Covers of groups

Here is the second of these discussions of elementary problems on finite groups with unexpected ramifications. This topic was suggested to me by Hamid Reza Dorbidi form Jiroft, Iran.

Let F be a finite set of finite groups. A finite group G is a cover of F if every group in F is isomorphic to a subgroup of G; it is minimal if no proper subgroup of G is a cover, and minimum if no group of smaller order than G is a cover. Covers certainly exist; for example, the direct product of the groups in F is a cover of F. But we would like more economical covers, that is, minimal or (especially) minimum covers. Note that any cover contains a minimal cover, but not necessarily a minimum one.

A paper on this topic just went on the arXiv, 2311.15652. I am not going to tell you about everything in that paper; I want to highlight some questions we cannot answer.

Perhaps the major question we can’t answer is this. It is clear that a set F can have only finitely many minimum covers. Which sets F have infinitely many minimal covers? Consider the set consisting of two cyclic groups of prime orders q and r. If (q,r) = (2,3), there are only three minimal covers, the cyclic group of order 6, S₃, and A₄. But if q and r are both greater than 5 there are infinitely many. So the question is quite subtle!

A set of nilpotent groups has a nilpotent minimum cover, and a set of p-groups has a minimum cover which is a p-group. But we do not know whether a set of abelian groups has a minimum cover which is abelian. We do, on the other hand, know the order of the smallest abelian group covering a set of abelian p-groups; I’ll describe this shortly.

Also, we do not know whether a set of p-groups can have infinitely many minimal covers, even if we strengthen the definition, and say that a cover is strongly minimal if it has no proper subgroup or quotient which is a cover.

We could also ask whether a set of soluble groups has a minimum cover which is soluble. But this question has a simple negative answer. The alternating group of degree 4 and the dihedral group of order 10 are both subgroups of A₅, but no other group of equal or smaller size contains both of them.

If F is a set consisting of two non-abelian simple groups, then a minimum cover is either their direct product or simple. Both cases can occur:

• the minimum cover of PSL(2,8) and A₅ is their direct product;
• the minimum cover of PSL(2,7) and A₆ is A₇.

This suggests a few questions:

1. The pairs of simple groups of the same order are all known. Can any of these be minimum covers of a set of two simple groups?
2. Do there exist three simple groups with the property that the product of the orders of the first two is the order of the third? If so, can both of these groups of the same order be minimum covers of the set consisting of the first two?
3. Is it the case that, for “most” pairs of simple groups, the unique minimum cover is their direct product?

Another type of question concerns minimum n-covers; these are minimum covers for the set of all groups of order n. Note that Cayley’s theorem shows that the symmetric group S_n is an n-cover, but usually not the smallest such. For example, if n is a power of a prime p (and n>2), then the Sylow p-subgroup of S_n is a smaller n-cover.

If n is a power of the prime p, then the order of a minimum n-cover is a power of p. But which power? In particular, is there an upper bound for the size of a minimum p^m-cover of the form p^F(m), where the function F is independent of p? If so, what is its order of magnitude? (We have a quadratic lower bound, but no upper bound since we do not even know whether such a function exists. The upper bound given by the Sylow p-subgroup of S_n is not of this form: it is p^{(pm−1)/(p−1)}. (This gives the right answer for 4, but for 8 it gives 128 whereas the correct value is 32.)

On the subject of small p-groups, we know that for odd primes p, the minimum size of a p³-cover is either p⁶ or p⁷, but do not know which it is. This question should perhaps be resolvable. I will sketch here the proof that there is no cover of order p⁵ for p > 3. Such a group would have nilpotency class at most 4, and so would be a regular p-group. In such a group, the elements of order 1 or p form a subgroup A. Since the group must contain both the elementary abelian group of order p³ and the non-abelian group of this order with exponent p, the order of A is at least p⁴. But G also contains the cyclic group C of order p³, and the intersection of A and C has order at most p, so their product has order at least p⁶.

Computation shows that there are minimal 27-covers of order 729; indeed there are 16 such groups.

Note that, if F is a set of abelian p-groups, then the order of the (unique) smallest abelian group containing them is known. The algorithm to find the group is as follows. Write each group in F as a sum of cyclic p-groups of non-increasing order, say of orders p^a(1), …, p^a(k). We can assume that we have the same value of k for all the groups, by adding “dummy” copies of the trivial group. Let b(j) be the maximum of all the values a(j) for fixed j, over all the groups in F. Then the smallest abelian group containing all the groups in F is the sum of cyclic groups of orders p^b(1), …, p^b(k); and this is its canonical form, since the bs will be non-increasing.

Using this, we can write down the order of the unique smallest abelian group containing all abelian groups of order p^m: It is p^Φ(m), where Φ(m) is the sum of the floor of m/k as k runs from 1 to m. The function Φ was studied by Dirichlet, and has many interpretations; its order of magnitude is roughly m log m, and its values appear as sequence A006218 in the On-Line Encyclopedia of Integer Sequences, where many references can be found.

We do not know whether there is a smaller non-abelian p-group which embeds all abelian groups of order p^m.