I have been looking for information about the Frattini subgroup of a finite group. Almost all the books dealing with this topic discuss this subgroup for $p$-groups.

I am actually willing to discuss the following questions:

Let $G$ be a finite group. Is the Frattini subgroup of $G$ abelian?

Why is the order of the Frattini factor divisible by each prime divisor of $|G|$?

  • $\begingroup$ No to the first question. Counterexample? Choose a finite non-abelian $\;p$-group with non-abelian maximal proper subgroup. For the second question: for any such prime divisor there is a maximal subgroup divisible by that prime... $\endgroup$ – DonAntonio Apr 23 '14 at 14:26
  • $\begingroup$ and it is enough to find only one maximal subgroup with order divisible by the prime? $\endgroup$ – user145150 Apr 23 '14 at 14:29
  • $\begingroup$ for get that: I misread that last part since, obviously, if the Frattini subgroup is trivial then it cannot be divided any any prime...yet you're talking of the Frattini factor...do you mean the quotient $\;G/\Phi(G)\;$ ? $\endgroup$ – DonAntonio Apr 23 '14 at 14:37
  • $\begingroup$ do you recommend any reference? $\endgroup$ – user145150 Apr 23 '14 at 14:44

Here is a proof that, for a finite group $G$, $|G/\Phi(G)|$ is divisible by all primes $p$ dividing $|G|$. It uses the Schur-Zassenhaus Theorem. I don't know whether that can be avoided.

First note that each Sylow $p$-subgroup $P$ of $\Phi(G)$ is normal in $G$. (So, in particular, $\Phi(G)$ is nilpotent.) To see that, we have $G = \Phi(G)N_G(P)$ by the Frattini Argument, and then by the fact that $\Phi(G)$ consists of the non-generators of $G$, we have $G = N_G(P)$.

Now if there is a prime $p$ dividing $|G|$ but not dividing $|G/\Phi(G)|$, then $\Phi(G)$ contains a Sylow $p$-subgroup $P$ of $G$ and $P \unlhd G$. So, by the Schur-Zassenhaus Theorem, $P$ has a complement $H$ in $G$. Let $M$ be a maximal subgroup of $G$ containing $H$. Then $p$ divides $|G:M|$ and hence $p$ divides $|G/\Phi(G)|$, contradiction.

  • $\begingroup$ Thanks, I did not find any reference for the proof $\endgroup$ – user145150 Apr 23 '14 at 14:54

To find a finite group with non-abelian Frattini subgroup, you'll need a group with a non-abelian maximal subgroup, but that does not suffice. The dihedral group of order 16 has a maximal subgroup isomorphic to the dihedral group of order 8, but the Frattini subgroup of a non-cyclic group of order 16 has order at most 4, so is abelian (it is Klein 4 in this case).

The smallest examples have order 64. One particular example is given by a matrix group:$$G=\left\langle \begin{bmatrix} 0 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \mod 4\right\rangle \leq \operatorname{GL}(2,\mathbb{Z}/4\mathbb{Z})$$ which is a semi-direct product of $C_4$ acting on $C_4 \times C_4$. Its Frattini subgroup is isomorphic to $C_2 \times D_8$. The only other possibility for a non-abelian Frattini subgroup of a group of order 64 is $C_2 \times Q_8$.

One reason books emphasize Frattini subgroups of $p$-groups is that they have a very nice definition there: $\Phi(G) = G^p [G,G]$. Hence calculations and theorems are much easier. For solvable groups, there is still some research into embedding properties related to the Frattini subgroup, and for non-solvable groups, there are significant questions left open.

I gave an answer you might be interested in, describing the groups that can occur as Frattini subgroups. In particular, some non-abelian groups are on the list. :-)

  • $\begingroup$ so if I could construct the frattini subgroup for a finite group, would that be something? $\endgroup$ – user145150 Apr 23 '14 at 15:51

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