SMMC 2022 A4

The Simon Marais Mathematics Competition happened last weekend. It is a maths competition for undergraduate students across Europe, Asia, Africa and Oceania. This post is about problem A4, which I submitted. I’ll talk a bit about where the problem came from, a generalisation, a conjecture and also provide a solution. The entire paper is available on the Marais website, and solutions should be put up there at some time in the near future.

Problem (SMMC 2022 A4)
Let $n$ be a positive integer, and let $q\geq 3$ be an odd integer such that every prime factor of $q$ is larger than $n$ . Prove that

$\frac{1}{n!(q-1)^n}\prod_{i=1}^n (q^i-1)$

is an integer that has no prime factor in common with $\displaystyle{\frac{q-1}{2}}$ .

Origins

Let $G=GL_n(\mathbb{F}_q)$ and let $N$ be the subgroup of monomial matrices (a matrix is a monomial matrix if and only if it has exactly one nonzero entry in each row and column). I show below in my solution that this question is equivalent to the fact that the integer $|G|/|N|$ is coprime to $(q-1)/2$ . Now why would I ever care about that?

This coprimality fact implies that the cohomology of $G$ with mod $(q-1)/2$ coefficients is isomorphic to the cohomology of $N$ with mod $(q-1)/2$ coefficients. And I was interested in these cohomology groups because the second cohomology group classifies central extensions, which is what I used to think about back in my PhD days. The group $N$ feels somewhat more “combinatorial” than $G$ , so it is nice to be able to pass information from $N$ to $G$ for free.

Generalisations (known and conjectural)

Let $G$ be a split reductive group over $\mathbb{F}_q$ , which I conflate with its $\mathbb{F}_q$ -points below in an abuse of notation. Let $T$ be a maximal split torus and $N$ its normaliser in $G$ . Then

$\frac{|G|}{|N|}=q^{|\Phi^+|}\prod_i \frac{q^{d_i}-1}{d_i(q-1)}.$

Here $\Phi^+$ is the set of positive roots and the collection of integers $\{d_i\}$ are the exponents of the Weyl group. Then the same argument as in my proof below shows that this fraction is an integer, relatively prime to $\frac{q-1}{2}$ .

If we remove the assumption that $G$ is split, then I suspect the same conclusion is satisfied, but there is an additional argument needed as the formula for the quotient has additional factors. I have not worked out this argument and really don’t want to resort to case by case arguments, so there is your conjecture (I expect we now need to say $T$ is a maximal torus containing a maximal split torus).

Solution

First we show that the fraction in the question is an integer. Since $q-1$ divides $q^d-1$ as a polynomial for all $d$ , the statement only depends on the residue class of $q$ modulo $n!$ . Since every prime factor of $q$ is greater than $n$ , $q$ is relatively prime to $n!$ . So by Dirichlet’s theorem on primes in arithmetic progressions, we may assume without loss of generality that $q$ is prime.

Let $G=GL_n(\mathbb{F}_q)$ and let $N$ be the subgroup of monomial matrices. Then

$|G|=q^{\frac{n(n-1)}{2}}\prod_{d=1}^n q^d-1 \qquad \mbox{and}\qquad |N|=n!(q-1)^n.$

By Lagrange’s theorem $|G|/|N|$ is an integer. Since $q$ is relatively prime to $|N|$ , we can further divide by the largest power of $q$ in $|G|$ and deduce that

$\frac{|G|}{q^{\frac{n(n-1)}{2}}|N|}=\frac{1}{n!(q-1)^n}\prod_{d=1}^n (q^d-1)$

is an integer.

Now let $p$ be a prime dividing $\frac{q-1}{2}$ and let $d$ be a positive integer. To conclude, it suffices to show that the fraction

$\frac{q^d-1}{d(q-1)}$

has zero $p$ -adic valuation. Write $q=1+2m$ , then by the binomial theorem,

$\frac{q^d-1}{d(q-1)}=\frac{1}{2dm}\sum_{i=1}^d {d \choose i}(2n)^i.$

Let $a=v_p(d)$ and $b=v_p(m)$ . Since $v_p(i!)=\lfloor\frac{i}{p}\rfloor+\lfloor\frac{i}{p^2}\rfloor+\lfloor\frac{i}{p^3}\rfloor+\cdots <\frac{i}{p-1}$ , we get

$v_p\left( {d \choose i}(2n)^i \right)\geq v_p\left(\frac{d(2n)^i}{i!}\right)>a-\frac{i}{p-1}+i(b+v_p(2)).$

We have the inequality $v_p(2)-\frac{1}{p-1}\geq \frac{v_p(2)-1}{2}$ , so

$v_p\left( {d \choose i}(2n)^i \right)>a+i\left(b+\frac{v_p(2)-1}{2}\right).$

For $i\geq 2$ , we therefore get

$v_p\left( {d \choose i}(2n)^i \right)>a+2(b-\frac{1}{2})=a+2b+v_p(2)-1\geq a+b+v_p(2)=v_p(2dm)$

as $b\geq 1$ from our assumption that $p$ divides $m$ .
Thus in our sum, the term with $i=1$ has a strictly smaller $p$ -adic valuation than every other term, so determines the $p$ -adic valuation of the sum, and we get

$v_p \left( \frac{q^d-1}{d(q-1)} \right)=v_p\left(\frac{2md}{2md}\right)=0,$

completing the proof.

SMMC 2022 A4

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