I am studying the standard model including the Higgs sector and electroweak interactions. Here, all of my terms have their usual meanings. Therefore my symmetry group is $SU(2)_L \times SU(2)_R \times U(1)_Y$. Here I have included $U(1)_Y$ within the larger group of $SU(2)_L \times SU(2)_R \approx O(4)$ as I want to describe electroweak interactions. Spontaneous symmetry breaking (SSB) and the Higgs mechanism reduce this group to $SU(2)_C \times U(1)_{EM}$. What I am finding confusing is understanding what happens to all of these generators when SSB occurs? In my understanding:

• $SU(2)_L$ has 3 generators, all of which are broken
• $SU(2)_R$ has 3 generators, but only $T^3_R$ is broken. The R subscript is just to note that this is a generator for the right-handed fermions
• $U(1)_Y$ has 1 generator which remains unbroken and is the gauge photon

In my understanding, the broken generators from $SU(2)_L \times SU(2)_R$ combine in some way to give us the three gauge bosons for weak interactions: $W^+, W^-, Z^0$, and the remaining two unbroken generators of $SU(2)_R$ are 'eaten' by the gauge bosons. So what are the generators for the remaining $SU(2)_C \times U(1)_{EM}$ symmetry group?

I understand the model when we just include $SU(2)_L \times U(1)_Y$ which results in just $U(1)_{EM}$ after SSB. What I am struggling to understand is the inclusion of the group $SU(2)_R$ and how all of this leads to the custodial symmetry? I also know that the different masses of the up and down quarks lead to an explicit custodial symmetry breaking, but it confuses me on how to incorporate this also.