groups – Random Permutations

Next summer, Jitendra Bajpai and I are organizing an LMS–HIMR Research School on “Growth and Expansion in Groups” here at the University of Warwick, 6–10 July 2026.

The school is aimed at graduate students and early-career researchers working in group theory (finite or infinite), especially any growth or diameter-related questions, or adjacent fields. There will be a series of minicourses on topics such as expansion in finite and classical groups, random walks and Poisson boundary theory, analytic approaches to growth, and word maps.

Minicourses will be given by Daniele Dona, Anna Erschler, Itay Glazer, and Noam Lifschitz. Additional plenary talks will be given by Laura Ciobanu, Ben Green, Harald Helfgott, and Michael Magee.

For further information see the Research School webpage here:

https://warwick.ac.uk/fac/sci/maths/research/events/2025-2026/lms-summer-school-2026/

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 ${C_3^4 \rtimes C_5^2 = (C_3^4 \rtimes C_5) \times C_5}$ .