I am studying quantum field theory using Srednicki's textbook. Problem 83.1 is:
Suppose that the color group is $G_C=SO(3)$ rather than $SU(3)$, and that each quark flavor is represented by a Dirac field in the 3 representation of $SO(3)$.
a) With $n$ flavors of massless quarks, what is the non-anomalous flavor symmetry group?
The answer is: Each Dirac field equals two left-handed Weyl fields. All $2n$ Weyl fields are in the 3 representation of $G_C=SO(3)$ (because it is real). So there is a $G_F=U(2n)$ flavor symmetry; the $U(1)$ is anomalous, leaving a non-anomalous flavor symmetry group $SU(2n)$.
My question is: Why does the 3 representation of $SO(3)$ contain a $U(2n)$ symmetry, and why does the $U(2n)$ symmetry break down into a $U(1)$ group and an $SU(2n)$ group?