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In a two-dimensional nondegenerate Hermitian form over a quadratic finite field, choose an orthonormal basis $v,w$ and $a$ with $a\bar{a}=-1$. Then $e=v+aw$ is an isotropic vector. Choose $g$ with $B(e,g)=1$, and choose $c$ with $c+\bar{c}=B(g,g)$. The vector $f=g-ce$ satisfies $B(e,f)=1$ and $B(f,f)=0$. Thus $e,f$ give the displayed matrix, including in characteristic two. Iterating on nondegenerate orthogonal complement for a sesquilinear form leaves only a unit-norm line in odd dimension.

A nondegenerate Hermitian form over a quadratic finite field has an orthonormal basis. If all diagonal values vanished, evaluating $B(v+aw,v+aw)$ would give $\mathrm{Tr}(\bar{a}B(v,w))=0$ for every $a$, contradicting quadratic finite-field norm and trace surjectivity and nondegeneracy. Thus some $v$ has nonzero norm; rescaling using the field norm makes it one. Its orthogonal complement for a sesquilinear form is nondegenerate, so induction completes the basis.

Here is Witt's lemma with the hypotheses appropriate to the classical groups. Let $V$ be finite-dimensional with a nonsingular alternating bilinear form, a nonsingular symmetric bilinear form in characteristic different from two, or a nonsingular Hermitian form with nontrivial involution. Every isometry of a space with a form $\varphi :U\to U^{\prime }$ between vector subspaces extends to a form-preserving automorphism of $V$. For quadratic forms, the same statement holds in every characteristic when the polar bilinear form of a quadratic form is nonsingular and the isometry of a space with a form preserves the quadratic form itself. The restricted forms on $U,U^{\prime }$ may be degenerate. We prove these versions, including the characteristic-two quadratic case needed later.

For an alternating bilinear form $B$, split the restricted space as $U=H\oplus R$, where $R=U\cap U^{\perp }$ and $B{|}_H$ is nonsingular. Choose a symplectic basis for $H$ and a basis $r_1,\dots ,r_t$ of $R$. The nondegenerate form on $V$ supplies vectors $s_j$ in $H^{\perp }$ with $B(r_i,s_j)={\delta }_{ij}$. Adding linear combinations of the $r_i$ to the $s_j$ makes $B(s_i,s_j)=0$: for each pair $i<j$ one may cancel its pairing by adding an appropriate multiple of $r_i$ to $s_j$, without disturbing the dual pairings or previously treated pairs. This uses no division by two. Thus $H$ together with the pairs $(r_i,s_i)$ is nonsingular and its orthogonal complement for a sesquilinear form has a symplectic basis.

Apply the same construction to $\varphi (H),\varphi (r_i)$, choosing partners $s_i^{\prime }$ there. The two completed bases have identical alternating Gram matrices and equal remaining dimensions. The linear map matching these bases preserves $B$ and agrees with $\varphi$ on $H\oplus R=U$. This proves the alternating version over an arbitrary field, including characteristic two.

For a quadratic form $Q$, define its polar bilinear form of a quadratic form by $B(x,y)=Q(x+y)-Q(x)-Q(y)$. Its symmetrypreserves $Q$. The zero-dimensional case is immediate. Induct on $\dim U$: choose $Q(t)\ne 0$ and a hyperplane $K\subseteq U\cap t^{\perp }$. Extending $\varphi {|}_K$ first reduces to $\varphi {|}_K=I$. Put $U=K\oplus Fu$, $v=\varphi (u)$, $d=u-v\in S=K^{\perp }$. If $Q(d)\ne 0$, $s_d$ completes the extension. Otherwise $B(u,d)=B(v,d)=0$. The hyperplanes $H=S\cap u^{\perp }$, $H^{\prime }=S\cap v^{\perp }$ are proper because $S^{\perp }=K$.

For $a\in S\setminus (H\cup H^{\prime })$ with $Q(a)\ne 0$, set $c=u-s_a(v)$. Thenand $s_a{s}_c$ fixes $K$ and sends $u$ to $v$. If no such $a$ exists and $|F|>2$, the identityforces $Q(z)=B(a,z)=0$. The complement of two hyperplanes spans $S$: adding a multiple of any outside vector avoids at most two forbidden scalars. Hence $d\perp S$, so $H=H^{\prime }$, forcing $Q{|}_S=0$, contrary to $t\in S$.

Over ${\mathbb{F}}_2$, choose $a$ outside both hyperplanes. In the remaining case $Q(a)=Q(d)=B(a,d)=0$, and $B(a,u)=B(a,v)=1$. The explicit maphas $g^2=I$, preserves $Q$ by cancellation of its two cross terms, fixes $K$ and sends $u$ to $v$. This closes the induction. The characteristic-independent symmetry argument and its binary exceptional step are also treated in Casselman's quadratic-form notes and Elman, Karpenko and Merkurjev, section 8.

