The Frattini subgroup of a finite group is the intersection of its maximal subgroups. It is well-known that the Frattini subgroup of a finite group is nilpotent. My question is whether a kind of converse is true:

Does every finite nilpotent group occur (up to isomorphism, of course) as the Frattini subgroup of some finite group?

The problem reduces immediately to the case of non-abelian groups of prime power order, since the Frattini subgroup of a direct product of finite groups is the direct product of the Frattini subgroups of the factors, and it is not hard to show that any finite abelian group is the Frattini subgroup of a finite (abelian) group. I had hoped this might kick off an induction, but I've not been able to do anything with it. The first interesting case seems to be the non-abelian groups of order eight. A computer search of groups up to order $2000$ (not including groups whose order is a multiple of $512$, however, which could not complete on hardware available to me), turned up nothing for these two groups. I am not certain what I would expect the answer to this question to be!


1 Answer 1


No. Gaschütz (1953) contains a wealth of information on the Frattini subgroup, including Satz 11 which says that $\Phi(H)$ is “nearly” abelian, in that it cannot have any serious inner automorphisms:

If $H$ is a finite group with $G \unlhd H$ and $G \leq \Phi(H)$, then $\operatorname{Inn(G)} \leq \Phi(\operatorname{Aut}(G))$.

This answers your question:

For $G=D_8$, $I=K_4 \not\leq \Phi(A) = \Phi(D_8) = C_2$. For $G=Q_8$, $I=K_4 \not\leq \Phi(A) = \Phi(S_4) = 1$.

This is actually exactly the criteria for a group $G$ to be the Frattini subgroup of any group, as shown in Eick (1997). However, there are some interesting results along the way:

Stitzing (1970) includes the theorem that if $G$ has a center of prime order and an abelian characteristic subgroup contained in the second center but properly containing the center, then $N$ cannot even be a $H$-normal subgroup of a Frattini subgroup of any group $H$. The dihedral group of order 8 satisfies this, as does the extra-special group of exponent $p^2$ and order $p^3$ for any odd prime $p$.

Hill-Parker (1973) show that if $G$ has size $p^n$ and nilpotency class greater than $n/2$, then $G$ is not an $H$-normal subgroup of a Frattini subgroup of any finite group $H$. In particular, amongst the groups of order $p^3$, only the abelian groups occur as normal subgroups of Frattini subgroups.

van der Waall (1974) handles all groups of order $p^4$. van der Waall–de Nijs (1995) handle order 32. Groups with cyclic centers were handled in Makan (1975).

Some people considered restricting $H$ more severely, for instance Hobby (1960) showed that some subgroups that cannot be derived subgroups of $p$-groups also cannot be Frattini subgroups of $p$-groups. This is generalized in Bechtell (1966) which gives a relativized Satz 11, and determines which groups of order $p^4$ can be Frattini subgroups of $p$-groups.

The “normal subgroup of $H$ contained in the Frattini” language may strike you as weird, and Allenby (1980) showed it is false generality: any such $G$ ($G\unlhd H$, $G \leq \Phi(H)$, $H$ finite) is already the Frattini subgroup of a finite group. This was also shown in Wright (1980).

Eick (1997) is a particularly nice paper in which the converse of Satz 11 is proven, in some sense completely classifying finite groups that occur as Frattini subgroups. In particular she shows that $$\{ \Phi(H) : H, \Phi(H) \text{ is finite } \} = \{ \Phi(H) : H \text{ is finite } \} \supsetneq \{ \Phi(H) : H \text{ is finite soluble } \} \supsetneq \{ \Phi(H) : H \text{ is finite nilpotent } \}$$


  • $\begingroup$ I believe Eick (1997) gives an explicit algorithm that, given a finite group $G$ satisfying Satz 11, realizes a finite group $H$ with $G \unlhd H$, $G \leq \Phi(H)$. Then Wright (1980) gives an effective procedure to find $K$ from $H$ with $\Phi(K)=G$. I believe this could be coded up in GAP, but I have not done so. $\endgroup$ Mar 19, 2014 at 4:32
  • $\begingroup$ Jack, excellent answer +1! And glad you mentioned one of my thesis advisors, Robert vd Waall! $\endgroup$ Mar 19, 2014 at 11:07
  • $\begingroup$ @JackSchmidt Thank you so much for a very helpful answer! I'm just starting to read those references I can access, but it seems there is some lovely mathematics behind the answer to this question. $\endgroup$
    – James
    Mar 20, 2014 at 0:35

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