Special linear group over a finite field

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

Over a field of odd characteristic, an involutory matrix has minimal polynomial dividing $(X-1)(X+1)$, so it is diagonalizable with eigenvalues in $\{1,-1\}$. In dimension two, determinant one requires either both eigenvalues one or both minus one. Hence the only element of order two in $S{L}_2$ is $-I$; the identity has order one.

A determinant-one matrix over ${\mathbb{F}}_2$ permutes the three nonzero vectors of ${\mathbb{F}}_2^2$. This group action is faithful because fixing two basis vectors forces the matrix to be the identity. There are $(4-1)(4-2)=6$ invertible matrices, all of determinant one since the only nonzero field element is one, so the action identifies the group with the symmetric group $S_3$. Reduction of integer determinant-one matrices modulo two is onto: $\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ and $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ induce two distinct transpositions, which generate $S_3$. Its kernel consists of matrices congruent to the identity modulo two, the level-two principal congruence subgroup.

The scalar center has order $\gcd (n,q-1)$. For $n>1$ the quotient is simple except at $(n,q)=(2,2),(2,3)$. Its faithful primitive projective-point action, abelian fixed-center transvections, elementary-matrix generation and perfectness give the Iwasawa simplicity lemma proof.

On the six-dimensional exterior square of ${\mathbb{F}}_2^4$, the Pfaffian $Q=x_{12}{x}_{34}+x_{13}{x}_{24}+x_{14}{x}_{23}$ is a nonsingular plus-type quadratic form. The action of $G{L}_4(2)$ preserves it and is faithful: fixing every decomposable exterior line fixes every two-space and then every one-space. The even-weight binary subset module on eight letters, modulo the all-one vector, gives another six-dimensional plus-type space with $Q=\mathrm{wt}(x)/2\mathrm{mod}2$. Coordinate permutations embed $S_8$ faithfully into its orthogonal group. That orthogonal group has order $40320$, while $G{L}_4(2)$ has order $20160$, identifying its index-two image with the alternating group $A_8$.

The special linear group $S{L}_2(5)$ has $24$ nonzero first columns and five second columns giving determinant one, hence order $120$; its centre is $\{I,-I\}$. In the quotient by this centre, the nonidentity conjugacy classes have sizes $15,20,12,12$, corresponding to projective element orders $2,3,5,5$. These arise from trace zero, paired traces $\pm 1$, and paired unipotent traces $\pm 2$. No proper union of these classes with the identity has size dividing $60$, so the quotient is simple. The simple groups with order a power of two times fifteen result identifies it with $A_5$.

Over ${\mathbb{F}}_5$, putThese determinant-one matrices satisfy $i^2=j^2=-I$, $ij=-ji$, and $r^3=I$. Conjugation by $r$ cycles $i,ij,j$. Thus $⟨i,j⟩$ is a quaternion group of order eight, normalized by an element of order three. Their generated subgroup has order $24$ in $S{L}_2(5)$, hence order $12$ after quotienting by $\{\pm I\}$. Its index-five coset action, together with simplicity and perfectness, identifies $PS{L}_2(5)$ with the alternating group $A_5$.

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.