In odd characteristic a symmetric bilinear form is recovered by the polarization identity for $Q(x)=B(x,x)/2$, so the quadratic proof also proves its extension theorem. For completeness, the Hermitian form case has a similarly explicit induction. Take $h$ linear in the first variable, with conjugation $a\mapsto \bar{a}$. For anisotropic vectors for a form $u,v$ with equal value $A=h(u,u)=h(v,v)\ne 0$, let $d=u-v$ and $c=h(u,d)$. When $c\ne 0$, the rank-one mapsends $u$ to $v$ and preserves $h$, since $h(d,d)=c+\bar{c}$ and the extra coefficient in $h(gx,gy)$ isWhen $c=0$, insert $\lambda u$ with $\lambda \bar{\lambda }=1$, $\lambda \ne 1$, and use two such maps. Such a scalar exists for a nontrivial involution: $\lambda =a/\bar{a}$ with $a\ne \bar{a}$ works. If the source vector subspace contains an anisotropic vector for a form, align it this way and induct in its nonsingular orthogonal complement for a sesquilinear form.

If instead every source vector is isotropic, the Hermitian polarization identity makes its whole restricted form zero. Choose $e$ in its basis and a partner $f$ pairing to one with $e$ and to zero with the remaining source basis. Make $h(f,f)=0$ by replacing $f$ by $f-te$, where $t+\bar{t}=h(f,f)$. The trace map is onto the fixed field: in odd characteristic divide by two; in characteristic two scale a nonzero trace. Choose $a+\bar{a}\ne 0$. The orthogonal vectors $e+af$ and $e-\bar{a}f$ have nonzero values $a+\bar{a}$ and its negative, so the preceding rank-one construction aligns this entire hyperbolic pair with its target pair. The remaining source basis lies in the pair's orthogonal complement for a sesquilinear form; induction finishes the Hermitian proof.

The hypotheses matter. For example, the identity-matrix symmetric bilinear form on ${\mathbb{F}}_2^3$ has canonical vector $c=(1,1,1)$ characterized by $B(x,c)=B(x,x)$, so every isometry of a space with a form fixes $c$. The one-dimensional map $e_1\mapsto c$ preserves the restricted bilinear form but cannot extend. Thus “all symmetric forms in characteristic two” is not a correct replacement for the quadratic or alternating versions proved above.

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.

For the degree-ten partition action of S6, take unordered partitions of six letters into two triples. There are $\left(\genfrac{}{}{0ex}{}{6}{3}\right)/2=10$. Fix $A\mid A^c$. Its stabilizer is $(S_3\times S_3)⋊S_2=S_3\wr S_2$, permuting each triple and interchanging them. The subgroup generated by a 3-cycle in each triple has order nine, so is a Sylow subgroup of $S_6$. Any normalizer preserves its two three-point orbits; conversely all the indicated within-triple permutations and the swap normalize it. This proves the asserted normalizer exactly.

For any different partition $C\mid C^c$, $|A\cap C|$ is one or two after excluding the identical partition. Interchanging $C$ and $C^c$ identifies these two possibilities. The stabilizer of $A\mid A^c$ is transitive on each fixed intersection pattern, hence on all other nine partitions. The degree-ten action is two-transitive. It is also faithful: two distinct partitions have a common refinement with two singleton cells and two cells of size two. Every unordered pair of letters occurs as those singleton cells. A permutation fixing every partition therefore fixes every pair and consequently every letter.

For the subgroup factorization from coset transitivity, use cosets $Hg$ (right cosets) and right multiplication by $K$. The orbit of $H$ is precisely $\{Hk:k\in K\}$. This is every coset if and only if every $g$ can be written $hk$, provingA natural $S_5$ fixing one letter is transitive on the ten partitions: the triple containing the fixed letter is determined by any two of the other five. Taking $H=S_3\wr S_2$, $K=S_5$ gives $S_6=HK$; inversion gives the requested order as well,

Now let $B$ be the binary nondegenerate alternating bilinear form. Its binary quadratic refinements form an affine space: the difference of two such forms is linear, so for a fixed $Q_0$ every one is uniquelyIn a symplectic basis each refinement isA two-dimensional summand with $(a_i,b_i)\ne (1,1)$ admits a basis of singular vectors paired to one and is hyperbolic; the $(1,1)$ summand is anisotropic. Two planes spanned by anisotropic vectors for a form together become two hyperbolic planes of a quadratic form: with anisotropic basis pairs $(e_1,f_1),(e_2,f_2)$, the singular vectors $e_1+e_2,f_1+e_2$ pair to one; their orthogonal complement for a sesquilinear form has singular paired basis $e_1+f_1+f_2,e_1+f_1+e_2+f_2$. Consequently every refinement is equivalent to either $m$ hyperbolic planes of a quadratic form or $(m-1)$ hyperbolic planes of a quadratic form and one anisotropic plane (all three nonzero vectors are anisotropic vectors for a form).

These two types are different. Indeed the character sum of each plane is $+2$ or $-2$, soThus there are exactly two symplectic orbits $Q^{+}(B),Q^{-}(B)$, the signs denoting $\mathrm{Arf}=0,1$. Counting the three coefficient pairs with product zero and the one with product one givesThe same character sum gives the corresponding numbers of singular vectors, including zero, in a fixed form of either type.

