# Research Updates: Boston–Shalev for conjugacy classes, growth in linear groups, and the (amazing) Kelley–Meka result

1. Boston–Shalev for conjugacy classes

Last week Daniele Garzoni and I uploaded to the arxiv a preprint on the Boston–Shalev conjecture for the conjugacy class weighting. The Boston–Shalev conjecture in its original form predicts that, in any finite simple group $G$, in any transitive action, the proportion of elements acting as derangements is at least some universal constant $c > 0$. This conjecture was proved by Fulman and Guralnick in a long series of papers. Daniele and I looked at conjugacy classes instead, and we found an analogous result to be true: the proportion of conjugacy classes containing derangements is at least some universal constant $c' > 0$.

Our proof depends on the correspondence between semisimple conjugacy classes in a group of Lie type and polynomials over a finite field possibly with certain restrictions: either symmetry or conjugate-symmetry. We studied these sets of polynomials from an “anatomical” perspective, and we needed to prove several nontrivial estimates, e.g., for

• the number of polynomials with a factor of a given degree (which is closely related the “multiplication table problem”),
• the number of polynomials with an even or odd number of irreducible factors,
• the number of polynomials with no factors of small degree,
• or the number of polynomials factorizing in a certain way (e.g., as $f = gg^*$, $g$ irreducible, $g^*$ the reciprocal polynomial).

For a particularly neat example, we found that, if the order of the ground field is odd, exactly half the self-reciprocal polynomials have an even number of irreducible factors — is there a simple proof of this fact?

2. Growth in Linear Groups

Yesterday Brendan Murphy, Endre Szabo, Laci Pyber, and I uploaded a substantial update to our preprint Growth in Linear Groups, in which we prove one general form of the “Helfgott–Lindenstrauss conjecture”. This conjecture asserts that if a symmetric subset $A$ of a general linear group $\mathrm{GL}_n(F)$ ($n$ bounded, $F$ an arbitrary field) exhibits bounded tripling, $|A^3| \le K|A|$, then $A$ suffers a precise structure: there are subgroup $H \trianglelefteq \Gamma \le \langle A \rangle$ such that $\Gamma / H$ is nilpotent of class at most $n-1$, $H$ is contained in a bounded power $A^{O_n(1)}$, and $A$ is covered by $K^{O_n(1)}$ cosets of $\Gamma$. Following prodding by the referee and others, we put a lot more work in and proved one additional property: $\Gamma$ can be taken to be normal in $\langle A \rangle$. This seemingly technical additional point is actually very subtle, and I strongly doubted whether it was true late into the project, more-or-less until we actually proved it.

We also added another significant “application”. This is not exactly an application of the result, but rather of the same toolkit. We showed that if $G \le \mathrm{GL}_n(F)$ (again $F$ an arbitrary field) is any finite subgroup which is $K(n)$-quasirandom, for some quantity $K(n)$ depending only on $n$, then the diameter of any Cayley graph of $G$ is polylogarithmic in the order of $|G|$ (that is, Babai’s conjecture holds for $G$). This was previously known for $G$ simple (Breuillard–Green–Tao, Pyber–Szabo, 2010). Our result establishes that it is only necessary that $G$ is sufficiently quasirandom. (There is a strong trend in asymptotic group theory of weakening results requiring simplicity to only requiring quasirandomness.)

The intention of our paper is more-or-less to “polish off” the theory of growth in bounded rank. By contrast, growth in high-rank simple groups is still poorly understood.

3. The Kelley–Meka result

Not my own work, but it cannot go unmentioned. There was a spectacular breakthrough in additive combinatorics last week. Kelley and Meka proved a Behrend-like upper bound for the density of a subset $A \subset \{1, \dots, n\}$ free of three-term arithmetic progressions (Roth’s theorem): the density of $A$ is bounded by $\exp(-c (\log n)^\beta)$ for some constants $c, \beta > 0$. Already there are other expositions of the method which are also worth looking at: see the notes by Bloom and Sisask and Green (to appear, possibly).

Until this work, density $1 /\log n$ was the “logarithmic barrier”, only very recently and barely overcome by Bloom and Sisask. Now that the logarithmic barrier has been completely smashed, it seems inevitable that the new barometer for progress on Roth’s theorem is the exponent $\beta$. Kelley and Meka obtain $\beta = 1/11$, while the Behrend construction shows $\beta \le 1/2$.

# Groups of order 2023

Since the current year commonly features in math competition problems, it seems useful to begin the year by noting that ${2023 = 7 \cdot 17^2}$.

The past couple years around this time I have set myself the further puzzle of working out the groups of that order. Usually it is not difficult, but occasionally it will be essentially impossible (e.g., in ${2048}$).

