4

This question is inspired by the cute answer to Order of general- and special linear groups over finite fields.

The formula $$ |\mathrm{GL}_n(\mathbb F_q)|=q^{\frac{n(n-1)}2}(q-1)(q^2-1)\cdots(q^n-1). $$ is obtained in that answer by simply but cleverly counting the number of linearly independent $n$-tuples of vectors in $\mathbb F_q^n$.

On the other hand, we know a lot about the structure of the group $\mathrm{GL}_n(\mathbb F_q)$: we can choose (in many ways) a maximal torus; let us take the one consisting of all invertible diagonal matrices, which is a subgroup of order $(q-1)^n$. We can next locate the unipotent radical of the corresponding Borel subgroup which in our case is the subgroup of all upper triangular matrices with $1$s along the main diagonal, thus has order $q^{\frac{n(n-1)}2}$. This accounts for $q^{\frac{n(n-1)}2}(q-1)^n$ elements.

How to account for the remaining factors $\frac{q^2-1}{q-1}$, $\frac{q^3-1}{q-1}$, ..., $\frac{q^n-1}{q-1}$? Do they also correspond to some subgroups that can be named, or maybe some explicitly describable conjugacy classes?

მამუკა ჯიბლაძე
  • 1,663
  • @ancientmathematician Of course, stupid me! Upper and lower triangulars are conjugate! I'll revise the question, thank you! – მამუკა ჯიბლაძე Mar 03 '23 at 10:11
  • 1
    Examples of subgroups of orders $q^{i} - 1$ may be seen as follows. I'll just do the case $i = n$. Consider the finite field $E = \operatorname{GF}(q^{n})$ as a vector space of dimension $n$ over $\operatorname{GF}(q) = \mathbb{F}{q}$. The multiplicative group $G$ of $E$, of order $q^{n} - 1$ acts on $E$ by multiplication, and the action is $\mathbb{F}{q}$-linear. – Andreas Caranti Mar 03 '23 at 10:32

1 Answers1

1

Let $F_q$ be a finite field of . $q$.

Let $T_n(F_q)$ the set of upper triangular matrices (the semi-dircet product of digoanl matrices and unipotent radical).

The quotient $GL(n,F_q)/ T_n(F_q)$ it the set of complete flags $0\subset E_1\subset E_2 \subset ..E_{n-1}\subset E_n=K^n$, where $E_1$ is a 1 dimensional subspace, $E_k$ a $k$-dimensional subsapce), so that for every $i$, $E_{i-1}$ is an hyperplane in $E_{i}$.

Then $q^i-1\over q-1$ appears as the number of $i-1$ dimensional subspace contained in a given $i$ dimensional space : indeed the number of hyperplane in a $d$ dimensionnel vector space over $F_q$ has cardinality $q^d-1\over q-1$ The product $1.{q^2-1\over q-1}... {q^{n-1}-1\over q-1}$ is of course the number of flags.

Thomas
  • 7,470
  • Excellent! So the formula is just $|B|\times|G/B|$ where $B$ is a Borel subgroup! And similarly $|\mathbf G_m|\times|\mathbf P^n|=|K^{n+1}|-1$ since $\mathbf P^n=(K^{n+1}\setminus{0})/\mathbf G_m$. By the way this also gives an interesting fact I never paid attention to: the complete flag variety $\mathbf{F}_n$ has the same cardinality as the product of projective spaces $\mathbf{P}^1\times\mathbf{P}^2\times\cdots\times\mathbf{P}^{n-1}.$ It is in fact most probably iteration of total spaces of projective bundles. – მამუკა ჯიბლაძე Mar 04 '23 at 07:13
  • By the way the last factor must be $\frac{q^n-1}{q-1}$ rather than $\frac{q^{n-1}-1}{q-1}$, right? – მამუკა ჯიბლაძე Mar 04 '23 at 07:19
  • Indeed, of $n=1$, $GL(n, F)=F^*$ has cardinality $q-1$, and $T(n,F)$ is reduce to the $(1)$ matrix. – Thomas Mar 06 '23 at 10:23