# How to break a irreducible representation into its subgroups

In Grand Unified Theories (though I'm sure this a general group theory result) people write the irreducible representations of a group (i.e., the gauge bosons) using a sum of irreducible representations of its subgroup (i.e., the unbroken groups after spontaneous symmetry breaking). For example for the ${\bf 24}$ of $SU(5)$ we write, $${\bf 24} = \left( {\mathbf{8}} , {\mathbf{1}} , {\mathbf{0}} \right) + \left( {\mathbf{1}} , {\mathbf{3}} , {\mathbf{0}} \right) + \left( {\mathbf{1}} , {\mathbf{1}} , {\mathbf{0}} \right) + \left( {\mathbf{3}} , {\mathbf{2}} , {\bf \frac{ 5 }{ 3}} \right) + \left( \bar{ {\mathbf{3}} } , {\mathbf{2}} , - {\bf \frac{ 5 }{ 3}} \right)$$ where the notation here is $\left( SU(3) , SU(2) , U(1) \right)$. Now I understand how to get the first 3 parts since they just arise from the fact that we have a combination of subgroups, but how do you derive the final two? Is this a trivial result or is there some technique (e.g. through Dynkin diagrams) that can be used to extract this for any representation?

I'm ideally looking for a practical technique to perform this breakdown and less for a formal derivation of why it works (unless the derivation is quick, which in my experience in group theory it rarely is).

One is the obvious traceless diagonal matrix diag(2,2,2,-3-3), un-normalized, acting like the identity in the respective 3- and 2-dim subspaces, so then the (1,1) singlet. The remaining 6+6 pieces hermitean conjugate to each other, are the off-diagonal 3x2 and 2x3 blocks, respectively, so then (3,2) and ($\overline{{\bf3}}$,2).