$SU(3)$ irreps decomposition in subgroup irreps On the notes of our professor there is written 
$$SU(3)\rightarrow SU(2) \times U(1)$$
so we can decompose the fundamental representation in
$$ \underline{3} \rightarrow (\underline{2},1/3) \oplus (\underline{1}, -2/3)$$
with first element in parenthesis belonging to $SU(2)$ and second term to $U(1)$.
And also
$$ \underline{6} \rightarrow (\underline{3},2/3) \oplus (\underline{2}, -1/3) \oplus (\underline{1}, -4/3), \quad \underline{3}^* \rightarrow (\underline{1},2/3) \oplus (\underline{2}, -1/3).$$
I don't really get what this means. I get $SU(2)$ is a subgroup ($su(2)$ is a subalgebra) but $U(1)$? there are generators that don't commute.
Note: we have not introducted roots, just weights as vectors of eigenvalues of base on the diagonal $T^3$ and $T^8$.
 A: I'll use boldface instead of your underscores to denote the dimensionality of the multiplet (rep).
I assume you are familiar with the Kronecker-product decomposition of the SU(3) triplet,
$$
{\mathbf 3}\otimes {\mathbf 3}={\mathbf 6}\oplus {\mathbf 3}^* , 
$$where, crucially, the first term in the decomposition is the symmetric part of the Kronecker product, while the second is the antisymmetrization  thereof.
Now consider the Gell-Mann su(3) matrices, and focus on a subgroup: $\lambda_1, \lambda_2,\lambda_3$ close into the su(2) algebra you are looking at. In addition, $\lambda_8$, commutes with these three, but is not the identity. 
So it is the generator of U(1), and your instructor changed its normalization by a $\sqrt 3$ so that its eigenvalues are 1/3,1/3, and -2/3, for the upper 2x2 block, the nontrivial su(2) subspace, and the 3-3 entry, respectively, on which, of course, the action of the su(2) is trivial. 
Summarizing, 
$$
{\mathbf 3}\to ({\mathbf 2},1/3)\oplus ({\mathbf 1},-2/3).
$$
Kronecker-square this and distribute the terms in each factor, all the while separating the symmetric part of the Kronecker product from the antisymmetric one.
Convince yourself the six symmetrized product states are 
$$
{\mathbf 3}\otimes_s {\mathbf 3}\to   ({\mathbf 2},1/3)\otimes_s   ({\mathbf 2},1/3)\oplus ({\mathbf 1},-2/3)\otimes _s ({\mathbf 1},-2/3)
\oplus ({\mathbf 2},1/3)\otimes_s ({\mathbf 1},-2/3)\\
= ({\mathbf 3},2/3)\oplus ({\mathbf 2},-1/3)\oplus ({\mathbf 1},-4/3 ). 
$$
The first term is the symmetric composition of two spin-1/2s to a spin 1. The U(1) charges add. So this is the  ${\mathbf 6}$.
The remaining three states of the antisymmetrization of the Kronecker product, since you can't antisymmetrize two singlets, are just
$$
{\mathbf 3}\otimes_a {\mathbf 3}\to  ({\mathbf 2},1/3)\otimes_a   ({\mathbf 2},1/3) \oplus ({\mathbf 2},1/3)\otimes_a ({\mathbf 1},-2/3)\\
=({\mathbf 1},2/3) \oplus ({\mathbf 2},-1/3), 
$$
where, this time, the first term is the spin singlet you get antisymmetrizing two spin 1/2s, and the second term is but the antisymmetric tensoring of a doublet with a singlet, the U(1) charges adding as usual. This cannot fail to be the ${\mathbf 3}^* $.
