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For distinct coordinates $i,j$, let $E_{ij}(t)=I+t{e}_{ij}$, where $e_{ij}$ is the matrix unit. For $t\ne 0$ this is a transvection: it fixes the hyperplane with $j$th coordinate zero pointwise, has determinant one and inverse $E_{ij}(-t)$. We show these elementary transvection matrices generate $SL(n,F)$ over any field.

Let $M\in SL(n,F)$. Row addition is left multiplication by an $E_{ij}(t)$. If the current diagonal pivot is zero, a nonzero entry occurs below it in the same remaining column, since the remaining square block is invertible; adding that row makes the pivot nonzero. Row additions then clear the entries below the pivot. Repeating produces an upper-triangular matrix with nonzero diagonal. Clearing the entries above the diagonal, starting from the last column, gives a diagonal matrix $D=\mathrm{diag}(a_1,\dots ,a_n)$ with ${\prod }_i{a}_i=1$. Thus it remains to realize determinant-one diagonal matrices using elementary transvections.

In a two-coordinate block putDirect multiplication givesThe diagonal matrix $D$ is the product, for $i=1,\dots ,n-1$, of these blocks on coordinates $i,n$, with entries $a_i,a_i^{-1}$. Their final $n$th entry is ${\prod }_{i<n}{a}_i^{-1}=a_n$. Therefore $D$, and hence $M$, is a product of elementary transvections. All row operations used are invertible transvections, and no division except by a nonzero pivot or $a_i$ has been made. This provesThe identity element $E_{ij}(0)$ can be omitted from the generating set. This is transvections generate the special linear group; ordinary row scaling alone would not prove it, because such a scaling can leave $SL(n,F)$.

Write $x_{ij}(t)=I+t{E}_{ij}$ for $i\ne j$ and $t\in {\mathbb{F}}_q$. When $t\ne 0$ these are transvections, and their inverses are $x_{ij}(-t)$. Row addition by such an elementary transvection matrix preserves the determinant. It also realizes signed interchanges and compensating diagonal scalings: in any two coordinate positions,Thus elimination can reduce any invertible matrix to diagonal form using only row additions and signed interchanges, all products of transvections. For a diagonal matrix with determinant one, $\mathrm{diag}(d_1,\dots ,d_n)$, use the displayed two-coordinate diagonal matrices in positions $(i,n)$ with parameter $d_i$, for $1\le i<n$. Their product has diagonal entries $d_1,\dots ,d_{n-1},{\prod }_{i<n}{d}_i^{-1}=d_n$. Hence the residual diagonal matrix is also a product of transvections, and reversing elimination provesThis proves transvections generate the special linear group, including the small fields.

For perfectness, use the convention $[x,y]=xy{x}^{-1}{y}^{-1}$ in the following calculations. If $n\ge 3$ and $i,j,k$ are distinct, direct matrix multiplication givesTaking $b=1$ puts every elementary generator in the commutator subgroup. Thus $S{L}_n(q)$ is perfect. When $n=2$ and $q>3$, choose $s\in {\mathbb{F}}_q^{\times }$ with $s^2\ne 1$; such an element exists because a degree-two polynomial has at most two roots and $q-1>2$. For $D=\mathrm{diag}(s,s^{-1})$,Both coefficients are nonzero, so varying $t$ supplies every upper and lower elementary generator as a commutator. ThereforeThe exceptions really fail: $S{L}_2(2)\cong S_3$ has a nontrivial sign quotient, and $S{L}_2(3)$ has quotient $PS{L}_2(3)\cong A_4$, which has the nontrivial Abelian quotient $A_4/V_4\cong C_3$. For the latter identification, the faithful projective action embeds the order-twelve group $PS{L}_2(3)$ as an index-two subgroup of $S_4$, necessarily $A_4$.

The Iwasawa simplicity lemma states that, in a faithful primitive group action, if a point stabilizer has an Abelian normal subgroup whose conjugates generate the whole group, then every nontrivial normal subgroup contains the commutator subgroup. In particular a nontrivial perfect group with this property is simple. The lemma is applied here, as allowed, without assuming the desired simplicity theorem.

Let $S{L}_n(q)$ act on the one-dimensional subspaces of ${\mathbb{F}}_q^n$. This action is two-transitive: a linear map taking two specified distinct lines to two others can be made determinant one by rescaling one of the target basis vectors, without changing either target line. Its kernel consists exactly of scalar matrices. Indeed fixing all lines makes a matrix diagonal in any fixed basis, and fixing the lines through $e_i+e_j$ makes all its diagonal entries equal. The scalar matrices in $S{L}_n(q)$ are $aI$ with $a^n=1$. They also form its group center, since a central matrix commutes with every elementary generator, forcing all off-diagonal entries to vanish and all diagonal entries to agree. Their number is $\gcd (n,q-1)$. Consequently the quotient $PS{L}_n(q)$ acts faithfully and primitively.

