The commutation relations of photon and gluon?

Question

In QED, the photon field has the following commutation relations: \begin{equation} [A^{\mu}(t,\vec{x}),A^{\nu}(t,\vec{y})]=0, \tag{1} \end{equation} where $A^{\mu}(t,\vec{x})$ is the photon filed. As for QCD, I can not find similar relations for gluon, can we have \begin{equation} [A_a^{\mu}(t,\vec{x}),A_a^{\nu}(t,\vec{y})]=0 \, \, \, \, \, (2) \end{equation} for gluon field with color index $a$? Or do we have the following equation \begin{equation} [A_a^{\mu}(t,\vec{x}),A_a^{\nu}(t,\vec{x})]=0 \, \, \, \, \, (3) \end{equation} for gluon at same $x$? From the definition of gluon field-strength tensor \begin{equation} F^{\mu \nu}=\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}-ig[A^{\mu},A^{\nu}], \, \, \, \, \, (4) \end{equation} where $A^{\mu}=A_a^{\mu}t^a$, the last term of Eq.(4) is often reexpressed as \begin{equation} [A^{\mu},A^{\nu}]=A_a^{\mu}A_b^{\nu}t^a t^b-A_b^{\nu}A_a^{\mu}t^bt^a=A_a^{\mu}A_b^{\nu} [t^a, t^b].\, \, \, \, \, (5) \end{equation} Does Eq.(5) implies that \begin{equation} A_a^{\mu}A_b^{\nu}=A_b^{\nu}A_a^{\mu} \, \, \, \, \, (6) \end{equation} thus gluon commutes at the same position. Could someone tell me which of the above equations are not correct, or how do we understand the commutation relations of gluon?