We know that no monopoles exist in the Weinberg-Salam model (a good explanation is provided in the answer to this previous question). Now let us assume that the initial $SU(2)\times U(1)$ gauge symmetry does not break down to $U(1)$ but down to $U(1)\times U(1)$.
Does this breaking yield any monopoles? Does the answer depend on whether the initial $U(1)$ symmetry is preserved after the breaking?
I think that if the two $U(1)$ symmetries after the breaking are just the initial $U(1)$ symmetry times the $U(1)$ subgroup of the $SU(2)$ symmetry, we could regard this as a breaking $SU(2)\rightarrow U(1)$, which is known to yield monopoles. If the two initial $U(1)$ symmetries mix and generate different final $U(1)$s, no monopoles exist just as in the standard Weinberg-Salam model. Is that correct?
Let me extend this question by another example. Consider the global flavor symmetry of two massless fermions, which breaks after generating one mass:
$$U(2)_V\times U(2)_A=SU(2)_V\times SU(2)_A\times U(1)_V\times U(1)_A\rightarrow U(1)_V\times U(1)_V\times U(1)_A.$$
Since the final two $U(1)_V$ symmetries correspond to fermion number conservation of individual flavors and the initial $U(1)_V$ symmetry corresponds to total fermion number conservation, the $U(1)_V$ symmetries mix and we get no global monopoles, right?