Fix a line $L=⟨v⟩$ and considerHere $vf$ is the rank-at-most-one map $z\mapsto f(z)v$. These matrices have determinant one, and $(vf)(vh)=0$, so $T_L$ is Abelian under addition of the functionals. If $gL=L$, then $gv=av$ and $g(I+vf)g^{-1}=I+av(f{g}^{-1})\in T_L$. Thus $T_L$ is normal in the line stabilizer. Its image in $PS{L}_n(q)$ is a nontrivial Abelian normal subgroup of the point stabilizer. Conjugates of these groups include every elementary transvection, hence generate $PS{L}_n(q)$. Perfectness passes to a quotient. The Iwasawa simplicity lemma therefore givesIn those exceptions the projective groups are $S_3$ and $A_4$, as above.

For $q=4$, counting columns gives $|S{L}_2(4)|=4(4^2-1)=60$. Its scalar group center is trivial, so $PS{L}_2(4)$ has order sixty. Its faithful action on the five points of the projective line embeds it in $S_5$. Because it is perfect, its sign homomorphism is trivial and its image lies in $A_5$. Equal orders then giveFor $q=5$, use an explicit subgroup to obtain a five-point action. In $S{L}_2(5)$ setAll three have determinant one. Matrix multiplication gives $i^2=j^2=-I$, $ij=-ji$, and $r^3=I$. The eight distinct matrices $\{\pm I,\pm i,\pm j,\pm ij\}$ form a quaternion group. Moreover,Thus $r$ normalizes that quaternion group and has order three. Their generated subgroup has order $24$ and contains the scalar group center $\{\pm I\}$. Its image in $PS{L}_2(5)$ has order twelve, hence index five since $|PS{L}_2(5)|=5(25-1)/2=60$. The resulting coset action is nontrivial and, by the simplicity already proved, faithful. Perfectness again puts its image in $A_5$, and equal orders implyThis is the quaternion construction of an index-five subgroup of PSL2 over F5; it avoids assuming a classification of groups of order sixty.

For the last comparison, the order of a general linear group over a finite field comes from choosing independent columns:The determinant map onto ${\mathbb{F}}_q^{\times }$ has kernel $S{L}_n(q)$, and the scalar quotient divides by $\gcd (n,q-1)$. HenceThe highest power of two dividing either order is $64$. Their Sylow $2$-subgroups are respectively the upper unitriangular groups $U_4(2)$ of order $2^6$ and the image of $U_3(4)$ of order $4^3$. The latter image is isomorphic to $U_3(4)$ because its intersection with the odd-order scalar group center is trivial.

The center of an upper unitriangular group can be computed directly here. Commuting $X\in U_4(2)$ with $I+E_{12}$ forces $X_{23}=X_{24}=0$; commuting with $I+E_{23}$ forces $X_{12}=X_{34}=0$; commuting with $I+E_{34}$ forces $X_{13}=0$. Only $X_{14}$ can remain, and those matrices commute with the whole group. Thus $Z(U_4(2))=\{I+t{E}_{14}:t\in {\mathbb{F}}_2\}$ has order two. In $U_3(4)$, commutation with $I+E_{12}$ and $I+E_{23}$ similarly forces $X_{23}=X_{12}=0$, leaving the arbitrary entry $X_{13}\in {\mathbb{F}}_4$. Its group center has order four. An isomorphism of the whole groups would send a Sylow subgroup to a Sylow subgroup and preserve its group center order, which is impossible. Therefore

Write $x_{ij}(a)=I+a{E}_{ij}$ for $i\ne j$. For $a\ne 0$ these are transvections, and for $a=0$ they are the identity. Left multiplication by $x_{ij}(a)$ adds $a$ times row $j$ to row $i$. We show that these elementary transvection matrices generate the special linear group.

Apply Gaussian elimination using only row additions. If a pivot is zero, invertibility ensures some lower entry in its column is nonzero; adding that row produces a nonzero pivot without interchanging or scaling rows. Clear entries below each pivot, and then above pivots in reverse order. This reduces any invertible matrix to a diagonal matrix $D=\mathrm{diag}(d_1,\dots ,d_n)$, with all $d_i\ne 0$, preserving determinant throughout. For a determinant-one matrix, ${\prod }_i{d}_i=1$.

A two-coordinate diagonal block is itself a product of elementary transvection matrices. On those two coordinates putThis formula is valid in characteristic two as well. For $n\ge 2$, multiply the diagonal blocks with entries $d_i,d_i^{-1}$ on coordinates $(i,n)$ for $1\le i<n$. Their product is $D$, because its last entry is ${\prod }_{i<n}{d}_i^{-1}=d_n$. Thus $D$ and the original matrix lie in the subgroup generated by transvections. Conversely every transvection has determinant one by part ii. HenceFor $n=1$, both the special linear group and the subgroup generated by the degenerate identity transvection are trivial. The argument also establishes transvections generate the special linear group over any field.