An asymptotic for the Hall–Paige conjecture

Problem 1 Let {G} be a group of order {n}. Is there a bijection {f \colon G \to G} such that the map {x \mapsto x f(x)} is also a bijection?

For example, if {G} has odd order, you can just take {f(x) = x}. Then {x f(x) = x^2}, and because every element has odd order this defines a bijection. If {G = \mathbf{F}_2^d}, then it suffices to find a linear map without an eigenvalue. Cyclic groups of even order never have complete mappings (option 1: stop reading and try to prove this as a diverting puzzle; option 2: read on, and it will be explained and become essential).

In general, a solution to Problem 1 is called a complete mapping. The definition of complete mapping is a little strange. Here is some motivation, if you want some. The construction of finite projective planes is a problem going back to Euler. You can think of a finite projective plane as a collection of {n-1} mutually orthogonal {n \times n} Latin squares, where two Latin squares are termed orthogonal if, when superimposed, the {n^2} pairs of symbols are distinct. The most familiar Latin square is the cyclic Latin square, which has entries {i + j \bmod n} for {0 \leq i, j < n}. More generally, for any group {G} of order {n}, the multiplication table of {G} is an {n \times n} Latin square. If you write down what it means for another square to be orthogonal to the {G}-based Latin square, you will find yourself re-inventing the definition of complete mapping.

Another motivation, of possibly broader interest, is simply that counting complete mappings turns out to be interesting and difficult, and requires us to hone some counting skills that, we hope, will be applicable elsewhere.

The concerted effort to answer Problem 1 began with Hall and Paige (1955). First, there is an obstruction, beginning with cyclic groups of even order. More generally, consider the abelianization {G^\textup{ab}} of {G}, and suppose all the elements of {G} have nontrivial sum in {G^\textup{ab}}. Then if {f} is a bijection the elements {x f(x)} will sum up to zero, so {x f(x)} cannot possibly be a bijection. Therefore in order for a complete mapping to exist, we must have {\prod G \in [G, G]}. Hall and Paige proved that this condition is equivalent to the Sylow 2-subgroups of {G} being trivial or noncyclic. Henceforth this condition will be called the Hall–Paige condition. Conversely, Hall and Paige conjectured that Problem 1 has a solution as long as {G} satisfies their condition.

Conjecture 2 (Hall–Paige) Every group satisfying the Hall–Paige condition has a complete mapping.

Progress on the Hall–Paige conjecture was slow. Hall and Paige (1955) proved the conjecture for solvable groups and symmetric and alternating groups, but progress on the full conjecture didn’t really get under way until the classification was established. For a complete history, see the book by Evans. Aschbacher (1990) proved various restrictions on the form of a minimal counterexample. The real breakthrough came in 2009 when Wilcox showed that a minimal counterexample would have to be simple, and furthermore could not be any of the simple groups of Lie type, which leaves only the Tits group and the 26 sporadic groups. At this point the problem is in principle a finite check, but still a mammoth one. Combining Wilcox’s technology with extensive computer algebra, Evans ruled out all of the remaining groups, with one exception: the fourth Janko group {J_4}, a group of order {86775571046077562880 \approx 9 \times 10^{19}}, and finally {J_4} was ruled out by Bray in work that remained unpublished until last year. Thus the Hall–Paige conjecture was proved. Although many people contributed, it is customary to attribute the final result to Wilcox, Evans, and Bray.

Meanwhile, another strand of study was developing. Suppose we take just the cyclic group {G = C_n}. We know there is at least one complete mapping as long as {n} is odd. But how many are there?

Problem 3 How many complete mappings does the cyclic group {C_n} have?

This problem can be compared with the well-known {n} queens puzzle: how many ways can you place {n} queens on an {n \times n} chessboard so that no two are attacking? If you make the chessboard toroidal, so that going off one edge brings you back on the opposite one, and if you only allow the queens to use one of the two diagonals, then you get an equivalent question.

Heuristically, there are {n!} bijections {f}, and for each the function {x\mapsto x f(x)} has a roughly {n!/n^n} chance of being a bijection, so you can guess that the number of complete mappings is about {n!^2 / n^n}; this conjecture is attributed to Vardi (1991) (in a weaker form) and Wanless (2011).

Conjecture 4 (Vardi–Wanless) The number of complete mappings of {C_n} is asymptotically {(e^{-1} + o(1))^n n!}.

