Sean Eberhard

Happy new year!

Updates from me: As of September I have taken up a new position at the University of Warwick. We’ve bought new wellies and we are enjoying the Warwickshire countryside, which has been pretty frosty so far. Mathematically, I am mentoring a postdoc and a new PhD student. As of the new year I have also just become an editor for Discrete Analysis.

Time for some numerical facts about the number ${2025}$ , and there are some good ones this year! First of all, ${2025}$ is a square: it is ${45^2}$ . Nice year to be ${45}$ ! The last square year was ${1936}$ and the next is ${2116}$ , so unless you are almost ${90}$ or younger than about ${10}$ this will be the only square year in your lifetime. Not only that, but ${45}$ is a triangular number, which makes 2025 a squared triangular number, so as observed on reddit and elsewhere we have

$\displaystyle 2025 = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)^2 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3.$

An even more profound and important discovery was made by the Alex Bellos of the Guardian: in fact ${2025 = (20 + 25)^2}$ . I will leave it to you to reflect on whether that’s as interesting.

My own personal nerdy exercise of the new year the past few years is to find all groups of order ${n}$ in year ${n}$ , where this year ${n = 2025 = 3^4 5^2}$ . This year is a step up in difficulty due to the presence of the large power ${3^4}$ .

The first step is to find the possibilities for the Sylow subgroups ${P}$ and ${Q}$ of order ${3^4}$ and ${5^2}$ respectively. There are just two groups of order ${5^2}$ , namely ${C_5^2}$ and ${C_{25}}$ , but actually ${15}$ different groups of order ${3^4}$ : ${5}$ abelian, ${6}$ of class ${2}$ , and ${4}$ of class ${3}$ . This is a calculation of Burnside, the first of three times Burnside will be mentioned in this post.

Next, I claim that every group ${G}$ of order ${2025}$ has the split structure either ${3^4 : 5^2}$ or ${5^2 : 3^4}$ , i.e., either ${P}$ or ${Q}$ is normal. This is a marginally more difficult version of a standard exercise in the use of Sylow’s theorem. By Burnside’s theorem, ${G}$ is solvable. By considering a minimal quotient of ${G}$ it follows that ${G}$ has a normal subgroup ${N}$ of index ${|G:N| \in \{3, 5\}}$ . Suppose ${|G:N| = 3}$ . Then ${Q \le N}$ , and since ${|N| = 3^3 5^2}$ and ${3, 3^2, 3^3 \not\equiv 1 \pmod 5}$ it follows from Sylow’s third theorem that ${Q}$ is the unique Sylow ${5}$ -subgroup of ${N}$ . Therefore ${Q}$ is characteristic in ${N}$ , so normal in ${G}$ . An analogous argument applies if ${|G:N| = 5}$ , since symmetrically ${5}$ mod ${3}$ has order ${2}$ .

Thus the problem reduces to classifying extensions of the form ${P \rtimes Q}$ and ${Q \rtimes P}$ . There are ${2 \cdot 15 = 30}$ possible direct products ${P \times Q}$ , so we may focus on the non-direct semidirect products.

