PSU(4,q) - Knowlpedia

For prime powers $q$ and fixed matrix size $4$ , $\operatorname{PSU}_{4}(q)$ is the finite group defined by

\operatorname{PSU}_{4}(q)=\operatorname{SU}_{4}(q)/\{\lambda I:\lambda^{q+1}=1,\ \lambda^{4}=1\}, \quad \operatorname{SU}_{4}(q)=\{A\in M_{4}(\mathbb F_{q^2}): (A^{(q)})^{\mathsf T}A=I,\ \det A=1\}.

Here $A^{(q)}$ is entrywise $q$ -th power. The matrix field has $q^2$ elements; its involution has fixed field of size $q$ . The scalar $\lambda$ belongs to $\mathbb F_{q^2}^\times$ ; multiplication is multiplication of scalar cosets.

Order and simplicity

Its order is

|\operatorname{PSU}_{4}(q)|=\frac{q^{6}(q^{2}-1)(q^{3}+1)(q^{4}-1)}{\gcd(4,q+1)}.

It is simple for every prime power q.

Parameter convention

The parameter $q$ is the size of the fixed field, while matrices have entries in $\mathbb F_{q^2}$ . The Hermitian form is $h(v,w)=\sum_i v_i^q w_i$ . All nondegenerate Hermitian forms of this dimension over the finite field give isomorphic groups. Type ${}^2A_{n-1}$ has absolute rank $n-1$ ; it is not a unitary group over the real or complex numbers.

References

Peter J. Cameron, Notes on Classical Groups, §§5.2–5.3, pp. 59–63; PSU(2,q), PSU(3,2), and the PSU(4,2) identification.
Robert A. Wilson, Classical groups (lecture notes), §3.1.7, pp. 5–6; §§3.2–3.4, pp. 7–17.