General linear group over a finite field

The group $G{L}_n({\mathbb{F}}_q)$ consists of invertible $n$ by $n$ matrices over the finite field ${\mathbb{F}}_q$.

For prime $p$, every order-$p$ element of $G{L}_2({\mathbb{F}}_p)$ is conjugate to the displayed nonidentity unipotent matrix. Its cyclic subgroup has vector group orbits of sizes one or $p$; counting all $p^2$ vectors implies at least $p$ fixed vectors. Choose a nonzero fixed vector as first basis vector. The matrix becomes triangular with first diagonal entry one; its other diagonal entry satisfies $d^p=1$ and hence $d=1$ in the prime finite field. The off-diagonal entry is nonzero and can be rescaled to one. All matrices involved have invertible changes of basis.

The general linear group over a finite field $G{L}_2({\mathbb{F}}_2)$ acts faithfully on the four vectors of ${\mathbb{F}}_2^2$, since a matrix fixing all vectors fixes the standard basis. The zero vector is fixed by every element; the induced action on the three nonzero vectors identifies this six-element group with $S_3$. Swapping the two coordinates fixes $(0,0)$ and $(1,1)$ and exchanges the other two vectors, so its four-point permutation is odd. This action therefore embeds the group into $S_4$ with nontrivial composite sign homomorphism.

Choosing linearly independent columns successively gives

The upper unitriangular group consists of upper triangular matrices with every diagonal entry equal to one. Over ${\mathbb{F}}_p$ it has order $p^{n(n-1)/2}$ and is a Sylow subgroup of $G{L}_n({\mathbb{F}}_p)$.

For a three-dimensional strictly upper triangular matrix $A$, the binomial theorem stops after $A^2$. Over a finite field of odd prime characteristic $p$, both coefficients $p$ and $\left(\genfrac{}{}{0ex}{}{p}{2}\right)$ vanish, so every nonidentity matrix in $U{T}_3({\mathbb{F}}_p)$ has order $p$. In characteristic two, a matrix with nonzero $A^2$ instead has order four.

The upper unitriangular group $U{T}_3({\mathbb{F}}_3)$ consists of $M(a,b,c)=\left(\begin{array}{ccc}1& a& c\\ 0& 1& b\\ 0& 0& 1\end{array}\right)$ for $a,b,c\in {\mathbb{F}}_3$. Its multiplication is $M(a,b,c)M(a^{\prime },b^{\prime },c^{\prime })=M(a+a^{\prime },b+b^{\prime },c+c^{\prime }+a{b}^{\prime })$. The elements $M(1,0,0)$ and $M(0,1,0)$ do not commute. Yet every element is $I+N$ with $N^3=0$, so characteristic three gives $(I+N{)}^3=I$. It is a nonabelian finite group of order $27$ whose nonidentity elements all have order three.

The group $PG{L}_2(F)=G{L}_2(F)/Z$ acts faithfully onby Möbius transformations. For $|F|=q$, this embeds $PG{L}_2(F)$ into the symmetric group $S_{q+1}$.

For any field $F$, a projective linear group element is uniquely determined by its images of three distinct points of the projective line, and any ordered distinct target triple is attainable. Given representatives $v_z,v_x$ forming a basis and $v_y=A{v}_z+B{v}_x$, the matrix with columns $A{v}_z,B{v}_x$ sends $(0,1,\infty )$ to $(x,y,z)$. Distinctness forces $A,B\ne 0$. The matrices realizing one projective map differ by a nonzero scalar; over ${\mathbb{F}}_p$ there are $p-1$ such matrices.

The projective line over a field $F$ is the set of one-dimensional subspaces of $F^2$ and may be identified with $F\cup \{\infty \}$.

The translations $x\mapsto x+b$ for $b\in {\mathbb{F}}_4$ form a Sylow $2$-subgroup of $PG{L}_2({\mathbb{F}}_4)$. Every nonidentity translation fixes infinity and exchanges the four finite points in two transpositions, so its action lies in $A_5$.

The determinant-one matrices form the normal subgroup $S{L}_n({\mathbb{F}}_q)=\ker (\det )$.

For a finite field of odd characteristic and nonzero $a,b$, let $U_a=\left(\begin{array}{cc}1& a\\ 0& 1\end{array}\right)$. An arbitrary determinant-one matrix $P=\left(\begin{array}{cc}r& s\\ t& u\end{array}\right)$ satisfies $P{U}_a=U_{b}P$ only if $t=0$ and $ar=bu$. The determinant condition $ru=1$ then gives $b/a=r^2$. Conversely $\mathrm{diag}(r,r^{-1})U_a\mathrm{diag}(r,r^{-1}{)}^{-1}=U_{a{r}^2}$. Thus nonzero upper unipotent matrices fall into two conjugacy classes under this square-class criterion. For example, $U_1$ and $U_3$ are conjugate over ${\mathbb{F}}_{11}$ because $5^2=3$, but not over ${\mathbb{F}}_5$. The criterion compares these particular upper unipotent representatives; it does not classify all determinant-one matrices.

A matrix $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ acts on the projective line by the Möbius transformation $x\mapsto (ax+b)/(cx+d)$, with the pole sent to infinity and infinity sent to $a/c$ when $c\ne 0$. The action is transitive because $\left(\begin{array}{cc}t& -1\\ 1& 0\end{array}\right)$ sends infinity to any finite $t$. The stabilizer subgroup of infinity consists of $\left(\begin{array}{cc}a& b\\ 0& a^{-1}\end{array}\right)$ with $a\ne 0$. The orbit-stabilizer theorem gives $|{\mathrm{SL}}_2({\mathbb{F}}_q)|=q(q^2-1)$. Its kernel is the scalar matrices of determinant one: a transformation fixing infinity has $c=0$, fixing zero then forces $b=0$, and fixing one gives $a=d$. Thus the kernel is $\{I,-I\}$, with these matrices coinciding in characteristic two.