The Frattini argument

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 = NG(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 Pg, 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 Pg = Pn. But then Pgn−1 = P, so h = gn−1 belongs to NG(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 = NG(P).N. But because N consists of non-generators, we can delete its elements one by one and obtain G = NG(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.

About Peter Cameron

I count all the things that need to be counted.
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5 Responses to The Frattini argument

  1. Tendai Shumba says:

    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!

      • Tendai Shumba says:

        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…

  2. Tendai Shumba says:

    Thanks, I have now seen those.

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