How are quadruple gluon vertices related to $SU(2)$ and $SU(3)$? I once read that the non-commutativity of the Lie Groups $SU(2)$ and $SU(3)$ is the reason that the weak and strong interactions have Feynman diagrams with quadruple vertices, where four gauge bosons interact.
Is that correct? Can somebody explain the reasoning behind this in more detail?
 A: Let us start with $U(1)$ electromagnetism and see why it does not have such interactions. The field strength tensor is given by $F_{\mu\nu}=\partial_\mu A_\nu - \partial_\nu A_\mu$, and the relevant part of the QED Lagrangian is proportional to $F_{\mu\nu}F^{\mu\nu}$. This means that the Lagrangian has only terms that are at most quadratic in the gauge field $A_\mu$. Therefore, as follows from Feynman rules, you cannot have more than two photon lines joining at a possible interaction vertex. 
In the case of nonabelian gauge theories like $SU(2)$ and $SU(3)$, the field strength is given schematically by $F^a_{\mu\nu}=\partial_\mu A_\nu^a - \partial_\nu A_\mu^a+g\;\epsilon^{abc}A_\mu^b A_\nu^c$, where I have included indices for the nonabelian symmetry transformations, and g represents a coupling constant. The additional term appears due to the nonabelian nature of the gauge group; it can also be written in terms of a commutator. As you can now see, squaring this term gives rise to fourth powers in the gauge field. This means, again according to Feynman rules, that vertices with four gauge-boson-lines are possible.   
A: General comment to the question (v3): 
Non-abelian YM [such as, e.g., YM with gauge group $SU(2)$ or $SU(3)$] has besides quartic gauge boson interactions also cubic gauge boson interactions, while abelian YM (aka. QED) has neither. 
This is because the Feynman-rules for the cubic (quartic) gauge boson vertices are linear (quadratic) in the Lie algebra structure constants $f_{abc}$, respectively. 
(Recall that by definition a Lie algebra is abelian iff all the structure constants $f_{abc}$ vanish.)
