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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.

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 $\{I,-I\}$.

Let $V$ have dimension $2m$ over the finite field ${\mathbb{F}}_q$, with a nondegenerate alternating bilinear form $B$. The symplectic group over a finite field isIt acts simply transitively on ordered symplectic bases: a linear map between two such bases is uniquely specified and preserves all pairings, hence the whole bilinear form.

Count a first hyperbolic pair $(e_1,f_1)$. There are $q^{2m}-1$ choices for $e_1\ne 0$. Nondegeneracy makes $B(e_1,-)$ a nonzero linear functional, so there are $q^{2m-1}$ choices of $f_1$ with $B(e_1,f_1)=1$. Their span is nondegenerate, and its symplectic orthogonal complement is a symplectic vector space of dimension $2m-2$. Recursing, with $|S{p}_0(q)|=1$, yieldsThe construction also proves the existence of the required symplectic bases, in characteristic two as well as odd characteristic.

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

A finite two-transitive group action is transitive on ordered pairs of distinct points. Equivalently, the action is transitive and a point stabilizer subgroup is transitive on the remaining points. Its rank of a transitive permutation group is two, and the orbit-stabilizer theorem gives $n(n-1)\mid |G|$ for degree $n$. Such an action is primitive: a block containing a point and another point must, by the transitivity of the stabilizer subgroup, contain every point.

These groups illustrate two quite different sources of symmetry. The structural theorem for finite two-transitive groups says that they have either a regular elementary-abelian normal subgroup, giving the affine case, or a nonabelian simple normal subgroup, giving the almost simple case. The full classification of the second case depends on the classification of finite simple groups; the useful point here is to describe the constructions, actions, and representative families rather than treat two-transitivity as an abstract property of a group without a specified action.

In the affine case let $V$ be a finite vector space over ${\mathbb{F}}_p$ and let $H\le GL(V)$ act transitively on $V\setminus \{0\}$. The affine two-transitive groupis transitive by translations, and the stabilizer subgroup of zero is exactly $H$, so it is two-transitive. Conversely, if a two-transitive group has an abelian minimal normal subgroup $V$, normal-subgroup orbits and primitivity make $V$ transitive. An abelian transitive faithful permutation subgroup is regular: an element fixing one point commutes with every element moving that point and hence fixes them all. Minimal normality makes $V$ elementary abelian, since its prime-primary subgroups and its subgroup of pth powers are characteristic. Conjugation by the stabilizer subgroup acts faithfully as linear automorphisms of $V$, and regularity gives $G=V⋊G_0$. This recovers precisely the construction above.

Many examples follow at once. The group $AG{L}_d(q)$ acts on the $q^d$ vectors of ${\mathbb{F}}_q^d$; $G{L}_d(q)$ sends any nonzero vector to any other by a basis map. For $d\ge 2$, the subgroup $AS{L}_d(q)$ also gives a two-transitive action, since the determinant of a basis map can be adjusted on a second basis vector while retaining the prescribed first vector. The affine group with linear part $S{p}_{2m}(q)$ gives another family: the nonzero-vector transitivity was proved using symplectic transvections in part 4(ii). Here the additive group of ${\mathbb{F}}_q^d$, for $q=p^f$, is an elementary abelian p-group of rank $fd$.

In particular $AG{L}_1(q)$ consists of $x\mapsto ax+b$ with $a\ne 0$. Given distinct $x,y$ and distinct $u,v$, the equations $ax+b=u$, $ay+b=v$ have the unique solution $a=(v-u)/(y-x)$, $b=u-ax$. Thus this action is sharply two-transitive, of degree $q$ and order $q(q-1)$. Adding field automorphisms gives $A\Gamma L_1(q)$, still two-transitive, with elements $x\mapsto a{x}^{p^i}+b$.

An almost simple group lies between a nonabelian simple group $T$ and its automorphism group. One must still specify an action: a group can have several permutation representations with different degrees or different transitivity properties. The natural actions of $S_n$ for $n\ge 2$ and $A_n$ for $n\ge 4$ are the most familiar examples of a two-transitive group action. For $n\ge 5$, the simple group $A_n$, a normal subgroup, places these groups in the almost simple case; the smaller two-transitive examples belong to the affine case. The symmetric group sends any ordered distinct pair to any other; for the alternating group one can adjust an odd map by swapping two letters outside the target pair. Thus both are two-transitive in the indicated ranges.

The projective linear examples are just as concrete. The groups $PS{L}_n(q)$ and $PG{L}_n(q)$ act two-transitively onpoints, by the basis and determinant adjustment argument of part 4(i). Their field-automorphism extensions are also two-transitive. On the projective line, $PG{L}_2(q)$ is sharply three-transitive: three distinct points determine a unique Möbius transformation. Its degree is $q+1$ and its order is $q(q^2-1)$. For even $q$, $PS{L}_2(q)=PG{L}_2(q)$; for odd $q$ the special projective group has half that order and remains two-transitive, but cannot be three-transitive. In projective dimension at least two, triples of distinct points may be collinear or noncollinear, so the natural action is not three-transitive.

