Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 7 / 3

A nonidentity transvection is a linear automorphism $\tau =I+N$ with $\mathrm{rank}N=1$ and $N^2=0$. Equivalently, it fixes a hyperplane pointwise and its displacement has image in a line lying in that hyperplane. The notation $T\setminus \{1\}$ allows the identity as a degenerate transvection; use that convention here, with $N=0$ also allowed.

For a nonidentity transvection, choose $d\ne 0$ spanning the image of $N$. There is a unique nonzero linear functional $f$ such that $N(v)=f(v)d$. Since $N^2(v)=f(v)f(d)d$, we have $f(d)=0$, so $d\in \ker f$. Conversely, this condition makes $(df{)}^2=0$ and gives inverse $I-df$. The identity is represented by $f=0$ and any $d\ne 0$. ThusWhen $f(d)=f^{\prime }(d)=0$, the two displacement maps have zero product, givingFor clarity use left actions of linear maps. Applying $g^{-1}$, then ${\tau }_{f,d}$, then $g$ givesWith the alternative conjugation notation ${\tau }^g=g^{-1}\tau g$, the answer is ${\tau }_{f,d}^g={\tau }_{f\circ g,g^{-1}(d)}$. In either convention the transformed direction lies in the kernel of the transformed functional.

For a nonidentity transvection ${\tau }_{f,d}$, take $e_1=d$ and choose $e_2$ with $f(e_2)=1$. Complete $e_1$ to a basis of $\ker f$ using $e_3,\dots ,e_n$. Then $e_1,e_2,e_3,\dots ,e_n$ is a basis of $V$, and the matrix of the transvection is $I+E_{12}$: it sends $e_2$ to $e_2+e_1$ and fixes the other basis vectors. Consequently all nonidentity transvections form one conjugacy class in $\mathrm{GL}(V)$. This matrix has determinant one, and conjugation preserves determinant, soFor $n\ge 3$, let $k$ be any GL-conjugator carrying one transvection to another. In an adapted basis for the first, the matrix $c=\mathrm{diag}(1,1,s,1,\dots ,1)$ centralizes it and has determinant $s$. Set $s=(\det k{)}^{-1}$. Then $kc\in \mathrm{SL}(V)$ gives the same conjugation as $k$. This proves the stronger transvection conjugacy over a finite field resultThe extra coordinate is essential to this determinant adjustment; no such claim is required in dimension 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.

Use the group commutator convention $[x,y]=x^{-1}{y}^{-1}xy$. If $n\ge 3$, choose distinct $i,j,k$. Multiplying elementary matrices, using $E_{ik}{E}_{kj}=E_{ij}$ and all reverse or repeated products here equal to zero, givesEvery elementary generator is therefore in the commutator subgroup. Part iii shows these generators generate $\mathrm{SL}(V)$, proving it is a perfect group.

It remains to handle $n=2$ and $q>3$. The polynomial $s^2-1$ has at most two roots, whereas ${\mathbb{F}}_q^{\times }$ has more than two elements, so choose $s\ne 0$ with $s^2\ne 1$. For $h=\mathrm{diag}(s,s^{-1})$, direct conjugation givesBoth coefficients are nonzero. As $t$ varies, these group commutators give every upper and lower elementary transvection, which generate ${\mathrm{SL}}_2({\mathbb{F}}_q)$. Thus the perfectness of the special linear group is established in all requested cases:The two pairs in the PDF are exclusions, not a literal logical disjunction. They really are exceptions: ${\mathrm{SL}}_2({\mathbb{F}}_2)\cong S_3$ has a nontrivial sign quotient. Also ${\mathrm{SL}}_2({\mathbb{F}}_3)$ maps onto ${\mathrm{PSL}}_2({\mathbb{F}}_3)\cong A_4$, which has the nontrivial abelian quotient $A_4/V_4\cong C_3$. For the latter identification, the faithful projective action has degree four and image of order 12, hence the index-two subgroup $A_4$ of $S_4$. Neither exceptional group is perfect.