# Are finite nilpotent groups the only finite groups with abelian Frattini quotient?

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.

• 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) Nov 1, 2017 at 16:01
• Indeed, this is well-known, see Theorem B here: emis.de/proceedings/Chicho2001/Otal.pdf Nov 1, 2017 at 16:15

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.

• – YCor
Nov 1, 2017 at 17:14

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 $W(N) \subset W(G)$.

• 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 $A$.
– YCor
Nov 2, 2017 at 14:56
• @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?). Nov 2, 2017 at 16:48
• 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.
– YCor
Nov 2, 2017 at 21:10
• Thanks for this. My question was derived from Galois theory, hence the groups I was considering were always finite. Nov 7, 2017 at 19:03