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The upper unitriangular group is generated by adjacent elementary transvection matrices $I+s{E}_{i,i+1}$. Commuting a matrix $X$ with each such generator forces every entry above the diagonal to vanish except possibly its $(1,n)$ entry: comparing $X{E}_{i,i+1}$ with $E_{i,i+1}X$ kills the entries above position $(i,i)$ in column $i$ and to the right of position $(i+1,i+1)$ in row $i+1$. The surviving matrices $I+t{E}_{1n}$ do commute with the entire group. In particular the group center has order $q$; it distinguishes $U_4(2)$ from $U_3(4)$ despite their common order $64$.

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

For $i\ne j$, put $x_{ij}(a)=I+a{E}_{ij}$. When $a\ne 0$ this is a transvection: it fixes the hyperplane $v_j=0$ pointwise and has rank-one difference from the identity. Its inverse is $x_{ij}(-a)$ and its determinant is one.

Left multiplication performs a row addition. For an invertible matrix, bring a nonzero entry into a pivot position by a signed interchange, then clear its column by row additions. A signed interchange on two coordinates is itself a product of row additions, becauseRepeating the pivot construction and then clearing above the pivots reduces any determinant-one matrix to a diagonal matrix of determinant one. FinallyA determinant-one diagonal matrix is a product of such two-coordinate matrices, using the last coordinate to compensate each of the first $n-1$ diagonal entries. Each displayed matrix is a product of elementary transvection matrices. Hence reversing the elimination provesThis argument works over every field, including characteristic two.

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.

Let $F={\mathbb{F}}_q$. The general linear group over a finite field consists of all invertible linear maps $F^n\to F^n$, and the special linear group over a finite field is its determinant-one subgroup. Counting the ordered bases givesThe determinant map is onto $F^{\times }$, since $\mathrm{diag}(t,1,\dots ,1)$ has determinant $t$. Its kernel is $S{L}_n(q)$, so

For $n>1$, commuting with all the elementary transvection matrices $I+t{E}_{ij}$ forces a matrix to be scalar. Therefore the center of a group of $S{L}_n(q)$ consists of $\lambda I$ with ${\lambda }^n=1$. The multiplicative group of a finite field is cyclic, so this center of a group has size $d=\gcd (n,q-1)$. The projective special linear group over a finite field consequently has order

Here is a proof of simplicity which isolates all the required ingredients. First we prove the Iwasawa simplicity lemma in the form needed. Suppose a group $H$ acts faithfully and primitively, is a perfect group, and has an abelian subgroup $U$ normal in a stabilizer subgroup $H_{\alpha }$, whose conjugates generate $H$. For a nontrivial normal subgroup $N$, its orbits form a block system. Faithfulness prevents all these orbits from being singletons, so primitivity makes $N$ transitive. Thus $H=N{H}_{\alpha }$. Modulo $N$, each conjugate of $U$ has exactly the same image as $U$, because we may write its conjugating element as $nh$ with $n\in N$ and $h\in H_{\alpha }$. The quotient $H/N$ is generated by this one abelian image and hence is abelian. Since $H$ is a perfect group, $H/N=1$. Thus every nontrivial normal subgroup is $H$.

Apply this to the action on projective points, that is, one-dimensional vector subspaces of $F^n$. Any ordered pair of distinct lines can be sent to any other by a linear map obtained by extending representatives to bases. Multiplying one target basis vector by a suitable nonzero scalar adjusts the determinant to one without changing either target line. Thus $S{L}_n(q)$ is two-transitive on these points. A linear map fixing every line is scalar: it is diagonal on a fixed basis, and fixing the lines spanned by $e_i+e_j$ makes all diagonal entries equal. The projective quotient therefore acts faithfully, two-transitively, and primitively.

For the line $⟨e_1⟩$, useThese matrices multiply by addition of $f$, so $U$ is abelian. Conjugating by a matrix preserving $⟨e_1⟩$ preserves $U$, so it is normal in that stabilizer subgroup. Its conjugates include every elementary transvection matrix $I+t{E}_{ij}$.

These elementary transvection matrices generate $S{L}_n(q)$ by row elimination. For completeness, row addition is elementary, signed row interchange is elementary, and compensating diagonal scalings are elementary because, in a two-dimensional coordinate block,Elimination using these operations reduces a determinant-one matrix to the identity.

It remains to show that these groups are perfect groups. In the following matrix identities use $[x,y]=xy{x}^{-1}{y}^{-1}$; the alternative group commutator convention generates the same commutator subgroup. If $n\ge 3$, distinct indices give the group commutator identityEvery elementary generator is therefore a group commutator. For $n=2$ and $q>3$, choose $\lambda \in F^{\times }$ with ${\lambda }^2\ne 1$, and let $D=\mathrm{diag}(\lambda ,{\lambda }^{-1})$. ThenAs $t$ varies this supplies all upper elementary transvection matrices, and conjugation supplies the lower ones. Thus $S{L}_n(q)$, and its projective quotient, are perfect in the asserted cases. The preceding simplicity argument now applies.

The two exceptions are genuine: the projective actions identify $PS{L}_2(2)$ with $S_3$ and $PS{L}_2(3)$ with $A_4$, of orders 6 and 12 respectively. The first contains its normal subgroup of order 3; the second contains the normal Klein four group. We have proved

The quotient of the special linear group by its scalar center, acting faithfully on projective space. It is a simple group for $n\ge 3$, or for $n=2$ and $|F|>3$. Elementary transvection matrices give generation and perfectness, and the Iwasawa simplicity lemma applies to the abelian subgroup of transvections with a fixed center.

Row additions reduce an invertible matrix to diagonal form without changing determinant. A determinant-one diagonal matrix is a product of diagonal two-coordinate blocks $\mathrm{diag}(a,a^{-1})$, each expressible as a product of six elementary transvection matrices. Therefore these matrices generate the special linear group over a field.

// Target: algebra.bigb