$$ is a maximal subgroup of $$G$$ not containing $$N$$, contradiction.

• This is a great answer. – the_fox Jan 28 at 18:05