= Solution
Write $x_{ij}(t)=I+tE_{ij}$ for $i\ne j$ and $t\in\mathbb F_q$. When $t\ne0$ 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,
$$
w(a)=\begin{pmatrix}1&a\\0&1\end{pmatrix}
\begin{pmatrix}1&0\\-a^{-1}&1\end{pmatrix}
\begin{pmatrix}1&a\\0&1\end{pmatrix}
=\begin{pmatrix}0&a\\-a^{-1}&0\end{pmatrix},\qquad
w(a)w(-1)=\begin{pmatrix}a&0\\0&a^{-1}\end{pmatrix}.
$$
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, $\operatorname{diag}(d_1,\ldots,d_n)$, use the displayed two-coordinate diagonal <matrices> in positions $(i,n)$ with parameter $d_i$, for $1\leq i<n$. Their product has diagonal entries $d_1,\ldots,d_{n-1},\prod_{i<n}d_i^{-1}=d_n$. Hence the residual diagonal <matrix> is also a product of <transvections>, and reversing elimination proves
$$
\boxed{SL_n(q)=\langle x_{ij}(t):i\ne j,\ t\in\mathbb F_q\rangle.}
$$
This proves <transvections generate the special linear group>, including the small fields.
For perfectness, use the convention $[x,y]=xyx^{-1}y^{-1}$ in the following calculations. If $n\geq3$ and $i,j,k$ are distinct, direct <matrix> multiplication gives
$$
[x_{ik}(a),x_{kj}(b)]=x_{ij}(ab).
$$
Taking $b=1$ puts every elementary generator in the <commutator subgroup>. Thus $SL_n(q)$ is <perfect>. When $n=2$ and $q>3$, choose $s\in\mathbb F_q^\times$ with $s^2\ne1$; such an element exists because a degree-two polynomial has at most two roots and $q-1>2$. For $D=\operatorname{diag}(s,s^{-1})$,
$$
[D,x_{12}(t)]=x_{12}((s^2-1)t),\qquad
[D,x_{21}(t)]=x_{21}((s^{-2}-1)t).
$$
Both coefficients are nonzero, so varying $t$ supplies every upper and lower elementary generator as a commutator. Therefore
$$
\boxed{SL_n(q)\text{ is perfect for }n\geq3\text{ or }(n=2,q>3).}
$$
The exceptions really fail: $SL_2(2)\cong S_3$ has a nontrivial sign quotient, and $SL_2(3)$ has quotient $PSL_2(3)\cong A_4$, 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> $PSL_2(3)$ 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 $SL_n(q)$ 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 $SL_n(q)$ 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 $PSL_n(q)$ acts faithfully and primitively.
Fix a line $L=\langle v\rangle$ and consider
$$
T_L=\{I+vf:f\in(\mathbb F_q^n)^*,\ f(v)=0\}.
$$
Here $vf$ 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 $g(I+vf)g^{-1}=I+av(fg^{-1})\in T_L$. Thus $T_L$ is normal in the line stabilizer. Its image in $PSL_n(q)$ is a nontrivial <Abelian> <normal subgroup> of the point stabilizer. Conjugates of these <groups> include every elementary <transvection>, hence generate $PSL_n(q)$. Perfectness passes to a quotient. The <Iwasawa simplicity lemma> therefore gives
$$
\boxed{PSL_n(q)\text{ is simple unless }(n,q)=(2,2),(2,3).}
$$
In those exceptions the projective <groups> are $S_3$ and $A_4$, as above.
For $q=4$, counting columns gives $|SL_2(4)|=4(4^2-1)=60$. Its scalar <group center> is trivial, so $PSL_2(4)$ 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 give
$$
\boxed{PSL_2(4)\cong A_5.}
$$
For $q=5$, use an explicit <subgroup> to obtain a five-point action. In $SL_2(5)$ set
$$
i=\begin{pmatrix}0&1\\4&0\end{pmatrix},\qquad
j=\begin{pmatrix}2&0\\0&3\end{pmatrix},\qquad
r=\begin{pmatrix}3&2\\1&1\end{pmatrix}.
$$
All 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,
$$
r i r^{-1}=ij,\qquad r(ij)r^{-1}=j,\qquad r j r^{-1}=i.
$$
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 $PSL_2(5)$ has order twelve, hence index five since $|PSL_2(5)|=5(25-1)/2=60$. The resulting <coset> action is nontrivial and, by the simplicity already proved, faithful. Perfectness again puts its image in $A_5$, and equal orders imply
$$
\boxed{PSL_2(5)\cong A_5.}
$$
This 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:
$$
|GL_n(q)|=\prod_{i=0}^{n-1}(q^n-q^i).
$$
The determinant map onto $\mathbb F_q^\times$ has kernel $SL_n(q)$, and the scalar quotient divides by $\gcd(n,q-1)$. Hence
$$
|PSL_4(2)|=(16-1)(16-2)(16-4)(16-8)=20160,
$$
$$
|PSL_3(4)|=\frac{(64-1)(64-4)(64-16)}{(4-1)\gcd(3,3)}=20160.
$$
The 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 $Z(U_4(2))=\{I+tE_{14}:t\in\mathbb F_2\}$ 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
$$
\boxed{|PSL_4(2)|=|PSL_3(4)|=20160,\qquad PSL_4(2)\not\cong PSL_3(4).}
$$
Back to article page
