Standard subgroups of finite general linear groups

The ground field and the group. Let $n\ge 1$ and let $q$ be a prime power, that is $q=p^m$ for a prime $p$ and an integer $m\ge 1$. There exists a field with exactly $q$ elements (For every prime $p$ and $n\ge 1$, a field with $p^n$ elements exists); fix one and call it ${\mathbb{F}}_q$, so that $\mathrm{\mid }{\mathbb{F}}_q\mathrm{\mid }=q$ (Finite fields and their order). Throughout, matrices have entries in ${\mathbb{F}}_q$: $M_n({\mathbb{F}}_q)=\{A:\{1,\dots ,n\}\times \{1,\dots ,n\}\to {\mathbb{F}}_q\},G:={\mathrm{GL}}_n({\mathbb{F}}_q),$ the second being the set of invertible matrices, which is a group under matrix multiplication (Invertible matrices and the general linear group ${\mathrm{GL}}_n(F)$, ${\mathrm{GL}}_n(F)$ is a group under matrix multiplication, including the trivial group ${\mathrm{GL}}_0(F)$). Fixed throughout are the identity matrix $I_n$ and the $i$-th standard basis vector $e_i$ of ${\mathbb{F}}_q^n$.

Indexing convention. On this page the rows and columns of a matrix are numbered $1,\dots ,n$ and a matrix is written $A=(a_{ij}{)}_{1\le i,j\le n}$; this is the usual relabelling of the $0$-indexed convention of Finite rectangular matrices over a commutative ring, their entries, rows and columns. We write $e_1,\dots ,e_n$ for the standard unit vectors of ${\mathbb{F}}_q^n$, so that $e_i$ has entry $1$ in position $i$ and entry $0$ elsewhere; this is the $1$-indexed relabelling of the ordered basis $e_0,\dots ,e_{n-1}$ of The standard list $e:n\to F^n$ with $e_i(i)=1_F$ and $e_i(j)=0_F$ for $j\mathrm{\ne }i$ is an ordered basis of $F^n$; hence ${\dim }_F{F}^n=n$, and $F^0$ is the zero space with basis $\mathrm{\varnothing }$ and dimension $0$. The standard flag of $V={\mathbb{F}}_q^n$ is the chain $V_0:=\{0\}⊊V_1:=⟨e_1⟩⊊V_2:=⟨e_1,e_2⟩⊊\cdots ⊊V_n:=⟨e_1,\dots ,e_n⟩=V,$ whose members are the standard coordinate subspaces; each $V_i$ has ${\dim }_{{\mathbb{F}}_q}{V}_i=i$ and $V_{i-1}⊊V_i$ for $1\le i\le n$, since $e_1,\dots ,e_i$ is a basis of $V_i$ and $e_i\in V_i\setminus V_{i-1}$.

Upper triangular, diagonal and unitriangular matrices. A matrix $A=(a_{ij})\in M_n({\mathbb{F}}_q)$ is upper triangular when $a_{ij}=0$ for all $i>j$ and diagonal when $a_{ij}=0$ for all $i\mathrm{\ne }j$ (Upper triangular, lower triangular and diagonal square matrices over a commutative ring). It is upper unitriangular, or simply unitriangular, when it is upper triangular and $a_{ii}=1$ for every $i$. Define the following subsets of $G$: $B:=\{b\in G:b\text{ is upper triangular}\},T:=\{t\in G:t\text{ is diagonal}\},U:=\{u\in G:u\text{ is unitriangular}\}\mathrm{.}$ $B$ is the standard Borel subgroup, $T$ the standard (diagonal) torus and $U$ the standard maximal unipotent subgroup of $G$.

