PSp(4,q) - Knowlpedia

For $n=2$ and prime powers $q$ , $\operatorname{PSp}_{4}(q)$ is the finite group

\operatorname{PSp}_{4}(q)=\operatorname{Sp}_{4}(q)/\{\lambda I:\lambda\in\mathbb F_{q}^\times,\lambda^2=1\}, \quad \operatorname{Sp}_{4}(q)=\{A\in\operatorname{GL}_{4}(\mathbb F_{q}):A^{\mathsf T}JA=J\},

where $\mathbb F_{q}$ is the field with $q$ elements. The matrix $J=\begin{pmatrix}0&I_{2}\\-I_{2}&0\end{pmatrix}$ gives the bilinear form $B(v,w)=v^{\mathsf T}Jw$ , which is alternating, meaning $B(v,v)=0$ for every $v$ , and nondegenerate. Multiplication is multiplication of scalar cosets.

Order and simplicity

Its order is

|\operatorname{PSp}_{4}(q)|=\frac{q^{4}(q^{2}-1)(q^{4}-1)}{\gcd(2,q-1)}.

It is simple exactly when q ≥ 3.

Parameter convention

Type $C_n$ has rank $n$ and matrix dimension $2n$ . The center has size $\gcd(2,q-1)$ , so the projective quotient does nothing when $q$ is even. Alternating means $B(v,v)=0$ , also in characteristic two.

References

Peter J. Cameron, Notes on Classical Groups, §§4.2–4.3, pp. 49–54; the Sp(4,2) action on six quadratic forms, p. 51.