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It is obvious that if the Frattini quotient of a finite group $G$ is abelian, then $G$ is abelian by nilpotent and that finite nilpotent groups have abelian Frattini quotient.

I wonder if there is an example of a non-nilpotent finite group with abelian Frattini quotient ?

A better thing would be to have a nice characterization of such finite groups.

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    $\begingroup$ By a theorem of Gaschütz, a finite group is nilpotent when its Frattini quotient is nilpotent. (Satz III.3.5 in Huppert's Endliche Gruppen I) $\endgroup$
    – Frieder Ladisch
    Commented Nov 1, 2017 at 16:01
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    $\begingroup$ Indeed, this is well-known, see Theorem B here: emis.de/proceedings/Chicho2001/Otal.pdf $\endgroup$
    – Luc Guyot
    Commented Nov 1, 2017 at 16:15
  • $\begingroup$ Even without the theorem of Gasch\"utz, the fact that the Frattini quotient of $G$ is Abelian implies that every maximal subgroup of $G$ contains the derived group$G^{\prime}$, and hence that every maximal subgroup of $G$ is normal, hence $G$ is nilpotent. $\endgroup$
    – Geoff Robinson
    Commented Oct 6 at 19:07

2 Answers 2

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This is well known, but here is a quick proof anyway.

Suppose that $G/\Phi(G)$ is nilpotent, and let $P \in {\rm Syl}_p(G)$. Then $P\Phi(G) \unlhd G$ so, by the Frattini Argument, $G = N_G(P)P\Phi(G)= N_G(P)\Phi(G)$. Hence $G = N_G(P)$ by the non-generator property of $\Phi(G)$. So all Sylow subgroups are normal, and $G$ is nilpotent.

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Here is my attempt to entertain those who knew the answer.

Question 1. Let $G$ be a finitely generated group such that $G/\Phi(G)$ is Abelian. Is $G$ nilpotent?

The answer is yes if we assume moreover that $G$ is linear (e.g., $G$ is finite). But the answer is no in general, as the the first Girgorchuk group $G_1$ is such that $[G_1, G_1] \subset \Phi(G_1)$. For more examples, see this MO post.

Let $W(G)$ denote the $N$-Frattini subgroup of $G$, that is the intersection of the maximal normal subgroups of $G$ when defined, $G$ otherwise. (See this post for results related to $W(G)$.) Note that $G/W(G)$ is Abelian for any soluble group $G$.

Question 2. Let $G$ be a finitely generated group such that $G/W(G)$ is Abelian. Is $G/\Phi(G)$ Abelian?

The answer is no, because some Sunic group of intermediate growth is a counter-example, see this preprint.

Question 3. Let $G$ be a finite group such that $G/W(G)$ is Abelian. Is $G$ soluble?

The answer is no because of the symetric group $S_n$ with $n \ge 5$. One may try to get a classification though. Possibly useful:

Lemma (Baer). Let $G$ be a group and let $N$ be a normal subgroup of $G$. Then $W(N)$ is a normal subgroup of $G$ and $W(N) \subset W(G)$.

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    $\begingroup$ Classifying finite groups such that $G/W(G)$ is abelian sounds a bit optimistic. For instance, $H\wr A$ is such a group for every finite group $H$ and any non-trivial finite solvable group $A$. $\endgroup$
    – YCor
    Commented Nov 2, 2017 at 14:56
  • $\begingroup$ @YCor Thanks for spotting the $G_1$-typo, it's corrected now. And indeed, "classification" is the wrong word. My intention was just to echo OP's question about the existence of a "nice characterization" of such groups. (By the way, are you assuming $H$ to be simple in your last comment?). $\endgroup$
    – Luc Guyot
    Commented Nov 2, 2017 at 16:48
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    $\begingroup$ No I didn't assume $H$ simple. The proof is as follows: in $H\wr A$, the normalizer of $H$ is normal and co-solvable. If $S$ is a simple quotient of $H\wr A$, the image of $H$ also has a normal co-solvable normalizer, hence is equal to $S$. So the image of $H$ is either $\{1\}$ or $S$. But it's the same for another conjugate copy of $H$ (and there's one since $A\neq 1$. Since these pairwise commute, it follows that all have trivial image. $\endgroup$
    – YCor
    Commented Nov 2, 2017 at 21:10
  • $\begingroup$ Thanks for this. My question was derived from Galois theory, hence the groups I was considering were always finite. $\endgroup$
    – Lior Bary-Soroker
    Commented Nov 7, 2017 at 19:03

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