Symplectic group over a finite field ($S{p}_{2m}(q)=\{g:g^{\mathsf{T}}Jg=J\}$)

This group preserves a nondegenerate alternating bilinear form on ${\mathbb{F}}_q^{2m}$. Choosing a nonzero vector and a partner of pairing one, then recursing on their orthogonal complement, gives $|S{p}_{2m}(q)|=q^{m^2}{\prod }_{i=1}^m(q^{2i}-1)$. Its center consists of scalars with ${\lambda }^2=1$.

For even n, the even-weight subspace of $F_2^n$ has radical $⟨1⟩$ for the dot product. Its quotient is a nondegenerate alternating bilinear form space of dimension n−2. The symmetric group acts by coordinate permutations. At n=6 this action is faithful and gives $S_6\cong S{p}_4(2)$.

For $q>3$, embedded perfect $S{L}_2(q)$ groups contain every symplectic transvection. For $q=3,m\ge 2$, four coefficient-one transvections on the lines of an isotropic plane multiply to the identity; their abelianized class is annihilated by 3 and 4. For $q=2,m\ge 3$, seven transvections on an isotropic three-space multiply to the identity; their class is annihilated by 2 and 7. Thus all nonexceptional cases are perfect.

This map preserves the alternating form and has inverse $T_{v,-c}$. Transvections along a fixed line form an abelian normal subgroup of its stabilizer. They generate the symplectic group by aligning successive symplectic basis pairs.

The scalar center has order $\gcd (2,q-1)$. The quotient is simple except at $(m,q)=(1,2),(1,3),(2,2)$. The last exception is $S{p}_4(2)\cong S_6$.