Can the gauge boson respective of a spontaneously broken generator remain massless in the context of the Higgs Mechanism? I'm studying a 3-3-1 model which is an extension of the standard model. The breaking $$SU(3)\times U(1) \to SU(2)\times U(1)'$$ occurs in a single step and through a single scalar VEV.
The problem is that implementing this model on Mathematica, evaluating the scalar covariant derivatives, assembling the vector mass matrix and diagonalizing it, I find that only 3 vector bosons acquire mass, not the 5 that would be expected realizing that $$\text{dim}[SU(3)\times U(1)]-\text{dim}[SU(2)\times U(1)']=5.$$
Assuming that my calculations are correct, is it possible that $$SU(3)\times U(1) \to SU(2)\times U(1)'$$ is indeed the breaking pattern caused by the vacuum of that single scalar even if the bosons remains massless or is the final symmetry strictly bigger than the one stated?
 A: TL;DR: No, that is not possible under normal circumstances.

*

*Let us for simplicity consider the spontaneous symmetry breaking (SSB) of the group
$$ G~=~ U(N+1) \quad \longrightarrow\quad H~=~ U(N), $$
i.e. there are $$
\dim_{\mathbb{R}} G-\dim_{\mathbb{R}} H~=~ (N+1)^2-N^2~=~2N+1$$ broken generators.


*At the Lie algebra level this corresponds to OP's example for $N=2$ and to electroweak SSB for $N=1$ because$^1$ $$u(N)\cong su(N)\oplus u(1), \qquad u(N\!+\!1)\cong su(N\!+\!1)\oplus u(1).$$
This is good enough to count DOFs.


*Let the scalar field $\Phi$ transform in the fundamental/defining representation $V\cong \mathbb{C}^{N+1}$ of $G$. Assume that it has a non-zero VEV $\Phi_0\neq 0$.
To be concrete, by a global $G$ transformation we may assume that
$$\Phi_0~\propto~ \begin{pmatrix}
0\cr \vdots \cr 0 \cr 1 \end{pmatrix}~\in~V.$$


*The stabilizer/isotropy subgroup is
$$H~\cong~\begin{pmatrix} H \cr & 1   \end{pmatrix}_{(N+1)\times (N+1)} ~ \subseteq~G. $$


*In the unitary gauge$^2$ we may assume that the scalar field (including quantum fluctuations) is of the form
$$\Phi ~\in~ \{0\}^N\times \mathbb{R}, $$
i.e. there is only 1 real physical Higgs boson. The remaining $2N+1$ fluctuations are eaten by gauge symmetry (along the broken directions).


*The mass terms for the gauge fields come from the Lagrangian term $|D_{\mu}\Phi_0|^2$. This makes precisely $2N+1$ components of the gauge fields $A_{\mu}\in \mathfrak{g}=u(N\!+\!1)$ massive, namely the ones in the last column.
--
$^1$ $SU(1)=\{1\}$ and $su(1)=\{0\}$ are singletons.
$^2$ The unitary gauge is here a partial gauge fixing condition that fixes the gauge symmetry along the broken directions. Ultimately we should also fix the $H$-gauge symmetry, but that's another story.
