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.

  • $\begingroup$ Nice question ! $\endgroup$ – the_fox Jan 27 at 20:29
  • $\begingroup$ 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 $G$. $\endgroup$ – the_fox Jan 27 at 20:39
  • $\begingroup$ 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? $\endgroup$ – Eric Towers Jan 27 at 21:28
  • $\begingroup$ @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. $\endgroup$ – Derek Holt Jan 27 at 21:32
  • $\begingroup$ @DerekHolt : That rejects the particular example, but not the proposed method for understanding the general failure of counterexamples. $\endgroup$ – 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.

  • 2
    $\begingroup$ This is a great answer. $\endgroup$ – the_fox Jan 28 at 18:05

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