Fix a form $Q$ of type $ϵ$. Completing the square by $x\mapsto x+v$ showsThe stabilizer of $Q$ is exactly $O(V,Q)$. It sends the parameter $v$ by its natural linear action or inverse action according to the chosen left/right convention. Other forms of the same type correspond to nonzero singular vectors $v$, and all forms of opposite type correspond to nonsingular vectors $v$. For two nonzero vectors of the same $Q$ value, the map between their spans is an isometry of a space with a form. Witt's lemma extends it, so the orthogonal stabilizer is transitive on each of these two vector sets. This proves the two-transitive binary quadratic-form actions: the symplectic group over a finite field is transitive on each type and its point stabilizer is transitive on all other members of that type.

In particular $O_{2m}^{-}(2)$ is transitive on the plus orbit. Identify that orbit with the cosets (right cosets) of a plus stabilizer, using $Q^g(x)=Q(gx)$ as a right action. The coset criterion then provesBoth genuine two-transitive actions require $m\ge 2$. At $m=1$ the plus orbit has three forms and the minus orbit is a singleton; transitivity and the factorization remain true, while two-transitive action on the singleton is only vacuous if one's convention permits degree one. No special simplicity assertion is needed for this argument. At $m=2$ the orbit sizes are ten and six, linking these actions with $S{p}_4(2)\cong S_6$ and the partition action above.

We construct a Hermitian hyperbolic plane over a finite field inside each successive two-dimensional piece. By the field norm surjectivity proved above, choose $a\in {\mathbb{F}}_{q^2}$ with $a\bar{a}=-1$. From two vectors in an orthonormal basis, form $e=v_1+a{v}_2$. Then $B(e,e)=1+a\bar{a}=0$, so $e$ is a nonzero isotropic vector. Nondegeneracy supplies $g$ with $B(e,g)=1$; rescaling the second argument achieves this normalization. In fact $g=v_1$ already works in this two-dimensional piece. By quadratic finite-field norm and trace surjectivity, choose $c$ with $c+\bar{c}=B(g,g)$ and put $f=g-ce$. Direct expansion givesThus $e,f$ are linearly independent, and their Gram matrix is $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, which is invertible even in characteristic two. Their span is nondegenerate, so its orthogonal complement for a sesquilinear form is again nondegenerate. Repeat the construction there. In even dimension this exhausts $V$; in odd dimension the final one-dimensional complement has a nonzero diagonal value, which can be normalized to one using the field norm. Calling its vector $d$, we obtainwith the $d$ terms present only in odd dimension. This proves all the requested pairings and the basis assertion.

The printed hint needs a qualification. If $X^2-X-1$ is irreducible over ${\mathbb{F}}_q$, its two roots are conjugate, so $\zeta +\bar{\zeta }=1$ and $\zeta \bar{\zeta }=-1$; the hinted vectors then do work. They need not work when it splits. For example, over ${\mathbb{F}}_{11}$ its roots are $4,8$, fixed by conjugation in ${\mathbb{F}}_{121}$. For either choice the proposed $e$ has norm $1+{\zeta }^2$, respectively $6$ or $10$, rather than zero. The norm-and-trace construction above proves the assertion without that restriction.

Take $B$ to be linear in its first argument and conjugate-linear in its second, with $\bar{a}=a^q$. For a Hermitian form over a quadratic finite field, $B(v,v)\in {\mathbb{F}}_q$. We first establish the quadratic finite-field norm and trace surjectivity needed for orthonormalization of a finite-field Hermitian form. The field norm $N(a)=a^{q+1}$ maps ${\mathbb{F}}_{q^2}^{\ast }$ onto ${\mathbb{F}}_q^{\ast }$ because the multiplicative group of a finite field is cyclic of order $q^2-1$. The field trace $\mathrm{Tr}(a)=a+a^q$ is an ${\mathbb{F}}_q$-linear map into ${\mathbb{F}}_q$. It is nonzero: a nonzero polynomial of degree $q$ cannot vanish at all $q^2$ elements. It is therefore onto, including in characteristic two.

There must be $v$ with $B(v,v)\ne 0$. Otherwise, for any $v,w,a$, expanding the diagonal value givesIf $B(v,w)\ne 0$, varying $a$ contradicts the surjectivity of the field trace. Thus all pairings would vanish, contrary to nondegeneracy. Choose $a$ with $N(a)=B(v,v{)}^{-1}$; then $v_1=av$ has norm one. Every $w$ decomposes uniquely asThe restricted Hermitian form over a quadratic finite field on the orthogonal complement for a sesquilinear form is nondegenerate: a vector there orthogonal to that complement is also orthogonal to $v_1$, hence to all of $V$, and is zero. Induction produces an orthonormal basis $v_1,\dots ,v_n$.

The unitary group over a finite field is the group of invertible ${\mathbb{F}}_{q^2}$-linear maps preserving $B$. In the orthonormal basis just constructed, its concise description isThe notation here uses the size of the matrix field; it is also frequently written $U_n(q)$. There is no determinant-one condition in this definition.