I know this question has been asked other times, but I am looking for a confirmation of the following.

1. When we say that the gauge group of the standard model is $G_{SM} = SU(3)_{c} \otimes SU(2)_{L} \otimes U(1)_{Y}$ we mean that the standard model is a $\textbf{gauge theory}$ with affine connections (gauge bosons) related to this group $\textbf{and}$ it has a $\textbf{global symmetry}$ $G_{SM}$?
2. If so, when we say that we have spontaneous symmetry breaking, we intend that the $\textbf{global}$ symmetry is broken, but the gauge structure remain the same? For example, when it is written that after SSB the group is $G_{SM}^{SSB} = SU(2) \otimes U(1)_{em}$ we mean that $G_{SM}^{SSB}$ is the only left global symmetry (and the underlying gauge structure remain the same)?
3. In the end, is the Higgs mechanism the same as the Goldstone mechanism with the difference that, because we have gauge freedom, we can reabsorb the massless Goldstone bosons "into" the gauge bosons, providing them with a mass?

I know this question has been asked other times, but I am looking for a confirmation of the following.

1. When we say that the gauge group of the standard model is $G_{SM} = SU(3)_{c} \times SU(2)_{L} \times U(1)_{Y}$ we mean that the standard model is a $\textbf{gauge theory}$ with affine connections (gauge bosons) related to this group $\textbf{and}$ it has a $\textbf{global symmetry}$ $G_{SM}$?
2. If so, when we say that we have spontaneous symmetry breaking, we intend that the $\textbf{global}$ symmetry is broken, but the gauge structure remain the same? For example, when it is written that after SSB the group is $G_{SM}^{SSB} = SU(2) \times U(1)_{em}$ we mean that $G_{SM}^{SSB}$ is the only left global symmetry (and the underlying gauge structure remain the same)?
3. In the end, is the Higgs mechanism the same as the Goldstone mechanism with the difference that, because we have gauge freedom, we can reabsorb the massless Goldstone bosons "into" the gauge bosons, providing them with a mass?