The *Frattini subgroup* of a finite group *G* can be defined in two equivalent ways:

- it is the intersection of all the maximal proper subgroups of
*G*; - it is the set of all
*non-generators*of*G*, that is, elements which can be dropped from any generating set (so*g*is a non-generator if, whenever*G*is generated by {*g*}∪*X*for some set*X*, then*G*is generated by*X*).

To see the equivalence, suppose first that *g* lies outside some maximal subgroup *M*. Then *G* is generated by *M* and *g* but not by *M*, so *g* is *not* a non-generator. Conversely, suppose that *g* is contained in every maximal subgroup, and let *X* be such that {*g*}∪*X* generates *G*. If *X* does not generate *G*, then it is contained in a maximal subgroup *M*; but *g* is also in *M*, so {*g*}∪*X* cannot generate *G*.

A little job I am doing at the moment requires me to remember and explain various facts about elementary group theory. One of these is:

**Theorem** *The Frattini subgroup of a finite group is nilpotent.*

I couldn’t remember the argument, so, following my standard practice, I went to bed last night thinking about it, and when I awoke I had the proof. It is, not so surprisingly, the *Frattini argument*.

My first permanent teaching job was at Bedford College, University of London (which now no longer exists, having fallen on hard times financially and been rescued by a merger with Royal Holloway College). I was there for a bit over a year in the mid-1970s. Paul Cohn was the head of department; he ran a study group which was reading a paper by Procesi about invariants of *n*-tuples of matrices. The participants were stuck on a certain point; I managed to notice that it was a ring-theoretic version of the Frattini argument. Paul was delighted that Procesi had adapted an argument by his compatriot Frattini.

It is an *argument*, not a *theorem*, and can be applied in a number of different contexts. But it appears in a pure form in the following theorem. (I discussed Sylow’s Theorem very recently here.)

**Theorem** *Let G be a finite group, N a normal subgroup of G, and P a Sylow p-subgroup of N. Then G = *N_{G}(*P*).*N*.

To prove this, we have to take an arbitrary element *g* of *G*, and factorise it as *hn*, where *h* belongs to the normaliser of *P* and *n* belongs to *N*.

Consider *P ^{g}*, the conjugate of

*P*by

*g*. This is contained in

*N*(since

*N*is a normal subgroup of

*G*), and has the same order as

*P*, so it is a Sylow

*p*-subgroup of

*N*. According to the second part of Sylow’s Theorem, it is conjugate to

*P*in

*N*; that is to say, there is an element

*n*in

*N*such that

*P*=

^{g}*P*. But then

^{n}*P*

^{gn−1}=

*P*, so

*h*=

*gn*

^{−1}belongs to N

_{G}(

*P*). Then

*g*=

*hn*as required.

Now we can prove the first theorem. Take *N* to be the Frattini subgroup of *G*: it is clearly normal in *G*. If *P* is any Sylow subgroup of *N*, then *G* = N_{G}(*P*).*N*. But because *N* consists of non-generators, we can delete its elements one by one and obtain *G* = N_{G}(*P*). Thus *P* is a normal subgroup of *G*, and *a fortiori* of *N*. But a finite group is nilpotent if and only if all its Sylow subgroups are normal; so we are done.

There is a typo, I believe, toward the end. Do you want N_G(P).P?

Thanks. Actually N_G(P).N, but not what I had written!

I actually noticed after I posted that it should be what you wrote here. On an unrelated note, where you note at the GRA workshop last week? I was looking forward to your summary.

Perhaps something strange is going on. I wrote notes about GRAW2, which you are not seeing for some reason…

Thanks, I have now seen those.