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(*G*)×*H*. 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 *g*∈*C*, *h*∈*D*, 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,g*α) for *g*∈*G*, 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.

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

nhas minimal generating sets of any cardinality between 2 andn−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 sizen−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.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).

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

(https://link.springer.com/article/10.1007/s00013-003-0788-y) which perhaps you already know.

I didn’t know that paper. But it was interesting reading Miller’s paper on the Frattini subgroup. He knew, for example, that a group which acts faithfully as a primitive permutation group has trivial Frattini subgroup.

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.