How does the underlying symmetry of QCD imply the allowance of a 4-gluon vertex? Quantum chromodynamics allows for a four-gluon vertex such as this, in a diagram

Such a vertex would never be allowed in quantum electrodynamics, which has an underlying U(1) gauge symmetry.
I know that quantum chromodynamics has an underlying SU(3) gauge symmetry, but how does this imply that a four-gluon vertex like what is shown above would be allowed?
 A: The QCD Lagrangian is,
$\mathcal{L}=\sum_{\alpha} \overline{\psi}_{q, a}\left(i \gamma^{\mu} \partial_{\mu} \delta_{a b}-g_{s} \gamma^{\mu} t_{a b}^{C} \mathcal{A}_{\mu}^{C}-m_{q} \delta_{a b}\right) \psi_{q, b}-\frac{1}{4} F_{\mu \nu}^{A} F^{A \mu \nu}$.
The field strength tensor is given by,
$F_{\mu \nu}^{A}=\partial_{\mu} \mathcal{A}_{\nu}^{A}-\partial_{\nu} \mathcal{A}_{\mu}^{A}-g_{s} f_{A B C} \mathcal{A}_{\mu}^{B} \mathcal{A}_{\nu}^{C}$,
where the $A_\mu^A$ terms represent the gluon field. It is then clear that $\frac{1}{4} F_{\mu \nu}^{A} F^{A \mu \nu}$ involves a product of four gluon field operators.  In the $U(1)$ case, the field strength tensor has no $-g_{s} f_{A B C} \mathcal{A}_{\mu}^{B} \mathcal{A}_{\nu}^{C}$ term, and so we do not get photon-photon interactions.
For more on this, see the PDG's review of QCD.
A: Let's compare the following train of thought for $U(1)$ vs $SU(3)$: 
U(1)


*

*$U(1)$ is an Abelian group. 

*This means, the commutator of two group elements is zero. 

*The field strength tensor that appears in the Lagrangian, $F^{\mu\nu}$, contains such a commutator: $$F_{\mu\nu}^{QED} = \partial_\mu A_\nu - \partial_\nu A_\mu + \text{i} g [A_\mu, A_\nu] = \partial_\mu A_\nu - \partial_\nu A_\mu$$

*In the Lagrangian, we have a term like $F_{\mu\nu} F^{\mu\nu}$. 

*This means that at most, we can have terms quadratic in $A_\mu$, which leads to the photon propagator and the fermion-photon interaction. 


SU(3)


*

*$SU(3)$ is a non-Abelian group. 

*This means, the commutator of two group elements is not zero. 

*The field strength tensor that appears in the Lagrangian, $F^{\mu\nu}$, contains such a commutator: $$F_{\mu\nu}^{QCD} = \partial_\mu A_\nu - \partial_\nu A_\mu + \text{i} g [A_\mu, A_\nu]$$

*In the Lagrangian, we have a term like $F_{\mu\nu} F^{\mu\nu}$. 

*This means that at most, we can have terms quartic in $A_\mu$. This means, $A^4$, $A^3$, $A^2$, $A$. Terms linear in $A$ lead to quark-gluon interaction, terms quadratic in $A$ lead to gluon propagators and the terms with $A^3$ and $A^4$ lead to the three-gluon and four-gluon interaction terms. 


And this is how the underlying symmetry of QCD implies the allowance of a 4-gluon vertex! 
