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.

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