9
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
6
+50
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.