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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 $\{I,-I\}$.

Replace the determinant condition by preservation of a nondegenerate alternating bilinear form $B$ on $F^{2m}$. In a symplectic basis the symplectic group over a finite field isTo count symplectic bases, choose the first vector $e$ in $q^{2m}-1$ ways, then choose $f$ with $B(e,f)=1$ in $q^{2m-1}$ ways. Their span is nondegenerate; its symplectic orthogonal complement has dimension $2m-2$ and supplies the rest of the basis. The resulting recurrence gives the finite symplectic group order

The elementary replacements are symplectic transvectionsExpanding $B(Tx,Ty)$ shows that the two cross terms cancel and the last term vanishes because $B(v,v)=0$. Thus these maps preserve $B$; their inverses are $T_{v,-c}$.

We need both generation and primitivity to use the argument from part (i). For nonzero vectors $v,w$ with $B(v,w)\ne 0$, the symplectic transvection with direction $w-v$ and parameter $1/B(v,w)$ sends $v$ to $w$. When $B(v,w)=0$, choose $z$ nonorthogonal to both and use two such steps. Such $z$ exists because two proper hyperplanes cannot cover a vector space: if the two linear forms are independent prescribe both values to be one, and if dependent prescribe either a nonzero value. Hence symplectic transvections act transitively on nonzero vectors.

They generate the whole symplectic group over a finite field. Indeed first use them to make a given symplectic transformation fix $e_1$. Its image $f^{\prime }$ of $f_1$ still satisfies $B(e_1,f^{\prime })=1$. A symplectic transvection with direction $f_1-f^{\prime }$ fixes $e_1$ and sends $f^{\prime }$ to $f_1$ if the pairing is nonzero. If that pairing is zero, pass through $f^{\prime }+c{e}_1$ for any $c\ne 0$: it has nonzero pairing with both partners, and both required directions lie in $e_1^{\perp }$. We have now fixed the first hyperbolic pair, and induction applies in its nondegenerate orthogonal complement. The final two-dimensional case is exactly the elementary generation of $S{L}_2(q)=S{p}_2(q)$ proved in part (i).

For a fixed projective point $⟨e⟩$, the symplectic transvections $T_{e,c}$ form an abelian subgroup normal in its stabilizer subgroup; conjugation scales $c$ by the square of the scalar multiplying $e$. Their conjugates supply all symplectic transvections. A central element commutes with every symplectic transvection, so it preserves every direction line and must be scalar. Conversely scalars commute with the whole group. The kernel of the projective action consists of scalar matrices, and preservation of $B$ restricts the scalar to ${\lambda }^2=1$. Thus the center of a group has size $\gcd (2,q-1)$ and the projective symplectic group over a finite field acts faithfully.

For $m\ge 2$ the stabilizer subgroup has precisely three orbits: its fixed point, the other points in $e^{\perp }$, and the points outside $e^{\perp }$. Transitivity on each follows by extending, respectively, an independent isotropic pair or a hyperbolic pair to a symplectic basis. WriteA nontrivial proper block through the fixed point would be a union of these stabilizer subgroup orbits, hence would have size either $h$ or $q^{2m-1}+1$. A block size divides $N$. But $N-qh=1$, so $h>1$ cannot divide $N$, and $N/2<q^{2m-1}+1<N$, excluding the other size. This proves primitivity. For $m=1$ the projective action is the two-transitive action of $PS{L}_2(q)$.

Here is a direct check of perfectness of finite symplectic groups, including the small fields where the diagonal group commutator needs replacement. If $q>3$, every symplectic transvection lies in an embedded $S{L}_2(q)$ on a hyperbolic plane containing its direction, and that subgroup is perfect by part (i). Since the symplectic transvections generate, the whole symplectic group over a finite field is perfect.

For $q=3$ and $m\ge 2$, all $T_{v,1}$ are conjugate, so their common image $z$ generates the abelianization and satisfies $3z=0$. Choose independent perpendicular vectors $e,f$. The four commuting symplectic transvections with directions $e,f,e+f,e-f$ and parameter one multiply to the identity: the sum of their rank-one matrices is zero in characteristic 3. Thus also $4z=0$, giving $z=0$.

For $q=2$ and $m\ge 3$, use the seven nonzero vectors of a three-dimensional isotropic subspace of a symplectic vector space. Their symplectic transvections commute, and their product is the identity. To see this explicitly, write a vector in three isotropic basis coordinates: in the sum of the outer products $v{v}^{\mathsf{T}}$ each diagonal coefficient occurs four times and each off-diagonal coefficient twice, all zero in characteristic 2. The common abelianization image $z$ therefore satisfies $2z=7z=0$. Again the group is perfect.

The simplicity argument from part (i) now applies to the projective symplectic group over a finite field. The remaining exceptional group besides $S{p}_2(2)$ and $PS{p}_2(3)$ is $S{p}_4(2)$. It is not simple: it is isomorphic to $S_6$. One can verify this identification without assuming it. On $F_2^4$ setThese are all 16 quadratic forms with polar form $B$; the Arf invariant of a quadratic form gives $\mathrm{Arf}(Q_a)=Q_0(a)$. There are six with invariant one. The symplectic group over a finite field permutes these six forms. Differences of their parameter vectors span $F_2^4$: writing $e_i,f_i$ for the symplectic basis, the six parameters areTheir pairwise differences include each of $e_1,e_2,f_1,f_2$. An element fixing all six forms therefore fixes all linear forms $B(a-b,-)$, hence is the identity. This gives a faithful action on six objects; the group order is 720, so its image is all of $S_6$. Consequently