# Unification of the electroweak theory

Can the electroweak theory be described by the spontaneous symmetry breaking of $SU(3)$ to $SU(2)\times U(1)$?

It is indeed possible to break $SU(3)$ to $SU(2) \times U(1)$. To see that we need to check that $SU(2)$ and $U(1)$ are subgroups of $SU(3)$. Its easy to see that $SU(2)$ is a subgroup since the first three Gell-mann matrices are given by, $$\lambda _i = \left( \begin{array}{cc} \sigma _i & 0 \\ 0 & 0 \end{array} \right) \quad (i = 1 ,2,3)$$ and since these are just the Pauli matrices we know they form a group. Furthermore, it is also well known that there is another Gell-mann matrix that commutes with these $3$, $$\lambda _8 \equiv \frac{1}{\sqrt{3}} \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 2 \end{array} \right)$$
To break $SU(3)$ into this subgroup one needs to use a scalar in the adjoint (matrix) representation. This choice will work as long as the VEV of the scalar commutes with the subgroups. To see why this is consider the kinetic term for the scalar, $$\mbox{Tr} D ^\mu \Phi ^\dagger D _\mu \Phi = \mbox{Tr} [ \Phi , T ^a ] [ \Phi , T ^b ] A _\mu ^a A ^{ b , \mu } + ...$$ where since $\Phi$ is in the adjoint representation we have, $D _\mu \Phi = \partial _\mu \Phi - i A ^a _\mu \left[ \Phi , T ^a \right]$ ($T ^a$ are the group generators and $A ^a$ is the vector field). We see that if $\Phi$ gains a VEV the condition that the vector field, $A _\mu ^a$ remain massless is that $T ^a$ commute with the VEV. If a gauge boson remains massless then its gauge symmetry is conserved.
With this in mind all we have to do is pick a VEV for the scalar that commutes with our subgroup. This will be the case for the VEV, $$\left\langle \Phi \right\rangle = v \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 2 \end{array} \right)$$
Note that while one can produce the $SU(3) \rightarrow SU(2) \times U(1)$ pattern this way, this is insufficient to reproduce the phenomenology of the SM. To do that one would need to fit the SM into multiplets of $SU(3)$ which you couldn't do without introducing new fields. For more discussion on this point or any of the above I encourage you to look at a treatment of grand unification.
In fact it is possible, see the paper Spontaneous Breaking of Symmetries by Li Fong Li. In general for the Adjoint Representation of $SU(n)$(the octet for $SU(3)$) you can have the following breaking (when $\lambda_2>0$, a parameter in the general potential): $SU(n) \rightarrow SU(l)\times SU(n-l)\times U(1), \;\; l= [\frac{1}{2}n]$