Commutation relation between su(N) and clifford algebra generators. Why does the $ \gamma_5 $ matrix commute with the generators of the $su(N)$ algebra?
In the case of the chiral symmetry from physics, [$Q_a$, $Q_b^5$] = $i \epsilon_{abc}Q^5_c$ where the $Q_a, Q^5_a$ form the $su(N)$ algebra.
In order for this to hold it must be so that $[T_a, \gamma_5]=0 $ , at least to my understanding.     
I am aware of the fact that the gamma matrices under a unitary transformation create a new set of, say, $\tilde{\gamma}$ which transform like the first set. I am not sure, however, how that is applicable in this case. 
The physical interpretation I have encountered in textbooks is "they correspond to different algebras", however I am not satisfied by this, I need a stronger argument. 
 A: A bit of Theoretical Physics in QCD
I think this is best answered from a physics perspective. Let $\mathcal{L}_{QCD}$ be a QCD Lagrangian that is invariant under the gauge transformation,
$$\psi(x) \to e^{-i\alpha}\psi(x)$$
If all masses are equal in $\mathcal{L}_{QCD}$ then there is also an invariance,
$$
\psi(x) \to e^{-i\alpha^aT_a}\psi(x)
$$
where $\psi$ becomes a column vector with quark flavour indices and $T_a$ is the $SU(N_f)$ generator. If all quark masses vanish then we also obtain a chiral symmetry,
$$
\psi(x) \to e^{-i\alpha^a\gamma^5T_a}\psi(x)
$$
The chiral projection as you will know is defined as $\psi_{{R},{L}}(x) = \frac{1}{2}(1 \pm \gamma^5)\psi(x)$ from this we can construct $\gamma^5\psi = \psi_R - \psi_L$
The Maths
To obtain the algebra, look at the infinitesimal form,
$$
e^{-i\alpha^a\gamma^5T_a} = 1 - i\alpha^a\gamma^5T_a + \cdots
$$
where $\alpha$ are angles and $T_a,\gamma^5$ are matrices. 
We can decompose the field into a linear combination $\psi=\psi_L+\psi_R$ and express a new Lagrangian $\mathcal{L}_{QCD} =\mathcal{L}_{QCD,R} + \mathcal{L}_{QCD,L}$ which are each invariant under separate transformations,
$$
\psi(x)_{L,R} \to e^{-i \alpha^a_{L,R} (T_{L,R})_a}\psi(x)_{L,R}
$$
These separate invariances must be preserved in $su(n_f)\times su(n_f)$ so we must have that,
$$
\psi(x) \to \psi(x)_R - i\alpha^a_R(T_R)_a \psi(x)_R + \psi_L - i\alpha^a_L(T_L)_a \psi_L + \cdots 
$$
and so we can compare terms,
$$
- i\alpha^a_R(T_R)_a \psi(x)_R - i\alpha^a_L(T_L)_a \psi_L 
=
- i\alpha^a\gamma^5T_a\psi(x)
$$
to see that we can project out left and right handedness of the generator and also that $\alpha^a(T_{R,L})_a\psi(x)_{R,L} = 0$
We can deduce that $[\gamma^5,T_a]\psi(x)=0$ since we can either project out the chirality of $T_a$ or $\psi(x)$ but whichever we choose, due to orthogonality, the result will be the same.
Very Hand-Wavy Summary
We can also deduce that chiral symmetry would not be conserved if the right-handed and left-handed $T_a$ gave non-zero contributions from both $\psi_R,\psi_L$ and since the whole point is to project $\psi_R$ and $\psi_L$ you can deduce that any way you look at it, even if we assume $[T_a,\gamma^5]\ne0$ alone, then the resultant action we demand once acting on $\psi$ renders it the same as $[T_a,\gamma^5]=0$
References


*

*Dirac, Majorana and Weyl fermions

*Conceptual Foundations of Modern Particle Physics

*Random Standard Model Lectures
