PSU(1,q) - Knowlpedia

For prime powers $q$ and fixed matrix size $1$ , $\operatorname{PSU}_{1}(q)$ is the finite group defined by

\operatorname{PSU}_{1}(q)=\operatorname{SU}_{1}(q)/\{\lambda I:\lambda^{q+1}=1,\ \lambda^{1}=1\}, \quad \operatorname{SU}_{1}(q)=\{A\in M_{1}(\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}_{1}(q)|=1.

It is the trivial group, so it is not simple under the nontrivial-group convention. The exact order is 1.

Parameter convention

The determinant-one condition forces the sole matrix entry to be 1. The resulting group has one element. This is a degenerate matrix-size case, not an additional irreducible Lie-type family.

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.