Why is the Symmetry Group for the Electroweak force $SU(2) times U(1)$ and not $U(2)$?

Let me first say that I'm a layman who's trying to understand group theory and gauge theory, so excuse me if my question doesn't make sense.

Before symmetry breaking, the Electroweak force has 4 degrees of freedom ($B^0$, $W^1$, $W^2$, and $W^3$ right?) and after symmetry breaking, we are left with the weak $SU(2)$ bosons ($W^+$, $W^-$, $Z$) and the photon. Why, then, is the symmetry group before Electroweak symmetry breaking $SU(2) \times U(1)$ and not just $U(2)$ (Since a Unitary group of $n$ dimension contains $n^2$ degrees of freedom?) What am I missing?

Answer

The correct electroweak gauge group is $SU(2)_L \times U(1)_Y$ where $Y$ denotes the weak hypercharge. After the Higgs field spontaneously breaks this exact symmetry, third generator of $SU(2)_L$ (weak isospin) and weak hypercharge combine to give the remaining unbroken $U(1)_{em}$.

Gauge bosons and fermions fall under different representations of this gauge group and transform non-trivially under the action of the group. For example $W_{1,2,3}$ transforms as a triplet under $SU(2)_L$. Left handed fermions transform as $SU(2)_L$ doublets. This simply means that under the action of a symmetry transformation, components of each representation are mixed in a special way depending on the representation of the gauge group. So quantum numbers of our fields determine how they transform and interact with each other. For the electroweak sector of the Standard Model, the correct properties are obtained with the gauge group $SU(2)_L \times U(1)_Y$ and not with $U(2)$.

Having said that in grand unified theories one starts with a larger gauge group that contains the SM gauge group and try to break it down to the SM gauge group.