Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 2 / 4 / d / Solution

Replace $F^n$ by a symplectic vector space $(V,B)$ of dimension $2m$, and elementary row operations by symplectic transvectionsThe cross terms in $B(Tx,Ty)$ cancel, and $B(v,v)=0$, so these maps preserve $B$. Their inverses are $T_{v,-c}$.

They generate $Sp(V)$ by the following symplectic basis elimination. If $B(u,v)\ne 0$, a single transvection with direction $v-u$ and parameter $1/B(u,v)$ sends $u$ to $v$. If $B(u,v)=0$ and $u\ne v$, choose $w$ with both $B(u,w)$ and $B(w,v)$ nonzero: the union of two proper hyperplanes does not exhaust a vector space, even over ${\mathbb{F}}_2$. Two transvections then suffice. Thus one can make a form-preserving linear map fix the first basis vector $e$. Next align its partner $f$, while keeping $e$ fixed. Two partners have difference in $e^{\perp }$, so the preceding single-transvection construction uses a direction in $e^{\perp }$ whenever their mutual pairing is nonzero. If it is zero, the intermediate partner $f+ae$, $a\ne 0$, has nonzero pairing with both. All these transvections fix $e$. Once both $e,f$ are fixed, recurse on their nondegenerate symplectic orthogonal complement. This proves generation over every field.

Use the action on the points of projective space, whose kernel is the scalar center $Z=\{\lambda I:{\lambda }^2=1\}$. For $m=1$ this is the $S{L}_2$ case. For $m\ge 2$ the action is primitive, although it is generally not two-transitive. The stabilizer of a line $L$ has two orbits on other lines: those orthogonal to $L$ and those not orthogonal to $L$. This follows by extending a hyperbolic pair, or an independent isotropic pair, to a symplectic basis. A block containing $L$ and a second line $M$ contains the entire corresponding orbit of the stabilizer of $L$. The stabilizer of $M$ then supplies a line of the other type: choose a vector orthogonal to one of $L,M$ and not to the other. Nondegeneracy and independence guarantee such a vector, also in dimension four. Hence the block contains both orbits and is the whole set.

The transvections along $L$ form an abelian normal subgroup of its stabilizer; their conjugates generate the group. Thus the Iwasawa simplicity lemma applies as soon as perfectness is established. For $|F|>3$, every transvection lies in an embedded $S{L}_2(F)$ acting on a hyperbolic plane, and 4(c) proves that embedded group perfect.

For the remaining fields, the perfectness of finite symplectic groups has a short direct proof. Over ${\mathbb{F}}_3$ with $m\ge 2$, all parameter-one transvections have the same class $t$ in the abelianization, and $3t=0$. In an isotropic plane with basis $u,v$, the four directions $u,v,u+v,u-v$ give commuting transvections whose product is $1$: their rank-one terms add to zero, sinceover ${\mathbb{F}}_3$. Thus $4t=0$, giving $t=0$. Parameter-two transvections are inverses, so all generators vanish in the abelianization. Over ${\mathbb{F}}_2$ with $m\ge 3$, use the seven nonzero vectors of an isotropic three-space. Their commuting transvections multiply to $1$, since each diagonal coordinate of ${\sum }_{0\ne v}v{v}^{\mathsf{T}}$ occurs four times and each off-diagonal coordinate twice. Their common abelianized class satisfies $2t=7t=0$, and again the group is perfect.

Consequently the projective symplectic group is simple, with the necessary exceptionsThe first two are the linear exceptions; the last is $S{p}_4(2)\cong S_6$, proved in 6(c). The word “projective” matters: the unquotiented symplectic group may have the nontrivial scalar center $\{I,-I\}$.