## Integrals of groups

Everyone who has studied mathematics knows what the derivative and integral of a function are. The derivative measures rate of change, the integral (the inverse operation) measures area under a curve. They are inverse operations; and two functions have the same integral if and only if they differ by a constant.

In a group, the commutator is a function of two variables which is the identity if and only if the two arguments commute. The commutator subgroup of a group is the subgroup generated by the commutators; so it is the identity subgroup if and only if the group is abelian (all pairs of elements commute).

The commutator subgroup is also known as the derived group. I don’t know who introduced this terminology; it suggests an analogy between group theory and calculus which would never have occurred to me. (There is a closer analogy with differentiation in the relation between a Lie group and its Lie algebra, but that is a different story.)

On one of my visits to Lisbon, João Araújo and Francesco Matucci introduced me to what sounded like an interesting game: if forming the derived group is the analogue of differentiation in calculus, what can we say about integration? We will say that H is an integral of G if G is isomorphic to the derived subgroup of H.

The first thing that can be said is that a basic fact from calculus fails rather spectacularly here. The analogue of a constant function is an abelian group (one whose derived group is trivial); the analogue of adding a constant should presumably be taking the direct product with an abelian group. But, for example, among the integrals of the cyclic group of order 3 we have the dihedral group of order 6 and both non-abelian groups of order 27, which certainly do not differ by an abelian direct factor!

Nevertheless, we have been playing the game for a while, and have just posted a preprint on the arXiv.

The first thing we can say, which finite group theorists will appreciate, is that, if a finite group G has an integral, then it has a finite integral. For suppose that H is an integral of G. Choose pairs of elements of H whose commutators generate G; the subgroup of H generated by these elements is also an integral of G. So we may assume that H is finitely generated. Now conjugate elements of H lie in the same coset of the derived group; so the conjugacy classes in H have size bounded by |G|; that is, H is a BFC-group. In a finitely generated BFC-group, the centre has finite index, and so is finitely generated; so the centre is the direct product of a finite group and a torsion-free group. Factoring out the torsion-free group we obtain a finite group whose derived group is G.

Every abelian group is integrable. However, not every group is integrable. For example, for every even number n greater than 4, the dihedral group of order n has no integral. Of the two non-abelian groups of order p3, where p is prime, one is integrable and the other is not.

How do we decide whether a group is integrable? This leads to an interesting question to which we don’t have an answer:

Problem: Find a good upper bound for the order of the smallest integral of an integrable group of order n.

It is clear that there is such a bound. For there are only finitely many groups of order n up to isomorphism; we can take the largest order of the smallest integral of an integrable group in the collection. However, just knowing that a bound exists is no use for the obvious algorithm for testing integrability, viz., check all groups whose orders are multiples of n up to f(n), where f is the above function, to see whether any of them has derived group isomorphic to G.

The smallest integrals of the cyclic group of order 2 are the two non-abelian groups of order 8, and we conjecture that the function f satisfies f(n) ≤ n3 for all n. We can show that an abelian group of order n has integral of order at most n1+o(1), though finding an exact formula for the smallest integral is quite tricky; the right answer could conceivably be Cn for some constant C.

A result which may be folklore characterises the set of positive integers n such that every group of order n is abelian: they are precisely the integers n which are cube-free and for which there do not exist primes p and q such that either

• p and q divide n, and q divides p−1; or
• p2 and q divide n, and q divides p+1.

A proof by Robin Chapman can be found here or here.

Since every abelian group is integrable, one could ask for a similar characterisation of the set of numbers n for which every group of order n is integrable. This question is answered in our paper. The answer is almost the same as above; the second condition on the primes p and q simply has to be deleted. Thus, for example, there exist non-abelian groups of order 75, but every group of this order is integrable.

An interesting problem here is: does the density of the set of such numbers exist, and if so, what is it? Of the first hundred million positive integers, there are 32,261,534 for which every group of that order is integrable.

There are groups which can be integrated arbitrarily many times (that is, which are the nth derived group of some other group, for all n). For example, if we take the group of strictly upper triangular (2n+1)×(2n+1) matrices over a field F, the nth derived group is isomorphic to the additive group of F. However, we do not know an answer to the following question:

Problem: For which groups G does there exist a strictly ascending sequence G=G0<G1<… with each group the derived group of the following one?

If G is a perfect group (equal to its derived group), then of course such a sequence exists with the inequality replaced by equality or non-strict inequality (all terms equal to G); but with the strict inequality, we have no examples. Such a group must have non-trivial centre, since we show that for groups with trivial centre the sequence must terminate after finitely many steps.

All of this can be viewed in a different light, as an example of what might be termed “inverse group theory”. There are various constructions which start from a group and produce another group: very often the new group is a subgroup (for example, the derived group, Frattini subgroup, Fitting subgroup, centre), but it may be a quotient (soluble residual, derived quotient) or even something else (Schur multiplier or other cohomology group). I would like to say “functor”, but of course not all of these constructions are functorial in the sense of category theory.

For each of these constructions, there is an inverse problem: Given a group G, is there a group H such that the derived group (or Frattini subgroup, or soluble residual, or Schur multiplier) of H is G?

Some of these problems are trivial. For two examples,

• The centre of a group is abelian. But every abelian group is the centre of a group (for example, itself).
• The Fitting subgroup of a group is nilpotent. But every nilpotent group is the Fitting subgroup of a group (for example, itself).

We have by no means explored all the examples listed. But it appears that the question is non-trivial, and has interesting aspects, for derived group and for Frattini subgroup. Let me give one problem related to the latter.

True or false? If the (finite) p-group G is the Frattini subgroup of a group, then it is the Frattini subgroup of a finite p-group.

I mention in conclusion that not every p-group is the Frattini subgroup of a p-group. I count all the things that need to be counted.
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### 2 Responses to Integrals of groups

1. Peter Cameron says:

Now a good lesson in how the arXiv works.

When the paper appeared this morning, three people (Alireza Abdollahi, Michael Kinyon, and Avinoam Mann) had written to us telling us things that we hadn’t known before. As a result, the next version of the paper will be much improved. So thank you all!

In particular, our problem about the infinite ascending sequence of finite groups is answered in the negative, as a consequence of a theorem of Bernhard Neumann (my mathematical father’s father). Most mortifying to me, this is a paper which I have cited in another publication! Either I didn’t read it carefully enough, or my brain has started to leak (or more likely both!)

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