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 Any comments appreciated, as always!

Symmetric intersecting families

Jeff Kahn, Bhargav Narayanan, Sophie Spirkl, and I have just uploaded to the arxiv our note On symmetric intersecting families of vectors. We are interested in problems in extremal set theory with symmetry constraints. Often in extremal set theory it is useful to “compress”, which you can think of as trying to reduce your arbitrary system to a simpler system, one that may be easier to understand. With symmetry constraints, the added challenge is that your compression method should also satisfy those symmetry constraints.

Our particular focus is symmetric intersecting families. A family {\mathcal{A} \subset 2^{[n]}} of subsets {A \subset [n]} (where {[n] = \{1, \dots, n\}}) is termed intersecting if for any {A, B \in \mathcal{A}}, we have {A \cap B \neq \emptyset}. More generally {\mathcal{A}} is termed {r}-wise intersecting if for any {A_1, \dots, A_r \in \mathcal{A}} we have {A_1 \cap A_2 \cap \cdots \cap A_r \neq \emptyset}. How large can an {r}-wise intersecting family {\mathcal{A} \subset 2^{[n]}} be? Trivially, it can be as large as {2^{n-1}}: just take all sets containing some point {1 \in [n]}. But what if we demand that {\mathcal{A}} have a transitive symmetry group? If {r=2} we can just take the family of all sets of size greater than {n/2}, but if {r \geq 3} the situation is unclear. This question was answered by Frankl in 1981 for {r \geq 4}, and answered for {r = 3} only recently by Ellis and Narayanan.

The Ellis–Narayanan proof can be understood, at least if you squint, as a compression argument. Define the {p}-biased measure {\mu_p} on {2^{[n]}} by

\displaystyle  \mu_p(\{A\}) = p^{|A|} (1-p)^{n-|A|}.

Put another way, {\mu_p(\mathcal{A})} is the probability that {A \in \mathcal{A}} if {A} is chosen randomly by flipping {n} independent {p}-biased coins. We want to show that {\mu_{1/2}(\mathcal{A}) = o(1)} under the assumption that {\mathcal{A}} is {3}-wise intersecting and symmetric. This is not obvious, but what is obvious is that

\displaystyle  \mu_{2/3}(\mathcal{A}) \leq 2/3.

This is a bit like the basic observation that an intersecting family can have density at most {1/2}, because for each set {A} you can have only either {A} or {A^c}. For the above inequality, define three sets {A_1, A_2, A_3} by independently selecting two of them for each {x \in [n]} and including {x} in those two sets. Then {A_1, A_2, A_3} are coupled random variables, each individually having the {\mu_{2/3}} distribution, such that {A_1 \cap A_2 \cap A_3 = \emptyset}. The above inequality follows. Now although it is difficult to compress {\mathcal{A}} while preserving symmetry, we can “compress our counting method” {\mu_{1/2}} up to {\mu_{2/3}}. A sharp threshold theorem of Friedgut and Kalai (relying on the BKKKL theorem, from the analysis of boolean functions) implies that, under symmetry, the function {p \mapsto \mu_p(\mathcal{A})} has a sharp threshold: it changes from {\varepsilon} to {1-\varepsilon} over an interval of length {O(\log \varepsilon^{-1} / \log n)}. Since {\mu_{2/3}(\mathcal{A}) \leq 2/3}, this implies that {\mu_{1/2}(\mathcal{A})} must be small; in fact {\mu_{1/2}(\mathcal{A}) \leq n^{-c}} for some constant {c>0}.

In our new paper we looked at a further variant of this problem. Suppose {\mathcal{A}} is a subset of {[3]^n} which is intersecting (just {2}-wise now) in the sense that for any two vectors {x, y \in \mathcal{A}} there is an {i\in [n]} such that {x_i = y_i}. Again, assume that {\mathcal{A}} has a transitive group of symmetries. Must it be the case that {\mathcal{A}} has size {o(3^n)}? The problem is that there is no obvious {p}-biased measure to move around this time, but we got around this problem with the following cute device: Embed {[3]} into

\displaystyle  W = \{\{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}\}

in the obvious way, and replace {\mathcal{A}} with its up-closure in {W^n}. Let {\mu_0} be the uniform measure on the bottom level and {\mu_1} the uniform measure on the top level. Then {\mu_0(\mathcal{A})} is the thing we are interested in, and {\mu_{1/2}} is the uniform measure on {W^n}. Again it is obvious that {\mu_{1/2}(\mathcal{A}) \leq 1/2}, so we are done if we can prove a sharp threshold theorem for {\mu_p(\mathcal{A})}, and that’s what we did. We again get a bound of the form {n^{-c}}.

That bound {n^{-c}} is a little unsatisfying actually. We expect the truth to look like {\exp(-cn^\delta)} for some {\delta > 0}. Going back to the problem of {r}-wise intersecting families in {[2]^n}, this was proved by Cameron, Frankl, and Kantor for {r\geq 4}, and there are plenty of examples to show you can’t do better for any {r \geq 3}, but the upper bound for {r=3} remains mysterious. But at least the new proofs tell us where to look: we want a {3}-wise intersecting family with small total influence, or we want to prove that that’s impossible.