First consider the extensions of the form ${Q \rtimes_\alpha P}$ where ${\alpha : P \to \mathrm{Aut}(Q)}$ is a nontrivial homomorphism. In particular ${|\mathrm{Aut}(Q)|}$ should be divisible by ${3}$ . Since ${|\mathrm{Aut}(C_{25})| = 20}$ it follows that ${Q = C_5^2}$ and ${\mathrm{Aut}(Q) \cong \mathrm{GL}_2(5)}$ . Note that ${\mathrm{GL}_2(5)}$ has order ${480}$ and its quotient ${\mathrm{PGL}_2(5)}$ is isomorphic to ${S_5}$ . Therefore the Sylow ${3}$ -subgroups of ${\mathrm{GL}_2(5)}$ have order ${3}$ , they are all conjugate, and moreover the nontrivial automorphism of ${C_3}$ is also realized an inner automorphism of ${\mathrm{GL}_2(5)}$ (since that is true in ${S_5}$ ). As discussed last year, we often have ${P \rtimes_\alpha Q \cong P \rtimes_\beta Q}$ for different ${\alpha, \beta}$ , and we really only care about equivalence classes of homomorphism ${\alpha \in \mathrm{Hom}(P, \mathrm{Aut}(Q))}$ up to the natural action of ${\mathrm{Aut}(P) \times \mathrm{Aut}(Q)}$ (where ${\mathrm{Aut}(P)}$ acts by precomposition and ${\mathrm{Aut}(Q)}$ acts by conjugation). Since the image of ${\alpha}$ must be contained in a Sylow ${3}$ -subgroup of ${\mathrm{Aut}(Q)}$ , we get a unique equivalence class ${[\alpha]}$ for each automorphism class of maximal subgroups of ${P}$ . We now have to go through the list of possibilities for the Sylow ${3}$ -subgroup ${P}$ and count the automorphism classes of index- ${3}$ subgroups. This can be done in GAP with a bit of torment, which I will spare you, but the result is that each of the groups of order ${3^4}$ have ${1}$ , ${2}$ , or ${3}$ classes of maximal subgroup, and altogether there are ${31}$ essentially different pairs ${(P, M)}$ where ${P}$ is a group of order ${3^4}$ and ${M < P}$ is a maximal subgroup, and thus there are ${31}$ isomorphism classes of extensions of the form ${C_5^2 \rtimes P}$ .

It is similar with extensions of the form ${P \rtimes Q}$ . In this case ${|\mathrm{Aut}(P)|}$ has to be divisible by ${5}$ , and I claim this happens only for ${P = C_3^4}$ . Indeed, otherwise, ${P / \Phi(P) \cong C_3^d}$ for some ${d \le 3}$ and therefore ${Q}$ acts trivially on ${P / \Phi(P)}$ , which implies by a lemma of Burnside that ${Q}$ acts trivially on ${P}$ (see (24.1) in Aschbacher’s Finite Group Theory). Now ${\mathrm{GL}_4(3)}$ has order- ${5}$ Sylow ${5}$ -subgroups, so similarly to the previous paragraph we find that the number of classes of homomorphisms ${\alpha : Q \to \mathrm{Aut}(P)}$ is just the number of automorphism classes of maximal subgroups of ${Q}$ . Both ${C_{25}}$ and ${C_5^2}$ have a unique such class, so the number of nontrivial extensions of the form ${P \rtimes Q}$ is just ${2}$ : there is one of the form ${C_3^4 \rtimes C_{25}}$ and one of the form ${C_3^4 \rtimes C_5^2 = (C_3^4 \rtimes C_5) \times C_5}$ .

Thus altogether there are ${30 + 31 + 2 = 63}$ groups of order ${2025}$ ! Happy new year!

Recall that a Moore graph (of diameter 2) is a regular graph of girth ${5}$ and diameter ${2}$ . Equivalently, it is a triangle-free graph such that any two nonadjacent vertices have a unique common neighbour. There are famously few Moore graphs: only the 5-cycle of degree 2, the Petersen graph of degree 3, and the Hoffman–Singleton graph of degree 7, and possibly a graph (or even several graphs) of degree ${57}$ and order ${3250}$ (the “missing Moore graph”). This is the Hoffman–Singleton theorem.

The Hoffman–Singleton theorem being too mysterious, it is tempting to try to find a weakening of the definition of a Moore graph that allows a few more graphs but still finitely many. One hopes that the “finitely many” part would have some more direct combinatorial proof (as opposed to the algebraic voodoo that goes into Hoffman–Singleton). Such a conjecture was recently posed by Wood and stated in a paper of Devillers, Kamcev, McKay, O Cathain, Royle, Van de Voorde, Wanless, and Wood.

Conjecture 1 For every integer ${t \ge 2}$ there are only finitely many triangle-free ${K_{2,t}}$ -free graph with diameter ${2}$ apart from stars.

For ${t = 2}$ this is immediate from Hoffman–Singleton. The authors (Devillers, Kamcev, …) find almost ${6}$ million examples of such graphs with ${t=3}$ , but “nothing suggestive of an infinite class”. However, the following construction (which arose in conversations with Padraig O Cathain) disproves the conjecture for ${t = 7}$ .

