## On the Frattini subgroup

I wrote earlier about the Frattini subgroup of a group. It can be defined in either of two ways (as the set of non-generators of a group, the elements which can be dropped from any generating set containing them; or as the intersection of all the maximal subgroups), and the proof that it is nilpotent is a good illustration of the Frattini argument.

As you will see from the preceding post, I have been thinking about the Frattini subgroup recently. What I propose to do here is to tell you something about it which I don’t recall meeting when I was a student (and so perhaps you didn’t either), and also to throw some light on two other things: how research is done, and how mathematics was written a century ago.

In the course of what I was doing, I began to wonder whether the Frattini subgroup of the direct product G×H is the direct product of the Frattini subgroups of G and H. GAP confirmed that this was so in a couple of examples. In the past, I would have had to go to the library and look through books, and probably lug huge volumes of Mathematical Reviews down from the shelves. Now I can’t go to the library, but I do have the internet. So I googled the question. I found a fairly recent paper determining exactly when this holds, for arbitrary (possibly infinite) direct products of arbitrary (possibly infinite) groups. In the bibliography was a paper by G. A. Miller from 1915 which, it was claimed, proved the result for finite direct products of finite groups.

The paper was in the Transactions of the American Mathematical Society, long enough ago that it is freely available, so I downloaded a copy. It is only seven pages long, but does not contain a single theorem; it just consists of text, with a few important points italicised. A quick glance didn’t lead me to the theorem quoted, although he deals with the Frattini subgroup of a Sylow subgroup of the symmetric group, so has to handle direct products.

This is how mathematics was done in those days. I think the change in style was not too abrupt; in Burnside’s book, he often rambles on for several pages, but at least he states a theorem at the end of his ramblings. I think that today we would not allow our students to write like that, and journals would not accept our papers if we did it ourselves.

So I had to prove it for myself, to be sure it was right. I will also explain why finiteness of the groups is needed.

I will use the definition as intersection of maximal subgroups, though Miller uses the other definition of the Frattini subgroup.

Consider G×H. If A is a maximal subgroup of G, then A×H is a maximal subgroup of G×H; the intersection of all these is Φ(GH. Similarly the other way around. So certainly Φ(G)×Φ(H) is an intersection of some maximal subgroups, so contains Φ(G×H). We have to show that intersecting this with other maximal subgroups does not make it any smaller.

So we have to think about arbitrary maximal subgroups of a direct product G×H. In fact, it is known (I did learn this as a student) that all subgroups of the direct product are produced in the following way. Let A and C be subgroups of G, with A a normal subgroup of C; and similarly B and D in H. Suppose that the factor groups C/A and D/B are isomorphic; let α be an isomorphism between them. Now consider the set of all pairs (g,h) in G×H such that gC, hD, and (Ag)α = Bh. This is a subgroup, and every subgroup has this form.

It is easy to see that, in order for the subgroup in the previous paragraph to be maximal, we have to have C = G, D = H, and the isomorphic quotients G/A and H/B simple.

Now we can see why the formula for the Frattini subgroup of a direct product can fail for infinite groups. Let G be an infinite simple group which has no maximal subgroups. Then Φ(G) = G. But G×G has a diagonal subgroup consisting of all (g,gα) for gG, which is maximal; so Φ(G×G) is not equal to G×G.

But the Frattini subgroup of a finite simple group G is trivial; for every finite group has a maximal subgroup, so its Frattini subgroup is proper, and it is normal, so is trivial if G is simple.

Thus, for a maximal subgroup of the type we are considering, A contains Φ(G) and B contains Φ(H), so our maximal subgroup contains Φ(G)×Φ(H), as required.

## About Peter Cameron

I count all the things that need to be counted.
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### 5 Responses to On the Frattini subgroup

1. A postscript to the above. Miller has quite a lot to say about minimal generating sets of a group (those where no element can be dropped). He states, as a kind of throwaway remark, that the symmetric group of degree n has minimal generating sets of any cardinality between 2 and n−1 inclusive.
Now one of my all-time favourite theorems is Julius Whiston’s result that there is no minimal generating set larger than n−1 (and indeed there is no larger independent set, and an independent set of size n−1 is a generating set).
Philippe Cara and I went on to determine all the minimal generating sets of size n−1. There are two types, both related to trees; the simpler to describe consists of all the transpositions corresponding to the edges of a tree.

2. S. Srinivas Rau says:

E.Schenkman’s “Group Theory” presents the theorem of Dlab & Korinek(1960); Frattini (Product of finitely generated groups)=Product of Frattini groups. The only possible exception is a simple group with no maximal subgroup ( existence unknown in the 1960s).My query:analogue for radical of sum of two algebras/Lie algebras.(Frattini subgroup is the analogue of the radical of an algebra).

3. Thanks for this lovely post. Reading this, I became curious about which finite groups have trivial Frattini subgroup. I immediately found a rather nice paper of L.-C. Kappe & J. Kirtland
Here is a slightly different (and hopefully correct) proof of the inclusion $\Phi(G)\times\Phi(H)\le \Phi(G\times H)$.
Let $G$ be a finite group and let $S$ be a subset of $G$. Recall that if $G=\langle \Phi(G), S\rangle$, then $G=\langle S\rangle$. (Otherwise, $\langle S\rangle < G$ and so there exists a maximal subgroup $M$ of $G$ containing $\langle S\rangle$. But now $G=\langle \Phi(G), S\rangle \le M <G$, which is absurd.) Call this fact the \textit{Fundamental Frattini Property} (FFP). In particular, note that if $H$ is a subgroup of $G$, then the equality $G=\Phi(G)H$ implies that $G=H$.
Now, write $D=G\times H$. Identify $G$ and $H$ as subgroups of $D$ as usual. Let $M$ be a maximal subgroup of $D$. If $\Phi(G)\not\le M$, then $D=M\Phi(G)$. (Note: $G$ normalizes $\Phi(G)$ and $H$ centralizes $\Phi(G)$. Consequently $\Phi(G)$ is normal in $D$, and so $\Phi(G)M$ is actually a subgroup of $D$.) By Dedekind's Lemma, $G=G\cap D=G\cap M\Phi(G)=\Phi(G)(G\cap M).$
Thus $G=\Phi(G)(G\cap M)$, and so $G=(G\cap M)$ by the FFP. But now, $G\le M$, against our hypothesis that $\Phi(G)\not\le M$. We conclude that $\Phi(G)\le M$. A similar argument shows that $\Phi(H)\le M$. Thus $\Phi(G)\Phi(H)\le M$. As $M$ was an arbitrary maximal subgroup of $D$, the inclusion $\Phi(G)\Phi(H)\le\Phi(D)$ holds.