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.

Infinite dimensional vector spaces

This is a quick note to prove that two bases of an infinite dimensional vector space have the same cardinality. We freely use the axiom of choice and other standard facts about cardinalities of infinite sets. We will in fact prove the following:

Theorem: Let V be a vector space with basis \{v_i\}_{i\in I} with I an infinite set. Let \{w_j\}_{j\in J} be a linearly independent subset of V. (e.g. a basis of a subspace). Then |J|\leq |I|.

To prove this, WLOG J is a basis of V (by extending \{w_j\}_{j\in J} to a basis of V if necessary). For all i\in I, write

    \[v_i=\sum_j c_{ij}w_j.\]

Let E\subset I\times J be the set of pairs (i,j) with c_{ij}\neq 0. Then E\to I has finite fibres, since the sum above is finite, and E\to J is surjective, since the v_i lie in the span of the w_j with j in the image of V, but also the v_i generate V. Since I is assumed infinite, this is enough to prove that |I|\geq |J|, as required.