This year is not very interesting actually, group-theoretically. This is an abelian year in the sense that all groups of order ${2023}$ are abelian, and the only possibilities are ${C_{2023}}$ and ${C_7 \times C_{17} \times C_{17}}$. This is an easy consequence of Sylow’s theorem. Let ${G}$ be a group of order ${2023}$. The number of Sylow ${7}$-subgroups must divide ${17^2}$ and be ${\equiv 1 \pmod 7}$, so it must be ${1}$, and similarly the number of Sylow ${17}$-subgroups must also be ${1}$, so ${G \cong C_7 \times H}$ where ${H}$ has order ${17^2}$. A group of order ${p^2}$ must be abelian (standard exercise), so ${H \cong C_{289}}$ or ${C_{17} \times C_{17}}$.

To make it more challenging, can we actually characterize the positive integers ${n}$ such that all groups of order ${n}$ are abelian? Call such an integer abelian.

Proposition 1 Let ${n}$ be a positive integer. Then ${n}$ is abelian if and only if
(1) ${n}$ is cube-free,
(2) if ${p, q}$ are distinct primes such that ${pq \mid n}$ then ${p}$ does not divide ${q-1}$,
(3) if ${p, q}$ are distinct primes such that ${pq^2 \mid n}$ then ${p}$ does not divide ${q+1}$.

Let us first show that the conditions are necessary. Suppose (1) fails. Then ${p^3 \mid n}$ for some prime ${p}$. There is a nonabelian group ${H}$ of order ${p^3}$ (see extraspecial groups), so ${H \times C_{n / p^3}}$ is a nonabelian group of order ${n}$. Similarly if (2) fails then there is a nonabelian semidirect product ${C_q : C_p}$, so we can again form a nonabelian group of order ${n}$. Finally if (3) fails then note that ${\mathrm{Aut}(C_q^2) \cong \mathrm{GL}_2(q)}$, and ${\mathrm{GL}_2(q)}$ contains an element of order ${q^2 - 1}$ (since the finite field of order ${q^2}$ has a primitive root), so we have another nonabelian semidirect product ${(C_q \times C_q) : C_p}$.

Now we show that the conditions are sufficient. We use induction on ${n}$. Let ${G}$ be a group of order ${n}$, where ${n}$ satisfies conditions (1), (2), and (3). Assume first ${G}$ is not simple. Let ${N}$ be a maximal normal subgroup of ${G}$. Then ${|N|}$ and ${|G/N|}$ are both integers smaller than ${n}$ satisfying (1), (2), and (3), so ${N}$ and ${G/N}$ are abelian by induction. Since all subgroups of ${G/N}$ are normal and ${N}$ was chosen to be maximal normal, ${G/N}$ must be cyclic of prime order, say ${p}$. Let ${P \le G}$ be a Sylow ${p}$-subgroup and let ${H \le N}$ be the product of the Sylow ${q}$-subgroups for ${q \ne p}$. Then it follows that ${G}$ is the semidirect product ${G = H : P}$.

Next consider the conjugation action of ${P}$ on ${H}$. Let ${Q \le H}$ be a Sylow ${q}$-subgroup. Then ${Q}$ is isomorphic to one of ${C_q, C_{q^2}, C_q \times C_q}$, and ${\mathrm{Aut}(Q)}$ is respectively isomorphic to one of ${C_{q-1}, C_{q(q-1)}, \mathrm{GL}_2(q)}$. Note that ${|\mathrm{GL}_2(q)| = q (q-1)^2(q+1)}$. By the given conditions, none of these has order divisible by ${p}$, so ${P}$ acts trivially on ${Q}$. Hence ${P}$ acts trivially on ${H}$ and ${G = H \times P}$ is abelian, as claimed.

To finish we must rule out the possibility that ${G}$ is simple. This seems to require some nontrivial group theory. One lazy method is to note that condition (2) implies that ${n}$ is odd (or a power of ${2}$), so we are done by the Feit–Thompson theorem. An easier argument is to use Burnside’s transfer theorem, which states that if ${P}$ is a Sylow ${p}$-subgroup and ${C_G(P) = N_G(P)}$ then ${G}$ has a normal subgroup of order ${|G:P|}$ (in particular ${G}$ is not simple). In our case, let ${p}$ be the smallest prime dividing ${n}$ and let ${P}$ be a Sylow ${p}$-subgroup. Let ${C = C_G(P)}$ and ${N = N_G(P)}$. Then ${N / C}$ is isomorphic to a subgroup of ${\mathrm{Aut}(P)}$. As before ${\mathrm{Aut}(P)}$ is one of ${C_{p-1}}$, ${C_{p(p-1)}}$, or ${\mathrm{GL}_2(p)}$. The orders of these groups have no prime factor larger than ${p+1}$. Since ${p}$ is the smallest prime dividing ${n}$ and ${p > 2}$, it follows that ${N / C}$ is trivial, so we are done.

It is also possible to characterize nilpotent numbers in general, and even solvable numbers. See Pakianathan, J., & Shankar, K. (2000). Nilpotent Numbers. The American Mathematical Monthly, 107(7), 631–634. https://doi.org/10.2307/2589118.