Now the conference is over.

On the last morning, Marty Isaacs posed an interesting problem. Let *p* be a prime, and *n* a positive integer. Is there an infinite group in which exactly *n* elements are not *p*th powers? This was during a talk in which he told us several things about finite groups with this property, with complete proofs: notably, either such a group has order at most *n*^{2}, or it is a sharply 2-transitive group of order *n*(*n*+1), where *n* is a power of *p*.

I gave the last talk of the meeting. Contrary to my usual practice, I will tell you a bit about it, for two reasons. First, this is joint work with Colva Roney-Dougal; we don’t have a preprint yet, but I regard it as public following my talk, and so worth publicising. The second reason will emerge later.

The aim of the project is to say something about generating sets for finite groups. In the case of an elementary abelian *p*-group, the minimal generating sets are just bases in a vector space, and the theory is so easy that we teach it to all mathematics students. By the Burnside basis theorem, arbitrary *p*-groups are essentially no worse. A set generates a *p*-group *P* if and only if the corresponding cosets generate *P*/Φ(*P*), an elementary abelian group (where Φ(*P*) is the Frattini subgroup); so we just have a vector space “blown up” by the order of the Frattini subgroup. But for arbitrary groups, things are more complicated.

One gadget which has been much studied recently is the *generating graph* of a group *G*, in which group elements *x* and *y* are joined if and only if {*x,y*} = *G*. Of course, this is the null graph unless *G* is 2-generated; but, by a miracle which we don’t understand, at least every finite simple group is 2-generated.

When I first learned about this, I calculated the generating graph of *A*_{5}, a group whose order is equal to the age of P-cubed. I found to my astonishment that the automorphism group of the graph has order 23482733690880. But it is not a new simple group; the reason is not hard to find. Elements of order 3 or 5 which generate the same cyclic subgroup have the same neighbour set in the graph, and so can be permuted arbitrarily. There is a normal subgroup of order (2!)^{10}(4!)^{6} doing this, and the quotient is just *S*_{5}, the automorphism group of *A*_{5}.

So we can reduce the graph by the “g-equivalence relation”, in which two vertices are equivalent if they have the same neighbours. The resulting smaller graph shares many important properties (including clique number and chromatic number) with the original one, and is likely to have a more sensible automorphism group.

We were led to another equivalence relation, the “m-equivalence relation”, in which two elements are equivalent if they lie in the same maximal subgroups of the group *G*. It is clear that this is a refinement of g-equivalence, but is non-trivial for any finite group. So now I can state one of our conjectures:

*Conjecture*: In a finite simple group (or maybe even a 2-generated almost simple group), two elements are g-equivalent if and only if they are m-equivalent.

This is false in general, even for 2-generated groups. For *S*_{4}, for example, a double transposition lies in no 2-element generating set, so they are g-equivalent to the identity, but not m-equivalent: there are 14 g-equivalence classes, but 15 m-equivalence classes. The numbers of m-equivalence classes in symmetric groups form an interesting sequence, beginning 1, 2, 5, 15, 67, 362, 1479, …

However, I learned later that Budapest is not a good place for marketing generating sets; they are readily available on street corners:

I have made a few edits to these posts, based on suggestions from Laci Babai.

P-cubed has given a positive answer to Marty’s question, with a construction involving an infinite sequence of free products with amalgamation.