Short version: Can someone give an example of an anisotropic Hermitian form over a field such that its corresponding projective unitary group is not simple?
Let $F$ be a (commutative, associative, unital) field and $\sigma$ a field automorphism with $\sigma \neq 1$ and $\sigma^2=1$. Let $V$ be a finite dimensional vector space with a function $\beta:V \times V \to F$ such that for all $u,v,w \in V$ and $\alpha \in F$ $$\begin{array}{rl} \beta(\alpha v,w) &=& \alpha \beta(v,w) \\ \beta(u+v,w) &=& \beta(u,w) + \beta(v,w) \\ \beta(u,v) &=& \sigma( \beta(v,u)) \\ \end{array}$$ $$\beta(v,v) = 0 \implies v = 0$$ and define $U(\beta) =\left\{ g \in \operatorname{GL}(V) : \beta(gv,gw) = \beta(v,w) \text{ for all } v,w \in V\right\}$ to be the unitary group of the anisotropic Hermitian form $\beta$.
For instance, $F=\mathbb{C}$, $V=F^n$ for any positive integer $n$, and $\beta(v,w) =\sum v_i \bar w_i$. Then $U(\beta)/Z(U(\beta))$ is simple unless $n=1$.
Can someone give a specific example of a $\beta$ with $n\geq 2$ where $U(\beta)$ has a non-central proper normal subgroup?
Is there some reasonable condition on the field $F$ so that no $U(\beta)$ has a non-central proper normal subgroup?
Finite fields don't have anisotropic Hermitian $\beta$. I am not sure if fields of positive characteristic do. At any rate for condition (2) I am looking for conditions that include fields where $\beta$ exists and $U(\beta)/Z(U(\beta))$ is simple for $n \geq 2$.