The projective special unitary group over a finite field $PS{U}_3(q)$, for $q\ge 3$, acts on the isotropic points of a space with a nondegenerate Hermitian form of dimension three over ${\mathbb{F}}_{q^2}$. Its degree and order areFor two distinct isotropic lines the Hermitian form pairing is nonzero, since a three-dimensional nondegenerate Hermitian space has maximal totally isotropic dimension one. Scale their representatives to form a hyperbolic pair and extend to a basis adapted to the Hermitian form. The corresponding isometry sends one ordered pair to another, and a norm-one scalar on the orthogonal one-dimensional complement adjusts its determinant without moving the two lines. Thus the special projective unitary action is two-transitive.

Two further families of simple groups have analogous rank-two actions:The Suzuki group of Lie type and the Ree group of type G2 have stabilizer subgroups containing subgroups acting regularly on all the other points in these actions; this gives two-transitivity. For example $Sz(8)$ has degree 65 and order 29120. These are exceptional Lie-type families, distinct from the preceding linear and unitary constructions.

The symplectic group over a finite field in characteristic two has a particularly illuminating pair of two-transitive actions which are not their projective-vector actions. For $m\ge 3$, $S{p}_{2m}(2)$ acts on the two classes of quadratic forms whose polar form is the given alternating bilinear form, distinguished by the Arf invariant of a quadratic form. Their degrees areTransitivity on each class follows by choosing symplectic bases giving the standard quadratic-form normal forms. Fix a form $Q$. Every other form with the same polar form is $Q_a(x)=Q(x)+B(a,x)$, and has the same Arf invariant of a quadratic form exactly when $Q(a)=0$. The stabilizer subgroup of $Q$, an orthogonal group over a finite field, is transitive on its nonzero isotropic vectors: extend such a vector to a hyperbolic pair and map adapted quadratic-form bases. It is therefore transitive on the other forms of the same type. This proves two-transitivity of both actions. In particular $S{p}_6(2)$ has actions of degrees 28 and 36. The degree-six action of $S{p}_4(2)\cong S_6$ in part 4(ii) is the smaller-dimensional precursor.

Finally the Mathieu groups supply exceptional examples with even higher transitivity:In particular all five natural actions are two-transitive. Together these examples exhibit affine spaces, projective and Hermitian geometries, symplectic quadratic forms, and exceptional finite simple groups as different sources of the same pair symmetry.

The organizing distinction is affine versus almost simple; the action, its degree, and its stabilizer subgroup are essential data. The explicitly verified affine, symmetric, alternating, projective, unitary, and symplectic actions already give many infinite families, while the exceptional families show why no single matrix construction covers every example.

A symplectic basis organizes a nondegenerate alternating bilinear form into independent pairs. Let $(V,\omega )$ be a finite-dimensional vector space over a field $K$, with $\omega$ a nondegenerate bilinear form satisfying $\omega (v,v)=0$ for every vector $v$. Its dimension is even, say $2n$, and a symplectic basis is an ordered basis $\mathcal{B}=(e_1,\dots ,e_n,f_1,\dots ,f_n)$ such thatHere ${\delta }_{ij}$ is the Kronecker delta. Each $(e_i,f_i)$ is a hyperbolic pair; different pairs have zero mutual pairing. These conventions apply in every characteristic. In characteristic two, $-1=1$, but the condition $\omega (v,v)=0$ is still essential: a symmetric bilinear form need not be an alternating bilinear form. The existence theorem and the characteristic caveat are given in Bilinear Forms by Keith Conrad (Keith Conrad, Bilinear Forms, §1, Theorem 1.6; §5, Theorems 5.1 and 5.4 and Definition 5.3).

The benefit is a common coordinate language for every symplectic vector space of the same finite dimension over the same field. The span of all $e_i$, and separately of all $f_i$, is a Lagrangian subspace; each paired span is a symplectic subspace. In Hamiltonian mechanics, the corresponding pairing of position and momentum coordinates describes phase space and canonical transformations. On a symplectic manifold, a symplectic basis describes one tangent space; Darboux coordinates additionally describe a whole local neighborhood.^[ref]

A symplectic vector space is a vector space equipped with a nondegenerate bilinear form $\omega$ that is an alternating bilinear form: $\omega (v,v)=0$ for every vector $v$. This implies $\omega (v,w)=-\omega (w,v)$. When the characteristic is two, that latter equation alone does not imply the alternating condition. A symplectic vector space of finite dimension has even dimension: splitting off a hyperbolic pair leaves a smaller nondegenerate bilinear form, and repeating gives a decomposition into two-dimensional planes.