Let ${p}$ be a prime congruent to ${11}$ mod ${12}$ . This condition ensures that ${-1}$ and ${-3}$ are quadratic nonresidues. Let ${G = \mathbf{F}_p^2}$ and let ${A = \{(x, \pm x^2) : x \in \mathbf{F}_p, x \ne 0\}}$ . Now consider the Cayley graph ${\mathrm{Cay}(G, A)}$ . Equivalently, this is the graph with vertex set $\mathbf{Z}/p\mathbf{Z} \times \mathbf{Z}/p\mathbf{Z}$ and $(x_1, y_1)$ joined to $(x_2, y_2)$ if and only if $y_1 - y_2 = \pm (x_1 - x_2)^2 \ne 0$ . We claim ${\mathrm{Cay}(G, A)}$ is triangle-free, ${K_{2,7}}$ -free, and diameter-2.

Claim 1: ${\mathrm{Cay}(G, A)}$ is triangle-free. Equivalently, ${A}$ is sum-free.

Proof: We must check that there do not exist nonzero ${x, y, z}$ such that ${(x, \pm x^2) + (y, \pm y^2) + (z, \pm z^2) = 0}$ . Equivalently, we must check that

$\displaystyle (x+y)^2 \pm x^2 \pm y^2 = 0$

implies ${xy(x+y) = 0}$ . There are four cases, depending on the two signs:

${(-, -)}$ : We get ${2xy = 0}$ .

${(-, +)}$ : We get get ${2(x+y)y = 0}$ .

${(+, -)}$ : We get ${2(x+y)x = 0}$ .

${(+, +)}$ : We get ${x^2 + xy + y^2 = 0}$ . Since the polynomial ${X^2 + X + 1}$ has discriminant ${-3}$ , which is a quadratic nonresidue, it follows that ${x = y = 0}$ .

Claim 2: ${\mathrm{Cay}(G, A)}$ is ${K_{2,7}}$ -free and diameter- ${2}$ . In fact, we have ${2 \le 1_A * 1_A(g) \le 6}$ for every nonzero ${g \notin A}$ .

Here ${1_A * 1_A(g)}$ is the number of solutions to ${a_1 + a_2 = g}$ with ${a_1, a_2 \in A}$ . It is an exercise to check that the first statement follows from the second, so we just prove the second.

Proof: Let ${g = (z, w)}$ be a nonzero point not in ${A}$ . First assume ${z \ne 0}$ . The value of ${1_A*1_A(g)}$ is the number of quadruples ${(x, y, \epsilon_1, \epsilon_2)}$ with ${x, y \in \mathbf{F}_p}$ nonzero and ${\epsilon_1, \epsilon_2 = \pm 1}$ such that

$\displaystyle x + y = z, \qquad \epsilon_1 x^2 + \epsilon_2 y^2 = w.$

For fixed ${z, w, \epsilon_1, \epsilon_2}$ is equivalent to count ${x \ne 0}$ such that

$\displaystyle \epsilon_1 x^2 + \epsilon_2 (x-z)^2 = w.$

If the signs are equal then this is a nontrivial quadratic equation, so there are at most ${2}$ solutions. If the signs are different then we get a linear equation

$\displaystyle 2xz - z^2 = \epsilon_1 w,$

which has a unique solution. If that solution is ${0}$ or ${z}$ then we get ${w = \pm z^2}$ , contrary to the assumption that ${g \notin A}$ .

Now assume ${z = 0}$ and ${w}$ is nonzero. Then ${1_A * 1_A(g)}$ is the number of triples ${(x, \epsilon_1, \epsilon_2)}$ with ${x}$ nonzero such that ${(\epsilon_1 + \epsilon_2) x^2 = w}$ . If the signs are different there are no solutions, since ${w}$ is nonzero. Otherwise the equation is ${2 \epsilon_1 x^2 = w}$ . Since ${-1}$ is a quadratic nonresidue, one of the choices of sign gives ${0}$ solutions and the other gives ${2}$ solutions. Therefore ${1_A * 1_A(g) = 2}$ .