This is where additive combinatorics enters. A few years ago Freddie Manners, Rudi Mrazović, and I started thinking about Problem 3, motivated by the observation that it can be thought of as requesting the number of solutions to

\displaystyle  \pi_1 + \pi_2 = \pi_3

(or additive triples) with {\pi_1, \pi_2, \pi_3 \in S}, where {S \subset C_n^n} is the set of bijections. The usual tool for counting additive triples (or solutions to any linear equation) is Fourier analysis; hence the problem reduces to understanding the Fourier transform of {S}. After barking up this tree (for quite a while), we were eventually able to prove the Vardi–Wanless conjecture. In fact we proved something even more precise: the solution to Problem 3 turns out to be asymptotically

\displaystyle  (e^{-1/2} + o(1)) n!^2/n^{n-1},

which differs from the Vardi–Wanless guess (though they never made a guess so precise) in two important respects: there is an extra factor of {n}, and a factor of {e^{-1/2}} (what the hell?).

Since the key to unlocking Problem 3 is Fourier analysis, it seems like we are using the abelianness of {C_n} in an absolutely essential way, but today Freddie and Rudi and I have a new announcement: we can count complete mappings in nonabelian groups too, and the asymptotic is essentially unchanged.

Theorem 5 Let {G} be a group of order {n} satisfying the Hall–Paige condition. Then the number of complete mappings of {G} is asymptotically

\displaystyle  (e^{-1/2} + o(1)) |G^\textup{ab}| n!^2 / n^n.

Here are some other highlights from our paper:

  1. The asymptotic above is begging for a heuristic explanation. We give one! It uses something called the principle of maximum entropy, which I personally can’t wait to use again. Basically, if {f} is a random bijection, {x \mapsto x f(x)} is more prone to collisions than a random function, and, heuristically, has a Gibbs distribution with a particular partition function, and a calculation shows that its probability of being a bijection is therefore smaller by a constant factor.
  2. We evaluate the next term in the asymptotic! We actually show that the number of complete mappings is

    \displaystyle  e^{-1/2} (1 + (1/3 + \textup{inv}(G)/4)/n + O(1/n^2)) |G^\textup{ab}| n!^2/n^n,

    where {\textup{inv}(G)} is the proportion of involutions in {G}. This verifies another conjecture of Wanless: that elementary abelian {2}-groups have the most complete mappings of any group of the same order. We can keep going, in principle, but really we would rather not: there is a combinatorial explosion in the number of “collision types” (in a sense we make precise), and to give the next term in the asymptotic you need to sum over all of these.

  3. Orthogonally, we can give effective estimates instead of asymptotics. For example, we can show that as long as {|G| > 10^5} and all nontrivial complex representations of {G} have degree at least {21}, then {G} has a positive number of complete mappings (in fact, within a constant factor of {n!^2/n^n} such). By combining a few such effective statements, we can cover all the simple sporadic groups, apart from the two smallest {M_{11}} and {M_{12}}. You can view this as a second-generation proof of the Evans–Bray contribution to the Hall–Paige conjecture. In fact, our proof covers all nonabelian finite simple groups except for some alternating groups, some {\textup{PSL}_2(q)'s}, and about ten other groups that we list.

The preprint is available at https://arxiv.org/abs/2003.01798. Any comments appreciated, as always!

Invariable generation of classical groups

Elements {g_1, \dots, g_k} in a group {G} invariably generate if they still generate after an adversary replaces them by conjugates. This is a function of conjugacy classes: we could say that conjugacy classes {\mathcal{C}_1, \dots, \mathcal{C}_k} in a group {G} invariably generate if {\langle g_1, \dots, g_k\rangle = G} whenever {g_i \in \mathcal{C}_i} for each {i}. This concept was invented by Dixon to quantify expected running time of the most standard algorithm for computing Galois groups: reduce modulo various primes {p}, get your Frobenius element {g_p}, and then try to infer what your Galois group is from the information that it contains {g_p} (which is defined only up to conjugacy) for each {p}. If {\textup{Gal}(f)} is secretly {G}, and you somehow know a priori that {\textup{Gal}(f) \leq G}, then the number of primes you need on average to prove that {\textup{Gal}(f) = G} is the expected number of elements it takes to invariably generate {G}.

For example, if {G = S_n}, then we know that four random elements invariably generate with positive probability, while three random elements almost surely (as {n\to\infty}) do not invariably generate. Therefore if {\textup{Gal}(f) = S_n} then it typically takes four primes to prove it.

A few days ago Eilidh McKemmie posted a paper on the arxiv which extends this result to finite classical groups: e.g., if {G} is {\textup{SL}_n(q)} then, for large enough constant {q} and {n\to\infty}, four random elements invariably generate with positive probability, but three do not. (The bounded-rank case is rather different in character, and I think two elements suffice.) The proof is pretty cool: invariable generation in {G} is related to invariable generation in the Weyl group, which is either {S_n} or {C_2 \wr S_n}, and we already understand invariable generation for these groups (using a small trick for the latter).

I believe the restriction to large enough constant {q} is a technical rather than essential problem. Assuming it can be overcome, we will be able to deduce the following rather clean statement: If {G} is a finite simple group then four random elements invariably generate {G} with probability bounded away from zero. Moreover, if the rank of {G} is unbounded then three random elements do not.