# What does Frattini length measure?

I have heard derived length, for example, described as a measure of "how non-commutative" the group is. An abelian group will have derived length $1$, whereas a non-solvable group will be so non-commutative that the even the fastest converging commutator series will continue infinitely.

Analogously, nilpotency class measures "how close to being abelian" a group is by finding the length of the slowest converging commutator series. A group of nilpotency class $2$ has essentially only one "twist" in its structure, and a group sufficiently far from being abelian cannot be completely unraveled by commutators without using bigger guns.

Then, between these two definitions, we've got Fitting length, which measures how far a group is from being nilpotent.

Of course, there is another series floating around out there.

The Frattini subgroup of a group $G$, denoted $\Phi(G)$, is the intersection of all maximal subgroups of $G$. (Equivalently, $\Phi(G)$ is the set of all non-generators of $G$.) We define the Frattini series by $\Phi_0(G)=G$ and $\Phi_{k+1}(G)=\Phi(\Phi_k(G))$. The smallest $n$ for which $\Phi_n(G)=1$ is the Frattini length of $G$.

So, intuitively, what does Frattini length measure?

• I don't believe the Frattini length is very helpful in understanding a group. A related idea that does seem helpful is the list of Frattini chief factors. K/N is Frattini if K/N is G-stable, G-irreducible, and not complemented in G/N. The Frattini subgroup is formed from the "bottom" Frattini factors, but there can be some above it, that is, another layer of a "Frattini series" that skips the complemented factors. – Jack Schmidt Nov 12 '12 at 0:29
• @JackSchmidt So, what does this extra Frattini layer tell us? I'm afraid I don't immediately understand what their intuitive meaning is. (Probably because I am not sure what it means for $K/N$ to be $G$-stable and $G$-irreducible.) Also, why do you say that Frattini length is not helpful for understanding a group? I'd happily accept an explanation of all this as an answer if you'd be willing to write one. – Alexander Gruber Nov 12 '12 at 1:39
• K/N is G-stable means K and N are normal subgroups of G with N ≤ K. G-irreducible means that there is no normal subgroup of G properly contained between N and K. In other words, K/N is a chief factor, and is Frattini if and only if it is not complemented. – Jack Schmidt Nov 12 '12 at 4:46
• Here's a paper I just found: groups with $\Phi(G)=1$ that could be helpful for understanding a part of this. – Alexander Gruber Feb 4 '14 at 13:46

## 2 Answers

A chief factor of a finite group $G$ is a pair $H/K$ such that $H \leq K$ are two normal subgroups of $G$ such that there is no other normal subgroup of $G$ properly contained between $H$ and $K$. Chief factors of finite groups come in two major flavors: elementary abelian $p$-groups and a direct product of copies of a fixed finite non-abelian simple group.

There are three kinds of chief factors: complemented, Frattini, and the rest. A complemented chief factor is one in which there is some $M/K$ such that $G=MH$ and $M \cap H = K$. A Frattini chief factor is one where $H/K \leq \Phi(G)/K$. A complemented factor can be abelian or not, but a Frattini factor must be abelian. The rest consist only of non-abelian chief factors, and I don't know of any examples offhand (they'd have to be pretty big).

If $H/K$ is a chief factor and $H \leq \Phi(G)$, then of cours $K \leq \Phi(G)$ and so $\Phi(G/K) = \Phi(G)/K$ and $H$ is a Frattini factor. Conversely, a chain of Frattini factors $H_1/H_2, H_2/H_3, \dots, H_n/1$ will all live in the Frattinis subgroup: $H_1 \leq \Phi(G)$. So the Frattini subgroup is the bottom-most section of Frattini chief factors.

The other Frattini chief factors make up other layers of Frattini-ness. So much like the $p,p'$-series of a finite $p$-solvable group, you can view a finite group as made up of Frattini and non-Frattini layers of chief factors.

The Frattini factors can often be ignored in a group. For instance, if none of the Frattini chief factors are $p$-groups, then the group has no elements of order $p$. The centralizer of the $p$-chief factors is equal to the centralizer of the non-Frattini $p$-chief factors. Frattini factors are covered by all maximal subgroups, but each maximal subgroup in a finite solvable group picks out a chief factor that it complements. The Frattini factors are sort of a formless ooze between the more rigid layers of the complemented factors. There is a formula for the number of generators of a group and the number of generating sequences of a group in terms of the non-Frattini chief factors of the group.

The Frattini subgroup of the Frattini subgroup is the part of the Frattini subgroup that really doesn't matter, even to the part of the original group that doesn't matter. The Frattini subgroup is nilpotent, so its Frattini subgroup immediately devolves into the exponent-p derived series, which is pretty dull.

• The module structure of elementary abelian sections of a finite group are very closely related to its Frattini/complemented nature. The module theoretic radical (maximal superfluous module) is the Frattini of the semi-direct product, and the cohomology of the module determines whether it is complemented (this is slightly tricky, as the G/H cohomology only tells you if it could be Frattini, the full cohomology tells you if it is Frattini). – Jack Schmidt Nov 12 '12 at 19:12
• I guess you accidently swapped the order of the set inclusion in your first sentence "$H/K$ such that $H \le K$", i.e. it should read "$H \ge K$" as the smaller normal subgroup stands conventionally on the right of the factor group. – StefanH Dec 29 '15 at 12:07
• Also I do not understand your sentence about the "major flavors" of chief factors. Should it be "cyclic groups of order $p$ or some non-abelian simple group" with no direct product? For if $A \times B \le H/K$ (or $A\times B$ is isomorphic to some subgroup of $H/K$) with $A,B \ne 1$, then we would have some nontrivial normal subgroups $A,B$, hence for example the inverse image $C$ of $A$ in $H/K$ is some normal subgroup with $K < C < H$ and $C / K \cong A$, i.e. we have some normal subgroup between $K$ and $H$? – StefanH Dec 29 '15 at 12:07

Perhaps we can see it the following way. We have the characterization of the Frattini subgroup of a group $\,G\,$:

$$x\in\Phi(G)\Longleftrightarrow\,\,\text{if for some}\,\,X\subset G\,\,,\,\langle\, X\,,\,x\,\rangle =G\Longrightarrow \langle\,X\,\rangle=G$$

as the non-generators of the group $\,G\,$.

Thus, if the series

$$\Phi(G)=:\Phi_0(G)\supset\Phi(\Phi(G))=:\Phi_1(G)\supset...\supset\Phi_n(G)=1$$

Now, what does it mean $\,\Phi(K)=1\,$? It means that any non-trivial element in $\,K\,$ is a non-redundant generator of some set of generators of $\,K\,$ . The only groups that I can think of right now fulfilling this are the elementary abelian ones, so perhaps we could say the Frattini Length of a group somehow measures how much of the original group we have to "scrap out" in order to be left with an abelian elementary one (or with any other kind of group for which any element is a non-redundant generator of some set of generators).

Just an idea...

• Here's a big list of stuff with Frattini length $1$. groupprops.subwiki.org/wiki/Special:Ask/… – Alexander Gruber Nov 11 '12 at 3:11
• If $K$ were nilpotent, then we could classify $K$ in the way that you did... – peoplepower Nov 11 '12 at 3:19
• Ok: the Baer-Specker group, with which (whom? Almost...) I had quite a struggle some 10 years ago while working on a decent proof that it isn't free abelian, makes it crystal clear that one can have finite Frattini length (even length one!) without having an elementary abelian "part inside". Most of the other examples I think have some rather hefty elementary abelian chunks within them. – DonAntonio Nov 11 '12 at 3:21