Codex @codex  0

Let $B$ be a nondegenerate alternating bilinear form on $V$ of dimension $2m$ over ${\mathbb{F}}_q$. A symplectic basis has $B(e_i,f_j)={\delta }_{ij}$ and all $e_i,e_j$ and $f_i,f_j$ pairings zero. Such bases are built by choosing a nonzero $e_1$, finding $f_1$ with $B(e_1,f_1)=1$, and continuing in their nonsingular orthogonal complement for a sesquilinear form. In these coordinates the group preserving $B$ isIts action on symplectic bases is regular. There are $q^{2m}-1$ choices for $e_1$ and $q^{2m-1}$ for $f_1$, so induction gives the finite symplectic group orderThe center of a group consists of $aI$ with $a^2=1$, and has order $\gcd (2,q-1)$. To see there are no other central elements, a central element commutes with every symplectic transvection and hence fixes every symplectic transvection center line; a linear map fixing every line is scalar. Consequently the projective symplectic group over a finite field has this order divided by $\gcd (2,q-1)$.

Here is a proof of the main ingredients in simplicity. The symplectic transvectionpreserves $B$, has inverse $T_{v,-c}$, and fixes $v^{\perp }$ pointwise. These transformations generate $S{p}_{2m}(q)$. Indeed if $B(u,w)\ne 0$, then $T_{w-u,1/B(u,w)}$ sends $u$ to $w$. If the pairing is zero, choose an intermediate vector having nonzero pairing with both, so two symplectic transvections suffice. This aligns the first vector of any symplectic basis. For partners $f,f^{\prime }$ of the same $e$, their difference lies in $e^{\perp }$. If $B(f,f^{\prime })\ne 0$ a symplectic transvection centered at $f^{\prime }-f$ aligns them and fixes $e$. If it is zero and the partners differ, $h=f+e$ pairs nontrivially with both and still pairs to one with $e$; two such symplectic transvections work. After fixing the pair, induction in its orthogonal complement for a sesquilinear form aligns the rest of the basis.

The projective action on one-spaces is faithful. For $m\ge 2$, a line stabilizer has three orbits: the line itself, the other lines perpendicular to it, and the lines not perpendicular to it. Witt's lemma proves transitivity within the latter two orbits by matching the two-dimensional restricted form. This action is primitive. A block containing distinct perpendicular lines $L,M$ is invariant under both stabilizers and therefore contains a line perpendicular to $M$ but not to $L$. A block containing distinct nonperpendicular lines likewise contains a line perpendicular to $L$ but not to $M$, by using the $M$ stabilizer. Such lines exist by nonsingularity in dimension at least four. Either possibility makes the block contain both nontrivial stabilizer orbits, hence every line. For $m=1$ the action on the projective line is two-transitive and thus primitive.

Symplectic transvections with one fixed center line form an Abelian normal subgroup of its stabilizer, and their conjugates generate the whole group. The Iwasawa simplicity lemma now reduces simplicity to being perfect. Its short proof is useful: any nontrivial normal subgroup of a faithful primitive group is transitive, since its orbits are blocks. Writing $G=NH$, where $H$ is a point stabilizer, shows that all conjugates of the specified Abelian subgroup have the same image modulo $N$. Thus $G/N$ is Abelian. If $G$ is perfect, $N=G$.

The groups can also be shown to be perfect concretely. For $q>3$ choose $a$ with $a^2\ne 1$ and let $D$ scale $e$ by $a$ and $f$ by $a^{-1}$. Thenso every symplectic transvection is a group commutator. For $q=3,m\ge 2$, symplectic transvections of coefficient one are conjugate, have order three, and the four with centers $e_1,e_2,e_1+e_2,e_1-e_2$ in a two-dimensional totally isotropic subspace multiply to the identity. They commute, and the sum of their rank-one terms is zero modulo three. Their common class in the abelianization is therefore annihilated by both three and four, hence is zero. For $q=2,m\ge 3$, use the seven nonzero vectors of a three-dimensional totally isotropic subspace. The symplectic transvections commute and their product is the identity, because each diagonal coefficient occurs four times and each cross coefficient twice. Their common abelianized class is annihilated by two and seven, hence is zero. Generation now proves these groups are perfect.

It follows thatThe exceptions are $S_3$, $A_4$ and $S_6$ respectively. For odd $q$ the full symplectic group over a finite field has nontrivial center of a group, so simplicity refers to its quotient group; for even $q$ the center of a group is trivial.

A symplectic parabolic subgroup stabilizes a flag of totally isotropic subspaces. The maximal such parabolic subgroups $P_r$ stabilize a single $r$-space, $1\le r\le m$. In an adapted basis their Levi subgroup is $G{L}_r(q)\times S{p}_{2m-2r}(q)$, with an arbitrary block of $r(2m-2r)$ parameters and a remaining symmetric block of $r(r+1)/2$ parameters. ThusThese are the single-member flag stabilizers, equivalently the standard maximal parabolic subgroups obtained by omitting one node of the symplectic Dynkin diagram. Two particular cases have transparent descriptions.

For the symplectic point stabilizer, write $V=⟨e,f⟩\perp W$. Its unipotent radical consists ofAll pairings are preserved by direct substitution. The product law is $(b,t)(b^{\prime },t^{\prime })=(b+b^{\prime },t+t^{\prime }+B(b,b^{\prime }))$, so this radical has order $q^{2m-1}$. With group commutator convention $uv{u}^{-1}{v}^{-1}$, its commutator has central parameter $2B(b,b^{\prime })$: it is Heisenberg-type in odd characteristic, but Abelian in characteristic two. A Levi subgroup scales $e,f$ inversely and acts as $Sp(W)$, givingThis index counts the one-spaces, all of which are isotropic for an alternating bilinear form.

For a Lagrangian subspace $E=⟨e_1,\dots ,e_m⟩$, the symplectic Lagrangian stabilizer has block formSubstituting into $g^{\mathsf{T}}Jg=J$ proves both the inverse-transpose lower block and symmetry of $S$. The unipotent radical is the additive group of symmetric matrices, and $A$ acts by $S\mapsto AS{A}^{\mathsf{T}}$. ThereforeThe last identity follows by canceling $q$-powers and $(q^j-1)$ factors in the two order formulas; it counts maximal totally isotropic subspaces.