The basic structure. $B$, $T$ and $U$ are subgroups of $G$, we have $T\le B$ and $U\le B$, and $T\cap U=\{I_n\}$. Here $I_n\in B\cap T\cap U$, and each of the three sets is closed under products and inverses: the product of two upper triangular (respectively diagonal, respectively unitriangular) matrices is upper triangular (respectively diagonal, respectively unitriangular), and the inverse of an invertible upper triangular matrix is upper triangular, with diagonal entries $a_{ii}^{-1}$ in the diagonal case. For instance, if $u=I_n+N$ with $N$ strictly upper triangular, then $N^n=0$ and $u^{-1}=I_n-N+N^2-\cdots +(-1{)}^{n-1}{N}^{n-1},$ which is again unitriangular; the displayed product is computed with the associative multiplication and distributivity of Matrix multiplication is associative, unital, distributive, and compatible with scalar multiplication, and $N^n=0$ because $N$ moves every vector at least one step up the flag $0\le ⟨e_1⟩\le ⟨e_1,e_2⟩\le \cdots \le {\mathbb{F}}_q^n$. The notation $U$ is consistent with "unipotent": every $u\in U$ satisfies $(u-I_n{)}^n=0$.

Every element of $B$ is a unique product $tu$. Let $b=(b_{ij})\in B$. Since $b$ is upper triangular and invertible, its determinant is the product $b_{11}{b}_{22}\cdots b_{nn}$ (The determinant of a triangular matrix is the product of its diagonal entries), so $b_{ii}\mathrm{\ne }0$ for every $i$; hence $t:=\mathrm{diag}(b_{11},\dots ,b_{nn})\in T,u:=t^{-1}b,$ where $t^{-1}=\mathrm{diag}(b_{11}^{-1},\dots ,b_{nn}^{-1})$. A product of a diagonal and an upper triangular matrix is upper triangular with diagonal entries $b_{ii}^{-1}{b}_{ii}=1$, so $u\in U$ and $b=tu$ with $t\in T$, $u\in U$. Conversely, if $b=t^{\mathrm{\prime }}{u}^{\mathrm{\prime }}$ with $t^{\mathrm{\prime }}\in T$ and $u^{\mathrm{\prime }}\in U$, then comparing diagonal entries gives $t^{\mathrm{\prime }}=t$, and then $u^{\mathrm{\prime }}=t^{\mathrm{\prime }-1}b=t^{-1}b=u$. So the multiplication map $T\times U⟶B,(t,u)⟼tu,$ is a bijection: every $b\in B$ has a unique factorisation $b=tu$ with $t\in T$ and $u\in U$.

Semidirect product structure. $U$ is normal in $B$: for $t\in T$ and $u\in U$ the conjugate $tu{t}^{-1}$ is again unitriangular, and for $g\in B$, written $g=t_0{u}_0$ with $t_0\in T$, $u_0\in U$, one gets $gu{g}^{-1}=t_0(u_{0}u{u}_0^{-1})t_0^{-1}\in U$ because $U$ is a subgroup and $T$ normalises $U$. Since $B=TU$, $T\cap U=\{I_n\}$ and $U⊴B$, the group $B$ is the internal semidirect product $B=T⋉U,$ with $T$ acting on $U$ by conjugation. In particular every $b\in B$ has the form $tu$ with $t\in T$, $u\in U$ and this expression is unique, which is the structure used throughout this page.

Two special cases. For $n=1$ all matrices are $1\times 1$, so $G=B=T={\mathbb{F}}_q^{\times }$ and $U=\{I_1\}$ is trivial. For $n=2$ the elements of $B$ are $\left(\begin{array}{cc}a& c\\ 0& d\end{array}\right)$ with $ad\mathrm{\ne }0$, and the factorisation reads $\left(\begin{array}{cc}a& c\\ 0& d\end{array}\right)=\left(\begin{array}{cc}a& 0\\ 0& d\end{array}\right)\left(\begin{array}{cc}1& a^{-1}c\\ 0& 1\end{array}\right)$.

Counting remark. Since an upper triangular matrix has the $n(n+1)\mathrm{/}2$ entries on and above the diagonal and an arbitrary assignment of these entries with all diagonal entries nonzero is invertible, $\mathrm{\mid }B\mathrm{\mid }=(q-1{)}^n{q}^{n(n-1)\mathrm{/}2}$; $a_{ij}$ with $i<j$ is arbitrary in $U$. The diagonal torus $T$ is abelian and isomorphic to $({\mathbb{F}}_q^{\times }{)}^n$, while $U$ has order $q^{n(n-1)\mathrm{/}2}$; the factorisation $B=T⋉U$ therefore refines $\mathrm{\mid }B\mathrm{\mid }=\mathrm{\mid }T\mathrm{\mid }\cdot \mathrm{\mid }U\mathrm{\mid }$.